{ "dataset": { "theorems": [ { "id": 0, "type": "example", "label": "Lebl-contfunc:0", "categories": [ "bounds", "numbers", "example" ], "title": "The set$\\Q$of rational numbers does not have the least-upper-bound property", "contents": [ "The set$\\Q$of rational numbers does not have the least-upper-bound property. The subset$\\{ x \\in \\Q : x^2 < 2 \\}$does not have a supremum in$\\Q$. We will see later (\\exampleref{example:sqrt2}) that the supremum is$\\sqrt{2}$, which is not rational% \\footnote{This is true for all other roots of 2, and interestingly, the fact that$\\sqrt[k]{2}$is never rational for$k > 1$implies no piano can ever be perfectly tuned in all keys. See, for example: \\url{https://youtu.be/1Hqm0dYKUx4}.}. Suppose$x \\in \\Q$such that$x^2 = 2$. Write$x=\\nicefrac{m}{n}$in lowest terms. So${(\\nicefrac{m}{n})}^2 = 2$or$m^2 = 2n^2$. Hence,$m^2$is divisible by 2, and so$m$is divisible by 2. Write$m = 2k$and so${(2k)}^2 = 2n^2$. Divide by 2 and note that$2k^2 = n^2$, and hence$n$is divisible by 2. But that is a contradiction as$\\nicefrac{m}{n}$is in lowest terms." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 1, "type": "example", "label": "Lebl-contfunc:1", "categories": [ "fields", "numbers", "example" ], "title": "The set$\\Q$of rational numbers is a field", "contents": [ "The set$\\Q$of rational numbers is a field. On the other hand$\\Z$is not a field, as it does not contain multiplicative inverses. For example, there is no$x \\in \\Z$such that$2x = 1$, so (M5) is not satisfied. You can check that (M5) is the only property that fails\\footnote{An algebraist would say that$\\Z$is an ordered ring, or perhaps a commutative ordered ring.}." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 2, "type": "proposition", "label": "Lebl-contfunc:prop:bordfield", "categories": [ "fields" ], "title": "Let$F$be an ordered field and$x,y,z,w \\in F$", "contents": [ "Let$F$be an ordered field and$x,y,z,w \\in F$. Then \\begin{enumerate}[(i)] \\item \\label{prop:bordfield:i} If$x > 0$, then$-x < 0$(and vice versa). \\item \\label{prop:bordfield:ii} If$x > 0$and$y < z$, then$xy < xz$. \\item \\label{prop:bordfield:iii} If$x < 0$and$y < z$, then$xy > xz$. \\item \\label{prop:bordfield:iv} If$x \\not= 0$, then$x^2 > 0$. \\item \\label{prop:bordfield:v} If$0 < x < y$, then$0 < \\nicefrac{1}{y} < \\nicefrac{1}{x}$. \\item \\label{prop:bordfield:vi} If$0 < x < y$, then$x^2 < y^2$. \\item \\label{prop:bordfield:vii} If$x \\leq y$and$z \\leq w$, then$x + z \\leq y + w$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "Let us prove \\ref{prop:bordfield:i}. The inequality $x > 0$ implies by item", "\\ref{defn:ordfield:i} of the definition of ordered fields that", "$x + (-x) > 0 + (-x)$. Apply the algebraic properties of fields to", "obtain $0 > -x$. The \\myquote{vice versa} follows by a similar calculation.", "For \\ref{prop:bordfield:ii}, note that $y < z$ implies", "$0 < z - y$ by", "item \\ref{defn:ordfield:i} of the definition of ordered fields.", "Apply item", "\\ref{defn:ordfield:ii} of the definition of ordered fields to obtain", "$0 < x(z-y)$. By algebraic properties, $0 < xz - xy$.", "Again by item", "\\ref{defn:ordfield:i} of the definition, $xy < xz$.", "Part \\ref{prop:bordfield:iii} is left as an exercise.", "To prove part \\ref{prop:bordfield:iv}, first suppose $x > 0$.", "By item \\ref{defn:ordfield:ii} of the definition of ordered fields,", "$x^2 > 0$ (use $y=x$). If $x < 0$, we use", "part \\ref{prop:bordfield:iii} of this proposition, where we plug in $y=x$ and", "$z=0$.", "To prove part \\ref{prop:bordfield:v}, notice that", "$\\nicefrac{1}{x}$ cannot be equal to zero (why?).", "Suppose $\\nicefrac{1}{x} < 0$,", "then $\\nicefrac{-1}{x} > 0$ by \\ref{prop:bordfield:i}.", "Apply part \\ref{defn:ordfield:ii} of the definition (as $x > 0$) to obtain", "$x(\\nicefrac{-1}{x}) > 0$ or $-1 > 0$, which contradicts $1 > 0$ by using part", "\\ref{prop:bordfield:i} again. Hence $\\nicefrac{1}{x} > 0$.", "Similarly, $\\nicefrac{1}{y} > 0$. Thus $(\\nicefrac{1}{x})(\\nicefrac{1}{y}) > 0$", "by definition of ordered field, and by part \\ref{prop:bordfield:ii},", "\\begin{equation*}", "(\\nicefrac{1}{x})(\\nicefrac{1}{y})x < (\\nicefrac{1}{x})(\\nicefrac{1}{y})y .", "\\end{equation*}", "By algebraic properties, $\\nicefrac{1}{y} < \\nicefrac{1}{x}$.", "Parts \\ref{prop:bordfield:vi} and \\ref{prop:bordfield:vii} are left as", "exercises." ], "refs": [ "defn:ordfield:i", "defn:ordfield:ii", "prop:bordfield:i", "prop:bordfield:ii", "prop:bordfield:iii", "prop:bordfield:iv", "prop:bordfield:v", "prop:bordfield:vi", "prop:bordfield:vii" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 3, "type": "proposition", "label": "Lebl-contfunc:3", "categories": [ "fields" ], "title": "Let$x,y \\in F$, where$F$is an ordered field", "contents": [ "Let$x,y \\in F$, where$F$is an ordered field. If$xy > 0$, then either both$x$and$y$are positive, or both are negative." ], "refs": [], "proofs": [ { "contents": [ "We show the contrapositive: If either one of $x$ or $y$ is zero, or", "if $x$ and $y$ have opposite signs, then $xy$ is not positive.", "If $x$ or $y$ is zero, then $xy$ is zero and hence not positive.", "Hence assume that $x$ and $y$ are nonzero and have", "opposite signs.", "Without loss of generality,", "suppose $x > 0$ and $y < 0$.", "Multiply $y < 0$ by $x$ to get $xy < 0x = 0$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 4, "type": "example", "label": "Lebl-contfunc:example:complexfield", "categories": [ "fields", "ordered sets", "numbers", "example" ], "title": "The reader may have heard about the \\emph{\\myindex{complex numbers}}, usually denoted by$\\C$", "contents": [ "The reader may have heard about the \\emph{\\myindex{complex numbers}}, usually denoted by$\\C$.\\glsadd{not:complex} That is,$\\C$is the set of numbers of the form$x + iy$, where$x$and$y$are real numbers, and$i$is the imaginary number, a number such that$i^2 = -1$. The reader may remember from algebra that$\\C$is also a field; however, it is not an ordered field. While one can make$\\C$into an ordered set in some way, it is not possible to put an order on$\\C$that would make it an ordered field: In every ordered field,$-1 < 0$and$x^2 > 0$for all nonzero$x$, but in$\\C$,$i^2 = -1$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 5, "type": "proposition", "label": "Lebl-contfunc:5", "categories": [ "fields", "bounds" ], "title": "Boundedness of Fields", "contents": [ "Let$F$be an ordered field with the least-upper-bound property. Let$A \\subset F$be a nonempty set that is bounded below. Then$\\inf\\, A$exists." ], "refs": [], "proofs": [ { "contents": [ "Let $B \\coloneqq \\{ -x : x \\in A \\}$. Let $b \\in F$ be a lower bound for $A$:", "If $x \\in A$, then $x \\geq b$", "and hence $-x \\leq -b$. So $-b$", "is an upper bound for $B$.", "Since $F$ has the least-upper-bound property, $c\\coloneqq\\sup\\, B$ exists, and $c \\leq -b$.", "As $y \\leq c$", "for all $y \\in B$, then $-c \\leq x$ for all $x \\in A$.", "So", "$-c$ is a lower bound for $A$. As $-c \\geq b$,", "the greatest lower bound of $A$ exists and equals $-c$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 6, "type": "theorem", "label": "Lebl-contfunc:6", "categories": [ "fields", "numbers", "uniqueness" ], "title": "Uniqueness of Fields", "contents": [ "There exists a unique\\footnote{Uniqueness is up to isomorphism, but we wish to avoid excessive use of algebra. For us, it is simply enough to assume that a set of real numbers exists. See Rudin~\\cite{Rudin:baby} for the construction and more details.} ordered field$\\R$with the \\hyperref[defn:lub]{least-upper-bound property} such that$\\Q \\subset \\R$." ], "refs": [ "defn:lub" ], "proofs": [], "ref_ids": [] }, { "id": 7, "type": "proposition", "label": "Lebl-contfunc:7", "categories": [], "title": "If$x \\in \\R$is such that$x \\leq \\epsilon$for all$\\epsilon \\in \\R$where$\\epsilon > 0$, then$x \\leq 0$", "contents": [ "If$x \\in \\R$is such that$x \\leq \\epsilon$for all$\\epsilon \\in \\R$where$\\epsilon > 0$, then$x \\leq 0$." ], "refs": [], "proofs": [ { "contents": [ "If $x > 0$, then $0 < \\nicefrac{x}{2} < x$ (why?). Take", "$\\epsilon = \\nicefrac{x}{2}$ to get a contradiction. Thus $x \\leq 0$." ], "refs": [], "ref_ids": [] }, { "contents": [ "Take the set", "$A \\coloneqq \\{ x \\in \\R : x^2 < 2 \\}$. We first show that $A$ is bounded above and", "nonempty. The inequality $x \\geq 2$ implies $x^2 \\geq 4$", "(see \\exerciseref{exercise:squareineq}).", "So if $x^2 < 2$, then $x < 2$. So $A$ is bounded above.", "As $1 \\in A$, the set $A$ is nonempty. The supremum, therefore, exists.", "Let $r \\coloneqq \\sup\\, A$. We will show that $r^2 = 2$ by showing", "that $r^2 \\geq 2$ and $r^2 \\leq 2$. This is the way analysts show", "equality, by showing two inequalities.", "We already know that $r \\geq 1 > 0$.", "%\\medskip", "In the following, it may seem we are pulling certain expressions out of", "a hat. When writing a proof such as this, we would, of course, come up with", "the expressions only after playing around with what we wish to prove. The", "order in which we write the proof is not necessarily the order in which we", "come up with the proof.", "Let us first show that $r^2 \\geq 2$.", "Take a positive number $s$ such that $s^2 < 2$. We wish to find an $h > 0$", "such that ${(s+h)}^2 < 2$.", "As $2-s^2 > 0$, we have $\\frac{2-s^2}{2s+1} > 0$.", "Choose an $h \\in \\R$ such that", "$0 < h < \\frac{2-s^2}{2s+1}$.", "Furthermore, assume $h < 1$. Estimate,", "\\begin{equation*}", "\\begin{aligned}", "{(s+h)}^2 - s^2 & = h(2s + h) \\\\", "& < h(2s+1) & & \\quad \\bigl(\\text{since } h < 1\\bigr) \\\\", "& < 2-s^2 & & \\quad \\bigl(\\text{since } h < \\tfrac{2-s^2}{2s+1} \\bigr) .", "\\end{aligned}", "\\end{equation*}", "Therefore, ${(s+h)}^2 < 2$. Hence $s+h \\in A$, but as $h > 0$,", "we have $s+h > s$. So $s < r = \\sup\\, A$. As $s$ was an arbitrary", "positive number such that $s^2 < 2$, it follows that $r^2 \\geq 2$.", "%\\medskip", "Now take a positive number $s$ such that", "$s^2 > 2$. We wish to find an $h > 0$ such that", "${(s-h)}^2 > 2$ and $s-h$ is still positive.", "As $s^2-2 > 0$, we have $\\frac{s^2-2}{2s} > 0$.", "Let $h \\coloneqq \\frac{s^2-2}{2s}$,", "and check $s-h=s-\\frac{s^2-2}{2s} = \\frac{s}{2}+\\frac{1}{s} > 0$.", "Estimate,", "\\begin{equation*}", "\\begin{aligned}", "s^2 - {(s-h)}^2 & = 2sh - h^2 \\\\", "& < 2sh & & \\quad \\bigl( \\text{since } h^2 > 0 \\text{ as } h \\not= 0 \\bigr) \\\\", "& = s^2-2 & & \\quad \\bigl( \\text{since } h = \\tfrac{s^2-2}{2s} \\bigr) .", "\\end{aligned}", "\\end{equation*}", "By subtracting $s^2$ from both sides and multiplying by $-1$, we find", "${(s-h)}^2 > 2$. Therefore, $s-h \\notin A$.", "Moreover, if $x \\geq s-h$,", "then $x^2 \\geq {(s-h)}^2 > 2$ (as $x > 0$ and $s-h > 0$) and so $x \\notin A$.", "Thus,", "$s-h$ is an upper bound for $A$. However, $s-h < s$, or in other", "words, $s > r = \\sup\\, A$. Hence, $r^2 \\leq 2$.", "\\medskip", "Together, $r^2 \\geq 2$ and $r^2 \\leq 2$ imply", "$r^2 = 2$. The existence part is finished. We still need to", "handle uniqueness. Suppose $s \\in \\R$ such that $s^2 = 2$ and $s > 0$.", "Thus, $s^2 = r^2$. However, if $0 < s < r$, then $s^2 < r^2$. Similarly,", "$0 < r < s$ implies $r^2 < s^2$. Hence, $s = r$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 8, "type": "example", "label": "Lebl-contfunc:example:sqrt2", "categories": [ "existence", "bounds", "example", "uniqueness" ], "title": "Uniqueness of Bounds", "contents": [ "Claim: \\emph{There exists a unique positive$r \\in \\R$such that$r^2 = 2$. We denote$r$by$\\sqrt{2}$.}", "\\begin{proof} Take the set$A \\coloneqq \\{ x \\in \\R : x^2 < 2 \\}$. We first show that$A$is bounded above and nonempty. The inequality$x \\geq 2$implies$x^2 \\geq 4$(see \\exerciseref{exercise:squareineq}). So if$x^2 < 2$, then$x < 2$. So$A$is bounded above. As$1 \\in A$, the set$A$is nonempty. The supremum, therefore, exists.", "Let$r \\coloneqq \\sup\\, A$. We will show that$r^2 = 2$by showing that$r^2 \\geq 2$and$r^2 \\leq 2$. This is the way analysts show equality, by showing two inequalities. We already know that$r \\geq 1 > 0$.", "%\\medskip In the following, it may seem we are pulling certain expressions out of a hat. When writing a proof such as this, we would, of course, come up with the expressions only after playing around with what we wish to prove. The order in which we write the proof is not necessarily the order in which we come up with the proof.", "Let us first show that$r^2 \\geq 2$. Take a positive number$s$such that$s^2 < 2$. We wish to find an$h > 0$such that${(s+h)}^2 < 2$. As$2-s^2 > 0$, we have$\\frac{2-s^2}{2s+1} > 0$. Choose an$h \\in \\R$such that$0 < h < \\frac{2-s^2}{2s+1}$. Furthermore, assume$h < 1$. Estimate,", "\\begin{equation*}\n\\begin{aligned}\n{(s+h)}^2 - s^2 & = h(2s + h) \\\\\n & < h(2s+1) & & \\quad \\bigl(\\text{since } h < 1\\bigr) \\\\\n & < 2-s^2 & & \\quad \\bigl(\\text{since } h < \\tfrac{2-s^2}{2s+1} \\bigr) .\n\\end{aligned}\n\\end{equation*}", "Therefore,${(s+h)}^2 < 2$. Hence$s+h \\in A$, but as$h > 0$, we have$s+h > s$. So$s < r = \\sup\\, A$. As$s$was an arbitrary positive number such that$s^2 < 2$, it follows that$r^2 \\geq 2$.", "%\\medskip", "Now take a positive number$s$such that$s^2 > 2$. We wish to find an$h > 0$such that${(s-h)}^2 > 2$and$s-h$is still positive. As$s^2-2 > 0$, we have$\\frac{s^2-2}{2s} > 0$. Let$h \\coloneqq \\frac{s^2-2}{2s}$, and check$s-h=s-\\frac{s^2-2}{2s} = \\frac{s}{2}+\\frac{1}{s} > 0$. Estimate,", "\\begin{equation*}\n\\begin{aligned}\ns^2 - {(s-h)}^2 & = 2sh - h^2 \\\\\n & < 2sh & & \\quad \\bigl( \\text{since } h^2 > 0 \\text{ as } h \\not= 0 \\bigr) \\\\\n & = s^2-2 & & \\quad \\bigl( \\text{since } h = \\tfrac{s^2-2}{2s} \\bigr) .\n\\end{aligned}\n\\end{equation*}", "By subtracting$s^2$from both sides and multiplying by$-1$, we find${(s-h)}^2 > 2$. Therefore,$s-h \\notin A$.", "Moreover, if$x \\geq s-h$, then$x^2 \\geq {(s-h)}^2 > 2$(as$x > 0$and$s-h > 0$) and so$x \\notin A$. Thus,$s-h$is an upper bound for$A$. However,$s-h < s$, or in other words,$s > r = \\sup\\, A$. Hence,$r^2 \\leq 2$.", "\\medskip", "Together,$r^2 \\geq 2$and$r^2 \\leq 2$imply$r^2 = 2$. The existence part is finished. We still need to handle uniqueness. Suppose$s \\in \\R$such that$s^2 = 2$and$s > 0$. Thus,$s^2 = r^2$. However, if$0 < s < r$, then$s^2 < r^2$. Similarly,$0 < r < s$implies$r^2 < s^2$. Hence,$s = r$. \\end{proof}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 9, "type": "theorem", "label": "Lebl-contfunc:thm:arch", "categories": [ "numbers", "named theorem" ], "title": "Archimedean Property", "contents": [ "%FIXMEevillayouthack \\pagebreak[3] \\leavevmode \\begin{enumerate}[(i)] \\item \\label{thm:arch:i} \\emph{(\\myindex{Archimedean property})}% \\footnote{Named after the Ancient Greek mathematician \\href{https://en.wikipedia.org/wiki/Archimedes}{Archimedes of Syracuse} (c.\\ 287 BC -- c.\\ 212 BC)\\@. This property is Axiom V from Archimedes' \\myquote{On the Sphere and Cylinder} 225 BC\\@.} If$x, y \\in \\R$and$x > 0$, then there exists an$n \\in \\N$such that\\begin{equation*} nx > y . \\end{equation*}\\item \\label{thm:arch:ii} \\emph{($\\Q$is dense in$\\R$\\index{density of rational numbers})} If$x, y \\in \\R$and$x < y$, then there exists an$r \\in \\Q$such that$x < r < y$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "Let us prove \\ref{thm:arch:i}. Divide through by $x$.", "Then \\ref{thm:arch:i} says that for every real number $t\\coloneqq \\nicefrac{y}{x}$,", "we can find $n \\in \\N$ such that $n > t$. In other words,", "\\ref{thm:arch:i} says that $\\N \\subset \\R$ is not bounded above.", "Suppose for contradiction that $\\N$ is bounded above. Let $b \\coloneqq \\sup \\N$.", "The number $b-1$ cannot possibly be an upper bound for $\\N$ as it is strictly", "less than $b$ (the least upper bound). Thus there exists an $m \\in \\N$ such that $m > b-1$.", "Add one to obtain $m+1 > b$, contradicting $b$ being an", "upper bound.", "\\begin{myfigureht}", "\\subimport*{figures/}{figdensofQ.pdf_t}", "\\caption{Idea of the proof of the density of $\\Q$: Find $n$ such that $y-x >", "\\nicefrac{1}{n}$, then take the least $m$ such that $\\nicefrac{m}{n} > x$.\\label{figdensofQ}}", "\\end{myfigureht}", "Let us tackle \\ref{thm:arch:ii}.", "See \\figureref{figdensofQ}", "for a picture of the idea behind the proof.", "First assume $x \\geq 0$.", "Note that $y-x > 0$.", "By \\ref{thm:arch:i}, there exists an $n \\in \\N$ such that", "\\begin{equation*}", "n(y-x) > 1", "\\qquad \\text{or} \\qquad", "y-x > \\nicefrac{1}{n}.", "\\end{equation*}", "Again by \\ref{thm:arch:i} the set", "$A \\coloneqq \\{ k \\in \\N : k > nx \\}$ is nonempty. By the", "\\hyperlink{wop:link}{well ordering property}", "of $\\N$, $A$ has a least element $m$, and as $m \\in A$,", "then $m > nx$.", "Divide through by $n$ to get $x < \\nicefrac{m}{n}$.", "As $m$ is the least", "element of $A$, $m-1 \\notin A$.", "If $m > 1$, then $m-1 \\in \\N$, but $m-1 \\notin A$ and so $m-1 \\leq nx$.", "If $m=1$,", "then $m-1 = 0$, and $m-1 \\leq nx$ still holds as $x \\geq 0$.", "In other words,", "\\begin{equation*}", "m-1 \\leq nx \\qquad \\text{or} \\qquad m \\leq nx+1 .", "\\end{equation*}", "On the other hand,", "from $n(y-x) > 1$ we obtain $ny > 1+nx$.", "Hence $ny > 1+nx \\geq m$, and therefore $y > \\nicefrac{m}{n}$.", "Putting everything together, we obtain $x < \\nicefrac{m}{n} < y$.", "So take $r = \\nicefrac{m}{n}$.", "Now assume $x < 0$. If $y > 0$, then just take $r=0$. If", "$y \\leq 0$, then $0 \\leq -y < -x$, and we", "find a rational $q$ such that $-y < q < -x$. Then take $r = -q$." ], "refs": [ "thm:arch:i", "thm:arch:ii" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 10, "type": "corollary", "label": "Lebl-contfunc:10", "categories": [ "consequence" ], "title": "", "contents": [ "$\\inf \\{ \\nicefrac{1}{n} : n \\in \\N \\} = 0$. See \\figureref{fig:oneovernset}." ], "refs": [], "proofs": [ { "contents": [ "Let $A \\coloneqq \\{ \\nicefrac{1}{n} : n \\in \\N \\}$. Obviously, $A$ is not empty.", "Furthermore,", "$\\nicefrac{1}{n} > 0$ for all $n \\in \\N$, so $0$ is a lower bound", "and $b \\coloneqq \\inf\\, A$ exists.", "As $0$ is a lower bound, then $b \\geq 0$.", "Take an arbitrary $a > 0$. By the", "\\hyperref[thm:arch:i]{Archimedean property}, there exists an $n$ such that", "$na > 1$, that is, $a > \\nicefrac{1}{n} \\in A$. Therefore,", "$a$ cannot be a lower bound for $A$. Hence $b=0$." ], "refs": [ "thm:arch:i" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 11, "type": "proposition", "label": "Lebl-contfunc:prop:supinfalg", "categories": [ "bounds" ], "title": "Boundedness of Real Numbers", "contents": [ "Let$A \\subset \\R$be nonempty. \\begin{enumerate}[(i)] \\item If$x \\in \\R$and$A$is bounded above, then$\\sup (x+A) = x + \\sup\\, A$. \\item If$x \\in \\R$and$A$is bounded below, then$\\inf (x+A) = x + \\inf\\, A$. \\item If$x > 0$and$A$is bounded above, then$\\sup (xA) = x ( \\sup\\, A )$. \\item If$x > 0$and$A$is bounded below, then$\\inf (xA) = x ( \\inf\\, A )$. \\item If$x < 0$and$A$is bounded below, then$\\sup (xA) = x ( \\inf\\, A )$. \\item If$x < 0$and$A$is bounded above, then$\\inf (xA) = x ( \\sup\\, A )$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "Let us only prove the first statement. The rest are left as exercises.", "Suppose $b$ is an upper bound for $A$. That is, $y \\leq b$ for all $y \\in A$.", "Then $x+y \\leq x+b$ for all $y \\in A$, and so $x+b$ is an upper", "bound for $x+A$. In particular, if $b = \\sup\\, A$, then", "\\begin{equation*}", "\\sup (x+A) \\leq x+b = x+ \\sup\\, A .", "\\end{equation*}", "The opposite inequality is similar: If $c$ is an upper bound for $x+A$,", "then $x+y \\leq c$", "for all $y \\in A$ and so $y \\leq c-x$ for all $y \\in A$.", "So $c-x$ is an upper bound for $A$.", "If $c = \\sup (x+A)$, then", "\\begin{equation*}", "\\sup\\, A \\leq c-x = \\sup (x+A) -x .", "\\end{equation*}", "The result follows." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 12, "type": "proposition", "label": "Lebl-contfunc:infsupineq:prop", "categories": [ "bounds" ], "title": "Boundedness of Real Numbers", "contents": [ "Let$A, B \\subset \\R$be nonempty sets such that$x \\leq y$whenever$x \\in A$and$y \\in B$. Then$A$is bounded above,$B$is bounded below, and$\\sup\\, A \\leq \\inf\\, B$." ], "refs": [], "proofs": [ { "contents": [ "Any $x \\in A$ is a lower bound for $B$.", "Therefore, $x \\leq \\inf\\, B$ for all $x \\in A$,", "so $\\inf\\, B$ is an upper bound for $A$.", "Hence, $\\sup\\, A \\leq \\inf\\, B$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 13, "type": "proposition", "label": "Lebl-contfunc:prop:existsxepsfromsup", "categories": [ "bounds" ], "title": "Boundedness of Real Numbers", "contents": [ "If$S \\subset \\R$is nonempty and bounded above, then for every$\\epsilon > 0$there exists an$x \\in S$such that$(\\sup\\, S) - \\epsilon < x \\leq \\sup\\, S$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 14, "type": "proposition", "label": "Lebl-contfunc:prop:absbas", "categories": [ "characterization" ], "title": "\\leavevmode \\begin{enumerate}[(i)] \\item \\label{prop:absbas:i}$\\abs{x} \\geq 0$, moreover,$\\abs{x}...", "contents": [ "\\leavevmode \\begin{enumerate}[(i)] \\item \\label{prop:absbas:i}$\\abs{x} \\geq 0$, moreover,$\\abs{x}=0$if and only if$x = 0$. \\item \\label{prop:absbas:ii}$\\abs{-x} = \\abs{x}$for all$x \\in \\R$. \\item \\label{prop:absbas:iii}$\\abs{xy} = \\abs{x}\\abs{y}$for all$x,y \\in \\R$. \\item \\label{prop:absbas:iv}$\\abs{x}^2 = x^2$for all$x \\in \\R$. \\item \\label{prop:absbas:v}$\\abs{x} \\leq y$if and only if$-y \\leq x \\leq y$. \\item \\label{prop:absbas:vi}$-\\abs{x} \\leq x \\leq \\abs{x}$for all$x \\in \\R$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "\\ref{prop:absbas:i}:", "First suppose $x \\geq 0$.", "Then $\\abs{x} = x \\geq 0$. Also, $\\abs{x} = x = 0$ if and only", "if $x=0$.", "On the other hand, if $x < 0$, then $\\abs{x} = -x > 0$, and $\\abs{x}$ is never zero.", "\\medskip", "\\ref{prop:absbas:ii}: If $x > 0$, then $-x < 0$ and so $\\abs{-x} = -(-x) = x =", "\\abs{x}$. Similarly when $x < 0$, or $x = 0$.", "\\medskip", "\\ref{prop:absbas:iii}:", "If $x$ or $y$ is zero, then the result is immediate. When $x$ and", "$y$ are both positive, then $\\abs{x}\\abs{y} = xy$. As $xy$ is also", "positive, $xy = \\abs{xy}$.", "If $x$ and $y$ are both negative,", "then $xy=(-x)(-y)$ is still positive and $\\abs{xy}=xy$. Also,", "$\\abs{x}\\abs{y} = (-x)(-y) = xy$.", "If $x > 0$ and $y < 0$, then $\\abs{x}\\abs{y} = x(-y) = -(xy)$. Now", "$xy$ is negative and $\\abs{xy} = -(xy)$. Similarly when", "$x < 0$ and $y > 0$.", "\\medskip", "\\ref{prop:absbas:iv}:", "Immediate if $x \\geq 0$. If $x < 0$, then $\\abs{x}^2 = {(-x)}^2 =", "x^2$.", "\\medskip", "\\ref{prop:absbas:v}:", "Suppose $\\abs{x} \\leq y$.", "If $x \\geq 0$, then $x \\leq y$.", "It follows that $y \\geq 0$, leading to $-y \\leq 0 \\leq x$.", "So $-y \\leq x \\leq y$ holds.", "If $x < 0$, then $\\abs{x} \\leq y$ means $-x \\leq y$.", "Negating both sides we get $x \\geq -y$.", "Again $y \\geq 0$ and so $y \\geq 0 > x$.", "Hence, $-y \\leq x \\leq y$.", "On the other hand, suppose $-y \\leq x \\leq y$ is true.", "If $x \\geq 0$, then $x \\leq y$ is equivalent to $\\abs{x} \\leq y$.", "If $x < 0$, then $-y \\leq x$ implies $(-x) \\leq y$,", "which is equivalent to $\\abs{x} \\leq y$.", "\\medskip", "\\ref{prop:absbas:vi}:", "Apply \\ref{prop:absbas:v} with $y = \\abs{x}$." ], "refs": [ "prop:absbas:i", "prop:absbas:ii", "prop:absbas:iii", "prop:absbas:iv", "prop:absbas:v", "prop:absbas:vi" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 15, "type": "proposition", "label": "Lebl-contfunc:15", "categories": [], "title": "$\\sabs{x+y} \\leq \\sabs{x}+\\sabs{y}$for all$x, y \\in \\R$", "contents": [ "$\\sabs{x+y} \\leq \\sabs{x}+\\sabs{y}$for all$x, y \\in \\R$." ], "refs": [], "proofs": [ { "contents": [ "\\propref{prop:absbas} gives", "$- \\sabs{x} \\leq x \\leq \\sabs{x}$ and", "$- \\sabs{y} \\leq y \\leq \\sabs{y}$. Add these two inequalities to obtain", "\\begin{equation*}", "- \\bigl(\\sabs{x}+\\sabs{y}\\bigr) \\leq x+y \\leq \\sabs{x}+ \\sabs{y} .", "\\end{equation*}", "Apply \\propref{prop:absbas} again to find", "$\\sabs{x+y} \\leq \\sabs{x}+\\sabs{y}$." ], "refs": [ "prop:absbas" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 16, "type": "corollary", "label": "Lebl-contfunc:16", "categories": [ "consequence" ], "title": "Triangle Inequality", "contents": [ "Let$x,y \\in \\R$. \\begin{enumerate}[(i)] \\item \\emph{(\\myindex{reverse triangle inequality})} ~$\\bigl\\lvert (\\abs{x}-\\abs{y}) \\bigr\\rvert \\leq \\abs{x-y}$. \\item$\\abs{x-y} \\leq \\abs{x}+\\abs{y}$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "Let us plug in $x=a-b$ and $y=b$ into the standard", "triangle inequality to obtain", "\\begin{equation*}", "\\abs{a} = \\abs{a-b+b} \\leq \\abs{a-b} + \\abs{b} ,", "\\end{equation*}", "or $\\abs{a}-\\abs{b} \\leq \\abs{a-b}$. Switching the roles of $a$ and $b$", "we find", "$\\abs{b}-\\abs{a} \\leq \\abs{b-a} = \\abs{a-b}$. Applying", "\\propref{prop:absbas}, we obtain the reverse triangle", "inequality.", "The second item in the corollary is obtained from the standard", "triangle inequality by just replacing $y$ with $-y$, and noting $\\abs{-y} =", "\\abs{y}$." ], "refs": [ "prop:absbas" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 17, "type": "corollary", "label": "Lebl-contfunc:17", "categories": [ "consequence" ], "title": "Let$x_1, x_2, \\ldots, x_n \\in \\R$", "contents": [ "Let$x_1, x_2, \\ldots, x_n \\in \\R$. Then\\begin{equation*} \\abs{x_1 + x_2 + \\cdots + x_n} \\leq \\abs{x_1} + \\abs{x_2} + \\cdots + \\abs{x_n} . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "We proceed by \\hyperref[induction:thm]{induction}.", "The conclusion holds trivially for $n=1$, and", "for $n=2$ it is the standard triangle inequality. Suppose the corollary", "holds for $n$. Take $n+1$ numbers $x_1,x_2,\\ldots,x_{n+1}$ and", "first use the standard triangle inequality, then the induction", "hypothesis", "\\begin{equation*}", "\\begin{split}", "\\sabs{x_1 + x_2 + \\cdots + x_n + x_{n+1}} & \\leq", "\\sabs{x_1 + x_2 + \\cdots + x_n} + \\sabs{x_{n+1}} \\\\", "& \\leq", "\\sabs{x_1} + \\sabs{x_2} + \\cdots + \\sabs{x_n} + \\sabs{x_{n+1}} . \\qedhere", "\\end{split}", "\\end{equation*}" ], "refs": [ "induction:thm" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 18, "type": "example", "label": "Lebl-contfunc:18", "categories": [ "example", "intervals" ], "title": "Triangle Inequality", "contents": [ "Find a number$M$such that$\\sabs{x^2-9x+1} \\leq M$for all$-1 \\leq x \\leq 5$.", "Using the triangle inequality, write", "\\begin{equation*}\n\\sabs{x^2-9x+1} \\leq \\sabs{x^2}+\\sabs{9x}+\\sabs{1}\n=\n\\sabs{x}^2+9\\sabs{x}+1 .\n\\end{equation*}", "The expression$\\sabs{x}^2+9\\sabs{x}+1$is largest when$\\abs{x}$is largest (why?). In the interval provided,$\\abs{x}$is largest when$x=5$and so$\\abs{x}=5$. One possibility for$M$is", "\\begin{equation*}\nM = 5^2+9(5)+1 = 71 .\n\\end{equation*}", "There are, of course, other$M$that work. The bound of 71 is much higher than it need be, but we didn't ask for the best possible$M$, just one that works." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 19, "type": "proposition", "label": "Lebl-contfunc:prop:funcsupinf", "categories": [ "bounds", "numbers" ], "title": "Boundedness of Real Numbers", "contents": [ "If$f \\colon D \\to \\R$and$g \\colon D \\to \\R$($D$nonempty) are bounded\\footnote{The boundedness hypothesis is for simplicity, it can be dropped if we allow for the extended real numbers.} functions and\\begin{equation*} f(x) \\leq g(x) \\qquad \\text{for all } x \\in D, \\end{equation*}then \\begin{equation} \\label{prop:funcsupinf:eq} \\sup_{x \\in D} f(x) \\leq \\sup_{x \\in D} g(x) \\qquad \\text{and} \\qquad \\inf_{x \\in D} f(x) \\leq \\inf_{x \\in D} g(x) . \\end{equation}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 20, "type": "proposition", "label": "Lebl-contfunc:prop:intervaldef", "categories": [ "characterization", "intervals" ], "title": "Characterization of Intervals via Equivalence", "contents": [ "A set$I \\subset \\R$is an interval if and only if$I$contains at least 2 points and for all$a,c \\in I$and$b \\in \\R$such that$a < b < c$, we have$b \\in I$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 21, "type": "theorem", "label": "Lebl-contfunc:21", "categories": [ "cardinality", "named theorem" ], "title": "Cantor's Theorem", "contents": [ "\\index{Cantor's theorem}$\\R$is uncountable." ], "refs": [], "proofs": [ { "contents": [ "Let $X \\subset \\R$ be a countably infinite", "subset such that for every pair of real numbers", "$a < b$, there is an $x \\in X$ such that $a < x < b$. Were $\\R$ countable,", "we could take $X = \\R$. We will show that $X$ is necessarily", "a proper subset, and so $X$ cannot equal $\\R$, and $\\R$ must be", "uncountable.", "As $X$ is countably infinite,", "there is a bijection from $\\N$ to $X$. We write $X$ as", "a sequence of real numbers $x_1, x_2, x_3,\\ldots$, such that", "each number in $X$", "is given by $x_n$ for some $n \\in \\N$.", "We inductively", "construct two sequences of real numbers $a_1,a_2,a_3,\\ldots$ and", "$b_1,b_2,b_3,\\ldots$. Let", "$a_1 \\coloneqq x_1$ and $b_1 \\coloneqq x_1+1$.", "Note that $a_1 < b_1$ and", "$x_1 \\notin (a_1,b_1)$.", "For some $k > 1$,", "suppose $a_1,a_2,\\ldots,a_{k-1}$ and $b_1,b_2,\\ldots,b_{k-1}$ have been", "defined,", "suppose", "$a_1 < a_2 < \\cdots < a_{k-1} < b_{k-1} < \\cdots < b_2 < b_1$,", "and suppose", "for each $j=1,2,\\ldots,k-1$,", "we have $x_\\ell \\notin (a_{j},b_{j})$", "for $\\ell=1,2,\\ldots,j$.", "\\begin{enumerate}[(i)]", "\\item Define $a_k \\coloneqq x_n$, where $n$ is the smallest $n \\in \\N$", "such that $x_n \\in (a_{k-1},b_{k-1})$. Such an $x_n$ exists by our", "assumption on $X$, and $n \\geq k$ by the assumption on $(a_{k-1},b_{k-1})$.", "\\item Next, define $b_k$ to be some real number in $(a_{k},b_{k-1})$.", "\\end{enumerate}", "Notice that %$a_k < b_k$ and", "$a_{k-1} < a_k < b_k < b_{k-1}$.", "Also notice that $(a_{k},b_{k})$ does not contain $x_k$ and hence", "does not contain $x_j$ for $j=1,2,\\ldots,k$.", "The two sequences are now defined.", "Claim: \\emph{$a_n < b_m$ for all $n$ and $m$ in $\\N$.} Proof: Let us first", "assume $n < m$. Then $a_n < a_{n+1} < \\cdots < a_{m-1} < a_m < b_m$.", "Similarly for $n > m$. The claim follows.", "Let $A \\coloneqq \\{ a_n : n \\in \\N \\}$ and $B \\coloneqq \\{ b_n : n \\in \\N \\}$.", "By \\propref{infsupineq:prop} and the claim above,", "\\begin{equation*}", "\\sup\\, A \\leq \\inf\\, B .", "\\end{equation*}", "Define $y \\coloneqq \\sup\\, A$. The number $y$ cannot be a member of $A$: If $y = a_n$", "for some $n$, then $y < a_{n+1}$, which is impossible.", "Similarly, $y$ cannot be a member of $B$. Therefore,", "$a_n < y$ for all $n\\in \\N$", "and $y < b_n$ for all $n\\in \\N$.", "In other words, for every $n \\in \\N$, we have $y \\in (a_n,b_n)$.", "By the construction of the sequence, $x_n \\not\\in (a_n,b_n)$,", "and so $y \\not= x_n$. As this was true for all $n \\in \\N$, we have that $y", "\\not\\in X$.", "We have constructed a real number $y$ that is not in $X$, and thus $X$", "is a proper subset of $\\R$.", "The sequence $x_1,x_2,\\ldots$ cannot contain all elements of $\\R$", "and thus $\\R$ is uncountable." ], "refs": [ "infsupineq:prop" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 22, "type": "proposition", "label": "Lebl-contfunc:prop:decimalprop", "categories": [ "decimals", "numbers" ], "title": "Uniqueness of Real Numbers", "contents": [ "\\leavevmode \\begin{enumerate}[(i)] \\item Every infinite sequence of digits$0.d_1d_2d_3\\ldots$represents a unique real number$x \\in [0,1]$, and\\begin{equation*} D_n \\leq x \\leq D_n+\\frac{1}{{10}^n} \\qquad \\text{for all } n \\in \\N. \\end{equation*}\\item For every$x \\in (0,1]$there exists an infinite sequence of digits$0.d_1d_2d_3\\ldots$that represents$x$. There exists a unique representation such that\\begin{equation*} D_n < x \\leq D_n+\\frac{1}{{10}^n} \\qquad \\text{for all } n \\in \\N. \\end{equation*}\\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "We start with the first item. Take an arbitrary infinite sequence of", "digits $0.d_1d_2d_3\\ldots$. Use the geometric sum formula to write", "\\begin{equation*}", "\\begin{split}", "D_n =", "\\frac{d_1}{10} +", "\\frac{d_2}{{10}^2} +", "\\frac{d_3}{{10}^3} +", "\\cdots +", "\\frac{d_n}{{10}^n}", "& \\leq", "\\frac{9}{10} +", "\\frac{9}{{10}^2} +", "\\frac{9}{{10}^3} +", "\\cdots +", "\\frac{9}{{10}^n}", "\\\\", "& =", "\\frac{9}{10}", "\\bigl( 1 + \\nicefrac{1}{10} +", "{(\\nicefrac{1}{10})}^2 + \\cdots +", "{(\\nicefrac{1}{10})}^{n-1} \\bigr)", "\\\\", "& =", "\\frac{9}{10}", "\\left(", "\\frac{1-{(\\nicefrac{1}{10})}^{n}}{1-\\nicefrac{1}{10}}", "\\right)", "= 1-{(\\nicefrac{1}{10})}^{n}", "< 1 .", "\\end{split}", "\\end{equation*}", "In particular, $D_n < 1$ for all $n$. A sum of nonnegative numbers is", "nonnegative so $D_n \\geq 0$, and hence", "\\begin{equation*}", "0 \\leq \\sup_{n\\in \\N} \\, D_n \\leq 1 .", "\\end{equation*}", "Therefore, $0.d_1d_2d_3\\ldots$ represents a unique number $x \\coloneqq \\sup_{n\\in", "\\N} D_n \\in [0,1]$.", "As $x$ is a supremum, then $D_n \\leq x$.", "Take $m \\in \\N$. If $m < n$, then $D_m - D_n \\leq 0$. If $m > n$, then", "computing as above", "\\begin{equation*}", "D_m - D_n", "=", "\\frac{d_{n+1}}{{10}^{n+1}} +", "\\frac{d_{n+2}}{{10}^{n+2}} +", "\\frac{d_{n+3}}{{10}^{n+3}} +", "\\cdots +", "\\frac{d_{m}}{{10}^m}", "\\leq", "\\frac{1}{{10}^{n}}", "\\bigl(", "1-{(\\nicefrac{1}{10})}^{m-n}", "\\bigr)", "<", "\\frac{1}{{10}^{n}} .", "\\end{equation*}", "Take the supremum over $m$ to find", "\\begin{equation*}", "x - D_n", "\\leq", "\\frac{1}{{10}^{n}} .", "\\end{equation*}", "We move on to the", "second item. Take any $x \\in (0,1]$.", "First let us tackle the existence.", "For convenience, let $D_0 \\coloneqq 0$.", "Then,", "$D_0 < x \\leq D_0 + {10}^{-0}$.", "Suppose we defined the digits $d_1,d_2,\\ldots,d_n$,", "and that", "$D_k < x \\leq D_k + {10}^{-k}$, for $k=0,1,2,\\ldots,n$. We need to define $d_{n+1}$.", "By the", "\\hyperref[thm:arch:i]{Archimedean property} of the real numbers,", "find an integer $j$ such that", "$x-D_n \\leq j {10}^{-(n+1)}$. Take the least such $j$ and obtain", "\\begin{equation} \\label{eq:theDnjineq}", "(j-1){10}^{-(n+1)} < x-D_n \\leq j {10}^{-(n+1)} .", "\\end{equation}", "Let $d_{n+1} \\coloneqq j-1$.", "As $D_n < x$,", "then $d_{n+1} = j-1 \\geq 0$. On the other hand, since", "$x-D_n \\leq {10}^{-n}$, we have that", "$j$ is at most 10, and therefore $d_{n+1} \\leq 9$.", "So $d_{n+1}$ is a", "decimal digit.", "Since $D_{n+1} = D_n + d_{n+1} {10}^{-(n+1)}$", "add $D_n$ to the inequality", "\\eqref{eq:theDnjineq} above:", "\\begin{multline*}", "D_{n+1} = D_n + (j-1){10}^{-(n+1)} < x", "\\leq", "D_n + j {10}^{-(n+1)} \\\\", "=", "D_n + (j-1) {10}^{-(n+1)} +", "{10}^{-(n+1)} = D_{n+1} + {10}^{-(n+1)} .", "\\end{multline*}", "And so", "$D_{n+1} < x \\leq D_{n+1} + {10}^{-(n+1)}$ holds.", "We inductively", "defined an infinite sequence of digits $0.d_1d_2d_3\\ldots$.", "Consider $D_{n} < x \\leq D_{n} + {10}^{-n}$.", "As $D_n < x$ for all $n$, then", "$\\sup \\{ D_n : n \\in \\N \\} \\leq x$.", "The second inequality for $D_n$ implies", "\\begin{equation*}", "x - \\sup \\{ D_m : m \\in \\N \\}", "\\leq", "x - D_n \\leq 10^{-n} .", "\\end{equation*}", "As the inequality holds for all $n$ and", "${10}^{-n}$ can be made arbitrarily small", "(see \\exerciseref{exercise:bnlimit}),", "we have $x \\leq \\sup \\{ D_m : m \\in \\N \\}$.", "Therefore, $\\sup \\{ D_m : m \\in \\N \\} = x$.", "What is left to show is the uniqueness.", "Suppose $0.e_1e_2e_3\\ldots$ is another representation of $x$.", "Let $E_n$ be the $n$-digit truncation of $0.e_1e_2e_3\\ldots$, and suppose", "$E_n < x \\leq E_n + {10}^{-n}$ for all $n \\in \\N$.", "Suppose for some $K \\in \\N$, $e_n = d_n$ for all $n < K$, so", "$D_{K-1} = E_{K-1}$. Then", "\\begin{equation*}", "E_K = D_{K-1} + e_K{10}^{-K} < x \\leq E_K + {10}^{-K} = D_{K-1} +", "e_K{10}^{-K} + {10}^{-K} .", "\\end{equation*}", "Subtracting $D_{K-1}$ and multiplying by ${10}^{K}$ we get", "\\begin{equation*}", "e_K < (x - D_{K-1}){10}^K \\leq e_K + 1 .", "\\end{equation*}", "Similarly,", "\\begin{equation*}", "d_K < (x - D_{K-1}){10}^K \\leq d_K + 1 .", "\\end{equation*}", "Hence, both $e_K$ and $d_K$ are the largest integer $j$", "such that $j < (x - D_{K-1}){10}^K$, and therefore $e_K = d_K$. That is,", "the representation is unique." ], "refs": [ "eq:theDnjineq", "thm:arch:i" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 23, "type": "theorem", "label": "Lebl-contfunc:23", "categories": [ "cardinality" ], "title": "The set$(0,1]$is uncountable", "contents": [ "The set$(0,1]$is uncountable." ], "refs": [], "proofs": [ { "contents": [ "Let $X \\coloneqq \\{ x_1,x_2,x_3,\\ldots \\}$ be any countable subset of real numbers in $(0,1]$.", "We will construct a real number not in $X$. Let", "\\begin{equation*}", "x_n = 0.d_1^nd_2^nd_3^n\\ldots", "\\end{equation*}", "be the unique representation from the proposition, that is, $d_j^n$ is the", "$j$th digit of the $n$th number. Let", "\\begin{equation*}", "e_n \\coloneqq", "\\begin{cases}", "1 & \\text{if } d_n^n \\not= 1, \\\\", "2 & \\text{if } d_n^n = 1.", "\\end{cases}", "\\end{equation*}", "Let $E_n$ be the $n$-digit truncation of $y = 0.e_1e_2e_3\\ldots$. Because", "all the digits are nonzero we get $E_n < E_{n+1} \\leq y$. Therefore,", "\\begin{equation*}", "E_n < y \\leq E_n + {10}^{-n}", "\\end{equation*}", "for all $n$, and the representation is the unique one for $y$ from", "the proposition. For every $n$, the $n$th digit", "of $y$ is different from the $n$th digit of $x_n$, so $y \\not= x_n$.", "Therefore $y \\notin X$, and as $X$ was an arbitrary countable subset,", "$(0,1]$ must be uncountable. See \\figureref{cantorexamplefig} for an", "example.", "\\begin{myfigureht}", "\\subimport*{figures/}{cantorexamplefig.pdf_t}", "\\caption{Example of Cantor diagonalization, the diagonal digits $d_n^n$", "marked.\\label{cantorexamplefig}}", "\\end{myfigureht}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 24, "type": "proposition", "label": "Lebl-contfunc:prop:rationaldecimal", "categories": [ "decimals", "numbers" ], "title": "If$x \\in (0,1]$is a rational number and $x = 0", "contents": [ "If$x \\in (0,1]$is a rational number and$x = 0.d_1d_2d_3\\ldots$, then the decimal digits eventually start repeating. That is, there are positive integers$N$and$P$, such that for all$n \\geq N$,$d_n = d_{n+P}$." ], "refs": [], "proofs": [ { "contents": [ "Suppose $x = \\nicefrac{p}{q}$ for positive integers $p$ and $q$.", "Suppose also that $x$ is a number with a unique representation, as", "otherwise we have seen above that both its representations are repeating,", "see also \\exerciseref{exercise:nonuniquedecimals}. This also means", "that $x \\not= 1$ so $p < q$.", "To compute the first digit, we take $10 p$ and divide by", "$q$. Let $d_1$ be the quotient, and the remainder $r_1$ is some integer", "between $0$ and $q-1$. That is, $d_1$ is the largest integer", "such that $d_1 q \\leq 10p$ and then $r_1 = 10p - d_1q$.", "As $p < q$, then $d_1 < 10$, so $d_1$ is a digit.", "Furthermore,", "\\begin{equation*}", "\\frac{d_1}{10} \\leq \\frac{p}{q} =", "\\frac{d_1}{10} + \\frac{r_1}{10q} \\leq \\frac{d_1}{10} +", "\\frac{1}{10} .", "\\end{equation*}", "The first inequality is strict since $x$ has a", "unique representation. That is, $d_1$ really is the first digit.", "What is left is $\\nicefrac{r_1}{(10q)}$. This is the same as computing the", "first digit of $\\nicefrac{r_1}{q}$.", "To compute $d_2$ divide $10 r_1$ by $q$, and so on.", "After computing $n-1$ digits, we have", "$\\nicefrac{p}{q} = D_{n-1} + \\nicefrac{r_{n-1}}{(10^{n-1} q)}$.", "To get the $n$th digit,", "divide $10 r_{n-1}$ by $q$", "to get quotient $d_n$, remainder $r_n$, and the inequalities", "\\begin{equation*}", "\\frac{d_n}{10} \\leq \\frac{r_{n-1}}{q} =", "\\frac{d_n}{10} + \\frac{r_n}{10q} \\leq \\frac{d_n}{10} +", "\\frac{1}{10} .", "\\end{equation*}", "Dividing by $10^{n-1}$ and adding $D_{n-1}$ we find", "\\begin{equation*}", "D_n \\leq D_{n-1} + \\frac{r_{n-1}}{10^{n-1} q} = \\frac{p}{q} \\leq D_n +", "\\frac{1}{10^n} .", "\\end{equation*}", "By uniqueness, we really have", "the $n$th digit $d_n$ from the construction.", "The new digit depends only the remainder from the previous", "step.", "There are at most $q$ possible remainders.", "Hence, the process must start repeating itself after at most $q$ steps,", "and so $P$ is at most $q$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 25, "type": "example", "label": "Lebl-contfunc:25", "categories": [ "decimals", "numbers", "example" ], "title": "The number \\begin{equation*} x = 0", "contents": [ "The number", "\\begin{equation*}\nx = 0.101001000100001000001\\ldots\n\\end{equation*}", "is irrational. That is, the digits are$n$zeros, then a one, then$n+1$zeros, then a one, and so on and so forth. The fact that$x$is irrational follows from the proposition; the digits never start repeating. For every$P$, if we go far enough, we find a 1 followed by at least$P+1$zeros." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 0, "type": "example", "label": "Lebl-contfunc:0", "categories": [ "convergence", "sequences", "example" ], "title": "Convergence of Sequences", "contents": [ "The constant sequence$1,1,1,1,\\ldots$converges to 1. For every$\\epsilon > 0$, pick$M = 1$. That is,$\\sabs{x_n-x} = \\sabs{1-1} < \\epsilon$for all$n$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 1, "type": "example", "label": "Lebl-contfunc:1", "categories": [ "convergence", "sequences", "example" ], "title": "Claim: \\emph{The sequence$\\{ \\nicefrac{1}{n} \\}_{n=1}^\\infty$is convergent and} \\begin{equation*}...", "contents": [ "Claim: \\emph{The sequence$\\{ \\nicefrac{1}{n} \\}_{n=1}^\\infty$is convergent and}", "\\begin{equation*}\n\\lim_{n\\to \\infty} \\frac{1}{n} = 0 .\n\\end{equation*}", "Proof: Given an$\\epsilon > 0$, find an$M \\in \\N$such that$0 < \\nicefrac{1}{M} < \\epsilon$(\\hyperref[thm:arch:i]{Archimedean property} at work). For all$n \\geq M$,", "\\begin{equation*}\n\\abs{x_n - x}\n=\n\\abs{\\frac{1}{n} - 0} = \\abs{\\frac{1}{n}} = \\frac{1}{n} \\leq \\frac{1}{M} < \\epsilon .\n\\end{equation*}" ], "refs": [ "thm:arch:i" ], "proofs": [ { "contents": [ "Given an$\\epsilon > 0$, find an$M \\in \\N$such that$0 < \\nicefrac{1}{M} < \\epsilon$(\\hyperref[thm:arch:i]{Archimedean property} at work). For all$n \\geq M$, \\begin{equation*}", "\\abs{x_n - x}", "=", "\\abs{\\frac{1}{n} - 0} = \\abs{\\frac{1}{n}} = \\frac{1}{n} \\leq \\frac{1}{M} < \\epsilon .", "\\end{equation*}" ], "refs": [ "thm:arch:i" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 2, "type": "example", "label": "Lebl-contfunc:2", "categories": [ "convergence", "sequences", "divergence", "example" ], "title": "Existence of Sequences", "contents": [ "The sequence$\\bigl\\{ {(-1)}^n \\bigr\\}_{n=1}^\\infty$is divergent. Proof: If there were a limit$x$, then for$\\epsilon = \\frac{1}{2}$we expect an$M$that satisfies the definition. Suppose such an$M$exists. Then for an even$n \\geq M$, we compute", "\\begin{equation*}\n\\nicefrac{1}{2} > \\abs{x_n - x} = \\abs{1 - x}\n\\qquad \\text{and} \\qquad\n\\nicefrac{1}{2} > \\abs{x_{n+1} - x} = \\abs{-1 - x} .\n\\end{equation*}", "And we obtain a contradiction", "\\begin{equation*}\n2 = \\abs{1 - x - (-1 -x)} \\leq\n\\abs{1 - x} + \\abs{-1 -x} < \\nicefrac{1}{2} + \\nicefrac{1}{2} = 1 .\n\\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "If there were a limit$x$, then for$\\epsilon = \\frac{1}{2}$we expect an$M$that satisfies the definition. Suppose such an$M$exists. Then for an even$n \\geq M$, we compute \\begin{equation*}", "\\nicefrac{1}{2} > \\abs{x_n - x} = \\abs{1 - x}", "\\qquad \\text{and} \\qquad", "\\nicefrac{1}{2} > \\abs{x_{n+1} - x} = \\abs{-1 - x} .", "\\end{equation*} And we obtain a contradiction \\begin{equation*}", "2 = \\abs{1 - x - (-1 -x)} \\leq", "\\abs{1 - x} + \\abs{-1 -x} < \\nicefrac{1}{2} + \\nicefrac{1}{2} = 1 .", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 3, "type": "proposition", "label": "Lebl-contfunc:prop:limisunique", "categories": [ "convergence", "sequences" ], "title": "Uniqueness of Sequences", "contents": [ "A convergent sequence has a unique limit." ], "refs": [], "proofs": [ { "contents": [ "%NOTE: should be word for word the same as 7.3.3", "Suppose $\\{ x_n \\}_{n=1}^\\infty$ has limits $x$ and $y$.", "Take an arbitrary $\\epsilon > 0$.", "From the definition find an $M_1$ such that for all $n \\geq M_1$,", "$\\abs{x_n-x} < \\nicefrac{\\epsilon}{2}$. Similarly, find an $M_2$", "such that for all $n \\geq M_2$, we have", "$\\abs{x_n-y} < \\nicefrac{\\epsilon}{2}$.", "Now take an $n$ such that $n \\geq M_1$ and also $n \\geq M_2$, and estimate", "\\begin{equation*}", "\\begin{split}", "\\abs{y-x}", "& =", "\\abs{x_n-x - (x_n -y)} \\\\", "& \\leq", "\\abs{x_n-x} + \\abs{x_n -y} \\\\", "& <", "\\frac{\\epsilon}{2} + \\frac{\\epsilon}{2} = \\epsilon .", "\\end{split}", "\\end{equation*}", "As $\\abs{y-x} < \\epsilon$ for all $\\epsilon > 0$, then $\\abs{y-x} = 0$", "and $y=x$. Hence the limit (if it exists) is unique." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 4, "type": "proposition", "label": "Lebl-contfunc:4", "categories": [ "convergence", "sequences", "boundedness" ], "title": "Boundedness of Sequences", "contents": [ "A convergent sequence$\\{ x_n \\}_{n=1}^\\infty$is bounded." ], "refs": [], "proofs": [ { "contents": [ "Suppose $\\{ x_n \\}_{n=1}^\\infty$ converges to $x$. Thus there exists an $M \\in \\N$", "such that for all $n \\geq M$, we have", "$\\abs{x_n - x} < 1$. For $n \\geq M$,", "\\begin{equation*}", "\\begin{split}", "\\abs{x_n} & = \\abs{x_n - x + x}", "\\\\", "& \\leq \\abs{x_n - x} + \\abs{x}", "\\\\", "& < 1 + \\abs{x} .", "\\end{split}", "\\end{equation*}", "The set $\\bigl\\{ \\abs{x_1}, \\abs{x_2}, \\ldots, \\abs{x_{M-1}} , 1+\\sabs{x} \\bigr\\}$", "is a finite set and hence let", "\\begin{equation*}", "B \\coloneqq \\max \\bigl\\{ \\abs{x_1}, \\abs{x_2}, \\ldots, \\abs{x_{M-1}}, 1+\\sabs{x} \\bigr\\} .", "\\end{equation*}", "Then for all $n \\in \\N$,", "\\begin{equation*}", "\\abs{x_n} \\leq B. \\qedhere", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 5, "type": "example", "label": "Lebl-contfunc:5", "categories": [ "convergence", "example" ], "title": "Convergence of Convergence", "contents": [ "Let us show$\\left\\{ \\frac{n^2+1}{n^2+n} \\right\\}_{n=1}^\\infty$converges and", "\\begin{equation*}\n\\lim_{n\\to\\infty} \\frac{n^2+1}{n^2+n} = 1 .\n\\end{equation*}", "Given$\\epsilon > 0$, find$M \\in \\N$such that$\\frac{1}{M} < \\epsilon$. Then for all$n \\geq M$,", "\\begin{equation*}\n\\begin{split}\n%\\abs{\\frac{n^2+1}{n^2+n} - 1} & =\n%\\abs{\\frac{n^2+1 - (n^2+n)}{n^2+n}} \\\\\n\\abs{\\frac{n^2+1}{n^2+n} - 1} =\n\\abs{\\frac{n^2+1 - (n^2+n)}{n^2+n}}\n& =\n\\abs{\\frac{1 - n}{n^2+n}} \\\\\n& =\n\\frac{n-1}{n^2+n} \\\\\n& \\leq \n\\frac{n}{n^2+n} \n =\n\\frac{1}{n+1} \\\\\n& \\leq \\frac{1}{n}\n\\leq \\frac{1}{M} < \\epsilon .\n\\end{split}\n\\end{equation*}", "Therefore,$\\lim_{n\\to\\infty} \\frac{n^2+1}{n^2+n} = 1$. This example shows that sometimes to get what you want, you must throw away some information to get a simpler estimate." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 6, "type": "theorem", "label": "Lebl-contfunc:thm:monotoneconv", "categories": [ "characterization", "convergence", "sequences", "boundedness" ], "title": "Monotone Convergence Theorem", "contents": [ "\\index{monotone convergence theorem}% A monotone sequence$\\{ x_n \\}_{n=1}^\\infty$is bounded if and only if it is convergent.", "Furthermore, if$\\{ x_n \\}_{n=1}^\\infty$is monotone increasing and bounded, then\\begin{equation*} \\lim_{n\\to \\infty} x_n = \\sup \\{ x_n : n \\in \\N \\} . \\end{equation*}If$\\{ x_n \\}_{n=1}^\\infty$is monotone decreasing and bounded, then\\begin{equation*} \\lim_{n\\to \\infty} x_n = \\inf \\{ x_n : n \\in \\N \\} . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Consider a monotone increasing sequence $\\{ x_n \\}_{n=1}^\\infty$. Suppose", "first the sequence is bounded,", "that is,", "the set $\\{ x_n : n \\in \\N \\}$ is bounded. Let", "\\begin{equation*}", "x \\coloneqq \\sup \\{ x_n : n \\in \\N \\} .", "\\end{equation*}", "Let $\\epsilon > 0$ be arbitrary. As $x$ is the supremum,", "there must be at least one $M \\in \\N$ such that $x_{M} > x-\\epsilon$.", "As $\\{ x_n \\}_{n=1}^\\infty$ is monotone increasing,", "then it is easy to see (by \\hyperref[induction:thm]{induction}) that", "$x_n \\geq x_{M}$ for all $n \\geq M$. Hence for all $n \\geq M$,", "\\begin{equation*}", "\\abs{x_n-x} = x-x_n \\leq x-x_{M} < \\epsilon .", "\\end{equation*}", "So $\\{ x_n \\}_{n=1}^\\infty$ converges to $x$.", "Therefore, a bounded monotone increasing sequence converges.", "For the other direction, we already proved that a convergent sequence", "is bounded.", "The proof for monotone decreasing sequences is left as an exercise." ], "refs": [ "induction:thm" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 7, "type": "example", "label": "Lebl-contfunc:7", "categories": [ "convergence", "boundedness", "sequences", "example", "tests" ], "title": "Boundedness of Sequences", "contents": [ "Take the sequence$\\bigl\\{ \\frac{1}{\\sqrt{n}} \\bigr\\}_{n=1}^\\infty$.", "The sequence is bounded below as$\\frac{1}{\\sqrt{n}} > 0$for all$n \\in \\N$. Let us show that it is monotone decreasing. We start with$\\sqrt{n+1} \\geq \\sqrt{n}$(why is that true?). From this inequality we obtain", "\\begin{equation*}\n\\frac{1}{\\sqrt{n+1}} \\leq \\frac{1}{\\sqrt{n}} .\n\\end{equation*}", "So the sequence is monotone decreasing and bounded below (hence bounded). Via \\thmref{thm:monotoneconv} we find that the sequence is convergent and", "\\begin{equation*}\n\\lim_{n\\to \\infty} \\frac{1}{\\sqrt{n}}\n=\n\\inf \\left\\{ \\frac{1}{\\sqrt{n}} : n \\in \\N \\right\\} .\n\\end{equation*}", "We already know that the infimum is greater than or equal to 0, as 0 is a lower bound. Take a number$b \\geq 0$such that$b \\leq \\frac{1}{\\sqrt{n}}$for all$n$. We square both sides to obtain", "\\begin{equation*}\nb^2 \\leq \\frac{1}{n} \\qquad \\text{for all } n \\in \\N.\n\\end{equation*}", "We have seen before that this implies that$b^2 \\leq 0$(a consequence of the \\hyperref[thm:arch:i]{Archimedean property}). As$b^2 \\geq 0$as well, we have$b^2 = 0$and so$b = 0$. Hence,$b=0$is the greatest lower bound, and$\\lim\\limits_{n\\to\\infty} \\frac{1}{\\sqrt{n}} = 0$." ], "refs": [ "thm:arch:i" ], "proofs": [], "ref_ids": [] }, { "id": 8, "type": "example", "label": "Lebl-contfunc:8", "categories": [ "series", "sequences", "boundedness", "example" ], "title": "Boundedness of Sequences", "contents": [ "A word of caution: Showing that a monotone sequence is bounded in order to use \\thmref{thm:monotoneconv} may be difficult. The sequence$\\{ 1 + \\nicefrac{1}{2} + \\cdots + \\nicefrac{1}{n} \\}_{n=1}^\\infty$is a monotone increasing sequence that grows slowly and in fact grows slower and slower as$n$gets larger. We will see, once we get to series, that this sequence has no upper bound and so does not converge. It is not at all obvious that this sequence has no upper bound." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 9, "type": "proposition", "label": "Lebl-contfunc:prop:supinfseq", "categories": [ "sequences", "boundedness" ], "title": "Boundedness of Sequences", "contents": [ "Let$S \\subset \\R$be a nonempty bounded set. Then there exist monotone sequences$\\{ x_n \\}_{n=1}^\\infty$and$\\{ y_n \\}_{n=1}^\\infty$such that$x_n, y_n \\in S$and\\begin{equation*} \\sup\\,S = \\lim_{n\\to \\infty} x_n \\qquad \\text{and} \\qquad \\inf\\,S = \\lim_{n\\to\\infty} y_n . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 10, "type": "proposition", "label": "Lebl-contfunc:10", "categories": [ "convergence", "sequences" ], "title": "Convergence of Sequences", "contents": [ "Let$\\{ x_n \\}_{n=1}^\\infty$be a sequence. Then the following statements are equivalent: \\begin{enumerate}[(i)] \\item \\label{prop:ktail:i} The sequence$\\{ x_n \\}_{n=1}^\\infty$converges. \\item \\label{prop:ktail:ii} The$K$-tail$\\{ x_{n+K} \\}_{n=1}^\\infty$converges for all$K \\in \\N$. \\item \\label{prop:ktail:iii} The$K$-tail$\\{ x_{n+K} \\}_{n=1}^\\infty$converges for some$K \\in \\N$. \\end{enumerate} Furthermore, if any (and hence all) of the limits exist, then for all$K \\in \\N$\\begin{equation*} \\lim_{n\\to \\infty} x_n = \\lim_{n \\to \\infty} x_{n+K} . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "It is clear that", "\\ref{prop:ktail:ii} implies \\ref{prop:ktail:iii}.", "We will therefore show first that", "\\ref{prop:ktail:i}", "implies", "\\ref{prop:ktail:ii},", "and then we will show that", "\\ref{prop:ktail:iii}", "implies", "\\ref{prop:ktail:i}. That is,", "%mbxSTARTIGNORE", "\\begin{equation*}", "\\begin{tikzcd}[row sep=0pt]", "& \\text{\\ref{prop:ktail:ii}} \\ar[dr, Rightarrow] & \\\\", "\\text{\\ref{prop:ktail:i}} \\ar[ur, Rightarrow, \"\\text{to prove}\"] & &", "\\text{\\ref{prop:ktail:iii}} \\ar[ll, Rightarrow, \"\\text{to prove}\"]", "\\end{tikzcd}", "\\end{equation*}", "%mbxENDIGNORE", "%mbxlatex \\begin{center}", "%mbxlatex \\subimport*{figures/}{tail_of_sequence_cd.pdf_t}", "%mbxlatex \\end{center}", "In the process we will also show that the limits are equal.", "We start with \\ref{prop:ktail:i} implies \\ref{prop:ktail:ii}.", "Suppose $\\{x_n \\}_{n=1}^\\infty$ converges to some $x \\in \\R$.", "Let $K \\in \\N$ be arbitrary, and", "define $y_n \\coloneqq x_{n+K}$. We wish to show that $\\{ y_n \\}_{n=1}^\\infty$ converges", "to $x$.", "Given an $\\epsilon > 0$, there exists an $M \\in \\N$ such that", "$\\abs{x-x_n} < \\epsilon$ for all $n \\geq M$.", "Note that $n \\geq M$ implies $n+K \\geq M$. Therefore, for", "all $n \\geq M$, we have", "\\begin{equation*}", "\\abs{x-y_n} = \\abs{x-x_{n+K}} < \\epsilon .", "\\end{equation*}", "Consequently, $\\{ y_n \\}_{n=1}^\\infty$ converges to $x$.", "Let us move to \\ref{prop:ktail:iii} implies \\ref{prop:ktail:i}.", "Let $K \\in \\N$ be given, define", "$y_n \\coloneqq x_{n+K}$, and suppose that $\\{ y_n \\}_{n=1}^\\infty$ converges to $x \\in \\R$.", "That is, given an $\\epsilon > 0$, there exists an $M' \\in \\N$ such that", "$\\abs{x-y_n} < \\epsilon$ for all $n \\geq M'$.", "Let $M \\coloneqq M'+K$. Then $n \\geq M$ implies $n-K \\geq M'$.", "Thus, whenever $n \\geq M$, we have", "\\begin{equation*}", "\\abs{x-x_n} = \\abs{x-y_{n-K}} < \\epsilon.", "\\end{equation*}", "Therefore, $\\{ x_n \\}_{n=1}^\\infty$ converges to $x$." ], "refs": [ "prop:ktail:i", "prop:ktail:ii", "prop:ktail:iii" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 11, "type": "proposition", "label": "Lebl-contfunc:prop:seqtosubseq", "categories": [ "convergence", "sequences" ], "title": "If$\\{ x_n \\}_{n=1}^\\infty$is a convergent sequence, then every subsequence$\\{ x_{n_i} \\}_{i=1}^\\i...", "contents": [ "If$\\{ x_n \\}_{n=1}^\\infty$is a convergent sequence, then every subsequence$\\{ x_{n_i} \\}_{i=1}^\\infty$is also convergent, and\\begin{equation*} \\lim_{n\\to \\infty} x_n = \\lim_{i\\to \\infty} x_{n_i} . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Suppose $\\lim_{n\\to \\infty} x_n = x$. So for every", "$\\epsilon > 0$, there is an $M \\in \\N$ such that for all $n \\geq M$,", "\\begin{equation*}", "\\abs{x_n - x} < \\epsilon .", "\\end{equation*}", "It is not hard to prove (do it!) by \\hyperref[induction:thm]{induction} that", "$n_i \\geq i$ for all $i \\in \\N$. Hence $i \\geq M$ implies $n_i \\geq M$. Thus,", "for all $i \\geq M$,", "\\begin{equation*}", "\\abs{x_{n_i} - x} < \\epsilon ,", "\\end{equation*}", "and we are done." ], "refs": [ "induction:thm" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 12, "type": "example", "label": "Lebl-contfunc:12", "categories": [ "convergence", "existence", "sequences", "example", "divergence" ], "title": "Convergence of Sequences", "contents": [ "Existence of a convergent subsequence does not imply convergence of the sequence itself. Take the sequence$0,1,0,1,0,1,\\ldots$. That is,$x_n = 0$if$n$is odd, and$x_n = 1$if$n$is even. The sequence$\\{ x_n \\}_{n=1}^\\infty$is divergent; however, the subsequence$\\{ x_{2i} \\}_{i=1}^\\infty$converges to 1 and the subsequence$\\{ x_{2i+1} \\}_{i=1}^\\infty$converges to 0. Compare \\propref{seqconvsubseqconv:prop}." ], "refs": [ "seqconvsubseqconv:prop" ], "proofs": [], "ref_ids": [] }, { "id": 13, "type": "lemma", "label": "Lebl-contfunc:13", "categories": [ "convergence", "sequences", "auxiliary result" ], "title": "Squeeze Lemma", "contents": [ "\\index{squeeze lemma} \\label{squeeze:lemma} Let$\\{ a_n \\}_{n=1}^\\infty$,$\\{ b_n \\}_{n=1}^\\infty$, and$\\{ x_n \\}_{n=1}^\\infty$be sequences such that\\begin{equation*} a_n \\leq x_n \\leq b_n \\quad \\text{for all } n \\in \\N. \\end{equation*}Suppose$\\{ a_n \\}_{n=1}^\\infty$and$\\{ b_n \\}_{n=1}^\\infty$converge and\\begin{equation*} \\lim_{n\\to \\infty} a_n = \\lim_{n\\to \\infty} b_n . \\end{equation*}Then$\\{ x_n \\}_{n=1}^\\infty$converges and\\begin{equation*} \\lim_{n\\to \\infty} x_n = \\lim_{n\\to \\infty} a_n = \\lim_{n\\to \\infty} b_n . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let $x \\coloneqq \\lim_{n\\to\\infty} a_n = \\lim_{n\\to\\infty} b_n$.", "Let $\\epsilon > 0$ be given.", "Find an $M_1$ such that for all $n \\geq M_1$, we have", "that $\\abs{a_n-x} < \\epsilon$, and an $M_2$", "such that for all $n \\geq M_2$,", "we have $\\abs{b_n-x} < \\epsilon$. Set $M \\coloneqq \\max \\{M_1, M_2 \\}$.", "Suppose $n \\geq M$.", "In particular,", "$x - a_n < \\epsilon$, or", "$x - \\epsilon < a_n$. Similarly, $b_n < x + \\epsilon$.", "Putting everything together, we find", "\\begin{equation*}", "x - \\epsilon < a_n \\leq x_n \\leq b_n < x + \\epsilon .", "\\end{equation*}", "In other words, $-\\epsilon < x_n-x < \\epsilon$ or $\\abs{x_n-x} < \\epsilon$.", "So $\\{x_n\\}_{n=1}^\\infty$ converges to $x$.", "See \\figureref{figsqueeze}." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 14, "type": "example", "label": "Lebl-contfunc:14", "categories": [ "convergence", "sequences", "example" ], "title": "Squeeze Lemma", "contents": [ "One application of the \\hyperref[squeeze:lemma]{squeeze lemma} is to compute limits of sequences using limits that we already know. For example, consider the sequence$\\bigl\\{ \\frac{1}{n\\sqrt{n}} \\bigr\\}_{n=1}^\\infty$. Since$\\sqrt{n} \\geq 1$for all$n \\in \\N$, we have", "\\begin{equation*}\n0 \\leq \\frac{1}{n\\sqrt{n}} \\leq \\frac{1}{n}\n\\end{equation*}", "for all$n \\in \\N$. We already know$\\lim_{n\\to\\infty} \\nicefrac{1}{n} = 0$. Hence, using the constant sequence$\\{ 0 \\}_{n=1}^\\infty$and the sequence$\\{ \\nicefrac{1}{n} \\}_{n=1}^\\infty$in the squeeze lemma, we conclude", "\\begin{equation*}\n\\lim_{n\\to\\infty} \\frac{1}{n\\sqrt{n}} = 0 .\n\\end{equation*}" ], "refs": [ "squeeze:lemma" ], "proofs": [], "ref_ids": [] }, { "id": 15, "type": "lemma", "label": "Lebl-contfunc:limandineq:lemma", "categories": [ "convergence", "sequences", "auxiliary result" ], "title": "Auxiliary Result for Sequences", "contents": [ "Let$\\{ x_n \\}_{n=1}^\\infty$and$\\{ y_n \\}_{n=1}^\\infty$be convergent sequences and\\begin{equation*} x_n \\leq y_n \\quad \\text{for all } n \\in \\N . \\end{equation*}Then\\begin{equation*} \\lim_{n\\to\\infty} x_n \\leq \\lim_{n\\to\\infty} y_n . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let $x \\coloneqq \\lim_{n\\to\\infty} x_n$ and $y \\coloneqq \\lim_{n\\to\\infty} y_n$.", "Let", "$\\epsilon > 0$ be given. Find an $M_1$ such that for all $n \\geq M_1$,", "we have $\\abs{x_n-x} < \\nicefrac{\\epsilon}{2}$. Find an $M_2$ such that", "for all $n \\geq M_2$, we have", "$\\abs{y_n-y} < \\nicefrac{\\epsilon}{2}$. In particular,", "for some $n \\geq \\max\\{ M_1, M_2 \\}$, we have", "$x-x_n < \\nicefrac{\\epsilon}{2}$ and", "$y_n-y < \\nicefrac{\\epsilon}{2}$. We add these inequalities to", "obtain", "\\begin{equation*}", "y_n-x_n+x-y < \\epsilon, \\qquad \\text{or} \\qquad", "y_n-x_n < y-x+ \\epsilon .", "\\end{equation*}", "Since $x_n \\leq y_n$, we have", "$0 \\leq y_n-x_n$ and hence $0 < y-x+ \\epsilon$.", "In other words,", "\\begin{equation*}", "x-y < \\epsilon .", "\\end{equation*}", "Because $\\epsilon > 0$ was arbitrary, we obtain", "$x-y \\leq 0$.", "Therefore, $x \\leq y$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 16, "type": "corollary", "label": "Lebl-contfunc:limandineq:cor", "categories": [ "consequence", "convergence", "sequences" ], "title": "\\leavevmode \\begin{enumerate}[(i)] \\item If$\\{ x_n \\}_{n=1}^\\infty$is a convergent sequence such ...", "contents": [ "\\leavevmode \\begin{enumerate}[(i)] \\item If$\\{ x_n \\}_{n=1}^\\infty$is a convergent sequence such that$x_n \\geq 0$for all$n \\in \\N$, then\\begin{equation*} \\lim_{n\\to\\infty} x_n \\geq 0. \\end{equation*}\\item Let$a,b \\in \\R$and let$\\{ x_n \\}_{n=1}^\\infty$be a convergent sequence such that\\begin{equation*} a \\leq x_n \\leq b \\quad \\text{for all } n \\in \\N . \\end{equation*}Then\\begin{equation*} a \\leq \\lim_{n\\to\\infty} x_n \\leq b. \\end{equation*}\\end{enumerate}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 17, "type": "proposition", "label": "Lebl-contfunc:prop:contalg", "categories": [ "convergence", "sequences" ], "title": "Convergence of Sequences", "contents": [ "Let$\\{ x_n \\}_{n=1}^\\infty$and$\\{ y_n \\}_{n=1}^\\infty$be convergent sequences. \\begin{enumerate}[(i)] \\item \\label{prop:contalg:i} The sequence$\\{ z_n \\}_{n=1}^\\infty$, where$z_n \\coloneqq x_n + y_n$, converges and\\begin{equation*} \\lim_{n \\to \\infty} (x_n + y_n) = \\lim_{n \\to \\infty} z_n = \\lim_{n \\to \\infty} x_n + \\lim_{n \\to \\infty} y_n . \\end{equation*}\\item \\label{prop:contalg:ii} The sequence$\\{ z_n \\}_{n=1}^\\infty$, where$z_n \\coloneqq x_n - y_n$, converges and\\begin{equation*} \\lim_{n \\to \\infty} (x_n - y_n) = \\lim_{n \\to \\infty} z_n = \\lim_{n \\to \\infty} x_n - \\lim_{n \\to \\infty} y_n . \\end{equation*}\\item \\label{prop:contalg:iii} The sequence$\\{ z_n \\}_{n=1}^\\infty$, where$z_n \\coloneqq x_n y_n$, converges and\\begin{equation*} \\lim_{n \\to \\infty} (x_n y_n) = \\lim_{n \\to \\infty} z_n = \\left( \\lim_{n \\to \\infty} x_n \\right) \\left( \\lim_{n \\to \\infty} y_n \\right) . \\end{equation*}\\item \\label{prop:contalg:iv} If$\\lim_{n\\to\\infty} y_n \\not= 0$and$y_n \\not= 0$for all$n \\in \\N$, then the sequence$\\{ z_n \\}_{n=1}^\\infty$, where$z_n \\coloneqq \\dfrac{x_n}{y_n}$, converges and\\begin{equation*} \\lim_{n \\to \\infty} \\frac{x_n}{y_n} = \\lim_{n \\to \\infty} z_n = \\frac{\\lim_{n \\to \\infty} x_n}{\\lim_{n \\to \\infty} y_n} . \\end{equation*}\\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "We start with \\ref{prop:contalg:i}.", "Suppose $\\{ x_n \\}_{n=1}^\\infty$ and $\\{ y_n \\}_{n=1}^\\infty$ are convergent sequences and", "write $z_n \\coloneqq x_n + y_n$. Let $x \\coloneqq \\lim_{n\\to\\infty} x_n$,", "$y \\coloneqq \\lim_{n\\to\\infty} y_n$, and $z \\coloneqq x+y$.", "Let $\\epsilon > 0$ be given.", "Find an $M_1$ such that for all $n \\geq M_1$,", "we have", "$\\abs{x_n - x} < \\nicefrac{\\epsilon}{2}$.", "Find an $M_2$ such that for all $n \\geq M_2$,", "we have", "$\\abs{y_n - y} < \\nicefrac{\\epsilon}{2}$. Take $M \\coloneqq \\max \\{ M_1, M_2 \\}$.", "For all $n \\geq M$, we have", "\\begin{equation*}", "\\begin{split}", "\\abs{z_n - z} &=", "\\abs{(x_n+y_n) - (x+y)} \\\\", "& =", "\\abs{x_n-x + y_n-y} \\\\", "& \\leq", "\\abs{x_n-x} + \\abs{y_n-y} \\\\", "& <", "\\frac{\\epsilon}{2} +", "\\frac{\\epsilon}{2}", "= \\epsilon.", "\\end{split}", "\\end{equation*}", "Therefore \\ref{prop:contalg:i} is proved.", "Proof of \\ref{prop:contalg:ii} is almost identical and is left as an", "exercise.", "Let us tackle", "\\ref{prop:contalg:iii}.", "Suppose again that $\\{ x_n \\}_{n=1}^\\infty$ and $\\{ y_n \\}_{n=1}^\\infty$ are convergent sequences and", "write $z_n \\coloneqq x_n y_n$. Let $x \\coloneqq \\lim_{n\\to\\infty} x_n$,", "$y \\coloneqq \\lim_{n\\to\\infty} y_n$, and $z \\coloneqq xy$.", "Let $\\epsilon > 0$ be given.", "Let $K \\coloneqq \\max\\{ \\abs{x}, \\abs{y}, \\nicefrac{\\epsilon}{3} , 1 \\}$.", "Find an $M_1$ such that for all $n \\geq M_1$,", "we have", "$\\abs{x_n - x} < \\frac{\\epsilon}{3K}$.", "Find an $M_2$ such that for all $n \\geq M_2$,", "we have", "$\\abs{y_n - y} < \\frac{\\epsilon}{3K}$. Take $M \\coloneqq \\max \\{ M_1, M_2 \\}$.", "For all $n \\geq M$, we have", "\\begin{equation*}", "\\begin{split}", "\\abs{z_n - z} &=", "\\abs{(x_ny_n) - (xy)} \\\\", "& =", "\\abs{(x_n-x+x)(y_n-y+y) - xy} \\\\", "& =", "\\abs{(x_n-x)y + x(y_n-y) +(x_n-x)(y_n-y)} \\\\", "& \\leq", "\\abs{(x_n-x)y} + \\abs{x(y_n - y)} +", "\\abs{(x_n-x)(y_n-y)} \\\\", "& =", "\\abs{x_n -x}\\abs{y} +", "\\abs{x}\\abs{y_n -y} +", "\\abs{x_n -x}\\abs{y_n -y}", "\\\\", "& <", "\\frac{\\epsilon}{3K} K +", "K \\frac{\\epsilon}{3K} +", "\\frac{\\epsilon}{3K}", "\\frac{\\epsilon}{3K}", "\\qquad \\qquad \\text{(now notice that } \\tfrac{\\epsilon}{3K} \\leq 1", "\\text{ and }", "K \\geq 1\\text{)}", "\\\\", "& \\leq", "\\frac{\\epsilon}{3} + \\frac{\\epsilon}{3} + \\frac{\\epsilon}{3}", "= \\epsilon .", "\\end{split}", "\\end{equation*}", "Finally, we examine", "\\ref{prop:contalg:iv}. Instead of proving", "\\ref{prop:contalg:iv} directly, we prove the following simpler claim:", "Claim: \\emph{If $\\{ y_n \\}_{n=1}^\\infty$ is a convergent sequence such that", "$\\lim_{n\\to\\infty} y_n \\not= 0$ and $y_n \\not= 0$ for all $n \\in \\N$, then", "$\\{ \\nicefrac{1}{y_n} \\}_{n=1}^\\infty$ converges and}", "\\begin{equation*}", "\\lim_{n\\to\\infty} \\frac{1}{y_n} = \\frac{1}{\\lim_{n\\to\\infty} y_n} .", "\\end{equation*}", "Once the claim is proved, we take the sequence", "$\\{ \\nicefrac{1}{y_n} \\}_{n=1}^\\infty$,", "multiply it by the sequence $\\{ x_n \\}_{n=1}^\\infty$ and apply item", "\\ref{prop:contalg:iii}.", "Proof of claim: Let $\\epsilon > 0$ be given.", "Let $y \\coloneqq \\lim_{n\\to\\infty} y_n$.", "As $\\abs{y} \\not= 0$, then", "$\\min \\left\\{ \\abs{y}^2\\frac{\\epsilon}{2}, \\, \\frac{\\abs{y}}{2} \\right\\} > 0$.", "Find an $M$ such that for all $n \\geq M$,", "we have", "\\begin{equation*}", "\\abs{y_n - y} < \\min \\left\\{ \\abs{y}^2\\frac{\\epsilon}{2}, \\, \\frac{\\abs{y}}{2}", "\\right\\} .", "\\end{equation*}", "For all $n \\geq M$, we have", "$\\abs{y - y_n} < \\nicefrac{\\abs{y}}{2}$, and so", "\\begin{equation*}", "\\abs{y} =", "\\abs{y - y_n + y_n } \\leq", "\\abs{y - y_n} + \\abs{ y_n } < \\frac{\\abs{y}}{2} + \\abs{y_n}.", "\\end{equation*}", "Subtracting $\\nicefrac{\\abs{y}}{2}$ from both sides we obtain", "$\\nicefrac{\\abs{y}}{2} < \\abs{y_n}$, or in other words,", "\\begin{equation*}", "\\frac{1}{\\abs{y_n}} < \\frac{2}{\\abs{y}} .", "\\end{equation*}", "We finish the proof of the claim:", "\\begin{equation*}", "\\begin{split}", "\\abs{\\frac{1}{y_n} - \\frac{1}{y}} &=", "\\abs{\\frac{y - y_n}{y y_n}} \\\\", "& =", "\\frac{\\abs{y - y_n}}{\\abs{y} \\abs{y_n}} \\\\", "& \\leq", "\\frac{\\abs{y - y_n}}{\\abs{y}} \\, \\frac{2}{\\abs{y}} \\\\", "& <", "\\frac{\\abs{y}^2 \\frac{\\epsilon}{2}}{\\abs{y}} \\, \\frac{2}{\\abs{y}}", "= \\epsilon .", "\\end{split}", "\\end{equation*}", "And we are done." ], "refs": [ "prop:contalg:i", "prop:contalg:ii", "prop:contalg:iii", "prop:contalg:iv" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 18, "type": "proposition", "label": "Lebl-contfunc:18", "categories": [ "convergence", "sequences" ], "title": "\\pagebreak[2] Let$\\{ x_n \\}_{n=1}^\\infty$be a convergent sequence such that$x_n \\geq 0$for all$n ...", "contents": [ "\\pagebreak[2] Let$\\{ x_n \\}_{n=1}^\\infty$be a convergent sequence such that$x_n \\geq 0$for all$n \\in \\N$. Then\\begin{equation*} \\lim_{n\\to\\infty} \\sqrt{x_n} = \\sqrt{ \\lim_{n\\to\\infty} x_n } . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let $\\{ x_n \\}_{n=1}^\\infty$ be a convergent sequence and let $x \\coloneqq", "\\lim_{n\\to\\infty} x_n$.", "As we just mentioned, $x \\geq 0$.", "First suppose $x=0$. Let $\\epsilon > 0$ be given.", "Then there is an $M$ such that for all $n \\geq M$, we have", "$x_n = \\abs{x_n} < \\epsilon^2$, or in other words, $\\sqrt{x_n} < \\epsilon$.", "Hence,", "\\begin{equation*}", "\\abs{\\sqrt{x_n} - \\sqrt{x}} =", "\\sqrt{x_n} < \\epsilon.", "\\end{equation*}", "Now suppose $x > 0$ (and hence $\\sqrt{x} > 0$).", "\\begin{equation*}", "\\begin{split}", "\\abs{\\sqrt{x_n}-\\sqrt{x}} &=", "\\abs{\\frac{x_n-x}{\\sqrt{x_n}+\\sqrt{x}}} \\\\", "&=", "\\frac{1}{\\sqrt{x_n}+\\sqrt{x}}", "\\abs{x_n-x} \\\\", "& \\leq", "\\frac{1}{\\sqrt{x}}", "\\abs{x_n-x} .", "\\end{split}", "\\end{equation*}", "We leave the rest of the proof to the reader." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 19, "type": "proposition", "label": "Lebl-contfunc:19", "categories": [ "convergence", "sequences" ], "title": "If$\\{ x_n \\}_{n=1}^\\infty$is a convergent sequence, then$\\{ \\abs{x_n} \\}_{n=1}^\\infty$is converge...", "contents": [ "If$\\{ x_n \\}_{n=1}^\\infty$is a convergent sequence, then$\\{ \\abs{x_n} \\}_{n=1}^\\infty$is convergent and\\begin{equation*} \\lim_{n\\to\\infty} \\abs{x_n} = \\abs{\\lim_{n\\to\\infty} x_n} . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "We simply note the reverse triangle inequality", "\\begin{equation*}", "\\big\\lvert \\abs{x_n} - \\abs{x} \\big\\rvert \\leq \\abs{x_n-x} .", "\\end{equation*}", "Hence if $\\abs{x_n -x}$ can be made arbitrarily small, so can", "$\\big\\lvert \\abs{x_n} - \\abs{x} \\big\\rvert$.", "Details are left to the reader." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 20, "type": "example", "label": "Lebl-contfunc:20", "categories": [ "convergence", "sequences", "boundedness", "example" ], "title": "Boundedness of Sequences", "contents": [ "Let$\\{ x_n \\}_{n=1}^\\infty$be defined by$x_1 \\coloneqq 2$and", "\\begin{equation*}\nx_{n+1} \\coloneqq x_n - \\frac{x_n^2-2}{2x_n} .\n\\end{equation*}", "We must first find out if this sequence is well-defined; we must show we never divide by zero. Then we must find out if the sequence converges. Only then can we attempt to find the limit.", "So let us prove that for all$n$$x_n$exists and$x_n > 0$(so the sequence is well-defined and bounded below). Let us show this by \\hyperref[induction:thm]{induction}. We know that$x_1 = 2 > 0$. For the induction step, suppose$x_n$exists and$x_n > 0$. Then", "\\begin{equation*}\nx_{n+1} = x_n - \\frac{x_n^2-2}{2x_n} =\n\\frac{2x_n^2 - x_n^2+2}{2x_n} =\n\\frac{x_n^2+2}{2x_n} .\n\\end{equation*}", "It is always true that$x_n^2+2 > 0$, and as$x_n > 0$, then$x_{n+1} = \\frac{x_n^2+2}{2x_n} > 0$.", "Next let us show that the sequence is monotone decreasing. If we show that$x_n^2-2 \\geq 0$for all$n$, then$x_{n+1} \\leq x_n$for all$n$. Obviously$x_1^2-2 = 4-2 = 2 > 0$. For an arbitrary$n$, we have", "\\begin{equation*}\nx_{n+1}^2-2 =\n{\\left( \\frac{x_n^2+2}{2x_n} \\right)}^2 - 2\n=\n\\frac{x_n^4+4x_n^2+4 - 8x_n^2}{4x_n^2}\n=\n\\frac{x_n^4-4x_n^2+4}{4x_n^2}\n=\n\\frac{{\\left( x_n^2-2 \\right)}^2}{4x_n^2} .\n\\end{equation*}", "Since squares are nonnegative,$x_{n+1}^2-2 \\geq 0$for all$n$. Therefore,$\\{ x_n \\}_{n=1}^\\infty$is monotone decreasing and bounded ($x_n > 0$for all$n$), and so the limit exists. It remains to find the limit.", "Write", "\\begin{equation*}\n2x_nx_{n+1} = x_n^2+2 .\n\\end{equation*}", "Since$\\{ x_{n+1} \\}_{n=1}^\\infty$is the 1-tail of$\\{ x_n \\}_{n=1}^\\infty$, it converges to the same limit. Let us define$x \\coloneqq \\lim_{n\\to\\infty} x_n$. Take the limit of both sides to obtain", "\\begin{equation*}\n2x^2 = x^2+2 ,\n\\end{equation*}", "or$x^2 = 2$. As$x_n > 0$for all$n$we get$x \\geq 0$, and therefore$x = \\sqrt{2}$." ], "refs": [ "induction:thm" ], "proofs": [], "ref_ids": [] }, { "id": 21, "type": "example", "label": "Lebl-contfunc:21", "categories": [ "convergence", "boundedness", "sequences", "example", "series" ], "title": "Boundedness of Sequences", "contents": [ "Suppose$x_1 \\coloneqq 1$and$x_{n+1} \\coloneqq x_n^2+x_n$. If we blindly assumed that the limit exists (call it$x$), then we would get the equation$x = x^2+x$, from which we might conclude$x=0$. However, it is not hard to show that$\\{ x_n \\}_{n=1}^\\infty$is unbounded and therefore does not converge.", "The thing to notice in this example is that the method still works, but it depends on the initial value$x_1$. If we set$x_1 \\coloneqq 0$, then the sequence converges and the limit really is 0. An entire branch of mathematics, called dynamics, deals precisely with these issues. See \\exerciseref{exercise:convergentinitialvalues}." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 22, "type": "proposition", "label": "Lebl-contfunc:convzero:prop", "categories": [ "convergence", "sequences" ], "title": "Convergence of Sequences", "contents": [ "Let$\\{ x_n \\}_{n=1}^\\infty$be a sequence. Suppose there is an$x \\in \\R$and a convergent sequence$\\{ a_n \\}_{n=1}^\\infty$such that\\begin{equation*} \\lim_{n\\to\\infty} a_n = 0 \\end{equation*}and\\begin{equation*} \\abs{x_n - x} \\leq a_n \\qquad \\text{for all $n \\in \\N$.} \\end{equation*}Then$\\{ x_n \\}_{n=1}^\\infty$converges and$\\lim\\limits_{n\\to\\infty} x_n = x$." ], "refs": [], "proofs": [ { "contents": [ "Let $\\epsilon > 0$ be given. Note that $a_n \\geq 0$", "for all $n$. Find an $M \\in \\N$ such that for", "all $n \\geq M$, we have", "$a_n = \\abs{a_n - 0} < \\epsilon$. Then, for all $n \\geq M$,", "we have", "\\begin{equation*}", "\\abs{x_n - x} \\leq a_n < \\epsilon . \\qedhere", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 23, "type": "proposition", "label": "Lebl-contfunc:23", "categories": [ "boundedness" ], "title": "Boundedness of Bounded Sequences", "contents": [ "Let$c > 0$. \\begin{enumerate}[(i)] \\item If$c < 1$, then\\begin{equation*} \\lim_{n\\to\\infty} c^n = 0. \\end{equation*}\\item If$c > 1$, then$\\{ c^n \\}_{n=1}^\\infty$is unbounded. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "First consider $c < 1$. As $c > 0$, then $c^n > 0$ for all $n \\in \\N$ by", "\\hyperref[induction:thm]{induction}. As $c < 1$, then $c^{n+1} < c^n$ for all $n$.", "So $\\{ c^n \\}_{n=1}^\\infty$ is", "a decreasing sequence that is bounded below. Hence, it is convergent.", "Let $x \\coloneqq \\lim_{n\\to\\infty} c^n$. The 1-tail $\\{ c^{n+1} \\}_{n=1}^\\infty$", "also converges to $x$. Taking the limit of both sides of $c^{n+1} = c \\cdot", "c^n$, we obtain $x = cx$, or $(1-c)x=0$. It follows that $x=0$ as $c \\not= 1$.", "Now consider $c > 1$.", "Let $B > 0$ be arbitrary.", "As $\\nicefrac{1}{c} < 1$, then $\\bigl\\{ {(\\nicefrac{1}{c})}^n \\bigr\\}_{n=1}^\\infty$", "converges to $0$.", "Hence for some large enough $n$, we get", "\\begin{equation*}", "\\frac{1}{c^n} =", "{\\left(\\frac{1}{c}\\right)}^n < \\frac{1}{B} .", "\\end{equation*}", "In other words, $c^n > B$, and $B$ is not an upper bound for", "$\\{ c^n \\}_{n=1}^\\infty$. As $B$ was arbitrary, $\\{ c^n \\}_{n=1}^\\infty$ is unbounded." ], "refs": [ "induction:thm" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 24, "type": "lemma", "label": "Lebl-contfunc:24", "categories": [ "convergence", "boundedness", "auxiliary result", "sequences", "tests", "divergence" ], "title": "Ratio Test", "contents": [ "\\index{ratio test for sequences} \\label{seq:ratiotest} Let$\\{ x_n \\}_{n=1}^\\infty$be a sequence such that$x_n \\not= 0$for all$n$and such that the limit\\begin{equation*} L \\coloneqq \\lim_{n\\to\\infty} \\frac{\\abs{x_{n+1}}}{\\abs{x_n}} \\qquad \\text{exists.} \\end{equation*}\\begin{enumerate}[(i)] \\item If$L < 1$, then$\\{ x_n \\}_{n=1}^\\infty$converges and$\\lim\\limits_{n\\to\\infty} x_n = 0$. \\item If$L > 1$, then$\\{ x_n \\}_{n=1}^\\infty$is unbounded (hence diverges). \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "Suppose $L < 1$.", "As", "$\\frac{\\abs{x_{n+1}}}{\\abs{x_n}} \\geq 0$ for all $n$, then $L \\geq 0$. Pick", "$r$ such that $L < r < 1$.", "We wish to compare the sequence $\\{ x_n \\}_{n=1}^\\infty$ to the sequence $\\{ r^n \\}_{n=1}^\\infty$. The idea is that", "while the ratio $\\frac{\\abs{x_{n+1}}}{\\abs{x_n}}$ is not going to be less than $L$ eventually,", "it will eventually be less than $r$, which is still less than 1.", "The intuitive idea of the proof is illustrated in \\figureref{figratseq}.", "\\begin{myfigureht}", "\\subimport*{figures/}{figratseq.pdf_t}", "\\caption{The short lines represent the ratios", "$\\frac{\\abs{x_{n+1}}}{\\abs{x_n}}$", "approaching $L < 1$.\\label{figratseq}}", "\\end{myfigureht}", "As $r-L > 0$, there exists an $M \\in \\N$ such that for", "all $n \\geq M$, we have", "\\begin{equation*}", "\\abs{\\frac{\\abs{x_{n+1}}}{\\abs{x_n}} - L} < r-L .", "\\end{equation*}", "Therefore, for $n \\geq M$,", "\\begin{equation*}", "\\frac{\\abs{x_{n+1}}}{\\abs{x_n}} - L < r-L", "\\qquad \\text{or} \\qquad", "\\frac{\\abs{x_{n+1}}}{\\abs{x_n}} < r .", "\\end{equation*}", "For $n > M$ (that is for $n \\geq M+1$)", "write", "\\begin{equation*}", "\\abs{x_n} =", "\\abs{x_M}", "\\frac{\\abs{x_{M+1}}}{\\abs{x_{M}}}", "\\frac{\\abs{x_{M+2}}}{\\abs{x_{M+1}}}", "\\cdots", "\\frac{\\abs{x_{n}}}{\\abs{x_{n-1}}}", "<", "\\abs{x_M}", "r r \\cdots r = \\abs{x_M} r^{n-M} = (\\abs{x_M} r^{-M}) r^n .", "\\end{equation*}", "The sequence $\\{ r^n \\}_{n=1}^\\infty$ converges to zero and hence", "$\\abs{x_M} r^{-M} r^n$ converges to zero. By \\propref{convzero:prop},", "the $M$-tail", "$\\{x_n\\}_{n=M+1}^\\infty$ converges to zero and therefore $\\{x_n\\}_{n=1}^\\infty$ converges to zero.", "Now suppose $L > 1$. Pick", "$r$ such that $1 < r < L$. As $L-r > 0$,", "there exists an $M \\in \\N$ such that for", "all $n \\geq M$", "\\begin{equation*}", "\\abs{\\frac{\\abs{x_{n+1}}}{\\abs{x_n}} - L} < L-r .", "\\end{equation*}", "Therefore,", "\\begin{equation*}", "\\frac{\\abs{x_{n+1}}}{\\abs{x_n}} > r .", "\\end{equation*}", "Again for $n > M$,", "write", "\\begin{equation*}", "\\abs{x_n} =", "\\abs{x_M}", "\\frac{\\abs{x_{M+1}}}{\\abs{x_{M}}}", "\\frac{\\abs{x_{M+2}}}{\\abs{x_{M+1}}}", "\\cdots", "\\frac{\\abs{x_{n}}}{\\abs{x_{n-1}}}", ">", "\\abs{x_M}", "r r \\cdots r = \\abs{x_M} r^{n-M} = (\\abs{x_M} r^{-M}) r^n .", "\\end{equation*}", "The sequence $\\{ r^n \\}_{n=1}^\\infty$ is unbounded (since $r > 1$), and so", "$\\{x_n\\}_{n=1}^\\infty$ cannot be bounded (if $\\abs{x_n} \\leq B$ for all $n$, then", "$r^n < \\frac{B}{\\abs{x_M}} r^{\\:M}$ for all $n > M$, which is impossible).", "Consequently, $\\{ x_n \\}_{n=1}^\\infty$ cannot converge." ], "refs": [ "convzero:prop" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 25, "type": "example", "label": "Lebl-contfunc:25", "categories": [ "convergence", "example" ], "title": "Convergence of Convergence", "contents": [ "A simple application of the lemma above is to prove", "\\begin{equation*}\n\\lim_{n\\to\\infty} \\frac{2^n}{n!} = 0 .\n\\end{equation*}", "Proof: Compute", "\\begin{equation*}\n\\frac{2^{n+1} / (n+1)!}{2^n/n!}\n=\n\\frac{2^{n+1}}{2^n}\\frac{n!}{(n+1)!}\n=\n\\frac{2}{n+1} .\n\\end{equation*}", "It is not hard to see that$\\bigl\\{ \\frac{2}{n+1} \\bigr\\}_{n=1}^\\infty$converges to zero. The conclusion follows by the lemma." ], "refs": [], "proofs": [ { "contents": [ "Compute \\begin{equation*}", "\\frac{2^{n+1} / (n+1)!}{2^n/n!}", "=", "\\frac{2^{n+1}}{2^n}\\frac{n!}{(n+1)!}", "=", "\\frac{2}{n+1} .", "\\end{equation*} It is not hard to see that$\\bigl\\{ \\frac{2}{n+1} \\bigr\\}_{n=1}^\\infty$converges to zero. The conclusion follows by the lemma." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 26, "type": "example", "label": "Lebl-contfunc:example:nto1overn", "categories": [ "convergence", "tests", "sequences", "example" ], "title": "Ratio Test", "contents": [ "A more complicated (and useful) application of the ratio test is to prove", "\\begin{equation*}\n\\lim_{n\\to\\infty} n^{1/n} = 1 .\n\\end{equation*}", "Proof: Let$\\epsilon > 0$be given. Consider the sequence$\\bigl\\{ \\frac{n}{{(1+\\epsilon)}^n} \\bigr\\}_{n=1}^\\infty$. Compute", "\\begin{equation*}\n\\frac{(n+1)/{(1+\\epsilon)}^{n+1}}{n/{(1+\\epsilon)}^{n}}\n=\n\\frac{n+1}{n} \\frac{1}{1+\\epsilon} .\n\\end{equation*}", "The limit of$\\frac{n+1}{n} = 1+\\frac{1}{n}$as$n \\to \\infty$is 1, and so", "\\begin{equation*}\n\\lim_{n\\to \\infty} \\frac{(n+1)/{(1+\\epsilon)}^{n+1}}{n/{(1+\\epsilon)}^{n}}\n=\n\\frac{1}{1+\\epsilon} < 1 .\n\\end{equation*}", "Therefore,$\\bigl\\{ \\frac{n}{{(1+\\epsilon)}^n} \\bigr\\}_{n=1}^\\infty$converges to 0. In particular, there exists an$M$such that for$n \\geq M$, we have$\\frac{n}{{(1+\\epsilon)}^n} < 1$, or$n < {(1+\\epsilon)}^n$, or$n^{1/n} < 1+\\epsilon$. As$n \\geq 1$, then$n^{1/n} \\geq 1$, and so$0 \\leq n^{1/n}-1 < \\epsilon$. Consequently,$\\lim\\limits_{n\\to\\infty} n^{1/n} = 1$." ], "refs": [], "proofs": [ { "contents": [ "Let$\\epsilon > 0$be given. Consider the sequence$\\bigl\\{ \\frac{n}{{(1+\\epsilon)}^n} \\bigr\\}_{n=1}^\\infty$. Compute \\begin{equation*}", "\\frac{(n+1)/{(1+\\epsilon)}^{n+1}}{n/{(1+\\epsilon)}^{n}}", "=", "\\frac{n+1}{n} \\frac{1}{1+\\epsilon} .", "\\end{equation*} The limit of$\\frac{n+1}{n} = 1+\\frac{1}{n}$as$n \\to \\infty$is 1, and so \\begin{equation*}", "\\lim_{n\\to \\infty} \\frac{(n+1)/{(1+\\epsilon)}^{n+1}}{n/{(1+\\epsilon)}^{n}}", "=", "\\frac{1}{1+\\epsilon} < 1 .", "\\end{equation*} Therefore,$\\bigl\\{ \\frac{n}{{(1+\\epsilon)}^n} \\bigr\\}_{n=1}^\\infty$converges to 0. In particular, there exists an$M$such that for$n \\geq M$, we have$\\frac{n}{{(1+\\epsilon)}^n} < 1$, or$n < {(1+\\epsilon)}^n$, or$n^{1/n} < 1+\\epsilon$. As$n \\geq 1$, then$n^{1/n} \\geq 1$, and so$0 \\leq n^{1/n}-1 < \\epsilon$. Consequently,$\\lim\\limits_{n\\to\\infty} n^{1/n} = 1$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 27, "type": "proposition", "label": "Lebl-contfunc:27", "categories": [ "convergence", "sequences", "boundedness" ], "title": "Boundedness of Sequences", "contents": [ "Let$\\{ x_n \\}_{n=1}^\\infty$be a bounded sequence. Let$a_n$and$b_n$be as in the definition above. \\begin{enumerate}[(i)] \\item The sequence$\\{ a_n \\}_{n=1}^\\infty$is bounded monotone decreasing and$\\{ b_n \\}_{n=1}^\\infty$is bounded monotone increasing. In particular,$\\liminf\\limits_{n\\to\\infty} x_n$and$\\limsup\\limits_{n\\to\\infty} x_n$exist. \\item$\\displaystyle \\limsup_{n \\to \\infty} x_n = \\inf \\{ a_n : n \\in \\N \\}$and$\\displaystyle \\liminf_{n \\to \\infty} x_n = \\sup \\{ b_n : n \\in \\N \\}$. \\item$\\displaystyle \\liminf_{n \\to \\infty} x_n \\leq \\limsup_{n \\to \\infty} x_n$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "Let us see why $\\{ a_n \\}_{n=1}^\\infty$ is a decreasing sequence. As $a_n$ is the least upper", "bound for $\\{ x_k : k \\geq n \\}$, it is also", "an upper bound for the subset $\\{ x_k : k \\geq n+1 \\}$. Therefore", "$a_{n+1}$, the least upper bound for", "$\\{ x_k : k \\geq n+1 \\}$, has to be less than or equal to $a_n$,", "the least upper bound for", "$\\{ x_k : k \\geq n) \\}$.", "That is,", "$a_n \\geq a_{n+1}$ for all $n$. Similarly (an exercise),", "$\\{ b_n \\}_{n=1}^\\infty$ is an increasing sequence.", "It is left as an exercise to show that", "if $\\{ x_n \\}_{n=1}^\\infty$ is bounded, then $\\{ a_n \\}_{n=1}^\\infty$ and", "$\\{ b_n \\}_{n=1}^\\infty$ must be bounded.", "The second item follows as the sequences", "$\\{ a_n \\}_{n=1}^\\infty$ and $\\{ b_n \\}_{n=1}^\\infty$ are monotone and bounded.", "For the third item, note that $b_n \\leq a_n$, as the $\\inf$ of a nonempty set", "is less than or equal to its $\\sup$. The sequences $\\{ a_n \\}_{n=1}^\\infty$", "and $\\{ b_n \\}_{n=1}^\\infty$", "converge to the limsup and the liminf respectively.", "Apply \\lemmaref{limandineq:lemma} to obtain", "\\begin{equation*}", "\\lim_{n\\to \\infty} b_n \\leq \\lim_{n\\to \\infty} a_n. \\qedhere", "\\end{equation*}" ], "refs": [ "limandineq:lemma" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 28, "type": "example", "label": "Lebl-contfunc:example:liminfsupex", "categories": [ "convergence", "sequences", "example" ], "title": "Limit of Sequences", "contents": [ "Let$\\{ x_n \\}_{n=1}^\\infty$be defined by", "\\begin{equation*}\nx_n \\coloneqq\n\\begin{cases}\n\\frac{n+1}{n} & \\text{if } n \\text{ is odd,} \\\\\n0 & \\text{if } n \\text{ is even.}\n\\end{cases}\n\\end{equation*}", "Let us compute the$\\liminf$and$\\limsup$of this sequence. See also \\figureref{sequence-limsupliminf_an_bn-example}. First the limit inferior:", "\\begin{equation*}\n\\liminf_{n\\to\\infty} x_n = \n\\lim_{n\\to\\infty}\n\\bigl(\n\\inf \\{ x_k : k \\geq n \\}\n\\bigr)\n=\n\\lim_{n\\to\\infty} 0 = 0 .\n\\end{equation*}", "For the limit superior, we write", "\\begin{equation*}\n\\limsup_{n\\to\\infty} x_n = \n\\lim_{n\\to\\infty}\n\\bigl(\n\\sup \\{ x_k : k \\geq n \\}\n\\bigr) .\n\\end{equation*}", "It is not hard to see that", "\\begin{equation*}\n\\sup \\{ x_k : k \\geq n \\} =\n\\begin{cases}\n\\frac{n+1}{n} & \\text{if } n \\text{ is odd,} \\\\\n\\frac{n+2}{n+1} & \\text{if } n \\text{ is even.}\n\\end{cases}\n\\end{equation*}", "We leave it to the reader to show that the limit is 1. That is,", "\\begin{equation*}\n\\limsup_{n\\to\\infty} x_n = 1 .\n\\end{equation*}", "Do note that the sequence$\\{ x_n \\}_{n=1}^\\infty$is not a convergent sequence. \\begin{myfigureht} \\includegraphics{figures/sequence-limsupliminf_an_bn-example} \\caption{First 20 terms of the sequence in \\exampleref{example:liminfsupex}. The marking is as in \\figureref{sequence-limsupliminf_an_bn}. \\label{sequence-limsupliminf_an_bn-example}} \\end{myfigureht}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 29, "type": "theorem", "label": "Lebl-contfunc:subseqlimsupinf:thm", "categories": [ "sequences", "boundedness" ], "title": "Boundedness of Sequences", "contents": [ "If$\\{ x_n \\}_{n=1}^\\infty$is a bounded sequence, then there exists a subsequence$\\{ x_{n_k} \\}_{k=1}^\\infty$such that\\begin{equation*} \\lim_{k\\to \\infty} x_{n_k} = \\limsup_{n \\to \\infty} x_n . \\end{equation*}Similarly, there exists a (perhaps different) subsequence$\\{ x_{m_k} \\}_{k=1}^\\infty$such that\\begin{equation*} \\lim_{k\\to \\infty} x_{m_k} = \\liminf_{n \\to \\infty} x_n . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Define $a_n \\coloneqq \\sup \\{ x_k : k \\geq n \\}$.", "Write", "$x \\coloneqq \\limsup_{n\\to\\infty} x_n = \\lim_{n\\to\\infty} a_n$.", "We define the subsequence inductively.", "Let $n_1 \\coloneqq 1$, and suppose $n_1,n_2,\\ldots,n_{k-1}$ are already defined for", "some $k \\geq 2$. Pick an $m \\geq n_{k-1} + 1$", "such that", "\\begin{equation*}", "a_{(n_{k-1}+1)} - x_m < \\frac{1}{k} .", "\\end{equation*}", "Such an $m$ exists", "as $a_{(n_{k-1}+1)}$ is a supremum of the", "set $\\{ x_\\ell : \\ell \\geq n_{k-1} + 1 \\}$ and hence there are elements", "of the sequence arbitrarily close (or even possibly equal) to the supremum.", "Set $n_{k} \\coloneqq m$. The subsequence $\\{ x_{n_k} \\}_{k=1}^\\infty$ is defined. Next, we", "must prove that it converges to $x$.", "For all $k \\geq 2$, we have", "$a_{(n_{k-1}+1)} \\geq a_{n_k}$ (why?) and $a_{n_{k}} \\geq x_{n_k}$.", "Therefore, for every $k \\geq 2$,", "\\begin{equation*}", "\\begin{split}", "\\abs{a_{n_k} - x_{n_k}} & =", "a_{n_k} - x_{n_k}", "\\\\", "& \\leq", "a_{(n_{k-1}+1)} - x_{n_k}", "\\\\", "& < \\frac{1}{k} .", "\\end{split}", "\\end{equation*}", "Let us show that $\\{ x_{n_k} \\}_{k=1}^\\infty$ converges to $x$.", "Note that the subsequence need not be monotone. Let $\\epsilon > 0$ be given.", "As $\\{ a_n \\}_{n=1}^\\infty$ converges to $x$, the subsequence", "$\\{ a_{n_k} \\}_{k=1}^\\infty$ converges to $x$.", "Thus there exists an $M_1 \\in \\N$", "such that for all $k \\geq M_1$, we have", "\\begin{equation*}", "\\abs{a_{n_k} - x} < \\frac{\\epsilon}{2} .", "\\end{equation*}", "Find an $M_2 \\in \\N$ such that", "\\begin{equation*}", "\\frac{1}{M_2} \\leq \\frac{\\epsilon}{2}.", "\\end{equation*}", "Take $M \\coloneqq \\max \\{M_1 , M_2 , 2 \\}$. For all $k \\geq M$,", "\\begin{equation*}", "\\begin{split}", "\\abs{x- x_{n_k}} & =", "\\abs{a_{n_k} - x_{n_k} + x - a_{n_k}}", "\\\\", "& \\leq \\abs{a_{n_k} - x_{n_k}} + \\abs{x - a_{n_k}}", "\\\\", "& < \\frac{1}{k} + \\frac{\\epsilon}{2}", "\\\\", "& \\leq \\frac{1}{M_2} + \\frac{\\epsilon}{2} \\leq \\frac{\\epsilon}{2} +", "\\frac{\\epsilon}{2} = \\epsilon .", "\\end{split}", "\\end{equation*}", "We leave the statement for $\\liminf$ as an exercise." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 30, "type": "proposition", "label": "Lebl-contfunc:liminfsupconv:prop", "categories": [ "characterization", "convergence", "sequences", "boundedness" ], "title": "Characterization of Sequences via Equivalence", "contents": [ "Let$\\{ x_n \\}_{n=1}^\\infty$be a bounded sequence. Then$\\{ x_n \\}_{n=1}^\\infty$converges if and only if\\begin{equation*} \\liminf_{n\\to \\infty} x_n = \\limsup_{n\\to \\infty} x_n. \\end{equation*}Furthermore, if$\\{ x_n \\}_{n=1}^\\infty$converges, then\\begin{equation*} \\lim_{n\\to \\infty} x_n = \\liminf_{n\\to \\infty} x_n = \\limsup_{n\\to \\infty} x_n. \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let $a_n$ and $b_n$ be as in \\defnref{liminflimsup:def}.", "In particular, for all $n \\in \\N$,", "\\begin{equation*}", "b_n \\leq x_n \\leq a_n .", "\\end{equation*}", "First suppose", "$\\liminf_{n\\to\\infty} x_n = \\limsup_{n\\to\\infty} x_n$.", "Then $\\{ a_n \\}_{n=1}^\\infty$ and $\\{ b_n \\}_{n=1}^\\infty$", "both converge to the same limit.", "By the squeeze lemma", "(\\lemmaref{squeeze:lemma}), $\\{ x_n \\}_{n=1}^\\infty$ converges and", "\\begin{equation*}", "\\lim_{n\\to \\infty} b_n", "=", "\\lim_{n\\to \\infty} x_n", "=", "\\lim_{n\\to \\infty} a_n .", "\\end{equation*}", "Now suppose $\\{ x_n \\}_{n=1}^\\infty$ converges to $x$.", "By \\thmref{subseqlimsupinf:thm},", "there exists a subsequence $\\{ x_{n_k} \\}_{k=1}^\\infty$", "converging to $\\limsup_{n\\to\\infty} x_n$.", "As $\\{ x_n \\}_{n=1}^\\infty$ converges to $x$,", "every subsequence converges to $x$ and", "so $\\limsup_{n\\to\\infty} x_n = \\lim_{k\\to\\infty} x_{n_k} = x$.", "Similarly, $\\liminf_{n\\to\\infty} x_n = x$." ], "refs": [ "squeeze:lemma" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 31, "type": "proposition", "label": "Lebl-contfunc:prop:subseqslimsupinf", "categories": [ "sequences", "boundedness" ], "title": "Boundedness of Sequences", "contents": [ "Suppose$\\{ x_n \\}_{n=1}^\\infty$is a bounded sequence and$\\{ x_{n_k} \\}_{k=1}^\\infty$is a subsequence. Then\\begin{equation*} \\liminf_{n\\to\\infty} x_n \\leq \\liminf_{k\\to\\infty} x_{n_k} \\leq \\limsup_{k\\to\\infty} x_{n_k} \\leq \\limsup_{n\\to\\infty} x_n . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "The middle inequality has been proved already. We will prove the third", "inequality, and leave the first inequality as an exercise.", "We want to prove that", "$\\limsup_{k\\to\\infty} x_{n_k} \\leq \\limsup_{n\\to\\infty} x_n$. Define", "$a_n \\coloneqq \\sup \\{ x_k : k \\geq n \\}$", "as usual.", "Also define", "$c_n \\coloneqq \\sup \\{ x_{n_k} : k \\geq n \\}$.", "It is not true that $\\{ c_n \\}_{n=1}^\\infty$ is necessarily a subsequence", "of $\\{ a_n \\}_{n=1}^\\infty$.", "However, as $n_k \\geq k$ for all $k$, we have", "$\\{ x_{n_k} : k \\geq n \\} \\subset \\{ x_k : k \\geq n \\}$.", "A supremum of a subset is less than or equal to the supremum of the", "set, and therefore", "\\begin{equation*}", "c_n \\leq a_n \\qquad \\text{for all $n$}.", "\\end{equation*}", "\\lemmaref{limandineq:lemma} gives", "\\begin{equation*}", "\\lim_{n\\to\\infty} c_n \\leq \\lim_{n\\to\\infty} a_n ,", "\\end{equation*}", "which is the desired conclusion." ], "refs": [ "limandineq:lemma" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 32, "type": "proposition", "label": "Lebl-contfunc:seqconvsubseqconv:prop", "categories": [ "characterization", "convergence", "sequences", "boundedness" ], "title": "Characterization of Sequences via Equivalence", "contents": [ "A bounded sequence$\\{ x_n \\}_{n=1}^\\infty$is convergent and converges to$x$if and only if every convergent subsequence$\\{ x_{n_k} \\}_{k=1}^\\infty$converges to$x$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 33, "type": "theorem", "label": "Lebl-contfunc:33", "categories": [ "convergence", "sequences", "boundedness" ], "title": "Boundedness of Sequences", "contents": [ "\\index{Bolzano--Weierstrass theorem}\\label{thm:bwseq} Suppose a sequence$\\{ x_n \\}_{n=1}^\\infty$of real numbers is bounded. Then there exists a convergent subsequence$\\{ x_{n_i} \\}_{i=1}^\\infty$." ], "refs": [], "proofs": [ { "contents": [ "\\thmref{subseqlimsupinf:thm} says that there exists", "a subsequence whose limit is $\\limsup_{n\\to\\infty} x_n$." ], "refs": [], "ref_ids": [] }, { "contents": [ "[Alternate proof of Bolzano--Weierstrass]", "As the sequence is bounded, then there exist two numbers $a_1 < b_1$", "such that $a_1 \\leq x_n \\leq b_1$ for all $n \\in \\N$.", "We will define a subsequence $\\{ x_{n_i} \\}_{i=1}^\\infty$ and two", "sequences $\\{ a_i \\}_{i=1}^\\infty$ and $\\{ b_i \\}_{i=1}^\\infty$, such that", "$\\{ a_i \\}_{i=1}^\\infty$ is monotone increasing, $\\{ b_i \\}_{i=1}^\\infty$ is monotone decreasing,", "$a_i \\leq x_{n_i} \\leq b_i$ and such that $\\lim_{i\\to\\infty} a_i =", "\\lim_{i\\to\\infty} b_i$. That", "$x_{n_i}$ converges then follows by the \\hyperref[squeeze:lemma]{squeeze lemma}.", "We define the sequences inductively. We will define the sequences so that", "for all $i$, we have $a_i < b_i$,", "and that $x_n \\in [a_i,b_i]$ for infinitely many $n \\in \\N$.", "We have already defined $a_1$ and $b_1$. We take $n_1 \\coloneqq 1$, that is", "$x_{n_1} = x_1$.", "Suppose that up to some $k \\in \\N$,", "we have defined the subsequence $x_{n_1}, x_{n_2}, \\ldots,", "x_{n_k}$, and the sequences $a_1,a_2,\\ldots,a_k$", "and $b_1,b_2,\\ldots,b_k$.", "Let $y \\coloneqq \\frac{a_k+b_k}{2}$.", "Clearly", "$a_k < y < b_k$. If there exist infinitely many $j \\in \\N$", "such that $x_j \\in [a_k,y]$, then set $a_{k+1} \\coloneqq a_k$, $b_{k+1}", "\\coloneqq y$,", "and pick $n_{k+1} > n_{k}$", "such that $x_{n_{k+1}} \\in [a_k,y]$. If there are not infinitely many", "$j$ such that", "$x_j \\in [a_k,y]$, then it must be true that there are infinitely many $j \\in", "\\N$ such that", "$x_j \\in [y,b_k]$. In this case pick $a_{k+1} \\coloneqq y$, $b_{k+1}", "\\coloneqq b_k$,", "and pick $n_{k+1} > n_{k}$", "such that $x_{n_{k+1}} \\in [y,b_k]$.", "We now have the sequences defined. What is left to prove is that", "$\\lim_{i\\to\\infty} a_i = \\lim_{i\\to\\infty} b_i$. The limits exist as the sequences", "are monotone. In the construction,", "$b_i - a_i$ is cut in half in each step. Therefore,", "$b_{i+1} - a_{i+1} = \\frac{b_i-a_i}{2}$. By", "\\hyperref[induction:thm]{induction},", "\\begin{equation*}", "b_i - a_i = \\frac{b_1-a_1}{2^{i-1}} .", "\\end{equation*}", "Let $x \\coloneqq \\lim_{i\\to\\infty} a_i$. As $\\{ a_i \\}_{i=1}^\\infty$ is monotone,", "\\begin{equation*}", "x = \\sup \\{ a_i : i \\in \\N \\} .", "\\end{equation*}", "Let $y \\coloneqq \\lim_{i\\to\\infty} b_i = \\inf \\{ b_i : i \\in \\N \\}$. Since $a_i < b_i$ for", "all $i$, then $x \\leq y$.", "As the sequences are monotone, then", "for all $i$, we have (why?)", "\\begin{equation*}", "y-x \\leq b_i-a_i = \\frac{b_1-a_1}{2^{i-1}} .", "\\end{equation*}", "Because $\\frac{b_1-a_1}{2^{i-1}}$ is arbitrarily small and $y-x \\geq 0$,", "we have $y-x = 0$. Finish by the \\hyperref[squeeze:lemma]{squeeze lemma}." ], "refs": [ "induction:thm", "squeeze:lemma" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 34, "type": "proposition", "label": "Lebl-contfunc:prop:unboundedmonotone", "categories": [ "sequences", "boundedness" ], "title": "Boundedness of Sequences", "contents": [ "Suppose$\\{ x_n \\}_{n=1}^\\infty$is a monotone unbounded sequence. Then \\begin{equation*} \\lim_{n \\to \\infty} x_n = \\begin{cases} \\infty & \\text{if } \\{ x_n \\}_{n=1}^\\infty \\text{ is increasing,}", "-\\infty & \\text{if } \\{ x_n \\}_{n=1}^\\infty \\text{ is decreasing.} \\end{cases} \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "The case of monotone increasing follows from", "\\exerciseref{exercise:infseqlimlims} part c) below.", "Suppose $\\{x_n\\}_{n=1}^\\infty$ is decreasing and unbounded.", "That the sequence is unbounded means that", "for every $K \\in \\R$, there is an $M \\in \\N$ such that $x_M < K$.", "By monotonicity, $x_n \\leq x_M < K$ for all $n \\geq M$.", "Therefore, $\\lim_{n\\to\\infty} x_n = -\\infty$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 35, "type": "example", "label": "Lebl-contfunc:35", "categories": [ "example" ], "title": "\\begin{equation*} \\lim_{n\\to \\infty} n = \\infty, \\qquad \\lim_{n\\to \\infty} n^2 = \\infty, \\qquad \\...", "contents": [ "\\begin{equation*}\n\\lim_{n\\to \\infty} n = \\infty,\n\\qquad\n\\lim_{n\\to \\infty} n^2 = \\infty,\n\\qquad\n\\lim_{n\\to \\infty} -n = -\\infty.\n\\end{equation*}", "We leave verification to the reader." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 36, "type": "proposition", "label": "Lebl-contfunc:36", "categories": [ "sequences", "boundedness" ], "title": "Boundedness of Sequences", "contents": [ "Let$\\{ x_n \\}_{n=1}^\\infty$be an unbounded sequence. Define$\\{ a_n \\}_{n=1}^\\infty$and$\\{ b_n \\}_{n=1}^\\infty$as above. Then$\\{ a_n \\}_{n=1}^\\infty$is decreasing, and$\\{ b_n \\}_{n=1}^\\infty$is increasing. If$a_n$is a real number for every$n$, then$\\limsup_{n\\to\\infty} x_n = \\lim_{n\\to\\infty} a_n$. If$b_n$is a real number for every$n$, then$\\liminf_{n\\to\\infty} x_n = \\lim_{n\\to\\infty} b_n$." ], "refs": [], "proofs": [ { "contents": [ "As before,", "$a_n = \\sup \\{ x_k : k \\geq n \\} \\geq \\sup \\{ x_k : k \\geq n+1 \\} =", "a_{n+1}$. So $\\{ a_n \\}_{n=1}^\\infty$ is decreasing. Similarly,", "$\\{ b_n \\}_{n=1}^\\infty$ is increasing.", "If the sequence $\\{ a_n \\}_{n=1}^\\infty$ is a sequence of real numbers, then", "$\\lim_{n\\to\\infty} a_n = \\inf \\{ a_n : n \\in \\N \\}$. This follows from", "\\thmref{thm:monotoneconv} if $\\{ a_n \\}_{n=1}^\\infty$ is bounded and", "\\propref{prop:unboundedmonotone} if $\\{a_n \\}_{n=1}^\\infty$", "is unbounded. We proceed similarly with $\\{ b_n \\}_{n=1}^\\infty$." ], "refs": [ "prop:unboundedmonotone" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 37, "type": "example", "label": "Lebl-contfunc:37", "categories": [ "example" ], "title": "Existence of Limit Superior", "contents": [ "Suppose$x_n \\coloneqq 0$for odd$n$and$x_n \\coloneqq n$for even$n$. Then$a_n = \\infty$for all$n$, since for every$M$, there exists an even$k$such that$x_k = k \\geq M$. On the other hand,$b_n = 0$for all$n$, as for every$n$, the set$\\{ b_k : k \\geq n \\}$consists of$0$and positive numbers. So,", "\\begin{equation*}\n\\lim_{n\\to \\infty} x_n \\quad \\text{does not exist},\n\\qquad\n\\limsup_{n\\to \\infty} x_n = \\infty ,\n\\qquad\n\\liminf_{n\\to \\infty} x_n = 0.\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 38, "type": "example", "label": "Lebl-contfunc:38", "categories": [ "named theorem", "sequences", "cauchy", "example" ], "title": "Cauchy Convergence Criterion", "contents": [ "The sequence$\\{ \\nicefrac{1}{n} \\}_{n=1}^\\infty$is a Cauchy sequence.", "Proof: Given$\\epsilon > 0$, find$M$such that$M > \\nicefrac{2}{\\epsilon}$. Then for$n,k \\geq M$, we have$\\nicefrac{1}{n} < \\nicefrac{\\epsilon}{2}$and$\\nicefrac{1}{k} < \\nicefrac{\\epsilon}{2}$. Therefore, for$n, k \\geq M$, we have", "\\begin{equation*}\n\\abs{\\frac{1}{n} - \\frac{1}{k}}\n\\leq\n\\abs{\\frac{1}{n}} + \\abs{\\frac{1}{k}}\n< \\frac{\\epsilon}{2} + \\frac{\\epsilon}{2} = \\epsilon.\n\\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Given$\\epsilon > 0$, find$M$such that$M > \\nicefrac{2}{\\epsilon}$. Then for$n,k \\geq M$, we have$\\nicefrac{1}{n} < \\nicefrac{\\epsilon}{2}$and$\\nicefrac{1}{k} < \\nicefrac{\\epsilon}{2}$. Therefore, for$n, k \\geq M$, we have \\begin{equation*}", "\\abs{\\frac{1}{n} - \\frac{1}{k}}", "\\leq", "\\abs{\\frac{1}{n}} + \\abs{\\frac{1}{k}}", "< \\frac{\\epsilon}{2} + \\frac{\\epsilon}{2} = \\epsilon.", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 39, "type": "example", "label": "Lebl-contfunc:39", "categories": [ "named theorem", "sequences", "cauchy", "example" ], "title": "Cauchy Convergence Criterion", "contents": [ "The sequence$\\bigl\\{ {(-1)}^n \\bigr\\}_{n=1}^\\infty$is not a Cauchy sequence.", "Proof: Given any$M \\in \\N$, take$n \\geq M$to be any even number, and let$k \\coloneqq n+1$. Then", "\\begin{equation*}\n\\begin{split}\n\\abs{{(-1)}^n - {(-1)}^k}\n& =\n\\abs{{(-1)}^n - {(-1)}^{n+1}}\n\\\\\n& =\n\\abs{1 - (-1)} = 2 .\n\\end{split}\n\\end{equation*}", "Therefore, for any$\\epsilon \\leq 2$the definition cannot be satisfied, and the sequence is not Cauchy." ], "refs": [], "proofs": [ { "contents": [ "Given any$M \\in \\N$, take$n \\geq M$to be any even number, and let$k \\coloneqq n+1$. Then \\begin{equation*}", "\\begin{split}", "\\abs{{(-1)}^n - {(-1)}^k}", "& =", "\\abs{{(-1)}^n - {(-1)}^{n+1}}", "\\\\", "& =", "\\abs{1 - (-1)} = 2 .", "\\end{split}", "\\end{equation*} Therefore, for any$\\epsilon \\leq 2$the definition cannot be satisfied, and the sequence is not Cauchy." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 40, "type": "example", "label": "Lebl-contfunc:40", "categories": [ "named theorem", "sequences", "cauchy", "example" ], "title": "Cauchy Convergence Criterion", "contents": [ "%The sequence$\\{ \\frac{n+1}{n} \\}_{n=1}^\\infty$is a Cauchy sequence. % %Proof: Given$\\epsilon > 0$, find$M$such that %$M > \\nicefrac{2}{\\epsilon}$. Then for$n,k \\geq M$, %we have$\\nicefrac{1}{n} < \\nicefrac{\\epsilon}{2}$%and %$\\nicefrac{1}{k} < \\nicefrac{\\epsilon}{2}$. Therefore, for$n, k \\geq M$, %we have %", "\\begin{equation*}\n%\\begin{split}\n%\\abs{\\frac{n+1}{n} - \\frac{k+1}{k}}\n%& =\n%\\abs{\\frac{k(n+1)-n(k+1)}{nk}}\n%\\\\\n%& =\n%\\abs{\\frac{kn+k-nk-n}{nk}}\n%\\\\\n%& =\n%\\abs{\\frac{k-n}{nk}}\n%\\\\\n%& \\leq\n%\\abs{\\frac{k}{nk}}\n%+\n%\\abs{\\frac{-n}{nk}}\n%\\\\\n%& = \\frac{1}{n} + \\frac{1}{k}\n%< \\frac{\\epsilon}{2} + \\frac{\\epsilon}{2} = \\epsilon .\n%\\end{split}\n%\\end{equation*}", "%" ], "refs": [], "proofs": [ { "contents": [ "Given$\\epsilon > 0$, find$M$such that %$M > \\nicefrac{2}{\\epsilon}$. Then for$n,k \\geq M$, %we have$\\nicefrac{1}{n} < \\nicefrac{\\epsilon}{2}$%and %$\\nicefrac{1}{k} < \\nicefrac{\\epsilon}{2}$. Therefore, for$n, k \\geq M$, %we have % \\begin{equation*}", "%\\begin{split}", "%\\abs{\\frac{n+1}{n} - \\frac{k+1}{k}}", "%& =", "%\\abs{\\frac{k(n+1)-n(k+1)}{nk}}", "%\\\\", "%& =", "%\\abs{\\frac{kn+k-nk-n}{nk}}", "%\\\\", "%& =", "%\\abs{\\frac{k-n}{nk}}", "%\\\\", "%& \\leq", "%\\abs{\\frac{k}{nk}}", "%+", "%\\abs{\\frac{-n}{nk}}", "%\\\\", "%& = \\frac{1}{n} + \\frac{1}{k}", "%< \\frac{\\epsilon}{2} + \\frac{\\epsilon}{2} = \\epsilon .", "%\\end{split}", "%\\end{equation*} %" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 41, "type": "proposition", "label": "Lebl-contfunc:41", "categories": [ "sequences", "boundedness", "named theorem" ], "title": "Cauchy Convergence Criterion", "contents": [ "If a sequence is Cauchy, then it is bounded." ], "refs": [], "proofs": [ { "contents": [ "Suppose $\\{ x_n \\}_{n=1}^\\infty$ is Cauchy. Pick an $M$ such that for all", "$n,k \\geq M$, we have $\\abs{x_n-x_k} < 1$. In particular,", "for all $n \\geq M$,", "\\begin{equation*}", "\\abs{x_n - x_M} < 1 .", "\\end{equation*}", "By the reverse triangle inequality,", "$\\abs{x_n} - \\abs{x_M} \\leq \\abs{x_n - x_M} < 1$. Hence for $n \\geq M$,", "\\begin{equation*}", "\\abs{x_n} < 1 + \\abs{x_M}.", "\\end{equation*}", "Let", "\\begin{equation*}", "B \\coloneqq \\max \\bigl\\{ \\abs{x_1}, \\abs{x_2}, \\ldots, \\abs{x_{M-1}}, 1+ \\abs{x_M} \\bigr\\} .", "\\end{equation*}", "Then $\\abs{x_n} \\leq B$ for all $n \\in \\N$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 42, "type": "theorem", "label": "Lebl-contfunc:42", "categories": [ "characterization", "convergence", "sequences", "named theorem" ], "title": "Cauchy Convergence Criterion", "contents": [ "A sequence of real numbers is Cauchy if and only if it converges." ], "refs": [], "proofs": [ { "contents": [ "Suppose $\\{ x_n \\}_{n=1}^\\infty$ converges to $x$,", "and", "let $\\epsilon > 0$ be given.", "Then there", "exists an $M$ such that for $n \\geq M$,", "\\begin{equation*}", "\\abs{x_n - x} < \\frac{\\epsilon}{2} .", "\\end{equation*}", "Hence for $n \\geq M$ and $k \\geq M$,", "\\begin{equation*}", "\\abs{x_n - x_k} =", "\\abs{x_n - x + x - x_k}", "\\leq \\abs{x_n-x} + \\abs{x-x_k} < \\frac{\\epsilon}{2} + \\frac{\\epsilon}{2} =", "\\epsilon .", "\\end{equation*}", "Alright, that direction was easy. Now suppose $\\{ x_n \\}_{n=1}^\\infty$ is Cauchy.", "We have shown that $\\{ x_n \\}_{n=1}^\\infty$ is bounded.", "For a bounded sequence, liminf and limsup exist, and this is", "where we use the", "\\hyperref[defn:lub]{least-upper-bound property}.", "If we show that", "\\begin{equation*}", "\\liminf_{n\\to \\infty} x_n = \\limsup_{n\\to\\infty} x_n ,", "\\end{equation*}", "then $\\{ x_n \\}_{n=1}^\\infty$ must be convergent by \\propref{liminfsupconv:prop}.", "Define $a \\coloneqq \\limsup_{n\\to\\infty} x_n$ and", "$b \\coloneqq \\liminf_{n\\to\\infty} x_n$.", "By \\thmref{subseqlimsupinf:thm}, there exist subsequences", "$\\{ x_{n_i} \\}_{i=1}^\\infty$ and", "$\\{ x_{m_i} \\}_{i=1}^\\infty$, such that", "\\begin{equation*}", "\\lim_{i\\to\\infty} x_{n_i} = a", "\\qquad \\text{and} \\qquad", "\\lim_{i\\to\\infty} x_{m_i} = b.", "\\end{equation*}", "Given an $\\epsilon > 0$,", "there exists an $M_1$ such that", "$\\abs{x_{n_i} - a} < \\nicefrac{\\epsilon}{3}$ for all $i \\geq M_1$", "and an $M_2$ such that", "$\\abs{x_{m_i} - b} < \\nicefrac{\\epsilon}{3}$ for all $i \\geq M_2$.", "There also exists an $M_3$", "such that", "$\\abs{x_n-x_k} < \\nicefrac{\\epsilon}{3}$", "for all $n,k \\geq M_3$.", "Let $M \\coloneqq \\max \\{ M_1, M_2, M_3 \\}$.", "If $i \\geq M$, then $n_i \\geq M$ and $m_i \\geq M$. Hence,", "\\begin{equation*}", "\\begin{split}", "\\abs{a-b} & =", "\\abs{a-x_{n_i}+x_{n_i}", "-x_{m_i}+x_{m_i}", "-b} \\\\", "& \\leq", "\\abs{a-x_{n_i}}", "+ \\abs{x_{n_i} -x_{m_i}}", "+ \\abs{x_{m_i} -b} \\\\", "& <", "\\frac{\\epsilon}{3}", "+", "\\frac{\\epsilon}{3}", "+", "\\frac{\\epsilon}{3}", "= \\epsilon .", "\\end{split}", "\\end{equation*}", "As $\\abs{a-b} < \\epsilon$ for all $\\epsilon > 0$, then $a=b$ and", "the sequence converges." ], "refs": [ "defn:lub", "liminfsupconv:prop" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 43, "type": "example", "label": "Lebl-contfunc:43", "categories": [ "series", "convergence", "sequences", "example" ], "title": "Convergence of Sequences", "contents": [ "The series", "\\begin{equation*}\n\\sum_{n=1}^\\infty \\frac{1}{2^n}\n\\end{equation*}", "converges and the limit is 1. That is,", "\\begin{equation*}\n\\sum_{n=1}^\\infty \\frac{1}{2^n} = \n\\lim_{k\\to\\infty} \\sum_{n=1}^k \\frac{1}{2^n} = \n1 .\n\\end{equation*}", "Proof: We need the equality", "\\begin{equation*}\n\\left( \\sum_{n=1}^k \\frac{1}{2^n} \\right)\n+ \\frac{1}{2^k}\n= 1 .\n\\end{equation*}", "The equality is immediate when$k=1$. The proof for general$k$follows by \\hyperref[induction:thm]{induction}, which we leave to the reader. See \\figureref{figcutseries} for an illustration. \\begin{myfigureht} \\subimport*{figures/}{figcutseries.pdf_t} \\caption{The equality$\\left( \\sum_{n=1}^k \\frac{1}{2^n} \\right) + \\frac{1}{2^k} = 1$illustrated for$k=3$.\\label{figcutseries}} \\end{myfigureht}", "Let$s_k$be the partial sum. We write", "\\begin{equation*}\n\\abs{\n1 - s_k \n}\n=\n\\abs{\n1 - \n\\sum_{n=1}^k \\frac{1}{2^n}\n}\n=\n\\abs{\\frac{1}{2^k}} = \n\\frac{1}{2^k} .\n\\end{equation*}", "The sequence$\\bigl\\{ \\frac{1}{2^k} \\bigr\\}_{k=1}^\\infty$, and therefore$\\bigl\\{ \\abs{1-s_k} \\bigr\\}_{k=1}^\\infty$, converges to zero. So,$\\{ s_k \\}_{k=1}^\\infty$converges to 1." ], "refs": [ "induction:thm" ], "proofs": [ { "contents": [ "We need the equality \\begin{equation*}", "\\left( \\sum_{n=1}^k \\frac{1}{2^n} \\right)", "+ \\frac{1}{2^k}", "= 1 .", "\\end{equation*} The equality is immediate when$k=1$. The proof for general$k$follows by \\hyperref[induction:thm]{induction}, which we leave to the reader. See \\figureref{figcutseries} for an illustration. \\begin{myfigureht} \\subimport*{figures/}{figcutseries.pdf_t} \\caption{The equality$\\left( \\sum_{n=1}^k \\frac{1}{2^n} \\right) + \\frac{1}{2^k} = 1$illustrated for$k=3$.\\label{figcutseries}} \\end{myfigureht} Let$s_k$be the partial sum. We write \\begin{equation*}", "\\abs{", "1 - s_k", "}", "=", "\\abs{", "1 -", "\\sum_{n=1}^k \\frac{1}{2^n}", "}", "=", "\\abs{\\frac{1}{2^k}} =", "\\frac{1}{2^k} .", "\\end{equation*} The sequence$\\bigl\\{ \\frac{1}{2^k} \\bigr\\}_{k=1}^\\infty$, and therefore$\\bigl\\{ \\abs{1-s_k} \\bigr\\}_{k=1}^\\infty$, converges to zero. So,$\\{ s_k \\}_{k=1}^\\infty$converges to 1." ], "refs": [ "induction:thm" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 44, "type": "proposition", "label": "Lebl-contfunc:geometric:prop", "categories": [ "convergence", "series" ], "title": "Convergence of Series", "contents": [ "Suppose$-1 < r < 1$. Then the \\emph{\\myindex{geometric series}}$\\sum_{n=0}^\\infty r^n$converges, and\\begin{equation*} \\sum_{n=0}^\\infty r^n = \\frac{1}{1-r} . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 45, "type": "proposition", "label": "Lebl-contfunc:45", "categories": [ "characterization", "convergence", "series" ], "title": "Characterization of Series via Equivalence", "contents": [ "Let$\\sum_{n=1}^\\infty x_n$be a series. Let$M \\in \\N$. Then\\begin{equation*} \\sum_{n=1}^\\infty x_n \\quad \\text{converges if and only if} \\quad \\sum_{n=M}^\\infty x_n \\quad \\text{converges.} \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "We look at partial sums of the two series (for $k \\geq M$)", "\\begin{equation*}", "\\sum_{n=1}^{k} x_n", "=", "\\left(", "\\sum_{n=1}^{M-1} x_n", "\\right)", "+", "\\sum_{n=M}^{k} x_n .", "\\end{equation*}", "Note that", "$\\sum_{n=1}^{M-1} x_n$ is a fixed number. Use", "\\propref{prop:contalg} to finish the proof." ], "refs": [ "prop:contalg" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 46, "type": "proposition", "label": "Lebl-contfunc:prop:cachyser", "categories": [ "characterization", "series", "named theorem" ], "title": "Cauchy Convergence Criterion", "contents": [ "The series$\\sum_{n=1}^\\infty x_n$is Cauchy if and only if for every$\\epsilon > 0$, there exists an$M \\in \\N$such that for every$n \\geq M$and every$k > n$,\\begin{equation*} \\abs{ \\sum_{i={n+1}}^k x_i } < \\epsilon . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 47, "type": "proposition", "label": "Lebl-contfunc:47", "categories": [ "series", "convergence", "sequences" ], "title": "Let$\\sum_{n=1}^\\infty x_n$be a convergent series", "contents": [ "Let$\\sum_{n=1}^\\infty x_n$be a convergent series. Then the sequence$\\{ x_n \\}_{n=1}^\\infty$is convergent and\\begin{equation*} \\lim_{n\\to\\infty} x_n = 0. \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let $\\epsilon > 0$ be given. As $\\sum_{n=1}^\\infty x_n$ is convergent, it is Cauchy.", "Thus we find an $M$ such that for every $n \\geq M$,", "\\begin{equation*}", "\\epsilon >", "\\abs{ \\sum_{i={n+1}}^{n+1} x_i }", "=", "\\abs{ x_{n+1} } .", "\\end{equation*}", "Hence for every $n \\geq M+1$, we have $\\abs{x_{n}} < \\epsilon$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 48, "type": "example", "label": "Lebl-contfunc:48", "categories": [ "series", "divergence", "example" ], "title": "If$r \\geq 1$or$r \\leq -1$, then the geometric series$\\sum_{n=0}^\\infty r^n$diverges", "contents": [ "If$r \\geq 1$or$r \\leq -1$, then the geometric series$\\sum_{n=0}^\\infty r^n$diverges.", "Proof:$\\abs{r^n} = \\abs{r}^n \\geq 1^n = 1$. The terms do not go to zero and the series cannot converge." ], "refs": [], "proofs": [ { "contents": [ "$\\abs{r^n} = \\abs{r}^n \\geq 1^n = 1$. The terms do not go to zero and the series cannot converge." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 49, "type": "example", "label": "Lebl-contfunc:example:harmonicseries", "categories": [ "boundedness", "sequences", "example", "series", "divergence" ], "title": "Boundedness of Sequences", "contents": [ "The series$\\sum_{n=1}^\\infty \\nicefrac{1}{n}$diverges (despite the fact that$\\lim_{n\\to\\infty} \\nicefrac{1}{n} = 0$). This is the famous \\emph{\\myindex{harmonic series}}% \\footnote{The divergence of the harmonic series was known long before the theory of series was made rigorous. The proof we give is the earliest proof and was given by \\href{https://en.wikipedia.org/wiki/Oresme}{Nicole Oresme} (1323?--1382).}.", "Proof: We will show that the sequence of partial sums is unbounded, and hence cannot converge. Write the partial sums$s_n$for$n = 2^k$as: \\begin{align*} s_1 & = 1 ,", "s_2 & = \\left( 1 \\right) + \\left( \\frac{1}{2} \\right) ,", "s_4 & = \\left( 1 \\right) + \\left( \\frac{1}{2} \\right) + \\left( \\frac{1}{3} + \\frac{1}{4} \\right) ,", "s_8 & = \\left( 1 \\right) + \\left( \\frac{1}{2} \\right) + \\left( \\frac{1}{3} + \\frac{1}{4} \\right) + \\left( \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{7} + \\frac{1}{8} \\right) ,", "& ~~ \\vdots", "s_{2^k} & = 1 + \\sum_{i=1}^k \\left( \\sum_{m=2^{i-1}+1}^{2^i} \\frac{1}{m} \\right) . \\end{align*} Notice$\\nicefrac{1}{3} + \\nicefrac{1}{4} \\geq \\nicefrac{1}{4} + \\nicefrac{1}{4} = \\nicefrac{1}{2}$and$\\nicefrac{1}{5} + \\nicefrac{1}{6} + \\nicefrac{1}{7} + \\nicefrac{1}{8} \\geq \\nicefrac{1}{8} + \\nicefrac{1}{8} + \\nicefrac{1}{8} + \\nicefrac{1}{8} = \\nicefrac{1}{2}$. More generally", "\\begin{equation*}\n\\sum_{m=2^{k-1}+1}^{2^k} \\frac{1}{m}\n\\geq\n\\sum_{m=2^{k-1}+1}^{2^k} \\frac{1}{2^k}\n=\n(2^{k-1}) \\frac{1}{2^k} = \\frac{1}{2} .\n\\end{equation*}", "Therefore,", "\\begin{equation*}\ns_{2^k} = \n1 + \n\\sum_{i=1}^k\n\\left(\n\\sum_{m=2^{i-1}+1}^{2^i} \\frac{1}{m}\n\\right) \n\\geq\n1 + \\sum_{i=1}^k \\frac{1}{2} = 1 + \\frac{k}{2} .\n\\end{equation*}", "As$\\{ \\nicefrac{k}{2} \\}_{k=1}^\\infty$is unbounded by the \\hyperref[thm:arch:i]{Archimedean property}, that means that$\\{ s_{2^k} \\}_{k=1}^\\infty$is unbounded, and therefore$\\{ s_n \\}_{n=1}^\\infty$is unbounded. Hence$\\{ s_n \\}_{n=1}^\\infty$diverges, and consequently$\\sum_{n=1}^\\infty \\nicefrac{1}{n}$diverges." ], "refs": [ "thm:arch:i" ], "proofs": [ { "contents": [ "We will show that the sequence of partial sums is unbounded, and hence cannot converge. Write the partial sums$s_n$for$n = 2^k$as: \\begin{align*} s_1 & = 1 , s_2 & = \\left( 1 \\right) + \\left( \\frac{1}{2} \\right) , s_4 & = \\left( 1 \\right) + \\left( \\frac{1}{2} \\right) + \\left( \\frac{1}{3} + \\frac{1}{4} \\right) , s_8 & = \\left( 1 \\right) + \\left( \\frac{1}{2} \\right) + \\left( \\frac{1}{3} + \\frac{1}{4} \\right) + \\left( \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{7} + \\frac{1}{8} \\right) , & ~~ \\vdots s_{2^k} & = 1 + \\sum_{i=1}^k \\left( \\sum_{m=2^{i-1}+1}^{2^i} \\frac{1}{m} \\right) . \\end{align*} Notice$\\nicefrac{1}{3} + \\nicefrac{1}{4} \\geq \\nicefrac{1}{4} + \\nicefrac{1}{4} = \\nicefrac{1}{2}$and$\\nicefrac{1}{5} + \\nicefrac{1}{6} + \\nicefrac{1}{7} + \\nicefrac{1}{8} \\geq \\nicefrac{1}{8} + \\nicefrac{1}{8} + \\nicefrac{1}{8} + \\nicefrac{1}{8} = \\nicefrac{1}{2}$. More generally \\begin{equation*}", "\\sum_{m=2^{k-1}+1}^{2^k} \\frac{1}{m}", "\\geq", "\\sum_{m=2^{k-1}+1}^{2^k} \\frac{1}{2^k}", "=", "(2^{k-1}) \\frac{1}{2^k} = \\frac{1}{2} .", "\\end{equation*} Therefore, \\begin{equation*}", "s_{2^k} =", "1 +", "\\sum_{i=1}^k", "\\left(", "\\sum_{m=2^{i-1}+1}^{2^i} \\frac{1}{m}", "\\right)", "\\geq", "1 + \\sum_{i=1}^k \\frac{1}{2} = 1 + \\frac{k}{2} .", "\\end{equation*} As$\\{ \\nicefrac{k}{2} \\}_{k=1}^\\infty$is unbounded by the \\hyperref[thm:arch:i]{Archimedean property}, that means that$\\{ s_{2^k} \\}_{k=1}^\\infty$is unbounded, and therefore$\\{ s_n \\}_{n=1}^\\infty$is unbounded. Hence$\\{ s_n \\}_{n=1}^\\infty$diverges, and consequently$\\sum_{n=1}^\\infty \\nicefrac{1}{n}$diverges." ], "refs": [ "thm:arch:i" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 50, "type": "proposition", "label": "Lebl-contfunc:50", "categories": [ "convergence", "series" ], "title": "\\index{linearity of series} Let$\\alpha \\in \\R$and$\\sum_{n=1}^\\infty x_n$and$\\sum_{n=1}^\\infty y_n...", "contents": [ "\\index{linearity of series} Let$\\alpha \\in \\R$and$\\sum_{n=1}^\\infty x_n$and$\\sum_{n=1}^\\infty y_n$be convergent series. Then \\begin{enumerate}[(i)] \\item$\\sum_{n=1}^\\infty \\alpha x_n$is a convergent series and\\begin{equation*} \\sum_{n=1}^\\infty \\alpha x_n = \\alpha \\sum_{n=1}^\\infty x_n . \\end{equation*}\\item$\\sum_{n=1}^\\infty ( x_n + y_n )$is a convergent series and\\begin{equation*} \\sum_{n=1}^\\infty ( x_n + y_n ) = \\left( \\sum_{n=1}^\\infty x_n \\right) + \\left( \\sum_{n=1}^\\infty y_n \\right) . \\end{equation*}\\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "For the first item,", "we simply write the $k$th partial sum", "\\begin{equation*}", "\\sum_{n=1}^k \\alpha x_n", "=", "\\alpha \\left( \\sum_{n=1}^k x_n \\right) .", "\\end{equation*}", "We look at the right-hand side and note that the constant multiple of", "a convergent sequence", "is convergent. Hence, we take the limit of both sides to obtain", "the result.", "For the second item we also look at the", "$k$th partial sum", "\\begin{equation*}", "\\sum_{n=1}^k ( x_n + y_n )", "=", "\\left( \\sum_{n=1}^k x_n \\right)", "+", "\\left( \\sum_{n=1}^k y_n \\right) .", "\\end{equation*}", "We look at the right-hand side and note that the sum of convergent sequences", "is convergent. Hence, we take the limit of both sides to obtain", "the proposition." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 51, "type": "proposition", "label": "Lebl-contfunc:51", "categories": [ "convergence", "boundedness", "characterization", "sequences", "series" ], "title": "Characterization of Sequences via Equivalence", "contents": [ "If$x_n \\geq 0$for all$n$, then$\\sum_{n=1}^\\infty x_n$converges if and only if the sequence of partial sums is bounded above." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 52, "type": "proposition", "label": "Lebl-contfunc:52", "categories": [ "convergence", "absolute", "series" ], "title": "Convergence of Series", "contents": [ "If the series$\\sum_{n=1}^\\infty x_n$converges absolutely, then it converges." ], "refs": [], "proofs": [ { "contents": [ "A series is convergent if and only if it is Cauchy. Hence", "suppose $\\sum_{n=1}^\\infty \\abs{x_n}$ is Cauchy. That is, for every $\\epsilon > 0$,", "there exists an $M$ such that for all $k \\geq M$ and all $n > k$, we have", "\\begin{equation*}", "\\sum_{i=k+1}^n \\abs{x_i}", "=", "\\abs{ \\sum_{i=k+1}^n \\abs{x_i} }", "<", "\\epsilon .", "\\end{equation*}", "We apply the triangle inequality for a finite sum to obtain", "\\begin{equation*}", "\\abs{ \\sum_{i=k+1}^n x_i }", "\\leq", "\\sum_{i=k+1}^n \\abs{x_i}", "<", "\\epsilon .", "\\end{equation*}", "Hence $\\sum_{n=1}^\\infty x_n$ is Cauchy, and therefore it converges." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 53, "type": "proposition", "label": "Lebl-contfunc:53", "categories": [ "series", "convergence", "tests", "divergence" ], "title": "Comparison Test", "contents": [ "\\index{comparison test for series} Let$\\sum_{n=1}^\\infty x_n$and$\\sum_{n=1}^\\infty y_n$be series such that$0 \\leq x_n \\leq y_n$for all$n \\in \\N$. \\begin{enumerate}[(i)] \\item If$\\sum_{n=1}^\\infty y_n$converges, then so does$\\sum_{n=1}^\\infty x_n$. \\item If$\\sum_{n=1}^\\infty x_n$diverges, then so does$\\sum_{n=1}^\\infty y_n$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "As the terms of the series are all nonnegative, the sequences of", "partial sums are both monotone increasing.", "Since $x_n \\leq y_n$ for all $n$, the partial sums", "satisfy for all $k$", "\\begin{equation} \\label{comptest:eq}", "\\sum_{n=1}^k x_n \\leq \\sum_{n=1}^k y_n .", "\\end{equation}", "If the series $\\sum_{n=1}^\\infty y_n$ converges, the partial sums for the series", "are bounded. Therefore, the right-hand side of \\eqref{comptest:eq}", "is bounded for all $k$; there exists some $B \\in \\R$ such that", "$\\sum_{n=1}^k y_n \\leq B$ for all $k$, and so", "\\begin{equation*}", "\\sum_{n=1}^k x_n \\leq \\sum_{n=1}^k y_n \\leq B.", "\\end{equation*}", "Hence the partial sums for $\\sum_{n=1}^\\infty x_n$", "are also bounded. Since the partial sums are a monotone increasing sequence", "they are convergent. The first item is thus proved.", "On the other hand if $\\sum_{n=1}^\\infty x_n$ diverges, the sequence of partial sums", "must be unbounded since it is monotone increasing. That is, the partial", "sums for $\\sum_{n=1}^\\infty x_n$ are eventually bigger than any real number. Putting this", "together with \\eqref{comptest:eq} we see that for every $B \\in", "\\R$, there is a $k$ such that", "\\begin{equation*}", "B \\leq \\sum_{n=1}^k x_n \\leq \\sum_{n=1}^k y_n .", "\\end{equation*}", "Hence the partial sums for $\\sum_{n=1}^\\infty y_n$ are also unbounded, and", "$\\sum_{n=1}^\\infty y_n$ also diverges." ], "refs": [ "comptest:eq" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 54, "type": "proposition", "label": "Lebl-contfunc:54", "categories": [ "characterization", "convergence", "tests", "series" ], "title": "Characterization of Series via Equivalence", "contents": [ "% \\index{p-series@$p$-series}\\index{p-test@$p$-test} For$p \\in \\R$, the series\\begin{equation*} \\sum_{n=1}^\\infty \\frac{1}{n^p} \\end{equation*}converges if and only if$p > 1$." ], "refs": [], "proofs": [ { "contents": [ "First suppose $p \\leq 1$.", "As $n \\geq 1$, we have", "$\\frac{1}{n^p} \\geq \\frac{1}{n}$. Since", "$\\sum_{n=1}^\\infty \\frac{1}{n}$ diverges,", "$\\sum_{n=1}^\\infty \\frac{1}{n^p}$ must diverge for all $p \\leq 1$ by the comparison test.", "Now suppose $p > 1$.", "We proceed as we did for the", "harmonic series, but instead of showing that the sequence", "of partial sums is unbounded, we show that it is bounded.", "The terms of the series are positive, so the sequence of partial sums", "is monotone increasing and converges if it is bounded", "above.", "Let $s_n$ denote the $n$th partial sum.", "\\begin{align*}", "s_1 & = 1 , \\\\", "s_3 & = \\left( 1 \\right) + \\left( \\frac{1}{2^p} + \\frac{1}{3^p} \\right) , \\\\", "s_7 & = \\left( 1 \\right) + \\left( \\frac{1}{2^p} + \\frac{1}{3^p} \\right) +", "\\left( \\frac{1}{4^p} + \\frac{1}{5^p} + \\frac{1}{6^p} + \\frac{1}{7^p} \\right) , \\\\", "& ~~ \\vdots \\\\", "s_{2^k - 1} &=", "1 +", "\\sum_{i=1}^{k-1}", "\\left(", "\\sum_{m=2^i}^{2^{i+1}-1} \\frac{1}{m^p}", "\\right) .", "\\end{align*}", "Instead of estimating from below, we estimate from above.", "As $p$ is positive, then $2^p < 3^p$, and hence", "$\\frac{1}{2^p} + \\frac{1}{3^p} <", "\\frac{1}{2^p} + \\frac{1}{2^p}$. Similarly,", "$\\frac{1}{4^p} + \\frac{1}{5^p} +", "\\frac{1}{6^p} + \\frac{1}{7^p} <", "\\frac{1}{4^p} + \\frac{1}{4^p} +", "\\frac{1}{4^p} + \\frac{1}{4^p}$. Therefore,", "for all $k \\geq 2$,", "\\begin{equation*}", "\\begin{split}", "s_{2^k-1}", "& =", "1+", "\\sum_{i=1}^{k-1}", "\\left(", "\\sum_{m=2^{i}}^{2^{i+1}-1} \\frac{1}{m^p}", "\\right)", "\\\\", "& <", "1+", "\\sum_{i=1}^{k-1}", "\\left(", "\\sum_{m=2^{i}}^{2^{i+1}-1} \\frac{1}{{(2^i)}^p}", "\\right)", "\\\\", "& =", "1+", "\\sum_{i=1}^{k-1}", "\\left(", "\\frac{2^i}{{(2^i)}^p}", "\\right)", "\\\\", "& =", "1+", "\\sum_{i=1}^{k-1}", "{\\left(", "\\frac{1}{2^{p-1}}", "\\right)}^i .", "\\end{split}", "\\end{equation*}", "As $p > 1$, then $\\frac{1}{2^{p-1}} < 1$.", "\\propref{geometric:prop} says that", "\\begin{equation*}", "\\sum_{i=1}^\\infty", "{\\left(", "\\frac{1}{2^{p-1}}", "\\right)}^i", "\\end{equation*}", "converges. Thus,", "\\begin{equation*}", "s_{2^k-1} <", "1+", "\\sum_{i=1}^{k-1}", "{\\left(", "\\frac{1}{2^{p-1}}", "\\right)}^i", "\\leq", "1+", "\\sum_{i=1}^\\infty", "{\\left(", "\\frac{1}{2^{p-1}}", "\\right)}^i .", "\\end{equation*}", "For every $n$ there is a $k \\geq 2$ such that $n \\leq 2^k-1$, and", "as $\\{ s_n \\}_{n=1}^\\infty$ is a monotone sequence, $s_n \\leq s_{2^k-1}$.", "So for all $n$,", "\\begin{equation*}", "s_n <", "1+", "\\sum_{i=1}^\\infty", "{\\left(", "\\frac{1}{2^{p-1}}", "\\right)}^i", "\\end{equation*}", "Thus the sequence of partial sums is bounded, and the series converges." ], "refs": [ "geometric:prop" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 55, "type": "example", "label": "Lebl-contfunc:55", "categories": [ "convergence", "tests", "series", "example" ], "title": "Comparison Test", "contents": [ "The series$\\sum_{n=1}^\\infty \\frac{1}{n^2+1}$converges.", "Proof: First,$\\frac{1}{n^2+1} < \\frac{1}{n^2}$for all$n \\in \\N$. The series$\\sum_{n=1}^\\infty \\frac{1}{n^2}$converges by the$p$-series test. Therefore, by the comparison test,$\\sum_{n=1}^\\infty \\frac{1}{n^2+1}$converges." ], "refs": [], "proofs": [ { "contents": [ "First,$\\frac{1}{n^2+1} < \\frac{1}{n^2}$for all$n \\in \\N$. The series$\\sum_{n=1}^\\infty \\frac{1}{n^2}$converges by the$p$-series test. Therefore, by the comparison test,$\\sum_{n=1}^\\infty \\frac{1}{n^2+1}$converges." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 56, "type": "proposition", "label": "Lebl-contfunc:56", "categories": [ "convergence", "absolute", "series", "tests", "divergence" ], "title": "Ratio Test", "contents": [ "\\index{ratio test for series} Let$\\sum_{n=1}^\\infty x_n$be a series,$x_n \\not= 0$for all$n$, and such that\\begin{equation*} L \\coloneqq \\lim_{n\\to\\infty} \\frac{\\abs{x_{n+1}}}{\\abs{x_n}} \\qquad \\text{exists.} \\end{equation*}\\begin{enumerate}[(i)] \\item If$L < 1$, then$\\sum_{n=1}^\\infty x_n$converges absolutely. \\item If$L > 1$, then$\\sum_{n=1}^\\infty x_n$diverges. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "If $L > 1$, then", "\\lemmaref{seq:ratiotest} says that the sequence $\\{ x_n \\}_{n=1}^\\infty$", "diverges. Since it is a necessary condition for the convergence of series", "that the terms go to zero, we know that $\\sum_{n=1}^\\infty x_n$ must diverge.", "Thus suppose $L < 1$.", "We will argue that $\\sum_{n=1}^\\infty \\abs{x_n}$ must converge.", "The proof is similar to that of \\lemmaref{seq:ratiotest}. Of course $L \\geq", "0$.", "Pick", "$r$ such that $L < r < 1$. As $r-L > 0$, there exists an $M \\in \\N$ such that for", "all $n \\geq M$,", "\\begin{equation*}", "\\abs{\\frac{\\abs{x_{n+1}}}{\\abs{x_n}} - L} < r-L .", "\\end{equation*}", "Therefore,", "\\begin{equation*}", "\\frac{\\abs{x_{n+1}}}{\\abs{x_n}} < r .", "\\end{equation*}", "For $n > M$ (that is for $n \\geq M+1$),", "write", "\\begin{equation*}", "\\abs{x_n} =", "\\abs{x_M}", "\\frac{\\abs{x_{M+1}}}{\\abs{x_{M}}}", "\\frac{\\abs{x_{M+2}}}{\\abs{x_{M+1}}}", "\\cdots", "\\frac{\\abs{x_{n}}}{\\abs{x_{n-1}}}", "<", "\\abs{x_M}", "r r \\cdots r = \\abs{x_M} r^{n-M} = (\\abs{x_M} r^{-M}) r^n .", "\\end{equation*}", "For $k > M$, write the partial sum as", "\\begin{equation*}", "\\begin{split}", "\\sum_{n=1}^k \\abs{x_n}", "& =", "\\left(\\sum_{n=1}^{M} \\abs{x_n} \\right)", "+", "\\left(\\sum_{n=M+1}^{k} \\abs{x_n} \\right)", "\\\\", "& <", "\\left(\\sum_{n=1}^{M} \\abs{x_n} \\right)", "+", "\\left(\\sum_{n=M+1}^{k}", "(\\abs{x_M} r^{-M}) r^n", "\\right)", "\\\\", "& =", "\\left(\\sum_{n=1}^{M} \\abs{x_n} \\right)", "+", "\\bigl(\\abs{x_M} r^{-M}\\bigr)", "\\left( \\sum_{n=M+1}^{k} r^n \\right) .", "\\end{split}", "\\end{equation*}", "As $0 < r < 1$, the geometric series", "$\\sum_{n=0}^{\\infty} r^n$ converges, so", "$\\sum_{n=M+1}^{\\infty} r^n$ converges as well. We take the", "limit as $k$ goes to infinity on the right-hand side above to obtain", "\\begin{equation*}", "\\begin{split}", "\\sum_{n=1}^k \\abs{x_n}", "& <", "\\left(\\sum_{n=1}^{M} \\abs{x_n} \\right)", "+", "\\bigl(\\abs{x_M} r^{-M}\\bigr)", "\\left( \\sum_{n=M+1}^{k} r^n \\right)", "\\\\", "& \\leq", "\\left(\\sum_{n=1}^{M} \\abs{x_n} \\right)", "+", "\\bigl(\\abs{x_M} r^{-M}\\bigr)", "\\left( \\sum_{n=M+1}^{\\infty} r^n \\right) .", "\\end{split}", "\\end{equation*}", "The right-hand side is a number that does not depend on $k$.", "Hence the sequence of partial sums of $\\sum_{n=1}^\\infty \\abs{x_n}$ is bounded", "and $\\sum_{n=1}^\\infty \\abs{x_n}$ is convergent. Thus $\\sum_{n=1}^\\infty x_n$ is", "absolutely convergent." ], "refs": [ "seq:ratiotest" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 57, "type": "example", "label": "Lebl-contfunc:57", "categories": [ "convergence", "absolute", "example", "series", "tests" ], "title": "Ratio Test", "contents": [ "The series", "\\begin{equation*}\n\\sum_{n=1}^\\infty \\frac{2^n}{n!}\n\\end{equation*}", "converges absolutely.", "Proof: We write", "\\begin{equation*}\n\\lim_{n\\to\\infty} \\frac{2^{(n+1)}/(n+1)!}{2^n / n!} =\n\\lim_{n\\to\\infty} \\frac{2}{n+1} = 0 .\n\\end{equation*}", "Therefore, the series converges absolutely by the ratio test." ], "refs": [], "proofs": [ { "contents": [ "We write \\begin{equation*}", "\\lim_{n\\to\\infty} \\frac{2^{(n+1)}/(n+1)!}{2^n / n!} =", "\\lim_{n\\to\\infty} \\frac{2}{n+1} = 0 .", "\\end{equation*} Therefore, the series converges absolutely by the ratio test." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 58, "type": "proposition", "label": "Lebl-contfunc:58", "categories": [ "series", "convergence", "absolute", "divergence" ], "title": "Convergence of Series", "contents": [ "Let$\\sum_{n=1}^\\infty x_n$be a series and let\\begin{equation*} L \\coloneqq \\limsup_{n\\to\\infty} {\\sabs{x_n}}^{1/n} . \\end{equation*}\\begin{enumerate}[(i)] \\item If$L < 1$, then$\\sum_{n=1}^\\infty x_n$converges absolutely. \\item If$L > 1$, then$\\sum_{n=1}^\\infty x_n$diverges. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "If $L > 1$, then there exists\\footnote{%", "In case $L=\\infty$, see \\exerciseref{exercise:extendsubseqlimsupinf}.", "Alternatively, note that if $L=\\infty$, then $\\bigl\\{ {\\sabs{x_{n}}}^{1/n}", "\\bigr\\}_{n=1}^\\infty$", "is unbounded, and thus so is", "$\\{ x_n \\}_{n=1}^\\infty$.}", "a subsequence $\\{ x_{n_k} \\}_{k=1}^\\infty$ such that", "$L = \\lim_{k\\to\\infty} {\\sabs{x_{n_k}}}^{1/n_k}$. Let", "$r$ be such that $L > r > 1$. There exists an $M$ such", "that for all $k \\geq M$, we have", "${\\sabs{x_{n_k}}}^{1/n_k} > r > 1$, or in other words", "$\\sabs{x_{n_k}} > r^{n_k} > 1$.", "The subsequence", "$\\{ \\sabs{x_{n_k}} \\}_{k=1}^\\infty$, and therefore also", "$\\{ \\sabs{x_{n}} \\}_{n=1}^\\infty$,", "cannot possibly converge to zero, and so the series diverges.", "Now suppose $L < 1$. Pick $r$ such that $L < r < 1$.", "By definition of limit supremum,", "there is an $M$ such that for all $n \\geq M$,", "\\begin{equation*}", "\\sup \\bigl\\{ {\\abs{x_k}}^{1/k} : k \\geq n \\bigr\\} < r .", "\\end{equation*}", "Therefore, for all $n \\geq M$,", "\\begin{equation*}", "{\\abs{x_n}}^{1/n} < r , \\qquad \\text{or in other words} \\qquad \\abs{x_n} < r^n .", "\\end{equation*}", "Let $k > M$, and estimate the $k$th partial sum:", "\\begin{equation*}", "\\sum_{n=1}^k \\abs{x_n} =", "\\left( \\sum_{n=1}^M \\abs{x_n} \\right) +", "\\left( \\sum_{n=M+1}^k \\abs{x_n} \\right)", "\\leq", "\\left( \\sum_{n=1}^M \\abs{x_n} \\right) +", "\\left( \\sum_{n=M+1}^k r^n \\right) .", "\\end{equation*}", "As $0 < r < 1$,", "the geometric series $\\sum_{n=M+1}^\\infty r^n$ converges to", "$\\frac{r^{M+1}}{1-r}$. As everything is positive,", "\\begin{equation*}", "\\sum_{n=1}^k \\abs{x_n}", "\\leq", "\\left( \\sum_{n=1}^M \\abs{x_n} \\right) +", "\\frac{r^{M+1}}{1-r} .", "\\end{equation*}", "Thus the sequence of partial sums of $\\sum_{n=1}^\\infty \\abs{x_n}$ is bounded, and", "the series converges. Therefore, $\\sum_{n=1}^\\infty x_n$ converges absolutely." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 59, "type": "proposition", "label": "Lebl-contfunc:59", "categories": [ "series", "convergence", "sequences" ], "title": "Convergence of Sequences", "contents": [ "Let$\\{ x_n \\}_{n=1}^\\infty$be a monotone decreasing sequence of positive real numbers such that$\\lim\\limits_{n\\to\\infty} x_n = 0$. Then\\begin{equation*} \\sum_{n=1}^\\infty {(-1)}^n x_n \\qquad \\text{converges.} \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let $s_m \\coloneqq \\sum_{k=1}^m {(-1)}^n x_n$ be the $m$th partial sum. Then", "\\begin{equation*}", "s_{2k} =", "\\sum_{n=1}^{2k} {(-1)}^n x_n", "=", "(-x_1 + x_2) + \\cdots + (-x_{2k-1} + x_{2k})", "=", "\\sum_{\\ell=1}^{k} (-x_{2\\ell-1} + x_{2\\ell}) .", "\\end{equation*}", "The sequence $\\{ x_n \\}_{n=1}^\\infty$ is decreasing, so $(-x_{2\\ell-1}+x_{2\\ell}) \\leq 0$", "for all $\\ell$.", "Thus, the subsequence $\\{ s_{2k} \\}_{k=1}^\\infty$ of partial sums", "is a decreasing sequence. Similarly, $(x_{2\\ell}-x_{2\\ell+1}) \\geq 0$, and so", "\\begin{equation*}", "s_{2k} = - x_1 + ( x_2 - x_3 ) + \\cdots + ( x_{2k-2} - x_{2k-1} ) + x_{2k}", "\\geq -x_1 .", "\\end{equation*}", "The intuition behind the bound", "$0 \\geq s_{2k} \\geq -x_1$ is illustrated in \\figureref{fig:seralternate}.", "\\begin{myfigureht}", "\\includegraphics{figures/ser-alternate}", "\\caption{Showing that $0 \\geq s_{2k} \\geq -x_1$ where $k=4$ for an alternating series.\\label{fig:seralternate}}", "\\end{myfigureht}", "As $\\{ s_{2k} \\}_{k=1}^\\infty$ is decreasing and bounded below, it converges.", "Let $a \\coloneqq \\lim_{k\\to\\infty} s_{2k}$.", "We wish to show that $\\lim_{m\\to\\infty} s_m = a$ (and not just for the subsequence).", "Given $\\epsilon > 0$, pick $M$ such that $\\abs{s_{2k}-a} <", "\\nicefrac{\\epsilon}{2}$ whenever $k \\geq M$.", "Since $\\lim_{n\\to\\infty} x_n = 0$, we also", "make $M$ possibly larger", "to obtain", "$x_{2k+1} < \\nicefrac{\\epsilon}{2}$ whenever $k \\geq M$.", "Suppose $m \\geq 2M+1$. If $m=2k$, then $k \\geq M+\\nicefrac{1}{2} \\geq M$ and", "$\\abs{s_{m}-a}=\\abs{s_{2k}-a} < \\nicefrac{\\epsilon}{2} < \\epsilon$.", "If $m=2k+1$, then also $k \\geq M$. Notice", "$s_{2k+1} = s_{2k} - x_{2k+1}$.", "Thus", "\\begin{equation*}", "\\abs{s_{m}-a} =", "\\abs{s_{2k+1}-a} =", "\\abs{s_{2k}-a - x_{2k+1}} \\leq", "\\abs{s_{2k}-a} + x_{2k+1} <", "\\nicefrac{\\epsilon}{2}+ \\nicefrac{\\epsilon}{2} = \\epsilon . \\qedhere", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 60, "type": "proposition", "label": "Lebl-contfunc:60", "categories": [ "convergence", "absolute", "series" ], "title": "Convergence of Series", "contents": [ "Let$\\sum_{n=1}^\\infty x_n$be an absolutely convergent series converging to a number$x$. Let$\\sigma \\colon \\N \\to \\N$be a bijection. Then$\\sum_{n=1}^\\infty x_{\\sigma(n)}$is absolutely convergent and converges to$x$." ], "refs": [], "proofs": [ { "contents": [ "Let $\\epsilon > 0$ be given. As $\\sum_{n=1}^\\infty x_n$ is absolutely convergent, take $M$ such that", "\\begin{equation*}", "\\abs{\\left(\\sum_{n=1}^M x_n \\right) - x} < \\frac{\\epsilon}{2}", "\\qquad \\text{and} \\qquad", "\\sum_{n=M+1}^\\infty \\abs{x_n} < \\frac{\\epsilon}{2} .", "\\end{equation*}", "As $\\sigma$ is a bijection,", "there exists a number $K$ such that for each", "$n \\leq M$, there exists $k \\leq K$ such that $\\sigma(k) = n$.", "In other words", "$\\{ 1,2,\\ldots,M \\} \\subset \\sigma\\bigl(\\{ 1,2,\\ldots,K \\} \\bigr)$.", "For $N \\geq K$, let $Q \\coloneqq \\max \\sigma\\bigl(\\{ 1,2,\\ldots,N \\}\\bigr)$.", "Compute", "\\begin{equation*}", "\\begin{split}", "\\abs{\\left( \\sum_{n=1}^N x_{\\sigma(n)} \\right) - x}", "& =", "\\abs{ \\left( \\sum_{n=1}^M x_n", "+", "\\sum_{\\substack{n=1\\\\\\sigma(n) > M}}^N x_{\\sigma(n)} \\right) - x}", "\\\\", "& \\leq", "\\abs{ \\left( \\sum_{n=1}^M x_n \\right) - x}", "+", "\\sum_{\\substack{n=1\\\\\\sigma(n) > M}}^N \\abs{x_{\\sigma(n)}}", "\\\\", "& \\leq", "\\abs{ \\left( \\sum_{n=1}^M x_n \\right) - x}", "+", "\\sum_{n=M+1}^Q \\abs{x_{n}}", "\\\\", "& < \\nicefrac{\\epsilon}{2} + \\nicefrac{\\epsilon}{2} = \\epsilon .", "\\end{split}", "\\end{equation*}", "So", "$\\sum_{n=1}^\\infty x_{\\sigma(n)}$ converges to $x$. To see that the convergence", "is absolute, we apply the argument above to $\\sum_{n=1}^\\infty \\abs{x_n}$ to show", "that $\\sum_{n=1}^\\infty \\abs{x_{\\sigma(n)}}$ converges." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 61, "type": "example", "label": "Lebl-contfunc:example:harmonsumanything", "categories": [ "convergence", "example", "rearrangements", "series", "absolute" ], "title": "Limit of Series", "contents": [ "Let us show that the alternating harmonic series$\\sum_{n=1}^\\infty \\frac{{(-1)}^{n+1}}{n}$, which does not converge absolutely, can be rearranged to converge to anything. The odd terms and the even terms diverge to plus infinity and minus infinity respectively (prove this!):", "\\begin{equation*}\n\\sum_{m=1}^\\infty \\frac{1}{2m-1} = \\infty, \\qquad \\text{and} \\qquad\n\\sum_{m=1}^\\infty \\frac{-1}{2m} = -\\infty .\n\\end{equation*}", "Let$a_n \\coloneqq \\frac{{(-1)}^{n+1}}{n}$for simplicity, let an arbitrary number$L \\in \\R$be given, and set$\\sigma(1) \\coloneqq 1$. Suppose we have defined$\\sigma(n)$for all$n \\leq N$. If", "\\begin{equation*}\n\\sum_{n=1}^N a_{\\sigma(n)} \\leq L ,\n\\end{equation*}", "then let$\\sigma(N+1) \\coloneqq k$be the smallest odd$k \\in \\N$that we have not used yet, that is,$\\sigma(n) \\not= k$for all$n \\leq N$. Otherwise, let$\\sigma(N+1) \\coloneqq k$be the smallest even$k$that we have not yet used.", "By construction,$\\sigma \\colon \\N \\to \\N$is one-to-one. It is also onto, because if we keep adding either odd (resp.\\ even) terms, eventually we pass$L$and switch to the evens (resp.\\ odds). So we switch infinitely many times.", "Finally, let$N$be the$N$where we just pass$L$and switch. For example, suppose we have just switched from odd to even (so we start subtracting), and let$N' > N$be where we first switch back from even to odd. Then", "\\begin{equation*}\nL + \\frac{1}{\\sigma(N)} \\geq \\sum_{n=1}^{N-1} a_{\\sigma(n)}\n> \\sum_{n=1}^{N'-1} a_{\\sigma(n)} > L- \\frac{1}{\\sigma(N')}.\n\\end{equation*}", "Similarly for switching in the other direction. Therefore, the sum up to$N'-1$is within$\\frac{1}{\\min \\{ \\sigma(N), \\sigma(N') \\}}$of$L$. As we switch infinitely many times,$\\sigma(N) \\to \\infty$and$\\sigma(N') \\to \\infty$. Hence", "\\begin{equation*}\n\\sum_{n=1}^\\infty a_{\\sigma(n)} = \n\\sum_{n=1}^\\infty \\frac{{(-1)}^{\\sigma(n)+1}}{\\sigma(n)} = L .\n\\end{equation*}", "Here is an example to illustrate the proof. Suppose$L=1.2$, then the order is", "\\begin{equation*}\n1+\\nicefrac{1}{3}-\\nicefrac{1}{2}+\\nicefrac{1}{5}+\\nicefrac{1}{7}+\\nicefrac{1}{9}-\\nicefrac{1}{4}+\\nicefrac{1}{11}+\\nicefrac{1}{13}-\\nicefrac{1}{6}\n+\\nicefrac{1}{15}+\\nicefrac{1}{17}+\\nicefrac{1}{19} - \\nicefrac{1}{8} + \\cdots .\n\\end{equation*}", "At this point we are no more than$\\nicefrac{1}{8}$from the limit. See \\figureref{fig:serrearrange}. \\begin{myfigureht} \\includegraphics{figures/ser-rearrange} \\caption{The first 14 partial sums of the rearrangement converging to$1.2$.\\label{fig:serrearrange}} \\end{myfigureht}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 62, "type": "theorem", "label": "Lebl-contfunc:62", "categories": [ "convergence", "absolute", "series" ], "title": "Convergence of Series", "contents": [ "Suppose$\\sum_{n=0}^\\infty a_n$and$\\sum_{n=0}^\\infty b_n$are two convergent series, converging to$A$and$B$respectively. Suppose at least one of the series converges absolutely. Define\\begin{equation*} c_n \\coloneqq a_0 b_n + a_1 b_{n-1} + \\cdots + a_n b_0 = \\sum_{i=0}^n a_i b_{n-i} . \\end{equation*}Then the series$\\sum_{n=0}^\\infty c_n$converges to$AB$." ], "refs": [], "proofs": [ { "contents": [ "Suppose $\\sum_{n=0}^\\infty a_n$ converges absolutely, and let $\\epsilon > 0$ be", "given.", "In this proof instead of picking complicated estimates just to make", "the final estimate come out as less than $\\epsilon$,", "let us simply obtain an estimate that depends on $\\epsilon$", "and can be made arbitrarily small.", "Write", "\\begin{equation*}", "A_m \\coloneqq \\sum_{n=0}^m a_n , \\qquad B_m \\coloneqq \\sum_{n=0}^m b_n .", "\\end{equation*}", "We rearrange the $m$th partial sum of $\\sum_{n=0}^\\infty c_n$:", "\\begin{equation*}", "\\begin{split}", "\\abs{\\left(\\sum_{n=0}^m c_n \\right) - AB}", "& =", "\\abs{\\left( \\sum_{n=0}^m \\sum_{i=0}^n a_i b_{n-i} \\right) - AB}", "\\\\", "& =", "\\abs{\\left( \\sum_{n=0}^m", "B_n a_{m-n} \\right) - AB}", "\\\\", "& =", "\\abs{\\left( \\sum_{n=0}^m", "( B_n - B ) a_{m-n} \\right)", "+ B A_m - AB}", "\\\\", "& \\leq", "\\left(", "\\sum_{n=0}^m", "\\abs{ B_n - B } \\abs{a_{m-n}}", "\\right)", "+", "\\abs{B}\\abs{A_m - A}", "\\end{split}", "\\end{equation*}", "We can surely make the second term on the right-hand side go to zero.", "The trick is to handle the first term.", "Pick $K$ such that for all $m \\geq K$, we have", "$\\abs{A_m - A} < \\epsilon$ and", "also", "$\\abs{B_m - B} < \\epsilon$. Finally,", "as $\\sum_{n=0}^\\infty a_n$ converges absolutely,", "make sure that $K$ is large enough such that", "for all $m \\geq K$,", "\\begin{equation*}", "\\sum_{n=K}^m \\abs{a_n} < \\epsilon .", "\\end{equation*}", "As $\\sum_{n=0}^\\infty b_n$ converges, then", "$B_{\\text{max}} \\coloneqq \\sup \\bigl\\{ \\abs{ B_n - B } : n = 0,1,2,\\ldots", "\\bigr\\}$", "is finite. Take $m \\geq 2K$.", "In particular $m-K+1 > K$. So", "\\begin{equation*}", "\\begin{split}", "\\sum_{n=0}^m", "\\abs{ B_n - B } \\abs{a_{m-n}}", "& =", "\\left(", "\\sum_{n=0}^{m-K}", "\\abs{ B_n - B } \\abs{a_{m-n}}", "\\right)", "+", "\\left(", "\\sum_{n=m-K+1}^m", "\\abs{ B_n - B } \\abs{a_{m-n}}", "\\right)", "\\\\", "& \\leq", "\\left(", "\\sum_{n=K}^m", "\\abs{a_{n}}", "\\right)", "B_{\\text{max}}", "+", "\\left(", "\\sum_{n=0}^{K-1}", "\\epsilon \\abs{a_{n}}", "\\right)", "\\\\", "& \\leq", "\\epsilon", "B_{\\text{max}}", "+", "\\epsilon", "\\left(", "\\sum_{n=0}^\\infty \\abs{a_{n}}", "\\right) .", "\\end{split}", "\\end{equation*}", "Therefore, for $m \\geq 2K$, we have", "\\begin{equation*}", "\\begin{split}", "\\abs{\\left(\\sum_{n=0}^m c_n \\right) - AB}", "& \\leq", "\\left(", "\\sum_{n=0}^m", "\\abs{ B_n - B } \\abs{a_{m-n}}", "\\right)", "+", "\\abs{B}\\abs{A_m - A}", "\\\\", "& \\leq", "\\epsilon", "B_{\\text{max}}", "+", "\\epsilon", "\\left(", "\\sum_{n=0}^\\infty \\abs{a_{n}}", "\\right)", "+", "\\abs{B}\\epsilon", "=", "\\epsilon", "\\left(", "B_{\\text{max}}", "+", "\\left(", "\\sum_{n=0}^\\infty \\abs{a_{n}}", "\\right)", "+", "\\abs{B}", "\\right) .", "\\end{split}", "\\end{equation*}", "The expression in the parenthesis on the right-hand side", "is a fixed number.", "Hence,", "we can make the right-hand side arbitrarily small by picking a small enough", "$\\epsilon> 0$. So $\\sum_{n=0}^\\infty c_n$ converges to $AB$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 63, "type": "example", "label": "Lebl-contfunc:63", "categories": [ "convergence", "absolute", "example", "series", "tests", "named theorem" ], "title": "Cauchy Convergence Criterion", "contents": [ "If both series are only conditionally convergent, the Cauchy product series need not even converge. Suppose we take$a_n = b_n = {(-1)}^n \\frac{1}{\\sqrt{n+1}}$. The series$\\sum_{n=0}^\\infty a_n = \\sum_{n=0}^\\infty b_n$converges by the alternating series test; however, it does not converge absolutely as can be seen from the$p$-test. Let us look at the Cauchy product.", "\\begin{equation*}\nc_n = \n{(-1)}^n\n\\left(\n\\frac{1}{\\sqrt{n+1}} + \n\\frac{1}{\\sqrt{2n}} + \n\\frac{1}{\\sqrt{3(n-1)}} + \\cdots +\n%\\frac{1}{\\sqrt{2n}} + \n\\frac{1}{\\sqrt{n+1}}\n\\right)\n=\n{(-1)}^n\n\\sum_{i=0}^n \\frac{1}{\\sqrt{(i+1)(n-i+1)}} .\n\\end{equation*}", "Therefore,", "\\begin{equation*}\n\\abs{c_n} \n=\n\\sum_{i=0}^n \\frac{1}{\\sqrt{(i+1)(n-i+1)}} \n\\geq\n\\sum_{i=0}^n \\frac{1}{\\sqrt{(n+1)(n+1)}} \n= 1 .\n\\end{equation*}", "The terms do not go to zero and hence$\\sum_{n=0}^\\infty c_n$cannot converge." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 64, "type": "example", "label": "Lebl-contfunc:ps:expex", "categories": [ "convergence", "absolute", "example", "series", "tests" ], "title": "Ratio Test", "contents": [ "The series", "\\begin{equation*}\n\\sum_{n=0}^\\infty \\frac{1}{n!} x^n\n\\end{equation*}", "is absolutely convergent for all$x \\in \\R$using the ratio test: For any$x \\in \\R$", "\\begin{equation*}\n\\lim_{n \\to \\infty}\n\\frac{\\bigl(1/(n+1)!\\bigr) \\, x^{n+1}}{(1/n!) \\, x^{n}}\n=\n\\lim_{n \\to \\infty}\n\\frac{x}{n+1}\n=\n0.\n\\end{equation*}", "Recall from calculus that this series converges to$e^x$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 65, "type": "example", "label": "Lebl-contfunc:ps:1kex", "categories": [ "convergence", "tests", "power_series", "example", "series", "absolute", "divergence" ], "title": "Alternating Series Test", "contents": [ "The series", "\\begin{equation*}\n\\sum_{n=1}^\\infty \\frac{1}{n} x^n\n\\end{equation*}", "converges absolutely for all$x \\in (-1,1)$via the ratio test:", "\\begin{equation*}\n\\lim_{n \\to \\infty}\n\\abs{\n\\frac{\\bigl(1/(n+1) \\bigr) \\, x^{n+1}}{(1/n) \\, x^{n}}\n}\n=\n\\lim_{n \\to \\infty}\n\\abs{x} \\frac{n}{n+1}\n=\n\\abs{x} < 1 .\n\\end{equation*}", "The series converges at$x=-1$, as$\\sum_{n=1}^\\infty \\frac{{(-1)}^n}{n}$converges by the alternating series test. But the power series does not converge absolutely at$x=-1$, because$\\sum_{n=1}^\\infty \\frac{1}{n}$does not converge. The series diverges at$x=1$. When$\\abs{x} > 1$, then the series diverges via the ratio test." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 66, "type": "example", "label": "Lebl-contfunc:ps:divergeex", "categories": [ "series", "tests", "divergence", "example" ], "title": "Root Test", "contents": [ "The series", "\\begin{equation*}\n\\sum_{n=1}^\\infty n^n x^n\n\\end{equation*}", "diverges for all$x \\not= 0$. Let us apply the root test", "\\begin{equation*}\n\\limsup_{n\\to\\infty}\n\\,\n\\abs{n^n x^n}^{1/n}\n=\n\\limsup_{n\\to\\infty}\n\\,\nn \\abs{x}\n= \\infty .\n\\end{equation*}", "Therefore, the series diverges for all$x \\not= 0$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 67, "type": "proposition", "label": "Lebl-contfunc:prop:powerserrealradius", "categories": [ "convergence", "power_series", "series", "absolute", "divergence" ], "title": "Existence of Series", "contents": [ "Let$\\sum_{n=0}^\\infty a_n {(x-x_0)}^n$be a power series. If the series is convergent, then either it converges absolutely at all$x \\in \\R$, or there exists a number$\\rho$, such that the series converges absolutely on the interval$(x_0-\\rho,x_0+\\rho)$and diverges when$x < x_0-\\rho$or$x > x_0+\\rho$." ], "refs": [], "proofs": [ { "contents": [ "Write", "\\begin{equation*}", "R \\coloneqq \\limsup_{n\\to\\infty} {\\abs{a_n}}^{1/n} .", "\\end{equation*}", "We apply the root test,", "\\begin{equation*}", "L = \\limsup_{n\\to\\infty} {\\abs{a_n {(x-x_0)}^n}}^{1/n}", "=", "\\abs{x-x_0} \\limsup_{n\\to\\infty} {\\abs{a_n}}^{1/n}", "=", "\\abs{x-x_0} R .", "\\end{equation*}", "If $R = \\infty$, then $L=\\infty$ for every $x \\not= x_0$, and", "the series diverges by the root test.", "On the other hand,", "if $R = 0$, then $L=0$ for every $x$,", "and the series converges absolutely for all $x$.", "Suppose $0 < R < \\infty$.", "The series", "converges absolutely if", "$1 > L = R \\abs{x-x_0}$,", "that is,", "\\begin{equation*}", "\\abs{x-x_0} < \\nicefrac{1}{R} .", "\\end{equation*}", "The series diverges when", "$1 < L = R \\abs{x-x_0}$,", "or", "\\begin{equation*}", "\\abs{x-x_0} > \\nicefrac{1}{R} .", "\\end{equation*}", "Letting $\\rho \\coloneqq \\nicefrac{1}{R}$ completes the proof." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 68, "type": "proposition", "label": "Lebl-contfunc:68", "categories": [ "series", "convergence", "power_series", "divergence" ], "title": "Convergence of Series", "contents": [ "Let$\\sum_{n=0}^\\infty a_n {(x-x_0)}^n$be a power series, and let\\begin{equation*} R \\coloneqq \\limsup_{n\\to\\infty} {\\abs{a_n}}^{1/n} . \\end{equation*}If$R = \\infty$, the power series is divergent. If$R=0$, then the power series converges everywhere. Otherwise, the radius of convergence$\\rho = \\nicefrac{1}{R}$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 69, "type": "proposition", "label": "Lebl-contfunc:69", "categories": [ "convergence", "power_series", "series" ], "title": "Let$\\sum_{n=0}^\\infty a_n {(x-x_0)}^n$and$\\sum_{n=0}^\\infty b_n {(x-x_0)}^n$be two convergent pow...", "contents": [ "Let$\\sum_{n=0}^\\infty a_n {(x-x_0)}^n$and$\\sum_{n=0}^\\infty b_n {(x-x_0)}^n$be two convergent power series with radius of convergence at least$\\rho > 0$and$\\alpha \\in \\R$. Then for all$x$such that$\\abs{x-x_0} < \\rho$, we have\\begin{equation*} \\left(\\sum_{n=0}^\\infty a_n {(x-x_0)}^n\\right) + \\left(\\sum_{n=0}^\\infty b_n {(x-x_0)}^n\\right) = \\sum_{n=0}^\\infty (a_n+b_n) {(x-x_0)}^n , \\end{equation*}\\begin{equation*} \\alpha \\left(\\sum_{n=0}^\\infty a_n {(x-x_0)}^n\\right) = \\sum_{n=0}^\\infty \\alpha a_n {(x-x_0)}^n , \\end{equation*}and\\begin{equation*} \\left(\\sum_{n=0}^\\infty a_n {(x-x_0)}^n\\right) \\, \\left(\\sum_{n=0}^\\infty b_n {(x-x_0)}^n\\right) = \\sum_{n=0}^\\infty c_n {(x-x_0)}^n , \\end{equation*}where$c_n = a_0b_n + a_1 b_{n-1} + \\cdots + a_n b_0$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 70, "type": "example", "label": "Lebl-contfunc:70", "categories": [ "power_series", "series", "example" ], "title": "Let us expand$\\frac{x}{1+2x+x^2}$as a power series around the origin ($x_0 = 0$) and find the rad...", "contents": [ "Let us expand$\\frac{x}{1+2x+x^2}$as a power series around the origin ($x_0 = 0$) and find the radius of convergence.", "Write$1+2x+x^2 = {(1+x)}^2 = {\\bigl(1-(-x)\\bigr)}^2$, and suppose$\\abs{x} < 1$. Compute", "\\begin{equation*}\n\\begin{split}\n\\frac{x}{1+2x+x^2}\n&=\nx \\,\n{\\left(\n\\frac{1}{1-(-x)}\n\\right)}^2\n\\\\\n&=\nx \\,\n{\\left( \n\\sum_{n=0}^\\infty {(-1)}^n x^n \n\\right)}^2\n\\\\\n&=\nx \\,\n\\left(\n\\sum_{n=0}^\\infty c_n x^n \n\\right)\n\\\\\n&=\n\\sum_{n=0}^\\infty c_n x^{n+1} .\n\\end{split}\n\\end{equation*}", "Using the formula for the product of series, we obtain$c_0 = 1$,$c_1 = -1 -1 = -2$,$c_2 = 1+1+1 = 3$, etc. Hence, for$\\abs{x} < 1$,", "\\begin{equation*}\n\\frac{x}{1+2x+x^2}\n=\n\\sum_{n=1}^\\infty {(-1)}^{n+1} n x^n .\n\\end{equation*}", "The radius of convergence is at least 1. We leave it to the reader to verify that the radius of convergence is exactly equal to 1." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 0, "type": "proposition", "label": "Lebl-contfunc:0", "categories": [ "characterization", "limits", "topology", "sequences" ], "title": "Characterization of Sequences via Equivalence", "contents": [ "Let$S \\subset \\R$. Then$x \\in \\R$is a cluster point of$S$if and only if there exists a convergent sequence of numbers$\\{ x_n \\}_{n=1}^\\infty$such that$x_n \\not= x$and$x_n \\in S$for all$n$, and$\\lim\\limits_{n\\to\\infty} x_n = x$." ], "refs": [], "proofs": [ { "contents": [ "First suppose $x$ is a cluster point of $S$.", "For every $n \\in \\N$, pick $x_n$ to be an arbitrary point of", "$(x-\\nicefrac{1}{n},x+\\nicefrac{1}{n}) \\cap S \\setminus \\{x\\}$, which", "is nonempty because $x$ is a cluster point of $S$.", "Then", "$x_n$ is within $\\nicefrac{1}{n}$ of $x$, that is,", "\\begin{equation*}", "\\abs{x-x_n} < \\nicefrac{1}{n} .", "\\avoidbreak", "\\end{equation*}", "As $\\{ \\nicefrac{1}{n} \\}_{n=1}^\\infty$ converges to zero,", "$\\{ x_n \\}_{n=1}^\\infty$ converges to $x$.", "On the other hand, if we start with a sequence of numbers", "$\\{ x_n \\}_{n=1}^\\infty$ in $S$", "converging to $x$ such that $x_n \\not= x$ for all $n$, then for every", "$\\epsilon > 0$ there is an $M$ such that, in particular, $\\abs{x_M - x} <", "\\epsilon$. That is, $x_M \\in (x-\\epsilon,x+\\epsilon) \\cap S \\setminus \\{x\\}$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 1, "type": "proposition", "label": "Lebl-contfunc:1", "categories": [ "limits", "topology", "sequences" ], "title": "Uniqueness of Functions", "contents": [ "Let$c$be a cluster point of$S \\subset \\R$and let$f \\colon S \\to \\R$be a function such that$f(x)$converges as$x$goes to$c$. Then the limit of$f(x)$as$x$goes to$c$is unique." ], "refs": [], "proofs": [ { "contents": [ "Let $L_1$ and $L_2$ be two numbers that both satisfy the definition.", "Take an $\\epsilon > 0$ and find a $\\delta_1 > 0$ such that", "$\\abs{f(x)-L_1} < \\nicefrac{\\epsilon}{2}$", "for all $x \\in S \\setminus \\{c\\}$ with $\\abs{x-c} < \\delta_1$.", "Also find $\\delta_2 > 0$ such that", "$\\abs{f(x)-L_2} < \\nicefrac{\\epsilon}{2}$", "for all $x \\in S \\setminus \\{c\\}$ with $\\abs{x-c} < \\delta_2$.", "Put $\\delta \\coloneqq \\min \\{ \\delta_1, \\delta_2 \\}$. Suppose $x \\in S$,", "$\\abs{x-c} < \\delta$, and $x \\not= c$. As $\\delta > 0$ and $c$ is a cluster", "point, such an $x$ exists. Then", "\\begin{equation*}", "\\abs{L_1 - L_2} =", "\\abs{L_1 - f(x) + f(x) - L_2} \\leq", "\\abs{L_1 - f(x)} + \\abs{f(x) - L_2} < \\frac{\\epsilon}{2} + \\frac{\\epsilon}{2}", "= \\epsilon.", "\\end{equation*}", "As $\\abs{L_1-L_2} < \\epsilon$ for arbitrary $\\epsilon > 0$, then", "$L_1 = L_2$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 2, "type": "example", "label": "Lebl-contfunc:2", "categories": [ "example" ], "title": "Consider$f \\colon \\R \\to \\R$defined by$f(x) \\coloneqq x^2$", "contents": [ "Consider$f \\colon \\R \\to \\R$defined by$f(x) \\coloneqq x^2$. Then for any$c \\in \\R$,", "\\begin{equation*}\n\\lim_{x\\to c} f(x) = \\lim_{x\\to c} x^2 = c^2 .\n\\end{equation*}", "Proof: Let$c \\in \\R$be fixed, and suppose$\\epsilon > 0$is given. Write", "\\begin{equation*}\n\\delta \\coloneqq \\min \\left\\{ 1 , \\, \\frac{\\epsilon}{2\\abs{c}+1} \\right\\} .\n\\end{equation*}", "Take$x \\not= c$such that$\\abs{x-c} < \\delta$. In particular,$\\abs{x-c} < 1$. By reverse triangle inequality,", "\\begin{equation*}\n\\abs{x}-\\abs{c} \\leq \\abs{x-c} < 1 .\n\\end{equation*}", "Adding$2\\abs{c}$to both sides, we obtain$\\abs{x} + \\abs{c} < 2\\abs{c} + 1$. Estimate", "\\begin{equation*}\n\\begin{split}\n\\abs{f(x) - c^2} &= \\abs{x^2-c^2} \\\\\n&= \\abs{(x+c)(x-c)} \\\\\n&= \\abs{x+c}\\abs{x-c} \\\\\n&\\leq (\\abs{x}+\\abs{c})\\abs{x-c} \\\\\n&< (2\\abs{c}+1)\\abs{x-c} \\\\\n&< (2\\abs{c}+1)\\frac{\\epsilon}{2\\abs{c}+1} = \\epsilon .\n\\end{split}\n\\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let$c \\in \\R$be fixed, and suppose$\\epsilon > 0$is given. Write \\begin{equation*}", "\\delta \\coloneqq \\min \\left\\{ 1 , \\, \\frac{\\epsilon}{2\\abs{c}+1} \\right\\} .", "\\end{equation*} Take$x \\not= c$such that$\\abs{x-c} < \\delta$. In particular,$\\abs{x-c} < 1$. By reverse triangle inequality, \\begin{equation*}", "\\abs{x}-\\abs{c} \\leq \\abs{x-c} < 1 .", "\\end{equation*} Adding$2\\abs{c}$to both sides, we obtain$\\abs{x} + \\abs{c} < 2\\abs{c} + 1$. Estimate \\begin{equation*}", "\\begin{split}", "\\abs{f(x) - c^2} &= \\abs{x^2-c^2} \\\\", "&= \\abs{(x+c)(x-c)} \\\\", "&= \\abs{x+c}\\abs{x-c} \\\\", "&\\leq (\\abs{x}+\\abs{c})\\abs{x-c} \\\\", "&< (2\\abs{c}+1)\\abs{x-c} \\\\", "&< (2\\abs{c}+1)\\frac{\\epsilon}{2\\abs{c}+1} = \\epsilon .", "\\end{split}", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 3, "type": "example", "label": "Lebl-contfunc:3", "categories": [ "limits", "example" ], "title": "Limit of Functions", "contents": [ "Define$f \\colon [0,1) \\to \\R$by", "\\begin{equation*}\nf(x) \\coloneqq \n\\begin{cases}\nx & \\text{if } x > 0 , \\\\\n1 & \\text{if } x = 0 .\n\\end{cases}\n\\end{equation*}", "Then$\\lim\\limits_{x\\to 0} f(x) = 0$, even though$f(0) = 1$. See \\figureref{fig:limvaldiff}. \\begin{myfigureht} \\includegraphics{figures/limvaldiff} \\caption{Function with a different limit and value at$0$.\\label{fig:limvaldiff}} \\end{myfigureht}", "Proof: Let$\\epsilon > 0$be given. Let$\\delta \\coloneqq \\epsilon$. For$x \\in [0,1)$,$x \\not= 0$, and$\\abs{x-0} < \\delta$, we get", "\\begin{equation*}\n\\abs{f(x) - 0} = \\abs{x} < \\delta = \\epsilon .\n\\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let$\\epsilon > 0$be given. Let$\\delta \\coloneqq \\epsilon$. For$x \\in [0,1)$,$x \\not= 0$, and$\\abs{x-0} < \\delta$, we get \\begin{equation*}", "\\abs{f(x) - 0} = \\abs{x} < \\delta = \\epsilon .", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 4, "type": "lemma", "label": "Lebl-contfunc:seqflimit:lemma", "categories": [ "characterization", "auxiliary result", "sequences", "topology" ], "title": "Characterization of Functions via Equivalence", "contents": [ "Let$S \\subset \\R$, let$c$be a cluster point of$S$, let$f \\colon S \\to \\R$be a function, and let$L \\in \\R$.", "Then$f(x) \\to L$as$x \\to c$if and only if for every sequence$\\{ x_n \\}_{n=1}^\\infty$such that$x_n \\in S \\setminus \\{c\\}$for all$n$, and such that$\\lim_{n\\to\\infty} x_n = c$, we have that the sequence$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$converges to$L$." ], "refs": [], "proofs": [ { "contents": [ "Suppose", "$f(x) \\to L$ as $x \\to c$, and $\\{ x_n \\}_{n=1}^\\infty$ is a sequence", "such that", "$x_n \\in S \\setminus \\{c\\}$ and", "$\\lim_{n\\to\\infty} x_n = c$.", "We wish to show that $\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$ converges to $L$.", "Let $\\epsilon > 0$ be given. Find a $\\delta > 0$ such that", "if $x \\in S \\setminus \\{c\\}$ and $\\abs{x-c} < \\delta$, then", "$\\abs{f(x) - L} < \\epsilon$. As", "$\\{ x_n \\}_{n=1}^\\infty$ converges to $c$, find an $M$ such that for $n \\geq M$,", "we have that $\\abs{x_n - c} < \\delta$. Therefore, for $n \\geq M$,", "\\begin{equation*}", "\\abs{f(x_n) - L} < \\epsilon .", "\\end{equation*}", "Thus $\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$ converges to $L$.", "For the other direction, we use proof by contrapositive. Suppose", "it is not true that $f(x) \\to L$ as $x \\to c$. The negation of the", "definition is that there exists an $\\epsilon > 0$ such that for every", "$\\delta > 0$ there exists an $x \\in S \\setminus \\{c\\}$, where", "$\\abs{x-c} < \\delta$", "and $\\abs{f(x)-L} \\geq \\epsilon$.", "Let us use $\\nicefrac{1}{n}$ for $\\delta$ in the statement above to", "construct a sequence $\\{ x_n \\}_{n=1}^\\infty$. We have", "that there exists an $\\epsilon > 0$ such that for every $n$,", "there exists a point $x_n \\in S \\setminus \\{c\\}$, where", "$\\abs{x_n-c} < \\nicefrac{1}{n}$", "and $\\abs{f(x_n)-L} \\geq \\epsilon$.", "The sequence $\\{ x_n \\}_{n=1}^\\infty$ just constructed converges to $c$, but", "the sequence $\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$ does not converge to $L$.", "And we are done." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 5, "type": "example", "label": "Lebl-contfunc:5", "categories": [ "limits", "sequences", "example" ], "title": "Convergence of Sequences", "contents": [ "$\\displaystyle \\lim_{x \\to 0} \\sin( \\nicefrac{1}{x} )$does not exist, but$\\displaystyle \\lim_{x \\to 0} x\\sin( \\nicefrac{1}{x} ) = 0$. See \\figureref{figsin1x}.", "\\begin{myfigureht} %left guy also used in 10.4 \\subimport*{figures/}{sin1x_xsin1x.pdf_t} \\caption{Graphs of$\\sin(\\nicefrac{1}{x})$and$x \\sin(\\nicefrac{1}{x})$. Note that the computer cannot properly graph$\\sin(\\nicefrac{1}{x})$near zero as it oscillates too fast.\\label{figsin1x}} \\end{myfigureht}", "Proof: We start with$\\sin(\\nicefrac{1}{x})$. Define a sequence by$x_n \\coloneqq \\frac{1}{\\pi n + \\nicefrac{\\pi}{2}}$. It is not hard to see that$\\lim_{n\\to\\infty} x_n = 0$. Furthermore,", "\\begin{equation*}\n\\sin ( \\nicefrac{1}{x_n} )\n=\n\\sin (\\pi n + \\nicefrac{\\pi}{2})\n= {(-1)}^n .\n\\end{equation*}", "Therefore,$\\bigl\\{ \\sin ( \\nicefrac{1}{x_n} ) \\bigr\\}_{n=1}^\\infty$does not converge. By \\lemmaref{seqflimit:lemma},$\\displaystyle \\lim_{x \\to 0} \\sin( \\nicefrac{1}{x} )$does not exist.", "Now consider$x\\sin(\\nicefrac{1}{x})$. Let$\\{ x_n \\}_{n=1}^\\infty$be a sequence such that$x_n \\not= 0$for all$n$, and such that$\\lim_{n\\to\\infty} x_n = 0$. Notice that$\\abs{\\sin(t)} \\leq 1$for all$t \\in \\R$. Therefore,", "\\begin{equation*}\n\\abs{x_n\\sin(\\nicefrac{1}{x_n})-0}\n=\n\\abs{x_n}\\abs{\\sin(\\nicefrac{1}{x_n})}\n\\leq\n\\abs{x_n} .\n\\end{equation*}", "As$x_n$goes to 0, then$\\abs{x_n}$goes to zero, and hence$\\bigl\\{ x_n\\sin(\\nicefrac{1}{x_n}) \\bigr\\}_{n=1}^\\infty$converges to zero. By \\lemmaref{seqflimit:lemma},$\\displaystyle \\lim_{x \\to 0} x\\sin( \\nicefrac{1}{x} ) = 0$." ], "refs": [ "seqflimit:lemma" ], "proofs": [ { "contents": [ "We start with$\\sin(\\nicefrac{1}{x})$. Define a sequence by$x_n \\coloneqq \\frac{1}{\\pi n + \\nicefrac{\\pi}{2}}$. It is not hard to see that$\\lim_{n\\to\\infty} x_n = 0$. Furthermore, \\begin{equation*}", "\\sin ( \\nicefrac{1}{x_n} )", "=", "\\sin (\\pi n + \\nicefrac{\\pi}{2})", "= {(-1)}^n .", "\\end{equation*} Therefore,$\\bigl\\{ \\sin ( \\nicefrac{1}{x_n} ) \\bigr\\}_{n=1}^\\infty$does not converge. By \\lemmaref{seqflimit:lemma},$\\displaystyle \\lim_{x \\to 0} \\sin( \\nicefrac{1}{x} )$does not exist. Now consider$x\\sin(\\nicefrac{1}{x})$. Let$\\{ x_n \\}_{n=1}^\\infty$be a sequence such that$x_n \\not= 0$for all$n$, and such that$\\lim_{n\\to\\infty} x_n = 0$. Notice that$\\abs{\\sin(t)} \\leq 1$for all$t \\in \\R$. Therefore, \\begin{equation*}", "\\abs{x_n\\sin(\\nicefrac{1}{x_n})-0}", "=", "\\abs{x_n}\\abs{\\sin(\\nicefrac{1}{x_n})}", "\\leq", "\\abs{x_n} .", "\\end{equation*} As$x_n$goes to 0, then$\\abs{x_n}$goes to zero, and hence$\\bigl\\{ x_n\\sin(\\nicefrac{1}{x_n}) \\bigr\\}_{n=1}^\\infty$converges to zero. By \\lemmaref{seqflimit:lemma},$\\displaystyle \\lim_{x \\to 0} x\\sin( \\nicefrac{1}{x} ) = 0$." ], "refs": [ "seqflimit:lemma" ], "ref_ids": [ 4 ] } ], "ref_ids": [ 4 ] }, { "id": 6, "type": "corollary", "label": "Lebl-contfunc:6", "categories": [ "consequence", "limits", "topology" ], "title": "Limit of Functions", "contents": [ "Let$S \\subset \\R$and let$c$be a cluster point of$S$. Suppose$f \\colon S \\to \\R$and$g \\colon S \\to \\R$are functions such that the limits of$f(x)$and$g(x)$as$x$goes to$c$both exist, and\\begin{equation*} f(x) \\leq g(x) \\qquad \\text{for all } x \\in S \\setminus \\{ c \\}. \\end{equation*}Then\\begin{equation*} \\lim_{x\\to c} f(x) \\leq \\lim_{x\\to c} g(x) . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Take $\\{ x_n \\}_{n=1}^\\infty$ be a sequence of numbers in $S \\setminus \\{ c \\}$", "that converges to $c$. Let", "\\begin{equation*}", "L_1 \\coloneqq \\lim_{x\\to c} f(x), \\qquad \\text{and} \\qquad L_2 \\coloneqq \\lim_{x\\to c} g(x) .", "\\end{equation*}", "\\lemmaref{seqflimit:lemma} says that $\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$ converges to", "$L_1$ and $\\bigl\\{ g(x_n) \\bigr\\}_{n=1}^\\infty$ converges to $L_2$. We also", "have $f(x_n) \\leq g(x_n)$ for all $n$.", "We obtain $L_1 \\leq L_2$ using", "\\lemmaref{limandineq:lemma}." ], "refs": [ "limandineq:lemma", "seqflimit:lemma" ], "ref_ids": [ 4 ] } ], "ref_ids": [] }, { "id": 7, "type": "corollary", "label": "Lebl-contfunc:fconstineq:cor", "categories": [ "consequence", "limits", "topology" ], "title": "Existence of Functions", "contents": [ "Let$S \\subset \\R$and let$c$be a cluster point of$S$. Suppose$f \\colon S \\to \\R$is a function such that the limit of$f(x)$as$x$goes to$c$exists. Suppose there are two real numbers$a$and$b$such that\\begin{equation*} a \\leq f(x) \\leq b \\qquad \\text{for all } x \\in S \\setminus \\{ c \\}. \\end{equation*}Then\\begin{equation*} a \\leq \\lim_{x\\to c} f(x) \\leq b . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 8, "type": "corollary", "label": "Lebl-contfunc:fsqueeze:cor", "categories": [ "consequence", "limits", "topology" ], "title": "Existence of Functions", "contents": [ "Let$S \\subset \\R$and let$c$be a cluster point of$S$. Suppose$f \\colon S \\to \\R$,$g \\colon S \\to \\R$, and$h \\colon S \\to \\R$are functions such that\\begin{equation*} f(x) \\leq g(x) \\leq h(x) \\qquad \\text{for all } x \\in S \\setminus \\{ c \\}. \\end{equation*}Suppose the limits of$f(x)$and$h(x)$as$x$goes to$c$both exist, and\\begin{equation*} \\lim_{x\\to c} f(x) = \\lim_{x\\to c} h(x) . \\end{equation*}Then the limit of$g(x)$as$x$goes to$c$exists and\\begin{equation*} \\lim_{x\\to c} g(x) = \\lim_{x\\to c} f(x) = \\lim_{x\\to c} h(x) . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 9, "type": "corollary", "label": "Lebl-contfunc:falg:cor", "categories": [ "consequence", "limits", "topology" ], "title": "Limit of Functions", "contents": [ "Let$S \\subset \\R$and let$c$be a cluster point of$S$. Suppose$f \\colon S \\to \\R$and$g \\colon S \\to \\R$are functions such that the limits of$f(x)$and$g(x)$as$x$goes to$c$both exist. Then \\begin{enumerate}[(i)] \\item$\\displaystyle \\lim_{x\\to c} \\bigl(f(x)+g(x)\\bigr) = \\left(\\lim_{x\\to c} f(x)\\right) + \\left(\\lim_{x\\to c} g(x)\\right) . $\\item$\\displaystyle \\lim_{x\\to c} \\bigl(f(x)-g(x)\\bigr) = \\left(\\lim_{x\\to c} f(x)\\right) - \\left(\\lim_{x\\to c} g(x)\\right) . $\\item$\\displaystyle \\lim_{x\\to c} \\bigl(f(x)g(x)\\bigr) = \\left(\\lim_{x\\to c} f(x)\\right) \\left(\\lim_{x\\to c} g(x)\\right) . $\\item \\label{falg:cor:iv} If$\\displaystyle \\lim_{x\\to c} g(x) \\not= 0$and$g(x) \\not= 0$for all$x \\in S \\setminus \\{ c \\}$, then\\begin{equation*} \\lim_{x\\to c} \\frac{f(x)}{g(x)} = \\frac{\\lim_{x\\to c} f(x)}{\\lim_{x\\to c} g(x)} . \\end{equation*}\\end{enumerate}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 10, "type": "corollary", "label": "Lebl-contfunc:fabs:cor", "categories": [ "consequence", "limits", "topology" ], "title": "Existence of Functions", "contents": [ "Let$S \\subset \\R$and let$c$be a cluster point of$S$. Suppose$f \\colon S \\to \\R$is a function such that the limit of$f(x)$as$x$goes to$c$exists. Then\\begin{equation*} \\lim_{x\\to c} \\abs{f(x)} = \\abs{\\lim_{x\\to c} f(x)}. \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 11, "type": "proposition", "label": "Lebl-contfunc:prop:limrest", "categories": [ "characterization", "topology" ], "title": "Let$S \\subset \\R$,$c \\in \\R$, and let$f \\colon S \\to \\R$be a function", "contents": [ "Let$S \\subset \\R$,$c \\in \\R$, and let$f \\colon S \\to \\R$be a function. Suppose$A \\subset S$is such that there is some$\\alpha > 0$such that$(A \\setminus \\{ c \\}) \\cap (c-\\alpha,c+\\alpha) = (S \\setminus \\{ c \\}) \\cap (c-\\alpha,c+\\alpha)$. \\begin{enumerate}[(i)] \\item The point$c$is a cluster point of$A$if and only if$c$is a cluster point of$S$. \\item Supposing$c$is a cluster point of$S$, then$f(x) \\to L$as$x \\to c$if and only if$f|_A(x) \\to L$as$x \\to c$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "First, let $c$ be a cluster point of $A$.", "Since $A \\subset S$, then if $( A \\setminus \\{ c\\} ) \\cap", "(c-\\epsilon,c+\\epsilon)$ is nonempty for every $\\epsilon > 0$,", "then $( S \\setminus \\{ c\\} ) \\cap", "(c-\\epsilon,c+\\epsilon)$ is nonempty for every $\\epsilon > 0$.", "Thus $c$ is a cluster point of $S$.", "Second, suppose $c$ is a cluster", "point of $S$. Then for $\\epsilon > 0$ such that $\\epsilon < \\alpha$", "we get that $( A \\setminus \\{ c\\} ) \\cap (c-\\epsilon,c+\\epsilon) =", "( S \\setminus \\{ c\\} ) \\cap (c-\\epsilon,c+\\epsilon)$, which is nonempty. This is true for all", "$\\epsilon < \\alpha$ and hence", "$( A \\setminus \\{ c\\} ) \\cap (c-\\epsilon,c+\\epsilon)$ must be nonempty for all", "$\\epsilon > 0$. Thus $c$ is a cluster point of $A$.", "Now suppose $c$ is a cluster point of $S$ and $f(x) \\to L$ as $x \\to c$. That is, for every $\\epsilon > 0$", "there is a $\\delta > 0$ such that if $x \\in S \\setminus \\{ c \\}$", "and $\\abs{x-c} < \\delta$, then $\\abs{f(x)-L} < \\epsilon$. Because $A \\subset S$,", "if $x \\in A \\setminus \\{ c \\}$, then $x \\in S \\setminus \\{ c \\}$,", "and hence $f|_A(x) \\to L$ as $x \\to c$.", "Finally, suppose $f|_A(x) \\to L$ as $x \\to c$ and let $\\epsilon > 0$ be", "given.", "There is a $\\delta' > 0$ such that if $x \\in A \\setminus \\{ c \\}$", "and $\\abs{x-c} < \\delta'$, then $\\bigl\\lvert f|_A(x)-L \\bigr\\rvert < \\epsilon$.", "Take $\\delta \\coloneqq \\min \\{ \\delta', \\alpha \\}$.", "Now suppose $x \\in S \\setminus \\{ c \\}$ and", "$\\abs{x-c} < \\delta$. As $\\abs{x-c} < \\alpha$, we find $x \\in A \\setminus \\{ c \\}$,", "and as $\\abs{x-c} < \\delta'$,", "we get $\\abs{f(x)-L} = \\bigl\\lvert f|_A(x)-L", "\\bigr\\rvert < \\epsilon$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 12, "type": "proposition", "label": "Lebl-contfunc:prop:onesidedlimits", "categories": [ "characterization", "topology" ], "title": "Characterization of Functions via Equivalence", "contents": [ "Let$S \\subset \\R$be such that$c$is a cluster point of both$S \\cap (-\\infty,c)$and$S \\cap (c,\\infty)$, let$f \\colon S \\to \\R$be a function, and let$L \\in \\R$. Then$c$is a cluster point of$S$and\\begin{equation*} \\lim_{x \\to c} f(x) = L \\qquad \\text{if and only if} \\qquad \\lim_{x \\to c^-} f(x) = \\lim_{x \\to c^+} f(x) = L . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 13, "type": "proposition", "label": "Lebl-contfunc:contbasic:prop", "categories": [ "limits", "characterization", "topology", "sequences", "continuity" ], "title": "Continuity of Functions", "contents": [ "Consider a function$f \\colon S \\to \\R$defined on a set$S \\subset \\R$and let$c \\in S$. Then: %\\begin{enumerate}[(i),itemsep=0.5\\itemsep,parsep=0.5\\parsep,topsep=0.5\\topsep,partopsep=0.5\\partopsep] \\begin{enumerate}[(i)] \\item \\label{contbasic:prop:i} If$c$is not a cluster point of$S$, then$f$is continuous at$c$. \\item \\label{contbasic:prop:ii} If$c$is a cluster point of$S$, then$f$is continuous at$c$if and only if the limit of$f(x)$as$x \\to c$exists and\\begin{equation*} \\lim_{x\\to c} f(x) = f(c) . \\end{equation*}\\item \\label{contbasic:prop:iii} The function$f$is continuous at$c$if and only if for every sequence$\\{ x_n \\}_{n=1}^\\infty$where$x_n \\in S$and$\\lim\\limits_{n\\to\\infty} x_n = c$, the sequence$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$converges to$f(c)$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "\\pagebreak[2]", "We start with \\ref{contbasic:prop:i}. Suppose $c$ is not a cluster point of", "$S$. Then there exists a $\\delta > 0$", "such that $S \\cap (c-\\delta,c+\\delta) = \\{ c \\}$.", "For any $\\epsilon > 0$, simply pick this given $\\delta$.", "The only $x \\in S$ such that $\\abs{x-c} < \\delta$ is $x=c$. Then", "$\\abs{f(x)-f(c)} = \\abs{f(c)-f(c)} = 0 < \\epsilon$.", "Let us move to \\ref{contbasic:prop:ii}.", "Suppose $c$ is a cluster point of $S$. Let us first suppose", "that $\\lim_{x\\to c} f(x) = f(c)$. Then for every $\\epsilon > 0$,", "there is a $\\delta > 0$ such that if $x \\in S \\setminus \\{ c \\}$", "and $\\abs{x-c} < \\delta$, then $\\abs{f(x)-f(c)} < \\epsilon$.", "Also $\\abs{f(c)-f(c)} = 0 < \\epsilon$, so the definition of continuity at", "$c$ is satisfied. On the other hand, suppose $f$ is continuous", "at $c$. For every $\\epsilon > 0$, there exists a $\\delta > 0$", "such that for $x \\in S$ where $\\abs{x-c} < \\delta$, we have", "$\\abs{f(x)-f(c)} < \\epsilon$. Then the statement is, of course, still true if", "$x \\in S \\setminus \\{ c \\} \\subset S$. Therefore, $\\lim_{x\\to c} f(x) =", "f(c)$.", "For \\ref{contbasic:prop:iii}, first suppose $f$ is continuous at $c$.", "Let $\\{ x_n \\}_{n=1}^\\infty$", "be a sequence such that $x_n \\in S$ and $\\lim_{n\\to\\infty} x_n = c$. Let $\\epsilon > 0$", "be given. Find a $\\delta > 0$ such that $\\abs{f(x)-f(c)} < \\epsilon$", "for all $x \\in S$ where $\\abs{x-c} < \\delta$. Find an $M \\in \\N$", "such that for $n \\geq M$, we have $\\abs{x_n-c} < \\delta$. Then for", "$n \\geq M$, we have that $\\abs{f(x_n)-f(c)} < \\epsilon$,", "so $\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$", "converges to $f(c)$.", "We prove the other direction of \\ref{contbasic:prop:iii} by contrapositive.", "Suppose $f$ is not", "continuous at $c$. Then there exists an $\\epsilon > 0$", "such that for every $\\delta > 0$, there exists an $x \\in S$", "such that $\\abs{x-c} < \\delta$ and $\\abs{f(x)-f(c)} \\geq \\epsilon$.", "Define a sequence $\\{ x_n \\}_{n=1}^\\infty$ as follows.", "Let $x_n \\in S$ be such that $\\abs{x_n-c} < \\nicefrac{1}{n}$", "and $\\abs{f(x_n)-f(c)} \\geq \\epsilon$.", "Now $\\{ x_n \\}_{n=1}^\\infty$ is", "a sequence in $S$ such that", "$\\lim_{n\\to\\infty} x_n = c$ and such that", "$\\abs{f(x_n)-f(c)} \\geq \\epsilon$ for all $n \\in \\N$.", "Thus $\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$", "does not converge to $f(c)$. It may or may not converge, but it definitely", "does not converge to $f(c)$." ], "refs": [ "contbasic:prop:i", "contbasic:prop:ii", "contbasic:prop:iii" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 14, "type": "example", "label": "Lebl-contfunc:14", "categories": [ "continuity", "sequences", "example" ], "title": "Continuity of Functions", "contents": [ "The function$f \\colon (0,\\infty) \\to \\R$defined by$f(x) \\coloneqq \\nicefrac{1}{x}$is continuous.", "Proof: Fix$c \\in (0,\\infty)$. Let$\\{ x_n \\}_{n=1}^\\infty$be a sequence in$(0,\\infty)$such that$\\lim_{n\\to\\infty} x_n = c$. Then", "\\begin{equation*}\nf(c) = \\frac{1}{c}\n=\n\\frac{1}{\\lim_{n\\to\\infty} x_n}\n=\n\\lim_{n \\to \\infty} \\frac{1}{x_n}\n=\n\\lim_{n \\to \\infty} f(x_n) .\n\\end{equation*}", "Thus$f$is continuous at$c$. As$f$is continuous at all$c \\in (0,\\infty)$,$f$is continuous." ], "refs": [], "proofs": [ { "contents": [ "Fix$c \\in (0,\\infty)$. Let$\\{ x_n \\}_{n=1}^\\infty$be a sequence in$(0,\\infty)$such that$\\lim_{n\\to\\infty} x_n = c$. Then \\begin{equation*}", "f(c) = \\frac{1}{c}", "=", "\\frac{1}{\\lim_{n\\to\\infty} x_n}", "=", "\\lim_{n \\to \\infty} \\frac{1}{x_n}", "=", "\\lim_{n \\to \\infty} f(x_n) .", "\\end{equation*} Thus$f$is continuous at$c$. As$f$is continuous at all$c \\in (0,\\infty)$,$f$is continuous." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 15, "type": "proposition", "label": "Lebl-contfunc:15", "categories": [ "continuity" ], "title": "Continuity of Continuous Functions", "contents": [ "Let$f \\colon \\R \\to \\R$be a \\emph{\\myindex{polynomial}}. That is,\\begin{equation*} f(x) = a_d x^d + a_{d-1} x^{d-1} + \\cdots + a_1 x + a_0 , \\end{equation*}for some constants$a_0, a_1, \\ldots, a_d$. Then$f$is continuous." ], "refs": [], "proofs": [ { "contents": [ "Fix $c \\in \\R$.", "Let $\\{ x_n \\}_{n=1}^\\infty$ be a sequence such that", "$\\lim_{n\\to\\infty} x_n = c$. Then", "\\begin{equation*}", "\\begin{split}", "f(c) &=", "a_d c^d + a_{d-1} c^{d-1} + \\cdots + a_1 c + a_0", "\\\\", "&=", "a_d {\\left(\\lim_{n\\to\\infty} x_n\\right)}^d", "+ a_{d-1} {\\left(\\lim_{n\\to\\infty} x_n\\right)}^{d-1}", "+ \\cdots", "+ a_1 \\left(\\lim_{n\\to\\infty} x_n\\right) + a_0", "\\\\", "& =", "\\lim_{n \\to \\infty}", "\\left(", "a_d x_n^d + a_{d-1} x_n^{d-1} + \\cdots + a_1 x_n + a_0", "\\right)", "=", "\\lim_{n \\to \\infty}", "f(x_n) .", "\\avoidbreak", "\\end{split}", "\\end{equation*}", "Thus $f$ is continuous at $c$. As $f$ is continuous at all $c \\in \\R$,", "$f$ is continuous." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 16, "type": "proposition", "label": "Lebl-contfunc:contalg:prop", "categories": [ "continuity" ], "title": "Continuity of Functions", "contents": [ "Let$f \\colon S \\to \\R$and$g \\colon S \\to \\R$be functions continuous at$c \\in S$. \\begin{enumerate}[(i)] \\item The function$h \\colon S \\to \\R$defined by$h(x) \\coloneqq f(x)+g(x)$is continuous at$c$. \\item The function$h \\colon S \\to \\R$defined by$h(x) \\coloneqq f(x)-g(x)$is continuous at$c$. \\item The function$h \\colon S \\to \\R$defined by$h(x) \\coloneqq f(x)g(x)$is continuous at$c$. \\item If$g(x)\\not=0$for all$x \\in S$, the function$h \\colon S \\to \\R$given by$h(x) \\coloneqq \\frac{f(x)}{g(x)}$is continuous at$c$. \\end{enumerate}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 17, "type": "example", "label": "Lebl-contfunc:sincos:example", "categories": [ "continuity", "sequences", "example" ], "title": "Continuity of Functions", "contents": [ "The functions$\\sin(x)$and$\\cos(x)$are continuous. In the following computations we use the sum-to-product trigonometric identities. We also use the simple facts that$\\abs{\\sin(x)} \\leq \\abs{x}$,$\\abs{\\cos(x)} \\leq 1$, and$\\abs{\\sin(x)} \\leq 1$.", "\\begin{equation*}\n\\begin{split}\n\\abs{\\sin(x)-\\sin(c)} & =\n\\abs{\n2 \\sin \\left( \\frac{x-c}{2} \\right) \\cos \\left( \\frac{x+c}{2} \\right)\n}\n\\\\\n& =\n2\n\\abs{ \\sin \\left( \\frac{x-c}{2} \\right) }\n\\abs{ \\cos \\left( \\frac{x+c}{2} \\right) }\n\\\\\n& \\leq\n2\n\\abs{ \\sin \\left( \\frac{x-c}{2} \\right) }\n\\\\\n& \\leq\n2\n\\abs{ \\frac{x-c}{2} }\n= \\abs{x-c}\n\\end{split}\n\\end{equation*}", "\\begin{equation*}\n\\begin{split}\n\\abs{\\cos(x)-\\cos(c)} & =\n\\abs{\n-2 \\sin \\left( \\frac{x-c}{2} \\right) \\sin \\left( \\frac{x+c}{2} \\right)\n}\n\\\\\n& =\n2\n\\abs{ \\sin \\left( \\frac{x-c}{2} \\right) }\n\\abs{ \\sin \\left( \\frac{x+c}{2} \\right) }\n\\\\\n& \\leq\n2\n\\abs{ \\sin \\left( \\frac{x-c}{2} \\right) }\n\\\\\n& \\leq\n2\n\\abs{ \\frac{x-c}{2} }\n= \\abs{x-c}\n\\end{split}\n\\end{equation*}", "The claim that$\\sin$and$\\cos$are continuous follows by taking an arbitrary sequence$\\{ x_n \\}_{n=1}^\\infty$converging to$c$, or by applying the definition of continuity directly. Details are left to the reader." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 18, "type": "proposition", "label": "Lebl-contfunc:prop:compositioncont", "categories": [ "continuity" ], "title": "Continuity of Functions", "contents": [ "Let$A, B \\subset \\R$and$f \\colon B \\to \\R$and$g \\colon A \\to B$be functions. If$g$is continuous at$c \\in A$and$f$is continuous at$g(c)$, then$f \\circ g \\colon A \\to \\R$is continuous at$c$." ], "refs": [], "proofs": [ { "contents": [ "Let $\\{ x_n \\}_{n=1}^\\infty$ be a sequence in $A$ such that", "$\\lim_{n\\to\\infty} x_n = c$.", "As $g$ is continuous at $c$, we have $\\bigl\\{ g(x_n) \\bigr\\}_{n=1}^\\infty$ converges to $g(c)$.", "As $f$ is continuous at $g(c)$, we have $\\bigl\\{ f\\bigl(g(x_n)\\bigr)", "\\bigr\\}_{n=1}^\\infty$ converges", "to $f\\bigl(g(c)\\bigr)$.", "Thus $f \\circ g$ is continuous at $c$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 19, "type": "example", "label": "Lebl-contfunc:19", "categories": [ "continuity", "example", "intervals" ], "title": "Continuity of Functions", "contents": [ "Claim: \\emph{${\\bigl(\\sin(\\nicefrac{1}{x})\\bigr)}^2$is a continuous function on$(0,\\infty)$.}", "Proof: The function$\\nicefrac{1}{x}$is continuous on$(0,\\infty)$and$\\sin(x)$is continuous on$(0,\\infty)$(actually on$\\R$, but$(0,\\infty)$is the range for$\\nicefrac{1}{x}$). Hence the composition$\\sin(\\nicefrac{1}{x})$is continuous. Also,$x^2$is continuous on the interval$[-1,1]$(the range of$\\sin$). Thus the composition${\\bigl(\\sin(\\nicefrac{1}{x})\\bigr)}^2$is continuous on$(0,\\infty)$." ], "refs": [], "proofs": [ { "contents": [ "The function$\\nicefrac{1}{x}$is continuous on$(0,\\infty)$and$\\sin(x)$is continuous on$(0,\\infty)$(actually on$\\R$, but$(0,\\infty)$is the range for$\\nicefrac{1}{x}$). Hence the composition$\\sin(\\nicefrac{1}{x})$is continuous. Also,$x^2$is continuous on the interval$[-1,1]$(the range of$\\sin$). Thus the composition${\\bigl(\\sin(\\nicefrac{1}{x})\\bigr)}^2$is continuous on$(0,\\infty)$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 20, "type": "proposition", "label": "Lebl-contfunc:20", "categories": [ "continuity", "sequences" ], "title": "Continuity of Functions", "contents": [ "Let$f \\colon S \\to \\R$be a function and$c \\in S$. Suppose there exists a sequence$\\{ x_n \\}_{n=1}^\\infty$,$x_n \\in S$for all$n$, and$\\lim_{n\\to\\infty} x_n = c$such that$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$does not converge to$f(c)$. Then$f$is discontinuous at$c$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 21, "type": "example", "label": "Lebl-contfunc:example:jumpdiscont", "categories": [ "continuity", "sequences", "example" ], "title": "Continuity of Functions", "contents": [ "The function$f \\colon \\R \\to \\R$defined by", "\\begin{equation*}\nf(x) \\coloneqq \n\\begin{cases}\n-1 & \\text{if } x < 0, \\\\\n1 & \\text{if } x \\geq 0\n\\end{cases}\n\\end{equation*}", "is not continuous at 0.", "Proof: Consider$\\{ \\nicefrac{-1}{n} \\}_{n=1}^\\infty$, which converges to 0. Then$f(\\nicefrac{-1}{n}) = -1$for every$n$, and so$\\lim_{n\\to\\infty} f(\\nicefrac{-1}{n}) = -1$, but$f(0) = 1$. Thus the function is not continuous at 0. See \\figureref{fig:jumpdiscont}.", "\\begin{myfigureht} \\includegraphics{figures/jumpdiscont} \\caption{Jump discontinuity. The values of$f(\\nicefrac{-1}{n})$and$f(0)$are marked as black dots.\\label{fig:jumpdiscont}} \\end{myfigureht}", "Notice that$f(\\nicefrac{1}{n}) = 1$for all$n \\in \\N$. Hence,$\\lim_{n\\to\\infty} f(\\nicefrac{1}{n}) = f(0) = 1$. So$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$may converge to$f(0)$for some specific sequence$\\{ x_n \\}_{n=1}^\\infty$going to 0, despite the function being discontinuous at 0.", "Finally, consider$f\\Bigl(\\frac{{(-1)}^n}{n}\\Bigr) = {(-1)}^n$. This sequence diverges." ], "refs": [], "proofs": [ { "contents": [ "Consider$\\{ \\nicefrac{-1}{n} \\}_{n=1}^\\infty$, which converges to 0. Then$f(\\nicefrac{-1}{n}) = -1$for every$n$, and so$\\lim_{n\\to\\infty} f(\\nicefrac{-1}{n}) = -1$, but$f(0) = 1$. Thus the function is not continuous at 0. See \\figureref{fig:jumpdiscont}. \\begin{myfigureht} \\includegraphics{figures/jumpdiscont} \\caption{Jump discontinuity. The values of$f(\\nicefrac{-1}{n})$and$f(0)$are marked as black dots.\\label{fig:jumpdiscont}} \\end{myfigureht} Notice that$f(\\nicefrac{1}{n}) = 1$for all$n \\in \\N$. Hence,$\\lim_{n\\to\\infty} f(\\nicefrac{1}{n}) = f(0) = 1$. So$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$may converge to$f(0)$for some specific sequence$\\{ x_n \\}_{n=1}^\\infty$going to 0, despite the function being discontinuous at 0. Finally, consider$f\\Bigl(\\frac{{(-1)}^n}{n}\\Bigr) = {(-1)}^n$. This sequence diverges." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 22, "type": "example", "label": "Lebl-contfunc:22", "categories": [ "continuity", "sequences", "example" ], "title": "Continuity of Functions", "contents": [ "For an extreme example, take the so-called \\emph{\\myindex{Dirichlet function}}\\footnote{Named after the German mathematician \\href{https://en.wikipedia.org/wiki/Peter_Gustav_Lejeune_Dirichlet}{Johann Peter Gustav Lejeune Dirichlet} (1805--1859).}.", "\\begin{equation*}\nf(x) \\coloneqq\n\\begin{cases}\n1 & \\text{if } x \\text{ is rational,} \\\\\n0 & \\text{if } x \\text{ is irrational.}\n\\end{cases}\n\\avoidbreak\n\\end{equation*}", "The function$f$is discontinuous at all$c \\in \\R$.", "Proof: If$c$is rational, take a sequence$\\{ x_n \\}_{n=1}^\\infty$of irrational numbers such that$\\lim_{n\\to\\infty} x_n = c$(why can we?). Then$f(x_n) = 0$and so$\\lim_{n\\to\\infty} f(x_n) = 0$, but$f(c) = 1$. If$c$is irrational, take a sequence of rational numbers$\\{ x_n \\}_{n=1}^\\infty$that converges to$c$(why can we?). Then$\\lim_{n\\to\\infty} f(x_n) = 1$, but$f(c) = 0$." ], "refs": [], "proofs": [ { "contents": [ "If$c$is rational, take a sequence$\\{ x_n \\}_{n=1}^\\infty$of irrational numbers such that$\\lim_{n\\to\\infty} x_n = c$(why can we?). Then$f(x_n) = 0$and so$\\lim_{n\\to\\infty} f(x_n) = 0$, but$f(c) = 1$. If$c$is irrational, take a sequence of rational numbers$\\{ x_n \\}_{n=1}^\\infty$that converges to$c$(why can we?). Then$\\lim_{n\\to\\infty} f(x_n) = 1$, but$f(c) = 0$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 23, "type": "example", "label": "Lebl-contfunc:popcornfunction:example", "categories": [ "continuity", "sequences", "example" ], "title": "Continuity of Functions", "contents": [ "Define$f \\colon (0,1) \\to \\R$as", "\\begin{equation*}\nf(x) \\coloneqq \n\\begin{cases}\n\\nicefrac{1}{k} & \\text{if } x=\\nicefrac{m}{k}, \\text{ where } m,k \\in \\N\n\\text{ and have no common divisors (lowest terms),} \\\\\n0 & \\text{if } x \\text{ is irrational.}\n\\end{cases}\n\\end{equation*}", "See the graph of$f$in \\figureref{popcornfig}. We claim that$f$is continuous at all irrational$c$and discontinuous at all rational$c$. \\begin{myfigureht} \\includegraphics{figures/popcornfig} \\caption{Graph of the \\myquote{popcorn function.}\\label{popcornfig}} \\end{myfigureht}", "Proof: Let$c = \\nicefrac{m}{k}$be rational and in lowest terms. Take a sequence of irrational numbers$\\{ x_n \\}_{n=1}^\\infty$such that$\\lim_{n\\to\\infty} x_n = c$. Then$\\lim_{n\\to\\infty} f(x_n) = \\lim_{n\\to\\infty} 0 = 0$, but$f(c) = \\nicefrac{1}{k} \\not= 0$. So$f$is discontinuous at$c$.", "Now let$c$be irrational, so$f(c) = 0$. Take a sequence$\\{ x_n \\}_{n=1}^\\infty$in$(0,1)$such that$\\lim_{n\\to\\infty} x_n = c$. Given$\\epsilon > 0$, find$K \\in \\N$such that$\\nicefrac{1}{K} < \\epsilon$by the \\hyperref[thm:arch:i]{Archimedean property}. If$\\nicefrac{m}{k} \\in (0,1)$and$m,k \\in \\N$, then$0 < m < k$. So there are only finitely many rational numbers in$(0,1)$whose denominator$k$in lowest terms is less than$K$. As$\\lim_{n\\to\\infty} x_n = c$, every number not equal to$c$can appear at most finitely many times in$\\{ x_n \\}_{n=1}^\\infty$. Hence, there is an$M$such that for$n \\geq M$, all the numbers$x_n$that are rational have a denominator larger than or equal to$K$. Thus for$n \\geq M$,", "\\begin{equation*}\n\\abs{f(x_n) - 0} = f(x_n) \\leq \\nicefrac{1}{K} < \\epsilon .\n\\end{equation*}", "Therefore,$f$is continuous at irrational$c$." ], "refs": [ "thm:arch:i" ], "proofs": [ { "contents": [ "Let$c = \\nicefrac{m}{k}$be rational and in lowest terms. Take a sequence of irrational numbers$\\{ x_n \\}_{n=1}^\\infty$such that$\\lim_{n\\to\\infty} x_n = c$. Then$\\lim_{n\\to\\infty} f(x_n) = \\lim_{n\\to\\infty} 0 = 0$, but$f(c) = \\nicefrac{1}{k} \\not= 0$. So$f$is discontinuous at$c$. Now let$c$be irrational, so$f(c) = 0$. Take a sequence$\\{ x_n \\}_{n=1}^\\infty$in$(0,1)$such that$\\lim_{n\\to\\infty} x_n = c$. Given$\\epsilon > 0$, find$K \\in \\N$such that$\\nicefrac{1}{K} < \\epsilon$by the \\hyperref[thm:arch:i]{Archimedean property}. If$\\nicefrac{m}{k} \\in (0,1)$and$m,k \\in \\N$, then$0 < m < k$. So there are only finitely many rational numbers in$(0,1)$whose denominator$k$in lowest terms is less than$K$. As$\\lim_{n\\to\\infty} x_n = c$, every number not equal to$c$can appear at most finitely many times in$\\{ x_n \\}_{n=1}^\\infty$. Hence, there is an$M$such that for$n \\geq M$, all the numbers$x_n$that are rational have a denominator larger than or equal to$K$. Thus for$n \\geq M$, \\begin{equation*}", "\\abs{f(x_n) - 0} = f(x_n) \\leq \\nicefrac{1}{K} < \\epsilon .", "\\end{equation*} Therefore,$f$is continuous at irrational$c$." ], "refs": [ "thm:arch:i" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 24, "type": "example", "label": "Lebl-contfunc:example:removablediscont", "categories": [ "continuity", "example" ], "title": "Continuity of Functions", "contents": [ "Define$g \\colon \\R \\to \\R$by$g(x) \\coloneqq 0$if$x \\not= 0$and$g(0) \\coloneqq 1$. Then$g$is not continuous at zero, but continuous everywhere else (why?). The point$x=0$is called a \\emph{\\myindex{removable discontinuity}}. That is because if we would change the definition of$g$, by insisting that$g(0)$be$0$, we would obtain a continuous function. On the other hand, let$f$be the function of \\exampleref{example:jumpdiscont}. Then$f$does not have a removable discontinuity at$0$. No matter how we would define$f(0)$the function would still fail to be continuous. The difference is that$\\lim_{x\\to 0} g(x)$exists while$\\lim_{x\\to 0} f(x)$does not.", "We stay with this example to show another phenomenon. Let$A \\coloneqq \\{ 0 \\}$, then$g|_A$is continuous (why?), while$g$is not continuous on$A$. Similarly, if$B \\coloneqq \\R \\setminus \\{0 \\}$, then$g|_B$is also continuous, and$g$is in fact continuous on$B$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 25, "type": "lemma", "label": "Lebl-contfunc:25", "categories": [ "continuity", "auxiliary result", "boundedness", "intervals" ], "title": "Continuity of Functions", "contents": [ "A continuous function$f \\colon [a,b] \\to \\R$is bounded." ], "refs": [], "proofs": [ { "contents": [ "We prove the claim by contrapositive. Suppose $f$ is not bounded.", "Then for each", "$n \\in \\N$, there is an $x_n \\in [a,b]$, such that", "\\begin{equation*}", "\\abs{f(x_n)} \\geq n .", "\\end{equation*}", "The sequence $\\{ x_n \\}_{n=1}^\\infty$ is bounded as $a \\leq x_n \\leq b$.", "By the \\hyperref[thm:bwseq]{Bolzano--Weierstrass theorem},", "there is a convergent subsequence $\\{ x_{n_i} \\}_{i=1}^\\infty$.", "Let $x \\coloneqq \\lim_{i\\to\\infty} x_{n_i}$.", "Since $a \\leq x_{n_i} \\leq b$ for all $i$, then $a \\leq x \\leq b$.", "The sequence $\\bigl\\{ f(x_{n_i}) \\bigr\\}_{i=1}^\\infty$ is not bounded", "as", "$\\abs{f(x_{n_i})} \\geq n_i \\geq i$.", "Thus $f$ is not continuous at $x$ as", "\\begin{equation*}", "f(x)", "=", "f\\Bigl( \\lim_{i\\to\\infty} x_{n_i} \\Bigr) ,", "\\qquad \\text{but} \\qquad", "\\lim_{i\\to\\infty} f(x_{n_i}) \\enspace \\text{does not exist.} \\qedhere", "\\end{equation*}" ], "refs": [ "named:Bolzano", "named:Weierstrass", "thm:bwseq" ], "ref_ids": [ 32 ] } ], "ref_ids": [] }, { "id": 26, "type": "theorem", "label": "Lebl-contfunc:26", "categories": [ "continuity", "intervals" ], "title": "Extreme Value Theorem", "contents": [ "\\index{minimum-maximum theorem}% \\index{maximum-minimum theorem}% \\index{extreme value theorem}% A continuous function$f \\colon [a,b] \\to \\R$achieves both an absolute minimum and an absolute maximum on$[a,b]$." ], "refs": [ "named:Extreme Value" ], "proofs": [ { "contents": [ "The lemma says that $f$ is bounded, so", "the set $f\\bigl([a,b]\\bigr) = \\bigl\\{ f(x) : x \\in [a,b] \\bigr\\}$ has a supremum and an infimum.", "There exist sequences", "in the set $f\\bigl([a,b]\\bigr)$ that approach its supremum and its infimum.", "That is, there are sequences", "$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$ and $\\bigl\\{ f(y_n)", "\\bigr\\}_{n=1}^\\infty$, where $x_n$ and $y_n$ are in $[a,b]$,", "such that", "\\begin{equation*}", "\\lim_{n\\to\\infty} f(x_n) = \\inf f\\bigl([a,b]\\bigr) \\qquad \\text{and} \\qquad", "\\lim_{n\\to\\infty} f(y_n) = \\sup f\\bigl([a,b]\\bigr).", "\\end{equation*}", "We are not done yet; we need to find where the minima and the maxima are.", "The problem is that the sequences $\\{ x_n \\}_{n=1}^\\infty$ and", "$\\{ y_n \\}_{n=1}^\\infty$ need not converge.", "We know $\\{ x_n \\}_{n=1}^\\infty$ and $\\{ y_n \\}_{n=1}^\\infty$ are bounded", "(their elements belong to a bounded interval $[a,b]$).", "Apply the", "\\hyperref[thm:bwseq]{Bolzano--Weierstrass theorem}", "to find", "convergent subsequences", "$\\{ x_{n_i} \\}_{i=1}^\\infty$ and", "$\\{ y_{m_i} \\}_{i=1}^\\infty$. Let", "\\begin{equation*}", "x \\coloneqq \\lim_{i\\to\\infty} x_{n_i}", "\\qquad \\text{and} \\qquad", "y \\coloneqq \\lim_{i\\to\\infty} y_{m_i}.", "\\end{equation*}", "As $a \\leq x_{n_i} \\leq b$ for all $i$, we have $a \\leq x \\leq b$.", "Similarly, $a \\leq y \\leq b$. So $x$ and $y$ are in $[a,b]$.", "A limit of a subsequence is the same as the limit of the", "sequence, and we can take a limit past the continuous function $f$:", "\\begin{equation*}", "\\inf f\\bigl([a,b]\\bigr) = \\lim_{n\\to\\infty} f(x_n)", "= \\lim_{i\\to\\infty} f(x_{n_i}) =", "f \\Bigl( \\lim_{i\\to\\infty} x_{n_i} \\Bigr) = f(x) .", "\\end{equation*}", "Similarly,", "\\begin{equation*}", "\\sup f\\bigl([a,b]\\bigr) = \\lim_{n\\to\\infty} f(y_n)", "= \\lim_{i\\to\\infty} f(y_{m_i}) =", "f \\Bigl( \\lim_{i\\to\\infty} y_{m_i} \\Bigr) = f(y) .", "\\end{equation*}", "Hence, $f$ achieves an absolute minimum at $x$ and", "an absolute maximum at $y$." ], "refs": [ "named:Bolzano", "named:Weierstrass", "thm:bwseq" ], "ref_ids": [ 32 ] } ], "ref_ids": [] }, { "id": 27, "type": "example", "label": "Lebl-contfunc:27", "categories": [ "example", "intervals" ], "title": "The function$f(x) \\coloneqq x^2+1$defined on the interval$[-1,2]$achieves a minimum at$x=0$when$f...", "contents": [ "The function$f(x) \\coloneqq x^2+1$defined on the interval$[-1,2]$achieves a minimum at$x=0$when$f(0) = 1$. It achieves a maximum at$x=2$where$f(2) = 5$. Do note that the domain of definition matters. If we instead took the domain to be$[-10,10]$, then$f$would no longer have a maximum at$x=2$. Instead, the maximum would be achieved at either$x=10$or$x=-10$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 28, "type": "example", "label": "Lebl-contfunc:28", "categories": [ "example", "boundedness", "intervals" ], "title": "Boundedness of Functions", "contents": [ "The function$f \\colon \\R \\to \\R$defined by$f(x) \\coloneqq x$achieves neither a minimum, nor a maximum. So it is important that we are looking at a bounded interval." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 29, "type": "example", "label": "Lebl-contfunc:29", "categories": [ "boundedness", "intervals", "topology", "example", "continuity" ], "title": "Continuity of Functions", "contents": [ "The function$f \\colon (0,1) \\to \\R$defined by$f(x) \\coloneqq \\nicefrac{1}{x}$achieves neither a minimum, nor a maximum. It is continuous, but$(0,1)$is not closed. The values of the function are unbounded as we approach$0$. Also as we approach$x=1$, the values of the function approach$1$, but$f(x) > 1$for all$x \\in (0,1)$. There is no$x \\in (0,1)$such that$f(x) = 1$. So it is important that we are looking at a closed interval." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 30, "type": "example", "label": "Lebl-contfunc:30", "categories": [ "continuity", "topology", "boundedness", "example" ], "title": "Continuity of Functions", "contents": [ "Continuity is important. Define$f \\colon [0,1] \\to \\R$by$f(x) \\coloneqq \\nicefrac{1}{x}$for$x > 0$and let$f(0) \\coloneqq 0$. The function does not achieve a maximum. The domain$[0,1]$is closed and bounded, but the problem is that the function is not continuous at 0." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 31, "type": "lemma", "label": "Lebl-contfunc:IVT:lemma", "categories": [ "continuity", "auxiliary result", "intervals" ], "title": "Continuity of Functions", "contents": [ "Let$f \\colon [a,b] \\to \\R$be a continuous function. Suppose$f(a) < 0$and$f(b) > 0$. Then there exists a number$c \\in (a,b)$such that$f(c) = 0$." ], "refs": [], "proofs": [ { "contents": [ "We define two sequences $\\{ a_n \\}_{n=1}^\\infty$", "and $\\{ b_n \\}_{n=1}^\\infty$ inductively:", "\\begin{enumerate}[(i)]", "\\item Let $a_1 \\coloneqq a$ and $b_1 \\coloneqq b$.", "\\item If $f\\left(\\frac{a_n+b_n}{2}\\right) \\geq 0$, let $a_{n+1} \\coloneqq a_n$ and", "$b_{n+1} \\coloneqq \\frac{a_n+b_n}{2}$.", "\\item If $f\\left(\\frac{a_n+b_n}{2}\\right) < 0$, let $a_{n+1} \\coloneqq \\frac{a_n+b_n}{2}$ and", "$b_{n+1} \\coloneqq b_n$.", "\\end{enumerate}", "\\begin{myfigureht}", "\\includegraphics{figures/bisect}", "\\caption{Finding roots (bisection method).\\label{bisectfig}}", "\\end{myfigureht}", "See \\figureref{bisectfig} for an example defining the first five steps.", "If $a_n < b_n$, then $a_n < \\frac{a_n+b_n}{2} < b_n$. So", "$a_{n+1} < b_{n+1}$.", "Thus by \\hyperref[induction:thm]{induction} $a_n < b_n$ for all $n$.", "Furthermore, $a_n \\leq a_{n+1}$ and", "$b_n \\geq b_{n+1}$ for all $n$, that is, the sequences are monotone.", "As $a_n < b_n \\leq b_1 = b$ and", "$b_n > a_n \\geq a_1 = a$ for all $n$,", "the sequences are also bounded. Therefore, the", "sequences converge.", "Let $c \\coloneqq \\lim_{n\\to\\infty} a_n$ and $d \\coloneqq \\lim_{n\\to\\infty} b_n$,", "where also $a \\leq c \\leq d \\leq b$. We need", "to show that $c=d$.", "Notice", "\\begin{equation*}", "b_{n+1} - a_{n+1} = \\frac{b_n-a_n}{2}.", "\\end{equation*}", "By \\hyperref[induction:thm]{induction},", "\\begin{equation*}", "b_n - a_n = \\frac{b_1-a_1}{2^{n-1}} = 2^{1-n} (b-a) .", "\\end{equation*}", "As $2^{1-n}(b-a)$ converges to zero, we take the limit as $n$ goes to", "infinity to get", "\\begin{equation*}", "d-c = \\lim_{n\\to\\infty} (b_n - a_n) =", "\\lim_{n\\to\\infty} 2^{1-n} (b-a) = 0.", "\\end{equation*}", "In other words, $d=c$.", "By construction, for all $n$,", "\\begin{equation*}", "f(a_n) < 0", "\\qquad \\text{and} \\qquad", "f(b_n) \\geq 0 .", "\\end{equation*}", "Since", "$\\lim_{n\\to\\infty} a_n = \\lim_{n\\to\\infty} b_n = c$", "and $f$ is continuous at $c$, we may take", "limits in those inequalities:", "\\begin{equation*}", "f(c) = \\lim_{n\\to\\infty} f(a_n) \\leq 0", "\\qquad \\text{and} \\qquad", "f(c) = \\lim_{n\\to\\infty} f(b_n) \\geq 0 .", "\\end{equation*}", "As $f(c) \\geq 0$ and", "$f(c) \\leq 0$, we conclude $f(c) = 0$.", "Thus also $c \\not=a$ and $c \\not= b$, so", "$a < c < b$." ], "refs": [ "induction:thm" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 32, "type": "theorem", "label": "Lebl-contfunc:IVT:thm", "categories": [ "continuity", "named theorem", "intervals" ], "title": "Bolzano's Intermediate Value Theorem", "contents": [ "\\index{Bolzano's theorem} \\index{Bolzano's intermediate value theorem} \\index{intermediate value theorem} Let$f \\colon [a,b] \\to \\R$be a continuous function. Suppose$y \\in \\R$is such that$f(a) < y < f(b)$or$f(a) > y > f(b)$. Then there exists a$c \\in (a,b)$such that$f(c) = y$." ], "refs": [ "named:Bolzano" ], "proofs": [ { "contents": [ "If $f(a) < y < f(b)$, then define $g(x) \\coloneqq f(x)-y$. Then", "$g(a) < 0$ and $g(b) > 0$, and we apply \\lemmaref{IVT:lemma}", "to $g$ to find $c$. If $g(c) = 0$, then $f(c) = y$.", "Similarly, if $f(a) > y > f(b)$, then define $g(x) \\coloneqq y-f(x)$.", "Again, $g(a) < 0$ and $g(b) > 0$, and we apply \\lemmaref{IVT:lemma} to", "find $c$.", "As before, if $g(c) = 0$, then $f(c) = y$." ], "refs": [ "IVT:lemma", "named:Bolzano" ], "ref_ids": [ 31, 32 ] } ], "ref_ids": [] }, { "id": 33, "type": "example", "label": "Lebl-contfunc:33", "categories": [ "example" ], "title": "%\\index{bisection method} The polynomial$f(x) \\coloneqq x^3-2x^2+x-1$has a real root in$(1,2)$", "contents": [ "%\\index{bisection method} The polynomial$f(x) \\coloneqq x^3-2x^2+x-1$has a real root in$(1,2)$. We simply notice that$f(1) = -1$and$f(2) = 1$. Hence there must exist a point$c \\in (1,2)$such that$f(c) = 0$. To find a better approximation of the root we follow the proof of \\lemmaref{IVT:lemma}. We look at 1.5 and find that$f(1.5) = -0.625$. Therefore, there is a root of the polynomial in$(1.5,2)$. Next we look at 1.75 and note that$f(1.75) \\approx -0.016$. Hence there is a root of$f$in$(1.75,2)$. Next we look at 1.875 and find that$f(1.875) \\approx 0.44$, thus there is a root in$(1.75,1.875)$. We follow this procedure until we gain sufficient precision. In fact, the root is at$c \\approx 1.7549$." ], "refs": [ "IVT:lemma", "named:Bolzano" ], "proofs": [], "ref_ids": [ 31, 32 ] }, { "id": 34, "type": "proposition", "label": "Lebl-contfunc:34", "categories": [], "title": "Let$f(x)$be a polynomial of odd degree", "contents": [ "Let$f(x)$be a polynomial of odd degree. Then$f$has a real root." ], "refs": [], "proofs": [ { "contents": [ "Suppose $f$ is a polynomial of odd degree $d$. We write", "\\begin{equation*}", "f(x) = a_d x^d + a_{d-1} x^{d-1} + \\cdots + a_1 x + a_0 ,", "\\end{equation*}", "where $a_d \\not= 0$. We divide by $a_d$ to obtain a", "\\emph{\\myindex{monic polynomial}}\\footnote{The word \\emph{monic} means that", "the coefficient of $x^d$ is 1.}", "\\begin{equation*}", "g(x) \\coloneqq x^d + b_{d-1} x^{d-1} + \\cdots + b_1 x + b_0 ,", "\\end{equation*}", "where $b_k = \\nicefrac{a_k}{a_d}$.", "Let us show that $g(n)$ is", "positive for some large $n \\in \\N$.", "We first compare the highest order term with the rest:", "\\begin{equation*}", "\\begin{split}", "\\abs{\\frac{b_{d-1} n^{d-1} + \\cdots + b_1 n + b_0}{n^d}}", "& =", "\\frac{\\abs{b_{d-1} n^{d-1} + \\cdots + b_1 n + b_0}}{n^d}", "\\\\", "& \\leq", "\\frac{\\abs{b_{d-1}} n^{d-1} + \\cdots + \\abs{b_1} n + \\abs{b_0}}{n^d}", "\\\\", "& \\leq", "\\frac{\\abs{b_{d-1}} n^{d-1} + \\cdots + \\abs{b_1} n^{d-1} + \\abs{b_0} n^{d-1}}{n^d}", "\\\\", "& =", "\\frac{n^{d-1}\\bigl(\\abs{b_{d-1}} + \\cdots + \\abs{b_1} + \\abs{b_0}\\bigr)}{n^d}", "\\\\", "& =", "\\frac{1}{n}", "\\bigl(\\abs{b_{d-1}} + \\cdots + \\abs{b_1} + \\abs{b_0}\\bigr) .", "\\end{split}", "\\end{equation*}", "Therefore,", "\\begin{equation*}", "\\lim_{n\\to\\infty} \\frac{b_{d-1} n^{d-1} + \\cdots + b_1 n + b_0}{n^d}", "= 0 .", "\\end{equation*}", "Thus there exists an $M \\in \\N$ such that", "\\begin{equation*}", "\\abs{\\frac{b_{d-1} M^{d-1} + \\cdots + b_1 M + b_0}{M^d}} < 1 ,", "\\end{equation*}", "which implies", "\\begin{equation*}", "-(b_{d-1} M^{d-1} + \\cdots + b_1 M + b_0) < M^d .", "\\end{equation*}", "Therefore, $g(M) > 0$.", "Next, consider $g(-n)$ for $n \\in \\N$. By a similar argument,", "there exists a $K \\in \\N$ such that", "$b_{d-1} {(-K)}^{d-1} + \\cdots + b_1 (-K) + b_0 < K^d$", "and therefore $g(-K) < 0$ (see \\exerciseref{exercise:odddegnegativeK}).", "In the proof,", "make sure you use the fact that $d$ is odd.", "In particular, if $d$ is odd, then ${(-n)}^d = -(n^d)$.", "We appeal to the intermediate value theorem to find a", "$c \\in (-K,M)$, such that $g(c) = 0$. As $g(x) = \\frac{f(x)}{a_d}$,", "then $f(c) = 0$, and the proof is done." ], "refs": [ "named:Bolzano" ], "ref_ids": [ 32 ] } ], "ref_ids": [] }, { "id": 35, "type": "example", "label": "Lebl-contfunc:35", "categories": [ "existence", "example", "named theorem" ], "title": "Bolzano's Intermediate Value Theorem", "contents": [ "You may recall how hard we worked in \\exampleref{example:sqrt2} to show that$\\sqrt{2}$exists. With Bolzano's theorem, we can prove the existence$k$th root of any positive number$y > 0$without any effort. We claim that for any$k \\in \\N$and any$y > 0$, there exists a number$x > 0$such that$x^k = y$.", "Proof: If$y=1$, then it is clear, so assume$y\\not= 1$. Let$f(x) \\coloneqq x^k - y$. We notice$f(0) = -y < 0$. If$y < 1$, then$f(1) = 1^k -y > 0$. If$y > 1$, then$f(y) = y^k-y = y(y^{k-1}-1) > 0$. In either case, apply Bolzano's theorem to find an$x > 0$such that$f(x) = 0$, or in other words$x^k = y$." ], "refs": [ "named:Bolzano" ], "proofs": [ { "contents": [ "If$y=1$, then it is clear, so assume$y\\not= 1$. Let$f(x) \\coloneqq x^k - y$. We notice$f(0) = -y < 0$. If$y < 1$, then$f(1) = 1^k -y > 0$. If$y > 1$, then$f(y) = y^k-y = y(y^{k-1}-1) > 0$. In either case, apply Bolzano's theorem to find an$x > 0$such that$f(x) = 0$, or in other words$x^k = y$." ], "refs": [ "named:Bolzano" ], "ref_ids": [ 32 ] } ], "ref_ids": [ 32 ] }, { "id": 36, "type": "example", "label": "Lebl-contfunc:36", "categories": [ "continuity", "example", "intervals" ], "title": "Intermediate Value Theorem", "contents": [ "Interestingly, there exist discontinuous functions with the intermediate value property. The function", "\\begin{equation*}\nf(x) \\coloneqq\n\\begin{cases}\n\\sin(\\nicefrac{1}{x}) & \\text{if } x \\not= 0, \\\\\n0 & \\text{if } x=0,\n\\end{cases}\n\\end{equation*}", "is not continuous at$0$; however,$f$has the intermediate value property: Whenever$a < b$and$y$is such that$f(a) < y < f(b)$or$f(a) > y > f(b)$, there exists a$c \\in (a,b)$such that$f(c) = y$. See \\figureref{figsin1x} for a graph of$\\sin(\\nicefrac{1}{x})$. Proof is left as \\exerciseref{exercise:meanvaluepropsin1x}." ], "refs": [ "named:Bolzano" ], "proofs": [], "ref_ids": [ 32 ] }, { "id": 37, "type": "corollary", "label": "Lebl-contfunc:cor:imageofinterval", "categories": [ "boundedness", "intervals", "topology", "consequence", "continuity" ], "title": "Continuity of Continuous Functions", "contents": [ "If$f \\colon [a,b] \\to \\R$is continuous, then the direct image$f\\bigl([a,b]\\bigr)$is a closed and bounded interval or a single number." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 38, "type": "example", "label": "Lebl-contfunc:38", "categories": [ "continuity", "example" ], "title": "Continuity of Continuous Functions", "contents": [ "$f \\colon [0,1] \\to \\R$defined by$f(x) \\coloneqq x^2$is uniformly continuous.", "Proof: Note that$0 \\leq x,c \\leq 1$. Then", "\\begin{equation*}\n\\sabs{x^2-c^2} = \\sabs{x+c}\\sabs{x-c}\n\\leq \\bigl(\\sabs{x}+\\sabs{c}\\bigr) \\sabs{x-c}\n\\leq (1+1)\\sabs{x-c} .\n\\end{equation*}", "Therefore, given$\\epsilon > 0$, let$\\delta \\coloneqq \\nicefrac{\\epsilon}{2}$. If$\\sabs{x-c} < \\delta$, then$\\sabs{x^2-c^2} < \\epsilon$.", "\\medskip", "On the other hand,$g \\colon \\R \\to \\R$defined by$g(x) \\coloneqq x^2$is not uniformly continuous.", "Proof: Suppose it is uniformly continuous, then for every$\\epsilon > 0$, there would exist a$\\delta > 0$such that if$\\sabs{x-c} < \\delta$, then$\\sabs{x^2 -c^2} < \\epsilon$. Take$x > 0$and let$c \\coloneqq x+\\nicefrac{\\delta}{2}$. Write", "\\begin{equation*}\n\\epsilon >\n\\sabs{x^2-c^2} = \\sabs{x+c}\\sabs{x-c}\n=\n(2x+\\nicefrac{\\delta}{2})\\nicefrac{\\delta}{2} \n\\geq \n\\delta x .\n\\end{equation*}", "Therefore,$x < \\nicefrac{\\epsilon}{\\delta}$for all$x > 0$, which is a contradiction." ], "refs": [], "proofs": [ { "contents": [ "Note that$0 \\leq x,c \\leq 1$. Then \\begin{equation*}", "\\sabs{x^2-c^2} = \\sabs{x+c}\\sabs{x-c}", "\\leq \\bigl(\\sabs{x}+\\sabs{c}\\bigr) \\sabs{x-c}", "\\leq (1+1)\\sabs{x-c} .", "\\end{equation*} Therefore, given$\\epsilon > 0$, let$\\delta \\coloneqq \\nicefrac{\\epsilon}{2}$. If$\\sabs{x-c} < \\delta$, then$\\sabs{x^2-c^2} < \\epsilon$. \\medskip On the other hand,$g \\colon \\R \\to \\R$defined by$g(x) \\coloneqq x^2$is not uniformly continuous. Proof: Suppose it is uniformly continuous, then for every$\\epsilon > 0$, there would exist a$\\delta > 0$such that if$\\sabs{x-c} < \\delta$, then$\\sabs{x^2 -c^2} < \\epsilon$. Take$x > 0$and let$c \\coloneqq x+\\nicefrac{\\delta}{2}$. Write \\begin{equation*}", "\\epsilon >", "\\sabs{x^2-c^2} = \\sabs{x+c}\\sabs{x-c}", "=", "(2x+\\nicefrac{\\delta}{2})\\nicefrac{\\delta}{2}", "\\geq", "\\delta x .", "\\end{equation*} Therefore,$x < \\nicefrac{\\epsilon}{\\delta}$for all$x > 0$, which is a contradiction." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 39, "type": "example", "label": "Lebl-contfunc:39", "categories": [ "continuity", "characterization", "example" ], "title": "Characterization of Functions via Equivalence", "contents": [ "The function$f \\colon (0,1) \\to \\R$defined by$f(x) \\coloneqq \\nicefrac{1}{x}$is not uniformly continuous.", "Proof: Given$\\epsilon > 0$, then$\\epsilon > \\abs{\\nicefrac{1}{x}-\\nicefrac{1}{y}}$holds if and only if", "\\begin{equation*}\n\\epsilon >\n\\abs{\\nicefrac{1}{x}-\\nicefrac{1}{y}}\n=\n\\frac{\\abs{y-x}}{\\abs{xy}} \n=\n\\frac{\\abs{y-x}}{xy} ,\n\\end{equation*}", "or", "\\begin{equation*}\n\\abs{x-y} < xy \\epsilon .\n\\end{equation*}", "Suppose$\\epsilon < 1$, and we wish to see if a small$\\delta > 0$would work. If$x \\in (0,1)$and$y = x+\\nicefrac{\\delta}{2} \\in (0,1)$, then$\\abs{x-y} = \\nicefrac{\\delta}{2} < \\delta$. We plug$y$into the inequality above to get$\\nicefrac{\\delta}{2} < x \\bigl( x+\\nicefrac{\\delta}{2} \\bigr) \\epsilon < x$. If the definition of uniform continuity is satisfied, then the inequality$\\nicefrac{\\delta}{2} < x$holds for all$x > 0$. But then$\\delta \\leq 0$. Therefore, no single$\\delta > 0$works for all points." ], "refs": [], "proofs": [ { "contents": [ "Given$\\epsilon > 0$, then$\\epsilon > \\abs{\\nicefrac{1}{x}-\\nicefrac{1}{y}}$holds if and only if \\begin{equation*}", "\\epsilon >", "\\abs{\\nicefrac{1}{x}-\\nicefrac{1}{y}}", "=", "\\frac{\\abs{y-x}}{\\abs{xy}}", "=", "\\frac{\\abs{y-x}}{xy} ,", "\\end{equation*} or \\begin{equation*}", "\\abs{x-y} < xy \\epsilon .", "\\end{equation*} Suppose$\\epsilon < 1$, and we wish to see if a small$\\delta > 0$would work. If$x \\in (0,1)$and$y = x+\\nicefrac{\\delta}{2} \\in (0,1)$, then$\\abs{x-y} = \\nicefrac{\\delta}{2} < \\delta$. We plug$y$into the inequality above to get$\\nicefrac{\\delta}{2} < x \\bigl( x+\\nicefrac{\\delta}{2} \\bigr) \\epsilon < x$. If the definition of uniform continuity is satisfied, then the inequality$\\nicefrac{\\delta}{2} < x$holds for all$x > 0$. But then$\\delta \\leq 0$. Therefore, no single$\\delta > 0$works for all points." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 40, "type": "theorem", "label": "Lebl-contfunc:unifcont:thm", "categories": [ "continuity", "intervals" ], "title": "Continuity of Functions", "contents": [ "Let$f \\colon [a,b] \\to \\R$be a continuous function. Then$f$is uniformly continuous." ], "refs": [], "proofs": [ { "contents": [ "We prove the statement by contrapositive.", "Suppose $f$ is not uniformly continuous. We will prove", "that there is some", "$c \\in [a,b]$ where $f$ is not continuous. Let us negate", "the definition of uniformly continuous.", "There exists an $\\epsilon > 0$", "such that for every $\\delta > 0$, there exist points $x, y$ in $[a,b]$ with", "$\\abs{x-y} < \\delta$ and $\\abs{f(x)-f(y)} \\geq \\epsilon$.", "So for the $\\epsilon > 0$ above,", "we find sequences $\\{ x_n \\}_{n=1}^\\infty$ and $\\{ y_n \\}_{n=1}^\\infty$ such that", "$\\abs{x_n-y_n} < \\nicefrac{1}{n}$ and such that $\\abs{f(x_n)-f(y_n)} \\geq", "\\epsilon$. By", "\\hyperref[thm:bwseq]{Bolzano--Weierstrass},", "there exists a convergent subsequence", "$\\{ x_{n_k} \\}_{k=1}^\\infty$. Let $c \\coloneqq \\lim_{k\\to\\infty} x_{n_k}$.", "As $a \\leq x_{n_k} \\leq b$ for all $k$, we have $a \\leq c \\leq b$. Estimate", "\\begin{equation*}", "\\sabs{y_{n_k} - c} =", "\\sabs{y_{n_k} - x_{n_k} + x_{n_k} - c} \\leq", "\\sabs{y_{n_k} - x_{n_k}}", "+", "\\sabs{x_{n_k}-c}", "<", "\\nicefrac{1}{n_k}", "+", "\\sabs{x_{n_k}-c} .", "\\end{equation*}", "As $\\nicefrac{1}{n_k}$ and $\\abs{x_{n_k}-c}$ both go to zero when", "$k$ goes to infinity, $\\{ y_{n_k} \\}_{k=1}^\\infty$ converges and the limit", "is $c$. We now show that $f$ is not continuous at $c$.", "Estimate", "\\begin{equation*}", "\\begin{split}", "\\abs{f(x_{n_k}) - f(c)} & =", "\\abs{f(x_{n_k}) - f(y_{n_k}) + f(y_{n_k}) - f(c)} \\\\", "& \\geq", "\\abs{f(x_{n_k}) - f(y_{n_k})} - \\abs{f(y_{n_k}) - f(c)} \\\\", "& \\geq", "\\epsilon - \\abs{f(y_{n_k})-f(c)} .", "\\end{split}", "\\end{equation*}", "Or in other words,", "\\begin{equation*}", "\\abs{f(x_{n_k})-f(c)}", "+", "\\abs{f(y_{n_k})-f(c)} \\geq", "\\epsilon .", "\\end{equation*}", "At least one of the sequences $\\bigl\\{ f(x_{n_k}) \\bigr\\}_{k=1}^\\infty$ or", "$\\bigl\\{ f(y_{n_k}) \\bigr\\}_{k=1}^\\infty$ cannot converge to $f(c)$, otherwise the", "left-hand side of the inequality would go to zero while the right-hand side is positive.", "Thus $f$ cannot be continuous at $c$." ], "refs": [ "named:Bolzano", "named:Weierstrass", "thm:bwseq" ], "ref_ids": [ 32 ] } ], "ref_ids": [] }, { "id": 41, "type": "lemma", "label": "Lebl-contfunc:unifcauchycauchy:lemma", "categories": [ "continuity", "sequences", "auxiliary result" ], "title": "Continuity of Functions", "contents": [ "Let$S \\subset \\R$and let$f \\colon S \\to \\R$be a uniformly continuous function. Let$\\{ x_n \\}_{n=1}^\\infty$be a Cauchy sequence in$S$. Then$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$is Cauchy." ], "refs": [], "proofs": [ { "contents": [ "Let $\\epsilon > 0$ be given. There is a $\\delta > 0$ such that", "$\\abs{f(x)-f(y)} < \\epsilon$ whenever $x,y \\in S$ and $\\abs{x-y} < \\delta$. Find an $M", "\\in \\N$ such that for all $n, k \\geq M$, we have $\\abs{x_n-x_k} < \\delta$.", "Then for all $n, k \\geq M$, we have $\\abs{f(x_n)-f(x_k)} < \\epsilon$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 42, "type": "proposition", "label": "Lebl-contfunc:context:prop", "categories": [ "continuity", "limits", "characterization", "intervals" ], "title": "Characterization of Functions via Equivalence", "contents": [ "A function$f \\colon (a,b) \\to \\R$is uniformly continuous if and only if the limits\\begin{equation*} L_a \\coloneqq \\lim_{x \\to a} f(x) \\qquad \\text{and} \\qquad L_b \\coloneqq \\lim_{x \\to b} f(x) \\end{equation*}exist and the function$\\widetilde{f} \\colon [a,b] \\to \\R$defined by \\begin{equation*} \\widetilde{f}(x) \\coloneqq \\begin{cases} f(x) & \\text{if } x \\in (a,b),", "L_a & \\text{if } x = a,", "L_b & \\text{if } x = b \\end{cases} \\end{equation*} is continuous." ], "refs": [], "proofs": [ { "contents": [ "One direction is quick. If $\\widetilde{f}$ is continuous, then", "it is uniformly continuous by \\thmref{unifcont:thm}. As $f$ is the", "restriction of $\\widetilde{f}$ to $(a,b)$, $f$ is also uniformly continuous", "(exercise).", "Now suppose $f$ is uniformly continuous. We must first show", "that the limits $L_a$ and $L_b$ exist. Let us concentrate on $L_a$.", "Take $\\{ x_n \\}_{n=1}^\\infty$ in $(a,b)$ such that", "$\\lim_{n\\to\\infty} x_n = a$.", "The sequence $\\{ x_n \\}_{n=1}^\\infty$ is Cauchy, so by", "\\lemmaref{unifcauchycauchy:lemma}", "the sequence $\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$ is Cauchy and thus convergent.", "Let $L_1 \\coloneqq \\lim_{n\\to\\infty} f(x_n)$. Take another sequence", "$\\{ y_n \\}_{n=1}^\\infty$ in $(a,b)$ such that $\\lim_{n\\to\\infty} y_n = a$. By the same reasoning", "we get $L_2 \\coloneqq \\lim_{n\\to\\infty} f(y_n)$. If we show that $L_1 = L_2$, then", "the limit $L_a = \\lim_{x\\to a} f(x)$ exists. Let $\\epsilon > 0$ be given.", "Find $\\delta > 0$ such that $\\abs{x-y} < \\delta$ implies $\\abs{f(x)-f(y)} <", "\\nicefrac{\\epsilon}{3}$. Find $M \\in \\N$ such that for", "$n \\geq M$, we have $\\abs{a-x_n} < \\nicefrac{\\delta}{2}$,", "$\\abs{a-y_n} < \\nicefrac{\\delta}{2}$,", "$\\abs{f(x_n)-L_1} < \\nicefrac{\\epsilon}{3}$, and", "$\\abs{f(y_n)-L_2} < \\nicefrac{\\epsilon}{3}$. Then for $n \\geq M$,", "\\begin{equation*}", "\\abs{x_n-y_n} =", "\\abs{x_n-a+a-y_n} \\leq", "\\abs{x_n-a}+\\abs{a-y_n} < \\nicefrac{\\delta}{2} + \\nicefrac{\\delta}{2} =", "\\delta.", "\\end{equation*}", "So", "\\begin{equation*}", "\\begin{split}", "\\abs{L_1-L_2} &=", "\\abs{L_1-f(x_n)+f(x_n)-f(y_n)+f(y_n)-L_2} \\\\", "& \\leq", "\\abs{L_1-f(x_n)}+\\abs{f(x_n)-f(y_n)}+\\abs{f(y_n)-L_2} \\\\", "& \\leq", "\\nicefrac{\\epsilon}{3} + \\nicefrac{\\epsilon}{3} + \\nicefrac{\\epsilon}{3}", "=", "\\epsilon .", "\\end{split}", "\\end{equation*}", "Therefore, $L_1 = L_2$.", "Thus $L_a$ exists. To show that $L_b$ exists is left as an exercise.", "If $L_a = \\lim_{x\\to a} f(x)$", "exists, then $\\lim_{x\\to a} \\widetilde{f}(x)$ exists", "and equals $L_a$", "(see \\propref{prop:limrest}). Similarly for $L_b$.", "Hence $\\widetilde{f}$ is continuous at $a$ and $b$.", "And since $f$ is continuous at $c \\in (a,b)$, then", "$\\widetilde{f}$ is continuous at $c \\in (a,b)$ (\\propref{prop:limrest} again)." ], "refs": [ "prop:limrest", "unifcauchycauchy:lemma" ], "ref_ids": [ 11, 41 ] } ], "ref_ids": [] }, { "id": 43, "type": "proposition", "label": "Lebl-contfunc:43", "categories": [ "continuity" ], "title": "Continuity of Functions", "contents": [ "A Lipschitz continuous function is uniformly continuous." ], "refs": [], "proofs": [ { "contents": [ "Let $f \\colon S \\to \\R$ be a function and let $K$ be a constant such that", "$\\abs{f(x)-f(y)} \\leq K \\abs{x-y}$", "for all $x, y$ in $S$.", "Let $\\epsilon > 0$ be given. Take $\\delta \\coloneqq", "\\nicefrac{\\epsilon}{K}$.", "For all $x$ and $y$ in $S$ such that", "$\\abs{x-y} < \\delta$,", "\\begin{equation*}", "\\abs{f(x)-f(y)} \\leq K \\abs{x-y} < K \\delta = K \\frac{\\epsilon}{K} =", "\\epsilon .", "\\end{equation*}", "Therefore, $f$ is uniformly continuous." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 44, "type": "example", "label": "Lebl-contfunc:44", "categories": [ "continuity", "example" ], "title": "Continuity of Functions", "contents": [ "The functions$\\sin(x)$and$\\cos(x)$are Lipschitz continuous. In \\exampleref{sincos:example} we have seen the following two inequalities.", "\\begin{equation*}\n\\abs{\\sin(x)-\\sin(y)} \n\\leq \\abs{x-y}\n\\qquad \\text{and} \\qquad\n\\abs{\\cos(x)-\\cos(y)}\n\\leq \\abs{x-y} .\n\\end{equation*}", "Hence sine and cosine are Lipschitz continuous with$K=1$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 45, "type": "example", "label": "Lebl-contfunc:45", "categories": [ "continuity", "example" ], "title": "Continuity of Functions", "contents": [ "The function$f \\colon [1,\\infty) \\to \\R$defined by$f(x) \\coloneqq \\sqrt{x}$is Lipschitz continuous. Proof:", "\\begin{equation*}\n\\abs{\\sqrt{x}-\\sqrt{y}} = \n\\abs{\\frac{x-y}{\\sqrt{x}+\\sqrt{y}}}\n=\n\\frac{\\abs{x-y}}{\\sqrt{x}+\\sqrt{y}} .\n\\end{equation*}", "As$x \\geq 1$and$y \\geq 1$, we see that$\\frac{1}{\\sqrt{x}+\\sqrt{y}} \\leq \\frac{1}{2}$. Therefore,", "\\begin{equation*}\n\\abs{\\sqrt{x}-\\sqrt{y}} = \n\\abs{\\frac{x-y}{\\sqrt{x}+\\sqrt{y}}}\n\\leq\n\\frac{1}{2}\n\\abs{x-y}.\n\\end{equation*}", "On the other hand,$g \\colon [0,\\infty) \\to \\R$defined by$g(x) \\coloneqq \\sqrt{x}$is not Lipschitz continuous. Proof: Suppose for all$x,y \\in [0,\\infty)$,", "\\begin{equation*}\n\\abs{\\sqrt{x}-\\sqrt{y}} \n\\leq\nK \\abs{x-y} ,\n\\end{equation*}", "for some$K$. Set$y=0$to obtain$\\sqrt{x} \\leq K x$. If$K > 0$, then for$x > 0$we get$\\nicefrac{1}{K} \\leq \\sqrt{x}$or$\\nicefrac{1}{K^2} \\leq x$. This cannot possibly be true for all$x > 0$. Thus no such$K > 0$exists and$g$is not Lipschitz continuous. See \\figureref{fig:sqrtgraph} and note how secant lines would be more and more vertical as we get closer to$x=0$. \\begin{myfigureht} \\includegraphics{figures/sqrtgraph} \\caption{Graph of$\\sqrt{x}$and some secant lines through$(0,0)$and$(x,\\sqrt{x})$.\\label{fig:sqrtgraph}} \\end{myfigureht}", "The last example$g$is a function that is uniformly continuous but not Lipschitz continuous. To see that$\\sqrt{x}$is uniformly continuous as a function on$[0,\\infty)$, note that it is uniformly continuous when restricted to$[0,1]$by \\thmref{unifcont:thm}. It is also Lipschitz (and so uniformly continuous) when restricted to$[1,\\infty)$. It is not hard (exercise) to show that this means that$\\sqrt{x}$is a uniformly continuous function on$[0,\\infty)$." ], "refs": [], "proofs": [ { "contents": [ "\\begin{equation*}", "\\abs{\\sqrt{x}-\\sqrt{y}} =", "\\abs{\\frac{x-y}{\\sqrt{x}+\\sqrt{y}}}", "=", "\\frac{\\abs{x-y}}{\\sqrt{x}+\\sqrt{y}} .", "\\end{equation*} As$x \\geq 1$and$y \\geq 1$, we see that$\\frac{1}{\\sqrt{x}+\\sqrt{y}} \\leq \\frac{1}{2}$. Therefore, \\begin{equation*}", "\\abs{\\sqrt{x}-\\sqrt{y}} =", "\\abs{\\frac{x-y}{\\sqrt{x}+\\sqrt{y}}}", "\\leq", "\\frac{1}{2}", "\\abs{x-y}.", "\\end{equation*} On the other hand,$g \\colon [0,\\infty) \\to \\R$defined by$g(x) \\coloneqq \\sqrt{x}$is not Lipschitz continuous. Proof: Suppose for all$x,y \\in [0,\\infty)$, \\begin{equation*}", "\\abs{\\sqrt{x}-\\sqrt{y}}", "\\leq", "K \\abs{x-y} ,", "\\end{equation*} for some$K$. Set$y=0$to obtain$\\sqrt{x} \\leq K x$. If$K > 0$, then for$x > 0$we get$\\nicefrac{1}{K} \\leq \\sqrt{x}$or$\\nicefrac{1}{K^2} \\leq x$. This cannot possibly be true for all$x > 0$. Thus no such$K > 0$exists and$g$is not Lipschitz continuous. See \\figureref{fig:sqrtgraph} and note how secant lines would be more and more vertical as we get closer to$x=0$. \\begin{myfigureht} \\includegraphics{figures/sqrtgraph} \\caption{Graph of$\\sqrt{x}$and some secant lines through$(0,0)$and$(x,\\sqrt{x})$.\\label{fig:sqrtgraph}} \\end{myfigureht} The last example$g$is a function that is uniformly continuous but not Lipschitz continuous. To see that$\\sqrt{x}$is uniformly continuous as a function on$[0,\\infty)$, note that it is uniformly continuous when restricted to$[0,1]$by \\thmref{unifcont:thm}. It is also Lipschitz (and so uniformly continuous) when restricted to$[1,\\infty)$. It is not hard (exercise) to show that this means that$\\sqrt{x}$is a uniformly continuous function on$[0,\\infty)$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 46, "type": "proposition", "label": "Lebl-contfunc:liminfty:unique", "categories": [ "limits" ], "title": "Uniqueness of Limits", "contents": [ "The limit at$\\infty$or$-\\infty$as defined above is unique if it exists." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 47, "type": "example", "label": "Lebl-contfunc:47", "categories": [ "limits", "example" ], "title": "Limit of Limits", "contents": [ "Let$f(x) \\coloneqq \\frac{1}{\\abs{x}+1}$. Then", "\\begin{equation*}\n\\lim_{x\\to \\infty} f(x) = 0 \\qquad \\text{and} \\qquad\n\\lim_{x\\to -\\infty} f(x) = 0 .\n\\end{equation*}", "Proof: Let$\\epsilon > 0$be given. Find$M > 0$large enough so that$\\frac{1}{M+1} < \\epsilon$. If$x \\geq M$, then$0 < \\frac{1}{\\abs{x}+1} = \\frac{1}{x+1} \\leq \\frac{1}{M+1} < \\epsilon$. The first limit follows. The proof for$-\\infty$is left to the reader." ], "refs": [], "proofs": [ { "contents": [ "Let$\\epsilon > 0$be given. Find$M > 0$large enough so that$\\frac{1}{M+1} < \\epsilon$. If$x \\geq M$, then$0 < \\frac{1}{\\abs{x}+1} = \\frac{1}{x+1} \\leq \\frac{1}{M+1} < \\epsilon$. The first limit follows. The proof for$-\\infty$is left to the reader." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 48, "type": "example", "label": "Lebl-contfunc:48", "categories": [ "continuity", "limits", "sequences", "example" ], "title": "Continuity of Functions", "contents": [ "Let$f(x) \\coloneqq \\sin(\\pi x)$. Then$\\lim_{x\\to\\infty} f(x)$does not exist. To prove this fact note that if$x = 2n+\\nicefrac{1}{2}$for some$n \\in \\N$, then$f(x)=1$, while if$x = 2n+\\nicefrac{3}{2}$, then$f(x)=-1$. So they cannot both be within a small$\\epsilon$of a single real number.", "Be careful not to confuse continuous limits with limits of sequences. We could say", "\\begin{equation*}\n\\lim_{n \\to \\infty} \\sin(\\pi n) = 0, \\qquad \\text{but} \\qquad\n\\lim_{x \\to \\infty} \\sin(\\pi x) \\enspace \\text{does not exist}.\n\\end{equation*}", "Of course the notation is ambiguous: Are we thinking of the sequence$\\bigl\\{ \\sin (\\pi n) \\bigr\\}_{n=1}^\\infty$or the function$\\sin(\\pi x)$of a real variable? We are simply using the convention that$n \\in \\N$, while$x \\in \\R$. When the notation is not clear, it is good to explicitly mention where the variable lives, or what kind of limit are you using. If there is possibility of confusion, one can write, for example,", "\\begin{equation*}\n\\lim_{\\substack{n \\to \\infty\\\\n \\in \\N}} \\sin(\\pi n) .\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 49, "type": "lemma", "label": "Lebl-contfunc:seqflimitinf:lemma", "categories": [ "characterization", "auxiliary result", "limits", "sequences" ], "title": "Characterization of Functions via Equivalence", "contents": [ "Suppose$f \\colon S \\to \\R$is a function,$\\infty$is a cluster point of$S \\subset \\R$, and$L \\in \\R$. Then\\begin{equation*} \\lim_{x\\to\\infty} f(x) = L \\qquad \\text{if and only if} \\qquad \\lim_{n\\to\\infty} f(x_n) = L \\end{equation*}for all sequences$\\{ x_n \\}_{n=1}^\\infty$in$S$such that$\\lim\\limits_{n\\to\\infty} x_n = \\infty$." ], "refs": [], "proofs": [ { "contents": [ "First suppose $f(x) \\to L$ as $x \\to \\infty$.", "Given an $\\epsilon > 0$, there exists an $M$ such that for all $x \\geq M$,", "we have $\\abs{f(x)-L} < \\epsilon$.", "Let $\\{ x_n \\}_{n=1}^\\infty$", "be a sequence in $S$ such that $\\lim_{n\\to\\infty} x_n = \\infty$. Then there exists an", "$N$ such that for all $n \\geq N$, we have $x_n \\geq M$. And thus", "$\\abs{f(x_n)-L} < \\epsilon$.", "We prove the converse by contrapositive. Suppose $f(x)$ does", "not go to $L$ as $x \\to \\infty$.", "This means that there exists an $\\epsilon > 0$,", "such that for every $n \\in \\N$, there exists an $x \\in S$, $x \\geq n$, let", "us call it $x_n$, such that $\\abs{f(x_n)-L} \\geq \\epsilon$.", "Consider the sequence $\\{ x_n \\}_{n=1}^\\infty$. Clearly", "$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$ does not converge to $L$. It remains to note", "that $\\lim_{n\\to\\infty} x_n = \\infty$, because $x_n \\geq n$ for all $n$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 50, "type": "example", "label": "Lebl-contfunc:50", "categories": [ "limits", "example" ], "title": "Limit of Limits", "contents": [ "Let us show that$\\lim\\limits_{x \\to \\infty} \\frac{1+x^2}{1+x} = \\infty$.", "Proof: For$x \\geq 1$, we have", "\\begin{equation*}\n\\frac{1+x^2}{1+x} \\geq \n\\frac{x^2}{x+x} = \n\\frac{x}{2} .\n\\end{equation*}", "Given$N \\in \\R$, take$M = \\max \\{ 2N+1 , 1 \\}$. If$x \\geq M$, then$x \\geq 1$and$\\nicefrac{x}{2} > N$. So", "\\begin{equation*}\n\\frac{1+x^2}{1+x} \\geq \n\\frac{x}{2} > N .\n\\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "For$x \\geq 1$, we have \\begin{equation*}", "\\frac{1+x^2}{1+x} \\geq", "\\frac{x^2}{x+x} =", "\\frac{x}{2} .", "\\end{equation*} Given$N \\in \\R$, take$M = \\max \\{ 2N+1 , 1 \\}$. If$x \\geq M$, then$x \\geq 1$and$\\nicefrac{x}{2} > N$. So \\begin{equation*}", "\\frac{1+x^2}{1+x} \\geq", "\\frac{x}{2} > N .", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 51, "type": "proposition", "label": "Lebl-contfunc:prop:inflimcompositions", "categories": [ "topology" ], "title": "Suppose$f \\colon A \\to B$,$g \\colon B \\to \\R$,$A, B \\subset \\R$,$a \\in \\R \\cup \\{ -\\infty, \\infty...", "contents": [ "Suppose$f \\colon A \\to B$,$g \\colon B \\to \\R$,$A, B \\subset \\R$,$a \\in \\R \\cup \\{ -\\infty, \\infty\\}$is a cluster point of$A$, and$b \\in \\R \\cup \\{ -\\infty, \\infty\\}$is a cluster point of$B$. Suppose\\begin{equation*} \\lim_{x \\to a} f(x) = b\\qquad \\text{and} \\qquad \\lim_{y \\to b} g(y) = c \\end{equation*}for some$c \\in \\R \\cup \\{ -\\infty, \\infty \\}$. If$b \\in B$, then suppose$g(b) = c$. Then\\begin{equation*} \\lim_{x \\to a} g\\bigl(f(x)\\bigr) = c . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 52, "type": "example", "label": "Lebl-contfunc:52", "categories": [ "example" ], "title": "Let$h(x) \\coloneqq e^{-x^2+x}$", "contents": [ "Let$h(x) \\coloneqq e^{-x^2+x}$. Then", "\\begin{equation*}\n\\lim_{x\\to \\infty} h(x) = 0 .\n\\end{equation*}", "Proof: The claim follows once we know", "\\begin{equation*}\n\\lim_{x\\to \\infty} -x^2+x = -\\infty\n\\end{equation*}", "and", "\\begin{equation*}\n\\lim_{y\\to -\\infty} e^y = 0 ,\n\\end{equation*}", "which is usually proved when the exponential function is defined." ], "refs": [], "proofs": [ { "contents": [ "The claim follows once we know \\begin{equation*}", "\\lim_{x\\to \\infty} -x^2+x = -\\infty", "\\end{equation*} and \\begin{equation*}", "\\lim_{y\\to -\\infty} e^y = 0 ,", "\\end{equation*} which is usually proved when the exponential function is defined." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 53, "type": "proposition", "label": "Lebl-contfunc:prop:monotlimits", "categories": [ "topology", "monotonicity" ], "title": "Let$S \\subset \\R$,$c \\in \\R$,$f \\colon S \\to \\R$be increasing, and$g \\colon S \\to \\R$be decreasing", "contents": [ "Let$S \\subset \\R$,$c \\in \\R$,$f \\colon S \\to \\R$be increasing, and$g \\colon S \\to \\R$be decreasing. If$c$is a cluster point of$S \\cap (-\\infty,c)$, then\\begin{equation*} \\lim_{x \\to c^-} f(x) = \\sup \\{ f(x) : x < c, x \\in S \\} \\qquad \\text{and} \\qquad \\lim_{x \\to c^-} g(x) = \\inf \\{ g(x) : x < c, x \\in S \\} . \\end{equation*}If$c$is a cluster point of$S \\cap (c,\\infty)$, then\\begin{equation*} \\lim_{x \\to c^+} f(x) = \\inf \\{ f(x) : x > c, x \\in S \\} \\qquad \\text{and} \\qquad \\lim_{x \\to c^+} g(x) = \\sup \\{ g(x) : x > c, x \\in S \\} . \\end{equation*}If$\\infty$is a cluster point of$S$, then\\begin{equation*} \\lim_{x \\to \\infty} f(x) = \\sup \\{ f(x) : x \\in S \\} \\qquad \\text{and} \\qquad \\lim_{x \\to \\infty} g(x) = \\inf \\{ g(x) : x \\in S \\} . \\end{equation*}If$-\\infty$is a cluster point of$S$, then\\begin{equation*} \\lim_{x \\to -\\infty} f(x) = \\inf \\{ f(x) : x \\in S \\} \\qquad \\text{and} \\qquad \\lim_{x \\to -\\infty} g(x) = \\sup \\{ g(x) : x \\in S \\} . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let us assume $f$ is increasing, and we will show the first", "equality. The rest of the proof is very similar and is left as an", "exercise.", "Let $a \\coloneqq \\sup \\{ f(x) : x < c, x \\in S \\}$. If $a = \\infty$,", "then given an $M \\in \\R$, there exists an $x_M \\in S$, $x_M < c$, such that $f(x_M) > M$.", "As $f$ is increasing, $f(x) \\geq f(x_M) > M$ for all $x \\in S$ with $x > x_M$.", "Take $\\delta \\coloneqq c-x_M > 0$ to obtain the definition of the limit going to", "infinity.", "Next suppose $a < \\infty$.", "Let $\\epsilon > 0$ be given. Because $a$ is the supremum and", "$S \\cap (-\\infty,c)$ is nonempty, $a \\in \\R$ and", "there exists an", "$x_\\epsilon \\in S$,", "$x_\\epsilon < c$,", "such that $f(x_\\epsilon) > a-\\epsilon$. As $f$ is increasing,", "if $x \\in S$ and $x_\\epsilon < x < c$, we have", "$a-\\epsilon < f(x_\\epsilon) \\leq f(x) \\leq a$. Let", "$\\delta \\coloneqq c-x_\\epsilon$. Then for $x \\in S \\cap (-\\infty,c)$", "with $\\abs{x-c} < \\delta$,", "we have $\\abs{f(x)-a} < \\epsilon$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 54, "type": "corollary", "label": "Lebl-contfunc:cor:continterval", "categories": [ "monotonicity", "intervals", "characterization", "consequence", "continuity" ], "title": "Characterization of Continuous Functions via Equivalence", "contents": [ "If$I \\subset \\R$is an interval and$f \\colon I \\to \\R$is monotone and not constant, then$f(I)$is an interval if and only if$f$is continuous." ], "refs": [], "proofs": [ { "contents": [ "Without loss of generality, suppose $f$ is increasing.", "First suppose $f$ is continuous. Take two points", "$f(x_1), f(x_2)$ in $f(I)$ and suppose", "$f(x_1) < f(x_2)$.", "As $f$ is increasing, then $x_1 < x_2$. By the", "\\hyperref[IVT:thm]{intermediate value theorem},", "given $y$ with $f(x_1) < y < f(x_2)$, we find", "a $c \\in (x_1,x_2) \\subset I$ such that $f(c) = y$, so $y \\in f(I)$.", "Hence, $f(I)$ is an interval.", "Let us prove the reverse direction by contrapositive.", "Suppose $f$ is not continuous at $c \\in I$,", "and that $c$ is not an endpoint of $I$.", "Let", "\\begin{equation*}", "a \\coloneqq \\lim_{x \\to c^-} f(x) = \\sup \\bigl\\{ f(x) : x \\in I, x < c \\bigr\\} ,", "\\qquad", "b \\coloneqq \\lim_{x \\to c^+} f(x) = \\inf \\bigl\\{ f(x) : x \\in I, x > c \\bigr\\} .", "\\end{equation*}", "As $c$ is a discontinuity, $a < b$.", "If $x < c$, then $f(x) \\leq a$, and", "if $x > c$, then $f(x) \\geq b$. Therefore", "no point", "in $(a,b) \\setminus \\bigl\\{ f(c) \\bigr\\}$ is in $f(I)$.", "There exists $x_1 \\in I$ with $x_1 < c$, so", "$f(x_1) \\leq a$, and there exists $x_2 \\in I$ with $x_2 > c$,", "so $f(x_2) \\geq b$. Both $f(x_1)$ and $f(x_2)$ are in $f(I)$,", "but there are points in between them that are not in $f(I)$.", "So $f(I)$ is not an interval. See \\figureref{fig:figinccont}.", "When $c \\in I$ is an endpoint, the proof is similar and is left as an exercise." ], "refs": [ "IVT:thm", "named:Bolzano" ], "ref_ids": [ 32 ] } ], "ref_ids": [] }, { "id": 55, "type": "corollary", "label": "Lebl-contfunc:cor:monotcountcont", "categories": [ "consequence", "monotonicity", "intervals" ], "title": "Monotonicity of Monotone Functions", "contents": [ "Let$I \\subset \\R$be an interval and$f \\colon I \\to \\R$be monotone. Then$f$has at most countably many discontinuities." ], "refs": [], "proofs": [ { "contents": [ "Let $E \\subset I$ be the set of all discontinuities", "that are not endpoints of $I$. As there are", "only two endpoints, it is enough to show that $E$ is countable.", "Without loss of generality, suppose $f$ is increasing.", "We will define an injection $h \\colon E \\to \\Q$.", "For each $c \\in E$,", "both one-sided limits of $f$ exist as $c$ is not an endpoint.", "Let", "\\begin{equation*}", "a \\coloneqq \\lim_{x \\to c^-} f(x) = \\sup \\bigl\\{ f(x) : x \\in I, x < c \\bigr\\} ,", "\\qquad", "b \\coloneqq \\lim_{x \\to c^+} f(x) = \\inf \\bigl\\{ f(x) : x \\in I, x > c \\bigr\\} .", "\\end{equation*}", "As $c$ is a discontinuity, $a < b$.", "There exists a rational number $q \\in (a,b)$, so let $h(c) \\coloneqq q$.", "Suppose $d \\in E$ is another discontinuity.", "If $d > c$, there", "exist an $x \\in I$ with $c < x < d$, and so $\\lim_{x \\to d^-} f(x) \\geq b$.", "Hence the rational number we choose for $h(d)$ is different from $q$,", "since $q=h(c) < b$ and $h(d) > b$.", "Similarly if $d < c$. After making such a choice for", "every element of $E$, we have a", "one-to-one (injective) function into $\\Q$. Therefore, $E$ is countable." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 56, "type": "example", "label": "Lebl-contfunc:example:countdiscont", "categories": [ "boundedness", "monotonicity", "intervals", "topology", "example", "continuity" ], "title": "Continuity of Functions", "contents": [ "By$\\lfloor x \\rfloor$denote the largest integer less than or equal to$x$. Define$f \\colon [0,1] \\to \\R$by", "\\begin{equation*}\nf(x) \\coloneqq\nx +\n\\sum_{n=0}^{\\lfloor 1/(1-x) \\rfloor}\n2^{-n} ,\n\\end{equation*}", "for$x < 1$and$f(1) \\coloneqq 3$. It is an exercise to show that$f$is strictly increasing, bounded, and has a discontinuity at all points$1-\\nicefrac{1}{k}$for$k \\in \\N$. In particular, there are countably many discontinuities, but the function is bounded and defined on a closed bounded interval. See \\figureref{fig:countdiscont}. \\begin{myfigureht} \\includegraphics{figures/increasing-discont-fig} \\caption{Strictly increasing function on$[0,1]$with countably many discontinuities.\\label{fig:countdiscont}} \\end{myfigureht}", "Similarly, one can find an example of a monotone function discontinuous on a dense set such as the rational numbers. See the exercises." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 57, "type": "proposition", "label": "Lebl-contfunc:prop:invcont", "categories": [ "continuity", "monotonicity", "intervals" ], "title": "Continuity of Continuous Functions", "contents": [ "If$I \\subset \\R$is an interval and$f \\colon I \\to \\R$is strictly monotone, then the inverse$f^{-1} \\colon f(I) \\to I$is continuous." ], "refs": [], "proofs": [ { "contents": [ "Let us suppose $f$ is strictly increasing. The proof is almost", "identical for a strictly decreasing function.", "Since $f$ is strictly increasing, so is $f^{-1}$. That is, if $f(x) <", "f(y)$, then we must have $x < y$ and therefore", "$f^{-1}\\bigl(f(x)\\bigr) < f^{-1}\\bigl(f(y)\\bigr)$.", "Take $c \\in f(I)$.", "If $c$ is not a cluster point of $f(I)$, then $f^{-1}$ is continuous at $c$", "automatically. So let $c$ be a cluster point of $f(I)$.", "Suppose both of the following one-sided limits exist:", "\\begin{align*}", "x_0 & \\coloneqq \\lim_{y \\to c^-} f^{-1}(y) =", "\\sup \\bigl\\{ f^{-1}(y) : y < c, y \\in f(I) \\bigr\\}", "=", "\\sup \\bigl\\{ x \\in I : f(x) < c \\bigr\\} , \\\\", "x_1 & \\coloneqq \\lim_{y \\to c^+} f^{-1}(y) =", "\\inf \\bigl\\{ f^{-1}(y) : y > c, y \\in f(I) \\bigr\\}", "=", "\\inf \\bigl\\{ x \\in I : f(x) > c \\bigr\\} .", "\\end{align*}", "We have $x_0 \\leq x_1$ as $f^{-1}$ is increasing.", "For all $x \\in I$ where $x > x_0$, we have $f(x) \\geq c$. As $f$ is strictly increasing,", "we must have $f(x) > c$ for all $x \\in I$ where $x > x_0$. Therefore,", "\\begin{equation*}", "\\{ x \\in I : x > x_0 \\} \\subset \\bigl\\{ x \\in I : f(x) > c \\bigr\\}.", "\\end{equation*}", "The infimum of the left-hand set is $x_0$, and the infimum of the right-hand", "set is $x_1$, so we obtain $x_0 \\geq x_1$.", "So $x_1 = x_0$, and $f^{-1}$ is continuous at $c$.", "If one of the one-sided limits does not exist, the argument is similar", "and is left as an exercise." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 58, "type": "example", "label": "Lebl-contfunc:58", "categories": [ "continuity", "example", "intervals" ], "title": "Continuity of Functions", "contents": [ "The proposition does not require$f$itself to be continuous. Let$f \\colon \\R \\to \\R$be defined by", "\\begin{equation*}\nf(x) \\coloneqq\n\\begin{cases}\nx & \\text{if } x < 0, \\\\\nx+1 & \\text{if } x \\geq 0. \\\\\n\\end{cases}\n\\end{equation*}", "The function$f$is not continuous at$0$. The image of$I = \\R$is the set$(-\\infty,0)\\cup [1,\\infty)$, not an interval. Then$f^{-1} \\colon (-\\infty,0)\\cup [1,\\infty) \\to \\R$can be written as", "\\begin{equation*}\nf^{-1}(y) =\n\\begin{cases}\ny & \\text{if } y < 0, \\\\\ny-1 & \\text{if } y \\geq 1. \n\\end{cases}\n\\end{equation*}", "It is not difficult to see that$f^{-1}$is a continuous function. See \\figureref{invcontfig} for the graphs. \\begin{myfigureht} \\subimport*{figures/}{invcontfigAB.pdf_t} \\caption{Graph of$f$on the left and$f^{-1}$on the right.\\label{invcontfig}} \\end{myfigureht}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 0, "type": "example", "label": "Lebl-contfunc:0", "categories": [ "example", "derivatives" ], "title": "Let$f(x) \\coloneqq x^2$defined on the whole real line", "contents": [ "Let$f(x) \\coloneqq x^2$defined on the whole real line. Let$c \\in \\R$be arbitrary. We find that if$x \\not=c$,", "\\begin{equation*}\n\\frac{x^2-c^2}{x-c} =\n\\frac{(x+c)(x-c)}{x-c} =\n(x+c) .\n\\end{equation*}", "Therefore,", "\\begin{equation*}\nf'(c) = \n\\lim_{x\\to c} \\frac{x^2-c^2}{x-c} =\n\\lim_{x\\to c} (x+c) = 2c.\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 1, "type": "example", "label": "Lebl-contfunc:1", "categories": [ "example", "derivatives" ], "title": "Let$f(x) \\coloneqq ax + b$for numbers$a, b \\in \\R$", "contents": [ "Let$f(x) \\coloneqq ax + b$for numbers$a, b \\in \\R$. Let$c \\in \\R$be arbitrary. For$x \\not=c$,", "\\begin{equation*}\n\\frac{f(x)-f(c)}{x-c} =\n\\frac{a(x-c)}{x-c} = a .\n\\end{equation*}", "Therefore,", "\\begin{equation*}\nf'(c) =\n\\lim_{x\\to c} \n\\frac{f(x)-f(c)}{x-c} =\n\\lim_{x\\to c} \na = a.\n\\end{equation*}", "In fact, every differentiable function \\myquote{infinitesimally} behaves like the affine function$ax + b$. You can guess many results and formulas for derivatives, if you work them out for affine functions first." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 2, "type": "example", "label": "Lebl-contfunc:2", "categories": [ "example", "derivatives" ], "title": "The function$f(x) \\coloneqq \\sqrt{x}$is differentiable for$x > 0$", "contents": [ "The function$f(x) \\coloneqq \\sqrt{x}$is differentiable for$x > 0$. To see this fact, fix$c > 0$, and suppose$x \\not= c$and$x > 0$. Compute", "\\begin{equation*}\n\\frac{\\sqrt{x}-\\sqrt{c}}{x-c}\n=\n\\frac{\\sqrt{x}-\\sqrt{c}}{(\\sqrt{x}-\\sqrt{c})(\\sqrt{x}+\\sqrt{c})}\n=\n\\frac{1}{\\sqrt{x}+\\sqrt{c}} .\n\\end{equation*}", "Therefore,", "\\begin{equation*}\nf'(c) =\n\\lim_{x\\to c}\n\\frac{\\sqrt{x}-\\sqrt{c}}{x-c}\n=\n\\lim_{x\\to c}\n\\frac{1}{\\sqrt{x}+\\sqrt{c}}\n=\n\\frac{1}{2\\sqrt{c}} .\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 3, "type": "example", "label": "Lebl-contfunc:3", "categories": [ "example", "derivatives" ], "title": "The function$f(x) \\coloneqq \\abs{x}$is not differentiable at the origin", "contents": [ "The function$f(x) \\coloneqq \\abs{x}$is not differentiable at the origin. When$x > 0$,", "\\begin{equation*}\n\\frac{\\abs{x}-\\abs{0}}{x-0} =\n\\frac{x-0}{x-0} = 1 .\n\\end{equation*}", "When$x < 0$,", "\\begin{equation*}\n\\frac{\\abs{x}-\\abs{0}}{x-0} =\n\\frac{-x-0}{x-0} = -1 .\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 4, "type": "proposition", "label": "Lebl-contfunc:4", "categories": [ "continuity", "derivatives" ], "title": "Continuity of Derivatives", "contents": [ "Let$f \\colon I \\to \\R$be differentiable at$c \\in I$, then it is continuous at$c$." ], "refs": [], "proofs": [ { "contents": [ "We know the limits", "\\begin{equation*}", "\\lim_{x\\to c}\\frac{f(x)-f(c)}{x-c} = f'(c)", "\\qquad", "\\text{and}", "\\qquad", "\\lim_{x\\to c}(x-c) = 0", "\\end{equation*}", "exist. Furthermore,", "\\begin{equation*}", "f(x)-f(c) =", "\\left( \\frac{f(x)-f(c)}{x-c} \\right) (x-c) .", "\\end{equation*}", "Therefore, the limit of $f(x)-f(c)$ exists and", "\\begin{equation*}", "\\lim_{x\\to c} \\bigl( f(x)-f(c) \\bigr) =", "\\left(\\lim_{x\\to c} \\frac{f(x)-f(c)}{x-c} \\right)", "\\left(\\lim_{x\\to c} (x-c) \\right) =", "f'(c) \\cdot 0 = 0.", "\\end{equation*}", "Hence $\\lim\\limits_{x\\to c} f(x) = f(c)$, and $f$ is continuous at $c$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 5, "type": "proposition", "label": "Lebl-contfunc:5", "categories": [ "derivatives" ], "title": "\\index{linearity of the derivative} Let$I$be an interval, let$f \\colon I \\to \\R$and$g \\colon I \\t...", "contents": [ "\\index{linearity of the derivative} Let$I$be an interval, let$f \\colon I \\to \\R$and$g \\colon I \\to \\R$be differentiable at$c \\in I$, and let$\\alpha \\in \\R$. \\begin{enumerate}[(i)] \\item Define$h \\colon I \\to \\R$by$h(x) \\coloneqq \\alpha f(x)$. Then$h$is differentiable at$c$and$h'(c) = \\alpha f'(c)$. \\item Define$h \\colon I \\to \\R$by$h(x) \\coloneqq f(x) + g(x)$. Then$h$is differentiable at$c$and$h'(c) = f'(c) + g'(c)$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "First, let $h(x) \\coloneqq \\alpha f(x)$.", "For $x \\in I$, $x \\not= c$,", "\\begin{equation*}", "\\frac{h(x)-h(c)}{x-c} =", "\\frac{\\alpha f(x) - \\alpha f(c)}{x-c}", "=", "\\alpha \\frac{f(x) - f(c)}{x-c} .", "\\end{equation*}", "The limit as $x$ goes to $c$ exists on the right-hand side", "by \\corref{falg:cor}. We get", "\\begin{equation*}", "\\lim_{x\\to c}\\frac{h(x)-h(c)}{x-c} =", "\\alpha \\lim_{x\\to c} \\frac{f(x) - f(c)}{x-c} .", "\\end{equation*}", "Therefore, $h$ is differentiable at $c$,", "and the derivative is computed as given.", "Next, define $h(x) \\coloneqq f(x)+g(x)$.", "For $x \\in I$, $x \\not= c$, we have", "\\begin{equation*}", "\\frac{h(x)-h(c)}{x-c} =", "\\frac{\\bigl(f(x) + g(x)\\bigr) - \\bigl(f(c) + g(c)\\bigr)}{x-c}", "=", "\\frac{f(x) - f(c)}{x-c}", "+", "\\frac{g(x) - g(c)}{x-c} .", "\\end{equation*}", "The limit as $x$ goes to $c$ exists on the right-hand side", "by \\corref{falg:cor}. We get", "\\begin{equation*}", "\\lim_{x\\to c}\\frac{h(x)-h(c)}{x-c} =", "\\lim_{x\\to c} \\frac{f(x) - f(c)}{x-c}", "+", "\\lim_{x\\to c}\\frac{g(x) - g(c)}{x-c} .", "\\end{equation*}", "Therefore, $h$ is differentiable at $c$,", "and the derivative is computed as given." ], "refs": [ "falg:cor" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 6, "type": "proposition", "label": "Lebl-contfunc:6", "categories": [ "product_rule", "derivatives" ], "title": "Product Rule", "contents": [ "\\index{product rule} Let$I$be an interval, let$f \\colon I \\to \\R$and$g \\colon I \\to \\R$be functions differentiable at$c$. If$h \\colon I \\to \\R$is defined by\\begin{equation*} h(x) \\coloneqq f(x) g(x) , \\end{equation*}then$h$is differentiable at$c$and\\begin{equation*} h'(c) = f(c) g'(c) + f'(c) g(c) . \\end{equation*}" ], "refs": [ "named:Product Rule" ], "proofs": [], "ref_ids": [] }, { "id": 7, "type": "proposition", "label": "Lebl-contfunc:7", "categories": [ "quotient_rule", "derivatives" ], "title": "Quotient Rule", "contents": [ "\\index{quotient rule} Let$I$be an interval, let$f \\colon I \\to \\R$and$g \\colon I \\to \\R$be differentiable at$c$and$g(x) \\not= 0$for all$x \\in I$. If$h \\colon I \\to \\R$is defined by\\begin{equation*} h(x) \\coloneqq \\frac{f(x)}{g(x)}, \\end{equation*}then$h$is differentiable at$c$and\\begin{equation*} h'(c) = \\frac{f'(c) g(c) - f(c) g'(c)}{{\\bigl(g(c)\\bigr)}^2} . \\end{equation*}" ], "refs": [ "named:Quotient Rule" ], "proofs": [], "ref_ids": [] }, { "id": 8, "type": "proposition", "label": "Lebl-contfunc:8", "categories": [ "chain_rule", "derivatives" ], "title": "Chain Rule", "contents": [ "\\index{chain rule} Let$I_1, I_2$be intervals, let$g \\colon I_1 \\to I_2$be differentiable at$c \\in I_1$, and$f \\colon I_2 \\to \\R$be differentiable at$g(c)$. If$h \\colon I_1 \\to \\R$is defined by\\begin{equation*} h(x) \\coloneqq (f \\circ g) (x) = f\\bigl(g(x)\\bigr) , \\end{equation*}then$h$is differentiable at$c$and\\begin{equation*} h'(c) = f'\\bigl(g(c)\\bigr)g'(c) . \\end{equation*}" ], "refs": [ "named:Chain Rule" ], "proofs": [ { "contents": [ "Let $d \\coloneqq g(c)$. Define", "$u \\colon I_2 \\to \\R$ and $v \\colon I_1 \\to \\R$ by", "\\begin{equation*}", "u(y) \\coloneqq", "\\begin{cases}", "\\frac{f(y) - f(d)}{y-d} & \\text{if } y \\not=d, \\\\", "f'(d) & \\text{if } y = d,", "\\end{cases}", "\\qquad", "v(x) \\coloneqq", "\\begin{cases}", "\\frac{g(x) - g(c)}{x-c} & \\text{if } x \\not=c, \\\\", "g'(c) & \\text{if } x = c.", "\\end{cases}", "\\end{equation*}", "Because $f$ is differentiable at $d = g(c)$, we find that", "$u$ is continuous at $d$. Similarly, $v$ is continuous at $c$.", "For any $x$ and $y$,", "\\begin{equation*}", "f(y)-f(d) = u(y) (y-d)", "\\qquad \\text{and} \\qquad", "g(x)-g(c) = v(x) (x-c) .", "\\end{equation*}", "Plug in to obtain", "\\begin{equation*}", "h(x)-h(c)", "=", "f\\bigl(g(x)\\bigr)-f\\bigl(g(c)\\bigr)", "=", "u\\bigl( g(x) \\bigr) \\bigl(g(x)-g(c)\\bigr)", "=", "u\\bigl( g(x) \\bigr) \\bigl(v(x) (x-c)\\bigr) .", "\\end{equation*}", "Therefore, if $x \\not= c$,", "\\begin{equation} \\label{eq:chainruleeq}", "\\frac{h(x)-h(c)}{x-c}", "=", "u\\bigl( g(x) \\bigr) v(x) .", "\\end{equation}", "By continuity of $u$ and $v$ at $d$ and $c$ respectively, we find", "$\\lim_{y \\to d} u(y)", "= f'(d) = f'\\bigl(g(c)\\bigr)$ and", "$\\lim_{x \\to c} v(x) = g'(c)$.", "The function $g$ is continuous at $c$, and so $\\lim_{x \\to c} g(x) = g(c)$.", "Hence the limit of", "the right-hand side of \\eqref{eq:chainruleeq}", "as $x$ goes to $c$", "exists and is equal to $f'\\bigl(g(c)\\bigr) g'(c)$. Thus $h$", "is differentiable at $c$ and $h'(c) = f'\\bigl(g(c)\\bigr)g'(c)$." ], "refs": [ "eq:chainruleeq" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 9, "type": "lemma", "label": "Lebl-contfunc:relminmax:lemma", "categories": [ "auxiliary result", "extrema", "derivatives" ], "title": "Auxiliary Result for Derivatives", "contents": [ "Suppose$f \\colon (a,b) \\to \\R$is differentiable at$c \\in (a,b)$, and$f$has a relative minimum or a relative maximum at$c$. Then$f'(c) = 0$." ], "refs": [], "proofs": [ { "contents": [ "Suppose $c$ is a", "relative maximum of $f$. That is, there is a $\\delta > 0$ such", "that for every $x \\in (a,b)$ where", "$\\abs{x-c} < \\delta$, we have $f(x)-f(c) \\leq 0$.", "Consider the difference", "quotient. If $c < x < c+\\delta$, then", "\\begin{equation*}", "\\frac{f(x)-f(c)}{x-c} \\leq 0 ,", "\\end{equation*}", "and if $c-\\delta < y < c$, then", "\\begin{equation*}", "\\frac{f(y)-f(c)}{y-c} \\geq 0 .", "\\end{equation*}", "See \\figureref{fig:critpt} for an illustration.", "\\begin{myfigureht}", "\\includegraphics{figures/critpt}", "\\caption{Slopes of secants at a relative maximum.\\label{fig:critpt}}", "\\end{myfigureht}", "As $a < c < b$, there exist", "sequences $\\{ x_n \\}_{n=1}^\\infty$ and", "$\\{ y_n \\}_{n=1}^\\infty$ in $(a,b)$ and within $\\delta$ of $c$,", "such that $x_n > c$, and", "$y_n < c$ for all $n \\in \\N$, and such that", "$\\lim_{n\\to\\infty} x_n = \\lim_{n\\to\\infty} y_n = c$.", "Since $f$", "is differentiable at $c$,", "\\begin{equation*}", "0 \\geq \\lim_{n\\to\\infty} \\frac{f(x_n)-f(c)}{x_n-c}", "=", "f'(c)", "=", "\\lim_{n\\to\\infty} \\frac{f(y_n)-f(c)}{y_n-c} \\geq 0.", "\\end{equation*}", "We are done with a maximum.", "For a minimum, consider", "the function $-f$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 10, "type": "theorem", "label": "Lebl-contfunc:thm:rolle", "categories": [ "mean_value", "continuity", "named theorem", "derivatives" ], "title": "Rolle's Theorem", "contents": [ "\\index{Rolle's theorem} Let$f \\colon [a,b] \\to \\R$be a continuous function differentiable on$(a,b)$such that$f(a) = f(b)$. Then there exists a$c \\in (a,b)$such that$f'(c) = 0$." ], "refs": [ "named:Rolle" ], "proofs": [ { "contents": [ "As $f$ is continuous on $[a,b]$, it attains an absolute minimum and an", "absolute", "maximum in $[a,b]$. We wish to apply \\lemmaref{relminmax:lemma}, and", "so we need to find some $c \\in (a,b)$ where $f$ attains a minimum or a", "maximum.", "Write $K \\coloneqq f(a) = f(b)$.", "If there exists an $x$ such that $f(x) > K$, then the absolute", "maximum is bigger than $K$ and hence occurs at some $c \\in (a,b)$, and", "therefore $f'(c) = 0$. On the other hand, if there exists an $x$", "such that $f(x) < K$, then the absolute minimum occurs at some", "$c \\in (a,b)$, and so $f'(c) = 0$. If there is no $x$ such that", "$f(x) > K$ or", "$f(x) < K$, then $f(x) = K$ for all $x$ and then", "$f'(x) = 0$ for all $x \\in [a,b]$, so any $c \\in (a,b)$ works." ], "refs": [ "relminmax:lemma" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 11, "type": "theorem", "label": "Lebl-contfunc:thm:mvt", "categories": [ "mean_value", "continuity", "derivatives" ], "title": "Mean Value Theorem", "contents": [ "\\index{mean value theorem} Let$f \\colon [a,b] \\to \\R$be a continuous function differentiable on$(a,b)$. Then there exists a point$c \\in (a,b)$such that\\begin{equation*} f(b)-f(a) = f'(c)(b-a) . \\end{equation*}" ], "refs": [ "named:Mean Value" ], "proofs": [ { "contents": [ "Define the", "function $g \\colon [a,b] \\to \\R$ by", "\\begin{equation*}", "g(x) \\coloneqq f(x)-f(b)-\\frac{f(b)-f(a)}{b-a}(x-b) .", "\\end{equation*}", "The function $g$ is differentiable on $(a,b)$,", "continuous on $[a,b]$, such that $g(a) = 0$ and $g(b) = 0$. Thus there exists", "a", "$c \\in (a,b)$ such that $g'(c) = 0$, that is,", "\\begin{equation*}", "0 = g'(c) = f'(c)-\\frac{f(b)-f(a)}{b-a} .", "\\end{equation*}", "In other words,", "$f(b)-f(a) = f'(c)(b-a)$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 12, "type": "theorem", "label": "Lebl-contfunc:thm:cauchymvt", "categories": [ "mean_value", "continuity", "derivatives" ], "title": "Mean Value Theorem", "contents": [ "\\index{Cauchy's mean value theorem} Let$f \\colon [a,b] \\to \\R$and$\\varphi \\colon [a,b] \\to \\R$be continuous functions differentiable on$(a,b)$. Then there exists a point$c \\in (a,b)$such that\\begin{equation*} \\bigl(f(b)-f(a)\\bigr)\\varphi'(c) = f'(c)\\bigl(\\varphi(b)-\\varphi(a)\\bigr) . \\end{equation*}" ], "refs": [ "named:Mean Value" ], "proofs": [], "ref_ids": [ 11 ] }, { "id": 13, "type": "proposition", "label": "Lebl-contfunc:prop:derzeroconst", "categories": [ "derivatives" ], "title": "Let$I$be an interval and let$f \\colon I \\to \\R$be a differentiable function such that$f'(x) = 0$f...", "contents": [ "Let$I$be an interval and let$f \\colon I \\to \\R$be a differentiable function such that$f'(x) = 0$for all$x \\in I$. Then$f$is constant." ], "refs": [], "proofs": [ { "contents": [ "Take arbitrary $x,y \\in I$ with $x < y$.", "As $I$ is an interval, $[x,y] \\subset I$.", "Then $f$ restricted to $[x,y]$ satisfies the hypotheses", "of the \\hyperref[thm:mvt]{mean value theorem}.", "Therefore, there is a $c \\in (x,y)$ such that", "\\begin{equation*}", "f(y)-f(x) = f'(c)(y-x).", "\\end{equation*}", "As $f'(c) = 0$, we have $f(y) = f(x)$. Hence,", "the function is constant." ], "refs": [ "named:Mean Value", "thm:mvt" ], "ref_ids": [ 11 ] } ], "ref_ids": [] }, { "id": 14, "type": "proposition", "label": "Lebl-contfunc:incdecdiffprop", "categories": [ "characterization", "derivatives" ], "title": "Let$I$be an interval and let$f \\colon I \\to \\R$be a differentiable function", "contents": [ "Let$I$be an interval and let$f \\colon I \\to \\R$be a differentiable function. %\\begin{enumerate}[(i),itemsep=0.5\\itemsep,parsep=0.5\\parsep,topsep=0.5\\topsep,partopsep=0.5\\partopsep] \\begin{enumerate}[(i)] \\item$f$is increasing if and only if$f'(x) \\geq 0$for all$x \\in I$. \\item$f$is decreasing if and only if$f'(x) \\leq 0$for all$x \\in I$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "Let us prove the first item. Suppose $f$ is increasing.", "For all $x,c \\in I$ with $x \\neq c$,", "\\begin{equation*}", "\\frac{f(x)-f(c)}{x-c} \\geq 0 .", "\\end{equation*}", "Taking a limit as $x$ goes to $c$, we see that $f'(c) \\geq 0$.", "For the other direction, suppose $f'(x) \\geq 0$ for all $x \\in I$.", "Take any $x, y \\in I$ where $x < y$, and note that $[x,y] \\subset I$.", "By the \\hyperref[thm:mvt]{mean value theorem}, there is some $c \\in (x,y)$ such that", "\\begin{equation*}", "f(y)-f(x) = f'(c)(y-x) .", "\\end{equation*}", "As $f'(c) \\geq 0$ and $y-x > 0$, then $f(y) - f(x) \\geq 0$ or $f(x) \\leq", "f(y)$, and so", "$f$ is increasing.", "We leave the second item, decreasing $f$, to the reader as exercise." ], "refs": [ "named:Mean Value", "thm:mvt" ], "ref_ids": [ 11 ] } ], "ref_ids": [] }, { "id": 15, "type": "proposition", "label": "Lebl-contfunc:incdecdiffstrictprop", "categories": [ "derivatives" ], "title": "Let$I$be an interval and let$f \\colon I \\to \\R$be a differentiable function", "contents": [ "Let$I$be an interval and let$f \\colon I \\to \\R$be a differentiable function. \\begin{enumerate}[(i)] \\item \\label{incdecdiffstrictprop:i} If$f'(x) > 0$for all$x \\in I$, then$f$is strictly increasing. \\item \\label{incdecdiffstrictprop:ii} If$f'(x) < 0$for all$x \\in I$, then$f$is strictly decreasing. \\end{enumerate}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 16, "type": "proposition", "label": "Lebl-contfunc:firstderminmaxtest", "categories": [ "continuity", "extrema", "derivatives" ], "title": "Continuity of Derivatives", "contents": [ "Let$f \\colon (a,b) \\to \\R$be continuous. Let$c \\in (a,b)$and suppose$f$is differentiable on$(a,c)$and$(c,b)$. \\begin{enumerate}[(i)] \\item If$f'(x) \\leq 0$whenever$x \\in (a,c)$and$f'(x) \\geq 0$whenever$x \\in (c,b)$, then$f$has an absolute minimum at$c$. \\item If$f'(x) \\geq 0$whenever$x \\in (a,c)$and$f'(x) \\leq 0$whenever$x \\in (c,b)$, then$f$has an absolute maximum at$c$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "We prove the first item and leave the second to the reader.", "Take $x \\in (a,c)$", "and a sequence $\\{ y_n\\}_{n=1}^\\infty$ such that $x < y_n < c$ for all $n$", "and $\\lim_{n\\to\\infty} y_n = c$.", "By the preceding proposition,", "$f$ is decreasing on $(a,c)$ so $f(x) \\geq f(y_n)$ for all $n$.", "As $f$ is", "continuous at $c$, we take the limit to get", "$f(x) \\geq f(c)$.", "Similarly, take $x \\in (c,b)$", "and $\\{ y_n\\}_{n=1}^\\infty$ a sequence such that $c < y_n < x$ and", "$\\lim_{n\\to\\infty} y_n = c$.", "The function is increasing on $(c,b)$ so $f(x) \\geq f(y_n)$ for all $n$.", "By continuity of $f$, we get", "$f(x) \\geq f(c)$. Thus $f(x) \\geq f(c)$ for all", "$x \\in (a,b)$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 17, "type": "proposition", "label": "Lebl-contfunc:prop:endpointderivative", "categories": [ "continuity", "derivatives" ], "title": "Continuity of Derivatives", "contents": [ "\\leavevmode \\begin{enumerate}[(i)] \\item Suppose$f \\colon [a,b) \\to \\R$is continuous, differentiable in$(a,b)$, and$\\lim_{x \\to a} f'(x) = L$. Then$f$is differentiable at$a$and$f'(a) = L$. \\item Suppose$f \\colon (a,b] \\to \\R$is continuous, differentiable in$(a,b)$, and$\\lim_{x \\to b} f'(x) = L$. Then$f$is differentiable at$b$and$f'(b) = L$. \\end{enumerate}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 18, "type": "theorem", "label": "Lebl-contfunc:thm:darboux", "categories": [ "derivatives" ], "title": "Darboux's Theorem", "contents": [ "\\index{Darboux's theorem} Let$f \\colon [a,b] \\to \\R$be differentiable. Suppose$y \\in \\R$is such that$f'(a) < y < f'(b)$or$f'(a) > y > f'(b)$. Then there exists a$c \\in (a,b)$such that$f'(c) = y$." ], "refs": [], "proofs": [ { "contents": [ "Suppose", "$f'(a) < y < f'(b)$.", "Define", "\\begin{equation*}", "g(x) \\coloneqq yx - f(x) .", "\\end{equation*}", "The function $g$ is continuous on $[a,b]$, and so $g$ attains a maximum at some $c \\in", "[a,b]$.", "The function $g$ is also differentiable on $[a,b]$.", "Compute $g'(x) = y-f'(x)$. Thus $g'(a) > 0$. As the derivative is", "the limit of difference quotients and is positive, there must be some", "difference quotient that is positive. That is, there must exist", "an $x > a$ such that", "\\begin{equation*}", "\\frac{g(x)-g(a)}{x-a} > 0 ,", "\\end{equation*}", "or $g(x) > g(a)$. Thus $g$", "cannot possibly have a maximum at $a$. Similarly, as $g'(b) < 0$,", "we find an $x < b$ (a different $x$) such that", "$\\frac{g(x)-g(b)}{x-b} < 0$ or that $g(x) > g(b)$, thus", "$g$ cannot possibly have a maximum at $b$.", "Therefore, $c \\in (a,b)$,", "and \\lemmaref{relminmax:lemma} applies: As $g$ attains a maximum", "at $c$ we find $g'(c) = 0$", "and so $f'(c) = y$.", "Similarly, if $f'(a) > y > f'(b)$, consider $g(x) \\coloneqq f(x)- yx$." ], "refs": [ "relminmax:lemma" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 19, "type": "example", "label": "Lebl-contfunc:baddifffunc:example", "categories": [ "continuity", "extrema", "example", "derivatives" ], "title": "Continuity of Derivatives", "contents": [ "Let$f \\colon \\R \\to \\R$be the function defined by", "\\begin{equation*}\nf(x) \\coloneqq\n\\begin{cases}\n{\\bigl( x \\sin(\\nicefrac{1}{x}) \\bigr)}^2 & \\text{if } x \\not= 0, \\\\\n0 & \\text{if } x = 0.\n\\end{cases}\n\\end{equation*}", "We claim that$f$is differentiable everywhere, but$f' \\colon \\R \\to \\R$is not continuous at the origin. Furthermore,$f$has a minimum at$0$, but the derivative changes sign infinitely often near the origin. See \\figureref{fig:nonc1diff}. \\begin{myfigureht} \\subimport*{figures/}{nonc1diff_full.pdf_t} \\caption{A function with a discontinuous derivative. The function$f$is on the left and$f'$is on the right. Notice that$f(x) \\leq x^2$on the left graph.\\label{fig:nonc1diff}} \\end{myfigureht}", "Proof: It is immediate from the definition that$f$has an absolute minimum at$0$; we know$f(x) \\geq 0$for all$x$and$f(0) = 0$.", "For$x \\not= 0$,$f$is differentiable and the derivative is$2 \\sin (\\nicefrac{1}{x}) \\bigl( x \\sin (\\nicefrac{1}{x}) - \\cos(\\nicefrac{1}{x}) \\bigr)$. As an exercise, show that for$x_n = \\frac{4}{(8n+1)\\pi}$, we have$\\lim_{n\\to\\infty} f'(x_n) = -1$, and for$y_n = \\frac{4}{(8n+3)\\pi}$, we have$\\lim_{n\\to\\infty} f'(y_n) = 1$. So$f'$cannot be continuous at$0$no matter what$f'(0)$is.", "Let us show that$f$is differentiable at$0$and$f'(0)=0$. For$x \\not= 0$,", "\\begin{equation*}\n\\abs{\\frac{f(x)-f(0)}{x-0} - 0}\n=\n\\abs{\\frac{x^2 \\sin^2(\\nicefrac{1}{x})}{x}}\n=\n\\abs{x \\sin^2(\\nicefrac{1}{x})}\n\\leq\n\\abs{x} .\n\\end{equation*}", "And, of course, as$x$tends to zero,$\\abs{x}$tends to zero, and hence$\\abs{\\frac{f(x)-f(0)}{x-0} - 0}$goes to zero. Therefore,$f$is differentiable at 0 and the derivative at 0 is 0. A key point in the calculation above is that$\\abs{f(x)} \\leq x^2$, see also Exercises \\ref{exercise:bndmuldiff} and \\ref{exercise:diffsqueeze}." ], "refs": [ "exercise:bndmuldiff", "exercise:diffsqueeze" ], "proofs": [ { "contents": [ "It is immediate from the definition that$f$has an absolute minimum at$0$; we know$f(x) \\geq 0$for all$x$and$f(0) = 0$. For$x \\not= 0$,$f$is differentiable and the derivative is$2 \\sin (\\nicefrac{1}{x}) \\bigl( x \\sin (\\nicefrac{1}{x}) - \\cos(\\nicefrac{1}{x}) \\bigr)$. As an exercise, show that for$x_n = \\frac{4}{(8n+1)\\pi}$, we have$\\lim_{n\\to\\infty} f'(x_n) = -1$, and for$y_n = \\frac{4}{(8n+3)\\pi}$, we have$\\lim_{n\\to\\infty} f'(y_n) = 1$. So$f'$cannot be continuous at$0$no matter what$f'(0)$is. Let us show that$f$is differentiable at$0$and$f'(0)=0$. For$x \\not= 0$, \\begin{equation*}", "\\abs{\\frac{f(x)-f(0)}{x-0} - 0}", "=", "\\abs{\\frac{x^2 \\sin^2(\\nicefrac{1}{x})}{x}}", "=", "\\abs{x \\sin^2(\\nicefrac{1}{x})}", "\\leq", "\\abs{x} .", "\\end{equation*} And, of course, as$x$tends to zero,$\\abs{x}$tends to zero, and hence$\\abs{\\frac{f(x)-f(0)}{x-0} - 0}$goes to zero. Therefore,$f$is differentiable at 0 and the derivative at 0 is 0. A key point in the calculation above is that$\\abs{f(x)} \\leq x^2$, see also Exercises \\ref{exercise:bndmuldiff} and \\ref{exercise:diffsqueeze}." ], "refs": [ "exercise:bndmuldiff", "exercise:diffsqueeze" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 20, "type": "theorem", "label": "Lebl-contfunc:20", "categories": [ "continuity", "taylor_series", "named theorem", "derivatives" ], "title": "Taylor's Theorem", "contents": [ "\\index{Taylor's theorem} \\label{thm:taylor} Suppose$f \\colon [a,b] \\to \\R$is a function with$n$continuous derivatives on$[a,b]$and such that$f^{(n+1)}$exists on$(a,b)$. Given distinct points$x_0$and$x$in$[a,b]$, we can find a point$c$between$x_0$and$x$such that\\begin{equation*} f(x)=P_{n}^{x_0}(x)+\\frac{f^{(n+1)}(c)}{(n+1)!}{(x-x_0)}^{n+1} . \\end{equation*}" ], "refs": [ "named:Taylor" ], "proofs": [ { "contents": [ "Find a number $M_{x,x_0}$ (depending on $x$ and $x_0$) solving the equation", "\\begin{equation*}", "f(x)=P_{n}^{x_0}(x)+M_{x,x_0}{(x-x_0)}^{n+1} .", "\\end{equation*}", "Define a function $g(s)$ by", "\\begin{equation*}", "g(s) \\coloneqq f(s)-P_n^{x_0}(s)-M_{x,x_0}{(s-x_0)}^{n+1} .", "\\end{equation*}", "We compute", "the $k$th derivative at $x_0$ of the Taylor polynomial", "${(P_n^{x_0})}^{(k)}(x_0) = f^{(k)}(x_0)$ for", "$k=0,1,2,\\ldots,n$ (the zeroth derivative of a function is the function", "itself). Therefore,", "\\begin{equation*}", "g(x_0) = g'(x_0) = g''(x_0) = \\cdots = g^{(n)}(x_0) = 0 .", "\\end{equation*}", "In particular, $g(x_0) = 0$.", "On the other hand $g(x) = 0$. By the", "\\hyperref[thm:mvt]{mean value theorem}", "there exists an $x_1$ between $x_0$ and $x$ such that $g'(x_1) = 0$.", "Applying the \\hyperref[thm:mvt]{mean value theorem}", "to $g'$ we obtain that there exists", "$x_2$ between $x_0$ and $x_1$ (and therefore between $x_0$ and $x$)", "such that $g''(x_2) = 0$. We repeat the", "argument $n+1$ times to obtain a number $x_{n+1}$ between $x_0$ and $x_n$", "(and therefore between $x_0$ and $x$) such that $g^{(n+1)}(x_{n+1}) = 0$.", "Let $c \\coloneqq x_{n+1}$.", "We compute the $(n+1)$th derivative of $g$ to find", "\\begin{equation*}", "g^{(n+1)}(s) = f^{(n+1)}(s)-(n+1)!\\,M_{x,x_0} .", "\\end{equation*}", "Plugging in $c$ for $s$ we obtain $M_{x,x_0} = \\frac{f^{(n+1)}(c)}{(n+1)!}$, and", "we are done." ], "refs": [ "named:Mean Value", "named:Taylor", "thm:mvt" ], "ref_ids": [ 11, 20 ] } ], "ref_ids": [] }, { "id": 21, "type": "proposition", "label": "Lebl-contfunc:21", "categories": [ "continuity", "extrema", "derivatives" ], "title": "Continuity of Derivatives", "contents": [ "\\index{second derivative test} Suppose$f \\colon (a,b) \\to \\R$is twice continuously differentiable,$x_0 \\in (a,b)$,$f'(x_0) = 0$and$f''(x_0) > 0$. Then$f$has a strict relative minimum at$x_0$." ], "refs": [], "proofs": [ { "contents": [ "As $f''$ is continuous, there exists a $\\delta > 0$", "such that $f''(c) > 0$ for all $c \\in (x_0-\\delta,x_0+\\delta)$,", "see \\exerciseref{exercise:positivecontneigh}.", "Take $x \\in (x_0-\\delta,x_0+\\delta)$, $x \\not= x_0$.", "Taylor's theorem says that for some $c$ between $x_0$ and $x$,", "\\begin{equation*}", "f(x)", "=", "f(x_0) + f'(x_0) (x-x_0) +", "\\frac{f''(c)}{2}{(x-x_0)}^{2}", "=", "f(x_0) + \\frac{f''(c)}{2}{(x-x_0)}^{2} .", "\\end{equation*}", "As $f''(c) > 0$, and ${(x-x_0)}^2 > 0$, then $f(x) > f(x_0)$." ], "refs": [ "named:Taylor" ], "ref_ids": [ 20 ] } ], "ref_ids": [] }, { "id": 22, "type": "lemma", "label": "Lebl-contfunc:lemma:ift", "categories": [ "auxiliary result", "continuity", "inverse_functions", "derivatives" ], "title": "Continuity of Derivatives", "contents": [ "Let$I,J \\subset \\R$be intervals. If$f \\colon I \\to J$is strictly monotone (hence one-to-one), onto ($f(I) = J$), differentiable at$x_0 \\in I$, and$f'(x_0) \\not= 0$, then the inverse$f^{-1}$is differentiable at$y_0 = f(x_0)$and\\begin{equation*} (f^{-1})'(y_0) = \\frac{1}{f'\\bigl( f^{-1}(y_0) \\bigr)} = \\frac{1}{f'(x_0)} . \\end{equation*}If$f$is continuously differentiable and$f'$is never zero, then$f^{-1}$is continuously differentiable." ], "refs": [], "proofs": [ { "contents": [ "By \\propref{prop:invcont}, $f$ has a continuous inverse. For convenience", "call the inverse $g \\colon J \\to I$.", "Let $x_0,y_0$ be as in the statement. For $x \\in I$, write $y \\coloneqq f(x)$.", "If $x \\not= x_0$, and so $y \\not= y_0$, we find", "\\begin{equation*}", "\\frac{g(y)-g(y_0)}{y-y_0} =", "\\frac{g\\bigl(f(x)\\bigr)-g\\bigl(f(x_0)\\bigr)}{f(x)-f(x_0)} =", "\\frac{x-x_0}{f(x)-f(x_0)} .", "\\end{equation*}", "See \\figureref{inversefig} for the geometric idea.", "\\begin{myfigureht}", "\\subimport*{figures/}{inversefigAB.pdf_t}", "\\caption{Interpretation of the derivative of the inverse", "function.\\label{inversefig}}", "\\end{myfigureht}", "Let", "\\begin{equation*}", "Q(x) \\coloneqq", "\\begin{cases}", "\\frac{x-x_0}{f(x)-f(x_0)} & \\text{if } x \\neq x_0, \\\\", "\\frac{1}{f'(x_0)} & \\text{if } x = x_0 \\quad \\text{(notice that }", "f'(x_0) \\neq 0 \\text{)}.", "\\end{cases}", "\\end{equation*}", "As $f$ is differentiable at $x_0$,", "\\begin{equation*}", "\\lim_{x \\to x_0} Q(x) =", "\\lim_{x \\to x_0}", "\\frac{x-x_0}{f(x)-f(x_0)}", "=", "\\frac{1}{f'(x_0)}", "=", "Q(x_0) ,", "\\end{equation*}", "that is, $Q$ is continuous at $x_0$.", "As $g(y)$ is continuous at $y_0$,", "the composition $Q\\bigl(g(y)\\bigr) = \\frac{g(y)-g(y_0)}{y-y_0}$", "is continuous at $y_0$ by", "\\propref{prop:compositioncont}.", "Therefore,", "\\begin{equation*}", "\\frac{1}{f'\\bigl(g(y_0)\\bigr)}", "= Q\\bigl(g(y_0)\\bigr)", "= \\lim_{y \\to y_0} Q\\bigl(g(y)\\bigr)", "= \\lim_{y \\to y_0} \\frac{g(y)-g(y_0)}{y-y_0} .", "\\end{equation*}", "So $g$ is differentiable at $y_0$ and $g'(y_0) =", "\\frac{1}{f'\\left(\\vphantom{1^1_1}g(y_0)\\right)}$.", "If $f'$ is continuous and nonzero at all $x \\in I$,", "then the lemma applies at all $x \\in I$. As $g$ is also", "continuous (it is differentiable), the derivative $g'(y) =", "\\frac{1}{f'\\left(\\vphantom{1^1_1}g(y)\\right)}$ must be continuous." ], "refs": [ "prop:compositioncont", "prop:invcont" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 23, "type": "theorem", "label": "Lebl-contfunc:23", "categories": [ "continuity", "inverse_functions", "derivatives" ], "title": "Inverse Function Theorem", "contents": [ "\\index{inverse function theorem} Let$f \\colon (a,b) \\to \\R$be a continuously differentiable function,$x_0 \\in (a,b)$a point where$f'(x_0) \\not= 0$. Then there exists an open interval$I \\subset (a,b)$with$x_0 \\in I$, the restriction$f|_{I}$is injective with a continuously differentiable inverse$g \\colon J \\to I$defined on an interval$J \\coloneqq f(I)$, and\\begin{equation*} g'(y) = \\frac{1}{f'\\bigl( g(y) \\bigr)} \\qquad \\text{for all } y \\in J. \\end{equation*}" ], "refs": [ "named:Inverse Function" ], "proofs": [ { "contents": [ "Without loss of generality, suppose $f'(x_0) > 0$. As $f'$ is", "continuous, there must exist an open interval $I = (x_0-\\delta,x_0+\\delta)$", "such that $f'(x) > 0$ for all $x \\in I$. See", "\\exerciseref{exercise:positivecontneigh}.", "By \\propref{incdecdiffstrictprop}, $f$ is strictly increasing", "on $I$, and hence the restriction $f|_{I}$ is bijective onto $J: = f(I)$.", "As $f$ is continuous,", "\\corref{cor:continterval}", "(or directly via the", "\\hyperref[IVT:thm]{intermediate value theorem})", "implies that", "$f(I)$ is an interval.", "Now apply \\lemmaref{lemma:ift}." ], "refs": [ "IVT:thm", "cor:continterval", "incdecdiffstrictprop", "lemma:ift", "named:Inverse Function" ], "ref_ids": [ 23 ] } ], "ref_ids": [] }, { "id": 24, "type": "corollary", "label": "Lebl-contfunc:24", "categories": [ "consequence", "continuity", "derivatives" ], "title": "Uniqueness of Derivatives", "contents": [ "Given$n \\in \\N$and$x \\geq 0$, there exists a unique number$y \\geq 0$(denoted$x^{1/n} \\coloneqq y$), such that$y^n = x$. Furthermore, the function$g \\colon (0,\\infty) \\to (0,\\infty)$defined by$g(x) \\coloneqq x^{1/n}$is continuously differentiable and\\begin{equation*} g'(x) = \\frac{1}{nx^{(n-1)/n}} = \\frac{1}{n} \\, x^{(1-n)/n} , \\end{equation*}using the convention$x^{m/n} \\coloneqq {(x^{1/n})}^{m}$." ], "refs": [], "proofs": [ { "contents": [ "For $x=0$, the existence of a unique root is trivial.", "Let $f \\colon (0,\\infty) \\to (0,\\infty)$ be defined by $f(y) \\coloneqq y^n$.", "The function $f$ is continuously differentiable", "and $f'(y) = ny^{n-1}$, see \\exerciseref{exercise:diffofxn}.", "For $y > 0$, the derivative $f'$ is strictly positive,", "and so again by \\propref{incdecdiffstrictprop}, $f$ is strictly", "increasing (this can also be proved directly) and hence injective.", "Suppose $M$ and $\\epsilon$ are such that", "$M > 1$ and $1 > \\epsilon > 0$.", "Then", "$f(M) = M^n \\geq M$ and", "$f(\\epsilon) = \\epsilon^n \\leq \\epsilon$.", "For every $x$ with $\\epsilon < x < M$,", "we have, by the", "\\hyperref[IVT:thm]{intermediate value theorem}, that $x \\in", "f\\bigl( [\\epsilon,M] \\bigr) \\subset", "f\\bigl( (0,\\infty) \\bigr)$. As $M$ and $\\epsilon$ were arbitrary, $f$ is onto", "$(0,\\infty)$, and hence $f$ is bijective.", "Let $g$ be the inverse of $f$, and we obtain", "the existence and uniqueness of positive", "$n$th roots. \\lemmaref{lemma:ift} says that $g$ has a continuous", "derivative and $g'(x) =", "\\frac{1}{f'\\left(\\vphantom{1^1_1}g(x)\\right)} = \\frac{1}{n {(x^{1/n})}^{n-1}}$." ], "refs": [ "IVT:thm", "incdecdiffstrictprop", "lemma:ift", "named:Inverse Function" ], "ref_ids": [ 23 ] } ], "ref_ids": [] }, { "id": 25, "type": "example", "label": "Lebl-contfunc:25", "categories": [ "inverse_functions", "example", "derivatives" ], "title": "Inverse Function Theorem", "contents": [ "The corollary provides a good example of where the inverse function theorem gives us an interval smaller than$(a,b)$. Take$f \\colon \\R \\to \\R$defined by$f(x) \\coloneqq x^2$. Then$f'(x_0) \\not= 0$as long as$x_0 \\not= 0$. If$x_0 > 0$, we can take$I=(0,\\infty)$, but no larger." ], "refs": [ "named:Inverse Function" ], "proofs": [], "ref_ids": [ 23 ] }, { "id": 26, "type": "example", "label": "Lebl-contfunc:26", "categories": [ "continuity", "example", "derivatives" ], "title": "Continuity of Derivatives", "contents": [ "Another useful example is$f(x) \\coloneqq x^3$. The function$f \\colon \\R \\to \\R$is one-to-one and onto, so$f^{-1}(y) = y^{1/3}$exists on the entire real line including zero and negative$y$. The function$f$has a continuous derivative, but$f^{-1}$has no derivative at the origin. The point is that$f'(0) = 0$. See \\figureref{cubecuberootfig} for a graph, notice the vertical tangent on the cube root at the origin. See also \\exerciseref{exercise:oddroot}. \\begin{myfigureht} \\includegraphics{figures/cubecuberoot} \\caption{Graphs of$x^3$and$x^{1/3}$.\\label{cubecuberootfig}} \\end{myfigureht}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 0, "type": "proposition", "label": "Lebl-contfunc:sumulbound:prop", "categories": [ "techniques", "inequalities", "functions", "series" ], "title": "Boundedness of Partitions", "contents": [ "Let$f \\colon [a,b] \\to \\R$be a bounded function. Let$m, M \\in \\R$be such that for all$x \\in [a,b]$, we have$m \\leq f(x) \\leq M$. Then for every partition$P$of$[a,b]$, \\begin{equation} \\label{sumulbound:eq} m(b-a) \\leq L(P,f) \\leq U(P,f) \\leq M(b-a) . \\end{equation}" ], "refs": [], "proofs": [ { "contents": [ "Let $P$ be a partition of $[a,b]$. Note that $m \\leq m_i$ for all $i$ and", "$M_i \\leq M$ for all $i$. Also $m_i \\leq M_i$ for all $i$. Finally,", "$\\sum_{i=1}^n \\Delta x_i = (b-a)$. Therefore,", "\\begin{multline*}", "m(b-a) =", "m \\left( \\sum_{i=1}^n \\Delta x_i \\right)", "=", "\\sum_{i=1}^n m \\Delta x_i", "\\leq", "\\sum_{i=1}^n m_i \\Delta x_i", "\\leq", "\\\\", "\\leq", "\\sum_{i=1}^n M_i \\Delta x_i", "\\leq", "\\sum_{i=1}^n M \\Delta x_i", "=", "M \\left( \\sum_{i=1}^n \\Delta x_i \\right)", "=", "M(b-a) .", "\\end{multline*}", "Hence we get \\eqref{sumulbound:eq}. In particular, the set of lower and", "upper sums are bounded sets." ], "refs": [ "sumulbound:eq" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 1, "type": "example", "label": "Lebl-contfunc:example:dirichletfunc", "categories": [ "techniques", "series", "example" ], "title": "Take the \\myindex{Dirichlet function}$f \\colon [0,1] \\to \\R$, where$f(x) \\coloneqq 1$if$x \\in \\Q$...", "contents": [ "Take the \\myindex{Dirichlet function}$f \\colon [0,1] \\to \\R$, where$f(x) \\coloneqq 1$if$x \\in \\Q$and$f(x) \\coloneqq 0$if$x \\notin \\Q$. Then", "\\begin{equation*}\n\\underline{\\int_0^1} f = 0 \\qquad \\text{and} \\qquad\n\\overline{\\int_0^1} f = 1 .\n\\end{equation*}", "The reason is that for any partition$P$and every$i$, we have$m_i = \\inf \\bigl\\{ f(x) : x \\in [x_{i-1},x_i] \\bigr\\} = 0$and$M_i = \\sup \\bigl\\{ f(x) : x \\in [x_{i-1},x_i] \\bigr\\} = 1$. Thus", "\\begin{equation*}\nL(P,f) = \\sum_{i=1}^n 0 \\cdot \\Delta x_i = 0 , \\quad \\text{and} \\quad\nU(P,f) = \\sum_{i=1}^n 1 \\cdot \\Delta x_i = \\sum_{i=1}^n \\Delta x_i = 1 .\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 2, "type": "proposition", "label": "Lebl-contfunc:prop:refinement", "categories": [ "techniques", "functions", "inequalities" ], "title": "Boundedness of Partitions", "contents": [ "Let$f \\colon [a,b] \\to \\R$be a bounded function, and let$P$be a partition of$[a,b]$. Let$\\widetilde{P}$be a refinement of$P$. Then\\begin{equation*} L(P,f) \\leq L(\\widetilde{P},f) \\qquad \\text{and} \\qquad U(\\widetilde{P},f) \\leq U(P,f) . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "The tricky part of this proof is to get the notation correct.", "Let $\\widetilde{P} = \\{ \\widetilde{x}_0, \\widetilde{x}_1, \\ldots,", "\\widetilde{x}_{\\ell} \\}$ be", "a refinement of", "$P = \\{ x_0, x_1, \\ldots, x_n \\}$. Then", "$x_0 = \\widetilde{x}_0$ and", "$x_n = \\widetilde{x}_{\\ell}$. In fact, there are integers", "$k_0 < k_1 < \\cdots < k_n$ such that $x_i = \\widetilde{x}_{k_i}$ for", "$i=0,1,2,\\ldots,n$.", "Let $\\Delta \\widetilde{x}_q \\coloneqq \\widetilde{x}_q - \\widetilde{x}_{q-1}$ for", "$q=0,1,2,\\ldots,\\ell$.", "See \\figureref{fig:refinement}.", "We get", "\\begin{equation*}", "\\Delta x_i", "=", "x_i - x_{i-1} =", "\\widetilde{x}_{k_i} - \\widetilde{x}_{k_{i-1}} =", "\\sum_{q=k_{i-1}+1}^{k_i}", "\\widetilde{x}_{q} - \\widetilde{x}_{q-1}", "=", "\\sum_{q=k_{i-1}+1}^{k_i} \\Delta \\widetilde{x}_q .", "\\end{equation*}", "\\begin{myfigureht}", "\\subimport*{figures/}{figrefinement.pdf_t}", "\\caption{Refinement of a subinterval. Notice $\\Delta x_i =", "\\Delta \\widetilde{x}_{q-2} +", "\\Delta \\widetilde{x}_{q-1} +", "\\Delta \\widetilde{x}_{q}$,", "and also", "$k_{i-1}+1 = q-2$ and", "$k_{i} = q$.\\label{fig:refinement}}", "\\end{myfigureht}", "Let $m_i$ be as before and correspond to the partition $P$.", "Let $\\widetilde{m}_q \\coloneqq \\inf \\bigl\\{ f(x) : \\widetilde{x}_{q-1} \\leq x \\leq", "\\widetilde{x}_q \\bigr\\}$.", "Now, $m_i \\leq \\widetilde{m}_q$ for $k_{i-1} < q \\leq k_i$. Therefore,", "\\begin{equation*}", "m_i \\Delta x_i", "=", "m_i \\sum_{q=k_{i-1}+1}^{k_i} \\Delta \\widetilde{x}_q", "=", "\\sum_{q=k_{i-1}+1}^{k_i} m_i \\Delta \\widetilde{x}_q", "\\leq", "\\sum_{q=k_{i-1}+1}^{k_i} \\widetilde{m}_q \\Delta \\widetilde{x}_q .", "\\end{equation*}", "So", "\\begin{equation*}", "L(P,f) =", "\\sum_{i=1}^n m_i \\Delta x_i", "\\leq", "\\sum_{i=1}^n \\,", "\\sum_{q=k_{i-1}+1}^{k_i} \\widetilde{m}_q \\Delta \\widetilde{x}_q", "=", "\\sum_{q=1}^{\\ell}", "\\widetilde{m}_q \\Delta \\widetilde{x}_q = L(\\widetilde{P},f).", "\\end{equation*}", "The proof of $U(\\widetilde{P},f) \\leq U(P,f)$ is left as an exercise." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 3, "type": "proposition", "label": "Lebl-contfunc:intulbound:prop", "categories": [ "functions", "inequalities" ], "title": "Boundedness of Bounded Functions", "contents": [ "Let$f \\colon [a,b] \\to \\R$be a bounded function. Let$m, M \\in \\R$be such that for all$x \\in [a,b]$, we have$m \\leq f(x) \\leq M$. Then \\begin{equation} \\label{intulbound:eq} m(b-a) \\leq \\underline{\\int_a^b} f \\leq \\overline{\\int_a^b} f \\leq M(b-a) . \\end{equation}" ], "refs": [], "proofs": [ { "contents": [ "By \\propref{sumulbound:prop}, for every partition $P$,", "\\begin{equation*}", "m(b-a) \\leq L(P,f) \\leq U(P,f) \\leq M(b-a).", "\\end{equation*}", "The inequality", "$m(b-a) \\leq L(P,f)$ implies $m(b-a) \\leq \\underline{\\int_a^b} f$.", "The inequality", "$U(P,f) \\leq M(b-a)$ implies $\\overline{\\int_a^b} f \\leq M(b-a)$.", "The middle inequality in", "\\eqref{intulbound:eq} is the main point of the proposition.", "Let $P_1, P_2$ be partitions of $[a,b]$. Define", "$\\widetilde{P} \\coloneqq P_1 \\cup P_2$.", "The set $\\widetilde{P}$ is a partition of $[a,b]$, which", "is a refinement of $P_1$ and a refinement of $P_2$.", "By \\propref{prop:refinement},", "$L(P_1,f) \\leq L(\\widetilde{P},f)$ and", "$U(\\widetilde{P},f) \\leq U(P_2,f)$. So", "\\begin{equation*}", "L(P_1,f) \\leq L(\\widetilde{P},f) \\leq U(\\widetilde{P},f) \\leq U(P_2,f) .", "\\end{equation*}", "In other words, for two arbitrary partitions $P_1$ and $P_2$, we have", "$L(P_1,f) \\leq U(P_2,f)$.", "Recall \\propref{infsupineq:prop}, and take the supremum and", "infimum over all partitions:", "\\begin{multline*}", "\\underline{\\int_a^b} f =", "\\sup \\, \\bigl\\{ L(P,f) : P \\text{ a partition of } [a,b] \\bigr\\}", "\\\\", "\\leq", "\\inf \\, \\bigl\\{ U(P,f) : P \\text{ a partition of } [a,b] \\bigr\\}", "=", "\\overline{\\int_a^b} f . \\qedhere", "\\end{multline*}" ], "refs": [ "infsupineq:prop", "intulbound:eq", "prop:refinement", "sumulbound:prop" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 4, "type": "proposition", "label": "Lebl-contfunc:intbound:prop", "categories": [ "integration", "named theorem" ], "title": "Riemann Integral", "contents": [ "Let$f \\colon [a,b] \\to \\R$be a Riemann integrable function. Let$m, M \\in \\R$be such that$m \\leq f(x) \\leq M$for all$x \\in [a,b]$. Then\\begin{equation*} m(b-a) \\leq \\int_a^b f \\leq M(b-a) . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 5, "type": "example", "label": "Lebl-contfunc:5", "categories": [ "integration", "inequalities", "example" ], "title": "We integrate constant functions using \\propref{intulbound:prop}", "contents": [ "We integrate constant functions using \\propref{intulbound:prop}. If$f(x) \\coloneqq c$for some constant$c$, then we take$m = M = c$. In inequality \\eqref{intulbound:eq} all the inequalities must be equalities. Thus$f$is integrable on$[a,b]$and$\\int_a^b f = c(b-a)$." ], "refs": [ "intulbound:eq", "intulbound:prop" ], "proofs": [], "ref_ids": [] }, { "id": 6, "type": "example", "label": "Lebl-contfunc:6", "categories": [ "applications", "techniques", "integration", "example", "series", "inequalities", "named theorem" ], "title": "Riemann Integral", "contents": [ "Let$f \\colon [0,2] \\to \\R$be defined by", "\\begin{equation*}\nf(x) \\coloneqq\n\\begin{cases}\n1 & \\text{if } x < 1,\\\\\n\\nicefrac{1}{2} & \\text{if } x = 1,\\\\\n0 & \\text{if } x > 1.\n\\end{cases}\n\\end{equation*}", "We claim$f$is Riemann integrable and$\\int_0^2 f = 1$.", "Proof: Let$0 < \\epsilon < 1$be arbitrary. Let$P \\coloneqq \\{0, 1-\\epsilon, 1+\\epsilon, 2\\}$be a partition. We use the notation from the definition of the Darboux sums. Then \\begin{align*} m_1 &= \\inf \\bigl\\{ f(x) : x \\in [0,1-\\epsilon] \\bigr\\} = 1 , & M_1 &= \\sup \\bigl\\{ f(x) : x \\in [0,1-\\epsilon] \\bigr\\} = 1 ,", "m_2 &= \\inf \\bigl\\{ f(x) : x \\in [1-\\epsilon,1+\\epsilon] \\bigr\\} = 0 , & M_2 &= \\sup \\bigl\\{ f(x) : x \\in [1-\\epsilon,1+\\epsilon] \\bigr\\} = 1 ,", "m_3 &= \\inf \\bigl\\{ f(x) : x \\in [1+\\epsilon,2] \\bigr\\} = 0 , & M_3 &= \\sup \\bigl\\{ f(x) : x \\in [1+\\epsilon,2] \\bigr\\} = 0 . \\end{align*} Furthermore,$\\Delta x_1 = 1-\\epsilon$,$\\Delta x_2 = 2\\epsilon$, and$\\Delta x_3 = 1-\\epsilon$. See \\figureref{darbouxfigstep}. \\begin{myfigureht} \\includegraphics{figures/darbouxfigstep} \\caption{Darboux sums for the step function.$L(P,f)$is the area of the shaded rectangle,$U(P,f)$is the area of both rectangles, and$U(P,f)-L(P,f)$is the area of the unshaded rectangle.\\label{darbouxfigstep}} \\end{myfigureht}", "We compute \\begin{align*} & L(P,f) = \\sum_{i=1}^3 m_i \\Delta x_i = 1 \\cdot (1-\\epsilon) + 0 \\cdot 2\\epsilon + 0 \\cdot (1-\\epsilon) = 1-\\epsilon ,", "& U(P,f) = \\sum_{i=1}^3 M_i \\Delta x_i = 1 \\cdot (1-\\epsilon) + 1 \\cdot 2\\epsilon + 0 \\cdot (1-\\epsilon) = 1+\\epsilon . \\end{align*} Thus,", "\\begin{equation*}\n\\overline{\\int_0^2} f - \n\\underline{\\int_0^2} f\n\\leq\nU(P,f) - L(P,f)\n=\n(1+\\epsilon)\n- (1-\\epsilon) = 2 \\epsilon .\n\\end{equation*}", "By \\propref{intulbound:prop},$\\underline{\\int_0^2} f \\leq \\overline{\\int_0^2} f$. As$\\epsilon$was arbitrary,$\\overline{\\int_0^2} f = \\underline{\\int_0^2} f$. So$f$is Riemann integrable. Finally,", "\\begin{equation*}\n1-\\epsilon = L(P,f) \\leq \\int_0^2 f \\leq U(P,f) =\n1+\\epsilon.\n\\end{equation*}", "Hence,$\\bigl\\lvert \\int_0^2 f - 1 \\bigr\\rvert \\leq \\epsilon$. As$\\epsilon$was arbitrary, we conclude$\\int_0^2 f = 1$." ], "refs": [ "intulbound:prop" ], "proofs": [ { "contents": [ "Let$0 < \\epsilon < 1$be arbitrary. Let$P \\coloneqq \\{0, 1-\\epsilon, 1+\\epsilon, 2\\}$be a partition. We use the notation from the definition of the Darboux sums. Then \\begin{align*} m_1 &= \\inf \\bigl\\{ f(x) : x \\in [0,1-\\epsilon] \\bigr\\} = 1 , & M_1 &= \\sup \\bigl\\{ f(x) : x \\in [0,1-\\epsilon] \\bigr\\} = 1 , m_2 &= \\inf \\bigl\\{ f(x) : x \\in [1-\\epsilon,1+\\epsilon] \\bigr\\} = 0 , & M_2 &= \\sup \\bigl\\{ f(x) : x \\in [1-\\epsilon,1+\\epsilon] \\bigr\\} = 1 , m_3 &= \\inf \\bigl\\{ f(x) : x \\in [1+\\epsilon,2] \\bigr\\} = 0 , & M_3 &= \\sup \\bigl\\{ f(x) : x \\in [1+\\epsilon,2] \\bigr\\} = 0 . \\end{align*} Furthermore,$\\Delta x_1 = 1-\\epsilon$,$\\Delta x_2 = 2\\epsilon$, and$\\Delta x_3 = 1-\\epsilon$. See \\figureref{darbouxfigstep}. \\begin{myfigureht} \\includegraphics{figures/darbouxfigstep} \\caption{Darboux sums for the step function.$L(P,f)$is the area of the shaded rectangle,$U(P,f)$is the area of both rectangles, and$U(P,f)-L(P,f)$is the area of the unshaded rectangle.\\label{darbouxfigstep}} \\end{myfigureht} We compute \\begin{align*} & L(P,f) = \\sum_{i=1}^3 m_i \\Delta x_i = 1 \\cdot (1-\\epsilon) + 0 \\cdot 2\\epsilon + 0 \\cdot (1-\\epsilon) = 1-\\epsilon , & U(P,f) = \\sum_{i=1}^3 M_i \\Delta x_i = 1 \\cdot (1-\\epsilon) + 1 \\cdot 2\\epsilon + 0 \\cdot (1-\\epsilon) = 1+\\epsilon . \\end{align*} Thus, \\begin{equation*}", "\\overline{\\int_0^2} f -", "\\underline{\\int_0^2} f", "\\leq", "U(P,f) - L(P,f)", "=", "(1+\\epsilon)", "- (1-\\epsilon) = 2 \\epsilon .", "\\end{equation*} By \\propref{intulbound:prop},$\\underline{\\int_0^2} f \\leq \\overline{\\int_0^2} f$. As$\\epsilon$was arbitrary,$\\overline{\\int_0^2} f = \\underline{\\int_0^2} f$. So$f$is Riemann integrable. Finally, \\begin{equation*}", "1-\\epsilon = L(P,f) \\leq \\int_0^2 f \\leq U(P,f) =", "1+\\epsilon.", "\\end{equation*} Hence,$\\bigl\\lvert \\int_0^2 f - 1 \\bigr\\rvert \\leq \\epsilon$. As$\\epsilon$was arbitrary, we conclude$\\int_0^2 f = 1$." ], "refs": [ "intulbound:prop" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 7, "type": "proposition", "label": "Lebl-contfunc:7", "categories": [ "functions", "techniques", "integration", "inequalities", "named theorem" ], "title": "Riemann Integral", "contents": [ "Let$f \\colon [a,b] \\to \\R$be a bounded function. Then$f$is Riemann integrable if for every$\\epsilon > 0$, there exists a partition$P$of$[a,b]$such that\\begin{equation*} U(P,f) - L(P,f) < \\epsilon . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "If for every $\\epsilon > 0$ such a $P$ exists, then", "\\begin{equation*}", "0 \\leq", "\\overline{\\int_a^b} f -", "\\underline{\\int_a^b} f", "\\leq", "U(P,f) - L(P,f) < \\epsilon .", "\\end{equation*}", "Therefore,", "$\\overline{\\int_a^b} f = \\underline{\\int_a^b} f$, and $f$ is integrable." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 8, "type": "example", "label": "Lebl-contfunc:8", "categories": [ "functions", "techniques", "integration", "example", "series" ], "title": "Continuity of Riemann Integration", "contents": [ "Let us show$\\frac{1}{1+x}$is integrable on$[0,b]$for all$b > 0$. We will see later that continuous functions are integrable, but let us demonstrate how we do it directly.", "Let$\\epsilon > 0$be given. Take$n \\in \\N$and let$x_i \\coloneqq \\nicefrac{ib}{n}$form the partition$P \\coloneqq \\{ x_0,x_1,\\ldots,x_n \\}$of$[0,b]$. Then$\\Delta x_i = \\nicefrac{b}{n}$for all$i$. As$f$is decreasing, for every subinterval$[x_{i-1},x_i]$,", "\\begin{equation*}\nm_i = \\inf \\left\\{ \\frac{1}{1+x} : x \\in [x_{i-1},x_i] \\right\\} = \\frac{1}{1+x_i} ,\n\\quad\nM_i = \\sup \\left\\{ \\frac{1}{1+x} : x \\in [x_{i-1},x_i] \\right\\} =\n\\frac{1}{1+x_{i-1}} .\n\\end{equation*}", "Then \\begin{multline*} U(P,f)-L(P,f) = \\sum_{i=1}^n \\Delta x_i (M_i-m_i) = \\frac{b}{n} \\sum_{i=1}^n \\left( \\frac{1}{1+\\nicefrac{(i-1)b}{n}} - \\frac{1}{1+\\nicefrac{ib}{n}} \\right) =", "= \\frac{b}{n} \\left( \\frac{1}{1+\\nicefrac{0b}{n}} - \\frac{1}{1+\\nicefrac{nb}{n}} \\right) = \\frac{b^2}{n(b+1)} . \\end{multline*} The sum telescopes, the terms successively cancel each other, something we have seen before. Picking$n$to be such that$\\frac{b^2}{n(b+1)} < \\epsilon$, the proposition is satisfied, and the function is integrable." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 9, "type": "lemma", "label": "Lebl-contfunc:lemma:darbouxadd", "categories": [ "auxiliary result", "functions", "inequalities" ], "title": "Boundedness of Bounded Functions", "contents": [ "Suppose$a < b < c$and$f \\colon [a,c] \\to \\R$is a bounded function. Then\\begin{equation*} \\underline{\\int_a^c} f = \\underline{\\int_a^b} f + \\underline{\\int_b^c} f \\quad \\text{and} \\quad \\overline{\\int_a^c} f = \\overline{\\int_a^b} f + \\overline{\\int_b^c} f . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "If we have partitions $P_1 = \\{ x_0,x_1,\\ldots,x_k \\}$", "of $[a,b]$ and $P_2 = \\{ x_k, x_{k+1}, \\ldots, x_n \\}$ of $[b,c]$,", "then the set $P \\coloneqq P_1 \\cup P_2 = \\{ x_0, x_1, \\ldots, x_n \\}$ is", "a partition of $[a,c]$. We find", "\\begin{equation*}", "L(P,f) =", "\\sum_{i=1}^n m_i \\Delta x_i", "=", "\\sum_{i=1}^k m_i \\Delta x_i", "+", "\\sum_{i=k+1}^n m_i \\Delta x_i", "=", "L(P_1,f) + L(P_2,f) .", "\\end{equation*}", "When we take the supremum of the right-hand side over all $P_1$ and $P_2$,", "we are taking a supremum of the left-hand side", "over all partitions $P$ of $[a,c]$ that contain $b$. If $Q$ is a partition", "of $[a,c]$ and $P = Q \\cup \\{ b \\}$, then $P$ is a refinement of $Q$", "and so $L(Q,f) \\leq L(P,f)$. Therefore, taking a supremum only over the $P$", "that contain $b$ is sufficient to find the supremum of $L(P,f)$", "over all partitions $P$, see \\exerciseref{exercise:dominatingb}.", "Finally, recall \\exerciseref{exercise:supofsum}", "to compute", "\\begin{equation*}", "\\begin{split}", "\\underline{\\int_a^c} f", "& =", "\\sup \\, \\bigl\\{ L(P,f) : P \\text{ a partition of } [a,c] \\bigr\\}", "\\\\", "& =", "\\sup \\, \\bigl\\{ L(P,f) : P \\text{ a partition of } [a,c], b \\in P \\bigr\\}", "\\\\", "& =", "\\sup \\, \\bigl\\{ L(P_1,f) + L(P_2,f) :", "P_1 \\text{ a partition of } [a,b], P_2 \\text{ a partition of } [b,c] \\bigr\\}", "\\\\", "& =", "\\sup \\, \\bigl\\{ L(P_1,f) : P_1 \\text{ a partition of } [a,b] \\bigr\\}", "+", "\\sup \\, \\bigl\\{ L(P_2,f) : P_2 \\text{ a partition of } [b,c] \\bigr\\}", "\\\\", "&=", "\\underline{\\int_a^b} f + \\underline{\\int_b^c} f .", "\\end{split}", "\\end{equation*}", "Similarly, for $P$, $P_1$, and $P_2$ as above, we obtain", "\\begin{equation*}", "U(P,f) =", "\\sum_{i=1}^n M_i \\Delta x_i", "=", "\\sum_{i=1}^k M_i \\Delta x_i", "+", "\\sum_{i=k+1}^n M_i \\Delta x_i", "=", "U(P_1,f) + U(P_2,f) .", "\\end{equation*}", "We wish to take the infimum on the right", "over all $P_1$ and $P_2$, and so we are taking the infimum", "over all partitions $P$ of $[a,c]$ that contain $b$. If $Q$ is a partition", "of $[a,c]$ and $P = Q \\cup \\{ b \\}$, then $P$ is a refinement of $Q$", "and so $U(Q,f) \\geq U(P,f)$. Therefore, taking an infimum only over the $P$", "that contain $b$ is sufficient to find the infimum of $U(P,f)$ for", "all $P$.", "We obtain", "\\begin{equation*}", "\\overline{\\int_a^c} f", "=", "\\overline{\\int_a^b} f + \\overline{\\int_b^c} f . \\qedhere", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 10, "type": "proposition", "label": "Lebl-contfunc:10", "categories": [ "integration", "characterization", "named theorem" ], "title": "Riemann Integral", "contents": [ "Let$a < b < c$. A function$f \\colon [a,c] \\to \\R$is Riemann integrable if and only if$f$is Riemann integrable on$[a,b]$and$[b,c]$. If$f$is Riemann integrable, then\\begin{equation*} \\int_a^c f = \\int_a^b f + \\int_b^c f . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Suppose $f \\in \\sR\\bigl([a,c]\\bigr)$, then it is bounded and", "$\\overline{\\int_a^c} f =", "\\underline{\\int_a^c} f =", "\\int_a^c f$. The lemma gives", "\\begin{equation*}", "\\int_a^c f", "=", "\\underline{\\int_a^c} f", "=", "\\underline{\\int_a^b} f + \\underline{\\int_b^c} f", "\\leq", "\\overline{\\int_a^b} f + \\overline{\\int_b^c} f", "=", "\\overline{\\int_a^c} f", "=", "\\int_a^c f .", "\\end{equation*}", "Thus the inequality is an equality,", "\\begin{equation*}", "\\underline{\\int_a^b} f + \\underline{\\int_b^c} f", "=", "\\overline{\\int_a^b} f + \\overline{\\int_b^c} f .", "\\end{equation*}", "As we also know", "$\\underline{\\int_a^b} f \\leq \\overline{\\int_a^b} f$", "and", "$\\underline{\\int_b^c} f \\leq \\overline{\\int_b^c} f$, we", "conclude", "\\begin{equation*}", "\\underline{\\int_a^b} f", "=", "\\overline{\\int_a^b} f", "\\qquad \\text{and} \\qquad", "\\underline{\\int_b^c} f", "=", "\\overline{\\int_b^c} f .", "\\end{equation*}", "Thus $f$ is Riemann integrable on $[a,b]$ and $[b,c]$ and the desired formula", "holds.", "Now assume $f$ is Riemann integrable on $[a,b]$ and on $[b,c]$.", "Again it is bounded, and the lemma gives", "\\begin{equation*}", "\\underline{\\int_a^c} f", "=", "\\underline{\\int_a^b} f + \\underline{\\int_b^c} f", "=", "\\int_a^b f + \\int_b^c f", "=", "\\overline{\\int_a^b} f + \\overline{\\int_b^c} f", "=", "\\overline{\\int_a^c} f .", "\\end{equation*}", "Therefore, $f$ is Riemann integrable on $[a,c]$, and the integral is computed", "as indicated." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 11, "type": "corollary", "label": "Lebl-contfunc:intsubcor", "categories": [ "consequence" ], "title": "If$f \\in \\sR\\bigl([a,b]\\bigr)$and$[c,d] \\subset [a,b]$, then the restriction$f|_{[c,d]}$is in$\\sR...", "contents": [ "If$f \\in \\sR\\bigl([a,b]\\bigr)$and$[c,d] \\subset [a,b]$, then the restriction$f|_{[c,d]}$is in$\\sR\\bigl([c,d]\\bigr)$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 12, "type": "proposition", "label": "Lebl-contfunc:12", "categories": [], "title": "\\index{linearity of the integral}\\label{prop:integrallinear} Let$f$and$g$be in$\\sR\\bigl([a,b]\\big...", "contents": [ "\\index{linearity of the integral}\\label{prop:integrallinear} Let$f$and$g$be in$\\sR\\bigl([a,b]\\bigr)$and$\\alpha \\in \\R$. \\begin{enumerate}[(i)] \\item$\\alpha f$is in$\\sR\\bigl([a,b]\\bigr)$and\\begin{equation*} \\int_a^b \\alpha f(x) \\,dx = \\alpha \\int_a^b f(x) \\,dx . \\end{equation*}\\item$f+g$is in$\\sR\\bigl([a,b]\\bigr)$and\\begin{equation*} \\int_a^b \\bigl( f(x)+g(x) \\bigr) \\,dx = \\int_a^b f(x) \\,dx + \\int_a^b g(x) \\,dx . \\end{equation*}\\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "\\pagebreak[2]", "Let us prove the first item for $\\alpha \\geq 0$.", "Let $P$ be a partition of $[a,b]$, and", "$m_i \\coloneqq \\inf \\bigl\\{ f(x) : x \\in [x_{i-1},x_i] \\bigr\\}$ as usual.", "As $\\alpha \\geq 0$, the multiplication by $\\alpha$ moves", "past the infimum,", "\\begin{equation*}", "\\inf \\bigl\\{ \\alpha f(x) : x \\in [x_{i-1},x_i] \\bigr\\}", "=", "\\alpha \\inf \\bigl\\{ f(x) : x \\in [x_{i-1},x_i] \\bigr\\} = \\alpha m_i .", "\\end{equation*}", "Therefore,", "\\begin{equation*}", "L(P,\\alpha f) =", "\\sum_{i=1}^n \\alpha m_i \\Delta x_i = \\alpha \\sum_{i=1}^n m_i \\Delta x_i = \\alpha", "L(P,f).", "\\end{equation*}", "Similarly,", "\\begin{equation*}", "U(P,\\alpha f) = \\alpha U(P,f) .", "\\end{equation*}", "Again, as $\\alpha \\geq 0$, we", "may move multiplication by $\\alpha$ past the supremum. Hence,", "\\begin{equation*}", "\\begin{split}", "\\underline{\\int_a^b} \\alpha f(x)\\,dx & =", "\\sup \\, \\bigl\\{ L(P,\\alpha f) : P \\text{ a partition of } [a,b] \\bigr\\}", "\\\\", "& =", "\\sup \\, \\bigl\\{ \\alpha L(P,f) : P \\text{ a partition of } [a,b] \\bigr\\}", "\\\\", "& =", "\\alpha \\,", "\\sup \\, \\bigl\\{ L(P,f) : P \\text{ a partition of } [a,b] \\bigr\\}", "\\\\", "& =", "\\alpha", "\\underline{\\int_a^b} f(x)\\,dx .", "\\end{split}", "\\end{equation*}", "Similarly, we show", "\\begin{equation*}", "\\overline{\\int_a^b} \\alpha f(x)\\,dx", "=", "\\alpha", "\\overline{\\int_a^b} f(x)\\,dx .", "\\end{equation*}", "The conclusion now follows for $\\alpha \\geq 0$.", "To finish the proof of the first item (for $\\alpha < 0$), we need to show", "that $-f$ is Riemann integrable and", "$\\int_a^b - f(x)\\,dx =", "-", "\\int_a^b f(x)\\,dx$. The proof of this fact is left as", "\\exerciseref{exercise:proofoflinpropparti}.", "The proof of the second item is left as", "\\exerciseref{exercise:proofoflinproppartii}.", "It is not difficult, but it is not as", "trivial as it may appear at first glance." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 13, "type": "proposition", "label": "Lebl-contfunc:prop:upperlowerlinineq", "categories": [ "inequalities" ], "title": "Boundedness of Bounded Functions", "contents": [ "Let$f \\colon [a,b] \\to \\R$and$g \\colon [a,b] \\to \\R$be bounded functions. Then\\begin{equation*} %\\overline{\\int_a^b} \\bigl(f(x)+g(x)\\bigr)\\,dx \\leq %\\overline{\\int_a^b}f(x)\\,dx+\\overline{\\int_a^b}g(x)\\,dx \\overline{\\int_a^b} (f+g) \\leq \\overline{\\int_a^b}f+\\overline{\\int_a^b}g , \\qquad \\text{and} \\qquad \\underline{\\int_a^b} (f+g) \\geq \\underline{\\int_a^b}f+\\underline{\\int_a^b}g %\\underline{\\int_a^b} \\bigl(f(x)+g(x)\\bigr)\\,dx \\geq %\\underline{\\int_a^b}f(x)\\,dx+\\underline{\\int_a^b}g(x)\\,dx . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 14, "type": "proposition", "label": "Lebl-contfunc:14", "categories": [ "inequalities" ], "title": "Boundedness of Bounded Functions", "contents": [ "\\index{monotonicity of the integral} Let$f \\colon [a,b] \\to \\R$and$g \\colon [a,b] \\to \\R$be bounded, and$f(x) \\leq g(x)$for all$x \\in [a,b]$. Then\\begin{equation*} \\underline{\\int_a^b} f \\leq \\underline{\\int_a^b} g \\qquad \\text{and} \\qquad \\overline{\\int_a^b} f \\leq \\overline{\\int_a^b} g . \\end{equation*}Moreover, if$f$and$g$are in$\\sR\\bigl([a,b]\\bigr)$, then\\begin{equation*} \\int_a^b f \\leq \\int_a^b g . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let $P = \\{ x_0, x_1, \\ldots, x_n \\}$ be a partition of $[a,b]$. Then", "let", "\\begin{equation*}", "m_i \\coloneqq \\inf \\, \\bigl\\{ f(x) : x \\in [x_{i-1},x_i] \\bigr\\}", "\\qquad \\text{and} \\qquad", "\\widetilde{m}_i \\coloneqq \\inf \\, \\bigl\\{ g(x) : x \\in [x_{i-1},x_i] \\bigr\\} .", "\\end{equation*}", "As $f(x) \\leq g(x)$, then $m_i \\leq \\widetilde{m}_i$.", "Therefore,", "\\begin{equation*}", "L(P,f)", "=", "\\sum_{i=1}^n m_i \\Delta x_i", "\\leq", "\\sum_{i=1}^n \\widetilde{m}_i \\Delta x_i", "=", "L(P,g) .", "\\end{equation*}", "We take the supremum over all $P$ (see \\propref{prop:funcsupinf}) to obtain", "\\begin{equation*}", "\\underline{\\int_a^b} f", "\\leq", "\\underline{\\int_a^b} g .", "\\end{equation*}", "Similarly, we obtain the same conclusion for the upper integrals.", "Finally,", "if $f$ and $g$ are Riemann integrable all the integrals are equal,", "and the conclusion follows." ], "refs": [ "prop:funcsupinf" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 15, "type": "lemma", "label": "Lebl-contfunc:lemma:contint", "categories": [ "auxiliary result", "functions" ], "title": "Continuity of Continuous Functions", "contents": [ "If$f \\colon [a,b] \\to \\R$is a continuous function, then$f \\in \\sR\\bigl([a,b]\\bigr)$." ], "refs": [], "proofs": [ { "contents": [ "As $f$ is continuous on a closed bounded interval, it is", "bounded and", "uniformly continuous.", "Given $\\epsilon > 0$, find a $\\delta > 0$ such that", "$\\abs{x-y} < \\delta$ implies $\\abs{f(x)-f(y)} < \\frac{\\epsilon}{b-a}$.", "Let $P = \\{ x_0, x_1, \\ldots, x_n \\}$", "be a partition of $[a,b]$ such that $\\Delta x_i < \\delta$ for all $i = 1,2,", "\\ldots, n$. For example,", "take $n$ such that $\\frac{b-a}{n} < \\delta$, and", "let $x_i \\coloneqq \\frac{i}{n}(b-a) + a$.", "Then for all $x, y \\in [x_{i-1},x_i]$, we have", "$\\abs{x-y} \\leq \\Delta x_i < \\delta$, and so", "\\begin{equation*}", "f(x)-f(y) \\leq \\abs{f(x)-f(y)} < \\frac{\\epsilon}{b-a} .", "\\end{equation*}", "As $f$ is continuous on $[x_{i-1},x_i]$, it attains a maximum and a minimum", "on this interval.", "Let $x$ be a point where $f$ attains the maximum and $y$ be a point", "where $f$ attains the minimum. Then $f(x) = M_i$", "and $f(y) = m_i$ in the notation from the definition of the integral.", "Therefore,", "\\begin{equation*}", "M_i-m_i = f(x)-f(y) <", "\\frac{\\epsilon}{b-a} .", "\\end{equation*}", "And so", "\\begin{equation*}", "\\begin{split}", "\\overline{\\int_a^b} f -", "\\underline{\\int_a^b} f", "& \\leq", "U(P,f) - L(P,f)", "\\\\", "& =", "\\left(", "\\sum_{i=1}^n", "M_i \\Delta x_i", "\\right)", "-", "\\left(", "\\sum_{i=1}^n", "m_i \\Delta x_i", "\\right)", "\\\\", "& =", "\\sum_{i=1}^n", "(M_i-m_i) \\Delta x_i", "\\\\", "& <", "\\frac{\\epsilon}{b-a}", "\\sum_{i=1}^n", "\\Delta x_i", "\\\\", "& =", "\\frac{\\epsilon}{b-a} (b-a)", "= \\epsilon .", "\\end{split}", "\\end{equation*}", "As $\\epsilon > 0$ was arbitrary,", "\\begin{equation*}", "\\overline{\\int_a^b} f = \\underline{\\int_a^b} f ,", "\\end{equation*}", "and $f$ is Riemann integrable on $[a,b]$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 16, "type": "lemma", "label": "Lebl-contfunc:lemma:boundedimpriemann", "categories": [ "auxiliary result", "functions", "inequalities" ], "title": "Boundedness of Bounded Functions", "contents": [ "Let$f \\colon [a,b] \\to \\R$be a bounded function,$\\{ a_n \\}_{n=1}^\\infty$and$\\{b_n \\}_{n=1}^\\infty$be sequences such that$a < a_n < b_n < b$for all$n$, with$\\lim_{n\\to\\infty} a_n = a$and$\\lim_{n\\to\\infty} b_n = b$. Suppose$f \\in \\sR\\bigl([a_n,b_n]\\bigr)$for all$n$. Then$f \\in \\sR\\bigl([a,b]\\bigr)$and\\begin{equation*} \\int_a^b f = \\lim_{n \\to \\infty} \\int_{a_n}^{b_n} f . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let $M > 0$ be a real number such that $\\abs{f(x)} \\leq M$.", "As $(b-a) \\geq (b_n-a_n)$,", "\\begin{equation*}", "-M(b-a) \\leq", "-M(b_n-a_n) \\leq", "\\int_{a_n}^{b_n} f", "\\leq", "M(b_n-a_n) \\leq", "M(b-a) .", "\\end{equation*}", "Therefore, the sequence of numbers", "$\\bigl\\{ \\int_{a_n}^{b_n} f \\bigr\\}_{n=1}^\\infty$ is bounded and by", "\\hyperref[thm:bwseq]{Bolzano--Weierstrass}", "has a convergent subsequence indexed by $n_k$. Let us call", "$L$ the limit of the subsequence", "$\\bigl\\{ \\int_{a_{n_k}}^{b_{n_k}} f \\bigr\\}_{k=1}^\\infty$.", "\\lemmaref{lemma:darbouxadd} says that", "the lower and upper integral are additive", "and the hypothesis says that", "$f$ is integrable on $[a_{n_k},b_{n_k}]$.", "Therefore", "\\begin{equation*}", "\\underline{\\int_a^b} f", "=", "\\underline{\\int_a^{a_{n_k}}} f", "+", "\\int_{a_{n_k}}^{b_{n_k}} f", "+", "\\underline{\\int_{b_{n_k}}^b} f", "\\geq", "-M(a_{n_k}-a)", "+", "\\int_{a_{n_k}}^{b_{n_k}} f", "-", "M(b-b_{n_k}) .", "\\end{equation*}", "We take the limit as $k$ goes to $\\infty$ on the right-hand side,", "\\begin{equation*}", "\\underline{\\int_a^b} f", "\\geq", "-M\\cdot 0", "+", "L", "-", "M\\cdot 0", "= L .", "\\end{equation*}", "Next we use additivity of the upper integral,", "\\begin{equation*}", "\\overline{\\int_a^b} f", "=", "\\overline{\\int_a^{a_{n_k}}} f", "+", "\\int_{a_{n_k}}^{b_{n_k}} f", "+", "\\overline{\\int_{b_{n_k}}^b} f", "\\leq", "M(a_{n_k}-a)", "+", "\\int_{a_{n_k}}^{b_{n_k}} f", "+", "M(b-b_{n_k}) .", "\\end{equation*}", "We take the same subsequence", "$\\{ \\int_{a_{n_k}}^{b_{n_k}} f \\}_{k=1}^\\infty$ and take the limit", "to obtain", "\\begin{equation*}", "\\overline{\\int_a^b} f", "\\leq", "M\\cdot 0", "+", "L", "+", "M\\cdot 0", "= L .", "\\end{equation*}", "Thus $\\overline{\\int_a^b} f = \\underline{\\int_a^b} f = L$", "and hence $f$ is Riemann integrable and $\\int_a^b f = L$.", "In particular, no matter what", "subsequence we chose,", "the $L$ is the same number.", "To prove the final statement of the lemma we use", "\\propref{seqconvsubseqconv:prop}. We have shown that every convergent", "subsequence", "$\\bigl\\{ \\int_{a_{n_k}}^{b_{n_k}} f \\bigr\\}_{k=1}^\\infty$ converges to $L = \\int_a^b f$.", "Therefore, the sequence", "$\\bigl\\{ \\int_{a_n}^{b_n} f \\bigr\\}_{n=1}^\\infty$ is convergent and converges to $\\int_a^b f$." ], "refs": [ "lemma:darbouxadd", "seqconvsubseqconv:prop", "thm:bwseq" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 17, "type": "theorem", "label": "Lebl-contfunc:17", "categories": [ "functions", "inequalities" ], "title": "Boundedness of Bounded Functions", "contents": [ "Let$f \\colon [a,b] \\to \\R$be a bounded function with finitely many discontinuities. Then$f \\in \\sR\\bigl([a,b]\\bigr)$." ], "refs": [], "proofs": [ { "contents": [ "We divide the interval into finitely many intervals $[a_i,b_i]$", "so that $f$ is continuous", "on the interior $(a_i,b_i)$. If $f$ is continuous on $(a_i,b_i)$,", "then it is continuous and hence integrable on $[c_i,d_i]$ whenever $a_i < c_i < d_i < b_i$. By", "\\lemmaref{lemma:boundedimpriemann},", "the restriction", "of $f$ to $[a_i,b_i]$ is integrable.", "By additivity of the integral (and \\hyperref[induction:thm]{induction}),", "$f$ is integrable on the union of the intervals." ], "refs": [ "induction:thm", "lemma:boundedimpriemann" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 18, "type": "proposition", "label": "Lebl-contfunc:18", "categories": [ "integration", "named theorem" ], "title": "Riemann Integral", "contents": [ "Let$f \\colon [a,b] \\to \\R$be Riemann integrable. Let$g \\colon [a,b] \\to \\R$be such that$f(x) = g(x)$for all$x \\in [a,b] \\setminus S$, where$S$is a finite set. Then$g$is Riemann integrable and\\begin{equation*} \\int_a^b g = \\int_a^b f. \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "[Sketch of proof]", "Using additivity of the integral, split the interval $[a,b]$ into", "smaller intervals such that $f(x) = g(x)$ holds for all $x$ except at the", "endpoints (details are left to the reader).", "Therefore, without loss of generality suppose $f(x) = g(x)$ for", "all $x \\in (a,b)$. The proof follows by \\lemmaref{lemma:boundedimpriemann},", "and is left as \\exerciseref{exercise:changeendpointsintegral}." ], "refs": [ "lemma:boundedimpriemann" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 19, "type": "proposition", "label": "Lebl-contfunc:prop:monotoneintegrable", "categories": [], "title": "Monotonicity of Integration Theory", "contents": [ "Let$f \\colon [a,b] \\to \\R$be a monotone function. Then$f \\in \\sR\\bigl([a,b]\\bigr)$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 20, "type": "theorem", "label": "Lebl-contfunc:thm:FTCv1", "categories": [ "functions" ], "title": "Continuity of Continuous Functions", "contents": [ "Let$F \\colon [a,b] \\to \\R$be a continuous function, differentiable on$(a,b)$. Let$f \\in \\sR\\bigl([a,b]\\bigr)$be such that$f(x) = F'(x)$for$x \\in (a,b)$. Then\\begin{equation*} \\int_a^b f = F(b)-F(a) . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let $P = \\{ x_0, x_1, \\ldots, x_n \\}$ be a partition of $[a,b]$.", "For each interval $[x_{i-1},x_i]$, use the", "\\hyperref[thm:mvt]{mean value theorem} to find a", "$c_i \\in (x_{i-1},x_i)$ such that", "\\begin{equation*}", "f(c_i) \\Delta x_i = F'(c_i) (x_i - x_{i-1}) = F(x_i) - F(x_{i-1}) .", "\\end{equation*}", "See \\figureref{fig:fundthmfig}, and", "notice that the area of the", "$i$th rectangle is", "$F(x_{i+1})-F(x_{i-2})$", "for all three rectangles.", "The idea is that by taking smaller and smaller subintervals", "we prove that this area is the integral of $f$.", "\\begin{myfigureht}", "\\includegraphics{figures/fundthmfig}", "\\caption{Mean value theorem on subintervals of a partition", "approximating the area under the curve.\\label{fig:fundthmfig}}", "\\end{myfigureht}", "Using the notation from the definition of the integral,", "$m_i \\leq f(c_i) \\leq M_i$, and multiplying by $\\Delta x_i$ gets", "\\begin{equation*}", "m_i \\Delta x_i \\leq F(x_i) - F(x_{i-1}) \\leq M_i \\Delta x_i .", "\\end{equation*}", "We sum over $i = 1,2, \\ldots, n$ to get", "\\begin{equation*}", "\\sum_{i=1}^n m_i \\Delta x_i", "\\leq \\sum_{i=1}^n \\bigl(F(x_i) - F(x_{i-1}) \\bigr)", "\\leq \\sum_{i=1}^n M_i \\Delta x_i .", "\\end{equation*}", "In the middle sum, all the terms except the first and last cancel", "and we end up with $F(x_n)-F(x_0) = F(b)-F(a)$. The sums on the left", "and on the right are the lower and the upper sum respectively. So", "\\begin{equation*}", "L(P,f) \\leq F(b)-F(a) \\leq U(P,f) .", "\\end{equation*}", "We take the supremum of $L(P,f)$ over all partitions $P$ and the left inequality", "yields", "\\begin{equation*}", "\\underline{\\int_a^b} f \\leq F(b)-F(a) .", "\\end{equation*}", "Similarly, taking", "the infimum of $U(P,f)$ over all partitions $P$ yields", "\\begin{equation*}", "F(b)-F(a) \\leq \\overline{\\int_a^b} f .", "\\end{equation*}", "As $f$ is Riemann integrable, we have", "\\begin{equation*}", "\\int_a^b f =", "\\underline{\\int_a^b} f \\leq F(b)-F(a) \\leq \\overline{\\int_a^b} f", "= \\int_a^b f .", "\\end{equation*}", "The inequalities must be equalities and we are done." ], "refs": [ "named:Mean Value", "thm:mvt" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 21, "type": "example", "label": "Lebl-contfunc:21", "categories": [ "calculus", "example" ], "title": "Fundamental Theorem of Calculus", "contents": [ "To compute", "\\begin{equation*}\n\\int_0^1 x^2 \\,dx ,\n\\end{equation*}", "we notice$x^2$is the derivative of$\\frac{x^3}{3}$. The fundamental theorem says", "\\begin{equation*}\n\\int_0^1 x^2 \\,dx =\n\\frac{1^3}{3}\n-\n\\frac{0^3}{3}\n= \\frac{1}{3}.\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 22, "type": "theorem", "label": "Lebl-contfunc:thm:FTCv2", "categories": [ "integration", "named theorem" ], "title": "Riemann Integral", "contents": [ "Let$f \\colon [a,b] \\to \\R$be a Riemann integrable function. Define\\begin{equation*} F(x) \\coloneqq \\int_a^x f . \\end{equation*}First,$F$is continuous on$[a,b]$. Second, if$f$is continuous at$c \\in [a,b]$, then$F$is differentiable at$c$and$F'(c) = f(c)$." ], "refs": [], "proofs": [ { "contents": [ "As $f$ is bounded, there is an $M > 0$", "such that $\\abs{f(x)} \\leq M$ for all $x \\in [a,b]$. Suppose $x,y \\in [a,b]$", "with $x > y$. Then", "\\begin{equation*}", "\\abs{F(x)-F(y)} =", "\\abs{\\int_a^x f - \\int_a^y f}", "=", "\\abs{\\int_y^x f}", "\\leq", "M\\abs{x-y} .", "\\end{equation*}", "By symmetry, the same also holds if $x < y$.", "So $F$ is Lipschitz continuous and hence continuous.", "Now suppose $f$ is continuous at $c$.", "Let $\\epsilon > 0$ be given. Let $\\delta > 0$ be such that", "for $x \\in [a,b]$,", "$\\abs{x-c} < \\delta$ implies $\\abs{f(x)-f(c)} < \\epsilon$.", "In particular,", "for such $x$, we have", "\\begin{equation*}", "f(c)-\\epsilon < f(x) < f(c) + \\epsilon.", "\\end{equation*}", "Thus if $x > c$, then", "\\begin{equation*}", "\\bigl(f(c)-\\epsilon\\bigr) (x-c) \\leq \\int_c^x f \\leq", "\\bigl(f(c) + \\epsilon\\bigr)(x-c).", "\\end{equation*}", "When $c > x$, then the inequalities are reversed. Therefore,", "assuming $x \\not= c$, we get", "\\begin{equation*}", "f(c)-\\epsilon", "\\leq", "\\frac{\\int_c^{x} f}{x-c}", "\\leq", "f(c)+\\epsilon .", "\\end{equation*}", "As", "\\begin{equation*}", "\\frac{F(x)-F(c)}{x-c}", "=", "\\frac{\\int_a^{x} f - \\int_a^{c} f}{x-c}", "=", "\\frac{\\int_c^{x} f}{x-c} ,", "\\end{equation*}", "we have", "\\begin{equation*}", "\\abs{\\frac{F(x)-F(c)}{x-c} - f(c)} \\leq \\epsilon .", "\\end{equation*}", "The result follows. It is left to the reader to see why is it OK that we", "just have a non-strict inequality." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 23, "type": "theorem", "label": "Lebl-contfunc:23", "categories": [ "calculus" ], "title": "Change of Variables Theorem", "contents": [ "\\index{change of variables theorem} Let$g \\colon [a,b] \\to \\R$be a continuously differentiable function, let$f \\colon [c,d] \\to \\R$be continuous, and suppose$g\\bigl([a,b]\\bigr) \\subset [c,d]$. Then\\begin{equation*} \\int_a^b f\\bigl(g(x)\\bigr)\\, g'(x)\\, dx = \\int_{g(a)}^{g(b)} f(s)\\, ds . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "As $g$, $g'$, and $f$ are continuous, $f\\bigl(g(x)\\bigr)\\,g'(x)$", "is a continuous function of $[a,b]$, therefore it is Riemann integrable.", "Similarly, $f$ is integrable on every subinterval of $[c,d]$.", "Define $F \\colon [c,d] \\to \\R$ by", "\\begin{equation*}", "F(y) \\coloneqq \\int_{g(a)}^{y} f(s)\\,ds .", "\\end{equation*}", "By the second form of the fundamental", "theorem of calculus", "(see \\remarkref{remark:fundthmbase} and \\exerciseref{secondftc:exercise}),", "$F$ is a differentiable function and $F'(y) = f(y)$. Apply the chain", "rule,", "\\begin{equation*}", "\\bigl( F \\circ g \\bigr)' (x) =", "F'\\bigl(g(x)\\bigr) g'(x)", "=", "f\\bigl(g(x)\\bigr) g'(x) .", "\\end{equation*}", "Note that $F\\bigl(g(a)\\bigr) = 0$ and", "use the first form of the fundamental theorem", "to obtain", "\\begin{multline*}", "\\qquad %to center things more", "\\int_{g(a)}^{g(b)} f(s)\\,ds = F\\bigl(g(b)\\bigr) = F\\bigl(g(b)\\bigr)-F\\bigl(g(a)\\bigr)", "\\\\", "=", "\\int_a^b", "\\bigl( F \\circ g \\bigr)' (x) \\,dx", "=", "\\int_a^b", "f\\bigl(g(x)\\bigr) g'(x)", "\\,dx .", "\\qquad %to center things more", "\\qedhere", "\\end{multline*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 24, "type": "example", "label": "Lebl-contfunc:24", "categories": [ "example" ], "title": "The derivative of$\\sin(x)$is$\\cos(x)$", "contents": [ "The derivative of$\\sin(x)$is$\\cos(x)$. Using$g(x) \\coloneqq x^2$, we solve", "\\begin{equation*}\n\\int_0^{\\sqrt{\\pi}} x \\cos(x^2) \\, dx = \\int_0^\\pi \\frac{\\cos(s)}{2} \\, ds\n=\n\\frac{1}{2}\n\\int_0^\\pi \\cos(s) \\, ds\n=\n\\frac{\n\\sin(\\pi) - \\sin(0)\n}{2}\n=\n0 .\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 25, "type": "example", "label": "Lebl-contfunc:25", "categories": [ "improper", "integration", "example", "inequalities", "named theorem" ], "title": "Riemann Integral", "contents": [ "Consider", "\\begin{equation*}\n\\int_{-1}^{1} \\frac{\\ln \\abs{x}}{x} \\,dx .\n\\end{equation*}", "It may be tempting to take$g(x) \\coloneqq \\ln \\abs{x}$. Compute$g'(x) = \\nicefrac{1}{x}$and try to write", "\\begin{equation*}\n\\int_{g(-1)}^{g(1)} s \\,ds = \n\\int_{0}^{0} s \\,ds = 0. \n\\end{equation*}", "This \\myquote{solution} is incorrect, and it does not say that we can solve the given integral. First problem is that$\\frac{\\ln \\abs{x}}{x}$is not continuous on$[-1,1]$. It is not defined at 0, and cannot be made continuous by defining a value at 0. Second,$\\frac{\\ln \\abs{x}}{x}$is not even Riemann integrable on$[-1,1]$(it is unbounded). The integral we wrote down simply does not make sense. Finally,$g$is not continuous on$[-1,1]$, let alone continuously differentiable." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 26, "type": "proposition", "label": "Lebl-contfunc:26", "categories": [], "title": "Uniqueness of Trigonometric Functions", "contents": [ "There exists a unique function$L \\colon (0,\\infty) \\to \\R$such that \\begin{enumerate}[(i)] \\item \\label{it:log:i}$L(1) = 0$. \\item \\label{it:log:ii}$L$is differentiable and$L'(x) = \\nicefrac{1}{x}$. \\item \\label{it:log:iii}$L$is strictly increasing, bijective, and\\begin{equation*} \\lim_{x\\to 0} L(x) = -\\infty , \\qquad \\text{and} \\qquad \\lim_{x\\to \\infty} L(x) = \\infty . \\end{equation*}\\item \\label{it:log:iv}$L(xy) = L(x)+L(y)$for all$x,y \\in (0,\\infty)$. \\item \\label{it:log:v} If$q$is a rational number and$x > 0$, then$L(x^q) = q L(x)$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "To prove existence, we define a candidate and show it satisfies", "all the properties. Let", "\\begin{equation*}", "L(x) \\coloneqq \\int_1^x \\frac{1}{t}\\,dt .", "\\end{equation*}", "Obviously, \\ref{it:log:i} holds. Property \\ref{it:log:ii} holds", "via the second form of the fundamental theorem of calculus", "(\\thmref{thm:FTCv2}).", "To prove property \\ref{it:log:iv},", "we change variables $u=yt$ to obtain", "\\begin{equation*}", "L(x) =", "\\int_1^{x} \\frac{1}{t}\\,dt", "=", "\\int_y^{xy} \\frac{1}{u}\\,du", "=", "\\int_1^{xy} \\frac{1}{u}\\,du", "-", "\\int_1^{y} \\frac{1}{u}\\,du", "=", "L(xy)-L(y) .", "\\end{equation*}", "Let us prove \\ref{it:log:iii}.", "Property \\ref{it:log:ii} together with the fact that $L'(x) = \\nicefrac{1}{x} > 0$", "for $x > 0$, implies that $L$", "is strictly increasing and hence one-to-one.", "Let us show $L$ is onto.", "As $\\nicefrac{1}{t} \\geq \\nicefrac{1}{2}$ when $t \\in [1,2]$,", "\\begin{equation*}", "L(2) = \\int_1^2 \\frac{1}{t} \\,dt \\geq \\nicefrac{1}{2} .", "\\end{equation*}", "By induction, \\ref{it:log:iv} implies that for $n \\in \\N$,", "\\begin{equation*}", "L(2^n) = L(2) + L(2) + \\cdots + L(2) = n L(2) .", "\\end{equation*}", "Given $y > 0$,", "by the \\hyperref[thm:arch:i]{Archimedean property} of the real numbers", "(notice $L(2) > 0$), there is an $n \\in \\N$ such that", "$L(2^n) > y$. The", "\\hyperref[IVT:thm]{intermediate value theorem}", "gives an $x_1 \\in (1,2^n)$ such that $L(x_1) = y$. Thus", "$(0,\\infty)$ is in the image of $L$.", "As $L$ is increasing, $L(x) > y$ for all $x > 2^n$, and so", "\\begin{equation*}", "\\lim_{x\\to\\infty} L(x) = \\infty .", "\\end{equation*}", "Next", "$0 = L(\\nicefrac{x}{x}) = L(x) + L(\\nicefrac{1}{x})$, and", "so $L(x) = - L(\\nicefrac{1}{x})$. Using $x=2^{-n}$, we obtain", "as above that $L$ achieves all negative numbers. And", "\\begin{equation*}", "\\lim_{x \\to 0} L(x) =", "\\lim_{x \\to 0} -L(\\nicefrac{1}{x})", "=", "\\lim_{x \\to \\infty} -L(x)", "= - \\infty .", "\\end{equation*}", "In the limits, note that only $x > 0$ are in the domain of $L$.", "Let us prove \\ref{it:log:v}.", "Fix $x > 0$.", "As above, \\ref{it:log:iv} implies", "$L(x^n) = n L(x)$", "for all $n \\in \\N$.", "We already found that", "$L(x) = - L(\\nicefrac{1}{x})$,", "so $L(x^{-n}) = - L(x^n) = -n L(x)$. Then for $m \\in \\N$", "\\begin{equation*}", "L(x) = L\\Bigl({(x^{1/m})}^m\\Bigr) = m L\\bigl(x^{1/m}\\bigr) .", "\\end{equation*}", "Putting everything together for $n \\in \\Z$ and $m \\in \\N$, we have", "$L(x^{n/m}) = n L(x^{1/m}) = (\\nicefrac{n}{m}) L(x)$.", "Uniqueness follows using properties \\ref{it:log:i} and", "\\ref{it:log:ii}. Via the first form of the", "fundamental theorem of calculus (\\thmref{thm:FTCv1}),", "\\begin{equation*}", "L(x) = \\int_1^x \\frac{1}{t}\\,dt", "\\end{equation*}", "is the unique function such that $L(1) = 0$ and $L'(x) = \\nicefrac{1}{x}$." ], "refs": [ "IVT:thm", "it:log:i", "it:log:ii", "it:log:iii", "it:log:iv", "it:log:v", "thm:arch:i" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 27, "type": "proposition", "label": "Lebl-contfunc:27", "categories": [], "title": "Uniqueness of Trigonometric Functions", "contents": [ "There exists a unique function$E \\colon \\R \\to (0,\\infty)$such that \\begin{enumerate}[(i)] \\item \\label{it:exp:i}$E(0) = 1$. \\item \\label{it:exp:ii}$E$is differentiable and$E'(x) = E(x)$. \\item \\label{it:exp:iii}$E$is strictly increasing, bijective, and\\begin{equation*} \\lim_{x\\to -\\infty} E(x) = 0 , \\qquad \\text{and} \\qquad \\lim_{x\\to \\infty} E(x) = \\infty . \\end{equation*}\\item \\label{it:exp:iv}$E(x+y) = E(x)E(y)$for all$x,y \\in \\R$. \\item \\label{it:exp:v} If$q \\in \\Q$, then$E(qx) = {E(x)}^q$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "Again, we prove existence of such a function by defining a candidate", "and proving that it satisfies all the properties.", "The $L = \\ln$ defined above is invertible. Let $E$ be the", "inverse function of $L$. Property \\ref{it:exp:i} is immediate.", "Property \\ref{it:exp:ii} follows", "via the inverse function theorem, in particular", "via \\lemmaref{lemma:ift}: $L$~satisfies", "all the hypotheses of the lemma, and hence", "\\begin{equation*}", "E'(x) = \\frac{1}{L'\\bigl(E(x)\\bigr)} = E(x) .", "\\end{equation*}", "Let us look at property \\ref{it:exp:iii}.", "The function $E$ is strictly increasing since", "$E'(x) = E(x) > 0$. As $E$ is the inverse of $L$, it must also", "be bijective.", "To find the limits, we use that", "$E$ is strictly increasing and onto $(0,\\infty)$.", "For every $M > 0$, there is an $x_0$ such that", "$E(x_0) = M$ and $E(x) \\geq M$ for all $x \\geq x_0$.", "Similarly, for every $\\epsilon > 0$, there is", "an $x_0$ such that $E(x_0) = \\epsilon$ and", "$E(x) < \\epsilon$ for all $x < x_0$.", "Therefore,", "\\begin{equation*}", "\\lim_{n\\to -\\infty} E(x) = 0 , \\qquad \\text{and} \\qquad", "\\lim_{n\\to \\infty} E(x) = \\infty .", "\\end{equation*}", "To prove property \\ref{it:exp:iv}, we use the corresponding", "property for the logarithm.", "Take $x, y \\in \\R$.", "As $L$ is bijective, find $a$ and $b$ such that $x = L(a)$ and $y = L(b)$. Then", "\\begin{equation*}", "E(x+y) =", "E\\bigl(L(a)+L(b)\\bigr) =", "E\\bigl(L(ab)\\bigr) = ab = E(x)E(y) .", "\\end{equation*}", "Property \\ref{it:exp:v} also follows from the corresponding property of $L$.", "Given $x \\in \\R$, let $a$ be such that $x = L(a)$ and", "\\begin{equation*}", "E(qx) = E\\bigl(qL(a)\\bigr)", "=", "E\\bigl(L(a^q)\\bigr) = a^q = {E(x)}^q .", "\\end{equation*}", "Uniqueness follows from", "\\ref{it:exp:i} and", "\\ref{it:exp:ii}.", "Let $E$ and $F$", "be two functions satisfying", "\\ref{it:exp:i} and \\ref{it:exp:ii}.", "\\begin{equation*}", "\\frac{d}{dx} \\Bigl( F(x)E(-x) \\Bigr)", "=", "F'(x)E(-x) - E'(-x)F(x)", "=", "F(x)E(-x) - E(-x)F(x) = 0 .", "\\end{equation*}", "Therefore, by \\propref{prop:derzeroconst},", "$F(x)E(-x) = F(0)E(-0) = 1$ for all $x \\in \\R$.", "Next, $1 = E(0) = E(x-x) = E(x)E(-x)$.", "%Doing the computation with $F = E$,", "%we obtain $E(x)E(-x) = 1$.", "Then", "\\begin{equation*}", "0 = 1-1 = F(x)E(-x) - E(x)E(-x) = \\bigl(F(x)-E(x)\\bigr) E(-x) .", "\\end{equation*}", "%Since $E(x)E(-x) = 1$,", "Finally, $E(-x) \\not= 0$\\footnote{%", "$E$ is a function into $(0,\\infty)$ after all.", "However, $E(-x) \\neq 0$ also follows", "from $E(x)E(-x) = 1$. Therefore, we can prove uniqueness of $E$", "given \\ref{it:exp:i} and \\ref{it:exp:ii}, even for functions $E \\colon \\R", "\\to \\R$.}", "for all $x \\in \\R$.", "So", "$F(x)-E(x) = 0$ for all $x$, and we are done." ], "refs": [ "it:exp:i", "it:exp:ii", "it:exp:iii", "it:exp:iv", "it:exp:v", "lemma:ift", "named:Inverse Function", "prop:derzeroconst" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 28, "type": "proposition", "label": "Lebl-contfunc:28", "categories": [], "title": "Let$x, y \\in \\R$", "contents": [ "Let$x, y \\in \\R$. \\begin{enumerate}[(i)] \\item$\\exp(xy) = {\\bigl(\\exp(x)\\bigr)}^y$. \\item If$x > 0$, then$\\ln(x^y) = y \\ln (x)$. \\end{enumerate}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 29, "type": "proposition", "label": "Lebl-contfunc:29", "categories": [ "series", "improper" ], "title": "p-test for Improper Integrals", "contents": [ "% \\index{p-test for integrals@$p$-test for integrals} \\label{impropriemann:ptest} The improper integral\\begin{equation*} \\int_1^\\infty \\frac{1}{x^p} \\,dx \\end{equation*}converges to$\\frac{1}{p-1}$if$p > 1$and diverges if$0 < p \\leq 1$.", "The improper integral\\begin{equation*} \\int_0^1 \\frac{1}{x^p} \\,dx \\end{equation*}converges to$\\frac{1}{1-p}$if$0 < p < 1$and diverges if$p \\geq 1$." ], "refs": [], "proofs": [ { "contents": [ "The proof follows by application of the", "\\hyperref[thm:FTCv1]{fundamental theorem of calculus}.", "Let us do the proof for $p > 1$ for the infinite right endpoint and", "leave the rest to the reader. Hint: You should handle $p=1$", "separately.", "Suppose $p > 1$. Then", "using the fundamental theorem,", "\\begin{equation*}", "\\int_1^b \\frac{1}{x^p} \\,dx", "=", "\\int_1^b x^{-p} \\,dx", "=", "\\frac{b^{-p+1}}{-p+1}", "-", "\\frac{1^{-p+1}}{-p+1}", "=", "\\frac{-1}{(p-1)b^{p-1}}", "+", "\\frac{1}{p-1} .", "\\end{equation*}", "As $p > 1$, then $p-1 > 0$. Take the limit as $b \\to \\infty$", "to obtain that $\\frac{1}{b^{p-1}}$ goes to 0. The result follows." ], "refs": [ "thm:FTCv1" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 30, "type": "proposition", "label": "Lebl-contfunc:impropriemann:tail", "categories": [ "integration", "characterization", "named theorem" ], "title": "Riemann Integral", "contents": [ "Let$f \\colon [a,\\infty) \\to \\R$be a function that is Riemann integrable on$[a,b]$for all$b > a$. For every$b > a$, the integral$\\int_b^\\infty f$converges if and only if$\\int_a^\\infty f$converges, in which case\\begin{equation*} \\int_a^\\infty f = \\int_a^b f + \\int_b^\\infty f . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let $c > b$. Then", "\\begin{equation*}", "\\int_a^c f", "=", "\\int_a^b f +", "\\int_b^c f .", "\\end{equation*}", "Taking the limit $c \\to \\infty$ finishes the proof." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 31, "type": "proposition", "label": "Lebl-contfunc:impropriemann:possimp", "categories": [ "integration", "characterization", "named theorem" ], "title": "Riemann Integral", "contents": [ "Suppose$f \\colon [a,\\infty) \\to \\R$is nonnegative ($f(x) \\geq 0$for all$x$) and$f$is Riemann integrable on$[a,b]$for all$b > a$. \\begin{enumerate}[(i)] \\item\\begin{equation*} \\int_a^\\infty f = \\sup \\left\\{ \\int_a^x f : x \\geq a \\right\\} . \\end{equation*}\\item Suppose$\\{ x_n \\}_{n=1}^\\infty$is a sequence with$\\lim_{n\\to\\infty} x_n = \\infty$. Then$\\int_a^\\infty f$converges if and only if$\\lim_{n\\to\\infty} \\int_a^{x_n} f$exists, in which case\\begin{equation*} \\int_a^\\infty f = \\lim_{n\\to\\infty} \\int_a^{x_n} f . \\end{equation*}\\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "We start with the first item.", "As $f$ is nonnegative,", "$\\int_a^x f$ is increasing as a function of $x$.", "If the supremum is infinite, then for every $M \\in \\R$", "we find $N$ such that $\\int_a^N f \\geq M$. As $\\int_a^x f$", "is increasing, $\\int_a^x f \\geq M$ for all $x \\geq N$. So", "$\\int_a^\\infty f$ diverges to infinity.", "Next suppose the supremum is finite, say", "$A \\coloneqq \\sup \\left\\{ \\int_a^x f : x \\geq a \\right\\}$.", "For every $\\epsilon > 0$, we find an $N$ such that", "$A - \\int_a^N f < \\epsilon$. As $\\int_a^x f$ is increasing,", "then", "$A - \\int_a^x f < \\epsilon$ for all $x \\geq N$ and hence", "$\\int_a^\\infty f$ converges to $A$.", "Let us look at the second item.", "If $\\int_a^\\infty f$ converges, then every sequence $\\{ x_n \\}_{n=1}^\\infty$ going to", "infinity works. The trick is", "proving the other direction. Suppose $\\{ x_n \\}_{n=1}^\\infty$ is such that", "$\\lim_{n\\to\\infty} x_n = \\infty$ and", "\\begin{equation*}", "\\lim_{n\\to\\infty} \\int_a^{x_n} f = A", "\\end{equation*}", "converges. Given $\\epsilon > 0$, pick $N$ such that for", "all $n \\geq N$, we have", "$A - \\epsilon < \\int_a^{x_n} f < A + \\epsilon$.", "Because $\\int_a^x f$ is increasing as a function of $x$, we have that for all", "$x \\geq x_N$", "\\begin{equation*}", "A - \\epsilon < \\int_a^{x_N} f \\leq \\int_a^x f .", "\\end{equation*}", "As $\\{ x_n \\}_{n=1}^\\infty$ goes to $\\infty$, then for any given", "$x$, there is an $x_m$ such that $m \\geq N$ and $x \\leq x_m$. Then", "\\begin{equation*}", "\\int_a^{x} f \\leq \\int_a^{x_m} f < A + \\epsilon .", "\\end{equation*}", "In particular, for all $x \\geq x_N$, we have", "$\\abs{\\int_a^{x} f - A} < \\epsilon$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 32, "type": "proposition", "label": "Lebl-contfunc:32", "categories": [ "integration", "named theorem", "improper" ], "title": "Riemann Integral", "contents": [ "% \\index{comparison test for improper integrals} Let$f \\colon [a,\\infty) \\to \\R$and$g \\colon [a,\\infty) \\to \\R$be functions that are Riemann integrable on$[a,b]$for all$b > a$. Suppose that for all$x \\geq a$,\\begin{equation*} \\abs{f(x)} \\leq g(x) . \\end{equation*}\\begin{enumerate}[(i)] \\item If$\\int_a^\\infty g$converges, then$\\int_a^\\infty f$converges, and in this case$\\abs{\\int_a^\\infty f} \\leq \\int_a^\\infty g$. \\item If$\\int_a^\\infty f$diverges, then$\\int_a^\\infty g$diverges. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "We start with the first item.", "For every $b$ and $c$, such that $a \\leq b \\leq c$, we have", "$-g(x) \\leq f(x) \\leq g(x)$, and so", "\\begin{equation*}", "\\int_b^c -g \\leq \\int_b^c f \\leq \\int_b^c g .", "\\end{equation*}", "In other words, $\\abs{\\int_b^c f} \\leq \\int_b^c g$.", "Let $\\epsilon > 0$ be given. Because", "of \\propref{impropriemann:tail},", "\\begin{equation*}", "\\int_a^\\infty g =", "\\int_a^b g +", "\\int_b^\\infty g .", "\\end{equation*}", "As $\\int_a^b g$ goes to", "$\\int_a^\\infty g$ as $b$ goes to infinity,", "$\\int_b^\\infty g$ goes to 0 as $b$ goes to infinity. Choose $B$", "such that", "\\begin{equation*}", "\\int_B^\\infty g < \\epsilon .", "\\end{equation*}", "As $g$ is nonnegative, if $B \\leq b < c$, then", "$\\int_b^c g < \\epsilon$ as well.", "Let $\\{ x_n \\}_{n=1}^\\infty$ be a sequence going to infinity. Let $M$ be such that", "$x_n \\geq B$ for all $n \\geq M$. Take $n, m \\geq M$,", "with $x_n \\leq x_m$,", "\\begin{equation*}", "\\abs{\\int_a^{x_m} f - \\int_a^{x_n} f}", "=", "\\abs{\\int_{x_n}^{x_m} f}", "\\leq \\int_{x_n}^{x_m} g < \\epsilon .", "\\end{equation*}", "Therefore, the sequence $\\bigl\\{ \\int_a^{x_n} f \\bigr\\}_{n=1}^\\infty$ is Cauchy and hence converges.", "We need to show that the limit is unique. Suppose $\\{ x_n \\}_{n=1}^\\infty$ is a sequence", "converging to infinity such that", "$\\bigl\\{ \\int_a^{x_n} f \\bigr\\}_{n=1}^\\infty$ converges to $L_1$, and $\\{", "y_n \\}_{n=1}^\\infty$ is a sequence", "converging to infinity such that", "$\\bigl\\{ \\int_a^{y_n} f \\bigr\\}_{n=1}^\\infty$ converges to $L_2$. Then there must be some $n$ such", "that", "$\\babs{\\int_a^{x_n} f - L_1} < \\epsilon$ and", "$\\babs{\\int_a^{y_n} f - L_2} < \\epsilon$. We can also suppose $x_n \\geq B$", "and $y_n \\geq B$. Then", "\\begin{equation*}", "\\abs{L_1 - L_2} \\leq", "\\abs{L_1 - \\int_a^{x_n} f}", "+", "\\abs{\\int_a^{x_n} f- \\int_a^{y_n} f}", "+", "\\abs{\\int_a^{y_n} f - L_2}", "<", "\\epsilon", "+", "\\abs{\\int_{x_n}^{y_n} f}", "+", "\\epsilon", "<", "3 \\epsilon.", "\\end{equation*}", "As $\\epsilon > 0$ was arbitrary, $L_1 = L_2$, and hence", "$\\int_a^\\infty f$ converges.", "Above we have shown that $\\abs{\\int_a^c f} \\leq \\int_a^c g$ for all $c > a$.", "By taking the limit $c \\to \\infty$, the first item is proved.", "The second item is simply a contrapositive of the first item." ], "refs": [ "impropriemann:tail" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 33, "type": "example", "label": "Lebl-contfunc:33", "categories": [ "example", "improper" ], "title": "Comparison Test for Improper Integrals", "contents": [ "The improper integral", "\\begin{equation*}\n\\int_0^\\infty \\frac{\\sin(x^2)(x+2)}{x^3+1} \\,dx\n\\end{equation*}", "converges.", "Proof: Observe we simply need to show that the integral converges when going from 1 to infinity. For$x \\geq 1$we obtain", "\\begin{equation*}\n\\abs{\\frac{\\sin(x^2)(x+2)}{x^3+1}}\n\\leq\n\\frac{x+2}{x^3+1}\n\\leq \\frac{x+2}{x^3} \\leq\n\\frac{x+2x}{x^3} \\leq \\frac{3}{x^2} .\n\\end{equation*}", "Then", "\\begin{equation*}\n\\int_1^\\infty \\frac{3}{x^2}\\,dx\n=\n3 \\int_1^\\infty \\frac{1}{x^2}\\,dx\n%=\n%\\lim_{c\\to\\infty} \\int_1^c \\frac{3}{x^2} \\,dx\n=\n3 .\n\\end{equation*}", "So using the comparison test and the tail test, the original integral converges." ], "refs": [], "proofs": [ { "contents": [ "Observe we simply need to show that the integral converges when going from 1 to infinity. For$x \\geq 1$we obtain \\begin{equation*}", "\\abs{\\frac{\\sin(x^2)(x+2)}{x^3+1}}", "\\leq", "\\frac{x+2}{x^3+1}", "\\leq \\frac{x+2}{x^3} \\leq", "\\frac{x+2x}{x^3} \\leq \\frac{3}{x^2} .", "\\end{equation*} Then \\begin{equation*}", "\\int_1^\\infty \\frac{3}{x^2}\\,dx", "=", "3 \\int_1^\\infty \\frac{1}{x^2}\\,dx", "%=", "%\\lim_{c\\to\\infty} \\int_1^c \\frac{3}{x^2} \\,dx", "=", "3 .", "\\end{equation*} So using the comparison test and the tail test, the original integral converges." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 34, "type": "example", "label": "Lebl-contfunc:34", "categories": [ "convergence", "example", "improper" ], "title": "Comparison Test for Improper Integrals", "contents": [ "You should be careful when doing formal manipulations with improper integrals. The integral", "\\begin{equation*}\n\\int_2^\\infty \\frac{2}{x^2-1}\\,dx\n\\end{equation*}", "converges via the comparison test using$\\nicefrac{1}{x^2}$again. However, if you succumb to the temptation to write", "\\begin{equation*}\n\\frac{2}{x^2-1} = \n\\frac{1}{x-1}\n-\n\\frac{1}{x+1} \n\\end{equation*}", "and try to integrate each part separately, you will not succeed. It is \\emph{not} true that you can split the improper integral in two; you cannot split the limit.", "\\begin{equation*}\n\\begin{split}\n\\int_2^\\infty \\frac{2}{x^2-1} \\,dx &=\n\\lim_{b\\to \\infty} \\int_2^b \\frac{2}{x^2-1} \\,dx\n\\\\\n&=\n\\lim_{b\\to \\infty}\n\\left(\n\\int_2^b \\frac{1}{x-1}\\,dx\n-\n\\int_2^b \\frac{1}{x+1}\\,dx\n\\right)\n\\\\\n&\\not=\n\\int_2^\\infty \\frac{1}{x-1}\\,dx\n-\n\\int_2^\\infty \\frac{1}{x+1}\\,dx .\n\\end{split}\n\\end{equation*}", "The last line in the computation does not even make sense. Both of the integrals diverge to infinity, since we can apply the comparison test appropriately with$\\nicefrac{1}{x}$. We get$\\infty - \\infty$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 35, "type": "example", "label": "Lebl-contfunc:35", "categories": [ "example" ], "title": "\\begin{equation*} \\int_{-\\infty}^\\infty \\frac{1}{1+x^2} \\, dx = \\lim_{a \\to -\\infty} \\, \\lim_{b \\...", "contents": [ "\\begin{equation*}\n\\int_{-\\infty}^\\infty \\frac{1}{1+x^2} \\, dx\n=\n\\lim_{a \\to -\\infty} \\, \\lim_{b \\to \\infty} \\,\n\\int_{a}^b \\frac{1}{1+x^2} \\, dx\n=\n\\lim_{a \\to -\\infty} \\, \\lim_{b \\to \\infty}\n\\bigl( \\arctan(b) - \\arctan(a) \\bigr)\n=\n\\pi .\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 36, "type": "proposition", "label": "Lebl-contfunc:36", "categories": [ "characterization", "integration", "inequalities", "improper" ], "title": "Characterization of Riemann Integration via Equivalence", "contents": [ "Suppose$f \\colon \\R \\to \\R$is integrable on every bounded interval$[a,b]$. Then\\begin{equation*} \\lim_{a \\to -\\infty} \\, \\lim_{b \\to \\infty} \\, \\int_a^b f \\quad \\text{converges} \\qquad \\text{if and only if} \\qquad \\lim_{b \\to \\infty} \\, \\lim_{a \\to -\\infty} \\, \\int_a^b f \\quad \\text{converges,} \\end{equation*}in which case the two expressions are equal. If either of the expressions converges, then the improper integral converges and\\begin{equation*} \\lim_{a\\to\\infty} \\int_{-a}^a f = \\int_{-\\infty}^\\infty f . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Without loss of generality, assume $a < 0$ and $b > 0$. Suppose", "the first expression converges. Then", "\\begin{equation*}", "\\begin{split}", "\\lim_{a \\to -\\infty} \\, \\lim_{b \\to \\infty} \\, \\int_a^b f", "& =", "\\lim_{a \\to -\\infty} \\, \\lim_{b \\to \\infty}", "\\left(", "\\int_a^0 f", "+", "\\int_0^b f", "\\right)", "=", "\\left(", "\\lim_{a \\to -\\infty}", "\\int_a^0 f", "\\right)", "+", "\\left(", "\\lim_{b \\to \\infty}", "\\int_0^b f", "\\right) \\\\", "& =", "\\lim_{b \\to \\infty}", "\\left(", "\\left(", "\\lim_{a \\to -\\infty}", "\\int_a^0 f", "\\right)", "+", "\\int_0^b f", "\\right)", "=", "\\lim_{b \\to \\infty} \\,", "\\lim_{a \\to -\\infty}", "\\left(", "\\int_a^0 f", "+", "\\int_0^b f", "\\right) .", "\\end{split}", "\\end{equation*}", "Similar computation shows the other direction. Therefore, if", "either expression converges, then the improper integral converges", "and", "\\begin{multline*}", "\\int_{-\\infty}^\\infty f", "=", "\\lim_{a \\to -\\infty} \\, \\lim_{b \\to \\infty} \\, \\int_a^b f", "=", "\\left(", "\\lim_{a \\to -\\infty}", "\\int_a^0 f", "\\right)", "+", "\\left(", "\\lim_{b \\to \\infty}", "\\int_0^b f", "\\right)", "\\\\", "=", "\\left(", "\\lim_{a \\to \\infty}", "\\int_{-a}^0 f", "\\right)", "+", "\\left(", "\\lim_{a \\to \\infty}", "\\int_0^a f", "\\right)", "=", "\\lim_{a \\to \\infty}", "\\left(", "\\int_{-a}^0 f", "+", "\\int_0^a f", "\\right)", "=", "\\lim_{a \\to \\infty}", "\\int_{-a}^a f .", "\\end{multline*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 37, "type": "example", "label": "Lebl-contfunc:37", "categories": [ "convergence", "example", "improper" ], "title": "Limit of Improper Integrals", "contents": [ "On the other hand, you must be careful to take the limits independently before you know convergence. Let$f(x) = \\frac{x}{\\abs{x}}$for$x \\not= 0$and$f(0) = 0$. If$a < 0$and$b > 0$, then", "\\begin{equation*}\n\\int_{a}^b f\n=\n\\int_{a}^0 f\n+\n\\int_{0}^b f\n=\na+b .\n\\end{equation*}", "For every fixed$a < 0$, the limit as$b \\to \\infty$is infinite. So even the first limit does not exist, and the improper integral$\\int_{-\\infty}^\\infty f$does not converge. On the other hand, if$a > 0$, then", "\\begin{equation*}\n\\int_{-a}^{a} f\n=\n(-a)+a = 0 .\n\\end{equation*}", "Therefore,", "\\begin{equation*}\n\\lim_{a\\to\\infty}\n\\int_{-a}^{a} f\n= 0 .\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 38, "type": "example", "label": "Lebl-contfunc:38", "categories": [ "applications", "convergence", "example", "series", "calculus", "examples", "inequalities", "improper" ], "title": "Fundamental Theorem of Calculus", "contents": [ "An example to keep in mind for improper integrals is the so-called \\emph{\\myindex{sinc function}}% \\footnote{Shortened from Latin: \\emph{sinus cardinalis}}. This function comes up quite often in both pure and applied mathematics. Define", "\\begin{equation*}\n\\operatorname{sinc}(x) \\coloneqq\n\\begin{cases}\n\\frac{\\sin(x)}{x} & \\text{if } x \\not= 0 , \\\\\n1 & \\text{if } x = 0 .\n\\end{cases}\n\\end{equation*}", "\\begin{myfigureht} \\includegraphics{figures/sincfig} \\caption{The sinc function.\\label{figsinc}} \\end{myfigureht}", "It is not difficult to show that the sinc function is continuous at zero, but that is not important right now. What is important is that", "\\begin{equation*}\n\\int_{-\\infty}^\\infty \\operatorname{sinc}(x) \\,dx = \\pi ,\n\\qquad \\text{while} \\qquad\n\\int_{-\\infty}^\\infty \\abs{\\operatorname{sinc}(x)} \\,dx = \\infty .\n\\end{equation*}", "The integral of the sinc function is a continuous analogue of the alternating harmonic series$\\sum_{n=1}^\\infty \\nicefrac{{(-1)}^n}{n}$, while the absolute value is like the regular harmonic series$\\sum_{n=1}^\\infty \\nicefrac{1}{n}$. In particular, the fact that the integral converges must be done directly rather than using comparison test.", "We will not prove the first statement exactly. Let us simply prove that the integral of the sinc function converges, but we will not worry about the exact limit. Because$\\frac{\\sin(-x)}{-x} = \\frac{\\sin(x)}{x}$, it is enough to show that", "\\begin{equation*}\n\\int_{2\\pi}^\\infty \\frac{\\sin(x)}{x}\\,dx\n\\end{equation*}", "converges. We also avoid$x=0$this way to make our life simpler.", "For every$n \\in \\N$, we have that for$x \\in [\\pi 2n, \\pi (2n+1)]$,", "\\begin{equation*}\n\\frac{\\sin(x)}{\\pi (2n+1)}\n\\leq\n\\frac{\\sin(x)}{x}\n\\leq\n\\frac{\\sin(x)}{\\pi 2n} ,\n\\end{equation*}", "as$\\sin(x) \\geq 0$. For$x \\in [\\pi (2n+1), \\pi (2n+2)]$,", "\\begin{equation*}\n\\frac{\\sin(x)}{\\pi (2n+1)}\n\\leq\n\\frac{\\sin(x)}{x}\n\\leq\n\\frac{\\sin(x)}{\\pi (2n+2)} ,\n\\end{equation*}", "as$\\sin(x) \\leq 0$.", "Via the fundamental theorem of calculus,", "\\begin{equation*}\n\\frac{2}{\\pi (2n+1)}\n=\n\\int_{\\pi 2n}^{\\pi (2n+1)}\n\\frac{\\sin(x)}{\\pi (2n+1)}\n\\,dx\n\\leq\n\\int_{\\pi 2n}^{\\pi (2n+1)}\n\\frac{\\sin(x)}{x}\n\\,dx\n\\leq\n\\int_{\\pi 2n}^{\\pi (2n+1)}\n\\frac{\\sin(x)}{\\pi 2n}\n\\,dx\n=\n\\frac{1}{\\pi n} .\n\\end{equation*}", "Similarly,", "\\begin{equation*}\n\\frac{-2}{\\pi (2n+1)}\n\\leq\n\\int_{\\pi (2n+1)}^{\\pi (2n+2)}\n\\frac{\\sin(x)}{x}\n\\,dx\n\\leq\n\\frac{-1}{\\pi (n+1)} .\n\\end{equation*}", "Adding the two together we find", "\\begin{equation*}\n0\n=\n\\frac{2}{\\pi (2n+1)}\n+\n\\frac{-2}{\\pi (2n+1)}\n\\leq\n\\int_{2\\pi n}^{2\\pi (n+1)}\n\\frac{\\sin(x)}{x}\n\\,dx\n\\leq\n\\frac{1}{\\pi n} \n+\n\\frac{-1}{\\pi (n+1)} \n=\n\\frac{1}{\\pi n(n+1)} .\n\\end{equation*}", "See \\figureref{fig:sincbound}. \\begin{myfigureht} \\includegraphics{figures/sincbound} \\caption{Bound of$\\int_{2\\pi n}^{2\\pi (n+1)} \\frac{\\sin(x)}{x} \\,dx$using the shaded integral (signed area$\\frac{1}{\\pi n} + \\frac{-1}{\\pi (n+1)}$).\\label{fig:sincbound}} \\end{myfigureht}", "For$k \\in \\N$,", "\\begin{equation*}\n\\int_{2\\pi}^{2k\\pi} \\frac{\\sin(x)}{x} \\,dx\n=\n\\sum_{n=1}^{k-1}\n\\int_{2\\pi n}^{2\\pi (n+1)} \\frac{\\sin(x)}{x} \\,dx \n\\leq\n\\sum_{n=1}^{k-1}\n\\frac{1}{\\pi n(n+1)} .\n\\end{equation*}", "We find the partial sums of a series with positive terms. The series converges as$\\sum_{n=1}^\\infty \\frac{1}{\\pi n (n+1)}$is a convergent series. Thus as a sequence,", "\\begin{equation*}\n\\lim_{k\\to \\infty} \\int_{2\\pi}^{2k\\pi} \\frac{\\sin(x)}{x} \\,dx\n=L \\leq\n\\sum_{n=1}^{\\infty}\n\\frac{1}{\\pi n(n+1)} < \\infty .\n\\end{equation*}", "Let$M > 2\\pi$be arbitrary, and let$k \\in \\N$be the largest integer such that$2k\\pi \\leq M$. For$x \\in [2k\\pi,M]$, we have$\\frac{-1}{2k\\pi} \\leq \\frac{\\sin(x)}{x} \\leq \\frac{1}{2k\\pi}$, and so", "\\begin{equation*}\n\\abs{\\int_{2k\\pi}^{M} \\frac{\\sin(x)}{x} \\,dx } \\leq\n\\frac{M-2k\\pi}{2k\\pi} \\leq \\frac{1}{k} .\n\\end{equation*}", "As$k$is the largest$k$such that$2k\\pi \\leq M$, then as$M\\in \\R$goes to infinity, so does$k \\in \\N$.", "Then", "\\begin{equation*}\n\\int_{2\\pi}^M \\frac{\\sin(x)}{x}\\,dx\n=\n\\int_{2\\pi}^{2k\\pi} \\frac{\\sin(x)}{x} \\,dx\n+\n\\int_{2k\\pi}^{M} \\frac{\\sin(x)}{x} \\,dx .\n\\end{equation*}", "As$M$goes to infinity, the first term on the right-hand side goes to$L$, and the second term on the right-hand side goes to zero. Hence", "\\begin{equation*}\n\\int_{2\\pi}^\\infty \\frac{\\sin(x)}{x} \\,dx = L .\n%\\leq \\sum_{n=1}^{\\infty}\n%\\frac{1}{\\pi n(n+1)} < \\infty .\n\\end{equation*}", "The double-sided integral of sinc also exists as noted above. We leave the other statement---that the integral of the absolute value of the sinc function diverges---as an exercise." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 39, "type": "proposition", "label": "Lebl-contfunc:39", "categories": [ "characterization", "series" ], "title": "Integral Test for Series", "contents": [ "\\index{integral test for series} Suppose$f \\colon [k,\\infty) \\to \\R$is a decreasing nonnegative function where$k \\in \\Z$. Then\\begin{equation*} \\sum_{n=k}^\\infty f(n) \\quad \\text{converges} \\qquad \\text{if and only if} \\qquad \\int_k^\\infty f \\quad \\text{converges}. \\end{equation*}In this case\\begin{equation*} \\int_k^\\infty f \\leq \\sum_{n=k}^\\infty f(n) \\leq f(k)+ \\int_k^\\infty f . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let $\\ell, m \\in \\Z$ be such that $m > \\ell \\geq k$.", "Because $f$ is decreasing, we have", "$\\int_{n}^{n+1} f \\leq f(n) \\leq \\int_{n-1}^{n} f$. Therefore,", "\\begin{equation} \\label{impropriemann:eqseries}", "\\int_\\ell^m f", "=", "\\sum_{n=\\ell}^{m-1} \\int_{n}^{n+1} f", "\\leq", "\\sum_{n=\\ell}^{m-1} f(n)", "\\leq", "f(\\ell) +", "\\sum_{n=\\ell+1}^{m-1} \\int_{n-1}^{n} f", "\\leq", "f(\\ell)+", "\\int_\\ell^{m-1} f .", "\\end{equation}", "Suppose first that $\\int_k^\\infty f$ converges and", "let $\\epsilon > 0$ be given.", "As before, since $f$ is positive, then there exists", "an $L \\in \\N$ such that if $\\ell \\geq L$, then", "$\\int_\\ell^{m} f < \\nicefrac{\\epsilon}{2}$ for all $m \\geq \\ell$.", "The function", "$f$ must decrease to zero (why?), so make $L$ large enough so that", "for $\\ell \\geq L$, we have $f(\\ell) < \\nicefrac{\\epsilon}{2}$.", "Thus, for $m > \\ell \\geq L$, we have", "via \\eqref{impropriemann:eqseries},", "\\begin{equation*}", "\\sum_{n=\\ell}^{m} f(n)", "\\leq", "f(\\ell)+", "\\int_\\ell^{m} f < \\nicefrac{\\epsilon}{2} + \\nicefrac{\\epsilon}{2} = \\epsilon .", "\\end{equation*}", "The series is therefore Cauchy and thus converges. The estimate in the", "proposition is obtained by letting $m$ go to infinity in", "\\eqref{impropriemann:eqseries} with $\\ell = k$.", "Conversely, suppose $\\int_k^\\infty f$ diverges.", "As $f$ is positive, then by", "\\propref{impropriemann:possimp},", "the sequence $\\{ \\int_k^m f \\}_{m=k}^\\infty$ diverges to infinity.", "Using", "\\eqref{impropriemann:eqseries} with $\\ell = k$, we find", "\\begin{equation*}", "\\int_k^m f", "\\leq", "\\sum_{n=k}^{m-1} f(n) .", "\\end{equation*}", "As the left-hand side goes to infinity as $m \\to \\infty$, so does the", "right-hand side." ], "refs": [ "impropriemann:eqseries", "impropriemann:possimp" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 40, "type": "example", "label": "Lebl-contfunc:40", "categories": [ "calculus", "series", "example", "inequalities" ], "title": "Fundamental Theorem of Calculus", "contents": [ "The integral test can be used not only to show that a series converges, but to estimate its sum to arbitrary precision. Let us show$\\sum_{n=1}^\\infty \\frac{1}{n^2}$exists and estimate its sum to within 0.01. As this series is the$p$-series for$p=2$, we already proved it converges (let us pretend we do not know that), but we only roughly estimated its sum.", "The fundamental theorem of calculus says that for all$k \\in \\N$,", "\\begin{equation*}\n\\int_{k}^\\infty \\frac{1}{x^2}\\,dx = \\frac{1}{k} .\n\\end{equation*}", "In particular, the series must converge. But we also have", "\\begin{equation*}\n\\frac{1}{k} = \\int_k^\\infty \\frac{1}{x^2}\\,dx\n\\leq\n\\sum_{n=k}^\\infty \\frac{1}{n^2}\n\\leq\n\\frac{1}{k^2}\n+\n\\int_k^\\infty \\frac{1}{x^2}\\,dx\n=\n\\frac{1}{k^2}\n+\n\\frac{1}{k} .\n\\end{equation*}", "Adding the partial sum up to$k-1$we get", "\\begin{equation*}\n\\frac{1}{k} + \\sum_{n=1}^{k-1} \\frac{1}{n^2}\n\\leq\n\\sum_{n=1}^\\infty \\frac{1}{n^2}\n\\leq\n\\frac{1}{k^2}\n+\n\\frac{1}{k} + \\sum_{n=1}^{k-1} \\frac{1}{n^2} .\n\\end{equation*}", "In other words,$\\nicefrac{1}{k} + \\sum_{n=1}^{k-1} \\nicefrac{1}{n^2}$is an estimate for the sum to within$\\nicefrac{1}{k^2}$. Therefore, if we wish to find the sum to within 0.01, we note$\\nicefrac{1}{{10}^2} = 0.01$. We obtain", "\\begin{equation*}\n1.6397\\ldots\n\\approx\n\\frac{1}{10} + \\sum_{n=1}^{9} \\frac{1}{n^2}\n\\leq\n\\sum_{n=1}^\\infty \\frac{1}{n^2}\n\\leq\n\\frac{1}{100}\n+\n\\frac{1}{10} + \\sum_{n=1}^{9} \\frac{1}{n^2}\n\\approx\n1.6497\\ldots .\n\\end{equation*}", "The actual sum is$\\nicefrac{\\pi^2}{6} \\approx 1.6449\\ldots$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 0, "type": "example", "label": "Lebl-contfunc:0", "categories": [ "convergence", "functions", "example" ], "title": "Convergence of Pointwise Convergence", "contents": [ "On$[-1,1]$, the sequence of functions defined by$f_n(x) \\coloneqq x^{2n}$converges pointwise to$f \\colon [-1,1] \\to \\R$, where", "\\begin{equation*}\nf(x) =\n\\begin{cases}\n1 & \\text{if } x=-1 \\text{ or } x=1, \\\\\n0 & \\text{otherwise.}\n\\end{cases}\n\\avoidbreak\n\\end{equation*}", "See \\figureref{x2nfig}.", "\\begin{myfigureht} \\includegraphics{figures/x2nfig} \\caption{Graphs of$f_1$,$f_2$,$f_3$, and$f_8$for$f_n(x) \\coloneqq x^{2n}$.\\label{x2nfig}} \\end{myfigureht}", "To see this is so, first take$x \\in (-1,1)$. Then$0 \\leq x^2 < 1$. We have seen before that", "\\begin{equation*}\n\\abs{x^{2n} - 0} = {(x^2)}^n \\to 0 \\quad \\text{as} \\quad n \\to \\infty .\n\\end{equation*}", "Therefore,$\\lim_{n\\to\\infty}f_n(x) = 0$.", "When$x = 1$or$x=-1$, then$x^{2n} = 1$for all$n$and hence$\\lim_{n\\to\\infty}f_n(x) = 1$. For all other$x$, the sequence$\\bigl\\{ f_n(x) \\bigr\\}_{n=1}^\\infty$does not converge." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 1, "type": "example", "label": "Lebl-contfunc:example:geomsumptconv", "categories": [ "example" ], "title": "Limit of Pointwise Convergence", "contents": [ "We write", "\\begin{equation*}\n\\sum_{k=0}^\\infty x^k\n\\end{equation*}", "to denote the limit of the functions", "\\begin{equation*}\nf_n(x) \\coloneqq \\sum_{k=0}^n x^k .\n\\end{equation*}", "When studying series, we saw that for$(-1,1)$the$f_n$converge pointwise to", "\\begin{equation*}\n\\frac{1}{1-x} .\n\\end{equation*}", "The subtle point here is that while$\\frac{1}{1-x}$is defined for all$x \\not=1$, and$f_n$are defined for all$x$(even at$x=1$), convergence only happens on$(-1,1)$. Therefore, when we write", "\\begin{equation*}\nf(x) \\coloneqq \\sum_{k=0}^\\infty x^k\n\\end{equation*}", "we mean that$f$is defined on$(-1,1)$and is the pointwise limit of the partial sums." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 2, "type": "example", "label": "Lebl-contfunc:2", "categories": [ "example" ], "title": "Existence of Pointwise Convergence", "contents": [ "Let$f_n(x) \\coloneqq \\sin(nx)$. Then$f_n$does not converge pointwise to any function on any interval. It may converge at certain points, such as when$x=0$or$x=\\pi$. It is left as an exercise that in any interval$[a,b]$, there exists an$x$such that$\\sin(xn)$does not have a limit as$n$goes to infinity. See \\figureref{fig:nonconvsinxn}. \\begin{myfigureht} \\includegraphics{figures/nonconvsinxn} \\caption{Graphs of$\\sin(nx)$for$n=1,2,\\ldots,10$, with higher$n$in lighter gray.% \\label{fig:nonconvsinxn}} \\end{myfigureht}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 3, "type": "proposition", "label": "Lebl-contfunc:ptwsconv:prop", "categories": [ "convergence", "characterization" ], "title": "Characterization of Pointwise Convergence via Equivalence", "contents": [ "Let$f_n \\colon S \\to \\R$and$f \\colon S \\to \\R$be functions. Then$\\{ f_n \\}_{n=1}^\\infty$converges pointwise to$f$if and only if for every$x \\in S$and every$\\epsilon > 0$, there exists an$N \\in \\N$such that\\begin{equation*} \\abs{f_n(x)-f(x)} < \\epsilon \\quad \\text{for all } n \\geq N . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 4, "type": "proposition", "label": "Lebl-contfunc:4", "categories": [ "convergence", "functions" ], "title": "Convergence of Pointwise Convergence", "contents": [ "Let$\\{ f_n \\}_{n=1}^\\infty$be a sequence of functions$f_n \\colon S \\to \\R$. If$\\{ f_n \\}_{n=1}^\\infty$converges uniformly to$f \\colon S \\to \\R$, then$\\{ f_n \\}_{n=1}^\\infty$converges pointwise to$f$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 5, "type": "example", "label": "Lebl-contfunc:5", "categories": [ "convergence", "functions", "example", "continuity" ], "title": "Continuity of Pointwise Convergence", "contents": [ "The functions$f_n(x) \\coloneqq x^{2n}$do not converge uniformly on$[-1,1]$, even though they converge pointwise. To see this, suppose for contradiction that the convergence is uniform. For$\\epsilon \\coloneqq \\nicefrac{1}{2}$, there would have to exist an$N$such that$x^{2N} = \\abs{x^{2N} - 0} < \\nicefrac{1}{2}$for all$x \\in (-1,1)$(as$f_n(x)$converges to 0 on$(-1,1)$). But that means that for every sequence$\\{ x_k \\}_{k=1}^\\infty$in$(-1,1)$such that$\\lim_{k\\to\\infty} x_k = 1$, we have$x_k^{2N} < \\nicefrac{1}{2}$for all$k$. On the other hand,$x^{2N}$is a continuous function of$x$(it is a polynomial). Therefore, we obtain a contradiction", "\\begin{equation*}\n1 = 1^{2N} = \\lim_{k\\to\\infty} x_k^{2N} \\leq \\nicefrac{1}{2} .\n\\end{equation*}", "However, if we restrict our domain to$[-a,a]$where$0 < a < 1$, then$\\{ f_n \\}_{n=1}^\\infty$converges uniformly to 0 on$[-a,a]$. Note that$a^{2n} \\to 0$as$n \\to \\infty$. Given$\\epsilon > 0$, pick$N \\in \\N$such that$a^{2n} < \\epsilon$for all$n \\geq N$. If$x \\in [-a,a]$, then$\\abs{x} \\leq a$. So for all$n \\geq N$and all$x \\in [-a,a]$,", "\\begin{equation*}\n\\abs{x^{2n}} = \\abs{x}^{2n} \\leq a^{2n} < \\epsilon .\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 6, "type": "proposition", "label": "Lebl-contfunc:6", "categories": [ "convergence", "characterization" ], "title": "Characterization of Uniform Convergence via Equivalence", "contents": [ "A sequence of bounded functions$f_n \\colon S \\to \\R$converges uniformly to$f \\colon S \\to \\R$if and only if\\begin{equation*} \\lim_{n\\to\\infty} \\snorm{f_n - f}_S = 0 . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "First suppose", "$\\lim_{n\\to\\infty} \\norm{f_n - f}_S = 0$. Let $\\epsilon > 0$ be", "given. There exists an $N$ such that", "for $n \\geq N$, we have $\\snorm{f_n - f}_S < \\epsilon$. As $\\norm{f_n-f}_S$", "is the supremum of $\\abs{f_n(x)-f(x)}$, we see that for all $x \\in S$,", "we have $\\abs{f_n(x)-f(x)} \\leq \\snorm{f_n - f}_S < \\epsilon$.", "On the other hand, suppose $\\{ f_n \\}_{n=1}^\\infty$ converges uniformly to $f$.", "Let $\\epsilon > 0$ be given. Then find $N$ such that", "for all $n \\geq N$, we have", "$\\abs{f_n(x)-f(x)} < \\epsilon$ for all $x \\in S$.", "Taking the supremum over $x \\in S$, we see that", "$\\snorm{f_n - f}_S \\leq \\epsilon$. Hence $\\lim_{n\\to\\infty} \\snorm{f_n-f}_S = 0$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 7, "type": "example", "label": "Lebl-contfunc:7", "categories": [ "convergence", "example" ], "title": "Convergence of Uniform Convergence", "contents": [ "Let$f_n \\colon [0,1] \\to \\R$be defined by$f_n(x) \\coloneqq \\frac{nx+ \\sin(nx^2)}{n}$. We claim$\\{ f_n \\}_{n=1}^\\infty$converges uniformly to$f(x) \\coloneqq x$. Let us compute:", "\\begin{equation*}\n\\begin{split}\n\\norm{f_n-f}_{[0,1]}\n& =\n\\sup \\left\\{ \\abs{\\frac{nx+ \\sin(nx^2)}{n} - x} : x \\in [0,1] \\right\\}\n\\\\\n& =\n\\sup \\left\\{ \\frac{\\abs{\\sin(nx^2)}}{n} : x \\in [0,1] \\right\\}\n\\\\\n& \\leq\n\\sup \\bigl\\{ \\nicefrac{1}{n} : x \\in [0,1] \\bigr\\}\n\\\\\n& = \\nicefrac{1}{n}.\n\\end{split}\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 8, "type": "proposition", "label": "Lebl-contfunc:prop:uniformcauchy", "categories": [ "norms", "convergence", "characterization" ], "title": "Characterization of Uniform Convergence via Equivalence", "contents": [ "Let$f_n \\colon S \\to \\R$be bounded functions. Then$\\{ f_n \\}_{n=1}^\\infty$is Cauchy in the uniform norm if and only if there exists an$f \\colon S \\to \\R$and$\\{ f_n \\}_{n=1}^\\infty$converges uniformly to$f$." ], "refs": [], "proofs": [ { "contents": [ "First suppose $\\{ f_n \\}_{n=1}^\\infty$ is Cauchy in the uniform norm.", "Let us define $f$. Fix $x$.", "The sequence $\\bigl\\{ f_n(x) \\bigr\\}_{n=1}^\\infty$ is Cauchy because", "\\begin{equation*}", "\\abs{f_m(x)-f_k(x)}", "\\leq", "\\norm{f_m-f_k}_S .", "\\end{equation*}", "Thus $\\bigl\\{ f_n(x) \\bigr\\}_{n=1}^\\infty$ converges to some real number. Define $f \\colon S", "\\to \\R$ by", "\\begin{equation*}", "f(x) \\coloneqq \\lim_{n \\to \\infty} f_n(x) .", "\\end{equation*}", "The sequence", "$\\{ f_n \\}_{n=1}^\\infty$ converges pointwise to $f$. To show that the convergence", "is uniform, let $\\epsilon > 0$ be given. Find an $N$ such that", "for all $m, k \\geq N$, we have", "$\\norm{f_m-f_k}_S < \\nicefrac{\\epsilon}{2}$. In other words, for", "all $x$, we have", "$\\abs{f_m(x)-f_k(x)} < \\nicefrac{\\epsilon}{2}$. For any fixed $x$, take the limit", "as $k$ goes to infinity. Then $\\abs{f_m(x)-f_k(x)}$", "goes to $\\abs{f_m(x)-f(x)}$.", "Consequently for all $x$,", "\\begin{equation*}", "\\abs{f_m(x)-f(x)} \\leq \\nicefrac{\\epsilon}{2} < \\epsilon .", "\\end{equation*}", "Hence, $\\{ f_n \\}_{n=1}^\\infty$ converges uniformly.", "Next, we prove the other direction.", "Suppose $\\{ f_n \\}_{n=1}^\\infty$ converges uniformly to", "$f$. Given $\\epsilon > 0$, find $N$ such that for all $n \\geq N$,", "we have $\\abs{f_n(x)-f(x)} < \\nicefrac{\\epsilon}{4}$ for all $x \\in S$.", "Therefore, for all $m, k \\geq N$ and all $x$,", "\\begin{multline*}", "\\abs{f_m(x)-f_k(x)} =", "\\abs{f_m(x)-f(x)+f(x)-f_k(x)}", "\\\\", "\\leq", "\\abs{f_m(x)-f(x)}+\\abs{f(x)-f_k(x)} < \\nicefrac{\\epsilon}{4} +", "\\nicefrac{\\epsilon}{4} = \\nicefrac{\\epsilon}{2} .", "\\end{multline*}", "Take the supremum over all $x$ to obtain", "\\begin{equation*}", "\\norm{f_m-f_k}_S \\leq \\nicefrac{\\epsilon}{2} < \\epsilon . \\qedhere", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 9, "type": "example", "label": "Lebl-contfunc:9", "categories": [ "continuity", "example" ], "title": "Continuity of Pointwise Convergence", "contents": [ "Define$f_n \\colon [0,1] \\to \\R$as", "\\begin{equation*}\nf_n(x) \\coloneqq\n\\begin{cases}\n1-nx & \\text{if } x < \\nicefrac{1}{n},\\\\\n0 & \\text{if } x \\geq \\nicefrac{1}{n}.\n\\end{cases}\n\\end{equation*}", "See \\figureref{contconvcntr:fig}.", "\\begin{myfigureht} \\includegraphics{figures/contconvcntr} \\caption{Graph of$f_n(x)$.% \\label{contconvcntr:fig}} \\end{myfigureht}", "Each function$f_n$is continuous. Fix an$x \\in (0,1]$. If$n \\geq \\nicefrac{1}{x}$, then$x \\geq \\nicefrac{1}{n}$. Therefore for$n \\geq \\nicefrac{1}{x}$, we have$f_n(x) = 0$, and so", "\\begin{equation*}\n\\lim_{n \\to \\infty} f_n(x) = 0.\n\\end{equation*}", "On the other hand, if$x=0$, then", "\\begin{equation*}\n\\lim_{n \\to \\infty} f_n(0) = \n\\lim_{n \\to \\infty} 1 = 1.\n\\end{equation*}", "Thus the pointwise limit of$f_n$is the function$f \\colon [0,1] \\to \\R$defined by", "\\begin{equation*}\nf(x) \\coloneqq\n\\begin{cases}\n1 & \\text{if } x = 0,\\\\\n0 & \\text{if } x > 0.\n\\end{cases}\n\\end{equation*}", "The function$f$is not continuous at 0." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 10, "type": "theorem", "label": "Lebl-contfunc:10", "categories": [ "continuity", "functions" ], "title": "Continuity of Uniform Convergence", "contents": [ "Suppose$S \\subset \\R$. Let$\\{ f_n \\}_{n=1}^\\infty$be a sequence of continuous functions$f_n \\colon S \\to \\R$converging uniformly to$f \\colon S \\to \\R$. Then$f$is continuous." ], "refs": [], "proofs": [ { "contents": [ "Let $x \\in S$ be fixed. Let $\\{ x_n \\}_{n=1}^\\infty$ be a sequence in $S$", "converging to $x$.", "Let $\\epsilon > 0$ be given.", "As $\\{ f_k \\}_{k=1}^\\infty$ converges uniformly to $f$, we find a $k \\in \\N$ such that", "\\begin{equation*}", "\\abs{f_k(y)-f(y)} < \\nicefrac{\\epsilon}{3}", "\\end{equation*}", "for all $y \\in S$. As $f_k$ is continuous at $x$,", "we find an $N \\in \\N$ such that for all $m \\geq N$,", "\\begin{equation*}", "\\abs{f_k(x_m)-f_k(x)} < \\nicefrac{\\epsilon}{3} .", "\\end{equation*}", "Thus for all", "$m \\geq N$,", "\\begin{equation*}", "\\begin{split}", "\\abs{f(x_m)-f(x)}", "& =", "\\abs{f(x_m)-f_k(x_m)+f_k(x_m)-f_k(x)+f_k(x)-f(x)}", "\\\\", "& \\leq", "\\abs{f(x_m)-f_k(x_m)}+", "\\abs{f_k(x_m)-f_k(x)}+", "\\abs{f_k(x)-f(x)}", "\\\\", "& <", "\\nicefrac{\\epsilon}{3} +", "\\nicefrac{\\epsilon}{3} +", "\\nicefrac{\\epsilon}{3} = \\epsilon .", "\\end{split}", "\\end{equation*}", "Therefore, $\\bigl\\{ f(x_m) \\bigr\\}_{m=1}^\\infty$ converges to $f(x)$,", "and consequently $f$ is continuous at $x$.", "As $x$ was arbitrary, $f$ is continuous everywhere." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 11, "type": "example", "label": "Lebl-contfunc:11", "categories": [ "continuity", "example" ], "title": "Fundamental Theorem of Calculus", "contents": [ "Define$f_n \\colon [0,1] \\to \\R$as", "\\begin{equation*}\nf_n(x) \\coloneqq\n\\begin{cases}\n0 & \\text{if } x = 0,\\\\\nn-n^2x & \\text{if } 0 < x < \\nicefrac{1}{n},\\\\\n0 & \\text{if } x \\geq \\nicefrac{1}{n}.\n\\end{cases}\n\\avoidbreak\n\\end{equation*}", "See \\figureref{intconvcntr:fig}.", "\\begin{myfigureht} \\includegraphics{figures/intconvcntr} \\caption{Graph of$f_n(x)$.% \\label{intconvcntr:fig}} \\end{myfigureht}", "Each$f_n$is Riemann integrable (it is continuous on$(0,1]$and bounded), and the fundamental theorem of calculus says that", "\\begin{equation*}\n\\int_0^1 f_n =\n\\int_0^{\\nicefrac{1}{n}} (n-n^2x)\\,dx = \\nicefrac{1}{2} .\n\\end{equation*}", "Let us compute the pointwise limit of$\\{ f_n \\}_{n=1}^\\infty$. Fix an$x \\in (0,1]$. For$n \\geq \\nicefrac{1}{x}$, we have$x \\geq \\nicefrac{1}{n}$and so$f_n(x) = 0$. Hence,", "\\begin{equation*}\n\\lim_{n \\to \\infty} f_n(x) = 0.\n\\end{equation*}", "We also have$f_n(0) = 0$for all$n$. So the pointwise limit of$\\{ f_n \\}_{n=1}^\\infty$is the zero function. In summary,", "\\begin{equation*}\n\\nicefrac{1}{2} =\n\\lim_{n\\to\\infty}\n\\int_0^1 f_n (x)\\,dx\n\\not=\n\\int_0^1\n\\left(\n\\lim_{n\\to\\infty}\nf_n(x)\\right)\\,dx\n=\n\\int_0^1 0\\,dx = 0 .\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 12, "type": "theorem", "label": "Lebl-contfunc:integralinterchange:thm", "categories": [], "title": "Let$\\{ f_n \\}_{n=1}^\\infty$be a sequence of Riemann integrable functions$f_n \\colon [a,b] \\to \\R$...", "contents": [ "Let$\\{ f_n \\}_{n=1}^\\infty$be a sequence of Riemann integrable functions$f_n \\colon [a,b] \\to \\R$converging uniformly to$f \\colon [a,b] \\to \\R$. Then$f$is Riemann integrable, and\\begin{equation*} \\int_a^b f = \\lim_{n\\to\\infty} \\int_a^b f_n . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let $\\epsilon > 0$ be given.", "As $f_n$ goes to $f$ uniformly, we find an $M \\in \\N$ such that", "for all $n \\geq M$, we have", "$\\abs{f_n(x)-f(x)} < \\frac{\\epsilon}{2(b-a)}$ for all $x \\in [a,b]$.", "In particular, by reverse triangle inequality,", "$\\abs{f(x)} < \\frac{\\epsilon}{2(b-a)} + \\abs{f_n(x)}$ for all $x$.", "Hence $f$ is bounded,", "as $f_n$ is bounded.", "Note that $f_n$ is integrable and compute", "\\begin{equation*}", "\\begin{split}", "\\overline{\\int_a^b} f", "-", "\\underline{\\int_a^b} f", "& =", "\\overline{\\int_a^b} \\bigl( f(x) - f_n(x) + f_n(x) \\bigr)\\,dx", "-", "\\underline{\\int_a^b} \\bigl( f(x) - f_n(x) + f_n(x) \\bigr)\\,dx", "\\\\", "& \\leq", "\\overline{\\int_a^b} \\bigl( f(x) - f_n(x) \\bigr)\\,dx + \\overline{\\int_a^b} f_n(x) \\,dx", "-", "\\underline{\\int_a^b} \\bigl( f(x) - f_n(x) \\bigr)\\,dx - \\underline{\\int_a^b} f_n(x) \\,dx", "\\\\", "& =", "\\overline{\\int_a^b} \\bigl( f(x) - f_n(x) \\bigr)\\,dx + \\int_a^b f_n(x) \\,dx", "-", "\\underline{\\int_a^b} \\bigl( f(x) - f_n(x) \\bigr)\\,dx - \\int_a^b f_n(x) \\,dx", "\\\\", "& =", "\\overline{\\int_a^b} \\bigl( f(x) - f_n(x) \\bigr)\\,dx", "-", "\\underline{\\int_a^b} \\bigl( f(x) - f_n(x) \\bigr)\\,dx", "\\\\", "& \\leq", "\\frac{\\epsilon}{2(b-a)} (b-a) +", "\\frac{\\epsilon}{2(b-a)} (b-a) = \\epsilon .", "\\end{split}", "\\end{equation*}", "The first inequality is \\propref{prop:upperlowerlinineq}.", "The second inequality follows by \\propref{intulbound:prop} and", "the fact that for all $x \\in [a,b]$, we have", "$\\frac{-\\epsilon}{2(b-a)} < f(x)-f_n(x) < \\frac{\\epsilon}{2(b-a)}$.", "As $\\epsilon > 0$ was arbitrary, $f$ is Riemann integrable.", "Finally, we compute $\\int_a^b f$. We apply \\propref{intbound:prop}", "in the calculation. Again, for all $n \\geq M$ (where $M$ is the same as above),", "\\begin{equation*}", "\\begin{split}", "\\abs{\\int_a^b f - \\int_a^b f_n} & =", "\\abs{ \\int_a^b \\bigl(f(x) - f_n(x)\\bigr)\\,dx}", "\\\\", "& \\leq", "\\frac{\\epsilon}{2(b-a)} (b-a) = \\frac{\\epsilon}{2} < \\epsilon .", "\\end{split}", "\\end{equation*}", "Therefore, $\\bigl\\{ \\int_a^b f_n \\bigr\\}_{n=1}^\\infty$ converges to $\\int_a^b f$." ], "refs": [ "intbound:prop", "intulbound:prop", "prop:upperlowerlinineq" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 13, "type": "example", "label": "Lebl-contfunc:13", "categories": [ "integrals", "convergence", "example", "derivatives" ], "title": "Existence of Uniform Convergence", "contents": [ "Suppose we wish to compute", "\\begin{equation*}\n\\lim_{n\\to\\infty} \\int_0^1 \\frac{nx+ \\sin(nx^2)}{n} \\,dx .\n\\end{equation*}", "It is impossible to compute the integrals for any particular$n$using calculus as$\\sin(nx^2)$has no closed-form antiderivative. However, we can compute the limit. We have shown before that$\\frac{nx+ \\sin(nx^2)}{n}$converges uniformly on$[0,1]$to$x$. By \\thmref{integralinterchange:thm}, the limit exists and", "\\begin{equation*}\n\\lim_{n\\to\\infty} \\int_0^1 \\frac{nx+ \\sin(nx^2)}{n} \\,dx\n=\n\\int_0^1\nx \\,dx = \\nicefrac{1}{2} .\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 14, "type": "example", "label": "Lebl-contfunc:14", "categories": [ "convergence", "example" ], "title": "Convergence of Pointwise Convergence", "contents": [ "If convergence is only pointwise, the limit need not even be Riemann integrable. On$[0,1]$define", "\\begin{equation*}\nf_n(x) \\coloneqq\n\\begin{cases}\n1 & \\text{if } x = \\nicefrac{p}{q} \\text{ in lowest terms and } q \\leq n, \\\\\n0 & \\text{otherwise.}\n\\end{cases}\n\\end{equation*}", "Each function$f_n$differs from the zero function at finitely many points; there are only finitely many fractions in$[0,1]$with denominator less than or equal to$n$. So$f_n$is integrable and$\\int_0^1 f_n = \\int_0^1 0 = 0$. It is an easy exercise to show that$\\{ f_n \\}_{n=1}^\\infty$converges pointwise to the \\myindex{Dirichlet function}", "\\begin{equation*}\nf(x) \\coloneqq\n\\begin{cases}\n1 & \\text{if } x \\in \\Q, \\\\\n0 & \\text{otherwise,}\n\\end{cases}\n\\end{equation*}", "which is not Riemann integrable." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 15, "type": "example", "label": "Lebl-contfunc:15", "categories": [ "convergence", "example" ], "title": "Boundedness of Pointwise Convergence", "contents": [ "In fact, if the convergence is only pointwise, the limit of bounded functions is not even necessarily bounded. Define$f_n \\colon [0,1] \\to \\R$by", "\\begin{equation*}\nf_n(x) \\coloneqq\n\\begin{cases}\n0 & \\text{if } x < \\nicefrac{1}{n},\\\\\n\\nicefrac{1}{x} & \\text{else.}\n\\end{cases}\n\\end{equation*}", "For every$n$we get that$\\abs{f_n(x)} \\leq n$for all$x \\in [0,1]$so the functions are bounded. However,$\\{ f_n \\}_{n=1}^\\infty$converges pointwise to the unbounded function", "\\begin{equation*}\nf(x) \\coloneqq\n\\begin{cases}\n0 & \\text{if } x = 0,\\\\\n\\nicefrac{1}{x} & \\text{else.}\n\\end{cases}\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 16, "type": "example", "label": "Lebl-contfunc:16", "categories": [ "convergence", "example", "derivatives" ], "title": "Convergence of Pointwise Convergence", "contents": [ "Let$f_n(x) \\coloneqq \\frac{\\sin(nx)}{n}$. Then$f_n$converges uniformly to 0. See \\figureref{fig:conv1nsinxn}. The derivative of the limit is 0. But$f_n'(x) = \\cos(nx)$, which does not converge even pointwise, for example$f_n'(\\pi) = {(-1)}^n$. Moreover,$f_n'(0) = 1$for all$n$, which does converge, but not to$0$. \\begin{myfigureht} \\includegraphics{figures/conv1nsinxn} \\caption{Graphs of$\\frac{\\sin(nx)}{n}$for$n=1,2,\\ldots,10$, with higher$n$in lighter gray.% \\label{fig:conv1nsinxn}} \\end{myfigureht}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 17, "type": "example", "label": "Lebl-contfunc:exercise:badconvergenceder", "categories": [ "convergence", "continuity", "example", "derivatives" ], "title": "Continuity of Pointwise Convergence", "contents": [ "Let$f_n(x) \\coloneqq \\frac{1}{1+nx^2}$. If$x \\not= 0$, then$\\lim_{n \\to \\infty} f_n(x) = 0$, but$\\lim_{n \\to \\infty} f_n(0) = 1$. Hence,$\\{ f_n \\}_{n=1}^\\infty$converges pointwise to a function that is not continuous at$0$. We compute", "\\begin{equation*}\nf_n'(x) %= \\frac{d}{dx} \\frac{1}{1+n x^2}\n= \\frac{-2 n x}{(1+ n x^2)^2} .\n\\end{equation*}", "For every$x$,$\\lim_{n\\to\\infty} f_n'(x) = 0$, so the derivatives converge pointwise to 0, but the reader can check that the convergence is not uniform on any interval containing$0$. The limit of$f_n$is not differentiable at$0$---it is not even continuous at$0$. See \\figureref{fig:convder1over1pnxsq}. \\begin{myfigureht} \\includegraphics{figures/convder1over1pnxsq} \\caption{Graphs of$\\frac{1}{1+nx^2}$and its derivative for$n=1,2,\\ldots,10$, with higher$n$in lighter gray.% \\label{fig:convder1over1pnxsq}} \\end{myfigureht}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 18, "type": "theorem", "label": "Lebl-contfunc:thm:dersconverge", "categories": [ "convergence", "functions", "continuity" ], "title": "Continuity of Uniform Convergence", "contents": [ "Let$I$be a bounded interval and let$f_n \\colon I \\to \\R$be continuously differentiable functions. Suppose$\\{ f_n' \\}_{n=1}^\\infty$converges uniformly to$g \\colon I \\to \\R$, and suppose$\\bigl\\{ f_n(c) \\bigr\\}_{n=1}^\\infty$is a convergent sequence for some$c \\in I$. Then$\\{ f_n \\}_{n=1}^\\infty$converges uniformly to a continuously differentiable function$f \\colon I \\to \\R$, and$f' = g$." ], "refs": [], "proofs": [ { "contents": [ "Define $f(c) \\coloneqq \\lim_{n\\to \\infty} f_n(c)$.", "As $f_n'$ are continuous and hence Riemann integrable,", "then", "via the fundamental theorem of calculus, we find that for $x \\in I$,", "\\begin{equation*}", "f_n(x) = f_n(c) + \\int_c^x f_n' .", "\\end{equation*}", "As $\\{ f_n' \\}_{n=1}^\\infty$ converges uniformly on $I$, it converges uniformly", "on $[c,x]$ (or $[x,c]$ if $x < c$).", "Thus, the limit as $n \\to \\infty$ on the right-hand side exists.", "Define $f$ at the remaining points (where $x\\neq c$) by this limit:", "\\begin{equation*}", "f(x) \\coloneqq", "\\lim_{n\\to\\infty} f_n(c) + \\lim_{n\\to\\infty} \\int_c^x f_n'", "=", "f(c) + \\int_c^x g .", "\\end{equation*}", "The function $g$ is continuous, being the uniform limit of continuous", "functions. Hence $f$ is differentiable and $f'(x) = g(x)$ for all $x \\in I$", "by the second form of the fundamental theorem.", "It remains to prove", "uniform convergence.", "Suppose $I$ has a lower bound $a$ and upper bound $b$.", "Let $\\epsilon > 0$ be given. Take $M$", "such that for all $n \\geq M$, we have", "$\\abs{f(c)-f_n(c)} < \\nicefrac{\\epsilon}{2}$", "and", "$\\abs{g(x)-f_n'(x)} < \\frac{\\epsilon}{2(b-a)}$", "for all $x \\in I$. Then", "\\begin{equation*}", "\\begin{split}", "\\abs{f(x) - f_n(x)} & =", "\\abs{\\left(f(c) + \\int_c^x g\\right) - \\left( f_n(c) + \\int_c^x f_n' \\right)}", "\\\\", "& \\leq", "\\abs{f(c) - f_n(c)} + \\abs{\\int_c^x g - \\int_c^x f_n'}", "\\\\", "& =", "\\abs{f(c) - f_n(c)} + \\abs{\\int_c^x \\bigl(g(s) - f_n'(s)\\bigr) \\, ds}", "\\\\", "& <", "\\frac{\\epsilon}{2}", "+", "\\frac{\\epsilon}{2(b-a)}", "(b-a)", "=\\epsilon. \\qedhere", "\\end{split}", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 19, "type": "proposition", "label": "Lebl-contfunc:19", "categories": [ "convergence", "functions", "series", "continuity" ], "title": "Continuity of Pointwise Convergence", "contents": [ "Let$\\sum_{n=0}^\\infty c_n {(x-a)}^n$be a convergent power series with a radius of convergence$\\rho$, where$0 < \\rho \\leq \\infty$. Then the series converges uniformly in$[a-r,a+r]$whenever$0 < r < \\rho$.", "In particular, the series converges (pointwise) to a continuous function on$(a-\\rho,a+\\rho)$if$\\rho < \\infty$, or on$\\R$if$\\rho = \\infty$." ], "refs": [], "proofs": [ { "contents": [ "Let $I \\coloneqq (a-\\rho,a+\\rho)$ if $\\rho < \\infty$,", "or let $I \\coloneqq \\R$ if $\\rho= \\infty$.", "Take $0 < r < \\rho$.", "The series converges absolutely for every $x \\in I$,", "in particular if $x = a+r$.", "So $\\sum_{n=0}^\\infty \\abs{c_n} r^n$ converges.", "Given $\\epsilon >0$, find $M$ such that for all $k \\geq M$,", "\\begin{equation*}", "\\sum_{n=k+1}^\\infty \\abs{c_n} {r}^n < \\epsilon .", "\\end{equation*}", "For all $x \\in [a-r,a+r]$ and all $m > k$,", "\\begin{multline*}", "\\abs{\\sum_{n=0}^m c_n {(x-a)}^n -", "\\sum_{n=0}^k c_n {(x-a)}^n}", "=", "\\abs{\\sum_{n=k+1}^m c_n {(x-a)}^n}", "\\\\", "\\leq", "\\sum_{n=k+1}^m \\abs{c_n} {\\abs{x-a}}^n", "\\leq", "\\sum_{n=k+1}^m \\abs{c_n} {r}^n", "\\leq", "\\sum_{n=k+1}^\\infty \\abs{c_n} {r}^n", "<\\epsilon.", "\\end{multline*}", "The partial sums are therefore uniformly Cauchy on $[a-r,a+r]$ and", "hence converge uniformly on that set.", "Moreover, the partial sums are polynomials, which are", "continuous, and so their uniform limit on $[a-r,a+r]$", "is a continuous function.", "As $r < \\rho$ was arbitrary, the limit function", "is continuous on all of $I$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 20, "type": "corollary", "label": "Lebl-contfunc:20", "categories": [ "consequence", "series" ], "title": "Limit of Power Series", "contents": [ "Let$\\sum_{n=0}^\\infty c_n {(x-a)}^n$be a convergent power series with a radius of convergence$0 < \\rho \\leq \\infty$. Let$I \\coloneqq (a-\\rho,a+\\rho)$if$\\rho < \\infty$or$I \\coloneqq \\R$if$\\rho= \\infty$. Let$f \\colon I \\to \\R$be the limit. Then\\begin{equation*} \\int_a^x f = \\sum_{n=1}^\\infty \\frac{c_{n-1}}{n} {(x-a)}^{n} , \\end{equation*}where the radius of convergence of this series is at least$\\rho$." ], "refs": [], "proofs": [ { "contents": [ "Take $0 < r < \\rho$.", "The partial sums $\\sum_{n=0}^k c_n {(x-a)}^n$ converge uniformly on $[a-r,a+r]$.", "For every fixed $x \\in [a-r,a+r]$, the convergence is also uniform", "on $[a,x]$ (or $[x,a]$ if $x < a$).", "Hence,", "\\begin{equation*}", "\\int_a^x f =", "\\int_a^x \\lim_{k\\to\\infty} \\sum_{n=0}^k c_n {(s-a)}^n \\, ds", "=", "\\lim_{k\\to\\infty}", "\\int_a^x \\sum_{n=0}^k c_n {(s-a)}^n \\, ds", "=", "\\lim_{k\\to\\infty}", "\\sum_{n=1}^{k+1} \\frac{c_{n-1}}{n} {(x-a)}^{n} . \\qedhere", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 21, "type": "corollary", "label": "Lebl-contfunc:cor:differentiatepowerser", "categories": [ "consequence", "functions", "series" ], "title": "Limit of Power Series", "contents": [ "Let$\\sum_{n=0}^\\infty c_n {(x-a)}^n$be a convergent power series with a radius of convergence$0 < \\rho \\leq \\infty$. Let$I \\coloneqq (a-\\rho,a+\\rho)$if$\\rho < \\infty$or$I \\coloneqq \\R$if$\\rho= \\infty$. Let$f \\colon I \\to \\R$be the limit. Then$f$is a differentiable function, and\\begin{equation*} f'(x) = \\sum_{n=0}^\\infty (n+1) c_{n+1} {(x-a)}^{n} , \\end{equation*}where the radius of convergence of this series is$\\rho$." ], "refs": [], "proofs": [ { "contents": [ "Take $0 < r < \\rho$.", "The series converges uniformly on", "$[a-r,a+r]$,", "but we need uniform convergence of the derivative.", "Let", "\\begin{equation*}", "R \\coloneqq \\limsup_{n \\to \\infty} \\abs{c_n}^{1/n} .", "\\end{equation*}", "As the series is convergent $R < \\infty$, and", "the radius of convergence is $\\nicefrac{1}{R}$ (or $\\infty$ if $R=0$).", "Let $\\epsilon > 0$ be given. In \\exampleref{example:nto1overn},", "we saw $\\lim_{n\\to\\infty} n^{1/n} = 1$.", "Hence there exists an $N$ such that for all $n \\geq N$, we have", "$n^{1/n} < 1+\\epsilon$.", "So", "\\begin{equation*}", "R =", "\\limsup_{n \\to \\infty}", "\\abs{c_n}^{1/n}", "\\leq", "\\limsup_{n \\to \\infty}", "\\abs{n c_n}^{1/n}", "\\leq", "(1+\\epsilon)", "\\limsup_{n \\to \\infty}", "\\abs{c_n}^{1/n}", "=", "(1+\\epsilon)R .", "\\end{equation*}", "As $\\epsilon$ was arbitrary, $\\limsup_{n \\to \\infty} \\abs{n c_n}^{1/n} = R$.", "Therefore, $\\sum_{n=1}^\\infty n c_{n} {(x-a)}^{n}$ has radius of", "convergence $\\rho$. By dividing by $(x-a)$, we find", "$\\sum_{n=0}^\\infty (n+1) c_{n+1} {(x-a)}^{n}$ has radius of convergence", "$\\rho$ as well.", "Consequently, the partial sums", "$\\sum_{n=0}^k (n+1) c_{n+1} {(x-a)}^{n}$,", "which are derivatives of the partial sums", "$\\sum_{n=0}^{k+1} c_{n} {(x-a)}^{n}$,", "converge uniformly on $[a-r,a+r]$. Furthermore,", "the series clearly converges at $x=a$.", "We may thus apply \\thmref{thm:dersconverge}, and", "we are done as $r < \\rho$ was arbitrary." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 22, "type": "example", "label": "Lebl-contfunc:example:exponentialbypowerseries", "categories": [ "series", "example" ], "title": "We could have used this result to define the exponential function", "contents": [ "We could have used this result to define the exponential function. That is, the power series", "\\begin{equation*}\nf(x) \\coloneqq \\sum_{n=0}^\\infty \\frac{x^n}{n!}\n\\end{equation*}", "has radius of convergence$\\rho=\\infty$. Furthermore,$f(0) = 1$, and by differentiating term by term, we find that$f'(x) = f(x)$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 23, "type": "example", "label": "Lebl-contfunc:23", "categories": [ "convergence", "example", "derivatives" ], "title": "Convergence of Limits and Operations", "contents": [ "The series", "\\begin{equation*}\n\\sum_{n=1}^\\infty n x^n\n\\end{equation*}", "converges to$\\frac{x}{{(1-x)}^2}$on$(-1,1)$.", "Proof: On$(-1,1)$,$\\sum_{n=0}^\\infty x^n$converges to$\\frac{1}{1-x}$. The derivative$\\sum_{n=1}^\\infty n x^{n-1}$then converges on the same interval to$\\frac{1}{{(1-x)}^2}$. Multiplying by$x$obtains the result." ], "refs": [], "proofs": [ { "contents": [ "On$(-1,1)$,$\\sum_{n=0}^\\infty x^n$converges to$\\frac{1}{1-x}$. The derivative$\\sum_{n=1}^\\infty n x^{n-1}$then converges on the same interval to$\\frac{1}{{(1-x)}^2}$. Multiplying by$x$obtains the result." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 24, "type": "theorem", "label": "Lebl-contfunc:24", "categories": [ "existence", "continuity", "named theorem", "uniqueness" ], "title": "Picard's Existence and Uniqueness Theorem", "contents": [ "% \\index{existence and uniqueness theorem}\\index{Picard's theorem}% \\label{thm:fs:picard} Let$I, J \\subset \\R$be closed bounded intervals, let$I^\\circ$and$J^\\circ$be their interiors% \\footnote{By interior of$[a,b]$, we mean$(a,b)$.}, and let$(x_0,y_0) \\in I^\\circ \\times J^\\circ$. Suppose$F \\colon I \\times J \\to \\R$is continuous and Lipschitz in the second variable, that is, there exists an$L \\in \\R$such that\\begin{equation*} \\abs{F(x,y) - F(x,z)} \\leq L \\abs{y-z} \\qquad \\text{for all } y,z \\in J, x \\in I . \\end{equation*}Then there exists an$h > 0$such that$[x_0-h,x_0+h] \\subset I$and a unique differentiable function$f \\colon [x_0 - h, x_0 + h] \\to J \\subset \\R$such that \\begin{equation} \\label{picard:diffeq} f'(x) = F\\bigl(x,f(x)\\bigr) \\qquad \\text{and} \\qquad f(x_0) = y_0. \\end{equation}" ], "refs": [], "proofs": [ { "contents": [ "\\pagebreak[2]", "Suppose we could find a solution $f$. Using the fundamental", "theorem of calculus we integrate the equation", "$f'(x) = F\\bigl(x,f(x)\\bigr)$, $f(x_0) = y_0$, and write \\eqref{picard:diffeq}", "as the integral equation", "\\begin{equation} \\label{picard:inteq}", "f(x) = y_0 + \\int_{x_0}^x F\\bigl(t,f(t)\\bigr)\\,dt .", "\\end{equation}", "The idea of our proof is that we try to plug in approximations to", "a solution to the right-hand side of \\eqref{picard:inteq} to get better approximations on the", "left-hand side of \\eqref{picard:inteq}. We hope that in the end the sequence", "converges and solves", "\\eqref{picard:inteq} and hence \\eqref{picard:diffeq}.", "The technique below is called \\emph{\\myindex{Picard iteration}},", "and the individual functions $f_k$ are called the", "\\emph{Picard iterates}\\index{Picard iterate}.", "Without loss of generality, suppose $x_0 = 0$ (exercise below).", "Another", "exercise tells us that $F$ is bounded as it is continuous.", "Therefore pick some $M > 0$ so that", "$\\abs{F(x,y)} \\leq M$ for all $(x,y) \\in I\\times J$.", "Pick $\\alpha > 0$ such that", "$[-\\alpha,\\alpha] \\subset I$ and $[y_0-\\alpha, y_0 + \\alpha] \\subset J$.", "Define", "\\begin{equation*}", "h \\coloneqq \\min \\left\\{ \\alpha, \\frac{\\alpha}{M+L\\alpha} \\right\\} .", "\\end{equation*}", "Observe", "$[-h,h] \\subset I$.", "Set $f_0(x) \\coloneqq y_0$.", "We define $f_k$ inductively.", "Assuming $f_{k-1}([-h,h]) \\subset [y_0-\\alpha,y_0+\\alpha]$,", "we see", "$F\\bigl(t,f_{k-1}(t)\\bigr)$ is", "a well-defined function of $t$ for $t \\in [-h,h]$.", "Further if $f_{k-1}$ is continuous", "on $[-h,h]$, then", "$F\\bigl(t,f_{k-1}(t)\\bigr)$ is", "continuous as", "a function of $t$ on $[-h,h]$ (left as an exercise).", "Define", "\\begin{equation*}", "f_k(x) \\coloneqq y_0+ \\int_{0}^x F\\bigl(t,f_{k-1}(t)\\bigr)\\,dt ,", "\\end{equation*}", "and $f_k$ is continuous on $[-h,h]$ by the fundamental theorem of calculus.", "To see that $f_k$ maps $[-h,h]$ to $[y_0-\\alpha,y_0+\\alpha]$, we compute for", "$x \\in [-h,h]$", "\\begin{equation*}", "\\abs{f_k(x) - y_0} =", "\\abs{\\int_{0}^x F\\bigl(t,f_{k-1}(t)\\bigr)\\,dt }", "\\leq", "M\\abs{x}", "\\leq", "Mh", "\\leq", "M", "\\frac{\\alpha}{M+L\\alpha}", "\\leq \\alpha .", "\\end{equation*}", "We next define $f_{k+1}$ using $f_k$ and so on.", "Thus we have inductively defined a sequence $\\{ f_k \\}_{k=1}^\\infty$ of functions.", "We need to show that it converges to a function $f$ that solves", "the equation~\\eqref{picard:inteq} and therefore~\\eqref{picard:diffeq}.", "We wish to show that the sequence $\\{ f_k \\}_{k=1}^\\infty$ converges uniformly", "to some function on $[-h,h]$. First, for $t \\in [-h,h]$,", "we have the following", "useful bound", "\\begin{equation*}", "\\abs{F\\bigl(t,f_{n}(t)\\bigr) -", "F\\bigl(t,f_{k}(t)\\bigr)}", "\\leq", "L \\abs{f_n(t)-f_k(t)}", "\\leq", "L \\norm{f_n-f_k}_{[-h,h]} ,", "\\end{equation*}", "where $\\norm{f_n-f_k}_{[-h,h]}$ is the uniform norm, that", "is the supremum of $\\abs{f_n(t)-f_k(t)}$ for $t \\in [-h,h]$.", "Now note that $\\abs{x} \\leq h \\leq \\frac{\\alpha}{M+L\\alpha}$.", "Therefore", "\\begin{equation*}", "\\begin{split}", "\\abs{f_n(x) - f_k(x)}", "& =", "\\abs{\\int_{0}^x F\\bigl(t,f_{n-1}(t)\\bigr)\\,dt", "-", "\\int_{0}^x F\\bigl(t,f_{k-1}(t)\\bigr)\\,dt}", "\\\\", "& =", "\\abs{\\int_{0}^x", "\\Bigl(", "F\\bigl(t,f_{n-1}(t)\\bigr)-", "F\\bigl(t,f_{k-1}(t)\\bigr)", "\\Bigr)", "\\,dt}", "\\\\", "& \\leq", "L\\norm{f_{n-1}-f_{k-1}}_{[-h,h]}", "\\abs{x}", "\\\\", "& \\leq", "\\frac{L\\alpha}{M+L\\alpha}", "\\norm{f_{n-1}-f_{k-1}}_{[-h,h]} .", "\\end{split}", "\\end{equation*}", "Let $C \\coloneqq \\frac{L\\alpha}{M+L\\alpha}$ and note that $C < 1$.", "Taking supremum on the left-hand side we get", "\\begin{equation*}", "\\norm{f_n-f_k}_{[-h,h]} \\leq C \\norm{f_{n-1}-f_{k-1}}_{[-h,h]} .", "\\end{equation*}", "Without loss of generality,", "suppose $n \\geq k$. Then by \\hyperref[induction:thm]{induction} we can show", "\\begin{equation*}", "\\norm{f_n-f_k}_{[-h,h]} \\leq C^{k} \\norm{f_{n-k}-f_{0}}_{[-h,h]} .", "\\end{equation*}", "For $x \\in [-h,h]$, we have", "\\begin{equation*}", "\\abs{f_{n-k}(x)-f_{0}(x)}", "=", "\\abs{f_{n-k}(x)-y_0}", "\\leq \\alpha .", "\\end{equation*}", "Therefore,", "\\begin{equation*}", "\\norm{f_n-f_k}_{[-h,h]} \\leq C^{k} \\norm{f_{n-k}-f_{0}}_{[-h,h]} \\leq C^{k} \\alpha .", "\\end{equation*}", "As $C < 1$, $\\{f_n\\}_{n=1}^\\infty$ is uniformly Cauchy and by", "\\propref{prop:uniformcauchy} we obtain that $\\{ f_n \\}_{n=1}^\\infty$", "converges uniformly on $[-h,h]$ to some function $f \\colon [-h,h] \\to \\R$.", "The function $f$ is the uniform limit of continuous functions and therefore", "continuous. Furthermore, since $f_n\\bigl([-h,h]\\bigr) \\subset", "[y_0-\\alpha,y_0+\\alpha]$ for all $n$,", "then $f\\bigl([-h,h]\\bigr) \\subset [y_0-\\alpha,y_0+\\alpha]$", "(why?).", "We now need to show that $f$ solves \\eqref{picard:inteq}.", "First, as before we notice", "\\begin{equation*}", "\\abs{F\\bigl(t,f_{n}(t)\\bigr) -", "F\\bigl(t,f(t)\\bigr)}", "\\leq", "L \\abs{f_n(t)-f(t)}", "\\leq", "L \\norm{f_n-f}_{[-h,h]} .", "\\end{equation*}", "As $\\norm{f_n-f}_{[-h,h]}$ converges to 0, then", "$F\\bigl(t,f_n(t)\\bigr)$ converges uniformly to $F\\bigl(t,f(t)\\bigr)$", "for $t \\in [-h,h]$. Hence, for $x \\in [-h,h]$", "the convergence is uniform %(why?)", "for $t \\in [0,x]$ (or $[x,0]$ if $x < 0$). Therefore,", "\\begin{align*}", "y_0", "+", "\\int_0^{x}", "F(t,f(t)\\bigr)\\,dt", "& =", "y_0", "+", "\\int_0^{x}", "F\\bigl(t,\\lim_{n\\to\\infty} f_n(t)\\bigr)\\,dt", "& &", "\\\\", "& =", "y_0", "+", "\\int_0^{x}", "\\lim_{n\\to\\infty} F\\bigl(t,f_n(t)\\bigr)\\,dt", "& & \\text{(by continuity of } F\\text{)}", "\\\\", "& =", "\\lim_{n\\to\\infty}", "\\left(", "y_0", "+", "\\int_0^{x}", "F\\bigl(t,f_n(t)\\bigr)\\,dt", "\\right)", "& & \\text{(by uniform convergence)}", "\\\\", "& =", "\\lim_{n\\to\\infty}", "f_{n+1}(x)", "=", "f(x) .", "& &", "\\end{align*}", "We apply the fundamental theorem of calculus (\\thmref{thm:FTCv2}) to show that", "$f$ is differentiable and its derivative is $F\\bigl(x,f(x)\\bigr)$. It is obvious", "that $f(0) = y_0$.", "Finally, what is left to do is to show uniqueness. Suppose $g \\colon [-h,h]", "\\to J \\subset \\R$ is another solution.", "As before we use the fact that", "$\\abs{F\\bigl(t,f(t)\\bigr) - F\\bigl(t,g(t)\\bigr)} \\leq L \\norm{f-g}_{[-h,h]}$.", "Then", "\\begin{equation*}", "\\begin{split}", "\\abs{f(x)-g(x)}", "& =", "\\abs{", "y_0", "+", "\\int_0^{x}", "F\\bigl(t,f(t)\\bigr)\\,dt", "-", "\\left(", "y_0", "+", "\\int_0^{x}", "F\\bigl(t,g(t)\\bigr)\\,dt", "\\right)", "}", "\\\\", "& =", "\\abs{", "\\int_0^{x}", "\\Bigl(", "F\\bigl(t,f(t)\\bigr)", "-", "F\\bigl(t,g(t)\\bigr)", "\\Bigr)", "\\,dt", "}", "\\\\", "& \\leq", "L\\norm{f-g}_{[-h,h]}\\abs{x}", "\\leq", "Lh\\norm{f-g}_{[-h,h]}", "\\leq", "\\frac{L\\alpha}{M+L\\alpha}\\norm{f-g}_{[-h,h]} .", "\\end{split}", "\\end{equation*}", "As", "before, $C = \\frac{L\\alpha}{M+L\\alpha} < 1$. By taking supremum over $x \\in", "[-h,h]$ on the", "left-hand side we obtain", "\\begin{equation*}", "\\norm{f-g}_{[-h,h]} \\leq C \\norm{f-g}_{[-h,h]} .", "\\end{equation*}", "This is only possible if $\\norm{f-g}_{[-h,h]} = 0$. Therefore, $f=g$, and the", "solution is unique." ], "refs": [ "induction:thm", "picard:diffeq", "picard:inteq", "prop:uniformcauchy" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 25, "type": "example", "label": "Lebl-contfunc:example:picardexponential", "categories": [ "existence", "example", "differential_equations", "series", "named theorem", "uniqueness" ], "title": "Picard's Existence and Uniqueness Theorem", "contents": [ "Consider", "\\begin{equation*}\nf'(x) = f(x), \\qquad f(0) = 1 .\n\\end{equation*}", "That is, we suppose$F(x,y) = y$, and we are looking for a function$f$such that$f'(x) = f(x)$. Let us forget for the moment that we solved this equation in \\sectionref{sec:logandexp}. See also \\figureref{fig:exp} for a plot of both the equation, showing the slope$F(x,y)=y$at each point, and the solution, the exponential, that satisfies$f(0)=1$.", "We pick any$I$that contains 0 in the interior. We pick an arbitrary$J$that contains 1 in its interior. We can use$L = 1$. The theorem guarantees an$h > 0$such that there exists a unique solution$f \\colon [-h,h] \\to \\R$. This solution is usually denoted by", "\\begin{equation*}\ne^x \\coloneqq f(x) .\n\\end{equation*}", "We leave it to the reader to verify that by picking$I$and$J$large enough the proof of the theorem guarantees that we are able to pick$\\alpha$such that we get any$h$we want as long as$h < \\nicefrac{1}{2}$. We omit the calculation. Of course, we know %(though we have not proved) this function exists as a function for all$x$, so an arbitrary$h$ought to work, but the theorem only provides$h < \\nicefrac{1}{2}$.", "By same reasoning as above, no matter what$x_0$and$y_0$are, the proof guarantees an arbitrary$h$as long as$h < \\nicefrac{1}{2}$. Fix such an$h$. We get a unique function defined on$[x_0-h,x_0+h]$. After defining the function on$[-h,h]$we find a solution on the interval$[0,2h]$and notice that the two functions must coincide on$[0,h]$by uniqueness. We thus iteratively construct the exponential for all$x \\in \\R$. Therefore, Picard's theorem could be used to prove the existence and uniqueness of the exponential.", "Let us compute the Picard iterates. We start with the constant function$f_0(x) \\coloneqq 1$. Then \\begin{align*} f_1(x) & = 1 + \\int_0^x f_0(s)\\,ds = 1+x,", "f_2(x) & = 1 + \\int_0^x f_1(s)\\,ds = 1 + \\int_0^x (1+s)\\,ds = 1 + x + \\frac{x^2}{2},", "f_3(x) & = 1 + \\int_0^x f_2(s)\\,ds = 1 + \\int_0^x \\left(1+ s + \\frac{s^2}{2} \\right)\\,ds = 1 + x + \\frac{x^2}{2} + \\frac{x^3}{6} . \\end{align*} We recognize the beginning of the Taylor series for the exponential. See \\figureref{fig:exppicard}. \\begin{myfigureht} \\includegraphics{figures/exppicardfig} \\caption{The exponential (solid line) together with$f_0$,$f_1$,$f_2$,$f_3$(dashed).\\label{fig:exppicard}} \\end{myfigureht}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 26, "type": "example", "label": "Lebl-contfunc:26", "categories": [ "differential_equations", "example", "derivatives" ], "title": "Consider the equation \\begin{equation*} f'(x) = {\\bigl(f(x)\\bigr)}^2 \\qquad \\text{and} \\qquad f(0)=1", "contents": [ "Consider the equation", "\\begin{equation*}\nf'(x) = {\\bigl(f(x)\\bigr)}^2 \\qquad \\text{and} \\qquad f(0)=1.\n\\end{equation*}", "From elementary differential equations we know", "\\begin{equation*}\nf(x) = \\frac{1}{1-x}\n\\end{equation*}", "is the solution. The solution is only defined on$(-\\infty,1)$. That is, we are able to use$h < 1$, but never a larger$h$. The function that takes$y$to$y^2$is not Lipschitz as a function on all of$\\R$. As we approach$x=1$from the left, the solution becomes larger and larger. The derivative of the solution grows as$y^2$, and so the$L$required has to be larger and larger as$y_0$grows. If we apply the theorem with$x_0$close to 1 and$y_0 = \\frac{1}{1-x_0}$we find that the$h$that the proof guarantees is smaller and smaller as$x_0$approaches~1.", "The$h$from the proof is not the best$h$. By picking$\\alpha$correctly, the proof of the theorem guarantees$h=1-\\nicefrac{\\sqrt{3}}{2} \\approx 0.134$(we omit the calculation) for$x_0=0$and$y_0=1$, even though we saw above that any$h < 1$should work." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 27, "type": "example", "label": "Lebl-contfunc:27", "categories": [ "differential_equations", "continuity", "example" ], "title": "Uniqueness of Continuity of Functions", "contents": [ "Consider the equation", "\\begin{equation*}\nf'(x) = 2 \\sqrt{\\abs{f(x)}}, \\qquad f(0) = 0 .\n\\end{equation*}", "The function$F(x,y) = 2 \\sqrt{\\abs{y}}$is continuous, but not Lipschitz in$y$(why?). The equation does not satisfy the hypotheses of the theorem. The function", "\\begin{equation*}\nf(x) =\n\\begin{cases}\nx^2 & \\text{if } x \\geq 0,\\\\\n-x^2 & \\text{if } x < 0,\n\\end{cases}\n\\end{equation*}", "is a solution, but$f(x) = 0$is also a solution. A solution exists, but is not unique." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 28, "type": "example", "label": "Lebl-contfunc:28", "categories": [ "differential_equations", "continuity", "example", "derivatives" ], "title": "Darboux's Theorem", "contents": [ "Consider$y' = \\varphi(x)$where$\\varphi(x) \\coloneqq 0$if$x \\in \\Q$and$\\varphi(x)\\coloneqq 1$if$x \\not\\in \\Q$. In other words, the$F(x,y) = \\varphi(x)$is discontinuous. The equation has no solution regardless of the initial conditions. A solution would have derivative$\\varphi$, but$\\varphi$does not have the intermediate value property at any point (why?). No solution exists by \\hyperref[thm:darboux]{Darboux's theorem}." ], "refs": [ "thm:darboux" ], "proofs": [], "ref_ids": [] }, { "id": 0, "type": "example", "label": "Lebl-contfunc:0", "categories": [ "metric_spaces", "example" ], "title": "The set of real numbers$\\R$is a metric space with the metric \\begin{equation*} d(x,y) \\coloneqq \\...", "contents": [ "The set of real numbers$\\R$is a metric space with the metric", "\\begin{equation*}\nd(x,y) \\coloneqq \\abs{x-y} .\n\\end{equation*}", "Items \\ref{metric:pos}--\\ref{metric:com} of the definition are easy to verify. The triangle inequality \\ref{metric:triang} follows immediately from the standard triangle inequality for real numbers:", "\\begin{equation*}\nd(x,z) = \\abs{x-z} = \n\\abs{x-y+y-z} \\leq\n\\abs{x-y}+\\abs{y-z} =\nd(x,y)+ d(y,z) .\n\\end{equation*}", "This metric is the \\emph{\\myindex{standard metric on$\\R$}}. If we talk about$\\R$as a metric space without mentioning a specific metric, we mean this particular metric." ], "refs": [ "metric:com", "metric:pos", "metric:triang" ], "proofs": [], "ref_ids": [] }, { "id": 1, "type": "example", "label": "Lebl-contfunc:1", "categories": [ "applications", "example" ], "title": "We can also put a different metric on the set of real numbers", "contents": [ "We can also put a different metric on the set of real numbers. For example, take the set of real numbers$\\R$together with the metric", "\\begin{equation*}\nd(x,y) \\coloneqq\n\\frac{\\abs{x-y}}{\\abs{x-y}+1} .\n\\end{equation*}", "Items \\ref{metric:pos}--\\ref{metric:com} are again easy to verify. The triangle inequality \\ref{metric:triang} is a little bit more difficult. Note that$d(x,y) = \\varphi(\\abs{x-y})$where$\\varphi(t) = \\frac{t}{t+1}$and$\\varphi$is an increasing function (positive derivative, see \\figureref{fig:tovertp1}). Hence", "\\begin{equation*}\n\\begin{split}\nd(x,z) = \\varphi(\\abs{x-z})\n& = \n\\varphi(\\abs{x-y+y-z}) \\\\\n& \\leq\n\\varphi(\\abs{x-y}+\\abs{y-z})\n\\\\\n& =\n\\frac{\\abs{x-y}+\\abs{y-z}}{\\abs{x-y}+\\abs{y-z}+1} \\\\\n& =\n\\frac{\\abs{x-y}}{\\abs{x-y}+\\abs{y-z}+1} +\n\\frac{\\abs{y-z}}{\\abs{x-y}+\\abs{y-z}+1}\n\\\\\n& \\leq\n\\frac{\\abs{x-y}}{\\abs{x-y}+1} +\n\\frac{\\abs{y-z}}{\\abs{y-z}+1} =\nd(x,y)+ d(y,z) .\n\\end{split}\n\\end{equation*}", "The function$d$is thus a metric, and gives an example of a nonstandard metric on$\\R$. With this metric,$d(x,y) < 1$for all$x,y \\in \\R$. That is, every two points are less than 1 unit apart. \\begin{myfigureht} \\includegraphics{figures/tovertp1graph} \\caption{Graph of$\\frac{t}{t+1}$for positive$t$with an asymptote at 1.\\label{fig:tovertp1}} \\end{myfigureht}" ], "refs": [ "metric:com", "metric:pos", "metric:triang" ], "proofs": [], "ref_ids": [] }, { "id": 2, "type": "lemma", "label": "Lebl-contfunc:2", "categories": [ "auxiliary result" ], "title": "Auxiliary Result for Metric Spaces", "contents": [ "Suppose$x =(x_1,x_2,\\ldots,x_n) \\in \\R^n$,$y =(y_1,y_2,\\ldots,y_n) \\in \\R^n$. Then\\begin{equation*} {\\biggl( \\sum_{k=1}^n x_k y_k \\biggr)}^2 \\leq \\biggl(\\sum_{k=1}^n x_k^2 \\biggr) \\biggl(\\sum_{k=1}^n y_k^2 \\biggr) . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "A square of a real number is nonnegative. Hence a sum of squares is", "nonnegative:", "\\begin{equation*}", "\\begin{split}", "0 & \\leq", "\\sum_{k=1}^n \\sum_{\\ell=1}^n {(x_k y_\\ell - x_\\ell y_k)}^2", "\\\\", "& =", "\\sum_{k=1}^n \\sum_{\\ell=1}^n \\bigl( x_k^2 y_\\ell^2 + x_\\ell^2 y_k^2 - 2 x_k", "x_\\ell y_k", "y_\\ell \\bigr)", "\\\\", "& =", "\\biggl( \\sum_{k=1}^n x_k^2 \\biggr)", "\\biggl( \\sum_{\\ell=1}^n y_\\ell^2 \\biggr)", "+", "\\biggl( \\sum_{k=1}^n y_k^2 \\biggr)", "\\biggl( \\sum_{\\ell=1}^n x_\\ell^2 \\biggr)", "-", "2", "\\biggl( \\sum_{k=1}^n x_k y_k \\biggr)", "\\biggl( \\sum_{\\ell=1}^n x_\\ell y_\\ell \\biggr) .", "\\end{split}", "\\end{equation*}", "We relabel and divide by 2 to obtain precisely what we wanted,", "\\begin{equation*}", "0 \\leq", "\\biggl( \\sum_{k=1}^n x_k^2 \\biggr)", "\\biggl( \\sum_{k=1}^n y_k^2 \\biggr)", "-", "{\\biggl( \\sum_{k=1}^n x_k y_k \\biggr)}^2 .", "\\qedhere", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 3, "type": "example", "label": "Lebl-contfunc:3", "categories": [ "completeness", "example" ], "title": "Let us construct the standard metric\\index{standard metric on$\\R^n$} for$\\R^n$", "contents": [ "Let us construct the standard metric\\index{standard metric on$\\R^n$} for$\\R^n$. Define", "\\begin{equation*}\nd(x,y) \\coloneqq\n\\sqrt{\n{(x_1-y_1)}^2 + \n{(x_2-y_2)}^2 + \n\\cdots +\n{(x_n-y_n)}^2\n} =\n\\sqrt{\n\\sum_{k=1}^n\n{(x_k-y_k)}^2 \n} .\n\\end{equation*}", "For$n=1$, the real line, this metric agrees with what we defined above. For$n > 1$, the only tricky part of the definition to check, as before, is the triangle inequality. It is less messy to work with the square of the metric. In the following estimate, note the use of the Cauchy--Schwarz inequality.", "\\begin{equation*}\n\\begin{split}\n{\\bigl(d(x,z)\\bigr)}^2 & =\n\\sum_{k=1}^n\n{(x_k-z_k)}^2 \n\\\\\n& =\n\\sum_{k=1}^n\n{(x_k-y_k+y_k-z_k)}^2 \n\\\\\n& =\n\\sum_{k=1}^n\n\\Bigl(\n{(x_k-y_k)}^2+{(y_k-z_k)}^2 + 2(x_k-y_k)(y_k-z_k)\n\\Bigr)\n\\\\\n& =\n\\sum_{k=1}^n\n{(x_k-y_k)}^2\n+\n\\sum_{k=1}^n\n{(y_k-z_k)}^2 \n+\n2\n\\sum_{k=1}^n\n(x_k-y_k)(y_k-z_k)\n\\\\\n& \\leq\n\\sum_{k=1}^n\n{(x_k-y_k)}^2\n+\n\\sum_{k=1}^n\n{(y_k-z_k)}^2 \n+\n2\n\\sqrt{\n\\sum_{k=1}^n\n{(x_k-y_k)}^2\n\\sum_{k=1}^n\n{(y_k-z_k)}^2\n}\n\\\\\n& =\n{\\left(\n\\sqrt{\n\\sum_{k=1}^n\n{(x_k-y_k)}^2\n}\n+\n\\sqrt{\n\\sum_{k=1}^n\n{(y_k-z_k)}^2 \n}\n\\right)}^2\n=\n{\\bigl( d(x,y) + d(y,z) \\bigr)}^2 .\n\\end{split}\n\\end{equation*}", "Because the square root is an increasing function, the inequality is preserved when we take the square root of both sides, and we obtain the triangle inequality." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 4, "type": "example", "label": "Lebl-contfunc:4", "categories": [ "convergence", "metric_spaces", "example" ], "title": "Limit of Metric Spaces", "contents": [ "The set of complex numbers$\\C$is the set of numbers$z = x+iy$, where$x$and$y$are in$\\R$. By imposing$i^2 = -1$, we make$\\C$into a field. For the purposes of taking limits, the set$\\C$is regarded as the metric space$\\R^2$, where$z=x+iy \\in \\C$corresponds to$(x,y) \\in \\R^2$. For$z=x+iy$define the \\emph{\\myindex{complex modulus}}\\index{modulus} by$\\sabs{z} \\coloneqq \\sqrt{x^2+y^2}$. Then for two complex numbers$z_1 = x_1 + iy_1$and$z_2 = x_2 + iy_2$, the distance is", "\\begin{equation*}\nd(z_1,z_2) = \\sqrt{{(x_1-x_2)}^2+ {(y_1-y_2)}^2} = \\sabs{z_1-z_2}.\n\\end{equation*}", "Furthermore, when working with complex numbers it is often convenient to write the metric in terms of the so-called \\emph{\\myindex{complex conjugate}}: The conjugate of$z=x+iy$is$\\bar{z} \\coloneqq x-iy$. Then${\\sabs{z}}^2 = x^2 +y^2 = z\\bar{z}$, and so${\\sabs{z_1-z_2}}^2 = (z_1-z_2)\\overline{(z_1-z_2)}$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 5, "type": "example", "label": "Lebl-contfunc:5", "categories": [ "applications", "metric_spaces", "example" ], "title": "An example to keep in mind is the so-called \\emph{\\myindex{discrete metric}}", "contents": [ "An example to keep in mind is the so-called \\emph{\\myindex{discrete metric}}. For any set$X$, define", "\\begin{equation*}\nd(x,y) \\coloneqq\n\\begin{cases}\n1 & \\text{if } x \\not= y, \\\\\n0 & \\text{if } x = y.\n\\end{cases}\n\\end{equation*}", "That is, all points are equally distant from each other. When$X$is a finite set, we can draw a diagram, see for example \\figureref{fig:msdiscmetric}. Of course, in the diagram the distances are not the normal euclidean distances in the plane. Things become subtle when$X$is an infinite set such as the real numbers. \\begin{myfigureht} \\subimport*{figures/}{msdiscmetric.pdf_t} \\caption{Sample discrete metric space$\\{ a,b,c,d,e \\}$, the distance between any two points is$1$.\\label{fig:msdiscmetric}} \\end{myfigureht}", "While this particular example may seldom come up in practice, it gives a useful \\myquote{smell test.} If you make a statement about metric spaces, try it with the discrete metric. To show that$(X,d)$is indeed a metric space is left as an exercise." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 6, "type": "example", "label": "Lebl-contfunc:example:msC01", "categories": [ "applications", "topology", "example", "metric_spaces", "continuity", "named theorem" ], "title": "Picard's Theorem", "contents": [ "Let$C\\bigl([a,b],\\R\\bigr)$\\glsadd{not:contfuncs} be the set of continuous real-valued functions on the interval$[a,b]$. Define the metric on$C\\bigl([a,b],\\R\\bigr)$as", "\\begin{equation*}\nd(f,g) \\coloneqq \\sup_{x \\in [a,b]} \\abs{f(x)-g(x)} .\n\\end{equation*}", "Let us check the properties. First,$d(f,g)$is finite as$\\abs{f(x)-g(x)}$is a continuous function on a closed bounded interval$[a,b]$, and so is bounded. It is clear that$d(f,g) \\geq 0$, it is the supremum of nonnegative numbers. If$f = g$, then$\\abs{f(x)-g(x)} = 0$for all$x$, and hence$d(f,g) = 0$. Conversely, if$d(f,g) = 0$, then for every$x$, we have$\\abs{f(x)-g(x)} \\leq d(f,g) = 0$, and hence$f(x) = g(x)$for all$x$, and so$f=g$. That$d(f,g) = d(g,f)$is equally trivial. To show the triangle inequality we use the standard triangle inequality;", "\\begin{equation*}\n\\begin{split}\nd(f,g) & =\n\\sup_{x \\in [a,b]} \\abs{f(x)-g(x)} =\n\\sup_{x \\in [a,b]} \\abs{f(x)-h(x)+h(x)-g(x)}\n\\\\\n& \\leq\n\\sup_{x \\in [a,b]} \\bigl( \\abs{f(x)-h(x)}+\\abs{h(x)-g(x)} \\bigr)\n\\\\\n& \\leq\n\\sup_{x \\in [a,b]} \\abs{f(x)-h(x)}+\n\\sup_{x \\in [a,b]} \\abs{h(x)-g(x)} = d(f,h) + d(h,g) .\n\\end{split}\n\\end{equation*}", "When treating$C\\bigl([a,b],\\R\\bigr)$as a metric space without mentioning a metric, we mean this particular metric. Notice that$d(f,g) = \\norm{f-g}_{[a,b]}$, the uniform norm of \\defnref{def:unifnorm}.", "This example may seem esoteric at first, but it turns out that working with spaces such as$C\\bigl([a,b],\\R\\bigr)$is really the meat of a large part of modern analysis. Treating sets of functions as metric spaces allows us to abstract away a lot of the grubby detail and prove powerful results such as \\hyperref[thm:fs:picard]{Picard's theorem} with less work." ], "refs": [ "named:Picard", "thm:fs:picard" ], "proofs": [], "ref_ids": [] }, { "id": 7, "type": "example", "label": "Lebl-contfunc:7", "categories": [ "applications", "metric_spaces", "example" ], "title": "Another useful example of a metric space is the sphere with a metric usually called the \\emph{\\my...", "contents": [ "Another useful example of a metric space is the sphere with a metric usually called the \\emph{\\myindex{great circle distance}}. Let$S^2$be the \\myindex{unit sphere}\\index{sphere} in$\\R^3$, that is$S^2 \\coloneqq \\{ x \\in \\R^3 : x_1^2+x_2^2+x_3^2 = 1 \\}$. Take$x$and$y$in$S^2$, draw a line through the origin and$x$, and another line through the origin and$y$, and let$\\theta$be the angle that the two lines make. Then define$d(x,y) \\coloneqq \\theta$. See \\figureref{fig:spheremetric}. The law of cosines from vector calculus says$d(x,y) = \\arccos(x_1y_1+x_2y_2+x_3y_3)$. It is relatively easy to see that this function satisfies the first three properties of a metric. Triangle inequality is harder to prove, and requires a bit more trigonometry and linear algebra than we wish to indulge in right now, so let us leave it without proof.", "\\begin{myfigureht} \\subimport*{figures/}{spheremetric.pdf_t} \\caption{The great circle distance on the unit sphere.\\label{fig:spheremetric}} \\end{myfigureht}", "This distance is the shortest distance between points on a sphere if we are allowed to travel on the sphere only. It is easy to generalize to arbitrary diameters. If we take a sphere of radius$r$, we let the distance be$d(x,y) \\coloneqq r \\theta$. As an example, this is the standard distance you would use if you compute a distance on the surface of the earth, such as computing the distance a plane travels from London to Los Angeles." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 8, "type": "proposition", "label": "Lebl-contfunc:8", "categories": [ "metric_spaces" ], "title": "Let$(X,d)$be a metric space and$Y \\subset X$", "contents": [ "Let$(X,d)$be a metric space and$Y \\subset X$. Then the restriction$d|_{Y \\times Y}$is a metric on$Y$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 9, "type": "example", "label": "Lebl-contfunc:9", "categories": [ "metric_spaces", "example" ], "title": "Take the metric space$\\R$with the standard metric", "contents": [ "Take the metric space$\\R$with the standard metric. For$x \\in \\R$and$\\delta > 0$,", "\\begin{equation*}\nB(x,\\delta) = (x-\\delta,x+\\delta) \\qquad \\text{and} \\qquad\nC(x,\\delta) = [x-\\delta,x+\\delta] .\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 10, "type": "example", "label": "Lebl-contfunc:10", "categories": [ "metric_spaces", "example" ], "title": "Be careful when working on a subspace", "contents": [ "Be careful when working on a subspace. Consider the metric space$[0,1]$as a subspace of$\\R$. Then in$[0,1]$,", "\\begin{equation*}\nB(0,\\nicefrac{1}{2}) = B_{[0,1]}(0,\\nicefrac{1}{2})\n= \\bigl\\{ y \\in [0,1] : \\abs{0-y} < \\nicefrac{1}{2} \\bigr\\}\n= [0,\\nicefrac{1}{2}) .\n\\end{equation*}", "This is different from$B_{\\R}(0,\\nicefrac{1}{2}) = (\\nicefrac{-1}{2},\\nicefrac{1}{2})$. The important thing to keep in mind is which metric space we are working in." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 11, "type": "example", "label": "Lebl-contfunc:11", "categories": [ "metric_spaces", "topology", "example" ], "title": "The set$(0,\\infty) \\subset \\R$is open: Given any$x \\in (0,\\infty)$, let$\\delta \\coloneqq x$", "contents": [ "The set$(0,\\infty) \\subset \\R$is open: Given any$x \\in (0,\\infty)$, let$\\delta \\coloneqq x$. Then$B(x,\\delta) = (0,2x) \\subset (0,\\infty)$.", "The set$[0,\\infty) \\subset \\R$is closed: Given$x \\in (-\\infty,0) = [0,\\infty)^c$, let$\\delta \\coloneqq -x$. Then$B(x,\\delta) = (-2x,0) \\subset (-\\infty,0) = [0,\\infty)^c$.", "The set$[0,1) \\subset \\R$is neither open nor closed. First, every ball in$\\R$around$0$,$B(0,\\delta) = (-\\delta,\\delta)$, contains negative numbers and hence is not contained in$[0,1)$. So$[0,1)$is not open. Second, every ball in$\\R$around$1$,$B(1,\\delta) = (1-\\delta,1+\\delta)$, contains numbers strictly less than 1 and greater than 0 (e.g.\\$1-\\nicefrac{\\delta}{2}$as long as$\\delta < 2$). Thus$[0,1)^c = \\R \\setminus [0,1)$is not open, and$[0,1)$is not closed.", "If$(X,d)$is any metric space, and$x \\in X$is a point, then$\\{ x \\}$is closed (exercise). On the other hand,$\\{ x \\}$may or may not be open depending on$X$. The set$\\{ 0 \\} \\subset \\R$is not open as$B(0,\\delta)$contains nonzero numbers for every$\\delta > 0$. If$X=\\{ x \\}$, then$\\{ x \\}$is open." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 12, "type": "proposition", "label": "Lebl-contfunc:prop:topology:open", "categories": [ "metric_spaces", "topology" ], "title": "Let$(X,d)$be a metric space", "contents": [ "Let$(X,d)$be a metric space. \\begin{enumerate}[(i)] \\item \\label{topology:openi}$\\emptyset$and$X$are open. \\item \\label{topology:openii} If$V_1, V_2, \\ldots, V_k$are open subsets of$X$, then\\begin{equation*} \\bigcap_{j=1}^k V_j \\end{equation*}is also open. That is, a finite intersection of open sets is open. \\item \\label{topology:openiii} If$\\{ V_\\lambda \\}_{\\lambda \\in I}$is an arbitrary collection of open subsets of$X$, then\\begin{equation*} \\bigcup_{\\lambda \\in I} V_\\lambda \\end{equation*}is also open. That is, a union of open sets is open. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "The sets $\\emptyset$ and $X$ are obviously open in $X$.", "Let us prove \\ref{topology:openii}.", "If $x \\in \\bigcap_{j=1}^k V_j$, then $x \\in V_j$ for all $j$.", "As $V_j$ are all open, for every $j$ there exists a $\\delta_j > 0$", "such that $B(x,\\delta_j) \\subset V_j$. Take $\\delta \\coloneqq \\min \\{", "\\delta_1,\\delta_2,\\ldots,\\delta_k \\}$ and notice $\\delta > 0$. We have", "$B(x,\\delta) \\subset B(x,\\delta_j) \\subset V_j$ for every $j$ and so", "$B(x,\\delta) \\subset \\bigcap_{j=1}^k V_j$. Consequently the intersection is open.", "Let us prove \\ref{topology:openiii}.", "If $x \\in \\bigcup_{\\lambda \\in I} V_\\lambda$, then $x \\in V_\\lambda$ for some", "$\\lambda \\in I$.", "As $V_\\lambda$ is open, there exists a $\\delta > 0$", "such that $B(x,\\delta) \\subset V_\\lambda$. But then", "$B(x,\\delta) \\subset \\bigcup_{\\lambda \\in I} V_\\lambda$,", "and so the union is open." ], "refs": [ "topology:openii", "topology:openiii" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 13, "type": "example", "label": "Lebl-contfunc:13", "categories": [ "topology", "example" ], "title": "Notice the difference between items \\ref{topology:openii} and \\ref{topology:openiii}", "contents": [ "Notice the difference between items \\ref{topology:openii} and \\ref{topology:openiii}. Item \\ref{topology:openii} is not true for an arbitrary intersection. For instance,$\\bigcap_{n=1}^\\infty (\\nicefrac{-1}{n},\\nicefrac{1}{n}) = \\{ 0 \\}$, which is not open." ], "refs": [ "topology:openii", "topology:openiii" ], "proofs": [], "ref_ids": [] }, { "id": 14, "type": "proposition", "label": "Lebl-contfunc:prop:topology:closed", "categories": [ "metric_spaces", "topology" ], "title": "%FIXMEevillayouthack \\pagebreak[2] Let$(X,d)$be a metric space", "contents": [ "%FIXMEevillayouthack \\pagebreak[2] Let$(X,d)$be a metric space. \\begin{enumerate}[(i)] \\item \\label{topology:closedi}$\\emptyset$and$X$are closed. \\item \\label{topology:closedii} If$\\{ E_\\lambda \\}_{\\lambda \\in I}$is an arbitrary collection of closed subsets of$X$, then\\begin{equation*} \\bigcap_{\\lambda \\in I} E_\\lambda \\end{equation*}is also closed. That is, an intersection of closed sets is closed. \\item \\label{topology:closediii} If$E_1, E_2, \\ldots, E_k$are closed subsets of$X$, then\\begin{equation*} \\bigcup_{j=1}^k E_j \\end{equation*}is also closed. That is, a finite union of closed sets is closed. \\end{enumerate}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 15, "type": "proposition", "label": "Lebl-contfunc:prop:topology:ballsopenclosed", "categories": [ "metric_spaces", "topology" ], "title": "Let$(X,d)$be a metric space,$x \\in X$, and$\\delta > 0$", "contents": [ "Let$(X,d)$be a metric space,$x \\in X$, and$\\delta > 0$. Then$B(x,\\delta)$is open and$C(x,\\delta)$is closed." ], "refs": [], "proofs": [ { "contents": [ "Let $y \\in B(x,\\delta)$. Let $\\alpha \\coloneqq \\delta-d(x,y)$. As $\\alpha", "> 0$, consider $z \\in B(y,\\alpha)$. Then", "\\begin{equation*}", "d(x,z) \\leq d(x,y) + d(y,z) < d(x,y) + \\alpha = d(x,y) + \\delta-d(x,y) =", "\\delta .", "\\end{equation*}", "Therefore, $z \\in B(x,\\delta)$ for every $z \\in B(y,\\alpha)$. So", "$B(y,\\alpha) \\subset B(x,\\delta)$, and so", "$B(x,\\delta)$ is open. See \\figureref{fig:ballisopen}.", "\\begin{myfigureht}", "\\subimport*{figures/}{ballisopen.pdf_t}", "\\caption{Proof that $B(x,\\delta)$ is open: $B(y,\\alpha) \\subset", "B(x,\\delta)$ with the triangle inequality illustrated.\\label{fig:ballisopen}}", "\\end{myfigureht}", "The proof that $C(x,\\delta)$ is closed is left as an exercise." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 16, "type": "proposition", "label": "Lebl-contfunc:prop:topology:intervals:openclosed", "categories": [ "topology" ], "title": "Let$a, b$be two real numbers,$a < b$", "contents": [ "Let$a, b$be two real numbers,$a < b$. Then$(a,b)$,$(a,\\infty)$, and$(-\\infty,b)$are open in$\\R$. Also$[a,b]$,$[a,\\infty)$, and$(-\\infty,b]$are closed in$\\R$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 17, "type": "proposition", "label": "Lebl-contfunc:prop:topology:subspaceopen", "categories": [ "characterization", "metric_spaces", "topology" ], "title": "Characterization of Metric Spaces via Equivalence", "contents": [ "Suppose$(X,d)$is a metric space, and$Y \\subset X$. Then$U \\subset Y$is open in$Y$(in the subspace topology) if and only if there exists an open set$V \\subset X$(so open in$X$) such that$V \\cap Y = U$." ], "refs": [], "proofs": [ { "contents": [ "Suppose $V \\subset X$ is open and $V \\cap Y = U$.", "Let $x \\in U$.", "As $V$ is open and $x \\in V$, there", "exists a $\\delta > 0$ such that $B_X(x,\\delta) \\subset V$.", "Then", "\\begin{equation*}", "B_Y(x,\\delta) = B_X(x,\\delta) \\cap Y \\subset V \\cap Y = U .", "\\end{equation*}", "So $U$ is open in $Y$.", "The proof of the opposite direction, that is, that if $U \\subset Y$", "is open in the subspace topology there exists a $V$ is left as", "\\exerciseref{exercise:mssubspace}." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 18, "type": "proposition", "label": "Lebl-contfunc:prop:topology:subspacesame", "categories": [ "characterization", "metric_spaces", "topology" ], "title": "Suppose$(X,d)$is a metric space,$V \\subset X$is open, and$E \\subset X$is closed", "contents": [ "Suppose$(X,d)$is a metric space,$V \\subset X$is open, and$E \\subset X$is closed. \\begin{enumerate}[(i)] \\item \\label{prop:topology:subspacesame:i}$U \\subset V$is open in the subspace topology if and only if$U$is open in$X$. \\item \\label{prop:topology:subspacesame:ii}$F \\subset E$is closed in the subspace topology if and only if$F$is closed in$X$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "We prove", "\\ref{prop:topology:subspacesame:i}", "and leave", "\\ref{prop:topology:subspacesame:ii} as an exercise.", "If $U \\subset V$ is open in the subspace topology, by", "\\propref{prop:topology:subspaceopen}, there is a set $W \\subset X$", "open in $X$ such that $U = W \\cap V$. Intersection of two open sets", "is open so $U$ is open in $X$.", "Now suppose $U$ is open in $X$. Then $U = U \\cap V$. So", "$U$ is open in $V$ again by \\propref{prop:topology:subspaceopen}." ], "refs": [ "prop:topology:subspaceopen", "prop:topology:subspacesame:i", "prop:topology:subspacesame:ii" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 19, "type": "proposition", "label": "Lebl-contfunc:19", "categories": [ "characterization", "connectedness", "metric_spaces", "topology" ], "title": "Characterization of Metric Spaces via Equivalence", "contents": [ "Let$(X,d)$be a metric space. A nonempty set$S \\subset X$is disconnected if and only if there exist open sets$U_1$and$U_2$in$X$such that$U_1 \\cap U_2 \\cap S = \\emptyset$,$U_1 \\cap S \\not= \\emptyset$,$U_2 \\cap S \\not= \\emptyset$, and\\begin{equation*} S = \\bigl( U_1 \\cap S \\bigr) \\cup \\bigl( U_2 \\cap S \\bigr) . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "First suppose $S$ is disconnected: There are", "nonempty disjoint $S_1$ and $S_2$ that are", "open in $S$ and $S = S_1 \\cup S_2$.", "\\propref{prop:topology:subspaceopen} says there exist $U_1$ and $U_2$", "that are open in $X$ such that $U_1 \\cap S = S_1$ and", "$U_2 \\cap S = S_2$.", "For the other direction start with the $U_1$ and $U_2$.", "Then $U_1 \\cap S$ and $U_2 \\cap S$ are open in $S$ by", "\\propref{prop:topology:subspaceopen}.", "Via the discussion before the proposition, $S$ is disconnected." ], "refs": [ "prop:topology:subspaceopen" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 20, "type": "example", "label": "Lebl-contfunc:20", "categories": [ "connectedness", "example" ], "title": "Suppose$S \\subset \\R$and there are$x,y,z$such that$x < z < y$with$x,y \\in S$and$z \\notin S$", "contents": [ "Suppose$S \\subset \\R$and there are$x,y,z$such that$x < z < y$with$x,y \\in S$and$z \\notin S$. Claim: \\emph{$S$is disconnected}. Proof: Notice", "\\begin{equation*}\n\\bigl( (-\\infty,z) \\cap S \\bigr)\n\\cup\n\\bigl( (z,\\infty) \\cap S \\bigr)\n= S .\n\\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Notice \\begin{equation*}", "\\bigl( (-\\infty,z) \\cap S \\bigr)", "\\cup", "\\bigl( (z,\\infty) \\cap S \\bigr)", "= S .", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 21, "type": "proposition", "label": "Lebl-contfunc:21", "categories": [ "connectedness", "characterization" ], "title": "Characterization of Connectedness via Equivalence", "contents": [ "A nonempty set$S \\subset \\R$is connected if and only if$S$is an interval or a single point." ], "refs": [], "proofs": [ { "contents": [ "Suppose $S$ is connected. If $S$ is a single point,", "then we are done. So suppose $x < y$ and $x,y \\in S$. If $z \\in \\R$ is such", "that $x < z < y$, then $(-\\infty,z) \\cap S$ is nonempty and $(z,\\infty) \\cap", "S$ is nonempty. The two sets are disjoint. As", "$S$ is connected, we must have they their union is not $S$, so $z \\in S$.", "By \\propref{prop:intervaldef}, $S$ is an interval.", "If $S$ is a single point, it is connected.", "Therefore, suppose $S$ is an interval.", "Consider open subsets $U_1$ and $U_2$ of $\\R$ such that", "$U_1 \\cap S$ and $U_2 \\cap S$ are nonempty, and", "$S =", "\\bigl( U_1 \\cap S \\bigr)", "\\cup", "\\bigl( U_2 \\cap S \\bigr)$. We will show that $U_1 \\cap S$", "and $U_2 \\cap S$ contain a common point, so they are not disjoint,", "proving that $S$ is connected.", "Suppose $x \\in U_1 \\cap S$", "and $y \\in U_2 \\cap S$. Without loss of generality, assume $x < y$.", "As $S$ is an interval, $[x,y] \\subset S$.", "Note that $U_2 \\cap [x,y] \\not= \\emptyset$, and", "let $z \\coloneqq \\inf (U_2 \\cap [x,y])$.", "We wish to show that $z \\in U_1$.", "If $z = x$, then $z \\in U_1$.", "If $z > x$,", "then for every $\\epsilon > 0$, the ball $B(z,\\epsilon) =", "(z-\\epsilon,z+\\epsilon)$ contains points of $[x,y]$ not in $U_2$,", "as $z$ is the infimum of $U_2 \\cap [x,y]$.", "So $z \\notin U_2$ as $U_2$ is open.", "Therefore, $z \\in U_1$ as every point of $[x,y]$ is in", "$U_1$ or $U_2$.", "As $U_1$ is open,", "$B(z,\\delta) \\subset U_1$ for a small enough $\\delta > 0$.", "As $z$ is the infimum of the nonempty set $U_2 \\cap [x,y]$,", "there must exist some $w \\in U_2 \\cap [x,y]$", "such that $w \\in [z,z+\\delta) \\subset B(z,\\delta) \\subset U_1$.", "Therefore, $w \\in U_1 \\cap U_2 \\cap [x,y]$.", "So $U_1 \\cap S$ and $U_2 \\cap S$ are not disjoint, and", "$S$ is connected.", "See \\figureref{fig:intervalcon}." ], "refs": [ "prop:intervaldef" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 22, "type": "example", "label": "Lebl-contfunc:22", "categories": [ "applications", "connectedness", "topology", "example", "metric_spaces" ], "title": "Oftentimes a ball$B(x,\\delta)$is connected, but this is not necessarily true in every metric space", "contents": [ "Oftentimes a ball$B(x,\\delta)$is connected, but this is not necessarily true in every metric space. For a simplest example, take a two point space$\\{ a, b\\}$with the discrete metric. Then$B(a,2) = \\{ a , b \\}$, which is not connected as$B(a,1) = \\{ a \\}$and$B(b,1) = \\{ b \\}$are open and disjoint." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 23, "type": "proposition", "label": "Lebl-contfunc:23", "categories": [ "metric_spaces", "topology" ], "title": "Let$(X,d)$be a metric space and$A \\subset X$", "contents": [ "Let$(X,d)$be a metric space and$A \\subset X$. The closure$\\widebar{A}$is closed, and$A \\subset \\widebar{A}$. Furthermore, if$A$is closed, then$\\widebar{A} = A$." ], "refs": [], "proofs": [ { "contents": [ "The closure is an intersection of closed sets, so $\\widebar{A}$ is closed.", "There is at least one closed set containing $A$, namely $X$ itself,", "so $A \\subset \\widebar{A}$.", "If $A$ is closed, then $A$ is a closed set that contains $A$.", "So $\\widebar{A} \\subset A$, and thus $A = \\widebar{A}$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 24, "type": "example", "label": "Lebl-contfunc:24", "categories": [ "topology", "example" ], "title": "The closure of$(0,1)$in$\\R$is$[0,1]$", "contents": [ "The closure of$(0,1)$in$\\R$is$[0,1]$. Proof: If$E$is closed and contains$(0,1)$, then$E$contains$0$and$1$(why?). Thus$[0,1] \\subset E$. But$[0,1]$is also closed. Hence, the closure$\\overline{(0,1)} = [0,1]$." ], "refs": [], "proofs": [ { "contents": [ "If$E$is closed and contains$(0,1)$, then$E$contains$0$and$1$(why?). Thus$[0,1] \\subset E$. But$[0,1]$is also closed. Hence, the closure$\\overline{(0,1)} = [0,1]$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 25, "type": "example", "label": "Lebl-contfunc:25", "categories": [ "metric_spaces", "topology", "example" ], "title": "Be careful to notice what ambient metric space you are working with", "contents": [ "Be careful to notice what ambient metric space you are working with. If$X = (0,\\infty)$, then the closure of$(0,1)$in$(0,\\infty)$is$(0,1]$. Proof: Similarly as above,$(0,1]$is closed in$(0,\\infty)$(why?). Any closed set$E$that contains$(0,1)$must contain 1 (why?). Therefore,$(0,1] \\subset E$, and hence$\\overline{(0,1)} = (0,1]$when working in$(0,\\infty)$." ], "refs": [], "proofs": [ { "contents": [ "Similarly as above,$(0,1]$is closed in$(0,\\infty)$(why?). Any closed set$E$that contains$(0,1)$must contain 1 (why?). Therefore,$(0,1] \\subset E$, and hence$\\overline{(0,1)} = (0,1]$when working in$(0,\\infty)$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 26, "type": "proposition", "label": "Lebl-contfunc:prop:msclosureappr", "categories": [ "metric_spaces", "characterization" ], "title": "Characterization of Metric Spaces via Equivalence", "contents": [ "Let$(X,d)$be a metric space and$A \\subset X$. Then$x \\in \\widebar{A}$if and only if for every$\\delta > 0$,$B(x,\\delta) \\cap A \\not=\\emptyset$." ], "refs": [], "proofs": [ { "contents": [ "Let us prove the two contrapositives.", "Let us show that $x \\notin \\widebar{A}$ if and only if there exists", "a $\\delta > 0$ such that $B(x,\\delta) \\cap A = \\emptyset$.", "First suppose $x \\notin \\widebar{A}$. We know $\\widebar{A}$ is", "closed. Thus there is a $\\delta > 0$ such that", "$B(x,\\delta) \\subset \\widebar{A}^c$. As $A \\subset \\widebar{A}$ we", "see that $B(x,\\delta) \\subset \\widebar{A}^c \\subset A^c$ and hence", "$B(x,\\delta) \\cap A = \\emptyset$.", "On the other hand, suppose there is a $\\delta > 0$ such that", "$B(x,\\delta) \\cap A = \\emptyset$.", "In other words,", "$A \\subset {B(x,\\delta)}^c$.", "As", "${B(x,\\delta)}^c$ is a closed set, as $x \\not \\in {B(x,\\delta)}^c$,", "and as $\\widebar{A}$ is the intersection", "of closed sets containing $A$, we have $x \\notin \\widebar{A}$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 27, "type": "example", "label": "Lebl-contfunc:27", "categories": [ "example" ], "title": "Suppose$A \\coloneqq (0,1]$and$X \\coloneqq \\R$", "contents": [ "Suppose$A \\coloneqq (0,1]$and$X \\coloneqq \\R$. Then$\\widebar{A}=[0,1]$,$A^\\circ = (0,1)$, and$\\partial A = \\{ 0, 1 \\}$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 28, "type": "example", "label": "Lebl-contfunc:28", "categories": [ "example" ], "title": "Consider$X \\coloneqq \\{ a, b \\}$with the discrete metric, and let$A \\coloneqq \\{ a \\}$", "contents": [ "Consider$X \\coloneqq \\{ a, b \\}$with the discrete metric, and let$A \\coloneqq \\{ a \\}$. Then$\\widebar{A} = A^\\circ = A$and$\\partial A = \\emptyset$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 29, "type": "proposition", "label": "Lebl-contfunc:29", "categories": [ "metric_spaces", "topology" ], "title": "Let$(X,d)$be a metric space and$A \\subset X$", "contents": [ "Let$(X,d)$be a metric space and$A \\subset X$. Then$A^\\circ$is open and$\\partial A$is closed." ], "refs": [], "proofs": [ { "contents": [ "Given $x \\in A^\\circ$, there is a $\\delta > 0$ such that $B(x,\\delta)", "\\subset A$. If $z \\in B(x,\\delta)$, then as open balls are open,", "there is an $\\epsilon > 0$ such that $B(z,\\epsilon) \\subset B(x,\\delta)", "\\subset A$. So $z \\in A^\\circ$. Therefore, $B(x,\\delta) \\subset", "A^\\circ$, and so $A^\\circ$ is open.", "As $A^\\circ$ is open, then", "$\\partial A = \\widebar{A} \\setminus A^\\circ = \\widebar{A} \\cap", "{(A^\\circ)}^c$ is closed." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 30, "type": "proposition", "label": "Lebl-contfunc:30", "categories": [ "metric_spaces", "characterization" ], "title": "Characterization of Metric Spaces via Equivalence", "contents": [ "Let$(X,d)$be a metric space and$A \\subset X$. Then$x \\in \\partial A$if and only if for every$\\delta > 0$,$B(x,\\delta) \\cap A$and$B(x,\\delta) \\cap A^c$are both nonempty." ], "refs": [], "proofs": [ { "contents": [ "Suppose $x \\in \\partial A = \\widebar{A} \\setminus A^\\circ$ and", "let $\\delta > 0$ be arbitrary.", "By \\propref{prop:msclosureappr}, $B(x,\\delta)$ contains", "a point of $A$. If $B(x,\\delta)$ contained no points of $A^c$,", "then $x$ would be in $A^\\circ$. Hence $B(x,\\delta)$ contains a point of", "$A^c$ as well.", "Let us prove the other direction by contrapositive.", "Suppose $x \\notin \\partial A$, so $x \\notin \\widebar{A}$ or $x \\in A^\\circ$.", "If $x \\notin \\widebar{A}$, then", "$B(x,\\delta) \\subset \\widebar{A}^c$", "for some $\\delta > 0$ as $\\widebar{A}$ is closed.", "So $B(x,\\delta) \\cap A$ is empty, because $\\widebar{A}^c \\subset", "A^c$.", "If $x \\in A^\\circ$, then", "$B(x,\\delta) \\subset A$ for some $\\delta > 0$,", "so $B(x,\\delta) \\cap A^c$ is empty." ], "refs": [ "prop:msclosureappr" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 31, "type": "corollary", "label": "Lebl-contfunc:31", "categories": [ "consequence", "metric_spaces" ], "title": "Let$(X,d)$be a metric space and$A \\subset X$", "contents": [ "Let$(X,d)$be a metric space and$A \\subset X$. Then$\\partial A = \\widebar{A} \\cap \\widebar{A^c}$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 32, "type": "proposition", "label": "Lebl-contfunc:prop:mslimisunique", "categories": [ "convergence", "metric_spaces" ], "title": "Uniqueness of Metric Spaces", "contents": [ "A convergent sequence in a metric space has a unique limit." ], "refs": [], "proofs": [ { "contents": [ "%NOTE: should be word for word the same as 2.1.6", "Suppose $\\{ x_n \\}_{n=1}^\\infty$ has limits $x$ and $y$.", "Take an arbitrary $\\epsilon > 0$.", "From the definition find an $M_1$ such that for all $n \\geq M_1$,", "$d(x_n,x) < \\nicefrac{\\epsilon}{2}$. Similarly find an $M_2$", "such that for all $n \\geq M_2$, we have", "$d(x_n,y) < \\nicefrac{\\epsilon}{2}$. Now take an $n$ such that", "$n \\geq M_1$ and also $n \\geq M_2$, and estimate", "\\begin{equation*}", "\\begin{split}", "d(y,x)", "& \\leq", "d(y,x_n) + d(x_n,x) \\\\", "& <", "\\frac{\\epsilon}{2} + \\frac{\\epsilon}{2} = \\epsilon .", "\\end{split}", "\\end{equation*}", "As $d(y,x) < \\epsilon$ for all $\\epsilon > 0$, then $d(x,y) = 0$", "and $y=x$. Hence the limit (if it exists) is unique." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 33, "type": "proposition", "label": "Lebl-contfunc:prop:msconvbound", "categories": [ "convergence", "metric_spaces" ], "title": "Boundedness of Metric Spaces", "contents": [ "A convergent sequence in a metric space is bounded." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 34, "type": "proposition", "label": "Lebl-contfunc:prop:msconvifa", "categories": [ "convergence", "metric_spaces" ], "title": "Characterization of Metric Spaces via Equivalence", "contents": [ "A sequence$\\{ x_n \\}_{n=1}^\\infty$in a metric space$(X,d)$converges to$p \\in X$if and only if there exists a sequence$\\{ a_n \\}_{n=1}^\\infty$of real numbers such that\\begin{equation*} d(x_n,p) \\leq a_n \\quad \\text{for all } n \\in \\N, \\qquad \\text{and} \\qquad \\lim_{n\\to\\infty} a_n = 0. \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 35, "type": "proposition", "label": "Lebl-contfunc:prop:mssubseq", "categories": [ "convergence", "metric_spaces" ], "title": "Convergence of Metric Spaces", "contents": [ "Let$\\{ x_n \\}_{n=1}^\\infty$be a sequence in a metric space$(X,d)$. \\begin{enumerate}[(i)] \\item If$\\{ x_n \\}_{n=1}^\\infty$converges to$p \\in X$, then every subsequence$\\{ x_{n_k} \\}_{k=1}^\\infty$converges to$p$. \\item If for some$K \\in \\N$the$K$-tail$\\{ x_n \\}_{n=K+1}^\\infty$converges to$p \\in X$, then$\\{ x_n \\}_{n=1}^\\infty$converges to$p$. \\end{enumerate}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 36, "type": "example", "label": "Lebl-contfunc:36", "categories": [ "convergence", "continuity", "metric_spaces", "example" ], "title": "Characterization of Metric Spaces via Equivalence", "contents": [ "Take$C\\bigl([a,b],\\R\\bigr)$be the set of continuous functions with the metric being the uniform norm. We saw that we obtain a metric space. If we look at the definition of convergence, we notice that it is identical to uniform convergence. That is,$\\{ f_n \\}_{n=1}^\\infty$converges uniformly if and only if it converges in the metric space sense." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 37, "type": "proposition", "label": "Lebl-contfunc:prop:msconveuc", "categories": [ "convergence", "characterization" ], "title": "Characterization of Convergence via Equivalence", "contents": [ "Let$\\{ x_m \\}_{m=1}^\\infty$be a sequence in$\\R^n$, where$x_m = \\bigl(x_{m,1},x_{m,2},\\ldots,x_{m,n}\\bigr) \\in \\R^n$. Then$\\{ x_m \\}_{m=1}^\\infty$converges if and only if$\\{ x_{m,k} \\}_{m=1}^\\infty$converges for every$k=1,2,\\ldots,n$, in which case\\begin{equation*} \\lim_{m\\to\\infty} x_m = \\Bigl( \\lim_{m\\to\\infty} x_{m,1}, \\lim_{m\\to\\infty} x_{m,2}, \\ldots, \\lim_{m\\to\\infty} x_{m,n} \\Bigr) . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Suppose", "$\\{ x_m \\}_{m=1}^\\infty$ converges to", "$y = (y_1,y_2,\\ldots,y_n) \\in \\R^n$.", "Given $\\epsilon > 0$, there exists an $M$ such that for all", "$m \\geq M$, we have", "\\begin{equation*}", "d(y,x_m) < \\epsilon.", "\\end{equation*}", "Fix some $k=1,2,\\ldots,n$. For all $m \\geq M$,", "\\begin{equation*}", "\\bigl\\lvert y_k - x_{m,k} \\bigr\\rvert", "=", "\\sqrt{{\\bigl(y_k - x_{m,k} \\bigr)}^2}", "\\leq", "\\sqrt{\\sum_{\\ell=1}^n {\\bigl(y_\\ell-x_{m,\\ell}\\bigr)}^2}", "= d(y,x_m) < \\epsilon .", "\\end{equation*}", "Hence the sequence $\\{ x_{m,k} \\}_{m=1}^\\infty$ converges to $y_k$.", "For the other direction, suppose", "$\\{ x_{m,k} \\}_{m=1}^\\infty$ converges to $y_k$ for every $k=1,2,\\ldots,n$.", "Given $\\epsilon > 0$, pick an $M$ such that if $m \\geq M$, then", "$\\bigl\\lvert y_k-x_{m,k} \\bigr\\rvert < \\nicefrac{\\epsilon}{\\sqrt{n}}$ for all", "$k=1,2,\\ldots,n$. Then", "\\begin{equation*}", "d(y,x_m)", "=", "\\sqrt{\\sum_{k=1}^n {\\bigl(y_k-x_{m,k}\\bigr)}^2}", "<", "\\sqrt{\\sum_{k=1}^n {\\left(\\frac{\\epsilon}{\\sqrt{n}}\\right)}^2}", "=", "\\sqrt{\\sum_{k=1}^n \\frac{{\\epsilon^2}}{n}}", "= \\epsilon .", "\\end{equation*}", "That is, the sequence $\\{ x_m \\}_{m=1}^\\infty$ converges to", "$y = (y_1,y_2,\\ldots,y_n) \\in \\R^n$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 38, "type": "example", "label": "Lebl-contfunc:38", "categories": [ "convergence", "metric_spaces", "characterization", "example" ], "title": "Characterization of Metric Spaces via Equivalence", "contents": [ "As we said, the set$\\C$of complex numbers$z = x+iy$is considered as the metric space$\\R^2$. The proposition says that the sequence$\\{ z_n \\}_{n=1}^\\infty = \\{ x_n + iy_n \\}_{n=1}^\\infty$converges to$z = x+iy$if and only if$\\{ x_n \\}_{n=1}^\\infty$converges to$x$and$\\{ y_n \\}_{n=1}^\\infty$converges to$y$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 39, "type": "proposition", "label": "Lebl-contfunc:prop:msconvtopo", "categories": [ "characterization", "convergence", "metric_spaces", "topology" ], "title": "Characterization of Metric Spaces via Equivalence", "contents": [ "Let$(X,d)$be a metric space and$\\{x_n\\}_{n=1}^\\infty$a sequence in$X$. Then$\\{ x_n \\}_{n=1}^\\infty$converges to$p \\in X$if and only if for every open neighborhood$U$of$p$, there exists an$M \\in \\N$such that for all$n \\geq M$, we have$x_n \\in U$." ], "refs": [], "proofs": [ { "contents": [ "Suppose $\\{ x_n \\}_{n=1}^\\infty$ converges to $p$. Let $U$ be an open neighborhood", "of $p$, then there exists an $\\epsilon > 0$ such that $B(p,\\epsilon) \\subset", "U$. As the sequence converges, find an $M \\in \\N$ such that for all $n \\geq", "M$, we have $d(p,x_n) < \\epsilon$, or in other words $x_n \\in B(p,\\epsilon)", "\\subset U$.", "Let us prove the other direction. Given $\\epsilon > 0$, let $U \\coloneqq", "B(p,\\epsilon)$ be the neighborhood of $p$. Then there is an $M \\in \\N$", "such that for $n \\geq M$, we have $x_n \\in U = B(p,\\epsilon)$, or in other", "words, $d(p,x_n) < \\epsilon$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 40, "type": "proposition", "label": "Lebl-contfunc:prop:msclosedlim", "categories": [ "convergence", "metric_spaces", "topology" ], "title": "Convergence of Metric Spaces", "contents": [ "Let$(X,d)$be a metric space,$E \\subset X$a closed set, and$\\{ x_n \\}_{n=1}^\\infty$a sequence in$E$that converges to some$p \\in X$. Then$p \\in E$." ], "refs": [], "proofs": [ { "contents": [ "Let us prove the contrapositive.", "Suppose $\\{ x_n \\}_{n=1}^\\infty$ is a sequence in $X$ that converges to $p \\in E^c$.", "As $E^c$ is open, \\propref{prop:msconvtopo} says that there is", "an $M$ such that for all $n \\geq M$,", "$x_n \\in E^c$. So $\\{ x_n \\}_{n=1}^\\infty$ is not a sequence in $E$." ], "refs": [ "prop:msconvtopo" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 41, "type": "proposition", "label": "Lebl-contfunc:prop:msclosureapprseq", "categories": [ "convergence", "metric_spaces", "characterization" ], "title": "Characterization of Metric Spaces via Equivalence", "contents": [ "Let$(X,d)$be a metric space and$A \\subset X$. Then$p \\in \\widebar{A}$if and only if there exists a sequence$\\{ x_n \\}_{n=1}^\\infty$of elements in$A$such that$\\lim\\limits_{n\\to\\infty} x_n = p$." ], "refs": [], "proofs": [ { "contents": [ "Let $p \\in \\widebar{A}$. For every $n \\in \\N$,", "\\propref{prop:msclosureappr} guarantees", "a point $x_n \\in B(p,\\nicefrac{1}{n}) \\cap A$.", "As $d(p,x_n) < \\nicefrac{1}{n}$, we have $\\lim_{n\\to\\infty} x_n = p$.", "For the other direction, see \\exerciseref{exercise:reverseclosedseq}." ], "refs": [ "prop:msclosureappr" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 42, "type": "proposition", "label": "Lebl-contfunc:42", "categories": [ "convergence", "metric_spaces", "completeness" ], "title": "A convergent sequence in a metric space is Cauchy", "contents": [ "A convergent sequence in a metric space is Cauchy." ], "refs": [], "proofs": [ { "contents": [ "Suppose $\\{ x_n \\}_{n=1}^\\infty$ converges to $p$.", "Given $\\epsilon > 0$, there is an $M$ such that for all $n \\geq M$,", "we have $d(p,x_n) < \\nicefrac{\\epsilon}{2}$. Hence", "for all $n,k \\geq M$, we have", "$d(x_n,x_k) \\leq d(x_n,x) + d(x,x_k) < \\nicefrac{\\epsilon}{2} +", "\\nicefrac{\\epsilon}{2} = \\epsilon$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 43, "type": "proposition", "label": "Lebl-contfunc:43", "categories": [ "metric_spaces", "completeness" ], "title": "The space$\\R^n$with the standard metric is a complete metric space", "contents": [ "The space$\\R^n$with the standard metric is a complete metric space." ], "refs": [], "proofs": [ { "contents": [ "Let $\\{ x_m \\}_{m=1}^\\infty$ be a Cauchy sequence", "in $\\R^n$, where $x_m = \\bigl(x_{m,1},x_{m,2},\\ldots,x_{m,n}\\bigr) \\in \\R^n$.", "As the sequence is Cauchy, given $\\epsilon > 0$, there exists an $M$ such that for all", "$i,j \\geq M$,", "\\begin{equation*}", "d(x_i,x_j) < \\epsilon.", "\\end{equation*}", "Fix some $k=1,2,\\ldots,n$. For $i,j \\geq M$,", "\\begin{equation*}", "\\bigl\\lvert x_{i,k} - x_{j,k} \\bigr\\rvert", "=", "\\sqrt{{\\bigl(x_{i,k} - x_{j,k}\\bigr)}^2}", "\\leq", "\\sqrt{\\sum_{\\ell=1}^n {\\bigl(x_{i,\\ell}-x_{j,\\ell}\\bigr)}^2}", "= d(x_i,x_j) < \\epsilon .", "\\end{equation*}", "Hence the sequence $\\{ x_{m,k} \\}_{m=1}^\\infty$ is Cauchy. As $\\R$ is", "complete the sequence converges; there exists a $y_k \\in \\R$ such that", "$y_k = \\lim_{m\\to\\infty} x_{m,k}$.", "Write $y = (y_1,y_2,\\ldots,y_n) \\in \\R^n$.", "By \\propref{prop:msconveuc}, $\\{ x_m \\}_{m=1}^\\infty$ converges", "to $y \\in \\R^n$, and hence $\\R^n$ is complete." ], "refs": [ "prop:msconveuc" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 44, "type": "proposition", "label": "Lebl-contfunc:prop:CabRcomplete", "categories": [ "continuity", "metric_spaces", "completeness" ], "title": "Continuity of Metric Spaces", "contents": [ "The space of continuous functions$C\\bigl([a,b],\\R\\bigr)$with the uniform norm as metric is a complete metric space." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 45, "type": "proposition", "label": "Lebl-contfunc:prop:closedcomplete", "categories": [ "metric_spaces", "topology", "completeness" ], "title": "Suppose$(X,d)$is a complete metric space and$E \\subset X$is closed", "contents": [ "Suppose$(X,d)$is a complete metric space and$E \\subset X$is closed. Then$E$is a complete metric space with the subspace metric." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 46, "type": "example", "label": "Lebl-contfunc:46", "categories": [ "compactness", "metric_spaces", "topology", "example" ], "title": "Boundedness of Metric Spaces", "contents": [ "Let$\\R$be the metric space with the standard metric.", "The set$\\R$is not compact. Proof: For$j \\in \\N$, let$U_j \\coloneqq (-j,j)$. Any$x \\in \\R$is in some$U_j$(by the \\hyperref[thm:arch:i]{Archimedean property}), so we have an open cover. Suppose we have a finite subcover$\\R \\subset U_{j_1} \\cup U_{j_2} \\cup \\cdots \\cup U_{j_m}$, and suppose$j_1 < j_2 < \\cdots < j_m$. Then$\\R \\subset U_{j_m}$, but that is a contradiction as$j_m \\in \\R$on one hand and$j_m \\notin U_{j_m} = (-j_m,j_m)$on the other.", "The set$(0,1) \\subset \\R$is also not compact. Proof: Take the sets$U_{j} \\coloneqq (\\nicefrac{1}{j},1-\\nicefrac{1}{j})$for$j=3,4,5,\\ldots$. As above$(0,1) = \\bigcup_{j=3}^\\infty U_j$. And similarly as above, if there exists a finite subcover, then there is one$U_j$such that$(0,1) \\subset U_j$, which again leads to a contradiction.", "The set$\\{ 0 \\} \\subset \\R$is compact. Proof: Given an open cover$\\{ U_{\\lambda} \\}_{\\lambda \\in I}$, there must exist a$\\lambda_0$such that$0 \\in U_{\\lambda_0}$as it is a cover. But then$U_{\\lambda_0}$gives a finite subcover.", "We will prove below that$[0,1]$, and in fact every closed and bounded interval$[a,b]$, is compact." ], "refs": [ "thm:arch:i" ], "proofs": [ { "contents": [ "For$j \\in \\N$, let$U_j \\coloneqq (-j,j)$. Any$x \\in \\R$is in some$U_j$(by the \\hyperref[thm:arch:i]{Archimedean property}), so we have an open cover. Suppose we have a finite subcover$\\R \\subset U_{j_1} \\cup U_{j_2} \\cup \\cdots \\cup U_{j_m}$, and suppose$j_1 < j_2 < \\cdots < j_m$. Then$\\R \\subset U_{j_m}$, but that is a contradiction as$j_m \\in \\R$on one hand and$j_m \\notin U_{j_m} = (-j_m,j_m)$on the other. The set$(0,1) \\subset \\R$is also not compact. Proof: Take the sets$U_{j} \\coloneqq (\\nicefrac{1}{j},1-\\nicefrac{1}{j})$for$j=3,4,5,\\ldots$. As above$(0,1) = \\bigcup_{j=3}^\\infty U_j$. And similarly as above, if there exists a finite subcover, then there is one$U_j$such that$(0,1) \\subset U_j$, which again leads to a contradiction. The set$\\{ 0 \\} \\subset \\R$is compact. Proof: Given an open cover$\\{ U_{\\lambda} \\}_{\\lambda \\in I}$, there must exist a$\\lambda_0$such that$0 \\in U_{\\lambda_0}$as it is a cover. But then$U_{\\lambda_0}$gives a finite subcover. We will prove below that$[0,1]$, and in fact every closed and bounded interval$[a,b]$, is compact." ], "refs": [ "thm:arch:i" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 47, "type": "proposition", "label": "Lebl-contfunc:47", "categories": [ "compactness", "metric_spaces", "topology" ], "title": "Boundedness of Metric Spaces", "contents": [ "Let$(X,d)$be a metric space. If$K \\subset X$is compact, then$K$is closed and bounded." ], "refs": [], "proofs": [ { "contents": [ "First, we prove that a compact set is bounded.", "Fix $p \\in X$. We have the open cover", "\\begin{equation*}", "K \\subset \\bigcup_{n=1}^\\infty B(p,n) = X .", "\\end{equation*}", "If $K$ is compact, then there exists some set of indices", "$n_1 < n_2 < \\ldots < n_m$ such that", "\\begin{equation*}", "K \\subset \\bigcup_{j=1}^m B(p,n_j) = B(p,n_m) .", "\\end{equation*}", "As $K$ is contained in a ball, $K$ is bounded.", "See the left-hand side of \\figureref{fig:compactbndclosed}.", "Next, we show a set that is not closed is not compact. Suppose", "$\\widebar{K} \\not= K$, that is, there is a point $x \\in \\widebar{K}", "\\setminus K$.", "If $y \\not= x$, then", "$y \\notin C(x,\\nicefrac{1}{n})$", "for $n \\in \\N$", "such that $\\nicefrac{1}{n} < d(x,y)$.", "Furthermore, $x \\notin K$, so", "\\begin{equation*}", "K \\subset \\bigcup_{n=1}^\\infty {C(x,\\nicefrac{1}{n})}^c .", "\\end{equation*}", "A closed ball is closed, so its complement ${C(x,\\nicefrac{1}{n})}^c$ is open, and", "we have an open cover.", "If we take any", "finite collection of indices $n_1 < n_2 < \\ldots < n_m$, then", "\\begin{equation*}", "\\bigcup_{j=1}^m {C(x,\\nicefrac{1}{n_j})}^c", "=", "{C(x,\\nicefrac{1}{n_m})}^c", "\\end{equation*}", "As $x$ is in the closure of $K$,", "then", "$C(x,\\nicefrac{1}{n_m}) \\cap K \\not= \\emptyset$. So there is no", "finite subcover and $K$ is not compact.", "See the right-hand side of \\figureref{fig:compactbndclosed}." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 48, "type": "lemma", "label": "Lebl-contfunc:ms:lebesgue", "categories": [ "convergence", "auxiliary result", "topology", "metric_spaces", "named theorem" ], "title": "Lebesgue Covering Lemma", "contents": [ "\\index{Lebesgue covering lemma} Let$(X,d)$be a metric space and$K \\subset X$. Suppose every sequence in$K$has a subsequence convergent in$K$. Given an open cover$\\{ U_\\lambda \\}_{\\lambda \\in I}$of$K$, there exists a$\\delta > 0$such that for every$x \\in K$, there exists a$\\lambda \\in I$with$B(x,\\delta) \\subset U_\\lambda$." ], "refs": [ "named:Lebesgue" ], "proofs": [ { "contents": [ "We prove the lemma by contrapositive.", "If the conclusion is not true, then", "there is", "an open cover $\\{ U_\\lambda \\}_{\\lambda \\in I}$ of $K$ with", "the following property.", "For every $n \\in \\N$, there exists an $x_n \\in K$ such that", "$B(x_n,\\nicefrac{1}{n})$ is not a subset of any $U_\\lambda$.", "Take any $x \\in K$. There is", "a $\\lambda \\in I$ such that $x \\in U_\\lambda$. As $U_\\lambda$ is open,", "there is an $\\epsilon > 0$", "such that $B(x,\\epsilon) \\subset U_\\lambda$. Take $M$ such that", "$\\nicefrac{1}{M} < \\nicefrac{\\epsilon}{2}$. If $y \\in", "B(x,\\nicefrac{\\epsilon}{2})$ and $n \\geq M$, then", "\\begin{equation*}", "B(y,\\nicefrac{1}{n}) \\subset", "B(y,\\nicefrac{1}{M}) \\subset", "B(y,\\nicefrac{\\epsilon}{2}) \\subset B(x,\\epsilon)", "\\subset U_\\lambda ,", "\\end{equation*}", "where", "$B(y,\\nicefrac{\\epsilon}{2}) \\subset B(x,\\epsilon)$", "follows by triangle inequality.", "See \\figureref{fig:lebesguedelta}.", "Thus $y \\not= x_n$.", "In other words, for all $n \\geq M$, $x_n \\notin B(x,\\nicefrac{\\epsilon}{2})$.", "The sequence cannot have a subsequence converging to $x$. As $x \\in K$ was", "arbitrary we are done." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 49, "type": "theorem", "label": "Lebl-contfunc:thm:mscompactisseqcpt", "categories": [ "convergence", "compactness", "metric_spaces" ], "title": "Characterization of Metric Spaces via Equivalence", "contents": [ "Let$(X,d)$be a metric space. Then$K \\subset X$is compact if and only if every sequence in$K$has a subsequence converging to a point in$K$." ], "refs": [], "proofs": [ { "contents": [ "Claim: \\emph{Let $K \\subset X$ be a subset of $X$ and", "$\\{ x_n \\}_{n=1}^\\infty$ a sequence in $K$. Suppose that for each $x \\in K$,", "there is a ball $B(x,\\alpha_x)$ for some $\\alpha_x > 0$ such that", "$x_n \\in B(x,\\alpha_x)$ for only finitely many $n \\in \\N$.", "Then $K$ is not compact.}", "Proof of the claim:", "Notice", "\\begin{equation*}", "K \\subset \\bigcup_{x \\in K} B(x,\\alpha_x) .", "\\end{equation*}", "Any finite collection of these balls contains at most finitely many", "elements of $\\{ x_n \\}_{n=1}^\\infty$,", "and so there must be an $x_n \\in K$", "not in their union. Hence, $K$ is not compact and the claim is", "proved.", "So suppose that $K$ is compact and $\\{ x_n \\}_{n=1}^\\infty$ is a sequence in $K$.", "Then there exists an $x \\in K$ such that", "for all $\\delta > 0$,", "$B(x,\\delta)$ contains $x_n$ for infinitely many $n \\in \\N$.", "We define the subsequence inductively.", "The ball $B(x,1)$ contains some $x_k$, so let $n_1 \\coloneqq k$.", "Suppose $n_{j-1}$ is defined.", "There must exist a $k > n_{j-1}$", "such that $x_k \\in B(x,\\nicefrac{1}{j})$. Define", "$n_j \\coloneqq k$.", "We now posses a subsequence $\\{ x_{n_j} \\}_{j=1}^\\infty$.", "Since", "$d(x,x_{n_j}) < \\nicefrac{1}{j}$, \\propref{prop:msconvifa} says", "$\\lim_{j\\to\\infty} x_{n_j} = x$.", "For the other direction, suppose every sequence in~$K$", "has a", "subsequence converging in~$K$.", "Take", "an open cover $\\{ U_\\lambda \\}_{\\lambda \\in I}$ of $K$.", "Using the Lebesgue covering lemma above, find a $\\delta > 0$", "such that for every $x \\in K$, there is a $\\lambda \\in I$ with", "$B(x,\\delta) \\subset U_\\lambda$.", "Pick $x_1 \\in K$ and find $\\lambda_1 \\in I$ such that $B(x_1,\\delta) \\subset", "U_{\\lambda_1}$.", "If $K \\subset U_{\\lambda_1}$, we stop as we have found a", "finite subcover.", "Otherwise, there must be", "a point $x_2 \\in K \\setminus U_{\\lambda_1}$.", "Note that $d(x_2,x_1) \\geq \\delta$.", "There must exist some $\\lambda_2 \\in I$ such that", "$B(x_2,\\delta) \\subset U_{\\lambda_2}$.", "We work inductively. Suppose $\\lambda_{n-1}$ is defined.", "Either", "$U_{\\lambda_1} \\cup", "U_{\\lambda_2} \\cup \\cdots \\cup", "U_{\\lambda_{n-1}}$ is a finite cover of $K$, in which case we", "stop, or", "there must be", "a point $x_n \\in K \\setminus \\bigl( U_{\\lambda_1} \\cup", "U_{\\lambda_2} \\cup \\cdots \\cup", "U_{\\lambda_{n-1}}\\bigr)$.", "Note that $d(x_n,x_j) \\geq \\delta$ for all $j = 1,2,\\ldots,n-1$.", "Next, there must be some $\\lambda_n \\in I$", "such that $B(x_n,\\delta) \\subset U_{\\lambda_n}$.", "See \\figureref{fig:seqcompactiscompact}.", "\\begin{myfigureht}", "\\subimport*{figures/}{seqcompactiscompact.pdf_t}", "\\caption{Covering $K$ by $U_{\\lambda}$. The points", "$x_1,x_2,x_3,x_4$,", "the three sets", "$U_{\\lambda_1}$,", "$U_{\\lambda_2}$,", "$U_{\\lambda_2}$,", "and", "the first three balls", "of radius $\\delta$ are drawn.\\label{fig:seqcompactiscompact}}", "\\end{myfigureht}", "Either at some point we obtain a finite subcover of $K$,", "or we obtain an", "infinite", "sequence $\\{ x_n \\}_{n=1}^\\infty$ as above.", "For contradiction, suppose that", "there is no finite subcover and we have the sequence $\\{ x_n \\}_{n=1}^\\infty$.", "For all $n$ and $k$, $n \\not= k$,", "we have $d(x_n,x_k) \\geq \\delta$.", "So no subsequence of $\\{ x_n \\}_{n=1}^\\infty$ is Cauchy.", "Hence, no subsequence of $\\{ x_n \\}_{n=1}^\\infty$ is convergent,", "which is a contradiction." ], "refs": [ "named:Lebesgue", "prop:msconvifa" ], "ref_ids": [ 48 ] } ], "ref_ids": [] }, { "id": 50, "type": "example", "label": "Lebl-contfunc:example:intervalcompact", "categories": [ "convergence", "compactness", "topology", "example" ], "title": "Boundedness of Compactness", "contents": [ "\\thmref{thm:bwseq}, the Bolzano--Weierstrass theorem for sequences of real numbers, says that every bounded sequence in$\\R$has a convergent subsequence. Therefore, every sequence in a closed interval$[a,b] \\subset \\R$has a convergent subsequence. The limit is also in$[a,b]$as limits preserve non-strict inequalities. Hence a closed bounded interval$[a,b] \\subset \\R$is (sequentially) compact." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 51, "type": "proposition", "label": "Lebl-contfunc:prop:closedsubsetcompact", "categories": [ "compactness", "metric_spaces", "topology" ], "title": "Let$(X,d)$be a metric space and let$K \\subset X$be compact", "contents": [ "Let$(X,d)$be a metric space and let$K \\subset X$be compact. If$E \\subset K$is a closed set, then$E$is compact." ], "refs": [], "proofs": [ { "contents": [ "Let $\\{ x_n \\}_{n=1}^\\infty$ be a sequence in $E$. It is also a sequence in $K$.", "Therefore, it has a convergent subsequence $\\{ x_{n_j} \\}_{j=1}^\\infty$ that converges to", "some $x \\in K$. As $E$ is closed the limit of a sequence in $E$ is also in $E$", "and so $x \\in E$. Thus $E$ must be compact." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 52, "type": "theorem", "label": "Lebl-contfunc:52", "categories": [ "compactness", "topology" ], "title": "Boundedness of Compactness", "contents": [ "% \\index{Heine--Borel theorem} \\label{thm:msbw} A closed bounded subset$K \\subset \\R^n$is compact." ], "refs": [], "proofs": [ { "contents": [ "For $\\R = \\R^1$, suppose $K \\subset \\R$ is closed and bounded.", "Then $K \\subset [a,b]$ for some closed and bounded interval,", "which is compact by \\exampleref{example:intervalcompact}.", "As $K$ is a closed subset of a compact set,", "it is compact by \\propref{prop:closedsubsetcompact}.", "We carry out the proof for $n=2$ and leave arbitrary $n$ as an exercise.", "As $K \\subset \\R^2$ is bounded, there exists a set", "$B=[a,b]\\times[c,d] \\subset \\R^2$ such that $K \\subset B$. We will show", "that $B$ is compact. Then $K$, being a closed subset of a compact $B$, is", "also compact.", "Let $\\bigl\\{ (x_k,y_k) \\bigr\\}_{k=1}^\\infty$ be a sequence in $B$. That is,", "$a \\leq x_k \\leq b$ and", "$c \\leq y_k \\leq d$ for all $k$. A bounded sequence of real numbers", "has a convergent", "subsequence so there is a subsequence $\\{ x_{k_j} \\}_{j=1}^\\infty$", "that is convergent. The subsequence", "$\\{ y_{k_j} \\}_{j=1}^\\infty$ is also a bounded sequence so there exists", "a subsequence", "$\\{ y_{k_{j_i}} \\}_{i=1}^\\infty$ that is convergent. A subsequence of a", "convergent sequence is still convergent, so", "$\\{ x_{k_{j_i}} \\}_{i=1}^\\infty$ is convergent.", "Let", "\\begin{equation*}", "x \\coloneqq \\lim_{i\\to\\infty} x_{k_{j_i}}", "\\qquad \\text{and} \\qquad", "y \\coloneqq \\lim_{i\\to\\infty} y_{k_{j_i}} .", "\\end{equation*}", "By \\propref{prop:msconveuc},", "$\\bigl\\{ (x_{k_{j_i}},y_{k_{j_i}}) \\bigr\\}_{i=1}^\\infty$ converges to $(x,y)$.", "Furthermore, as $a \\leq x_k \\leq b$ and", "$c \\leq y_k \\leq d$ for all $k$, we know that $(x,y) \\in B$." ], "refs": [ "prop:closedsubsetcompact", "prop:msconveuc" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 53, "type": "example", "label": "Lebl-contfunc:53", "categories": [ "applications", "compactness", "topology", "characterization", "example", "metric_spaces", "completeness", "named theorem" ], "title": "Lebesgue Covering Lemma", "contents": [ "The discrete metric provides interesting counterexamples again. Let$(X,d)$be a metric space with the discrete metric, that is,$d(x,y) = 1$if$x \\not= y$. Suppose$X$is an infinite set. Then \\begin{enumerate}[(i)] \\item$(X,d)$is a complete metric space. \\item Any subset$K \\subset X$is closed and bounded. \\item A subset$K \\subset X$is compact if and only if it is a finite set. \\item The conclusion of the Lebesgue covering lemma is always satisfied, e.g. with$\\delta = \\nicefrac{1}{2}$, even for noncompact$K \\subset X$. \\end{enumerate} The proofs of the statements above are either trivial or are relegated to the exercises below." ], "refs": [ "named:Lebesgue" ], "proofs": [], "ref_ids": [ 48 ] }, { "id": 54, "type": "proposition", "label": "Lebl-contfunc:prop:contiscont", "categories": [ "convergence", "continuity", "metric_spaces", "characterization" ], "title": "Characterization of Metric Spaces via Equivalence", "contents": [ "Let$(X,d_X)$and$(Y,d_Y)$be metric spaces. Then$f \\colon X \\to Y$is continuous at$c \\in X$if and only if for every sequence$\\{ x_n \\}_{n=1}^\\infty$in$X$converging to$c$, the sequence$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$converges to$f(c)$." ], "refs": [], "proofs": [ { "contents": [ "Suppose $f$ is continuous at $c$. Let $\\{ x_n \\}_{n=1}^\\infty$ be a", "sequence in $X$ converging to $c$. Given $\\epsilon > 0$,", "there is a $\\delta > 0$ such that $d_X(x,c) < \\delta$ implies", "$d_Y\\bigl(f(x),f(c)\\bigr) < \\epsilon$. So take $M$ such that", "for all $n \\geq M$, we have $d_X(x_n,c) < \\delta$, then", "$d_Y\\bigl(f(x_n),f(c)\\bigr) < \\epsilon$. Hence $\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$", "converges to $f(c)$.", "On the other hand, suppose $f$ is not continuous at $c$.", "Then there exists an $\\epsilon > 0$,", "such that for every $n \\in \\N$ there exists an $x_n \\in X$,", "with", "$d_X(x_n,c) < \\nicefrac{1}{n}$ such that $d_Y\\bigl(f(x_n),f(c)\\bigr) \\geq", "\\epsilon$. Then $\\{ x_n \\}_{n=1}^\\infty$ converges to $c$, but", "$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$", "does not converge to $f(c)$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 55, "type": "example", "label": "Lebl-contfunc:55", "categories": [ "convergence", "continuity", "example" ], "title": "Continuity of Continuity", "contents": [ "Suppose$f \\colon \\R^2 \\to \\R$is a polynomial. That is,", "\\begin{equation*}\nf(x,y) =\n\\sum_{j=0}^d\n\\sum_{k=0}^{d-j}\na_{jk}\\,x^jy^k =\na_{0\\,0} + a_{1\\,0} \\, x +\na_{0\\,1} \\, y+ \na_{2\\,0} \\, x^2+ \na_{1\\,1} \\, xy+ \na_{0\\,2} \\, y^2+ \\cdots +\na_{0\\,d} \\, y^d ,\n\\end{equation*}", "for some$d \\in \\N$(the degree) and$a_{jk} \\in \\R$. We claim$f$is continuous. Let$\\bigl\\{ (x_n,y_n) \\bigr\\}_{n=1}^\\infty$be a sequence in$\\R^2$that converges to$(x,y) \\in \\R^2$. We proved that this means$\\lim_{n\\to\\infty} x_n = x$and$\\lim_{n\\to\\infty} y_n = y$. By \\propref{prop:contalg},", "\\begin{equation*}\n\\lim_{n\\to\\infty}\nf(x_n,y_n) =\n\\lim_{n\\to\\infty}\n\\sum_{j=0}^d\n\\sum_{k=0}^{d-j}\na_{jk} \\, x_n^jy_n^k \n=\n\\sum_{j=0}^d\n\\sum_{k=0}^{d-j}\na_{jk} \\, x^jy^k\n=\nf(x,y) .\n\\end{equation*}", "So$f$is continuous at$(x,y)$, and as$(x,y)$was arbitrary$f$is continuous everywhere. Similarly, a polynomial in$n$variables is continuous." ], "refs": [ "prop:contalg" ], "proofs": [], "ref_ids": [] }, { "id": 56, "type": "example", "label": "Lebl-contfunc:56", "categories": [ "convergence", "characterization", "example", "metric_spaces", "continuity" ], "title": "Continuity of Metric Spaces", "contents": [ "Let$X$be a metric space and$f \\colon X \\to \\C$a complex-valued function. Write$f(p) = g(p) + i \\, h(p)$, where$g \\colon X \\to \\R$and$h \\colon X \\to \\R$are the real and imaginary parts of$f$. Then$f$is continuous at$c \\in X$if and only if its real and imaginary parts are continuous at~$c$. This fact follows because$\\bigl\\{ f(p_n) = g(p_n) + i \\, h(p_n) \\bigr\\}_{n=1}^\\infty$converges to$f(p) = g(p) + i \\, h(p)$if and only if$\\bigl\\{ g(p_n) \\bigr\\}_{n=1}^\\infty$converges to$g(p)$and$\\bigl\\{ h(p_n) \\bigr\\}_{n=1}^\\infty$converges to$h(p)$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 57, "type": "lemma", "label": "Lebl-contfunc:lemma:continuouscompact", "categories": [ "auxiliary result", "compactness", "metric_spaces", "continuity" ], "title": "Continuity of Metric Spaces", "contents": [ "Let$(X,d_X)$and$(Y,d_Y)$be metric spaces and$f \\colon X \\to Y$a continuous function. If$K \\subset X$is a compact set, then$f(K)$is a compact set." ], "refs": [], "proofs": [ { "contents": [ "A sequence in $f(K)$ can be written as", "$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$, where", "$\\{ x_n \\}_{n=1}^\\infty$ is a sequence in $K$. The set $K$ is compact and", "therefore there is a subsequence", "$\\{ x_{n_j} \\}_{j=1}^\\infty$ that converges to some $x \\in K$.", "By continuity,", "\\begin{equation*}", "\\lim_{j\\to\\infty} f(x_{n_j}) = f(x) \\in f(K) .", "\\end{equation*}", "So every sequence in $f(K)$ has a subsequence convergent to", "a point in $f(K)$, and $f(K)$ is compact by \\thmref{thm:mscompactisseqcpt}." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 58, "type": "theorem", "label": "Lebl-contfunc:58", "categories": [ "compactness", "metric_spaces", "continuity" ], "title": "Continuity of Metric Spaces", "contents": [ "Let$(X,d)$be a nonempty compact metric space and let$f \\colon X \\to \\R$be continuous. Then$f$is bounded and in fact$f$achieves an absolute minimum and an absolute maximum on$X$." ], "refs": [], "proofs": [ { "contents": [ "As $X$ is compact and $f$ is continuous,", "$f(X) \\subset \\R$ is compact. Hence $f(X)$ is closed", "and bounded. In particular,", "$\\sup f(X) \\in f(X)$ and", "$\\inf f(X) \\in f(X)$, because both the sup and the inf", "can be achieved by sequences in $f(X)$ and $f(X)$ is closed.", "Therefore, there is some $x \\in X$ such that $f(x) = \\sup f(X)$", "and some $y \\in X$ such that $f(y) = \\inf f(X)$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 59, "type": "lemma", "label": "Lebl-contfunc:lemma:mstopocontloc", "categories": [ "auxiliary result", "characterization", "topology", "metric_spaces", "continuity" ], "title": "Characterization of Metric Spaces via Equivalence", "contents": [ "Let$(X,d_X)$and$(Y,d_Y)$be metric spaces. A function$f \\colon X \\to Y$is continuous at$c \\in X$if and only if for every open neighborhood$U$of$f(c)$in$Y$, the set$f^{-1}(U)$contains an open neighborhood of$c$in$X$. See \\figureref{fig:mscontfuncpt}." ], "refs": [], "proofs": [ { "contents": [ "First suppose that $f$ is continuous at $c$.", "Let $U$ be an open neighborhood of $f(c)$", "in $Y$, then $B_Y\\bigl(f(c),\\epsilon\\bigr) \\subset U$ for some $\\epsilon >", "0$. By continuity of $f$, there exists a $\\delta > 0$", "such that whenever $x$ is such that $d_X(x,c) < \\delta$, then", "$d_Y\\bigl(f(x),f(c)\\bigr) < \\epsilon$. In other words,", "\\begin{equation*}", "B_X(c,\\delta) \\subset f^{-1}\\bigl(B_Y\\bigl(f(c),\\epsilon\\bigr)\\bigr) \\subset", "f^{-1}(U) ,", "\\end{equation*}", "and $B_X(c,\\delta)$ is an open neighborhood of $c$.", "For the other direction,", "let $\\epsilon > 0$ be given. If", "$f^{-1}\\bigl(B_Y\\bigl(f(c),\\epsilon\\bigr)\\bigr)$ contains an open", "neighborhood $W$ of $c$, it contains a ball. That is, there is some $\\delta > 0$", "such that", "\\begin{equation*}", "B_X(c,\\delta) \\subset W \\subset f^{-1}\\bigl(B_Y\\bigl(f(c),\\epsilon\\bigr)\\bigr) .", "\\end{equation*}", "That means precisely that if $d_X(x,c) < \\delta$,", "then $d_Y\\bigl(f(x),f(c)\\bigr) < \\epsilon$.", "So $f$ is continuous at~$c$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 60, "type": "theorem", "label": "Lebl-contfunc:thm:mstopocont", "categories": [ "characterization", "continuity", "metric_spaces", "topology" ], "title": "Characterization of Metric Spaces via Equivalence", "contents": [ "Let$(X,d_X)$and$(Y,d_Y)$be metric spaces. A function$f \\colon X \\to Y$is continuous if and only if for every open$U \\subset Y$,$f^{-1}(U)$is open in$X$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 61, "type": "example", "label": "Lebl-contfunc:61", "categories": [ "continuity", "topology", "example" ], "title": "Continuity of Continuity", "contents": [ "Let$f \\colon X \\to Y$be a continuous function. \\thmref{thm:mstopocont} tells us that if$E \\subset Y$is closed, then$f^{-1}(E) = X \\setminus f^{-1}(E^c)$is also closed. Therefore, if we have a continuous function$f \\colon X \\to \\R$, then the \\emph{\\myindex{zero set}} of$f$, that is,$f^{-1}(0) = \\bigl\\{ x \\in X : f(x) = 0 \\bigr\\}$, is closed. We have just proved the most basic result in \\emph{\\myindex{algebraic geometry}}, the study of zero sets of polynomials: The zero set of a polynomial is closed.", "Similarly, the set where$f$is nonnegative,$f^{-1}\\bigl( [0,\\infty) \\bigr) = \\bigl\\{ x \\in X : f(x) \\geq 0 \\bigr\\}$, is closed. On the other hand, the set where$f$is positive,$f^{-1}\\bigl( (0,\\infty) \\bigr) = \\bigl\\{ x \\in X : f(x) > 0 \\bigr\\}$, is open." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 62, "type": "theorem", "label": "Lebl-contfunc:thm:Xcompactfunifcont", "categories": [ "compactness", "metric_spaces", "continuity" ], "title": "Continuity of Metric Spaces", "contents": [ "Let$(X,d_X)$and$(Y,d_Y)$be metric spaces. Suppose$f \\colon X \\to Y$is continuous and$X$is compact. Then$f$is uniformly continuous." ], "refs": [], "proofs": [ { "contents": [ "Let $\\epsilon > 0$ be given. For each $c \\in X$, pick $\\delta_c > 0$ such that", "$d_Y\\bigl(f(x),f(c)\\bigr) < \\nicefrac{\\epsilon}{2}$", "whenever", "$x \\in B(c,\\delta_c)$.", "%$d_X(x,c) < \\delta_c$.", "The balls", "$B(c,\\delta_c)$ cover $X$, and the space $X$ is compact.", "Apply the \\hyperref[ms:lebesgue]{Lebesgue covering lemma} to obtain a", "$\\delta > 0$ such that for every $x \\in X$, there is a $c \\in X$", "for which $B(x,\\delta) \\subset B(c,\\delta_c)$.", "Suppose $p, q \\in X$ where $d_X(p,q) < \\delta$.", "Find a $c \\in X$ such that $B(p,\\delta) \\subset B(c,\\delta_c)$.", "Then $q \\in B(c,\\delta_c)$. By the triangle inequality", "and the definition of $\\delta_c$,", "\\begin{equation*}", "d_Y\\bigl(f(p),f(q)\\bigr)", "\\leq", "d_Y\\bigl(f(p),f(c)\\bigr)", "+", "d_Y\\bigl(f(c),f(q)\\bigr)", "<", "\\nicefrac{\\epsilon}{2}+", "\\nicefrac{\\epsilon}{2} = \\epsilon . \\qedhere", "\\end{equation*}" ], "refs": [ "ms:lebesgue", "named:Lebesgue" ], "ref_ids": [ 48 ] } ], "ref_ids": [] }, { "id": 63, "type": "proposition", "label": "Lebl-contfunc:prop:integralcontcont", "categories": [ "continuity" ], "title": "Continuity of Continuity", "contents": [ "If$f \\colon [a,b] \\times [c,d] \\to \\R$is continuous, then$g \\colon [c,d] \\to \\R$defined by\\begin{equation*} g(y) \\coloneqq \\int_a^b f(x,y) \\,dx \\qquad \\text{is continuous}. \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Fix $y \\in [c,d]$ and", "let $\\epsilon > 0$ be given.", "As $f$ is continuous on $[a,b] \\times [c,d]$, which is compact, $f$", "is uniformly continuous.", "In particular, there exists a $\\delta > 0$ such that", "whenever $z \\in [c,d]$ and", "$\\abs{z-y} < \\delta$, we have", "$\\abs{f(x,z)-f(x,y)} < \\frac{\\epsilon}{b-a}$ for all $x \\in [a,b]$.", "So suppose $\\abs{z-y} < \\delta$. Then", "\\begin{multline*}", "\\abs{", "g(z)-", "g(y)", "}", "=", "\\abs{", "\\int_a^b", "f(x,z) \\,dx", "-", "\\int_a^b", "f(x,y) \\,dx", "}", "\\\\", "=", "\\abs{", "\\int_a^b", "\\bigl(", "f(x,z) - f(x,y)", "\\bigr)", "\\,dx", "}", "\\leq", "(b-a)", "\\frac{\\epsilon}{b-a}", "= \\epsilon . \\qedhere", "\\end{multline*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 64, "type": "example", "label": "Lebl-contfunc:64", "categories": [ "applications", "continuity", "metric_spaces", "example" ], "title": "Continuity of Metric Spaces", "contents": [ "Useful examples of uniformly continuous functions are again the so-called \\emph{Lipschitz continuous}% \\index{Lipschitz continuous!in a metric space}% \\index{function!Lipschitz} functions. That is, if$(X,d_X)$and$(Y,d_Y)$are metric spaces, then$f \\colon X \\to Y$is called Lipschitz or$K$-Lipschitz if there exists a$K \\in \\R$such that", "\\begin{equation*}\nd_Y\\bigl(f(p),f(q)\\bigr) \\leq K \\, d_X(p,q)\n\\qquad \\text{for all } p,q \\in X.\n\\end{equation*}", "A Lipschitz function is uniformly continuous: Take$\\delta = \\nicefrac{\\epsilon}{K}$. A function can be uniformly continuous but not Lipschitz, as we already saw:$\\sqrt{x}$on$[0,1]$is uniformly continuous but not Lipschitz.", "It is worth mentioning that, if a function is Lipschitz, it tends to be easiest to simply show it is Lipschitz even if we are only interested in knowing continuity." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 65, "type": "proposition", "label": "Lebl-contfunc:prop:mslimitisunique", "categories": [ "convergence", "metric_spaces" ], "title": "Uniqueness of Metric Spaces", "contents": [ "Let$(X,d_X)$and$(Y,d_Y)$be metric spaces,$S \\subset X$,$p \\in X$a cluster point of$S$, and let$f \\colon S \\to Y$be a function such that$f(x)$converges as$x$goes to$p$. Then the limit of$f(x)$as$x$goes to$p$is unique." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 66, "type": "lemma", "label": "Lebl-contfunc:ms:seqflimit:lemma", "categories": [ "convergence", "metric_spaces", "characterization", "auxiliary result" ], "title": "Characterization of Metric Spaces via Equivalence", "contents": [ "Let$(X,d_X)$and$(Y,d_Y)$be metric spaces,$S \\subset X$,$p \\in X$a cluster point of$S$, and let$f \\colon S \\to Y$be a function.", "Then$f(x)$converges to$L \\in Y$as$x$goes to$p$if and only if for every sequence$\\{ x_n \\}_{n=1}^\\infty$in$S \\setminus \\{p\\}$such that$\\lim_{n\\to\\infty} x_n = p$, the sequence$\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty$converges to$L$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 67, "type": "theorem", "label": "Lebl-contfunc:67", "categories": [ "fixed_points", "metric_spaces", "completeness", "named theorem" ], "title": "Banach Fixed Point Theorem", "contents": [ "% [Contraction mapping principle\\index{contraction mapping principle} or \\myindex{Banach fixed point theorem}\\index{fixed point theorem}% \\footnote{Named after the Polish mathematician \\href{https://en.wikipedia.org/wiki/Stefan_Banach}{Stefan Banach} (1892--1945) who first stated the theorem in 1922.}] \\label{thm:contr} Let$(X,d)$be a nonempty complete metric space and$\\varphi \\colon X \\to X$a contraction. Then$\\varphi$has a unique fixed point." ], "refs": [ "named:Banach" ], "proofs": [ { "contents": [ "Pick $x_0 \\in X$.", "Define a sequence $\\{ x_n \\}_{n=1}^\\infty$ by $x_{n+1} \\coloneqq \\varphi(x_n)$. Then", "\\begin{equation*}", "d(x_{n+1},x_n) = d\\bigl(\\varphi(x_n),\\varphi(x_{n-1})\\bigr)", "\\leq k d(x_n,x_{n-1}) .", "%\\leq \\cdots", "%\\leq k^n d(x_1,x_0) .", "\\end{equation*}", "Repeating $n$ times, we get $d(x_{n+1},x_n) \\leq k^n d(x_1,x_0)$.", "For $m > n$,", "\\begin{equation*}", "\\begin{split}", "d(x_m,x_n)", "& \\leq \\sum_{i=n}^{m-1} d(x_{i+1},x_i) \\\\", "& \\leq \\sum_{i=n}^{m-1} k^i d(x_1,x_0) \\\\", "& = k^n d(x_1,x_0) \\sum_{i=0}^{m-n-1} k^i \\\\", "& \\leq k^n d(x_1,x_0) \\sum_{i=0}^{\\infty} k^i", "= k^n d(x_1,x_0) \\frac{1}{1-k} .", "\\end{split}", "\\end{equation*}", "In particular, the sequence is Cauchy (why?). Since $X$ is complete,", "we let $x \\coloneqq \\lim_{n\\to\\infty} x_n$, and we claim that $x$", "is our unique fixed point.", "Fixed point? The function $\\varphi$ is a contraction,", "so it is Lipschitz continuous:", "\\begin{equation*}", "\\varphi(x) = \\varphi\\Bigl( \\lim_{n\\to\\infty} x_n\\Bigr) = \\lim_{n\\to\\infty}", "\\varphi(x_n) =", "\\lim_{n\\to\\infty} x_{n+1} = x .", "\\end{equation*}", "Unique? Let $x$ and $y$ be fixed points.", "\\begin{equation*}", "d(x,y) = d\\bigl(\\varphi(x),\\varphi(y)\\bigr) \\leq k\\, d(x,y) .", "\\end{equation*}", "As $k < 1$, the inequality means that $d(x,y) = 0$, and hence $x=y$. The theorem is", "proved." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 68, "type": "theorem", "label": "Lebl-contfunc:68", "categories": [ "continuity", "topology", "named theorem" ], "title": "Picard's Theorem", "contents": [ "% \\index{existence and uniqueness theorem}\\index{Picard's theorem} Let$I, J \\subset \\R$be closed and bounded intervals, let$I^\\circ$and$J^\\circ$be their interiors, and let$(x_0,y_0) \\in I^\\circ \\times J^\\circ$. Suppose$F \\colon I \\times J \\to \\R$is continuous and Lipschitz in the second variable, that is, there exists an$L \\in \\R$such that\\begin{equation*} \\abs{F(x,y) - F(x,z)} \\leq L \\abs{y-z} \\qquad \\text{for all } y,z \\in J, x \\in I . \\end{equation*}Then there exists an$h > 0$such that$[x_0-h,x_0+h] \\subset I$and a unique differentiable function$f \\colon [x_0 - h, x_0 + h] \\to J \\subset \\R$such that\\begin{equation*} f'(x) = F\\bigl(x,f(x)\\bigr) \\qquad \\text{and} \\qquad f(x_0) = y_0. \\end{equation*}" ], "refs": [ "named:Picard" ], "proofs": [ { "contents": [ "Without loss of generality, assume $x_0 =0$ (exercise).", "As $I \\times J$ is compact and", "$F$ is continuous, $F$ is bounded.", "So find an $M > 0$ such that", "$\\abs{F(x,y)} \\leq M$ for all $(x,y) \\in I\\times J$.", "Pick $\\alpha > 0$ such that", "$[-\\alpha,\\alpha] \\subset I$ and $[y_0-\\alpha, y_0 + \\alpha] \\subset J$.", "Let", "\\begin{equation*}", "h \\coloneqq \\min \\left\\{ \\alpha, \\frac{\\alpha}{M+L\\alpha} \\right\\} .", "\\end{equation*}", "Note $[-h,h] \\subset I$. Let", "\\begin{equation*}", "Y \\coloneqq \\bigl\\{ f \\in C\\bigl([-h,h],\\R\\bigr) : f\\bigl([-h,h]\\bigr) \\subset J \\bigr\\} .", "\\end{equation*}", "That is, $Y$ is the set of continuous functions on $[-h,h]$ with values in", "$J$, in other words,", "exactly those functions where $F\\bigl(x,f(x)\\bigr)$ makes sense.", "It is left as an exercise to show that $Y$ is a closed subset of $C\\bigl([-h,h],\\R\\bigr)$", "(because $J$ is closed).", "The space $C\\bigl([-h,h],\\R\\bigr)$ is complete, and", "a closed subset of a complete metric space is a complete metric space with", "the subspace metric, see \\propref{prop:closedcomplete}. So $Y$ with the", "subspace metric is a complete metric space.", "We will write $d(f,g) = \\snorm{f-g}_{[-h,h]}$ for this metric.", "Define a mapping", "$T \\colon Y \\to C\\bigl([-h,h],\\R\\bigr)$ by", "\\begin{equation*}", "T(f)(x)", "\\coloneqq", "y_0 + \\int_0^x F\\bigl(t,f(t)\\bigr)\\,dt .", "\\end{equation*}", "It is an exercise to check that", "$T$ is well-defined, and that for $f \\in Y$, $T(f)$ really is in $C\\bigl([-h,h],\\R\\bigr)$.", "Let $f \\in Y$ and $\\abs{x} \\leq h$.", "As $F$ is bounded by $M$, we have", "\\begin{equation*}", "\\begin{split}", "\\abs{T(f)(x) - y_0}", "&= \\abs{\\int_0^x F\\bigl(t,f(t)\\bigr)\\,dt} \\\\", "& \\leq", "\\abs{x}M \\leq hM \\leq \\frac{\\alpha M}{M+ L\\alpha} \\leq \\alpha .", "\\end{split}", "\\end{equation*}", "So $T(f)\\bigl([-h,h]\\bigr) \\subset [y_0-\\alpha,y_0+\\alpha] \\subset J$, and", "$T(f) \\in Y$. In other words, $T(Y) \\subset Y$. From now on,", "we consider $T$ as a mapping of $Y$ to $Y$.", "We claim $T \\colon Y \\to Y$ is a contraction. First, for $x \\in [-h,h]$", "and $f,g \\in Y$, we have", "\\begin{equation*}", "\\abs{F\\bigl(x,f(x)\\bigr) - F\\bigl(x,g(x)\\bigr)} \\leq", "L\\abs{f(x)- g(x)} \\leq L \\, d(f,g) .", "\\end{equation*}", "Therefore,", "\\begin{equation*}", "\\begin{split}", "\\abs{T(f)(x) - T(g)(x)}", "&= \\abs{\\int_0^x \\Bigl( F\\bigl(t,f(t)\\bigr) - F\\bigl(t,g(t)\\bigr)\\Bigr)\\,dt} \\\\", "& \\leq \\abs{x} L \\, d(f,g)", "\\leq h L\\, d(f,g)", "\\leq \\frac{L\\alpha}{M+L\\alpha} \\, d(f,g) .", "\\end{split}", "\\end{equation*}", "We chose $M > 0$ and so", "$\\frac{L\\alpha}{M+L\\alpha} < 1$.", "Take supremum over $x \\in [-h,h]$ of the left-hand side above to obtain", "$d\\bigl(T(f),T(g)\\bigr) \\leq \\frac{L\\alpha}{M+L\\alpha} \\, d(f,g)$,", "that is, $T$ is a contraction.", "The fixed point theorem (\\thmref{thm:contr})", "gives a unique $f \\in Y$ such that $T(f) = f$.", "In other words,", "\\begin{equation*} %\\label{equation:msinteqpicard}", "f(x) = y_0 + \\int_0^x F\\bigl(t,f(t)\\bigr)\\,dt .", "\\end{equation*}", "Clearly, $f(0) = y_0$.", "By the fundamental theorem of calculus (\\thmref{thm:FTCv2}),", "$f$ is differentiable and its derivative is", "$F\\bigl(x,f(x)\\bigr)$.", "Differentiable functions are continuous, so", "$f$ is the unique differentiable $f \\colon [-h,h] \\to J$", "such that", "$f'(x) = F\\bigl(x,f(x)\\bigr)$ and $f(0) = y_0$." ], "refs": [ "prop:closedcomplete" ], "ref_ids": [] } ], "ref_ids": [ 6 ] }, { "id": 0, "type": "example", "label": "Lebl-contfunc:0", "categories": [ "vector_spaces", "example" ], "title": "The set$\\R^n$is a vector space, addition and multiplication by a scalar is done componentwise: If...", "contents": [ "The set$\\R^n$is a vector space, addition and multiplication by a scalar is done componentwise: If$a \\in \\R$,$v = (v_1,v_2,\\ldots,v_n) \\in \\R^n$, and$w = (w_1,w_2,\\ldots,w_n) \\in \\R^n$, then \\begin{align*} & v+w \\coloneqq (v_1,v_2,\\ldots,v_n) + (w_1,w_2,\\ldots,w_n) = (v_1+w_1,v_2+w_2,\\ldots,v_n+w_n) ,", "& a v \\coloneqq a (v_1,v_2,\\ldots,v_n) = (a v_1, a v_2,\\ldots, a v_n) . \\end{align*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 1, "type": "example", "label": "Lebl-contfunc:1", "categories": [ "vector_spaces", "example" ], "title": "A trivial example of a vector space is$X \\coloneqq \\{ 0 \\}$", "contents": [ "A trivial example of a vector space is$X \\coloneqq \\{ 0 \\}$. The operations are defined in the obvious way:$0 + 0 \\coloneqq 0$and$a0 \\coloneqq 0$. A zero vector must always exist, so all vector spaces are nonempty sets, and this$X$is the smallest possible vector space." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 2, "type": "example", "label": "Lebl-contfunc:2", "categories": [ "continuity", "vector_spaces", "example" ], "title": "Continuity of Vector Spaces", "contents": [ "The space$C([0,1],\\R)$of continuous functions on the interval$[0,1]$is a vector space. For two functions$f$and$g$in$C([0,1],\\R)$and$a \\in \\R$, we make the obvious definitions of$f+g$and$af$:", "\\begin{equation*}\n(f+g)(x) \\coloneqq f(x) + g(x), \\qquad (af) (x) \\coloneqq a\\bigl(f(x)\\bigr) .\n\\end{equation*}", "The 0 is the function that is identically zero. We leave it as an exercise to check that all the vector space conditions are satisfied. The space$C^1([0,1],\\R)$of continuously differentiable functions is a subspace of$C([0,1],\\R)$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 3, "type": "example", "label": "Lebl-contfunc:3", "categories": [ "vector_spaces", "example", "implicit_inverse" ], "title": "The space of polynomials$c_0 + c_1 t + c_2 t^2 + \\cdots + c_m t^m$(of arbitrary degree~$m$) is a ...", "contents": [ "The space of polynomials$c_0 + c_1 t + c_2 t^2 + \\cdots + c_m t^m$(of arbitrary degree~$m$) is a vector space, denoted by$\\R[t]$\\glsadd{not:realpoly} (coefficients are real and the variable is$t$). The operations are defined in the same way as for functions above. Suppose there are two polynomials, one of degree$m$and one of degree$n$. Assume$n \\geq m$for simplicity. Then \\begin{multline*} (c_0 + c_1 t + c_2 t^2 + \\cdots + c_m t^m) + (d_0 + d_1 t + d_2 t^2 + \\cdots + d_n t^n) =", "(c_0+d_0) + (c_1+d_1) t + (c_2 + d_2) t^2 + \\cdots + (c_m+d_m) t^m + d_{m+1} t^{m+1} + \\cdots + d_n t^n \\end{multline*} and", "\\begin{equation*}\na(c_0 + c_1 t + c_2 t^2 + \\cdots + c_m t^m)\n=\n(ac_0) + (ac_1) t + (ac_2) t^2 + \\cdots + (ac_m) t^m .\n\\end{equation*}", "Despite what it looks like,$\\R[t]$is not equivalent to$\\R^n$for any$n$. In particular, it is not \\myquote{finite-dimensional.} We will make this notion precise in just a little bit. One can make a finite-dimensional vector subspace by restricting the degree. For example, if$\\sP_n$is the set of polynomials of degree$n$or less, then$\\sP_n$is a finite-dimensional vector space, and we could identify it with$\\R^{n+1}$.", "Above, the variable$t$is really just a formal placeholder. By setting$t$equal to a real number, we obtain a function. So the space$\\R[t]$can be thought of as a subspace of$C(\\R,\\R)$. If we restrict the range of$t$to$[0,1]$,$\\R[t]$can be identified with a subspace of$C([0,1],\\R)$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 4, "type": "proposition", "label": "Lebl-contfunc:4", "categories": [ "vector_spaces" ], "title": "For$S \\subset X$to be a vector subspace of a vector space$X$, we only need to check: %If$X$is a v...", "contents": [ "For$S \\subset X$to be a vector subspace of a vector space$X$, we only need to check: %If$X$is a vector space, for$S \\subset X$%to be a vector subspace, we only need to check \\begin{enumerate}[1)] \\item$0 \\in S$. \\item$S$is closed under addition: If$x,y \\in S$, then$x+y \\in S$. \\item$S$is closed under scalar multiplication: If$x \\in S$and$a \\in \\R$, then$ax \\in S$. \\end{enumerate}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 5, "type": "example", "label": "Lebl-contfunc:5", "categories": [ "example" ], "title": "Let$Y \\coloneqq \\bigl\\{ (1,1) \\bigr\\} \\subset \\R^2$", "contents": [ "Let$Y \\coloneqq \\bigl\\{ (1,1) \\bigr\\} \\subset \\R^2$. Then", "\\begin{equation*}\n\\spn(Y)\n=\n\\bigl\\{ (x,x) \\in \\R^2 : x \\in \\R \\bigr\\} .\n\\end{equation*}", "That is,$\\spn(Y)$is the line through the origin and the point$(1,1)$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 6, "type": "example", "label": "Lebl-contfunc:example:vecspr2span", "categories": [ "convexity", "vector_spaces", "example" ], "title": "Let$Y \\coloneqq \\bigl\\{ (1,1), (0,1) \\bigr\\} \\subset \\R^2$", "contents": [ "Let$Y \\coloneqq \\bigl\\{ (1,1), (0,1) \\bigr\\} \\subset \\R^2$. Then", "\\begin{equation*}\n\\spn(Y)\n=\n\\R^2 ,\n\\end{equation*}", "as every point$(x,y) \\in \\R^2$can be written as a linear combination", "\\begin{equation*}\n(x,y) = x (1,1) + (y-x) (0,1) .\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 7, "type": "example", "label": "Lebl-contfunc:7", "categories": [ "vector_spaces", "example" ], "title": "Let$Y \\coloneqq \\{ 1, t, t^2, t^3, \\ldots \\} \\subset \\R[t]$, and$E \\coloneqq \\{ 1, t^2, t^4, t^6,...", "contents": [ "Let$Y \\coloneqq \\{ 1, t, t^2, t^3, \\ldots \\} \\subset \\R[t]$, and$E \\coloneqq \\{ 1, t^2, t^4, t^6, \\ldots \\} \\subset \\R[t]$. The span of$Y$is all polynomials,", "\\begin{equation*}\n\\spn(Y) = \\R[t] .\n\\end{equation*}", "The span of$E$is the set of polynomials with even powers of$t$only." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 8, "type": "proposition", "label": "Lebl-contfunc:8", "categories": [ "vector_spaces" ], "title": "Let$X$be a vector space and$Y \\subset X$is a subset", "contents": [ "Let$X$be a vector space and$Y \\subset X$is a subset. Then the set$\\spn(Y)$is a vector subspace of$X$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 9, "type": "proposition", "label": "Lebl-contfunc:9", "categories": [ "vector_spaces" ], "title": "Uniqueness of Vector Spaces", "contents": [ "Suppose a vector space$X$has basis$B = \\{ x_1, x_2, \\ldots, x_n \\}$. Then every$y \\in X$has a unique representation of the form\\begin{equation*} y = \\sum_{k=1}^n a_k \\, x_k \\end{equation*}for some scalars$a_1, a_2, \\ldots, a_n$." ], "refs": [], "proofs": [ { "contents": [ "As $X$ is the span of $B$,", "every $y \\in X$ is a linear combination of elements of $B$.", "Suppose", "\\begin{equation*}", "y = \\sum_{k=1}^n a_k \\, x_k = \\sum_{k=1}^n b_k \\, x_k .", "\\end{equation*}", "Then", "\\begin{equation*}", "\\sum_{k=1}^n (a_k-b_k) x_k = 0 .", "\\end{equation*}", "By linear independence of the basis, $a_k = b_k$ for all $k$, and so", "the representation is unique." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 10, "type": "proposition", "label": "Lebl-contfunc:mv:dimprop", "categories": [ "characterization", "vector_spaces" ], "title": "Characterization of Vector Spaces via Equivalence", "contents": [ "\\pagebreak[2] Let$X$be a vector space and$d$a nonnegative integer. \\begin{enumerate}[(i)] \\item \\label{mv:dimprop:i} If$X$is spanned by$d$vectors, then$\\dim \\, X \\leq d$. \\item \\label{mv:dimprop:ii} If$T$is a linearly independent set and$v \\in X \\setminus \\spn (T)$, then$T \\cup \\{ v \\}$is linearly independent. \\item \\label{mv:dimprop:iii}$\\dim \\, X = d$if and only if$X$has a basis of$d$vectors. In particular,$\\dim \\, \\R^n = n$. \\item \\label{mv:dimprop:iv} If$Y \\subset X$is a vector subspace and$\\dim \\, X = d$, then$\\dim \\, Y \\leq d$. \\item \\label{mv:dimprop:v} If$\\dim \\, X = d$and a set$T$of$d$vectors spans$X$, then$T$is linearly independent. \\item \\label{mv:dimprop:vi} If$\\dim \\, X = d$and a set$T$of$m$vectors is linearly independent, then there is a set$S$of$d-m$vectors such that$T \\cup S$is a basis of$X$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "All statements hold trivially when $d=0$, so assume $d \\geq 1$.", "We start with \\ref{mv:dimprop:i}.", "Suppose $S \\coloneqq \\{ x_1 , x_2, \\ldots, x_d \\}$ spans $X$, and", "$T \\coloneqq \\{ y_1, y_2, \\ldots, y_m \\}$ is a linearly independent", "subset of $X$. We wish to show that $m \\leq d$.", "As $S$ spans $X$,", "write", "\\begin{equation*}", "y_1 = \\sum_{k=1}^d a_{k,1} x_k ,", "\\end{equation*}", "for some numbers $a_{1,1},a_{2,1},\\ldots,a_{d,1}$.", "One of the", "$a_{k,1}$ is nonzero, otherwise $y_1$ would be zero.", "Without loss of generality, suppose $a_{1,1} \\not= 0$. Solve", "\\begin{equation*}", "x_1 = \\frac{1}{a_{1,1}} y_1 - \\sum_{k=2}^d \\frac{a_{k,1}}{a_{1,1}} x_k .", "\\end{equation*}", "In particular, $\\{ y_1 , x_2, \\ldots, x_d \\}$ spans $X$, since $x_1$ can be", "obtained from $\\{ y_1 , x_2, \\ldots, x_d \\}$. Therefore, there are some numbers", "for some numbers $a_{1,2},a_{2,2},\\ldots,a_{d,2}$, such that", "\\begin{equation*}", "y_2 = a_{1,2} y_1 + \\sum_{k=2}^d a_{k,2} x_k .", "\\end{equation*}", "As $T$ is linearly independent---and so $\\{ y_1, y_2 \\}$ is linearly", "independent---one of the $a_{k,2}$", "for $k \\geq 2$ must be nonzero. Without loss of generality suppose", "$a_{2,2} \\not= 0$. Solve", "\\begin{equation*}", "x_2 = \\frac{1}{a_{2,2}} y_2 - \\frac{a_{1,2}}{a_{2,2}} y_1 - \\sum_{k=3}^d", "\\frac{a_{k,2}}{a_{2,2}} x_k .", "\\avoidbreak", "\\end{equation*}", "In particular,", "$\\{ y_1 , y_2, x_3, \\ldots, x_d \\}$ spans $X$.", "We continue this procedure. If $m < d$, we are done. Suppose", "$m \\geq d$.", "After $d$ steps, we obtain that", "$\\{ y_1 , y_2, \\ldots, y_d \\}$ spans $X$. Any", "other vector $v$ in $X$ is a linear combination of", "$\\{ y_1 , y_2, \\ldots, y_d \\}$ and hence cannot be in $T$ as $T$ is", "linearly independent. So $m = d$.", "We continue with \\ref{mv:dimprop:ii}.", "Suppose", "$T = \\{x_1,x_2,\\ldots,x_m\\}$ is linearly independent,", "does not span $X$, and $v \\in X \\setminus \\spn (T)$.", "Suppose", "$a_1 x_1 + a_2 x_2 + \\cdots + a_m x_m + a_{m+1} v = 0$", "for some scalars $a_1,a_2,\\ldots,a_{m+1}$.", "If $a_{m+1} \\not=0$, then $v$ would be a linear combination of $T$,", "so $a_{m+1} = 0$. Then, as $T$ is linearly independent, $a_1=a_2=\\cdots=a_m = 0$.", "So $T \\cup \\{ v \\}$ is linearly independent.", "We move to \\ref{mv:dimprop:iii}.", "If $\\dim \\, X = d$,", "then there must exist some linearly independent set $T$ of $d$ vectors,", "and $T$ must span $X$, otherwise we could choose a larger set of linearly", "independent vectors via \\ref{mv:dimprop:ii}.", "So we have a basis of $d$ vectors.", "On the other hand, if we have a basis of $d$ vectors,", "the dimension is at least $d$ as a basis is linearly independent.", "A basis also spans $X$, and so", "by \\ref{mv:dimprop:i} we know that dimension is at most $d$.", "Hence the dimension of $X$ must equal $d$.", "The \\myquote{in particular} follows by noting that", "$\\{ e_1, e_2, \\ldots, e_n \\}$ is a basis of $\\R^n$.", "To see \\ref{mv:dimprop:iv},", "suppose $Y \\subset X$ is a vector subspace,", "where $\\dim \\, X = d$. As $X$ cannot contain $d+1$ linearly independent", "vectors, neither can $Y$.", "For \\ref{mv:dimprop:v}, suppose $T$ is a set of $m$ vectors", "that is linearly dependent", "and spans $X$.", "We will show that $m > d$.", "One of the", "vectors is a linear combination of the others. If we remove it", "from $T$, we obtain a set of $m-1$ vectors that still span $X$.", "Hence", "$d = \\dim \\, X \\leq m-1$ by~\\ref{mv:dimprop:i}.", "For \\ref{mv:dimprop:vi} suppose $T = \\{ x_1, x_2, \\ldots, x_m \\}$ is", "a linearly independent set. First, $m \\leq d$ by definition of dimension.", "If $m=d$, the set $T$ must span $X$ as in the proof of \\ref{mv:dimprop:iii},", "otherwise we could add another vector to $T$.", "If $m < d$, $T$ cannot span $X$ by \\ref{mv:dimprop:iii}.", "So find $v$ not in the span of~$T$.", "Via \\ref{mv:dimprop:ii}, the set $T \\cup \\{ v \\}$ is", "a linearly independent set of $m+1$ elements.", "Therefore, we repeat this procedure $d-m$ times to", "find a set of $d$ linearly independent vectors.", "Again, they must span $X$, otherwise we could add yet another vector." ], "refs": [ "mv:dimprop:i", "mv:dimprop:ii", "mv:dimprop:iii", "mv:dimprop:iv", "mv:dimprop:v", "mv:dimprop:vi" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 11, "type": "proposition", "label": "Lebl-contfunc:prop:LXYvs", "categories": [ "vector_spaces" ], "title": "Let$X$,$Y$, and$Z$be vector spaces", "contents": [ "Let$X$,$Y$, and$Z$be vector spaces. \\begin{enumerate}[(i)] \\item If$A \\in L(X,Y)$, then$A0 = 0$. \\item If$A,B \\in L(X,Y)$, then$A+B \\in L(X,Y)$. \\item If$A \\in L(X,Y)$and$a \\in \\R$, then$aA \\in L(X,Y)$. \\item If$A \\in L(Y,Z)$and$B \\in L(X,Y)$, then$AB \\in L(X,Z)$. \\item If$A \\in L(X,Y)$is invertible, then$A^{-1} \\in L(Y,X)$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "We leave the first four items as a quick exercise,", "\\exerciseref{exercise:LXYvs}.", "Let us prove the last item.", "Let $a \\in \\R$ and $y \\in Y$. As $A$ is onto, then there is an", "$x \\in X$ such that $y = Ax$.", "As it is also one-to-one, $A^{-1}(Az) = z$ for all $z \\in X$.", "So", "\\begin{equation*}", "A^{-1}(ay)", "=", "A^{-1}(aAx)", "=", "A^{-1}\\bigl(A(ax)\\bigr)", "= ax", "= aA^{-1}(y).", "\\end{equation*}", "Similarly, let $y_1,y_2 \\in Y$ and $x_1, x_2 \\in X$ be such that", "$Ax_1 = y_1$ and", "$Ax_2 = y_2$, then", "\\begin{equation*}", "A^{-1}(y_1+y_2)", "=", "A^{-1}(Ax_1+Ax_2)", "=", "A^{-1}\\bigl(A(x_1+x_2)\\bigr)", "= x_1+x_2", "= A^{-1}(y_1) + A^{-1}(y_2). \\qedhere", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 12, "type": "proposition", "label": "Lebl-contfunc:mv:lindefonbasis", "categories": [ "vector_spaces" ], "title": "If$A \\in L(X,Y)$is linear, then it is completely determined by its values on a basis of$X$", "contents": [ "If$A \\in L(X,Y)$is linear, then it is completely determined by its values on a basis of$X$. Furthermore, if$B$is a basis of$X$, then every function$\\widetilde{A} \\colon B \\to Y$extends to a linear function$A$on$X$." ], "refs": [], "proofs": [ { "contents": [ "Let $\\{ x_1, x_2, \\ldots, x_n \\}$ be a basis of $X$, and let", "$y_k \\coloneqq A x_k$. Every $x \\in X$ has a unique representation", "\\begin{equation*}", "x = \\sum_{k=1}^n b_k \\, x_k", "\\end{equation*}", "for some numbers $b_1,b_2,\\ldots,b_n$. By linearity,", "\\begin{equation*}", "Ax =", "A\\sum_{k=1}^n b_k x_k", "=", "\\sum_{k=1}^n b_k \\, Ax_k", "=", "\\sum_{k=1}^n b_k \\, y_k .", "\\end{equation*}", "The \\myquote{furthermore} follows by setting $y_k \\coloneqq \\widetilde{A}(x_k)$,", "and then for", "$x = \\sum_{k=1}^n b_k \\, x_k$,", "defining the extension as", "$A(x) \\coloneqq \\sum_{k=1}^n b_k \\, y_k$. The function is well-defined by", "uniqueness of the representation of $x$.", "We leave it to the reader to check that $A$ is linear." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 13, "type": "proposition", "label": "Lebl-contfunc:mv:prop:lin11onto", "categories": [ "characterization", "vector_spaces" ], "title": "Characterization of Vector Spaces via Equivalence", "contents": [ "If$X$is a finite-dimensional vector space and$A \\in L(X)$, then$A$is one-to-one if and only if it is onto." ], "refs": [], "proofs": [ { "contents": [ "Let $\\{ x_1,x_2,\\ldots,x_n \\}$ be a basis for $X$.", "First suppose $A$ is one-to-one. Let $c_1,c_2,\\ldots,c_n$ be such that", "\\begin{equation*}", "0 =", "\\sum_{k=1}^n c_k \\, Ax_k =", "A\\sum_{k=1}^n c_k \\, x_k", ".", "\\end{equation*}", "As $A$ is one-to-one,", "the only vector that is taken to 0 is 0 itself.", "Hence,", "\\begin{equation*}", "0 =", "\\sum_{k=1}^n c_k \\, x_k,", "\\end{equation*}", "and so $c_k = 0$ for all $k$ as $\\{ x_1,x_2,\\ldots,x_n \\}$ is a basis.", "So $\\{ Ax_1, Ax_2, \\ldots, Ax_n \\}$ is linearly independent.", "By \\propref{mv:dimprop}", "and the fact that the dimension is $n$, we conclude", "$\\{ Ax_1, Ax_2, \\ldots, Ax_n \\}$ spans $X$. Consequently, $A$ is onto,", "as any $y \\in X$ can be written as", "\\begin{equation*}", "y = \\sum_{k=1}^n a_k \\, Ax_k =", "A\\sum_{k=1}^n a_k \\, x_k .", "\\avoidbreak", "\\end{equation*}", "For the other direction, suppose $A$ is onto.", "Suppose that for some", "$c_1,c_2,\\ldots,c_n$,", "\\begin{equation*}", "0 = A\\sum_{k=1}^n c_k \\, x_k =", "\\sum_{k=1}^n c_k \\, Ax_k .", "\\end{equation*}", "As $A$ is determined by the action on the basis,", "$\\{ Ax_1, Ax_2, \\ldots, Ax_n \\}$ spans $X$.", "So by \\propref{mv:dimprop}, the set is linearly independent,", "and $c_k = 0$ for all $k$. In other words, if $Ax = 0$, then $x=0$.", "Thus, $A$ is one-to-one." ], "refs": [ "mv:dimprop" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 14, "type": "proposition", "label": "Lebl-contfunc:prop:LXYfinitedim", "categories": [ "vector_spaces" ], "title": "If$X$and$Y$are finite-dimensional vector spaces, then$L(X,Y)$is also finite-dimensional", "contents": [ "If$X$and$Y$are finite-dimensional vector spaces, then$L(X,Y)$is also finite-dimensional." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 15, "type": "proposition", "label": "Lebl-contfunc:15", "categories": [ "convexity" ], "title": "Let$x \\in \\R^n$and$r > 0$", "contents": [ "Let$x \\in \\R^n$and$r > 0$. The ball$B(x,r) \\subset \\R^n$%(using the standard metric on$\\R^n$) is convex." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 16, "type": "example", "label": "Lebl-contfunc:16", "categories": [ "convexity", "vector_spaces", "example" ], "title": "A convex combination is, in particular, a linear combination", "contents": [ "A convex combination is, in particular, a linear combination. So every vector subspace$V$of a vector space$X$is convex." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 17, "type": "example", "label": "Lebl-contfunc:17", "categories": [ "convexity", "continuity", "vector_spaces", "example" ], "title": "Continuity of Vector Spaces", "contents": [ "%A somewhat more complicated example is given by the following. Let$C([0,1],\\R)$be the vector space of continuous real-valued functions on$\\R$. Let$V \\subset C([0,1],\\R)$be the set of those$f$such that", "\\begin{equation*}\n\\int_0^1 f(x)\\,dx \\leq 1 \\qquad \\text{and} \\qquad\nf(x) \\geq 0 \\enspace \\text{for all } x \\in [0,1] .\n\\end{equation*}", "Then$V$is convex. Take$t \\in [0,1]$, and note that if$f,g \\in V$, then$(1-t) f(x) + t g(x) \\geq 0$for all$x$. Furthermore,", "\\begin{equation*}\n\\int_0^1 \\bigl((1-t)f(x) + t g(x)\\bigr) \\,dx\n=\n(1-t) \\int_0^1 f(x) \\,dx\n+ t \\int_0^1 g(x) \\,dx \\leq 1 .\n\\end{equation*}", "Note that$V$is not a vector subspace of$C([0,1],\\R)$. The function$f(x) \\coloneqq 1$is in$V$, but$2f$and$-f$is not." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 18, "type": "proposition", "label": "Lebl-contfunc:prop:intersectionconvex", "categories": [ "convexity", "vector_spaces" ], "title": "The intersection of two convex sets is convex", "contents": [ "The intersection of two convex sets is convex. In fact, if$\\{ C_\\lambda \\}_{\\lambda \\in I}$is an arbitrary collection of convex sets in a vector space, then\\begin{equation*} C \\coloneqq \\bigcap_{\\lambda \\in I} C_\\lambda \\qquad \\text{is convex.} \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "If $x, y \\in C$, then $x,y \\in C_\\lambda$ for all", "$\\lambda \\in I$, and hence if $t \\in [0,1]$, then $(1-t)x + ty \\in", "C_\\lambda$ for all $\\lambda \\in I$. Therefore, $(1-t)x + ty \\in C$ and $C$", "is convex." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 19, "type": "example", "label": "Lebl-contfunc:19", "categories": [ "convexity", "example" ], "title": "The convex hull of$\\{ 0, 1 \\}$in$\\R$is$[0,1]$", "contents": [ "The convex hull of$\\{ 0, 1 \\}$in$\\R$is$[0,1]$. Proof: A convex set containing 0 and 1 must contain$[0,1]$, so$[0,1] \\subset \\operatorname{co}(\\{0,1\\})$. The set$[0,1]$is convex and contains$\\{0,1\\}$, so$\\operatorname{co}(\\{0,1\\}) \\subset [0,1]$." ], "refs": [], "proofs": [ { "contents": [ "A convex set containing 0 and 1 must contain$[0,1]$, so$[0,1] \\subset \\operatorname{co}(\\{0,1\\})$. The set$[0,1]$is convex and contains$\\{0,1\\}$, so$\\operatorname{co}(\\{0,1\\}) \\subset [0,1]$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 20, "type": "proposition", "label": "Lebl-contfunc:20", "categories": [ "convexity", "vector_spaces" ], "title": "Let$X,Y$be vector spaces,$A \\in L(X,Y)$, and let$C \\subset X$be convex", "contents": [ "Let$X,Y$be vector spaces,$A \\in L(X,Y)$, and let$C \\subset X$be convex. Then$A(C)$is convex." ], "refs": [], "proofs": [ { "contents": [ "Take two points $p,q \\in A(C)$. Pick $u,v \\in C$ such that", "$Au = p$ and $Av=q$. As $C$ is convex, then", "$(1-t)u+t v \\in C$", "for all $t \\in [0,1]$, so", "\\begin{equation*}", "(1-t)p+t q", "=", "(1-t)Au+tAv", "=", "A\\bigl((1-t)u+tv\\bigr)", "\\in A(C) . \\qedhere", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 21, "type": "theorem", "label": "Lebl-contfunc:21", "categories": [ "norms", "characterization" ], "title": "Characterization of Norms via Equivalence", "contents": [ "Let$x, y \\in \\R^n$, then\\begin{equation*} \\sabs{x \\cdot y} \\leq \\snorm{x} \\, \\snorm{y} = \\sqrt{x\\cdot x}\\, \\sqrt{y\\cdot y}, \\avoidbreak \\end{equation*}with equality if and only if$x = \\lambda y$or$y = \\lambda x$for some$\\lambda \\in \\R$." ], "refs": [], "proofs": [ { "contents": [ "If $x=0$ or $y = 0$, then the theorem holds trivially.", "So assume $x\\not= 0$ and $y \\not= 0$.", "If $x$ is a scalar multiple of $y$, that is, $x = \\lambda y$ for some", "$\\lambda \\in \\R$, then the theorem holds with equality:", "\\begin{equation*}", "\\sabs{ x \\cdot y } =", "\\sabs{\\lambda y \\cdot y} = \\sabs{\\lambda} \\, \\sabs{y\\cdot y} =", "\\sabs{\\lambda} \\, \\snorm{y}^2 = \\snorm{\\lambda y} \\, \\snorm{y}", "= \\snorm{x} \\, \\snorm{y} .", "\\end{equation*}", "Fixing $x$ and $y$,", "$\\snorm{x+ty}^2$", "is a quadratic polynomial", "as a function of $t$:", "\\begin{equation*}", "\\snorm{x+ty}^2 =", "(x+ty) \\cdot (x+ty) =", "x \\cdot x + x \\cdot ty + ty \\cdot x + ty \\cdot ty", "=", "\\snorm{x}^2 + 2t(x \\cdot y) + t^2 \\snorm{y}^2 .", "\\end{equation*}", "If $x$ is not a scalar multiple of $y$, then", "$\\snorm{x+ty}^2 > 0$ for all $t$. So the polynomial $\\snorm{x+ty}^2$", "is never zero.", "Elementary algebra says that the discriminant must be negative:", "\\begin{equation*}", "4 {(x \\cdot y)}^2 - 4 \\snorm{x}^2\\snorm{y}^2 < 0.", "\\end{equation*}", "In other words, ${(x \\cdot y)}^2 < \\snorm{x}^2\\snorm{y}^2$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 22, "type": "proposition", "label": "Lebl-contfunc:prop:finitedimpropnormfin", "categories": [ "norms", "continuity", "vector_spaces" ], "title": "Continuity of Vector Spaces", "contents": [ "Let$X$and$Y$be normed vector spaces,$A \\in L(X,Y)$, and$X$is finite-dimensional. Then$\\snorm{A} < \\infty$, and$A$is uniformly continuous (Lipschitz with constant$\\snorm{A}$)." ], "refs": [], "proofs": [ { "contents": [ "As we said we only prove the proposition for euclidean spaces, so suppose", "that $X = \\R^n$ and the norm is the standard euclidean norm.", "The general case is left as an exercise.", "Let $\\{ e_1,e_2,\\ldots,e_n \\}$ be the standard basis of $\\R^n$.", "Write $x \\in \\R^n$, with $\\snorm{x} = 1$, as", "\\begin{equation*}", "x = \\sum_{k=1}^n c_k \\, e_k .", "\\end{equation*}", "Since $e_k \\cdot e_\\ell = 0$ whenever $k\\not=\\ell$ and $e_k \\cdot e_k = 1$,", "we have $c_k = x \\cdot e_k$. By Cauchy--Schwarz,", "\\begin{equation*}", "\\sabs{c_k} = \\sabs{ x \\cdot e_k }", "\\leq \\snorm{x} \\, \\snorm{e_k} = 1 .", "\\end{equation*}", "Then", "\\begin{equation*}", "\\snorm{Ax} =", "\\norm{\\sum_{k=1}^n c_k \\, Ae_k}", "\\leq", "\\sum_{k=1}^n \\sabs{c_k} \\, \\snorm{Ae_k}", "\\leq", "\\sum_{k=1}^n \\snorm{Ae_k} .", "\\end{equation*}", "The right-hand side does not depend on $x$. We found", "a finite upper bound for $\\snorm{Ax}$ independent of $x$, so $\\snorm{A} < \\infty$.", "Take normed vector spaces $X$ and $Y$, and $A \\in L(X,Y)$ with", "$\\snorm{A} < \\infty$.", "For $v,w \\in X$,", "\\begin{equation*}", "\\snorm{Av - Aw} =", "\\snorm{A(v-w)} \\leq \\snorm{A} \\, \\snorm{v-w} .", "\\end{equation*}", "As $\\snorm{A} < \\infty$, then the inequality above says that", "$A$ is Lipschitz with constant $\\snorm{A}$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 23, "type": "proposition", "label": "Lebl-contfunc:prop:finitedimpropnorm", "categories": [ "norms", "vector_spaces" ], "title": "\\pagebreak[3] Let$X$,$Y$, and$Z$be finite-dimensional normed vector spaces\\footnote{If we strike ...", "contents": [ "\\pagebreak[3] Let$X$,$Y$, and$Z$be finite-dimensional normed vector spaces\\footnote{If we strike the \\myquote{In particular} part and interpret the algebra with infinite operator norms properly, namely decree that$0$times$\\infty$is 0, then this result also holds for infinite-dimensional spaces.}. \\begin{enumerate}[(i)] \\item \\label{item:finitedimpropnorm:i} If$A,B \\in L(X,Y)$and$c \\in \\R$, then\\begin{equation*} \\snorm{A+B} \\leq \\snorm{A}+\\snorm{B}, \\qquad \\snorm{cA} = \\sabs{c} \\, \\snorm{A} . \\end{equation*}In particular, the operator norm is a norm on the vector space$L(X,Y)$. \\item \\label{item:finitedimpropnorm:ii} If$A \\in L(X,Y)$and$B \\in L(Y,Z)$, then\\begin{equation*} \\snorm{BA} \\leq \\snorm{B} \\, \\snorm{A} . \\end{equation*}\\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "First, since all the spaces are finite-dimensional, then all the operator", "norms are finite, and the statements make sense to begin with.", "For \\ref{item:finitedimpropnorm:i}, let $x \\in X$ be arbitrary. Then", "\\begin{equation*}", "\\snorm{(A+B)x} =", "\\snorm{Ax+Bx} \\leq", "\\snorm{Ax}+\\snorm{Bx} \\leq", "\\snorm{A} \\, \\snorm{x}+\\snorm{B} \\,\\snorm{x} =", "\\bigl(\\snorm{A}+\\snorm{B}\\bigr) \\snorm{x} .", "\\end{equation*}", "So $\\snorm{A+B} \\leq \\snorm{A}+\\snorm{B}$.", "Similarly,", "\\begin{equation*}", "\\snorm{(cA)x} =", "\\sabs{c} \\, \\snorm{Ax} \\leq \\bigl(\\sabs{c} \\,\\snorm{A}\\bigr) \\snorm{x} .", "\\end{equation*}", "Thus $\\snorm{cA} \\leq \\sabs{c} \\, \\snorm{A}$. Next,", "\\begin{equation*}", "\\sabs{c} \\, \\snorm{Ax}", "=", "\\snorm{cAx} \\leq \\snorm{cA} \\, \\snorm{x} .", "\\end{equation*}", "Hence $\\sabs{c} \\, \\snorm{A} \\leq \\snorm{cA}$.", "For \\ref{item:finitedimpropnorm:ii}, write", "\\begin{equation*}", "\\snorm{BAx} \\leq \\snorm{B} \\, \\snorm{Ax} \\leq \\snorm{B} \\, \\snorm{A} \\, \\snorm{x} .", "\\qedhere", "\\end{equation*}" ], "refs": [ "item:finitedimpropnorm:i", "item:finitedimpropnorm:ii" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 24, "type": "proposition", "label": "Lebl-contfunc:prop:finitedimpropinv", "categories": [ "norms", "continuity", "vector_spaces" ], "title": "Continuity of Vector Spaces", "contents": [ "Let$X$be a finite-dimensional normed vector space. Let$GL(X) \\subset L(X)$\\glsadd{not:GLX} be the set of invertible linear operators.\\footnote{$GL(X)$is called the \\emph{\\myindex{general linear group}}, that is where the acronym GL comes from.} \\begin{enumerate}[(i)] \\item \\label{finitedimpropinv:i} If$A \\in GL(X)$,$B \\in L(X)$, and \\begin{equation} \\label{eqcontineq} \\snorm{A-B} < \\frac{1}{\\snorm{A^{-1}}}, \\end{equation} then$B \\in GL(X)$, that is,$B$is invertible. In particular,$GL(X)$is open. \\item \\label{finitedimpropinv:ii}$A \\mapsto A^{-1}$is a continuous function on$GL(X)$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "Let us prove \\ref{finitedimpropinv:i}. We know something about $A^{-1}$", "and $A-B$; they are linear operators.", "So apply them to a vector:", "\\begin{equation*}", "A^{-1}(A-B)x", "=", "x-A^{-1}Bx .", "\\end{equation*}", "Therefore,", "\\begin{equation*}", "\\begin{split}", "\\snorm{x}", "& =", "\\snorm{A^{-1} (A-B)x + A^{-1}Bx}", "\\\\", "& \\leq", "\\snorm{A^{-1}}\\,\\snorm{A-B}\\, \\snorm{x} + \\snorm{A^{-1}}\\,\\snorm{Bx} .", "\\end{split}", "\\end{equation*}", "Assume $x \\neq 0$ and so $\\snorm{x} \\neq 0$.", "Using \\eqref{eqcontineq}, we obtain", "\\begin{equation*}", "\\snorm{x} < \\snorm{x} + \\snorm{A^{-1}} \\, \\snorm{Bx} .", "\\end{equation*}", "Thus $\\snorm{Bx} \\not= 0$ for all $x \\not= 0$, and consequently", "$Bx \\not= 0$ for all $x \\not= 0$. So", "$B$ is one-to-one; if $Bx = By$, then $B(x-y) = 0$, so $x=y$.", "As $B$ is a one-to-one linear mapping from $X$ to $X$, which is finite-dimensional,", "it is also onto by \\propref{mv:prop:lin11onto}.", "Therefore, $B$ is invertible.", "It follows that, in particular, $GL(X)$ is open.", "Let us prove \\ref{finitedimpropinv:ii}.", "We must show that the inverse is continuous.", "Fix a $A \\in GL(X)$. Let $B$ be near $A$,", "specifically $\\snorm{A-B} < \\frac{1}{2 \\snorm{A^{-1}}}$.", "Then \\eqref{eqcontineq} is satisfied and $B$ is invertible.", "A similar computation as above (using $B^{-1}y$ instead of $x$) gives", "\\begin{equation*}", "\\snorm{B^{-1}y} \\leq", "\\snorm{A^{-1}} \\, \\snorm{A-B} \\, \\snorm{B^{-1}y} + \\snorm{A^{-1}} \\, \\snorm{y}", "\\leq", "\\frac{1}{2} \\snorm{B^{-1}y} + \\snorm{A^{-1}}\\,\\snorm{y} ,", "\\end{equation*}", "or", "\\begin{equation*}", "\\snorm{B^{-1}y} \\leq", "2\\snorm{A^{-1}}\\,\\snorm{y} .", "\\end{equation*}", "So", "$", "\\snorm{B^{-1}} \\leq 2 \\snorm{A^{-1}}", "$.", "Now", "\\begin{equation*}", "A^{-1}(A-B)B^{-1} =", "A^{-1}(AB^{-1}-I) =", "B^{-1}-A^{-1} ,", "\\end{equation*}", "and", "\\begin{equation*}", "\\snorm{B^{-1}-A^{-1}} =", "\\snorm{A^{-1}(A-B)B^{-1}} \\leq", "\\snorm{A^{-1}}\\,\\snorm{A-B}\\,\\snorm{B^{-1}}", "\\leq", "2\\snorm{A^{-1}}^2", "\\snorm{A-B} .", "\\end{equation*}", "Therefore, as $B$ tends to $A$, $\\snorm{B^{-1}-A^{-1}}$ tends to 0, and", "so the inverse operation is a continuous function at $A$." ], "refs": [ "eqcontineq", "finitedimpropinv:i", "finitedimpropinv:ii", "mv:prop:lin11onto" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 25, "type": "proposition", "label": "Lebl-contfunc:prop:matrixcont", "categories": [ "norms", "continuity", "matrices", "characterization" ], "title": "Characterization of Matrices and Determinants via Equivalence", "contents": [ "The topology (the set of open sets) on$L(\\R^n,\\R^m)$is the same whether we consider$L(\\R^n,\\R^m)$as a metric space using the operator norm, or the euclidean metric of$\\R^{nm}$.", "In particular, let$S$be a metric space and let$\\pi \\colon L(\\R^n,\\R^m) \\to \\R^{nm}$identify an operator with the$nm$-tuple of entries of the corresponding matrix. Then$f \\colon S \\to L(\\R^n,\\R^m)$is continuous if and only if$\\pi \\circ f \\colon S \\to \\R^{nm}$is continuous. Similarly for$g \\colon L(\\R^n,\\R^m) \\to S$and$g \\circ \\pi^{-1} \\colon \\R^{nm} \\to S$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 26, "type": "proposition", "label": "Lebl-contfunc:26", "categories": [ "continuity", "matrices", "vector_spaces" ], "title": "Continuity of Vector Spaces", "contents": [ "\\pagebreak[0] \\leavevmode \\begin{enumerate}[(i)] \\item \\label{prop:det:i}$\\det(I) = 1$. \\item \\label{prop:det:ii} For every$j=1,2,\\ldots,n$, the function$x_j \\mapsto \\det\\bigl([x_1 ~~~ x_2 ~~~ \\cdots ~~~ x_n ]\\bigr)$is linear. \\item \\label{prop:det:iii} If two columns of a matrix are interchanged, then the determinant changes sign. \\item \\label{prop:det:iv} If two columns of$A$are equal, then$\\det(A) = 0$. \\item \\label{prop:det:v} If a column is zero, then$\\det(A) = 0$. \\item \\label{prop:det:vi}$A \\mapsto \\det(A)$is a continuous function on$L(\\R^n)$. \\item \\label{prop:det:vii} $\\det\\left( \\left[\\begin{smallmatrix} a & b", "c &d \\end{smallmatrix}\\right] \\right) = ad-bc$, and $\\det \\bigl( [a] \\bigr) = a$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "We go through the proof quickly, as you have likely seen it before.", "%", "Item \\ref{prop:det:i} is trivial. For \\ref{prop:det:ii}, note that each term in the definition of the", "determinant contains exactly one factor from each column.", "%", "Item \\ref{prop:det:iii} follows as switching two columns is switching the", "two corresponding numbers in every element in $S_n$. Hence, all the signs", "are changed.", "Item \\ref{prop:det:iv} follows because if two columns are equal, and we", "switch them, we get", "the same matrix back. So item \\ref{prop:det:iii} says the determinant must be 0.", "%", "Item \\ref{prop:det:v} follows because the product in each term in the definition includes", "one element from the zero column.", "Item \\ref{prop:det:vi} follows as $\\det$ is a polynomial in the entries of the matrix", "and hence continuous (as a function of the entries of the matrix).", "A function defined on", "matrices is continuous in the operator norm if and only if it is", "continuous as a function of the entries (\\propref{prop:matrixcont}).", "Finally, item \\ref{prop:det:vii} is a direct computation." ], "refs": [ "prop:det:i", "prop:det:ii", "prop:det:iii", "prop:det:iv", "prop:det:v", "prop:det:vi", "prop:det:vii", "prop:matrixcont" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 27, "type": "proposition", "label": "Lebl-contfunc:27", "categories": [ "characterization" ], "title": "Characterization of Several Variables and Derivatives via Equivalence", "contents": [ "If$A$and$B$are$n$-by-$n$matrices, then$\\det(AB) = \\det(A)\\det(B)$. Furthermore,$A$is invertible if and only if$\\det(A) \\not= 0$and in this case,$\\det(A^{-1}) = \\frac{1}{\\det(A)}$." ], "refs": [], "proofs": [ { "contents": [ "Let $b_1,b_2,\\ldots,b_n$ be the columns of $B$. Then", "\\begin{equation*}", "AB = [ Ab_1 \\quad Ab_2 \\quad \\cdots \\quad Ab_n ] .", "\\end{equation*}", "That is, the columns of $AB$ are", "$Ab_1,Ab_2,\\ldots,Ab_n$.", "Let $b_{j,k}$ denote the elements of $B$ and", "$a_j$ the columns of $A$.", "By linearity of the determinant,", "\\begin{equation*}", "\\begin{split}", "\\det(AB) & =", "\\det \\bigl([ Ab_1 \\quad Ab_2 \\quad \\cdots \\quad Ab_n ] \\bigr) =", "\\det \\left(\\left[ \\sum_{j=1}^n b_{j,1} a_j \\quad Ab_2 \\quad \\cdots \\quad Ab_n \\right]\\right) \\\\", "& =", "\\sum_{j=1}^n", "b_{j,1}", "\\det \\bigl([ a_j \\quad Ab_2 \\quad \\cdots \\quad Ab_n ]\\bigr) \\\\", "& =", "\\sum_{1 \\leq j_1,j_2,\\ldots,j_n \\leq n}", "b_{j_1,1}", "b_{j_2,2}", "\\cdots", "b_{j_n,n}", "\\det \\bigl([ a_{j_1} \\quad a_{j_2} \\quad \\cdots \\quad a_{j_n} ]\\bigr) \\\\", "& =", "\\left(", "\\sum_{(j_1,j_2,\\ldots,j_n) \\in S_n}", "b_{j_1,1}", "b_{j_2,2}", "\\cdots", "b_{j_n,n}", "\\operatorname{sgn}(j_1,j_2,\\ldots,j_n)", "\\right)", "\\det \\bigl([ a_{1} \\quad a_{2} \\quad \\cdots \\quad a_{n} ]\\bigr) .", "\\end{split}", "\\end{equation*}", "In the last equality, we sum over the elements of $S_n$", "instead of all $n$-tuples for integers between 1 and $n$,", "because", "when two columns in the determinant are the same, then the", "determinant is zero. Reordering the columns to the", "original ordering to obtains the sgn.", "The conclusion that $\\det(AB) = \\det(A)\\det(B)$", "follows by recognizing that the expression in parentheses above is the determinant of $B$.", "We obtain this by plugging in $A=I$.", "The expression we get for the determinant of $B$ has rows and columns", "swapped, so as a bonus, we have also just proved that the determinant of", "a matrix and its transpose are equal.", "Let us prove the", "\\myquote{Furthermore.}", "If $A$ is invertible,", "then $A^{-1}A = I$.", "Consequently $\\det(A^{-1})\\det(A) = \\det(A^{-1}A) = \\det(I) = 1$.", "If $A$ is not invertible, then it is not one-to-one, and so $A$ takes", "some nonzero vector to zero.", "In other words, the columns of $A$ are linearly dependent.", "Suppose", "\\begin{equation*}", "\\sum_{k=1}^n \\gamma_k\\, a_k = 0 ,", "\\end{equation*}", "where not all $\\gamma_k$ are equal to 0.", "Without loss of generality, suppose $\\gamma_1\\neq 0$.", "Take", "\\begin{equation*}", "B \\coloneqq", "\\begin{bmatrix}", "\\gamma_1 & 0 & 0 & \\cdots & 0 \\\\", "\\gamma_2 & 1 & 0 & \\cdots & 0 \\\\", "\\gamma_3 & 0 & 1 & \\cdots & 0 \\\\", "\\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\", "\\gamma_n & 0 & 0 & \\cdots & 1", "\\end{bmatrix} .", "\\end{equation*}", "Using the definition of the determinant (there is only a single", "permutation $\\sigma$ for which $\\prod_{i=1}^n b_{i,\\sigma_i}$ is nonzero)", "we find $\\det(B) = \\gamma_1 \\not= 0$.", "Then", "$\\det(AB) = \\det(A)\\det(B) = \\gamma_1\\det(A)$.", "The first column of $AB$ is zero, and hence $\\det(AB) = 0$. We conclude", "$\\det(A) = 0$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 28, "type": "proposition", "label": "Lebl-contfunc:28", "categories": [ "matrices", "vector_spaces" ], "title": "Determinant is independent of the basis: If$A$and$B$are$n$-by-$n$matrices and$B$is invertible, th...", "contents": [ "Determinant is independent of the basis: If$A$and$B$are$n$-by-$n$matrices and$B$is invertible, then\\begin{equation*} \\det(A) = \\det(B^{-1}AB) . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "$\\det(B^{-1}AB) =", "\\det(B^{-1})\\det(A)\\det(B) =", "\\frac{1}{\\det(B)}\\det(A)\\det(B) = \\det(A)$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 29, "type": "proposition", "label": "Lebl-contfunc:prop:elemmatrixdecomp", "categories": [ "matrices" ], "title": "Existence of Matrices and Determinants", "contents": [ "Let$T$be an$n$-by-$n$invertible matrix. Then there exists a finite sequence of elementary matrices$E_1, E_2, \\ldots, E_k$such that\\begin{equation*} T = E_1 E_2 \\cdots E_k , \\end{equation*}and\\begin{equation*} \\det(T) = \\det(E_1)\\det(E_2)\\cdots \\det(E_k) . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 30, "type": "proposition", "label": "Lebl-contfunc:30", "categories": [ "norms" ], "title": "Let$U \\subset \\R^n$be an open subset and$f \\colon U \\to \\R^m$a function", "contents": [ "Let$U \\subset \\R^n$be an open subset and$f \\colon U \\to \\R^m$a function. Suppose$x \\in U$and there exist$A,B \\in L(\\R^n,\\R^m)$such that\\begin{equation*} \\lim_{h \\to 0} \\frac{\\snorm{f(x+h)-f(x) - Ah}}{\\snorm{h}} = 0 \\qquad \\text{and} \\qquad \\lim_{h \\to 0} \\frac{\\snorm{f(x+h)-f(x) - Bh}}{\\snorm{h}} = 0 . \\end{equation*}Then$A=B$." ], "refs": [], "proofs": [ { "contents": [ "Suppose $h \\in \\R^n$, $h \\not= 0$. Compute", "\\begin{equation*}", "\\begin{split}", "\\frac{\\snorm{(A-B)h}}{\\snorm{h}} & =", "\\frac{\\snorm{-\\bigl(f(x+h)-f(x) - Ah\\bigr) + f(x+h)-f(x) - Bh}}{\\snorm{h}} \\\\", "& \\leq", "\\frac{\\snorm{f(x+h)-f(x) - Ah}}{\\snorm{h}} + \\frac{\\snorm{f(x+h)-f(x) -", "Bh}}{\\snorm{h}} .", "\\end{split}", "\\end{equation*}", "So", "$\\frac{\\snorm{(A-B)h}}{\\snorm{h}} \\to 0$ as $h \\to 0$. Given", "$\\epsilon > 0$, for all nonzero $h$ in some $\\delta$-ball around", "the origin we have", "\\begin{equation*}", "\\epsilon >", "\\frac{\\snorm{(A-B)h}}{\\snorm{h}}", "=", "\\norm{(A-B)\\frac{h}{\\snorm{h}}} .", "\\end{equation*}", "For any given $v \\in \\R^n$ with $\\snorm{v}=1$,", "if $h = (\\nicefrac{\\delta}{2}) \\, v$, then $\\snorm{h} < \\delta$", "and $\\frac{h}{\\snorm{h}} = v$.", "So $\\snorm{(A-B)v} < \\epsilon$. Taking the supremum over all $v$ with", "$\\snorm{v} = 1$, we get the operator norm", "$\\snorm{A-B} \\leq \\epsilon$. As $\\epsilon > 0$", "was arbitrary, $\\snorm{A-B} = 0$, or in other words $A = B$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 31, "type": "example", "label": "Lebl-contfunc:31", "categories": [ "norms", "vector_spaces", "example" ], "title": "If$f(x) = Ax$for a linear mapping$A$, then$f'(x) = A$: \\begin{equation*} \\frac{\\snorm{f(x+h)-f(x)...", "contents": [ "If$f(x) = Ax$for a linear mapping$A$, then$f'(x) = A$:", "\\begin{equation*}\n\\frac{\\snorm{f(x+h)-f(x) - Ah}}{\\snorm{h}}\n=\n\\frac{\\snorm{A(x+h)-Ax - Ah}}{\\snorm{h}}\n=\n\\frac{0}{\\snorm{h}} = 0 .\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 32, "type": "example", "label": "Lebl-contfunc:32", "categories": [ "norms", "matrices", "example", "derivatives", "continuity" ], "title": "Existence of Derivatives", "contents": [ "Let$f \\colon \\R^2 \\to \\R^2$be defined by", "\\begin{equation*}\nf(x,y) = \\bigl(f_1(x,y),f_2(x,y)\\bigr) \\coloneqq (1+x+2y+x^2,2x+3y+xy).\n\\end{equation*}", "Let us show that$f$is differentiable at the origin and compute the derivative directly using the definition. If the derivative exists, it is in$L(\\R^2,\\R^2)$, so it can be represented by a$2$-by-$2$matrix $\\left[\\begin{smallmatrix}a&b", "c&d\\end{smallmatrix}\\right]$. Suppose $h = (h_1,h_2)$. We need the following expression to go to zero. \\begin{multline*} \\frac{\\snorm{ f(h_1,h_2)-f(0,0) - (ah_1 +bh_2 , ch_1+dh_2)} }{\\snorm{(h_1,h_2)}} =", "\\frac{\\sqrt{ {\\bigl((1-a)h_1 + (2-b)h_2 + h_1^2\\bigr)}^2 + {\\bigl((2-c)h_1 + (3-d)h_2 + h_1h_2\\bigr)}^2}}{\\sqrt{h_1^2+h_2^2}} . \\end{multline*} If we choose$a=1$,$b=2$,$c=2$,$d=3$, the expression becomes", "\\begin{equation*}\n\\frac{\\sqrt{\nh_1^4 + h_1^2h_2^2}}{\\sqrt{h_1^2+h_2^2}}\n=\n\\sabs{h_1}\n\\frac{\\sqrt{\nh_1^2 + h_2^2}}{\\sqrt{h_1^2+h_2^2}}\n= \\sabs{h_1} .\n\\end{equation*}", "This expression does indeed go to zero as$h \\to 0$. The function$f$is differentiable at the origin and the derivative$f'(0)$is represented by the matrix $\\left[\\begin{smallmatrix}1&2", "2&3\\end{smallmatrix}\\right]$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 33, "type": "proposition", "label": "Lebl-contfunc:33", "categories": [ "continuity" ], "title": "Continuity of Continuity and Differentiability", "contents": [ "Let$U \\subset \\R^n$be open and$f \\colon U \\to \\R^m$be differentiable at$p \\in U$. Then$f$is continuous at$p$." ], "refs": [], "proofs": [ { "contents": [ "Another way to write the differentiability of $f$ at $p$ is to consider", "\\begin{equation*}", "r(h) \\coloneqq f(p+h)-f(p) - f'(p) h .", "\\end{equation*}", "The function $f$ is differentiable at $p$ if", "$\\frac{\\snorm{r(h)}}{\\snorm{h}}$ goes to zero as $h \\to 0$,", "so", "$r(h)$ itself goes to zero. The mapping $h \\mapsto f'(p) h$", "is a linear mapping between finite-dimensional spaces, hence continuous", "and $f'(p) h \\to 0$ as $h \\to 0$. Thus,", "$f(p+h)$ must go to $f(p)$ as $h \\to 0$. That is, $f$ is continuous at $p$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 34, "type": "proposition", "label": "Lebl-contfunc:34", "categories": [ "continuity" ], "title": "Suppose$U \\subset \\R^n$is open,$f \\colon U \\to \\R^m$and$g \\colon U \\to \\R^m$are differentiable at...", "contents": [ "Suppose$U \\subset \\R^n$is open,$f \\colon U \\to \\R^m$and$g \\colon U \\to \\R^m$are differentiable at$p \\in U$, and$\\alpha \\in \\R$. Then the functions$f+g$and$\\alpha f$are differentiable at$p$,\\begin{equation*} (f+g)'(p) = f'(p) + g'(p) , \\qquad \\text{and} \\qquad (\\alpha f)'(p) = \\alpha f'(p) . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let $h \\in \\R^n$, $h \\not= 0$. Then", "\\begin{multline*}", "\\frac{\\norm{f(p+h)+g(p+h)-\\bigl(f(p)+g(p)\\bigr) - \\bigl(f'(p) + g'(p)\\bigr)h}}{\\snorm{h}}", "\\\\", "\\leq", "\\frac{\\norm{f(p+h)-f(p) - f'(p)h}}{\\snorm{h}}", "+", "\\frac{\\norm{g(p+h)-g(p) - g'(p)h}}{\\snorm{h}} ,", "\\end{multline*}", "and", "\\begin{equation*}", "\\frac{\\norm{\\alpha f(p+h) - \\alpha f(p) - \\alpha f'(p)h}}{\\snorm{h}}", "=", "\\sabs{\\alpha} \\frac{\\norm{f(p+h))-f(p) - f'(p)h}}{\\snorm{h}} .", "\\end{equation*}", "The limits as $h$ goes to zero of the right-hand sides are zero by", "hypothesis. The result follows." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 35, "type": "theorem", "label": "Lebl-contfunc:35", "categories": [ "continuity", "functions", "chain_rule" ], "title": "Chain Rule", "contents": [ "\\index{chain rule} Let$U \\subset \\R^n$and$V \\subset \\R^m$be open sets,$f \\colon U \\to \\R^m$be differentiable at$p \\in U$,$f(U) \\subset V$, and let$g \\colon V \\to \\R^\\ell$be differentiable at$f(p)$. Then$F \\colon U \\to \\R^{\\ell}$defined by\\begin{equation*} F(x) \\coloneqq g\\bigl(f(x)\\bigr) \\end{equation*}is differentiable at$p$, and\\begin{equation*} F'(p) = g'\\bigl(f(p)\\bigr) f'(p) . \\end{equation*}" ], "refs": [ "named:Chain Rule" ], "proofs": [ { "contents": [ "Let $A \\coloneqq f'(p)$ and $B \\coloneqq g'\\bigl(f(p)\\bigr)$. Take a nonzero $h \\in \\R^n$", "and write $q \\coloneqq f(p)$, $k \\coloneqq f(p+h)-f(p)$. Let", "\\begin{equation*}", "r(h) \\coloneqq f(p+h)-f(p) - A h . %= k - Ah.", "\\end{equation*}", "Then $r(h) = k-Ah$ or $Ah = k-r(h)$, and $f(p+h) = q+k$.", "We look at the quantity we need to go", "to zero:", "\\begin{equation*}", "\\begin{split}", "\\frac{\\snorm{F(p+h)-F(p) - BAh}}{\\snorm{h}}", "& =", "\\frac{\\snorm{g\\bigl(f(p+h)\\bigr)-g\\bigl(f(p)\\bigr) - BAh}}{\\snorm{h}}", "\\\\", "& =", "\\frac{\\snorm{g(q+k)-g(q) - B\\bigl(k-r(h)\\bigr)}}{\\snorm{h}}", "\\\\", "& \\leq", "\\frac", "{\\snorm{g(q+k)-g(q) - Bk}}", "{\\snorm{h}}", "+", "\\snorm{B}", "\\frac", "{\\snorm{r(h)}}", "{\\snorm{h}}", "\\\\", "& =", "\\frac", "{\\snorm{g(q+k)-g(q) - Bk}}", "{\\snorm{k}}", "\\frac", "{\\snorm{f(p+h)-f(p)}}", "{\\snorm{h}}", "+", "\\snorm{B}", "\\frac", "{\\snorm{r(h)}}", "{\\snorm{h}} .", "\\end{split}", "\\end{equation*}", "First, $\\snorm{B}$ is a constant and $f$ is differentiable at $p$,", "so", "the term $\\snorm{B}\\frac{\\snorm{r(h)}}{\\snorm{h}}$ goes to 0.", "Next, because $f$ is continuous at $p$,", "$k$ goes to 0 as $h$ goes to 0.", "Thus", "$\\frac", "{\\snorm{g(q+k)-g(q) - Bk}}", "{\\snorm{k}}$ goes to 0, because $g$ is differentiable at $q$.", "Finally,", "\\begin{equation*}", "\\frac", "{\\snorm{f(p+h)-f(p)}}", "{\\snorm{h}}", "\\leq", "\\frac", "{\\snorm{f(p+h)-f(p)-Ah}}", "{\\snorm{h}}", "+", "\\frac", "{\\snorm{Ah}}", "{\\snorm{h}}", "\\leq", "\\frac", "{\\snorm{f(p+h)-f(p)-Ah}}", "{\\snorm{h}}", "+", "\\snorm{A} .", "\\end{equation*}", "As $f$ is differentiable at $p$,", "for small enough $h$, the quantity", "$\\frac{\\snorm{f(p+h)-f(p)-Ah}}{\\snorm{h}}$ is bounded. Hence, the", "term", "$", "\\frac", "{\\snorm{f(p+h)-f(p)}}", "{\\snorm{h}}", "$", "stays bounded as $h$ goes to 0. Therefore,", "$\\frac{\\snorm{F(p+h)-F(p) - BAh}}{\\snorm{h}}$ goes to zero, and", "$F'(p) = BA$, which is what was claimed." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 36, "type": "proposition", "label": "Lebl-contfunc:mv:prop:jacobianmatrix", "categories": [ "continuity", "matrices", "higher_order", "derivatives" ], "title": "Let$U \\subset \\R^n$be open and let$f \\colon U \\to \\R^m$be differentiable at$p \\in U$", "contents": [ "Let$U \\subset \\R^n$be open and let$f \\colon U \\to \\R^m$be differentiable at$p \\in U$. Then all the partial derivatives at$p$exist and, in terms of the standard bases of$\\R^n$and$\\R^m$,$f'(p)$is represented by the matrix \\begin{equation*} \\begin{bmatrix} \\frac{\\partial f_1}{\\partial x_1}(p) & \\frac{\\partial f_1}{\\partial x_2}(p) & \\ldots & \\frac{\\partial f_1}{\\partial x_n}(p)", "[6pt] \\frac{\\partial f_2}{\\partial x_1}(p) & \\frac{\\partial f_2}{\\partial x_2}(p) & \\ldots & \\frac{\\partial f_2}{\\partial x_n}(p)", "\\vdots & \\vdots & \\ddots & \\vdots", "\\frac{\\partial f_m}{\\partial x_1}(p) & \\frac{\\partial f_m}{\\partial x_2}(p) & \\ldots & \\frac{\\partial f_m}{\\partial x_n}(p) \\end{bmatrix} . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Fix a $j$ and note that for nonzero $h$,", "\\begin{equation*}", "\\begin{split}", "\\norm{\\frac{f(p+h e_j)-f(p)}{h} - f'(p) \\, e_j} & =", "\\norm{\\frac{f(p+h e_j)-f(p) - f'(p) \\, h e_j}{h}} \\\\", "& =", "\\frac{\\snorm{f(p+h e_j)-f(p) - f'(p) \\, h e_j}}{\\snorm{h e_j}} .", "\\end{split}", "\\end{equation*}", "As $h$ goes to 0, the right-hand side goes to zero by", "differentiability of $f$. Hence,", "\\begin{equation*}", "\\lim_{h \\to 0}", "\\frac{f(p+h e_j)-f(p)}{h} = f'(p) \\, e_j .", "\\end{equation*}", "The limit is in $\\R^m$.", "Represent $f$ in components", "$f = (f_1,f_2,\\ldots,f_m)$.", "Taking a limit in $\\R^m$", "is the same as taking the limit in each component separately.", "So for every $k$,", "the partial derivative", "\\begin{equation*}", "\\frac{\\partial f_k}{\\partial x_j} (p)", "=", "\\lim_{h \\to 0}", "\\frac{f_k(p+h e_j)-f_k(p)}{h}", "\\end{equation*}", "exists and is equal to the $k$th component of $f'(p)\\, e_j$, which is the", "$j$th column of $f'(p)$,", "and we are done." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 37, "type": "lemma", "label": "Lebl-contfunc:lemma:mvtmv", "categories": [ "norms", "continuity", "auxiliary result" ], "title": "Continuity of Continuity and Differentiability", "contents": [ "If$\\varphi \\colon [a,b] \\to \\R^n$is differentiable on$(a,b)$and continuous on$[a,b]$, then there exists a$t_0 \\in (a,b)$such that\\begin{equation*} \\snorm{\\varphi(b)-\\varphi(a)} \\leq (b-a) \\snorm{\\varphi'(t_0)} . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "By the mean value theorem on the scalar-valued function", "$t \\mapsto \\bigl(\\varphi(b)-\\varphi(a) \\bigr) \\cdot \\varphi(t)$,", "where the dot is the dot product, we obtain", "a $t_0 \\in (a,b)$ such that", "\\begin{equation*}", "\\begin{split}", "\\snorm{\\varphi(b)-\\varphi(a)}^2", "& =", "\\bigl( \\varphi(b)-\\varphi(a) \\bigr)", "\\cdot", "\\bigl( \\varphi(b)-\\varphi(a) \\bigr)", "\\\\", "& =", "\\bigl(\\varphi(b)-\\varphi(a) \\bigr) \\cdot \\varphi(b) -", "\\bigl(\\varphi(b)-\\varphi(a) \\bigr) \\cdot \\varphi(a)", "\\\\", "& =", "(b-a)", "\\bigl(\\varphi(b)-\\varphi(a) \\bigr) \\cdot \\varphi'(t_0) ,", "\\end{split}", "\\end{equation*}", "where we treat $\\varphi'$ as a vector in $\\R^n$ by the abuse of", "notation we mentioned in the previous section.", "If we think of $\\varphi'(t)$ as a vector, then by", "\\exerciseref{exercise:normonedim},", "$\\snorm{\\varphi'(t)}_{L(\\R,\\R^n)} = \\snorm{\\varphi'(t)}_{\\R^n}$.", "That is, the euclidean norm of the vector is the same as the operator norm", "of $\\varphi'(t)$.", "By the Cauchy--Schwarz inequality", "\\begin{equation*}", "\\snorm{\\varphi(b)-\\varphi(a)}^2", "=", "(b-a)\\bigl(\\varphi(b)-\\varphi(a) \\bigr) \\cdot \\varphi'(t_0)", "\\leq", "(b-a)", "\\snorm{\\varphi(b)-\\varphi(a)} \\, \\snorm{\\varphi'(t_0)} . \\qedhere", "\\end{equation*}" ], "refs": [ "named:Mean Value" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 38, "type": "proposition", "label": "Lebl-contfunc:mv:prop:convexlip", "categories": [ "convexity", "continuity", "norms" ], "title": "Let$U \\subset \\R^n$be a convex open set,$f \\colon U \\to \\R^m$be a differentiable function, and an...", "contents": [ "Let$U \\subset \\R^n$be a convex open set,$f \\colon U \\to \\R^m$be a differentiable function, and an$M$be such that\\begin{equation*} \\snorm{f'(p)} \\leq M \\qquad \\text{for all } p \\in U. \\end{equation*}Then$f$is Lipschitz with constant$M$, that is,\\begin{equation*} \\snorm{f(p)-f(q)} \\leq M \\snorm{p-q} \\qquad \\text{for all } p,q \\in U. \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Fix $p$ and $q$ in $U$ and note that", "$(1-t)p+tq \\in U$ for all $t \\in [0,1]$", "by convexity.", "Next", "\\begin{equation*}", "\\frac{d}{dt} \\Bigl[f\\bigl((1-t)p+tq\\bigr)\\Bigr]", "=", "f'\\bigl((1-t)p+tq\\bigr) (q-p) .", "\\end{equation*}", "By \\lemmaref{lemma:mvtmv}, there is some", "$t_0 \\in (0,1)$ such that", "\\begin{equation*}", "\\begin{split}", "\\snorm{f(p)-f(q)} & \\leq", "\\norm{\\frac{d}{dt} \\Big|_{t=t_0} \\Bigl[ f\\bigl((1-t)p+tq\\bigr) \\Bigr] }", "\\\\", "& \\leq", "\\norm{f'\\bigl((1-t_0)p+t_0q\\bigr)} \\, \\snorm{q-p} \\leq", "M \\snorm{q-p} . \\qedhere", "\\end{split}", "\\end{equation*}" ], "refs": [ "lemma:mvtmv", "named:Mean Value" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 39, "type": "example", "label": "Lebl-contfunc:39", "categories": [ "convexity", "norms", "example", "derivatives", "continuity" ], "title": "Boundedness of Derivatives", "contents": [ "If$U$is not convex the proposition is not true: Consider the set", "\\begin{equation*}\nU \\coloneqq \\bigl\\{ (x,y) : 0.5 < x^2+y^2 < 2 \\bigr\\}\n\\setminus \\bigl\\{ (x,0) : x < 0 \\bigr\\} .\n\\end{equation*}", "For$(x,y) \\in U$, let$f(x,y)$be the angle that the line from the origin to$(x,y)$makes with the positive$x$axis. We even have a formula for$f$:", "\\begin{equation*}\nf(x,y) = 2 \\operatorname{arctan}\\left( \\frac{y}{x+\\sqrt{x^2+y^2}}\\right) .\n\\end{equation*}", "Think a spiral staircase with room in the middle. See \\figureref{mv:fignonlip}.", "\\begin{myfigureht} \\subimport*{figures/}{nonlip_full.pdf_t} \\caption{A non-Lipschitz function with uniformly bounded derivative.\\label{mv:fignonlip}} \\end{myfigureht}", "The function is differentiable, and the derivative is bounded on$U$, which is not hard to see. Now think of what happens near where the negative$x$-axis cuts the annulus in half. As we approach this cut from positive$y$,$f(x,y)$approaches$\\pi$. From negative$y$,$f(x,y)$approaches$-\\pi$. So for small$\\epsilon > 0$,$\\sabs{f(-1,\\epsilon)-f(-1,-\\epsilon)}$approaches$2\\pi$, but$\\snorm{(-1,\\epsilon)-(-1,-\\epsilon)} = 2\\epsilon$, which is arbitrarily small. The conclusion of the proposition does not hold for this nonconvex$U$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 40, "type": "corollary", "label": "Lebl-contfunc:40", "categories": [ "consequence", "continuity" ], "title": "If$U \\subset \\R^n$is open and connected,$f \\colon U \\to \\R^m$is differentiable, and$f'(x) = 0$for...", "contents": [ "If$U \\subset \\R^n$is open and connected,$f \\colon U \\to \\R^m$is differentiable, and$f'(x) = 0$for all$x \\in U$, then$f$is constant." ], "refs": [], "proofs": [ { "contents": [ "For any given $x \\in U$, there is a ball $B(x,\\delta) \\subset U$. The ball", "$B(x,\\delta)$ is convex. Since", "$\\snorm{f'(y)} \\leq 0$ for all $y \\in B(x,\\delta)$, then by the proposition,", "$\\snorm{f(x)-f(y)} \\leq 0 \\snorm{x-y} = 0$. So $f(x) = f(y)$ for all $y \\in", "B(x,\\delta)$.", "Therefore, $f^{-1}(c)$ is open for all $c \\in \\R^m$.", "Suppose $c_0 \\in \\R^m$ is such that", "$f^{-1}(c_0) \\not= \\emptyset$.", "As $f$ is also continuous,", "the two sets", "\\begin{equation*}", "U' = f^{-1}(c_0), \\qquad U'' = f^{-1}\\bigl(\\R^m\\setminus\\{c_0\\}\\bigr)", "\\end{equation*}", "are open and disjoint, and further $U = U' \\cup U''$. As $U'$ is nonempty", "and $U$ is connected, then", "$U'' = \\emptyset$. So $f(x) = c_0$ for all $x \\in U$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 41, "type": "proposition", "label": "Lebl-contfunc:mv:prop:contdiffpartials", "categories": [ "continuity", "characterization", "higher_order", "derivatives" ], "title": "Characterization of Derivatives via Equivalence", "contents": [ "Let$U \\subset \\R^n$be open and$f \\colon U \\to \\R^m$. The function$f$is continuously differentiable if and only if the partial derivatives$\\frac{\\partial f_k}{\\partial x_j}$exist for all$k$and$j$and are continuous." ], "refs": [], "proofs": [ { "contents": [ "We proved that if $f$ is differentiable, then", "the partial derivatives exist. The partial", "derivatives are the entries of the matrix representing $f'(x)$. If", "$f' \\colon U \\to L(\\R^n,\\R^m)$ is continuous, then the entries are", "continuous, and hence the partial derivatives are continuous.", "To prove the opposite direction,", "suppose the partial derivatives exist and are continuous.", "Fix $x \\in U$. If we show that $f'(x)$ exists we are done, because", "the entries of the matrix representing $f'(x)$ are the partial", "derivatives and if the entries are continuous functions,", "the matrix-valued function $f'$ is continuous.", "We do induction on dimension. First,", "the conclusion is true when $n=1$", "(exercise, note that $f$ is vector-valued).", "In this case, $f'(x)$ is essentially the derivative of", "\\volIref{\\chapterref*{vI-der:chapter}}{\\chapterref{der:chapter}}.", "Suppose the conclusion is true for $\\R^{n-1}$.", "That is,", "if we restrict to the first $n-1$ variables, the function is differentiable.", "When taking the partial derivatives in $x_1$ through $x_{n-1}$,", "it does not matter if we consider $f$ or $f$ restricted to the set where", "$x_n$ is fixed.", "In the following, by a slight abuse of notation,", "we think of $\\R^{n-1}$ as a subset of $\\R^n$, that is, the set in $\\R^n$ where $x_n = 0$.", "In other words, we identify the vectors $(x_1,x_2,\\ldots,x_{n-1})$ and", "$(x_1,x_2,\\ldots,x_{n-1},0)$.", "Fix $p \\in U$ and let", "\\begin{equation*}", "A \\coloneqq", "\\begin{bmatrix}", "\\frac{\\partial f_1}{\\partial x_1}(p)", "& \\ldots &", "\\frac{\\partial f_1}{\\partial x_n}(p)", "\\\\", "\\vdots & \\ddots & \\vdots", "\\\\", "\\frac{\\partial f_m}{\\partial x_1}(p)", "& \\ldots &", "\\frac{\\partial f_m}{\\partial x_n}(p)", "\\end{bmatrix} ,", "\\qquad", "A' \\coloneqq", "\\begin{bmatrix}", "\\frac{\\partial f_1}{\\partial x_1}(p)", "& \\ldots &", "\\frac{\\partial f_1}{\\partial x_{n-1}}(p)", "\\\\", "\\vdots & \\ddots & \\vdots", "\\\\", "\\frac{\\partial f_m}{\\partial x_1}(p)", "& \\ldots &", "\\frac{\\partial f_m}{\\partial x_{n-1}}(p)", "\\end{bmatrix} ,", "\\qquad", "v \\coloneqq", "\\begin{bmatrix}", "\\frac{\\partial f_1}{\\partial x_n}(p)", "\\\\", "\\vdots", "\\\\", "\\frac{\\partial f_m}{\\partial x_n}(p)", "\\end{bmatrix} .", "\\end{equation*}", "Let $\\epsilon > 0$ be given. By the induction hypothesis, there", "is a $\\delta > 0$ such that", "for every $h' \\in \\R^{n-1}$ with $\\snorm{h'} < \\delta$, we have", "\\begin{equation*}", "\\frac{\\snorm{f(p+h') - f(p) - A' h'}}{\\snorm{h'}} < \\epsilon .", "\\end{equation*}", "By continuity of the partial derivatives, suppose $\\delta$ is small", "enough so that", "\\begin{equation*}", "\\abs{\\frac{\\partial f_k}{\\partial x_n}(p+h)", "- \\frac{\\partial f_k}{\\partial x_n}(p)} < \\epsilon", "\\end{equation*}", "for all $k$ and all $h \\in \\R^n$ with $\\snorm{h} < \\delta$.", "Suppose $h = h' + t e_n$ is a vector in $\\R^n$, where $h' \\in \\R^{n-1}$,", "$t \\in \\R$, such that", "$\\snorm{h} < \\delta$. Then $\\snorm{h'} \\leq \\snorm{h} < \\delta$.", "Note that $Ah = A' h' + tv$.", "\\begin{equation*}", "\\begin{split}", "\\snorm{f(p+h) - f(p) - Ah}", "& = \\snorm{f(p+h' + t e_n) - f(p+h') - tv + f(p+h') - f(p) - A' h'}", "\\\\", "& \\leq \\snorm{f(p+h' + t e_n) - f(p+h') -tv} + \\snorm{f(p+h') - f(p) -", "A' h'}", "\\\\", "& \\leq \\snorm{f(p+h' + t e_n) - f(p+h') -tv} + \\epsilon \\snorm{h'} .", "\\end{split}", "\\end{equation*}", "As all the partial derivatives exist, by the mean value theorem,", "for each $k$ there is some $\\theta_k \\in [0,t]$ (or $[t,0]$ if $t < 0$), such that", "\\begin{equation*}", "f_k(p+h' + t e_n) - f_k(p+h') =", "t \\frac{\\partial f_k}{\\partial x_n}(p+h'+\\theta_k e_n).", "\\end{equation*}", "We have $\\snorm{h'+\\theta_k e_n} \\leq \\snorm{h} < \\delta$,", "and so we can finish the estimate", "\\begin{equation*}", "\\begin{split}", "\\snorm{f(p+h) - f(p) - Ah}", "& \\leq \\snorm{f(p+h' + t e_n) - f(p+h') -tv} + \\epsilon \\snorm{h'}", "\\\\", "& \\leq \\sqrt{\\sum_{k=1}^m {\\left(t\\frac{\\partial f_k}{\\partial x_n}(p+h'+\\theta_k e_n) -", "t \\frac{\\partial f_k}{\\partial x_n}(p)\\right)}^2} + \\epsilon \\snorm{h'}", "\\\\", "& \\leq \\sqrt{m}\\, \\epsilon \\sabs{t} + \\epsilon \\snorm{h'}", "\\\\", "& \\leq (\\sqrt{m}+1)\\epsilon \\snorm{h} . \\qedhere", "\\end{split}", "\\end{equation*}" ], "refs": [ "named:Mean Value" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 42, "type": "corollary", "label": "Lebl-contfunc:42", "categories": [ "consequence", "continuity" ], "title": "Continuity of Continuity and Differentiability", "contents": [ "A polynomial$p \\colon \\R^n \\to \\R$in several variables\\begin{equation*} p(x_1,x_2,\\ldots,x_n) = \\sum_{0 \\leq j_1+j_2+\\cdots+j_n \\leq d} c_{j_1,j_2,\\ldots,j_n} \\, x_1^{j_1} x_2^{j_2} \\cdots x_n^{j_n} \\end{equation*}is continuously differentiable." ], "refs": [], "proofs": [ { "contents": [ "Consider the partial derivative of $p$ in the $x_n$ variable.", "Write $p$ as", "\\begin{equation*}", "p(x) = \\sum_{j=0}^d p_j(x_1,\\ldots,x_{n-1}) \\, x_n^j ,", "\\end{equation*}", "where $p_j$ are polynomials in one less variable.", "Then", "\\begin{equation*}", "\\frac{\\partial p}{\\partial x_n}(x)", "= \\sum_{j=1}^d p_j(x_1,\\ldots,x_{n-1}) \\, j x_n^{j-1} ,", "\\end{equation*}", "which is again a polynomial.", "So the partial derivatives of polynomials exist and are again polynomials.", "By the continuity of algebraic operations, polynomials are continuous functions.", "Therefore $p$ is continuously differentiable." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 43, "type": "theorem", "label": "Lebl-contfunc:43", "categories": [ "continuity", "functions", "implicit_inverse" ], "title": "Inverse Function Theorem", "contents": [ "\\index{inverse function theorem} \\label{thm:inverse} Let$U \\subset \\R^n$be an open set and let$f \\colon U \\to \\R^n$be a continuously differentiable function. Suppose$p \\in U$and$f'(p)$is invertible (that is,$J_f(p) \\not=0$). Then there exist open sets$V, W \\subset \\R^n$such that$p \\in V \\subset U$,$f(V) = W$, and$f|_V$is one-to-one. Hence a function$g \\colon W \\to V$exists such that$g(y) \\coloneqq (f|_V)^{-1}(y)$. Furthermore,$g$is continuously differentiable and\\begin{equation*} g'(y) = {\\bigl(f'(x)\\bigr)}^{-1}, \\qquad \\text{for all } x \\in V, y = f(x). \\end{equation*}" ], "refs": [ "named:Inverse Function" ], "proofs": [ { "contents": [ "Write $A = f'(p)$. As $f'$ is continuous, there is an open ball", "$V$ centered at $p$ such that", "\\begin{equation*}", "\\snorm{A-f'(x)} < \\frac{1}{2\\snorm{A^{-1}}}", "\\qquad \\text{for all } x \\in V.", "\\end{equation*}", "Consequently, the derivative $f'(x)$ is invertible for all $x \\in V$", "by \\propref{prop:finitedimpropinv}.", "Given $y \\in \\R^n$, define $\\varphi_y \\colon V \\to \\R^n$ by", "\\begin{equation*}", "\\varphi_y (x) \\coloneqq x + A^{-1}\\bigl(y-f(x)\\bigr) .", "\\end{equation*}", "As $A^{-1}$ is one-to-one,", "$\\varphi_y(x) = x$ ($x$ is a fixed point) if only if", "$y-f(x) = 0$, or in other words $f(x)=y$. Using the chain rule we obtain", "\\begin{equation*}", "\\varphi_y'(x) = I - A^{-1} f'(x) = A^{-1} \\bigl( A-f'(x) \\bigr) .", "\\end{equation*}", "So for $x \\in V$, we have", "\\begin{equation*}", "\\snorm{\\varphi_y'(x)} \\leq \\snorm{A^{-1}} \\, \\snorm{A-f'(x)} < \\nicefrac{1}{2} .", "\\end{equation*}", "As $V$ is a ball, it is convex. Hence", "\\begin{equation*}", "\\snorm{\\varphi_y(x_1)-\\varphi_y(x_2)} \\leq \\frac{1}{2} \\snorm{x_1-x_2}", "\\qquad", "\\text{for all } x_1,x_2 \\in V.", "\\end{equation*}", "In other words, $\\varphi_y$ is a contraction defined on $V$, though we so far", "do not know what is the range of $\\varphi_y$. We cannot yet", "apply the fixed", "point theorem, but we can say that $\\varphi_y$", "has at most one fixed point in $V$:", "If $\\varphi_y(x_1) = x_1$ and", "$\\varphi_y(x_2) = x_2$, then", "$\\snorm{x_1-x_2} = \\snorm{\\varphi_y(x_1)-\\varphi_y(x_2)} \\leq", "\\frac{1}{2} \\snorm{x_1-x_2}$, so $x_1 = x_2$.", "That is, there exists at most one $x \\in V$", "such that $f(x) = y$, and so $f|_V$ is one-to-one.", "Let $W \\coloneqq f(V)$ and let $g \\colon W \\to V$ be the inverse of $f|_V$.", "We need to show that $W$ is open. Take a $y_0 \\in W$.", "There is a unique $x_0 \\in V$ such that $f(x_0) = y_0$.", "Let $r > 0$ be small enough such that the closed ball $C(x_0,r) \\subset V$", "(such $r > 0$ exists as $V$ is open).", "Suppose $y$ is such that", "\\begin{equation*}", "\\snorm{y-y_0} <", "\\frac{r}{2\\snorm{A^{-1}}} .", "\\end{equation*}", "If we show that $y \\in W$, then we have shown that $W$ is open.", "If $x_1 \\in", "C(x_0,r)$, then", "\\begin{equation*}", "\\begin{split}", "\\snorm{\\varphi_y(x_1)-x_0}", "& \\leq", "\\snorm{\\varphi_y(x_1)-\\varphi_y(x_0)} +", "\\snorm{\\varphi_y(x_0)-x_0} \\\\", "& \\leq", "\\frac{1}{2}\\snorm{x_1-x_0} +", "\\snorm{A^{-1}(y-y_0)} \\\\", "& \\leq", "\\frac{1}{2}r +", "\\snorm{A^{-1}} \\, \\snorm{y-y_0} \\\\", "& <", "\\frac{1}{2}r +", "\\snorm{A^{-1}}", "\\frac{r}{2\\snorm{A^{-1}}} = r .", "\\end{split}", "\\end{equation*}", "So $\\varphi_y$ takes $C(x_0,r)$ into $B(x_0,r) \\subset C(x_0,r)$. It is a", "contraction on $C(x_0,r)$ and $C(x_0,r)$ is complete (closed subset of $\\R^n$", "is complete).", "Apply the contraction mapping principle to obtain a fixed point $x$,", "i.e.\\ $\\varphi_y(x) = x$. That is, $f(x) = y$, and $y \\in", "f\\bigl(C(x_0,r)\\bigr) \\subset f(V) = W$. Therefore, $W$ is open.", "Next we need to show that $g$ is continuously differentiable and compute", "its derivative. First, let us show that it is differentiable.", "Let $y \\in W$ and $k \\in \\R^n$, $k\\not= 0$, such that $y+k \\in W$.", "Because $f|_V$ is a one-to-one and onto mapping of $V$ onto $W$,", "there are unique", "$x \\in V$ and $h \\in \\R^n$, $h \\not= 0$ and $x+h \\in V$, such that", "$f(x) = y$ and $f(x+h) = y+k$.", "In other words, $g(y) = x$ and $g(y+k) = x+h$. See", "\\figureref{fig:inversefuncRn2}.", "\\begin{myfigureht}", "\\subimport*{figures/}{inversefuncRn2.pdf_t}", "\\caption{Proving that $g$ is differentiable.\\label{fig:inversefuncRn2}}", "\\end{myfigureht}", "We can still", "squeeze some information from the fact that $\\varphi_y$ is a contraction.", "\\begin{equation*}", "\\varphi_y(x+h)-\\varphi_y(x) = h + A^{-1} \\bigl( f(x)-f(x+h) \\bigr) = h - A^{-1} k .", "\\end{equation*}", "So", "\\begin{equation*}", "\\snorm{h-A^{-1}k} = \\snorm{\\varphi_y(x+h)-\\varphi_y(x)} \\leq", "\\frac{1}{2}\\snorm{x+h-x} = \\frac{\\snorm{h}}{2}.", "\\end{equation*}", "By the inverse triangle inequality, $\\snorm{h} - \\snorm{A^{-1}k} \\leq", "\\frac{1}{2}\\snorm{h}$.", "So", "\\begin{equation*}", "\\snorm{h} \\leq 2 \\snorm{A^{-1}k} \\leq 2 \\snorm{A^{-1}} \\, \\snorm{k}.", "\\end{equation*}", "In particular, as $k$ goes to 0, so does $h$.", "As $x \\in V$, then $f'(x)$ is invertible.", "Let $B \\coloneqq \\bigl(f'(x)\\bigr)^{-1}$, which is what we think the derivative of", "$g$ at $y$ is. Then", "\\begin{equation*}", "\\begin{split}", "\\frac{\\snorm{g(y+k)-g(y)-Bk}}{\\snorm{k}}", "& =", "\\frac{\\snorm{h-Bk}}{\\snorm{k}}", "\\\\", "& =", "\\frac{\\snorm{h-B\\bigl(f(x+h)-f(x)\\bigr)}}{\\snorm{k}}", "\\\\", "& =", "\\frac{\\snorm{B\\bigl(f(x+h)-f(x)-f'(x)h\\bigr)}}{\\snorm{k}}", "\\\\", "& \\leq", "\\snorm{B}", "\\frac{\\snorm{h}}{\\snorm{k}}\\,", "\\frac{\\snorm{f(x+h)-f(x)-f'(x)h}}{\\snorm{h}}", "\\\\", "& \\leq", "2\\snorm{B} \\, \\snorm{A^{-1}}", "\\frac{\\snorm{f(x+h)-f(x)-f'(x)h}}{\\snorm{h}} .", "\\end{split}", "\\end{equation*}", "As $k$ goes to 0, so does $h$. So the right-hand side goes to 0 as $f$ is", "differentiable, and hence", "the left-hand side also goes to 0. And", "$B$ is precisely what we wanted $g'(y)$ to be.", "We have $g$ is differentiable, let us show it is $C^1(W)$.", "The function $g \\colon W \\to V$ is continuous (it is differentiable),", "$f'$ is a continuous function from $V$", "to $L(\\R^n)$, and $X \\mapsto X^{-1}$ is a continuous function on", "the set of invertible operators.", "As", "$g'(y) = {\\bigl( f'\\bigl(g(y)\\bigr)\\bigr)}^{-1}$ is the composition", "of these three", "continuous functions, it is continuous." ], "refs": [ "named:Chain Rule", "prop:finitedimpropinv" ], "ref_ids": [ 35 ] } ], "ref_ids": [] }, { "id": 44, "type": "corollary", "label": "Lebl-contfunc:cor:IVTopenmap", "categories": [ "consequence", "continuity" ], "title": "Continuity of Continuity and Differentiability", "contents": [ "Suppose$U \\subset \\R^n$is open and$f \\colon U \\to \\R^n$is a continuously differentiable mapping such that$f'(x)$is invertible for all$x \\in U$. Then for every open set$V \\subset U$, the set$f(V)$is open ($f$is said to be an \\emph{\\myindex{open mapping}})." ], "refs": [], "proofs": [ { "contents": [ "Without loss of generality, suppose $U=V$.", "For each $y \\in f(V)$, pick $x \\in f^{-1}(y)$ (there could be more", "than one such point), then by the inverse function theorem there is a", "neighborhood of $x$ in $V$ that maps onto a neighborhood of $y$. Hence", "$f(V)$ is open." ], "refs": [ "named:Inverse Function" ], "ref_ids": [ 43 ] } ], "ref_ids": [] }, { "id": 45, "type": "example", "label": "Lebl-contfunc:45", "categories": [ "example", "implicit_inverse" ], "title": "The theorem, and the corollary, is not true if$f'(x)$is not invertible for some~$x$", "contents": [ "The theorem, and the corollary, is not true if$f'(x)$is not invertible for some~$x$. For example, the map$f(x,y) \\coloneqq (x,xy)$, maps$\\R^2$onto the set$\\R^2 \\setminus \\bigl\\{ (0,y) : y \\neq 0 \\bigr\\}$, which is neither open nor closed. In fact,$f^{-1}(0,0) = \\bigl\\{ (0,y) : y \\in \\R \\bigr\\}$. This bad behavior only occurs on the$y$-axis, everywhere else the function is locally invertible. If we avoid the$y$-axis,$f$is even one-to-one." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 46, "type": "example", "label": "Lebl-contfunc:46", "categories": [ "continuity", "implicit_inverse", "example", "derivatives" ], "title": "Just because$f'(x)$is invertible everywhere does not mean that$f$is one-to-one", "contents": [ "Just because$f'(x)$is invertible everywhere does not mean that$f$is one-to-one. It is \\myquote{locally} one-to-one, but perhaps not \\myquote{globally.} Consider$f \\colon \\R^2 \\setminus \\bigl\\{ (0,0) \\bigr\\} \\to \\R^2 \\setminus \\bigl\\{ (0,0) \\bigr\\}$defined by$f(x,y) \\coloneqq (x^2-y^2,2xy)$. It is left to the reader to verify the following statements. The map$f$is differentiable and the derivative is invertible. On the other hand,$f$is 2-to-1 globally: For every$(a,b)$that is not the origin, there are exactly two solutions to$x^2-y^2=a$and$2xy=b$($f$is also onto). Notice that once you show that there is at least one solution, replacing$x$and$y$with$-x$and$-y$we obtain another solution." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 47, "type": "proposition", "label": "Lebl-contfunc:47", "categories": [], "title": "Uniqueness of Several Variables and Derivatives", "contents": [ "Let$A = [A_x~A_y] \\in L(\\R^{n+m},\\R^m)$and suppose$A_y$is invertible. If$B = - {(A_y)}^{-1} A_x$, then\\begin{equation*} 0 = A ( x, Bx) = A_x x + A_y Bx . \\end{equation*}Furthermore,$y=Bx$is the unique$y \\in \\R^m$such that$A(x,y) = 0$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 48, "type": "theorem", "label": "Lebl-contfunc:48", "categories": [ "implicit_inverse", "functions", "derivatives" ], "title": "Implicit Function Theorem", "contents": [ "\\index{implicit function theorem} \\label{thm:implicit} Let$U \\subset \\R^{n+m}$be an open set and let$f \\colon U \\to \\R^m$be a$C^1(U)$mapping. Let$(p,q) \\in U$be a point such that$f(p,q) = 0$and such that\\begin{equation*} \\frac{\\partial(f_1,\\ldots,f_m)}{\\partial(y_1,\\ldots,y_m)} (p,q) \\neq 0 . \\end{equation*}Then there exists an open set$W \\subset \\R^n$with$p \\in W$, an open set$W' \\subset \\R^m$with$q \\in W'$, where$W \\times W' \\subset U$, and a$C^1(W)$map$g \\colon W \\to W'$, with$g(p) = q$, and for all$x \\in W$, the point$g(x)$is the unique point in$W'$such that\\begin{equation*} f\\bigl(x,g(x)\\bigr) = 0 . \\end{equation*}Furthermore, if$A = [ A_x ~ A_y ] = f'(p,q)$, then\\begin{equation*} g'(p) = -{(A_y)}^{-1}A_x . \\end{equation*}" ], "refs": [ "named:Implicit Function" ], "proofs": [ { "contents": [ "Define $F \\colon U \\to \\R^{n+m}$ by $F(x,y) \\coloneqq \\bigl(x,f(x,y)\\bigr)$.", "It is clear that $F$ is $C^1$, and we want to show that its derivative", "at $(p,q)$ is invertible.", "Let us compute the derivative. The quotient", "\\begin{equation*}", "\\frac{\\snorm{f(p+h,q+k) - f(p,q) - A_x h - A_y k}}{\\snorm{(h,k)}}", "\\end{equation*}", "goes to zero as $\\snorm{(h,k)} = \\sqrt{\\snorm{h}^2+\\snorm{k}^2}$ goes to zero.", "But then so does", "\\begin{multline*}", "\\frac{\\snorm{F(p+h,q+k)-F(p,q) - (h,A_x h+A_y k)}}{\\snorm{(h,k)}}", "\\\\", "\\begin{aligned}", "& =", "\\frac{\\snorm{\\bigl(h,f(p+h,q+k)-f(p,q)\\bigr) - (h,A_x h+A_y", "k)}}{\\snorm{(h,k)}}", "\\\\", "& =", "\\frac{\\snorm{f(p+h,q+k) - f(p,q) - A_x h - A_y k}}{\\snorm{(h,k)}} .", "\\end{aligned}", "\\end{multline*}", "So the derivative of $F$ at $(p,q)$ takes $(h,k)$ to $(h,A_x h+A_y k)$.", "In block matrix form, it is", "$\\left[\\begin{smallmatrix}I & 0\\\\A_x & A_y\\end{smallmatrix}\\right]$. If", "$(h,A_x h+A_y k) = (0,0)$, then $h=0$, and so $A_y k = 0$. As $A_y$ is", "one-to-one, $k=0$. Thus $F'(p,q)$ is one-to-one, and hence invertible.", "We apply the inverse function theorem.", "That is, there exists an open set $V \\subset \\R^{n+m}$ with", "$F(p,q) = (p,0) \\in V$,", "and a $C^1$", "mapping $G \\colon V \\to \\R^{n+m}$, such that $F\\bigl(G(x,s)\\bigr) = (x,s)$ for", "all $(x,s) \\in V$, $G$ is one-to-one, and $G(V)$ is open. % (where $x \\in \\R^n$ and $s \\in \\R^m$).", "Write $G = (G_1,G_2)$ (the first $n$ and the next $m$ components of $G$).", "Then", "\\begin{equation*}", "F\\bigl(G_1(x,s),G_2(x,s)\\bigr) = \\Bigl(G_1(x,s),f\\bigl(G_1(x,s),G_2(x,s) \\bigr)\\Bigr)", "= (x,s) .", "\\end{equation*}", "So $x = G_1(x,s)$ and $f\\bigl(G_1(x,s),G_2(x,s)\\bigr) = f\\bigl(x,G_2(x,s)\\bigr) = s$.", "Plugging in $s=0$, we obtain", "\\begin{equation*}", "f\\bigl(x,G_2(x,0)\\bigr) = 0 .", "\\end{equation*}", "As the set $G(V)$ is open and $(p,q) \\in G(V)$,", "there exist some open sets", "$\\widetilde{W}$ and $W'$ such that $\\widetilde{W} \\times W' \\subset G(V)$ with $p", "\\in \\widetilde{W}$ and", "$q \\in W'$.", "Take $W \\coloneqq \\bigl\\{ x \\in \\widetilde{W} : G_2(x,0) \\in W' \\bigr\\}$.", "The function that takes $x$ to $G_2(x,0)$ is continuous and therefore $W$", "is open.", "Define", "$g \\colon W \\to \\R^m$ by $g(x) \\coloneqq G_2(x,0)$, which is the $g$ in the theorem.", "The fact that $g(x)$ is the unique point in $W'$ follows because $W \\times", "W' \\subset G(V)$ and $G$ is one-to-one.", "Next, differentiate", "\\begin{equation*}", "x\\mapsto f\\bigl(x,g(x)\\bigr)", "\\end{equation*}", "at $p$,", "which is the zero map, so its derivative is zero.", "Using the chain rule,", "\\begin{equation*}", "0 = A\\bigl(h,g'(p)h\\bigr) = A_xh + A_yg'(p)h", "\\end{equation*}", "for all $h \\in \\R^{n}$,", "and we obtain the desired derivative for $g$." ], "refs": [ "named:Chain Rule", "named:Inverse Function" ], "ref_ids": [ 35, 43 ] } ], "ref_ids": [] }, { "id": 49, "type": "example", "label": "Lebl-contfunc:49", "categories": [ "matrices", "example", "derivatives" ], "title": "Consider the set given by$x^2+y^2-{(z+1)}^3 = -1$and$e^x+e^y+e^z = 3$near the point$(0,0,0)$", "contents": [ "Consider the set given by$x^2+y^2-{(z+1)}^3 = -1$and$e^x+e^y+e^z = 3$near the point$(0,0,0)$. It is the zero set of the mapping", "\\begin{equation*}\nf(x,y,z) = \\bigl(x^2+y^2-{(z+1)}^3+1,e^x+e^y+e^z-3\\bigr) ,\n\\end{equation*}", "whose derivative is", "\\begin{equation*}\nf' =\n\\begin{bmatrix}\n2x & 2y & -3{(z+1)}^2 \\\\\ne^x & e^y & e^z\n\\end{bmatrix} .\n\\end{equation*}", "The matrix", "\\begin{equation*}\n\\begin{bmatrix}\n2(0) & -3{(0+1)}^2 \\\\\ne^0 & e^0\n\\end{bmatrix}\n=\n\\begin{bmatrix}\n0 & -3 \\\\\n1 & 1\n\\end{bmatrix}\n\\end{equation*}", "is invertible. Hence near$(0,0,0)$, we can solve for$y$and$z$as$C^1$functions of$x$such that for$x$near$0$,", "\\begin{equation*}\nx^2+y(x)^2-{\\bigl(z(x)+1\\bigr)}^3 = -1,\n\\qquad\ne^x+e^{y(x)}+e^{z(x)} = 3 .\n\\end{equation*}", "In other words, near the origin the set of solutions is a smooth curve in$\\R^3$that goes through the origin. The theorem does not tell us how to find$y(x)$and$z(x)$explicitly, it just tells us they exist." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 50, "type": "proposition", "label": "Lebl-contfunc:mv:prop:swapders", "categories": [ "derivatives" ], "title": "Suppose$U \\subset \\R^n$is open and$f \\colon U \\to \\R$is a$C^2$function, and$\\ell$and$m$are two in...", "contents": [ "Suppose$U \\subset \\R^n$is open and$f \\colon U \\to \\R$is a$C^2$function, and$\\ell$and$m$are two integers from$1$to$n$. Then\\begin{equation*} \\frac{\\partial^2 f}{\\partial x_m \\partial x_{\\ell}} = \\frac{\\partial^2 f}{\\partial x_{\\ell} \\partial x_m} . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Fix a $p \\in U$, and let $e_{\\ell}$ and $e_m$ be the standard basis vectors.", "Pick two positive numbers $s$ and $t$ small enough so that", "$p+s_0e_{\\ell} +t_0e_m \\in U$ whenever", "$0 < s_0 \\leq s$ and $0 < t_0 \\leq t$. This can be done as $U$ is open and so", "contains a small open ball (or a box if you wish) around $p$.", "Use the mean value theorem on the function", "\\begin{equation*}", "\\tau \\mapsto f(p+se_{\\ell} + \\tau e_m)-f(x + \\tau e_m) ,", "\\end{equation*}", "on the interval $[0,t]$", "to find a $t_0 \\in (0,t)$", "such that", "\\begin{equation*}", "\\frac{f(p+se_{\\ell} + te_m)- f(p+t e_m) - f(p+s e_{\\ell})+f(p)}{t}", "=", "\\frac{\\partial f}{\\partial x_m}(p + s e_{\\ell} + t_0 e_m)", "-", "\\frac{\\partial f}{\\partial x_m}(p + t_0 e_m) .", "\\end{equation*}", "Similarly, there exists a number $s_0 \\in (0,s)$ such that", "\\begin{equation*}", "\\frac{\\frac{\\partial f}{\\partial x_m}(p + s e_{\\ell} + t_0 e_m)", "-", "\\frac{\\partial f}{\\partial x_m}(p + t_0 e_m)}{s}", "=", "\\frac{\\partial^2 f}{\\partial x_{\\ell} \\partial x_m}(p + s_0 e_{\\ell} + t_0 e_m) .", "\\end{equation*}", "In other words,", "\\begin{equation*}", "g(s,t) \\coloneqq", "\\frac{f(p+se_{\\ell} + te_m)- f(p+t e_m) - f(p+s e_{\\ell})+f(p)}{st}", "=", "\\frac{\\partial^2 f}{\\partial x_{\\ell} \\partial x_m}(p + s_0 e_{\\ell} + t_0 e_m) .", "\\end{equation*}", "\\begin{myfigureht}", "\\subimport*{figures/}{der2orderflip.pdf_t}", "\\caption{Using the mean value theorem to estimate", "a second order partial derivative by", "a certain difference quotient.\\label{fig:der2orderflip}}", "\\end{myfigureht}", "See \\figureref{fig:der2orderflip}.", "The $s_0$ and $t_0$ depend on $s$ and $t$,", "but $0 < s_0 < s$ and", "$0 < t_0 < t$.", "Let the domain of the function $g$ be the set $(0,\\epsilon) \\times", "(0,\\epsilon)$ for some small $\\epsilon > 0$.", "As $(s,t) \\in (0,\\epsilon) \\times (0,\\epsilon)$ goes to $(0,0)$,", "$(s_0,t_0)$ also goes to $(0,0)$.", "By continuity of the second partial derivatives,", "\\begin{equation*}", "\\lim_{(s,t) \\to (0,0)} g(s,t) =", "\\frac{\\partial^2 f}{\\partial x_{\\ell} \\partial x_m}(p) .", "\\end{equation*}", "Now reverse the roles of $s$ and $t$ (and $\\ell$ and $m$). Start with", "the function $\\sigma \\mapsto f(p+\\sigma e_{\\ell} + te_m)-f(p + \\sigma", "e_{\\ell})$", "find an $s_1 \\in (0,s)$ such that", "\\begin{equation*}", "\\frac{f(p+ se_{\\ell} + te_m )- f(p+s e_{\\ell}) - f(p+t e_m)+f(p)}{s}", "=", "\\frac{\\partial f}{\\partial x_{\\ell}}(p + s_1 e_{\\ell} + t e_m)", "-", "\\frac{\\partial f}{\\partial x_{\\ell}}(p + s_1 e_{\\ell}) .", "\\end{equation*}", "Find a $t_1 \\in (0,t)$ such that", "\\begin{equation*}", "\\frac{\\frac{\\partial f}{\\partial x_{\\ell}}(p + s_1 e_{\\ell} + t e_m)", "-", "\\frac{\\partial f}{\\partial x_{\\ell}}(p + s_1 e_{\\ell})}{t}", "=", "\\frac{\\partial^2 f}{\\partial x_m \\partial x_{\\ell}}(p + s_1 e_{\\ell} + t_1 e_m) .", "\\end{equation*}", "So $g(s,t) = \\frac{\\partial^2 f}{\\partial x_m \\partial", "x_{\\ell}}(p + s_1 e_{\\ell} + t_1 e_m)$ for the same $g$ as above.", "As before,", "\\begin{equation*}", "\\lim_{(s,t) \\to (0,0)} g(s,t) =", "\\frac{\\partial^2 f}{\\partial x_m \\partial x_{\\ell}}(p) .", "\\end{equation*}", "Therefore, the two partial derivatives are equal." ], "refs": [ "named:Mean Value" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 0, "type": "theorem", "label": "Lebl-contfunc:0", "categories": [], "title": "Continuity of Path Integration", "contents": [ "Suppose$f \\colon [a,b] \\times [c,d] \\to \\R$is a continuous function, such that$\\frac{\\partial f}{\\partial y}$exists for all$(x,y) \\in [a,b] \\times [c,d]$and is continuous. Define$g \\colon [c,d] \\to \\R$by\\begin{equation*} g(y) \\coloneqq \\int_a^b f(x,y) \\,dx . \\end{equation*}Then$g$is continuously differentiable and\\begin{equation*} g'(y) = \\int_a^b \\frac{\\partial f}{\\partial y}(x,y) \\,dx . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Fix $y \\in [c,d]$ and let $\\epsilon > 0$ be given.", "As $\\frac{\\partial f}{\\partial y}$ is continuous on $[a,b] \\times [c,d]$ it", "is uniformly continuous. In particular, there exists $\\delta > 0$ such that", "whenever $y_1 \\in [c,d]$ with", "$\\abs{y_1-y} < \\delta$ and all $x \\in [a,b]$, we have", "\\begin{equation*}", "\\abs{\\frac{\\partial f}{\\partial y}(x,y_1)-\\frac{\\partial f}{\\partial y}(x,y)} < \\epsilon .", "\\end{equation*}", "Suppose $h$ is such that $y+h \\in [c,d]$ and $\\abs{h} < \\delta$.", "Fix $x$ for a moment", "and apply the mean value theorem to find a $y_1$ between $y$ and $y+h$ such that", "\\begin{equation*}", "\\frac{f(x,y+h)-f(x,y)}{h}", "=", "\\frac{\\partial f}{\\partial y}(x,y_1) .", "\\end{equation*}", "As $\\abs{y_1-y} \\leq \\abs{h} < \\delta$,", "\\begin{equation*}", "\\abs{", "\\frac{f(x,y+h)-f(x,y)}{h}", "-", "\\frac{\\partial f}{\\partial y}(x,y)", "}", "=", "\\abs{", "\\frac{\\partial f}{\\partial y}(x,y_1)", "-", "\\frac{\\partial f}{\\partial y}(x,y)", "}", "< \\epsilon .", "\\end{equation*}", "The argument worked for every $x \\in [a,b]$ (different $y_1$ may have been", "used). Thus, as a function of", "$x$", "\\begin{equation*}", "x \\mapsto \\frac{f(x,y+h)-f(x,y)}{h}", "\\qquad", "\\text{converges uniformly to}", "\\qquad", "x \\mapsto \\frac{\\partial f}{\\partial y}(x,y)", "\\qquad", "\\text{as } h \\to 0 .", "\\end{equation*}", "We defined uniform convergence for sequences although the idea is the", "same. You may replace $h$ with a sequence of nonzero", "numbers $\\{ h_n \\}_{n=1}^\\infty$", "converging to $0$ such that $y+h_n \\in [c,d]$ and let $n \\to \\infty$.", "Consider the difference quotient of $g$,", "\\begin{equation*}", "\\frac{g(y+h)-g(y)}{h}", "=", "\\frac{\\int_a^b f(x,y+h) \\,dx -", "\\int_a^b f(x,y) \\,dx }{h}", "=", "\\int_a^b \\frac{f(x,y+h)-f(x,y)}{h} \\,dx .", "\\end{equation*}", "Uniform convergence implies the limit can be taken underneath the integral.", "So", "\\begin{equation*}", "\\lim_{h\\to 0}", "\\frac{g(y+h)-g(y)}{h}", "=", "\\int_a^b", "\\lim_{h\\to 0}", "\\frac{f(x,y+h)-f(x,y)}{h} \\,dx", "=", "\\int_a^b", "\\frac{\\partial f}{\\partial y}(x,y) \\,dx .", "\\end{equation*}", "Then $g'$ is continuous on $[c,d]$ by", "\\volIref{\\propref*{vI-prop:integralcontcont} from volume I}{\\propref{prop:integralcontcont}} mentioned above." ], "refs": [ "prop:integralcontcont" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 1, "type": "example", "label": "Lebl-contfunc:1", "categories": [ "example" ], "title": "Let \\begin{equation*} f(y) = \\int_0^1 \\sin(x^2-y^2) \\,dx", "contents": [ "Let", "\\begin{equation*}\nf(y) = \\int_0^1 \\sin(x^2-y^2) \\,dx .\n\\end{equation*}", "Then", "\\begin{equation*}\nf'(y) = \\int_0^1 -2y\\cos(x^2-y^2) \\,dx .\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 2, "type": "example", "label": "Lebl-contfunc:example:counterexamplediffunder", "categories": [ "differentiation", "integration", "example" ], "title": "Continuity of Integration", "contents": [ "Consider", "\\begin{equation*}\n\\int_0^{1} \\frac{x-1}{\\ln(x)} \\,dx .\n\\end{equation*}", "The function under the integral extends to be continuous on$[0,1]$, and hence the integral exists, see \\exerciseref{exercise:counterexamplediffunder}. Trouble is finding it. We introduce a parameter$y$and define a function:", "\\begin{equation*}\ng(y) \\coloneqq \\int_0^{1} \\frac{x^y-1}{\\ln(x)} \\,dx .\n\\end{equation*}", "The function$\\frac{x^y-1}{\\ln(x)}$also extends to a continuous function of$x$and$y$for$(x,y) \\in [0,1] \\times [0,1]$(also part of the exercise). See \\figureref{fig:diffunderexample}. \\begin{myfigureht} \\includegraphics{figures/diffunderexample} \\caption{The graph$z= \\frac{x^y-1}{\\ln(x)}$on$[0,1] \\times [0,1]$.\\label{fig:diffunderexample}} \\end{myfigureht}", "Hence,$g$is a continuous function on$[0,1]$and$g(0) = 0$. For every$\\epsilon > 0$, the$y$derivative of the integrand,$x^y$, is continuous on$[0,1] \\times [\\epsilon,1]$. Therefore, for$y >0$, we may differentiate under the integral sign,", "\\begin{equation*}\ng'(y) =\n\\int_0^{1} \\frac{\\ln(x) x^y}{\\ln(x)} \\,dx \n=\n\\int_0^{1} x^y \\,dx =\n\\frac{1}{y+1} .\n\\end{equation*}", "We need to figure out$g(1)$given that$g'(y) = \\frac{1}{y+1}$and$g(0) = 0$. Elementary calculus says that$g(1) = \\int_0^1 g'(y)\\,dy = \\ln(2)$. Thus,", "\\begin{equation*}\n\\int_0^{1} \\frac{x-1}{\\ln(x)} \\,dx = \\ln(2).\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 3, "type": "example", "label": "Lebl-contfunc:mv:example:unitsquarepath", "categories": [ "paths", "example" ], "title": "Let$\\gamma \\colon [0,4] \\to \\R^2$be defined by \\begin{equation*} \\gamma(t) \\coloneqq \\begin{cases...", "contents": [ "Let$\\gamma \\colon [0,4] \\to \\R^2$be defined by", "\\begin{equation*}\n\\gamma(t) \\coloneqq\n\\begin{cases}\n(t,0) & \\text{if } t \\in [0,1],\\\\\n(1,t-1) & \\text{if } t \\in (1,2],\\\\\n(3-t,1) & \\text{if } t \\in (2,3],\\\\\n(0,4-t) & \\text{if } t \\in (3,4].\n\\end{cases}\n\\end{equation*}", "\\begin{myfigureht} \\includegraphics{figures/squarepath} \\caption{The path$\\gamma$traversing the unit square.\\label{fig:squarepath}} \\end{myfigureht}", "The path$\\gamma$is the unit square traversed counterclockwise. See \\figureref{fig:squarepath}. It is a piecewise smooth path. For example,$\\gamma|_{[1,2]}(t) = (1,t-1)$and so$(\\gamma|_{[1,2]})'(t) = (0,1) \\not= 0$. Similarly for the other 3 sides. Notice that$(\\gamma|_{[1,2]})'(1) = (0,1)$,$(\\gamma|_{[0,1]})'(1) = (1,0)$, but$\\gamma^{\\:\\prime}(1)$does not exist. At the corners$\\gamma$is not differentiable. The path$\\gamma$is a simple closed path, as$\\gamma|_{[0,4)}$is one-to-one and$\\gamma(0)=\\gamma(4)$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 4, "type": "example", "label": "Lebl-contfunc:4", "categories": [ "paths", "differentiation", "example" ], "title": "Continuity of Path Integrals", "contents": [ "The condition$\\gamma^{\\:\\prime}(t) \\not= 0$means that the image$\\gamma\\bigl([a,b]\\bigr)$has no \\myquote{corners} where$\\gamma$is smooth. Consider", "\\begin{equation*}\n\\gamma(t) \\coloneqq\n\\begin{cases}\n(t^2,0) & \\text{if } t < 0,\\\\\n(0,t^2) & \\text{if } t \\geq 0.\n\\end{cases}\n\\end{equation*}", "See \\figureref{fig:cornersmoothpath}. It is left for the reader to check that$\\gamma$is continuously differentiable, yet the image$\\gamma(\\R) = \\bigl\\{ (x,y) \\in \\R^2 : (x,y) = (s,0) \\text{ or } (x,y) = (0,s) \\text{ for some } s \\geq 0 \\bigr\\}$has a \\myquote{corner} at the origin. And that is because$\\gamma^{\\:\\prime}(0) = (0,0)$. More complicated examples with, say, infinitely many corners exist, see the exercises. \\begin{myfigureht} \\includegraphics{figures/cornersmoothpath} \\caption{``Smooth'' path with a corner if we allow zero derivative. The points corresponding to several values of$t$are marked with dots.\\label{fig:cornersmoothpath}} \\end{myfigureht}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 5, "type": "example", "label": "Lebl-contfunc:5", "categories": [ "paths", "example" ], "title": "Continuity of Path Integrals", "contents": [ "A graph of a continuously differentiable function$f \\colon [a,b] \\to \\R$is a smooth path. Define$\\gamma \\colon [a,b] \\to \\R^2$by", "\\begin{equation*}\n\\gamma(t) \\coloneqq \\bigl(t,f(t)\\bigr) .\n\\end{equation*}", "Then$\\gamma^{\\:\\prime}(t) = \\bigl( 1 , f'(t) \\bigr)$, which is never zero, and$\\gamma\\bigl([a,b]\\bigr)$is the graph of$f$.", "There are other ways of parametrizing the path. That is, there are different paths with the same image. The function$t \\mapsto (1-t)a+tb$, takes the interval$[0,1]$to$[a,b]$. Define$\\alpha \\colon [0,1] \\to \\R^2$by", "\\begin{equation*}\n\\alpha(t) \\coloneqq \\bigl((1-t)a+tb,f((1-t)a+tb)\\bigr) .\n\\end{equation*}", "Then$\\alpha'(t) = \\bigl( b-a ,~ (b-a)f'((1-t)a+tb) \\bigr)$, which is never zero. As sets,$\\alpha\\bigl([0,1]\\bigr) = \\gamma\\bigl([a,b]\\bigr) = \\bigl\\{ (x,y) \\in \\R^2 : x \\in [a,b] \\text{ and } f(x) = y \\bigr\\}$, which is just the graph of$f$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 6, "type": "proposition", "label": "Lebl-contfunc:prop:reparamapiecewisesmooth", "categories": [ "paths", "parametrization" ], "title": "If$\\gamma \\colon [a,b] \\to \\R^n$is a piecewise smooth path, and$\\gamma \\circ h \\colon [c,d] \\to \\...", "contents": [ "If$\\gamma \\colon [a,b] \\to \\R^n$is a piecewise smooth path, and$\\gamma \\circ h \\colon [c,d] \\to \\R^n$is a piecewise smooth reparametrization, then$\\gamma \\circ h$is a piecewise smooth path." ], "refs": [], "proofs": [ { "contents": [ "Assume that $h$ preserves orientation, that is, $h$ is strictly", "increasing.", "If $h \\colon [c,d] \\to [a,b]$ gives a piecewise smooth reparametrization,", "then for some partition", "$r_0 = c < r_1 < r_2 < \\cdots < r_\\ell = d$, the restriction", "$h|_{[r_{j-1},r_j]}$ is continuously differentiable with a positive", "derivative.", "Let $t_0 = a < t_1 < t_2 < \\cdots < t_k = b$ be the partition from the", "definition of piecewise smooth for $\\gamma$ together with the", "points $\\{ h(r_0), h(r_1), h(r_2), \\ldots, h(r_\\ell) \\}$.", "Let $s_j \\coloneqq h^{-1}(t_j)$. Then", "$s_0 = c < s_1 < s_2 < \\cdots < s_k = d$", "is a partition that includes (is a refinement of) the", "$\\{ r_0,r_1,\\ldots,r_\\ell \\}$.", "If $\\tau \\in [s_{j-1},s_j]$, then $h(\\tau) \\in [t_{j-1},t_j]$", "since $h(s_{j-1}) = t_{j-1}$,", "$h(s_{j}) = t_j$, and", "$h$ is strictly increasing.", "Also $h|_{[s_{j-1},s_j]}$ is continuously differentiable, and", "$\\gamma|_{[t_{j-1},t_j]}$ is also continuously differentiable.", "Then", "\\begin{equation*}", "(\\gamma \\circ h)|_{[s_{j-1},s_{j}]} (\\tau)", "=", "\\gamma|_{[t_{j-1},t_{j}]} \\bigl( h|_{[s_{j-1},s_j]}(\\tau) \\bigr) .", "\\end{equation*}", "The function", "$(\\gamma \\circ h)|_{[s_{j-1},s_{j}]}$ is therefore continuously", "differentiable and", "by the chain rule", "\\begin{equation*}", "\\bigl( (\\gamma \\circ h)|_{[s_{j-1},s_{j}]} \\bigr) ' (\\tau)", "=", "\\bigl( \\gamma|_{[t_{j-1},t_{j}]} \\bigr)' \\bigl( h(\\tau) \\bigr)", "(h|_{[s_{j-1},s_j]})'(\\tau) \\not= 0 .", "\\end{equation*}", "Consequently, $\\gamma \\circ h$ is a piecewise smooth path.", "The proof for orientation reversing $h$ is left as an exercise." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 7, "type": "example", "label": "Lebl-contfunc:7", "categories": [ "forms", "example" ], "title": "\\begin{equation*} \\omega(x,y) \\coloneqq \\frac{-y}{x^2+y^2} \\,dx + \\frac{x}{x^2+y^2} \\,dy \\end{equ...", "contents": [ "\\begin{equation*}\n\\omega(x,y) \\coloneqq \\frac{-y}{x^2+y^2} \\,dx + \\frac{x}{x^2+y^2} \\,dy\n\\end{equation*}", "is a one-form defined on$\\R^2 \\setminus \\{ (0,0) \\}$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 8, "type": "proposition", "label": "Lebl-contfunc:mv:prop:pathconcat", "categories": [ "paths", "forms" ], "title": "Let$\\gamma \\colon [a,c] \\to \\R^n$be a piecewise smooth path, and$b \\in (a,c)$", "contents": [ "Let$\\gamma \\colon [a,c] \\to \\R^n$be a piecewise smooth path, and$b \\in (a,c)$. Define the piecewise smooth paths$\\alpha \\coloneqq \\gamma|_{[a,b]}$and$\\beta \\coloneqq \\gamma|_{[b,c]}$. Let$\\omega$be a one-form defined on$\\gamma\\bigl([a,c]\\bigr)$. Then\\begin{equation*} \\int_{\\gamma} \\omega = \\int_{\\alpha} \\omega + \\int_{\\beta} \\omega . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 9, "type": "example", "label": "Lebl-contfunc:example:mv:irrotoneformint", "categories": [ "paths", "integration", "forms", "example", "parametrization" ], "title": "Let the one-form$\\omega$and the path$\\gamma \\colon [0,2\\pi] \\to \\R^2$be defined by \\begin{equatio...", "contents": [ "Let the one-form$\\omega$and the path$\\gamma \\colon [0,2\\pi] \\to \\R^2$be defined by", "\\begin{equation*}\n\\omega(x,y) \\coloneqq \\frac{-y}{x^2+y^2} \\,dx + \\frac{x}{x^2+y^2} \\,dy,\n\\qquad\n\\gamma(t) \\coloneqq \\bigl(\\cos(t),\\sin(t)\\bigr) .\n\\end{equation*}", "Then", "\\begin{equation*}\n\\begin{split}\n\\int_{\\gamma} \\omega\n& =\n\\int_0^{2\\pi}\n\\Biggl(\n\\frac{-\\sin(t)}{{\\bigl(\\cos(t)\\bigr)}^2+{\\bigl(\\sin(t)\\bigr)}^2}\n\\bigl(-\\sin(t)\\bigr)\n+\n\\frac{\\cos(t)}{{\\bigl(\\cos(t)\\bigr)}^2+{\\bigl(\\sin(t)\\bigr)}^2}\n\\bigl(\\cos(t)\\bigr)\n\\Biggr) \\, dt\n\\\\\n& =\n\\int_0^{2\\pi}\n1 \\, dt\n= 2\\pi .\n\\end{split}\n\\end{equation*}", "Next, parametrize the same curve as$\\alpha \\colon [0,1] \\to \\R^2$defined by$\\alpha(t) \\coloneqq \\bigl(\\cos(2\\pi t),\\sin(2 \\pi t)\\bigr)$, that is,$\\alpha$is a smooth reparametrization of$\\gamma$. Then", "\\begin{equation*}\n\\begin{split}\n\\int_{\\alpha} \\omega\n& =\n\\int_0^{1}\n\\Biggl(\n\\frac{-\\sin(2\\pi t)}{{\\bigl(\\cos(2\\pi t)\\bigr)}^2+{\\bigl(\\sin(2\\pi t)\\bigr)}^2}\n\\bigl(-2\\pi \\sin(2\\pi t)\\bigr)\n\\\\\n& \\phantom{=\\int_0^1\\Biggl(~}\n+\n\\frac{\\cos(2 \\pi t)}{{\\bigl(\\cos(2 \\pi t)\\bigr)}^2+{\\bigl(\\sin(2 \\pi t)\\bigr)}^2}\n\\bigl(2 \\pi \\cos(2 \\pi t)\\bigr)\n\\Biggr) \\, dt\n\\\\\n& =\n\\int_0^{1}\n2\\pi \\, dt\n= 2\\pi .\n\\end{split}\n\\end{equation*}", "Finally, reparametrize with$\\beta \\colon [0,2\\pi] \\to \\R^2$as$\\beta(t) \\coloneqq \\bigl(\\cos(-t),\\sin(-t)\\bigr)$. Then", "\\begin{equation*}\n\\begin{split}\n\\int_{\\beta} \\omega\n& =\n\\int_0^{2\\pi}\n\\Biggl(\n\\frac{-\\sin(-t)}{{\\bigl(\\cos(-t)\\bigr)}^2+{\\bigl(\\sin(-t)\\bigr)}^2}\n\\bigl(\\sin(-t)\\bigr)\n+\n\\frac{\\cos(-t)}{{\\bigl(\\cos(-t)\\bigr)}^2+{\\bigl(\\sin(-t)\\bigr)}^2}\n\\bigl(-\\cos(-t)\\bigr)\n\\Biggr) \\, dt\n\\\\\n& =\n\\int_0^{2\\pi}\n(-1) \\, dt\n= -2\\pi .\n\\end{split}\n\\end{equation*}", "The path$\\alpha$is an orientation preserving reparametrization of$\\gamma$, and the integrals are the same. The path$\\beta$is an orientation reversing reparametrization of$\\gamma$and the integral is minus the original. See \\figureref{fig:circlepathrepar}. \\begin{myfigureht} \\includegraphics{figures/circlepathrepar} \\caption{A circular path reparametrized in two different ways. The arrow indicates the orientation of$\\gamma$and$\\alpha$. The path$\\beta$traverses the circle in the opposite direction.\\label{fig:circlepathrepar}} \\end{myfigureht}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 10, "type": "proposition", "label": "Lebl-contfunc:mv:prop:pathintrepararam", "categories": [ "paths", "parametrization", "forms" ], "title": "Let$\\gamma \\colon [a,b] \\to \\R^n$be a piecewise smooth path and$\\gamma \\circ h \\colon [c,d] \\to \\...", "contents": [ "Let$\\gamma \\colon [a,b] \\to \\R^n$be a piecewise smooth path and$\\gamma \\circ h \\colon [c,d] \\to \\R^n$a piecewise smooth reparametrization. Suppose$\\omega$is a one-form defined on the set$\\gamma\\bigl([a,b]\\bigr)$. Then \\begin{equation*} \\int_{\\gamma \\circ h} \\omega = \\begin{cases} \\int_{\\gamma} \\omega & \\text{if } h \\text{ preserves orientation,}", "-\\int_{\\gamma} \\omega & \\text{if } h \\text{ reverses orientation.} \\end{cases} \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Assume first that $\\gamma$ and $h$ are both smooth.", "Write $\\omega = \\omega_1 \\, dx_1 + \\omega_2 \\, dx_2 + \\cdots +", "\\omega_n \\, dx_n$.", "Suppose that $h$ is orientation preserving. Use", "the change of variables formula for the Riemann integral:", "\\begin{equation*}", "\\begin{split}", "\\int_{\\gamma} \\omega", "& =", "\\int_a^b", "\\left(", "\\sum_{j=1}^n", "\\omega_j\\bigl(\\gamma(t)\\bigr) \\gamma_j^{\\:\\prime}(t)", "\\right) dt", "\\\\", "& =", "\\int_c^d", "\\left(", "\\sum_{j=1}^n", "\\omega_j\\Bigl(\\gamma\\bigl(h(\\tau)\\bigr)\\Bigr) \\gamma_j^{\\:\\prime}\\bigl(h(\\tau)\\bigr)", "\\right) h'(\\tau) \\, d\\tau", "\\\\", "& =", "\\int_c^d", "\\left(", "\\sum_{j=1}^n", "\\omega_j\\Bigl(\\gamma\\bigl(h(\\tau)\\bigr)\\Bigr) (\\gamma_j \\circ h)'(\\tau)", "\\right) d\\tau", "=", "\\int_{\\gamma \\circ h} \\omega .", "\\end{split}", "\\end{equation*}", "If $h$ is orientation reversing, it swaps the order of the limits on the", "integral and introduces a minus sign.", "The details, along with finishing the proof for piecewise smooth", "paths, is left as \\exerciseref{mv:exercise:pathpiece}." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 11, "type": "example", "label": "Lebl-contfunc:11", "categories": [ "paths", "example" ], "title": "Take$\\gamma \\colon [0,2\\pi] \\to \\R^2$given by$\\gamma(t) \\coloneqq \\bigl(\\cos(t),\\sin(t)\\bigr)$, a...", "contents": [ "Take$\\gamma \\colon [0,2\\pi] \\to \\R^2$given by$\\gamma(t) \\coloneqq \\bigl(\\cos(t),\\sin(t)\\bigr)$, and$\\beta \\colon [0,2\\pi] \\to \\R^2$by$\\beta(t) \\coloneqq \\bigl(\\cos(2t),\\sin(2t)\\bigr)$. Notice that$\\gamma\\bigl([0,2\\pi]\\bigr) = \\beta\\bigl([0,2\\pi]\\bigr)$; we travel around the same curve, the unit circle. But$\\gamma$goes around the unit circle once in the counter clockwise direction, and$\\beta$goes around the unit circle twice (in the same direction). See \\figureref{fig:circlepathrepar2}. \\begin{myfigureht} \\includegraphics{figures/circlepathrepar2} \\caption{Circular path traversed once by$\\gamma \\colon [0,2\\pi] \\to \\R^2$and twice by$\\beta \\colon [0,2\\pi] \\to \\R^2$.\\label{fig:circlepathrepar2}} \\end{myfigureht}", "Compute \\begin{align*} & \\int_{\\gamma} -y\\, dx + x\\,dy = \\int_0^{2\\pi} \\Bigl( \\bigl(-\\sin(t) \\bigr) \\bigl(-\\sin(t) \\bigr) + \\cos(t) \\cos(t) \\Bigr) dt = 2 \\pi,", "& \\int_{\\beta} -y\\, dx + x\\,dy = \\int_0^{2\\pi} \\Bigl( \\bigl(-\\sin(2t) \\bigr) \\bigl(-2\\sin(2t) \\bigr) + \\cos(t) \\bigl(2\\cos(t)\\bigr) \\Bigr) dt = 4 \\pi. \\end{align*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 12, "type": "proposition", "label": "Lebl-contfunc:mv:prop:lineintrepararam", "categories": [ "paths", "parametrization" ], "title": "Continuity of Path Integrals", "contents": [ "Let$\\gamma \\colon [a,b] \\to \\R^n$be a piecewise smooth path and$\\gamma \\circ h \\colon [c,d] \\to \\R^n$a piecewise smooth reparametrization. Suppose$f$is a continuous function defined on the set$\\gamma\\bigl([a,b]\\bigr)$. Then\\begin{equation*} \\int_{\\gamma \\circ h} f\\, ds = \\int_{\\gamma} f\\, ds . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Suppose $h$ is orientation preserving and that $\\gamma$ and $h$", "are both smooth. Then", "\\begin{equation*}", "\\begin{split}", "\\int_{\\gamma} f \\, ds", "& =", "\\int_a^b", "f\\bigl(\\gamma(t)\\bigr) \\snorm{\\gamma^{\\:\\prime}(t)} \\, dt", "\\\\", "& =", "\\int_c^d", "f\\Bigl(\\gamma\\bigl(h(\\tau)\\bigr)\\Bigr)", "\\snorm{\\gamma^{\\:\\prime}\\bigl(h(\\tau)\\bigr)} h'(\\tau) \\, d\\tau", "\\\\", "& =", "\\int_c^d", "f\\Bigl(\\gamma\\bigl(h(\\tau)\\bigr)\\Bigr)", "\\snorm{\\gamma^{\\:\\prime}\\bigl(h(\\tau)\\bigr) h'(\\tau)} \\, d\\tau", "\\\\", "& =", "\\int_c^d", "f\\bigl((\\gamma \\circ h)(\\tau)\\bigr) \\snorm{(\\gamma \\circ h)'(\\tau)} \\, d\\tau", "\\\\", "& =", "\\int_{\\gamma \\circ h} f \\, ds .", "\\end{split}", "\\end{equation*}", "If $h$ is orientation reversing it swaps the order of the limits on the", "integral, but you also have to introduce a minus sign in order", "to take $h'$ inside the norm.", "The details, along with finishing the proof for piecewise smooth", "paths is left to the reader as \\exerciseref{mv:exercise:linepiece}." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 13, "type": "example", "label": "Lebl-contfunc:13", "categories": [ "paths", "parametrization", "example" ], "title": "Let$f(x,y) \\coloneqq x$", "contents": [ "Let$f(x,y) \\coloneqq x$. Let$C \\subset \\R^2$be half of the unit circle for$x \\geq 0$. We wish to compute", "\\begin{equation*}\n\\int_C f \\, ds .\n\\end{equation*}", "Parametrize the curve$C$via$\\gamma \\colon [\\nicefrac{-\\pi}{2},\\nicefrac{\\pi}{2}] \\to \\R^2$defined as$\\gamma(t) \\coloneqq \\bigl(\\cos(t),\\sin(t)\\bigr)$. Then$\\gamma^{\\:\\prime}(t) = \\bigl(-\\sin(t),\\cos(t)\\bigr)$, and", "\\begin{equation*}\n\\int_C f \\, ds =\n\\int_\\gamma f \\, ds\n=\n\\int_{-\\pi/2}^{\\pi/2} \\cos(t) \\sqrt{ {\\bigl(-\\sin(t)\\bigr)}^2 + \n{\\bigl(\\cos(t)\\bigr)}^2 } \\, dt\n=\n\\int_{-\\pi/2}^{\\pi/2} \\cos(t) \\, dt = 2.\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 14, "type": "example", "label": "Lebl-contfunc:14", "categories": [ "paths", "parametrization", "example" ], "title": "Let$x,y \\in \\R^n$be two points and write$[x,y]$as the straight line segment between the two point...", "contents": [ "Let$x,y \\in \\R^n$be two points and write$[x,y]$as the straight line segment between the two points$x$and$y$. Parametrize$[x,y]$by$\\gamma(t) \\coloneqq (1-t)x + ty$for$t$running between$0$and$1$. See \\figureref{fig:straightpath}. Then$\\gamma^{\\:\\prime}(t) = y-x$, and therefore", "\\begin{equation*}\n\\ell\\bigl([x,y]\\bigr)\n=\n\\int_{[x,y]} ds\n=\n\\int_0^1 \\snorm{y-x} \\, dt\n=\n\\snorm{y-x} .\n\\end{equation*}", "The length of$[x,y]$is the standard euclidean distance between$x$and$y$, justifying the name. \\begin{myfigureht} \\includegraphics{figures/straightpath} \\caption{Straight path between$x$and$y$parametrized by$(1-t)x + ty$.\\label{fig:straightpath}} \\end{myfigureht}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 15, "type": "example", "label": "Lebl-contfunc:15", "categories": [ "paths", "independence", "integration", "example" ], "title": "Let$\\gamma \\colon [0,1] \\to \\R^2$be the path$\\gamma(t) \\coloneqq (t,0)$going from$(0,0)$to$(1,0)$", "contents": [ "Let$\\gamma \\colon [0,1] \\to \\R^2$be the path$\\gamma(t) \\coloneqq (t,0)$going from$(0,0)$to$(1,0)$. Let$\\beta \\colon [0,1] \\to \\R^2$be the path$\\beta(t) \\coloneqq \\bigl(t,(1-t)t\\bigr)$also going between the same points. Then \\begin{align*} & \\int_\\gamma y \\, dx = \\int_0^1 \\gamma_2(t) \\gamma_1^{\\:\\prime}(t) \\, dt = \\int_0^1 0 (1) \\, dt = 0 ,", "& \\int_\\beta y \\, dx = \\int_0^1 \\beta_2(t) \\beta_1'(t) \\, dt = \\int_0^1 (1-t)t(1) \\, dt = \\frac{1}{6} . \\end{align*} The integral of$y\\,dx$is not path independent. In particular,$\\int_{(0,0)}^{(1,0)} y\\,dx$does not make sense." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 16, "type": "proposition", "label": "Lebl-contfunc:mv:prop:pathinddf", "categories": [ "paths", "characterization", "forms", "independence", "domains" ], "title": "Characterization of Path Integrals via Equivalence", "contents": [ "Let$U \\subset \\R^n$be a path connected open set and$\\omega$a one-form defined on$U$. Then$\\int_x^y \\omega$is path independent (for all$x,y \\in U$) if and only if there exists a continuously differentiable$f \\colon U \\to \\R$such that$\\omega = df$.", "In fact, if such an$f$exists, then for every pair of points$x,y \\in U$\\begin{equation*} \\int_{x}^y \\omega = f(y)-f(x) . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "First suppose that the integral is path independent. Pick $p \\in U$. Since", "$U$ is path connected, there exists a path from $p$ to every $x \\in U$.", "Define", "\\begin{equation*}", "f(x) \\coloneqq \\int_{p}^x \\omega .", "\\end{equation*}", "Write $\\omega = \\omega_1 \\,dx_1 + \\omega_2 \\,dx_2 + \\cdots + \\omega_n \\,dx_n$.", "We wish to show that for every $j = 1,2,\\ldots,n$, the", "partial derivative $\\frac{\\partial f}{\\partial x_j}$ exists", "and is equal to $\\omega_j$.", "Let $e_j$ be an arbitrary standard basis vector, and $h$ a nonzero real", "number. Compute", "\\begin{equation*}", "\\frac{f(x+h e_j) - f(x)}{h} =", "\\frac{1}{h} \\left( \\int_{p}^{x+he_j} \\omega - \\int_{p}^x \\omega \\right)", "=", "\\frac{1}{h} \\int_{x}^{x+he_j} \\omega ,", "\\end{equation*}", "which follows by \\propref{mv:prop:pathconcat} and path independence as", "$\\int_{p}^{x+he_j} \\omega =", "\\int_{p}^{x} \\omega +", "\\int_{x}^{x+he_j} \\omega$, because we pick a path from $p$ to", "$x+he_j$ that also happens to pass through $x$, and then we cut this path in", "two, see \\figureref{fig:pathindantider}.", "\\begin{myfigureht}", "\\subimport*{figures/}{pathindantider.pdf_t}", "\\caption{Using path independence in computing the partial", "derivative.\\label{fig:pathindantider}}", "\\end{myfigureht}", "Since $U$ is open, suppose $h$ is so small so that all points of distance", "$\\abs{h}$ or", "less from $x$ are in $U$.", "As the integral is path independent,", "pick the simplest path possible from $x$ to $x+he_j$, that is", "$\\gamma(t) \\coloneqq x+t he_j$ for $t \\in [0,1]$. The path is in $U$.", "Notice $\\gamma^{\\:\\prime}(t) = h e_j$", "has only one nonzero component and that is the $j$th component, which is", "$h$. Therefore,", "\\begin{equation*}", "\\frac{1}{h} \\int_{x}^{x+he_j} \\omega", "=", "\\frac{1}{h} \\int_{\\gamma} \\omega", "=", "\\frac{1}{h} \\int_0^1 \\omega_j(x+the_j) h \\, dt", "=", "\\int_0^1 \\omega_j(x+the_j) \\, dt .", "\\end{equation*}", "We wish to take the limit as $h \\to 0$. The function $\\omega_j$ is", "continuous at $x$. Given $\\epsilon > 0$, suppose $h$ is small enough so that", "$\\abs{\\omega_j(x)-\\omega_j(y)} < \\epsilon$ whenever $\\snorm{x-y} \\leq \\abs{h}$.", "Thus,", "$\\abs{\\omega_j(x+the_j)-\\omega_j(x)} < \\epsilon$ for all $t \\in [0,1]$,", "and we estimate", "\\begin{equation*}", "\\abs{\\int_0^1 \\omega_j(x+the_j) \\, dt - \\omega_j(x)}", "=", "\\abs{\\int_0^1 \\bigl( \\omega_j(x+the_j) - \\omega_j(x) \\bigr) \\, dt}", "\\leq", "\\epsilon .", "\\end{equation*}", "That is,", "\\begin{equation*}", "\\lim_{h\\to 0}\\frac{f(x+h e_j) - f(x)}{h} = \\omega_j(x) .", "\\end{equation*}", "All partials of $f$ exist and are equal to $\\omega_j$, which are continuous", "functions. Thus, $f$ is continuously differentiable, and furthermore", "$df = \\omega$.", "For the other direction,", "suppose a continuously differentiable $f$ exists such that $df = \\omega$.", "Take a smooth", "path", "$\\gamma \\colon [a,b] \\to U$ such that $\\gamma(a) = x$ and", "$\\gamma(b) = y$. Then", "\\begin{equation*}", "\\begin{split}", "\\int_\\gamma df", "& =", "\\int_a^b", "\\biggl(", "\\frac{\\partial f}{\\partial x_1}\\bigl(\\gamma(t)\\bigr) \\gamma_1^{\\:\\prime}(t)+", "\\frac{\\partial f}{\\partial x_2}\\bigl(\\gamma(t)\\bigr) \\gamma_2^{\\:\\prime}(t)+ \\cdots +", "\\frac{\\partial f}{\\partial x_n}\\bigl(\\gamma(t)\\bigr) \\gamma_n^{\\:\\prime}(t)", "\\biggr) \\, dt", "\\\\", "& =", "\\int_a^b", "\\frac{d}{dt} \\Bigl[ f\\bigl(\\gamma(t)\\bigr) \\Bigr]\\, dt", "\\\\", "& = f(y)-f(x) .", "\\end{split}", "\\end{equation*}", "The value of the integral only depends on $x$ and $y$, not the", "path taken. Therefore the integral is path independent.", "We leave checking this fact for a piecewise smooth path as an exercise." ], "refs": [ "mv:prop:pathconcat" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 17, "type": "proposition", "label": "Lebl-contfunc:17", "categories": [ "paths", "domains", "characterization", "forms" ], "title": "Characterization of Path Integrals via Equivalence", "contents": [ "Let$U \\subset \\R^n$be a path connected open set and$\\omega$a one-form defined on$U$. Then$\\omega = df$for some continuously differentiable$f \\colon U \\to \\R$if and only if\\begin{equation*} \\int_{\\gamma} \\omega = 0 \\qquad \\text{for every piecewise smooth closed path } \\gamma \\colon [a,b] \\to U. \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Suppose $\\omega = df$ and let $\\gamma$ be a piecewise smooth", "closed path.", "Since $\\gamma(a) = \\gamma(b)$ for a closed path,", "the previous proposition says", "\\begin{equation*}", "\\int_{\\gamma} \\omega = f\\bigl(\\gamma(b)\\bigr) - f\\bigl(\\gamma(a)\\bigr) = 0 .", "\\end{equation*}", "Now suppose that for every piecewise smooth closed path $\\gamma$, $\\int_{\\gamma} \\omega = 0$.", "Let $x,y$ be two points in $U$ and let $\\alpha \\colon [0,1] \\to U$ and", "$\\beta \\colon [0,1] \\to U$ be two piecewise smooth paths with $\\alpha(0) = \\beta(0) = x$", "and $\\alpha(1) = \\beta(1) = y$. See \\figureref{fig:twopaths}.", "\\begin{myfigureht}", "\\subimport*{figures/}{twopaths.pdf_t}", "\\caption{Two paths from $x$ to $y$.\\label{fig:twopaths}}", "\\end{myfigureht}", "Define $\\gamma \\colon [0,2] \\to U$ by", "\\begin{equation*}", "\\gamma(t) \\coloneqq", "\\begin{cases}", "\\alpha(t) & \\text{if } t \\in [0,1], \\\\", "\\beta(2-t) & \\text{if } t \\in (1,2].", "\\end{cases}", "\\end{equation*}", "This path is piecewise smooth. This is due to the fact that", "$\\gamma|_{[0,1]}(t) = \\alpha(t)$ and", "$\\gamma|_{[1,2]}(t) = \\beta(2-t)$ (note especially $\\gamma(1) = \\alpha(1) =", "\\beta(2-1)$).", "It is also closed as $\\gamma(0) = \\alpha(0) = \\beta(0) = \\gamma(2)$.", "So", "\\begin{equation*}", "0 = \\int_{\\gamma} \\omega = \\int_{\\alpha} \\omega - \\int_{\\beta} \\omega .", "\\end{equation*}", "This follows first by \\propref{mv:prop:pathconcat}, and then noticing that", "the second part is $\\beta$ traveled backwards so that we get minus the", "$\\beta$ integral. Thus the integral of $\\omega$ on $U$ is path independent." ], "refs": [ "mv:prop:pathconcat" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 18, "type": "theorem", "label": "Lebl-contfunc:18", "categories": [ "domains", "forms" ], "title": "Continuity of Differential Forms", "contents": [ "Let$U \\subset \\R^n$be a star-shaped domain and$\\omega$a continuously differentiable one-form defined on$U$. That is, if\\begin{equation*} \\omega = \\omega_1 \\,dx_1 + \\omega_2 \\,dx_2 + \\cdots + \\omega_n \\,dx_n , \\end{equation*}then$\\omega_1,\\omega_2,\\ldots,\\omega_n$are continuously differentiable functions. Suppose that for every$j$and$k$\\begin{equation*} \\frac{\\partial \\omega_j}{\\partial x_k} = \\frac{\\partial \\omega_k}{\\partial x_j} , \\avoidbreak \\end{equation*}then there exists a twice continuously differentiable function$f \\colon U \\to \\R$such that$df = \\omega$." ], "refs": [], "proofs": [ { "contents": [ "Suppose $U$ is a star-shaped domain with respect to $p=(p_1,p_2,\\ldots,p_n) \\in U$.", "Given $x = (x_1,x_2,\\ldots,x_n) \\in U$, define the path $\\gamma \\colon [0,1] \\to U$ as", "$\\gamma(t) \\coloneqq (1-t)p + tx$, so $\\gamma^{\\:\\prime}(t) = x-p$. Let", "\\begin{equation*}", "f(x) \\coloneqq \\int_{\\gamma} \\omega", "=", "\\int_0^1", "\\left(", "\\sum_{k=1}^n", "\\omega_k \\bigl((1-t)p + tx \\bigr) \\, (x_k-p_k)", "\\right) dt .", "\\end{equation*}", "We differentiate in $x_j$ under the integral, which is allowed as", "everything, including the partials, is continuous:", "\\begin{equation*}", "\\begin{split}", "\\frac{\\partial f}{\\partial x_j}(x) & =", "\\int_0^1", "\\left(", "\\left(", "\\sum_{k=1}^n", "\\frac{\\partial \\omega_k}{\\partial x_j} \\bigl((1-t)p + tx \\bigr) \\, t", "(x_k-p_k)", "\\right)", "+", "\\omega_j \\bigl((1-t)p + tx \\bigr)", "\\right)", "dt", "\\\\", "& =", "\\int_0^1", "\\left(", "\\left(", "\\sum_{k=1}^n", "\\frac{\\partial \\omega_j}{\\partial x_k} \\bigl((1-t)p + tx \\bigr) \\, t", "(x_k-p_k)", "\\right)", "+", "\\omega_j \\bigl((1-t)p + tx \\bigr)", "\\right) dt", "\\\\", "& =", "\\int_0^1", "\\frac{d}{dt}", "\\Bigl[", "t \\omega_j\\bigl((1-t)p + tx \\bigr)", "\\Bigr]", "\\,", "dt", "\\\\", "&= \\omega_j(x) .", "\\end{split}", "\\end{equation*}", "And this is precisely what we wanted." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 19, "type": "example", "label": "Lebl-contfunc:19", "categories": [ "paths", "independence", "integration", "example" ], "title": "Without some hypothesis on$U$the theorem is not true", "contents": [ "Without some hypothesis on$U$the theorem is not true. Let", "\\begin{equation*}\n\\omega(x,y) \\coloneqq \\frac{-y}{x^2+y^2} \\,dx + \\frac{x}{x^2+y^2} \\,dy\n\\end{equation*}", "be defined on$\\R^2 \\setminus \\{ 0 \\}$. Then", "\\begin{equation*}\n\\frac{\\partial}{\\partial y} \\left[ \n\\frac{-y}{x^2+y^2} \\right] =\n\\frac{y^2-x^2}{{(x^2+y^2)}^2}\n=\n\\frac{\\partial}{\\partial x} \\left[ \n\\frac{x}{x^2+y^2} \\right] .\n\\end{equation*}", "However, there is no$f \\colon \\R^2 \\setminus \\{ 0 \\} \\to \\R$such that$df = \\omega$. In \\exampleref{example:mv:irrotoneformint} we integrated from$(1,0)$to$(1,0)$along the unit circle counterclockwise, that is$\\gamma(t) = \\bigl(\\cos(t),\\sin(t)\\bigr)$for$t \\in [0,2\\pi]$, and we found the integral to be$2\\pi$. We would have gotten$0$if the integral was path independent, or in other words if there would exist an$f$such that$df = \\omega$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 0, "type": "proposition", "label": "Lebl-contfunc:mv:sumulbound:prop", "categories": [ "domains" ], "title": "Boundedness of Rectangles and Partitions", "contents": [ "Suppose$R \\subset \\R^n$is a closed rectangle and$f \\colon R \\to \\R$is a bounded function. Let$m, M \\in \\R$be such that for all$x \\in R$, we have$m \\leq f(x) \\leq M$. Then for every partition$P$of~$R$,\\begin{equation*} m \\, V(R) \\leq L(P,f) \\leq U(P,f) \\leq M\\, V(R) . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Let $P$ be a partition of $R$. For all $i$, we have", "$m \\leq m_i \\leq M_i \\leq M$. Also $\\sum_{i=1}^N V(R_i) = V(R)$. Therefore,", "\\begin{multline*}", "m \\, V(R) =", "m \\left( \\sum_{i=1}^N V(R_i) \\right)", "=", "\\sum_{i=1}^N m \\, V(R_i)", "\\leq", "\\sum_{i=1}^N m_i \\, V(R_i)", "\\leq", "\\\\", "\\leq", "\\sum_{i=1}^N M_i \\, V(R_i)", "\\leq", "\\sum_{i=1}^N M \\,V(R_i)", "=", "M \\left( \\sum_{i=1}^N V(R_i) \\right)", "=", "M \\,V(R) . \\qedhere", "\\end{multline*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 1, "type": "proposition", "label": "Lebl-contfunc:mv:prop:refinement", "categories": [ "domains" ], "title": "Boundedness of Rectangles and Partitions", "contents": [ "Suppose$R \\subset \\R^n$is a closed rectangle,$P$is a partition of$R$, and$\\widetilde{P}$is a refinement of$P$. If$f \\colon R \\to \\R$is bounded, then\\begin{equation*} L(P,f) \\leq L(\\widetilde{P},f) \\qquad \\text{and} \\qquad U(\\widetilde{P},f) \\leq U(P,f) . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "We prove the first inequality, and the second follows similarly.", "Let $R_1,R_2,\\ldots,R_N$ be the subrectangles of $P$", "and", "$\\widetilde{R}_1,\\widetilde{R}_2,\\ldots,\\widetilde{R}_{\\widetilde{N}}$ be the", "subrectangles of", "$\\widetilde{R}$.", "Let $I_k$ be the set of all indices $j$ such that $\\widetilde{R}_j \\subset R_k$.", "For example, in figures \\ref{mv:figrect} and", "\\ref{mv:figrectpart}, $I_4 = \\{ 6, 7, 8, 9 \\}$ as", "$R_4 =", "\\widetilde{R}_6 \\cup \\widetilde{R}_7 \\cup", "\\widetilde{R}_8 \\cup \\widetilde{R}_9$.", "Then,", "\\begin{equation*}", "R_k = \\bigcup_{j \\in I_k} \\widetilde{R}_j,", "\\qquad", "V(R_k) = \\sum_{j \\in I_k} V(\\widetilde{R}_j).", "\\end{equation*}", "Let $m_j \\coloneqq \\inf \\bigl\\{ f(x) : x \\in R_j \\bigr\\}$, and", "$\\widetilde{m}_j \\coloneqq \\inf \\bigl\\{ f(x) : \\in \\widetilde{R}_j \\bigr\\}$ as usual.", "If $j \\in I_k$, then $m_k \\leq \\widetilde{m}_j$. Then", "\\begin{equation*}", "L(P,f) =", "\\sum_{k=1}^N m_k V(R_k)", "=", "\\sum_{k=1}^N \\sum_{j\\in I_k} m_k V(\\widetilde{R}_j)", "\\leq", "\\sum_{k=1}^N \\sum_{j\\in I_k} \\widetilde{m}_j V(\\widetilde{R}_j)", "=", "\\sum_{j=1}^{\\widetilde{N}} \\widetilde{m}_j V(\\widetilde{R}_j) = L(\\widetilde{P},f) . \\qedhere", "\\end{equation*}" ], "refs": [ "mv:figrect", "mv:figrectpart" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 2, "type": "proposition", "label": "Lebl-contfunc:mv:intulbound:prop", "categories": [ "domains" ], "title": "Boundedness of Rectangles and Partitions", "contents": [ "Let$R \\subset \\R^n$be a closed rectangle and$f \\colon R \\to \\R$a bounded function. Let$m, M \\in \\R$be such that for all$x \\in R$, we have$m \\leq f(x) \\leq M$. Then \\begin{equation} \\label{mv:intulbound:eq} m \\, V(R) \\leq \\underline{\\int_R} f \\leq \\overline{\\int_R} f \\leq M \\, V(R). \\end{equation}" ], "refs": [], "proofs": [ { "contents": [ "For every partition $P$, via \\propref{mv:sumulbound:prop},", "\\begin{equation*}", "m\\,V(R) \\leq L(P,f) \\leq U(P,f) \\leq M\\,V(R).", "\\end{equation*}", "Taking supremum of $L(P,f)$ and infimum of $U(P,f)$ over all partitions $P$,", "we obtain the first and the last inequality in", "\\eqref{mv:intulbound:eq}.", "The key inequality in", "\\eqref{mv:intulbound:eq}", "is the middle one.", "Let $P=(P_1,P_2,\\ldots,P_n)$ and", "$Q=(Q_1,Q_2,\\ldots,Q_n)$", "be partitions of $R$. Define", "$\\widetilde{P} = ( \\widetilde{P}_1,\\widetilde{P}_2,\\ldots,\\widetilde{P}_n )$", "by letting", "$\\widetilde{P}_k \\coloneqq P_k \\cup Q_k$ for every $k$.", "Then $\\widetilde{P}$ is a partition of $R$,", "and $\\widetilde{P}$ is a refinement of $P$ and also a refinement of $Q$.", "By \\propref{mv:prop:refinement},", "$L(P,f) \\leq L(\\widetilde{P},f)$ and", "$U(\\widetilde{P},f) \\leq U(Q,f)$. Therefore,", "\\begin{equation*}", "L(P,f) \\leq L(\\widetilde{P},f) \\leq U(\\widetilde{P},f) \\leq U(Q,f) .", "\\end{equation*}", "In other words, for two arbitrary partitions $P$ and $Q$, we have", "$L(P,f) \\leq U(Q,f)$.", "Via \\volIref{\\propref*{vI-infsupineq:prop} from volume I}{\\propref{infsupineq:prop}},", "we obtain", "\\begin{equation*}", "\\sup \\, \\bigl\\{ L(P,f) : P \\text{ a partition of } R \\bigl\\}", "\\leq", "\\inf \\, \\bigl\\{ U(P,f) : P \\text{ a partition of } R \\bigl\\} .", "\\end{equation*}", "In other words, $\\underline{\\int_R} f \\leq \\overline{\\int_R} f$." ], "refs": [ "infsupineq:prop", "mv:intulbound:eq", "mv:prop:refinement", "mv:sumulbound:prop" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 3, "type": "proposition", "label": "Lebl-contfunc:mv:intbound:prop", "categories": [ "integration", "domains" ], "title": "Let$f \\colon R \\to \\R$be a Riemann integrable function on a closed rectangle$R \\subset \\R^n$", "contents": [ "Let$f \\colon R \\to \\R$be a Riemann integrable function on a closed rectangle$R \\subset \\R^n$. Let$m, M \\in \\R$be such that$m \\leq f(x) \\leq M$for all$x \\in R$. Then\\begin{equation*} m \\, V(R) \\leq \\int_{R} f \\leq M \\, V(R) . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 4, "type": "example", "label": "Lebl-contfunc:4", "categories": [ "integration", "example" ], "title": "A constant function is Riemann integrable", "contents": [ "A constant function is Riemann integrable. Proof: Suppose$f(x) = c$for all$x \\in R$. Then", "\\begin{equation*}\nc \\, V(R) \\leq \\underline{\\int_R} f \\leq \\overline{\\int_R} f \\leq c\\, V(R) .\n\\end{equation*}", "So$f$is integrable, and furthermore$\\int_R f = c\\,V(R)$." ], "refs": [], "proofs": [ { "contents": [ "Suppose$f(x) = c$for all$x \\in R$. Then \\begin{equation*}", "c \\, V(R) \\leq \\underline{\\int_R} f \\leq \\overline{\\int_R} f \\leq c\\, V(R) .", "\\end{equation*} So$f$is integrable, and furthermore$\\int_R f = c\\,V(R)$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 5, "type": "proposition", "label": "Lebl-contfunc:mv:intlinearity:prop", "categories": [ "domains" ], "title": "\\index{linearity of the integral} Let$R \\subset \\R^n$be a closed rectangle and let$f$and$g$be in$...", "contents": [ "\\index{linearity of the integral} Let$R \\subset \\R^n$be a closed rectangle and let$f$and$g$be in$\\sR(R)$and$\\alpha \\in \\R$. \\begin{enumerate}[(i)] \\item$\\alpha f$is in$\\sR(R)$and\\begin{equation*} \\int_R \\alpha f = \\alpha \\int_R f . \\end{equation*}\\item$f+g$is in$\\sR(R)$and\\begin{equation*} \\int_R (f+g) = \\int_R f + \\int_R g . \\end{equation*}\\end{enumerate}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 6, "type": "proposition", "label": "Lebl-contfunc:6", "categories": [ "domains" ], "title": "\\index{monotonicity of the integral} Let$R \\subset \\R^n$be a closed rectangle, let$f$and$g$be in$...", "contents": [ "\\index{monotonicity of the integral} Let$R \\subset \\R^n$be a closed rectangle, let$f$and$g$be in$\\sR(R)$, and suppose$f(x) \\leq g(x)$for all$x \\in R$. Then\\begin{equation*} \\int_R f \\leq \\int_R g . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 7, "type": "proposition", "label": "Lebl-contfunc:mv:prop:upperlowerepsilon", "categories": [ "domains", "characterization" ], "title": "Characterization of Rectangles and Partitions via Equivalence", "contents": [ "Let$R \\subset \\R^n$be a closed rectangle and$f \\colon R \\to \\R$a bounded function. Then$f \\in \\sR(R)$if and only if for every$\\epsilon > 0$, there exists a partition$P$of$R$such that\\begin{equation*} U(P,f) - L(P,f) < \\epsilon . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "First, if $f$ is integrable, then the supremum of $L(P,f)$ and", "infimum of $U(Q,f)$ over all partitions $P$ and $Q$ are equal and hence the", "infimum of $U(P,f)-L(Q,f)$ is zero. Taking a common refinement", "$\\widetilde{P}$ of $P$ and $Q$ we find", "$U(\\widetilde{P},f)-L(\\widetilde{P},f) \\leq U(P,f)-L(Q,f)$.", "Hence the infimum of $U(P,f)-L(P,f)$ over all partitions $P$ is zero, and", "so for every $\\epsilon > 0$, there must be some partition $P$ such that", "$U(P,f) - L(P,f) < \\epsilon$.", "For the other direction, given an $\\epsilon > 0$ find $P$ such that", "$U(P,f) - L(P,f) < \\epsilon$.", "\\begin{equation*}", "\\overline{\\int_R} f -", "\\underline{\\int_R} f", "\\leq", "U(P,f) - L(P,f)", "< \\epsilon .", "\\end{equation*}", "As $\\overline{\\int_R} f \\geq \\underline{\\int_R} f$ and the above holds for", "every $\\epsilon > 0$, we conclude", "$\\overline{\\int_R} f = \\underline{\\int_R} f$ and $f \\in \\sR(R)$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 8, "type": "proposition", "label": "Lebl-contfunc:mv:prop:integralsmallerset", "categories": [ "integration", "domains" ], "title": "Let$S \\subset \\R^n$be a closed rectangle", "contents": [ "Let$S \\subset \\R^n$be a closed rectangle. If$f \\colon S \\to \\R$is integrable and$R \\subset S$is a closed rectangle, then$f$is integrable on$R$." ], "refs": [], "proofs": [ { "contents": [ "Given $\\epsilon > 0$, find a partition $P=(P_1,\\ldots,P_n)$", "of $S$ such that", "$U(P,f)-L(P,f) < \\epsilon$. By making a refinement of $P$", "if necessary,", "assume that the endpoints of $R$ are in $P$. That is,", "if $R = [a_1,b_1] \\times [a_2,b_2] \\times \\cdots \\times [a_n,b_n]$,", "then $a_i,b_i \\in P_i$.", "Let $\\widetilde{P} = (\\widetilde{P}_1,\\ldots,\\widetilde{P}_n)$", "be the partition of $R$ given by $\\widetilde{P}_i = P_i \\cap [a_i,b_i]$.", "Subrectangles of $\\widetilde{P}$ are subrectangles of~$P$, that", "is, $R$ is a union of subrectangles of~$P$.", "Divide the subrectangles of $P$ into two collections:", "Let $R_1,R_2\\ldots,R_K$ be the subrectangles of $P$ that are", "also subrectangles of $\\widetilde{P}$ and", "let $R_{K+1},\\ldots, R_N$ be the rest.", "See \\figureref{fig:figrectsubrect}.", "Let $m_k$ and $M_k$ be the infimum and supremum", "of $f$ on $R_k$ as usual. Then,", "\\begin{equation*}", "\\begin{split}", "\\epsilon & >", "U(P,f)-L(P,f)", "=", "\\sum_{k=1}^K (M_k-m_k) V(R_k)", "+", "\\sum_{k=K+1}^N (M_k-m_k) V(R_k)", "\\\\", "&", "\\geq", "\\sum_{k=1}^K (M_k-m_k) V(R_k)", "=", "U(\\widetilde{P},f|_R)-L(\\widetilde{P},f|_R) .", "\\end{split}", "\\end{equation*}", "Therefore, $f|_R$ is integrable." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 9, "type": "proposition", "label": "Lebl-contfunc:prop:diameterrectangle", "categories": [ "domains" ], "title": "If a rectangle$R \\subset \\R^n$has longest side at most$\\alpha$, then for all$x,y \\in R$, \\begin{e...", "contents": [ "If a rectangle$R \\subset \\R^n$has longest side at most$\\alpha$, then for all$x,y \\in R$,\\begin{equation*} \\snorm{x-y} \\leq \\sqrt{n} \\, \\alpha . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "\\begin{equation*}", "\\begin{split}", "\\snorm{x-y}", "& =", "\\sqrt{", "{(x_1-y_1)}^2", "+", "{(x_2-y_2)}^2", "+ \\cdots +", "{(x_n-y_n)}^2", "}", "\\\\", "& \\leq", "\\sqrt{", "{(b_1-a_1)}^2", "+", "{(b_2-a_2)}^2", "+ \\cdots +", "{(b_n-a_n)}^2", "}", "\\\\", "& \\leq", "\\sqrt{", "{\\alpha}^2", "+", "{\\alpha}^2", "+ \\cdots +", "{\\alpha}^2", "}", "=", "\\sqrt{n} \\, \\alpha . \\qedhere", "\\end{split}", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 10, "type": "theorem", "label": "Lebl-contfunc:mv:thm:contintrect", "categories": [ "domains" ], "title": "Continuity of Rectangles and Partitions", "contents": [ "Let$R \\subset \\R^n$be a closed rectangle. If$f \\colon R \\to \\R$is continuous, then$f \\in \\sR(R)$." ], "refs": [], "proofs": [ { "contents": [ "The proof is analogous to the one-variable proof with some complications.", "The set $R$ is a closed and bounded subset of $\\R^n$, and hence compact. So", "$f$ is uniformly continuous", "by \\volIref{\\thmref*{vI-thm:Xcompactfunifcont} from volume I}{\\thmref{thm:Xcompactfunifcont}}.", "Let $\\epsilon > 0$ be given. Find a $\\delta > 0$ such that", "$\\snorm{x-y} < \\delta$ implies $\\sabs{f(x)-f(y)} < \\frac{\\epsilon}{V(R)}$.", "Let $P$ be a partition of $R$, such that longest side of every subrectangle", "is strictly less than $\\frac{\\delta}{\\sqrt{n}}$.", "If $x, y \\in R_k$ for a subrectangle $R_k$ of $P$, then,", "by the proposition,", "$\\snorm{x-y} < \\sqrt{n} \\frac{\\delta}{\\sqrt{n}} = \\delta$. Therefore,", "\\begin{equation*}", "f(x)-f(y) \\leq \\sabs{f(x)-f(y)} < \\frac{\\epsilon}{V(R)} .", "\\end{equation*}", "As $f$ is continuous on $R_k$, which is compact, $f$ attains a maximum and a minimum", "on this subrectangle.", "Let $x$ be a point where $f$ attains the maximum and $y$ be a point", "where $f$ attains the minimum. Then $f(x) = M_k$", "and $f(y) = m_k$ in the notation from the definition of the integral.", "Thus,", "\\begin{equation*}", "M_k-m_k = f(x)-f(y) <", "\\frac{\\epsilon}{V(R)} .", "\\end{equation*}", "And so", "\\begin{equation*}", "\\begin{split}", "U(P,f) - L(P,f)", "& =", "\\left(", "\\sum_{k=1}^N", "M_k V(R_k)", "\\right)", "-", "\\left(", "\\sum_{k=1}^N", "m_k V(R_k)", "\\right)", "\\\\", "& =", "\\sum_{k=1}^N", "(M_k-m_k) V(R_k)", "\\\\", "& <", "\\frac{\\epsilon}{V(R)}", "\\sum_{k=1}^N", "V(R_k)", "= \\epsilon.", "\\end{split}", "\\avoidbreak", "\\end{equation*}", "Via application of \\propref{mv:prop:upperlowerepsilon}, we find that $f \\in", "\\sR(R)$." ], "refs": [ "mv:prop:upperlowerepsilon", "thm:Xcompactfunifcont" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 11, "type": "example", "label": "Lebl-contfunc:11", "categories": [ "example" ], "title": "Continuity of Multivariable Integration", "contents": [ "The function$f \\colon \\R^2 \\to \\R$defined by", "\\begin{equation*}\nf(x,y) \\coloneqq\n\\begin{cases}\n-x{(x^2+y^2-1)}^2 & \\text{if } \\sqrt{x^2+y^2} \\leq 1, \\\\\n0 & \\text{else},\n\\end{cases}\n\\end{equation*}", "is continuous and its support is the closed unit disc$C(0,1) = \\bigl\\{ (x,y) : \\sqrt{x^2 + y^2} \\leq 1 \\bigr\\}$, which is a compact set, so$f$has compact support. Note that the function is zero on the entire$y$-axis and on the unit circle, but all points that lie in the closed unit disc are still within the support as they are in the closure of points where$f$is nonzero. See \\figureref{fig:compsup}. \\begin{myfigureht} \\subimport*{figures/}{compsup_full.pdf_t} \\caption{Function with compact support (left), the support is the closed unit disc (right).\\label{fig:compsup}} \\end{myfigureht}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 12, "type": "example", "label": "Lebl-contfunc:12", "categories": [ "example" ], "title": "Continuity of Multivariable Integration", "contents": [ "Let$B(0,1) \\subset \\R^2$be the unit disc. The function$f \\colon B(0,1) \\to \\R$defined by", "\\begin{equation*}\nf(x,y) \\coloneqq\n\\begin{cases}\n0 & \\text{if } \\sqrt{x^2+y^2} > \\nicefrac{1}{2}, \\\\\n\\nicefrac{1}{2} - \\sqrt{x^2+y^2} & \\text{if } \\sqrt{x^2+y^2} \\leq \\nicefrac{1}{2},\n\\end{cases}\n\\end{equation*}", "is continuous on$B(0,1)$and its support is the smaller closed ball$C(0,\\nicefrac{1}{2})$. As that is a compact set,$f$has compact support.", "The function$g \\colon B(0,1) \\to \\R$defined by", "\\begin{equation*}\ng(x,y) \\coloneqq\n\\begin{cases}\n0 & \\text{if } x \\leq 0, \\\\\nx & \\text{if } x > 0,\n\\end{cases}\n\\end{equation*}", "is continuous on$B(0,1)$, but its support is the set$\\bigl\\{ (x,y) \\in B(0,1) : x \\geq 0 \\bigr\\}$. In particular,$g$~is not compactly supported." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 13, "type": "example", "label": "Lebl-contfunc:13", "categories": [ "domains", "example" ], "title": "Continuity of Domains and Boundaries", "contents": [ "The continuous function$f \\colon B(0,1) \\to \\R$given by$f(x,y) \\coloneqq \\sin\\bigl(\\frac{1}{1-x^2-y^2}\\bigr)$does not have compact support; as$f$is not constantly zero on any neighborhood of every point in$B(0,1)$, the support is the entire disc$B(0,1)$. The function does not extend as above to a continuous function on$\\R^2$. In fact, it is not difficult to show that$f$cannot be extended in any way whatsoever to be continuous on all of$\\R^2$(the boundary of the disc is the problem)." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 14, "type": "proposition", "label": "Lebl-contfunc:mv:prop:rectanglessupp", "categories": [ "domains" ], "title": "Continuity of Rectangles and Partitions", "contents": [ "Suppose$f \\colon \\R^n \\to \\R$is a continuous function with compact support. If$R$and$S$are closed rectangles such that$\\operatorname{supp}(f) \\subset R$and$\\operatorname{supp}(f) \\subset S$, then\\begin{equation*} \\int_S f = \\int_R f . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "As $f$ is continuous, it is automatically integrable on the rectangles $R$, $S$, and $R", "\\cap S$.", "Then \\exerciseref{mv:zerooutside} says", "$\\int_S f = \\int_{S \\cap R} f = \\int_R f$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 15, "type": "example", "label": "Lebl-contfunc:15", "categories": [ "integration", "domains", "example" ], "title": "Define \\begin{equation*} f(x,y) \\coloneqq \\begin{cases} 1 & \\text{if } x=\\nicefrac{1}{2} \\text{ a...", "contents": [ "Define", "\\begin{equation*}\nf(x,y) \\coloneqq \n\\begin{cases}\n1 & \\text{if } x=\\nicefrac{1}{2} \\text{ and } y \\in \\Q, \\\\\n0 & \\text{otherwise.}\n\\end{cases}\n\\end{equation*}", "Then$f$is Riemann integrable on$R \\coloneqq [0,1]^2$and$\\int_R f = 0$. Moreover,$\\int_0^1 \\int_0^1 f(x,y) \\, dx \\, dy = 0$. However,", "\\begin{equation*}\n\\int_0^1 f(\\nicefrac{1}{2},y) \\, dy\n\\end{equation*}", "does not exist, so strictly speaking,$\\int_0^1 \\int_0^1 f(x,y) \\, dy \\, dx$does not make sense. See \\figureref{fig:fubinibad}.", "\\begin{myfigureht} \\subimport*{figures/}{fubinibad.pdf_t} \\caption{Left:$[0,1]^2$with the line$x=\\nicefrac{1}{2}$marked dotted and$\\int_0^1 f(x,y) \\, dx$marked as gray solid line for a generic$y$. Center: Similar picture but$\\int_0^1 f(x,y) \\, dy$marked for some$x \\not= \\nicefrac{1}{2}$. Right: The three different rectangles in the partition used to integrate$f$in different grays.\\label{fig:fubinibad}} \\end{myfigureht}", "Proof: We start with integrability of$f$. Consider the partition of$[0,1]^2$where the partition in the$x$direction is$\\{ 0, \\nicefrac{1}{2}-\\epsilon, \\nicefrac{1}{2}+\\epsilon,1\\}$and in the$y$direction$\\{ 0, 1 \\}$. The corresponding subrectangles are", "\\begin{equation*}\nR_1 \\coloneqq [0, \\nicefrac{1}{2}-\\epsilon] \\times [0,1],\n\\qquad\nR_2 \\coloneqq [\\nicefrac{1}{2}-\\epsilon,\n\\nicefrac{1}{2}+\\epsilon] \\times [0,1],\n\\qquad\nR_3 \\coloneqq [\\nicefrac{1}{2}+\\epsilon,1] \\times [0,1] .\n\\end{equation*}", "We have$m_1 = M_1 = 0$,$m_2 =0$,$M_2 = 1$, and$m_3 = M_3 = 0$. Therefore,", "\\begin{equation*}\nL(P,f) = \nm_1 V(R_1)\n+\nm_2 V(R_2)\n+\nm_3 V(R_3)\n=\n0 (\\nicefrac{1}{2}-\\epsilon)\n+\n0 (2\\epsilon)\n+\n0 (\\nicefrac{1}{2}-\\epsilon) = 0 ,\n\\end{equation*}", "and", "\\begin{equation*}\nU(P,f) = \nM_1 V(R_1)\n+\nM_2 V(R_2)\n+\nM_3 V(R_3)\n=\n0 (\\nicefrac{1}{2}-\\epsilon)\n+\n1 (2\\epsilon)\n+\n0 (\\nicefrac{1}{2}-\\epsilon) = 2 \\epsilon .\n\\end{equation*}", "The upper and lower sums are arbitrarily close and the lower sum is always zero, so the function is integrable and$\\int_R f = 0$.", "For every fixed$y$, the function that takes$x$to$f(x,y)$is zero except perhaps at a single point$x=\\nicefrac{1}{2}$. Such a function is integrable and$\\int_0^1 f(x,y) \\, dx = 0$. Therefore,$\\int_0^1 \\int_0^1 f(x,y) \\, dx \\, dy = 0$. However, if$x=\\nicefrac{1}{2}$, the function that takes$y$to$f(\\nicefrac{1}{2},y)$is the nonintegrable function that is 1 on the rationals and 0 on the irrationals. See \\volIref{\\exampleref*{vI-example:dirichletfunc} from volume I}{\\exampleref{example:dirichletfunc}}." ], "refs": [], "proofs": [ { "contents": [ "We start with integrability of$f$. Consider the partition of$[0,1]^2$where the partition in the$x$direction is$\\{ 0, \\nicefrac{1}{2}-\\epsilon, \\nicefrac{1}{2}+\\epsilon,1\\}$and in the$y$direction$\\{ 0, 1 \\}$. The corresponding subrectangles are \\begin{equation*}", "R_1 \\coloneqq [0, \\nicefrac{1}{2}-\\epsilon] \\times [0,1],", "\\qquad", "R_2 \\coloneqq [\\nicefrac{1}{2}-\\epsilon,", "\\nicefrac{1}{2}+\\epsilon] \\times [0,1],", "\\qquad", "R_3 \\coloneqq [\\nicefrac{1}{2}+\\epsilon,1] \\times [0,1] .", "\\end{equation*} We have$m_1 = M_1 = 0$,$m_2 =0$,$M_2 = 1$, and$m_3 = M_3 = 0$. Therefore, \\begin{equation*}", "L(P,f) =", "m_1 V(R_1)", "+", "m_2 V(R_2)", "+", "m_3 V(R_3)", "=", "0 (\\nicefrac{1}{2}-\\epsilon)", "+", "0 (2\\epsilon)", "+", "0 (\\nicefrac{1}{2}-\\epsilon) = 0 ,", "\\end{equation*} and \\begin{equation*}", "U(P,f) =", "M_1 V(R_1)", "+", "M_2 V(R_2)", "+", "M_3 V(R_3)", "=", "0 (\\nicefrac{1}{2}-\\epsilon)", "+", "1 (2\\epsilon)", "+", "0 (\\nicefrac{1}{2}-\\epsilon) = 2 \\epsilon .", "\\end{equation*} The upper and lower sums are arbitrarily close and the lower sum is always zero, so the function is integrable and$\\int_R f = 0$. For every fixed$y$, the function that takes$x$to$f(x,y)$is zero except perhaps at a single point$x=\\nicefrac{1}{2}$. Such a function is integrable and$\\int_0^1 f(x,y) \\, dx = 0$. Therefore,$\\int_0^1 \\int_0^1 f(x,y) \\, dx \\, dy = 0$. However, if$x=\\nicefrac{1}{2}$, the function that takes$y$to$f(\\nicefrac{1}{2},y)$is the nonintegrable function that is 1 on the rationals and 0 on the irrationals. See \\volIref{\\exampleref*{vI-example:dirichletfunc} from volume I}{\\exampleref{example:dirichletfunc}}." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 16, "type": "theorem", "label": "Lebl-contfunc:mv:fubinivA", "categories": [ "integration", "domains", "named theorem" ], "title": "Fubini's Theorem", "contents": [ "\\index{Fubini's theorem} Let$R \\times S \\subset \\R^n \\times \\R^m$be a closed rectangle and$f \\colon R \\times S \\to \\R$be integrable. The functions$g \\colon R \\to \\R$and$h \\colon R \\to \\R$defined by\\begin{equation*} g(x) \\coloneqq \\underline{\\int_S} f_x \\qquad \\text{and} \\qquad h(x) \\coloneqq \\overline{\\int_S} f_x \\end{equation*}are integrable on$R$and\\begin{equation*} \\int_R g = \\int_R h = \\int_{R \\times S} f . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "A partition of $R \\times S$ is a concatenation of a partition of $R$ and a", "partition of~$S$. That is, write a partition of $R \\times S$", "as $(P,P') = (P_1,P_2,\\ldots,P_n,P'_1,P'_2,\\ldots,P'_m)$,", "where", "$P = (P_1,P_2,\\ldots,P_n)$ and", "$P' = (P'_1,P'_2,\\ldots,P'_m)$ are partitions of $R$ and $S$ respectively.", "Let", "$R_1,R_2,\\ldots,R_N$ be the subrectangles of $P$ and", "$R'_1,R'_2,\\ldots,R'_K$ be the subrectangles of $P'$.", "The subrectangles of $(P,P')$ are", "$R_i \\times R'_j$ where $1 \\leq i \\leq N$ and $1 \\leq j \\leq K$.", "Let", "\\begin{equation*}", "m_{i,j} \\coloneqq", "\\inf_{(x,y) \\in R_i \\times R'_j} f(x,y) .", "\\end{equation*}", "Notice that", "$V(R_i \\times R'_j) = V(R_i)V(R'_j)$ and hence", "\\begin{equation*}", "L\\bigl((P,P'),f\\bigr) =", "\\sum_{i=1}^N", "\\sum_{j=1}^K", "m_{i,j} \\, V(R_i \\times R'_j)", "=", "\\sum_{i=1}^N", "\\left(", "\\sum_{j=1}^K", "m_{i,j} \\, V(R'_j) \\right) V(R_i) .", "\\end{equation*}", "Define", "\\begin{equation*}", "m_j(x) \\coloneqq \\inf_{y \\in R'_j} f(x,y) = \\inf_{y \\in R'_j} f_x(y) .", "\\end{equation*}", "For $x \\in R_i$, we have $m_{i,j} \\leq m_j(x)$, and therefore,", "\\begin{equation*}", "\\sum_{j=1}^K", "m_{i,j} \\, V(R'_j)", "\\leq \\sum_{j=1}^K m_j(x) \\, V(R'_j) = L(P',f_x) \\leq", "\\underline{\\int_S} f_x = g(x) .", "\\end{equation*}", "The inequality holds for all $x \\in R_i$, and so", "\\begin{equation*}", "\\sum_{j=1}^K", "m_{i,j} \\, V(R'_j)", "\\leq \\inf_{x \\in R_i} g(x) .", "\\end{equation*}", "We obtain", "\\begin{equation*}", "L\\bigl((P,P'),f\\bigr)", "\\leq", "\\sum_{j=1}^N", "\\left(", "\\inf_{x \\in R_j} g(x)", "\\right) V(R_j) = L(P,g) .", "\\end{equation*}", "Similarly, $U\\bigl((P,P'),f) \\geq U(P,h)$, and the proof of this inequality is", "left as an exercise.", "Putting the two inequalities together with the fact that $g(x) \\leq h(x)$ for all $x$,", "\\begin{equation*}", "L\\bigl((P,P'),f\\bigr)", "\\leq", "L(P,g) \\leq", "U(P,g) \\leq", "U(P,h) \\leq", "U\\bigl((P,P'),f\\bigr) .", "\\end{equation*}", "Since $f$ is integrable, it must be that $g$ is integrable as", "\\begin{equation*}", "U(P,g) - L(P,g)", "\\leq", "U\\bigl((P,P'),f\\bigr) -", "L\\bigl((P,P'),f\\bigr) ,", "\\end{equation*}", "and we can make the right-hand side arbitrarily small.", "As for any partition we have", "$L\\bigl((P,P'),f\\bigr) \\leq L(P,g) \\leq U\\bigl((P,P'),f\\bigr)$, we have", "$\\int_R g = \\int_{R \\times S} f$.", "Likewise,", "\\begin{equation*}", "L\\bigl((P,P'),f\\bigr)", "\\leq", "L(P,g) \\leq", "L(P,h) \\leq", "U(P,h) \\leq", "U\\bigl((P,P'),f\\bigr) ,", "\\end{equation*}", "and hence", "\\begin{equation*}", "U(P,h) - L(P,h)", "\\leq", "U\\bigl((P,P'),f\\bigr) -", "L\\bigl((P,P'),f\\bigr) .", "\\end{equation*}", "As $f$ is integrable, so is $h$.", "Moreover, $L\\bigl((P,P'),f\\bigr) \\leq L(P,h) \\leq U\\bigl((P,P'),f\\bigr)$", "implies", "$\\int_R h = \\int_{R \\times S} f$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 17, "type": "theorem", "label": "Lebl-contfunc:mv:fubinivB", "categories": [ "integration", "domains", "named theorem" ], "title": "Fubini's Theorem", "contents": [ "\\index{Fubini's theorem} \\pagebreak[2] Let$R \\times S \\subset \\R^n \\times \\R^m$be a closed rectangle and$f \\colon R \\times S \\to \\R$be integrable. The functions$g \\colon S \\to \\R$and$h \\colon S \\to \\R$defined by\\begin{equation*} g(y) \\coloneqq \\underline{\\int_R} f^y \\qquad \\text{and} \\qquad h(y) \\coloneqq \\overline{\\int_R} f^y \\end{equation*}are integrable on$S$and\\begin{equation*} \\int_S g = \\int_S h = \\int_{R \\times S} f . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 18, "type": "proposition", "label": "Lebl-contfunc:mv:prop:ballsnull", "categories": [ "characterization", "measure" ], "title": "Characterization of Measure Theory via Equivalence", "contents": [ "Let$\\delta > 0$be given. A set$S \\subset \\R^n$is of measure zero if and only if for every$\\epsilon > 0$, there exists a sequence of open balls$\\{ B_k \\}_{k=1}^\\infty$, where the radius of$B_k$is$r_k < \\delta$, and such that\\begin{equation*} S \\subset \\bigcup_{k=1}^\\infty B_k \\qquad \\text{and} \\qquad \\sum_{k=1}^\\infty r_k^n < \\epsilon. \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "If $C$ is a closed cube (rectangle with all sides equal) of side $s$,", "then $C$ is contained in a closed ball of radius", "$\\sqrt{n}\\, s$ by \\propref{prop:diameterrectangle}, and hence", "in an open ball of radius $2 \\sqrt{n}\\, s$.", "Suppose $R$ is a rectangle of positive volume.", "Let $s > 0$ be a number less than the smallest side of $R$ and", "such that $2\\sqrt{n} \\, s < \\delta$.", "If each side of $R$ is an integer multiple of $s$, then $R$ is", "contained in a union of closed cubes", "$C_1, C_2, \\ldots, C_m$ of side $s$ such that", "$\\sum_{k=1}^m V(C_k) = V(R)$. So suppose the sides", "of $R$ are not integer multiples of $s$.", "Consider a side of length $(\\ell+\\alpha) s$, for", "an integer $\\ell$ and $0 \\leq \\alpha < 1$. As $s$ is less than the smallest", "side, $\\ell \\geq 1$, and so $(\\ell+\\alpha)s \\leq 2\\ell s$.", "Increasing this side to $2\\ell s$,", "and similarly increasing every side of $R$, we obtain a new larger", "rectangle of volume at most $2^n$ times larger, whose sides are", "multiples of~$s$. See \\figureref{fig:nullrectcube}.", "Thus $R$ is contained in a union of", "closed cubes $C_1, C_2, \\ldots, C_m$ of side $s$ such that", "\\begin{equation*}", "\\sum_{k=1}^m V(C_k) \\leq 2^n V(R) .", "\\end{equation*}", "\\begin{myfigureht}", "\\subimport*{figures/}{nullrectcube.pdf_t}", "\\caption{Covering a rectangle by cubes of total size at most $2^n V(R)$.\\label{fig:nullrectcube}}", "\\end{myfigureht}", "So suppose that $S$ is a null set and", "there exist open rectangles $\\{ R_j \\}_{j=1}^\\infty$ whose union contains $S$ and such that", "\\eqref{mv:eq:nullR} is true. Choose closed", "cubes $\\{ C_k \\}_{k=1}^\\infty$ with $C_k$ of side $s_k$ as above that cover", "all the rectangles $\\{ R_j \\}_{j=1}^\\infty$ and so that", "\\begin{equation*}", "\\sum_{k=1}^\\infty s_k^n =", "\\sum_{k=1}^\\infty V(C_k) \\leq", "2^n \\sum_{j=1}^\\infty V(R_j)", "< 2^n \\epsilon.", "\\end{equation*}", "Covering each $C_k$ with a ball $B_k$ of radius $r_k = 2\\sqrt{n} \\, s_k < \\delta$,", "we obtain", "\\begin{equation*}", "\\sum_{k=1}^\\infty r_k^n", "=", "\\sum_{k=1}^\\infty {(2\\sqrt{n})}^n s_k^n", "<", "{(4\\sqrt{n})}^n \\epsilon .", "\\end{equation*}", "As $S \\subset\\bigcup_{j} R_j \\subset \\bigcup_{k} C_k \\subset \\bigcup_{k}", "B_k$ and ${(4\\sqrt{n})}^n \\epsilon$ can be arbitrarily small,", "the forward direction follows.", "For the other direction, suppose $S$ is covered by balls $B_j$", "of radii $r_j$, such that $\\sum_{j=1}^\\infty r_j^n < \\epsilon$,", "as in the statement of the proposition.", "Each $B_j$ is contained in an open cube $R_j$ of side $2r_j$.", "So $V(R_j) = {(2 r_j)}^n = 2^n r_j^n$. Therefore,", "\\begin{equation*}", "S \\subset \\bigcup_{j=1}^\\infty R_j \\qquad \\text{and} \\qquad", "\\sum_{j=1}^\\infty V(R_j)", "\\leq", "\\sum_{j=1}^\\infty 2^n r_j^n < 2^n \\epsilon. \\qedhere", "\\end{equation*}" ], "refs": [ "mv:eq:nullR", "prop:diameterrectangle" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 19, "type": "example", "label": "Lebl-contfunc:19", "categories": [ "domains", "measure", "example" ], "title": "The set$\\Q^n \\subset \\R^n$of points with rational coordinates is of measure zero", "contents": [ "The set$\\Q^n \\subset \\R^n$of points with rational coordinates is of measure zero.", "Proof: The set$\\Q^n$is countable, so write it as a sequence$q_1,q_2,\\ldots$. For each$q_j$, find an open rectangle$R_j$with$q_j \\in R_j$and$V(R_j) < \\epsilon 2^{-j}$. Then", "\\begin{equation*}\n\\Q^n \\subset \\bigcup_{j=1}^\\infty R_j \\qquad \\text{and} \\qquad\n\\sum_{j=1}^\\infty V(R_j) <\n\\sum_{j=1}^\\infty \\epsilon 2^{-j} = \\epsilon .\n\\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "The set$\\Q^n$is countable, so write it as a sequence$q_1,q_2,\\ldots$. For each$q_j$, find an open rectangle$R_j$with$q_j \\in R_j$and$V(R_j) < \\epsilon 2^{-j}$. Then \\begin{equation*}", "\\Q^n \\subset \\bigcup_{j=1}^\\infty R_j \\qquad \\text{and} \\qquad", "\\sum_{j=1}^\\infty V(R_j) <", "\\sum_{j=1}^\\infty \\epsilon 2^{-j} = \\epsilon .", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 20, "type": "proposition", "label": "Lebl-contfunc:20", "categories": [ "measure" ], "title": "A countable union of measure zero sets is of measure zero", "contents": [ "A countable union of measure zero sets is of measure zero." ], "refs": [], "proofs": [ { "contents": [ "Suppose", "\\begin{equation*}", "S = \\bigcup_{j=1}^\\infty S_j ,", "\\end{equation*}", "where $S_j$ are all measure zero sets. Let $\\epsilon > 0$ be given.", "For each $j$,", "there exists a sequence of open rectangles $\\{ R_{j,k} \\}_{k=1}^\\infty$", "such that", "\\begin{equation*}", "S_j \\subset \\bigcup_{k=1}^\\infty R_{j,k}", "\\qquad \\text{and} \\qquad", "\\sum_{k=1}^\\infty V(R_{j,k}) < 2^{-j} \\epsilon .", "\\end{equation*}", "Then", "\\begin{equation*}", "S \\subset \\bigcup_{j=1}^\\infty \\bigcup_{k=1}^\\infty R_{j,k} .", "\\end{equation*}", "All $V(R_{j,k})$ are nonnegative, so the sum over all $j$ and $k$", "can be done by summing first over the $k$ and then over the $j$,", "see", "\\volIref{\\exerciseref*{vI-exercise:tonellifubiniforsums} in volume I}{\\exerciseref{exercise:tonellifubiniforsums}}.", "In particular, as", "\\begin{equation*}", "\\sum_{j=1}^\\infty \\sum_{k=1}^\\infty V(R_{j,k}) <", "\\sum_{j=1}^\\infty 2^{-j} \\epsilon = \\epsilon . \\qedhere", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 21, "type": "example", "label": "Lebl-contfunc:mv:example:planenull", "categories": [ "domains", "measure", "example" ], "title": "Suppose$n \\in \\N$,$k=1,2,\\ldots,n$, and$c \\in \\R$", "contents": [ "Suppose$n \\in \\N$,$k=1,2,\\ldots,n$, and$c \\in \\R$. Then$P \\coloneqq \\{ x \\in \\R^n : x_k = c \\}$is of measure zero. Note that if$n \\geq 2$, then$P$is uncountable.", "Proof: First fix$s \\in \\N$and consider", "\\begin{equation*}\nP_s \\coloneqq \\bigl\\{ x \\in \\R^n : x_k = c \\text{ and } \\sabs{x_j} \\leq s\n\\text{ for all } j\\not=k \\bigr\\} .\n\\end{equation*}", "Given any$\\epsilon > 0$define the open rectangle", "\\begin{equation*}\nR \\coloneqq \\bigl\\{ x \\in \\R^n : c-\\epsilon < x_k < c+\\epsilon \\text{ and } \\sabs{x_j} < s+1\n\\text{ for all } j\\not=k \\bigr\\} .\n\\end{equation*}", "Clearly,$P_s \\subset R$. Furthermore,", "\\begin{equation*}\nV(R) = 2\\epsilon {\\bigl(2(s+1)\\bigr)}^{n-1} .\n\\end{equation*}", "As$s$is fixed,$V(R)$can be arbitrarily small by picking$\\epsilon$small enough. So$P_s$is of measure zero.", "Next", "\\begin{equation*}\nP = \\bigcup_{j=1}^\\infty P_j\n\\end{equation*}", "and a countable union of measure zero sets is of measure zero." ], "refs": [], "proofs": [ { "contents": [ "First fix$s \\in \\N$and consider \\begin{equation*}", "P_s \\coloneqq \\bigl\\{ x \\in \\R^n : x_k = c \\text{ and } \\sabs{x_j} \\leq s", "\\text{ for all } j\\not=k \\bigr\\} .", "\\end{equation*} Given any$\\epsilon > 0$define the open rectangle \\begin{equation*}", "R \\coloneqq \\bigl\\{ x \\in \\R^n : c-\\epsilon < x_k < c+\\epsilon \\text{ and } \\sabs{x_j} < s+1", "\\text{ for all } j\\not=k \\bigr\\} .", "\\end{equation*} Clearly,$P_s \\subset R$. Furthermore, \\begin{equation*}", "V(R) = 2\\epsilon {\\bigl(2(s+1)\\bigr)}^{n-1} .", "\\end{equation*} As$s$is fixed,$V(R)$can be arbitrarily small by picking$\\epsilon$small enough. So$P_s$is of measure zero. Next \\begin{equation*}", "P = \\bigcup_{j=1}^\\infty P_j", "\\end{equation*} and a countable union of measure zero sets is of measure zero." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 22, "type": "example", "label": "Lebl-contfunc:22", "categories": [ "domains", "example" ], "title": "If$a < b$, then$m^*\\bigl([a,b]\\bigr) = b-a$", "contents": [ "If$a < b$, then$m^*\\bigl([a,b]\\bigr) = b-a$.", "Proof: In$\\R$, open rectangles are open intervals. Since$[a,b] \\subset (a-\\epsilon,b+\\epsilon)$for all$\\epsilon > 0$, we have$m^*\\bigl([a,b]\\bigr) \\leq b-a$.", "The other inequality is harder. Suppose$\\bigl\\{ (a_j,b_j) \\bigr\\}_{j=1}^\\infty$are open intervals such that", "\\begin{equation*}\n[a,b] \\subset \\bigcup_{j=1}^\\infty (a_j,b_j) .\n\\end{equation*}", "We wish to bound$\\sum_{j=1}^\\infty (b_j-a_j)$from below. Since$[a,b]$is compact, finitely many of the open intervals still cover$[a,b]$. As throwing out some of the intervals only makes the sum smaller, we only need to consider the finite number of intervals covering$[a,b]$. If$(a_i,b_i) \\subset (a_j,b_j)$, then we throw out$(a_i,b_i)$as well. The intervals that are left have distinct left endpoints, and whenever$a_j < a_i < b_j$, then$b_j < b_i$. Therefore,$[a,b] \\subset \\bigcup_{j=1}^k (a_j,b_j)$for some$k$, and we assume that the intervals are sorted such that$a_1 < a_2 < \\cdots < a_k$. As$(a_2,b_2)$is not contained in$(a_1,b_1)$, since$a_j > a_2$for all$j > 2$, and since the intervals must contain every point in$[a,b]$, we find that$a_2 < b_1$, or in other words$a_1 < a_2 < b_1 < b_2$. Similarly$a_j < a_{j+1} < b_j < b_{j+1}$for all$j$. Furthermore,$a_1 < a$and$b_k > b$. See \\figureref{fig:measureinterval} for a sample configuration. As$b_j-a_j > a_{j+1}-a_j$, we obtain", "\\begin{equation*}\n\\sum_{j=1}^k (b_j-a_j)\n\\geq\n\\sum_{j=1}^{k-1} (a_{j+1}-a_j)\n+\n(b_k-a_k)\n=\nb_k-a_1 > b-a .\n\\end{equation*}", "So$m^*\\bigl([a,b]\\bigr) \\geq b-a$. \\begin{myfigureht} \\subimport*{figures/}{measureinterval.pdf_t} \\caption{Open intervals covering$[a,b]$which satisfy$a_j < a_{j+1} < b_j < b_{j+1}$for all$j$.\\label{fig:measureinterval}} \\end{myfigureht}" ], "refs": [], "proofs": [ { "contents": [ "In$\\R$, open rectangles are open intervals. Since$[a,b] \\subset (a-\\epsilon,b+\\epsilon)$for all$\\epsilon > 0$, we have$m^*\\bigl([a,b]\\bigr) \\leq b-a$. The other inequality is harder. Suppose$\\bigl\\{ (a_j,b_j) \\bigr\\}_{j=1}^\\infty$are open intervals such that \\begin{equation*}", "[a,b] \\subset \\bigcup_{j=1}^\\infty (a_j,b_j) .", "\\end{equation*} We wish to bound$\\sum_{j=1}^\\infty (b_j-a_j)$from below. Since$[a,b]$is compact, finitely many of the open intervals still cover$[a,b]$. As throwing out some of the intervals only makes the sum smaller, we only need to consider the finite number of intervals covering$[a,b]$. If$(a_i,b_i) \\subset (a_j,b_j)$, then we throw out$(a_i,b_i)$as well. The intervals that are left have distinct left endpoints, and whenever$a_j < a_i < b_j$, then$b_j < b_i$. Therefore,$[a,b] \\subset \\bigcup_{j=1}^k (a_j,b_j)$for some$k$, and we assume that the intervals are sorted such that$a_1 < a_2 < \\cdots < a_k$. As$(a_2,b_2)$is not contained in$(a_1,b_1)$, since$a_j > a_2$for all$j > 2$, and since the intervals must contain every point in$[a,b]$, we find that$a_2 < b_1$, or in other words$a_1 < a_2 < b_1 < b_2$. Similarly$a_j < a_{j+1} < b_j < b_{j+1}$for all$j$. Furthermore,$a_1 < a$and$b_k > b$. See \\figureref{fig:measureinterval} for a sample configuration. As$b_j-a_j > a_{j+1}-a_j$, we obtain \\begin{equation*}", "\\sum_{j=1}^k (b_j-a_j)", "\\geq", "\\sum_{j=1}^{k-1} (a_{j+1}-a_j)", "+", "(b_k-a_k)", "=", "b_k-a_1 > b-a .", "\\end{equation*} So$m^*\\bigl([a,b]\\bigr) \\geq b-a$. \\begin{myfigureht} \\subimport*{figures/}{measureinterval.pdf_t} \\caption{Open intervals covering$[a,b]$which satisfy$a_j < a_{j+1} < b_j < b_{j+1}$for all$j$.\\label{fig:measureinterval}} \\end{myfigureht}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 23, "type": "proposition", "label": "Lebl-contfunc:mv:prop:compactnull", "categories": [ "domains", "measure" ], "title": "Suppose$E \\subset \\R^n$is a compact set of measure zero", "contents": [ "Suppose$E \\subset \\R^n$is a compact set of measure zero. Then for every$\\epsilon > 0$, there exist finitely many open rectangles$R_1,R_2,\\ldots,R_k$such that\\begin{equation*} E \\subset R_1 \\cup R_2 \\cup \\cdots \\cup R_k \\qquad \\text{and} \\qquad \\sum_{j=1}^k V(R_j) < \\epsilon. \\end{equation*}Moreover, for every$\\epsilon > 0$and every$\\delta > 0$, there exist finitely many open balls$B_1,B_2,\\ldots,B_{\\ell}$of radii$r_1,r_2,\\ldots,r_{\\ell} < \\delta$such that\\begin{equation*} E \\subset B_1 \\cup B_2 \\cup \\cdots \\cup B_{\\ell} \\qquad \\text{and} \\qquad \\sum_{j=1}^{\\ell} r_j^n < \\epsilon. \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "As $E$ is of measure zero,", "there exists a sequence of open rectangles $\\{ R_j \\}_{j=1}^\\infty$ such that", "\\begin{equation*}", "E \\subset \\bigcup_{j=1}^\\infty R_j", "\\qquad \\text{and} \\qquad", "\\sum_{j=1}^\\infty V(R_j) < \\epsilon.", "\\end{equation*}", "By compactness, there are finitely", "many of these rectangles that still contain $E$. That is, there is some $k$ such", "that", "$E \\subset R_1 \\cup R_2 \\cup \\cdots \\cup R_k$. Hence", "\\begin{equation*}", "\\sum_{j=1}^k V(R_j) \\leq", "\\sum_{j=1}^\\infty V(R_j) < \\epsilon.", "\\end{equation*}", "The proof that we can choose balls instead of rectangles is left as an", "exercise." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 24, "type": "example", "label": "Lebl-contfunc:example:cantor", "categories": [ "measure", "example" ], "title": "So that the reader is not under the impression that there are only few measure zero sets and that...", "contents": [ "So that the reader is not under the impression that there are only few measure zero sets and that these sets are uncomplicated, here is an uncountable, compact, measure zero subset of$[0,1]$, which contains no intervals. Any$x \\in [0,1]$can be expanded in ternary:", "\\begin{equation*}\nx = \\sum_{n=1}^\\infty d_n 3^{-n} ,\n\\qquad \\text{where } d_n=0, 1, \\text{ or } 2.\n\\end{equation*}", "See \\volIref{\\sectionref*{vI-sec:decimals} in volume I\\@, in particular \\exerciseref*{vI-exercise:decimalpropbaseb}}{\\sectionref{sec:decimals}, in particular \\exerciseref{exercise:decimalpropbaseb}}. Define the \\emph{\\myindex{Cantor set}}$C$as", "\\begin{equation*}\nC \\coloneqq \\Bigl\\{ x \\in [0,1] : x = \\sum_{n=1}^\\infty d_n 3^{-n},\n\\text{ where } d_n = 0 \\text{ or } d_n = 2 \\text{ for all } n \\Bigr\\} .\n\\end{equation*}", "That is,$x$is in$C$if it has a ternary expansion in only$0$s and$2$s. If$x$has two expansions, as long as one of them does not have any$1$s, then$x$is in$C$. Define$C_0 \\coloneqq [0,1]$and", "\\begin{equation*}\nC_k \\coloneqq \\Bigl\\{ x \\in [0,1] : x = \\sum_{n=1}^\\infty d_n 3^{-n},\n\\text{ where } d_n = 0 \\text{ or } d_n = 2 \\text{ for all } n=1,2,\\ldots,k \\Bigr\\} .\n\\end{equation*}", "Clearly,", "\\begin{equation*}\nC = \\bigcap_{k=1}^\\infty C_k .\n\\end{equation*}", "See \\figureref{fig:cantor}.", "We leave as an exercise to prove: \\begin{enumerate}[(i)] \\item Each$C_k$is a finite union of closed intervals. It is obtained by taking$C_{k-1}$, and from each closed interval removing the \\myquote{middle third.} \\item Each$C_k$is closed, and so$C$is closed. \\item$m^*(C_k) =1 - \\sum_{n=1}^k \\frac{2^n}{3^{n+1}}$. \\item Hence,$m^*(C) = 0$. \\item The set$C$is in one-to-one correspondence with$[0,1]$, in other words,$C$is uncountable. \\end{enumerate} \\begin{myfigureht} \\subimport*{figures/}{cantorfig.pdf_t} \\caption{Cantor set construction.\\label{fig:cantor}} \\end{myfigureht}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 25, "type": "lemma", "label": "Lebl-contfunc:lemma:ballmapder", "categories": [ "auxiliary result" ], "title": "Continuity of Multivariable Integration", "contents": [ "Suppose$U \\subset \\R^n$is an open set,$B \\subset U$is an open (resp.\\ closed) ball of radius at most$r$,$f \\colon U \\to \\R^n$is continuously differentiable, and suppose$\\snorm{f'(x)} \\leq M$for all$x \\in B$. Then$f(B) \\subset B'$, where$B'$is an open (resp.\\ closed) ball of radius at most$Mr$." ], "refs": [], "proofs": [ { "contents": [ "Suppose $B$ is open.", "As the ball $B$ is convex,", "\\propref{mv:prop:convexlip} says that $\\snorm{f(x)-f(y)} \\leq M \\snorm{x-y}$ for", "all $x,y \\in B$.", "So if $\\snorm{x-y} < r$, then $\\snorm{f(x)-f(y)} < Mr$.", "In other words, if", "$B=B(y,r)$, then $f(B) \\subset B\\bigl(f(y),M r \\bigr)$.", "If $B$ is closed, then $\\overline{B(y,r)} = B$.", "As $f$ is continuous,", "$f(B) = f\\bigl(\\overline{B(y,r)}\\bigr) \\subset", "\\overline{f\\bigl(B(y,r)\\bigr)} \\subset \\overline{B\\bigl(f(y),M r", "\\bigr)}$, as $f(\\widebar{A}) \\subset \\overline{f(A)}$ for any set~$A$.", "% and any continuous $f$ (limit of a sequence in $A$ goes to a limit of a sequence in $f(A)$)." ], "refs": [ "mv:prop:convexlip" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 26, "type": "proposition", "label": "Lebl-contfunc:prop:imagenull", "categories": [ "measure" ], "title": "Continuity of Measure Theory", "contents": [ "Suppose$U \\subset \\R^n$is open and$f \\colon U \\to \\R^n$is continuously differentiable. If$E \\subset U$is a measure zero set, then$f(E)$is measure zero." ], "refs": [], "proofs": [ { "contents": [ "We prove the proposition for a compact $E$ and leave the general case as", "an exercise.", "Suppose $E$ is compact and of measure zero.", "First, we will replace $U$ by a smaller open set to make $\\snorm{f'(x)}$ bounded.", "At each point $x \\in", "E$ pick an open ball $B(x,r_x)$ such that the closed ball $C(x,r_x) \\subset", "U$. By compactness, we only need to take finitely", "many points $x_1,x_2,\\ldots,x_q$ to cover $E$ with the balls $B(x_j,r_{x_j})$. Define", "\\begin{equation*}", "U' \\coloneqq \\bigcup_{j=1}^q B(x_j,r_{x_j}), \\qquad", "K \\coloneqq \\bigcup_{j=1}^q C(x_j,r_{x_j}).", "\\end{equation*}", "We have $E \\subset U' \\subset K \\subset U$. The set $K$, being a finite", "union of compact sets, is compact.", "The function that takes $x$ to $\\snorm{f'(x)}$ is continuous, and therefore", "there exists an $M > 0$ such that $\\snorm{f'(x)} \\leq M$ for all $x \\in K$.", "So without loss of generality, we may replace $U$ by $U'$ and from now on", "suppose that $\\snorm{f'(x)} \\leq M$ for all $x \\in U$.", "At each $x \\in E$, take the maximum radius $\\delta_x$", "such that $B(x,\\delta_x) \\subset U$ (we may assume $U \\not= \\R^n$).", "Let $\\delta \\coloneqq \\inf_{x\\in E} \\delta_x$.", "We want to show that $\\delta > 0$.", "Take a sequence $\\{ x_j \\}_{j=1}^\\infty$ in $E$ so that $\\delta_{x_j} \\to \\delta$.", "As $E$ is compact, we can pick the sequence to be convergent to some $y \\in", "E$. Once $\\snorm{x_j-y} < \\frac{\\delta_y}{2}$, then", "$\\delta_{x_j} > \\frac{\\delta_y}{2}$ by the triangle inequality.", "Thus, $\\delta > 0$.", "Given $\\epsilon > 0$, there exist balls $B_1,B_2,\\ldots,B_k$ of radii", "$r_1,r_2,\\ldots,r_k < \\nicefrac{\\delta}{2}$ such that", "\\begin{equation*}", "E \\subset B_1 \\cup B_2 \\cup \\cdots \\cup B_k", "\\qquad \\text{and} \\qquad", "\\sum_{j=1}^k r_j^n < \\epsilon.", "\\end{equation*}", "We can assume that each ball contains a point of $E$ and so", "the balls are contained in $U$.", "Suppose $B_1', B_2', \\ldots, B_k'$ are the balls of radius", "$Mr_1, Mr_2, \\ldots, Mr_k$ from", "\\lemmaref{lemma:ballmapder}, such that $f(B_j) \\subset B_j'$ for all $j$.", "Then,", "\\begin{equation*}", "f(E) \\subset f(B_1) \\cup f(B_2) \\cup \\cdots \\cup f(B_k)", "\\subset B_1' \\cup B_2' \\cup \\cdots \\cup B_k'", "\\qquad \\text{and} \\qquad", "\\sum_{j=1}^k {(Mr_j)}^n", "< M^n \\epsilon. \\qedhere", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 27, "type": "proposition", "label": "Lebl-contfunc:27", "categories": [ "characterization" ], "title": "Characterization of Multivariable Integration via Equivalence", "contents": [ "\\pagebreak[2] A function$f \\colon D \\to \\R$is continuous at$x \\in D$if and only if$o(f,x) = 0$." ], "refs": [], "proofs": [ { "contents": [ "First suppose that $f$ is continuous at $x \\in D$. Given $\\epsilon > 0$,", "there exists a $\\delta > 0$ such that for $y \\in B_D(x,\\delta)$,", "we have $\\sabs{f(x)-f(y)} < \\epsilon$. Therefore, if $y_1,y_2 \\in", "B_D(x,\\delta)$, then", "\\begin{equation*}", "f(y_1)-f(y_2) =", "\\bigl(f(y_1)-f(x)\\bigr)-\\bigl(f(y_2)-f(x)\\bigr) < \\epsilon + \\epsilon = 2 \\epsilon .", "\\end{equation*}", "Take the supremum over $y_1$ and $y_2$ to find", "\\begin{equation*}", "o(f,x,\\delta) =", "\\sup_{y_1,y_2 \\in B_D(x,\\delta)} \\bigl(f(y_1)-f(y_2)\\bigr)", "\\leq", "2 \\epsilon .", "\\end{equation*}", "As $o(x,f) \\leq o(f,x,\\delta) \\leq 2\\epsilon$, and $\\epsilon > 0$ was arbitrary,", "$o(x,f) = 0$.", "On the other hand, suppose $o(x,f) = 0$. Given $\\epsilon > 0$,", "find a $\\delta > 0$ such that $o(f,x,\\delta) < \\epsilon$. If", "$y \\in B_D(x,\\delta)$, then", "\\begin{equation*}", "\\sabs{f(x)-f(y)}", "\\leq", "\\sup_{y_1,y_2 \\in B_D(x,\\delta)} \\bigl(f(y_1)-f(y_2)\\bigr)", "=", "o(f,x,\\delta) < \\epsilon. \\qedhere", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 28, "type": "proposition", "label": "Lebl-contfunc:prop:seclosed", "categories": [], "title": "Let$D \\subset \\R^n$be closed,$f \\colon D \\to \\R$, and$\\epsilon > 0$", "contents": [ "Let$D \\subset \\R^n$be closed,$f \\colon D \\to \\R$, and$\\epsilon > 0$. The set$\\bigl\\{ x \\in D : o(f,x) \\geq \\epsilon \\bigr\\}$is closed." ], "refs": [], "proofs": [ { "contents": [ "Equivalently, we want to show that", "$G \\coloneqq \\bigl\\{ x \\in D : o(f,x) < \\epsilon \\bigr\\}$ is open in the subspace topology.", "Consider $x \\in G$.", "As $\\inf_{\\delta > 0} o(f,x,\\delta) < \\epsilon$, find a $\\delta > 0$ such", "that", "\\begin{equation*}", "o(f,x,\\delta) < \\epsilon .", "\\end{equation*}", "Take any $\\xi \\in B_D(x,\\nicefrac{\\delta}{2})$. Notice that", "$B_D(\\xi,\\nicefrac{\\delta}{2}) \\subset B_D(x,\\delta)$. Therefore,", "\\begin{equation*}", "o(f,\\xi,\\nicefrac{\\delta}{2}) =", "\\sup_{y_1,y_2 \\in B_D(\\xi,\\nicefrac{\\delta}{2})} \\bigl(f(y_1)-f(y_2)\\bigr)", "\\leq", "\\sup_{y_1,y_2 \\in B_D(x,\\delta)} \\bigl(f(y_1)-f(y_2)\\bigr) = o(f,x,\\delta) <", "\\epsilon .", "\\end{equation*}", "So $o(f,\\xi) < \\epsilon$ as well. As this is true for all $\\xi \\in", "B_D(x,\\nicefrac{\\delta}{2})$, we get that $G$ is open in the subspace", "topology, and $D \\setminus G$ is closed as claimed." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 29, "type": "theorem", "label": "Lebl-contfunc:29", "categories": [ "domains", "integration", "characterization", "measure" ], "title": "Characterization of Riemann Integration via Equivalence", "contents": [ "% \\index{Riemann--Lebesgue theorem}% \\index{Lebesgue--Vitali theorem}% Let$R \\subset \\R^n$be a closed rectangle and$f \\colon R \\to \\R$bounded. Then$f$is Riemann integrable if and only if the set of discontinuities of$f$is of measure zero." ], "refs": [], "proofs": [ { "contents": [ "Let $S \\subset R$ be the set of discontinuities of $f$, that is,", "$S = \\bigl\\{ x \\in R : o(f,x) > 0 \\bigr\\}$.", "Suppose $S$ is a measure zero set: $m^*(S) = 0$.", "The trick to proving that $f$ is integrable is to isolate the bad set", "into a small set of subrectangles of a partition. A partition has", "finitely many subrectangles, so we need", "compactness. If $S$ were closed, then it would be compact and we could cover", "it by finitely many small rectangles. Unfortunately, $S$", "itself is not closed in general, but the following set is.", "Given $\\epsilon > 0$, define", "\\begin{equation*}", "S_\\epsilon \\coloneqq \\bigl\\{ x \\in R : o(f,x) \\geq \\epsilon \\bigr\\} .", "\\end{equation*}", "By \\propref{prop:seclosed}, $S_\\epsilon$ is closed,", "and as it is also a subset of the bounded $R$,", "$S_\\epsilon$ is compact. Moreover,", "$S_\\epsilon \\subset S$ and $S$ is of measure zero,", "so $S_\\epsilon$ is of measure zero.", "Via \\propref{mv:prop:compactnull}, finitely many open rectangles", "$O_1,O_2,\\ldots,O_k$ cover $S_\\epsilon$ and", "$\\sum_{j=1}^\\infty V(O_j) < \\epsilon$.", "The set $T \\coloneqq R \\setminus ( O_1 \\cup \\cdots \\cup O_k )$ is closed, bounded,", "and so compact. As $o(f,x) < \\epsilon$ for all $x \\in T$,", "for each $x \\in T$, there is a $\\delta > 0$ such that", "$o(f,x,\\delta) < \\epsilon$, so there exists a small closed rectangle", "$T_x \\subset B(x,\\delta)$ with $x$ in the interior of $T_x$, such that", "\\begin{equation*}", "\\sup_{y\\in T_x} f(y) - \\inf_{y\\in T_x} f(y) < \\epsilon.", "\\end{equation*}", "The interiors of the rectangles $T_x$ cover $T$. As $T$ is compact,", "finitely many such rectangles $T_1, T_2, \\ldots, T_m$", "cover $T$.", "Construct a partition $P$ out of the", "endpoints of the rectangles $T_1,T_2,\\ldots,T_m$", "and $O_1,O_2,\\ldots,O_k$ (ignoring those that", "are outside the endpoints of $R$). The", "subrectangles $R_1,R_2,\\ldots,R_p$ of $P$ are", "such that every $R_j$ is contained in $T_\\ell$ for some~$\\ell$", "or the closure of $O_\\ell$ for some~$\\ell$. Order", "the rectangles so that $R_1,R_2,\\ldots,R_q$ are those", "that are contained in some $T_\\ell$, and $R_{q+1},R_{q+2},\\ldots,R_{p}$", "are the rest.", "See \\figureref{fig:nullsetintegrate}.", "So", "\\begin{equation*}", "\\sum_{j=1}^q V(R_j) \\leq V(R)", "\\qquad \\text{and} \\qquad", "\\sum_{j=q+1}^p V(R_j)", "\\leq", "\\sum_{\\ell=1}^k V(O_\\ell)", "< \\epsilon .", "\\end{equation*}", "The second estimate holds because the $R_j$ that are subsets of $\\widebar{O}_\\ell$", "give a partition of $\\widebar{O}_\\ell$ and hence their volumes sum to", "$V(O_\\ell)$.", "Let $m_j$ and $M_j$ be the inf and sup of $f$ over $R_j$ as usual.", "If $R_j \\subset T_\\ell$ for some $\\ell$, then $M_j-m_j < \\epsilon$.", "Let $B \\in \\R$ be such that", "$\\sabs{f(x)} \\leq B$ for all $x \\in R$, so $M_j-m_j \\leq 2B$ over all", "rectangles. Then", "\\begin{equation*}", "\\begin{split}", "U(P,f)-L(P,f)", "& =", "\\sum_{j=1}^p (M_j-m_j) V(R_j)", "\\\\", "& =", "\\left(", "\\sum_{j=1}^q (M_j-m_j) V(R_j)", "\\right)", "+", "\\left(", "\\sum_{j=q+1}^p (M_j-m_j) V(R_j)", "\\right)", "\\\\", "& <", "\\left(", "\\sum_{j=1}^q \\epsilon\\, V(R_j)", "\\right)", "+", "\\left(", "\\sum_{j=q+1}^p 2 B\\, V(R_j)", "\\right)", "\\\\", "& <", "\\epsilon\\, V(R)", "+", "2B \\epsilon = \\epsilon \\bigl(V(R)+2B\\bigr) .", "\\end{split}", "\\avoidbreak", "\\end{equation*}", "We can make the right-hand side as small as we want,", "and hence $f$ is integrable.", "\\begin{myfigureht}", "\\subimport*{figures/}{nullsetintegrate.pdf_t}", "\\caption{A rectangle $R$ with $S_\\epsilon$ marked as thick black line, and the", "$O_\\ell$ as shaded rectangles. The partition is given by the dotted", "lines. Note how the $R_j$ partition the $O_\\ell$.\\label{fig:nullsetintegrate}}", "\\end{myfigureht}", "\\medskip", "\\pagebreak[2]", "For the other direction, suppose $f$ is Riemann integrable", "on $R$.", "Let $S$ be the set of discontinuities of $f$ again. Consider", "the sequence of sets", "\\begin{equation*}", "S_{1/k} = \\bigl\\{ x \\in R : o(f,x) \\geq \\nicefrac{1}{k} \\bigr\\}.", "\\end{equation*}", "Fix a $k \\in \\N$.", "Given an $\\epsilon > 0$, find a partition $P$ with subrectangles", "$R_1,R_2,\\ldots,R_p$ such that", "\\begin{equation*}", "U(P,f)-L(P,f) =", "\\sum_{j=1}^p (M_j-m_j) V(R_j)", "< \\epsilon .", "\\end{equation*}", "Suppose $R_1,R_2,\\ldots,R_p$ are ordered so that", "the interiors of $R_1,R_2,\\ldots,R_{q}$ intersect $S_{1/k}$,", "while the interiors of $R_{q+1},R_{q+2},\\ldots,R_p$", "are disjoint from $S_{1/k}$.", "Let $R_j^\\circ$ denote the interior of $R_j$.", "Suppose $j \\leq q$ and consider", "$x \\in R_j^\\circ \\cap S_{1/k}$.", "Let $\\delta > 0$ be small enough so that $B(x,\\delta) \\subset R_j$.", "As $x \\in S_{1/k}$, we get $o(f,x,\\delta) \\geq o(f,x) \\geq \\nicefrac{1}{k}$,", "which, along with $B(x,\\delta) \\subset R_j$, implies", "$M_j-m_j \\geq \\nicefrac{1}{k}$.", "Then", "\\begin{equation*}", "\\epsilon >", "\\sum_{j=1}^p (M_j-m_j) V(R_j)", "\\geq", "\\sum_{j=1}^q (M_j-m_j) V(R_j)", "\\geq", "\\frac{1}{k}", "\\sum_{j=1}^q V(R_j) .", "\\end{equation*}", "In other words,", "$\\sum_{j=1}^q V(R_j) < k \\epsilon$.", "Let $G$ be the set of all boundaries of all the subrectangles", "of $P$. The set $G$ is of measure zero (it can be covered by", "finitely many sets from \\exampleref{mv:example:planenull}).", "We find", "\\begin{equation*}", "S_{1/k} \\subset R_1^\\circ \\cup R_2^\\circ \\cup \\cdots \\cup R_q^\\circ \\cup G .", "\\end{equation*}", "As $G$ can also be covered by open rectangles arbitrarily small volume,", "$S_{1/k}$ must be of measure zero. As", "\\begin{equation*}", "S = \\bigcup_{k=1}^\\infty S_{1/k}", "\\end{equation*}", "and a countable union of measure zero sets is of measure zero,", "$S$ is of measure zero." ], "refs": [ "mv:prop:compactnull", "prop:seclosed" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 30, "type": "corollary", "label": "Lebl-contfunc:cor:closednessofriemannintegrable", "categories": [ "consequence", "integration", "domains" ], "title": "Continuity of Riemann Integration", "contents": [ "\\pagebreak[2] Let$R \\subset \\R^n$be a closed rectangle. Let$\\sR(R)$be the set of Riemann integrable functions on$R$. Then \\begin{enumerate}[(i)] \\item$\\sR(R)$is a real algebra: If$f,g \\in \\sR(R)$and$a \\in \\R$, then$af \\in \\sR(R)$,$f+g \\in \\sR(R)$and$fg \\in \\sR(R)$. \\item If$f,g \\in \\sR(R)$and\\begin{equation*} \\varphi(x) \\coloneqq \\max \\bigl\\{ f(x) , g(x) \\bigr\\} , \\qquad \\psi(x) \\coloneqq \\min \\bigl\\{ f(x) , g(x) \\bigr\\} , \\end{equation*}then$\\varphi, \\psi \\in \\sR(R)$. \\item If$f \\in \\sR(R)$, then$\\sabs{f} \\in \\sR(R)$, where$\\sabs{f}(x) \\coloneqq \\sabs{f(x)}$. \\item If$R' \\subset \\R^n$is another closed rectangle,$U \\subset \\R^n$and$U' \\subset \\R^n$are open sets such that$R \\subset U$and$R' \\subset U'$,$g \\colon U \\to U'$is continuously differentiable, bijective,$g^{-1}$is continuously differentiable,$g(R) \\subset R'$, and$f \\in \\sR(R')$, then the composition$f \\circ g$is Riemann integrable on$R$. \\end{enumerate}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 31, "type": "proposition", "label": "Lebl-contfunc:31", "categories": [ "characterization", "measure", "domains", "jordan", "named theorem" ], "title": "Jordan Measurability Theorem", "contents": [ "A bounded set$S \\subset \\R^n$is Jordan measurable if and only if the boundary$\\partial S$is a measure zero set." ], "refs": [], "proofs": [ { "contents": [ "Suppose $R$ is a closed rectangle such that $S$ is", "contained in the interior of $R$.", "If $x \\in \\partial S$, then for every $\\delta > 0$,", "the sets $S \\cap B(x,\\delta)$ (where $\\chi_S$ is 1) and", "the sets $(R \\setminus S) \\cap B(x,\\delta)$ (where $\\chi_S$ is 0) are", "both nonempty. So $\\chi_S$ is not continuous at $x$.", "If $x$ is either in the interior of $S$ or in the complement of the closure", "$\\widebar{S}$, then $\\chi_S$ is either identically 1 or identically 0", "in a whole neighborhood of $x$ and hence $\\chi_S$ is continuous at $x$.", "Therefore, the set of discontinuities of $\\chi_S$ is precisely the", "boundary $\\partial S$. The proposition follows." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 32, "type": "proposition", "label": "Lebl-contfunc:prop:jordanmeas", "categories": [ "jordan", "named theorem" ], "title": "Jordan Measurability Theorem", "contents": [ "Suppose$S$and$T$are bounded Jordan measurable sets. Then \\begin{enumerate}[(i)] \\item The closure$\\widebar{S}$is Jordan measurable. \\item The interior$S^\\circ$is Jordan measurable. \\item$S \\cup T$is Jordan measurable. \\item$S \\cap T$is Jordan measurable. \\item$S \\setminus T$is Jordan measurable. \\end{enumerate}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 33, "type": "proposition", "label": "Lebl-contfunc:33", "categories": [ "jordan", "named theorem" ], "title": "Jordan Measurability Theorem", "contents": [ "If$S \\subset \\R^n$is Jordan measurable, then$V(S) = m^*(S)$." ], "refs": [], "proofs": [ { "contents": [ "Given $\\epsilon > 0$,", "let $R$ be a closed rectangle that contains $S$. Let $P$ be a partition", "of $R$ such that", "\\begin{equation*}", "U(P,\\chi_S) \\leq \\left( \\int_R \\chi_S \\right) + \\epsilon = V(S) + \\epsilon", "\\qquad \\text{and} \\qquad", "L(P,\\chi_S) \\geq \\left( \\int_R \\chi_S \\right) - \\epsilon = V(S)-\\epsilon.", "\\end{equation*}", "Let $R_1,R_2,\\ldots,R_k$ be all the subrectangles of $P$ such that $\\chi_S$ is not", "identically zero on each $R_j$. That is, there is some point $x \\in R_j$ such", "that $x \\in S$ (i.e.\\ $\\chi_S(x)=1$). Let $O_j$ be an open rectangle such that $R_j \\subset O_j$", "and $V(O_j) < V(R_j) + \\nicefrac{\\epsilon}{k}$. Notice that $S \\subset", "\\bigcup_j O_j$. Then", "\\begin{equation*}", "U(P,\\chi_S) = \\sum_{j=1}^k V(R_j) >", "\\left(\\sum_{j=1}^k V(O_j)\\right) - \\epsilon \\geq m^*(S) - \\epsilon .", "\\end{equation*}", "As", "$U(P,\\chi_S) \\leq V(S) + \\epsilon$, then", "$m^*(S) - \\epsilon \\leq V(S) + \\epsilon$, or in other words", "$m^*(S) \\leq V(S)$.", "Let $R'_1,R'_2,\\ldots,R'_\\ell$ be all the subrectangles of $P$ such that", "$\\chi_S$ is identically one on each $R'_j$. In other words,", "these are the subrectangles contained in $S$.", "The interiors", "of the subrectangles $R'^\\circ_j$ are disjoint and", "$V(R'^\\circ_j) = V(R'_j)$. Via", "\\exerciseref{exercise:outermeasureofsumofrectangles},", "\\begin{equation*}", "m^*\\Bigl(\\bigcup_{j=1}^\\ell R'^\\circ_j\\Bigr)", "=", "\\sum_{j=1}^\\ell", "V(R'^\\circ_j) .", "\\end{equation*}", "Hence", "\\begin{equation*}", "m^*(S) \\geq", "m^*\\Bigl(\\bigcup_{j=1}^\\ell R'_j\\Bigr)", "\\geq", "m^*\\Bigl(\\bigcup_{j=1}^\\ell R'^\\circ_j\\Bigr)", "=", "\\sum_{j=1}^\\ell", "V(R'^\\circ_j)", "=", "\\sum_{j=1}^\\ell", "V(R'_j)", "=", "L(P,f) \\geq V(S) - \\epsilon .", "\\end{equation*}", "Therefore $m^*(S) \\geq V(S)$ as well." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 34, "type": "proposition", "label": "Lebl-contfunc:34", "categories": [ "integration", "jordan", "named theorem" ], "title": "Jordan Measurability Theorem", "contents": [ "If$S \\subset \\R^n$is a bounded Jordan measurable set and$f \\colon S \\to \\R$is a bounded continuous function, then$f$is integrable on$S$." ], "refs": [], "proofs": [ { "contents": [ "Define the function $\\widetilde{f}$ as above for some closed rectangle $R$ with $S", "\\subset R$. If $x \\in R \\setminus \\widebar{S}$, then $\\widetilde{f}$", "is identically zero in a neighborhood of $x$. Similarly if $x$ is in the", "interior of $S$, then $\\widetilde{f} = f$ on a neighborhood of $x$", "and $f$ is continuous at $x$. Therefore, $\\widetilde{f}$ is only ever", "possibly discontinuous at $\\partial S$, which is a set of measure zero,", "and we are finished." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 35, "type": "proposition", "label": "Lebl-contfunc:prop:jordanintbasic", "categories": [ "integration", "jordan", "named theorem" ], "title": "Jordan Measurability Theorem", "contents": [ "Suppose$S \\subset \\R^n$is a bounded Jordan measurable set and$f \\colon S \\to \\R$and$g \\colon S \\to \\R$are Riemann integrable on$S$, and$\\alpha \\in \\R$. Then \\begin{enumerate}[(i)] \\item If$f = 0$almost everywhere, then$\\int_S f = 0$. \\item If$f = g$almost everywhere, then$\\int_S f = \\int_S g$. \\item$f+g$is Riemann integrable on$S$and$\\int_S (f+g) = \\int_S f + \\int_S g$. \\item$\\alpha f$is Riemann integrable on$S$and$\\int_S \\alpha f = \\alpha \\int_S f$. \\item If$f(x) \\leq g(x)$for almost every$x$, then$\\int_S f \\leq \\int_S g$. \\end{enumerate}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 36, "type": "proposition", "label": "Lebl-contfunc:prop:jordanintadd", "categories": [ "integration", "named theorem" ], "title": "Jordan Measurability Theorem", "contents": [ "Suppose$A \\subset \\R^n$and$B \\subset \\R^n$are disjoint bounded Jordan measurable sets and$f \\colon A \\cup B \\to \\R$is such that the restrictions$f|_A$and$f|_B$are Riemann integrable on$A$and$B$respectively. Then$f$is Riemann integrable on$A \\cup B$and\\begin{equation*} \\int_{A \\cup B} f = \\int_A f + \\int_B f . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 37, "type": "proposition", "label": "Lebl-contfunc:prop:intovertypeIset", "categories": [ "jordan", "integration", "domains", "named theorem" ], "title": "Jordan Measurability Theorem", "contents": [ "Let$f \\colon [a,b] \\to \\R$and$g \\colon [a,b] \\to \\R$be continuous functions and such that for all$x \\in (a,b)$,$f(x) < g(x)$. Let\\begin{equation*} U \\coloneqq \\bigl\\{ (x,y) \\in \\R^2 : a < x < b \\text{ and } f(x) < y < g(x) \\bigr\\} . \\end{equation*}See \\figureref{fig:typeIdomain}. Then$U$is Jordan measurable, and if$\\varphi \\colon U \\to \\R$is Riemann integrable on$U$, then\\begin{equation*} \\int_U \\varphi = \\int_a^b \\int_{f(x)}^{g(x)} \\varphi(x,y) \\, dy \\, dx . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 38, "type": "proposition", "label": "Lebl-contfunc:prop:imagejordanmeas", "categories": [ "change of variables", "jordan", "named theorem" ], "title": "Jordan Measurability Theorem", "contents": [ "Suppose$U \\subset \\R^n$is open and$S \\subset U$is a compact Jordan measurable set. Suppose$g \\colon U \\to \\R^n$is a one-to-one continuously differentiable mapping such that the Jacobian determinant$J_g$is never zero on$S$. Then$g(S)$is bounded and Jordan measurable." ], "refs": [], "proofs": [ { "contents": [ "Let $T \\coloneqq g(S)$. %As $S \\subset \\R^n$ is closed and bounded it is compact.", "By", "\\volIref{\\lemmaref*{vI-lemma:continuouscompact} from volume I}{\\lemmaref{lemma:continuouscompact}},", "the set", "$T$ is also compact and so closed and bounded.", "We claim $\\partial T \\subset g(\\partial S)$. Suppose the claim is proved.", "As $S$ is Jordan measurable, then", "$\\partial S$ is measure zero. Then $g(\\partial S)$ is measure", "zero by \\propref{prop:imagenull}. As $\\partial T \\subset g(\\partial", "S)$, then $T$ is Jordan measurable.", "It is therefore left to prove the claim.", "As $T$ is closed, $\\partial T \\subset T$.", "Suppose $y \\in \\partial T$, then there must exist an", "$x \\in S$", "such that $g(x) = y$, and by hypothesis $J_g(x) \\not= 0$.", "We use the inverse function theorem (\\thmref{thm:inverse}). We find", "a neighborhood $V \\subset U$ of $x$ and an open set $W$ such that", "the restriction $f|_V$ is a one-to-one and onto function from $V$ to $W$", "with a continuously differentiable inverse. In particular, $g(x) = y \\in W$.", "As $y \\in \\partial T$, there exists a sequence $\\{ y_k \\}_{k=1}^\\infty$ in $W$ with", "$\\lim_{k\\to\\infty} y_k = y$ and $y_k \\notin T$. As $g|_V$ is invertible and in", "particular has a continuous inverse, there exists", "a sequence $\\{ x_k \\}_{k=1}^\\infty$ in $V$ such that $g(x_k) = y_k$ and", "$\\lim_{k\\to\\infty} x_k = x$.", "Since $y_k \\notin T = g(S)$, clearly $x_k \\notin S$. Since $x \\in S$, we", "conclude that $x \\in \\partial S$. The claim is proved, $\\partial T \\subset", "g(\\partial S)$." ], "refs": [ "prop:imagenull", "thm:inverse" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 39, "type": "proposition", "label": "Lebl-contfunc:39", "categories": [ "domains", "jordan", "named theorem" ], "title": "Jordan Measurability Theorem", "contents": [ "Let$U \\subset \\R^2$be a bounded domain with piecewise smooth boundary. Then$U$is Jordan measurable." ], "refs": [], "proofs": [ { "contents": [ "We must show that", "$\\partial U$ is a null set. As $\\partial U$ is a finite", "union of piecewise smooth paths, which", "are finite unions of smooth paths, we need only show that", "a smooth path in $\\R^2$ is a null set.", "Let $\\gamma \\colon [a,b] \\to \\R^2$ be a smooth path.", "It is enough to show that", "$\\gamma\\bigl((a,b)\\bigr)$ is a null set, as adding", "the points $\\gamma(a)$ and $\\gamma(b)$,", "to a null set still results in a null set.", "Define", "\\begin{equation*}", "f \\colon (a,b) \\times (-1,1) \\to \\R^2,", "\\qquad \\text{as} \\qquad", "f(x,y) \\coloneqq \\gamma(x) .", "\\end{equation*}", "The set $(a,b) \\times \\{ 0 \\}$ is a null set in $\\R^2$ and", "$\\gamma\\bigl((a,b)\\bigr) = f\\bigl( (a,b) \\times \\{ 0 \\} \\bigr)$.", "By \\propref{prop:imagenull},", "$\\gamma\\bigl((a,b)\\bigr)$ is a null set in $\\R^2$", "and so", "$\\gamma\\bigl([a,b]\\bigr)$ is a null set, and", "so finally $\\partial U$ is a null set." ], "refs": [ "prop:imagenull" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 40, "type": "theorem", "label": "Lebl-contfunc:40", "categories": [ "orientation", "domains", "named theorem" ], "title": "Green's Theorem", "contents": [ "\\index{Green's theorem} Suppose$U \\subset \\R^2$is a bounded domain with piecewise smooth boundary with the boundary positively oriented. Suppose$P$and$Q$are continuously differentiable functions defined on some open set that contains the closure$\\widebar{U}$. Then\\begin{equation*} \\int_{\\partial U} P \\, dx + Q\\, dy = \\int_{U} \\left(\\frac{\\partial Q}{\\partial x} - \\frac{\\partial P}{\\partial y} \\right) . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "[Proof of Green's theorem for $U$ of type III]", "Let $f,g,h,k$ be the functions defined above.", "Using \\propref{prop:intovertypeIset},", "$U$ is Jordan measurable and as $U$ is of type I\\@, then", "\\begin{equation*}", "\\begin{split}", "\\int_U", "\\left(- \\frac{\\partial P}{\\partial y} \\right)", "& =", "\\int_a^b \\int_{g(x)}^{f(x)}", "\\left(- \\frac{\\partial P}{\\partial y} (x,y) \\right)", "\\, dy \\, dx", "\\\\", "& =", "\\int_a^b \\Bigl(", "- P\\bigl(x,f(x)\\bigr) +", "P\\bigl(x,g(x)\\bigr)", "\\Bigr) \\, dx", "\\\\", "& =", "\\int_a^b P\\bigl(x,g(x)\\bigr) \\, dx", "-", "\\int_a^b P\\bigl(x,f(x)\\bigr) \\, dx .", "\\end{split}", "\\end{equation*}", "We integrate $P\\,dx$ along the boundary.", "The one-form $P\\,dx$ integrates to zero", "along the straight vertical lines in the boundary. Therefore it is", "only", "integrated along the top and along the bottom. As a parameter,", "$x$ runs from left to right. If we use the parametrizations that take $x$", "to $\\bigl(x,f(x)\\bigr)$ and to", "$\\bigl(x,g(x)\\bigr)$ we recognize path integrals above. However the second", "path integral is in the wrong direction; the top should be going right to", "left, and so we must switch orientation.", "\\begin{equation*}", "\\int_{\\partial U} P \\, dx", "=", "\\int_a^b P\\bigl(x,g(x)\\bigr) \\, dx", "+", "\\int_b^a P\\bigl(x,f(x)\\bigr) \\, dx", "=", "\\int_U", "\\left(- \\frac{\\partial P}{\\partial y} \\right) .", "\\end{equation*}", "Similarly, $U$ is also of type II\\@. The form $Q\\,dy$ integrates to zero along", "horizontal lines. So", "\\begin{equation*}", "\\int_U", "\\frac{\\partial Q}{\\partial x}", "=", "\\int_c^d \\int_{k(y)}^{h(y)}", "\\frac{\\partial Q}{\\partial x}(x,y)", "\\, dx \\, dy", "=", "\\int_a^b \\Bigl(", "Q\\bigl(y,h(y)\\bigr)", "-", "Q\\bigl(y,k(y)\\bigr)", "\\Bigr) \\, dx", "=", "\\int_{\\partial U} Q \\, dy .", "\\end{equation*}", "Putting the two computations together we obtain", "\\begin{equation*}", "\\int_{\\partial U} P\\, dx + Q \\, dy", "=", "\\int_{\\partial U} P\\, dx + \\int_{\\partial U} Q \\, dy", "=", "\\int_U", "\\Bigl(-\\frac{\\partial P}{\\partial y}\\Bigr)", "+", "\\int_U", "\\frac{\\partial Q}{\\partial x}", "=", "\\int_U", "\\Bigl(", "\\frac{\\partial Q}{\\partial x}", "-\\frac{\\partial P}{\\partial y}", "\\Bigr) . \\qedhere", "\\end{equation*}" ], "refs": [ "prop:intovertypeIset" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 41, "type": "example", "label": "Lebl-contfunc:41", "categories": [ "example", "orientation", "vector fields", "domains", "named theorem" ], "title": "Green's Theorem", "contents": [ "Suppose$P(x,y) = \\frac{-y}{x^2+y^2}$,$Q(x,y) = \\frac{x}{x^2+y^2}$. If we think of$(P,Q)$as a vector, so that we have a so-called \\emph{\\myindex{vector field}},$(P,Q)$is called the \\emph{\\myindex{vortex vector field}}, as it gives the velocity of particles traveling in a vortex around the origin. Variations on this vector field come up often in applications. Suppose that$\\gamma$is a path that goes counterclockwise around a rectangle whose interior contains the origin. We claim", "\\begin{equation*}\n\\int_{\\gamma} \\frac{-y}{x^2+y^2} \\, dx + \\frac{x}{x^2+y^2} \\, dy = 2 \\pi .\n\\end{equation*}", "First we draw a circle$C$of radius$r > 0$centered at the origin such that the entire circle is within$\\gamma$and oriented clockwise. Consider$U$to be the domain between$\\gamma$and$C$. See \\figureref{fig:vortexbox}. The integral around$\\partial U$is the integral around$\\gamma$plus the integral around$C$. Now$U$is not a domain of type III, so we cannot just apply the version of Green's theorem we actually proved. However, if we cut the box along the axis as shown in the figure with dashed lines, the four resulting domains, let us call them$U_1,U_2,U_3,U_4$, are of type III. The dashed lines are oriented in opposite directions for the two$U_j$that share them, and so when we integrate along both, the integrals cancel. That is, \\begin{multline*} \\int_{\\partial U} P\\, dx + Q \\, dy =", "\\int_{\\partial U_1} P\\, dx + Q \\, dy + \\int_{\\partial U_2} P\\, dx + Q \\, dy + \\int_{\\partial U_3} P\\, dx + Q \\, dy + \\int_{\\partial U_4} P\\, dx + Q \\, dy . \\end{multline*} Now we can apply Green's theorem to every$U_j$. We leave it to the reader to verify that outside of the origin,$\\frac{\\partial Q}{\\partial x} - \\frac{\\partial P}{\\partial y} = 0$. So", "\\begin{equation*}\n\\int_{\\partial U_j} P\\, dx + Q \\, dy \n=\n\\int_{U_j} \\left( \\frac{\\partial Q}{\\partial x} - \\frac{\\partial P}{\\partial\ny} \\right)\n=\n\\int_{U_j} 0 = 0 .\n\\end{equation*}", "Next we notice that", "\\begin{equation*}\n\\int_C P \\, dx + Q \\, dy\n+\n\\int_{\\gamma} P \\, dx + Q \\, dy\n=\n\\int_{\\partial U} P \\, dx + Q \\, dy = 0 .\n\\end{equation*}", "So the integral around$C$is minus the integral around$\\gamma$. The integral around$C$is easy to compute as on$C$we have$x^2+y^2 = r^2$, so$P(x,y) = \\frac{-y}{r^2}$and$Q(x,y) = \\frac{x}{r^2}$. We leave it to the reader to compute", "\\begin{equation*}\n\\int_C P\\, dx + Q \\, dy\n=\n\\int_C \\frac{-y}{r^2}\\, dx + \\frac{x}{r^2} \\, dy\n=\n- 2 \\pi .\n\\end{equation*}", "The claim follows. \\begin{myfigureht} \\subimport*{figures/}{vortexbox.pdf_t} \\caption{Changing the box integral to an integral around a small circle around the origin. The domain$U$is the entire shaded area between the circle and the box.\\label{fig:vortexbox}} \\end{myfigureht}", "We remark that if$\\gamma$would not contain the origin,$\\int_\\gamma P\\,dx+Q\\,dy = 0$, as we could just apply Green's to$\\gamma$. So this integral can detect whether the origin is inside$\\gamma$or not." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 42, "type": "example", "label": "Lebl-contfunc:42", "categories": [ "harmonic", "example", "named theorem" ], "title": "Green's Theorem", "contents": [ "Suppose$U \\subset \\R^2$is open and$f \\colon U \\to \\R$is harmonic, that is,$f$is twice continuously differentiable and satisfies the \\emph{\\myindex{Laplace equation}},$\\frac{\\partial^2 f}{\\partial x^2} + \\frac{\\partial^2 f}{\\partial y^2} = 0$. Harmonic functions are, for instance, the steady state heat distribution, or the electric potential between charges. We will prove one of the most fundamental properties of these functions.", "Let$D_r \\coloneqq B(p,r)$be a disc such that its closure$\\overline{D_r} = C(p,r) \\subset U$. Write$p = (x_0,y_0)$. We orient$\\partial D_r$positively. See \\exerciseref{green:balltype3orient}. Then via Green's and differentiation under the integral,", "\\begin{equation*}\n\\begin{split}\n0\n& =\n\\frac{1}{2\\pi r}\n\\int_{D_r}\n\\left(\n\\frac{\\partial^2 f}{\\partial x^2} +\n\\frac{\\partial^2 f}{\\partial y^2}\n\\right)\n\\\\\n& \n=\n\\frac{1}{2\\pi r}\n\\int_{\\partial D_r}\n- \\frac{\\partial f}{\\partial y} \\, dx + \n\\frac{\\partial f}{\\partial x} \\, dy\n\\\\\n&\n=\n\\frac{1}{2\\pi r}\n\\int_0^{2\\pi}\n\\biggl(\n- \\frac{\\partial f}{\\partial y} \\bigl(x_0+r\\cos(t),y_0+r\\sin(t)\\bigr) \\bigl(-r\\sin(t)\\bigr)\n\\\\\n& \\hspace{1.2in}\n+ \\frac{\\partial f}{\\partial x} \\bigl(x_0+r\\cos(t),y_0+r\\sin(t)\\bigr) r\\cos(t)\n\\biggr) \\, dt\n\\\\\n&\n=\n\\frac{d}{dr}\n\\left[\n\\frac{1}{2\\pi}\n\\int_0^{2\\pi}\nf\\bigl(x_0+r\\cos(t),y_0+r\\sin(t)\\bigr) \\, dt\n\\right] .\n\\end{split}\n\\end{equation*}", "Let$g(r) \\coloneqq \\frac{1}{2\\pi} \\int_0^{2\\pi} f\\bigl(x_0+r\\cos(t),y_0+r\\sin(t)\\bigr) \\, dt$for$r \\geq 0$(small enough). The function is continuous at$r=0$(exercise), and we have just proved that$g'(r) = 0$for all$r > 0$. Therefore,$g(0) = g(r)$for all$r > 0$, and", "\\begin{equation*}\ng(r) = g(0) = \n\\frac{1}{2\\pi}\n\\int_0^{2\\pi}\nf\\bigl(x_0+0\\cos(t),y_0+0\\sin(t)\\bigr) \\, dt\n=\nf(x_0,y_0).\n\\end{equation*}", "We proved the \\emph{\\myindex{mean value property}} of harmonic functions:", "\\begin{equation*}\nf(x_0,y_0) = \n\\frac{1}{2\\pi}\n\\int_0^{2\\pi}\nf\\bigl(x_0+r\\cos(t),y_0+r\\sin(t)\\bigr) \\, dt \n=\n\\frac{1}{2\\pi r}\n\\int_{\\partial D_r} f \\, ds .\n\\end{equation*}", "That is, for a harmonic function, the value at$p = (x_0,y_0)$equals the average of its values over a circle of any radius$r$centered at$(x_0,y_0)$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 43, "type": "proposition", "label": "Lebl-contfunc:prop:volrectdet", "categories": [ "domains", "jordan", "named theorem" ], "title": "Jordan Measurability Theorem", "contents": [ "Suppose$R \\subset \\R^n$is a rectangle and$A \\colon \\R^n \\to \\R^n$is linear. Then$A(R)$is Jordan measurable and$V\\bigl(A(R)\\bigr) = \\sabs{\\det (A)} \\, V(R)$." ], "refs": [], "proofs": [ { "contents": [ "It is enough to prove for elementary matrices. The proof is left as an", "exercise." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 44, "type": "theorem", "label": "Lebl-contfunc:44", "categories": [ "integration", "jordan", "named theorem" ], "title": "Jordan Measurability Theorem", "contents": [ "Suppose$U \\subset \\R^n$is open,$S \\subset U$is a compact Jordan measurable set, and$g \\colon U \\to \\R^n$is a one-to-one continuously differentiable mapping, such that$J_g$is never zero on$S$. Suppose$f \\colon g(S) \\to \\R$is Riemann integrable. Then$f \\circ g$is Riemann integrable on$S$and\\begin{equation*} \\int_{g(S)} f(x) \\, dx = \\int_S f\\bigl(g(x)\\bigr) \\, \\sabs{J_g(x)} \\, dx . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "The set $S$ can be covered by finitely many closed rectangles", "$P_1,P_2,\\ldots,P_k$, whose", "interiors do not overlap such that each $P_j \\subset U$", "(\\exerciseref{mv:changeofvarcoverbyrects}).", "Proving the theorem for $P_j \\cap S$ instead of $S$ is enough.", "Define $f(y) \\coloneqq 0$ for all $y \\notin g(S)$.", "The new $f$ is still Riemann integrable since $g(S)$ is Jordan measurable.", "We can now replace the integrals over $S$ with integrals over the whole", "rectangle.", "We therefore assume that $S$ is equal to a rectangle $R$.", "Let $\\epsilon > 0$ be given.", "For every $x \\in R$, let", "\\begin{equation*}", "W_x \\coloneqq \\bigl\\{ y \\in U : \\snorm{g'(x)-g'(y)} < \\nicefrac{\\epsilon}{2} \\bigr\\} .", "\\end{equation*}", "By \\exerciseref{mv:changeofvarWxopen},", "$W_x$ is open.", "As $x \\in W_x$ for every $x$, it is an open cover.", "By the Lebesgue covering lemma", "(\\volIref{\\lemmaref*{vI-ms:lebesgue} from volume I}{\\lemmaref{ms:lebesgue}}),", "there exists a $\\delta > 0$ such that", "for every $y \\in R$, there is an $x$ such that $B(y,\\delta) \\subset W_x$.", "In other words, if $P$ is a rectangle of maximum side length less", "than $\\frac{\\delta}{\\sqrt{n}}$ and $y \\in P$, then $P \\subset", "B(y,\\delta) \\subset W_x$. By triangle inequality,", "$\\snorm{g'(\\xi)-g'(\\eta)} < \\epsilon$ for all $\\xi, \\eta \\in P$.", "Let $R_1,R_2,\\ldots,R_N$ be subrectangles partitioning $R$ such that", "the maximum side of every $R_j$ is less than", "$\\frac{\\delta}{\\sqrt{n}}$.", "We also make sure that the minimum side length is at least", "$\\frac{\\delta}{2\\sqrt{n}}$, which we can do if $\\delta$ is", "sufficiently small relative to the sides of $R$ (\\exerciseref{mv:changeofvarrectside}).", "Consider some $R_j$ and some fixed $x_j \\in R_j$.", "First suppose $x_j=0$, $g(0) = 0$, and $g'(0) = I$.", "For any given $y \\in R_j$,", "apply the fundamental theorem of calculus", "to the function $t \\mapsto g(ty)$ to find", "$g(y) = \\int_0^1 g'(ty)y \\,dt$. As the", "side of $R_j$ is at most $\\frac{\\delta}{\\sqrt{n}}$,", "then $\\snorm{y} \\leq \\delta$. So", "\\begin{equation*}", "\\snorm{g(y)-y} =", "\\norm{\\int_0^1 \\bigl(g'(ty) y - y\\bigr) \\,dt} \\leq", "\\int_0^1 \\snorm{g'(ty) y - y} \\,dt \\leq", "\\snorm{y} \\int_0^1 \\snorm{g'(ty) - I} \\,dt", "\\leq", "\\delta \\epsilon .", "\\end{equation*}", "Therefore, $g(R_j) \\subset \\widetilde{R}_j$, where", "$\\widetilde{R}_j$ is a rectangle obtained from", "$R_j$ by extending by", "$\\delta \\epsilon$ on all sides. See \\figureref{changeofvarssq:fig}.", "\\begin{myfigureht}", "\\subimport*{figures/}{changeofvarssq.pdf_t}", "\\caption{Image of $R_j$ under $g$ lies inside", "$\\widetilde{R}_j$. A sample point $y \\in R_j$ (on the boundary of $R_j$ in fact) is marked", "and $g(y)$ must lie within with a radius of $\\delta\\epsilon$", "(also marked).\\label{changeofvarssq:fig}}", "\\end{myfigureht}", "If the sides of $R_j$ are $s_1,s_2,\\ldots,s_n$, then", "$V(R_j) = s_1 s_2 \\cdots s_n$. Recall $\\delta \\leq 2\\sqrt{n} \\, s_j$.", "Thus,", "\\begin{equation*}", "\\begin{split}", "V(\\widetilde{R}_j) & =", "(s_1+2\\delta \\epsilon )", "(s_2+2\\delta \\epsilon )", "\\cdots", "(s_n+2\\delta \\epsilon )", "\\\\", "& \\leq", "(s_1+4 \\sqrt{n}\\,s_1 \\epsilon )", "(s_2+4 \\sqrt{n}\\,s_2 \\epsilon )", "\\cdots", "(s_n+4 \\sqrt{n}\\,s_n \\epsilon )", "\\\\", "& =", "s_1 (1+4 \\sqrt{n}\\, \\epsilon )", "\\,", "s_2 (1+4 \\sqrt{n}\\, \\epsilon )", "\\cdots", "s_n (1+4 \\sqrt{n}\\, \\epsilon )", "=", "V(R_j) \\, {(1+4\\sqrt{n} \\, \\epsilon)}^n .", "\\end{split}", "\\end{equation*}", "In other words,", "\\begin{equation*}", "V\\bigl(g(R_j)\\bigr) \\leq V(\\widetilde{R}_j) \\leq V(R_j) \\, {(1+4\\sqrt{n} \\, \\epsilon)}^n .", "\\end{equation*}", "Next, suppose $A \\coloneqq g'(0)$ is not necessarily the identity.", "Write $g = A \\circ \\widetilde{g}$ where $\\widetilde{g}'(0) = I$.", "By \\propref{prop:volrectdet},", "$V\\bigl(A(R_j)\\bigr) = \\sabs{\\det(A)} \\, V(R_j)$, and hence", "\\begin{equation*}", "\\begin{split}", "V\\bigl(g(R_j)\\bigr) & \\leq", "\\sabs{\\det(A)} \\, V(R_j) \\, {(1+4\\sqrt{n} \\, \\epsilon)}^n \\\\", "& =", "\\sabs{J_g(0)} \\, V(R_j) \\, {(1+4\\sqrt{n} \\, \\epsilon)}^n .", "\\end{split}", "\\end{equation*}", "Translation does not change volume, and therefore", "for every $R_j$, and $x_j \\in R_j$, including when $x_j \\not= 0$ and $g(x_j)", "\\not= 0$, we find", "\\begin{equation*}", "V\\bigl(g(R_j)\\bigr) \\leq", "\\sabs{J_g(x_j)} \\, V(R_j) \\, {(1+4\\sqrt{n} \\, \\epsilon)}^n .", "\\end{equation*}", "Write $f$ as", "$f = f_+ - f_-$ for two nonnegative Riemann integrable", "functions $f_+$ and $f_-$:", "\\begin{equation*}", "f_+(x) \\coloneqq \\max \\bigl\\{ f(x) , 0 \\bigr\\}, \\qquad", "f_-(x) \\coloneqq \\max \\bigl\\{ -f(x) , 0 \\bigr\\} .", "\\end{equation*}", "So, if we prove the theorem for a nonnegative $f$,", "we obtain the theorem for arbitrary $f$.", "Therefore, suppose that", "$f(y) \\geq 0$ for all $y \\in R$.", "For a small enough", "$\\delta > 0$, we have", "\\begin{equation*}", "\\begin{split}", "\\epsilon + \\int_R f\\bigl(g(x)\\bigr) \\, \\sabs{J_g(x)} \\, dx", "& \\geq", "\\sum_{j=1}^N \\biggl(\\sup_{x \\in R_j} f\\bigl(g(x)\\bigr) \\, \\sabs{J_g(x)} \\biggr) \\, V(R_j)", "\\\\", "& \\geq", "\\sum_{j=1}^N \\biggl(\\sup_{x \\in R_j} f\\bigl(g(x)\\bigr) \\biggr) \\, \\sabs{J_g(x_j)} \\, V(R_j)", "\\\\", "& \\geq", "\\sum_{j=1}^N \\biggl(\\sup_{y \\in g(R_j)} f(y) \\biggr) \\,", "V\\bigl(g(R_j)\\bigr)", "\\frac{1}{{(1+4\\sqrt{n} \\, \\epsilon)}^n}", "\\\\", "& \\geq", "\\sum_{j=1}^N \\left(\\int_{g(R_j)}f(y) \\,dy \\right)", "\\frac{1}{{(1+4\\sqrt{n} \\, \\epsilon)}^n}", "\\\\", "& =", "\\frac{1}{{(1+4\\sqrt{n} \\, \\epsilon)}^n}", "\\int_{g(R)} f(y) \\,dy .", "\\end{split}", "\\end{equation*}", "The last equality follows because the overlaps of the rectangles", "are their boundaries, which are of measure zero, and hence the image", "of their boundaries is also measure zero.", "Let $\\epsilon$ go to zero to find", "\\begin{equation*}", "\\int_R f\\bigl(g(x)\\bigr) \\, \\sabs{J_g(x)} \\, dx \\geq \\int_{g(R)} f(y) \\,dy .", "\\end{equation*}", "By adding this result for several rectangles covering an $S$ we obtain the", "result for an arbitrary bounded Jordan measurable $S \\subset U$,", "and nonnegative integrable function $f$:", "\\begin{equation*}", "\\int_S f\\bigl(g(x)\\bigr) \\, \\sabs{J_g(x)} \\, dx \\geq \\int_{g(S)} f(y) \\,dy .", "\\end{equation*}", "Recall that $g^{-1}$ exists and $g^{-1}\\bigl(g(S)\\bigr) = S$.", "Also, $1 = J_{g\\circ g^{-1}} = J_g\\bigl(g^{-1}(y)\\bigr) \\,J_{g^{-1}}(y)$ for $y \\in g(S)$.", "So", "\\begin{equation*}", "\\begin{split}", "\\int_{g(S)} f(y) \\, dy", "& =", "\\int_{g(S)} f\\bigl(g\\bigl(g^{-1}(y)\\bigr)\\bigr) \\,", "\\sabs{J_g\\bigl(g^{-1}(y)\\bigr)} \\, \\sabs{J_{g^{-1}}(y)} \\, dy", "\\\\", "& \\geq", "\\int_{g^{-1}(g(S))} f\\bigl(g(x)\\bigr) \\, \\sabs{J_g(x)} \\, dx", "=", "\\int_{S} f\\bigl(g(x)\\bigr) \\, \\sabs{J_g(x)} \\, dx .", "\\end{split}", "\\end{equation*}", "The conclusion of the theorem holds", "for all nonnegative $f$ and as we", "mentioned above, it thus holds for all Riemann integrable $f$." ], "refs": [ "prop:volrectdet" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 0, "type": "proposition", "label": "Lebl-contfunc:prop:continuityofcomplex", "categories": [ "complex analysis" ], "title": "Suppose$\\{ z_n \\}_{n=1}^\\infty$,$\\{ w_n \\}_{n=1}^\\infty$are sequences of complex numbers convergi...", "contents": [ "Suppose$\\{ z_n \\}_{n=1}^\\infty$,$\\{ w_n \\}_{n=1}^\\infty$are sequences of complex numbers converging to$z$and$w$respectively. Then \\begin{enumerate}[(i)] \\item$\\displaystyle \\lim_{n\\to \\infty} z_n + w_n = z + w$. \\item$\\displaystyle \\lim_{n\\to \\infty} z_n w_n = z w$. \\item Assuming$w_n \\not= 0$for all$n$and$w\\not= 0$,$\\displaystyle \\lim_{n\\to \\infty} \\frac{z_n}{w_n} = \\frac{z}{w}$. \\item$\\displaystyle \\lim_{n\\to \\infty} \\sabs{z_n} = \\sabs{z}$. \\item$\\displaystyle \\lim_{n\\to \\infty} \\bar{z}_n = \\bar{z}$. \\end{enumerate}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 1, "type": "proposition", "label": "Lebl-contfunc:prop:cachysercomplex", "categories": [ "series" ], "title": "Existence of Complex Numbers", "contents": [ "The complex series$\\sum_{n=1}^\\infty z_n$is Cauchy if for every$\\epsilon > 0$, there exists an$M \\in \\N$such that for every$n \\geq M$and every$k > n$, we have\\begin{equation*} \\abs{ \\sum_{j={n+1}}^k z_j } < \\epsilon . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 2, "type": "proposition", "label": "Lebl-contfunc:prop:absconvmeansconv", "categories": [ "series" ], "title": "Convergence of Complex Numbers", "contents": [ "If a complex series$\\sum_{n=1}^\\infty z_n$converges absolutely, then it converges." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 3, "type": "example", "label": "Lebl-contfunc:3", "categories": [ "continuity", "example" ], "title": "Continuity of Continuity", "contents": [ "The functions$f_n \\colon \\R \\to \\R$,", "\\begin{equation*}\nf_n(x) \\coloneqq \\frac{1}{1+nx^2},\n\\end{equation*}", "are continuous and converge pointwise to the discontinuous function", "\\begin{equation*}\nf(x) \\coloneqq \n\\begin{cases}\n1 & \\text{if } x=0, \\\\\n0 & \\text{else.}\n\\end{cases}\n\\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 4, "type": "proposition", "label": "Lebl-contfunc:prop:uniformconvbounded", "categories": [], "title": "Boundedness of Analysis", "contents": [ "Let$X$be a set and$(Y,d)$a metric space. If$f_n \\colon X \\to Y$are bounded functions and converge uniformly to$f \\colon X \\to Y$, then$f$is bounded." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 5, "type": "proposition", "label": "Lebl-contfunc:prop:unifcauchymetric", "categories": [ "series" ], "title": "Convergence of Convergence", "contents": [ "Let$X$be a set,$(Y,d)$be a metric space, and$f_n \\colon X \\to Y$be functions. If$\\{ f_n \\}_{n=1}^\\infty$converges uniformly, then$\\{f_n\\}_{n=1}^\\infty$is uniformly Cauchy. Conversely, if$\\{f_n\\}_{n=1}^\\infty$is uniformly Cauchy and$(Y,d)$is Cauchy-complete, then$\\{f_n\\}_{n=1}^\\infty$converges uniformly." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 6, "type": "theorem", "label": "Lebl-contfunc:thm:weiermtest", "categories": [ "series" ], "title": "Convergence of Convergence", "contents": [ "Let$X$be a set. Suppose$f_n \\colon X \\to \\C$are functions and$M_n > 0$numbers such that\\begin{equation*} \\sabs{f_n(x)}\\leq M_n \\quad \\text{for all } x \\in X, \\qquad \\text{and} \\qquad \\sum_{n=1}^\\infty M_n \\quad \\text{converges}. \\end{equation*}Then\\begin{equation*} \\sum_{n=1}^\\infty f_n(x) \\quad \\text{converges uniformly}. \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Suppose $\\sum_{n=1}^\\infty M_n$ converges. Given $\\epsilon > 0$,", "we have that the partial sums of $\\sum_{n=1}^\\infty M_n$ are Cauchy so", "there is an $N$ such that for all $m, n \\geq N$ with $m \\geq n$, we have", "\\begin{equation*}", "\\sum_{k=n+1}^m M_k < \\epsilon .", "\\end{equation*}", "We estimate a Cauchy difference of the partial", "sums of the functions", "\\begin{equation*}", "\\abs{\\sum_{k=n+1}^m f_k(x)} \\leq", "\\sum_{k=n+1}^m \\sabs{f_k(x)} \\leq", "\\sum_{k=n+1}^m M_k < \\epsilon .", "\\end{equation*}", "The series converges by \\propref{prop:cachysercomplex}.", "The convergence is uniform, as $N$ does not depend on $x$.", "Indeed, for all $n \\geq N$,", "\\begin{equation*}", "\\abs{\\sum_{k=1}^\\infty f_k(x) - \\sum_{k=1}^n f_k(x)} \\leq", "\\abs{\\sum_{k=n+1}^\\infty f_k(x)} \\leq \\epsilon .", "\\qedhere", "\\end{equation*}" ], "refs": [ "prop:cachysercomplex" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 7, "type": "example", "label": "Lebl-contfunc:example:sinnsqfourier", "categories": [ "fourier analysis", "series", "example" ], "title": "Convergence of Fourier Series", "contents": [ "The series", "\\begin{equation*}\n\\sum_{n=1}^\\infty \\frac{\\sin(nx)}{n^2}\n\\end{equation*}", "converges uniformly on$\\R$. See \\figureref{fig:fouriersern2}. This series is a Fourier series, and we will see more of these in a later section. Proof: The series converges uniformly because$\\sum_{n=1}^\\infty \\frac{1}{n^2}$converges and", "\\begin{equation*}\n\\abs{\\frac{\\sin(nx)}{n^2}} \\leq \n\\frac{1}{n^2} .\n\\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "The series converges uniformly because$\\sum_{n=1}^\\infty \\frac{1}{n^2}$converges and \\begin{equation*}", "\\abs{\\frac{\\sin(nx)}{n^2}} \\leq", "\\frac{1}{n^2} .", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 8, "type": "example", "label": "Lebl-contfunc:8", "categories": [ "series", "example" ], "title": "Boundedness of Power Series", "contents": [ "The series", "\\begin{equation*}\n\\sum_{n=0}^\\infty \\frac{x^n}{n!} \n\\end{equation*}", "converges uniformly on every bounded interval. This series is a power series that we will study shortly. Proof: Take the interval$[-r,r] \\subset \\R$(every bounded interval is contained in some$[-r,r]$). The series$\\sum_{n=0}^\\infty \\frac{r^n}{n!}$converges by the ratio test, so$\\sum_{n=0}^\\infty \\frac{x^n}{n!}$converges uniformly on$[-r,r]$as", "\\begin{equation*}\n\\abs{\\frac{x^n}{n!} } \\leq \n\\frac{r^n}{n!} .\n\\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Take the interval$[-r,r] \\subset \\R$(every bounded interval is contained in some$[-r,r]$). The series$\\sum_{n=0}^\\infty \\frac{r^n}{n!}$converges by the ratio test, so$\\sum_{n=0}^\\infty \\frac{x^n}{n!}$converges uniformly on$[-r,r]$as \\begin{equation*}", "\\abs{\\frac{x^n}{n!} } \\leq", "\\frac{r^n}{n!} .", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 9, "type": "proposition", "label": "Lebl-contfunc:prop:uniformswitch", "categories": [ "series" ], "title": "Existence of Convergence", "contents": [ "Let$(X,d_X)$and$(Y,d_Y)$be metric spaces, and suppose$(Y,d_Y)$is Cauchy-complete. Suppose$f_n \\colon X \\to Y$converge uniformly to$f \\colon X \\to Y$. Let$\\{ x_k \\}_{k=1}^\\infty$be a sequence in$X$and$x \\coloneqq \\lim_{k \\to \\infty} x_k$. Suppose\\begin{equation*} a_n \\coloneqq \\lim_{k \\to \\infty} f_n(x_k) \\end{equation*}exists for all$n$. Then$\\{a_n\\}_{n=1}^\\infty$converges and\\begin{equation*} \\lim_{k \\to \\infty} f(x_k) = \\lim_{n\\to\\infty} a_n . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "First we show that $\\{ a_n \\}_{n=1}^\\infty$ converges. As", "$\\{ f_n \\}_{n=1}^\\infty$ converges uniformly it is uniformly Cauchy.", "Let $\\epsilon > 0$ be given. There is", "an $M$ such that for all $m,n \\geq M$, we have", "\\begin{equation*}", "d_Y\\bigl(f_n(x_k),f_m(x_k)\\bigr) < \\epsilon \\qquad \\text{for all } k .", "\\end{equation*}", "Note that", "$d_Y(a_n,a_m) \\leq", "d_Y\\bigl(a_n,f_n(x_k)\\bigr) +", "d_Y\\bigl(f_n(x_k),f_m(x_k)\\bigr) +", "d_Y\\bigl(f_m(x_k),a_m\\bigr)$ and take the limit as $k \\to \\infty$ to find", "\\begin{equation*}", "d_Y(a_n,a_m) \\leq \\epsilon .", "\\end{equation*}", "Hence $\\{a_n\\}_{n=1}^\\infty$ is Cauchy and converges since $Y$ is complete. Write", "$a \\coloneqq \\lim_{k \\to \\infty} a_n$.", "Find a $k \\in \\N$ such that", "\\begin{equation*}", "d_Y\\bigl(f_k(p),f(p)\\bigr) < \\nicefrac{\\epsilon}{3}", "\\end{equation*}", "for all $p \\in X$. Assume $k$ is large enough", "so that", "\\begin{equation*}", "d_Y(a_k,a) < \\nicefrac{\\epsilon}{3} .", "\\end{equation*}", "Find an $N \\in \\N$ such that for $m \\geq N$,", "\\begin{equation*}", "d_Y\\bigl(f_k(x_m),a_k\\bigr) < \\nicefrac{\\epsilon}{3} .", "\\end{equation*}", "Then for", "$m \\geq N$,", "\\begin{equation*}", "d_Y\\bigl(f(x_m),a\\bigr)", "\\leq", "d_Y\\bigl(f(x_m),f_k(x_m)\\bigr)", "+", "d_Y\\bigl(f_k(x_m),a_k\\bigr)", "+", "d_Y\\bigl(a_k,a\\bigr)", "<", "\\nicefrac{\\epsilon}{3} +", "\\nicefrac{\\epsilon}{3} +", "\\nicefrac{\\epsilon}{3} = \\epsilon . \\qedhere", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 10, "type": "corollary", "label": "Lebl-contfunc:cor:metricuniformcontinuous", "categories": [ "consequence", "continuity", "series" ], "title": "Continuity of Convergence", "contents": [ "Let$X$and$Y$be metric spaces. If$f_n \\colon X \\to Y$are continuous functions such that$\\{ f_n \\}_{n=1}^\\infty$converges uniformly to$f \\colon X \\to Y$, then$f$is continuous." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 11, "type": "corollary", "label": "Lebl-contfunc:11", "categories": [ "consequence", "topology" ], "title": "Let$(X,d)$be a compact metric space", "contents": [ "Let$(X,d)$be a compact metric space. Then$C(X,\\C)$is a Cauchy-complete metric space." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 12, "type": "example", "label": "Lebl-contfunc:12", "categories": [ "continuity", "fourier analysis", "series", "example" ], "title": "Continuity of Fourier Series", "contents": [ "By \\exampleref{example:sinnsqfourier} the Fourier series", "\\begin{equation*}\n\\sum_{n=1}^\\infty \\frac{\\sin(nx)}{n^2}\n\\end{equation*}", "converges uniformly and hence is continuous by \\corref{cor:metricuniformcontinuous} (as is visible in \\figureref{fig:fouriersern2})." ], "refs": [ "cor:metricuniformcontinuous" ], "proofs": [], "ref_ids": [] }, { "id": 13, "type": "proposition", "label": "Lebl-contfunc:prop:complexlimitswapintegral", "categories": [ "integration", "series" ], "title": "Convergence of Convergence", "contents": [ "Suppose$f_n \\colon [a,b] \\to \\C$are Riemann integrable and suppose that$\\{ f_n \\}_{n=1}^\\infty$converges uniformly to$f \\colon [a,b] \\to \\C$. Then$f$is Riemann integrable and\\begin{equation*} \\int_a^b f = \\lim_{n\\to \\infty} \\int_a^b f_n . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 14, "type": "corollary", "label": "Lebl-contfunc:14", "categories": [ "consequence", "integration", "series" ], "title": "Convergence of Convergence", "contents": [ "\\pagebreak[2] Suppose$f_n \\colon [a,b] \\to \\C$are Riemann integrable and suppose that\\begin{equation*} \\sum_{n=1}^\\infty f_n(x) \\end{equation*}converges uniformly. Then the series is Riemann integrable on$[a,b]$and\\begin{equation*} \\int_a^b \\sum_{n=1}^\\infty f_n(x) \\,dx = \\sum_{n=1}^\\infty \\int_a^b f_n(x) \\,dx \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 15, "type": "example", "label": "Lebl-contfunc:15", "categories": [ "integration", "fourier analysis", "example", "series", "named theorem" ], "title": "Weierstrass Approximation Theorem", "contents": [ "Let us show how to integrate a Fourier series.", "\\begin{equation*}\n\\int_{0}^x \\sum_{n=1}^\\infty \\frac{\\cos(nt)}{n^2} \\,dt\n=\n\\sum_{n=1}^\\infty \\int_{0}^x \\frac{\\cos(nt)}{n^2}\\,dt\n=\n\\sum_{n=1}^\\infty \\frac{\\sin(nx)}{n^3}\n\\end{equation*}", "The swapping of integral and sum is possible because of uniform convergence, which we have proved before using the Weierstrass$M$-test (\\thmref{thm:weiermtest})." ], "refs": [ "thm:weiermtest" ], "proofs": [], "ref_ids": [] }, { "id": 16, "type": "theorem", "label": "Lebl-contfunc:thm:dersconvergecomplex", "categories": [ "continuity", "series", "differentiability" ], "title": "Continuity of Convergence", "contents": [ "Let$I \\subset \\R$be a bounded interval and let$f_n \\colon I \\to \\C$be continuously differentiable functions. Suppose$\\{ f_n' \\}_{n=1}^\\infty$converges uniformly to$g \\colon I \\to \\C$, and suppose$\\{ f_n(c) \\}_{n=1}^\\infty$is a convergent sequence for some$c \\in I$. Then$\\{ f_n \\}_{n=1}^\\infty$converges uniformly to a continuously differentiable function$f \\colon I \\to \\C$, and$f' = g$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 17, "type": "example", "label": "Lebl-contfunc:17", "categories": [ "example", "series", "differentiability", "continuity", "named theorem" ], "title": "Weierstrass Approximation Theorem", "contents": [ "There exist continuous nowhere differentiable functions. Such functions are often called \\emph{Weierstrass functions}\\index{Weierstrass function}, although this particular one, essentially due to Takagi\\footnote{\\href{https://en.wikipedia.org/wiki/Teiji_Takagi}{Teiji Takagi} (1875--1960) was a Japanese mathematician.}, is a different example than what Weierstrass gave. Define", "\\begin{equation*}\n\\varphi(x) \\coloneqq \\sabs{x} \\qquad \\text{for } x \\in [-1,1] .\n\\end{equation*}", "Extend$\\varphi$to all of$\\R$by making it 2-periodic: Decree that$\\varphi(x) = \\varphi(x+2)$. The function$\\varphi \\colon \\R \\to \\R$is continuous, in fact,$\\sabs{\\varphi(x)-\\varphi(y)} \\leq \\sabs{x-y}$(why?). See \\figureref{fig:triangwave}. \\begin{myfigureht} \\includegraphics{figures/triangwave} \\caption{The 2-periodic function$\\varphi$.\\label{fig:triangwave}} \\end{myfigureht}", "As$\\sum_{n=0}^\\infty {\\left(\\frac{3}{4}\\right)}^n$converges and$\\sabs{\\varphi(x)} \\leq 1$for all$x$, by the$M$-test (\\thmref{thm:weiermtest}),", "\\begin{equation*}\nf(x) \\coloneqq \\sum_{n=0}^\\infty \n{\\left(\\frac{3}{4}\\right)}^n \\varphi(4^n x)\n\\end{equation*}", "converges uniformly and hence is continuous. See \\figureref{fig:nowherediff}.", "\\begin{myfigureht} \\includegraphics{figures/nowherediff} \\caption{Plot of the nowhere differentiable function$f$.\\label{fig:nowherediff}} \\end{myfigureht}", "We claim$f \\colon \\R \\to \\R$is nowhere differentiable. Fix$x$, and we will show$f$is not differentiable at$x$. Define", "\\begin{equation*}\n\\delta_m \\coloneqq \\pm \\frac{1}{2} 4^{-m} ,\n\\avoidbreak\n\\end{equation*}", "where the sign is chosen so that there is no integer between$4^m x$and$4^m(x+\\delta_m) = 4^m x \\pm \\frac{1}{2}$.", "We want to look at the difference quotient", "\\begin{equation*}\n\\frac{f(x+\\delta_m)-f(x)}{\\delta_m}\n=\n\\sum_{n=0}^\\infty \n{\\left(\\frac{3}{4}\\right)}^n\n\\frac{\\varphi\\bigl(4^n(x+\\delta_m)\\bigr)-\\varphi(4^nx)}{\\delta_m} .\n\\end{equation*}", "Fix$m$for a moment. Consider the expression inside the series:", "\\begin{equation*}\n\\gamma_{n} \\coloneqq\n\\frac{\\varphi\\bigl(4^n(x+\\delta_m)\\bigr)-\\varphi(4^nx)}{\\delta_m} .\n\\end{equation*}", "If$n > m$, then$4^n\\delta_m$is an even integer. As$\\varphi$is 2-periodic we get that$\\gamma_n = 0$.", "As there is no integer between$4^m(x+\\delta_m) = 4^m x\\pm\\nicefrac{1}{2}$and$4^m x$, then on this interval$\\varphi(t) = \\pm t + \\ell$for some integer$\\ell$. In particular,$\\abs{\\varphi\\bigl(4^m(x+ \\delta_m)\\bigr)-\\varphi(4^mx)} = \\abs{4^mx\\pm\\nicefrac{1}{2}-4^mx} = \\nicefrac{1}{2}$. Therefore,", "\\begin{equation*}\n\\sabs{\\gamma_m} =\n\\abs{\n\\frac{\\varphi\\bigl(4^m(x+\\delta_m)\\bigr)-\\varphi(4^mx)}{\\pm (\\nicefrac{1}{2}) 4^{-m}}\n}\n= 4^m .\n\\end{equation*}", "Similarly, suppose$n < m$. Since$\\sabs{\\varphi(s) -\\varphi(t)} \\leq \\sabs{s-t}$,", "\\begin{equation*}\n\\sabs{\\gamma_n} =\n\\abs{\\frac{\\varphi\\bigl(4^nx\\pm(\\nicefrac{1}{2})4^{n-m}\\bigr)-\\varphi(4^nx)}{\\pm\n(\\nicefrac{1}{2}) 4^{-m}}}\n\\leq\n\\abs{\\frac{\\pm(\\nicefrac{1}{2})4^{n-m}}{\\pm (\\nicefrac{1}{2}) 4^{-m}}} = 4^n\n.\n\\end{equation*}", "And so", "\\begin{equation*}\n\\begin{split}\n\\abs{\n\\frac{f(x+\\delta_m)-f(x)}{\\delta_m}\n}\n%& =\n%\\abs{\n%\\sum_{n=0}^\\infty \n%{\\left(\\frac{3}{4}\\right)}^n\n%\\frac{\\varphi\\bigl(4^n(x+\\delta_m)\\bigr)-\\varphi(4^nx)}{\\delta_m}\n%}\n=\n\\abs{\n\\sum_{n=0}^\\infty \n{\\left(\\frac{3}{4}\\right)}^n\n\\gamma_n\n}\n& =\n\\abs{\n\\sum_{n=0}^m \n{\\left(\\frac{3}{4}\\right)}^n\n\\gamma_n\n}\n\\\\\n& \\geq\n\\abs{\n{\\left(\\frac{3}{4}\\right)}^m\n\\gamma_m}\n-\n\\abs{\n\\sum_{n=0}^{m-1} \n{\\left(\\frac{3}{4}\\right)}^n\n\\gamma_n\n}\n\\\\\n& \\geq\n3^m\n-\n\\sum_{n=0}^{m-1} \n3^n\n=\n3^m\n-\n\\frac{3^{m}-1}{3-1}\n=\n\\frac{3^m +1}{2} .\n\\end{split}\n\\end{equation*}", "As$m \\to \\infty$, we have$\\delta_m \\to 0$, but$\\frac{3^m+1}{2}$goes to infinity. So$f$cannot be differentiable at~$x$." ], "refs": [ "thm:weiermtest" ], "proofs": [], "ref_ids": [] }, { "id": 18, "type": "proposition", "label": "Lebl-contfunc:18", "categories": [ "topology", "series" ], "title": "Existence of Power Series", "contents": [ "\\pagebreak[2] Let$\\sum_{n=0}^\\infty c_n {(z-a)}^n$be a power series. There exists a$\\rho \\in [0,\\infty]$such that \\begin{enumerate}[(i)] \\item If$\\rho = 0$, then the series diverges. \\item If$\\rho = \\infty$, then the series converges absolutely for all$z \\in \\C$. \\item If$0 < \\rho < \\infty$, then the series converges absolutely on$B(a,\\rho)$, and diverges when$\\sabs{z-a} > \\rho$. \\end{enumerate} Furthermore, if$0 < r < \\rho$, then the series converges uniformly on the closed ball$C(a,r)$." ], "refs": [], "proofs": [ { "contents": [ "We use the real version of this proposition,", "\\volIref{\\propref*{vI-prop:powerserrealradius} in volume I}{\\propref{prop:powerserrealradius}}.", "Let", "\\begin{equation*}", "R \\coloneqq \\limsup_{n\\to\\infty} \\sqrt[n]{\\sabs{c_n}} .", "\\end{equation*}", "If $R = 0$, then", "$\\sum_{n=0}^\\infty \\sabs{c_n} \\, \\sabs{z-a}^n$ converges for all $z$.", "If $R = \\infty$, then", "$\\sum_{n=0}^\\infty \\sabs{c_n} \\, \\sabs{z-a}^n$ converges only at $z=a$.", "Otherwise, let $\\rho \\coloneqq \\nicefrac{1}{R}$ and", "$\\sum_{n=0}^\\infty \\sabs{c_n} \\, \\sabs{z-a}^n$ converges when", "$\\sabs{z-a} < \\rho$, and diverges (in fact the terms of the series", "do not go to zero) when $\\sabs{z-a} > \\rho$.", "To prove the ``Furthermore,'' suppose", "$0 < r < \\rho$ and $z \\in C(a,r)$. Then", "consider the partial sums", "\\begin{equation*}", "\\abs{\\sum_{n=0}^k c_n {(z-a)}^n}", "\\leq", "\\sum_{n=0}^k \\sabs{c_n} \\sabs{z-a}^n", "\\leq", "\\sum_{n=0}^k \\sabs{c_n} r^n . \\qedhere", "\\end{equation*}" ], "refs": [ "prop:powerserrealradius" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 19, "type": "example", "label": "Lebl-contfunc:19", "categories": [ "series", "example" ], "title": "We list some series we already know: \\begin{align*} & & & \\sum_{n=0}^\\infty z^n & & \\text{has rad...", "contents": [ "We list some series we already know: \\begin{align*} & & & \\sum_{n=0}^\\infty z^n & & \\text{has radius of convergence } 1. & &", "& & & \\sum_{n=0}^\\infty \\frac{1}{n!} z^n & & \\text{has radius of convergence } \\infty. & &", "& & & \\sum_{n=0}^\\infty n^n z^n & & \\text{has radius of convergence } 0. & & \\end{align*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 20, "type": "example", "label": "Lebl-contfunc:20", "categories": [ "characterization", "series", "example", "complex analysis" ], "title": "Characterization of Analytic Functions via Equivalence", "contents": [ "Note the difference between$\\frac{1}{1-z}$and its power series. Let us expand$\\frac{1}{1-z}$as power series around a point$a \\not= 1$. Let$c \\coloneqq \\frac{1}{1-a}$, then", "\\begin{equation*}\n\\frac{1}{1-z} = \n\\frac{c}{1-c(z-a)}\n=\nc\n\\sum_{n=0}^\\infty c^{n} {(z-a)}^n\n=\n\\sum_{n=0}^\\infty \\left( \\frac{1}{{(1-a)}^{n+1}} \\right) {(z-a)}^n .\n\\end{equation*}", "The series$\\sum_{n=0}^\\infty c^n {(z-a)}^n$converges if and only if the series on the right-hand side converges and", "\\begin{equation*}\n\\limsup_{n\\to\\infty}\n\\sqrt[n]{\\sabs{c^n}} = \\sabs{c}\n= \\frac{1}{\\sabs{1-a}} .\n\\end{equation*}", "The radius of convergence of the power series is$\\sabs{1-a}$, that is the distance from$1$to$a$. The function$\\frac{1}{1-z}$has a power series representation around every$a\\not= 1$and so is analytic in$\\C \\setminus \\{ 1 \\}$. The domain of the function is bigger than the region of convergence of the power series representing the function at any point." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 21, "type": "proposition", "label": "Lebl-contfunc:21", "categories": [ "continuity", "series", "complex analysis" ], "title": "Continuity of Analytic Functions", "contents": [ "If\\begin{equation*} f(z) \\coloneqq \\sum_{n=0}^\\infty c_n {(z-a)}^n \\end{equation*}is convergent in$B(a,\\rho)$for some$\\rho > 0$, then$f \\colon B(a,\\rho) \\to \\C$is continuous. In particular, analytic functions are continuous." ], "refs": [], "proofs": [ { "contents": [ "For $z_0 \\in B(a,\\rho)$, pick $r < \\rho$ such that $z_0 \\in B(a,r)$.", "On $B(a,r)$ the", "partial sums (which are continuous) converge uniformly,", "and so the limit $f|_{B(a,r)}$ is continuous.", "Any sequence converging to", "$z_0$ has some tail that is completely in the open ball $B(a,r)$,", "hence $f$ is continuous at $z_0$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 22, "type": "theorem", "label": "Lebl-contfunc:thm:fubiniforsums", "categories": [ "series", "complex analysis" ], "title": "Convergence of Complex Numbers", "contents": [ "Let$\\{ a_{k,m} \\}_{k=1,m=1}^\\infty$be a double sequence of complex numbers and suppose that for every$k$the series\\begin{equation*} \\sum_{m=1}^\\infty \\sabs{a_{k,m}} \\qquad \\text{converges} \\end{equation*}and furthermore that\\begin{equation*} \\sum_{k=1}^\\infty \\left( \\sum_{m=1}^\\infty \\sabs{a_{k,m}} \\right) \\qquad \\text{converges}. \\end{equation*}Then\\begin{equation*} \\sum_{k=1}^\\infty \\left( \\sum_{m=1}^\\infty a_{k,m} \\right) = \\sum_{m=1}^\\infty \\left( \\sum_{k=1}^\\infty a_{k,m} \\right) , \\end{equation*}where all the series involved converge." ], "refs": [], "proofs": [ { "contents": [ "Let $E$ be the set $\\{ \\nicefrac{1}{n} : n \\in \\N \\} \\cup \\{ 0 \\}$,", "and treat it as a metric space with the metric inherited from $\\R$.", "Define the sequence of functions $f_k \\colon E \\to \\C$", "by", "\\begin{equation*}", "f_k(\\nicefrac{1}{n}) \\coloneqq \\sum_{m=1}^n a_{k,m}", "\\qquad", "\\text{and}", "\\qquad", "f_k(0) \\coloneqq \\sum_{m=1}^\\infty a_{k,m} .", "\\end{equation*}", "As the series converges, each $f_k$ is continuous at $0$", "(since 0 is the only cluster point, they are continuous at every point of", "$E$, but we don't need that).", "For all $x \\in E$, we have", "\\begin{equation*}", "\\sabs{f_k(x)} \\leq \\sum_{m=1}^\\infty \\sabs{a_{k,m}} .", "\\end{equation*}", "As $\\sum_k \\sum_m \\sabs{a_{k,m}}$ converges (and does not depend on", "$x$), we know that", "\\begin{equation*}", "\\sum_{k=1}^n f_k(x)", "\\end{equation*}", "converges uniformly on $E$. Define", "\\begin{equation*}", "g(x) \\coloneqq \\sum_{k=1}^\\infty f_k(x) ,", "\\end{equation*}", "which is, therefore, a continuous function at $0$.", "So", "\\begin{equation*}", "\\begin{split}", "\\sum_{k=1}^\\infty \\left( \\sum_{m=1}^\\infty a_{k,m} \\right)", "& =", "\\sum_{k=1}^\\infty f_k(0)", "= g(0)", "= \\lim_{n\\to\\infty} g(\\nicefrac{1}{n}) \\\\", "&=", "\\lim_{n\\to\\infty}\\sum_{k=1}^\\infty f_k(\\nicefrac{1}{n})", "=", "\\lim_{n\\to\\infty}\\sum_{k=1}^\\infty \\sum_{m=1}^n a_{k,m} \\\\", "&=", "\\lim_{n\\to\\infty}\\sum_{m=1}^n \\sum_{k=1}^\\infty a_{k,m}", "=", "\\sum_{m=1}^\\infty \\left( \\sum_{k=1}^\\infty a_{k,m} \\right) . \\qedhere", "\\end{split}", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 23, "type": "theorem", "label": "Lebl-contfunc:23", "categories": [ "series", "complex analysis" ], "title": "\\index{Taylor's theorem!real-analytic}% \\label{thm:tayloranal} Let \\begin{equation*} f(x) \\colone...", "contents": [ "\\index{Taylor's theorem!real-analytic}% \\label{thm:tayloranal} Let\\begin{equation*} f(x) \\coloneqq \\sum_{k=0}^\\infty a_k x^k \\end{equation*}be a power series converging in$(-\\rho,\\rho)$for some$\\rho > 0$. Given any$a \\in (-\\rho,\\rho)$, and$x$such that$\\sabs{x-a} < \\rho-\\sabs{a}$, we have\\begin{equation*} f(x) = \\sum_{k=0}^\\infty \\frac{f^{(k)}(a)}{k!} {(x-a)}^{k} . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "Given $a$ and $x$ as in the theorem,", "write", "\\begin{equation*}", "\\begin{split}", "f(x) &= \\sum_{k=0}^\\infty a_k {\\bigl((x-a)+a\\bigr)}^k \\\\", "&= \\sum_{k=0}^\\infty a_k \\sum_{m=0}^k \\binom{k}{m} a^{k-m} {(x-a)}^m .", "\\end{split}", "\\end{equation*}", "Define $c_{k,m} \\coloneqq a_k \\binom{k}{m} a^{k-m}$ if $m \\leq k$ and $0$ if $m >", "k$. Then", "\\begin{equation} \\label{eq:tsproof}", "f(x) = \\sum_{k=0}^\\infty \\, \\sum_{m=0}^\\infty c_{k,m} {(x-a)}^m .", "\\end{equation}", "Let us show that the double sum converges absolutely.", "\\begin{equation*}", "\\begin{split}", "\\sum_{k=0}^\\infty \\, \\sum_{m=0}^\\infty \\abs{ c_{k,m} {(x-a)}^m}", "& = \\sum_{k=0}^\\infty \\, \\sum_{m=0}^k \\abs{ a_k \\binom{k}{m} a^{k-m} {(x-a)}^m }", "\\\\", "& = \\sum_{k=0}^\\infty \\sabs{a_k} \\sum_{m=0}^k \\binom{k}{m} \\sabs{a}^{k-m}", "{\\sabs{x-a}}^m \\\\", "& = \\sum_{k=0}^\\infty \\sabs{a_k} {\\bigl(\\sabs{x-a}+\\sabs{a}\\bigr)}^k ,", "\\end{split}", "\\end{equation*}", "and this series converges as long as", "$(\\sabs{x-a}+\\sabs{a}) < \\rho$ or in other words if", "$\\sabs{x-a} < \\rho-\\sabs{a}$.", "Using \\thmref{thm:fubiniforsums},", "swap the order of summation in \\eqref{eq:tsproof}, and", "the following series converges when $\\sabs{x-a} < \\rho-\\sabs{a}$:", "\\begin{equation*}", "f(x) =", "\\sum_{k=0}^\\infty \\, \\sum_{m=0}^\\infty c_{k,m} {(x-a)}^m", "=", "\\sum_{m=0}^\\infty", "\\left( \\sum_{k=0}^\\infty", "c_{k,m} \\right) {(x-a)}^m .", "\\end{equation*}", "The formula in terms of derivatives at $a$ follows by", "differentiating the series to obtain \\eqref{eq:formulaforpscoeffs}." ], "refs": [ "eq:formulaforpscoeffs", "eq:tsproof", "thm:fubiniforsums" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 24, "type": "corollary", "label": "Lebl-contfunc:cor:powerseranalytic", "categories": [ "consequence", "series", "approximation" ], "title": "Existence of Power Series", "contents": [ "For every$a \\in \\C$, if$\\sum_{k=0}^\\infty c_k {(z-a)}^k$converges to$f(z)$in$B(a,\\rho)$and$b \\in B(a,\\rho)$, then there exists a power series$\\sum_{k=0}^\\infty d_k {(z-b)}^k$that converges to$f(z)$in$B(b,\\rho-\\sabs{b-a})$." ], "refs": [], "proofs": [ { "contents": [ "Without loss of generality assume that $a=0$. We can rotate to assume that $b$ is real, but", "since that is harder to picture, let us do it explicitly.", "Let $\\alpha \\coloneqq \\frac{\\bar{b}}{\\sabs{b}}$.", "Notice that", "\\begin{equation*}", "\\abs{\\nicefrac{1}{\\alpha}} = \\sabs{\\alpha}", "= 1 .", "\\end{equation*}", "Therefore the series", "$\\sum_{k=0}^\\infty c_k {(\\nicefrac{z}{\\alpha})}^k =", "\\sum_{k=0}^\\infty c_k \\alpha^{-k} {z}^k$", "converges to $f(\\nicefrac{z}{\\alpha})$ in $B(0,\\rho)$.", "When $z=x$ is real", "we apply \\thmref{thm:tayloranal} at $\\sabs{b}$ and get", "a series that converges", "to $f(\\nicefrac{z}{\\alpha})$ on $B(\\sabs{b},\\rho-\\sabs{b})$.", "That is, there is a convergent series", "\\begin{equation*}", "f(\\nicefrac{z}{\\alpha}) =", "\\sum_{k=0}^\\infty a_k {\\bigl(z - \\sabs{b}\\bigr)}^k .", "\\end{equation*}", "Using $\\alpha b = \\sabs{b}$, we find", "\\begin{equation*}", "f(z) = f(\\nicefrac{\\alpha z}{\\alpha}) =", "\\sum_{k=0}^\\infty a_k {(\\alpha z - \\sabs{b})}^k", "=", "\\sum_{k=0}^\\infty a_k\\alpha^k {\\bigl(z - \\nicefrac{\\sabs{b}}{\\alpha}\\bigr)}^k", "=", "\\sum_{k=0}^\\infty a_k\\alpha^k {(z - b)}^k ,", "\\end{equation*}", "and this series converges for all $z$ such that", "$\\bigl\\lvert \\alpha z-\\sabs{b}\\bigr\\rvert < \\rho-\\sabs{b}$", "or $\\sabs{z - b} < \\rho-\\sabs{b}$." ], "refs": [ "thm:tayloranal" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 25, "type": "lemma", "label": "Lebl-contfunc:25", "categories": [ "auxiliary result", "series", "complex analysis" ], "title": "Auxiliary Result for Complex Numbers", "contents": [ "Suppose$f(z) = \\sum_{k=0}^\\infty a_k z^k$is a convergent power series and$\\{ z_n \\}_{n=1}^\\infty$is a sequence of nonzero complex numbers converging to 0, such that$f(z_n) = 0$for all$n$. Then$a_k = 0$for every~$k$." ], "refs": [], "proofs": [ { "contents": [ "By continuity we know $f(0) = 0$ so $a_0 = 0$.", "Suppose there exists some nonzero $a_k$.", "Let $m$ be the smallest $m$ such that $a_m \\not= 0$. Then", "\\begin{equation*}", "f(z) = \\sum_{k=m}^\\infty a_k z^k =", "z^m \\sum_{k=m}^\\infty a_k z^{k-m} =", "z^m \\sum_{k=0}^\\infty a_{k+m} z^{k} .", "\\end{equation*}", "Write $g(z) = \\sum_{k=0}^\\infty a_{k+m} z^{k}$ (this series converges in", "on the same set as $f$). $g$ is continuous and $g(0) = a_m \\not= 0$. Thus", "there exists some $\\delta > 0$ such that $g(z) \\not= 0$ for all $z \\in", "B(0,\\delta)$. As $f(z) = z^m g(z)$, the only point in $B(0,\\delta)$ where", "$f(z) = 0$ is when $z=0$, but this contradicts the assumption", "that $f(z_n) = 0$ for all $n$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 26, "type": "theorem", "label": "Lebl-contfunc:thm:identityanalytic", "categories": [ "topology", "complex analysis" ], "title": "Identity Theorem", "contents": [ "% \\index{identity theorem}% Let$U \\subset \\C$be open and connected. If$f \\colon U \\to \\C$and$g \\colon U \\to \\C$are analytic functions that are equal on a set$E \\subset U$, and$E$has a cluster point in$U$, then$f(z) = g(z)$for all$z \\in U$." ], "refs": [], "proofs": [ { "contents": [ "Without loss of generality suppose $E$ is the set of all points $z \\in U$ such that", "$g(z)=f(z)$. Note that $E$ must be closed as $f$ and $g$ are continuous.", "Suppose $E$ has a cluster point. Without loss of generality assume that $0$", "is this cluster point. Near $0$,", "we have the expansions", "\\begin{equation*}", "f(z) = \\sum_{k=0}^\\infty a_k {z}^k", "\\qquad", "\\text{and}", "\\qquad", "g(z) = \\sum_{k=0}^\\infty b_k {z}^k ,", "\\end{equation*}", "which converge in some ball $B(0,\\rho)$. Therefore the series", "\\begin{equation*}", "0 = f(z)-g(z) =", "\\sum_{k=0}^\\infty (a_k-b_k) z^k", "\\end{equation*}", "converges in $B(0,\\rho)$. As $0$ is a cluster point of $E$, there", "is a sequence of nonzero points $\\{ z_n \\}_{n=1}^\\infty$ such that", "$f(z_n) -g(z_n) = 0$. Hence, by the lemma above", "$a_k = b_k$ for all $k$. Therefore, $B(0,\\rho) \\subset E$.", "Thus the set of cluster points of $E$ is open. The set of cluster points", "of $E$ is also closed: A limit of cluster points of $E$ is in $E$", "as it is closed, and it is clearly a cluster point of $E$.", "As $U$ is connected, the set of cluster points of $E$", "is equal to $U$, or in other words $E = U$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 27, "type": "proposition", "label": "Lebl-contfunc:27", "categories": [ "complex analysis" ], "title": "Let$z,w \\in \\C$be complex numbers", "contents": [ "Let$z,w \\in \\C$be complex numbers. Then\\begin{equation*} e^{z+w} = e^z e^w. \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "We already know that", "the equality", "$e^{x+y} = e^x e^y$ holds for all", "real numbers $x$ and $y$.", "For every fixed $y \\in \\R$, consider the expressions as", "functions of $x$ and apply the identity theorem", "(\\thmref{thm:identityanalytic}) to get that", "$e^{z+y} = e^ze^y$ for all $z \\in \\C$. Fixing an arbitrary $z \\in \\C$,", "we get", "$e^{z+y} = e^ze^y$ for all $y \\in \\R$. Again by the identity theorem", "$e^{z+w} = e^z e^w$", "for all $w \\in \\C$." ], "refs": [ "thm:identityanalytic" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 28, "type": "proposition", "label": "Lebl-contfunc:28", "categories": [], "title": "Existence of Analysis", "contents": [ "The sine and cosine functions have the following properties: \\begin{enumerate}[(i)] \\item For all$z \\in \\C$,\\index{Euler's formula}\\begin{equation*} e^{iz} = \\cos(z) + i\\sin(z) \\qquad \\text{(Euler's formula)}. \\end{equation*}\\item$\\cos(0) = 1$,$\\sin(0) = 0$. \\item For all$z \\in \\C$,\\begin{equation*} \\cos(-z) = \\cos(z), \\qquad \\sin(-z) = -\\sin(z). \\end{equation*}\\item For all$z \\in \\C$,\\begin{equation*} \\cos(z) = \\sum_{k=0}^\\infty \\frac{{(-1)}^k}{(2k)!} z^{2k} , \\qquad \\sin(z) = \\sum_{k=0}^\\infty \\frac{{(-1)}^k}{(2k+1)!} z^{2k+1} . \\end{equation*}\\item For all$x \\in \\R$\\begin{equation*} \\cos(x) = \\Re (e^{ix}) \\qquad\\text{and}\\qquad \\sin(x) = \\Im (e^{ix}) . \\end{equation*}\\item For all$z \\in \\C$,\\begin{equation*} {\\bigl( \\cos(z) \\bigr)}^2 + {\\bigl( \\sin(z) \\bigr)}^2 = 1 . \\end{equation*}\\item For all$x \\in \\R$,\\begin{equation*} \\sabs{\\sin(x)} \\leq 1, \\qquad \\sabs{\\cos(x)} \\leq 1 . \\end{equation*}\\item For all$x \\in \\R$,\\begin{equation*} \\frac{d}{dx} \\bigl[ \\cos(x) \\bigr] = -\\sin(x) \\qquad \\text{and} \\qquad \\frac{d}{dx} \\bigl[ \\sin(x) \\bigr] = \\cos(x) . \\end{equation*}\\item For all$x \\geq 0$,\\begin{equation*} \\sin(x) \\leq x . \\end{equation*}\\item There exists an$x > 0$such that$\\cos(x) = 0$. We define \\glsadd{not:pi}\\begin{equation*} \\pi \\coloneqq 2 \\, \\inf \\{ x > 0 : \\cos(x) = 0 \\} . \\end{equation*}\\item For all$z \\in \\C$,\\begin{equation*} e^{2\\pi i} = 1 \\qquad \\text{and} \\qquad e^{z + i 2\\pi} = e^z. \\end{equation*}\\item Sine and cosine are$2\\pi$-periodic and not periodic with any smaller period. That is,$2\\pi$is the smallest number such that for all$z \\in \\C$,\\begin{equation*} \\sin(z+2\\pi) = \\sin(z) \\qquad \\text{and} \\qquad \\cos(z+2\\pi) = \\cos(z) . \\end{equation*}\\item The function$x \\mapsto e^{ix}$is a bijective map from$[0,2\\pi)$onto the set of$z \\in \\C$such that$\\sabs{z} = 1$. \\end{enumerate}" ], "refs": [], "proofs": [ { "contents": [ "The first three items follow directly from the definition.", "The computation of the power series for both is left as an exercise.", "As complex conjugate is a continuous function, the definition", "of $e^z$ implies", "$\\overline{e^z} = e^{\\bar{z}}$. If", "$x$ is real,", "\\begin{equation*}", "\\overline{e^{ix}} = e^{-ix} .", "\\end{equation*}", "Thus for real $x$,", "$\\cos(x) =", "\\frac{e^{ix}-e^{-ix}}{2} =", "\\frac{e^{ix}-\\overline{e^{ix}}}{2} =", "\\Re (e^{ix})$", "and similarly $\\sin(x) = \\Im (e^{ix})$.", "For real $x$, we compute", "\\begin{equation*}", "1 = e^{ix} e^{-ix}", "= e^{ix} \\, \\overline{e^{ix}}", "= \\sabs{e^{ix}}^2", "= \\babs{\\cos(x) + i \\sin(x)}^2", "= {\\bigl( \\cos(x) \\bigr)}^2 + {\\bigl( \\sin(x) \\bigr)}^2 .", "\\end{equation*}", "A slightly more complicated computation shows this fact for", "complex numbers, see \\exerciseref{exercise:cossinidentity}.", "In particular, is $e^{ix}$ is unimodular for real $x$;", "the values lie on the unit circle.", "A square of a real number is always nonnegative:", "\\begin{equation*}", "{\\bigl(\\sin(x)\\bigr)}^2 = 1-{\\bigl(\\cos(x)\\bigr)}^2 \\leq 1 .", "\\end{equation*}", "So $\\sabs{\\sin(x)} \\leq 1$ and similarly", "$\\sabs{\\cos(x)} \\leq 1$.", "We leave the computation of the derivatives to the reader as exercises.", "Let us prove that $\\sin(x) \\leq x$ for $x \\geq 0$.", "Consider", "$f(x) \\coloneqq x-\\sin(x)$ and differentiate:", "\\begin{equation*}", "f'(x) = \\frac{d}{dx} \\bigl[ x - \\sin(x) \\bigr]", "=", "1 -\\cos(x) \\geq 0 ,", "\\end{equation*}", "for all $x \\in \\R$ as $\\sabs{\\cos(x)} \\leq 1$.", "In other words, $f$ is increasing and $f(0) = 0$.", "So $f$ must be nonnegative when $x \\geq 0$ and hence, $\\sin(x) \\geq x$.", "Next, we claim there exists a positive $x$ such that $\\cos(x) = 0$.", "As $\\cos(0) = 1 > 0$, $\\cos(x) > 0$", "for $x$ near $0$. Namely,", "there is some", "$y > 0$, such that $\\cos(x) > 0$ on $[0,y)$.", "Then $\\sin(x)$ is strictly", "increasing on $[0,y)$. As $\\sin(0) = 0$, then", "$\\sin(x) > 0$ for $x \\in (0,y)$. Take $a \\in (0,y)$. By", "the mean value theorem, there is a $c \\in (a,y)$ such that", "\\begin{equation*}", "2 \\geq \\cos(a)-\\cos(y) = \\sin(c)(y-a) \\geq \\sin(a)(y-a) .", "\\end{equation*}", "As $a \\in (0,y)$, then $\\sin(a) > 0$ and so", "\\begin{equation*}", "y \\leq \\frac{2}{\\sin(a)} + a .", "\\end{equation*}", "Hence there is some largest $y$ such that $\\cos(x) > 0$ in $[0,y)$,", "and let $y$ be the largest such number.", "By continuity, $\\cos(y) = 0$.", "In fact, $y$ is the", "smallest positive $y$ such that $\\cos(y) = 0$. As mentioned,", "$\\pi$ is defined to be $2y$.", "As $\\cos(\\nicefrac{\\pi}{2}) = 0$, then", "${\\bigl(\\sin(\\nicefrac{\\pi}{2})\\bigr)}^2 = 1$.", "As $\\sin$ is positive on $(0,\\nicefrac{\\pi}{2})$, we have", "$\\sin(\\nicefrac{\\pi}{2}) = 1$.", "Hence,", "\\begin{equation*}", "e^{i \\pi /2} = i ,", "\\end{equation*}", "and by the law of exponents,", "\\begin{equation*}", "e^{i \\pi} = -1 ,", "\\qquad", "e^{i 2\\pi} = 1 .", "\\end{equation*}", "So $e^{i2\\pi} = 1 = e^0$. The law of exponents also says", "\\begin{equation*}", "e^{z+i2\\pi} = e^z e^{i2\\pi} = e^z", "\\end{equation*}", "for all $z \\in \\C$. Immediately, we also obtain", "$\\cos(z+2\\pi) = \\cos(z)$ and $\\sin(z+2\\pi) = \\sin(z)$.", "So $\\sin$ and $\\cos$ are $2\\pi$-periodic.", "We claim that $\\sin$ and $\\cos$ are not periodic with a smaller period. It", "would suffice to show that if $e^{ix} = 1$ for the", "smallest positive $x$, then", "$x = 2\\pi$. Let $x$ be the smallest positive $x$ such that", "$e^{ix} = 1$.", "Of course, $x \\leq 2\\pi$.", "By the law of exponents,", "\\begin{equation*}", "{\\bigl(e^{ix/4}\\bigr)}^4 = 1 .", "\\end{equation*}", "If $e^{ix/4} = a+ib$, then", "\\begin{equation*}", "{(a+ib)}^4", "=a^4-6a^2b^2+b^4 + i\\bigl(4ab(a^2-b^2)\\bigr)", "=1 .", "\\end{equation*}", "Then either $a = 0$ or $a^2 = b^2$.", "As $\\nicefrac{x}{4} \\leq \\nicefrac{\\pi}{2}$, then $a = \\cos(\\nicefrac{x}{4}) \\geq 0$ and", "$b = \\sin(\\nicefrac{x}{4}) > 0$.", "If $a^2=b^2$, then", "$a^4-6a^2b^2+b^4 = -4a^4 < 0$ and in particular not equal to 1.", "Therefore $a=0$, in which case $\\nicefrac{x}{4} = \\nicefrac{\\pi}{2}$.", "Hence $2\\pi$ is the smallest period we could choose for $e^{ix}$", "and so also for $\\cos$ and $\\sin$.", "Finally, we wish to show that $e^{ix}$ is one-to-one and onto", "from the set $[0,2\\pi)$ to the set of $z \\in \\C$ such that", "$\\sabs{z} = 1$. Suppose $e^{ix} = e^{iy}$ and", "$x > y$. Then", "$e^{i(x-y)} = 1$, meaning $x-y$ is a multiple of $2\\pi$ and hence", "only one of them can live in $[0,2\\pi)$.", "To show onto, pick $(a,b) \\in \\R^2$ such that $a^2+b^2 = 1$.", "Suppose first that $a,b \\geq 0$. By the intermediate value theorem,", "there must exist an $x \\in [0,\\nicefrac{\\pi}{2}]$ such that", "$\\cos(x) = a$, and hence $b^2 = \\bigl(\\sin(x)\\bigr)^2$. As", "$b$ and $\\sin(x)$ are nonnegative, $b = \\sin(x)$.", "Since $-\\sin(x)$ is the derivative of $\\cos(x)$", "and $\\cos(-x) = \\cos(x)$, then $\\sin(x) < 0$ for $x \\in [\\nicefrac{-\\pi}{2},0)$.", "Using the same reasoning, we obtain that", "if $a > 0$ and $b \\leq 0$, we can find an $x$ in $[\\nicefrac{-\\pi}{2},0)$,", "and by periodicity,", "$x \\in [\\nicefrac{3\\pi}{2},2\\pi)$ such that $\\cos(x) = a$ and $\\sin(x)=b$.", "Multiplying by $-1$ is the same as multiplying by $e^{i\\pi}$ or", "$e^{-i\\pi}$. So we can always assume that $a \\geq 0$ (details are left", "as exercise)." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 29, "type": "lemma", "label": "Lebl-contfunc:lemma:polyalwaysgetssmaller", "categories": [ "auxiliary result", "series" ], "title": "Existence of Complex Numbers", "contents": [ "Let$\\epsilon > 0$, let$p(z)$be a nonconstant complex polynomial, or more generally a nonconstant power series converging in$B(z_0,\\epsilon)$, and suppose$p(z_0) \\not= 0$. Then there exists a$w \\in B(z_0,\\epsilon)$such that$\\sabs{p(w)} < \\sabs{p(z_0)}$." ], "refs": [], "proofs": [ { "contents": [ "We prove this lemma for a polynomial and leave the general case as", "\\exerciseref{exercise:minprinciple}.", "Without loss of generality assume that $z_0 = 0$ and $p(0) = 1$. Write", "\\begin{equation*}", "p(z) = 1+a_kz^k + a_{k+1}z^{k+1} + \\cdots + a_d z^d ,", "\\end{equation*}", "where $a_k \\not= 0$. Pick $t$ such that $a_k e^{ikt} = -\\sabs{a_k}$, which", "we can do by the discussion on trigonometric functions. Suppose", "$r > 0$ is small enough such that", "$1-r^k \\sabs{a_k} > 0$. We have", "\\begin{equation*}", "p(r e^{it}) =", "1-r^k \\sabs{a_k} + r^{k+1}a_{k+1}e^{i(k+1)t} + \\cdots + r^{d}a_{d}e^{idt} .", "\\end{equation*}", "So", "\\begin{equation*}", "\\begin{split}", "\\abs{", "p(r e^{it}) } - \\abs{", "r^{k+1}a_{k+1}e^{i(k+1)t} + \\cdots + r^{d}a_{d}e^{idt}", "}", "& \\leq", "\\abs{", "p(r e^{it})", "- r^{k+1}a_{k+1}e^{i(k+1)t} - \\cdots - r^{d}a_{d}e^{idt}", "}", "\\\\", "& =", "\\abs{", "1-r^k \\sabs{a_k}", "}", "=", "1-r^k \\sabs{a_k} .", "\\end{split}", "\\end{equation*}", "In other words,", "\\begin{equation*}", "\\abs{", "p(r e^{it}) }", "\\leq", "1-r^k \\left( \\sabs{a_k}", "-", "r", "\\abs{", "a_{k+1}e^{i(k+1)t} + \\cdots + r^{d-k-1}a_{d}e^{idt}", "}", "\\right) .", "\\end{equation*}", "For small enough $r$, the expression in the parentheses is positive", "as $\\sabs{a_k} > 0$. Hence, $\\abs{p(re^{it})} < 1 = p(0)$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 30, "type": "theorem", "label": "Lebl-contfunc:30", "categories": [ "topology", "complex analysis" ], "title": "Maximum Modulus Principle", "contents": [ "\\index{maximum modulus principle}% \\index{maximum principle!analytic functions}% \\label{thm:maxprinciple} If$U \\subset \\C$is open and connected,$f \\colon U \\to \\C$is analytic, and$\\sabs{f(z)}$attains a relative maximum at$z_0 \\in U$, then$f$is constant." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 31, "type": "lemma", "label": "Lebl-contfunc:31", "categories": [ "auxiliary result" ], "title": "Existence of Complex Numbers", "contents": [ "Let$p(z)$be a nonconstant complex polynomial. Then for an$M > 0$, there exists an$R > 0$such that$\\sabs{p(z)} \\geq M$whenever$\\sabs{z} \\geq R$." ], "refs": [], "proofs": [ { "contents": [ "Write $p(z) = a_0 + a_1 z + \\cdots + a_d z^d$ and suppose that $d \\geq 1$", "and $a_d \\not= 0$.", "Suppose $\\sabs{z} \\geq R$ (so also $\\sabs{z}^{-1} \\leq R^{-1}$).", "We estimate:", "\\begin{equation*}", "\\begin{split}", "\\sabs{p(z)}", "& \\geq", "\\sabs{a_d z^d} -", "\\sabs{a_0} - \\sabs{a_1 z} - \\cdots - \\sabs{a_{d-1} z^{d-1} }", "\\\\", "& =", "\\sabs{z}^d \\bigl(", "\\sabs{a_d} -", "\\sabs{a_0} \\, \\sabs{z}^{-d} -", "\\sabs{a_1} \\, \\sabs{z}^{-d+1} - \\cdots - \\sabs{a_{d-1}} \\, \\sabs{z}^{-1}", "\\bigr)", "\\\\", "& \\geq", "R^d \\bigl(\\sabs{a_d} -", "\\sabs{a_0}R^{-d} - \\sabs{a_1}R^{1-d} - \\cdots - \\sabs{a_{d-1}}R^{-1} \\bigr)", ".", "\\end{split}", "\\end{equation*}", "Then the expression in parentheses is eventually positive for large enough", "$R$. In particular, for large enough $R$ we get that this expression", "is greater than", "$\\frac{\\sabs{a_d}}{2}$, and so", "\\begin{equation*}", "\\sabs{p(z)}", "\\geq", "R^d \\frac{\\sabs{a_d}}{2} .", "\\end{equation*}", "Therefore,", "we can pick $R$ large enough to be bigger than a given $M$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 32, "type": "theorem", "label": "Lebl-contfunc:32", "categories": [], "title": "Fundamental Theorem of Algebra", "contents": [ "\\index{fundamental theorem of algebra}% Let$p(z)$be a nonconstant complex polynomial, then there exists a$z_0 \\in \\C$such that$p(z_0) = 0$." ], "refs": [], "proofs": [ { "contents": [ "Let $\\mu \\coloneqq \\inf \\bigl\\{ \\sabs{p(z)} : z \\in \\C \\bigr\\}$. Find an $R$ such that", "for all $z$ with $\\sabs{z} \\geq R$, we have $\\sabs{p(z)} \\geq \\mu+1$.", "Therefore, every $z$ with $\\sabs{p(z)}$ close to $\\mu$ must be in the", "closed ball $C(0,R) = \\bigl\\{ z \\in \\C : \\sabs{z} \\leq R \\bigr\\}$. As $\\sabs{p(z)}$", "is a continuous real-valued function, it achieves its minimum", "on the compact set $C(0,R)$ (closed and bounded) and this minimum must", "be $\\mu$. So there is a $z_0 \\in C(0,R)$ such that $\\sabs{p(z_0)} = \\mu$.", "As that is a minimum of $\\sabs{p(z)}$ on $\\C$, then by the first lemma", "above, we have $\\sabs{p(z_0)} = 0$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 33, "type": "example", "label": "Lebl-contfunc:33", "categories": [ "continuity", "example" ], "title": "Continuity of Continuity", "contents": [ "There exist sequences of continuous functions on$[0,1]$that are uniformly bounded but contain no subsequence converging even pointwise. Let us state without proof that$f_n(x) \\coloneqq \\sin (2\\pi n x)$is one such sequence. Below we will show that there must always exist a subsequence converging at countably many points, but$[0,1]$is uncountable." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 34, "type": "example", "label": "Lebl-contfunc:34", "categories": [ "continuity", "series", "example" ], "title": "Continuity of Convergence", "contents": [ "The sequence$f_n(x) \\coloneqq x^n$of continuous functions on$[0,1]$is uniformly bounded, but contains no subsequence that converges uniformly, although the sequence converges pointwise (to a discontinuous function)." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 35, "type": "example", "label": "Lebl-contfunc:35", "categories": [ "series", "example", "approximation" ], "title": "Boundedness of Convergence", "contents": [ "The sequence$\\{ f_n \\}_{n=1}^\\infty$of functions in$C\\bigl([0,1],\\R\\bigr)$given by$f_n(x) \\coloneqq \\frac{n^3x}{1+n^4x^2}$converges pointwise to the zero function (obvious at$x=0$, and for$x > 0$, we have$\\frac{n^3x}{1+n^4x^2} \\leq \\frac{1}{nx}$). As for each$x$,$\\{f_n(x)\\}_{n=1}^\\infty$converges to 0, it is bounded so$\\{ f_n \\}_{n=1}^\\infty$is pointwise bounded.", "Via calculus, we find that the maximum of$f_n$on$[0,1]$occurs at the critical point$x=\\nicefrac{1}{n^2}$:", "\\begin{equation*}\n\\snorm{f_n}_{[0,1]}\n=\nf_n\\left(\\nicefrac{1}{n^2}\\right)\n= \\nicefrac{n}{2} .\n\\end{equation*}", "So$\\lim_{n\\to\\infty} \\snorm{f_n}_{[0,1]} = \\infty$, and this sequence is not uniformly bounded." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 36, "type": "proposition", "label": "Lebl-contfunc:prop:subsequenceoncountableX", "categories": [ "series" ], "title": "Boundedness of Convergence", "contents": [ "Let$X$be a countable set and$f_n \\colon X \\to \\C$give a pointwise bounded sequence of functions. Then$\\{ f_n \\}_{n=1}^\\infty$has a subsequence that converges pointwise." ], "refs": [], "proofs": [ { "contents": [ "Let $x_1,x_2,x_3,\\ldots$ be an enumeration of the elements of $X$.", "The sequence $\\{ f_n(x_1) \\}_{n=1}^\\infty$ is bounded and hence", "we have a subsequence of $\\{ f_n \\}_{n=1}^{\\infty}$, which we denote by", "$\\{ f_{1,k} \\}_{k=1}^\\infty$,", "such that", "$\\{ f_{1,k}(x_1) \\}_{k=1}^\\infty$ converges.", "Next $\\{ f_{1,k}(x_2) \\}_{k=1}^\\infty$ is bounded and so", "$\\{ f_{1,k} \\}_{k=1}^\\infty$ has a subsequence", "$\\{ f_{2,k} \\}_{k=1}^\\infty$ such that", "$\\{ f_{2,k}(x_2) \\}_{k=1}^\\infty$ converges. Note that", "$\\{ f_{2,k}(x_1) \\}_{k=1}^\\infty$ is still convergent.", "In general, we have a sequence $\\{ f_{m,k} \\}_{k=1}^\\infty$,", "which is a subsequence of $\\{ f_{m-1,k} \\}_{k=1}^\\infty$,", "such that $\\{ f_{m,k}(x_j) \\}_{k=1}^\\infty$ converges for $j=1,2,\\ldots, m$.", "We let $\\{ f_{m+1,k} \\}_{k=1}^\\infty$ be a subsequence of", "$\\{ f_{m,k} \\}_{k=1}^\\infty$", "such that", "$\\{ f_{m+1,k}(x_{m+1}) \\}_{k=1}^\\infty$ converges (and hence it converges for all", "$x_j$ for $j=1,2,\\ldots,m+1$). Rinse and repeat.", "If $X$ is finite, we are done as the process stops at some point.", "If $X$ is countably infinite,", "we pick the sequence", "$\\{ f_{k,k} \\}_{k=1}^\\infty$.", "This is a subsequence of the original sequence $\\{ f_n \\}_{n=1}^\\infty$.", "For every $m$, the tail $\\{ f_{k,k} \\}_{k=m}^\\infty$ is a subsequence of", "$\\{ f_{m,k} \\}_{k=1}^\\infty$", "and hence for any $m$ the sequence $\\{ f_{k,k}(x_m) \\}_{k=1}^\\infty$ converges." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 37, "type": "proposition", "label": "Lebl-contfunc:37", "categories": [ "topology", "continuity", "series" ], "title": "Continuity of Convergence", "contents": [ "Suppose$(X,d)$is a compact metric space,$f_n \\in C(X,\\C)$, and$\\{ f_n \\}_{n=1}^\\infty$converges uniformly, then$\\{ f_n \\}_{n=1}^\\infty$is uniformly equicontinuous." ], "refs": [], "proofs": [ { "contents": [ "Let $\\epsilon > 0$ be given.", "As $\\{ f_n \\}_{n=1}^\\infty$ converges uniformly, there is an $N \\in \\N$ such that for", "all $n \\geq N$", "\\begin{equation*}", "\\sabs{f_n(x)-f_N(x)} < \\nicefrac{\\epsilon}{3} \\qquad \\text{for all } x \\in X.", "\\end{equation*}", "As $X$ is compact, every continuous function is uniformly continuous.", "So $\\{ f_1,f_2,\\ldots,f_N \\}$ is a finite set of uniformly continuous", "functions. And so, as we mentioned above, the set is uniformly equicontinuous.", "Hence there is a $\\delta > 0$ such that", "\\begin{equation*}", "\\sabs{f_j(x)-f_j(y)} < \\nicefrac{\\epsilon}{3} < \\epsilon", "\\end{equation*}", "whenever $d(x,y) < \\delta$ and $1 \\leq j \\leq N$.", "Take $n > N$. For $d(x,y) < \\delta$, we have", "\\begin{equation*}", "\\sabs{f_n(x)-f_n(y)}", "\\leq", "\\sabs{f_n(x)-f_N(x)}", "+", "\\sabs{f_N(x)-f_N(y)}", "+", "\\sabs{f_N(y)-f_n(y)}", "<", "\\nicefrac{\\epsilon}{3}", "+", "\\nicefrac{\\epsilon}{3}", "+", "\\nicefrac{\\epsilon}{3}", "=\\epsilon . \\qedhere", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 38, "type": "proposition", "label": "Lebl-contfunc:38", "categories": [ "topology" ], "title": "Existence of Analysis", "contents": [ "A compact metric space$(X,d)$contains a countable dense subset, that is, there exists a countable$D \\subset X$such that$\\widebar{D} = X$." ], "refs": [], "proofs": [ { "contents": [ "For each $n \\in \\N$ there are finitely many", "balls of radius $\\nicefrac{1}{n}$ that cover $X$ (as $X$ is compact). That is,", "for every $n$, there exists", "a finite set of points $x_{n,1},x_{n,2},\\ldots,x_{n,k_n}$ such that", "\\begin{equation*}", "X= \\bigcup_{j=1}^{k_n} B(x_{n,j},\\nicefrac{1}{n}) .", "\\end{equation*}", "Let $D \\coloneqq \\bigcup_{n=1}^\\infty \\{ x_{n,1},x_{n,2},\\ldots,x_{n,k_n} \\}$.", "The set $D$ is countable as it is a countable union of finite sets.", "For every $x \\in X$", "and every $\\epsilon > 0$, there exists an $n$ such that", "$\\nicefrac{1}{n} < \\epsilon$ and an $x_{n,j} \\in D$ such that", "\\begin{equation*}", "x \\in B(x_{n,j},\\nicefrac{1}{n}) \\subset B(x_{n,j},\\epsilon) .", "\\end{equation*}", "Hence $x \\in \\widebar{D}$, so $\\widebar{D} = X$, and $D$ is dense." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 39, "type": "theorem", "label": "Lebl-contfunc:39", "categories": [ "topology", "continuity", "series" ], "title": "Continuity of Convergence", "contents": [ "\\index{Arzel\\`a--Ascoli theorem} \\label{thm:arzelaascoli} Let$(X,d)$be a compact metric space, and let$\\{ f_n \\}_{n=1}^\\infty$be pointwise bounded and uniformly equicontinuous sequence of functions$f_n \\in C(X,\\C)$. Then$\\{f_n\\}_{n=1}^\\infty$is uniformly bounded and$\\{ f_n \\}_{n=1}^\\infty$contains a uniformly convergent subsequence." ], "refs": [], "proofs": [ { "contents": [ "We first show that the sequence is uniformly bounded.", "By uniform equicontinuity,", "there is a $\\delta > 0$", "such that", "for all $x \\in X$ and all $n \\in \\N$,", "\\begin{equation*}", "B(x,\\delta) \\subset f_n^{-1}\\bigl(B(f_n(x),1)\\bigr) .", "\\end{equation*}", "The space $X$ is compact, so there exist $x_1,x_2,\\ldots,x_k$", "such that", "\\begin{equation*}", "X = \\bigcup_{j=1}^k B(x_j,\\delta) .", "\\end{equation*}", "As $\\{ f_n \\}_{n=1}^\\infty$ is pointwise bounded there exist $M_1,M_2,\\ldots,M_k$", "such that for $j=1,2,\\ldots,k$,", "\\begin{equation*}", "\\sabs{f_n(x_j)} \\leq M_j \\qquad \\text{for all } n.", "\\end{equation*}", "Let $M \\coloneqq 1+ \\max \\{ M_1,M_2,\\ldots,M_k \\}$. Given any", "$x \\in X$, there is a $j$ such that $x \\in B(x_j,\\delta)$. Therefore,", "for all $n$, we have", "$x \\in f_n^{-1}\\bigl(B(f_n(x_j),1)\\bigr)$, or in other words", "\\begin{equation*}", "\\sabs{f_n(x)-f_n(x_j)} < 1 .", "\\end{equation*}", "By the reverse triangle inequality,", "\\begin{equation*}", "\\sabs{f_n(x)} < 1+ \\sabs{f_n(x_j)} \\leq 1+M_j \\leq M .", "\\end{equation*}", "As $x$ was arbitrary, $\\{f_n\\}_{n=1}^\\infty$ is uniformly bounded.", "Next, pick a countable dense subset $D \\subset X$.", "By \\propref{prop:subsequenceoncountableX}, we find", "a subsequence $\\{ f_{n_j} \\}_{j=1}^\\infty$ that converges pointwise on $D$.", "Write $g_j \\coloneqq f_{n_j}$ for simplicity.", "The sequence $\\{ g_n \\}_{n=1}^\\infty$ is", "uniformly equicontinuous.", "Let $\\epsilon > 0$ be given, then there exists a $\\delta > 0$", "such that for all $x \\in X$ and all $n \\in \\N$,", "\\begin{equation*}", "B(x,\\delta) \\subset g_n^{-1}\\bigl(B(g_n(x),\\nicefrac{\\epsilon}{3})\\bigr).", "\\end{equation*}", "By density of $D$ and because $\\delta$ is fixed, every $x \\in X$ is in $B(y,\\delta)$", "for some $y \\in D$. By compactness of $X$,", "there is a finite subset $\\{ x_1,x_2,\\ldots,x_k \\} \\subset D$", "such that", "\\begin{equation*}", "X = \\bigcup_{j=1}^k B(x_j,\\delta) .", "\\end{equation*}", "As $\\{ x_1,x_2,\\ldots,x_k \\}$", "is a finite set and $\\{ g_n \\}_{n=1}^\\infty$", "converges pointwise on $D$, there exists a single $N$ such that for", "all $n,m \\geq N$,", "\\begin{equation*}", "\\sabs{g_n(x_j)-g_m(x_j)} < \\nicefrac{\\epsilon}{3}", "\\qquad \\text{for all } j=1,2,\\ldots,k.", "\\end{equation*}", "Let $x \\in X$ be arbitrary. There is some $j$ such that", "$x \\in B(x_j,\\delta)$ and so for all $\\ell \\in \\N$,", "\\begin{equation*}", "\\sabs{g_\\ell(x)-g_\\ell(x_j)} < \\nicefrac{\\epsilon}{3}.", "\\end{equation*}", "So for $n,m \\geq N$,", "\\begin{equation*}", "\\begin{split}", "\\sabs{g_n(x)-g_m(x)} & \\leq", "\\sabs{g_n(x)-g_n(x_j)} +", "\\sabs{g_n(x_j)-g_m(x_j)} +", "\\sabs{g_m(x_j)-g_m(x)}", "\\\\", "& <", "\\nicefrac{\\epsilon}{3} +", "\\nicefrac{\\epsilon}{3} +", "\\nicefrac{\\epsilon}{3} = \\epsilon .", "\\end{split}", "\\end{equation*}", "Hence, $\\{ g_n \\}_{n=1}^\\infty$ is uniformly Cauchy. By completeness of $\\C$,", "it is uniformly convergent.", "%FIXME: reference?" ], "refs": [ "prop:subsequenceoncountableX" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 40, "type": "corollary", "label": "Lebl-contfunc:40", "categories": [ "consequence", "continuity", "topology" ], "title": "Continuity of Continuity", "contents": [ "Let$(X,d)$be a compact metric space. Let$S \\subset C(X,\\C)$be a closed, bounded and uniformly equicontinuous set. Then$S$is compact." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 41, "type": "corollary", "label": "Lebl-contfunc:41", "categories": [ "consequence", "series", "differentiability" ], "title": "Boundedness of Convergence", "contents": [ "Suppose$\\{ f_n \\}_{n=1}^\\infty$is a sequence of differentiable functions on$[a,b]$,$\\{ f_n' \\}_{n=1}^\\infty$is uniformly bounded, and there is an$x_0 \\in [a,b]$such that$\\{ f_n(x_0) \\}_{n=1}^\\infty$is bounded. Then there exists a uniformly convergent subsequence$\\{ f_{n_j} \\}_{j=1}^\\infty$." ], "refs": [], "proofs": [ { "contents": [ "The trick is to use the mean value theorem. If $M$ is the uniform bound on", "$\\{ f_n' \\}_{n=1}^\\infty$, then by the mean value theorem for every $n$", "\\begin{equation*}", "\\sabs{f_n(x)-f_n(y)} \\leq M \\sabs{x-y} \\qquad \\text{for all } x,y \\in X.", "\\end{equation*}", "All the $f_n$ are Lipschitz with the same constant and hence", "the sequence is", "uniformly equicontinuous.", "Suppose $\\sabs{f_n(x_0)} \\leq M_0$ for all $n$.", "For all $x \\in [a,b]$,", "\\begin{equation*}", "\\sabs{f_n(x)} \\leq \\sabs{f_n(x_0)}+ \\sabs{f_n(x)-f_n(x_0)} \\leq M_0+ M \\sabs{x-x_0}", "\\leq M_0 + M(b-a) .", "\\end{equation*}", "So $\\{ f_n \\}_{n=1}^\\infty$ is uniformly bounded.", "We apply \\hyperref[thm:arzelaascoli]{Arzel\\`a--Ascoli} to find the subsequence." ], "refs": [ "thm:arzelaascoli" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 42, "type": "theorem", "label": "Lebl-contfunc:42", "categories": [ "continuity", "named theorem", "approximation" ], "title": "Weierstrass Approximation Theorem", "contents": [ "\\index{Weierstrass approximation theorem} If$f \\colon [a,b] \\to \\C$is continuous, then there exists a sequence$\\{ p_n \\}_{n=1}^\\infty$of polynomials converging to$f$uniformly on$[a,b]$. Furthermore, if$f$is real-valued, we can find$p_n$with real coefficients." ], "refs": [], "proofs": [ { "contents": [ "For $x \\in [0,1]$,", "define", "\\begin{equation*}", "g(x) \\coloneqq f\\bigl((b-a)x+a\\bigr)-f(a) - x\\bigl(f(b)-f(a)\\bigr) .", "\\end{equation*}", "If we prove the theorem for $g$ and find the sequence", "$\\{ p_n \\}_{n=1}^\\infty$ for $g$,", "it is proved for $f$ as we simply", "composed with an invertible affine function and added an affine", "function to $f$: We reverse the process and apply that to our", "$p_n$, to obtain polynomials approximating $f$.", "The function $g$ is defined on $[0,1]$ and $g(0)=g(1)=0$.", "For simplicity, assume that", "$g$ is defined on $\\R$ by letting", "$g(x) \\coloneqq 0$ if $x < 0$ or $x > 1$. This extended $g$ is continuous.", "Define", "\\begin{equation*}", "c_n \\coloneqq {\\left( \\int_{-1}^1 {(1-x^2)}^n\\,dx \\right)}^{-1} ,", "\\qquad", "q_n(x) \\coloneqq c_n (1-x^2)^n .", "\\end{equation*}", "The choice of $c_n$ is", "so that $\\int_{-1}^1 q_n(x)\\,dx = 1$.", "See \\figureref{fig:weierqn}.", "\\begin{myfigureht}", "\\includegraphics{figures/weierqn}", "\\caption{Plot of the approximate delta functions $q_n$ on $[-1,1]$ for", "$n=5,10,15,20,\\ldots,100$ with higher $n$ in lighter shade.\\label{fig:weierqn}}", "\\end{myfigureht}", "The functions $q_n$ are peaks around 0 (ignoring what happens outside", "of $[-1,1]$) that get narrower and taller as $n$ increases,", "while the area underneath is always 1.", "A classic approximation idea", "is to do a \\emph{\\myindex{convolution}} integral with peaks like this:", "For", "for $x \\in [0,1]$, let", "\\begin{equation*}", "p_n(x) \\coloneqq \\int_{0}^1 g(t)q_n(x-t) \\,dt \\quad \\left( = \\int_{-\\infty}^\\infty", "g(t)q_n(x-t) \\,dt \\right) .", "\\end{equation*}", "The idea of this convolution is that we do a \\myquote{weighted average} of the", "function $g$ around the point $x$ using $q_n$ as the weight.", "See \\figureref{fig:approxdeltaconv}.", "\\begin{myfigureht}", "\\includegraphics{figures/approxdeltaconv}", "\\caption{For $x=0.3$, the plot of $q_{100}(x-t)$ (light gray peak centered", "at $x$), some continuous function", "$g(t)$ (the jagged line) and the product $g(t)q_{100}(x-t)$ (the bold line).\\label{fig:approxdeltaconv}}", "\\end{myfigureht}", "As $q_n$ is a narrow peak, the integral", "mostly sees the values of $g$ that are", "close to $x$ and it does the weighted average of them.", "When the peak gets narrower, we compute this average closer to $x$", "and we expect the result to get closer to the value of $g(x)$. Really, we are", "approximating what is called a delta function\\footnote{The delta function", "is not actually a function,", "it is a \\myquote{thing} that should give", "\\myquote{$\\int_{-\\infty}^\\infty g(t) \\delta(x-t) \\, dt = g(x)$.}}", "(don't worry if you have not", "heard of this concept),", "and functions like $q_n$ are often called", "\\emph{approximate delta functions}\\index{approximate delta function}.", "We could do this with any set of polynomials that look like narrower", "and narrower peaks near zero. These just happen to be the simplest ones.", "We only need this behavior on $[-1,1]$ as the convolution sees nothing", "further than this as $g$ is zero outside $[0,1]$.", "Because $q_n$ is a polynomial, we write", "\\begin{equation*}", "q_n(x-t) = a_0(t) + a_1(t)\\,x + \\cdots + a_{2n}(t)\\, x^{2n} ,", "\\end{equation*}", "where $a_k(t)$ are polynomials in $t$, and hence", "integrable functions.", "So", "\\begin{equation*}", "\\begin{split}", "p_n(x) & =", "\\int_{0}^1 g(t)q_n(x-t) \\,dt", "\\\\", "&=", "\\left(", "\\int_0^1", "g(t)", "a_0(t)\\,dt", "\\right)", "+", "\\left(", "\\int_0^1", "g(t)", "a_1(t)\\,dt", "\\right)", "\\,", "x", "+", "\\cdots", "+", "\\left(", "\\int_0^1", "g(t)", "a_{2n}(t)\\,dt", "\\right)", "\\,", "x^{2n} .", "\\end{split}", "\\end{equation*}", "In other words, $p_n$ is a polynomial%", "\\footnote{%", "Do note that the functions $a_j$ depend on $n$, so the coefficients of $p_n$", "change as $n$ changes.}", "in $x$.", "If $g(t)$ is real-valued, then the functions $g(t)a_j(t)$ are", "real-valued and $p_n$ has real coefficients,", "proving the \\myquote{furthermore} part of the theorem.", "We still need to prove that $\\{ p_n \\}_{n=1}^\\infty$ converges to $g$.", "We start with estimating the size of~$c_n$.", "For $x \\in [0,1]$, we have that $1-x \\leq 1-x^2$. We estimate", "\\begin{equation*}", "\\begin{split}", "c_n^{-1} = \\int_{-1}^1 {(1-x^2)}^n \\, dx", "& = 2\\int_0^1 {(1-x^2)}^n \\, dx \\\\", "& \\geq 2\\int_0^{1} {(1-x)}^n \\, dx", "= \\frac{2}{n+1} .", "\\end{split}", "\\end{equation*}", "So $c_n \\leq \\frac{n+1}{2} \\leq n$.", "Let us see how small $q_n$ is if we ignore some small interval around the origin,", "where the peak is.", "Given any $\\delta > 0$, $\\delta < 1$,", "we have that for all", "$x$ such that $\\delta \\leq \\sabs{x} \\leq 1$,", "\\begin{equation*}", "q_n(x) \\leq c_n {(1-\\delta^2)}^n \\leq n{(1-\\delta^2)}^n ,", "\\end{equation*}", "because $q_n$ is increasing on $[-1,0]$ and decreasing on $[0,1]$.", "By the ratio test,", "$n{(1-\\delta^2)}^n$ goes to 0 as $n$ goes to infinity.", "The function $q_n$ is even, $q_n(t) = q_n(-t)$, and $g$", "is zero outside of $[0,1]$.", "So for $x \\in [0,1]$,", "\\begin{equation*}", "p_n(x) =", "\\int_{0}^1 g(t)q_n(x-t) \\, dt", "=", "\\int_{-x}^{1-x} g(x+t)q_n(-t) \\, dt", "=", "\\int_{-1}^{1} g(x+t)q_n(t) \\, dt .", "\\end{equation*}", "Let $\\epsilon > 0$ be given.", "As $[-1,2]$ is compact and $g$ is continuous on $[-1,2]$, we have that $g$ is uniformly continuous.", "Pick $0 < \\delta < 1$ such that if", "$\\sabs{x-y} < \\delta$ (and $x,y \\in [-1,2]$), then", "\\begin{equation*}", "\\sabs{g(x)-g(y)} < \\frac{\\epsilon}{2} .", "\\end{equation*}", "Let $M$ be such that $\\sabs{g(x)} \\leq M$ for all $x$. Let $N$ be", "such that for all $n \\geq N$,", "\\begin{equation*}", "4M n{(1-\\delta^2)}^n < \\frac{\\epsilon}{2} .", "\\end{equation*}", "Note that", "$\\int_{-1}^1 q_n(t) \\, dt = 1$ and $q_n(t) \\geq 0$ on $[-1,1]$. So for $n", "\\geq N$ and every $x \\in [0,1]$,", "\\begin{align*}", "\\sabs{p_n(x)-g(x)} & =", "\\abs{\\int_{-1}^1 g(x+t)q_n(t) \\, dt", "-g(x)\\int_{-1}^1 q_n(t) \\, dt} \\\\", "& =", "\\abs{\\int_{-1}^1 \\bigl(g(x+t)-g(x)\\bigr)q_n(t) \\, dt}", "\\displaybreak[0]\\\\", "& \\leq", "\\int_{-1}^1 \\sabs{g(x+t)-g(x)} q_n(t) \\, dt", "\\displaybreak[0]\\\\", "& =", "\\int_{-1}^{-\\delta} \\sabs{g(x+t)-g(x)} q_n(t) \\, dt", "\\quad +", "\\int_{-\\delta}^{\\delta} \\sabs{g(x+t)-g(x)} q_n(t) \\, dt", "\\\\", "& \\phantom{\\leq} +", "\\int_{\\delta}^1 \\sabs{g(x+t)-g(x)} q_n(t) \\, dt", "\\displaybreak[0]\\\\", "& \\leq", "2M", "\\int_{-1}^{-\\delta} q_n(t) \\, dt", "\\quad", "+", "\\quad", "\\frac{\\epsilon}{2}", "\\int_{-\\delta}^{\\delta} q_n(t) \\, dt", "\\quad", "+", "\\quad", "2M", "\\int_{\\delta}^1 q_n(t) \\, dt", "\\\\", "& \\leq", "2M n{(1-\\delta^2)}^n(1-\\delta)", "\\quad", "+", "\\quad", "\\frac{\\epsilon}{2}", "\\quad", "+", "\\quad", "2M n{(1-\\delta^2)}^n(1-\\delta) \\\\", "& <", "4M n{(1-\\delta^2)}^n", "+", "\\frac{\\epsilon}{2}", "< \\epsilon . \\qedhere", "\\end{align*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 43, "type": "corollary", "label": "Lebl-contfunc:43", "categories": [ "consequence" ], "title": "The metric spaces$C\\bigl([a,b],\\R\\bigr)$and$C\\bigl([a,b],\\C\\bigr)$each contain a countable dense ...", "contents": [ "The metric spaces$C\\bigl([a,b],\\R\\bigr)$and$C\\bigl([a,b],\\C\\bigr)$each contain a countable dense subset." ], "refs": [], "proofs": [ { "contents": [ "Without loss of generality, consider only", "$C\\bigl([a,b],\\R\\bigr)$", "(why?).", "Real polynomials are dense in $C\\bigl([a,b],\\R\\bigr)$ by Weierstrass.", "If we show that", "every real polynomial can be approximated by polynomials with rational", "coefficients, we are done.", "Indeed, there are only countably many", "rational numbers and so there are only countably many polynomials with", "rational coefficients (a countable union of countable sets is countable).", "Further without loss of generality, suppose $[a,b]=[0,1]$. Let", "\\begin{equation*}", "p(x) \\coloneqq \\sum_{k=0}^n a_k\\, x^k", "\\end{equation*}", "be a polynomial of degree $n$ where $a_k \\in \\R$. Given $\\epsilon > 0$, pick $b_k \\in \\Q$", "such that $\\sabs{a_k-b_k} < \\frac{\\epsilon}{n+1}$. Then", "if we let", "\\begin{equation*}", "q(x) \\coloneqq \\sum_{k=0}^n b_k \\, x^k ,", "\\end{equation*}", "we have", "\\begin{equation*}", "\\sabs{p(x)-q(x)}", "=", "\\abs{\\sum_{k=0}^n (a_k-b_k) x^k}", "\\leq", "\\sum_{k=0}^n \\sabs{a_k-b_k} x^k", "\\leq", "\\sum_{k=0}^n \\sabs{a_k-b_k}", "<", "\\sum_{k=0}^n \\frac{\\epsilon}{n+1} = \\epsilon . \\qedhere", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 44, "type": "corollary", "label": "Lebl-contfunc:44", "categories": [ "consequence", "series" ], "title": "Convergence of Convergence", "contents": [ "Let$[-a,a]$be an interval. Then there is a sequence of real polynomials$\\{ p_n \\}_{n=1}^\\infty$that converges uniformly to$\\sabs{x}$on$[-a,a]$and such that$p_n(0) = 0$for all$n$." ], "refs": [], "proofs": [ { "contents": [ "As $f(x) \\coloneqq \\sabs{x}$ is continuous and real-valued", "on $[-a,a]$, the Weierstrass theorem gives a sequence of", "real polynomials $\\{ \\widetilde{p}_n \\}_{n=1}^\\infty$ that converges to $f$", "uniformly on $[-a,a]$.", "Let", "\\begin{equation*}", "p_n(x) \\coloneqq \\widetilde{p}_n(x) - \\widetilde{p}_n(0) .", "\\end{equation*}", "Obviously $p_n(0) = 0$.", "Given $\\epsilon > 0$, let $N$ be such that", "for $n \\geq N$, we have", "$\\bigl\\lvert\\widetilde{p}_n(x)-\\sabs{x}\\big\\rvert <", "\\nicefrac{\\epsilon}{2}$ for all $x \\in [-a,a]$.", "In particular,", "$\\sabs{\\widetilde{p}_n(0)} < \\nicefrac{\\epsilon}{2}$.", "Then for $n \\geq N$,", "\\begin{equation*}", "\\bigl\\lvert p_n(x)-\\sabs{x} \\bigr\\rvert", "=", "\\bigl\\lvert \\widetilde{p}_n(x) - \\widetilde{p}_n(0) - \\sabs{x} \\bigr\\rvert", "\\leq", "\\bigl\\lvert \\widetilde{p}_n(x) - \\sabs{x} \\bigr\\rvert + \\sabs{\\widetilde{p}_n(0)} <", "\\nicefrac{\\epsilon}{2} + \\nicefrac{\\epsilon}{2} = \\epsilon . \\qedhere", "\\end{equation*}" ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 45, "type": "proposition", "label": "Lebl-contfunc:prop:closureofalgebra", "categories": [ "topology" ], "title": "Suppose$X$is a compact metric space", "contents": [ "Suppose$X$is a compact metric space. If$\\sA \\subset C(X,\\C)$is an algebra, then the closure$\\widebar{\\sA}$is also an algebra. Similarly for a real algebra in$C(X,\\R)$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 46, "type": "example", "label": "Lebl-contfunc:46", "categories": [ "example" ], "title": "Given any$X \\subset \\R$(or$X \\subset \\C$), the set$\\sP$of polynomials in one variable separates p...", "contents": [ "Given any$X \\subset \\R$(or$X \\subset \\C$), the set$\\sP$of polynomials in one variable separates points and vanishes at no point on$X$. That is,$1 \\in \\sP$, so it vanishes at no point. And for$x,y \\in X$,$x\\not= y$, take$f(t) \\coloneqq t$. Then$f(x) = x \\not= y = f(y)$. So$\\sP$separates points." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 47, "type": "example", "label": "Lebl-contfunc:47", "categories": [ "example" ], "title": "The set of functions of the form \\begin{equation*} f(t) = a_0 + \\sum_{n=1}^k a_n \\cos(nt) \\end{eq...", "contents": [ "The set of functions of the form", "\\begin{equation*}\nf(t) = a_0 + \\sum_{n=1}^k a_n \\cos(nt)\n\\end{equation*}", "is an algebra, which follows by the identity$\\cos(mt)\\cos(nt) = \\frac{\\cos((n+m) t)}{2}+ \\frac{\\cos((n-m) t)}{2}$. The algebra vanishes at no point as it contains a constant function. It does not separate points if the domain is an interval$[-a,a]$, as$f(-t) = f(t)$for all~$t$. It does separate points if the domain is$[0,\\pi]$;$\\cos(t)$is one-to-one on$[0,\\pi]$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 48, "type": "example", "label": "Lebl-contfunc:48", "categories": [ "continuity", "example" ], "title": "Continuity of Continuity", "contents": [ "The set$\\sP$of real polynomials with no constant term is an algebra that vanishes at the origin. Clearly, any function in the closure of$\\sP$also vanishes at the origin, so the closure of$\\sP$cannot be$C\\bigl([0,1],\\R\\bigr)$.", "Similarly, the set of constant functions is an algebra that does not separate points. Uniform limits of constants are constants, so we also do not obtain all continuous functions." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 49, "type": "proposition", "label": "Lebl-contfunc:prop:SWinterpolate", "categories": [], "title": "Suppose$\\sA$is an algebra of complex-valued functions on a set$X$that separates points and vanish...", "contents": [ "Suppose$\\sA$is an algebra of complex-valued functions on a set$X$that separates points and vanishes at no point. Suppose$x,y$are distinct points of$X$, and$c,d \\in \\C$. Then there is an$f \\in \\sA$such that\\begin{equation*} f(x) = c, \\qquad f(y) = d . \\avoidbreak \\end{equation*}If$\\sA$is a real algebra, the conclusion holds for$c,d \\in \\R$." ], "refs": [], "proofs": [ { "contents": [ "There must exist an $g,h,k \\in \\sA$", "such that", "$g(x) \\not= g(y)$, $h(x) \\not= 0$, $k(y) \\not= 0$.", "Let", "\\begin{equation*}", "\\begin{split}", "f & \\coloneqq", "c", "\\frac{\\bigl(g - g(y)\\bigr)h}{\\bigl(g(x)-g(y)\\bigr)h(x) } +", "d", "\\frac{\\bigl(g - g(x)\\bigr)k}{\\bigl(g(y)-g(x)\\bigr)k(y)}", "\\\\", "& =", "c", "\\frac{gh - g(y)h}{g(x)h(x)-g(y)h(x) } +", "d", "\\frac{gk - g(x)k}{g(y)k(y)-g(x)k(y)} .", "\\end{split}", "\\end{equation*}", "We are not dividing by zero (clear from the first formula).", "Also by the first formula, $f(x) = c$ and $f(y) = d$.", "By the second formula, $f \\in \\sA$ (as $\\sA$ is an algebra)." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 50, "type": "theorem", "label": "Lebl-contfunc:thm:SWreal", "categories": [ "continuity", "topology", "named theorem" ], "title": "Weierstrass Approximation Theorem", "contents": [ "% \\index{Stone--Weierstrass!real version}% Let$X$be a compact metric space and$\\sA$a real algebra of real-valued continuous functions on$X$, such that$\\sA$separates points and vanishes at no point. Then the closure$\\widebar{\\sA} = C(X,\\R)$." ], "refs": [], "proofs": [ { "contents": [ "The function $f$ is bounded (continuous on a compact set), so there is an $M$", "such that $\\sabs{f(x)} \\leq M$ for all $x \\in X$.", "Let $\\epsilon > 0$ be given. By the corollary to the Weierstrass theorem,", "there exists a real polynomial $c_1 y + c_2 y^2 + \\cdots+ c_n y^n$", "(vanishing at $y=0$) such that", "\\begin{equation*}", "\\abs{\\sabs{y} - \\sum_{k=1}^n c_k y^k} < \\epsilon", "\\end{equation*}", "for all $y \\in [-M,M]$.", "Because $\\widebar{\\sA}$ is an algebra and because there is no constant term in the", "polynomial,", "\\begin{equation*}", "\\sum_{k=1}^n c_k f^k \\in \\widebar{\\sA} .", "\\end{equation*}", "As $\\sabs{f(x)} \\leq M$, then for all $x \\in X$", "\\begin{equation*}", "\\abs{\\sabs{f(x)} - \\sum_{k=1}^n c_k {\\bigl(f(x)\\bigr)}^k}", "< \\epsilon .", "\\end{equation*}", "So $\\sabs{f}$ is in the closure of $\\widebar{\\sA}$, which is itself closed.", "In other words, $\\sabs{f} \\in \\widebar{\\sA}$." ], "refs": [], "ref_ids": [] }, { "contents": [ "Write:", "\\begin{equation*}", "\\max(f,g) = \\frac{f+g}{2} + \\frac{\\sabs{f-g}}{2} ,", "\\qquad\\text{and}\\qquad", "\\min(f,g) = \\frac{f+g}{2} - \\frac{\\sabs{f-g}}{2} .", "\\end{equation*}", "As $\\widebar{\\sA}$ is an algebra we are done." ], "refs": [], "ref_ids": [] }, { "contents": [ "Fix $f$, $x$, and $\\epsilon$.", "By \\propref{prop:SWinterpolate}, for every $y \\in X$, find an $h_y \\in", "\\sA$ such that", "\\begin{equation*}", "h_y(x) = f(x), \\qquad h_y(y)=f(y) .", "\\end{equation*}", "As $h_y$ and $f$ are continuous, the function $h_y-f$ is continuous,", "and the set", "\\begin{equation*}", "U_y \\coloneqq", "\\bigl\\{ t \\in X : h_y(t) > f(t) -\\epsilon \\bigr\\}", "=", "{(h_y-f)}^{-1} \\bigl( (-\\epsilon,\\infty) \\bigr)", "\\end{equation*}", "is open (it is the inverse image of an open set by a continuous function).", "Furthermore $y \\in U_y$. So the sets $U_y$ cover $X$.", "The space $X$ is compact,", "so there exist finitely many", "points $y_1,y_2,\\ldots,y_n$ in $X$ such", "that", "\\begin{equation*}", "X = \\bigcup_{k=1}^n U_{y_k} .", "\\end{equation*}", "Let", "\\begin{equation*}", "g_x \\coloneqq \\max(h_{y_1},h_{y_2},\\ldots,h_{y_n}) .", "\\end{equation*}", "By Claim 2, $g_x \\in \\widebar{\\sA}$. See \\figureref{fig:stonegx}.", "Moreover,", "\\begin{equation*}", "g_x(t) > f(t) -\\epsilon", "\\end{equation*}", "for all $t \\in X$, since for every $t$, there is a $y_k$ such that", "$t \\in U_{y_k}$, and so", "$h_{y_k}(t) > f(t) -\\epsilon$.", "Finally, $h_y(x) = f(x)$ for all $y \\in X$, so", "$g_x(x) = f(x)$." ], "refs": [ "prop:SWinterpolate" ], "ref_ids": [] }, { "contents": [ "For every $x \\in X$, find the function $g_x$ as in Claim 3.", "Let", "\\begin{equation*}", "V_x \\coloneqq \\bigl\\{ t \\in X : g_x(t) < f(t) + \\epsilon \\bigr\\}.", "\\end{equation*}", "The sets $V_x$ are open as $g_x$ and $f$ are continuous.", "As $g_x(x) = f(x)$, then $x \\in V_x$. So the sets $V_x$ cover $X$.", "By compactness of $X$,", "there", "are finitely many points $x_1,x_2,\\ldots,x_n$ such that", "\\begin{equation*}", "X = \\bigcup_{k=1}^n V_{x_k} .", "\\end{equation*}", "Let", "\\begin{equation*}", "\\varphi \\coloneqq \\min(g_{x_1},g_{x_2},\\ldots,g_{x_n}) .", "\\end{equation*}", "By Claim 2, $\\varphi \\in \\widebar{\\sA}$. Similarly as before (same argument as in", "Claim 3), for all $t \\in X$,", "\\begin{equation*}", "\\varphi(t) < f(t) + \\epsilon .", "\\end{equation*}", "Since all the $g_x$ satisfy $g_x(t) > f(t) - \\epsilon$ for all $t \\in X$,", "$\\varphi(t) > f(t) - \\epsilon$ as well.", "Hence, for all $t$,", "\\begin{equation*}", "-\\epsilon < \\varphi(t) - f(t) < \\epsilon ,", "\\end{equation*}", "which is the desired conclusion." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 51, "type": "example", "label": "Lebl-contfunc:51", "categories": [ "example" ], "title": "The functions of the form \\begin{equation*} f(t) = \\sum_{k=1}^n c_k \\, e^{kt}, \\end{equation*} fo...", "contents": [ "The functions of the form", "\\begin{equation*}\nf(t) = \\sum_{k=1}^n c_k \\, e^{kt},\n\\end{equation*}", "for$c_k \\in \\R$, are dense in$C\\bigl([a,b],\\R\\bigr)$. Such functions are a real algebra, which follows from$e^{kt} e^{\\ell t} = e^{(k+\\ell)t}$. They separate points as$e^t$is one-to-one. As$e^t > 0$for all$t$, the algebra does not vanish at any point." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 52, "type": "example", "label": "Lebl-contfunc:52", "categories": [ "fourier analysis", "example", "series", "continuity", "named theorem" ], "title": "Weierstrass Approximation Theorem", "contents": [ "We mentioned that the set of all functions of the form", "\\begin{equation*}\na_0 +\n\\sum_{n=1}^N a_n \\cos(nt)\n\\end{equation*}", "is an algebra. When considered on$[0,\\pi]$, it separates points and vanishes nowhere so \\hyperref[thm:SWreal]{Stone--Weierstrass} applies. As for polynomials, you \\emph{do not} want to conclude that every continuous function on$[0,\\pi]$has a uniformly convergent Fourier cosine series, that is, that every continuous function can be written as", "\\begin{equation*}\na_0 +\n\\sum_{n=1}^\\infty a_n \\cos(nt) .\n\\end{equation*}", "That is \\emph{not true}! There exist continuous functions whose Fourier series does not converge even pointwise let alone uniformly. See \\sectionref{sec:fourier}." ], "refs": [ "thm:SWreal" ], "proofs": [], "ref_ids": [] }, { "id": 53, "type": "theorem", "label": "Lebl-contfunc:thm:SWcomplex", "categories": [ "continuity", "topology", "named theorem" ], "title": "Weierstrass Approximation Theorem", "contents": [ "% \\index{Stone--Weierstrass!complex version}% Let$X$be a compact metric space and$\\sA$an algebra of complex-valued continuous functions on$X$, such that$\\sA$separates points, vanishes at no point, and is self-adjoint. Then the closure$\\widebar{\\sA} = C(X,\\C)$." ], "refs": [], "proofs": [ { "contents": [ "Suppose $\\sA_\\R \\subset \\sA$ is the set of the real-valued elements of", "$\\sA$.", "For $f \\in \\sA$, write $f = u+iv$ where $u$ and $v$ are real-valued.", "Then", "\\begin{equation*}", "u = \\frac{f+\\bar{f}}{2}, \\qquad", "v = \\frac{f-\\bar{f}}{2i} .", "\\end{equation*}", "So $u, v \\in \\sA$ as $\\sA$ is a self-adjoint algebra, and since they are", "real-valued $u, v \\in \\sA_\\R$.", "If $x \\not= y$, then find an $f \\in \\sA$ such that $f(x) \\not= f(y)$. If $f", "= u+iv$, then it is obvious that either $u(x) \\not= u(y)$ or $v(x) \\not=", "v(y)$. So $\\sA_\\R$ separates points.", "%", "Similarly, for every $x$ find $f \\in \\sA$ such that $f(x) \\not= 0$. If $f", "= u+iv$, then either $u(x) \\not= 0$ or $v(x) \\not= 0$.", "So $\\sA_\\R$ vanishes at no point.", "%", "The set $\\sA_\\R$ is a real algebra, and satisfies the hypotheses of the", "\\hyperref[thm:SWreal]{real Stone--Weierstrass theorem}.", "Given any $f = u+iv \\in C(X,\\C)$,", "we find $g,h \\in \\sA_\\R$ such that", "$\\sabs{u(t)-g(t)} < \\nicefrac{\\epsilon}{2}$ and", "$\\sabs{v(t)-h(t)} < \\nicefrac{\\epsilon}{2}$ for all $t \\in X$.", "Next, $g+i h \\in \\sA$, and", "\\begin{multline*}", "\\abs{f(t) - \\bigl(g(t)+ih(t)\\bigr)} =", "\\abs{u(t)+iv(t) - \\bigl(g(t)+ih(t)\\bigr)} \\\\", "\\leq", "\\sabs{u(t)-g(t)}+\\sabs{v(t)-h(t)} < \\nicefrac{\\epsilon}{2} +", "\\nicefrac{\\epsilon}{2} = \\epsilon", "\\end{multline*}", "for all $t \\in X$.", "So $\\widebar{\\sA} = C(X,\\C)$." ], "refs": [ "thm:SWreal" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 54, "type": "example", "label": "Lebl-contfunc:54", "categories": [ "topology", "example", "approximation", "continuity", "named theorem" ], "title": "Weierstrass Approximation Theorem", "contents": [ "Any continuous$F \\colon [0,1] \\times [0,1] \\to \\C$can be approximated uniformly by functions of the form", "\\begin{equation*}\n\\sum_{j=1}^n f_j(x) g_j(y) ,\n\\end{equation*}", "where$f_j \\colon [0,1] \\to \\C$and$g_j \\colon [0,1] \\to \\C$are continuous.", "Proof: It is not hard to see that the functions of the above form are a complex algebra. It is equally easy to show that they vanish nowhere, separate points, and the algebra is self-adjoint. As$[0,1] \\times [0,1]$is compact, \\hyperref[thm:SWcomplex]{Stone--Weierstrass} obtains the result." ], "refs": [ "thm:SWcomplex" ], "proofs": [ { "contents": [ "It is not hard to see that the functions of the above form are a complex algebra. It is equally easy to show that they vanish nowhere, separate points, and the algebra is self-adjoint. As$[0,1] \\times [0,1]$is compact, \\hyperref[thm:SWcomplex]{Stone--Weierstrass} obtains the result." ], "refs": [ "thm:SWcomplex" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 55, "type": "proposition", "label": "Lebl-contfunc:55", "categories": [ "fourier analysis" ], "title": "Characterization of Trigonometric Functions via Equivalence", "contents": [ "A trigonometric polynomial$f(x) = \\sum_{n=-N}^N c_n\\, e^{inx}$is real-valued for real$x$if and only if$c_{-m} = \\overline{c_m}$for all$m=-N,\\ldots,N$." ], "refs": [], "proofs": [ { "contents": [ "If $f(x)$ is real-valued, that is $\\overline{f(x)} = f(x)$, then", "\\begin{equation*}", "\\overline{c_m}", "=", "\\overline{", "\\frac{1}{2\\pi} \\int_{-\\pi}^\\pi", "f(x) e^{-imx} \\, dx", "}", "=", "\\frac{1}{2\\pi} \\int_{-\\pi}^\\pi", "\\overline{", "f(x) e^{-imx} } \\, dx", "=", "\\frac{1}{2\\pi} \\int_{-\\pi}^\\pi", "f(x) e^{imx} \\, dx", "= c_{-m} .", "\\end{equation*}", "The complex conjugate goes inside the integral because the integral is", "done on real and imaginary parts separately.", "On the other hand, if", "$c_{-m} = \\overline{c_m}$, then", "\\begin{equation*}", "\\overline{c_{-m}\\, e^{-imx}+ c_{m}\\, e^{imx}}", "=", "\\overline{c_{-m}}\\, e^{imx}+ \\overline{c_{m}}\\, e^{-imx}", "=", "c_{m}\\, e^{imx}+ c_{-m}\\, e^{-imx} ,", "\\end{equation*}", "which is real valued. Also $c_0 = \\overline{c_0}$, so", "$c_0$ is real.", "By pairing up the terms, we obtain that $f$ has to be real-valued." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 56, "type": "proposition", "label": "Lebl-contfunc:56", "categories": [], "title": "If \\begin{equation*} \\sum_{n=-N}^N c_n \\, e^{inx} = 0 \\end{equation*} for all$x \\in [-\\pi,\\pi]$, ...", "contents": [ "If\\begin{equation*} \\sum_{n=-N}^N c_n \\, e^{inx} = 0 \\end{equation*}for all$x \\in [-\\pi,\\pi]$, then$c_n = 0$for all$n$." ], "refs": [], "proofs": [ { "contents": [ "The result follows immediately from the integral formula for $c_n$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 57, "type": "example", "label": "Lebl-contfunc:57", "categories": [ "integration", "fourier analysis", "example", "series", "approximation", "continuity" ], "title": "Continuity of Fourier Series", "contents": [ "Consider the step function$h(x)$so that$h(x) \\coloneqq 1$on$[0,\\pi]$and$h(x) \\coloneqq -1$on$(-\\pi,0)$, extended periodically to a$2\\pi$-periodic function. With a little bit of calculus, we compute the coefficients:", "\\begin{equation*}\n\\hat{h}(0) = \\frac{1}{2\\pi} \\int_{-\\pi}^\\pi h(x) \\, dx = 0,\n\\qquad\n\\hat{h}(n) = \\frac{1}{2\\pi} \\int_{-\\pi}^\\pi h(x) e^{-inx} \\, dx = \\frac{i\\bigl(\n(-1)^n-1 \\bigr)}{\\pi n} \\quad \\text{for } n \\geq 1 .\n\\end{equation*}", "A little bit of simplification leads to", "\\begin{equation*}\ns_N(h;x) =\n\\sum_{n=-N}^N \\hat{h}(n) \\,e^{inx} \n=\n\\sum_{n=1}^N \\frac{2\\bigl(1-(-1)^n\\bigr)}{\\pi n} \\sin(n x) .\n\\avoidbreak\n\\end{equation*}", "See the left hand graph in \\figureref{fig:fourierheavicsaw} for a graph of$h$and several symmetric partial sums.", "For a second example, consider the function$g(x) \\coloneqq \\sabs{x}$on$[-\\pi,\\pi]$and then extended to a$2\\pi$-periodic function. Computing the coefficients, we find", "\\begin{equation*}\n\\hat{g}(0) = \\frac{1}{2\\pi} \\int_{-\\pi}^\\pi g(x) \\, dx = \\frac{\\pi}{2},\n\\qquad\n\\hat{g}(n) = \\frac{1}{2\\pi} \\int_{-\\pi}^\\pi g(x) e^{-inx} \\, dx\n= \\frac{(-1)^n-1}{\\pi n^2} \\quad \\text{for } n \\geq 1 .\n\\end{equation*}", "A little simplification yields", "\\begin{equation*}\ns_N(g;x) =\n\\sum_{n=-N}^N \\hat{g}(n) \\,e^{inx} \n=\n\\frac{\\pi}{2} + \n\\sum_{n=1}^N \\frac{2\\bigl((-1)^n-1\\bigr)}{\\pi n^2} \\cos(n x) .\n\\end{equation*}", "See the right hand graph in \\figureref{fig:fourierheavicsaw}.", "\\begin{myfigureht} \\subimport*{figures/}{fourierheavi_csaw.pdf_t} \\caption{The functions$h$and$g$in bold, with several symmetric partial sums in gray.\\label{fig:fourierheavicsaw}} \\end{myfigureht}", "Note that for both$f$and$g$, the even coefficients (except$\\hat{g}(0)$) happen to vanish, but that is not really important. What is important is convergence. First, at the discontinuity at$x=0$, we find$s_N(h;0) = 0$for all$N$, so$s_N(h;0)$converges to a different number from$h(0)$(at a nice enough jump discontinuity, the limit is the average of the two-sided limits, see the exercises). That should not be surprising; the coefficients are computed by an integral, and integration does not notice if the value of a function changes at a single point. We should remark, however, that we are not guaranteed that in general the Fourier series converges to the function even at a point where the function is continuous. We will prove convergence if the function is at least Lipschitz.", "What is really important is how fast the coefficients go to zero. For the discontinuous$h$, the coefficients$\\hat{h}(n)$go to zero approximately like$\\nicefrac{1}{n}$. On the other hand, for the continuous$g$, the coefficients$\\hat{g}(n)$go to zero approximately like$\\nicefrac{1}{n^2}$. The Fourier coefficients \\myquote{see} the discontinuity in some sense.", "Do note that continuity in this setting is the continuity of the periodic extension, that is, we include the endpoints$\\pm \\pi$. So the function$f(x) = x$defined on$(-\\pi,\\pi]$and extended periodically would be discontinuous at the endpoints$\\pm\\pi$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 58, "type": "proposition", "label": "Lebl-contfunc:58", "categories": [ "continuity", "fourier analysis", "series" ], "title": "Continuity of Fourier Series", "contents": [ "Let$\\sum_{n=-\\infty}^\\infty c_n\\, e^{inx}$be a Fourier series, and$C$,$\\alpha > 1$constants such that\\begin{equation*} \\sabs{c_n} \\leq \\frac{C}{\\sabs{n}^\\alpha} \\qquad \\text{for all } n \\in \\Z \\setminus \\{ 0 \\}. \\avoidbreak \\end{equation*}Then the series converges (absolutely and uniformly) to a continuous function on$\\R$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 59, "type": "proposition", "label": "Lebl-contfunc:59", "categories": [ "fourier analysis", "series", "approximation", "differentiability", "continuity" ], "title": "Continuity of Fourier Series", "contents": [ "Let$\\sum_{n=-\\infty}^\\infty c_n\\, e^{inx}$be a Fourier series, and$C$,$\\alpha > 2$constants such that\\begin{equation*} \\sabs{c_n} \\leq \\frac{C}{\\sabs{n}^\\alpha} \\qquad \\text{for all } n \\in \\Z \\setminus \\{ 0 \\}. \\avoidbreak \\end{equation*}Then the series converges to a continuously differentiable function on$\\R$." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 60, "type": "theorem", "label": "Lebl-contfunc:thm:l2bestapprox", "categories": [ "integration" ], "title": "Suppose$f$is a Riemann integrable function on$[a,b]$", "contents": [ "Suppose$f$is a Riemann integrable function on$[a,b]$. Let$\\{ \\varphi_n \\}_{n=1}^\\infty$be an orthonormal system on$[a,b]$and suppose\\begin{equation*} f(x) \\sim \\sum_{n=1}^\\infty c_n \\varphi_n(x) . \\end{equation*}If\\begin{equation*} s_k (x) \\coloneqq \\sum_{n=1}^k c_n \\varphi_n(x) \\quad\\text{and}\\quad p_k (x) \\coloneqq \\sum_{n=1}^k d_n \\varphi_n(x) \\end{equation*}for some other sequence$\\{ d_n \\}_{n=1}^\\infty$, then\\begin{equation*} \\int_a^b \\sabs{f(x)-s_k(x)}^2 \\, dx = \\snorm{f-s_k}_2^2 \\leq \\snorm{f-p_k}_2^2 = \\int_a^b \\sabs{f(x)-p_k(x)}^2 \\, dx \\end{equation*}with equality only if$d_n = c_n$for all$n=1,2,\\ldots,k$." ], "refs": [], "proofs": [ { "contents": [ "Let us write", "\\begin{equation*}", "\\int_a^b \\sabs{f-p_k}^2", "=", "\\int_a^b \\sabs{f}^2", "-", "\\int_a^b f \\widebar{p_k}", "-", "\\int_a^b \\widebar{f} p_k", "+", "\\int_a^b \\sabs{p_k}^2 .", "\\end{equation*}", "Now", "\\begin{equation*}", "\\int_a^b f \\widebar{p_k}", "=", "\\int_a^b f \\sum_{n=1}^k \\overline{d_n} \\overline{\\varphi_n}", "=", "\\sum_{n=1}^k \\overline{d_n} \\int_a^b f \\, \\overline{\\varphi_n}", "=", "\\sum_{n=1}^k \\overline{d_n} c_n ,", "\\end{equation*}", "and", "\\begin{equation*}", "\\int_a^b \\sabs{p_k}^2", "=", "\\int_a^b", "\\sum_{n=1}^k d_n \\varphi_n", "\\sum_{m=1}^k \\overline{d_m} \\overline{\\varphi_m}", "=", "\\sum_{n=1}^k", "\\sum_{m=1}^k", "d_n", "\\overline{d_m}", "\\int_a^b", "\\varphi_n", "\\overline{\\varphi_m}", "=", "\\sum_{n=1}^k", "\\sabs{d_n}^2 .", "\\end{equation*}", "So", "\\begin{equation*}", "\\begin{split}", "\\int_a^b \\sabs{f-p_k}^2", "& =", "\\int_a^b \\sabs{f}^2", "-", "\\sum_{n=1}^k \\overline{d_n} c_n", "-", "\\sum_{n=1}^k d_n \\overline{c_n}", "+", "\\sum_{n=1}^k", "\\sabs{d_n}^2", "\\\\", "& =", "\\int_a^b \\sabs{f}^2", "-", "\\sum_{n=1}^k \\sabs{c_n}^2", "+", "\\sum_{n=1}^k", "\\sabs{d_n-c_n}^2 .", "\\end{split}", "\\end{equation*}", "This is minimized precisely when $d_n = c_n$." ], "refs": [], "ref_ids": [] } ], "ref_ids": [] }, { "id": 61, "type": "theorem", "label": "Lebl-contfunc:thm:bessels", "categories": [ "integration" ], "title": "Suppose$f$is a Riemann integrable function on$[a,b]$", "contents": [ "Suppose$f$is a Riemann integrable function on$[a,b]$. Let$\\{ \\varphi_n \\}_{n=1}^\\infty$be an orthonormal system on$[a,b]$and suppose\\begin{equation*} f(x) \\sim \\sum_{n=1}^\\infty c_n \\varphi_n(x) . \\end{equation*}Then\\begin{equation*} \\sum_{n=1}^\\infty \\sabs{c_n}^2 \\leq \\int_a^b \\sabs{f}^2 = \\snorm{f}_2^2 . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 62, "type": "theorem", "label": "Lebl-contfunc:thm:fourierlocalization", "categories": [ "integration" ], "title": "Let$x$be fixed and let$f$be a$2\\pi$-periodic function Riemann integrable on$[-\\pi,\\pi]$", "contents": [ "Let$x$be fixed and let$f$be a$2\\pi$-periodic function Riemann integrable on$[-\\pi,\\pi]$. Suppose there exist$\\delta > 0$and$M$such that\\begin{equation*} \\sabs{f(x+t)-f(x)} \\leq M \\sabs{t} \\end{equation*}for all$t \\in (-\\delta,\\delta)$, then\\begin{equation*} \\lim_{N \\to \\infty} s_N(f;x) = f(x) . \\end{equation*}" ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 63, "type": "corollary", "label": "Lebl-contfunc:cor:fourierpiecewisesmooth", "categories": [ "consequence", "continuity", "integration" ], "title": "Continuity of Integration Theory", "contents": [ "Let$f$be a$2\\pi$-periodic function Riemann integrable on$[-\\pi,\\pi]$. Suppose there exist$x\\in \\R$and$\\delta > 0$such that$f$is continuous piecewise smooth on$[x-\\delta,x+\\delta]$, then\\begin{equation*} \\lim_{N \\to \\infty} s_N(f;x) = f(x) . \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "[Proof of \\thmref{thm:fourierlocalization}]", "For all $N$,", "\\begin{equation*}", "\\frac{1}{2\\pi} \\int_{-\\pi}^\\pi D_N = 1 .", "\\end{equation*}", "Write", "\\begin{equation*}", "\\begin{split}", "s_N(f;x)-f(x) & =", "\\frac{1}{2\\pi} \\int_{-\\pi}^\\pi f(x-t) D_N(t) \\, dt", "-", "f(x)", "\\frac{1}{2\\pi} \\int_{-\\pi}^\\pi D_N(t) \\, dt", "\\\\", "& =", "\\frac{1}{2\\pi} \\int_{-\\pi}^\\pi \\bigl( f(x-t) - f(x) \\bigr) D_N(t) \\, dt", "\\\\", "& =", "\\frac{1}{2\\pi} \\int_{-\\pi}^\\pi \\frac{f(x-t) - f(x)}{\\sin(\\nicefrac{t}{2})} \\sin\\bigl(", "(N+\\nicefrac{1}{2})t \\bigr) \\, dt .", "\\end{split}", "\\end{equation*}", "By the hypotheses,", "for small nonzero $t$,", "\\begin{equation*}", "\\abs{ \\frac{f(x-t) - f(x)}{\\sin(\\nicefrac{t}{2})} }", "\\leq", "\\frac{M\\sabs{t}}{\\sabs{\\sin(\\nicefrac{t}{2})}} .", "\\end{equation*}", "As $\\sin(\\theta) = \\theta + h(\\theta)$ where $\\frac{h(\\theta)}{\\theta} \\to", "0$ as $\\theta \\to 0$,", "we notice that", "$\\frac{M\\sabs{t}}{\\sabs{\\sin(\\nicefrac{t}{2})}}$ is continuous at the", "origin.", "Hence,", "$\\frac{f(x-t) - f(x)}{\\sin(\\nicefrac{t}{2})}$, as a function of $t$,", "is bounded near the origin.", "As $t=0$ is the only place on $[-\\pi,\\pi]$ where the denominator vanishes,", "it is the only place where there could be a problem.", "So, the function is bounded near $t=0$ and clearly", "Riemann integrable on any interval not including $0$, and thus", "it is Riemann integrable on $[-\\pi,\\pi]$.", "We use the trigonometric identity", "\\begin{equation*}", "\\sin\\bigl( (N+\\nicefrac{1}{2})t \\bigr)", "=", "\\cos(\\nicefrac{t}{2}) \\sin(Nt) +", "\\sin(\\nicefrac{t}{2}) \\cos(Nt) ,", "\\end{equation*}", "to compute", "\\begin{multline*}", "\\frac{1}{2\\pi} \\int_{-\\pi}^\\pi \\frac{f(x-t) - f(x)}{\\sin(\\nicefrac{t}{2})} \\sin\\bigl(", "(N+\\nicefrac{1}{2})t \\bigr) \\, dt", "= \\\\", "\\frac{1}{2\\pi} \\int_{-\\pi}^\\pi", "\\left( \\frac{f(x-t) - f(x)}{\\sin(\\nicefrac{t}{2})}", "\\cos (\\nicefrac{t}{2}) \\right) \\sin (Nt) \\, dt", "+", "\\frac{1}{2\\pi} \\int_{-\\pi}^\\pi \\bigl( f(x-t) - f(x) \\bigr)", "\\cos (Nt) \\, dt .", "\\end{multline*}", "As functions of $t$,", "$\\frac{f(x-t) - f(x)}{\\sin(\\nicefrac{t}{2})} \\cos (\\nicefrac{t}{2})$", "and", "$\\bigl( f(x-t) - f(x) \\bigr)$ are bounded Riemann integrable functions", "and so their Fourier coefficients go to zero by \\thmref{thm:bessels}. So the two", "integrals on the right-hand side, which compute the Fourier coefficients", "for the real version of the Fourier series go to 0 as $N$ goes to infinity.", "This is because $\\sin(Nt)$ and $\\cos(Nt)$ are also orthonormal systems", "with respect to the same inner product.", "Hence $s_N(f;x)-f(x)$ goes to 0, that is, $s_N(f;x)$ goes to $f(x)$." ], "refs": [ "thm:bessels", "thm:fourierlocalization" ], "ref_ids": [] } ], "ref_ids": [] }, { "id": 64, "type": "corollary", "label": "Lebl-contfunc:64", "categories": [ "integration", "characterization", "topology", "consequence", "series" ], "title": "Characterization of Convergence via Equivalence", "contents": [ "Suppose$f$is a$2\\pi$-periodic function, Riemann integrable on$[-\\pi,\\pi]$. If$J$is an open interval and$f(x) = 0$for all$x \\in J$, then$\\lim\\limits_{N\\to\\infty} s_N(f;x) = 0$for all$x \\in J$.", "In particular, if$f$and$g$are$2\\pi$-periodic functions, Riemann integrable on$[-\\pi,\\pi]$,$J$an open interval, and$f(x) = g(x)$for all$x \\in J$, then for all$x \\in J$, the sequence$\\bigl\\{ s_N(f;x) \\bigr\\}_{N=1}^\\infty$converges if and only if$\\bigl\\{ s_N(g;x) \\bigr\\}_{N=1}^\\infty$converges." ], "refs": [], "proofs": [], "ref_ids": [] }, { "id": 65, "type": "theorem", "label": "Lebl-contfunc:65", "categories": [ "integration", "named theorem" ], "title": "Parseval's Theorem", "contents": [ "\\index{Parseval's theorem} Let$f$and$g$be$2\\pi$-periodic functions, Riemann integrable on$[-\\pi,\\pi]$with\\begin{equation*} f(x) \\sim \\sum_{n=-\\infty}^\\infty c_n \\,e^{inx} \\qquad \\text{and} \\qquad g(x) \\sim \\sum_{n=-\\infty}^\\infty d_n \\,e^{inx} . \\end{equation*}Then\\begin{equation*} \\lim_{N\\to\\infty} \\snorm{f-s_N(f)}_2^2 = \\lim_{N\\to\\infty} \\frac{1}{2\\pi} \\int_{-\\pi}^\\pi \\sabs{f(x)-s_N(f;x)}^2 \\, dx =0 . \\end{equation*}Also\\begin{equation*} \\langle f , g \\rangle = \\frac{1}{2\\pi} \\int_{-\\pi}^\\pi f(x) \\overline{g(x)}\\, dx = \\sum_{n=-\\infty}^\\infty c_n \\overline{d_n} , \\end{equation*}and\\begin{equation*} \\snorm{f}_2^2 = \\frac{1}{2\\pi} \\int_{-\\pi}^\\pi \\sabs{f(x)}^2 \\, dx = \\sum_{n=-\\infty}^\\infty \\sabs{c_n}^2. \\end{equation*}" ], "refs": [], "proofs": [ { "contents": [ "There exists (exercise)", "a continuous $2\\pi$-periodic function $h$ such that", "\\begin{equation*}", "\\snorm{f-h}_2 < \\epsilon .", "\\end{equation*}", "Via \\hyperref[thm:SWcomplex]{Stone--Weierstrass},", "approximate $h$ with a trigonometric polynomial", "uniformly. That is, there is a trigonometric polynomial $P(x)$", "such that", "$\\sabs{h(x) - P(x)} < \\epsilon$ for all $x$.", "Hence", "\\begin{equation*}", "\\snorm{h-P}_2", "=", "\\sqrt{", "\\frac{1}{2\\pi}", "\\int_{-\\pi}^{\\pi}", "\\sabs{h(x)-P(x)}^2", "\\,", "dx", "}", "\\leq \\epsilon.", "\\end{equation*}", "If $P$ is of degree $N_0$, then for all $N \\geq N_0$ ,", "\\begin{equation*}", "\\snorm{h-s_N(h)}_2 \\leq \\snorm{h-P}_2 \\leq \\epsilon ,", "\\end{equation*}", "as $s_N(h)$ is the best approximation for $h$ in $L^2$ (\\thmref{thm:l2bestapprox}).", "By the inequality leading up to Bessel,", "\\begin{equation*}", "\\snorm{s_N(h)-s_N(f)}_2", "=", "\\snorm{s_N(h-f)}_2", "\\leq", "\\snorm{h-f}_2 \\leq \\epsilon .", "\\end{equation*}", "The $L^2$ norm satisfies the triangle inequality (exercise).", "Thus, for all $N \\geq N_0$,", "\\begin{equation*}", "\\snorm{f-s_N(f)}_2", "\\leq", "\\snorm{f-h}_2", "+", "\\snorm{h-s_N(h)}_2", "+", "\\snorm{s_N(h)-s_N(f)}_2", "\\leq 3\\epsilon .", "\\end{equation*}", "Hence, the first claim follows.", "Next,", "\\begin{equation*}", "\\langle s_N(f) , g \\rangle", "=", "\\frac{1}{2\\pi}", "\\int_{-\\pi}^\\pi", "s_N(f;x) \\overline{g(x)} \\, dx", "=", "\\sum_{n=-N}^N", "c_n", "\\frac{1}{2\\pi}", "\\int_{-\\pi}^\\pi", "e^{inx}", "\\overline{g(x)} \\, dx", "=", "\\sum_{n=-N}^N", "c_n", "\\overline{d_n} .", "\\end{equation*}", "We need the Schwarz (or Cauchy--Schwarz or Cauchy--Bunyakovsky--Schwarz)", "inequality for $L^2$, that is,", "\\begin{equation*}", "{\\abs{\\int_a^b f\\bar{g}}}^2", "\\leq", "\\left( \\int_a^b \\sabs{f}^2 \\right)", "\\left( \\int_a^b \\sabs{g}^2 \\right) .", "\\end{equation*}", "Its proof is left as an exercise; it is not much", "different from the finite-dimensional version.", "So", "\\begin{equation*}", "\\begin{split}", "\\abs{\\int_{-\\pi}^\\pi f\\bar{g} - \\int_{-\\pi}^\\pi s_N(f)\\bar{g}}", "& =", "\\abs{\\int_{-\\pi}^\\pi (f- s_N(f))\\bar{g}} \\\\", "%& \\leq", "%\\int_{-\\pi}^\\pi \\sabs{f- s_N(f)}\\, \\sabs{g} \\\\", "& \\leq", "{\\left(\\int_{-\\pi}^\\pi \\sabs{f- s_N(f)}^2 \\right)}^{1/2}", "{\\left( \\int_{-\\pi}^\\pi \\sabs{g}^2 \\right)}^{1/2} .", "\\end{split}", "\\end{equation*}", "The right-hand side goes to 0 as $N$ goes to infinity by the first", "claim of the theorem.", "That is, as $N$ goes to infinity, $\\langle s_N(f),g \\rangle$", "goes to $\\langle f,g \\rangle$, and", "the second claim is proved. The last claim in the theorem follows by using", "$g=f$." ], "refs": [ "thm:SWcomplex", "thm:l2bestapprox" ], "ref_ids": [] } ], "ref_ids": [] } ] } }