Advanced Calculus (I)

# Advanced Calculus (I) `2021 fall` 史習偉大大ㄉ高等微積分一...只有後半段... \ 前面的我有時間再慢慢補... ## Overview [TOC] ## Point-Set Topology of Metric Spaces ### Compact Sets --- ### Connected Sets Let $(M,d)$ be a metric space and $A\subseteq M$. \ Let $u,v\subseteq M$ be two nonempty open sets. \ We say that ==$u$ and $v$ separates $A$== if $\begin{cases} (1)\; A\subseteq u\cup v \\ (2)\; A\cap u \neq \emptyset \\ (3)\; A\cap v \neq \emptyset \\ (4)\; A\cap u\cap v =\emptyset \end{cases}$. And $A$ is said to be ==separated==, or ==disconnected==, if there exist such $u$ and $v$. \ Otherwise, $A$ is said to be ==connected==. --- Let $(M,d)$ be a metric space and $A\subseteq M$. \ Then $A$ is disconnected **iff** $\exists\,A_1,A_2 \subseteq M$ such that $\begin{cases} A_1,A_2 \neq \emptyset \\ A_1\cup A_2 =A \\ \bar{A}_1\cup A_2= A_1\cup\bar{A}_2 = \emptyset \end{cases}$ :::spoiler proof: > $\implies:$ \ > If $A$ is disconnected, then there exist two nonempty open sets $u$ and $v$ separating $A$. \ > Select $A_1=A\cap u$ and $A_2=A\cap v$. \ > By (2), $A_1\neq\emptyset$. By (3), $A_2\neq\emptyset$. \ > By (1), $A_1\cup A_2= (A\cap u)\cup(A\cap v)= A\cap(u\cup v)= A$. \ > By (4), $A_1\subseteq v^c$. \ > Since $v^c$ is closed, $\overline{A}_1\subseteq \overline{v^c}= v^c$. \ > $\Rightarrow$ $\overline{A}_1\cap v=\emptyset$ \ > $\Rightarrow$ $\overline{A}_1\cap A_2=\emptyset$ \ > Similarly, $A_1\cap \overline{A}_2=\emptyset$. > > $\impliedby:$ \ > $\exists\,A_1,A_2$ such that $\dots$ \ > Let $u=\bar{A}_{2}^{\,c}$ and $v=\bar{A}_{1}^{\,c}$. > Since $A_2\subseteq \bar{A}_1^{\,c}$ and $A_1\subseteq \bar{A}_2^{\,c}$, $v$ and $u$ are both nonempty. \ > Hence $A=A_1\cup A_2\subseteq u\cup v$. \ > Since $A_1\cap\bar{A}_1^{\,c}=\emptyset$ and $A_2\cap \bar{A}_2^{\,c}$, we have $A_1\cap v=\emptyset$ and $A_2\cap u=\emptyset$. \ > Then > $$\begin{aligned} > A\cap u\cap v &= (A_1\cap u\cap v)\cup(A_2\cap u\cap v) \\ > &= \emptyset > \end{aligned}$$ > Thus $A$ can be separated by $u$ and $v$, hence $A$ is disconnected. ::: --- $A\subseteq\mathbb{R}$ is connected ***iff*** for $x,y\in A$ with $x<z<y$, $z\in A$. (interval) :::spoiler proof: > $\implies:$ \ > Suppose $x<z<y$ and $x,y\in A$. \ > Assume $z\notin A$. \ > Let $A_1=(-\infty,z)\cap A$ and $A_2=(z,\infty)\cap A$. \ > Since $x\in A_1$ and $y\in A_2$, $A_1$, $A_2$ are nonempty. \ > Since $z\notin A$, $A=A_1\cup A_2$. \ > We have $\bar{A_1}\subseteq (-\infty,z]$ and $\bar{A}_2\subseteq [z,\infty)$. \ > $\Rightarrow$ $\bar{A}_1\cap A_2=\emptyset=A_1\cap\bar{A}_2$ \ > Then $A$ is disconnected. $\quad(\rightarrow\leftarrow)$. \ > Thus $z\in A$. > > $\impliedby:$ \ > Assume $A$ disconneted. \ > $\exists\,A_1,A_2\subseteq\mathbb{R}$ such that (i) $A_1,A_2\neq\emptyset$, (ii) $A=A_1\cup A_2$, > (iii) $\bar{A}_1\cap A_2= A_1\cap\bar{A}_2=\emptyset$. \ > By (i), $\exists\,x\in A_1, y\in A_2$. \ > By (iii), $x\neq y$. WLOG, assume $x<y$. \ > Let $z=\sup([x,y]\cap A_1)$. \ > Then $z\in \overline{A}_1$ and thus $z\notin A_2$. \ > - If $z\notin A_1$, then since $z\notin A_2$, $z\notin A$. $\quad(\rightarrow\leftarrow)$ > - If $z\in A_1$, then $z\notin \bar{A}_2$. \ > $\Rightarrow$ $\exists\,r>0$ such that $(z,z+r)\cap A_2=\emptyset$ \ > $\Rightarrow$ $x<z+\frac{r}{2}<y$, $\;z+\frac{r}{2}\notin A_1$, and $z+\frac{r}{2}\notin A_2$ \ > $\Rightarrow$ $z+\frac{r}{2}\notin A$ $\quad(\rightarrow\leftarrow)$ \ > Therefore $A$ is connected. ::: --- ### Subspace Topology Let $(M,d)$ be a metric space and $N\subseteq M$. \ The topology of $(N,d)$ is called the ==subspace topology== of $(M,d)$. \ $A\subseteq N$ could be open in $(N,d)$ but not open in $(M,d)$. \ For balls, let $x\in N$, $B_N(x,r)=B_M(x,r)\cap N$. --- Let $(M,d)$ be a metric space and $N\subseteq M$. \ $v\subseteq N$ is open in $(N,d)$ **iff** $\exists\,u\subseteq M$ is open in $(M,d)$ such that $v=u\cap N$. :::spoiler proof: > $\implies:$\ > Since $v$ is open in $(N,d)$, $\forall\,x\in v$, $\exists\,r_x>0$ such that $B_N(x,r_x)\subseteq v$. \ > Then $v\subseteq \displaystyle\bigcup_{x\in v}B_N(x,r_x)\subseteq v$. \ > Define $u=\displaystyle\bigcup_{x\in v}B_M(x,r_x)$. \ > $u$ is open in $(M,d)$. \ > Then > $$\begin{aligned} > u\cap N &= \bigcup_{x\in v}B_M(x,r_x) \cap N \\ > &= \bigcup_{x\in v}\left(B_M(x,r_x)\cap N\right) \\ > &=\bigcup_{x\in v}B_N(x,r_x) = v > \end{aligned}$$ > > $\impliedby:$\ > Suppose $x\in v =u\cap N$. \ > Since $u$ is open in $(M,d)$, $\exists\,\delta_x>0$ such that $B_M(x,\delta_x)\subseteq u$. \ > Then > $$\begin{aligned} > B_N(x,\delta_x) &= B_M(x,\delta_x)\cap N \\ > &\subseteq u\cap N = v > \end{aligned}$$ > Hence $v$ is open in $(N,d)$. ::: :::success There is similar result for closed sets. ::: --- Let $(M,d)$ be a metric space and $N\subseteq M$. \ $A\subseteq M$ is said to be ==open relative to $N$== if $A\cap N$ is open in $(M,d)$. \ If $A$ is open $(M,d)$, then A must be open relative to $N$. :::success There are similiar definitions for closed and compact. ::: --- Let $(M,d)$ be a metric space and $K\subseteq N\subseteq M$. \ $K$ is compact in $(M,d)$ ***iff*** $K$ is compact in $(N,d)$. proof: --- --- ## Mappings and Limits Let $(M,d)$ and $(N,\rho)$ be metric spaces and $A \subseteq M$.\ For $x_0 \in \overline{A}$, we say $y_0 \in N$ is the ==limit== of some function $f:A\to N$ at $x_0$ if $\forall\,\varepsilon >0,\exists\,\delta>0$ such that $\rho(f(x),y_0)<\varepsilon$ whenever $x\in A$ and $d(x,x_0)<\delta$, denoted by $\displaystyle\lim_{ \substack{x\to x_0}\\ x\in A }=y_0$. --- Let $(M,d)$ and $(N,\rho)$ be metric spaces, $A \subseteq M$, and $f:A\to N$ be a map.\ For $x_0 \in \overline{A}$, $\displaystyle\lim_{x\to x_0}f(x)=y$ ***iff*** for every sequence $\{x_k\}_{k=1}^\infty\subseteq A$ converging to $x_0$ in $(M,d)$, the sequence $\{f(x_k)\}_{k=1}^\infty$ converges to $y_0$ in $(N,\rho)$. :::spoiler proof: > $\implies$ :\ > $\forall\,\epsilon>0,\;\exists\,\delta>0$ such that $\rho(f(x),y_0)<\epsilon$ when $x\in A$ and $d(x,x_0)<\delta$.\ > Let $\{x_k\}_{k=1}^\infty\subseteq A$ converging to $x_0$.\ > $\Rightarrow$ $\exists\, K \in \mathbb{N}$ such that $d(x_k,x_0)<\delta$ whenever $k\geq K$.\ > $\Rightarrow$ $\rho(f(x_k),y_0)<\epsilon$ whenever $k\geq K$.\ > $\Rightarrow$ $\displaystyle\lim_{k\to\infty}=y_0$.\ > $\impliedby$ :\ > Prove by contradiction.\ > Assume $\displaystyle\lim_{x\to x_0}f(x)\neq y$.\ > Then $\exists\,\epsilon>0$ such that $\forall\,\delta>0$, $\exists\, x_\delta \in A$ with $d(x_\delta,x_0)<\delta$ but $\rho\left(f(x_\delta),y_0\right)\geq\epsilon$.\ > Let $\rho=\frac{1}{k}$, then $\exists\,\{x_k\}_{k=1}^\infty\subseteq A$ such that $d(x_k,x_0)<\frac{1}{k}$ but $\rho\left(f(x_0),y_0\right)\geq\epsilon$.\ > In this case, $\displaystyle\lim_{k\to\infty}x_k=x_0$ but $\displaystyle\lim_{k\to\infty}f(x_k)\neq y_0$. $\quad( \to\gets)$ ::: --- Let $(M,d)$ and $(N,\rho)$ be metric spaces, $A \subseteq M$, and $f:A\to N$ be a map.\ $f$ is said to be ==continuous== at $x_0$ if $\displaystyle\lim_{\substack{x\to x_0\\ x\in A}}f(x)=f(x_0)$.\ (note: $x$ may be an isolated point)\ If $f$ is continuous at every point of $A$, then $f$ is said to be continuous at $A$. --- Let $(M,d)$ and $(N,\rho)$ be metric spaces, $A \subseteq M$, and $f:A\to N$ be a map.\ Then the following statements are equivalent :\ (1) $f$ is continuous on $A$.\ (2) For every open set $v\subseteq N$, $\; f^{-1}(v)\subseteq A$ is open relative to $A$.\ (3) For every closed set $E\subseteq N$, $\; f^{-1}(E)\subseteq A$ is closed relative to $A$. :::spoiler proof: > #### (1)$\implies$(2) : > If $f^{-1}(v)=\emptyset$, then it is tivially open relative to $A$.\ > Consider $f^{-1}(v)\neq \emptyset$.\ > Suppose $x_0 \in f^{-1}(v)$.\ > $\Rightarrow$ $f(x_0)\in v$ > $\Rightarrow$ $\exists\,\epsilon>0$ such that $B_N(f(x_0),\epsilon)\subseteq v$\ > Since $f$ is continuous at $x_0$, $\exists\,\delta_{x_0}>0$ such that $f\left(B_M(x_0,\delta_{x_0})\cap A\right) \subseteq B_N(f(x_0),\epsilon) \subseteq v$.\ > Then $B_M(x_0,\delta_{x_0})\cap A \subseteq f^{-1}(v)$.\ > Since $f$ is continuous on $A$, for every $x\in f^{-1}(v)$, $\exists\,\delta_x>0$ such that $f\left(B_M(x_0,\delta_{x_0})\cap A\right) \subseteq v$, and hence $B_M(x,\delta_x)\cap A \subseteq f^{-1}(v)$.\ > Define $u=\displaystyle\bigcup_{x\in f^{-1}(v)}B_M(x,\delta_x)$.\ > Then $u$ is open in $M$ and $f^{-1}(v)\subseteq u\cap A$.\ > On the other side, > \begin{align} > f(u\cap A) &= f\left( \displaystyle\bigcup_{x\in f^{-1}(v)} (B_M(x,\delta_x)\cap A) \right)\\ > &= \displaystyle\bigcup_{x\in f^{-1}(v)}f(B_M(x,\delta_x)\cap A)\\ > &\subseteq v > \end{align} > $\Rightarrow$ $u\cap A\subseteq f^{-1}(v)$. > > #### (2)$\implies$(1) : > Suppose $x\in A$.\ > For given $\epsilon>0$, $B_N(f(x),\epsilon)$ is open in $N$.\ > By the hypothesis, there is open $u\subseteq M$ satisfying $u\cap A = f^{-1}(B_N(f(x),\epsilon))$.\ > $\Rightarrow$ $x\in u$ and $\exists\,\delta_x>0$ such that $B_M(x,\delta_x)\subseteq u$\ > Then $f(B_M(x,\delta_x)\cap A)\subseteq f(u\cap A)= B_N(f(x),\epsilon)$.\ > For $y\in B_M(x,\delta_x)\cap A$, we have $\rho(f(x),f(y))<\epsilon$.\ > Hence $f$ is continuous at $x$. > > #### (2)$\implies$(3) : > $E^c$ is open in $N$.\ > By the hypothesis, $f^{-1}(E^c)$ is open relative to $A$ and there exists an open $u\subseteq M$ such that $f^{-1}(E^c)=u\cap A$.\ > Then $u^c$ is closed in $M$ and $A\cap u^c= A\backslash (u\cap A)= A\backslash f^{-1}(E^c)= f^{-1}(E)$, > which is closed relative to $A$. > #### (3)$\implies$(2) : > (skip) ::: --- Let $(M,d)$ and $(N,\rho)$ be metric spaces, $A \subseteq M$.\ If $f:M\to N$ is a function, then its restriction $f|_A$ is continuous on $A$. --- ### Operations on Continuous Maps Let $(M,d)$ be a metric space, $(V,\|\cdot\|)$ be normed vector space, and $A\subseteq M$.\ Let $f,g:A\to V$ and $h:A\to \mathbb{R}$ be maps.\ $\forall\,x\in A$, $\forall\,\alpha\in\mathbb{R}$, define \begin{split} &(1)\; (f\pm g)(x)=f(x)\pm g(x)\\ &(2)\; (\alpha f)(x)=\alpha f(x)\\ &(3)\; (hf)(x)=h(x)f(x)\\ &(4)\; (\frac{f}{h})(x)=\frac{1}{h(x)}f(x),\; h(x)\neq 0 \end{split} Suppose $x_0\in \bar{A}$, $\displaystyle\lim_{x\to x_0}f(x)=v$, $\displaystyle\lim_{x\to x_0}g(x)=w$, $\displaystyle\lim_{x\to x_0}h(x)=c$.\ Then \begin{split} &(5) \displaystyle\lim_{x\to x_0}(f\pm g)(x)=v+w\\ &(6) \displaystyle\lim_{x\to x_0}(hf)(x)=cv\\ &(7) \displaystyle\lim_{x\to x_0}(\frac{f}{h})(x)=\frac{v}{c}\; \text{if }c\neq 0 \end{split} Also there are similar results for continuity.\ Let $(M,d),(N,\rho),(P,\gamma)$ be metric spaces and $f:M\to N$, $g:N\to P$.\ If $f$ is continuous at $x_0$ and $g$ is continuous at $f(x_0)$, then $g\circ f$ is continuous at $x_0$. --- ### Uniformly Continuous Let $(M,d)$ and $(N,\rho)$ be metric spaces, $A\subseteq M$ and $f:A\to N$.\ $f$ is ==uniformly continuous== on $A$ if $\forall\,\epsilon>0$, $\exists\,\delta>0$ such that $\rho(f(x),f(y))<\epsilon$ whenever $x,y\in A$ and $d(x,y)<\delta$.\ e.g.\ $f(x)=\frac{1}{x}$ is continuous on $(0,\infty)$ and uniformly continuous on $[a,\infty),\,a>0$. --- Let $A\subseteq \mathbb{R}$ and $f:A\to \mathbb{R}$.\ $f$ is ==Lipschitz function== if $\exists\,K>0$ such that $\left|f(x)-f(y)\right|\leq K\left|x-y\right|\;\forall\,x,y\in A$.\ Let $(M,d)$ and $(N,\rho)$ be metric spaces and $f:M\to N$.\ $f$ is ==Lipschitz function== if $\exists\,K>0$ such that $\rho(f(x),f(y))\leq K d(x,y)\;\forall\,x,y\in M$. --- Let $A\subseteq \mathbb{R}$ and $f:A\to \mathbb{R}$.\ $f$ is ==Hölder continuous with exponents $\alpha$== if $\exists\, K>0$ and $0<\alpha\leq 1$ such that $\left|f(x)-f(y)\right|\leq K\left|x-y\right|^\alpha \;\forall\,x,y\in A$. --- Suppose $f:\mathbb{R}\to\mathbb{R}$. $f$ is differentiable and $\exists\, K>0$ such that $|f'(x)|<K$ for all $x\in\mathbb{R}$ \ $\implies$ $f$ is Lipschitz \ $\implies$ $f$ is Hölder continuous with some $0<\alpha<1$ \ $\implies$ $f$ is uniformly continuous --- Let $(M,d)$ and $(N,\rho)$ be metric spaces, $A\subseteq M$ and $f:A\to N$.\ $f$ is uniformly continuous on $A$ **iff** for any two sequences $\left\{x_n\right\}_{n=1}^\infty$, $\left\{y_n\right\}_{n=1}^\infty \,\subseteq A$, $\displaystyle\lim_{n\to\infty}d(x,y)=0\implies\displaystyle\lim_{n\to\infty}\rho(f(x_n),f(y_n))=0$. :::spoiler proof: > $\implies$: \ > Prove by contradiction. \ > Assume $\displaystyle\lim_{n\to\infty}d(x_n,y_n)=0$ and $\displaystyle\lim_{n\to\infty}\rho(f(x_n),f(y_n))\neq 0$ for some $\{x_n\}_{n=1}^\infty, \{y_n\}_{n=1}^\infty \subseteq A$. \ > $\Rightarrow$ $\exists\,\epsilon>0$ such that $\forall\, > k\in\mathbb{N}$, $\exists\,n_k\geq k$ such that $\rho(f\left(x_{n_k}),f(y_{n_k})\right)\geq\epsilon$. \ > Since $f$ is uniformly continuous, $\exists\,\delta>0$ such that $\forall\, x,y\in A$ with $d(x,y)<\delta$, we have $\rho(f(x),f(y))<\epsilon$. $\quad(\rightarrow\leftarrow)$ \ > $\impliedby$: \ > Prove by contradiction. \ > $\exists\,\epsilon>0$ such that $\forall\, n\in\mathbb{N}$, $\exists\,x_n,y_n\in A$ with $d(x_n,y_n)<\frac{1}{n}$ but $\rho(f(x_n),f(y_n))\geq\epsilon$. \ > Then for the sequences $\{x_n\}_{n=1}^\infty,\{y_n\}_{n=1}^\infty \subseteq A$ with $\displaystyle\lim_{n\to\infty}d(x_n,y_n)=0$, $\displaystyle\lim_{n\to\infty}\rho(f(x_n),f(y_n))\neq 0$. $\quad(\rightarrow\leftarrow)$ ::: --- Suppose that $\forall\,\epsilon>0$, $\exists\,\delta(x,\epsilon)>0$ such that $\rho(f(x),f(y))<\epsilon$ whenever $d(x,y)<\delta(x,\epsilon)$.\ Define $\delta_f(\epsilon)=\displaystyle\inf_{x\in A}\delta(x,\epsilon)$. \ If $\delta_f(\epsilon)>0\;\forall\,\epsilon>0$, then $f$ is uniformly continuous on $A$. --- Let $(M,d)$ and $(N,\rho)$ be metric spaces, $A\subseteq M$, and $f:A\to N$ be uniformly continuous.\ If $\left\{x_n\right\}_{n=1}^\infty \subseteq A$ is a Cauchy sequence in $(M,d)$, then $\left\{f(x_n)\right\}_{n=1}^\infty$ is also a Cauchy sequence in $(N,\rho)$. :::spoiler proof: > $\forall\,\epsilon>0$, $\exists\,\delta>0$ such that $\rho(f(x),f(y))<\epsilon$ whenever $x,y\in A$ and $d(x,y)<\delta$. \ > Since $\{x_n\}_{n=1}^\infty \subseteq A$ is Cauchy in $(M,d)$, there exists $L\in\mathbb{N}$ such that $d(x_m,x_n)<\delta$ whenever $m,n\geq L$. \ > Therefore for $m,n\geq L$, we have $\rho(f(x_m),f(x_n))<\epsilon$. \ > Thus $\{f(x_n)\}_{n=1}^\infty$ is Cauchy in $(N,\rho)$. ::: --- ### Continuous Extension Let $(M,d)$ and $(N,\rho)$ be two metric spaces, $A\subseteq M$, and $f:A\to N$ be uniformly comitnuous.\ If $(N,\rho)$ is complete, then $f$ has a unique extension to a continuous function on $\bar{A}$, that is, $\exists\, g:\bar{A}\to N$ such that 1. $g$ is uniformly continuous on $\bar{A}$ 2. $g(x)=f(x)\;\forall\,x\in A$ 3. If there is $h:\bar{A}\to N$ satisfying 1. and 2., then $g=h$ on $\bar{A}$. :::spoiler proof: > We need to define the value of $g$ on $\bar{A}\backslash A$. \ > Suppose $x\in\bar{A}\backslash A$. \ > Then there exists $\left\{x_n\right\}_{n=1}^\infty \subseteq A$ converging to $x$. \ > Hence $\left\{x_n\right\}_{n=1}^\infty \subseteq A$ is Cauchy in $M$. \ > Since $f$ is uniformly continuous on $A$, $\left\{f(x_n)\right\}_{n=1}^\infty \subseteq A$ is a Cauchy sequence in $(N,\rho)$. \ > Since $(N,\rho)$ is complete, $\left\{f(x_n)\right\}_{n=1}^\infty$ converges in $(N,\rho)$, say $\displaystyle\lim_{n\to\infty}f(x_n)=y_x$. \ > Define $g(x) = \begin{cases} > f(x),\quad &x\in A \\ > y_x,\quad &x\in\bar{A}\backslash A > \end{cases}$ \ > To check that $g(x)$ is well-defined, consider another sequence $\left\{z_n\right\}_{n=1}^\infty \subseteq A$ converging to $x$. \ > Then $d(x_n,z_n)\to 0$ as $n\to\infty$. \ > Since $f$ is uniformly conitnuous on $A$, $\displaystyle\lim_{n\to\infty}\rho(f(x_n),f(z_n))=0$. \ > Therefore $\displaystyle\lim_{n\to\infty}f(x_n)=y_x=\displaystyle\lim_{n\to\infty}f(z_n)$. \ > Hence $g$ is well-defined on $\bar{A}$. \ > Check condition 1. :\ > Suppose $\epsilon>0$. \ > Since $f$ is uniformly continuous on $A$, $\exists\,\delta>0$ such that if $x, y\in A$ with $d(x,y)<\delta$, then $\rho(f(x),f(y))<\epsilon$. \ > For $u,v\in \bar{A}$ with $d(u,v)<\frac{\delta}{3}$, by the definition of $g$, $\exists\,\{u_n\}_{n=1}^{\infty}, \{v_n\}_{n=1}^{\infty}\subseteq A$ such that $\displaystyle\lim_{n\to\infty}u_n=u$ and $\displaystyle\lim_{n\to\infty}v_n=v$. \ > $\Rightarrow$ $\exists\, u_m, v_k$ with $d(u,u_m)<\frac{\delta}{3}$ and $d(v,v_k)<\frac{\delta}{3}$ such that > $\rho(f(u),f(u_m))<\frac{\epsilon}{3}$ and $\rho(f(v),f(v_k))<\frac{\epsilon}{3}$. \ > Then $$\begin{aligned} > d(u_m,v_k) &\leq d(u_m,u)+d(u,v)+d(v,v_k) \\ > &<\frac{\delta}{3}+\frac{\delta}{3}+\frac{\delta}{3} = \delta > \end{aligned}$$ > This implies $\rho(f(u_m),f(v_k))<\frac{\epsilon}{3}$ and hence > $$\begin{aligned} > \rho(g(u),g(v)) &\leq \rho(g(u),g(u_m))+\rho(g(u_m),g(v_k))+\rho(g(v_k),g(v)) \\ > &< \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3} \\ > &= \epsilon > \end{aligned}$$ > Hence $g$ is uniformly continuous on $\bar{A}$. \ > check condition 3. : \ > Suppose $h:\bar{A}\to N$ satisfying 1. and 2. . \ > Let $x\in\bar{A}\backslash A$. \ > Given $\epsilon>0$, $\exists\, \delta_1,\delta_2>0$ such that $d(x,y)<\delta_1 \implies \rho(g(x),g(y))<\frac{\epsilon}{2}$ > and $d(x,y)<\delta_2 \implies \rho(h(x),h(y))<\frac{\epsilon}{2}$. \ > Since $x\in \bar{A}\backslash A$, $\exists\, y\in A$ such that $d(x,y)<\min(\delta_1,\delta_2)$.\ > Then > $$\begin{aligned} > \rho(g(x),h(x)) &\leq \rho(g(x),g(y))+\rho(g(y),h(y))+\rho(h(y),h(x))\\ > &< \frac{\epsilon}{2}+0+\frac{\epsilon}{2}\\ > &=\epsilon > \end{aligned}$$ > Therefore $g(x)=h(x)\;\forall\, x\in\bar{A}$. ::: --- Let $(M,d)$ and $(N,\rho)$ be metric spaces, $A\subseteq M$ and $f:A\to N$ be a continuous map. \ If $K\subseteq A$ is compact, then $f(K)$ is compact in $(N,\rho)$. :::spoiler proof: > Suppose that $\{U_\alpha\}_{\alpha\in I}$ is an open cover of $f(K)$. \ > For every $\alpha\in I$, since $f:A\to N$ is continuous and $U_\alpha$ is open in $(N,\rho)$, $f^{-1}(U_\alpha)$ is open in $A$. \ > Then for each $\alpha\in I$, there exists $V_\alpha$ which is open in $M$ such that $f^{-1}(U_\alpha)=V_\alpha\cap A$. \ > Since $f(K)\subseteq \displaystyle\bigcup_{\alpha\in I}U_\alpha$ and $K\subseteq A$, $\{V_\alpha\}_{\alpha \in I}$ is an open cover of $K$. \ > Since $K$ is compact, there exists a finite subcover $\{V_{\alpha_i}\}_{i=1}^n$ such that > $K\subseteq \displaystyle\bigcup_{i=1}^{n}V_{\alpha_i} \cap A = \displaystyle\bigcup_{i=1}^{n}f^{-1}(U_{\alpha_i})$. \ > Then $f(K)\subseteq \displaystyle\bigcup_{i=1}^{n} U_{\alpha_i}$ and thus $f(K)$ is compact in $(N,\rho)$. ::: --- Let $(M,d)$ be a metric space and $K\subseteq M$ be compact. \ If $f:M\to\mathbb{R}$ is continuous, then $f$ attains its maximum and minimum in $K$. \ That is, $\exists\,x_0,x_1\in K$ such that $f(x_0)=\displaystyle\max_{x\in K}f(x)$ and $f(x_1)=\displaystyle\min_{x\in K}f(x)$. :::spoiler proof: > Since $f$ is continuous and $K$ is compact, $f(K)$ is compact in $\mathbb{R}$. \ > Hence, $f(K)$ is closed and bounded. \ > Then $f$ attains its extreme values in $K$. ::: :::info This provides the metric space version of extreme value theorem. ::: --- Let $(M,d)$ be a metric space, $K\subseteq M$ be compact and $f:K\to\mathbb{R}$ be a continuous map. \ Then the set $\big\{ x\in K\mid f(x)=\displaystyle\max_{x\in K}f(x)\big\}$ is nonempty and compact. --- Let $E$ be a noncompact set in $\mathbb{R}$. \ Then 1. there exists a continuous function on $E$ which is unbounded. 2. there exists a bounded and continuous function on $E$ which has no maximum. --- Let $(M,d)$ and $(N,\rho)$ be metric spaces, $A\subseteq M$, and $f:A\to N$ be a map. \ If $K\subseteq A$ is compact and $f$ is continuous, then $f$ is uniformly continuous on $K$. :::spoiler proof: > Since $f$ is continuous on $K$, given $\varepsilon>0$, for every $x\in K$, > $\exists\,\delta_x>0$ such that $\rho(f(x),f(y))<\frac{\varepsilon}{2}$ whenever $y\in K$ and $d(x,y)<\delta_x$. \ > Since $K$ is compact and $K\subseteq \displaystyle\bigcup_{x\in K}B(x,\frac{\delta_x}{2})$, > there exist $x_1,\dotsc,x_L\in K$ such that $K\subseteq \displaystyle\bigcup_{i=1}^L B(x_i,\frac{\delta_{x_i}}{2})$. \ > Select $\delta=\displaystyle\min_{1\leq i\leq L} \frac{\delta_{x_i}}{2}$. \ > Suppose $u,v\in K$ and $d(u,v)<\delta$. \ > Since $K\subseteq \displaystyle\bigcup_{i=1}^L B(x_i,\frac{\delta_{x_i}}{2})$, > there is $1\leq l\leq L$ such that $u\in B(x_l,\frac{\delta_{x_l}}{2})$. \ > Then > $$\begin{aligned} > d(v,x_l) &\leq d(v,u)+d(u,x_l) \\ > &< \delta+\frac{1}{2}\delta_{x_l} \\ > &\leq \delta_{x_l} > \end{aligned}$$ > Therefore $v\in B(x_l,\delta_{x_l})$ and we have > $$\begin{aligned} > \rho(f(u),f(v)) &\leq \rho(f(u),f(x_l))+\rho(f(x_l),f(v)) \\ > &< \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon > \end{aligned}$$ > Hence $d(u,v)<\delta\implies \rho(f(u),f(v))<\varepsilon$. \ > Thus $f$ is uniformly continuous on $K$. ::: --- Let $(M,d)$ and $(N,\rho)$ be two metric spaces, $K\subseteq M$ be compact, and $f:K\to N$ be one-to-one and continuous function. \ Then the inverse function $f^{-1}:f(K)\to K$ is continuous. :::spoiler proof: > To prove that $f^{-1}$ is continuous, > we can prove that for every closed $E\subseteq M$, the preimage $\left(f^{-1}\right)^{-1}(E)$ is closed relative to $f(K)$. \ > Since $E\subseteq M$ is closed and $K$ is compact in $M$, $E\cap K$ is compact in $M$. \ > Since $f$ is continous, $f(E\cap K)$ is compact on $N$, and hence is closed in $N$. \ > Since $f$ is one-to-one, we have $f(E\cap K)=\left(f^{-1}\right)^{-1}(E\cap K)=\left(f^{-1}\right)^{-1}(E)$ is closed in $f(K)$. \ > Therefore, $f^{-1}$ is continuous on $f(K)$. ::: counterexample for noncompact $K$: - $f(t)=(\cos(t),\sin(t))$ on $[\,0,2\pi)$ leads to discontinuous $f^{-1}$. --- ### Continuous Maps on Connected Sets and Path Connected Sets Let $(M,d)$ be a metric space and $A\subseteq M$. \ $A$ is said to be ==path connected== if for every pair of points $x,y\in A$, they can be joined by a *path* in $M$. That is, there exists a continuous map $\phi:[0,1]\to A$ such that $\phi(0)=x$ and $\phi(1)=y$. --- Let $V$ be a vector space and $A\subseteq V$.\ $A$ is called ==convex== if for all $x,y\in A$, the *line segment* joining $x$ and $y$, denoted by $\overline{xy}$, lies in $A$. :::danger Notice the difference between path and line segment. ::: --- Any open or closed ball in a vector space $V$ is convex. --- A convex set in a normed space is path connected by taking $\phi(t)=(1-t)x+ty$. --- Let $A,B\subseteq M$ be path connected. \ If there exists $a\in A$ and $b\in B$ and a path in $A\cup B$ joining $a$ and $b$, then $A\cup B$ is path connected. --- Under a metric space, \ path-connected $\implies$ connected :::spoiler proof: > Let $(M,d)$ be a metric space and $A\subseteq M$. \ > To prove by contradiction, assume that $A$ is disconnected. \ > Then there exist open sets $u$ and $v$ in $M$ such that (i) $A\subseteq u\cup v$ > (ii) $A\cap u\neq\emptyset$ (iii) $A\cap v\neq\emptyset$ (iv) $A\cap u\cap v=\emptyset$. \ > By (ii) and (iii), choose $x\in A\cap u$ and $y\in A\cap v$. \ > Since $A$ is path-connected, there exists a continuous map $\phi:[0,1]\to A$ > such that $\phi(0)=x$ and $\phi(1)=y$. \ > Then the sets $w_1= \phi^{-1}(u)$ and $w_2= \phi^{-1}(v)$ are open relative to $[0,1]$. \ > By (i), $[0,1]\subseteq w_1\cup w_2$. \ > Since $0\in w_1$ and $1\in w_2$, we have $[0,1]\cap w_1\neq\emptyset$ and $[0,1]\cap w_2\neq\emptyset$. \ > By (iv), $[0,1]\cap w_1\cap w_2 = \phi^{-1}(u)\cap \phi^{-1}(v)=\emptyset$. \ > Then $w_1$ and $w_2$ separate $[0,1]$. $\quad (\rightarrow\leftarrow)$ ::: :::success The converse is false. \ For example, $A=\left\{(x,\sin\frac{1}{x})\mid x\in(0,1)\right\}\cup ({0}\times[-1,1])$ is connected but not path-connected. ::: --- Let $(M,d)$ and $(N,\rho)$ be two metric spaces, $A\subseteq M$, and $f:A\to N$ be a continuous map. 1. If $A$ is connected, then $f(A)$ is connected. 2. If $A$ is path-connected, then $f(A)$ is path-connected. :::spoiler proof: > (1.): \ > To prove by contradiction, assume that $f(A)$ is disconnected. \ > Then there exist open sets $u$ and $v$ in $N$ such that (i) $f(A)\subseteq u\cup v$ (ii) $f(A)\cap u\neq\emptyset$ > (iii) $f(A)\cap v\neq \emptyset$ (iv) $f(A)\cap u\cap v=\emptyset$. \ > Since $f$ is continuous and $u,v$ are open in $N$, $f^{-1}(u)$ and $f^{-1}(v)$ are open relative to $A$. \ > Then there are $w_1$, $w_2$ open in $M$ such that $f^{-1}(u)=A\cap w_1$ and $f^{-1}(v)=A\cap w_2$. \ > By (i), $A\subseteq f^{-1}(u)\cup f^{-1}(v) \subseteq w_1\cup w_2$. \ > By (ii) and (iii), $A\cap w_1\neq\emptyset$ and $A\cap w_2\neq\emptyset$. \ > By (iv), $A\cap w_1\cap w_2=\emptyset$. \ > Hence, $A$ is disconnected. $\quad(\rightarrow\leftarrow)$ \ > (2.): \ > Suppose $y_1,y_2\in f(A)$. \ > $\Rightarrow$ $\exists\,x_1,x_2\in A$ such that $y_1=f(x_1)$ and $y_2=f(x_2)$. \ > Since $A$ is path-connected, there exists a continuous map $\phi:[0,1]\to A$ such that > $\phi(0)=x_1$ and $\phi(1)=x_2$. \ > Define $\psi(t)=f(\phi(t))$. \ > Clearly, $\psi:[0,1]\to f(A)$. \ > Since $f$ and $\phi$ are continuous, $\psi$ is continuous on $[0,1]$. \ > Then $\psi(0)=f(\phi(0))=f(x_1)=y_1$ and $\psi(1)=f(\phi(1))=f(x_2)=y_2$. \ > Hence, $\psi$ is a path in $f(A)$ joining $y_1$ and $y_2$. \ > Since $y_1$ and $y_2$ are arbitrary two points in $f(A)$, $f(A)$ is path-connected. ::: --- Let $V$ be a vector space and $\phi:[0,1]\to V$ be a continuous map. \ $\phi$ is said to be ==piecewise linear== if $\exists\,t_0,t_1,t_2,\dotsc,t_n\in[0,1]$ with $0=t_0<t_1<\cdots<t_n=1$ such that $\phi$ is a linear map on each $[t_{i-1},t_i]$. --- Let $V$ be a vector space and $x,y,z\in V$. \ If there are piecewise linear mapping $\phi_1,\phi_2:[0,1]\to V$ such that $\phi_1$ joins $x$ and $y$ and $\phi_2$ joins $y$ and $z$, then there exists a piecewise linear mapping $\phi:[0,1]\to V$ which joins $x$ and $z$. --- Let $G$ be a connected and open set in a vector space $V$. \ Then for any $x,y\in G$, there exists a piecewise linear mapping joining $x$ and $y$. :::spoiler proof: > Suppose $x\in G$. \ > Define $G_1= \left\{z\in G\mid \text{there exists a piecewise linear mapping joining } x \text{ and } z\right\}$. \ > To prove $G=G_1$, we want to show that $G_1$ is nonempty, open and closed in $G$. \ > Since $x\in G_1$, $G_1$ is nonempty. \ > Claim 1: $G_1$ is open \ > Suppose $z\in G_1$. \ > Since $G$ is open, $\exists\,\delta>0$ such that $B(z,\delta)\subseteq G$. \ > Since $B(z,\delta)$ is convex, for any point $z_1\in B(z,\delta)$, > there exists a piecewise linear mapping joining $z$ and $z_1$. \ > Then there is a piecewise linear mapping joining $x$ and $z_1$. \ > Hence $z_1\in G_1$. \ > Therefore, $B(z,\delta)\subseteq G_1$ and $G_1$ is open. \ > Claim 2: $G_1$ is closed \ > Suppose $w\in G\backslash G_1$. \ > Then there exists no piecewise linear mapping joining $x$ and $w$. \ > Since $G$ is open, $\exists\,r>0$ such that $B(w,r)\subseteq G$. \ > For any point $w_1\in B(w,r)$, there exists a piecewise linear mapping joining $w$ and $w_1$. \ > To prove $w_1\notin G_1$ by contradiction, assume $w_1\in G_1$. \ > Then there is a piecewise linear mapping joining $x$ and $w_1$. \ > $\Rightarrow$ $w\in G_1$ $\quad(\rightarrow\leftarrow)$ \ > Hence $B(w,r)\subseteq G\backslash G_1$ and $G\backslash G_1$ is open. \ > $\Rightarrow$ $G_1$ is closed. \ > Finally, since $G$ is connected and $G_1$ is a nonempty, open and closed subset of $G$, we have $G_1=G$. ::: --- |domain|function|codomain| |:-:|:-:|:-:| |compact|conitnuous|**compact**| |compact|continuous|**uniformly continuous**| |connected|continuous|**connected**| |path-connected|continuous|**path-connected**| --- ## Pointwise and Uniform Convergence Let $(M,d)$ and $(N,\rho)$ be two metric spaces, $A\subseteq M$, and $f,f_k:A\to N$ be maps $\forall\,k\in\mathbb{N}$. \ The sequence of maps $\{f_k\}_{k=1}^\infty$ is said to ==converge pointwise to $f$ on $A$== if $\displaystyle\lim_{k\to\infty}\rho(f_k(a),f(a))=0$ for every $a\in A$. \ That is, $\forall\,\varepsilon>0$ and $\forall\,a\in A$, $\exists\, N_{\varepsilon,a}>0$ such that $\rho(f_k(a),f(a))<\varepsilon$ whenever $k\geq N_{\varepsilon,a}$. :::danger Notice that these $N$ also depend on $a$. ::: --- Let $(M,d)$ and $(N,\rho)$ be two metric spaces, $A\subseteq M$, and $f,f_k:A\to N$ be maps $\forall\,N\in\mathbb{N}$. \ The sequence of maps $\{f_k\}_{k=1}^\infty$ is said to ==converge uniformly to $f$ on $A$== if $\forall\,\varepsilon>0$, $\exists\,N_\varepsilon>0$ such that $\rho(f_k(a),f(a))<\varepsilon$ whenever $a\in A$ and $k\geq N_\varepsilon$. :::spoiler Example: $f,f_k: [0,1]\to\mathbb{R}$ by $f_k(x)=x^k$ and $f(x)=\begin{cases} 0, &x\in[0,1) \\ 1, &x=1 \end{cases}$. \ Then 1. $f_k\to f$ pointwise 2. $\{f_k\}_{k=1}^\infty$ does not converge uniformly to $f$ on $[0,1]$. 3. For $0<a<1$, $f_k\to f$ uniformly on $[0,a]$ > 2: \ > Let $\varepsilon=\frac{1}{2}$. \ > Choose $x_N=\sqrt[N]{\frac{2}{3}}$ $\;\forall\,N\in\mathbb{N}$. \ > Then $\left|f_N(x_N)-f(x_N)\right|= \left|\frac{2}{3}-0\right|> \varepsilon$. > 3: \ > Fix $0<a<1$. \ > Given $\epsilon>0$, choose $N\in\mathbb{N}$ such that $N>\dfrac{\ln \epsilon}{\ln a}$. \ > Since $\ln a<0$, $a^N<\epsilon$. \ > For every $x\in[0,a]$ and $k\geq N$, $\left|f_k(x)-f(x)\right|= |x^k-0|\leq a^k\leq a^N< \epsilon$. \ > Hence, $f_k\to f$ uniformly on $[0,a]$. ::: --- Let $(M,d)$ and $(N,\rho)$ be two metric spaces, $A\subseteq M$, and $f,f_k:A\to N$ be maps $\forall\,k\in\mathbb{N}$. \ If $f_k\to f$ uniformly, then $f_k\to f$ pointwise. --- Let $(M,d)$ and $(N,\rho)$ be two metric spaces, $A\subseteq M$, and $f_k:A\to N$ be maps $\forall\,k\in\mathbb{N}$. \ Suppose that $(N,\rho)$ is complete. \ Then $\{f_k\}_{k=1}^\infty$ converges uniformly on $A$ ***iff*** $\forall\,\varepsilon>0$, $\exists\,L\in\mathbb{N}$ such that $\rho(f_m(x),f_n(x))<\varepsilon$ whenever $x\in A$ and $m,n\geq L$. :::spoiler proof: > $\implies:$ \ > Let $f:A\to N$ be a map where $f_k\to f$ uniformly on A. \ > Given $\epsilon>0$, $\exists\,L\in\mathbb{N}$ such that $\rho(f_k(x),f(x))<\frac{\epsilon}{2}$ > whenever $x\in A$ and $k\geq L$. \ > For $m,n\geq L$ and $x\in A$, > $$\begin{aligned} > \rho(f_m(x),f_n(x)) &\leq \rho(f_m(x),f(x))+\rho(f(x),f_n(x)) \\ > &< \frac{\epsilon}{2}+\frac{\epsilon}{2} =\epsilon > \end{aligned}$$ > $\impliedby:$ \ > Let $\{f_k\}_{k=1}^\infty$ be a sequence of maps on $A$ satisfying the Cauchy criterion. \ > Fix $a\in A$, $\{f_k(a)\}_{k=1}^\infty$ is a Cauchy sequence in $N$. \ > Since $(N,\rho)$ is complete, $\exists\,y_a\in N$ such that $\displaystyle\lim_{k\to\infty}f_k(a)=y_a$ in $N$. \ > By the same argument, $\forall\,x\in A$, there exists such $y_x$. \ > Define $f:A\to N$ by $f(x)=y_x$. \ > Then $f_k\to f$ pointwise on $A$. \ > Given $\epsilon>0$, by the Cauchy criterion, there exists $L\in\mathbb{N}$ such that > $\rho(f_m(x),f_n(x))<\frac{\epsilon}{2}$ whenever $x\in A$ and $m,n\geq L$. \ > Since $f_k\to f$ pointwise on $A$, for every $x\in A$, $\exists\,L_x\in\mathbb{N}$ with $L_x\geq L$ such that > $\rho(f_m(x),f(x))<\frac{\epsilon}{2}$ whenever $m>L_x$. \ > Hence for every $x\in A$ and $m\geq L$, we choose $m_x\geq L_x$. \ > Then > $$\begin{aligned} > \rho(f_m(x),f(x)) &\leq \rho(f_m(x),f_{m_x}(x))+\rho(f_{m_x}(x),f(x)) \\ > &< \frac{\epsilon}{2}+\frac{\epsilon}{2} =\epsilon > \end{aligned}$$ > Therefore $f_k\to f$ uniformly on $A$. ::: --- Let $(M,d)$ and $(N,\rho)$ be two metric spaces, $A\subseteq M$ and $f_k:A\to N$ be a sequence of continuous map converging to $f:A\to N$ uniformly on $A$. \ Then $f$ is continuous on $A$. :::spoiler proof: > Since $f_k\to f$ uniformly on $A$, for given $\varepsilon>0$, there exists $L\in\mathbb{N}$ such that for every $x\in A$ and $k\geq L$, > $$\rho(f_k(x),f(x))<\frac{\varepsilon}{3}$$ > Since $f_L$ is continuous on $A$, $\forall\,a\in A$, $\exists\,\delta_a>0$ such that $\rho(f_L(x),f_L(a))<\frac{\varepsilon}{3}$ > whenever $x\in A$ and $d(x,a)<\delta_a$. \ > Hence, for every $x\in A$ and $d(x,a)<\delta_a$, > $$\begin{aligned} > \rho(f(x),f(a)) &\leq \rho(f(x),f_L(x))+\rho(f_L(x),f_L(a))+\rho(f_L(a),f(a)) \\ > &<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3} \\ > &=\varepsilon > \end{aligned}$$ > Thus $f$ is continuous at $a$. \ > $f$ is continuous on $A$. ::: :::success If $f_k$ continuous but $f$ not continuous, then $f_k \not\to f$ uniformly. ::: :::info If $f_k\to f$ uniformly on $A$ and $a\in\overline{A}$, then $$\lim_{x\to a}\left(\lim_{k\to\infty}f_k(x)\right) = \lim_{k\to\infty}\left(\lim_{x\to a}f_k(x)\right)$$ ::: --- Let $f:[a,b]\to\mathbb{R}$ and $P=\{a=t_0<t_1<t_2<\cdots<t_n=b\}$ be a partition of $[a,b]$. \ The upper and lower sums of $P$ for $f$ are $U(P,f)=\displaystyle\sum_{i=1}^n M_i(t_i-t_{i-1})$ and $L(P,f)=\displaystyle\sum_{i=1}^n m_i(t_i-t_{i-1})$ where $M_i=\displaystyle\sup_{t\in [t_{i-1},t_i]}f(t)$ and $m_i=\displaystyle\inf_{t\in[t_{i-1},t_i]}f(t)$. \ Define $\underline{\int_a^b} f(x)\,dx=\displaystyle\sup_P L(P,f)$ and $\overline{\int_a^b}f(x)\,dx=\displaystyle\inf_{P} U(P,f)$. \ We have 1. $L(P,f)\leq U(P,f)$ 2. If $P_1$ is a refinement of $P$, then $L(P,f)\leq L(P_1,f)\leq U(P_1,f)\leq U(P,f)$. 3. For any two partitions $P_1$ and $P_2$, $L(P_1,f)\leq U(P_2,f)$. 4. $\underline{\int_a^b} f(x)\,dx \leq \overline{\int_a^b} f(x)\,dx$ If $\underline{\int_a^b}f(x)\,dx = \overline{\int_a^b}f(x)\,dx$, we say $f$ is ==integrable== on $[a,b]$ and denoted by $\int_a^b f(x)\,dx$. --- A function $f$ is integrable on $[a,b]$ ***iff*** $\forall\,\varepsilon>0$, there exists a partition $P$ of $[a,b]$ such that $U(P,f)-L(P,f)<\varepsilon$. --- If $f:[a,b]\to\mathbb{R}$ is piecewise continuous, then $f$ is integrable. --- Let $f_k:[a,b]\to\mathbb{R}$ be a sequence of integrable functions which converge uniformly to $f$ on $[a,b]$. \ Then $f$ is integrable on $[a,b]$ and $\displaystyle\lim_{k\to \infty}\int_a^b f_k(x)\,dx = \int_a^b f(x)\,dx = \int_a^b\lim_{k\to\infty}f_k(x)\,dx$. :::spoiler proof: > Since $\{f_k\}_{k=1}^\infty$ converges uniformly to $f$ on $[a,b]$, for given $\varepsilon>0$, > $\exists\,N\in\mathbb{N}$ such that $\left|f_k(x)-f(x)\right|<\varepsilon$ whenever $x\in [a,b]$ and $k\geq N$. \ > Since $f_N$ is integrable on $[a,b]$, there exists a partition $P$ of $[a,b]$ such that $U(P,f_N)-L(P,f_N)<\varepsilon$. \ > Let $M_i=\displaystyle\sup_{t\in[t_{i-1},t_i]}f(t)$, $m_i=\displaystyle\inf_{t\in[t_{i-1},t_i]}f(t)$, > and $M_i^{(N)}=\displaystyle\sup_{t\in[t_{i-1},t_i]}f_N(t)$, $m_i^{(N)}=\displaystyle\inf_{t\in[t_{i-1},t_i]}f_N(t)$. \ > Then > $$\begin{aligned} > \left|M_i-M_i^{(N)}\right| &\leq \sup_{t\in[t_{i-1},t_i]} \left|f(t)-f_N(t)\right|\leq\varepsilon \\ > \left|m_i-m_i^{(N)}\right| &\leq \sup_{t\in[t_{i-1},t_i]} \left|f(t)-f_N(t)\right|\leq\varepsilon > \end{aligned}$$ > And we have > $$\begin{aligned} > U(P,f)-L(P,f) &\leq \left|U(P,f)-U(P,f_N)\right|+\left|U(P,f_N)-L(P,f_N)\right|+\left|L(P,f_N)-L(P,f)\right| \\ > &<\sum_{i=1}^n \left|M_i-M_i^{(N)}\right|(t_i-t_{i-1})+\varepsilon+\sum_{i=1}^n \left|m_i-m_i^{(N)}\right|(t_i-t_{i-1}) \\ > &\leq2\varepsilon\sum_{i=1}^n(t_i-t_{i-1})+\varepsilon \\ > &=[2(b-a)+1]\varepsilon > \end{aligned}$$ > Hence, $f$ is integrable on $[a,b]$. \ > Moreover, for $k\geq N$, > $$\begin{aligned} > \left|\int_a^b f(x)\,dx-\int_a^b f_k(x)\,dx\right| &=\left|\int_a^b f(x)-f_k(x)\,dx\right| \\ > &\leq \int_a^b\left|f(x)-f_k(x)\right|\,dx \\ > &<\int_a^b\varepsilon\,dx =\varepsilon(b-a) > \end{aligned}$$ > Therefore $\displaystyle\int_a^b f(x)\,dx = \lim_{k\to\infty}\int_a^b f_k(x)\,dx$. ::: --- Suppose that $f_k\to f$ pointwise and $\int f_k\,dx\to\int f\,dx$. \ It cannot imply that $f_k\to f$ uniformly. :::spoiler counterexample: > The set $\mathbb{Q}\cap[0,1]$ is countable. Write $\mathbb{Q}\cap[0,1]=\{q_k\mid k\in \mathbb{N}\}$. \ > Define $f_k(x):[0,1]\to\mathbb{R}$ by $f_k(x)=\begin{cases} > 1, &x\in\{q_1,q_2,\dotsc q_k\} \\ > 0, &\text{otherwise} > \end{cases}$ and $f(x)=\begin{cases} > 1, &x\in\mathbb{Q}\cap[0,1] \\ > 0, &\text{otherwise} > \end{cases}$. \ > Then $f_k\to f$ pointwise on $[0,1]$. \ > Every $f_k(x)$ is integrable on $[0,1]$, but $f$ is not integrable on $[0,1]$. \ > Hence $\{f_k\}_{k=1}^\infty$ does not converge uniformly to $f$ on $[0,1]$. ::: --- Let $I\subseteq \mathbb{R}$ be a finite interval, $f_k:I\to\mathbb{R}$ be a sequence of differentiable functions. \ Suppose that $\{f_k(a)\}_{k=1}^\infty$ converges for some $a\in I$ and $\{f_k'\}_{k=1}^\infty$ converges uniformly to a function $\mathrm{g}$ on $I$. \ Then 1. $\{f_k\}_{k=1}^\infty$ converges uniformly to some function $f$ on $I$. 2. The limit function $f$ is differentiable on $I$ and $f'(x)=\mathrm{g}(x)$ for all $x\in I$. \ That is, $\mathrm{g}(x)=\displaystyle\lim_{k\to\infty}\left(\frac{d}{dx}f_k(x)\right)= f'(x)= \frac{d}{dx}\left(\lim_{k\to\infty}f_k(x)\right)$. :::spoiler proof: > For $x\in I$, by the F.T.C., $f_k(x)=f_k(a)+\int_a^x f_k'(t)\,dt$. \ > Since $\{f_k(a)\}_{k=1}^\infty$ converges, by the previous theorem, $\{\int_a^x f_k'(t)\,dt\}_{k=1}^\infty$ converges for every $x\in I$. \ > Then $\forall\,x\in I$, $\{f_k(x)\}_{k=1}^\infty$ converges, and we can define $f(x)=\displaystyle\lim_{k\to\infty}\left(f_k(a)+\int_a^x f_k'(t)\,dt\right)$. \ > Hence $f(a)=\displaystyle\lim_{k\to\infty}f_k(a)$ and $f(x)-f(a)=\displaystyle\lim_{k\to\infty}\int_a^xf_k'(t)\,dt= \int_a^x\displaystyle\lim_{k\to\infty}f_k'(t)\,dt= \int_a^x \mathrm{g}(t)\,dt$. \ > $\Rightarrow$ $f'(x)=\mathrm{g}(x)$ \ > Now we want to check $f_k\to f$ uniformly on $I$. \ > Since $\displaystyle\lim_{k\to\infty}f_k(a)=f(a)$, given $\varepsilon>0$, $\exists\,N_1\in\mathbb{N}$ such that $|f_k(a)-f(a)|<\frac{\varepsilon}{2}$ whenever $k\geq N_1$. \ > Since $f_k'(x)\to \mathrm{g}(x)$ uniformly on $I$, $\exists\,N_2\in\mathbb{N}$ such that $\left|f_k'(x)-\mathrm{g}(x)\right|<\dfrac{\varepsilon}{2|I|}$ whenever $x\in I$ and $k\geq N_2$. \ > Therefore, for $k\geq \max(N_1,N_2)$, > $$\begin{aligned} > \left|f(x)-f_k(x)\right| &= \left|\left[f(a)+\int_a^x \mathrm{g}(t)\,dt\right]-\left[f_k(a)+\int_a^xf_k'(t)\,dt\right]\right| \\ > &\leq|f(a)-f_k(a)|+\int_a^x\left|\mathrm{g}(t)-f_k'(t)\right|\,dt \\ > &<\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon > \end{aligned}$$ > Thus $f_k\to f$ uniformly on $I$. ::: --- Suppose $f_k$ differentiable and $f_k\to f$ uniformly. \ It cannot imply $f$ differentialble. :::spoiler counterexample: $$f_k(x)=\begin{cases} \frac{k}{2}x^2, &|x|\leq\frac{1}{k} \\ |x|-\frac{1}{2k}, &\frac{1}{k}\leq|x|\leq 1 \end{cases} \text{ and } f(x)=|x|$$ ::: --- #### Dini's Theorem Suppose that $K$ is compact and \ (a) $f_n:K\to\mathbb{R}$ is continuous on $K$ $\;\forall\, n\in\mathbb{N}$. \ (b) $f_n\to f$ pointwise on $K$ where $f$ is continuous \ (c\) $f_n(x)\leq f_{n+1}(x)$ $\;\forall\,n\in\mathbb{N}$ and $x\in K$. Then $f_n\to f$ uniformly on $K$. :::spoiler proof: > Define $\mathrm{g}_n=f-f_n$ $\;\forall\,n\in\mathbb{N}$. \ > Then, \ > (d) $\mathrm{g}_n$ is continuous on $K$ \ > (e) $\mathrm{g}_n\to \mathbf{0}$ pointwise on $K$ \ > (g) $\mathrm{g}_n(x)\geq \mathrm{g}_{n+1}(x)$ $\;\forall\,n\in\mathbb{N}$ and $x\in K$. \ > Our goal is to show $\mathrm{g}_n\to \mathbf{0}$ uniformly on $K$. \ > Given $\varepsilon>0$, define $K_n=\{x\in K\mid \mathrm{g}_n(x)\geq\varepsilon\}$. \ > For each $n\in\mathbb{N}$, since $\mathrm{g}_n$ is continuous on $K$, $K_n=\mathrm{g}_n^{-1}\big([\varepsilon,\infty)\big)$ is closed in $K$. Then $K_n$ is compact. \ > Moreover, by (g), $K_{n}\supseteq K_{n+1}$ $\forall\,n\in\mathbb{N}$. \ > Fix $x\in K$, since $\mathrm{g}_n(x)\to 0$ as $n\to \infty$, $x\notin K_n$ as $n$ is sufficiently large. \ > That implies $\displaystyle\bigcap_{n=1}^\infty K_n =\emptyset$. \ > According to this proposition > > If $\{K_n\}_{n=1}^\infty$ is a sequence of nonempty compact sets in a metric space and $K_n\supseteq K_{n+1}$ $\;\forall\,n\in\mathbb{N}$, then $\displaystyle\bigcap_{n=1}^\infty K_n\neq \emptyset$. > > there exists some $N\in\mathbb{N}$ such that $K_N=\emptyset$. > > That is, if $n\geq N$, $\mathrm{g}_n(x)<\varepsilon$ for every $x\in K$. \ > Hence $f_n\to f$ uniformly on $K$. ::: :::success - The result is true if condition (c\) is replaced by $f_n(x)\geq f_{n+1}$. - The compactness is necessary. You may consider $f_n(x)=\dfrac{1}{nx+1}$ on $(0,1)$. - The monotonicity is necessary. ::: --- ### Series of Functions Let $(M,d)$ be a metric space and $(V,\|\cdot\|)$ be a normed space, $A\subseteq M$,and $\mathrm{g}_k,\mathrm{g}:A\to V$ be functions. \ (1) We say that the series $\displaystyle\sum_{k=1}^\infty \mathrm{g}_k$ ==converges pointwise to $\mathrm{g}$ on $A$== if the sequence of partial sum $\{S_n\}_{n=1}^\infty$ given by $S_n(x)=\displaystyle\sum_{k=1}^n \mathrm{g}_k(x)$ converges pointwise to $\mathrm{g}$ on $A$. \ (2) We say that $\displaystyle\sum_{k=1}^\infty \mathrm{g}_k$ ==converge to $\mathrm{g}$ uniformly on $A$== if $\{S_n\}_{n=1}^\infty$ converges to $\mathrm{g}$ uniformly on $A$. :::spoiler Example: For the geometric series $\displaystyle\sum_{k=0}^\infty x^k$, $$S_n(x)=\sum_{k=0}^n x^k= \begin{cases} \frac{1-x^{n+1}}{1-x} &\text{if } x\neq 1 \\ n+1 &\text{if } x=1 \end{cases}$$ 1. For $x\in(-1,1)$, $S_n\to\frac{1}{1-x}$. \ $\Rightarrow$ $\sum_{k=0}^\infty x^k \to \frac{1}{1-x}$ pointwise on $(-1,1)$. 2. For $x\in(-\infty,1]\cup[1,\infty)$, $\{S_n\}_{n=1}^\infty$ diverges. \ $\Rightarrow$ $\sum_{k=0}^\infty x^k$ diverges. 3. Let $a\in(0,1)$ and $\mathrm{g}(x)=\frac{1}{1-x}$, for $x\in[-a,a]$, $$|S_n(x)-\mathrm{g}(x)|=\left|\frac{1-x^{n+1}}{1-x}-\frac{1}{1-x}\right|= \left|\frac{x^{n+1}}{1-x}\right|\leq \frac{|a|^{n+1}}{1-a}\to 0\text{ as }n\to\infty$$ Given $\varepsilon>0$, choose $N\in\mathbb{N}$ such that $\dfrac{|a|^{n+1}}{1-a}<\varepsilon$ whenever $n\geq N$. \ Then $|S_n(x)-\mathrm{g}(x)|<\varepsilon$ whenever $x\in[-a,a]$ and $n\geq N$. \ Hence $\sum_{k=0}^\infty x^k$ converges uniformly on $[-a,a]$. 4. $\displaystyle\sum_{k=0}^\infty x^k$ does not converge uniformly on $(-1,1)$ since $\displaystyle\sup_{x\in(-1,1)}|S_n(x)-\mathrm{g}(x)|=\infty$. ::: --- **Cauchy Criterion** \ Let $(M,d)$ be a metric space and $(V,\|\cdot\|)$ be a normed space, $A\subseteq M$, and $\mathrm{g}_k:A\to V$ be functions. \ If $\sum_{k=1}^\infty \mathrm{g}_k$ converges uniformly on $A$, then $\forall\,\varepsilon>0$, $\exists\,N\in\mathbb{N}$ such that $\left\|\sum_{k=m+1}^n \mathrm{g}_k(x)\right\|<\varepsilon$ whenever $x\in A$ and $n>m\geq N$. \ In addition, if $(V,\|\cdot\|)$ is a Banach space, then the converse holds. :::spoiler proof: > $\implies$: \ > Let $\sum_{k=1}^\infty \mathrm{g}_k$ converge uniformly to $\mathrm{g}$ on $A$. \ > Let $S_n=\sum_{k=1}^n \mathrm{g}_k$ be the partial sum. \ > For $\varepsilon>0$, $\exists\,N\in\mathbb{N}$ such that $\|S_n(x)-\mathrm{g}(x)\|<\frac{\varepsilon}{2}$ > whenever $x\in A$ and $n\geq N$. \ > Then for $n>m\geq N$, > $$\begin{aligned} > \left\|\sum_{k=m+1}^n \mathrm{g}_k(x)\right\| &= \|S_n(x)-S_m(x)\| \\ > &\leq \|S_n(x)-\mathrm{g}(x)\|+\|\mathrm{g}(x)-S_m(x)\| \\ > &<\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon > \end{aligned}$$ > for every $x\in A$. > > $\impliedby \text{ (in Banach space)}$: \ > For $x\in A$, given $\varepsilon>0$, $\exists\,N_{\varepsilon,x}\in\mathbb{N}$ such that > $\|S_n(x)-S_m(x)\|=\|\sum_{k=m+1}^n \mathrm{g}_k(x)\|< \varepsilon$ whenever $n>m\geq N_{\varepsilon,x}$. \ > Hence $\{S_n(x)\}_{n=1}^\infty$ is a Cauchy sequence in $V$. \ > Since $(V,\|\cdot\|)$ is complete, $\{S_n(x)\}_{n=1}^\infty$ converges in $V$. \ > Since $x\in A$ is arbitrary, $S_n\to \mathrm{g}=\sum_{k=1}^\infty \mathrm{g}_k$ pointwise on $A$. \ > To check that $S_n\to \mathrm{g}$ uniformly on $A$, given $\varepsilon>0$, \ > $\exists\,N\in\mathbb{N}$ such that $\|S_n(x)-S_m(x)\|<\frac{\varepsilon}{2}$ whenever $x\in A$ and $n>m\geq N$. \ > Since $S_n\to \mathrm{g}$ pointwise on $A$, $\forall\,x\in A$, $\exists\,m_x>N$ such that > $\|S_{m_x}(x)-\mathrm{g}(x)\|<\frac{\varepsilon}{2}$. \ > Then for $x\in A$ and $n\geq N$, > $$\begin{aligned} > \|S_n(x)-\mathrm{g}(x)\| &\leq \|S_n(x)-S_{m_x}(x)\|+\|S_{m_x}(x)-\mathrm{g}(x)\| \\ > &< \frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon > \end{aligned}$$ > Therefore, $S_n\to \mathrm{g}$ uniformly on $A$. ::: --- If $\sum_{k=1}^\infty \mathrm{g}_k$ converges uniformly on $A$, then $\mathrm{g}_k$ converges to $\mathbf{0}$ uniformly on $A$. --- Let $(M,d)$ be a metric space and $(V,\|\cdot\|)$ be a normed space, $A\subseteq M$, and $\mathrm{g}_k,\mathrm{g}:A\to V$ be functions. \ If $\mathrm{g}_k:A\to V$ are continuous and $\sum_{k=1}^\infty \mathrm{g}_k$ converges to $\mathrm{g}$ uniformly on $A$, then $\mathrm{g}$ is continuous. --- If $\mathrm{g}_k:[a,b]\to\mathbb{R}$ is integrable on $[a,b]$ and $\mathrm{g}=\sum_{k=1}^\infty \mathrm{g}_k$ converges uniformly on $[a,b]$, then $$\begin{aligned} \int_a^b \mathrm{g}(x)\,dx = \sum_{k=1}^\infty \int_a^b \mathrm{g}_k(x)\,dx \end{aligned}$$ --- #### Weirestrass M-test Let $(M,d)$ be a metric space and $(V,\|\cdot\|)$ be a Banach space, $A\subseteq M$, and $\mathrm{g}_k:A\to V$ be functions. \ Suppose that there exists $M_k>0$ such that $\displaystyle\sup_{x\in A}\|\mathrm{g}_k(x)\|\leq M_k$ whenever $k\in\mathbb{N}$ and $\sum_{k=1}^\infty M_k$ converges. \ Then $\displaystyle\sum_{k=1}^\infty \mathrm{g}_k :A\to V$ converges uniformly and absolutely on $A$. \ That is, $\displaystyle\sum_{k=1}^\infty \|\mathrm{g}_k\|: A\to\mathbb{R}$ converges uniformly on $A$. :::spoiler proof: > We intend to show that $\{S_n\}_{n=1}^\infty$ given by $S_n=\sum_{k=1}^n \mathrm{g}_k$ satisfies the Cauchy criterion. \ > Since $\sum_{k=1}^\infty M_k$ converges, given $\varepsilon>0$, $\exists\,N\in\mathbb{N}$ such that > $\sum_{k=m+1}^n M_k<\varepsilon$ whenever $n>m\geq N$. \ > Thus > $$\begin{aligned} > \sup_{x\in A}\|S_n(x)-S_m(x)\| &=\sup_{x\in A}\big\|\sum_{k=m+1}^n \mathrm{g}_k(x)\big\| \\ > &\leq \sup_{x\in A}\sum_{k=m+1}^n \|\mathrm{g}_k(x)\| \\ > &\leq \sum_{k=m+1}^n \sup_{x\in A} \|\mathrm{g}_k(x)\| \\ > &\leq \sum_{k=m+1}^n M_k \\ > &< \varepsilon > \end{aligned}$$ > Hence $\{S_n\}_{n=1}^\infty$ converges uniformly on $A$. > > Consider the sequence of functions $\|\mathrm{g}_k\|:A\to\mathbb{R}$. \ > Let $T_n(x)=\sum_{k=1}^n \|\mathrm{g}_k(x)\|$. \ > For $n>m\geq N$, > $$\begin{aligned} > \sup_{x\in A}|T_n(x)-T_m(x)| &= \sup_{x\in A}\sum_{k=m+1}^n \big\|\mathrm{g}_k(x)\big\| \\ > &\leq \sum_{k=m+1}^n \sup_{x\in A}\|\mathrm{g}_k(x)\| \\ > &\leq \sum_{k=m+1}^n M_k \\ > &<\varepsilon > \end{aligned}$$ > Therefore $\{T_n\}_{n=1}^\infty$ converges uniformly on $A$ and hence $\sum_{k=m+1}^\infty \|\mathrm{g}_k(x)\|$ converges uniformly on $A$. ::: --- :::spoiler Example of a continuous and nowhere differentiable function In $(\mathbb{R},|\cdot|)$, define $\phi(x)=|x|$ on $[-1,1]$ and extend $\phi$ to a $2$-periodic function (still called $\phi$). \ Then for $s,t\in\mathbb{R}$, $|\phi(s)-\phi(t)|\leq |s-t|$ ... (a). \ Define $f(x)=\displaystyle\sum_{n=0}^\infty \left(\frac{3}{4}\right)^n \phi\left(4^n x\right)$. \ $\left(\frac{3}{4}\right)^n\phi(4^n x)$ is continuous for every $n\in\mathbb{N}$. \ Since $0\leq \phi(x)\leq 1$, by M-test, the series converges uniformly on $\mathbb{R}$. \ Then $f$ is continuous on $\mathbb{R}$. To prove that $f$ is nowhere differentiable, suppose $x\in \mathbb{R}$. \ We want to show that $\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ does not exist. \ Fix $m\in\mathbb{N}$ and let $\delta_m=\pm\frac{1}{2}\cdot 4^{-m}$ where the sign is chosen such that $\mathbb{Z}\cap \big(4^m x, 4^m(x+\delta_m)\big)=\emptyset$ or $\mathbb{Z}\cap\big(4^m(x+\delta_m),4^m x\big)=\emptyset$. \ Then the interval would contain no interger. Define $\displaystyle r_n=\frac{\phi(4^n(x+\delta_m))-\phi(4^n x)}{\delta_m}$. \ When $n>m$, $4^n \delta_m= \pm\frac{1}{2} 4^{n-m}$ is an even interger. \ Since $\phi$ is periodic by $2$, $\phi(4^n(x+\delta_m))-\phi(4^n x)=0$ and $r_n=0$. \ When $0\leq n\leq m$, by (a), $|r_n|=\displaystyle\frac{\left|4^n(x+\delta_m)-4^n x\right|}{\delta_m} =4^n$ and $\phi(4^m(x+\delta_m))-\phi(4^m x)= \pm r^m\delta_m$. \ $\Rightarrow$ $|r_m|=4^m$ Thus $$\begin{aligned} \left|\frac{f(x+\delta_m)-f(x)}{\delta_m}\right| &= \left|\sum_{n=0}^m\left(\frac{3}{4}\right)^n r_n\right| \\ &\geq \left(\frac{3}{4}\right)^m |r_m| - \left|\sum_{n=0}^{m-1}\left(\frac{3}{4}\right)^n r_n\right| \\ &\geq 3^m-\sum_{n=0}^{m-1}3^n \\ &=\frac{1}{2}(3^m+1) \to\infty \text{ as }m\to\infty \end{aligned}$$ As $m\to\infty$, $\delta_m\to 0$, we have $\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ does not exist and hence $f$ is not differentiable at $x$. ::: --- ||domain|$f_k$|convergence|$f$|range| |:-:|:-:|:-:|:-:|:-:|:-:| |||continuous|uniformly|**continuous**| ||$[a,b]$|integrable **$\displaystyle\int\lim_{k\to\infty}=\lim_{k\to\infty}\int$**|uniformly||| |$\text{Dini}$|compact|continuous monotone|pointwise **uniformly**|continuous|$\mathbb{R}$| ||$[a,b]$|integrable monotone **$\displaystyle\int\lim_{k\to\infty}=\lim_{k\to\infty}\int$**|pointwise|integrable|$\mathbb{R}$| ||$[a,b]$|integrable bounded **$\displaystyle\int\lim_{k\to\infty}=\lim_{k\to\infty}\int$**|pointwise|integrable|$\mathbb{R}$| - where *integrable* means *Riemann integrable* - where *monotone* means $f_n(x)\leq f_{n+1}(x) \,\vee\, f_n(x)\geq f_{n+1}(x) \quad \forall\,n\in\mathbb{N}, x\in D_{f_n}$ --- ### Power Series We called a series of the form $$\sum_{k=0}^\infty c_k(x-a)^k$$ a ==power series centered at $a$== for some $\{c_k\}_{k=0}^\infty \subseteq\mathbb{R}$ and $a\in\mathbb{R}$. If $a=0$, it is called a ==Maclaurin series==. --- Let $\sum_{k=0}^\infty c_k(x-a)^k$ be a power series in $\mathbb{R}$. \ Suppose that the series converges at some $b\neq a$. \ Define $h=|b-a|$. \ Then the series converges on $(a-h,a+h)$. \ Moreover, the series converges uniformly on $[\alpha,\beta]$ if $[\alpha,\beta]\subseteq (a-h,a+h)$. :::spoiler proof: > WLOG, let $a=0$ and $\sum_{k=0}^\infty c_k x^k$ converge at $b\neq 0$. \ > Then $h=|b|$. \ > Since $\sum_{k=0}^\infty c_k b^k$ converges, $\displaystyle\lim_{k\to\infty}|c_k b^k| =\lim_{k\to\infty}|c_k|h^k =0$. \ > $\Rightarrow$ $\exists\,N\in\mathbb{N}$ such that $|c_k|h^k<1$ whenever $k\geq N$ \ > $\Rightarrow$ $|c_k|<\frac{1}{h^k}$ for every $k\geq N$ > > For $x_0\in(-h,h)$, > $$\begin{aligned} > \sum_{k=N}^\infty |c_k x_0^k| &= \sum_{k=N}^\infty |c_k||x_0|^k \\ > &\leq \sum_{k=N}^\infty \frac{1}{h^k}|x_0|^k \\ > &=\sum_{k=N}^\infty \left(\frac{|x_0|}{h}\right)^k \\ > &< \infty \qquad \text{(geometric series)} > \end{aligned}$$ > Then $\sum_{k=0}^\infty c_k x_0^k$ would converge absolutely and hence converges. \ > Thus $\sum_{k=0}^\infty c_k x^k$ converges pointwise on $(-h,h)$. > > Suppose $[\alpha,\beta]\subseteq(-h,h)$. \ > Since $(-h,h)$ is open, choose $\delta>0$ such that $[\alpha,\beta]\subseteq(-h+\delta,h-\delta)\subseteq(-h,h)$. \ > Then for $x_0\in[\alpha,\beta]$, $\frac{|x_0|}{h}< 1-\frac{\delta}{h}$. \ > Hence, for $\varepsilon>0$ and $n>m\geq N$, > $$\begin{aligned} > \left|\sum_{k=m+1}^n c_k x_0^k\right|&\leq\sum_{k=m+1}^n \left|c_kx_0^k\right| \\ > &\leq\sum_{k=m+1}^n |c_k|h^k(1-\frac{\delta}{h})^k \\ > &<\varepsilon \quad\text{ as }m,n \text{ sufficiently large} > \end{aligned}$$ > Therefore, $\sum_{k=0}^\infty c_k x^k$ converges uniformly on $[\alpha,\beta]$. ::: --- $R>0$ is called the ==radius of convergence== of the power series $\sum_{k=0}^\infty c_k(x-a)^k\subseteq\mathbb{R}$ if the series converges for every $x\in(a-R,a+R)$ and diverges for all $x\in(-\infty,a-R)\cup(a+R,\infty)$. $R=\displaystyle\sup\left\{r\geq 0 \;\middle\vert\; \sum_{k=0}^\infty c_k(x-a)^k \text{ converges in } [a-r,a+r]\right\}$ --- By ratio test, consider the series $\sum_{k=0}^\infty x_k$, $$\begin{aligned} \limsup_{k\to\infty}\left|\frac{x_{k+1}}{x_k}\right|<1 &\implies \text{ the series converges} \\ \liminf_{k\to\infty}\left|\frac{x_{k+1}}{x_k}\right|>1 &\implies \text{ the series diverges} \end{aligned}$$ --- Let $\sum_{k=0}^\infty c_k(x-a)^k\subseteq\mathbb{R}$ be a power series. \ If $\displaystyle\lim_{k\to\infty}\left|\frac{c_k}{c_{k+1}}\right|$ converges, then the radius of convergence of the power series is $\displaystyle\lim_{k\to\infty}\left|\frac{c_k}{c_{k+1}}\right|$. :::spoiler proof: > For $x\neq a$, let $c_k(x-a)^k=b_k$. \ > Then $\left|\frac{b_{k+1}}{b_k}\right|= \left|\frac{c_{k+1}}{c_k}\right||x-a|$. \ > Consider > $$\begin{aligned} > \limsup_{k\to\infty}\left|\frac{c_{k+1}}{c_k}\right||x-a|<1 &\iff |x-a|<\frac{1}{\displaystyle\limsup_{k\to\infty}\left|\frac{c_{k+1}}{c_k}\right|} = \liminf_{k\to\infty}\left|\frac{c_k}{c_{k+1}}\right| \\ > \liminf_{k\to\infty}\left|\frac{c_{k+1}}{c_k}\right||x-a|>1 &\iff |x-a|<\frac{1}{\displaystyle\liminf_{k\to\infty}\left|\frac{c_{k+1}}{c_k}\right|} = \limsup_{k\to\infty}\left|\frac{c_k}{c_{k+1}}\right| > \end{aligned}$$ > > If $\displaystyle\lim_{k\to\infty}\left|\frac{c_k}{c_{k+1}}\right|$ converges, then > $$\lim_{k\to\infty}\left|\frac{c_k}{c_{k+1}}\right|= \limsup_{k\to\infty}\left|\frac{c_k}{c_{k+1}}\right|= \liminf_{k\to\infty}\left|\frac{c_k}{c_{k+1}}\right|= R$$ ::: --- Let $\sum_{k=0}^\infty c_k(x-a)^k \subseteq\mathbb{R}$ be a power series with the radius of convergence $R$, and $[\alpha,\beta]\subseteq(a-R,a+R)$. \ Then the series $\sum_{k=0}^\infty (k+1)c_{k+1}(x-a)^k$ converges pointwise on $(a-R,a+R)$ and converges uniformly on $[\alpha,\beta]$. :::warning updating... ::: proof: Select $h>0$ such that $[\alpha,\beta]\subseteq[a-h,a+h]\subseteq(a-R,a+R)$. \ Then $r=\frac{|x-a|}{h}<1$. \ Since $a+h\in(a-R,a+R)$, $\sum_{k=0}^\infty \big((a+h)-a\big)^k=\sum_{k=0}^\infty c_k h^k$ converges. \ $\Rightarrow$ $\displaystyle\lim_{k\to\infty}|c_k h^k|=0$ \ $\Rightarrow$ $\exists\,N\in\mathbb{N}$ such that $|c_k h^k|<1$ whenever $k\geq N$. \ Then $$\begin{aligned} \sum_{k=N}^\infty \left|(k+1)c_k(x-a)^k\right| &= \sum_{k=N}^\infty (k+1)|c_k h^k|\left(\frac{|x-a|}{h}\right)^k \\ &< \sum_{k=N}^\infty (k+1)r^k \\ &<\infty \quad( r>1) \end{aligned}$$ By Weirestrass M-test, $\sum_{k=0}^\infty (k+1)c_{k+1}(x-a)^k$ converges uniformly on $[\alpha,\beta]$. Suppose $x\in (a-R,a+R)$. \ Choose $\delta>0$ such that $x\in[a-R+\delta,a+R-\delta]$. \ Since $\sum_{k=0}^\infty (k+1)c_{k+1}(x-a)^k$ converges uniformly on $[\alpha,\beta]$, it converges pointwise at $x$. \ Hence $\sum_{k=0}^\infty (k+1)c_{k+1}(x-a)^k$ converges pointwise on $(a-R,a+R)$. --- Let $\sum_{k=0}^\infty c_k(x-a)^k \subseteq\mathbb{R}$ be a power series with the radius of convergence $R$, and $[\alpha,\beta]\subseteq(a-R,a+R)$. \ Then the series is differentiable on $(a-R,a+R)$. \ Moreover, $$\frac{d}{dx}\sum_{k=0}^\infty c_k(x-a)^k = \sum_{k=0}^\infty \frac{d}{dx} c_k(x-a)^k = \sum_{k=1}^\infty kc_k(x-a)^{k-1}$$ :::success Since the differentiation is still a power series, it is also differentiable. \ By mathematical induction, a power series is infinitely differentiable. ::: --- Let $\sum_{k=0}^\infty c_k(x-a)^k$ be a power series with radius of convergence $R$ and $[\alpha,\beta]\subseteq(a-R,a+R)$. \ Then the power series $\sum_{k=0}^\infty c_k(x-a)^k$ is integrable on $[\alpha,\beta]$. \ Moreover, $$\int_\alpha^\beta \sum_{k=0}^\infty c_k(x-a)^k\,dx = \sum_{k=0}^\infty \int_\alpha^\beta c_k(x-a)^k\,dx$$ --- :::spoiler Example 1 The function $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$ converges on $\mathbb{R}$. \ Then $$\begin{aligned} \int_0^t e^x\,dx &= \int_0^t \sum_{k=0}^\infty \frac{x^k}{k!}\,dx \\ &=\sum_{k=0}^\infty \int_0^t \frac{x^k}{k!}\,dx \\ &=\left. \sum_{k=0}^\infty \frac{x^{k+1}}{(k+1)!} \right|_0^t \\ &=\sum_{k=1}^\infty \frac{t^k}{k!} \\ &=\left(\sum_{k=0}^\infty \frac{t^k}{k!}\right) -1 = e^t-1 \end{aligned}$$ ::: :::spoiler Example 2 The series $\sum_{k=1}^\infty \frac{x^k}{k}$ converges on $(-1,1)$. \ Hence, for $x\in(-1,1)$, $$\begin{aligned} \frac{d}{dx}\sum_{k=1}^\infty \frac{x^k}{k} &= \sum_{k=1}^\infty x^{k-1} \\ &= \sum_{k=0}^\infty x^k =\frac{1}{1-x} \end{aligned}$$ For $x\in (-1,1)$, $$\begin{aligned} \sum_{k=1}^\infty \frac{x^k}{k} &= \int_0^x \frac{d}{dt}\sum_{k=1}^\infty \frac{t^k}{k}\,dt \quad\text{ by F.T.C.} \\ &=\int_0^x \frac{1}{1-t}\,dt = -\ln(1-x) \end{aligned}$$ Moreover, we may consider $x=-1$. \ $\sum_{k=1}^\infty \frac{(-1)^k}{k}$ converges by alternating series test. \ We want to show that $\displaystyle\lim_{n\to\infty}\sum_{k=1}^n \frac{(-1)^k}{k} =-\ln 2$. \ First, $$\frac{d}{dt}\sum_{k=1}^n \frac{t^k}{k} = \sum_{k=0}^{n-1} t^k = \frac{1-t^n}{1-t} = \frac{1}{1-t}-\frac{t^n}{1-t}$$ and $$\sum_{k=1}^n \frac{(-1)^k}{k} = \int_{-1}^0 \frac{d}{dt}\sum_{k=1}^n \frac{t^k}{k}\,dt = \int_{-1}^0\frac{1}{1-t}\,dt-\int_{-1}^0\frac{t^n}{1-t}\,dt$$ Then $$\begin{aligned} \left|\sum_{k=1}^n \frac{(-1)^k}{k}-(-\ln 2)\right| &= \left|\int_{-1}^0\frac{1}{1-t}\,dt-(-\ln 2)\right| + \left|\int_{-1}^0\frac{t^n}{1-t}\,dt\right| \\ &< 0+\left|\int_{-1}^0 t^n\,dt\right| \\ &=\frac{1}{n+1} \quad \to 0 \text{ as } n\to\infty \end{aligned}$$ Therefore $\sum_{k=1}^\infty \frac{(-1)^k}{k}=-\ln 2$ and the series converges on $[-1,1)$. ::: :::spoiler Example 3 Find a function y satisfying $y''(x)+y(x)=0$. Solution (without complete verification): \ Suppose that a solution in the form of power series at $0$, say $y(x)=\sum_{k=0}^\infty c_k x^k$. \ Assume that the series converges on some $(-\delta,\delta)$. \ Then $y'(x)=\sum_{k=1}^\infty kc_{k}x^{k-1}$ and $y''(x)= \sum_{k=2}^\infty k(k-1)c_k x^{k-2}$. \ From the formula, we obtain $$\begin{aligned} 0 &= \sum_{k=2}^\infty k(k-1)c_k x^{k-2} + \sum_{k=0}^\infty c_k x^k \\ &=\sum_{k=0}^\infty \left[(k+2)(k+1)c_{k+2}+c_k\right] x^k \end{aligned}$$ The coefficients satisfy $(k+2)(k+1)c_{k+1}+c_k=0$ for $k=0,1,2,\dotsc$ and thus the recurrence ralation $c_{k+2}=-\frac{c_k}{(k+1)(k+2)}$. Hence we have $c_k=\begin{cases} \frac{(-1)^{k/2}}{k!}c_0 &k\text{ is even} \\ \frac{(-1)^{\lfloor k/2\rfloor}}{k!}c_1 &k\text{ is odd} \end{cases}$. \ Therefore, $$\begin{aligned} y &= c_0\left[1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots+\frac{(-1)^n}{(2n)!}x^{2n}+\cdots\right] + c_1\left[x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots+\frac{(-1)^n}{(2n+1)!}+\cdots\right] \\ &= c_0 \cos x +c_1 \sin x \end{aligned}$$ ::: --- ### Taylor Series Suppose that $f$ is a function such that $f'(a),f''(a),\dotsc,f^{(n)}(a)$ exist. \ Define $P_{n,a,f}(x)= c_0+c_1(x-a)+c_2(c-a)^2+\cdots+c_n(x-a)^n$ where $c_k=\frac{f^{(k)}(a)}{k!}$. \ $P_{n,a,f}$ is called the ==Taylor polynomial== of degree $n$ for $f$ at $a$. --- Suppose that $f$ is a function such that $f'(a)$, $f''(a)$,$\dotsc$,$f^{(n)}(a)$ exist. \ Then $$\lim_{x\to a}\frac{f(x)-P_{n,a}(x)}{(x-a)^n}=0$$ :::spoiler proof: > For $x\neq a$, > $$\frac{f(x)-P_{n,a}(x)}{(x-a)^n}= \frac{f(x)-\sum_{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}(x-a)^k}{(x-a)^n} - \frac{f^{(n)}(a)}{n!}$$ > Let $Q(x)=f(x)-\sum_{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}(x-a)^k$ and $\mathrm{g}(x)=(x-a)^n$.\ > Then for $1\leq i\leq n-1$, > $$Q^{(i)}(x)= f^{(i)}(x)-f^{(i)}(a)-\frac{f^{(i+1)}(a)}{1!}\,(x-a)-\cdots-\frac{f^{(n-1)}(a)\,(x-a)^{n-1-i}}{(n-1-i)!}$$ > Hence $\displaystyle\lim_{x\to a}Q^{(i)}(x)=0$ for $i=0,1,2,\dotsc,n-1$. \ > On the other hand, $g^{(i)}(x)=n(n-1)\cdots(n-i+1)(x-a)^{n-i}$ and hence $\displaystyle\lim_{x\to a}g^{(i)}(x)=0$ for $i=0,1,2,\dotsc,n-1$. \ > Use L'Hôpital's Rule $n-1$ times, > $$\begin{aligned} > \lim_{x\to a}\frac{f(x)-P_{n,a}(x)}{(x-a)^n} &= \lim_{x\to a}\frac{Q(x)}{g(x)}-\frac{f^{(n)}(a)}{n!} \\ > &= \lim_{x\to a} \frac{Q'(x)}{g'(x)}-\frac{f^{(n)}(a)}{n!} \\ > &\quad\quad\vdots \\ > &=\lim_{x\to a}\frac{Q^{(n-1)}(x)}{g^{(n-1)}(x)} - \frac{f^{(n)}(a)}{n!} \\ > &=\lim_{x\to a} \frac{f^{(n-1)}(x)-f^{(n-1)}(a)}{n!\,(x-a)} - \frac{f^{(n)}(a)}{n!} \\ > &=0 > \end{aligned}$$ ::: --- Let $P$, $Q$ be two polynomials in $(x-a)$ of degree less than or equal to $n$. \ Suppose that $P$ and $Q$ are equal up to order $n$ at $a$.\ Then $P=Q$. --- Suppose that $f$ has first derivative at $a$ and $P$ is a polynomial on $(x-a)$ of degree less than or equal to $n$ which equals $f$ up to order $n$ at $a$. \ Then $P(x)=P_{n,a,f}(x)$. --- Let $f$ be a $n+1$ times differentiable function. \ We define the remainder term $R_{n,a}(x)$ by $R_{n,a}(x)=f(x)-P_{n,a}(x)$. \ Then $$R_{n,a}(x)=\int_a^x \frac{f^{(n+1)}(t)}{n!}\,(x-t)^n\,dt$$ :::spoiler proof: > $$\begin{aligned} > f(x) &=\underbrace{f(a)}_{P_{0,a}(x)} + \underbrace{\int_a^x f'(t)\,dt}_{R_{0,a}(x)} \quad (\text{F.T.C.}) \\ > &=f(a)+\left.f'(t)\cdot t\right|_a^x-\int_a^x f''(t)\,dt \quad (\text{integration by parts}) \\ > &=f(a)+f'(x)\cdot x-f'(a)\cdot a-\int_a^xf''(t)\,t\,dt \\ > &=f(a)+f'(a)(x-a)-f'(a)\cdot x+f'(x)\cdot x-\int_a^x f''(t)\,t\,dt \\ > &=f(a)+f'(a)(x-a)+x\int_a^x f''(t)\,dt-\int_a^x f''(t)\,t\,dt \\ > &=\underbrace{f(a)+f'(a)(x-a)}_{P_{1,a}(x)} + \underbrace{\int_a^x f''(t)(x-t)\,dt}_{R_{1,a}(x)} \\ > &=f(a)+f'(a)(x-a)-\left.f''(t)\frac{(x-t)^2}{2}\right|_a^x+\int_a^x \frac{f'''(t)}{2}(x-t)^2\,dt \quad (\text{integration by parts})\\ > &=\underbrace{f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2}_{P_{2,a}(x)} + \underbrace{\int_a^x \frac{f'''(t)}{2}(x-t)^2\,dt}_{R_{2,a}(x)} > \end{aligned}$$ > By induction, if $f^{(n+1)}$ is continuous on $[a,x]$, then > $$R_{n,a}(x)=\int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n\,dt$$ ::: --- Let $f$ be a $n+1$ times differentiable function on $[a,x]$ and $R_{n,a}(x)$ be defined by $f(x)=f(a)+f'(a)\,(x-a)+\cdots+\frac{f^{(n)}(a)}{n!}\,(x-a)^n+R_{n,a}(x)$. \ Then - Cauchy form $$R_{n,a}(x)=\frac{f^{(n+1)}(\xi)}{n!}\,(x-\xi)(x-a) \text{ for some } \xi\in (a,x)$$ - Lagrange form $$R_{n,a}(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\,(x-a)^{n+1} \text{ for some } \xi\in (a,x)$$ - Integral form $$R_{n,a}(x)= \int_a^x \frac{(n+1)}{n!}\,(x-t)^n\,dt$$ The $\xi$ are usually different in Cauchy form and Lagrange form. \ $\xi$ depends on $a$ and $x$. In Lagrange form, if $|f^{(n+1)}(t)|<M$ for all $t\in[a,x]$, then $|R_{n,a}(x)|\leq M\frac{|x-a|^{n+1}}{(n+1)!}$. In intergral form, if $|f^{(n+1)}(t)|<M$ for all $t\in[a,x]$, then $|R_{n,a}(x)|\leq\frac{M}{n!}\left|\int_a^x (x-t)^n\,dt\right|\leq \frac{M}{(n+1)!}\left|\left.-(x-t)^{n+1}\right|_a^x\right|=\frac{M}{(n+1)!}\,|x-a|^{n+1}$. --- Let $f:(\alpha,\beta)\to\mathbb{R}$ be an infinitely differentiable function and $a\in (\alpha,\beta)$. 1. For every $n\in\mathbb{N}$ and $x\in (\alpha,\beta)$, $$f(x)=\sum_{k=0}^n \frac{f^{(k)}(a)}{k!}\,(x-a)^k+ \int_a^x\frac{f^{(n+1)}(t)}{n!}(x-t)^n\,dt$$ 2. Moreover, for some $0<h<\infty$ such that $(a-h,a+h)\subseteq(\alpha,\beta)$, suppose that $\exists\,M>0$ such that $|f^{(k)}(x)|\leq M$ for all $x\in (a-h,a+h)$ and $k\in\mathbb{N}$. \ Then $$f(x)=\sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!}\,(x-a)^k \text{ for all } x\in (a-h,a+h)$$ :::spoiler proof of 2: > Let $S_n(x)=\displaystyle\sum_{k=0}^n \frac{f^{(k)}(a)}{k!}\,(x-a)^k$. \ > For $x\in (a-h,a+h)$, > $$\begin{aligned} > |S_n(x)-f(x)|&\leq \left|\int_a^x\frac{f^{(n+1)}(t)}{n!}(x-t)^n\,dt \right| \\ > &\leq\frac{M}{h!}\, h^n \left|\int_a^x 1\,dt \right| \\ > &=\frac{M}{h!}\, h^n |x-a| \\ > &\leq \frac{M}{n!}\, h^{n+1} \quad \to 0 \text{ as } n\to\infty > \end{aligned}$$ > Hence $\{S_n(x)\}_{n=1}^\infty$ converges to $f(x)$ uniformly on $(a-h,a+h)$. \ > We obtain $f(x)=\displaystyle\sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!}$ uniformly on $(a-h,a+h)$. ::: ---