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

• Advanced Calculus (I)
  • Overview
  • Point-Set Topology of Metric Spaces
    • Compact Sets
    • Connected Sets
    • Subspace Topology
  • Mappings and Limits
    • Operations on Continuous Maps
    • Uniformly Continuous
    • Continuous Extension
    • Continuous Maps on Connected Sets and Path Connected Sets
  • Pointwise and Uniform Convergence
    • Series of Functions
    • Power Series
    • Taylor Series

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

$\{\begin{array}{l}(1)A\subseteq u\cup v\\ (2)A\cap u\ne \mathrm{\varnothing }\\ (3)A\cap v\ne \mathrm{\varnothing }\\ (4)A\cap u\cap v=\mathrm{\varnothing }\end{array}$.

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

$\mathrm{\exists }{A}_1,A_2\subseteq M$ such that

$\{\begin{array}{l}{A}_1,A_2\ne \mathrm{\varnothing }\\ A_1\cup A_2=A\\ {\overline{A}}_1\cup A_2=A_1\cup {\overline{A}}_2=\mathrm{\varnothing }\end{array}$

proof:

$⟹:$
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),

$A_1\ne \mathrm{\varnothing }$. By (3),

$A_2\ne \mathrm{\varnothing }$.
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,

${\bar{A}}_1\subseteq \overline{v^c}=v^c$.

$\Rightarrow$

${\bar{A}}_1\cap v=\mathrm{\varnothing }$

$\Rightarrow$

${\bar{A}}_1\cap A_2=\mathrm{\varnothing }$
Similarly,

$A_1\cap {\bar{A}}_2=\mathrm{\varnothing }$.

$⟸:$

$\mathrm{\exists }{A}_1,A_2$ such that

$\dots$
Let

$u={\overline{A}}_2^c$ and

$v={\overline{A}}_1^c$. Since

$A_2\subseteq {\overline{A}}_1^c$ and

$A_1\subseteq {\overline{A}}_2^c$,

$v$ and

$u$ are both nonempty.
Hence

$A=A_1\cup A_2\subseteq u\cup v$.
Since

$A_1\cap {\overline{A}}_1^c=\mathrm{\varnothing }$ and

$A_2\cap {\overline{A}}_2^c$, we have

$A_1\cap v=\mathrm{\varnothing }$ and

$A_2\cap u=\mathrm{\varnothing }$.
Then

$$\begin{array}{rl}A\cap u\cap v& =(A_1\cap u\cap v)\cup (A_2\cap u\cap v)\\ & =\mathrm{\varnothing }\end{array}$$ 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)

proof:

$⟹:$
Suppose

$x<z<y$ and

$x,y\in A$.
Assume

$z\notin A$.
Let

$A_1=(-\mathrm{\infty },z)\cap A$ and

$A_2=(z,\mathrm{\infty })\cap A$.
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

$\overline{A_1}\subseteq (-\mathrm{\infty },z]$ and

${\overline{A}}_2\subseteq [z,\mathrm{\infty })$.

$\Rightarrow$

${\overline{A}}_1\cap A_2=\mathrm{\varnothing }=A_1\cap {\overline{A}}_2$
Then

$A$ is disconnected.

$(\to \leftarrow )$.
Thus

$z\in A$.

$⟸:$
Assume

$A$ disconneted.

$\mathrm{\exists }{A}_1,A_2\subseteq \mathbb{R}$ such that (i)

$A_1,A_2\ne \mathrm{\varnothing }$, (ii)

$A=A_1\cup A_2$, (iii)

${\overline{A}}_1\cap A_2=A_1\cap {\overline{A}}_2=\mathrm{\varnothing }$.
By (i),

$\mathrm{\exists }x\in A_1,y\in A_2$.
By (iii),

$x\ne y$. WLOG, assume

$x<y$.
Let

$z=sup([x,y]\cap A_1)$.
Then

$z\in {\bar{A}}_1$ and thus

$z\notin A_2$. \

• If
  $z\notin A_1$, then since
  $z\notin A_2$,
  $z\notin A$.
  $(\to \leftarrow )$
• If
  $z\in A_1$, then
  $z\notin {\overline{A}}_2$.
  
  $\Rightarrow$
  $\mathrm{\exists }r>0$ such that
  $(z,z+r)\cap A_2=\mathrm{\varnothing }$
  
  $\Rightarrow$
  $x<z+\frac{r}{2}<y$,
  $z+\frac{r}{2}\notin A_1$, and
  $z+\frac{r}{2}\notin A_2$
  
  $\Rightarrow$
  $z+\frac{r}{2}\notin A$
  $(\to \leftarrow )$
  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

$\mathrm{\exists }u\subseteq M$ is open in

$(M,d)$ such that

$v=u\cap N$.

proof:

$⟹:$
Since

$v$ is open in

$(N,d)$,

$\mathrm{\forall }x\in v$,

$\mathrm{\exists }{r}_x>0$ 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

$$\begin{array}{rl}u\cap N& =\bigcup _{x\in v}{B}_M(x,r_x)\cap N\\ & =\bigcup _{x\in v}(B_M(x,r_x)\cap N)\\ & =\bigcup _{x\in v}{B}_N(x,r_x)=v\end{array}$$

$⟸:$
Suppose

$x\in v=u\cap N$.
Since

$u$ is open in

$(M,d)$,

$\mathrm{\exists }{\delta }_x>0$ such that

$B_M(x,{\delta }_x)\subseteq u$.
Then

$$\begin{array}{rl}{B}_N(x,{\delta }_x)& =B_M(x,{\delta }_x)\cap N\\ & \subseteq u\cap N=v\end{array}$$ Hence

$v$ is open in

$(N,d)$.

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

$N$.

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

$x_0\in \bar{A}$, we say

$y_0\in N$ is the limit of some function

$f:A\to N$ at

$x_0$ if

$\mathrm{\forall }\epsilon >0,\mathrm{\exists }\delta >0$ such that

$\rho (f(x),y_0)<\epsilon$ whenever

$x\in A$ and

$d(x,x_0)<\delta$, denoted by

$$\begin{gathered} {\displaystyle \underset{\begin{array}{c}x\to x_0\end{array} \\ x\in A}{lim}=y_0} \end{gathered}$$.

---

Let

$(M,d)$ and

$(N,\rho )$ be metric spaces,

$A\subseteq M$, and

$f:A\to N$ be a map.
For

$x_0\in \bar{A}$,

${\displaystyle \underset{x\to x_0}{lim}f(x)=y}$ iff for every sequence

$\{x_k{\}}_{k=1}^{\mathrm{\infty }}\subseteq A$ converging to

$x_0$ in

$(M,d)$, the sequence

$\{f(x_k){\}}_{k=1}^{\mathrm{\infty }}$ converges to

$y_0$ in

$(N,\rho )$.

proof:

$⟹$ :

$\mathrm{\forall }ϵ>0,\mathrm{\exists }\delta >0$ such that

$\rho (f(x),y_0)<ϵ$ when

$x\in A$ and

$d(x,x_0)<\delta$.
Let

$\{x_k{\}}_{k=1}^{\mathrm{\infty }}\subseteq A$ converging to

$x_0$.

$\Rightarrow$

$\mathrm{\exists }K\in \mathbb{N}$ such that

$d(x_k,x_0)<\delta$ whenever

$k\ge K$.

$\Rightarrow$

$\rho (f(x_k),y_0)<ϵ$ whenever

$k\ge K$.

$\Rightarrow$

${\displaystyle \underset{k\to \mathrm{\infty }}{lim}=y_0}$.

$⟸$ :
Prove by contradiction.
Assume

${\displaystyle \underset{x\to x_0}{lim}f(x)\ne y}$.
Then

$\mathrm{\exists }ϵ>0$ such that

$\mathrm{\forall }\delta >0$,

$\mathrm{\exists }{x}_{\delta }\in A$ with

$d(x_{\delta },x_0)<\delta$ but

$\rho (f(x_{\delta }),y_0)\ge ϵ$.
Let

$\rho =\frac{1}{k}$, then

$\mathrm{\exists }\{x_k{\}}_{k=1}^{\mathrm{\infty }}\subseteq A$ such that

$d(x_k,x_0)<\frac{1}{k}$ but

$\rho (f(x_0),y_0)\ge ϵ$.
In this case,

${\displaystyle \underset{k\to \mathrm{\infty }}{lim}{x}_k=x_0}$ but

${\displaystyle \underset{k\to \mathrm{\infty }}{lim}f(x_k)\ne y_0}$.

$(\to \leftarrow )$

---

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

${\displaystyle \underset{\begin{array}{c}x\to x_0\\ x\in A\end{array}}{lim}f(x)=f(x_0)}$.
(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

$A$.

proof:

(1)

$⟹$(2) :

If

$f^{-1}(v)=\mathrm{\varnothing }$, then it is tivially open relative to

$A$.
Consider

$f^{-1}(v)\ne \mathrm{\varnothing }$.
Suppose

$x_0\in f^{-1}(v)$.

$\Rightarrow$

$f(x_0)\in v$

$\Rightarrow$

$\mathrm{\exists }ϵ>0$ such that

$B_N(f(x_0),ϵ)\subseteq v$
Since

$f$ is continuous at

$x_0$,

$\mathrm{\exists }{\delta }_{x_0}>0$ such that

$f(B_M(x_0,{\delta }_{x_0})\cap A)\subseteq B_N(f(x_0),ϵ)\subseteq v$.
Then

$B_M(x_0,{\delta }_{x_0})\cap A\subseteq f^{-1}(v)$.
Since

$f$ is continuous on

$A$, for every

$x\in f^{-1}(v)$,

$\mathrm{\exists }{\delta }_x>0$ such that

$f(B_M(x_0,{\delta }_{x_0})\cap A)\subseteq v$, 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,

$$\begin{array}{rl}f(u\cap A)& =f\left({\displaystyle \bigcup _{x\in f^{-1}(v)}(B_M(x,{\delta }_x)\cap A)}\right)\\ & ={\displaystyle \bigcup _{x\in f^{-1}(v)}f(B_M(x,{\delta }_x)\cap A)}\\ & \subseteq v\end{array}$$

$\Rightarrow$

$u\cap A\subseteq f^{-1}(v)$.

(2)

$⟹$(1) :

Suppose

$x\in A$.
For given

$ϵ>0$,

$B_N(f(x),ϵ)$ is open in

$N$.
By the hypothesis, there is open

$u\subseteq M$ satisfying

$u\cap A=f^{-1}(B_N(f(x),ϵ))$.

$\Rightarrow$

$x\in u$ and

$\mathrm{\exists }{\delta }_x>0$ such that

$B_M(x,{\delta }_x)\subseteq u$
Then

$f(B_M(x,{\delta }_x)\cap A)\subseteq f(u\cap A)=B_N(f(x),ϵ)$.
For

$y\in B_M(x,{\delta }_x)\cap A$, we have

$\rho (f(x),f(y))<ϵ$.
Hence

$f$ is continuous at

$x$.

(2)

$⟹$(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

$A\cap u^c=A\mathrm{\setminus }(u\cap A)=A\mathrm{\setminus }{f}^{-1}(E^c)=f^{-1}(E)$, which is closed relative to

$A$.

(3)

$⟹$(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,

$(V,\Vert \cdot \Vert )$ be normed vector space, and

$A\subseteq M$.
Let

$f,g:A\to V$ and

$h:A\to \mathbb{R}$ be maps.

$\mathrm{\forall }x\in A$,

$\mathrm{\forall }\alpha \in \mathbb{R}$, define

$$\begin{array}{rl}& (1)(f\pm g)(x)=f(x)\pm g(x)\\ & (2)(\alpha f)(x)=\alpha f(x)\\ & (3)(hf)(x)=h(x)f(x)\\ & (4)(\frac{f}{h})(x)=\frac{1}{h(x)}f(x),h(x)\ne 0\end{array}$$ Suppose

$x_0\in \overline{A}$,

${\displaystyle \underset{x\to x_0}{lim}f(x)=v}$,

${\displaystyle \underset{x\to x_0}{lim}g(x)=w}$,

${\displaystyle \underset{x\to x_0}{lim}h(x)=c}$.
Then

$$\begin{array}{rl}& (5){\displaystyle \underset{x\to x_0}{lim}(f\pm g)(x)=v+w}\\ & (6){\displaystyle \underset{x\to x_0}{lim}(hf)(x)=cv}\\ & (7){\displaystyle \underset{x\to x_0}{lim}(\frac{f}{h})(x)=\frac{v}{c}\text{if }c\ne 0}\end{array}$$

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

$\mathrm{\forall }ϵ>0$,

$\mathrm{\exists }\delta >0$ such that

$\rho (f(x),f(y))<ϵ$ whenever

$x,y\in A$ and

$d(x,y)<\delta$.

e.g.

$f(x)=\frac{1}{x}$ is continuous on

$(0,\mathrm{\infty })$ and uniformly continuous on

$[a,\mathrm{\infty }),a>0$.

---

Let

$A\subseteq \mathbb{R}$ and

$f:A\to \mathbb{R}$.

$f$ is Lipschitz function if

$\mathrm{\exists }K>0$ such that

$|f(x)-f(y)|\le K|x-y|\mathrm{\forall }x,y\in A$.

Let

$(M,d)$ and

$(N,\rho )$ be metric spaces and

$f:M\to N$.

$f$ is Lipschitz function if

$\mathrm{\exists }K>0$ such that

$\rho (f(x),f(y))\le Kd(x,y)\mathrm{\forall }x,y\in M$.

---

Let

$A\subseteq \mathbb{R}$ and

$f:A\to \mathbb{R}$.

$f$ is Hölder continuous with exponents

$\alpha$ if

$\mathrm{\exists }K>0$ and

$0<\alpha \le 1$ such that

$|f(x)-f(y)|\le K{|x-y|}^{\alpha }\mathrm{\forall }x,y\in A$.

---

Suppose

$f:\mathbb{R}\to \mathbb{R}$.

$f$ is differentiable and

$\mathrm{\exists }K>0$ such that

$|f^{\prime }(x)|<K$ for all

$x\in \mathbb{R}$

$⟹$

$f$ is Lipschitz

$⟹$

$f$ is Hölder continuous with some

$0<\alpha <1$

$⟹$

$f$ 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

${\left\{x_n\right\}}_{n=1}^{\mathrm{\infty }}$,

${\left\{y_n\right\}}_{n=1}^{\mathrm{\infty }}\subseteq A$,

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}d(x,y)=0⟹{\displaystyle \underset{n\to \mathrm{\infty }}{lim}\rho (f(x_n),f(y_n))=0}}$.

proof:

$⟹$:
Prove by contradiction.
Assume

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}d(x_n,y_n)=0}$ and

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\rho (f(x_n),f(y_n))\ne 0}$ for some

$\{x_n{\}}_{n=1}^{\mathrm{\infty }},\{y_n{\}}_{n=1}^{\mathrm{\infty }}\subseteq A$.

$\Rightarrow$

$\mathrm{\exists }ϵ>0$ such that

$\mathrm{\forall }k\in \mathbb{N}$,

$\mathrm{\exists }{n}_k\ge k$ such that

$\rho (f(x_{n_k}),f(y_{n_k}))\ge ϵ$.
Since

$f$ is uniformly continuous,

$\mathrm{\exists }\delta >0$ such that

$\mathrm{\forall }x,y\in A$ with

$d(x,y)<\delta$, we have

$\rho (f(x),f(y))<ϵ$.

$(\to \leftarrow )$

$⟸$:
Prove by contradiction.

$\mathrm{\exists }ϵ>0$ such that

$\mathrm{\forall }n\in \mathbb{N}$,

$\mathrm{\exists }{x}_n,y_n\in A$ with

$d(x_n,y_n)<\frac{1}{n}$ but

$\rho (f(x_n),f(y_n))\ge ϵ$.
Then for the sequences

$\{x_n{\}}_{n=1}^{\mathrm{\infty }},\{y_n{\}}_{n=1}^{\mathrm{\infty }}\subseteq A$ with

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}d(x_n,y_n)=0}$,

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\rho (f(x_n),f(y_n))\ne 0}$.

$(\to \leftarrow )$

---

Suppose that

$\mathrm{\forall }ϵ>0$,

$\mathrm{\exists }\delta (x,ϵ)>0$ such that

$\rho (f(x),f(y))<ϵ$ whenever

$d(x,y)<\delta (x,ϵ)$.
Define

${\delta }_f(ϵ)={\displaystyle \underset{x\in A}{inf}\delta (x,ϵ)}$.
If

${\delta }_f(ϵ)>0\mathrm{\forall }ϵ>0$, 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

${\left\{x_n\right\}}_{n=1}^{\mathrm{\infty }}\subseteq A$ is a Cauchy sequence in

$(M,d)$, then

${\{f(x_n)\}}_{n=1}^{\mathrm{\infty }}$ is also a Cauchy sequence in

$(N,\rho )$.

proof:

$\mathrm{\forall }ϵ>0$,

$\mathrm{\exists }\delta >0$ such that

$\rho (f(x),f(y))<ϵ$ whenever

$x,y\in A$ and

$d(x,y)<\delta$.
Since

$\{x_n{\}}_{n=1}^{\mathrm{\infty }}\subseteq A$ is Cauchy in

$(M,d)$, there exists

$L\in \mathbb{N}$ such that

$d(x_m,x_n)<\delta$ whenever

$m,n\ge L$.
Therefore for

$m,n\ge L$, we have

$\rho (f(x_m),f(x_n))<ϵ$.
Thus

$\{f(x_n){\}}_{n=1}^{\mathrm{\infty }}$ 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

$\overline{A}$, that is,

$\mathrm{\exists }g:\overline{A}\to N$ such that

1. $g$ is uniformly continuous on
   $\overline{A}$
2. $g(x)=f(x)\mathrm{\forall }x\in A$
3. If there is
   $h:\overline{A}\to N$ satisfying 1. and 2., then
   $g=h$ on
   $\overline{A}$.

proof:

We need to define the value of

$g$ on

$\overline{A}\mathrm{\setminus }A$.
Suppose

$x\in \overline{A}\mathrm{\setminus }A$.
Then there exists

${\left\{x_n\right\}}_{n=1}^{\mathrm{\infty }}\subseteq A$ converging to

$x$.
Hence

${\left\{x_n\right\}}_{n=1}^{\mathrm{\infty }}\subseteq A$ is Cauchy in

$M$.
Since

$f$ is uniformly continuous on

$A$,

${\{f(x_n)\}}_{n=1}^{\mathrm{\infty }}\subseteq A$ is a Cauchy sequence in

$(N,\rho )$.
Since

$(N,\rho )$ is complete,

${\{f(x_n)\}}_{n=1}^{\mathrm{\infty }}$ converges in

$(N,\rho )$, say

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}f(x_n)=y_x}$.
Define

$g(x)=\{\begin{array}{ll}f(x),& x\in A\\ y_x,& x\in \overline{A}\mathrm{\setminus }A\end{array}$
To check that

$g(x)$ is well-defined, consider another sequence

${\left\{z_n\right\}}_{n=1}^{\mathrm{\infty }}\subseteq A$ converging to

$x$.
Then

$d(x_n,z_n)\to 0$ as

$n\to \mathrm{\infty }$.
Since

$f$ is uniformly conitnuous on

$A$,

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\rho (f(x_n),f(z_n))=0}$.
Therefore

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}f(x_n)=y_x={\displaystyle \underset{n\to \mathrm{\infty }}{lim}f(z_n)}}$.
Hence

$g$ is well-defined on

$\overline{A}$.
Check condition 1. :
Suppose

$ϵ>0$.
Since

$f$ is uniformly continuous on

$A$,

$\mathrm{\exists }\delta >0$ such that if

$x,y\in A$ with

$d(x,y)<\delta$, then

$\rho (f(x),f(y))<ϵ$.
For

$u,v\in \overline{A}$ with

$d(u,v)<\frac{\delta }{3}$, by the definition of

$g$,

$\mathrm{\exists }\{u_n{\}}_{n=1}^{\mathrm{\infty }},\{v_n{\}}_{n=1}^{\mathrm{\infty }}\subseteq A$ such that

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{u}_n=u}$ and

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{v}_n=v}$.

$\Rightarrow$

$\mathrm{\exists }{u}_m,v_k$ with

$d(u,u_m)<\frac{\delta }{3}$ and

$d(v,v_k)<\frac{\delta }{3}$ such that

$\rho (f(u),f(u_m))<\frac{ϵ}{3}$ and

$\rho (f(v),f(v_k))<\frac{ϵ}{3}$.
Then

$$\begin{array}{rl}d(u_m,v_k)& \le d(u_m,u)+d(u,v)+d(v,v_k)\\ & <\frac{\delta }{3}+\frac{\delta }{3}+\frac{\delta }{3}=\delta \end{array}$$ This implies

$\rho (f(u_m),f(v_k))<\frac{ϵ}{3}$ and hence

$$\begin{array}{rl}\rho (g(u),g(v))& \le \rho (g(u),g(u_m))+\rho (g(u_m),g(v_k))+\rho (g(v_k),g(v))\\ & <\frac{ϵ}{3}+\frac{ϵ}{3}+\frac{ϵ}{3}\\ & =ϵ\end{array}$$ Hence

$g$ is uniformly continuous on

$\overline{A}$.
check condition 3. :
Suppose

$h:\overline{A}\to N$ satisfying 1. and 2. .
Let

$x\in \overline{A}\mathrm{\setminus }A$.
Given

$ϵ>0$,

$\mathrm{\exists }{\delta }_1,{\delta }_2>0$ such that

$d(x,y)<{\delta }_1⟹\rho (g(x),g(y))<\frac{ϵ}{2}$ and

$d(x,y)<{\delta }_2⟹\rho (h(x),h(y))<\frac{ϵ}{2}$.
Since

$x\in \overline{A}\mathrm{\setminus }A$,

$\mathrm{\exists }y\in A$ such that

$d(x,y)<min({\delta }_1,{\delta }_2)$.
Then

$$\begin{array}{rl}\rho (g(x),h(x))& \le \rho (g(x),g(y))+\rho (g(y),h(y))+\rho (h(y),h(x))\\ & <\frac{ϵ}{2}+0+\frac{ϵ}{2}\\ & =ϵ\end{array}$$ Therefore

$g(x)=h(x)\mathrm{\forall }x\in \overline{A}$.

---

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

$(N,\rho )$.

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,

$\mathrm{\exists }{x}_0,x_1\in K$ such that

$f(x_0)={\displaystyle \underset{x\in K}{max}f(x)}$ and

$f(x_1)={\displaystyle \underset{x\in K}{min}f(x)}$.

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

$K$.

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

$\{x\in K\mid f(x)={\displaystyle \underset{x\in K}{max}f(x)\}}$ 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

$K$.

proof:

Since

$f$ is continuous on

$K$, given

$\epsilon >0$, for every

$x\in K$,

$\mathrm{\exists }{\delta }_x>0$ such that

$\rho (f(x),f(y))<\frac{\epsilon }{2}$ 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

$x_1,\dots ,x_L\in K$ such that

$K\subseteq {\displaystyle \bigcup _{i=1}^{L}B(x_i,\frac{{\delta }_{x_i}}{2})}$.
Select

$\delta ={\displaystyle \underset{1\le i\le L}{min}\frac{{\delta }_{x_i}}{2}}$.
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

$1\le l\le L$ such that

$u\in B(x_l,\frac{{\delta }_{x_l}}{2})$.
Then

$$\begin{array}{rl}d(v,x_l)& \le d(v,u)+d(u,x_l)\\ & <\delta +\frac{1}{2}{\delta }_{x_l}\\ & \le {\delta }_{x_l}\end{array}$$ Therefore

$v\in B(x_l,{\delta }_{x_l})$ and we have

$$\begin{array}{rl}\rho (f(u),f(v))& \le \rho (f(u),f(x_l))+\rho (f(x_l),f(v))\\ & <\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon \end{array}$$ Hence

$d(u,v)<\delta ⟹\rho (f(u),f(v))<\epsilon$.
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.

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

$\varphi :[0,1]\to A$ such that

$\varphi (0)=x$ and

$\varphi (1)=y$.

---

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

$A$.

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

$\varphi (t)=(1-t)x+ty$.

---

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

$⟹$ connected

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)

$A\cap u\ne \mathrm{\varnothing }$ (iii)

$A\cap v\ne \mathrm{\varnothing }$ (iv)

$A\cap u\cap v=\mathrm{\varnothing }$.
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

$\varphi :[0,1]\to A$ such that

$\varphi (0)=x$ and

$\varphi (1)=y$.
Then the sets

$w_1={\varphi }^{-1}(u)$ and

$w_2={\varphi }^{-1}(v)$ 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

$[0,1]\cap w_1\ne \mathrm{\varnothing }$ and

$[0,1]\cap w_2\ne \mathrm{\varnothing }$.
By (iv),

$[0,1]\cap w_1\cap w_2={\varphi }^{-1}(u)\cap {\varphi }^{-1}(v)=\mathrm{\varnothing }$.
Then

$w_1$ and

$w_2$ separate

$[0,1]$.

$(\to \leftarrow )$

The converse is false.
For example,

$A=\{(x,\sin \frac{1}{x})\mid x\in (0,1)\}\cup (0\times [-1,1])$ 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.

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)

$f(A)\cap u\ne \mathrm{\varnothing }$ (iii)

$f(A)\cap v\ne \mathrm{\varnothing }$ (iv)

$f(A)\cap u\cap v=\mathrm{\varnothing }$.
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),

$A\cap w_1\ne \mathrm{\varnothing }$ and

$A\cap w_2\ne \mathrm{\varnothing }$.
By (iv),

$A\cap w_1\cap w_2=\mathrm{\varnothing }$.
Hence,

$A$ is disconnected.

$(\to \leftarrow )$
(2.):
Suppose

$y_1,y_2\in f(A)$.

$\Rightarrow$

$\mathrm{\exists }{x}_1,x_2\in A$ such that

$y_1=f(x_1)$ and

$y_2=f(x_2)$.
Since

$A$ is path-connected, there exists a continuous map

$\varphi :[0,1]\to A$ such that

$\varphi (0)=x_1$ and

$\varphi (1)=x_2$.
Define

$\psi (t)=f(\varphi (t))$.
Clearly,

$\psi :[0,1]\to f(A)$.
Since

$f$ and

$\varphi$ are continuous,

$\psi$ is continuous on

$[0,1]$.
Then

$\psi (0)=f(\varphi (0))=f(x_1)=y_1$ and

$\psi (1)=f(\varphi (1))=f(x_2)=y_2$.
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

$\varphi :[0,1]\to V$ be a continuous map.

$\varphi$ is said to be piecewise linear if

$\mathrm{\exists }{t}_0,t_1,t_2,\dots ,t_n\in [0,1]$ with

$0=t_0<t_1<\cdots <t_n=1$ such that

$\varphi$ 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

${\varphi }_1,{\varphi }_2:[0,1]\to V$ such that

${\varphi }_1$ joins

$x$ and

$y$ and

${\varphi }_2$ joins

$y$ and

$z$, then there exists a piecewise linear mapping

$\varphi :[0,1]\to V$ 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

$y$.

proof:

Suppose

$x\in G$.
Define

$G_1=\{z\in G\mid \text{there exists a piecewise linear mapping joining }x\text{ and }z\}$.
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,

$\mathrm{\exists }\delta >0$ 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

$w\in G\mathrm{\setminus }{G}_1$.
Then there exists no piecewise linear mapping joining

$x$ and

$w$.
Since

$G$ is open,

$\mathrm{\exists }r>0$ 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

$w_1$.

$\Rightarrow$

$w\in G_1$

$(\to \leftarrow )$
Hence

$B(w,r)\subseteq G\mathrm{\setminus }{G}_1$ and

$G\mathrm{\setminus }{G}_1$ 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

$\mathrm{\forall }k\in \mathbb{N}$.
The sequence of maps

$\{f_k{\}}_{k=1}^{\mathrm{\infty }}$ is said to converge pointwise to

$f$ on

$A$ if

${\displaystyle \underset{k\to \mathrm{\infty }}{lim}\rho (f_k(a),f(a))=0}$ for every

$a\in A$.
That is,

$\mathrm{\forall }\epsilon >0$ and

$\mathrm{\forall }a\in A$,

$\mathrm{\exists }{N}_{\epsilon ,a}>0$ such that

$\rho (f_k(a),f(a))<\epsilon$ whenever

$k\ge N_{\epsilon ,a}$.

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

$\mathrm{\forall }N\in \mathbb{N}$.
The sequence of maps

$\{f_k{\}}_{k=1}^{\mathrm{\infty }}$ is said to converge uniformly to

$f$ on

$A$ if

$\mathrm{\forall }\epsilon >0$,

$\mathrm{\exists }{N}_{\epsilon }>0$ such that

$\rho (f_k(a),f(a))<\epsilon$ whenever

$a\in A$ and

$k\ge N_{\epsilon }$.

Example:

$f,f_k:[0,1]\to \mathbb{R}$ by

$f_k(x)=x^k$ and

$f(x)=\{\begin{array}{ll}0,& x\in [0,1)\\ 1,& x=1\end{array}$.
Then

1. $f_k\to f$ pointwise
2. $\{f_k{\}}_{k=1}^{\mathrm{\infty }}$ does not converge uniformly to
   $f$ on
   $[0,1]$.
3. For
   $0<a<1$,
   $f_k\to f$ uniformly on
   $[0,a]$

2:
Let

$\epsilon =\frac{1}{2}$.
Choose

$x_N=\sqrt[N]{\frac{2}{3}}$

$\mathrm{\forall }N\in \mathbb{N}$.
Then

$|f_N(x_N)-f(x_N)|=|\frac{2}{3}-0|>\epsilon$.

3:
Fix

$0<a<1$.
Given

$ϵ>0$, choose

$N\in \mathbb{N}$ such that

$N>{\displaystyle \frac{\ln ϵ}{\ln a}}$.
Since

$\ln a<0$,

$a^N<ϵ$.
For every

$x\in [0,a]$ and

$k\ge N$,

$|f_k(x)-f(x)|=|x^k-0|\le a^k\le a^N<ϵ$.
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

$\mathrm{\forall }k\in \mathbb{N}$.
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

$\mathrm{\forall }k\in \mathbb{N}$.
Suppose that

$(N,\rho )$ is complete.
Then

$\{f_k{\}}_{k=1}^{\mathrm{\infty }}$ converges uniformly on

$A$ iff

$\mathrm{\forall }\epsilon >0$,

$\mathrm{\exists }L\in \mathbb{N}$ such that

$\rho (f_m(x),f_n(x))<\epsilon$ whenever

$x\in A$ and

$m,n\ge L$.

proof:

$⟹:$
Let

$f:A\to N$ be a map where

$f_k\to f$ uniformly on A.
Given

$ϵ>0$,

$\mathrm{\exists }L\in \mathbb{N}$ such that

$\rho (f_k(x),f(x))<\frac{ϵ}{2}$ whenever

$x\in A$ and

$k\ge L$.
For

$m,n\ge L$ and

$x\in A$,

$$\begin{array}{rl}\rho (f_m(x),f_n(x))& \le \rho (f_m(x),f(x))+\rho (f(x),f_n(x))\\ & <\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ\end{array}$$

$⟸:$
Let

$\{f_k{\}}_{k=1}^{\mathrm{\infty }}$ be a sequence of maps on

$A$ satisfying the Cauchy criterion.
Fix

$a\in A$,

$\{f_k(a){\}}_{k=1}^{\mathrm{\infty }}$ is a Cauchy sequence in

$N$.
Since

$(N,\rho )$ is complete,

$\mathrm{\exists }{y}_a\in N$ such that

${\displaystyle \underset{k\to \mathrm{\infty }}{lim}{f}_k(a)=y_a}$ in

$N$.
By the same argument,

$\mathrm{\forall }x\in A$, there exists such

$y_x$.
Define

$f:A\to N$ by

$f(x)=y_x$.
Then

$f_k\to f$ pointwise on

$A$.
Given

$ϵ>0$, by the Cauchy criterion, there exists

$L\in \mathbb{N}$ such that

$\rho (f_m(x),f_n(x))<\frac{ϵ}{2}$ whenever

$x\in A$ and

$m,n\ge L$.
Since

$f_k\to f$ pointwise on

$A$, for every

$x\in A$,

$\mathrm{\exists }{L}_x\in \mathbb{N}$ with

$L_x\ge L$ such that

$\rho (f_m(x),f(x))<\frac{ϵ}{2}$ whenever

$m>L_x$.
Hence for every

$x\in A$ and

$m\ge L$, we choose

$m_x\ge L_x$.
Then

$$\begin{array}{rl}\rho (f_m(x),f(x))& \le \rho (f_m(x),f_{m_x}(x))+\rho (f_{m_x}(x),f(x))\\ & <\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ\end{array}$$ 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

$A$.

proof:

Since

$f_k\to f$ uniformly on

$A$, for given

$\epsilon >0$, there exists

$L\in \mathbb{N}$ such that for every

$x\in A$ and

$k\ge L$,

$$\rho (f_k(x),f(x))<\frac{\epsilon }{3}$$ Since

$f_L$ is continuous on

$A$,

$\mathrm{\forall }a\in A$,

$\mathrm{\exists }{\delta }_a>0$ such that

$\rho (f_L(x),f_L(a))<\frac{\epsilon }{3}$ whenever

$x\in A$ and

$d(x,a)<{\delta }_a$.
Hence, for every

$x\in A$ and

$d(x,a)<{\delta }_a$,

$$\begin{array}{rl}\rho (f(x),f(a))& \le \rho (f(x),f_L(x))+\rho (f_L(x),f_L(a))+\rho (f_L(a),f(a))\\ & <\frac{\epsilon }{3}+\frac{\epsilon }{3}+\frac{\epsilon }{3}\\ & =\epsilon \end{array}$$ Thus

$f$ is continuous at

$a$.

$f$ is continuous on

$A$.

If

$f_k$ continuous but

$f$ not continuous, then

$f_k\nrightarrow f$ uniformly.

If

$f_k\to f$ uniformly on

$A$ and

$a\in \bar{A}$, then

$$\underset{x\to a}{lim}(\underset{k\to \mathrm{\infty }}{lim}{f}_k(x))=\underset{k\to \mathrm{\infty }}{lim}(\underset{x\to a}{lim}{f}_k(x))$$

---

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

$M_i={\displaystyle \underset{t\in [t_{i-1},t_i]}{sup}f(t)}$ and

$m_i={\displaystyle \underset{t\in [t_{i-1},t_i]}{inf}f(t)}$.
Define

$\underset{―}{{\int }_a^b}f(x)dx={\displaystyle \underset{P}{sup}L(P,f)}$ and

$\overline{{\int }_a^b}f(x)dx={\displaystyle \underset{P}{inf}U(P,f)}$.
We have

1. $L(P,f)\le U(P,f)$
2. If
   $P_1$ is a refinement of
   $P$, then
   $L(P,f)\le L(P_1,f)\le U(P_1,f)\le U(P,f)$.
3. For any two partitions
   $P_1$ and
   $P_2$,
   $L(P_1,f)\le U(P_2,f)$.
4. $\underset{―}{{\int }_a^b}f(x)dx\le \overline{{\int }_a^b}f(x)dx$

If

$\underset{―}{{\int }_a^b}f(x)dx=\overline{{\int }_a^b}f(x)dx$, 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

$\mathrm{\forall }\epsilon >0$, there exists a partition

$P$ of

$[a,b]$ such that

$U(P,f)-L(P,f)<\epsilon$.

---

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

${\displaystyle \underset{k\to \mathrm{\infty }}{lim}{\int }_a^b{f}_k(x)dx={\int }_a^{b}f(x)dx={\int }_a^b\underset{k\to \mathrm{\infty }}{lim}{f}_k(x)dx}$.

proof:

Since

$\{f_k{\}}_{k=1}^{\mathrm{\infty }}$ converges uniformly to

$f$ on

$[a,b]$, for given

$\epsilon >0$,

$\mathrm{\exists }N\in \mathbb{N}$ such that

$|f_k(x)-f(x)|<\epsilon$ whenever

$x\in [a,b]$ and

$k\ge N$.
Since

$f_N$ is integrable on

$[a,b]$, there exists a partition

$P$ of

$[a,b]$ such that

$U(P,f_N)-L(P,f_N)<\epsilon$.
Let

$M_i={\displaystyle \underset{t\in [t_{i-1},t_i]}{sup}f(t)}$,

$m_i={\displaystyle \underset{t\in [t_{i-1},t_i]}{inf}f(t)}$, and

$M_i^{(N)}={\displaystyle \underset{t\in [t_{i-1},t_i]}{sup}{f}_N(t)}$,

$m_i^{(N)}={\displaystyle \underset{t\in [t_{i-1},t_i]}{inf}{f}_N(t)}$.
Then

$$\begin{array}{rl}|M_i-M_i^{(N)}|& \le \underset{t\in [t_{i-1},t_i]}{sup}|f(t)-f_N(t)|\le \epsilon \\ |m_i-m_i^{(N)}|& \le \underset{t\in [t_{i-1},t_i]}{sup}|f(t)-f_N(t)|\le \epsilon \end{array}$$ And we have

$$\begin{array}{rl}U(P,f)-L(P,f)& \le |U(P,f)-U(P,f_N)|+|U(P,f_N)-L(P,f_N)|+|L(P,f_N)-L(P,f)|\\ & <\sum _{i=1}^n|M_i-M_i^{(N)}|(t_i-t_{i-1})+\epsilon +\sum _{i=1}^n|m_i-m_i^{(N)}|(t_i-t_{i-1})\\ & \le 2\epsilon \sum _{i=1}^n(t_i-t_{i-1})+\epsilon \\ & =[2(b-a)+1]\epsilon \end{array}$$ Hence,

$f$ is integrable on

$[a,b]$.
Moreover, for

$k\ge N$,

$$\begin{array}{rl}|{\int }_a^{b}f(x)dx-{\int }_a^b{f}_k(x)dx|& =|{\int }_a^{b}f(x)-f_k(x)dx|\\ & \le {\int }_a^b|f(x)-f_k(x)|dx\\ & <{\int }_a^b\epsilon dx=\epsilon (b-a)\end{array}$$ Therefore

${\displaystyle {\int }_a^{b}f(x)dx=\underset{k\to \mathrm{\infty }}{lim}{\int }_a^b{f}_k(x)dx}$.

---

Suppose that

$f_k\to f$ pointwise and

$\int f_{k}dx\to \int fdx$.
It cannot imply that

$f_k\to f$ uniformly.

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

$f_k(x)=\{\begin{array}{ll}1,& x\in \{q_1,q_2,\dots q_k\}\\ 0,& \text{otherwise}\end{array}$ and

$f(x)=\{\begin{array}{ll}1,& x\in \mathbb{Q}\cap [0,1]\\ 0,& \text{otherwise}\end{array}$.
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

$\{f_k{\}}_{k=1}^{\mathrm{\infty }}$ 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

$\{f_k(a){\}}_{k=1}^{\mathrm{\infty }}$ converges for some

$a\in I$ and

$\{f_k^{\prime }{\}}_{k=1}^{\mathrm{\infty }}$ converges uniformly to a function

$\mathrm{g}$ on

$I$.
Then

1. $\{f_k{\}}_{k=1}^{\mathrm{\infty }}$ converges uniformly to some function
   $f$ on
   $I$.
2. The limit function
   $f$ is differentiable on
   $I$ and
   $f^{\prime }(x)=\mathrm{g}(x)$ for all
   $x\in I$.
   That is,
   $\mathrm{g}(x)={\displaystyle \underset{k\to \mathrm{\infty }}{lim}(\frac{d}{dx}{f}_k(x))=f^{\prime }(x)=\frac{d}{dx}(\underset{k\to \mathrm{\infty }}{lim}{f}_k(x))}$.

proof:

For

$x\in I$, by the F.T.C.,

$f_k(x)=f_k(a)+{\int }_a^x{f}_k^{\prime }(t)dt$.
Since

$\{f_k(a){\}}_{k=1}^{\mathrm{\infty }}$ converges, by the previous theorem,

$\{{\int }_a^x{f}_k^{\prime }(t)dt{\}}_{k=1}^{\mathrm{\infty }}$ converges for every

$x\in I$.
Then

$\mathrm{\forall }x\in I$,

$\{f_k(x){\}}_{k=1}^{\mathrm{\infty }}$ converges, and we can define

$f(x)={\displaystyle \underset{k\to \mathrm{\infty }}{lim}(f_k(a)+{\int }_a^x{f}_k^{\prime }(t)dt)}$.
Hence

$f(a)={\displaystyle \underset{k\to \mathrm{\infty }}{lim}{f}_k(a)}$ and

$f(x)-f(a)={\displaystyle \underset{k\to \mathrm{\infty }}{lim}{\int }_a^x{f}_k^{\prime }(t)dt={\int }_a^x{\displaystyle \underset{k\to \mathrm{\infty }}{lim}{f}_k^{\prime }(t)dt={\int }_a^x\mathrm{g}(t)dt}}$.

$\Rightarrow$

$f^{\prime }(x)=\mathrm{g}(x)$
Now we want to check

$f_k\to f$ uniformly on

$I$.
Since

${\displaystyle \underset{k\to \mathrm{\infty }}{lim}{f}_k(a)=f(a)}$, given

$\epsilon >0$,

$\mathrm{\exists }{N}_1\in \mathbb{N}$ such that

$|f_k(a)-f(a)|<\frac{\epsilon }{2}$ whenever

$k\ge N_1$.
Since

$f_k^{\prime }(x)\to \mathrm{g}(x)$ uniformly on

$I$,

$\mathrm{\exists }{N}_2\in \mathbb{N}$ such that

$|f_k^{\prime }(x)-\mathrm{g}(x)|<{\displaystyle \frac{\epsilon }{2|I|}}$ whenever

$x\in I$ and

$k\ge N_2$.
Therefore, for

$k\ge max(N_1,N_2)$,

$$\begin{array}{rl}|f(x)-f_k(x)|& =|[f(a)+{\int }_a^x\mathrm{g}(t)dt]-[f_k(a)+{\int }_a^x{f}_k^{\prime }(t)dt]|\\ & \le |f(a)-f_k(a)|+{\int }_a^x|\mathrm{g}(t)-f_k^{\prime }(t)|dt\\ & <\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon \end{array}$$ Thus

$f_k\to f$ uniformly on

$I$.

---

Suppose

$f_k$ differentiable and

$f_k\to f$ uniformly.
It cannot imply

$f$ differentialble.

counterexample:

$$f_k(x)=\{\begin{array}{ll}\frac{k}{2}{x}^2,& |x|\le \frac{1}{k}\\ |x|-\frac{1}{2k},& \frac{1}{k}\le |x|\le 1\end{array}\text{ and }f(x)=|x|$$

---

Dini's Theorem

Suppose that

$K$ is compact and
(a)

$f_n:K\to \mathbb{R}$ is continuous on

$K$

$\mathrm{\forall }n\in \mathbb{N}$.
(b)

$f_n\to f$ pointwise on

$K$ where

$f$ is continuous
(c)

$f_n(x)\le f_{n+1}(x)$

$\mathrm{\forall }n\in \mathbb{N}$ and

$x\in K$.

Then

$f_n\to f$ uniformly on

$K$.

proof:

Define

${\mathrm{g}}_n=f-f_n$

$\mathrm{\forall }n\in \mathbb{N}$.
Then,
(d)

${\mathrm{g}}_n$ is continuous on

$K$
(e)

${\mathrm{g}}_n\to \mathbf{0}$ pointwise on

$K$
(g)

${\mathrm{g}}_n(x)\ge {\mathrm{g}}_{n+1}(x)$

$\mathrm{\forall }n\in \mathbb{N}$ and

$x\in K$.
Our goal is to show

${\mathrm{g}}_n\to \mathbf{0}$ uniformly on

$K$.
Given

$\epsilon >0$, define

$K_n=\{x\in K\mid {\mathrm{g}}_n(x)\ge \epsilon \}$.
For each

$n\in \mathbb{N}$, since

${\mathrm{g}}_n$ is continuous on

$K$,

$K_n={\mathrm{g}}_n^{-1}([\epsilon ,\mathrm{\infty }))$ is closed in

$K$. Then

$K_n$ is compact.
Moreover, by (g),

$K_n\supseteq K_{n+1}$

$\mathrm{\forall }n\in \mathbb{N}$.
Fix

$x\in K$, since

${\mathrm{g}}_n(x)\to 0$ as

$n\to \mathrm{\infty }$,

$x\notin K_n$ as

$n$ is sufficiently large.
That implies

${\displaystyle \bigcap _{n=1}^{\mathrm{\infty }}{K}_n=\mathrm{\varnothing }}$.
According to this proposition

If

$\{K_n{\}}_{n=1}^{\mathrm{\infty }}$ is a sequence of nonempty compact sets in a metric space and

$K_n\supseteq K_{n+1}$

$\mathrm{\forall }n\in \mathbb{N}$, then

${\displaystyle \bigcap _{n=1}^{\mathrm{\infty }}{K}_n\ne \mathrm{\varnothing }}$.

there exists some

$N\in \mathbb{N}$ such that

$K_N=\mathrm{\varnothing }$.

That is, if

$n\ge N$,

${\mathrm{g}}_n(x)<\epsilon$ for every

$x\in K$.
Hence

$f_n\to f$ uniformly on

$K$.

• The result is true if condition (c) is replaced by
  $f_n(x)\ge f_{n+1}$.
• The compactness is necessary. You may consider
  $f_n(x)={\displaystyle \frac{1}{nx+1}}$ on
  $(0,1)$.
• The monotonicity is necessary.

---

Series of Functions

Let

$(M,d)$ be a metric space and

$(V,\Vert \cdot \Vert )$ be a normed space,

$A\subseteq M$,and

${\mathrm{g}}_k,\mathrm{g}:A\to V$ be functions.
(1) We say that the series

${\displaystyle \sum _{k=1}^{\mathrm{\infty }}{\mathrm{g}}_k}$ converges pointwise to

$\mathrm{g}$ on

$A$ if the sequence of partial sum

$\{S_n{\}}_{n=1}^{\mathrm{\infty }}$ 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

${\displaystyle \sum _{k=1}^{\mathrm{\infty }}{\mathrm{g}}_k}$ converge to

$\mathrm{g}$ uniformly on

$A$ if

$\{S_n{\}}_{n=1}^{\mathrm{\infty }}$ converges to

$\mathrm{g}$ uniformly on

$A$.

Example:

For the geometric series

${\displaystyle \sum _{k=0}^{\mathrm{\infty }}{x}^k}$,

$$S_n(x)=\sum _{k=0}^n{x}^k=\{\begin{array}{ll}\frac{1-x^{n+1}}{1-x}& \text{if }x\ne 1\\ n+1& \text{if }x=1\end{array}$$

1. For
   $x\in (-1,1)$,
   $S_n\to \frac{1}{1-x}$.
   
   $\Rightarrow$
   $\sum _{k=0}^{\mathrm{\infty }}{x}^k\to \frac{1}{1-x}$ pointwise on
   $(-1,1)$.
2. For
   $x\in (-\mathrm{\infty },1]\cup [1,\mathrm{\infty })$,
   $\{S_n{\}}_{n=1}^{\mathrm{\infty }}$ diverges.
   
   $\Rightarrow$
   $\sum _{k=0}^{\mathrm{\infty }}{x}^k$ diverges.
3. Let
   $a\in (0,1)$ and
   $\mathrm{g}(x)=\frac{1}{1-x}$, for
   $x\in [-a,a]$,
   $$|S_n(x)-\mathrm{g}(x)|=|\frac{1-x^{n+1}}{1-x}-\frac{1}{1-x}|=\left|\frac{x^{n+1}}{1-x}\right|\le \frac{|a{|}^{n+1}}{1-a}\to 0\text{ as }n\to \mathrm{\infty }$$ Given
   $\epsilon >0$, choose
   $N\in \mathbb{N}$ such that
   ${\displaystyle \frac{|a{|}^{n+1}}{1-a}}<\epsilon$ whenever
   $n\ge N$.
   Then
   $|S_n(x)-\mathrm{g}(x)|<\epsilon$ whenever
   $x\in [-a,a]$ and
   $n\ge N$.
   Hence
   $\sum _{k=0}^{\mathrm{\infty }}{x}^k$ converges uniformly on
   $[-a,a]$.
4. ${\displaystyle \sum _{k=0}^{\mathrm{\infty }}{x}^k}$ does not converge uniformly on
   $(-1,1)$ since
   ${\displaystyle \underset{x\in (-1,1)}{sup}|S_n(x)-\mathrm{g}(x)|=\mathrm{\infty }}$.

---

Cauchy Criterion
Let

$(M,d)$ be a metric space and

$(V,\Vert \cdot \Vert )$ be a normed space,

$A\subseteq M$, and

${\mathrm{g}}_k:A\to V$ be functions.
If

$\sum _{k=1}^{\mathrm{\infty }}{\mathrm{g}}_k$ converges uniformly on

$A$, then

$\mathrm{\forall }\epsilon >0$,

$\mathrm{\exists }N\in \mathbb{N}$ such that

$\Vert \sum _{k=m+1}^n{\mathrm{g}}_k(x)\Vert <\epsilon$ whenever

$x\in A$ and

$n>m\ge N$.
In addition, if

$(V,\Vert \cdot \Vert )$ is a Banach space, then the converse holds.

proof:

$⟹$:
Let

$\sum _{k=1}^{\mathrm{\infty }}{\mathrm{g}}_k$ converge uniformly to

$\mathrm{g}$ on

$A$.
Let

$S_n=\sum _{k=1}^n{\mathrm{g}}_k$ be the partial sum.
For

$\epsilon >0$,

$\mathrm{\exists }N\in \mathbb{N}$ such that

$\Vert S_n(x)-\mathrm{g}(x)\Vert <\frac{\epsilon }{2}$ whenever

$x\in A$ and

$n\ge N$.
Then for

$n>m\ge N$,

$$\begin{array}{rl}\Vert \sum _{k=m+1}^n{\mathrm{g}}_k(x)\Vert & =\Vert S_n(x)-S_m(x)\Vert \\ & \le \Vert S_n(x)-\mathrm{g}(x)\Vert +\Vert \mathrm{g}(x)-S_m(x)\Vert \\ & <\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon \end{array}$$ for every

$x\in A$.

$⟸\text{ (in Banach space)}$:
For

$x\in A$, given

$\epsilon >0$,

$\mathrm{\exists }{N}_{\epsilon ,x}\in \mathbb{N}$ such that

$\Vert S_n(x)-S_m(x)\Vert =\Vert \sum _{k=m+1}^n{\mathrm{g}}_k(x)\Vert <\epsilon$ whenever

$n>m\ge N_{\epsilon ,x}$.
Hence

$\{S_n(x){\}}_{n=1}^{\mathrm{\infty }}$ is a Cauchy sequence in

$V$.
Since

$(V,\Vert \cdot \Vert )$ is complete,

$\{S_n(x){\}}_{n=1}^{\mathrm{\infty }}$ converges in

$V$.
Since

$x\in A$ is arbitrary,

$S_n\to \mathrm{g}=\sum _{k=1}^{\mathrm{\infty }}{\mathrm{g}}_k$ pointwise on

$A$.
To check that

$S_n\to \mathrm{g}$ uniformly on

$A$, given

$\epsilon >0$,

$\mathrm{\exists }N\in \mathbb{N}$ such that

$\Vert S_n(x)-S_m(x)\Vert <\frac{\epsilon }{2}$ whenever

$x\in A$ and

$n>m\ge N$.
Since

$S_n\to \mathrm{g}$ pointwise on

$A$,

$\mathrm{\forall }x\in A$,

$\mathrm{\exists }{m}_x>N$ such that

$\Vert S_{m_x}(x)-\mathrm{g}(x)\Vert <\frac{\epsilon }{2}$.
Then for

$x\in A$ and

$n\ge N$,

$$\begin{array}{rl}\Vert S_n(x)-\mathrm{g}(x)\Vert & \le \Vert S_n(x)-S_{m_x}(x)\Vert +\Vert S_{m_x}(x)-\mathrm{g}(x)\Vert \\ & <\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon \end{array}$$ Therefore,

$S_n\to \mathrm{g}$ uniformly on

$A$.

---

If

$\sum _{k=1}^{\mathrm{\infty }}{\mathrm{g}}_k$ converges uniformly on

$A$, then

${\mathrm{g}}_k$ converges to

$\mathbf{0}$ uniformly on

$A$.

---

Let

$(M,d)$ be a metric space and

$(V,\Vert \cdot \Vert )$ 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

$\sum _{k=1}^{\mathrm{\infty }}{\mathrm{g}}_k$ 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

$\mathrm{g}=\sum _{k=1}^{\mathrm{\infty }}{\mathrm{g}}_k$ converges uniformly on

$[a,b]$, then

$$\begin{array}{r}{\int }_a^b\mathrm{g}(x)dx=\sum _{k=1}^{\mathrm{\infty }}{\int }_a^b{\mathrm{g}}_k(x)dx\end{array}$$

---

Weirestrass M-test

Let

$(M,d)$ be a metric space and

$(V,\Vert \cdot \Vert )$ 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

${\displaystyle \underset{x\in A}{sup}\Vert {\mathrm{g}}_k(x)\Vert \le M_k}$ whenever

$k\in \mathbb{N}$ and

$\sum _{k=1}^{\mathrm{\infty }}{M}_k$ converges.
Then

${\displaystyle \sum _{k=1}^{\mathrm{\infty }}{\mathrm{g}}_k:A\to V}$ converges uniformly and absolutely on

$A$.
That is,

${\displaystyle \sum _{k=1}^{\mathrm{\infty }}\Vert {\mathrm{g}}_k\Vert :A\to \mathbb{R}}$ converges uniformly on

$A$.

proof:

We intend to show that

$\{S_n{\}}_{n=1}^{\mathrm{\infty }}$ given by

$S_n=\sum _{k=1}^n{\mathrm{g}}_k$ satisfies the Cauchy criterion.
Since

$\sum _{k=1}^{\mathrm{\infty }}{M}_k$ converges, given

$\epsilon >0$,

$\mathrm{\exists }N\in \mathbb{N}$ such that

$\sum _{k=m+1}^n{M}_k<\epsilon$ whenever

$n>m\ge N$.
Thus

$$\begin{array}{rl}\underset{x\in A}{sup}\Vert S_n(x)-S_m(x)\Vert & =\underset{x\in A}{sup}\Vert \sum _{k=m+1}^n{\mathrm{g}}_k(x)\Vert \\ & \le \underset{x\in A}{sup}\sum _{k=m+1}^n\Vert {\mathrm{g}}_k(x)\Vert \\ & \le \sum _{k=m+1}^n\underset{x\in A}{sup}\Vert {\mathrm{g}}_k(x)\Vert \\ & \le \sum _{k=m+1}^n{M}_k\\ & <\epsilon \end{array}$$ Hence

$\{S_n{\}}_{n=1}^{\mathrm{\infty }}$ converges uniformly on

$A$.

Consider the sequence of functions

$\Vert {\mathrm{g}}_k\Vert :A\to \mathbb{R}$.
Let

$T_n(x)=\sum _{k=1}^n\Vert {\mathrm{g}}_k(x)\Vert$.
For

$n>m\ge N$,

$$\begin{array}{rl}\underset{x\in A}{sup}|T_n(x)-T_m(x)|& =\underset{x\in A}{sup}\sum _{k=m+1}^n\Vert {\mathrm{g}}_k(x)\Vert \\ & \le \sum _{k=m+1}^n\underset{x\in A}{sup}\Vert {\mathrm{g}}_k(x)\Vert \\ & \le \sum _{k=m+1}^n{M}_k\\ & <\epsilon \end{array}$$ Therefore

$\{T_n{\}}_{n=1}^{\mathrm{\infty }}$ converges uniformly on

$A$ and hence

$\sum _{k=m+1}^{\mathrm{\infty }}\Vert {\mathrm{g}}_k(x)\Vert$ converges uniformly on

$A$.

---

Example of a continuous and nowhere differentiable function

In

$(\mathbb{R},|\cdot |)$, define

$\varphi (x)=|x|$ on

$[-1,1]$ and extend

$\varphi$ to a

$2$-periodic function (still called

$\varphi$).
Then for

$s,t\in \mathbb{R}$,

$|\varphi (s)-\varphi (t)|\le |s-t|$ … (a).
Define

$f(x)={\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\left(\frac{3}{4}\right)}^n\varphi \left(4^{n}x\right)}$.

${\left(\frac{3}{4}\right)}^n\varphi (4^{n}x)$ is continuous for every

$n\in \mathbb{N}$.
Since

$0\le \varphi (x)\le 1$, 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

${\displaystyle \underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}}$ 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

$\mathbb{Z}\cap (4^{m}x,4^m(x+{\delta }_m))=\mathrm{\varnothing }$ or

$\mathbb{Z}\cap (4^m(x+{\delta }_m),4^{m}x)=\mathrm{\varnothing }$.
Then the interval would contain no interger.

Define

${\displaystyle r_n=\frac{\varphi (4^n(x+{\delta }_m))-\varphi (4^{n}x)}{{\delta }_m}}$.
When

$n>m$,

$4^n{\delta }_m=\pm \frac{1}{2}{4}^{n-m}$ is an even interger.
Since

$\varphi$ is periodic by

$2$,

$\varphi (4^n(x+{\delta }_m))-\varphi (4^{n}x)=0$ and

$r_n=0$.
When

$0\le n\le m$, by (a),

$|r_n|={\displaystyle \frac{|4^n(x+{\delta }_m)-4^{n}x|}{{\delta }_m}=4^n}$ and

$\varphi (4^m(x+{\delta }_m))-\varphi (4^{m}x)=\pm r^m{\delta }_m$.

$\Rightarrow$

$|r_m|=4^m$

Thus

$$\begin{array}{rl}\left|\frac{f(x+{\delta }_m)-f(x)}{{\delta }_m}\right|& =\left|\sum _{n=0}^m{\left(\frac{3}{4}\right)}^n{r}_n\right|\\ & \ge {\left(\frac{3}{4}\right)}^m|r_m|-\left|\sum _{n=0}^{m-1}{\left(\frac{3}{4}\right)}^n{r}_n\right|\\ & \ge 3^m-\sum _{n=0}^{m-1}{3}^n\\ & =\frac{1}{2}(3^m+1)\to \mathrm{\infty }\text{ as }m\to \mathrm{\infty }\end{array}$$ As

$m\to \mathrm{\infty }$,

${\delta }_m\to 0$, we have

${\displaystyle \underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}}$ does not exist and hence

$f$ is not differentiable at

$x$.

---

	domain	$f_k$	convergence	$f$	range
		continuous	uniformly	continuous
	$[a,b]$	integrable ${\displaystyle \int \underset{k\to \mathrm{\infty }}{lim}=\underset{k\to \mathrm{\infty }}{lim}\int }$	uniformly
$\text{Dini}$	compact	continuous monotone	pointwise uniformly	continuous	$\mathbb{R}$
	$[a,b]$	integrable monotone ${\displaystyle \int \underset{k\to \mathrm{\infty }}{lim}=\underset{k\to \mathrm{\infty }}{lim}\int }$	pointwise	integrable	$\mathbb{R}$
	$[a,b]$	integrable bounded ${\displaystyle \int \underset{k\to \mathrm{\infty }}{lim}=\underset{k\to \mathrm{\infty }}{lim}\int }$	pointwise	integrable	$\mathbb{R}$

• where integrable means Riemann integrable
• where monotone means
  $f_n(x)\le f_{n+1}(x)\vee f_n(x)\ge f_{n+1}(x)\mathrm{\forall }n\in \mathbb{N},x\in D_{f_n}$

---

Power Series

We called a series of the form

$$\sum _{k=0}^{\mathrm{\infty }}{c}_k(x-a{)}^k$$ a power series centered at

$a$ for some

$\{c_k{\}}_{k=0}^{\mathrm{\infty }}\subseteq \mathbb{R}$ and

$a\in \mathbb{R}$.

If

$a=0$, it is called a Maclaurin series.

---

Let

$\sum _{k=0}^{\mathrm{\infty }}{c}_k(x-a{)}^k$ be a power series in

$\mathbb{R}$.
Suppose that the series converges at some

$b\ne a$.
Define

$h=|b-a|$.
Then the series converges on

$(a-h,a+h)$.
Moreover, the series converges uniformly on

$[\alpha ,\beta ]$ if

$[\alpha ,\beta ]\subseteq (a-h,a+h)$.

proof:

WLOG, let

$a=0$ and

$\sum _{k=0}^{\mathrm{\infty }}{c}_k{x}^k$ converge at

$b\ne 0$.
Then

$h=|b|$.
Since

$\sum _{k=0}^{\mathrm{\infty }}{c}_k{b}^k$ converges,

${\displaystyle \underset{k\to \mathrm{\infty }}{lim}|c_k{b}^k|=\underset{k\to \mathrm{\infty }}{lim}|c_k|h^k=0}$.

$\Rightarrow$

$\mathrm{\exists }N\in \mathbb{N}$ such that

$|c_k|h^k<1$ whenever

$k\ge N$

$\Rightarrow$

$|c_k|<\frac{1}{h^k}$ for every

$k\ge N$

For

$x_0\in (-h,h)$,

$$\begin{array}{rl}\sum _{k=N}^{\mathrm{\infty }}|c_k{x}_0^k|& =\sum _{k=N}^{\mathrm{\infty }}|c_k||x_0{|}^k\\ & \le \sum _{k=N}^{\mathrm{\infty }}\frac{1}{h^k}|x_0{|}^k\\ & =\sum _{k=N}^{\mathrm{\infty }}{\left(\frac{|x_0|}{h}\right)}^k\\ & <\mathrm{\infty }\text{(geometric series)}\end{array}$$ Then

$\sum _{k=0}^{\mathrm{\infty }}{c}_k{x}_0^k$ would converge absolutely and hence converges.
Thus

$\sum _{k=0}^{\mathrm{\infty }}{c}_k{x}^k$ 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

$\epsilon >0$ and

$n>m\ge N$,

$$\begin{array}{rl}\left|\sum _{k=m+1}^n{c}_k{x}_0^k\right|& \le \sum _{k=m+1}^n\left|c_k{x}_0^k\right|\\ & \le \sum _{k=m+1}^n|c_k|h^k(1-\frac{\delta }{h}{)}^k\\ & <\epsilon \text{ as }m,n\text{ sufficiently large}\end{array}$$ Therefore,

$\sum _{k=0}^{\mathrm{\infty }}{c}_k{x}^k$ converges uniformly on

$[\alpha ,\beta ]$.

---

$R>0$ is called the radius of convergence of the power series

$\sum _{k=0}^{\mathrm{\infty }}{c}_k(x-a{)}^k\subseteq \mathbb{R}$ if the series converges for every

$x\in (a-R,a+R)$ and diverges for all

$x\in (-\mathrm{\infty },a-R)\cup (a+R,\mathrm{\infty })$.

$R={\displaystyle sup\{r\ge 0|\sum _{k=0}^{\mathrm{\infty }}{c}_k(x-a{)}^k\text{ converges in }[a-r,a+r]\}}$

---

By ratio test, consider the series

$\sum _{k=0}^{\mathrm{\infty }}{x}_k$,

$$\begin{array}{rl}\underset{k\to \mathrm{\infty }}{lim sup}\left|\frac{x_{k+1}}{x_k}\right|<1& ⟹\text{ the series converges}\\ \underset{k\to \mathrm{\infty }}{lim inf}\left|\frac{x_{k+1}}{x_k}\right|>1& ⟹\text{ the series diverges}\end{array}$$

---

Let

$\sum _{k=0}^{\mathrm{\infty }}{c}_k(x-a{)}^k\subseteq \mathbb{R}$ be a power series.
If

${\displaystyle \underset{k\to \mathrm{\infty }}{lim}\left|\frac{c_k}{c_{k+1}}\right|}$ converges, then the radius of convergence of the power series is

${\displaystyle \underset{k\to \mathrm{\infty }}{lim}\left|\frac{c_k}{c_{k+1}}\right|}$.

proof:

For

$x\ne a$, 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

$$\begin{array}{rl}\underset{k\to \mathrm{\infty }}{lim sup}\left|\frac{c_{k+1}}{c_k}\right||x-a|<1& ⟺|x-a|<\frac{1}{{\displaystyle \underset{k\to \mathrm{\infty }}{lim sup}\left|\frac{c_{k+1}}{c_k}\right|}}=\underset{k\to \mathrm{\infty }}{lim inf}\left|\frac{c_k}{c_{k+1}}\right|\\ \underset{k\to \mathrm{\infty }}{lim inf}\left|\frac{c_{k+1}}{c_k}\right||x-a|>1& ⟺|x-a|<\frac{1}{{\displaystyle \underset{k\to \mathrm{\infty }}{lim inf}\left|\frac{c_{k+1}}{c_k}\right|}}=\underset{k\to \mathrm{\infty }}{lim sup}\left|\frac{c_k}{c_{k+1}}\right|\end{array}$$

If

${\displaystyle \underset{k\to \mathrm{\infty }}{lim}\left|\frac{c_k}{c_{k+1}}\right|}$ converges, then

$$\underset{k\to \mathrm{\infty }}{lim}\left|\frac{c_k}{c_{k+1}}\right|=\underset{k\to \mathrm{\infty }}{lim sup}\left|\frac{c_k}{c_{k+1}}\right|=\underset{k\to \mathrm{\infty }}{lim inf}\left|\frac{c_k}{c_{k+1}}\right|=R$$

---

Let

$\sum _{k=0}^{\mathrm{\infty }}{c}_k(x-a{)}^k\subseteq \mathbb{R}$ be a power series with the radius of convergence

$R$, and

$[\alpha ,\beta ]\subseteq (a-R,a+R)$.
Then the series

$\sum _{k=0}^{\mathrm{\infty }}(k+1)c_{k+1}(x-a{)}^k$ converges pointwise on

$(a-R,a+R)$ and converges uniformly on

$[\alpha ,\beta ]$.

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

$a+h\in (a-R,a+R)$,

$\sum _{k=0}^{\mathrm{\infty }}((a+h)-a{)}^k=\sum _{k=0}^{\mathrm{\infty }}{c}_k{h}^k$ converges.

$\Rightarrow$

${\displaystyle \underset{k\to \mathrm{\infty }}{lim}|c_k{h}^k|=0}$

$\Rightarrow$

$\mathrm{\exists }N\in \mathbb{N}$ such that

$|c_k{h}^k|<1$ whenever

$k\ge N$.
Then

$$\begin{array}{rl}\sum _{k=N}^{\mathrm{\infty }}|(k+1)c_k(x-a{)}^k|& =\sum _{k=N}^{\mathrm{\infty }}(k+1)|c_k{h}^k|{\left(\frac{|x-a|}{h}\right)}^k\\ & <\sum _{k=N}^{\mathrm{\infty }}(k+1)r^k\\ & <\mathrm{\infty }(r>1)\end{array}$$ By Weirestrass M-test,

$\sum _{k=0}^{\mathrm{\infty }}(k+1)c_{k+1}(x-a{)}^k$ 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

$\sum _{k=0}^{\mathrm{\infty }}(k+1)c_{k+1}(x-a{)}^k$ converges uniformly on

$[\alpha ,\beta ]$, it converges pointwise at

$x$.
Hence

$\sum _{k=0}^{\mathrm{\infty }}(k+1)c_{k+1}(x-a{)}^k$ converges pointwise on

$(a-R,a+R)$.

---

Let

$\sum _{k=0}^{\mathrm{\infty }}{c}_k(x-a{)}^k\subseteq \mathbb{R}$ 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,

$$\frac{d}{dx}\sum _{k=0}^{\mathrm{\infty }}{c}_k(x-a{)}^k=\sum _{k=0}^{\mathrm{\infty }}\frac{d}{dx}{c}_k(x-a{)}^k=\sum _{k=1}^{\mathrm{\infty }}k{c}_k(x-a{)}^{k-1}$$

Since the differentiation is still a power series, it is also differentiable.
By mathematical induction, a power series is infinitely differentiable.

---

Let

$\sum _{k=0}^{\mathrm{\infty }}{c}_k(x-a{)}^k$ be a power series with radius of convergence

$R$ and

$[\alpha ,\beta ]\subseteq (a-R,a+R)$.
Then the power series

$\sum _{k=0}^{\mathrm{\infty }}{c}_k(x-a{)}^k$ is integrable on

$[\alpha ,\beta ]$.
Moreover,

$${\int }_{\alpha }^{\beta }\sum _{k=0}^{\mathrm{\infty }}{c}_k(x-a{)}^{k}dx=\sum _{k=0}^{\mathrm{\infty }}{\int }_{\alpha }^{\beta }{c}_k(x-a{)}^{k}dx$$

---

Example 1

The function

$e^x=\sum _{k=0}^{\mathrm{\infty }}\frac{x^k}{k!}$ converges on

$\mathbb{R}$.
Then

$$\begin{array}{rl}{\int }_0^t{e}^{x}dx& ={\int }_0^t\sum _{k=0}^{\mathrm{\infty }}\frac{x^k}{k!}dx\\ & =\sum _{k=0}^{\mathrm{\infty }}{\int }_0^t\frac{x^k}{k!}dx\\ & ={\sum _{k=0}^{\mathrm{\infty }}\frac{x^{k+1}}{(k+1)!}|}_0^t\\ & =\sum _{k=1}^{\mathrm{\infty }}\frac{t^k}{k!}\\ & =\left(\sum _{k=0}^{\mathrm{\infty }}\frac{t^k}{k!}\right)-1=e^t-1\end{array}$$

Example 2

The series

$\sum _{k=1}^{\mathrm{\infty }}\frac{x^k}{k}$ converges on

$(-1,1)$.
Hence, for

$x\in (-1,1)$,

$$\begin{array}{rl}\frac{d}{dx}\sum _{k=1}^{\mathrm{\infty }}\frac{x^k}{k}& =\sum _{k=1}^{\mathrm{\infty }}{x}^{k-1}\\ & =\sum _{k=0}^{\mathrm{\infty }}{x}^k=\frac{1}{1-x}\end{array}$$

For

$x\in (-1,1)$,

$$\begin{array}{rl}\sum _{k=1}^{\mathrm{\infty }}\frac{x^k}{k}& ={\int }_0^x\frac{d}{dt}\sum _{k=1}^{\mathrm{\infty }}\frac{t^k}{k}dt\text{ by F.T.C.}\\ & ={\int }_0^x\frac{1}{1-t}dt=-\ln (1-x)\end{array}$$

Moreover, we may consider

$x=-1$.

$\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^k}{k}$ converges by alternating series test.
We want to show that

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n\frac{(-1{)}^k}{k}=-\ln 2}$.
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

$$\begin{array}{rl}|\sum _{k=1}^n\frac{(-1{)}^k}{k}-(-\ln 2)|& =|{\int }_{-1}^0\frac{1}{1-t}dt-(-\ln 2)|+\left|{\int }_{-1}^0\frac{t^n}{1-t}dt\right|\\ & <0+\left|{\int }_{-1}^0{t}^{n}dt\right|\\ & =\frac{1}{n+1}\to 0\text{ as }n\to \mathrm{\infty }\end{array}$$ Therefore

$\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^k}{k}=-\ln 2$ and the series converges on

$[-1,1)$.

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

$y(x)=\sum _{k=0}^{\mathrm{\infty }}{c}_k{x}^k$.
Assume that the series converges on some

$(-\delta ,\delta )$.
Then

$y^{\prime }(x)=\sum _{k=1}^{\mathrm{\infty }}k{c}_k{x}^{k-1}$ and

$y^{″}(x)=\sum _{k=2}^{\mathrm{\infty }}k(k-1)c_k{x}^{k-2}$.
From the formula, we obtain

$$\begin{array}{rl}0& =\sum _{k=2}^{\mathrm{\infty }}k(k-1)c_k{x}^{k-2}+\sum _{k=0}^{\mathrm{\infty }}{c}_k{x}^k\\ & =\sum _{k=0}^{\mathrm{\infty }}[(k+2)(k+1)c_{k+2}+c_k]x^k\end{array}$$ The coefficients satisfy

$(k+2)(k+1)c_{k+1}+c_k=0$ for

$k=0,1,2,\dots$ and thus the recurrence ralation

$c_{k+2}=-\frac{c_k}{(k+1)(k+2)}$.

Hence we have

$c_k=\{\begin{array}{ll}\frac{(-1{)}^{k/2}}{k!}{c}_0& k\text{ is even}\\ \frac{(-1{)}^{\lfloor k/2\rfloor }}{k!}{c}_1& k\text{ is odd}\end{array}$.
Therefore,

$$\begin{array}{rl}y& =c_0[1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots +\frac{(-1{)}^n}{(2n)!}{x}^{2n}+\cdots ]+c_1[x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots +\frac{(-1{)}^n}{(2n+1)!}+\cdots ]\\ & =c_0\cos x+c_1\sin x\end{array}$$

---

Taylor Series

Suppose that

$f$ is a function such that

$f^{\prime }(a),f^{″}(a),\dots ,f^{(n)}(a)$ 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

$f^{\prime }(a)$,

$f^{″}(a)$,

$\dots$,

$f^{(n)}(a)$ exist.
Then

$$\underset{x\to a}{lim}\frac{f(x)-P_{n,a}(x)}{(x-a{)}^n}=0$$

proof:

For

$x\ne a$,

$$\frac{f(x)-P_{n,a}(x)}{(x-a{)}^n}=\frac{f(x)-\sum _{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}(x-a{)}^k}{(x-a{)}^n}-\frac{f^{(n)}(a)}{n!}$$ 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

$1\le i\le n-1$,

$$Q^{(i)}(x)=f^{(i)}(x)-f^{(i)}(a)-\frac{f^{(i+1)}(a)}{1!}(x-a)-\cdots -\frac{f^{(n-1)}(a)(x-a{)}^{n-1-i}}{(n-1-i)!}$$ Hence

${\displaystyle \underset{x\to a}{lim}{Q}^{(i)}(x)=0}$ for

$i=0,1,2,\dots ,n-1$.
On the other hand,

$g^{(i)}(x)=n(n-1)\cdots (n-i+1)(x-a{)}^{n-i}$ and hence

${\displaystyle \underset{x\to a}{lim}{g}^{(i)}(x)=0}$ for

$i=0,1,2,\dots ,n-1$.
Use L'Hôpital's Rule

$n-1$ times,

$$\begin{array}{rl}\underset{x\to a}{lim}\frac{f(x)-P_{n,a}(x)}{(x-a{)}^n}& =\underset{x\to a}{lim}\frac{Q(x)}{g(x)}-\frac{f^{(n)}(a)}{n!}\\ & =\underset{x\to a}{lim}\frac{Q^{\prime }(x)}{g^{\prime }(x)}-\frac{f^{(n)}(a)}{n!}\\ & ⋮\\ & =\underset{x\to a}{lim}\frac{Q^{(n-1)}(x)}{g^{(n-1)}(x)}-\frac{f^{(n)}(a)}{n!}\\ & =\underset{x\to a}{lim}\frac{f^{(n-1)}(x)-f^{(n-1)}(a)}{n!(x-a)}-\frac{f^{(n)}(a)}{n!}\\ & =0\end{array}$$

---

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

$$R_{n,a}(x)={\int }_a^x\frac{f^{(n+1)}(t)}{n!}(x-t{)}^{n}dt$$

proof:

$$\begin{array}{rl}f(x)& =\underset{P_{0,a}(x)}{\underbrace{f(a)}}+\underset{R_{0,a}(x)}{\underbrace{{\int }_a^x{f}^{\prime }(t)dt}}(\text{F.T.C.})\\ & =f(a)+{f^{\prime }(t)\cdot t|}_a^x-{\int }_a^x{f}^{″}(t)dt(\text{integration by parts})\\ & =f(a)+f^{\prime }(x)\cdot x-f^{\prime }(a)\cdot a-{\int }_a^x{f}^{″}(t)tdt\\ & =f(a)+f^{\prime }(a)(x-a)-f^{\prime }(a)\cdot x+f^{\prime }(x)\cdot x-{\int }_a^x{f}^{″}(t)tdt\\ & =f(a)+f^{\prime }(a)(x-a)+x{\int }_a^x{f}^{″}(t)dt-{\int }_a^x{f}^{″}(t)tdt\\ & =\underset{P_{1,a}(x)}{\underbrace{f(a)+f^{\prime }(a)(x-a)}}+\underset{R_{1,a}(x)}{\underbrace{{\int }_a^x{f}^{″}(t)(x-t)dt}}\\ & =f(a)+f^{\prime }(a)(x-a)-{f^{″}(t)\frac{(x-t{)}^2}{2}|}_a^x+{\int }_a^x\frac{f^{‴}(t)}{2}(x-t{)}^{2}dt(\text{integration by parts})\\ & =\underset{P_{2,a}(x)}{\underbrace{f(a)+f^{\prime }(a)(x-a)+\frac{f^{″}(a)}{2}(x-a{)}^2}}+\underset{R_{2,a}(x)}{\underbrace{{\int }_a^x\frac{f^{‴}(t)}{2}(x-t{)}^{2}dt}}\end{array}$$ 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

$f(x)=f(a)+f^{\prime }(a)(x-a)+\cdots +\frac{f^{(n)}(a)}{n!}(x-a{)}^n+R_{n,a}(x)$.
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

$|R_{n,a}(x)|\le M\frac{|x-a{|}^{n+1}}{(n+1)!}$.

In intergral form, if

$|f^{(n+1)}(t)|<M$ for all

$t\in [a,x]$, then

$|R_{n,a}(x)|\le \frac{M}{n!}|{\int }_a^x(x-t{)}^{n}dt|\le \frac{M}{(n+1)!}\left|{-(x-t{)}^{n+1}|}_a^x\right|=\frac{M}{(n+1)!}|x-a{|}^{n+1}$.

---

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

   $0<h<\mathrm{\infty }$ such that
   $(a-h,a+h)\subseteq (\alpha ,\beta )$, suppose that
   $\mathrm{\exists }M>0$ such that
   $|f^{(k)}(x)|\le M$ for all
   $x\in (a-h,a+h)$ and
   $k\in \mathbb{N}$.
   Then
   $$f(x)=\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(a)}{k!}(x-a{)}^k\text{ for all }x\in (a-h,a+h)$$

proof of 2:

Let

$S_n(x)={\displaystyle \sum _{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a{)}^k}$.
For

$x\in (a-h,a+h)$,

$$\begin{array}{rl}|S_n(x)-f(x)|& \le |{\int }_a^x\frac{f^{(n+1)}(t)}{n!}(x-t{)}^{n}dt|\\ & \le \frac{M}{h!}{h}^n\left|{\int }_a^{x}1dt\right|\\ & =\frac{M}{h!}{h}^n|x-a|\\ & \le \frac{M}{n!}{h}^{n+1}\to 0\text{ as }n\to \mathrm{\infty }\end{array}$$ Hence

$\{S_n(x){\}}_{n=1}^{\mathrm{\infty }}$ converges to

$f(x)$ uniformly on

$(a-h,a+h)$.
We obtain

$f(x)={\displaystyle \sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(a)}{k!}}$ uniformly on

$(a-h,a+h)$.

---