Advanced Calculus (II)

2022 spring

黃博峙大大的高等微積分二

我去成大修了習偉的一，傳送門在這裡！

Overview

• Advanced Calculus (II)
  • Overview
  • Space of Continuous Functions

Space of Continuous Functions

前面有些 overlap 的部份就跳過ㄌ，從這兒開始

Let

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

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

$A\subseteq M$. We denote

1. $\mathcal{C}(A;V)=\{f:A\to V\mid f\text{ is continuous on }A\}$
2. ${\mathcal{C}}_b(A;V)=\{f\in \mathcal{C}(A;V)\mid f\text{ is bounded on }A\}$

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Let

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

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

$A\subseteq M$.
Then

$B\subseteq {\mathcal{C}}_b(A;V)$ is said to be equi-continuous if

$\mathrm{\forall }ϵ>0$,

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

$\Vert f(x_1)-f(x_2)\Vert <ϵ$ whenever

$d(x_1,x_2)<\delta$,

$x_1,x_2\in A$, and

$f\in B$.

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In an equi-continuous set

$B$, every

$f\in B$ is uniformly continuous.

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Let

$B=\{f\in {\mathcal{C}}_b((0,1);\mathbb{R})\mid |f^{\prime }(x)|\le 1\mathrm{\forall }x\in (0,1)\}$.

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Let

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

$A\subseteq M$.
Then

$A$ is precompact if

$\bar{A}$ is compact.

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Let

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

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

$K\subseteq M$ be compact.
If

$B\subseteq \mathcal{C}(K;V)$ is pre-compact, then

$B$ is equi-continuous.

proof:

Proof by contradiction.
Suppose

$B$ not equi-continuous.

Then

$\mathrm{\exists }ϵ>0$ such that

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

$\mathrm{\exists }{x}_k,y_k\in K$ and

$f_k\in B$ such that

$d(x_k,y_k)<1/k$ but

$\Vert f_k(x_k)-f_k(y_k)\Vert \ge ϵ$.

Thus, we obtain

$\{x_k{\}}_{k=1}^{\mathrm{\infty }},\{y_k{\}}_{k=1}^{\mathrm{\infty }}\subseteq K$, and

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

Since

$B$ is pre-compact in

$(\mathcal{C}(K;V),\Vert \cdot {\Vert }_{\mathrm{\infty }})$, there exists

$\{f_{k_j}{\}}_{j=1}^{\mathrm{\infty }}\subseteq \{f_k{\}}_{k=1}^{\mathrm{\infty }}$ such that

${\displaystyle \underset{j\to \mathrm{\infty }}{lim}{f}_{k_j}=f}$ for some

$f\in (\mathcal{C}(K;V),\Vert \cdot {\Vert }_{\mathrm{\infty }})$.

Then

$f_{k_j}\to f$ uniformly on

$K$.

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

$\Vert f_{k_j}(x)-f(x)\Vert <{\displaystyle \frac{ϵ}{4}}$ whenever

$j\ge N_1$ and

$x\in K$.

Since

$f$ is continuous on the compact

$K$,

$f$ is uniformly continous on

$K$.

Hence

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

$\Vert f(x)-f(y)\Vert <\frac{ϵ}{4}$ whenever

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

Let

$N=max(N_1,\lfloor \frac{1}{\delta }\rfloor +1)$.

Then for

$j\ge N$,

$$d(x_{k_j},y_{k_j})<\frac{1}{k_j}\le \frac{1}{j}\le \frac{1}{N}\le \frac{1}{\lfloor \frac{1}{\delta }\rfloor +1}<\delta$$

This implies

$$\begin{array}{rl}ϵ& \le \Vert f_{k_j}(x_{k_j})-f_{k_j}(y_{k_j})\Vert \\ & \le \Vert f_{k_j}(x_{k_j})-f(x_{k_j})\Vert +\Vert f(x_{k_j})-f(y_{k_j})\Vert +\Vert f(y_{k_j})-f_{k_j}(y_{k_j})\Vert \\ & <\frac{3ϵ}{4}\end{array}$$

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Let

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

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

$K\subseteq M$ be compact.
If

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

$K$, then

$\{f_k{\}}_{k=1}^{\mathrm{\infty }}$ is equi-continuous.

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