---
# System prepended metadata

title: Advanced Calculus (II)

---

# 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](https://hackmd.io/@tinyynoob/SJvGWVbYt)！

## Overview

[TOC]


## 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, $(V,\|\cdot\|)$ be a normed vector space, and $A\subseteq M$. We denote
1. $\mathcal{C}(A;V)=\left\{f:A\to V \mid f \text{ is continuous on } A\right\}$
2. $\mathcal{C}_b(A;V) = \left\{f\in \mathcal{C}(A;V) \mid f \text{ is bounded on } A\right\}$

---

Let $(M,d)$ be a  metric space, $(V,\|\cdot\|)$ be a normed vector space, and $A\subseteq M$. \
Then $B\subseteq \mathcal{C}_b(A;V)$ is said to be ==equi-continuous== if $\forall\,\epsilon>0$, $\exists\,\delta>0$ such that $\|f(x_1)-f(x_2)\|<\epsilon$ whenever $d(x_1,x_2)<\delta$, $x_1,x_2\in A$, and $f\in B$.

---

In an equi-continuous set $B$, every $f\in B$ is uniformly continuous.

---

Let $B=\left\{f\in \mathcal{C}_b((0,1);\mathbb{R})\mid |f'(x)|\leq 1 \;\forall\,x\in(0,1)\right\}$.

---

Let $(M,d)$ be a metric space and $A\subseteq M$. \
Then $A$ is ==precompact== if $\overline{A}$ is compact.

---

Let $(M,d)$ be a  metric space, $(V,\|\cdot\|)$ 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 $\exists\,\epsilon>0$ such that $\forall\, k\in\mathbb{N}$, $\exists\,x_k,y_k\in K$ and $f_k\in B$ such that $d(x_k,y_k)<1/k$ but $\|f_k(x_k)-f_k(y_k)\|\geq\epsilon$.

Thus, we obtain $\{x_k\}_{k=1}^\infty, \{y_k\}_{k=1}^\infty\subseteq K$, and $\{f_k\}_{k=1}^\infty \subseteq B$.

Since $B$ is pre-compact in $(\mathcal{C}(K;V), \|\cdot\|_{\infty})$, there exists $\{f_{k_j}\}_{j=1}^\infty\subseteq \{f_k\}_{k=1}^\infty$ such that $\displaystyle\lim_{j\to\infty}f_{k_j}=f$ for some $f\in(\mathcal{C}(K;V), \|\cdot\|_{\infty})$.

Then $f_{k_j}\to f$ uniformly on $K$.

$\exists\,N_1\in\mathbb{N}$ such that $\|f_{k_j}(x)-f(x)\|<\displaystyle\frac{\epsilon}{4}$ whenever $j\geq N_1$ and $x\in K$.

Since $f$ is continuous on the compact $K$, $f$ is uniformly continous on $K$.

Hence $\exists\,\delta>0$ such that $\|f(x)-f(y)\|< \frac{\epsilon}{4}$ whenever $d(x,y)<\delta$.


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

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

This implies
$$\begin{aligned}
    \epsilon&\leq \|f_{k_j}(x_{k_j})-f_{k_j}(y_{k_j})\| \\
    &\leq \|f_{k_j}(x_{k_j})-f(x_{k_j})\| + \|f(x_{k_j})-f(y_{k_j})\| + \|f(y_{k_j})-f_{k_j}(y_{k_j})\| \\
    &< \frac{3\epsilon}{4}
\end{aligned}$$

---

Let $(M,d)$ be a metric space, $(V,\|\cdot\|)$ be a normed vector space, and $K\subseteq M$ be compact. \
If $\{f_k\}_{k=1}^\infty$ converges uniformly on $K$, then $\{f_k\}_{k=1}^\infty$ is equi-continuous.

---
