Basic question

1. What kind of “function” on a “space” should be considered?

   or What kind of “space” does the “functions” form or constitute?

2. What kind of linear operations between topological vector spaces

   or specific “space” of functions should one consider?

Definition

A metric space is a set $X$ equipped with a function $\rho : X\times X\rightarrow \R$ satisfying $\\$1. $\rho(x,y)\geq 0$ and “=” holds iff $x=y$ $\\$2. $\rho(x,y)=\rho(y,x)$ $\\$3. $\rho(x,z)\leq \rho(x,y)+\rho(y,z)$

Definition 1st category and 2nd category

$(Y,d)$ is a metric space, $E\subset Y,$ $\overline{E}$ is the closure of $E$

• If $\overline{E}$ has no interior point, then $E$ is called nowhere dense
• $\\$ A countable union of nowhere dense subsets is called a subset of the 1st category
• If is not 1st category, then it is called a subset of the 2nd category
• $\\$The complement of a subset of the 1st category is called a generic subset

Theorem

A complete metric space is of the $2nd$ category. i.e it can not be written as the union of countable many nowhere dense subsets

Definition

A metric space $(X,\rho)$ is called complete if every Cauchy sequence in $X$ converges to a point, that is $\lim\limits_{n\rightarrow \infin}\rho(x_n,x)=0$

Exercise

$(X,\rho)$ is a metric space, $E\subset X$

1. If $E$ is nowhere dense, then $E^c$ is dense
2. If $E_1,..E_n$ are nowhere dense, then $\mathop{\cup}\limits_{i=1}^{n}E_i$ is nowhere dense
3. If $E$ is nowhere dense in $X$ and $Y$ is open or the closure of an open subset of $X$, then $E\cap Y$ is nowhere dense in $Y$

Definition

$X$ is called bounded if $X=B(x_0,r)$ for some $x_0\in X$ and $r\in \R$, totally bounded if $\forall \varepsilon>0,\exist$ finitely many $x_1,..x_n$ such that $X=\mathop{\cup}\limits_{i=1}^{n}B(x_i,\varepsilon)$

Property

Let $(X,\rho)$ be a metric space, then TFAE

1. $X$ is compact
2. $X$ is sequentially compact
3. $X$ is complete and totally bounded

Theorem

Let $X$ be a complete metric space

1. $X$ is not of the $1st$ category. i.e. $X\neq \mathop{\cup}\limits_{n=1}^{\infin}E_n$, where each $E_n$ is nowhere dense
2. A generic subset (the complement of a subset of the $1st$ category) $Y$ of $X$ is dense

$C(X)=\{\text{ continuous functions on }X\}$, when $X$ is compact, $C(X)$ is a complete metric space.

Question: when $X$ is not compact, what kind of convergence or topology should we consider on $C(X)$

The completion of a metric space $(X,d)$

$(X_1,d_1),(X_2,d_2)$ are two metric spaces, a map $\varphi: X_1\rightarrow X_2$ is called an isometry if it respects the matrics. i.e $d_2(\varphi(x),\varphi(y))=d_1(x,y),\forall x,y\in X_1$.

If $\varphi$ is an isometry and a bijective map, we call $\varphi$ an isomorphism

If $\varphi$ is an isometry and $\varphi(x_1)$ is dense in $X_2$, we say $X_2$ is a completion of $X_1$, clearly, the completion of a metric space $(X,d)$ is unique up to isomorphism if exists

Existence of the completion of a metric space $(X,d)$

1. Consider $\widetilde{X}:=\{\text{ Cauchy sequences in } X\}/ \sim$ where two Cauchy sequences $\{x_n\},\{y_n\}$ are said two be equivalent if $d(x_n,y_n)\rightarrow 0$
2. Consider the map $\varphi : X\rightarrow \widetilde{X},x\mapsto[\{x,x,...\}]$ (the constant sequence). It is an isometry and $\varphi(x)$ is dense in $\widetilde{X}$. ($\tilde{d}(\varphi (x_m),\tilde{x})=\lim\limits_{n\rightarrow\infin} d(x_m,d_n)$ for $\tilde{x}=[\{x_n\}]$)

Example

$X$ is a complete metric space, $f_n\in C(X),f_n\xrightarrow{p.w} f$. Suppose for each point $x\in X,\lim\limits_{n\rightarrow\infin}f_n(x)$ converges and we denote the limit by $f(x)$. Then the discontinuities of $f$ is a subset of the $1st$ category. In other words, the points where $f$ is continuous is a generic subset

Claim

Let $\omega(x,r,g)=\mathop{Sup}\limits_{y,z\in B(x,r)}|g(y)-g(z)|$ and $\omega(x,g)=\lim\limits_{r\rightarrow 0^+}\omega(x,r,g)$

Suppose $f_n(x)\in C(X)$ and $f=\lim\limits_{p.w} f_n$, then $E_n: =\{x\in X\vert \omega (x,f)\geq \frac{1}{n}\}$ has no interior points

Lemma

For any open ball $B\subset X$ and $\forall \varepsilon >0$, there exist an open ball $B'\subset B$ and $N\in\N$ such that $|f_m(x)-f(x)|<\varepsilon$ for all $x\in B',m\geq N$

Theorem [s-s] 1.5 P163

$X=C[0,1]$ is a complete metric space, $\rho(f,g): =\mathop{max}\limits_{x\in [0,1]}|f(x)-g(x)|$. Then $Y=\{f\in C[0,1]\vert f \text{ is nowhere differentiable }\}$ is dense in $X$