AS ANA2 3

Problem 1

Let $A=\{z:z=e^{i\sin t},t\in \mathbb{R}\}$. $A$ is the image of the connected space $\mathbb{R}$ under the continuous map $t\mapsto e^{i\sin t}$. By Thm 144.4, $A$ is connected.

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Problem 2

Let $\phi :X\to \{\pm 1\}$ be any continuous function. For each $y\in Y$, $f^{-1}(y)$ is connected, so $\phi$ is constant on $f^{-1}(y)$ by Prp 329.1. This allows us to define $\psi :Y\to \{\pm 1\}$ by $\psi (y)=\phi (f^{-1}(y))$. Since $f$ is surjective, $X$ is compact, and $\phi =\psi \circ f$, $\psi$ is continuous by assignment 2, p4 . Let $C\subseteq Y$ be connected. Then, $\psi$ is constant on $C$. For all $x\in f^{-1}(C)$, $\phi (x)=\psi (f(x))=\psi (C)$. Thus, $\phi$ is constant on $f^{-1}(C)$. Since $\phi$ was chosen to be any arbitrary continuous map, it follows that $f^{-1}(C)$ is connected.

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Problem 3

Consider $\sin x\in C(\mathbb{R},\mathbb{R})$. If $f(x)\in \mathcal{B}$,

$$\begin{array}{rl}\Vert f-\sin {\Vert }_{\infty }& =\underset{x\in \mathbb{R}}{\sup }|f(x)-\sin x|\\ & =\underset{|x|>M}{\sup }|f(x)-\sin x|\\ & ⩾\frac{1}{2}.\end{array}$$

Thus, $\mathcal{B}$ is not dense in $C(\mathbb{R},\mathbb{R})$. This does not contradict the Stone-Weierstrass Theorem (Thm 315.6), since $\mathbb{R}$ is not compact.

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Problem 4

Consider $\mathcal{B}=\text{span}\{f\cdot g:f,g\in C([0,1])\}\subseteq C([0,1{]}^2,\mathbb{R})$. $\mathcal{B}$ is clearly unital. The map $((x,y)\mapsto x)\in \mathcal{B}$ separates $(x_1,y_1),(x_2,y_2)\in [0,1{]}^2$ if $x_1\ne x_2$, and $((x,y)\mapsto y)\in \mathcal{B}$ separates them if $x_1=x_2$. Thus, $\mathcal{B}$ is dense in $C([0,1{]}^2,\mathbb{R})$. Thus, there exists a sequence $\{b_n\}={\left\{{\sum }_{i=1}^{k_n}{f}_{i,n}\cdot g_{i,n}\right\}}_{n=1}^{\infty }\subseteq \mathcal{B}$ converging uniformly to $F$. Since $F$ is continuous (hence bounded), we have $b_{n}F\to F^2$ uniformly.

$$\begin{array}{rl}{\int }_0^1{\int }_0^1{F}^2(x,y)dxdy& ={\int }_0^1{\int }_0^1\underset{n\to \infty }{\lim }\left(\sum _{i=1}^{k_n}{f}_{i,n}\cdot g_{i,n}\right)(x,y)F(x,y)dxdy\\ & =\underset{n\to \infty }{\lim }\sum _{i=1}^{k_n}{\int }_0^1{\int }_0^1{f}_{i,n}(x)g_{i,n}(y)F(x,y)dxdy\\ & =0,\end{array}$$

by hypothesis.

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Class assignments

Exercise 349.1(Exercise).

$C_0(X)$ with the norm $\Vert \cdot {\Vert }_{\infty }$ is a Banach space.

Proof.

It is clear that$C_0(X)\subseteq C_b(X)$, which we know to be a Banach space by Rmk 163.6. Let $\{f_n\}$ be a sequence in $C_0(X)$ converging to $f\in C_b(X)$. Let $ϵ>0$. Let $N$ be such that $\Vert f_N-f\Vert <ϵ/2$. Let $K$ be a compact subset of $X$ such that $|f_N|<ϵ/2$ on $K^c$. It follows that $|f|<ϵ$ on $K^c$.□

Proposition 349.2(Proposition).

There exists continuous and surjective $f:\mathcal{C}\to [0,1]$.

Proof.

Define$f$ by

$$\begin{array}{r}f\left(\sum _{i=1}^{\infty }\frac{a_i}{3^i}\right)=\sum _{i=1}^{\infty }\frac{a_i}{2^{i+1}}.\end{array}$$

Let $x\in \mathcal{C}$ and $f(x)=y$. Let $ϵ>0$. Choose $n$ such that $1/2^n<ϵ$. Let $2\delta =1/3^n$. The ternary representations of all elements in $(x-\delta ,x+\delta )\cap \mathcal{C}$ will have their first $n$ digits fixed. If $x^{\prime }$ is such an element, then

$$|f(x)-f(x^{\prime })|⩽\sum _{i=n+1}^{\infty }\frac{1}{2^i}=\frac{1}{2^n}<ϵ.$$

Thus, $f$ is continuous.□

Note that $f$ is not injective, since $f(0.0\overline{2})=f(0.2)$.

Lemma 349.3(Lemma).

The inner product in continuous map from $H\times H$ to $\mathbb{C}$, that is, if $\{{\mathbf{x}}_n\}\to \mathbf{x}$ and $\{{\mathbf{y}}_n\}\to \mathbf{y}$, then $⟨{\mathbf{x}}_n,{\mathbf{y}}_n⟩\to ⟨\mathbf{x},\mathbf{y}⟩$.

Proof.

$$\begin{array}{rl}\left|⟨{\mathbf{x}}_n,{\mathbf{y}}_n⟩-⟨\mathbf{x},\mathbf{y}⟩\right|& ⩽|⟨{\mathbf{x}}_n,{\mathbf{y}}_n⟩-⟨{\mathbf{x}}_n,\mathbf{y}⟩|+|⟨{\mathbf{x}}_n,\mathbf{y}⟩-⟨\mathbf{x},\mathbf{y}⟩|\\ & =|⟨{\mathbf{x}}_n,{\mathbf{y}}_n-\mathbf{y}⟩|+|⟨{\mathbf{x}}_n-\mathbf{x},\mathbf{y}⟩|\\ & ⩽\Vert {\mathbf{x}}_n\Vert \Vert {\mathbf{y}}_n-\mathbf{y}\Vert +\Vert {\mathbf{x}}_n-\mathbf{x}\Vert \Vert \mathbf{y}\Vert \\ & \to 0\text{ as }n\to \infty .\end{array}$$ □

Proposition 349.4(Proposition).

Let $X$ be locally compact. There exists an isomorphism $\phi :C_0(X,\mathbb{R})\to \{f\in C(\overline{X},\mathbb{R}):f(\infty )=0\}$ (bijective linear norm preserving ring homomorphism).

Proof.

We have the obvious map $f\mapsto \phi (f)$ where

$$(\phi (f))(t)=\{\begin{array}{ll}f(t)& t\in X\\ 0& t=\infty .\end{array}$$

First, we have to show that $\phi (f)\in C(\overline{X})$. Let $O$ be an open subset of $\mathbb{R}$ not containing $0$. Then, $(\phi (f){)}^{-1}(O)=f^{-1}(O)$, which is open in $\overline{X}$ since it is open in $X$ and doesn’t contain $\infty$. Suppose $0\in O$. Let $ϵ>0$ be such that $(-ϵ,ϵ)\subseteq O$. There exists compact $K\subseteq X$ such that[^2] $f(K^c)\subseteq (-ϵ,ϵ)$, or $f^{-1}((-ϵ,ϵ{)}^c)\subseteq K$. Note that $f^{-1}(O{)}^c=f^{-1}(O^c)\subseteq f^{-1}((-ϵ,ϵ{)}^c)\subseteq K$. Since $O^c$ is closed and $f$ is continuous, $f^{-1}(O{)}^c$ is closed in $X$. Since it is a subset of a compact set, it follows that $f^{-1}(O{)}^c$ is compact. It follows that $(\phi (f){)}^{-1}(O)$ is open in $\overline{X}$.

The map is clearly injective, linear, norm preserving, and a ring homomorphism. We have to show surjectivity. Take any $g\in C(\overline{X},\mathbb{R})$ with $g(\infty )=0$. Put $f:=g{|}_X$. We need to check $f\in C_0(X,\mathbb{R})$, i.e. $f$ is continuous on $X$ (obvious) and $f$ vanishes at infinity: for every $ϵ>0$ there is a compact $K\subset X$ with $|f(x)|<\epsilon$ for all $x\notin K$.

Since $g$ is continuous at $\infty$ and $g(\infty )=0$ there exists an open neighborhood $U\subset \overline{X}$ of $\infty$ with $|g(y)|<\epsilon$ for all $y\in U$. By the definition of the one–point compactification, $K:=\overline{X}\setminus U$ is compact and $K\subset X$. For any $x\in X$ with $x\notin K$ we have $x\in U\cap X$, hence $|f(x)|=|g(x)|<\epsilon$. Finally, $\phi (f)$ is exactly the extension of $f$ that equals $0$ at $\infty$, so $\phi (f)=g$. Therefore $\phi$ is surjective.□