LEC ANA2 12

More generalizations of S-W

Theorem 324.1(Stone-Weierstrass, v4).

Let $X$ be a compact metric space, and let $\mathcal{A}\subseteq C(X,\mathbb{R})$ be a subalgebra which separates points. Then either $\overline{\mathcal{A}}=C(X,\mathbb{R})$ or $\overline{\mathcal{A}}={\mathcal{A}}_{x_0}$ for some $x_0$, where ${\mathcal{A}}_{x_0}$ is the set of all $f\in C(X,\mathbb{R})$ which vanish at $x_0$.

Proof.

If $1\in \overline{\mathcal{A}}$, then $\overline{\mathcal{A}}=C(X)$ by Theorem 315.6. So assume $1\notin \overline{\mathcal{A}}$. By Lemma 315.9, there exists $x_0\in X$ such that $f(x_0)=0$ for all $f\in \overline{\mathcal{A}}$. We will show that $\overline{\mathcal{A}}=\{f\in C(X):f(x_0)=0\}$.

Define $\tilde{\mathcal{A}}:=\{f+c1:c\in \mathbb{R},f\in \mathcal{A}\}$. Note that $\tilde{\mathcal{A}}$ is a unital subalgebra of $C(X,\mathbb{R})$. $\tilde{\mathcal{A}}$ separates points since $\mathcal{A}$ separates points. By Theorem 315.6, $\tilde{A}$ is dense in $C(X,\mathbb{R})$. Suppose $g\in C(X)$ such that $g(x_0)=0$. There exists a sequence $\{f_n+c_n\}\subseteq \tilde{A}$ converging uniformly to $g$. Since $f_n(0)=0$ for each $n$, we must have $\{c_n\}\to 0$. It follows that $\{f_n\}$ converges uniformly to $g$.□

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S-W for locally compact spaces

Definition 324.2(Definition).

$X$ is said to be locally compact if for all $x\in X$, there exists open $U\ni x$ such that $\overline{U}$ is compact. $\overline{U}$ may be called a compact neighborhood.

A normed linear space is locally compact iff it is finite dimensional. Open subsets of compact metric spaces are locally compact.

In an NLS, locally compact $⟺$ all closed balls are compact.

Definition 324.3(Definition).

Let $(X,\tau )$ be a locally compact metric space¹. Let $\overline{X}=X\cup \{\infty \}$. $U\subseteq \overline{X}$ is defined to be open if

1. $\infty \notin U$ and $U$ is open in $X$, or
2. $\infty \in U$ and $U^c$ is compact.

Let $\overline{\tau }$ denote the collection of subsets declared open.

Claim 324.4(Claim).

$(\overline{X},\overline{\tau })$ is a topological space.

Proof.

It is clear that $\overline{X},\varnothing \in \overline{\tau }$. Let ${\bigcup }_{\alpha \in I}{U}_{\alpha }$ be an arbitrary union of open sets in $\overline{\tau }$. If none of them contain $\infty$, then their union is open in $X$, and hence open in $\overline{X}$. If one of the $U_{\alpha }$‘s, say $U_{{\alpha }^{\prime }}$, contains $\infty$, rewrite the union as ${\left({\bigcap }_{\alpha \in I}(U_{\alpha }^c\cap X)\right)}^c$. Note that each $U_{\alpha }^c\cap X$ is closed in $X$, and $U_{{\alpha }^{\prime }}^c\cap X=U_{{\alpha }^{\prime }}^c$ is compact. It follows that the intersection is compact, so its complement is open in $\overline{X}$. Similar analysis shows that the open sets in $\overline{\tau }$ are closed under finite intersections.□

Claim 324.5(Claim).

$(\overline{X},\tilde{\tau })$ is compact.

Proof.

Let$\{U_{\alpha }{\}}_{\alpha \in I}$ cover $\overline{X}$. One of them must contain $\infty$, and hence their complement is compact, which then has a finite subcover.□

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Note the similarities to the two-point compactification performed in Exercise 15.1.

We will not prove the following fact:

Fact 324.6(Fact).

If $X$ is a locally compact and separable metric space then $(\overline{X},\overline{\tau })$ is metrizable.

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Definition 324.7(Definition).

Let $X$ be a locally compact metric space. Define

$$C_0(X):=\left\{f\in C(X):\forall ϵ>0\exists K\underset{\text{cpt}}{\subseteq }X\text{s.t.}|f(t)|<ϵ\forall t\in K^c\right\}$$

Exercise 324.8(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 Remark 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 $K^c$.□

It is now clear that $C_0(X)$ is an $\mathbb{R}$-algebra.

Proposition 324.9(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 that² $f(K^c)\subseteq (-ϵ,ϵ)$, 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. Showing that $\phi$ is onto takes some work, but it is very similar in nature to the preceding paragraph and I can’t be bothered.□

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Theorem 324.10(Stone-Weierstrass, v5).

Let $X$ be a locally compact metric space. Suppose $\mathcal{A}\subseteq C_0(X,\mathbb{R})$ is a subalgebra which separates points and does not vanish on $X$. Then $\mathcal{A}$ is dense in $C_0(X,\mathbb{R})$.

Proof.

Let$\phi$ be as in Proposition 9. Then, $\phi (\mathcal{A})\subseteq C(\overline{X},\mathbb{R})$ is a subalgebra which separates points. From Claim 5, Fact 6 and Theorem 1, $\overline{\phi (\mathcal{A})}=\{f\in C(\overline{X},\mathbb{R}):f(\infty )=0\}$. Since $\phi$ is an isometry between $C_0(X,\mathbb{R})$ and $\{f\in C(\overline{X},\mathbb{R}):f(\infty )=0\}$, it follows that $\overline{\mathcal{A}}=C_0(X,\mathbb{R})$.□

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Footnotes

1. Much of what we do here can be generalized to $X$ being a topological space with some additional hypotheses. ↩

2. The complements here are taken in $X$! ↩