proof of Stone-Weierstrass theorem

Let $\bar{\mathcal{A}}$ denote the closure of $\mathcal{A}$ in $C^{0}(X,\mathbb{R})$ according to the uniform convergence topology. We want to show that, if conditions 1 and 2 are satisfied, then $\bar{\mathcal{A}}=C^{0}(X,\mathbb{R})$ .

First, we shall show that, if $f\in\bar{\mathcal{A}}$ , then $|f|\in\bar{\mathcal{A}}$ . Since $f$ is a continuous function on a compact space $f$ must be bounded – there exists constants $a$ and $b$ such that $a\leq f\leq b$ . By the Weierstrass approximation theorem, for every $\epsilon>0$ , there exists a polynomial such that $|P(x)-|x||<\epsilon$ when $x\in[a,b]$ . (By the way, one does not need the full-blown Weierstrass approximation theorem to show that $P$ exists – see the entry “proof of Weierstrass approximation theorem” for an elementary construction of $P$ ) Define $g:X\to\mathbb{R}$ by $g(x)=P(f(x))$ . Since $\bar{\mathcal{A}}$ is an algebra, $g\in\bar{\mathcal{A}}$ . For all $x\in X$ , $|g(x)-|f(x)||<\epsilon$ . Since $\bar{\mathcal{A}}$ is closed under the uniform convergence topology, this implies that $|f|\in\bar{\mathcal{A}}$ .

A corollary of the fact just proven is that if $f,g\in\bar{\mathcal{A}}$ , then $\max(f,g)\in\bar{\mathcal{A}}$ and $\min(f,g)\in\bar{\mathcal{A}}$ . The reason for this is that one can write

\max(a,b)={1\over 2}\left(a+b+|a-b|\right)

\min(a,b)={1\over 2}\left(a+b-|a-b|\right)

Second, we shall show that, for every $f\in C^{0}(X,\mathbb{R})$ , every $x\in X$ , and every $\epsilon>0$ , there exists $g_{x}\in\bar{\mathcal{A}}$ such that $g_{x}\leq f+\epsilon$ and $g_{x}>f$ . By condition 1, if $y\neq x$ , there exists a function ${\tilde{h}}_{xy}\in\mathcal{A}$ such that ${\tilde{h}}_{xy}(x)\neq{\tilde{h}}_{xy}(y)$ . Define $h_{xy}$ by $h_{xy}(z)=p{\tilde{h}}_{xy}(z)+q$ , where the constants $p$ and $q$ have been chosen so that

h_{xy}(x)=f(x)+\frac{\epsilon}{2}

h_{xy}(y)=f(y)-\frac{\epsilon}{2}

By condition 2, $h_{xy}\in\mathcal{A}$ . (Note: This is the only place in the proof where condition 2 is used, but it is crucial since, otherwise, it might not be possible to construct a function which takes arbitrary preassigned values at two distinct points of $X$ . The necessity of condition 2 can be shown by a simple example: Suppose that $\mathcal{A}$ is the algebra of all continuous functions on $f$ which vanish at a point $O\in X$ . It is easy to see that this algebra satisfies all the hypotheses of the theorem except condition 2 and the conclusion of the theorem does not hold in this case.)

For every $y\neq x$ , define the set $U_{xy}$ as

U_{xy}=\{z\in X\mid h_{xy}(z)<f(z)+\epsilon\}

Since $f$ and $h_{xy}$ are continuous, $U_{xy}$ is an open set. Because $x\in U_{xy}$ and $y\in U_{xy}$ , $\left\{U_{xy}\mid y\in X\setminus\{x\}\right\}$ is an open cover of $X$ . By the definition of a compact space, there must exist a finite subcover. In other words, there exists a finite subset $\{y_{1},y_{2},\ldots,y_{n}\}\subset X$ such that $X=\bigcup_{m=0}^{n}U_{xy_{m}}$ . Define

g_{x}=\min(h_{xy_{1}},h_{xy_{2}},\ldots,h_{xy_{n}}).

By the corollary of the first part of the proof, $g_{x}\in\bar{\mathcal{A}}$ . By construction, $g_{x}(x)=f(x)+\epsilon/2$ and $g_{x}<f+\epsilon$ .

Third, we shall show that, for every $f\in C^{0}(X,\mathbb{R})$ and every $\epsilon>0$ , there exists a function $g\in\bar{\mathcal{A}}$ such that $f\leq g<f+\epsilon$ . This will complete the proof becauase it implies that $\bar{\mathcal{A}}=C^{0}(X,\mathbb{R})$ . For every $x\in X$ , define the set $V_{x}$ as

V_{x}=\{z\in X\mid g_{x}(z)>f(x)\}

where $g_{x}$ is defined as before. Since $f$ and $g_{x}$ are continuous, $V_{x}$ is an open set. Because $g_{x}(x)=f(x)+\epsilon/2>f(x)$ , $x\in V_{x}$ . Hence $\left\{V_{x}\mid x\right\}$ is an open cover of $X$ . By the definition of a compact space, there must exist a finite subcover. In other words, there exists a finite subset $\{x_{1},x_{2},\ldots x_{n}\}\subset X$ such that $X=\bigcup_{m=0}^{n}V_{x_{n}}$ . Define $g$ as

g(z)=\max\{g_{x_{1}}(z),g_{x_{2}}(z),\ldots,g_{x_{n}}(z)\}

By the corollary of the first part of the proof, $g\in\bar{\mathcal{A}}$ . By construction, $g>f$ . Since $g_{x}<f+\epsilon$ for every $x\in X$ , $g<f+\epsilon$ .

Title	proof of Stone-Weierstrass theorem
Canonical name	ProofOfStoneWeierstrassTheorem
Date of creation	2013-03-22 14:35:01
Last modified on	2013-03-22 14:35:01
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	18
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 46E15