# Contents

## Idea

The Stone–Weierstrass theorem says given a compact Hausdorff space $X$, one can uniformly approximate continuous functions $f:X\to ℝ$ by elements of any subalgebra that has enough elements to distinguish points. It is a far-reaching generalization of a classical theorem of Weierstrass, that real-valued continuous functions on a closed interval are uniformly approximable by polynomial functions.

## Precise statement

Let $X$ be a compact Hausdorff topological space; for a constructive version take $X$ to be a compact regular locale (see compactum). Recall that the algebra $C\left(X\right)$ of real-valued continuous functions $f:X\to ℝ$ is a commutative (real) Banach algebra with unit, under pointwise-defined addition and multiplication, and where the norm is the sup-norm

$\parallel f\parallel :={\mathrm{sup}}_{x\in X}\mid f\left(x\right)\mid$\|f\| := sup_{x \in X} |f(x)|

A subalgebra of $C\left(X\right)$ is a vector subspace $A\subseteq C\left(X\right)$ that is closed under the unit and algebra multiplication operations on $C\left(X\right)$. A Banach subalgebra is a subalgebra $A\subseteq C\left(X\right)$ which is closed as a subspace of the metric space $C\left(X\right)$ under the sup-norm metric. We say that $A\subseteq C\left(X\right)$ separates points if, given distinct points $x,y\in X$, there exists $f\in A$ such that $f\left(x\right)\ne f\left(y\right)$.

###### Theorem (Stone–Weierstrass)

A subalgebra inclusion $A\subseteq C\left(X\right)$ is dense if and only if it separates points. Equivalently, a Banach subalgebra inclusion $A\subseteq C\left(X\right)$ is the identity if and only if it separates points.

## Outline of proof

• The first step is the classical Weierstrass approximation, that polynomial functions $p\left(x\right):\left[a,b\right]\to ℝ$ are dense in $C\left(\left[a,b\right]\right)$; without loss of generality, take $a=-\frac{1}{4}$ and $b=\frac{1}{4}$. We use the method of constructing polynomials which approximate the identity of the convolution product (the Dirac distribution). Explicitly, consider the normalizations ${K}_{n}={s}_{n}/\parallel {s}_{n}{\parallel }_{1}$ where ${s}_{n}\left(x\right)=\left(1-{x}^{2}{\right)}^{n}$ over $\left[-1,1\right]$ and is compactly supported on $\left[-1,1\right]$, so that ${K}_{n}$ approximates the Dirac distribution concentrated at 0. Given $f:\left[-\frac{1}{4},\frac{1}{4}\right]\to ℝ$, extend to a function compactly supported on $\left[-\frac{1}{2},\frac{1}{2}\right]$. The convolution product

$\left({K}_{n}*f\right)\left(x\right)={\int }_{-1}^{1}{K}_{n}\left(x-y\right)f\left(y\right)dy$(K_n * f)(x) = \int_{-1}^1 K_n(x-y) f(y) d y

is polynomial (since by differentiating under the integral enough times, we eventually kill the convolving polynomial factor ${K}_{n}$), and one may verify that $\left({K}_{n}*f\right)$ converges to $f$ in sup norm (i.e., uniformly) as $n\to \infty$, and in particular when restricted over the interval $\left[-\frac{1}{4},\frac{1}{4}\right]$.

Now suppose given a Banach subalgebra $A\subseteq C\left(X\right)$.

• Given $f\in A\subseteq C\left(X\right)$, the values of $f$ are contained in some interval $\left[a,b\right]$. If polynomials ${p}_{n}$ converge to the absolute value function $\mid \cdot \mid :\left[a,b\right]\to ℝ$ in sup norm, then ${p}_{n}\left(f\right)\in A$ converges to $\mid f\mid \in C\left(X\right)$ in sup norm. Since $A$ is closed in $C\left(X\right)$ with respect to the sup-norm, it follows that $\mid f\mid \in A$.

• Next, $A$ is partially ordered by $f\le g$ if $f\left(x\right)\le g\left(x\right)$ for all $x\in X$, and we claim the poset $A$ is closed under binary meets and binary joins. For,

$f\vee g=\frac{1}{2}\left(\mid f-g\mid +\left(f+g\right)\right)$f \vee g = \frac1{2}(|f - g| + (f + g))

and $f\wedge g=-\left(\left(-f\right)\vee \left(-g\right)\right)$, and we showed in the last step that $A$ is closed under the operation $f↦\mid f\mid$.

Finally, suppose the Banach subalgebra $A$ separates points. Given $g\in C\left(X\right)$ and $\epsilon >0$, the last step is to show there exists $f\in A$ such that $\parallel f-g\parallel \le \epsilon$.

• ###### Lemma

Given $x\in X$, there exists ${f}_{x}\in A$ such that ${f}_{x}\left(x\right)=g\left(x\right)$ and $g\left(y\right)\le {f}_{x}\left(y\right)+\epsilon$ for all $y\in X$.

###### Proof

For each $y\in X$, $y\ne x$ we can choose $s\in A$ such that $s\left(x\right)\ne s\left(y\right)$, since $A$ separates points. Thus, given any $x,y$, there exist $s\in A$ and scalars $a$ and $b$ such that

$a+b\cdot s\left(x\right)=g\left(x\right),\phantom{\rule{2em}{0ex}}a+b\cdot s\left(y\right)=g\left(y\right)$a + b \cdot s(x) = g(x), \qquad a + b \cdot s(y) = g(y)

Denote $a\cdot 1+bs$ by ${f}_{x,y}$ (to indicate dependence on $x$ and $y$). For each $y$, choose a neighborhood ${V}_{y}$ so that $g\left(z\right)\le {f}_{x,y}\left(z\right)+\epsilon$ for all $z\in {V}_{y}$. Finitely many such neighborhoods ${V}_{{y}_{1}},\dots ,{V}_{{y}_{n}}$ cover $X$; let ${f}_{x}$ be the join of ${f}_{x,{y}_{1}},\dots ,{f}_{x,{y}_{n}}$. Then $g\left(y\right)\le {f}_{x}\left(y\right)+\epsilon$ for all $y\in X$.

• Given a choice of ${f}_{x}$ for each $x\in X$, as in the preceding lemma, we may choose a neighborhood ${V}_{x}$ such that ${f}_{x}\left(z\right)-\epsilon \le g\left(z\right)$ for all $z\in {V}_{x}$. Finitely many such neighborhoods ${V}_{{x}_{1}},\dots ,{V}_{{x}_{n}}$ cover $X$; let $f$ be the meet of ${f}_{{x}_{1}},\dots ,{f}_{{x}_{n}}$. Then

$f\left(y\right)-\epsilon \le g\left(y\right)\le f\left(y\right)+\epsilon$f(y) - \varepsilon \leq g(y) \leq f(y) + \varepsilon

for all $y\in X$, as was to be shown.

## Variations

There is a complex-valued version of Stone–Weierstrass. Let $C\left(X,ℂ\right)$ denote the commutative ${C}^{*}$-algebra of complex-valued functions $f:X\to ℂ$, where the star operation is pointwise-defined conjugation. A ${C}^{*}$-subalgebra is a subalgebra $A\subseteq C\left(X,ℂ\right)$ which is closed under the star operation.

###### Theorem

A ${C}^{*}$-subalgebra $A\subseteq C\left(X,ℂ\right)$ is dense if and only if it separates points.

There is also a locally compact version. Let $X$ be a locally compact Hausdorff space and let ${C}_{0}\left(X\right)$ be the space of (say real-valued) functions $f$ which “vanish at infinity”: for every $\epsilon >0$ there exists a compact set $K\subseteq X$ such that $\mid f\left(x\right)\mid <\epsilon$ for all $x$ outside $K$. (${C}_{0}\left(X\right)$ is no longer a Banach space, but it is locally convex and complete in its uniformity, and a Fréchet space if $X$ is second countable.) Under pointwise multiplication, ${C}_{0}\left(X\right)$ is a commutative algebra without unit. As before, we have a notion of subalgebra $A\subseteq {C}_{0}\left(X\right)$.

###### Theorem

$A\subseteq {C}_{0}\left(X\right)$ is dense if and only if it separates points and for no $x\in X$ is it true that every $f\in A$ vanishes at $x$.

## References

• B. Banaschewski, C. J. Mulvey, A constructive proof of the Stone-Weierstrass theorem, J. Pure Appl. Algebra 116 (1997), pp. 25–40, doi

Revised on July 23, 2011 13:39:18 by Urs Schreiber (89.204.137.100)