Stone-Weierstrass Theorem/Compact Space

Theorem

Let $T = \struct {X, \tau}$ be a compact topological space.

Let $\map C {X, \R}$ be the set of real-valued continuous functions on $T$.

Let $\times$ be the pointwise multiplication on $\map C {X, \R}$.

Let $\struct {\map C {X, \R}, \times}$ be the Banach algebra with respect to $\norm \cdot_\infty$.

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Let $\AA$ be a unital subalgebra of $\map C {X, \R}$.

Suppose that $\AA$ separates points of $X$, that is:

for distinct $p, q \in X$, there exists $h_{p q} \in \AA$ such that $\map {h_{p q} } p \ne \map {h_{p q} } q$.

Then the closure $\overline \AA$ of $\AA$ is equal to $\map C {X, \R}$.

Complex-Valued

Let $\map \CC {X, \C}$ be the set of complex-valued continuous functions on $X$.

Equip $\map \CC {X, \C}$ with pointwise addition, pointwise scalar multiplication and pointwise multiplication.

For each $f \in \map \CC {X, \C}$, define $\overline f : X \to \C$ by:

$\map {\overline f} x = \overline {\map f x}$

for each $x \in X$.

Let $\AA$ be a unital $\ast$-subalgebra of $\map \CC {X, \C}$.

Suppose that $\AA$ separates points of $X$, that is:

for distinct $p, q \in X$, there exists $h_{p q} \in \AA$ such that $\map {h_{p q} } p \ne \map {h_{p q} } q$.

Then the closure $\AA^-$ of $\AA$ is equal to $\map C {X, \C}$.

Proof

Let $\struct {C', \norm {\,\cdot\,}_{C'} }$ be the dual space of $\struct {\map C {X, \R}, \norm {\,\cdot\,}_\infty }$.

From Spanning Criterion of Normed Vector Space, it suffices to show that:

$\forall \ell \in C' : \ell \restriction_\AA = 0 \implies \ell = 0$

Let $B' \subseteq C'$ be the closed unit ball, that is:

$B' := \set {\ell \in C' : \norm \ell_{C'} \le 1}$

Let:

$U := \set {\ell \in B' : \ell \restriction_\AA = 0}$

Let $\map E U$ be the set of extreme points of $U$.

Lemma

$\map E U \setminus \set 0 = \O$

$\Box$

Let $w^\ast$ be the weak-$\star$ topology on $C'$.

Recall that $B'$ is $w^\ast$-compact by Banach-Alaoglu Theorem.

Thus $U$ is a $w^\ast$-compact convex subset of $C'$.

By Weak-* Topology is Hausdorff, $U$ is Hausdorff.

Thus by Krein-Milman Theorem we have:

$\map E U \ne \O$

Thus:

$\map E U = \set 0$

Therefore:

$U = \set 0$

$\blacksquare$

Also presented as

Some sources specifically define $T = \struct {X, \tau}$ to be a Hausdorff space.

However, we note that by hypothesis $\AA$ is such that it separates points.

Hence from Topological Space Separated by Mappings is Hausdorff it follows that $T$ is a fortiori a Hausdorff space.

Also see

Weierstrass Approximation Theorem, of which the Stone-Weierstrass Theorem is a generalization.

Source of Name

This entry was named for Marshall Harvey Stone and Karl Weierstrass.

Sources

2002: Peter D. Lax: Functional Analysis: $13.2$: The Krein-Milman Theorem