Stone-Weierstrass Theorem
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 $\map C {X, \R}$.
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