Stone-Weierstrass Theorem/Lemma

Lemma

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 $\map {h_{p q} } p \ne \map {h_{p q} } q$.

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

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 $U$.

Then:

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

Proof

Aiming for a contradiction, suppose there is an $\ell \in \map E U$ such that:

$(1):\quad \ell \ne 0$

Then:

$\norm \ell_{C'} = 1$

since:

$\ell = \norm \ell_{C'} \underbrace { \norm \ell_{C'}^{-1} \ell}_{\in U} + \paren {1 - \norm \ell_{C'} } \underbrace {0}_{\in U}$

Note that $X$ is Hausdorff by Topological Space Separated by Mappings is Hausdorff.

Thus we can apply Riesz-Kakutani Representation Theorem.

That is, there exists a signed measure $\mu$ on $X$ such that:

$\ds \forall f \in \map C {X, \R} : \map \ell f = \int_X f \rd \mu$

and:

$\map {\size \mu} X = 1$

Let $\map \supp \mu$ denote the support of $\mu$.

Consider any $g \in \AA$ such that:

$\forall x \in X : 0 < \map g x < 1$

Then we have:

$g \mu \in U$

Let:

$\ds a := \int_X g \rd \size {\mu}$

and:

$\ds b := \int_X \paren {1 - g} \rd \size {\mu}$

so that:

$a + b = 1$

As:

$\ds \mu = a \underbrace{\frac {g \mu} a}_{\in U} + b \underbrace {\frac {\paren {1 - g} \mu} b}_{\in U}$

we have:

$\ds \mu = \frac {g \mu} a$

Especially:

$\forall x \in \map \supp \mu : \map g x = a$

We claim that:

$\exists x_0 \in X : \map \supp \mu = \set {x_0}$

Aiming for a contradiction, suppose there are distinct $p, q \in \map \supp \mu$.

Then by hypothesis there exists $h_{p q} \in \AA$ such that:

$\map {h_{p q} } p \ne \map {h_{p q} } q$

Let $C,D > 0$ such that:

$C > \norm {h_{p q} }_\infty $

and:

$D > \norm {h_{p q} }_\infty + C$

Then:

$\ds g := \frac {h_{p q} + C} D$

must be constant on $\map \supp \mu$.

This is a contradiction.

$\Box$

Finally, we have:

\(\ds \forall f \in \map C {X, \R}: \, \)

\(\ds \map \ell f\)

\(=\)

\(\ds \map \mu {\set {x_0} } \map f {x_0}\)

as shown above

\(\ds \)

\(=\)

\(\ds \map \ell 1 \map f {x_0}\)

\(\ds \)

\(=\)

\(\ds 0\)

since the constant $1 \in \AA$

That is $\ell = 0$.

This is a contradiction to $(1)$.

$\blacksquare$