Stone-Weierstrass Theorem/Compact Space/Complex-Valued

Theorem

Let $X$ be a compact topological space.

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 $\BB$ be the set:

$\BB = \set {f \in \AA : f \in \map \CC {X, \R} }$

We show that $\BB$ is a algebra over $\R$.

From Bounded Continuous Functions on Topological Space form Banach Space, $\BB$ is a vector space over $\R$.

From Product of Continuous Functions on Topological Ring is Continuous, $\BB$ is an algebra over $\R$.

We show that:

$\BB = \set {\map \Re f : f \in \AA} = \set {\map \Im f : f \in \AA}$

Immediately we have:

$\set {f \in \AA : f \in \map \CC {X, \R} } \subseteq \set {\map \Re f : f \in \AA}$

since if $f \in \map \CC {X, \R}$, we have $\map \Re f = f$.

We also have:

$\set {f \in \AA : f \in \map \CC {X, \R} } \subseteq \set {\map \Im f : f \in \AA}$

since if $f \in \AA$, we have $i f \in \AA$ since $\AA$ is a subalgebra, and then $f = \map \Im {i f}$.

Towards proving the converse, for $f \in \AA$ write:

$\map \Re f = \dfrac {f + \overline f} 2$

from Sum of Complex Number with Conjugate.

Since $\AA$ is a $\ast$-subalgebra of $\map \CC {X, \C}$, we have $\map \Re f \in \AA$.

Similarly from Difference of Complex Number with Conjugate we hae:

$\map \Im f = \dfrac {f - \overline f} {2 i}$

Since $\AA$ is a $\ast$-subalgebra of $\map \CC {X, \C}$, we have $\map \Im f \in \AA$.

So for $f \in \AA$ we have $\map \Re f \in \AA \cap \map \CC {x, \R}$ and $\map \Im f \in \AA \cap \map \CC {X, \R}$.

Hence we conclude:

$\BB = \set {\map \Re f : f \in \AA} = \set {\map \Im f : f \in \AA}$

We prove that $\BB$ separates points.

We are given that $\AA$ separates points, hence:

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

Since $\map {h_{p q} } p \ne \map {h_{p q} } q$, we have either $\map \Re {\map {h_{p q} } p} \ne \map \Re {\map {h_{p q} } q}$ or $\map \Im {\map {h_{p q} } p} \ne \map \Im {\map {h_{p q} } q}$.

We have $\map \Re {h_{p q} }, \map \Im {h_{p q} } \in \BB$.

Hence $\BB$ separates points.

Hence by Stone-Weierstrass Theorem: Compact Space we have $\BB^- = \map \CC {X, \R}$.

Now let $f \in \map \CC {X, \C}$.

Write $f = \map \Re f + i \map \Im f$ with $\map \Re f, \map \Im f \in \map \CC {X, \R}$.

Let $\epsilon > 0$.

Since $\BB^- = \map \CC {X, \R}$, there exists $\phi, \psi \in \BB \subseteq \AA$ such that:

$\norm {\phi - \map \Re f}_\infty < \dfrac \epsilon 2$

and:

$\norm {\psi - \map \Im f}_\infty < \dfrac \epsilon 2$

We then have:

Since $\phi, \psi \in \AA$ and $\AA$ is a subalgebra, we have $\phi + i\psi \in \AA$.

Hence $\AA^- = \map \CC {X, \C}$.

$\blacksquare$