Stone-Weierstrass Theorem/Locally Compact Hausdorff Space

== Theorem ==
Let $X$ be a locally compact Hausdorff space.
Let $\struct {\map {\CC_0} {X, \R}, \norm {\, \cdot \,} }$ be the Banach algebra of real-valued continuous functions vanishing at infinity on $X$.
Let $\AA$ be a subalgebra of $\map {\CC_0} {X, \R}$ such that:
:$(1) \quad$ for each $x, y \in X$ with $x \ne y$ there exists $h_{x y} \in \AA$ such that $\map {h_{x y} } x \ne \map {h_{x y} } y$
:$(2) \quad$ for each $x \in X$ there exists $f_x \in \AA$ such that $\map {f_x} x \ne 0$. 
Then $\AA$ is everywhere dense in $\struct {\map {\CC_0} {X, \R}, \norm {\, \cdot \,} }$.
=== Complex-Valued ===
== Proof ==
Let $X^\ast = X \cup \set p$ be the Alexandroff extension of $X$. 
From Extension of Continuous Complex-Valued Function Vanishing at Infinity to Alexandroff Extension is Continuous, for each $f \in \map {\CC_0} {X, \R}$ we can define:
:$\map {f^\ast} x = \begin{cases}\map f x & x \in X \\ 0 & x = p\end{cases}$
for each $x \in X$ to obtain $f^\ast \in \map \CC {X^\ast, \R}$.
From Alexandroff Extension is Compact, $X^\ast$ is compact.
Define $\mathbf 1 : X^\ast \to \C$ by:
:$\map {\mathbf 1} x = 1$
for each $x \in X$.
From Constant Mapping is Continuous, we have $\mathbf 1 \in \map \CC {X^\ast, \R}$. 
Now define:
:$\BB = \set {f^\ast : f \in \AA} + \R \mathbf 1$
We show that $\BB$ is a subalgebra of $\map \CC {X^\ast, \R}$.
Since $\map \CC {X^\ast, \R}$ is an algebra, we have $\BB \subseteq \map \CC {X^\ast, \R}$.
Note that for $f, g \in \AA$ and $\lambda \in \R$ we have:
and:
for each $x \in X$.
Now consider:
:$f^\ast + t \mathbf 1 \in \BB$
and:
:$g^\ast + s \mathbf 1 \in \BB$
with $f, g \in \AA$.
Also let $\lambda \in \R$.
Then we have:
and:
Hence $\BB$ is a subalgebra of $\map \CC {X^\ast, \R}$.
We need to establish that $\BB$ separates points.
For $x, y \in X$ with $x \ne y$, we have:
:$\map {h_{x y} } x \ne \map {h_{x y} } y$
from $(1)$. 
Hence:
:$\map {h_{x y}^\ast} x \ne \map {h_{x y}^\ast} y$
We just need to show that $p$ is separated from other $y \in X$.
By condition $(2)$, there exists $f_x \in \AA$ with $\map {f_x} x \ne 0$. 
Then $\map {f_x^\ast} x \ne 0$. 
We then see $\map {f_x^\ast} p = 0 \ne \map {f_x^\ast} x$. 
Hence $\BB$ separates points.
Since $X^\ast$ is compact, we can apply Stone-Weierstrass Theorem: Compact Space to obtain:
:$\BB^- = \map \CC {X^\ast, \R}$.
Let $\epsilon > 0$ and $g \in \map {\CC_0} {X, \R}$. 
Then $g^\ast \in \map \CC {X^\ast, \R}$ as before.
So there exists $f \in \AA$ and $c \in \R$ such that:
:$\ds \sup_{x \mathop \in X^\ast} \size {\map {g^\ast} x - \map {f^\ast} x - c} < \frac \epsilon 2$
In particular:
:$\size {\map {g^\ast} p - \map {f^\ast} p - c} = \size c < \dfrac \epsilon 2$
Now we have from :
:$\ds \norm {g^\ast - f^\ast}_{\infty, X^\ast} \le \norm {g^\ast - f^\ast - c} + \size c < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$
where $\norm {\, \cdot \,}_{\infty, X^\ast}$ is the supremum norm on $\map \CC {X^\ast, \R}$. 
Note again that $\map {g^\ast} p - \map {f^\ast} p = 0$, so we also have:
:$\ds \norm {g - f} < \epsilon$
with $f \in \AA$. 
Since $\epsilon$ and $g$ were arbitrary, we have:
:$\AA^- = \map {\CC_0} {X, \R}$
Category:Stone-Weierstrass Theorem