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

Let $X$ be a locally compact Hausdorff space.

Let $\struct {\map {\CC_0} {X, \C}, \norm {\, \cdot \,} }$ be the Banach algebra of complex-valued continuous functions vanishing at infinity on $X$.

Let $\AA$ be a $\ast$-subalgebra of $\map {\CC_0} {X, \C}$ 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 $\struct {\map {\CC_0} {X, \C}, \norm {\, \cdot \,} }$.

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 Norm Axiom $\text N 3$: Triangle Inequality:

$\ds \norm {g^\ast - f^\ast}_{\infty, X^\ast} \le \norm {g^\ast - f^\ast - c} + \size c < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$

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:

$\AA^- = \map {\CC_0} {X, \R}$

$\blacksquare$