32.3 $C_{0}(X)$
Definition 32.3.1 ($C_{0}(X)$).label Let $X$ be a LCH space, then $C_{0}(X; \complex)$ equipped with pointwise operations and the uniform norm is a $C^{*}$-algebra.
Theorem 32.3.2.label Let $X$ be a LCH space, then the mapping
\[E: X \to \Omega(C_{0}(X)) \quad E(x)(f) = f(x)\]
is a homeomorphism. Under the identification $X = \Omega(C_{0}(X))$, the Gelfand transform is the identity.
Proof, [Theorem 7.4, Zhu93]. Let $X^{*} = X \sqcup \bracs{\infty}$ be the one-point compactification of $X$. For each $\phi \in \Omega(C_{0}(X))$, let
\[\phi^{*}: BC(X^{*}) \to \complex \quad f \mapsto \phi(f - f(\infty)) + f(\infty)\]
then for each $f, g \in BC(X^{*})$,
\begin{align*}fg&= (f - f(\infty))(g - g(\infty)) + f(\infty)(g - g(\infty)) + g(\infty)(f - f(\infty)) + f(\infty)g(\infty) \\ \phi^{*}(fg)&= \phi(f - f(\infty))\phi(g - g(\infty)) + f(\infty)\phi(g - g(\infty)) \\&+ g(\infty)(f - f(\infty)) + f(\infty)g(\infty) \\&= \braksn{\phi(f - f(\infty)) + f(\infty)}\braksn{\phi(g - g(\infty)) + g(\infty)}\\&= \phi^{*}(f)\phi^{*}(g)\end{align*}
so $\phi^{*} \in \Omega(BC(X^{*}))$. By Theorem 32.7.2, there exists $x \in X^{*}$ such that $\phi^{*}(f) = f(x)$ for all $f \in BC(X^{*})$. Since $\phi \ne 0$, $x \in X$, and $\phi = E(x)$.$\square$
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