Theorem: Gelfand-Naimark Theorem/Commutative Case

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a commutative $\text C^\ast$-algebra.
Let $\Phi_A$ be the spectrum of $A$.
Let $\struct {\map {\CC_0} {\Phi_A}, \overline \cdot, \norm {\, \cdot \,} }$ be the $\text C^\ast$-algebra of continuous complex-valued functions vanishing at infinity.
Define $G : A \to \map {\CC_0} {\Phi_A}$ by:
:$\map G a = \hat a$
where $\hat a$ is the Gelfand transform of $a$.
Then $G$ is a isometric $\ast$-algebra isomorphism.
That is, the commutative $\text C^\ast$-algebras are precisely $\struct {\map {\CC_0} X, \overline \cdot, \norm {\, \cdot \,} }$ for locally compact Hausdorff spaces $X$, up to isometric $\ast$-algebra isomorphism.

Proof:
First, from Spectrum of Commutative C*-Algebra is Non-Empty, we have $\Phi_A \ne \O$ and hence the Gelfand transform is indeed defined.
From Spectrum of Banach Algebra is Weak-* Locally Compact Hausdorff Space, $\Phi_A$ is a locally compact Hausdorff space, so we can understand $\map {\CC_0} {\Phi_A}$.
From the Gelfand Representation Theorem, $G$ is an algebra homomorphism with:
:$\norm {\map G a}_\infty = \map {r_A} a$
for each $a \in A$.
Since $A$ is commutative, we have:
:$a a^\ast = a^\ast a$
for each $a \in A$.
Hence every $a \in A$ is normal.
From Spectral Radius of Normal Element of C*-Algebra Equal to Norm, we have $\map {r_A} a = \norm a$.
Hence: 
:$\norm {\map G a}_\infty = \norm a$
and $G$ is a linear isometry.
It remains to show that $G$ is a $\ast$-algebra homomorphism and a surjection.
For each $\phi \in \Phi_A$, we have:
{{begin-eqn}}
{{eqn	| l = \map {\overline {\hat a} } \phi 
		| r = \overline {\map {\hat a} \phi} 
}}
{{eqn	| r = \overline {\map \phi a}
}}
{{eqn	| r = \map \phi {a^\ast} 
		| c = Character on C*-Algebra is *-Algebra Homomorphism
}}
{{eqn	| r = \map {\hat {\paren {a^\ast} } } \phi 
}}
{{end-eqn}}
Hence we have:
:$\overline {\map G a} = \map G {a^\ast}$
So $G$ is a $\ast$-algebra homomorphism.
From Image of Closed Set under Linear Isometry from Banach Space to Normed Vector Space is Closed, $G \sqbrk A$ is closed in $\map {\CC_0} {\Phi_A}$.
Hence from Image of *-Algebra under *-Algebra Homomorphism is *-Subalgebra, $G \sqbrk A$ is a $\ast$-subalgebra of $\map {\CC_0} {\Phi_A}$.
We look to apply Stone-Weierstrass Theorem: Locally Compact Hausdorff Space: Complex-Valued.
We need to show that:
:$(1) \quad$ for each $\phi, \psi \in \Phi_A$ with $\phi \ne \psi$ there exists $a \in A$ such that $\map {\hat a} \phi \ne \map {\hat a} \psi$. 
:$(2) \quad$ for each $\phi \in \Phi_A$ there exists $a \in A$ such that $\map {\hat a} \phi \ne 0$.
Both conditions are almost immediate.
To show $(1)$, note that if $\phi \ne \psi$ there exists $a \in A$ such that $\map \phi a \ne \map \psi a$, so $\map {\hat a} \phi \ne \map {\hat a} \psi$. 
Note that $(2)$ follows from the definition of a character: we have $\phi \ne {\mathbf 0}_{A^\ast}$ hence there exists $a \in A$ with $\map \phi a = \map {\hat a} \phi \ne 0$.
Hence by Stone-Weierstrass Theorem: Locally Compact Hausdorff Space: Complex-Valued, we have:
:$G \sqbrk A^- = \map {\CC_0} {\Phi_A}$
Since $G \sqbrk A$ is closed, we have:
:$G \sqbrk A^- = G \sqbrk A$
from Set is Closed iff Equals Topological Closure.
So:
:$G \sqbrk A = \map {\CC_0} {\Phi_A}$
and $G$ is surjection.
Hence $G$ is a isometric $\ast$-algebra isomorphism.
So every commutative $\text C^\ast$-algebra is isometrically $\ast$-algebra isomorphic to $\struct {\map {\CC_0} X, \overline \cdot, \norm {\, \cdot \,} }$ for some locally compact Hausdorff spaces $X$. 
Conversely, from Space of Complex-Valued Continuous Functions Vanishing at Infinity is C*-Algebra, $\struct {\map {\CC_0} X, \overline \cdot, \norm {\, \cdot \,} }$ is a commutative $\text C^\ast$-algebra for each locally compact Hausdorff space $X$. 
Each such $\text C^\ast$-algebra is commutative since $\C$ is a field from Complex Numbers form Field.
{{qed}}
Category:Gelfand-Naimark Theorem
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