Theorem 6.4.5 (Gelfand-Naimark).label Let $A$ be an abelian $C^{*}$-algebra then $A$ is semisimple as in Definition 6.2.1. Furthermore the Gelfand representation as in Theorem 6.3.11 $\scr{F}:A\to C_{\infty}(\OM(A))$ is an isometric isomorphism preserving the $*$-operation.
Proof. Since $A$ is abelian, every element $x\in A$ is normal so $\norm{x}{}=\norm{x}{sp}$ by Proposition 6.4.3. Hence $\scr{F}$ is isometric, showing that $A$ is semisimple. The image $\scr{F}(A)$ is complete therefore closed. Let $\om\in\OM(A)$ then for each $h\in A_{h}$
by Theorem 6.3.11 and Proposition 6.4.4. For every $x\in A$ we have $x=h+ik$ for some $h,k\in A_{h}$ and
since $*$-operation in $C_{\infty}(\OM(A))$ is complex conjugation, $\scr{F}:x\mto \ha{x}$ preserves the $*$-operation. Thus, $\scr{F}(A)$ is a self-adjoint, subalgebra of $C_{\infty}(\OM(A))$ that separates points of $\OM(A)$. By Stone-Weierstrass theorem, $\scr{F}(A)$ is dense in $C_{\infty}(\OM(A))$; hence $\scr{F}(A)=C_{\infty}(\OM)$ by closedness.$\square$