Commutative Gelfand Naimark

$\mathrm{\Delta }(A)$ is a nonempty compact Hausdorff space by [L1], so $C(\mathrm{\Delta }(A))$ is a complex Banach algebra under the supremum norm with pointwise operations.

Every element of $A$ is normal by [L2], so $r(a)=\mathrm{\parallel }a\mathrm{\parallel }$ for all $a$ by [L5].

$\mathrm{\Gamma }$ is a unital algebra homomorphism with $\mathrm{\parallel }\mathrm{\Gamma }(a){\mathrm{\parallel }}_{\mathrm{\infty }}=r(a)$, by [L4].

$\mathrm{\Gamma }(a^{\ast })(\chi )=\chi (a^{\ast })=\overline{\chi (a)}=\overline{\mathrm{\Gamma }(a)(\chi )}$ for every $\chi$, by [L3]; that is, $\mathrm{\Gamma }(a^{\ast })=\overline{\mathrm{\Gamma }(a)}$.

$\mathrm{\Gamma }$ is isometric: $\mathrm{\parallel }\mathrm{\Gamma }(a){\mathrm{\parallel }}_{\mathrm{\infty }}=r(a)=\mathrm{\parallel }a\mathrm{\parallel }$ for every $a$, by [step 1.2] and [step 1.3].

The range $B:=\mathrm{\Gamma }(A)$ is a unital self-adjoint complex subalgebra of $C(\mathrm{\Delta }(A))$: it is a subalgebra by [step 1.3] and contains the constant function one; it is self-adjoint by [step 1.4]; and it separates points, since for $\chi \mathrm{\ne }\psi$ there is $a\in A$ with $\chi (a)\mathrm{\ne }\psi (a)$, and then $\mathrm{\Gamma }(a)(\chi )\mathrm{\ne }\mathrm{\Gamma }(a)(\psi )$.

By [L6] applied to the compact Hausdorff space $\mathrm{\Delta }(A)$ and the unital point-separating self-adjoint function algebra $B$, the range $B$ is uniformly dense in $C(\mathrm{\Delta }(A))$.

The range $B=\mathrm{\Gamma }(A)$ is closed in $C(\mathrm{\Delta }(A))$: $\mathrm{\Gamma }$ is isometric by [step 2.1] and $A$ is complete, so $B$ is a complete subspace of the Banach space $C(\mathrm{\Delta }(A))$, hence closed.

A dense and closed subset equals the whole space, so $B=C(\mathrm{\Delta }(A))$ by [step 3.1] and [step 3.2]; together with [step 2.1], [step 1.3] and [step 1.4] this says that $\mathrm{\Gamma }$ is an isometric unital $\ast$-isomorphism onto $C(\mathrm{\Delta }(A))$.