> [!theorem] > > Let $\alg$ be a [[Commutative Banach Algebra|commutative]] unital [[C-Star Algebra|C* algebra]], then the [[Gelfand Transform|Gelfand transform]] $\Gamma: \alg \to C(\sigma(\alg))$ is an isometric $*$-isomorphism. > > If $f \in C(\sigma(\alg))$, denote $T_f \in \alg$ as the **inverse Gelfand transform**. > > *Proof*. Let $x \in \alg$ and $y = x^*$, then $y = y^*$ and > $ > \norm{y^{2^k}} = \norm{\paren{y^{2^{k - 1}}}^*y^{2^{k - 1}}} = \norm{y^{2^{k-1}}}^2 \quad \norm{y^{2^k}} = \norm{y}^{2^k} > $ > so $\norm{\wh y}_u = \norm{y}$. From here, > $ > \norm{x}^2 = \norm{y} = \norm{\wh y}_u = \norm{{\wh{x}}^2}_u = \norm{\wh x}_u^2 > $ > so $\Gamma$ is an isometry. Thus $\Gamma$ has closed range by the [[Method of Successive Approximations]]. From here, since $\Gamma$ has dense range, it is an isometric isomorphism. As $\Gamma x^* = \ol{\Gamma x}$, it is a $*$-isomorphism.