Theorem 7.4.13.label Let $A$ be an abelian Banach algebra then the Gelfand representation
\begin{align*}\scr{F}:A\to C_{\infty}(\OM(A))\end{align*}
is an isometry if and only if $\norm{x^2}{}=\norm{x}{}^{2}$ for all $x\in A$.
Proof. Suppose $\scr{F}$ is isometric then by Theorem 7.4.12
\begin{align*}\norm{x}{}=\norm{\ha{x}}{\infty}=\norm{x}{sp}\end{align*}
\begin{align*}\norm{x^2}{}=\norm{x^2}{sp}=\norm{x}{sp}^{2}=\norm{x}{}^{2}\end{align*}
Conversely if $\norm{x^2}{}=\norm{x}{}^{2}$ then by induction and Proposition 7.3.6 we have
\begin{align*}\norm{x}{sp}=\limit{n\to\infty}\norm{x^{2^n}}{}^{1/2^n}=\limit{n\to\infty}\norm{x}{}^{2^n/2^n}=\norm{x}{}\end{align*}
and by Theorem 7.4.12, $\scr{F}$ is isometric.$\square$
Post a Comment