Proposition 6.3.9.label Let $A$ be an abelian Banach algebra and $S^{*}$ be the unit ball of the conjugate space $A^{*}$ of $A$ then $\OM(A)\suf S^{*}$
Proof. If $\om\in\OM(A)$ then the kernel $\om\inv(0)$ is closed by Proposition 6.3.4. Recall any linear functional on a normed space is continuous if and only if its kernel is closed so $\om$ is continuous hence bounded. Consequently for each $x\in A$
\begin{align*}\abs{\om(x)}=\abs{\om(x^n)}^{1/n}\leq\norm{\om}{}^{1/n}\norm{x^n}{}^{1/n}&&\fall n\in\N^{+}\end{align*}
hence $\abs{\om(x)}\leq \limit{n\to\infty}\norm{\om}{}^{1/n}\norm{x^n}{}^{1/n}=\limit{n\to\infty}\norm{x^n}{}^{1/n}=\norm{x}{sp}\leq \norm{x}{}$$\square$