Theorem 9.4.12 (Gelfand Representation for Abelian Banach Algebra).label Let $A$ be an abelian Banach algebra, $C_{\infty}(\OM(A))$ is the abelian $C^{*}$-algebra of continuous functions on $\OM(A)$ vanishing at infinity. Then the Gelfand representation
is a homomorphism of abelian Banach algebras. If $A$ is unital then $\OM(A)$ is compact and $\spec{A}{x}=\ha{x}(\OM(A))$. If $A$ is not unital $\qspec{A}{x}=\ha{x}(\OM(A))\cup\curl{0}$. Hence in any case, for any $x\in A$
- (1)
$\norm{\ha{x}}{}=\norm{x}{sp}$
- (2)
$x\in G(A)\iff \scr{F}x\in G(C(\OM(A);\com))$
Proof. Since $\ha{x}:\OM(A)\to \com$ is defined by $\ha{x}(\om)=\om(x)$, $\ha{x}$ is $\s(A^{*},A)$-continuous on $\OM(A)$ so
is closed hence compact and we conclude $\ha{x}\in C_{\infty}(\OM(A))$. By definition $\scr{F}$ is linear and multiplicative, hence a homomorphism.
If $A$ is unital then $\frak{m}\define \curl{x-\lam\in A:\lam\in \spec{A}{x}}$ is a maximal regular ideal of $A$ hence
Conversely
Since the quasi spectrum always contain $0$, the assertion for nonunital $A$ follows from above. Thus we conclude
Lastly $x\in G(A)\iff$ the ideal generated by $x$ is proper $\iff$ there exists a maximal ideal containing $x$ $\iff$ there exists $\phi\in \OM(A)$ with $\phi(x)=0$.$\square$
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