Theorem 6.3.11: Gelfand Representation for Abelian Banach Algebra

Theorem 6.3.11 (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

\begin{align*}\scr{F}:A\to C_{\infty}(\OM(A))&&x\mto \ha{x}:\om\mto \om(x)\end{align*}

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,

\begin{align*}\norm{\ha{x}}{}=\norm{x}{sp}&&\fall x\in A\end{align*}

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

\begin{align*}\fall \e>0,\curl{\om\in \OM(A):\abs{\ha{x}(w)}\geq \e}\suf\OM'(A)\end{align*}

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

\begin{align*}0=\om_{\frak{m}}(x-\lam)=\om_{\frak{m}}(x)-\om_{\frak{m}}(\lam)=\om_{\frak{m}}(x)-\lam\end{align*}

Conversely

\begin{align*}\curl{\lam\in\com:\lam=\om(x)\text{ for some $\om\in \OM(A)$}}&=\curl{\lam\in \com:x-\lam\in\om\inv(0)\text{ for some $\om\in\OM(A)$}}\\&=\curl{\lam\in \com:x-\lam\text{ is not invertible}}=\spec{A}{x}\end{align*}

Since the quasi spectrum always contain $0$, the assertion for nonunital $A$ follows from above. Thus we conclude

\begin{align*}\norm{\ha{x}}{}=\SUP{\om\in\OM(A)}\abs{\om(x)}=\SUP{\lam\in\spec{A}{x}}\abs{\lam}=\norm{x}{sp}\end{align*}

$\square$

The Eigenspace

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The Eigenspace

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