60 lines
2.3 KiB
TeX
60 lines
2.3 KiB
TeX
\section{The Gelfand Transform}
|
|
\label{section:gelfand-transform}
|
|
|
|
\begin{definition}[Gelfand Transform]
|
|
\label{definition:gelfand-transform}
|
|
Let $A$ be a unital Banach algebra, then the \textbf{Gelfand transform} is the contractive homomorphism
|
|
\[
|
|
\Gamma = \Gamma_A: A \to C(\Omega(A); \complex) \quad (\Gamma_Ax)(\varphi) = \varphi(x)
|
|
\]
|
|
\end{definition}
|
|
|
|
\begin{remark}
|
|
\label{remark:gelfand-transform}
|
|
The Gelfand transform is limited in studying arbitrary Banach algebras, as they may admit no multiplicative functionals. However, these functionals come in abundance in the commutative case.
|
|
\end{remark}
|
|
|
|
|
|
\begin{proposition}
|
|
\label{proposition:gelfand-transform-gymnastics}
|
|
Let $A$ be a commutative unital Banach algebra and $x \in A$, then:
|
|
\begin{enumerate}
|
|
\item $\Gamma_A(1) = 1$.
|
|
\item $x \in G(A)$ if and only if $\Gamma_A x \in G(C(\Omega(A); \complex))$.
|
|
\item $(\Gamma_Ax)(\Omega(A)) = \sigma_A(x)$.
|
|
\item $\norm{\Gamma_Ax}_u = [x]_{sp}$.
|
|
\end{enumerate}
|
|
\end{proposition}
|
|
\begin{proof}[Proof, {{\cite[Theorem 1.1.13]{FollandHarmonic}}}. ]
|
|
(1): For each $\phi \in \Omega(A)$, $\phi(1) = 1$, so $\Gamma_A(1) = 1$.
|
|
|
|
(2): Since $A$ is commutative, $x \not\in G(A)$ if and only if the ideal generated by $x$ is proper, if and only if there exists a maximal ideal containing $x$, if and only if there exists $\phi \in \Omega(A)$ with $\phi(x) = 0$.
|
|
|
|
(3): By (2),
|
|
\[
|
|
(\Gamma_Ax)(\Omega(A)) = \sigma_{C(\Omega(A); \complex)}(\Gamma x) = \sigma_A(x)
|
|
\]
|
|
\end{proof}
|
|
|
|
\begin{proposition}
|
|
\label{proposition:gelfand-isometric}
|
|
Let $A$ be a commutative unital Banach algebra, then the following are equivalent:
|
|
\begin{enumerate}
|
|
\item For each $x \in A$, $\normn{x^2}_A = \norm{x}_A^2$.
|
|
\item $\Gamma_A$ is an isometry.
|
|
\end{enumerate}
|
|
\end{proposition}
|
|
\begin{proof}
|
|
(1) $\Rightarrow$ (2): For each $x \in A$, by the \hyperref[spectral radius formula]{proposition:spectral-radius-hadamard} and (4) of \autoref{proposition:gelfand-transform-gymnastics},
|
|
\[
|
|
\norm{\Gamma_A x}_u = [x]_{sp} = \norm{x}_A
|
|
\]
|
|
|
|
(2) $\Rightarrow$ (1): For each $x \in A$, by (4) of \autoref{proposition:gelfand-transform-gymnastics},
|
|
\[
|
|
\normn{x^2}_A \ge [x^2]_{sp} = \normn{\Gamma_A x^2}_u = \normn{\Gamma_A x}_u^2 = \normn{x}_A^2
|
|
\]
|
|
\end{proof}
|
|
|
|
|