\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}


