\section{Multiplicative Functionals}
\label{section:multiplicative-functional}

\begin{definition}[Multiplicative Functional]
\label{definition:multiplicative-functional}
    Let $A$ be a unital Banach algebra and $\phi \in A^*$, then $\phi$ is \textbf{multiplicative} if $\phi \ne 0$ and for each $x, y \in A$, $\phi(xy) = \phi(x)\phi(y)$. 
\end{definition}

\begin{proposition}
\label{proposition:multiplicative-unit}
    Let $A$ be a unital Banach algebra and $\phi \in A^*$ be a multiplicative functional, then
    \begin{enumerate}
        \item For each $x \in A$, $|\phi(x)| \le [x]_{sp} \le \norm{x}_A$. 
        \item $\norm{\phi}_{A^*} = 1$. 
        \item $\phi(G(A)) \subset \complex \setminus \bracs{0}$. 
    \end{enumerate}
\end{proposition}
\begin{proof}
    (3): For each $x \in G(A)$, $1 = \phi(xx^{-1}) = \phi(x)\phi(x^{-1}) \ne 0$. 
    
    (1): By (3), for every $\lambda \in \complex$ with $|\lambda| > [x]_{sp}$, $\lambda x \in G(A)$ and $\lambda - \phi(x) \ne 0$. Therefore $\phi(x) \in \ol{B(0, [x]_{sp})}$. 

    (2): For each $\lambda \in \complex$, $\phi(\lambda 1) = \lambda$, so $\norm{\phi}_{A^*} \le 1$. 
\end{proof}

\begin{proposition}
\label{proposition:multiplicative-less-unit}
    Let $A$ be a Banach algebra and $\phi \in A^*$ be a multiplicative functional, then $\norm{\phi}_{A^*} \le 1$. 
\end{proposition}
\begin{proof}
    Let $\tilde A$ be the unitisation of $A$, then by (U) of the \hyperref[unitisation]{definition:unitisation}, $\phi$ extends to a multiplicative functional $\tilde \phi$ on $\tilde A$. Therefore $\norm{\phi}_{A^*} \le \normn{\tilde \phi}_{{\tilde A}^*} = 1$. 
\end{proof}


\begin{definition}[Space of Multiplicative Linear Functionals]
\label{definition:multiplicative-linear-functional-space}
    Let $A$ be a Banach algebra, then $\Omega(A)$ is the \textbf{space of multiplicative linear functionals}, and with respect to the weak-* topology, 
    \begin{enumerate}
        \item If $A$ is unital, then $\Omega(A)$ is a compact Hausdorff space. 
        \item $\Omega(A) \cup \bracs{0}$ is a compact Hausdorff space, and $\Omega(A)$ is a LCH space.
    \end{enumerate}
\end{definition}
\begin{proof}
    (1): By \autoref{proposition:multiplicative-unit}, $\Omega(A)$ is a weak-* closed subset of $\bracsn{\phi \in A^*:\norm{\phi}_{A^*} = 1}$, so it is compact by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}.

    (2): By \autoref{proposition:multiplicative-less-unit}, $\Omega(A) \cup \bracs{0}$ is a weak-* closed subset of $\ol{B_{A^*}(0, 1)}$, so it is compact by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}.
\end{proof}

\begin{proposition}
\label{proposition:commutative-maximal-ideal-space}
    Let $A$ be a commutative unital Banach algebra, then the mapping
    \[
    \Omega(A) \to \cm(A) \quad \phi \mapsto \ker(\phi)
    \]

    is a bijection. 
\end{proposition}
\begin{proof}
    For each $\phi \in \cm(A)$, $\ker(\phi)$ is an ideal of codimension $1$, and must be maximal. 

    On the other hand, for each $I \in \Omega(A)$, the quotient $A/I$ is a field, so by the \hyperref[Gelfand-Mazur Theorem]{theorem:gelfand-mazur}, it is isomorphic to $\complex$. Thus the canonical projection $A \to A/I \iso \complex$ induces a multiplicative functional on $A$. 
\end{proof}

\begin{theorem}[Gleason-Kahane-Żelazko]
\label{theorem:gkz}
    Let $A$ be a unital Banach algebra and $\phi \in A^*$, then the following are equivalent:
    \begin{enumerate}
        \item $\phi$ is a multiplicative linear functional.
        \item $\phi(1) = 1$ and $\phi(G(A)) \subset \complex \setminus \bracs{0}$. 
    \end{enumerate}
\end{theorem}
\begin{proof}[Proof, {{\cite[Theorem 4.5]{Zhu}}}. ]
    (1) $\Rightarrow$ (2): \autoref{proposition:multiplicative-unit}. 

    (2) $\Rightarrow$ (1): Let $x \in A$ and $\lambda \in \complex$ with $|\lambda| > [x]_{sp}$, then $\lambda - x \in G(A)$ and $\phi(\lambda - x) \ne 0$. Therefore $\phi(x) \subset \ol{B(0, [x]_{sp})}$, and $\norm{\phi}_{A^*} = 1$. 

    Let $x \in \ker \phi$ with $\norm{x}_A \le 1$, and let
    \[
    f: \complex \to \complex \quad \lambda \mapsto \phi[\exp(\lambda x)] = \sum_{n = 0}^\infty \frac{\phi[(\lambda x)^n]}{n!}
    \]

    then
    \begin{enumerate}[label=(\alph*)]
        \item $f(0) = \phi(1) = 1$. 
        \item $Df(0) = \phi(x) = 0$. 
        \item Since $\norm{x}_A \le 1$, $|f| \le |\exp|$. As $\exp(\lambda x) \in G(A)$ for all $\lambda \in \complex$, $f(\lambda) \ne 0$ for all $\lambda \in \complex$. 
    \end{enumerate}

    By \autoref{proposition:entire-constant}, $f = 1$, and $\phi(x^2) = 0$, so for any $x \in \ker\phi$, $x^2 \in \ker\phi$ as well. 

    Now, for each $x, y \in A$, there exists $x_0, y_0 \in \ker \phi$ such that $x = x_0 + \phi(x)$ and $y = y_0 + \phi(x)$. In which case, 
    \begin{align*}
        \phi(xy) &= \phi(x_0y_0 + \phi(x)y_0 + \phi(y)x_0 + \phi(x)\phi(y)) \\
        &= \phi(x_0y_0) + \phi(x)\phi(y)
    \end{align*}

    In particular, 
    \[
    \phi(x^2) = \phi(x_0^2) + \phi(x)^2 = \phi(x)^2
    \]

    To conclude, let $x \in \ker\phi$, $y \in A$, then
    \begin{align*}
        \phi((x + y)^2) &= (\phi(x + y))^2 = \phi(x)^2 + 2\phi(x)(y) + \phi(y)^2 \\
        \phi(xy + yx) &= 2\phi(x)\phi(y)
    \end{align*}

    so $xy + yx \in \ker\phi$ as well. Finally, 
    \begin{align*}
        (xy + yx)^2 + (xy - yx)^2 &= 2(xyxy + yxyx) \\
        &= 2[x(yxy) + y(xyx)] \in \ker \phi
    \end{align*}

    and $(xy - yx)^2 \in \ker\phi$. Since
    \[
    (\phi(xy - yx))^2 = \phi((xy - yx)^2) = 0
    \]

    The commutator $xy - yx \in \ker \phi$ as well. Therefore $2xy = (xy + yx) - (xy - yx) \in \ker \phi$, $\ker \phi$ is an ideal, and $\phi$ is a homomorphism. 
\end{proof}


