From f4bcc76bd0a5bbdaedacde7daed0c8197b00091e Mon Sep 17 00:00:00 2001
From: Bokuan Li <bokuan.li@mail.mcgill.ca>
Date: Tue, 2 Jun 2026 20:21:07 -0400
Subject: [PATCH] Added the GKZ Theorem.

---
 src/op/banach/multiplicative.tex | 58 ++++++++++++++++++++++++++++++++
 1 file changed, 58 insertions(+)

diff --git a/src/op/banach/multiplicative.tex b/src/op/banach/multiplicative.tex
index d39cb00..350fa62 100644
--- a/src/op/banach/multiplicative.tex
+++ b/src/op/banach/multiplicative.tex
@@ -47,4 +47,62 @@
     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 I.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 $(xy - yx)^2 \in \ker\phi$ as well, and
+    \[
+    (\phi(xy - yx))^2 = \phi((xy - yx)^2) = 0
+    \]
+
+    Therefore $2xy = (xy + yx) - (xy - yx) \in \ker \phi$, $\ker \phi$ is an ideal, and $\phi$ is a homomorphism. 
+\end{proof}
+