Added basic facts about the Gelfand transform.

src/op/banach/gelfand.tex

@@ -0,0 +1,60 @@
+\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 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 $\Gamma_A$ is a contractive homomorphism. 
+        \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}}}. ]
+    (2): For each $\phi \in \Omega(A)$, $\phi(1) = 1$, so $\Gamma_A(1) = 1$. 
+
+    (3): 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 $\phi(x) = 0$. 
+
+    (4): By (1) and (3), 
+    \[
+    (\Gamma_Ax)(\Omega(A)) = \sigma_{C(\Omega(A); \complex)}(\Gamma x) = \sigma_A(x)
+    \]
+\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 (5) 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 (5) 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}
+
+

src/op/banach/index.tex

@@ -8,3 +8,4 @@
 \input{./spectrum.tex}
 \input{./fc.tex}
 \input{./multiplicative.tex}
+\input{./gelfand.tex}

src/op/banach/spectrum.tex

@@ -77,7 +77,7 @@
     Let $x \in A$. By \autoref{proposition:spectrum-non-empty}, there exists $\lambda \in \sigma_A(x)$. Since $\lambda \cdot 1 - x$ is not invertible and every non-zero element of $A$ is invertible, $x = \lambda \cdot 1$. Therefore the mapping $\complex \to A$ defined by $\lambda \mapsto \lambda \cdot 1$ is an isometric isomorphism. 
 \end{proof}
 
-\begin{proposition}
+\begin{proposition}[Spectral Radius Formula]
 \label{proposition:spectral-radius-hadamard}
     Let $A$ be a unital Banach algebra and $x \in A$, then $[\cdot]_{sp} = \limsup_{n \to \infty}\normn{x^n}_A^{1/n}$. 
 \end{proposition}
@@ -126,4 +126,24 @@
     Let $x \in A$ with $\sigma_A(x) \subset U$, and $\lambda \in U^c$. By \autoref{proposition:banach-algebra-inverse}, for any $y \in A$ with $\norm{y}_A \le \normn{(\lambda - x)^{-1}}_A^{-1}$, $\lambda - x - y \in G(A)$ as well. Since the mapping $\lambda \mapsto \normn{(\lambda - x)^{-1}}_A$ vanishes at infinity and $U^c$ is closed, $\delta = \inf_{\lambda \in U^c}\normn{(\lambda - x)^{-1}}_A^{-1} > 0$. Therefore for every $z \in B_A(x,\delta)$, $\sigma_A(x) \subset U$. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:commutative-spectrum-gymnastics}
+    Let $A$ be a unital Banach algebra and $x, y \in A$ with $x = y$, then
+    \begin{enumerate}
+        \item $\sigma_A(x + y) \subset \sigma_A(x) + \sigma_A(y)$.
+        \item $\sigma_A(xy) \subset \sigma_A(x)\sigma_A(y)$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    For any $z \in A$, denote $R(z) = \bracs{(\lambda - z)^{-1}|\lambda \in \sigma_A(z)}$. Let $B \subset A$ be the closed subalgebra generated by $1$, $x$, $y$, $R(x)$, $R(y)$, $R(xy)$, $R(x + y)$, then $B$ is a commutative algebra with $\sigma_B(x) = \sigma_A(x)$, $\sigma_B(y) = \sigma_A(y)$, $\sigma_B(xy) = \sigma_A(xy)$, and $\sigma_B(x + y) = \sigma_A(x + y)$. 
+
+    By \autoref{proposition:gelfand-transform-gymnastics}, for each $z \in B$, $(\Gamma_Bz)(\Omega(B)) = \sigma_B(z)$. Since for any $u, v \in C(\Omega(B); \complex)$, 
+    \begin{enumerate}
+        \item $\sigma_{C(\Omega(B); \complex)}(u + v) \subset \sigma_{C(\Omega(B); \complex)}(u) + \sigma_{C(\Omega(B); \complex)}(v)$.
+        \item $\sigma_{C(\Omega(B); \complex)}(uv) \subset \sigma_{C(\Omega(B); \complex)}(u)\sigma_{C(\Omega(B); \complex)}(v)$.
+    \end{enumerate}
+    
+    The above holds for $x$ and $y$ with respect to $\sigma_A$. 
+\end{proof}
+