Added basic facts about the Gelfand transform.

src/op/example/convolution.tex

@@ -0,0 +1,48 @@
+\section{$\ell^1(\integer)$}
+\label{section:convolution-algebra-integer}
+
+\begin{definition}[$\ell^1(\integer)$]
+\label{definition:convolution-algebra-integer}
+    Let $\ell^1(\integer)$ be the $\ell^1$ sequence space on $\integer$. For each $f, g \in \ell^1(\integer)$, let
+    \[
+    (f * g)(n) = \sum_{k \in \integer}f(k)g(n - k)
+    \]
+
+    then:
+    \begin{enumerate}
+        \item $\ell^1(\integer)$ is a commutative Banach algebra.
+        \item The multiplicative unit of $\ell^1(\integer)$ is $\delta_0 = \one_{\bracs{n = 0}}$. 
+    \end{enumerate}
+
+    The space $\ell^1(\integer)$ is the \textbf{convolution algebra} on $\integer$. 
+\end{definition}
+\begin{proof}
+    For each $f, g \in \ell^1(\integer)$, 
+    \begin{align*}
+        \normn{f * g}_{\ell^1(\integer)} &= \sum_{n \in \integer} \abs{\sum_{k \in \integer}f(k)g(n - k)} \\
+        &\le \sum_{n, k \in \integer}|f(k)| \cdot |g(n-k)| \le \sum_{k \in \integer}|f(k)| \cdot \sum_{n \in \integer}|g(n - k)| \\
+        &= \norm{f}_{\ell^1(\integer)} \cdot \norm{g}_{\ell^1(\integer)}
+    \end{align*}
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:convolution-integer-gelfand}
+    The Gelfand transform of $\ell^1(\integer)$ is not isometric. 
+\end{proposition}
+\begin{proof}
+    Let $f = \one_{\bracs{n = 1}} - \one_{\bracs{2 \le n \le 3}}$, then
+    \[
+    f^2(n) = \begin{cases}
+        -1 &n \in \bracs{1, 5} \\
+        -2 &n = 2 \\
+        -1 &n = 3 \\
+        2 &n = 4 \\
+        0 &n \not\in [1, 5] 
+    \end{cases}
+    \]
+
+    so $\normn{f^2}_{\ell^1(\integer)} = 7 < \normn{f}_{\ell^1(\integer)}^2$. By \autoref{proposition:gelfand-isometric}, the Gelfand transform is not isometric. 
+\end{proof}
+
+

src/op/example/index.tex

@@ -5,4 +5,5 @@
 \input{./bounded.tex}
 \input{./hardy.tex}
 \input{./disk.tex}
+\input{./convolution.tex}