Added basic facts about the Gelfand transform.

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}
+