Added the continuous functional calculus.

src/op/banach/spectrum.tex

@@ -146,4 +146,29 @@
     The above holds for $x$ and $y$ with respect to $\sigma_A$. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:spectrum-subalgebra-gymnastics}
+    Let $A$ be a unital Banach algebra, $B \subset A$ be a closed subalgebra containing $1$, and $x \in B$, then:
+    \begin{enumerate}
+        \item $\sigma_A(x) \subset \sigma_B(x)$. 
+        \item $\partial \sigma_B(x) \subset \sigma_A(x)$. 
+        \item $\sigma_B(x)$ is the union of $\sigma_A(x)$ and some bounded components of $\complex \setminus \sigma_A(x)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): $G(B) \subset G(A)$. 
+
+    (2): Let $\lambda \in \partial \sigma_B(x)$, then there exists $\seq{\lambda_n} \subset \complex \setminus \sigma_B(x)$ such that $\lambda_n - x \in G(B)$ for all $n \in \natp$, and $\lambda_n \to \lambda$ as $n \to \infty$. By \autoref{corollary:invertible-boundary-explode}, $\norm{(\lambda_n - x)^{-1}}_A \to \infty$ as $n \to \infty$. If $\lambda - x \in G(A)$, then $(\lambda_n - x)^{-1} \to (\lambda - x)^{-1}$ as $n \to \infty$. Thus $\norm{(\lambda - x)^{-1}}_A = \infty$, which is impossible. Therefore $\lambda - x \not\in G(A)$, and $\lambda \in \sigma_A(x)$. 
+\end{proof}
+
+\begin{theorem}["Runge's Theorem"]
+\label{theorem:spectrum-subalgebra-sufficiency}
+    Let $A$ be a unital Banach algebra, $x \in A$, $P \subset \complex \setminus \sigma_A(x)$ such that $P$ intersects every bounded component of $\complex \setminus \sigma_A(x)$, and $B \subset A$ be a closed algebra containing $1$, $x$, and $\bracsn{(\lambda - x)^{-1}|\lambda \in P}$, then $\sigma_A(x) = \sigma_B(x)$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 4.9]{MarcouxNotes}}}. ]
+    By construction, $P \subset \complex \setminus \sigma_B(x)$. In addition, for any polynomial $p \in \complex[z]$, $p(x) \in B$. Thus for every rational function $f \in \complex(z) \cap H(\complex_\infty \setminus (P \cup \bracs{\infty}); \complex)$, $f(x) \in B$. 
+
+    By \hyperref[Runge's Theorem]{theorem:runge}, $H(\complex_\infty \setminus (P \cup \bracs{\infty}); \complex)$ is dense in $H(\sigma_A(x); \complex)$. The continuity of the \hyperref[holomorphic functional calculus]{definition:holomorphic-functional-calculus} then implies that $f(x) \in B$ for all $f \in H(\sigma_A(x); \complex)$. In particular, $(\lambda - x)^{-1} \in B$ for all $\lambda \in \complex \setminus \sigma_A(x)$. Therefore $\sigma_B(x) \subset \sigma_A(x)$, and $\sigma_B(x) = \sigma_A(x)$ by \autoref{proposition:spectrum-subalgebra-gymnastics}.
+\end{proof}
+