Added basics of Banach algebras.

src/op/banach/spectrum.tex

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+\section{The Spectrum}
+\label{section:spectrum}
+
+\begin{definition}[Spectrum]
+\label{definition:spectrum}
+    Let $A$ be a unital Banach algebra and $x \in A$, then
+    \[
+    \sigma(x) = \sigma_A(x) = \bracs{\lambda \in \complex| \lambda - x \not\in G(A)}
+    \]
+
+    is the \textbf{spectrum} of $x$ in $A$, and its complement is the \textbf{resolvent set} of $x$. 
+\end{definition}
+
+\begin{definition}[Spectral Radius]
+\label{definition:spectral-radius}
+    Let $A$ be a unital Banach algebra and $x \in A$, then
+    \[
+    [x]_{sp} = \sup\bracs{|\lambda|: \lambda \in \sigma_A(x)}
+    \]
+
+    is the \textbf{spectral radius} of $x$. 
+\end{definition}
+
+\begin{definition}[Resolvent]
+\label{definition:resolvent}
+    Let $A$ be a unital Banach algebra and $x \in A$, then
+    \[
+    R_x: \complex \setminus \sigma_A(x) \to A \quad \lambda \mapsto \frac{1}{\lambda - x}
+    \]
+
+    is the \textbf{resolvent} function of $x$, which is holomorphic on $\complex \setminus \sigma_A(x)$.
+\end{definition}
+\begin{proof}
+    By \autoref{proposition:banach-algebra-inverse}, the mapping $y \mapsto y^{-1}$ is smooth on $G(A)$. Hence $R_x: \complex \setminus \sigma_A(x) \to A$ is holomorphic. 
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:spectrum-non-empty}
+    Let $A$ be a unital Banach algebra and $x \in A$, then $\sigma_A(x) \ne \emptyset$. 
+\end{proposition}
+\begin{proof}
+    Assume for contradiction that $\sigma_A(x) = \emptyset$, then by \autoref{definition:resolvent}, the resolvent
+    \[
+    R_x: \complex \setminus \sigma_A(x) \to A \quad \lambda \mapsto \frac{1}{\lambda - x}
+    \]
+
+    is an entire function. Since
+    \[
+    R_x(\lambda) = \frac{1}{\lambda - x} = \frac{\lambda^{-1}}{1 - \lambda^{-1}x}
+    \]
+
+    which tends to $0$ as $|\lambda| \to \infty$, $R_x \in H(\complex; A) \cap C_0(\complex; A)$. By \hyperref[Liouville's Theorem]{theorem:liouville}, $R_x = 0$, which is impossible. 
+\end{proof}
+
+\begin{theorem}[Gelfand-Naimark]
+\label{theorem:gelfand-naimark}
+    Let $A$ be a unital Banach algebra. If every non-zero element of $A$ is invertible, then $A$ is isometrically isomorphic to $\complex$. 
+\end{theorem}
+\begin{proof}
+    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}
+\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}
+\begin{proof}
+    Let $r = \limsup_{n \to \infty}\normn{x^n}_A^{1/n}$, then for any $\lambda \in \complex$ with $|\lambda| > r$, the series $\sum_{n = 0}^\infty \lambda^{-n-1}x^n$ converges absolutely, and to the inverse of $(x - \lambda)$. Therefore $r \ge [\cdot]_{sp}$. 
+
+    Let $D = \bracs{\lambda \in \complex|\ |\lambda| > [x]_{sp}}$, then since the series $\sum_{n = 0}^\infty \lambda^{-n-1}x^n$ is the expansion of $R_x$ at infinity, which is defined on $D$, it must converge on $D$. Let $\phi \in A^*$, then by the preceding discussion, the series $\sum_{n = 0}^\infty \lambda^{-n-1}\dpn{x^n, \phi}{A}$ converges on $D$. Thus for any $\lambda \in D$, 
+    \[
+    \sup_{n \in \natz}|\lambda^{-n-1} \dpn{x^n, \phi}{A}|^{1/n} < \infty
+    \]
+
+    By the \hyperref[Uniform Boundedness Principle]{theorem:uniform-boundedness}, 
+    \[
+    \sup_{n \in \natz}|\lambda|^{-n-1} \cdot \normn{x^n}_A < \infty
+    \]
+
+    Therefore $\limsup_{n \to \infty}\norm{x^n}_A^{1/n} \le [x]_{sp}$. 
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:spectrum-compact}
+    Let $A$ be a unital Banach algebra and $x \in A$, then $\sigma_A(x)$ is a compact subset of $B_\complex(0, \norm{x}_A)$. 
+\end{proposition}
+\begin{proof}
+    By \autoref{proposition:banach-algebra-inverse}, $G(A)$ and hence the resolvent set of $x$ is open, so $\sigma_A(x)$ is closed. By \autoref{proposition:spectral-radius-hadamard}, $[x]_{sp} \le \norm{x}_A$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:spectrum-product-gymnastics}
+    Let $A$ be a unital Banach algebra and $x, y \in A$, then:
+    \begin{enumerate}
+        \item $\sigma(xy) \cup \bracs{0} = \sigma(yx) \cup \bracs{0}$. 
+        \item $[xy]_{sp} = [yx]_{sp}$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $\lambda \in \complex \setminus \bracs{0}$, then by \autoref{proposition:swap-invertible}, $\lambda - xy \in G(A)$ if and only if $\lambda - yx \in G(A)$. 
+\end{proof}
+