Added some spectrum gruntwork.

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}[Spectral Radius Formula]
+\begin{proposition}[Beurling's 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}
@@ -128,7 +128,7 @@
 
 \begin{proposition}
 \label{proposition:commutative-spectrum-gymnastics}
-    Let $A$ be a unital Banach algebra and $x, y \in A$ with $x = y$, then
+    Let $A$ be a commutative 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 $\sigma_A(xy) \subset \sigma_A(x)\sigma_A(y)$.

src/op/c-star/index.tex

@@ -2,5 +2,6 @@
 \label{chap:c-star-algebras}
 
 \input{./involution.tex}
+\input{./unitary.tex}
 \input{./sa.tex}
 \input{./order.tex}

src/op/c-star/sa.tex

@@ -31,6 +31,91 @@
     If the above holds, then $x$ is \textbf{normal}. 
 \end{definition}
 
+\begin{theorem}
+\label{theorem:c-star-normal-spectral-radius}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then $\norm{x}_A = [x]_{sp}$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem II.8.1]{Zhu}}}. ]
+    First suppose that $x$ is self-adjoint. In this case, 
+    \begin{align*}
+        \normn{x^2}_A &= \normn{xx^*}_A = \norm{x}_A^2 \\
+        \normn{x^{2^n}}_A &= \norm{x}_A^{2^n}
+    \end{align*}
+
+    for all $n \in \natp$. Thus by the \hyperref[spectral radius formula]{proposition:spectral-radius-hadamard},
+    \[
+    [x]_{sp} = \limsup_{n \to \infty}\norm{x^{n}}_A^{1/n} \ge \limsup_{n \to \infty}\normn{x^{2^n}}_A^{1/2^n} = \norm{x}_A
+    \]
+
+    Now suppose that $x$ is only normal. Since $x$ and $x^*$ commute, $[xx^*]_{sp} \le [x]_{sp}[x^*]_{sp}$ by \autoref{proposition:commutative-spectrum-gymnastics}. Thus
+    \[
+    \norm{x}^2_A = \normn{xx^*}_A = [xx^*]_{sp} \le [x]_{sp}[x^*]_{sp} = [x]_{sp}^2
+    \]
+
+    by (5) of \autoref{proposition:c-star-algebra-gymnastics}. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:c-star-normal-spectral-radius-corollary}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then:
+    \begin{enumerate}
+        \item There exists $\lambda \in \sigma_A(x)$ such that $|\lambda| = \norm{x}_A$. 
+        \item If there exists $n \in \natp$ such that $x^n = 0$, then $x = 0$ as well. 
+    \end{enumerate}
+\end{corollary}
+\begin{proof}
+    (1): Since $\sigma_A(x)$ is compact, there exisst $\lambda \in \sigma_A(x)$ such that $|\lambda| = [x]_{sp}$. By \autoref{theorem:c-star-normal-spectral-radius}, $|\lambda| = [x]_{sp} = \norm{x}_A$. 
+
+    (2): By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-holomorphic}, 
+    \[
+    \bracs{\lambda^n| \lambda \in \sigma_A(x)} = \bracs{0}
+    \]
+
+    Thus $\sigma_A(x) = \bracs{0}$. By \autoref{theorem:c-star-normal-spectral-radius}, $\norm{x}_A = [x]_{sp} = 0$. 
+\end{proof}
+
+
+\begin{corollary}
+\label{corollary:c-star-unique-norm}
+    Let $A$ be a unital $C^*$-algebra over $\complex$, then for each $x \in A$, 
+    \[
+    \norm{x}_A^2 = \sup\bracs{|\lambda|\ | \lambda \in \sigma_A(x^*x)}
+    \]
+
+    In particular, there exists at most one norm on $A$ making it a $C^*$-algebra. 
+\end{corollary}
+\begin{proof}
+    Since $x^*x$ is self-adjoint, \autoref{theorem:c-star-normal-spectral-radius} implies that
+    \[
+    \norm{x}_A^2 = \norm{x^*x}_A = \sup\bracs{|\lambda|\ | \lambda \in \sigma_A(x^*x)}
+    \]
+
+    which depends only on the algebraic structure of $A$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:self-adjoint-spectrum}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be self-adjoint, then $\sigma_A(x) \subset \real$. 
+\end{proposition}
+\begin{proof}
+    Let
+    \[
+    y = \exp(ix) = \sum_{n = 0}^\infty \frac{i^nx^n}{n!}  
+    \]
+
+    then
+    \[
+    y^*= \sum_{n = 0}^\infty \frac{(-i)^n (x^*)^n}{n!} = \exp(-ix^*) 
+    \]
+
+    Since $x$ is normal, $y$ is also normal. By \autoref{proposition:functional-calculus-exp},,
+    \[
+    y^*y = \exp(-ix^* + ix) = \exp(-ix + ix) = 1
+    \]
+    
+    so $y$ is unitary. By \autoref{proposition:unitary-spectrum} and the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-holomorphic}, $\exp(i\sigma_A(x)) = \sigma_A(y) \subset \partial B_\complex(0, 1)$. Thus $i\sigma_A(x) \subset \bracs{\text{Re} = 0}$, and $\sigma_A(x) \subset \real$. 
+\end{proof}
+
 
 
 

src/op/c-star/unitary.tex

@@ -0,0 +1,55 @@
+\section{Unitary Elements}
+\label{section:unitary-c-star}
+
+\begin{definition}[Unitary]
+\label{definition:unitary-element}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$, then $x$ is \textbf{unitary} if $x \in G(A)$ and $x^* = x^{-1}$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:unitary-unit}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be unitary, then $\norm{x}_A = 1$. 
+\end{lemma}
+\begin{proof}
+    $\normn{x^2}_A = \norm{xx^*}_A = \norm{1}_A = 1$. 
+\end{proof}
+
+\begin{definition}[Unitarily Equivalent]
+\label{definition:unitary-equivalent}
+    Let $A$ be a unital $C^*$-algebra and $x, y \in A$, then $x$ and $y$ are \textbf{unitarily equivalent} if there exists a unitary element $u \in A$ such that $x = uyu^*$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:unitary-equivalent-same-stuff}
+    Let $A$ be a unital $C^*$-algebra and $x, y \in A$ be unitarily equivalent, then:
+    \begin{enumerate}
+        \item $\norm{x}_A = \norm{y}_A$. 
+        \item $\sigma_A(x) = \sigma_A(y)$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Let $u \in A$ be unitary such that $x = uyu^*$, then $\norm{x}_A \le \norm{u}_A\norm{x}_A\normn{u^*}_A$. By \autoref{lemma:unitary-unit} and (1) of \autoref{proposition:c-star-algebra-gymnastics}, $\norm{u}_A = \normn{u^*}_A = 1$, so $\norm{x}_A \le \norm{y}_A$. As the argument is symmetric, $\norm{x}_A = \norm{y}_A$. 
+
+    (2): Let $\lambda \in \complex$, then
+    \[
+    u(y - \lambda)u^* = uyu^* - \lambda uu^* = uyu^* - \lambda = x - \lambda
+    \]
+
+    Since $u, u^* \in G(A)$, $x - \lambda \in G(A)$ if and only if $y - \lambda \in G(A)$. Therefore $\sigma_A(x) = \sigma_A(y)$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:unitary-spectrum}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be unitary, then $\sigma_A(x) \subset \partial B_\complex(0, 1)$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition II.8.2]{Zhu}}}. ]
+    By \autoref{lemma:unitary-unit}, $\norm{x}_A = 1$, so $\sigma_A(x) \subset \ol{B_\complex(0, 1)}$. Thus
+    \[
+    \bracsn{\ol{\lambda}|\lambda \in \sigma_A(x)} = \sigma_A(x^*) = \sigma_A(x^{-1}) \subset \ol{\complex \setminus B_\complex(0, 1)}
+    \] 
+
+    by (4) of \autoref{proposition:c-star-algebra-gymnastics}. For any $\lambda \in \complex$, $\lambda \in \ol{B_\complex(0, 1)}$ and $\ol \lambda \in \ol{\complex \setminus B_\complex(0, 1)}$ if and only if $|\lambda| = 1$. Therefore $\sigma_A(x) \subset \partial B_\complex(0, 1)$. 
+\end{proof}
+
+
+