Added some spectrum gruntwork.

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}
+