\section{Self-Adjoint Elements}
\label{section:c-star-self-adjoint}

\begin{definition}[Self-Adjoint]
\label{definition:self-adjoint}
    Let $A$ be an involutive algebra over $\complex$ and $x \in A$, then $x$ is \textbf{self-adjoint} if $x = x^*$. The space $A_{sa} = \bracs{x \in A| x = x^*}$ is the \textbf{self-adjoint part} of $A$, and:
    \begin{enumerate}
        \item $A_{sa}$ is a $\real$ subspace of $A$. 
        \item $A = \complex(A_{sa})$ as a vector space.
        \item For each $x \in A$, let
        \[
        \text{Re}(x) = \frac{x + x^*}{2} \quad \text{Im}(x) = \frac{x - x^*}{2i}
        \]
        
        then $\text{Re}(x), \text{Im}(x) \in A_{sa}^2$ and $x = \text{Re}(x) + i\text{Im}(x)$.
        \item For each $x \in A$, $x^* = \text{Re}(x) - i\text{Im}(x)$. 
    \end{enumerate}
\end{definition}
\begin{proof}
    By \autoref{proposition:complex-conjugation-properties}. 
\end{proof}

\begin{definition}[Normal]
\label{definition:c-star-normal}
    Let $A$ be an involutive algebra over $\complex$ and $x \in A$, then the following are equivalent:
    \begin{enumerate}
        \item $\text{Re}(x)\text{Im}(x) = \text{Im}(x)\text{Re}(x)$. 
        \item $x^*x = xx^*$. 
    \end{enumerate}
    
    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}