\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{lemma}[Resolvent Equation]
\label{lemma:resolvent-equation}
    Let $A$ be a unital Banach algebra, $x \in A$, and $\lambda, \mu \in \complex \setminus \sigma_A(x)$, then
    \[
    R_x(\lambda) - R_x(\mu) = (\mu - \lambda)R_x(\lambda) R_x(\mu)
    \]
\end{lemma}
\begin{proof}
    \begin{align*}
        [R_x(\lambda) - R_x(\mu)](\mu - x) &= (\lambda - x)^{-1}(\mu - x) - 1 \\
        (\lambda - x)[R_x(\lambda) - R_x(\mu)](\mu - x)&= (\mu - x) - (\lambda - x) = \mu - \lambda \\
        R_x(\lambda) - R_x(\mu) &= (\mu - \lambda)R_x(\lambda)R_x(\mu)
    \end{align*}
\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-Mazur]
\label{theorem:gelfand-mazur}
    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}[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}
\begin{proof}[Proof, {{\cite[Theorem 1.8]{FollandHarmonic}}}. ]
    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}

\begin{proposition}
\label{proposition:spectrum-continuous}
    Let $A$ be a unital Banach algebra and $U \subset \complex$, then $\bracs{x \in A| \sigma_A(x) \subset U}$ is open. 
\end{proposition}
\begin{proof}[Proof, {{\cite[Proposition I.2.9]{Takesaki1}}}. ]
    Let $x \in A$ with $\sigma_A(x) \subset U$, and $\lambda \in U^c$. By \autoref{proposition:banach-algebra-inverse}, for any $y \in A$ with $\norm{y}_A \le \normn{(\lambda - x)^{-1}}_A^{-1}$, $\lambda - x - y \in G(A)$ as well. Since the mapping $\lambda \mapsto \normn{(\lambda - x)^{-1}}_A$ vanishes at infinity and $U^c$ is closed, $\delta = \inf_{\lambda \in U^c}\normn{(\lambda - x)^{-1}}_A^{-1} > 0$. Therefore for every $z \in B_A(x,\delta)$, $\sigma_A(x) \subset U$. 
\end{proof}

\begin{proposition}
\label{proposition:commutative-spectrum-gymnastics}
    Let $A$ be a 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 $\sigma_A(xy) \subset \sigma_A(x)\sigma_A(y)$.
    \end{enumerate}
\end{proposition}
\begin{proof}
    For any $z \in A$, denote $R(z) = \bracs{(\lambda - z)^{-1}|\lambda \in \sigma_A(z)}$. Let $B \subset A$ be the closed subalgebra generated by $1$, $x$, $y$, $R(x)$, $R(y)$, $R(xy)$, $R(x + y)$, then $B$ is a commutative algebra with $\sigma_B(x) = \sigma_A(x)$, $\sigma_B(y) = \sigma_A(y)$, $\sigma_B(xy) = \sigma_A(xy)$, and $\sigma_B(x + y) = \sigma_A(x + y)$. 

    By \autoref{proposition:gelfand-transform-gymnastics}, for each $z \in B$, $(\Gamma_Bz)(\Omega(B)) = \sigma_B(z)$. Since for any $u, v \in C(\Omega(B); \complex)$, 
    \begin{enumerate}
        \item $\sigma_{C(\Omega(B); \complex)}(u + v) \subset \sigma_{C(\Omega(B); \complex)}(u) + \sigma_{C(\Omega(B); \complex)}(v)$.
        \item $\sigma_{C(\Omega(B); \complex)}(uv) \subset \sigma_{C(\Omega(B); \complex)}(u)\sigma_{C(\Omega(B); \complex)}(v)$.
    \end{enumerate}
    
    The above holds for $x$ and $y$ with respect to $\sigma_A$. 
\end{proof}


