Added characterisation of positive linear functionals.
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\end{proof} \end{proof}
\begin{proposition}
\label{proposition:hermitian-functional-norm}
Let $E$ be a normed vector space over $\complex$, $*: E \to E$ be a complex conjugation map such that $\norm{x}_E = \normn{x^*}_E$ for all $x \in E$, and $\phi \in E^*$ be a Hermitian functional, then
\[
\norm{\phi}_{E^*} = \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}
\]
\end{proposition}
\begin{proof}
Since $\bracsn{x \in E|x = x^*} \subset E$, $\norm{\phi}_{E^*} \ge \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}$.
On the other hand, let $x \in E$ with $\norm{x}_E = 1$. Assume without loss of generality that $\dpn{x, \phi}{E} \in \real$, then
\begin{align*}
\dpn{x, \phi}{E} &= \dpn{\text{Re}(x), \phi}{E} + \underbrace{i\dpn{\text{Im}(x), \phi}{E}}_{\in \real} =
\dpn{\text{Re}(x), \phi}{E} \\
&\le \norm{\text{Re}(x)}_E \cdot \sup\bracsn{\dpn{y, \phi}{E}|y \in E, y = y^*, \norm{y}_E = 1}
\end{align*}
where $\norm{\text{Re}(x)}_E = \norm{{(x + x^*)}/{2}}_E \le \norm{x}_E$. As the above holds for all $x \in E$,
\[
\norm{\phi}_{E^*} \le \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}
\]
\end{proof}

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\begin{proposition} \begin{proposition}
\label{proposition:commutative-spectrum-gymnastics} \label{proposition:commutative-spectrum-gymnastics}
Let $A$ be a commutative 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 $xy = yx$, then
\begin{enumerate} \begin{enumerate}
\item $\sigma_A(x + y) \subset \sigma_A(x) + \sigma_A(y)$. \item $\sigma_A(x + y) \subset \sigma_A(x) + \sigma_A(y)$.
\item $\sigma_A(xy) \subset \sigma_A(x)\sigma_A(y)$. \item $\sigma_A(xy) \subset \sigma_A(x)\sigma_A(y)$.
@@ -161,7 +161,7 @@
(2): Let $\lambda \in \partial \sigma_B(x)$, then there exists $\seq{\lambda_n} \subset \complex \setminus \sigma_B(x)$ such that $\lambda_n - x \in G(B)$ for all $n \in \natp$, and $\lambda_n \to \lambda$ as $n \to \infty$. By \autoref{corollary:invertible-boundary-explode}, $\norm{(\lambda_n - x)^{-1}}_A \to \infty$ as $n \to \infty$. If $\lambda - x \in G(A)$, then $(\lambda_n - x)^{-1} \to (\lambda - x)^{-1}$ as $n \to \infty$. Thus $\norm{(\lambda - x)^{-1}}_A = \infty$, which is impossible. Therefore $\lambda - x \not\in G(A)$, and $\lambda \in \sigma_A(x)$. (2): Let $\lambda \in \partial \sigma_B(x)$, then there exists $\seq{\lambda_n} \subset \complex \setminus \sigma_B(x)$ such that $\lambda_n - x \in G(B)$ for all $n \in \natp$, and $\lambda_n \to \lambda$ as $n \to \infty$. By \autoref{corollary:invertible-boundary-explode}, $\norm{(\lambda_n - x)^{-1}}_A \to \infty$ as $n \to \infty$. If $\lambda - x \in G(A)$, then $(\lambda_n - x)^{-1} \to (\lambda - x)^{-1}$ as $n \to \infty$. Thus $\norm{(\lambda - x)^{-1}}_A = \infty$, which is impossible. Therefore $\lambda - x \not\in G(A)$, and $\lambda \in \sigma_A(x)$.
\end{proof} \end{proof}
\begin{theorem}["Runge's Theorem"] \begin{theorem}[Runge]
\label{theorem:spectrum-subalgebra-sufficiency} \label{theorem:spectrum-subalgebra-sufficiency}
Let $A$ be a unital Banach algebra, $x \in A$, $P \subset \complex \setminus \sigma_A(x)$ such that $P$ intersects every bounded component of $\complex \setminus \sigma_A(x)$, and $B \subset A$ be a closed algebra containing $1$, $x$, and $\bracsn{(\lambda - x)^{-1}|\lambda \in P}$, then $\sigma_A(x) = \sigma_B(x)$. Let $A$ be a unital Banach algebra, $x \in A$, $P \subset \complex \setminus \sigma_A(x)$ such that $P$ intersects every bounded component of $\complex \setminus \sigma_A(x)$, and $B \subset A$ be a closed algebra containing $1$, $x$, and $\bracsn{(\lambda - x)^{-1}|\lambda \in P}$, then $\sigma_A(x) = \sigma_B(x)$.
\end{theorem} \end{theorem}

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\input{./homomorphism.tex} \input{./homomorphism.tex}
\input{./gelfand.tex} \input{./gelfand.tex}
\input{./cont.tex} \input{./cont.tex}
\input{./order.tex} \input{./order.tex}
\input{./positive.tex}

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\section{Positive Linear Functionals}
\label{section:cstar-positive}
\begin{definition}[Positive Linear Functional]
\label{definition:cstar-positive-functional}
Let $A$ be a unital $C^*$-algebra and $\phi \in A^*$, then $\phi$ is \textbf{positive} if $\dpn{x, \phi}{A} \ge 0$ for all positive elements $x \in A$.
\end{definition}
\begin{theorem}
\label{theorem:cstar-positive-algebraic}
Let $A$ be a unital $C^*$-algebra and $\phi \in \hom(A; \complex)$, then the following are equivalent:
\begin{enumerate}
\item $\phi$ is a positive linear functional.
\item $\phi \in A^*$ with $\normn{\phi}_{A^*} = \dpn{1, \phi}{A}$.
\end{enumerate}
\end{theorem}
\begin{proof}
(1) $\Rightarrow$ (2): For any $x \in A_{sa}$ with $\norm{x}_A \le 1$, $\sigma_A(1 - x) \subset 1 - [-1, 1] = [0, 2]$ by \autoref{proposition:commutative-spectrum-gymnastics}. Thus $1 - x \ge 0$ by \autoref{proposition:positive-spectrum}, and $\dpn{x, \phi}{A} \le \dpn{1, \phi}{A}$. By \autoref{proposition:hermitian-functional-norm}, $\norm{\phi}_{A^*} \le \dpn{1, \phi}{A}$, so $\norm{\phi}_{A^*} = \dpn{1, \phi}{A}$.
(2) $\Rightarrow$ (1): For each $x \in A_{sa}$, by restricting to $A[x]$, assume without loss of generality that $A$ is commutative. By the \autoref{theorem:riesz-radon-c0}, $\phi$ takes the form of a Radon measure $\mu$ on $\Omega(A)$, and $\norm{\phi}_{A^*} = \norm{\mu}_{\text{var}}$. For each Borel set $E \in \cb_{\Omega(A)}$,
\begin{align*}
\norm{\mu}_{\text{var}} &= \int_{\Omega(A)} 1 d\mu = \mu(E) + \mu(\Omega(A) \setminus E) \\
&\le |\mu(E)| + |\mu(\Omega(A) \setminus E)| \le \norm{\mu}_{\text{var}}
\end{align*}
which is only possible if $\mu(E), \mu(\Omega(A) \setminus E) \ge 0$. As this holds for all $E \in \cb_{\Omega(A)}$, $\mu$ is positive, and $\phi$ then is a positive linear functional.
\end{proof}

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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: 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} \begin{enumerate}
\item $A_{sa}$ is a $\real$ subspace of $A$. \item $A_{sa}$ is a $\real$ subspace of $A$.
\item $A = \complex(A_{sa})$ as a vector space. \item $A = \complex(A_{sa})$, with equivalent norms.
\item For each $x \in A$, let \item For each $x \in A$, let
\[ \[
\text{Re}(x) = \frac{x + x^*}{2} \quad \text{Im}(x) = \frac{x - x^*}{2i} \text{Re}(x) = \frac{x + x^*}{2} \quad \text{Im}(x) = \frac{x - x^*}{2i}
@@ -64,7 +64,7 @@
\end{enumerate} \end{enumerate}
\end{corollary} \end{corollary}
\begin{proof} \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}, $|\lambda| = [x]_{sp} = \norm{x}_A$. (1): Since $\sigma_A(x)$ is compact, there exists $\lambda \in \sigma_A(x)$ 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}, (2): By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-holomorphic},
\[ \[