Added characterisation of positive linear functionals.

src/op/banach/spectrum.tex

@@ -128,7 +128,7 @@
 
 \begin{proposition}
 \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}
         \item $\sigma_A(x + y) \subset \sigma_A(x) + \sigma_A(y)$.
         \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 $\lambda \in \sigma_A(x)$. 
 \end{proof}
 
-\begin{theorem}["Runge's Theorem"]
+\begin{theorem}[Runge]
 \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 $\sigma_A(x) = \sigma_B(x)$. 
 \end{theorem}

src/op/c-star/index.tex

@@ -8,4 +8,5 @@
 \input{./homomorphism.tex}
 \input{./gelfand.tex}
 \input{./cont.tex}
-\input{./order.tex}
+\input{./order.tex}
+\input{./positive.tex}

src/op/c-star/positive.tex

@@ -0,0 +1,31 @@
+\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}
+
+
+

src/op/c-star/sa.tex

@@ -6,7 +6,7 @@
     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 $A = \complex(A_{sa})$, with equivalent norms. 
         \item For each $x \in A$, let
         \[
         \text{Re}(x) = \frac{x + x^*}{2} \quad \text{Im}(x) = \frac{x - x^*}{2i}
@@ -64,7 +64,7 @@
     \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}, $|\lambda| = [x]_{sp} = \norm{x}_A$. 
+    (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}, 
     \[