Added characterisation of positive linear functionals.

src/fa/order/index.tex

@@ -1,5 +1,7 @@
 \chapter{Order Structures}
 \label{chap:order-structure}
 
+\input{./order.tex}
+\input{./positive.tex}
 \input{./lattice.tex}
 \input{./norm.tex}

src/fa/order/lattice.tex

@@ -1,63 +1,6 @@
 \section{Vector Lattices}
 \label{section:vector-lattice}
 
-\begin{definition}[Ordered Vector Space]
-\label{definition:ordered-vector-space}
-    Let $E$ be a vector space over $\real$ and $\le$ be a partial order on $E$, then $(E, \le)$ is a \textbf{ordered vector space} if
-    \begin{enumerate}
-        \item[(LO1)] For any $x, y, z \in E$ with $x \le y$, $x + z \le y + z$.
-        \item[(LO2)] For any $x, y \in E$ and $\lambda > 0$, $x \le y$ implies that $\lambda x \le \lambda y$.
-    \end{enumerate}
-
-    
-\end{definition}
-
-\begin{proposition}
-\label{proposition:ordered-vector-space-properties}
-    Let $(E, \le)$ be an ordered vector space and $A, B \subset E$ such that $\sup(A)$ and $\sup (B)$ exist, then 
-    \begin{enumerate}
-        \item $\sup(A + B) = \sup(A) + \sup(B)$.
-        \item $\sup(A) = -\inf (-A)$
-    \end{enumerate}
-
-    
-\end{proposition}
-\begin{proof}
-    (1): For any $a \in A$ and $b \in B$, $a + b \le \sup(A) + \sup(B)$ by (LO1). Let $c \in E$ such that $c \ge a + b$ for all $a \in A$ and $b \in B$, then $c \ge a + \sup(B)$ for all $a \in A$, so $c \ge \sup(A) + \sup(B)$. Therefore $\sup(A) + \sup(B) = \sup(A + B)$.
-
-    (2): For any $a \in A$, $\sup(A) \ge a = -(-a)$, so $-\sup(A) \le \inf(-A)$, and $\sup(A) \ge -\inf(-A)$. On the other hand, for any $a \in A$, $-\inf(-A) \ge a$. Therefore $\sup(A) \le -\inf(-A)$. 
-\end{proof}
-
-
-\begin{definition}[Interval]
-\label{definition:ordered-vector-space-interval}
-    Let $(E, \le)$ be an ordered vector space and $x, y \in E$, then
-    \[
-    [x, y] = \bracs{z \in E| x \le z \le y}
-    \]
-
-    is the \textbf{order interval} with endpoints $x$ and $y$.
-\end{definition}
-
-\begin{definition}[Order Bounded]
-\label{definition:ordered-vector-space-bounded}
-    Let $(E, \le)$ be an ordered vector space and $A \subset E$, then $A$ is \textbf{order bounded} if there exists $x, y \in E$ such that $A \subset [x, y]$.
-\end{definition}
-
-\begin{definition}[Order Complete]
-\label{definition:order-vector-complete}
-    Let $(E, \le)$ be an ordered vector space, then $E$ is \textbf{order complete} if for any order bounded set $A \subset E$, $\sup (A)$ and $\inf (A)$ exist.
-\end{definition}
-
-\begin{definition}[Order Bounded Dual]
-\label{definition:order-bounded-dual}
-    Let $(E, \le)$ be an ordered vector space, then the space $E^b$ consisting of all linear functionals on $E$ that are bounded on order bounded sets is the \textbf{order bounded dual} of $E$.
-\end{definition}
-
-\begin{definition}[Order Dual]
-\label{definition:order-dual}
-    Let $(E, \le)$ be an ordered vector space and $\Phi^+ \in \hom(E; \real)$, then $\Phi^+$ is \textbf{positive} if for any $x \in E$ with $x \ge 0$, $\Phi^+(x) \ge 0$. The subspace $E^+ \subset \hom(E; \real)$ generated by the positive linear functionals on $E$ is the \textbf{order dual} of $E$. 
-\end{definition}
 
 
 \begin{definition}[Vector Lattice]
@@ -249,7 +192,7 @@
 \end{proof}
 
 
-\begin{proposition}[{{\cite[V.1.4]{SchaeferWolff}}}]
+\begin{proposition}
 \label{proposition:order-vector-dual}
     Let $(E, \le)$ be an ordered vector space. If for any $x, y \in E$ with $x, y \ge 0$, $[0, x] + [0, y] = [0, x + y]$, then:
     \begin{enumerate}
@@ -264,7 +207,7 @@
 
     
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[V.1.4]{SchaeferWolff}}}. ]
     (1): Let $C = \bracs{x \in E|x \ge 0}$, $\Phi^+ \in E^b$, and
     \[
     \Phi^+: C \to [0, \infty) \quad x \mapsto \sup(f([0, x]))

src/fa/order/order.tex

@@ -0,0 +1,56 @@
+\section{Ordered Vector Spaces}
+\label{section:ovs}
+
+\begin{definition}[Ordered Vector Space]
+\label{definition:ordered-vector-space}
+    Let $E$ be a vector space over $K \in \RC$ and $\le$ be a partial order on $E$, then $(E, \le)$ is a \textbf{ordered vector space} if
+    \begin{enumerate}
+        \item[(LO1)] For any $x, y, z \in E$ with $x \le y$, $x + z \le y + z$.
+        \item[(LO2)] For any $x, y \in E$ and $\lambda > 0$, $x \le y$ implies that $\lambda x \le \lambda y$.
+    \end{enumerate}
+
+    The set $C = \bracs{x \in E|x \ge 0}$ is the \textbf{positive cone} of $E$. 
+\end{definition}
+
+\begin{definition}[Ordered Topological Vector Space]
+\label{definition:ordered-tvs}
+    Let $(E, \le)$ be an ordered vector space over $K \in \RC$, and $\topo$ be a vector space topology on $E$, then the triple $(E, \topo, \le)$ is an \textbf{ordered topological vector space} if the positive cone $C = \bracs{x \in E|x \ge 0}$ is closed. 
+\end{definition}
+
+
+\begin{proposition}
+\label{proposition:ordered-vector-space-properties}
+    Let $(E, \le)$ be an ordered vector space and $A, B \subset E$ such that $\sup(A)$ and $\sup (B)$ exist, then 
+    \begin{enumerate}
+        \item $\sup(A + B) = \sup(A) + \sup(B)$.
+        \item $\sup(A) = -\inf (-A)$
+    \end{enumerate}
+
+    
+\end{proposition}
+\begin{proof}
+    (1): For any $a \in A$ and $b \in B$, $a + b \le \sup(A) + \sup(B)$ by (LO1). Let $c \in E$ such that $c \ge a + b$ for all $a \in A$ and $b \in B$, then $c \ge a + \sup(B)$ for all $a \in A$, so $c \ge \sup(A) + \sup(B)$. Therefore $\sup(A) + \sup(B) = \sup(A + B)$.
+
+    (2): For any $a \in A$, $\sup(A) \ge a = -(-a)$, so $-\sup(A) \le \inf(-A)$, and $\sup(A) \ge -\inf(-A)$. On the other hand, for any $a \in A$, $-\inf(-A) \ge a$. Therefore $\sup(A) \le -\inf(-A)$. 
+\end{proof}
+
+
+\begin{definition}[Interval]
+\label{definition:ordered-vector-space-interval}
+    Let $(E, \le)$ be an ordered vector space and $x, y \in E$, then
+    \[
+    [x, y] = \bracs{z \in E| x \le z \le y}
+    \]
+
+    is the \textbf{order interval} with endpoints $x$ and $y$.
+\end{definition}
+
+\begin{definition}[Order Bounded]
+\label{definition:ordered-vector-space-bounded}
+    Let $(E, \le)$ be an ordered vector space and $A \subset E$, then $A$ is \textbf{order bounded} if there exists $x, y \in E$ such that $A \subset [x, y]$.
+\end{definition}
+
+\begin{definition}[Order Complete]
+\label{definition:order-vector-complete}
+    Let $(E, \le)$ be an ordered vector space, then $E$ is \textbf{order complete} if for any order bounded set $A \subset E$, $\sup (A)$ and $\inf (A)$ exist.
+\end{definition}

src/fa/order/positive.tex

@@ -0,0 +1,34 @@
+\section{The Order Dual}
+\label{section:order-dual}
+
+
+\begin{definition}[Order Bounded Dual]
+\label{definition:order-bounded-dual}
+    Let $(E, \le)$ be an ordered vector space, then the space $E^b$ consisting of all linear functionals on $E$ that are bounded on order bounded sets is the \textbf{order bounded dual} of $E$.
+\end{definition}
+
+
+\begin{definition}[Order Dual]
+\label{definition:order-dual}
+    Let $(E, \le)$ be an ordered vector space over $K \in \RC$ and $\Phi^+ \in \hom(E; K)$, then $\Phi^+$ is \textbf{positive} if for any $x \in E$ with $x \ge 0$, $\text{Re}\dpn{x, \Phi^+}{E} \ge 0$. The subspace $E^+ \subset \hom(E; K)$ generated by the positive linear functionals on $E$ is the \textbf{order dual} of $E$. 
+\end{definition}
+
+
+\begin{theorem}[Bauer-Namioka]
+\label{theorem:bauer-namioka}
+    Let $E$ be an ordered vector space over $K \in \RC$ with positive cone $C$, $\topo$ be a vector space topology on $E$\footnote{The order and the topology need not to be compatible. }, $F \subset E$ be a subspace, and $\phi \in (F, \topo)^*$, then the following are equivalent:
+    \begin{enumerate}
+        \item There exists a continuous positive linear functional $\Phi \in (E, \topo)^*$ such that $\Phi|_F = \phi$. 
+        \item There exists $U \in \cn_\topo(0)$ convex such that
+        \[
+        \sup\bracs{\text{Re}\dpn{x, \phi}{F}|x \in F \cap (U - C)} < \infty
+        \]
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[V.5.4]{SchaeferWolff}}}. ]
+    (1) $\Rightarrow$ (2): Let $U = \bracs{\text{Re}(\Phi) < 1}$, then $U \in \cn_\topo(0)$ is convex. Since $\Phi$ is positive, for any $x \in U$ and $y \in C$, $\text{Re}\dpn{x - y, \Phi}{E} \le \text{Re}\dpn{x, \Phi}{E} < 1$. 
+
+    (2) $\Rightarrow$ (1): Assume without loss of generality that $K = \real$. Let $\alpha > 0$ such that $F \cap (U - C) \subset \bracs{\phi < \alpha}$. Since $U$ is convex and open, and $C$ is convex, $U - C \subset E$ is an open convex set, disjoint from the convex set $\bracs{\phi = \alpha}$. By the \autoref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-1}, there exists $\Phi \in E^*$ such that $U - C \subset \bracs{\Phi < \alpha}$, and $\bracs{\phi = \alpha} \subset \bracs{\Phi \ge \alpha}$. 
+
+    After rescaling, assume without loss of generality that $\bracs{\phi = \alpha} \subset \bracs{\Phi = \alpha}$, then $\Phi \in E^*$ is an extension of $\phi$. For each $x \in C$ and $\lambda > 0$, $-\lambda x \in U - C$, and $-\lambda\dpn{x, \Phi}{E} < \alpha$. As this holds for all $\lambda > 0$, $\dpn{x, \Phi}{E} \ge 0$, so $\Phi$ is the desired extension.
+\end{proof}

src/fa/tvs/complexify.tex

@@ -119,8 +119,7 @@
     The pair $(\complex(E), \iota)$ is the \textbf{complexification} of $E$ as a topological vector space, and
     \begin{enumerate}[start=4]
         \item If $E$ is locally convex, then so is $\complex(E)$. 
-        \item If $E$ is normed, then $\complex(E)$ is normable, and there exists a norm $\norm{\cdot}_{\complex(E)}: \complex(E) \to [0, \infty)$ such that $\iota: E \to \complex(E)$ is isometric. 
+        \item[(F)] For any topological vector space $F$ over $\real$ and continuous $\real$-linear map $T: E \to F$, there exists a unique continuous $\complex$-linear map $\complex(T): \complex(E) \to \complex(F)$ such that the following diagram commutes:
-        \item[(F)] For any vector space $F$ over $\real$ and continuous $\real$-linear map $T: E \to F$, there exists a unique continuous $\complex$-linear map $\complex(T): \complex(E) \to \complex(F)$ such that the following diagram commutes:
         \[
         \xymatrix{
         \mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & \mathbb{C}(F) \\
@@ -132,6 +131,8 @@
         \[
         \complex(T)(x + iy) = Tx + iTy
         \]
+
+        Moreover, if $E$ and $F$ are normed, then $\norm{\complex(T)}_{L(\complex(E); \complex(F))} = \norm{T}_{L(E; F)}$. 
     \end{enumerate}
     
 \end{definition}
@@ -141,25 +142,81 @@
     (U): By (U) of the \hyperref[complexification]{definition:complexification}, there exists a $\complex$-linear map $\complex(T): \complex(E) \to F$ such that the given diagram commutes. Since $T \circ \iota$ and $iT \circ \iota$ are continuous, $T$ is continuous by (U) of the \hyperref[direct sum]{definition:tvs-direct-sum}.
 
     (4): By \autoref{proposition:finite-lc-product}, the direct sum and product of finitely many locally convex spaces coincide. By \autoref{proposition:lc-projective-topology}, this topology is locally convex. 
+    
 
-    (5): Let $\norm{\cdot}_E: E \to [0, \infty)$ be the norm of $E$, and define
+    (F): Existence of $\complex(T)$ is given by (U) applied to $\iota \circ T$. 
+
+    For the isometry, 
+    \begin{align*}
+    \norm{\complex(T)}_{L(\complex(E); \complex(F))} &= \norm{\complex(T)}
+    \end{align*}
+    
+\end{proof}
+
+\begin{definition}[Complexification of Normed Spaces]
+\label{definition:complexification-of-normed-spaces}
+    Let $E$ be a normed vector space over $\real$, then
     \[
     \norm{\cdot}_{\complex(E)}: \complex(E) \to [0, \infty) \quad (x, y) \mapsto \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x + \sin(\theta)y}_E
     \]
 
-    then for any $\phi \in [0, 2\pi]$ and $x, y \in E$, 
+    is a norm on $\complex(E)$ such that the inclusion map $\iota: E \to \complex(E)$ is isometric. 
+    
+    Moreover, for any normed vector space $F$ over $\real$ and $T \in L(E; F)$, $\norm{\complex(T)}_{L(\complex(E); \complex(F))} = \norm{T}_{L(E; F)}$. 
+\end{definition}
+\begin{proof}
+    For any $\phi \in [0, 2\pi]$ and $x, y \in E$, 
     \begin{align*}
         \normn{e^{i \phi}(x, y)}_{\complex(E)} &= \normn{(\cos(\phi)x - \sin(\phi)y, \sin(\phi)x + \cos(\phi)y)}_{\complex(E)} \\
         &= \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta - \phi)x + \sin(\theta - \phi)y}_E \\
         &= \norm{(x, y)}_{\complex(E)}
     \end{align*}
 
-    so $\norm{(x, y)}_{\complex(E)}$ is a norm. For any $x \in E$, 
+    so $\norm{\cdot}_{\complex(E)}$ is a norm on $\complex(E)$. For any $x \in E$, 
     \[
     \norm{\iota x}_{\complex(E)} = \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x}_E = \norm{x}_E \\
     \]
 
     Therefore $\iota: E \to \complex(E)$ is isometric. 
 
-    (F): By (U) applied to $\iota \circ T$.
+    Now, let $F$ be a normed vector space over $\real$ and $T \in L(E; F)$, then
+    \begin{align*}
+        \norm{\complex(T)}_{L(\complex(E); \complex(F))} &= \sup\bracsn{\norm{\complex(T)(x,y)}_{\complex(F)}|(x, y) \in \complex(E), \norm{(x, y)}_{\complex(E)} = 1} \\
+        &\ge \sup\bracsn{\norm{Tx}|x \in E, \norm{x}_E = 1} = \norm{T}_{L(E; F)}
+    \end{align*}
+
+    On the other hand, let $(x, y) \in \complex(E)$, then there exists $\theta \in [0, 2\pi]$ such that
+    \begin{align*}
+    \normn{\complex(T)(x, y)}_{\complex(F)} &= \normn{(Tx, Ty)}_{\complex(F)} = \norm{\cos(\theta)Tx + \sin(\theta)Ty}_F \\
+    &\le \norm{T}_{L(E; F)} \cdot \norm{\cos(\theta)x + \sin(\theta)y}_E \\
+    &\le \norm{T}_{L(E; F)} \cdot \sup_{\phi \in [0, 2\pi]}\norm{\cos(\phi)x + \sin(\phi)y}_E \\
+    &= \norm{T}_{L(E; F)} \cdot \norm{(x, y)}_E
+    \end{align*}
+    
+    
 \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}
+

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/order.tex

@@ -58,8 +58,30 @@
 
 \begin{definition}[Absolute Value]
 \label{definition:absolute-value-c-star}
-    Let $A$ be a $C^*$-algebra and $x \in A$, then $|x| = \sqrt{x^*x}$ is the \textbf{absolute value} of $x$. If $x$ is self-adjoint, then $|x| = 
+    Let $A$ be a unital $C^*$-algebra and $x \in A$, then $|x| = \sqrt{x^*x}$ is the \textbf{absolute value} of $x$. 
 \end{definition}
 
+\begin{definition}[Positive and Negative Parts]
+\label{definition:positive-negative-cstar-algebra}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be self-adjoint, then there exists unique positive elements $x^+, x^- \in A$ such that
+    \begin{enumerate}
+        \item $x = x^+ - x^-$.
+        \item $x^+x^- = x^-x^+ = 0$. 
+    \end{enumerate}
+    
+    The pair $(x^+, x^-)$ are the \textbf{positive and negative parts} of $x$.
+\end{definition}
+\begin{proof}
+    Since $x$ is self-adjoint, $\sigma_A(x) \subset \real$ by \autoref{proposition:self-adjoint-spectrum}. Using the continuous functional calculus, existence is given by the functions $f^+(\lambda) = \lambda \vee 0$ and $f^-(\lambda) = \lambda \wedge 0$ and \autoref{proposition:positive-norm-inequality}. 
+
+    On the other hand, for each $p \in \real[z]$ with $p(0) = 0$, (2) implies that $p(x) = p(x^+) + p(-x^-)$. By the \hyperref[Stone-Weierstrass Theorem]{theorem:stone-weierstrass}, $f(x) = f(x^+) + f(-x^-)$ for all $f \in C(\real; \real)$ with $f(0) = 0$. In particular, (1) then implies that $f^+(x) = f+(x^+) + f^+(-x^-) = f^+(x^+) = x^+$, and likewise $f^-(x) = x^-$. Therefore the decomposition is given uniquely by the continuous functional calculus. 
+\end{proof}
+
+\begin{remark}
+\label{remark:positive-negative-cstar-algebra}
+    The condition in the sign decomposition that $x^+x^- = x^-x^+ = 0$ is essential. Otherwise I may use silly decompositions like $0 = 1 - 1$. 
+\end{remark}
+
+
 
 

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}, 
     \[