Minor cleanup.

src/op/c-star/state.tex

@@ -90,6 +90,34 @@
 \end{proof}
 
 
+\begin{corollary}
+\label{corollary:cstar-positive-property-probe}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then\footnote{The crude bound seems kind of tragic, but it wouldn't be true otherwise. }
+    \begin{align*}
+    \sigma_A(x) &\subset \bracs{\dpn{x, \phi}{A}|\phi \in P(A)} \\
+    &\subset \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} = \ol{\text{Conv}}(\sigma_A(x))
+    \end{align*}
+    
+
+    In particular, there exists $\phi \in P(A)$ such that $\norm{x}_A = |\dpn{x, \phi}{A}|$. 
+\end{corollary}
+\begin{proof}
+    Let $\lambda \in \sigma_A(x)$. By \autoref{proposition:gelfand-transform-gymnastics}, there exists $\phi \in \Omega(A[x])$ such that $\dpn{x, \phi}{A[x]} = \lambda$. By \autoref{proposition:multiplicative-pure-state}, $\phi \in P(A[x])$. The \hyperref[pure state extension theorem]{theorem:cstar-pure-state-extension} implies that there exists $\Phi \in P(A)$ such that $\Phi|_{A[x]} = \phi$. Thus $\Phi$ is a pure state with $\dpn{x, \Phi}{A} = \lambda$, and $ \sigma_A(x) \subset \bracs{\dpn{x, \Phi}{A}|\Phi \in P(A)}$. 
+
+    Let $\Phi \in S(A)$ and $\phi = \Phi|_{A[x]}$, then $\phi \in S(A[x])$ as well. By the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, the \hyperref[Spectral Theorem]{theorem:spectral-c-star}, and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, $\phi$ takes the form of a Radon probability measure $\mu$ on $\sigma_A(x)$. In which case,
+    \[
+    \dpn{x, \Phi}{A} = \dpn{x, \phi}{A[x]} = \int_{\sigma_A(x)}\lambda \mu(d\lambda) \in \ol{\text{Conv}}(\sigma_A(x))
+    \]
+
+    Finally, since $S(A)$ is compact and convex by \autoref{proposition:state-space-compact-convex}, 
+    \begin{align*}
+        \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} &= \ol{\text{Conv}}(\bracs{\dpn{x, \phi}{A}|\phi \in P(A)}) \\
+        &\subset \ol{\text{Conv}}(\sigma_A(x))
+    \end{align*}
+
+    by \autoref{proposition:compact-extensions} and \autoref{proposition:closure-of-image}. 
+\end{proof}
+
 \begin{theorem}
 \label{theorem:cstar-state-existence}
     Let $A$ be a unital $C^*$-algebra, $x \in A$, and $\lambda \in \sigma_A(x)$, then there exists $\phi \in S(A)$ such that $\dpn{x, \phi}{A} = \lambda$. 
@@ -112,8 +140,15 @@
         \item For each $x \in A$, $x = 0$ if and only if $\dpn{x, \phi}{A} = 0$ for all $\phi \in P(A)$. 
         \item The linear span of $P(A)$ is weak*-dense in $A^*$. 
     \end{enumerate}
+
+    Moreover, for any $x \in A$, 
+    \begin{enumerate}[start=2]
+        \item $x$ is self-adjoint if and only if $\dpn{x, \phi}{A} \in \real$ for all $\phi \in P(A)$. 
+        \item $x$ is positive if and only if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in P(A)$. 
+    \end{enumerate}
+    
 \end{corollary}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Theorem 13.9]{Zhu}}}. ]
     (1): Let $x \in A$ such that $\dpn{x, \phi}{A} = 0$ for all $\phi \in P(A)$. First suppose that $x$ is self-adjoint. By \autoref{theorem:cstar-state-existence}, $\sigma_A(x) = \bracs{0}$, and $\norm{x}_A = [x]_{sp} = 0$ by \autoref{theorem:c-star-normal-spectral-radius}. 
 
     Now suppose that $x$ is arbitrary. In this case, for each $\phi \in P(A)$, 
@@ -124,40 +159,13 @@
     because $\phi$ is Hermitian. Similarly, $\dpn{\text{Im}(x), \phi}{A} = 0$ as well. Thus $\text{Re}(x) = \text{Im}(x) = 0$, and $x = 0$ as well. 
 
     (2): Since the linear span of $P(A)$ separates points in $A$, it is weak*-dense in $A^*$ by \autoref{lemma:duality-dense}. 
-\end{proof}
-
-\begin{corollary}
-\label{corollary:cstar-positive-property-probe}
-    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then\footnote{The crude bound seems kind of tragic, but it wouldn't be true otherwise. }
-    \begin{align*}
-    \sigma_A(x) &\subset \bracs{\dpn{x, \phi}{A}|\phi \in P(A)} \\
-    &\subset \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} = \ol{\text{Conv}}(\sigma_A(x))
-    \end{align*}
-    
-
-    In particular, 
-    \begin{enumerate}
-        \item $x$ is self-adjoint if and only if $\dpn{x, \phi}{A} \in \real$ for all $\phi \in P(A)$. 
-        \item $x$ is positive if and only if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in P(A)$. 
-        \item There exists $\phi \in P(A)$ such that $\norm{x}_A = |\dpn{x, \phi}{A}|$. 
-    \end{enumerate}
-\end{corollary}
-\begin{proof}
-    Let $\lambda \in \sigma_A(x)$. By \autoref{proposition:gelfand-transform-gymnastics}, there exists $\phi \in \Omega(A[x])$ such that $\dpn{x, \phi}{A[x]} = \lambda$. By \autoref{proposition:multiplicative-pure-state}, $\phi \in P(A[x])$. The \hyperref[pure state extension theorem]{theorem:cstar-pure-state-extension} implies that there exists $\Phi \in P(A)$ such that $\Phi|_{A[x]} = \phi$. Thus $\Phi$ is a pure state with $\dpn{x, \Phi}{A} = \lambda$, and $ \sigma_A(x) \subset \bracs{\dpn{x, \Phi}{A}|\Phi \in P(A)}$. 
-
-    Let $\Phi \in S(A)$ and $\phi = \Phi|_{A[x]}$, then $\phi \in S(A[x])$ as well. By the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, the \hyperref[Spectral Theorem]{theorem:spectral-c-star}, and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, $\phi$ takes the form of a Radon probability measure $\mu$ on $\sigma_A(x)$. In which case,
-    \[
-    \dpn{x, \Phi}{A} = \dpn{x, \phi}{A[x]} = \int_{\sigma_A(x)}\lambda \mu(d\lambda) \in \ol{\text{Conv}}(\sigma_A(x))
-    \]
-
-    Finally, since $S(A)$ is compact and convex by \autoref{proposition:state-space-compact-convex}, 
-    \begin{align*}
-        \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} &= \ol{\text{Conv}}(\bracs{\dpn{x, \phi}{A}|\phi \in P(A)}) \\
-        &\subset \ol{\text{Conv}}(\sigma_A(x))
-    \end{align*}
-
-    by \autoref{proposition:compact-extensions} and \autoref{proposition:closure-of-image}. 
-
-    (1), (2): By \autoref{corollary:spectrum-characterisation-iff}. 
+
+    (3): Let $\phi \in P(A)$, then $\phi$ is Hermitian. If $x$ is self-adjoint, then $\dpn{x, \phi}{A} \in \real$. 
+
+    On the other hand, if $\dpn{x, \phi}{A} \in \real$, then $\dpn{x, \phi}{A} = \dpn{x^*, \phi}{A}$, and $\dpn{x - x^*, \phi}{A} =0 $. If this holds for all $\phi \in P(A)$, then $x - x^* = 0$ by (1), and $x$ is self-adjoint. 
+
+    (4): Let $\phi \in P(A)$, then $\phi$ is positive. Thus if $x$ is positive, $\dpn{x, \phi}{A} \ge 0$. 
+
+    On the other hand, if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in P(A)$, then $x$ is self-adjoint by (3). By \autoref{corollary:cstar-positive-property-probe}, $\sigma_A(x) \subset [0, \infty)$. As such, $x$ is positive by \autoref{corollary:spectrum-characterisation-iff}. 
 \end{proof}