Word massaging.

src/op/c-star/positive.tex

@@ -17,14 +17,15 @@
 \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 $E \in \cb_{\Omega(A)}$, 
+    (2) $\Rightarrow$ (1): For each $x \in A_{sa}$, $\sigma_A(x) = \sigma_{A[x]}(x)$ by \autoref{corollary:c-star-algebra-preserve-spectrum}, so $x \ge 0$ in $A$ if and only if $x \ge 0$ in $A[x]$ by \autoref{proposition:positive-spectrum}. In addition, the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark} implies that $x \ge 0$ if and only if $\Gamma_{A[x]}(x) \ge 0$. Moreover, since $1 \in A[x]$, the norm of $\phi$ alongside (2) is preserved by restricting to $A[x]$. By considering each commutative subalgebra, assume without loss of generality that $A$ is commutative. 
+    
+    By the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, $\phi$ takes the form of a complex 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}