Finished lecture 11.

src/op/c-star/cont.tex

@@ -49,8 +49,8 @@
         \]
 
         is continuous. 
+        \item Let $B$ be a unital $C^*$-algebra and $\Phi: A \to B$ be a unital *-homomorphism, then for any $x \in A$ and $f \in C(\sigma_A(x); \complex)$, $\Phi(f(x)) = f(\Phi(x))$.
     \end{enumerate}
-    
 \end{definition}
 \begin{proof}
     (1), (4), (5): By the \hyperref[Spectral Theorem]{theorem:spectral-c-star}, $\Omega(A[x])$ and $\sigma_A(x)$ may be identified. For each $f \in C(\sigma_A(x); \complex)$, define $f(x) = \Gamma^{-1}_{A[x]}(f)$, then the mapping $f \mapsto f(x)$ is a unital *-isomorphism by the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}. 
@@ -116,6 +116,30 @@
     \end{enumerate}
 \end{corollary}
 
-
+\begin{corollary}
+\label{corollary:sum-of-unitaries}
+    Let $A$ be a unital $C^*$-algebra, then
+    \begin{enumerate}
+        \item For any $x \in A_{sa}$, there exists a unitary element $u \in A$ such that $x = (u + u^*)/(2\norm{x}_A)$. 
+        \item For any $x \in A$, there exists unitary elements $u, v \in A$ such that $x = (u + u^*)/(\norm{x}_A) + i(v + v^*)/(2\norm{x}_A)$. 
+    \end{enumerate}
+\end{corollary}
+\begin{proof}
+    (1): Assume without loss of generality that $\norm{x}_A \le 1$. In which case, $\sigma(x) \subset [-1, 1]$, and $f(\lambda) = \lambda + i\sqrt{1 - \lambda^2}$ is defined and continuous on $[-1, 1]$. Furthermore, $|f(\lambda)| = 1$ for all $\lambda \in [-1, 1]$. Thus \autoref{corollary:normal-spectrum-identity} implies that $f(x)$ is unitary. Finally, since $f + \ol f = 2\text{Id}$, $x = (f(x) + \ol{f(x)})/(2\norm{x}_A)$.
+\end{proof}
+
+\begin{corollary}
+\label{corollary:star-homomorphism-continuous}
+    Let $A, B$ be unital $C^*$-algebras and $\Phi: A \to B$ be an injective unital *-homomorphism, then for each $x \in A$,
+    \begin{enumerate}
+        \item $\sigma_B(\Phi(x)) = \sigma_A(x)$. 
+        \item $\norm{\Phi(x)}_B = \norm{x}_A$.
+    \end{enumerate}
+\end{corollary}
+\begin{proof}[Proof, {{\cite[II.10.7]{Zhu}}}. ]
+    (1): Since $\Phi(G(A)) \subset G(B)$, $\sigma_B(\Phi(x)) \subset \sigma_A(x)$. If $\sigma_B(\Phi(x)) \subsetneq \sigma_A(x)$, then \hyperref[Urysohn's Lemma]{lemma:urysohn} implies that there exists $C(\sigma_A(x); \complex)$ such that $f|_{\sigma_B(\Phi(x))} = 0$ but $f \ne 0$. In which case, by (7) of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $\Phi(f(x)) = f(\Phi(x)) = 0$, which contradicts the fact that $\Phi$ is injective. 
+
+    (2): By \autoref{corollary:c-star-unique-norm}, $\Phi$ is isometric. 
+\end{proof}
 
 

src/op/c-star/order.tex

@@ -56,4 +56,10 @@
 \end{proof}
 
 
+\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| = 
+\end{definition}
+
+