Added more facts about the positive square root.

src/op/c-star/cont.tex

@@ -43,6 +43,12 @@
     \begin{enumerate}[start=3]
         \item Under the identification $\Omega(A[x]) = \sigma_A(x)$, for each $f \in C(\sigma_A(x); \complex)$, $f(x) = \Gamma^{-1}_{A[x]}(f)$.
         \item The mapping $f \mapsto f(x)$ is a unital *-isomorphism.  
+        \item For each $U \subset \complex$ open and $f \in C(U; \complex)$, the mapping
+        \[
+        \bracs{x \in A|\sigma_A(x) \subset U} \to A \quad x \mapsto f(x)
+        \]
+
+        is continuous. 
     \end{enumerate}
     
 \end{definition}
@@ -58,8 +64,58 @@
 
     (3): The mapping $f \mapsto f(x)$ is a *-homomorphism. 
 
+    (6): Fix $x \in A$ and let $K \in \cn_U(\sigma_A(x); \complex)$ be compact. By \autoref{proposition:spectrum-continuous}, there exists $r > 0$ such that $\sigma_A(y) \subset K$ for all $y \in B_A(x, r)$. 
+
+    Let $\eps > 0$. By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exists $p \in \complex[z, \ol z]$ such that $\sup_{\lambda \in K}|p(\lambda) - f(\lambda)| < \eps$. In which case, 
+    \begin{align*}
+    \norm{f(x) - f(y)}_A &\le \norm{f(x) - p(x)}_A + \norm{f(y) - p(y)}_A + \norm{p(x) - p(y)}_A \\
+    &\le \norm{p(x) - p(y)}_A + 2\eps
+    \end{align*}
+
+    for all $y \in B_A(x, r)$. Since $p$ is a continuous map on $A$, there exists $\delta \in (0, r)$ such that $\norm{p(x) - p(y)}_A < \eps$ for all $y \in B_A(x, \delta)$. Therefore $\norm{f(x) - f(y)}_A < 3\eps$ for all $y \in B_A(x, \delta)$. 
+
     (Uniqueness): By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}.
 \end{proof}
 
+\begin{theorem}[Spectral Mapping Theorem (Continuous)]
+\label{theorem:spectral-mapping-continuous}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then:
+    \begin{enumerate}
+        \item For every $f \in C(\sigma_A(x); \complex)$, $\sigma_A(f(x)) = f(\sigma_A(x))$. 
+        \item For every $f \in C(\sigma_A(x); \complex)$ and $g \in C(\sigma_A(f(x)); \complex)$, $(f \circ g)(x) = f(g(x))$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Since $f(x) \in A[x]$, by \autoref{proposition:gelfand-transform-gymnastics} and definition of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, 
+    \[
+    \sigma_A(f(x)) = \sigma_{A[x]}(f(x)) = \Gamma_{A[x]}(f(x))(\sigma_A(x)) = f(\sigma_A(x))
+    \]
+
+    (2): Firstly, the theorem holds directly if $f, g \in \complex[z, \ol z]$ are polynomials in $z$ and $\ol z$. 
+
+    Suppose that $g \in \complex[z, \ol z]$ but $f$ is arbitrary. By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exists $\seq{p_n} \subset \complex[z, \ol z]$ such that $p_n \to f$ uniformly on $\sigma_A(x)$. By property (6) of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, 
+    \[
+    (g \circ f)(x) = \limv{n}(g \circ f_n)(x) = \limv{n}g(f_n(x)) = \limv{n}g(f(x))
+    \]
+
+    Finally, suppose that both $f$ and $g$ are arbitary. By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exists $\seq{p_n} \subset \complex[z, \ol z]$ such that $p_n \to g$ uniformly on $\sigma_A(f(x))$. By continuity of the continuous functional calculus, 
+    \[
+    (g \circ f)(x) = \limv{n}(g_n \circ f)(x) = \limv{n}g_n(f(x)) = \limv{n}g(f(x))
+    \]
+\end{proof}
+
+
+
+\begin{corollary}
+\label{corollary:normal-spectrum-identity}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then:
+    \begin{enumerate}
+        \item $x$ is self-adjoint if and only if $\sigma(x) \subset \real$. 
+        \item $x$ is unitary if and only if $\sigma(x) \subset \partial B_\complex(0, 1)$. 
+        \item $x$ is a projection if and only if $\sigma(x) \subset \bracs{0, 1}$. 
+    \end{enumerate}
+\end{corollary}
+
+