Added the spectral mapping theorem.

src/op/banach/fc.tex

@@ -78,9 +78,7 @@
 
     and the mapping $f \mapsto f(x)$ is a homomorphism. 
 
-    (1): Since the constant $1$ function is the identity in $H(\sigma_A(x); \complex)$, $1(x) = 1$ by the homomorphism property. 
-
-    (2): Let $R > 0$ such that $\sigma_A(x) \subset B_\complex(0, R)$, then by \autoref{proposition:rs-complete} and \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula}, 
+    (1), (2): Let $R > 0$ such that $\sigma_A(x) \subset B_\complex(0, R)$, then by \autoref{proposition:rs-complete} and \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula}, 
     \[
         \text{Id}(x) = \frac{1}{2\pi i }\int_{\omega_{0, R}} \frac{1}{z - x}dz = \frac{1}{2\pi i }\sum_{n = 0}^\infty\int_{\omega_{0, R}}  z^{-n-1}x^n dz = 1
     \]
@@ -94,3 +92,36 @@
 
 \end{proof}
 
+\begin{theorem}[Spectral Mapping Theorem (Holomorphic)]
+\label{theorem:spectral-mapping-holomorphic}
+    Let $A$ be a unital Banach algebra and $x \in A$, then:
+    \begin{enumerate}
+        \item For each $f \in H(\sigma_A(x); \complex)$, $\sigma_A(f(x)) = f(\sigma_A(x))$. 
+        \item For each $f \in H(\sigma_A(x); \complex)$ and $g \in H(\sigma_A(f(x)); \complex)$, $(f \circ g)(x) = f(g(x))$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Let $\lambda \in \complex \setminus f(\sigma_A(x))$, then $1/(\lambda - f) \in H(\sigma_A(x); \complex)$. Since the holomorphic functional calculus is a homomorphism, $[1/(\lambda - f)](x) = (\lambda - f(x))^{-1}$. Thus $\sigma_A(f(x)) \subset f(\sigma_A(x))$. 
+
+    On the other hand, let $\lambda \in \sigma_A(x)$, then the function $h(\mu) = [f(\mu) - f(\lambda)]/(\mu - \lambda)$ extends to an element of $H(\sigma_A(x); \complex)$. Therefore by the homomorphism property, 
+    \[
+    f(x) - f(\lambda) = (x - \lambda)h(x) = h(x)(x - \lambda)
+    \]
+
+    Since $(x - \lambda) \not\in G(A)$, $f(x) - f(\lambda) \not\in G(A)$ as well. Therefore $\sigma_A(f(x)) \supset f(\sigma_A(x))$. 
+
+    (2): If additionally $f, g \in \complex(z)$, then the theorem holds by (1) and (2) of \autoref{definition:holomorphic-functional-calculus}. 
+
+    Now suppose that $f \in H(\sigma_A(x); \complex)$ is arbitrary and $g \in \complex(z)$. For each $\eps > 0$, by \autoref{proposition:spectrum-continuous} and continuity of the holomorphic functional calculus, there exists $U \in \cn_{H(\sigma_A(x); \complex)}(f)$ such that for each $h \in U$, 
+    \begin{enumerate}[label=(\alph*)]
+        \item $g \in H(\sigma_A(h(x)); \complex)$.
+        \item $\norm{g(f(x)) - g(h(x))}_A < \eps$. 
+    \end{enumerate}
+    
+    By \hyperref[Runge's Theorem]{corollary:runge-rational-approximation}, there exists $h \in U \cap \complex(z)$. Thus $\norm{g(f(x)) - (g \circ f)(x)}_A < \eps$ as well. As this holds for all $\eps > 0$, $g(f(x)) = (g \circ f)(x)$ for any $f \in H(\sigma_A(x); \complex)$ and $g \in \complex(x) \cap H(f(\sigma_A(x)); \complex)$. 
+    
+    By \hyperref[Runge's Theorem]{corollary:runge-rational-approximation} again and continuity of the holomorphic functional calculus, the above also holds for all $f \in H(\sigma_A(x); \complex)$ and $g \in H(f(\sigma_A(x)); \complex)$. 
+\end{proof}
+
+
+

src/op/banach/spectrum.tex

@@ -118,3 +118,12 @@
     (1): Let $\lambda \in \complex \setminus \bracs{0}$, then by \autoref{proposition:swap-invertible}, $\lambda - xy \in G(A)$ if and only if $\lambda - yx \in G(A)$. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:spectrum-continuous}
+    Let $A$ be a unital Banach algebra and $U \subset \complex$, then $\bracs{x \in A| \sigma_A(x) \subset U}$ is open. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition I.2.9]{Takesaki1}}}. ]
+    Let $x \in A$ with $\sigma_A(x) \subset U$, and $\lambda \in U^c$. By \autoref{proposition:banach-algebra-inverse}, for any $y \in A$ with $\norm{y}_A \le \normn{(\lambda - x)^{-1}}_A^{-1}$, $\lambda - x - y \in G(A)$ as well. Since the mapping $\lambda \mapsto \normn{(\lambda - x)^{-1}}_A$ vanishes at infinity and $U^c$ is closed, $\delta = \inf_{\lambda \in U^c}\normn{(\lambda - x)^{-1}}_A^{-1} > 0$. Therefore for every $z \in B_A(x,\delta)$, $\sigma_A(x) \subset U$. 
+\end{proof}
+
+