Added the continuous functional calculus.

src/op/c-star/cont.tex

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+\section{The Continuous Functional Calculus}
+\label{section:continuous-functional-calculus}
+
+\begin{theorem}[Spectral Theorem for $C^*$-Algebras]
+\label{theorem:spectral-c-star}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then the mapping
+    \[
+    \Phi: \Omega(A[x]) \to \sigma_A(x) \quad \Phi(\psi) = \Gamma_{A[x]}(x)(\psi) = \psi(x)
+    \]
+
+    is a homeomorphism. 
+\end{theorem}
+\begin{proof}
+    Firstly, $A[x]$ is commutative by \autoref{proposition:generated-subalgebra-dense}. Thus \autoref{corollary:c-star-algebra-preserve-spectrum} and (3) of \autoref{proposition:gelfand-transform-gymnastics} imply that
+    \[
+    \Phi(\Omega(A[x])) = \Gamma_{A[x]}(\Omega(A[x])) = \sigma_{A[x]}(x) = \sigma_A(x)
+    \]
+
+    and $\Phi$ is a surjection onto $\sigma_A(x)$. 
+
+    On the other hand, the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark} implies that $\Phi(x^*) = \ol{\Phi(x)}$, so since $A[x]$ is the smallest $C^*$-algebra containing $x$, any element $\psi \in \Omega(A[x])$ is uniquely determined by $\psi(x)$. Therefore $\Phi$ is injective. 
+
+    Finally, since $\Omega(A[x])$ is equipped with the weak* topology and $\Phi$ is the evaluation map at $x$, it is continuous. 
+
+    By \autoref{proposition:compact-hausdorff-homeomorphism}, $\Phi$ is a homeomorphism. 
+\end{proof}
+
+\begin{definition}[Continuous Functional Calculus]
+\label{definition:continuous-functional-calculus}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then there exists a unique continuous unital *-homomorphism
+    \[
+    C(\sigma_A(x); \complex) \to A[x] \quad f \mapsto f(x)
+    \]
+
+    such that:
+    \begin{enumerate}
+        \item $\one(x) = 1_A$. 
+        \item $\text{Id}(x) = x$.
+        \item $\overline{\text{Id}}(x) = x^*$.
+    \end{enumerate}
+
+    Moreover,
+    \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.  
+    \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}. 
+
+    (2): The identification 
+    \[
+    \Phi: \Omega(A[x]) \to \sigma_A(x) \quad \Phi(\psi) = \Gamma_{A[x]}(x)(\psi) = \psi(x)
+    \]
+
+    given by the Spectral Theorem implies that $\Gamma_{A[x]}(x) = \text{Id}$.
+
+    (3): The mapping $f \mapsto f(x)$ is a *-homomorphism. 
+
+    (Uniqueness): By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}.
+\end{proof}
+
+
+