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\section{The Borel Functional Calculus}
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\label{section:borel-functional-calculus}
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\begin{definition}[Spectral Measure]
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\label{definition:spectral-measure}
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Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, and $E: \cb_X \to B(H)$, then $E$ is a \textbf{spectral measure relative to $H$} if:
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\begin{enumerate}
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\item For each $B \in \cb_X$, $E(B)$ is an orthogonal projection.
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\item $E(\emptyset) = 0$, $E(X) = I_{B(H)}$.
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\item For each $B, C \in \cb_X$, $E(B \cap C) = E(B)E(C)$.
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\item For each $x, y \in H$, the mapping
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\[
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E_{x, y}: \cb_X \to \complex \quad B \mapsto \dpn{E(B)x, y}{H}
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\]
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is a complex Radon measure on $X$.
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\end{enumerate}
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\end{definition}
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\begin{definition}[Integration Against a Spectral Measure]
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\label{definition:spectral-measure-integral}
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Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, and $E: \cb_X \to B(H)$ be a spectral measure. Define
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\[
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I_E: C(X; \complex)^{**} \to B(H) \quad \phi \mapsto \int_X \phi dE
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\]
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where for each $x, y \in H$, $\angles{I_E(\phi) \cdot x, y}_{H} = \dpn{E_{x, y}, \phi}{C(X; \complex)^*}$, then:
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\begin{enumerate}
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\item $I_E$ is continuous from the weak*-topology on $C(X; \complex)^{**}$ to the weak operator topology on $B(H)$.
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\item $I_E$ is a unital *-homomorphism.
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\end{enumerate}
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For any $\phi \in C(X; \complex)^{**}$, $I_E(\phi) = \int_X \phi dE$ is the \textbf{integral} of $\phi$ with respect to $E$.
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\end{definition}
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\begin{proof}
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Firstly, let $x, y \in H$, $\seqf{B_j} \subset \cb_X$ be disjoint Borel sets, and $B = \bigsqcup_{j = 1}^n B_j$, then for each $1 \le i < j \le n$, $E(B_i)(H) \perp E(B_j)(H)$, so by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz} and the \hyperref[Pythagorean Theorem]{theorem:pythagoras},
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\begin{align*}
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\sum_{j = 1}^n |\dpn{E(B_j)x, y}{H}| &= \sum_{j = 1}^n |\dpn{E(B_j)x, E(B_j)y}{H}| \\
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&\le \sum_{j = 1}^n \norm{E(B_j)x}_H \norm{E(B_j)y}_H \\
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&\le \braks{\sum_{j = 1}^n \norm{E(B_j)x}_H^2}^{1/2} \cdot \braks{\sum_{j = 1}^n \norm{E(B_j)y}_H^2}^{1/2} \\
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&= \norm{E(B)x}_H \cdot \norm{E(B)y}_H \le \norm{x}_H \cdot \norm{y}_H
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\end{align*}
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As the above holds for all finite sequences of disjoint Borel sets, $\norm{E_{x, y}}_{C(X; \complex)^*} \le \norm{x}_H \norm{y}_H$. Thus for any $\phi \in C(X; \complex)^{**}$,
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\begin{align*}
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|\dpn{I_E(\phi) \cdot x, y}{H}| &= |\dpn{E_{x, y}, \phi}{C(X; \complex)^*}| \le \norm{E_{x, y}}_{C(X; \complex)^*} \cdot \norm{\phi}_{C(X; \complex)^{**}} \\
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&\le \norm{\phi}_{C(X; \complex)^{**}} \cdot \norm{x}_H \cdot \norm{y}_H
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\end{align*}
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Since the above holds for all $x, y \in H$, $I_E(\phi) \in B(H)$ with $\norm{I_E(\phi)}_{B(H)} \le \norm{\phi}_{C(X; \complex)^{**}}$.
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(1): For each $x, y \in H$, $E_{x, y} \in C(X; \complex)^*$. Since $\angles{\int \phi dE \cdot x, y}_{H} = \dpn{E_{x, y}, \phi}{C(X; \complex)^*}$ for every $\phi \in C(X; \complex)^{**}$, $I_E$ is continuous from the weak* topology on $C(X; \complex)^{**}$ to the weak operator topology on $B(H)$.
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(2): By \autoref{lemma:separable-metric-space-approx-identity}, the simple functions $\Sigma(X; \complex)$ are uniformly dense in the bounded Borel functions $B^\infty(X; \complex)$. Since
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\begin{enumerate}[label=(\roman*)]
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\item $I_E$ restricted to $\Sigma(X; \complex)$ is a *-homomorphism.
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\item Multiplication and conjugation are continuous in the uniform norm on $B^\infty(X; \complex)$
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\item Composition and transposition are continuous in the operator norm on $B(H)$
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\end{enumerate}
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the map $I_E$ restricted to $B^\infty(X; \complex)$ is a *-homomorphism by continuity. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(X; \complex) \subset B^\infty(X; \complex)$ is weak*-dense in $C(X; \complex)^{**}$. So as
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\begin{enumerate}[label=(\roman*)]
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\item $I_E$ restricted to $B^\infty(X; \complex)$ is a *-homomorphism.
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\item The involution $\phi \mapsto \ol \phi$ is weak*-continuous on $C(X; \complex)^{**}$.
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\item The transpose $T \mapsto T^*$ is weak-operator continuous on $B(H)$.
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\item The product $(\phi, \psi) \mapsto \phi \psi$ is separately weak*-continuous on $C(X; \complex)^{**}$.
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\item The composition $(S, T) \mapsto ST$ is separately weak-operator continuous on $B(H)$.
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\end{enumerate}
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the map $I_E$ is a *-homomorphism by (1). Finally, since $E(X) = I_{B(H)}$, $I_E$ is a unital *-homomorphism.
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\end{proof}
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\begin{theorem}[Spectral Theorem (I)]
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\label{theorem:spectral-theorem-vn-1}
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Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, then:
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\begin{enumerate}
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\item There exists a unique spectral measure $E: \cb_{\Omega(A)} \to B(H)$ such that
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\[
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T = \int_{\Omega(A)} \Gamma_A T dE \quad \forall T \in A
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\]
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\item Let $B \subset B(H)$ be the strong-operator closure of $A$ in $B(H)$, then
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\[
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B = \bracs{\int_{\Omega(A)} \phi dE \bigg | \phi \in C(\Omega(A); \complex)^{**}}
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\]
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\end{enumerate}
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The mapping $C(\Omega(A); \complex)^{**} \to B$ defined by $\phi \mapsto \int \phi dE$ is the \textbf{extended inverse Gelfand transform} of $A$.
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem 20.2]{Zhu}}}. ]
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(1): By the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, $\Gamma_A: A \to C(\Omega(A); \complex)$ is a unital *-isomorphism. For each $x, y \in H$, $\Gamma_A^{-1}$ induces a mapping
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\[
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E_{x, y}: C(\Omega(A); \complex) \to \complex \quad \dpn{f, E_{x, y}}{C(\Omega(A); \complex)} = \dpn{\Gamma_A^{-1}f \cdot x, y}{H}
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\]
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which, by the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, takes the form of a complex Radon measure on $\Omega(A)$. Thus by the uniqueness part of the Riesz Representation Theorem, such a spectral measure must be unique if it exists.
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Since $\norm{E_{x, y}}_{C(\Omega(A); \complex)^*} \le \norm{x}_H\norm{y}_H$ for all $x, y \in H$, $\bracsn{E_{x, y}|x, y \in H}$ induces a bounded linear map
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\[
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I: C(\Omega(A); \complex)^{**} \to B(H) \quad \dpn{I(\phi)x, y}{H} = \dpn{E_{x, y}, \phi}{C(\Omega(A); \complex)^*}
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\]
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with $I(f) = \Gamma_A^{-1}(f)$ for all $f \in C(\Omega(A); \complex)$.
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For any $C \in \cb_{\Omega(A)}$, $\one_C$ is a projection in $B^\infty(\Omega(A); \complex)$. So to see that
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\[
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E: \cb_{\Omega(A)} \to B(H) \quad \dpn{E(C)x, y}{H} = E_{x, y}(C)
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\]
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defines a spectral measure, it is sufficient to show that $I|_{B^\infty(\Omega(A); \complex)}$ is a *-homomorphism.
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Let $x, y \in H$, then as $\Gamma_A$ is a *-isomorphism, for any $f, g \in C(\Omega(A); \complex)$,
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\begin{align*}
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\dpn{fg, E_{x, y}}{C(\Omega(A); \complex)} &= \dpn{\Gamma_A^{-1}f \cdot \Gamma_A^{-1}g \cdot x, y}{H} \\
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&= \dpn{\Gamma_A^{-1}g \cdot x, (\Gamma_A^{-1}f)^* y}{H} = \dpn{g, E_{x, I(f)^*y}}{C(\Omega(A); \complex)}
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\end{align*}
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As the above holds for all $g \in C(\Omega(A); \complex)$, $fE_{x, y} = E_{x, I(f)^*y}$. Now, fix $\phi \in B^\infty(\Omega(A); \complex)$, then for every $f \in C(\Omega(A); \complex)$,
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\begin{align*}
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\dpn{E_{x, y}, \phi f}{C(\Omega(A); \complex)^*} &= \dpn{E_{x, I(f)^*y}, \phi}{C(\Omega(A); \complex)^*} = \dpn{I(\phi)x, I(f)^*y}{H} \\
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&= \dpn{I(f)I(\phi)x, y}{H} = \dpn{f, E_{I(\phi)x, y}}{C(\Omega(A); \complex)}
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\end{align*}
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so $\phi E_{x, y} = E_{I(\phi)x, y}$ for all $\phi \in B^\infty(\Omega(A); \complex)$. Thus for any $\phi, \psi \in B^\infty(\Omega(A); \complex)$,
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\begin{align*}
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\dpn{I(\phi \psi)x, y}{H} &= \dpn{E_{x, y}, \phi \psi}{C(\Omega(A); \complex)^*} = \dpn{E_{I(\psi) x, y}, \phi}{C(\Omega(A); \complex)^*} \\
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&= \dpn{I(\phi)I(\psi)x, y}{H}
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\end{align*}
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and $I|_{B^\infty(\Omega(A); \complex)}$ is a homomorphism\footnote{With the same amount of writing and considerably more mental gymnastics, it can be shown that $I$ is a *-homomorphism on the full space $C(\Omega(A); \complex)^{**}$. However, it is not needed to show that $E$ is a spectral measure, and the homomorphism property falls out at the end anyways.}.
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Finally, let $f \in C(\Omega(A); \real)$, then since $\Gamma_A$ is a *-isomorphism, $I(f) = \Gamma_A^{-1}(f)$ is self-adjoint. As such, for any $x \in H$, $\dpn{f, E_{x, x}}{C(\Omega(A); \complex)} = \dpn{I(f)x, x}{H} \in \real$, so $E_{x, x}$ is real-valued. Thus for any $\phi \in B^\infty(\Omega(A); \real)$ and $x \in H$, $\dpn{I(\phi)x, x}{H} = \dpn{E_{x, x}, \phi}{C(\Omega(A); \complex)^*} \in \real$ as well. Therefore $I(\phi)$ is self-adjoint, and $I|_{B^\infty(\Omega(A); \complex)}$ is a *-homomorphism.
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(2): By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(\Omega(A); \complex)$ is weak*-dense in $C(\Omega(A); \complex)^{**}$. Since the mapping $\phi \mapsto \int_{\Omega(A)}\phi dE$ is continuous from the weak* topology on $C(\Omega(A); \complex)^{**}$ to the weak operator topology on $B(H)$,
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\[
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B \supset \bracs{\int_{\Omega(A)} \phi dE \bigg | \phi \in C(\Omega(A); \complex)^{**}}
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\]
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by \autoref{proposition:closure-of-image}.
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On the other hand, by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, $\ol{B_{C(\Omega(A); \complex)^{**}}(0, 1)}$ is weak*-compact, so
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\[
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S := \bracs{\int_{\Omega(A)} \phi dE \bigg | \phi \in C(\Omega(A); \complex)^{**}, \norm{\phi}_{C(\Omega(A); \complex)^{**}} \le 1}
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\]
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is weak-operator compact. As $\Gamma_A: A \to C(\Omega(A); \complex)$ is an isometric isomorphism, $S \supset \ol{B_A(0, 1)}$. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $S \supset B_B(0, 1)$, and
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\[
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B = \bracs{\int_{\Omega(A)} \phi dE \bigg | \phi \in C(\Omega(A); \complex)^{**}}
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\]
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\end{proof}
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\begin{definition}[Borel Functional Calculus]
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\label{definition:borel-functional-calculus}
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Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $x \in A$ be normal, then there exists a unique continuous unital *-homomorphism
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Let $H$ be a complex Hilbert space, $T \in B(H)$ be normal, and $A \subset B(H)$ be the smallest von Neumann algebra acting on $H$ containing $T$ and $I$, then there exists a unique continuous unital *-homomorphism
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\[
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C(\sigma_A(x); \complex)^{**} \to A[x] \quad f \mapsto f(x)
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C(\sigma_{B(H)}(T); \complex)^{**} \to A \quad \phi \mapsto \phi(T)
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\]
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such that:
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\begin{enumerate}
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\item $\one(x) = 1_A$, $\text{Id}(x) = x$, and $\overline{\text{Id}}(x) = x^*$.
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\item The mapping $f \mapsto f(x)$ is continuous from the weak* topology on $C(\sigma_A(x); \complex)^{**}$ to the strong operator topology on $B(H)$.
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\item $\one(T) = I$, $\text{Id}(T) = T$, $\ol{Id}(T) = T^*$.
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\item The mapping $\phi \mapsto \phi(T)$ is continuous from the weak* topology on $C(\sigma_{B(H)}(T); \complex)^{**}$ to the weak operator topology on $C(\sigma_{B(H)}(T); \complex)$.
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\end{enumerate}
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Moreover, there exists a unique spectral measure $E: \sigma_{B(H)}(T) \to A$
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\end{definition}
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\begin{proof}
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Since the \autoref{definition:continuous-functional-calculus}
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By the \hyperref[Spectral Theorem]{theorem:spectral-theorem-vn-1} applied to $B(H)[T]$, there exists a unique spectral measure $E$ on $\sigma_{B(H)}(T)$ such that the mapping
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\[
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I_E: C(\sigma_{B(H)}(T); \complex)^{**} \to A \quad \phi \mapsto \int_{\sigma_A(T)} \phi dE
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\]
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extends the inverse Gelfand transform $\Gamma_{B(H)[T]}^{-1}: C(\sigma_{B(H)}(T); \complex) \to B(H)[T]$.
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For each $\phi \in C(\sigma_{B(H)}; \complex)^{**}$, let $\phi(T) = \int_{\sigma_{B(H)}(T)}\phi dE$, then the mapping $\phi \mapsto \phi(T)$ is continuous from the weak* topology on $C(\sigma_{B(H)}(T); \complex)^{**}$ to the weak operator topology on $C(\sigma_{B(H)}(T); \complex)$ by \autoref{definition:spectral-measure-integral}.
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Finally, by uniqueness of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, and (2), the mapping $\phi \mapsto \phi(T)$ is unique.
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\end{proof}
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