\section{The $L^\infty$ Functional Calculus} \label{section:borel-functional-calculus} \begin{definition}[Spectral Measure] \label{definition:spectral-measure} 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: \begin{enumerate} \item For each $B \in \cb_X$, $E(B)$ is an orthogonal projection. \item $E(\emptyset) = 0$, $E(X) = I_{B(H)}$. \item For each $B, C \in \cb_X$, $E(B \cap C) = E(B)E(C)$. \item For each $x, y \in H$, the mapping \[ E_{x, y}: \cb_X \to \complex \quad B \mapsto \dpn{E(B)x, y}{H} \] is a complex Radon measure on $X$. \end{enumerate} \end{definition} \begin{lemma} \label{lemma:spectral-measure-properties} Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, and $E: \cb_X \to B(H)$ be a spectral measure relative to $H$, then: \begin{enumerate} \item For each $x, y \in H$, $\norm{E_{x, y}}_{\text{var}} \le \norm{x}_H \norm{y}_H$. \item For each $x \in H$, $E_{x, x}$ is positive. \end{enumerate} Let $\mathscr{E} \subset M_R(X; \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, then \begin{enumerate}[start=2] \item For any $\mu \in \mathscr{E}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, $\nu \in \mathscr{E}$ as well. \item Let \[ J: B^\infty(X; \complex) \to \mathscr{E}^* \quad \dpn{\mu, J(f)}{\mathscr{E}} = \int_X f d\mu \] then $\mathscr{E}^*$ admits a unique weak*-continuous involution and a unique separately weak*-continuous product, such that $\mathscr{E}^*$ is a commutative unital $C^*$-algebra, and $J$ is a unital *-homomorphism. \end{enumerate} \end{lemma} \begin{proof} (1): 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}, \begin{align*} \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}| \\ &\le \sum_{j = 1}^n \norm{E(B_j)x}_H \norm{E(B_j)y}_H \\ &\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} \\ &= \norm{E(B)x}_H \cdot \norm{E(B)y}_H \le \norm{x}_H \cdot \norm{y}_H \end{align*} As the above holds for all finite sequences of disjoint Borel sets, $\norm{E_{x, y}}_{\text{var}} \le \norm{x}_H \norm{y}_H$. (2): For each $B \in \cb_X$, $E(B)$ is a projection, so $E_{x, x}(B) = \dpn{E(B)x, x}{H} \ge 0$. (3): For each $x, y \in H$ and $B, C \in \cb_X$, \[ \int_C \one_B dE_{x, y} = \dpn{E(C \cap B)x, y}{H} = \dpn{E(C)E(B)x, y}{H} = E_{E(B)x, y}(C) \] By linearity, $fdE_{x, y} \in \mathscr{E}$ for all $f \in \Sigma(X; \complex)$. For each $f \in \Sigma(X; \complex)$, the mapping $\mu \mapsto f d\mu$ is continuous in the total variation norm, so $fd\mu \in \mathscr{E}$ for all $\mu \in \mathscr{E}$ and $f \in \Sigma(X; \complex)$. By \autoref{proposition:lp-simple-dense}, $\Sigma(X; \complex)$ is dense in $L^1(\mu; \complex)$ for all $\mu \in \mathscr{E}$. Therefore $fd\mu \in \mathscr{E}$ for all $f \in L^1(\mu; \complex)$ and $\mu \in \mathscr{E}$. Finally, let $\mu \in \mathscr{E}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, then by the \hyperref[Radon-Nikodym Theorem]{theorem:lebesgue-radon-nikodym}, there exists $f \in L^1(\mu; \complex)$ such that $d\nu = f d\mu \in \mathscr{E}$. (4): By (3), for any $\mu \in \mathscr{E}$ and $f \in L^1(\mu; \complex)$, $fd\mu \in \mathscr{E}$ as well. By \autoref{proposition:measures-dual-algebra}, there exists a unique weak*-continuous involution and separately weak*-continuous product on $\mathscr{E}^*$ making $\mathscr{E}^*$ a commutative unital $C^*$-algebra, and $J|_{C(X; \complex)}$ a unital *-homomorphism. Since \begin{enumerate}[label=(\roman*)] \item $J$ is $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-$\sigma(\mathscr{E}^*, \mathscr{E})$ continuous. \item Conjugation on $B^\infty(X; \complex)$ is $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-continuous. \item Multiplication on $B^\infty(X; \complex)$ is separately $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-continuous. \end{enumerate} the mapping $J$ is a unital *-homomorphism. \end{proof} \begin{definition}[Integration Against Spectral Measure] \label{definition:spectral-measure-integral} Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, $E: \cb_X \to B(H)$ be a spectral measure relative to $H$, $\mathscr{E} \subset M_R(X; \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, and \[ J: B^\infty(X; \complex) \to \mathscr{E}^* \quad \dpn{\mu, J(f)}{\mathscr{E}} = \int_X f d\mu \] Then, $\mathscr{E}^*$ admits a unique weak*-continuous involution and a unique separately weak*-continuous product, such that $\mathscr{E}^*$ is a commutative unital $C^*$-algebra, and $J$ is a unital *-homomorphism. For each $\phi \in \mathscr{E}^*$, let $I_E(\phi) \in B(H)$ be the operator defined by \[ \dpn{I_E(\phi) \cdot x, y}{H} = \dpn{E_{x, y}, \phi}{\mathscr{E}} \quad \forall x, y \in H \] then \begin{enumerate} \item $I_E$ is a contraction from $\mathscr{E}^*$ to $B(H)$. \item $I_E$ is continuous from the weak*-topology on $\mathscr{E}^*$ to the weak operator topology on $B(H)$. \item $I_E$ is an injective unital *-homomorphism. \end{enumerate} For any $\phi \in \mathscr{E}^*$, $I_E(\phi) = \int_X \phi dE$ is the \textbf{integral} of $\phi$ with respect to $E$. \end{definition} \begin{proof} (1): Let $\phi \in \mathscr{E}^*$ and $x, y \in H$, then by \autoref{lemma:spectral-measure-properties}, \begin{align*} |\dpn{I_E(\phi) \cdot x, y}{H}| &= |\dpn{E_{x, y}, \phi}{\mathscr{E}}| \le \norm{E_{x, y}}_{\mathscr{E}} \cdot \norm{\phi}_{\mathscr{E}^{*}} \\ &\le \norm{\phi}_{\mathscr{E}^{*}} \cdot \norm{x}_H \cdot \norm{y}_H \end{align*} Since the above holds for all $x, y \in H$, $I_E(\phi) \in B(H)$ with $\norm{I_E(\phi)}_{B(H)} \le \norm{\phi}_{\mathscr{E}^{*}}$. (2): For each $x, y \in H$, $E_{x, y} \in \mathscr{E}$. Since $\angles{\int \phi dE \cdot x, y}_{H} = \dpn{E_{x, y}, \phi}{\mathscr{E}}$ for every $\phi \in \mathscr{E}^{*}$, $I_E$ is continuous from the weak* topology on $\mathscr{E}^{*}$ to the weak operator topology on $B(H)$. (3): 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 \begin{enumerate}[label=(\roman*)] \item $I_E$ restricted to $J(\Sigma(X; \complex))$ is a *-homomorphism. \item Multiplication and conjugation are continuous in the uniform norm on $B^\infty(X; \complex)$ \item Composition and adjunction are continuous in the operator norm on $B(H)$ \end{enumerate} the map $I_E$ restricted to $J(B^\infty(X; \complex))$ 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 $J(C(X; \complex))$ is weak*-dense in $\mathscr{E}^*$. As \begin{enumerate}[label=(\roman*)] \item $I_E$ restricted to $J(B^\infty(X; \complex))$ is a *-homomorphism. \item The involution $\phi \mapsto \ol \phi$ is weak*-continuous on $\mathscr{E}^{*}$. \item The adjunction $T \mapsto T^*$ is weak-operator continuous on $B(H)$. \item The product $(\phi, \psi) \mapsto \phi \psi$ is separately weak*-continuous on $\mathscr{E}^{*}$. \item The composition $(S, T) \mapsto ST$ is separately weak-operator continuous on $B(H)$. \end{enumerate} the map $I_E$ is a *-homomorphism by the weak* to weak-operator continuity established in (2). Since $E(X) = I_{B(H)}$, $I_E$ is a unital *-homomorphism. Finally, let $\phi \in \mathscr{E}^*$ with $I_E(\phi) = 0$, then $\dpn{I_E(\phi)x, y}{H} = \dpn{E_{x, y}, \phi}{\mathscr{E}} = 0$ for all $x, y \in H$. As $\mathscr{E}$ is the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, $\phi = 0$. Therefore $I_E$ is an injective unital *-homomorphism. \end{proof} \begin{theorem}[Spectral Theorem I] \label{theorem:spectral-theorem-vn-1} Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, then: \begin{enumerate} \item There exists a unique spectral measure $E: \cb_{\Omega(A)} \to B(H)$ such that\footnote{Omitting the natural map $C(\Omega(A); \complex) \to \mathscr{E}^*$. } \[ T = \int_{\Omega(A)} \Gamma_A T dE \quad \forall T \in A \] \item Let $B \subset B(H)$ be the strong-operator closure of $A$ in $B(H)$, and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, then \[ I_E: \mathscr{E}^* \to B \quad \phi \mapsto \int_{\Omega(A)}\phi dE \] is a *-isomorphism. \end{enumerate} The measure $E$ is the \textbf{spectral measure associated with $A$}, and the homomorphism $I_E$ is the \textbf{extended inverse Gelfand transform} of $A$. \end{theorem} \begin{proof}[Proof, {{\cite[Theorem 20.2]{Zhu}}}. ] (1): By the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, $\Gamma_A: A \to C(\Omega(A); \complex)$ is a *-isomorphism. For each $x, y \in H$, $\Gamma_A^{-1}$ induces a mapping \[ E_{x, y}: C(\Omega(A); \complex) \to \complex \quad \dpn{f, E_{x, y}}{C(\Omega(A); \complex)} = \dpn{\Gamma_A^{-1}f \cdot x, y}{H} \] 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. 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 \[ J_E: B^\infty(\Omega(A); \complex) \to B(H) \quad \dpn{J_E(\phi)x, y}{H} = \dpn{E_{x, y}, \phi}{C(\Omega(A); \complex)^*} \] with $J_E(f) = \Gamma_A^{-1}(f)$ for all $f \in C(\Omega(A); \complex)$. For any $C \in \cb_{\Omega(A)}$, $\one_C$ is a projection in $B^\infty(\Omega(A); \complex)$. So to see that \[ E: \cb_{\Omega(A)} \to B(H) \quad \dpn{E(C)x, y}{H} = E_{x, y}(C) \] defines a spectral measure, it is sufficient to show that $J_E$ is a *-homomorphism. Let $x, y \in H$, then as $\Gamma_A$ is a *-isomorphism, for any $f, g \in C(\Omega(A); \complex)$, \begin{align*} \dpn{fg, E_{x, y}}{C(\Omega(A); \complex)} &= \dpn{\Gamma_A^{-1}f \cdot \Gamma_A^{-1}g \cdot x, y}{H} \\ &= \dpn{\Gamma_A^{-1}g \cdot x, (\Gamma_A^{-1}f)^* y}{H} = \dpn{g, E_{x, J_E(f)^*y}}{C(\Omega(A); \complex)} \end{align*} As the above holds for all $g \in C(\Omega(A); \complex)$, $fE_{x, y} = E_{x, J_E(f)^*y}$. Now, fix $\phi \in B^\infty(\Omega(A); \complex)$, then for every $f \in C(\Omega(A); \complex)$, \begin{align*} \dpn{E_{x, y}, \phi f}{C(\Omega(A); \complex)^*} &= \dpn{E_{x, J_E(f)^*y}, \phi}{C(\Omega(A); \complex)^*} = \dpn{J_E(\phi)x, J_E(f)^*y}{H} \\ &= \dpn{J_E(f)J_E(\phi)x, y}{H} = \dpn{f, E_{J_E(\phi)x, y}}{C(\Omega(A); \complex)} \end{align*} so $\phi E_{x, y} = E_{J_E(\phi)x, y}$ for all $\phi \in B^\infty(\Omega(A); \complex)$. Thus for any $\phi, \psi \in B^\infty(\Omega(A); \complex)$, \begin{align*} \dpn{J_E(\phi \psi)x, y}{H} &= \dpn{E_{x, y}, \phi \psi}{C(\Omega(A); \complex)^*} = \dpn{E_{J_E(\psi) x, y}, \phi}{C(\Omega(A); \complex)^*} \\ &= \dpn{J_E(\phi)J_E(\psi)x, y}{H} \end{align*} and $J_E$ is a homomorphism. Finally, let $f \in C(\Omega(A); \real)$, then since $\Gamma_A$ is a *-isomorphism, $J_E(f) = \Gamma_A^{-1}(f)$ is self-adjoint. As such, for any $x \in H$, $\dpn{f, E_{x, x}}{C(\Omega(A); \complex)} = \dpn{J_E(f)x, x}{H} \in \real$, so $E_{x, x}$ is real-valued. Thus for any $\phi \in B^\infty(\Omega(A); \real)$ and $x \in H$, $\dpn{J_E(\phi)x, x}{H} = \dpn{E_{x, x}, \phi}{C(\Omega(A); \complex)^*} \in \real$ as well. Therefore $J_E(\phi)$ is self-adjoint, and $J_E$ is a *-homomorphism. (2): By \autoref{definition:spectral-measure-integral}, $I_E$ is an injective unital *-homomorphism, so it is sufficient to show that $I_E(\mathscr{E}^*) = B$. Let $J: C(\Omega(A); \complex) \to \mathscr{E}^*$ be defined by $\dpn{\mu, J(f)}{\mathscr{E}} = \int_{\Omega(A)}f d\mu$ for each $\mu \in \mathscr{E}$ and $f \in C(\Omega(A); \complex)$. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(\Omega(A); \complex)$ is weak*-dense in $C(\Omega(A); \complex)^{**}$, so $J(C(\Omega(A); \complex))$ is weak*-dense in $\mathscr{E}^*$. Since $I_E$ is continuous from the weak* topology on $\mathscr{E}^*$ to the weak operator topology on $B(H)$, $I_E(\mathscr{E}^*) \subset B$ by \autoref{proposition:closure-of-image}. On the other hand, by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, $\ol{B_{\mathscr{E}^*}(0, 1)}$ is weak*-compact, so $I_E(\ol{B_{\mathscr{E}^*}(0, 1)})$ is weak-operator compact. As $\Gamma_A: A \to C(\Omega(A); \complex)$ is an isometric isomorphism, $I_E(\ol{B_{\mathscr{E}^*}(0, 1)}) \supset \ol{B_A(0, 1)}$. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $I_E(\ol{B_{\mathscr{E}^*}(0, 1)}) \supset \ol{B_B(0, 1)}$, and $I_E(\mathscr{E}^*) = B$. \end{proof} \begin{theorem}[Spectral Theorem II] \label{theorem:spectral-theorem-vn-2} Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, $B$ be the von Neumann algebra generated by $A$, $E: \cb_{\Omega(A)} \to B(H)$ be the spectral measure associated with $A$, $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, and $\seqi{\xi} \subset H$ be a maximal family such that the subspaces $\bracsn{A\xi_i|i \in I}$ are mutually orthogonal, then: \begin{enumerate} \item For each $i \in I$, there exists a finite positive Radon measure $\mu_i$ on $\Omega(A)$ such that for every Borel set $C \in \cb_{\Omega(A)}$, $\mu_i(C) = 0$ if and only if $E_{x, y}(C) = 0$ for all $x, y \in \ol{A\xi_i}$. \item $\mathscr{E}$ is a quotient of $[l^1(I); L^1(\mu_i; \complex)]$, and $\mathscr{E}^*$ is a subspace of $[l^\infty(I); L^\infty(\mu_i; \complex)]$. \item There exists a unitary equivalence $U: H \to [l^2(I); L^2(\mu_i; \complex)]$ between $\mathscr{E}^*$ acting on $[l^2(I); L^2(\mu_i; \complex)]$ and $B$ acting on $H$, such that for each $i \in I$, $U|_{\ol{A\xi_i}}$ is an isometry onto the $i$-th factor of $[l^2(I); L^2(\mu_i; \complex)]$. \end{enumerate} \end{theorem} \begin{proof}[Proof, {{\cite[Theorem 1.47]{FollandHarmonic}}}. ] (1): Fix $ i \in I$ and let $\mu_i = E_{\xi_i, \xi_i}$, then for any $C \in \cb_{\Omega(A)}$ with $E_{x, y}(C) = 0$ for all $x, y \in \ol{A\xi_i}$, $\mu_i(C) = 0$. By (1) and (2) of \autoref{lemma:spectral-measure-properties}, $\mu_i$ is a finite positive Radon measure. By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1}, for each $S, T \in A$ and $C \in \cb_{\Omega(A)}$, \[ \dpn{E(C)S\xi_i, T\xi_i}{H} = \int_C \Gamma_AS \cdot \ol{\Gamma_AT} dE_{\xi_i, \xi_i} \] so $\Gamma_AS \cdot \ol{\Gamma_AT}dE_{\xi_i, \xi_i} = dE_{S\xi_i, T\xi_i} \ll \mu_i$. By (1) of \autoref{lemma:spectral-measure-properties} and completeness of $L^1(\mu; \complex)$, $\bracsn{E_{x, y}|x, y \in \ol{A\xi_i}}$ is absolutely continuous with respect to $\mu_i$. Therefore for any $C \in \cb_{\Omega(A)}$, $\mu_i(C) = 0$ if and only if $E_{x, y}(C) = 0$ for all $x, y \in \ol{A \xi_i}$. (2): Let \[ S: [l^1(I); L^1(\mu_i; \complex)] \to M_R(\Omega(A); \complex) \quad S(f) = \sum_{i \in I}f_i d\mu_i \] For any $i, j \in I$ with $i \ne j$, $x \in \ol{A\xi_i}$, $y \in \ol{A\xi_j}$, and $f \in C(\Omega(A); \complex)$, \[ \int_{\Omega(A)} f dE_{x, y} = \dpn{\Gamma_A^{-1}(f)x, y}{H} = 0 \] because $\ol{A\xi_i} \perp \ol{A\xi_j}$, so $E_{x, y} = 0$. By (1), $\bracsn{E_{x, y}|x, y \in \ol{A\xi_i}} \subset S([l^1(I); L^1(\mu_i; \complex)])$ for all $i \in I$. For each $i \in I$, let $P_i \in B(H)$ be the orthogonal projection of $H$ onto $\ol{A\xi_i}$, then as $\seqi{\xi}$ is maximal, $x = \sum_{i \in I}P_ix$ for all $x \in H$. Thus for any $x, y \in H$, \[ E_{x, y} = \sum_{i, j \in I}E_{P_ix, P_jy} = \sum_{i \in I}E_{P_ix, P_iy} \in S([l^1(I); L^1(\mu_i; \complex)]) \] so $\mathscr{E} \subset S([l^1(I); L^1(\mu_i; \complex)])$. On the other hand, for each $i \in I$, since $\mu_i$ is a Radon measure, $C(\Omega(A); \complex)$ is dense in $L^1(\mu_i; \complex)$ by \autoref{proposition:radon-cc-dense}. As \begin{align*} \mathscr{E} &\supset \bracsn{E_{x, y}|x, y \in \ol{A\xi_i}} \supset \bracsn{fdE_{\xi_i, \xi_i}|f \in C(\Omega(A); \complex)} \\ &= \bracsn{fd\mu_i|f \in C(\Omega(A); \complex)} \end{align*} and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ is closed, $\mathscr{E} \supset \bracsn{f d\mu_i|f \in L^1(\mu_i; \complex)}$. Finally, given that the above holds for all $i \in I$, $\mathscr{E} = S([l^1(I); L^1(\mu_i; \complex)])$. By \autoref{theorem:lp-sum-dual} and \autoref{theorem:lp-duality}, $[l^\infty(I); L^\infty(\mu_i; \complex)] = [l^1(I); L^1(\mu_i; \complex)]^*$, so $\mathscr{E}^*$ may be identified with its image under the adjoint of $S$. (3): Fix $i \in I$, then for any $S, T \in A$ with $S\xi_i = T\xi_i$, \[ \Gamma_AS dE_{\xi_i, \xi_i} = E_{S\xi_i, \xi_i} = E_{T\xi_i, \xi_i} = \Gamma_A T dE_{\xi_i, \xi_i} \] so $\Gamma_A S = \Gamma_A T$ $\mu_i$-almost everywhere. Thus the mapping \[ U_i: \ol{A\xi_i} \to L^2(\mu_i; \complex) \quad T\xi_i \mapsto \Gamma_AT \] is well-defined. Moreover, for any $S, T \in A$, \[ \dpn{S\xi_i, T\xi_i}{H} = \int \Gamma_AS \cdot \ol{\Gamma_A T} dE_{\xi_i, \xi_i} = \dpn{\Gamma_A S, \Gamma_A T}{L^2(\mu_i; \complex)} \] so $U_i$ extends into an isometry between $\ol{A\xi_i}$ and $L^2(\mu_i; \complex)$. For each $i \in I$, let $P_i \in B(H)$ be the orthogonal projection of $H$ onto $\ol{A\xi_i}$, then \[ U: H \to [l^2(I); L^2(\mu_i; \complex)] \quad (Ux)_i = U_i(P_ix) \] is an isometry between $H$ and $[l^2(I); L^2(\mu_i; \complex)]$ such that $U(Tx) = \Gamma_AT \cdot Ux$ for all $x \in H$ and $T \in A$. Finally, given that \begin{enumerate}[label=(\roman*)] \item By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(\Omega(A); \complex)$ is weak*-dense in $\mathscr{E}^*$. \item The weak* topology on $[l^\infty(I); L^\infty(\mu_i; \complex)]$ is equal to the weak operator topology of $[l^\infty(I); L^\infty(\mu_i; \complex)]$ acting on $[l^2(I); L^2(\mu_i; \complex)]$. \item $A$ is weak-operator dense in $B$. \item By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1}, the isomorphism $\phi \mapsto \int \phi dE$ is continuous from the weak* topology on $\mathscr{E}^*$ to the weak operator topology on $B$. \end{enumerate} the mapping $U$ is a unitary equivalence between $\mathscr{E}^*$ acting on $[l^2(I); L^2(\mu_i; \complex)]$ and $B$ acting on $H$. \end{proof} \begin{remark} \label{remark:spectral-theorem-vn-2} Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, $B$ be the von Neumann algebra generated by $A$, $E: \cb_{\Omega(A)} \to B(H)$ be the spectral measure associated with $A$, and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$. By \hyperref[Spectral Theorem II]{theorem:spectral-theorem-vn-2}, there exists a decomposable measure space $\Omega$, corresponding to a number of copies of $\Omega(A)$, such that $\mathscr{E}$ is a quotient of its $L^1$ space, $\mathscr{E}^*$ is a subspace of its $L^\infty$ space, and $H$ is isomorphic to its $L^2$ space. The preceding isomorphisms are all linked by a unitary equvalence between $B$ acting on $H$, and $\mathscr{E}^*$ acting on the $l^2$ direct sum. The complexity of $\Omega$, that is, the number of copies of $\Omega(A)$ that it contains, depends on two factors: \begin{enumerate} \item The complexity of the von Neumann algebra $B$: If $B$ is sufficiently complex, then $\mathscr{E}$ cannot be expressed as the $L^1$ space of a single measure on $\Omega(A)$. Instead, multiple copies of $\Omega(A)$ are needed to handle mutually singular measures with overlapping supports. For more details on this phenomenon, see \autoref{theorem:hilbert-measures-dual}. \item The size of the Hilbert space $H$ in comparision with $B$: If $H$ is extremely large, then a large number of vectors are required for $B$ to cover it. As such, many copies of $\Omega(A)$ are required to handle the complexity of $H$. \end{enumerate} More concretely, (1) manifests concretely as the size of the space $\mathscr{E}$, and (2) manifests as the size of the kernel of the mapping $L^1(\Omega) \to \mathscr{E}$. By limiting these two sources of complexity, it is possible to remove the need of multiple copies of $\Omega(A)$. In particular, \begin{enumerate} \item If $B$ admits a cyclic vector, then only one copy of $\Omega(A)$ is required for the construction in the Spectral Theorem \cite[Theorem 23.1]{Zhu}. \item If $H$ is separable, then at most countably many copies of $\Omega(A)$ are required for the construction in the Spectral Theorem. In which case, the measures can be summed to reduce the requirement to just one copy \cite[Page 24]{FollandHarmonic} \cite[Theorem 23.2]{Zhu}. \end{enumerate} \end{remark} \begin{definition}[$L^\infty$ Functional Calculus] \label{definition:linfty-functional-calculus} 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 \begin{enumerate} \item There exists a unique spectral measure $E: \cb_{\sigma_{B(H)}(T)} \to A$ such that \[ T = \int_{\sigma_{B(H)}(T)}\lambda E(d\lambda) \quad T^* = \int_{\sigma_{B(H)}(T)}\ol \lambda E(d\lambda) \] \item Let $\mathscr{E} \subset M_R(\sigma_{B(H)}(T); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, and $A \subset B(H)$ be the von Neumann algebra generated by $I$ and $T$, then \[ I_E: \mathscr{E}^* \to A \quad \phi \mapsto \phi(T) := \int_{\sigma_{B(H)}(T)}\phi dE \] is a *-isomorphism. \item $I_E: \mathscr{E}^* \to A$ is the unique weak* to weak-operator continuous unital *-homomorphism such that $I_E(\text{Id}) = T$. \end{enumerate} The spectral measure $E$ is the \textbf{resolution of the identity} for $T$, and the mapping $f \mapsto f(T)$ on $\mathscr{E}^*$ is the \textbf{$L^\infty$-functional calculus} of $T$. \end{definition} \begin{proof} (1), (2): By \hyperref[Spectral Theorem I]{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 \[ I_E: \mathscr{E}^{*} \to A \quad \phi \mapsto \int_{\sigma_{B(H)}(T)} \phi dE \] is a *-isomorphism that extends the inverse Gelfand transform $\Gamma_{B(H)[T]}^{-1}: C(\sigma_{B(H)}(T); \complex) \to B(H)[T]$. For each $\phi \in \mathscr{E}^{*}$, let $\phi(T) = \int_{\sigma_{B(H)}(T)}\phi dE$, then the mapping $\phi \mapsto \phi(T)$ is continuous from the weak* topology on $\mathscr{E}^{*}$ to the weak operator topology on $A$ by \autoref{definition:spectral-measure-integral}. (3): 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. \end{proof}