diff --git a/src/op/vn/fc.tex b/src/op/vn/fc.tex index db6e6f2..5097331 100644 --- a/src/op/vn/fc.tex +++ b/src/op/vn/fc.tex @@ -1,314 +1,5 @@ \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 For each $i \in I$, let $P_i: H \to \ol{A\xi_i}$ be the orthogonal projection onto $\ol{A\xi_i}$, then for any $x, y \in H$, $E_{x, y} = \sum_{i \in I}E_{P_ix, P_iy}$. - \item The natural map $C(\Omega(A); \complex) \to [l^\infty(I); L^\infty(\mu_i; \complex)]$ is injective. Equivalently, $\ol{\bigcup_{i \in I}\supp{\mu_i}} = \Omega(A)$. - \item The space $\mathscr{E}$ is a quotient of $[l^1(I); L^1(\mu_i; \complex)]$ under the mapping - \[ - \mathscr{M}: [l^1(I); L^1(\mu_i; \complex)] \to \mathscr{E} \quad f \mapsto \sum_{i \in I}f_id\mu_i - \] - - and $\mathscr{E}^*$ may be identified as a closed subspace of $[l^\infty(I); L^\infty(\mu_i; \complex)]$ through $\mathscr{M}^*$. - \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 $L^1(\mu_i; \complex)$, $\bracsn{E_{x, y}|x, y \in \ol{A\xi_i}}$ 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 $i, j \in I$ with $i \ne j$, $x \in \ol{A\xi_i}$, $y \in \ol{A\xi_j}$, and $f \in C(\Omega(A); \complex)$, then since $\ol{A\xi_i} \perp \ol{A\xi_j}$, - \[ - \int_{\Omega(A)} f dE_{x, y} = \dpn{\Gamma_A^{-1}(f)x, y}{H} = 0 - \] - - As the above holds for all $f \in C(\Omega(A); \complex)$, $E_{x, y} = 0$. - - Given that $\seqi{\xi}$ is maximal, $x = \sum_{i \in I}P_ix$ for all $x \in H$. Thus for any $x, y \in H$, - \[ - E_{x, y} = \sum_{i, j \in I}E_{P_ix, P_jy} = \sum_{i \in I}E_{P_ix, P_iy} \in \mathscr{M}([l^1(I); L^1(\mu_i; \complex)]) - \] - - (3): Let $T \in A$ with $\Gamma_A T = 0$ $\mu_i$-almost everywhere for all $i \in I$. By (1), $E_{P_ix, P_iy} \ll \mu_i$ for all $i \in I$. Thus for any $x, y \in H$, - \[ - \dpn{Tx, y}{H} = \int_{\Omega(A)}\Gamma_A T dE_{x, y} = \sum_{i \in I}\int_{\Omega(A)}\Gamma_A TdE_{P_ix, P_iy} = 0 - \] - - Therefore $C(\Omega(A); \complex)$ may be identified as a subspace of $[l^\infty(I); L^\infty(\mu_i; \complex)]$. - - (4): By (2), for each $x, y \in H$, $E_{x, y} = \sum_{i \in I}E_{P_ix, P_iy}$. By (1), $E_{P_ix, P_iy} \ll \mu_i$ for all $i \in I$, so $\mathscr{E} \subset \mathscr{M}([l^1(I); L^1(\mu_i; \complex)])$. - - 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 $i \in I$, $\mathscr{E} = \mathscr{M}([l^1(I); L^1(\mu_i; \complex)])$. 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 $\mathscr{M}^*$. - - (5): 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 $L^2(\mu_i; \complex)$. Thus the mapping - \[ - U: H \to [l^2(I); L^2(\mu_i; \complex)] \quad (Ux)_i = U_i(P_ix) - \] - - 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 equivalence 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 $H$ relative to $B$: 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 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 such that $B$ is isomorphic to an $L^\infty$ space on $\Omega(A)$ \cite[Page 24]{FollandHarmonic} \cite[Theorem 23.2]{Zhu}. - \end{enumerate} -\end{remark} - +\label{section:linfty-functional-calculus} @@ -345,3 +36,45 @@ \end{proof} +\begin{theorem} +\label{theorem:vn-projection-norm-dense} + Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then: + \begin{enumerate} + \item For each normal operator $T \in A$ and $f \in B^\infty(\sigma_{B(H)}(T); \complex)$ with $f(0) = 0$, $f(T) \in A$. + \item The linear span of projections in $A$ is norm dense in $A$. + \end{enumerate} +\end{theorem} +\begin{proof} + (1): Let $B \subset \sigma_{B(H)}(T) \setminus \bracs{0}$ be a Borel set. First suppose that $0 \not\in \ol{B}$. By \hyperref[Urysohn's Lemma]{lemma:urysohn}, there exists $f \in C(\sigma_{B(H)}(T); [0, 1])$ with $f(0) = 0$ and $f|_{\ol B} = 1$. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, there exists a net $\angles{g_\gamma}_{\gamma \in C} \subset C(\sigma_{B(H)}(T); \complex)$ such that $g_\gamma \to \one_B$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$. As $f\one_B = \one_B$, $fg_\gamma \to \one_B$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$ as well. + + By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, $h(T) \in A$ for all $h \in C(\sigma_{B(H)}(T); \complex)$ with $h(0) = 0$. In particular, $fg_\gamma(T) \in A$ for all $\gamma \in C$. Thus the \hyperref[$L^\infty$ functional calculus]{definition:linfty-functional-calculus} implies that $\one_B(T) \in A$ as well. + + If $B$ is arbitrary, then $\one_{B \setminus B_\complex(0, r)} \to \one_B$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$ as $r \downto 0$. As $\one_{B \setminus B_{\complex}(0, r)}(T) \in A$ for all $r > 0$, $\one_B(T) \in A$ as well. + + By linearity, $g(T) \in A$ for all $g \in \Sigma(\sigma_{B(H)}(T); \complex)$ with $g(0) = 0$. By \autoref{lemma:separable-metric-space-approx-identity}, $\bracsn{g \in \Sigma(\sigma_{B(H)}(T); \complex)|g(0) = 0}$ is uniformly dense in $\bracsn{f \in B^\infty(\sigma_{B(H)(T)}; \complex)|f(0) = 0}$. Therefore $f(T) \in A$ for all $f \in B^\infty(\sigma_{B(H)}(T); \complex)$ with $f(0) = 0$. +\end{proof} + +\begin{theorem} +\label{theorem:von-neumann-group-connected} + Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then: + \begin{enumerate} + \item $G(A)$ is path-connected in the norm topology. + \item The unitary group of $A$ is path-connected in the norm topology. + \item $I(A)$ is trivial. + \end{enumerate} +\end{theorem} +\begin{proof} + Using \autoref{theorem:existence-of-projections-vna} and after possibly shrinking $H$, assume without loss of generality that $I \in A$. + + After choosing and extending a branch of the complex logarithm, let $\phi: \complex \to \complex$ be a Borel measurable function such that: + \begin{enumerate}[label=(\roman*)] + \item $e^{\phi(z)} = z$ for all $z \in \complex \setminus \bracs{0}$. + \item For each $0 < r < R$, $\phi$ is bounded on the annulus $\ol{B(0, R)} \setminus B(0, r)$. + \end{enumerate} + + (1): Let $T \in G(A)$, then there exists $0 < r < R$ such that $\sigma_A(T) \subset \ol{B(0, R)} \setminus B(0, r)$. In which case, $\phi$ is a bounded Borel measurable function on $\sigma_A(T)$. By the \hyperref[Borel functional calculus]{definition:linfty-functional-calculus}, $\phi(T) \in A$ with $T = e^{\phi(T)}$. In which case, the path $t \mapsto e^{t\phi(T)}$ is a norm-continuous path in $G(A)$ from $I$ to $T$. + + (2): In particular, as $e^{t\phi}(\partial B(0, 1)) \subset \partial B(0, 1)$, the spectrum of $e^{t\phi}$ as an element in the domain of the $L^\infty$ functional calculus, is contained in $\partial B(0, 1)$. Thus if $T$ is unitary, then the path $t \mapsto e^{t\phi(T)}$ lies in the unitary group of $A$ by \autoref{corollary:spectrum-characterisation-iff}. +\end{proof} + + diff --git a/src/op/vn/index.tex b/src/op/vn/index.tex index 72555d3..ae6b71d 100644 --- a/src/op/vn/index.tex +++ b/src/op/vn/index.tex @@ -4,4 +4,5 @@ \input{./topologies.tex} \input{./cayley.tex} \input{./vn.tex} +\input{./spec.tex} \input{./fc.tex} \ No newline at end of file diff --git a/src/op/vn/spec.tex b/src/op/vn/spec.tex new file mode 100644 index 0000000..f4e17aa --- /dev/null +++ b/src/op/vn/spec.tex @@ -0,0 +1,311 @@ +\section{The Spectral Theorem} +\label{section:spectral-theorem} + +\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 $I \in A$, $B \subset B(H)$ 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 For each $i \in I$, let $P_i: H \to \ol{A\xi_i}$ be the orthogonal projection onto $\ol{A\xi_i}$, then for any $x, y \in H$, $E_{x, y} = \sum_{i \in I}E_{P_ix, P_iy}$. + \item The natural map $C(\Omega(A); \complex) \to [l^\infty(I); L^\infty(\mu_i; \complex)]$ is injective. Equivalently, $\ol{\bigcup_{i \in I}\supp{\mu_i}} = \Omega(A)$. + \item The space $\mathscr{E}$ is a quotient of $[l^1(I); L^1(\mu_i; \complex)]$ under the mapping + \[ + \mathscr{M}: [l^1(I); L^1(\mu_i; \complex)] \to \mathscr{E} \quad f \mapsto \sum_{i \in I}f_id\mu_i + \] + + and $\mathscr{E}^*$ may be identified as a closed subspace of $[l^\infty(I); L^\infty(\mu_i; \complex)]$ through $\mathscr{M}^*$. + \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 $L^1(\mu_i; \complex)$, $\bracsn{E_{x, y}|x, y \in \ol{A\xi_i}}$ 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 $i, j \in I$ with $i \ne j$, $x \in \ol{A\xi_i}$, $y \in \ol{A\xi_j}$, and $f \in C(\Omega(A); \complex)$, then since $\ol{A\xi_i} \perp \ol{A\xi_j}$, + \[ + \int_{\Omega(A)} f dE_{x, y} = \dpn{\Gamma_A^{-1}(f)x, y}{H} = 0 + \] + + As the above holds for all $f \in C(\Omega(A); \complex)$, $E_{x, y} = 0$. + + Given that $\seqi{\xi}$ is maximal, $x = \sum_{i \in I}P_ix$ for all $x \in H$. Thus for any $x, y \in H$, + \[ + E_{x, y} = \sum_{i, j \in I}E_{P_ix, P_jy} = \sum_{i \in I}E_{P_ix, P_iy} \in \mathscr{M}([l^1(I); L^1(\mu_i; \complex)]) + \] + + (3): Let $T \in A$ with $\Gamma_A T = 0$ $\mu_i$-almost everywhere for all $i \in I$. By (1), $E_{P_ix, P_iy} \ll \mu_i$ for all $i \in I$. Thus for any $x, y \in H$, + \[ + \dpn{Tx, y}{H} = \int_{\Omega(A)}\Gamma_A T dE_{x, y} = \sum_{i \in I}\int_{\Omega(A)}\Gamma_A TdE_{P_ix, P_iy} = 0 + \] + + Therefore $C(\Omega(A); \complex)$ may be identified as a subspace of $[l^\infty(I); L^\infty(\mu_i; \complex)]$. + + (4): By (2), for each $x, y \in H$, $E_{x, y} = \sum_{i \in I}E_{P_ix, P_iy}$. By (1), $E_{P_ix, P_iy} \ll \mu_i$ for all $i \in I$, so $\mathscr{E} \subset \mathscr{M}([l^1(I); L^1(\mu_i; \complex)])$. + + 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 $i \in I$, $\mathscr{E} = \mathscr{M}([l^1(I); L^1(\mu_i; \complex)])$. 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 $\mathscr{M}^*$. + + (5): 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 $L^2(\mu_i; \complex)$. Thus the mapping + \[ + U: H \to [l^2(I); L^2(\mu_i; \complex)] \quad (Ux)_i = U_i(P_ix) + \] + + 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 equivalence 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 $H$ relative to $B$: 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 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 such that $B$ is isomorphic to an $L^\infty$ space on $\Omega(A)$ \cite[Page 24]{FollandHarmonic} \cite[Theorem 23.2]{Zhu}. + \end{enumerate} +\end{remark} +