From a98ff324ac0834e5ff9e2f6c0470614d9e5dbd45 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Sat, 15 Aug 2026 16:10:55 -0400 Subject: [PATCH] Added another draft of the Borel functional calculus. --- src/op/vn/fc.tex | 117 ++++++++++++++++++++++++----------------------- 1 file changed, 61 insertions(+), 56 deletions(-) diff --git a/src/op/vn/fc.tex b/src/op/vn/fc.tex index cd7ef7c..d2583c2 100644 --- a/src/op/vn/fc.tex +++ b/src/op/vn/fc.tex @@ -23,7 +23,17 @@ 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 Let $\mathscr{E} \subset M_R(X; \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, then for any $\mu \in \mathscr{E}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, $\nu \in \mathscr{E}$ as well. + \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=1] + \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} @@ -45,6 +55,15 @@ 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 $d\nu = f d\mu \in \mathscr{E}$. + + (3): By (2), 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} @@ -73,16 +92,7 @@ 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} - ($C^*$ Structure of $\mathscr{E}^*$): By \autoref{lemma:spectral-measure-properties}, 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. - - (1): Let $\phi \in \mathscr{E}^*$ and $x, y \in H$, then + (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 @@ -119,20 +129,22 @@ \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 + \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 $B(H)$, then + \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 \[ - B = \bracs{\int_{\Omega(A)} \phi dE \bigg | \phi \in C(\Omega(A); \complex)^{**}} + I_E: \mathscr{E}^* \to B \quad \phi \mapsto \int_{\Omega(A)}\phi dE \] + + is a *-isomorphism. \end{enumerate} - The mapping $C(\Omega(A); \complex)^{**} \to B$ defined by $\phi \mapsto \int \phi dE$ is the \textbf{extended inverse Gelfand transform} of $A$. + The mapping $\mathscr{E}^* \to B$ defined by $\phi \mapsto \int \phi dE$ 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 unital *-isomorphism. For each $x, y \in H$, $\Gamma_A^{-1}$ induces a mapping + (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 \[ E_{x, y}: C(\Omega(A); \complex) \to \complex \quad \dpn{f, E_{x, y}}{C(\Omega(A); \complex)} = \dpn{\Gamma_A^{-1}f \cdot x, y}{H} \] @@ -141,84 +153,77 @@ 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 \[ - I_E: C(\Omega(A); \complex)^{**} \to B(H) \quad \dpn{I_E(\phi)x, y}{H} = \dpn{E_{x, y}, \phi}{C(\Omega(A); \complex)^*} + J_E: B^\infty(\Omega(A); \complex) \to B(H) \quad \dpn{J_E(\phi)x, y}{H} = \dpn{E_{x, y}, \phi}{C(\Omega(A); \complex)^*} \] - with $I_E(f) = \Gamma_A^{-1}(f)$ for all $f \in C(\Omega(A); \complex)$. + 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 $I_E|_{B^\infty(\Omega(A); \complex)}$ is a *-homomorphism. + 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, I_E(f)^*y}}{C(\Omega(A); \complex)} + &= \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 $g \in C(\Omega(A); \complex)$, $fE_{x, y} = E_{x, I_E(f)^*y}$. Now, fix $\phi \in B^\infty(\Omega(A); \complex)$, then for every $f \in C(\Omega(A); \complex)$, + 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, I_E(f)^*y}, \phi}{C(\Omega(A); \complex)^*} = \dpn{I_E(\phi)x, I_E(f)^*y}{H} \\ - &= \dpn{I_E(f)I_E(\phi)x, y}{H} = \dpn{f, E_{I_E(\phi)x, y}}{C(\Omega(A); \complex)} + \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 $\phi E_{x, y} = E_{I_E(\phi)x, y}$ for all $\phi \in B^\infty(\Omega(A); \complex)$. Thus for any $\phi, \psi \in B^\infty(\Omega(A); \complex)$, + 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{I_E(\phi \psi)x, y}{H} &= \dpn{E_{x, y}, \phi \psi}{C(\Omega(A); \complex)^*} = \dpn{E_{I_E(\psi) x, y}, \phi}{C(\Omega(A); \complex)^*} \\ - &= \dpn{I_E(\phi)I_E(\psi)x, y}{H} + \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 $I_E|_{B^\infty(\Omega(A); \complex)}$ is a homomorphism\footnote{With the same amount of writing and considerably more mental gymnastics, it can be shown that $I_E$ 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.}. + and $J_E$ is a homomorphism. - Finally, let $f \in C(\Omega(A); \real)$, then since $\Gamma_A$ is a *-isomorphism, $I_E(f) = \Gamma_A^{-1}(f)$ is self-adjoint. As such, for any $x \in H$, $\dpn{f, E_{x, x}}{C(\Omega(A); \complex)} = \dpn{I_E(f)x, x}{H} \in \real$, so $E_{x, x}$ is real-valued. Thus for any $\phi \in B^\infty(\Omega(A); \real)$ and $x \in H$, $\dpn{I_E(\phi)x, x}{H} = \dpn{E_{x, x}, \phi}{C(\Omega(A); \complex)^*} \in \real$ as well. Therefore $I_E(\phi)$ is self-adjoint, and $I_E|_{B^\infty(\Omega(A); \complex)}$ 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 \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 $B(H)$, - \[ - B \supset \bracs{\int_{\Omega(A)} \phi dE \bigg | \phi \in C(\Omega(A); \complex)^{**}} - \] - - by \autoref{proposition:closure-of-image}. + (2): By \autoref{definition:spectral-measure-integral}, $I_E$ is an injective unital *-homomorphism, so it is sufficient to show that $I_E(\mathscr{E}^*) = B$. - On the other hand, by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, $\ol{B_{C(\Omega(A); \complex)^{**}}(0, 1)}$ is weak*-compact, so - \[ - S := \bracs{\int_{\Omega(A)} \phi dE \bigg | \phi \in C(\Omega(A); \complex)^{**}, \norm{\phi}_{C(\Omega(A); \complex)^{**}} \le 1} - \] - - 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 - \[ - B = \bracs{\int_{\Omega(A)} \phi dE \bigg | \phi \in C(\Omega(A); \complex)^{**}} - \] + 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}, $I_E(\ol{B_{\mathscr{E}^*}(0, 1)}) \supset B_B(0, 1)$, and $I_E(\mathscr{E}^*) = B$. \end{proof} \begin{definition}[Borel Functional Calculus] \label{definition:borel-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 there exists a unique unital *-homomorphism - \[ - C(\sigma_{B(H)}(T); \complex)^{**} \to A \quad \phi \mapsto \phi(T) - \] - - such that: + 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 $\one(T) = I$, $\text{Id}(T) = T$, $\ol{\text{Id}}(T) = T^*$. - \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 $A$. - \end{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 + \[ + I_E: \mathscr{E}^* \to A \quad \phi \mapsto \phi(T) := \int_{\sigma_{B(H)}(T)}\phi dE + \] - Moreover, there exists a unique spectral measure $E: \cb_{\sigma_{B(H)}(T)} \to A$ such that $\phi(T) = \int_{\sigma_{B(H)}(T)}\phi dE$ for all $\phi \in C(\sigma_{B(H)}(T); \complex)^{**}$. + 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 mapping $f \mapsto f(T)$ on $\mathscr{E}^*$ is the \textbf{$L^\infty$-functional calculus} of $T$. \end{definition} \begin{proof} - 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 + (1), (2): 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 \[ I_E: C(\sigma_{B(H)}(T); \complex)^{**} \to A \quad \phi \mapsto \int_{\sigma_{B(H)}(T)} \phi dE \] - extends the inverse Gelfand transform $\Gamma_{B(H)[T]}^{-1}: C(\sigma_{B(H)}(T); \complex) \to B(H)[T]$. + 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 $\phi \in C(\sigma_{B(H)}(T); \complex)^{**}$, 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 $A$ by \autoref{definition:spectral-measure-integral}. - 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. + (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}