From 6cf96d9803b7b37f8ea39a61ca78abd0d612d917 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Sun, 16 Aug 2026 17:34:03 -0400 Subject: [PATCH] Fixed up spectral. --- src/op/vn/fc.tex | 62 +++++++++++++++++++++++++++++++++++------------- 1 file changed, 45 insertions(+), 17 deletions(-) diff --git a/src/op/vn/fc.tex b/src/op/vn/fc.tex index 5c34431..aff71c1 100644 --- a/src/op/vn/fc.tex +++ b/src/op/vn/fc.tex @@ -201,9 +201,9 @@ \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 $x, y \in \ol{A\xi_i}$. Moreover, the measures $\seqi{\mu}$ are mutually singular. - \item $\mathscr{E} = [l^1(I); L^1(\mu_i; \complex)]$ and $\mathscr{E}^* = [l^\infty(I); L^\infty(\mu_i; \complex)]$. - \item There exists a unitary equivalence $U: H \to [l^2(I); L^2(\mu_i; \complex)]$ between $[l^\infty(I); L^\infty(\mu_i; \complex)]$ acting on $[l^2(I); L^2(\mu_i; \complex)]$ and $B$ acting on $H$, such that for each $i \in I$, $U|_{A\xi_i}$ is an isometry onto the $i$-th factor of $[l^2(I); L^2(\mu_i; \complex)]$. + \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}}}. ] @@ -216,28 +216,33 @@ 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 $E_{x, y} = 0$. In particular, $\seqi{\mu}$ is a mutually singular family of measures, and $[l^1(I); L^1(\mu_i; \complex)]$ may be identified as a subspace of $M_R(\Omega(A); \complex)$. - - (2): For any $x, y \in H$, $\norm{E_{x, y}}_{\text{var}} \le \norm{x}_H \norm{y}_H$ by (1) of \autoref{lemma:spectral-measure-properties}. By (1), $\bracsn{E_{x, y}|x, y \in \ol{A\xi_i}} \subset [l^1(I); L^1(\mu_i; \complex)]$ 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 $x, y \in H$, + 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 $x, y \in H$, \[ - E_{x, y} = \sum_{i, j \in I}E_{P_ix, P_jy} = \sum_{i \in I}E_{P_ix, P_iy} \in [l^1(I); L^1(\mu_i; \complex)] + E_{x, y} = \sum_{i, j \in I}E_{P_ix, P_jy} = \sum_{i \in I}E_{P_ix, P_iy} \in S([l^1(I); L^1(\mu_i; \complex)]) \] - so $\mathscr{E} \subset [l^1(I); L^1(\mu_i; \complex)]$. + 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 - \[ - \mathscr{E} \supset \bracsn{E_{x, y}|x, y \in \ol{A\xi_i}} \supset \bracsn{fdE_{\xi_i, \xi_i}|f \in C(\Omega(A); \complex)} = \bracsn{fd\mu_i|f \in C(\Omega(A); \complex)} - \] + \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 $i \in I$, $\mathscr{E} = [l^1(I); L^1(\mu_i; \complex)]$. By \autoref{theorem:lp-sum-dual}, $[l^\infty(I); L^\infty(\mu_i; \complex)] = \mathscr{E}^*$. + 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 $S\xi_i = T\xi_i$, \[ @@ -265,15 +270,38 @@ 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 $[l^\infty(I); L^\infty(\mu_i; \complex)]$. + \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 $\mathscr{E}^* = [l^\infty(I); L^\infty(\mu_i; \complex)]$ to the weak operator topology on $B$. + \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 $[l^\infty(I); L^\infty(\mu_i; \complex)]$ acting on $[l^2(I); L^2(\mu_i; \complex)]$ and $B$ acting on $H$. + 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} @@ -297,12 +325,12 @@ \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 \[ - I_E: C(\sigma_{B(H)}(T); \complex)^{**} \to A \quad \phi \mapsto \int_{\sigma_{B(H)}(T)} \phi dE + I_E: \mathscr{E}^{*} \to A \quad \phi \mapsto \int_{\sigma_{B(H)}(T)} \phi dE \] 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 $\phi \in C(\sigma_{B(H)}(T); \complex)^{**}$, 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}. + 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}