From 6a53d4d10730cf1cff2806fdd1b03e32d01e6d02 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 11 Aug 2026 13:40:24 -0400 Subject: [PATCH] Added a continuity result in strong operator topology. --- src/fa/lc/beqtensor.tex | 39 ++++++++++++++++++++++++ src/fa/notation.tex | 1 + src/fa/tvs/bilinear.tex | 14 ++++----- src/op/vn/topologies.tex | 64 +++++++++++++++++++++++++++++++++++++++- src/op/vn/vn.tex | 14 +++++++++ 5 files changed, 124 insertions(+), 8 deletions(-) diff --git a/src/fa/lc/beqtensor.tex b/src/fa/lc/beqtensor.tex index c516ead..4e89132 100644 --- a/src/fa/lc/beqtensor.tex +++ b/src/fa/lc/beqtensor.tex @@ -58,5 +58,44 @@ is \textit{the} \textbf{injective cross norm} on $E \otimes_\eps F$. \end{definition} +\begin{definition}[Integral Bilinear Form] +\label{definition:integral-bilinear-form} + Let $E, F$ be locally convex spaces over $K \in \RC$ and $\lambda \in L^2(E, F; K)$ be a bilinear form, then $\lambda$ is \textbf{integral} if there exists equicontinuous subsets $S \subset E^*$ and $T \subset F^*$, and a Radon measure $\mu \in M_R(S \times T; K)$ such that + \[ + \lambda(x, y) = \int_{S \times T} \dpn{x, \phi}{E} \dpn{y, \psi}{F} \mu(d\phi, d\psi) + \] + + for all $x, y \in E$. + + The set $I(E, F)$ is the \textbf{space of integral bilinear forms} on $E$ and $F$. +\end{definition} + + +\begin{theorem} +\label{theorem:injective-dual-bilinear-form} + Let $E, F$ be locally convex spaces over $K \in \RC$, then $(E \wh \otimes_\eps F)^* = I(E, F)$. +\end{theorem} +\begin{proof} + Let $\lambda \in (E \wh \otimes_\eps F)^*$, then there exists equicontinuous subsets $S \subset E^*$ and $T \subset F^*$ such that for each $x \in E$ and $y \in F$, + \[ + |\lambda(x, y)| \le \sup_{\phi \in S}\sup_{\psi \in T} |\dpn{x, \phi}{E} \dpn{y, \psi}{F}| + \] + + For any $(x, y) \in E \times F$ and $(\phi, \psi) \in S \times T$, let $f_{xy}(\phi, \psi) = \dpn{x, \phi}{E} \dpn{y, \psi}{F}$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\Lambda \in C(S \times T; K)^*$ such that the following diagram commutes: + \[ + \xymatrix{ + & C(S \times T; K) \ar@{->}[rd]^{\Lambda} & \\ + E \times F \ar@{->}[ru]^{{(x, y) \mapsto f_{xy}}} \ar@{->}[rr]_{\lambda} & & K + } + \] + + Now, since $S$ and $T$ are equicontinuous, using the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, assume without loss of generality that $S$ and $T$ are weak*-compact. In which case, by the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, there exists $\mu \in M_R(S \times T; K)$ such that $\Lambda(f) = \int_{S \times T} f d\mu$ for all $f \in C(S \times T; K)$. Therefore + \[ + \lambda(x, y) = \Lambda(f_{xy}) = \int_{S \times T}f_{xy} d\mu = \int_{S \times T} \dpn{x, \phi}{E}\dpn{y, \psi}{F} d\mu + \] + + for all $(x, y) \in E \times F$, and $\lambda \in I(E; F)$. +\end{proof} + diff --git a/src/fa/notation.tex b/src/fa/notation.tex index 0e3c24f..c0137da 100644 --- a/src/fa/notation.tex +++ b/src/fa/notation.tex @@ -29,6 +29,7 @@ $E \,\wh{\otimes}_\pi F$ & Projective completion of $E$ and $F$. & \autoref{definition:projective-tensor-product} \\ $p \otimes q$ & Cross seminorm of $p$ and $q$. & \autoref{definition:cross-seminorm} \\ $N(E; F)$ & Nuclear mappings from $E$ to $F$. & \autoref{definition:nuclear-operator-normed} \\ + $I(E, F)$ & Integral bilinear forms on $E \times F$. & \autoref{definition:integral-bilinear-form} \\ % ---- Order Structures ---- $x \vee y$, $x \wedge y$ & $\sup$ and $\inf$ in vector lattice. & \autoref{definition:vector-lattice} \\ $|x|$ & Absolute value $x \vee (-x)$ in a vector lattice. & \autoref{definition:order-absolute-value} \\ diff --git a/src/fa/tvs/bilinear.tex b/src/fa/tvs/bilinear.tex index d418a4e..1b392fe 100644 --- a/src/fa/tvs/bilinear.tex +++ b/src/fa/tvs/bilinear.tex @@ -94,13 +94,13 @@ Now, let $S \in \sigma$, then $\bracsn{\lambda(x, \cdot)|x \in S}$ is equicontinuous by the $\sigma$-hypocontinuity of $\lambda$. For any $U \in \cn_0(G)$, there exists $V \in \cn_0(F)$ such that $\bigcup_{x \in S}\lambda(x, V \cap F_0) \subset U$. By \autoref{proposition:closure-of-image}, $\bigcup_{x \in S}\Lambda(x, \ol V) \subset \ol U$. Thus \autoref{proposition:tvs-good-neighbourhood-base} implies that $\bracsn{\Lambda(x, \cdot)|x \in S}$ is equicontinuous as well. - For each $x_0 \in \ol S$ and $y_0 \in F$, there exists $T \in \tau$ with $y_0 \in \ol T$. As $\Lambda|_{\ol S \times \ol T}$ is the unique continuous extension of $\lambda|_{S \times T}$, $\Lambda(x_0, \cdot)$ is a pointwise limit of elements of $\bracsn{\lambda(x, \cdot)|x \in S}$. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, - \begin{enumerate} - \item $\bracsn{\Lambda(x, \cdot)|x \in \ol S} \subset \ol{\bracsn{\Lambda(x, \cdot)|x \in S}}^{L_s(F; G)}$. - \item $\bracsn{\Lambda(x, \cdot)|x \in \ol S}$ is also equicontinuous. - \end{enumerate} - - so $\Lambda$ is $\ol \sigma$-hypocontinuous. Therefore $\Lambda$ is $(\ol \sigma, \ol \tau)$-hypocontinuous by symmetry. + For each $x_0 \in \ol S$ and $y_0 \in F$, there exists $T \in \tau$ with $y_0 \in \ol T$. As $\Lambda|_{\ol S \times \ol T}$ is the unique continuous extension of $\lambda|_{S \times T}$, $\Lambda(x_0, \cdot)$ is a pointwise limit of elements of $\bracsn{\lambda(x, \cdot)|x \in S}$. Thus + \[ + \bracsn{\Lambda(x, \cdot)|x \in \ol S} \subset \ol{\bracsn{\Lambda(x, \cdot)|x \in S}}^{L_s(F; G)} + \] + + + By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\bracsn{\Lambda(x, \cdot)|x \in \ol S}$ is also equicontinuous, so $\Lambda$ is $\ol \sigma$-hypocontinuous. Therefore $\Lambda$ is $(\ol \sigma, \ol \tau)$-hypocontinuous by symmetry. \end{proof} diff --git a/src/op/vn/topologies.tex b/src/op/vn/topologies.tex index 524c7b9..cc38866 100644 --- a/src/op/vn/topologies.tex +++ b/src/op/vn/topologies.tex @@ -16,6 +16,19 @@ Depending on the topology placed on $H \otimes H$, and the corresponding complet By \autoref{proposition:projective-tensor-product-dual} and the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}. \end{proof} +A natural topology consistent with the ultraweak topology would be the ultrastrong topology. + +\begin{definition}[Ultrastrong Topology] +\label{definition:bh-ultrastrong-topology} + Let $H$ be a complex Hilbert space. For each $x = \seq{x_n} \in L^2(\natp; H)$, let + \[ + \Phi_x: B(H) \to L^2(\natp; H) \quad (\Phi_xT)_n = Tx_n + \] + + then the \textbf{ultrastrong}/\textbf{$\sigma$-strong} topology on $B(H)$ is the topology generated by the maps $\bracsn{\Phi_x|x \in L^2(\natp; H)}$. +\end{definition} + + Seeing that $B(H)$ is a dual Banach space, the following fact is immediate: \begin{proposition} @@ -36,7 +49,6 @@ Now, a few facts about the more familiar operator topologies: \item Every bounded subset of $B(H)$ is relatively compact in the weak operator topology. \item The composition map $(S, T) \mapsto ST$ is separately continuous in the strong and weak operator topologies. \item The composition map $(S, T) \mapsto ST$ is left-hypocontinuous with respect to the strong operator topology and strong-operator bounded subsets of $B(H)$. - \item The adjoint map $T \mapsto T^*$ is continuous in the weak operator topology and the ultraweak topology. \end{enumerate} \end{proposition} \begin{proof} @@ -45,5 +57,55 @@ Now, a few facts about the more familiar operator topologies: (4): By the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, every strong-operator bounded subset of $B(H)$ is equicontinuous. \end{proof} +\begin{proposition} +\label{proposition:bh-adjoint-strong-continuous} + Let $H$ be a complex Hilbert space, then: + \begin{enumerate} + \item The adjoint map $T \mapsto T^*$ is continuous in the weak operator topology and the ultraweak topology. + \item $T \mapsto T^*$ restricted to the normal operators is continuous in the strong operator topology. + \item For any $f \in C(\complex; \complex)$, the mapping $T \mapsto f(T)$ restricted to any bounded set of normal operators is continuous in the strong operator topology. + \end{enumerate} +\end{proposition} +\begin{proof}[Proof, {{\cite[Section 19.1]{Zhu}}}. ] + (2): Let $S, T \in B(H)$, then + \begin{align*} + \normn{(S^* - T^*)x}_H^2 &= \normn{S^*x}_H^2 + \normn{T^*x}_H^2 - \dpn{x, ST^*x}{H} - \dpn{ST^*x, x}{H} \\ + &\le \normn{S^*x}_H^2 + \normn{T^*x}_H^2 - \dpn{x, TT^*x}{H} - \dpn{TT^*x, x}{H} \\ + &+ |\dpn{x, (T - S)T^*x}{H}| + |\dpn{(T - S)T^*x, x}{H}| \\ + &\le |\normn{S^*x}_H^2 - \normn{T^*x}_H^2| + 2\norm{x}_H\normn{(T - S)T^*x}_H + \end{align*} + Now, if $S$ and $T$ are normal, then $\normn{S^*x}_H = \norm{Sx}_H$ and $\norm{T^*x}_H = \norm{Tx}_H$, so + \begin{align*} + |\norm{S^*x}_H^2 - \norm{T^*x}_H^2| &= |\norm{Sx}_H^2 - \norm{Tx}_H^2| \\ + &\le \norm{(S - T)x}_H (\norm{Sx}_H + \norm{Tx}_H) \\ + &\le \norm{(S - T)x}_H (\norm{(S - T)x}_H + 2\norm{Tx}_H) + \end{align*} + + Therefore + \begin{align*} + \normn{(S^* - T^*)x}_H^2 &\le \norm{(S - T)x}_H (\norm{(S - T)x}_H + 2\norm{Tx}_H) \\ + &+ 2\norm{x}_H\normn{(T - S)T^*x}_H + \end{align*} + + and the adjoint map restricted to normal operators is continuous in the strong operator topology. + + (3): Let $S, T \in B_{B(H)}(0, 1)$ and $x \in H$ and $n \in \natp$, then + \begin{align*} + \normn{(S^n - T^n)x}_H &\le \sum_{k = 0}^{n-1}\normn{S^{n-1-k}(S - T)T^kx}_{H} \\ + &\le \sum_{k = 0}^{n - 1}\normn{(S - T)T^kx}_H + \end{align*} + + + so the mapping $T \mapsto T^n$ on $B_{B(H)}(0, 1)$ is continuous in the strong operator topology. By (2), the mapping $T \mapsto p(T, T^*)$ is strong-operator continuous for all $p \in \complex[z, \ol z]$. + + By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exist polynomials $p_n \in \complex[z, \ol z]$ such that $p_n \to f$ uniformly on $\ol{B_\complex(0, 1)}$. For any $T \in B_{B(H)}(0, 1)$, $x \in H$, and $n \in \natp$, + \begin{align*} + \norm{[f(T) - p_n(T)]x}_H &\le \norm{f(T) - p_n(T)}_{B(H)} \cdot \norm{x}_H \\ + &\le \norm{x}_H \cdot \sup_{z \in \ol{B_\complex(0, 1)}}|f(z) - p_n(z)| + \end{align*} + + + by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}. Thus $f$ is a uniform limit of strong-operator continuous functions on $B_{B(H)}(0, 1)$, and as such also strong-operator continuous by \autoref{proposition:uniform-limit-continuous}. +\end{proof} diff --git a/src/op/vn/vn.tex b/src/op/vn/vn.tex index 803b862..8ab09c1 100644 --- a/src/op/vn/vn.tex +++ b/src/op/vn/vn.tex @@ -141,6 +141,20 @@ Now, since $T \in A''$, $TP = PT$ as well, so $M$ is a reducing subspace for $T$. As $A$ is unital, $x \in M$, so $Tx \in M = \ol{\bracsn{Sx|S \in A}}$. \end{proof} +\begin{lemma}[Amplification] +\label{lemma:bh-amplification} + Let $H$ be a complex Hilbert space, $n \in \natp$, and + \[ + \pi: B(H) \to B(H^n) \quad [\pi(T)(x)]_n = Tx_n + \] + + then for each $\seqf{x_j} \subset H$, + \[ + \max_{1 \le j \le n}\norm{Tx_j}_H \le \norm{\pi(T)(x)}_{H^n} \le n \max_{1 \le j \le n}\norm{Tx_j}_H + \] +\end{lemma} + + \begin{theorem}[Von Neumann's Bicommutant Theorem] \label{theorem:bicommutant} Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a unital, self-adjoint subalgebra, then $A''$ is the strong-operator closure of $A$.