Added a continuity result in strong operator topology.

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}
+
 
 

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} \\

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}, 
+    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
-    \begin{enumerate}
+    \[
-        \item $\bracsn{\Lambda(x, \cdot)|x \in \ol S} \subset \ol{\bracsn{\Lambda(x, \cdot)|x \in S}}^{L_s(F; G)}$. 
+    \bracsn{\Lambda(x, \cdot)|x \in \ol S} \subset \ol{\bracsn{\Lambda(x, \cdot)|x \in S}}^{L_s(F; G)}
-        \item $\bracsn{\Lambda(x, \cdot)|x \in \ol S}$ is also equicontinuous. 
+    \]
-    \end{enumerate}
+    
-
+    
-    so $\Lambda$ is $\ol \sigma$-hypocontinuous. Therefore $\Lambda$ is $(\ol \sigma, \ol \tau)$-hypocontinuous by symmetry. 
+    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}
 
 

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}
 

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 $A$.