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}, 
-    \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}