Added bilinear forms.

src/fa/lc/continuous.tex

@@ -20,6 +20,7 @@
     $(4) \Rightarrow (3)$: Let $U \in \cn_F(0)$ be convex, circled, and radial, then its gauge $[\cdot]_U$ is a continuous seminorm on $F$ by \autoref{definition:locally-convex}. Thus there exists a continuous seminorm $[\cdot]_E$ such that $[Tx]_U \le [x]_E$. In which case, $V = \bracs{x \in E| [x]_E < 1} \in \cn_E(0)$ with $T(V) \subset U$. Therefore $T$ is continuous at $0$, and continuous by \autoref{definition:continuous-linear}.
 \end{proof}
 
+
 \begin{proposition}
 \label{proposition:tvs-convex-multilinear}
     Let $\seqf{E_j}$ and $F$ be locally convex spaces, and $T: \prod_{j = 1}^n E_j \to F$ be $n$-linear map, then the following are equivalent:

src/fa/tvs/bilinear.tex

@@ -0,0 +1,170 @@
+\section{Bilinear Mappings}
+\label{section:bilinear-tvs}
+
+
+
+\begin{theorem}
+\label{theorem:separate-joint-bilinear}
+    Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg$ be separately continuous bilinear maps from $E \times F$ to $G$. If one of the following holds:
+    \begin{enumerate}
+        \item[(B)] $E$ is Baire.
+        \item[(B')] $E$ is barrelled and $G$ is locally convex. 
+    \end{enumerate}
+
+    and that
+    \begin{enumerate}
+        \item[(M)] $E$ and $F$ are both metrisable. 
+        \item[(E)] For each $x \in E$, $\bracsn{\lambda(x, \cdot)|\lambda \in \alg} \subset L(F; G)$ is equicontinuous. 
+    \end{enumerate}
+
+    then $\alg$ is equicontinuous. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[III.5.1]{SchaeferWolff}}}. ]
+    Let $\seq{(x_n, y_n)} \subset E \times F$ and $\seq{\lambda_n} \subset \alg$ such that $(x_n, y_n) \to 0$ as $n \to \infty$. Since $\seq{y_n}$ is convergent, for each $n \in \natp$ and $x \in E$, $\bracsn{\lambda_n(x, y_n)|n \in \natp}$ is bounded by (E) and \autoref{proposition:equicontinuous-net}. By (B) or (B') and the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, $\bracsn{\lambda_n(\cdot, y_n)|n \in \natp}$ is equicontinuous, and $\lambda_n(x_n, y_n) \to 0$ as $n \to \infty$ by \autoref{proposition:equicontinuous-net}. By (M) and \autoref{proposition:equicontinuous-net}, $\alg$ is equicontinuous at $0$, and hence equicontinuous by \autoref{lemma:equicontinuous-bilinear}. 
+\end{proof}
+
+\begin{definition}[Hypocontinuity]
+\label{definition:hypocontinuity}
+    Let $E, F, G$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal of bounded sets, and $\lambda: E \times F \to G$ be a separately continuous bilinear map, then the following are equivalent:
+    \begin{enumerate}
+        \item For each $S \in \sigma$ and $V \in \cn_G(0)$, there exists $U \in \cn_F(0)$ such that $\lambda(S \times U) \subset V$. 
+        \item For each $S \in \sigma$, $\bracs{\lambda(x, \cdot)|x \in S} \subset F^*$ is equicontinuous. 
+    \end{enumerate}
+    
+    If the above holds, then $\lambda$ is \textbf{$\sigma$-hypocontinuous}. 
+
+    For any ideal $\tau \subset 2^F$ of bounded sets, $\lambda$ if \textbf{$(\sigma, \tau)$-hypocontinuous} if $\lambda$ is $\sigma$-hypocontinuous and $\tau$-hypocontinuous. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:separate-hypocontinuous}
+    Let $E, F, G$ be TVSs over $K \in \RC$, and $\lambda: E \times F \to G$ be a separately continuous bilinear map. 
+    If one of the following holds:
+    \begin{enumerate}
+        \item[(B)] $E$ is Baire.
+        \item[(B')] $E$ is barrelled and $G$ is locally convex. 
+    \end{enumerate}
+
+    then $\lambda$ is $B(E)$-hypocontinuous. 
+\end{proposition}
+\begin{proof}
+    Since $\lambda$ is separately continuous, the mapping
+    \[
+    E \to L(F; G) \quad x \mapsto \lambda(x, \cdot)
+    \]
+
+    is continuous with respect to the strong operator topology on $L(F; G)$. As such, for each $B \subset E$ bounded, $\bracs{\lambda(x, \cdot)|x \in B}$ is bounded in $L_s(F; G)$. By the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, $\bracs{\lambda(x, \cdot)|x \in B}$ is equicontinuous. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:hypocontinuous-restriction}
+    Let $E, F, G$ be TVSs over $K \in \RC$, $\sigma \subset B(E)$ and $\tau \subset B(F)$ be ideals of bounded sets, and $\lambda: E \times F \to G$ be a bilinear mapping. 
+    \begin{enumerate}
+        \item If $\lambda$ is $\sigma$-hypocontinuous, then $\lambda$ is continuous on $S \times F$ for all $S \in \sigma$. 
+        \item If $\lambda$ is $(\sigma, \tau)$-hypocontinuous, then $\lambda$ is uniformly continuous on $S \times T$ for all $S \in \sigma$ and $T \in \tau$. 
+    \end{enumerate}
+\end{proposition}
+% Omitted for obviousness
+
+\begin{proposition}
+\label{proposition:hypocontinuous-linear-extension}
+    Let $E, F, G$ be TVSs over $K \in \RC$, $E_0 \subset E$ and $F_0 \subset F$ be dense subspaces, and $\sigma \subset B(E_0)$ and $\tau \subset B(F_0)$ be ideals of bounded sets. Denote $\ol{\sigma}$ and $\ol \tau$ as the ideals generated by $\bracsn{\ol{S}|S \in \sigma}$ and $\bracsn{\ol{T}|T \in \tau}$, respectively. If
+    \begin{enumerate}[label=(\alph*)]
+        \item $G$ is a complete Hausdorff TVS. 
+        \item $\ol\sigma$ covers $E$ and $\ol{\tau}$ covers $F$. 
+    \end{enumerate}
+
+    Then, for any $(\sigma, \tau)$-hypocontinuous bilinear map $\lambda: E_0 \times F_0 \to G$, there exists a unique $\Lambda: E \times F \to G$ such that:
+    \begin{enumerate}
+        \item $\Lambda|_{E_0 \times F_0} = \lambda$. 
+        \item $\Lambda$ is bilinear and $(\ol\sigma, \ol\tau)$-hypocontinuous.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    By (2) of \autoref{proposition:hypocontinuous-restriction}, for each $S \in \sigma$ and $T \in \tau$, $\lambda|_{S \times T}$ is uniformly continuous. Thus (a) and \autoref{theorem:uniform-continuous-extension} imply that there exists a unique continuous extension of $\lambda$ to $\ol S \times \ol T$. By (b) and the \hyperref[gluing lemma]{lemma:glue-function}, there exists a unique $\Lambda: E \times F \to G$ such that:
+    \begin{enumerate}
+        \item $\Lambda|_{E_0 \times F_0} = \lambda$. 
+        \item[(2')] For each $S \in \sigma$ and $T \in \tau$, $\Lambda|_{\ol S \times \ol T}$ is continuous. 
+    \end{enumerate}
+
+    so the extension is unique by (1) of \autoref{proposition:hypocontinuous-restriction}. 
+
+    It remains to show that $\Lambda$ is bilinear and $(\ol \sigma, \ol \tau)$-hypocontinuous. To this end, observe that for each $x \in E_0$, $\lambda(x, \cdot) \in L(F_0; G)$, and extends to a unique element of $L(F; G)$ by the \hyperref[linear extension theorem]{theorem:linear-extension-theorem-tvs}. For each $S \in \sigma$ and $T \in \tau$, $\Lambda|_{\ol S \times \ol T}$ is the unique continuous extension of $\lambda|_{S \times T}$. As such, $\Lambda(x, \cdot)$ is the unique continuous extension of $\lambda(x, \cdot)$ to an element of $L(F; G)$, so $\Lambda(x, \cdot) \in L(F; G)$ for all $x \in E_0$. By symmetry, $\Lambda(\cdot, y) \in L(E; G)$ for all $y \in F_0$. 
+
+
+    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. 
+\end{proof}
+
+
+
+
+\begin{proposition}
+\label{proposition:multilinear-identify}
+	Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be a covering ideal, and $k \in \natp$, then
+	\begin{enumerate}
+        \item The map
+        \[
+        I: B_{\sigma}^k(E; B_{\sigma}(E; F)) \to B^{k+1}_{\sigma}(E; F)
+        \]
+
+        defined by
+        \[
+        (IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1})
+        \]
+
+        is an isomorphism.
+        \item The map
+        \[
+        I: \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} \to B^k_{\sigma}(E; F)
+        \]
+
+        defined by
+        \[
+        IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k)
+        \]
+
+        is an isomorphism.
+    \end{enumerate}
+
+    which allows the identification
+    \[
+    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} = B^k_{\sigma}(E; F)
+    \]
+
+    under the map $I$ in (2).
+\end{proposition}
+\begin{proof}
+    (1): To see that $I$ is surjective, let $T \in B_{\sigma}^{k+1}(E; F)$ and
+    \[
+    I^{-1}T: E \to B_{\sigma}(E; F) \quad x \mapsto T(x, \cdot)
+    \]
+
+    Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \sigma$. Since $\sigma$ is a covering ideal, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
+    \[
+    T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in \mathfrak{B}(F)
+    \]
+
+    by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\sigma}(E; F)$.
+
+    In addition, for any $S_1 \in \sigma$ and entourage $E(S_2, U)$ of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$, there exists $S \in \sigma$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in \mathfrak{B}(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\sigma}(E; F))$. Thus $I^{-1}T \in B^k_{\sigma}(E; B_{\sigma}(E; F))$.
+
+    It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \sigma$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.
+
+    On the other hand, let $S_1 \in \sigma$ and $E(S_2, U)$ be an entourage of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$. Let $S \in \sigma$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
+
+    (2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then
+    \[
+    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\sigma}(E; B_{\sigma}(E; F)) = B^{k+1}_{\sigma}(E; F)
+    \]
+
+    Thus (2) holds for all $k \in \natp$.
+\end{proof}
+

src/fa/tvs/equicontinuous.tex

@@ -88,28 +88,6 @@
     and $\lambda(x_0, y - y_0) \in U$ as well. Therefore $\alg$ is equicontinuous at $(x_0, y_0)$. 
 \end{proof}
 
-
-
-\begin{theorem}
-\label{theorem:separate-joint-bilinear}
-    Let $E, F, G$ be TVSs over $K \in \RC$ and $\alg$ be separately continuous bilinear maps from $E \times F$ to $G$. If one of the following holds:
-    \begin{enumerate}
-        \item[(B)] $E$ is Baire.
-        \item[(B')] $E$ is barrelled and $G$ is locally convex. 
-    \end{enumerate}
-
-    and that
-    \begin{enumerate}
-        \item[(M)] $E$ and $F$ are both metrisable. 
-        \item[(E)] For each $x \in E$, $\bracsn{\lambda(x, \cdot)|\lambda \in \alg} \subset L(F; G)$ is equicontinuous. 
-    \end{enumerate}
-
-    then $\alg$ is equicontinuous. 
-\end{theorem}
-\begin{proof}[Proof, {{\cite[III.5.1]{SchaeferWolff}}}. ]
-    Let $\seq{(x_n, y_n)} \subset E \times F$ and $\seq{\lambda_n} \subset \alg$ such that $(x_n, y_n) \to 0$ as $n \to \infty$. Since $\seq{y_n}$ is convergent, for each $n \in \natp$ and $x \in E$, $\bracsn{\lambda_n(x, y_n)|n \in \natp}$ is bounded by (E) and \autoref{proposition:equicontinuous-net}. By (B) or (B') and the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, $\bracsn{\lambda_n(\cdot, y_n)|n \in \natp}$ is equicontinuous, and $\lambda_n(x_n, y_n) \to 0$ as $n \to \infty$ by \autoref{proposition:equicontinuous-net}. By (M) and \autoref{proposition:equicontinuous-net}, $\alg$ is equicontinuous at $0$, and hence equicontinuous by \autoref{lemma:equicontinuous-bilinear}. 
-\end{proof}
-
 % TODO: Replace this with a more general version involving polars in the future. 
 \begin{theorem}[Banach-Alaoglu]
 \label{theorem:alaoglu}

src/fa/tvs/index.tex

@@ -14,4 +14,5 @@
 \input{./inductive.tex}
 \input{./vector-function.tex}
 \input{./space-of-linear.tex}
-\input{./equicontinuous.tex}
+\input{./equicontinuous.tex}
+\input{./bilinear.tex}

src/fa/tvs/space-of-linear.tex

@@ -20,69 +20,6 @@
 \end{proposition}
 % Proof omitted because it is obvious. 
 
-
-\begin{proposition}
-\label{proposition:multilinear-identify}
-	Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be a covering ideal, and $k \in \natp$, then
-	\begin{enumerate}
-        \item The map
-        \[
-        I: B_{\sigma}^k(E; B_{\sigma}(E; F)) \to B^{k+1}_{\sigma}(E; F)
-        \]
-
-        defined by
-        \[
-        (IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1})
-        \]
-
-        is an isomorphism.
-        \item The map
-        \[
-        I: \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} \to B^k_{\sigma}(E; F)
-        \]
-
-        defined by
-        \[
-        IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k)
-        \]
-
-        is an isomorphism.
-    \end{enumerate}
-
-    which allows the identification
-    \[
-    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} = B^k_{\sigma}(E; F)
-    \]
-
-    under the map $I$ in (2).
-\end{proposition}
-\begin{proof}
-    (1): To see that $I$ is surjective, let $T \in B_{\sigma}^{k+1}(E; F)$ and
-    \[
-    I^{-1}T: E \to B_{\sigma}(E; F) \quad x \mapsto T(x, \cdot)
-    \]
-
-    Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \sigma$. Since $\sigma$ is a covering ideal, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
-    \[
-    T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in \mathfrak{B}(F)
-    \]
-
-    by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\sigma}(E; F)$.
-
-    In addition, for any $S_1 \in \sigma$ and entourage $E(S_2, U)$ of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$, there exists $S \in \sigma$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in \mathfrak{B}(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\sigma}(E; F))$. Thus $I^{-1}T \in B^k_{\sigma}(E; B_{\sigma}(E; F))$.
-
-    It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \sigma$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.
-
-    On the other hand, let $S_1 \in \sigma$ and $E(S_2, U)$ be an entourage of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$. Let $S \in \sigma$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
-
-    (2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then
-    \[
-    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\sigma}(E; B_{\sigma}(E; F)) = B^{k+1}_{\sigma}(E; F)
-    \]
-
-    Thus (2) holds for all $k \in \natp$.
-\end{proof}
-
 \begin{definition}[Strong Operator Topology]
 \label{definition:strong-operator-topology}
     Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F^E$ is the \textbf{strong operator topology}.

src/op/vn/vn.tex

@@ -23,7 +23,7 @@
     
     
 \end{theorem}
-\begin{proof}[Proof, {{\cite[Theorem 17.1]{Zhu}}}. ]
+\begin{proof}[Proof, {{\cite[Section 17]{Zhu}}}. ]
     (1): After rescaling, assume without loss of generality that $-I \le T \le I$ for all $T \in \cf$. Since $\cf \subset A_{sa}$, $\norm{T}_{B(H)} = [T]_{sp} \le 1$ by \autoref{theorem:c-star-normal-spectral-radius}, where the spectral radius is taken with respect to $B(H)$. 
     
     Thus $\cf \subset \ol{B_{A}(0, 1)}$, and is relatively compact in the weak operator topology by the \hyperref[Banach-Alaoglu Theorem]{proposition:bh-ultraweak-bounded}. As such, $\bigcap_{T \in \cf}\ol{\bracs{S \in \cf|S \ge T}}^{\text{\small WOT}} \ne \emptyset$. Let $R \in \bigcap_{T \in \cf}\ol{\bracs{S \in \cf|S \ge T}}^{\text{\small WOT}}$. Since $A$ is strong-operator closed, so is $A_{sa}$ by \autoref{proposition:bh-operator-topologies-facts}. Thus for each $T \in A_{sa}$, $\bracs{S \in A_{sa}|S \ge T}$ is closed in the weak operator topology, and $R \in A_{sa}$ with $R \ge T$ for all $T \in \cf$.