Added bilinear forms.

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}.