Me when I forget to commit.

src/fa/tvs/bounded.tex

@@ -3,12 +3,12 @@
 
 \begin{definition}[Bounded]
 \label{definition:bounded}
-    Let $E$ be a TVS over $K \in \RC$ and $B \subset E$, then $B$ is \textbf{bounded} if for every $U \in \cn(0)$, there exists $\lambda \in K$ such that $\lambda U \supset B$.
+    Let $E$ be a TVS over $K \in \RC$ and $B \subset E$, then $B$ is \textbf{bounded} if for every $U \in \cn(0)$, there exists $\lambda \in K$ such that $\lambda U \supset B$. The collection $B(E)$ is the set of all bounded sets of $E$.
 \end{definition}
 
 \begin{proposition}[{{\cite[1.5.1]{SchaeferWolff}}}]
 \label{proposition:bounded-operations}
-    Let $E$ be a TVS over $K \in \RC$ and $A, B \subset E$ be bounded, then the following sets are bounded:
+    Let $E$ be a TVS over $K \in \RC$ and $A, B \in B(E)$, then the following sets are bounded:
     \begin{enumerate}
         \item Any $C \subset B$.
         \item The closure $\ol{B}$.

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

@@ -1,4 +1,4 @@
-\section{Spaces of Vector-Valued Maps}
+\section{Vector-Valued Function Spaces}
 \label{section:spaces-linear-map}
 
 \begin{proposition}[{{\cite[3.3.1]{SchaeferWolff}}}]
@@ -37,6 +37,65 @@
     for all $x \in S$.
 \end{proof}
 
+\begin{definition}[Space of Bounded Linear Maps]
+\label{definition:bounded-linear-map-space}
+    Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{S} \subset 2^E$ be an upward-directed system, and $k \in \nat$. The space $L_{\mathfrak{S}}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in B(F)$ for all $S \in \mathfrak{S}$, equipped with the $\bracsn{S^k| S \in \mathfrak{S}}$-uniform topology.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:multilinear-identify}
+	Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{S} \subset 2^E$ be an upward-directed system that contains all singletons, and $k \in \natp$, then
+	\begin{enumerate}
+        \item The map
+        \[
+        I: L_{\mathfrak{S}}^k(E; L_{\mathfrak{S}}(E; F)) \to L^{k+1}_{\mathfrak{S}}(E; F)
+        \]
+        defined by
+        \[
+        (IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1})
+        \]
+        is an isomorphism.
+        \item The map
+        \[
+        I: \underbrace{L_{\mathfrak{S}}(E; L_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} \to L^k_{\mathfrak{S}}(E; F)
+        \]
+        defined by
+        \[
+        IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k)
+        \]
+        is an isomorphism.
+    \end{enumerate}
+    which allows the identification
+    \[
+    \underbrace{L_{\mathfrak{S}}(E; L_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} = L^k_{\mathfrak{S}}(E; F)
+    \]
+    under the map $I$ in (2).
+\end{proposition}
+\begin{proof}
+    (1): To see that $I$ is surjective, let $T \in L_{\mathfrak{S}}^{k+1}(E; F)$ and
+    \[
+    I^{-1}T: E \to L_{\mathfrak{S}}(E; F) \quad x \mapsto T(x, \cdot)
+    \]
+    Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \mathfrak{S}$. Since $\mathfrak{S}$ is upward-directed and contains all singletons, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
+    \[
+    T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in B(F)
+    \]
+    by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in L_{\mathfrak{S}}(E; F)$.
+
+    In addition, for any $S_1 \in \mathfrak{S}$ and entourage $E(S_2, U)$ of $L_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$, there exists $S \in \mathfrak{S}$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in B(F)$, 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 $I^{-1}T(S^k) \in B(L_{\mathfrak{S}}(E; F))$. Thus $I^{-1}T \in L^k_{\mathfrak{S}}(E; L_{\mathfrak{S}}(E; F))$.
+
+    It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \mathfrak{S}$ 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 $S_1 \in \mathfrak{S}$ and $E(S_2, U)$ be an entourage of $L_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$. Let $S \in \mathfrak{S}$ 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
+    \[
+    \underbrace{L_{\mathfrak{S}}(E; L_{\mathfrak{S}}(E; \cdots)))}_{k+1 \text{ times}} = L^k_{\mathfrak{S}}(E; L_{\mathfrak{S}}(E; F)) = L^{k+1}_{\mathfrak{S}}(E; F)
+    \]
+    Thus (2) holds for all $k \in \natp$.
+\end{proof}
+
+
 
 \begin{definition}[Strong Operator Topology]
 \label{definition:strong-operator-topology}