Finished basic topologies on function spaces.

src/fa/tvs/bounded.tex

@@ -1,9 +1,6 @@
 \section{Bounded Sets}
 \label{section:bounded}
 
-\subsection{Bounded Sets}
-\label{subsection:tvs-bounded}
-
 \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 $\lambda U \supset B$.

src/fa/tvs/continuous.tex

@@ -84,41 +84,3 @@
     \end{enumerate}
     The uniformity $\fU$ and its induced topology are the \textbf{product uniformity/topology}, and $E$ equipped with $\fU$ is the \textbf{product TVS} of $\seqi{E}$.
 \end{definition}
-
-\begin{definition}[Uniform/Bounded Operator Topology]
-\label{definition:uniform-op-topo}
-    Let $E, F$ be TVSs over $K \in \RC$. For each bounded set $B \subset E$ and entourage $V \subset F \times F$, let
-    \[
-    U(B, V) = \bracs{(S, T) \in L(E; F) \times L(E; F)|(Sx, Tx) \in V \forall x \in B}
-    \]
-    then
-    \[
-    \fB = \bracs{U(B, V)| B \subset E \text{ bounded}, U \subset F \times F \text{ entourage}}
-    \]
-    is a fundamental system of entourages for a vector space uniformity on $L(E; F)$, which induces the \textbf{uniform/bounded operator topology} on $L(E; F)$.
-\end{definition}
-\begin{proof}
-    Firstly,
-    \begin{enumerate}
-        \item[(FB1)] Let $U(B, V), U(B', V') \in \fB$, then $U(B, V) \cap U(B', V') \supset U(B \cup B', V \cap V') \in \fB$.
-        \item[(UB1)] For any $U(B, V) \in \fB$, the diagonal is contained in $V$, so the diagonal is also contained in $U(B, V)$.
-        \item[(UB2)] For any $U(B, V) \in \fB$, there exists an entourage $W \subset F \times F$ such that $W \circ W \subset V$. So $U(B, W) \in \fB$ with
-        \[
-        U(B, W) \circ U(B, W) \subset U(B, W \circ W) \subset U(B, V)
-        \]
-    \end{enumerate}
-    By \ref{proposition:fundamental-entourage-criterion}, $\fB$ forms a fundamental system of entourages for a uniformity on $L(E; F)$.
-
-    TODO SCHAEFER WOLFF PAGE 79
-\end{proof}
-
-
-\begin{definition}[Strong Operator Topology]
-\label{definition:strong-op-topo}
-    Let $E, F$ be TVSs over $K \in \RC$, then the \textbf{strong operator topology} on $L(E; F)$ is the initial topology generated by the maps $\bracs{T \mapsto Tx| x \in E}$, where $F$ is equipped with the strong topology.
-\end{definition}
-
-\begin{definition}[Weak Operator Topology]
-\label{definition:weak-op-topo}
-    Let $E, F$ be TVSs over $K \in \RC$, then the \textbf{weak operator topology} on $L(E; F)$ is the initial topology generated by the maps $\bracs{T \mapsto Tx| x \in E}$, where $F$ is equipped with the weak topology.
-\end{definition}

src/fa/tvs/definition.tex

@@ -130,12 +130,12 @@
 \label{proposition:tvs-good-neighbourhood-base}
     Let $E$ be a topological vector space over $K \in \RC$, then
     \begin{enumerate}
-        \item $E$ admits a fundamental system of neighbourhoods at $0$ consisting of balanced and absorbing sets.
+        \item $E$ admits a fundamental system of neighbourhoods at $0$ consisting of circled and absorbing sets.
         \item The fundamental system of neighbourhoods in $(1)$ can be taken to be open or closed.
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    Firstly, (TVS2) implies that every neighbourhood of $0$ is balanced.
+    Firstly, (TVS2) implies that every neighbourhood of $0$ is circled.
 
     By \ref{proposition:uniform-neighbourhoods}, $E$ admits a fundamental system of neighbourhoods consisting of open sets or closed sets.
 

src/fa/tvs/index.tex

@@ -6,3 +6,4 @@
 \input{./src/fa/tvs/dual.tex}
 \input{./src/fa/tvs/continuous.tex}
 \input{./src/fa/tvs/completion.tex}
+\input{./src/fa/tvs/spaces-of-linear.tex}

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

@@ -0,0 +1,60 @@
+\section{Spaces of Vector-Valued Maps}
+\label{section:spaces-linear-map}
+
+\begin{proposition}[{{\cite[3.3.1]{SchaeferWolff}}}]
+\label{proposition:tvs-set-uniformity}
+    Let $T$ be a set, $\mathfrak{S} \subset 2^T$ be an upward-directed system of sets, $F$ be a TVS over $K \in \RC$, then
+    \begin{enumerate}
+        \item The $\mathfrak{S}$-uniformity on $F^T$ (\ref{definition:set-uniform}) is translation invariant.
+        \item The composition defined by
+        \[
+        T^F \times T^F \to T^F \quad (f + g)(x) = f(x) + g(x)
+        \]
+        is continuous.
+    \end{enumerate}
+    For any vector subspace $\cf \subset F^T$, the following are equivalent:
+    \begin{enumerate}
+        \item[(3)] The $\mathfrak{S}$-uniform topology on $\cf \subset F^E$ is a vector space topology.
+        \item[(4)] For each $S \in \mathfrak{S}$ and $f \in \cf$, $f(S) \subset E$ is bounded.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $U \subset F \times F$ be an entourage and $S \in \mathfrak{S}$. Using \ref{proposition:tvs-uniform}, assume without loss of generality that $U$ is translation-invariant. For any $(f, g) \in E(S, U)$, $h \in F^T$, and $x \in S$, $(f(x) + h(x), g(x) + h(x)) \in U$. Thus $(f + h, g + h) \in E(S, U)$ and $E(S, U)$ is translation-invariant.
+
+    (2): Let $f, g, f', g' \in \cf$, $S \in \mathfrak{S}$, and $U \subset F \times F$ be an entourage. By (TVS1), there exists an entourage $V \subset F \times F$ such that for any $x, x', y, y' \in F$ with $(x, x'), (y, y') \in V$, $(x + y, x' + y') \in U$. If $(f, g), (f', g') \in E(S, V) \cap \cf$, then for any $x \in S$, $(f(x) + g(x), f'(x) + g'(x)) \in U$, so $(f + g, f' + g') \in E(S, U) \cap \cf$.
+
+    (3) $\Rightarrow$ (4): Let $f \in \cf$, $S \in \mathfrak{S}$ and $U \subset F \times F$ be a symmetric entourage, then $E(S, U)(0)$ is a neighbourhood of $0$ with respect to the $\mathfrak{S}$-uniform topology. By (TVS2), there exists $\lambda > 0$ such that $f \in \lambda E(S, U)(0)$. In which case, for any $x \in S$, $\lambda^{-1}f(x) \in U(0)$ and $f(x) \in \lambda U(0)$. Thus $f(S) \subset \lambda U(0)$, and $f(S)$ is bounded.
+
+    (4) $\Rightarrow$ (3): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \mathfrak{S}$, then for any $x \in X$,
+    \begin{align*}
+    \lambda f(x) - \lambda' g(x) &= \lambda f(x) - \lambda' f(x) + \lambda' f(x) - \lambda' g(x) \\
+    &= (\lambda - \lambda')f(x) + \lambda' (f(x) - g(x))
+    \end{align*}
+    Let $U_0 \in \cn_F(0)$. By (TVS1) and \ref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn_F(0)$ circled such that $U + U \subset U_0$. By (TVS2), there exists $V \in \cn_F(0)$ such that $\lambda' V \subset U$. Since $f(S)$ is bounded, there exists $\eps > 0$ with $\eps f(S) \subset U$. In which case, if $\abs{\lambda - \lambda'} < \eps$ and $f(x) - g(x) \in V$ for all $x \in S$, then
+    \[
+    \lambda f(x) - \lambda' g(x) \in (\lambda - \lambda')f(S) + \lambda' V \subset U + U \subset U_0
+    \]
+    for all $x \in S$.
+\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 $L(E; F)$ is the \textbf{strong operator topology}.
+
+    The space $L_s(E; F)$ denotes $L(E; F)$ equipped with the strong operator topology.
+\end{definition}
+
+\begin{definition}[Weak Operator Topology]
+\label{definition:weak-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 $L(E; F_w)$ is the \textbf{weak operator topology}.
+
+    The space $L_w(E; F) = L_s(E; F_w)$ denotes $L(E; F)$ equipped with the weak operator topology.
+\end{definition}
+
+\begin{definition}[Bounded Convergence Topology]
+\label{definition:bounded-convergence-topology}
+    Let $E, F$ be TVSs over $K \in \RC$, $\fB \subset 2^E$ be the collection of bounded subsets of $E$, then the $\fB$-uniform topology on $L(E; F)$ is the \textbf{topology of bounded convergence}.
+
+    The space $L_b(E; F)$ denotes $L(E; F)$ equipped with the topology of bounded convergence.
+\end{definition}