Added facts about vector measures.

src/fa/lc/bornologic.tex

@@ -0,0 +1,56 @@
+\section{Bornologic Spaces}
+\label{section:bornologic}
+
+\begin{definition}[Bornologic Space]
+\label{definition:bornologic-space}
+    Let $E$ be a locally convex space, then the following are equivalent:
+    \begin{enumerate}
+        \item For any $U \subset E$ convex and balanced, if $U$ absorbs every bounded set of $E$, then $U \in \cn_E(0)$.
+        \item For any seminorm $\rho: E \to [0, \infty)$ that is bounded on all bounded sets of $E$, $\rho$ is continuous. 
+    \end{enumerate}
+    
+    If the above holds, then $E$ is a \textbf{bornologic space}.
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $B \subset E$ be bounded, then there exists $R > 0$ such that $\rho(B) \subset [0, R)$. In which case, 
+    \[
+    B \subset \bracs{\rho < R} = R\bracs{\rho < 1}
+    \]
+
+    By assumption, $\bracs{\rho < 1} \in \cn_E(0)$, so $\rho$ is continuous by \autoref{lemma:continuous-seminorm}.
+
+    (2) $\Rightarrow$ (1): Let $\rho$ be the \hyperref[gauge]{definition:gauge} of $U$, then for any $B \subset E$ bounded, there exists $R > 0$ such that $B \subset RU$. In which case, $\rho(B) \subset [0, R]$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:metrisable-bornologic}
+    Let $E$ be a metrisable locally convex space, then $E$ is bornologic.
+\end{proposition}
+\begin{proof}
+    Let $U \subset E$ be convex and balanced such that $U$ absorbs every bounded set of $E$. Let $\seq{U_n} \subset \cn^o(0)$ be a decreasing countable fundamental system of neighbourhoods at $0$. If $U_n \setminus nA \ne \emptyset$ for all $n \in \natp$, then there exists $\seq{x_n}$ such that $x_n \in U_n \setminus nA$ for all $n \in \natp$. In which case, $x_n \to 0$ as $n \to \infty$, so $\seq{x_n}$ is bounded. By assumption, there exists $n \in\natp$ such that $nA \supset \seq{x_n}$, which contradicts the fact that $\seq{x_n} \cap A = \emptyset$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:bornologic-bounded}
+    Let $E$ be a bornologic space, $F$ be a locally convex space, and $T \in \hom(E; F)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $T$ is continuous.
+        \item $T$ is bounded.
+    \end{enumerate}
+    
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): By \autoref{proposition:continuous-bounded}. 
+
+    (2) $\Rightarrow$ (1): Let $\rho: F \to [0, \infty)$ be a continuous seminorm, then $\rho \circ T$ is a seminorm on $E$ that is bounded on bounded sets. Since $E$ is bornologic, $\rho \circ T$ is continuous. Therefore $T$ is continuous by \autoref{proposition:tvs-convex-morphism}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:bornologic-continuous-complete}
+    Let $E$ be a bornologic space and $F$ be a complete Hausdorff locally convex space, then $L_b(E; F)$ is complete. In particular, $E^*$ equipped with the topology of bounded convergence is complete.
+\end{proposition}
+\begin{proof}
+    By \autoref{proposition:bornologic-bounded}, $L_b(E; F) = B(E; F)$. By \autoref{proposition:operator-space-completeness}, $B(E; F)$ is complete, so $L_b(E; F)$ is complete as well.
+\end{proof}
+
+

src/fa/lc/convex.tex

@@ -99,10 +99,16 @@
         \item $[\cdot]$ is uniformly continuous.
         \item $[\cdot]$ is continuous.
         \item $[\cdot]$ is continuous at $0$.
+        \item $\bracs{x \in E| [x] < 1} \in \cn_E(0)$.
     \end{enumerate}
 \end{lemma}
 \begin{proof}
-    $(3) \Rightarrow (1)$: Let $\eps > 0$, then there exists $V \in \cn(0)$ such that $[x] < \eps$ for all $x \in V$. In which case, for any $x, y \in E$ with $x - y \in V$, $\abs{[x] - [y]} \le [x - y] < \eps$. By \autoref{proposition:tvs-uniform}, $[\cdot]$ is uniformly continuous.
+    $(4) \Rightarrow (1)$: Let $x, y \in E$ and $r > 0$. If
+    \[
+    x - y \in \bracs{x \in E|[x] < r} = r\bracs{x \in E|[x] < 1} \in \cn_E(0)
+    \]
+    
+    then $[x - y] < r$.
 \end{proof}
 
 

src/fa/lc/index.tex

@@ -4,6 +4,7 @@
 
 \input{./convex.tex}
 \input{./continuous.tex}
+\input{./bornologic.tex}
 \input{./quotient.tex}
 \input{./projective.tex}
 \input{./inductive.tex}

src/fa/norm/absolute.tex

@@ -6,6 +6,22 @@
     Let $E$ be a normed space, then a series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$ \textbf{converges absolutely} if $\sum_{n \in \natp}\norm{x_n}_E < \infty$.
 \end{definition}
 
+\begin{lemma}
+\label{lemma:banach-absolute}
+    Let $E$ be a normed space, then the following are equivalent:
+    \begin{enumerate}
+        \item $E$ is a Banach space.
+        \item For any absolutely convergent series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$, there exists $x \in E$ such that $x = \sum_{n = 1}^\infty x_n$.
+    \end{enumerate}
+    
+\end{lemma}
+\begin{proof}
+    (2) $\Rightarrow$ (1): Let $\seq{x_n} \subset E$ be a Cauchy sequence, then there exists a subsequence ${n_k}$ such that $\norm{x_{n_{k+1}} - x_{n_k}} \le 2^{-k}$ for all $k \in \natp$. 
+    
+    For each $k \in \natp$, let $y_k = x_{n_{k+1}} - x_{n_k}$, then $\sum_{n = 1}^\infty y_k$ is absolutely convergent, and there exists $y \in E$ such that $y = \sum_{n = 1}^\infty y_n$. Let $x = x_{n_1} + y$, then $x_{n_k} \to x$ as $k \to \infty$. Since $\seq{x_n}$ is Cauchy, $x_n \to x$ as well by \autoref{lemma:cauchy-subnet}.
+\end{proof}
+
+
 \begin{definition}[Unconditional Convergence]
 \label{definition:unconditional-convergence}
     Let $E$ be a normed space, then a series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$ \textbf{converges unconditionally} if for any bijection $\sigma: \natp \to \natp$, 

src/fa/norm/normed.tex

@@ -31,6 +31,7 @@
     is a fundamental system of neighbourhoods at $0$ for $E$. Therefore $\norm{\cdot}_E$ induces the topology on $E$.
 \end{proof}
 
+
 \begin{theorem}[Successive Approximation]
 \label{theorem:successive-approximation}
     Let $E, F$ be normed spaces, $T \in L(E; F)$, $C \ge 0$, and $\gamma \in (0, 1)$. If for all $y \in F$, there exists $x \in E$ such that:

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

@@ -39,9 +39,53 @@
     for all $x \in S$.
 \end{proof}
 
+
+\begin{definition}[Space of Bounded Functions]
+\label{definition:bounded-function-space}
+    Let $T$ be a set, $E$ be a TVS over $K \in \RC$, and $f: T \to E$ be a function, then $f$ is \textbf{bounded} if $f(T)$ is a bounded subset of $E$. The set $B(T; E) \subset E^T$ is the space of all bounded functions from $T$ to $E$, and:
+    \begin{enumerate}
+        \item $B(T; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
+        \item $B(T; E)$ is a closed subset of $E^T$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $B(T; E)$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1): For any $f, g \in B(T; E)$ and $\lambda \in K$, $(\lambda f + g)(T) \subset (\lambda f)(T) + g(T)$, which is bounded by \autoref{proposition:bounded-operations}, so $B(T; E)$ is closed under addition and scalar multiplication. 
+    
+    Since $f(E)$ is bounded for all $f \in B(T; E)$, $B(T; E)$ forms a TVS over $K$ by \autoref{proposition:tvs-set-uniformity}.
+
+    (2): Let $f \in \ol{B(T; E)}$ and $U \in \cn_E(0)$ be circled, then there exists $g \in B(T; E)$ such that $(f - g)(E) \subset U$. Since $g \in B(T; E)$, there exists $\lambda \ge 0$ such that $g(E) \subset \lambda U$. In which case,
+    \[
+    f(E) \subset g(E) + (f - g)(E) \subset (\lambda + 1)U
+    \]
+    so $f \in B(T; E)$.
+
+    If $E$ is complete, then $E^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $B(T; E)$ is also complete by \autoref{proposition:complete-closed}.
+\end{proof}
+
+
+\begin{definition}[Space of Bounded Continuous Functions]
+\label{definition:bounded-continuous-function-space}
+    Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then $BC(X; E) = B(X; E) \cap C(X; E)$ is the space of bounded and continuous functions from $X$ to $E$, and
+    \begin{enumerate}
+        \item $BC(X; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
+        \item $BC(X; E)$ is a closed subspace of $C(X; E)$ and $B(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $BC(X; E)$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1): Since addition and scalar multiplication are continuous, $BC(X; E)$ is a subspace of $B(X; E)$, and hence a TVS over $K \in \RC$ by \autoref{definition:bounded-function-space}.
+
+    (2): Since $B(X; E)$ and $C(X; E)$ are both closed subspaces of $E^X$ by \autoref{definition:bounded-function-space}, $BC(X; E)$ is a closed subspace. 
+    
+    If $E$ is complete, then $B(X; E)$ and $C(X; E)$ are both complete under the uniform topology by \autoref{definition:bounded-function-space}. Therefore $BC(X; E)$ is also complete. 
+\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 $B_{\mathfrak{S}}^k(E; F)$ 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.
+
+    Let $\fB \subset 2^E$ be the collection of all bounded subsets of $E$, then $B_{\mathfrak{S}}(E; F) = B(E; F)$ is the \textbf{space of bounded linear maps} from $E$ to $F$. 
 \end{definition}
 
 \begin{proposition}
@@ -127,3 +171,35 @@
 
     The space $L_b(E; F)$ denotes $L(E; F)$ equipped with the topology of bounded convergence.
 \end{definition}
+
+\begin{proposition}
+\label{proposition:operator-space-completeness}
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, then:
+    \begin{enumerate}
+        \item $\hom(E; F)$ is a closed subspace of $F^E$ with respect to the product topology.
+        \item $B(E; F)$ is a closed subspace of $F^E$ with respect to the topology of bounded convergence. In particular, if $F$ is complete, then so is $B(E; F)$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For each $x, y \in E$ and $\lambda \in K$, the mappings
+    \[
+    \phi_{x, y}: F^E \to F \quad T \mapsto Tx + Ty - T(x + y)
+    \]
+
+    and
+    \[
+    \psi_{x, \lambda}: F^E \to F \quad T \mapsto T(\lambda x) - Tx
+    \]
+
+    are continuous with respect to the product topology. Since
+    \[
+    \hom(E; F) = \bigcap_{x, y \in E}\bracsn{\phi_{x, y} = 0} \cap \bigcap_{\substack{x \in E \\ \lambda \in K}}\bracsn{\psi_{x, \lambda} = 0}
+    \]
+
+    and $\bracs{0}$ is closed in $F$, $\hom(E; F)$ is a closed subspace of $F^E$. 
+
+    (2): By \autoref{definition:bounded-function-space} and (1), the space of bounded functions and the space of linear functions from $E$ to $F$ are closed subspaces of $F^E$ with respect to the topology of bounded convergence. Therefore $B(E; F)$ is also a closed subspace.
+
+\end{proof}
+
+