Added facts about vector measures.
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src/fa/lc/bornologic.tex
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src/fa/lc/bornologic.tex
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\section{Bornologic Spaces}
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\label{section:bornologic}
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\begin{definition}[Bornologic Space]
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\label{definition:bornologic-space}
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Let $E$ be a locally convex space, then the following are equivalent:
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\begin{enumerate}
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\item For any $U \subset E$ convex and balanced, if $U$ absorbs every bounded set of $E$, then $U \in \cn_E(0)$.
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\item For any seminorm $\rho: E \to [0, \infty)$ that is bounded on all bounded sets of $E$, $\rho$ is continuous.
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\end{enumerate}
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If the above holds, then $E$ is a \textbf{bornologic space}.
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\end{definition}
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\begin{proof}
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(1) $\Rightarrow$ (2): Let $B \subset E$ be bounded, then there exists $R > 0$ such that $\rho(B) \subset [0, R)$. In which case,
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\[
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B \subset \bracs{\rho < R} = R\bracs{\rho < 1}
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\]
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By assumption, $\bracs{\rho < 1} \in \cn_E(0)$, so $\rho$ is continuous by \autoref{lemma:continuous-seminorm}.
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(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, $\rho(B) \subset [0, R]$.
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\end{proof}
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\begin{proposition}
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\label{proposition:metrisable-bornologic}
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Let $E$ be a metrisable locally convex space, then $E$ is bornologic.
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\end{proposition}
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\begin{proof}
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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 $\seq{x_n} \cap A = \emptyset$.
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\end{proof}
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\begin{proposition}
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\label{proposition:bornologic-bounded}
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Let $E$ be a bornologic space, $F$ be a locally convex space, and $T \in \hom(E; F)$, then the following are equivalent:
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\begin{enumerate}
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\item $T$ is continuous.
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\item $T$ is bounded.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1) $\Rightarrow$ (2): By \autoref{proposition:continuous-bounded}.
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(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}.
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\end{proof}
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\begin{proposition}
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\label{proposition:bornologic-continuous-complete}
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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.
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\end{proposition}
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\begin{proof}
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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.
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\end{proof}
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@@ -99,10 +99,16 @@
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\item $[\cdot]$ is uniformly continuous.
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\item $[\cdot]$ is continuous.
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\item $[\cdot]$ is continuous at $0$.
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\item $\bracs{x \in E| [x] < 1} \in \cn_E(0)$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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$(3) \Rightarrow (1)$: 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.
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$(4) \Rightarrow (1)$: Let $x, y \in E$ and $r > 0$. If
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\[
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x - y \in \bracs{x \in E|[x] < r} = r\bracs{x \in E|[x] < 1} \in \cn_E(0)
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\]
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then $[x - y] < r$.
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\end{proof}
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@@ -4,6 +4,7 @@
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\input{./convex.tex}
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\input{./continuous.tex}
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\input{./bornologic.tex}
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\input{./quotient.tex}
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\input{./projective.tex}
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\input{./inductive.tex}
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