9.3 Bornologic Spaces
Definition 9.3.1 (Bornologic Space). Let $E$ be a locally convex space, then the following are equivalent:
For any $U \subset E$ convex and balanced, if $U$ absorbs every bounded set of $E$, then $U \in \cn_{E}(0)$.
For any seminorm $\rho: E \to [0, \infty)$ that is bounded on all bounded sets of $E$, $\rho$ is continuous.
If the above holds, then $E$ is a bornologic space.
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,
By assumption, $\bracs{\rho < 1}\in \cn_{E}(0)$, so $\rho$ is continuous by Lemma 9.1.9.
(2) $\Rightarrow$ (1): Let $\rho$ be the gauge of $U$, then for any $B \subset E$ bounded, there exists $R > 0$ such that $B \subset RU$. In which case, $\rho(B) \subset [0, R]$.$\square$
Proposition 9.3.2. Let $E$ be a metrisable locally convex space, then $E$ is bornologic.
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 $\seq{x_n}\cap A = \emptyset$.$\square$
Proposition 9.3.3. Let $E$ be a bornologic space, $F$ be a locally convex space, and $T \in \hom(E; F)$, then the following are equivalent:
$T$ is continuous.
$T$ is bounded.
Proof. (1) $\Rightarrow$ (2): By Proposition 8.5.3.
(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 Proposition 9.2.1.$\square$
Proposition 9.3.4. 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.
Proof. By Proposition 9.3.3, $L_{b}(E; F) = B(E; F)$. By Proposition 8.11.10, $B(E; F)$ is complete, so $L_{b}(E; F)$ is complete as well.$\square$