11.4 Bornological Spaces

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 11.1.9.

(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, $\rho(B) \subset [0, R]$.$\square$

Proof. (2) $\Rightarrow$ (3): Let $B \subset E$ be a bounded set, $\seq{x_n}\subset E$, and $\seq{\lambda_n}\subset K$ such that $\lambda_{n} \to 0$ as $n \to \infty$, then $\lambda_{n} x_{n} \to 0$ as $n \to \infty$. By sequential continuity of $T$, $T(\lambda_{n} x_{n}) \to 0$ as $n \to \infty$ as well. Thus $T(B)$ is also bounded.

(3) $\Rightarrow$ (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 bornological, $\rho \circ T$ is continuous. Therefore $T$ is continuous by Proposition 11.2.1.$\square$

Proof. Let $A \subset E$ be a convex and circled set that absorbs every bounded set of $E$, and $\seq{U_n}\subset \cn_{E}(0)$ be a decreasing fundamental system of neighbourhoods at $0$. If $A \not\in \cn_{E}(0)$, then $U_{n} \setminus nA \ne \emptyset$ for all $n \in \natp$. For each $n \in \natp$, let $x_{n} \in U_{n} \setminus nA$, then $x_{n} \to 0$ as $n \to \infty$, and $\seq{x_n}$ is bounded. However, since $A$ absorbs every bounded set of $E$, there exists $n \in \natp$ such that $nA \supset \seq{x_n}$, which contradicts the assumption that $A \not\in \cn_{E}(0)$.$\square$

Proof. Let $\rho: E \to [0, \infty)$ be a seminorm on $E$ that is bounded on all bounded sets. For each $i \in I$ and $B \subset E_{i}$ bounded, $T_{i}(B)$ is bounded by Proposition 11.4.2, and $\rho \circ T_{i}(B)$ is bounded by assumption. Thus for every $i \in I$, $\rho \circ T_{i}$ is continuous, so $\rho$ is continuous by (4) of Definition 11.7.1.$\square$

Proof. By Proposition 11.4.2, $L_{b}(E; F) = B(E; F)$. By Proposition 10.12.8, $B(E; F)$ is complete, so $L_{b}(E; F)$ is complete as well.$\square$

Proof. Let $\mathscr{T}$ be the collection of all topologies satisfying (1), then $\mathcal{T}\in \mathscr{T}$, so $\mathscr{T}\ne \emptyset$. Let $\mathcal{T}_{B}$ be the projective topology induced by $\mathscr{T}$.

(1): Since $\mathcal{T}_{B} \supset \mathcal{T}$, $\mathfrak{B}(E, \mathcal{T}) \supset \mathfrak{B}(E, \mathcal{T}_{B})$. Let $B \in (E, \mathcal{T})$ and $U \in \cn_{\mathcal{T}_B}(0)$, then there exists $\mathscr{S}\subset \mathscr{T}$ finite and $\bracsn{U_{\mathcal{S}}}_{\mathcal{S} \in \mathscr{S}}$ such that $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}}\subset U$. In which case, since $\mathfrak{B}(E, \mathcal{T}) = \mathfrak{B}(E, \mathcal{S})$ for all $\mathcal{S}\in \mathscr{S}$, $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}}$ absorbs $B$, so $U$ absorbs $B$ as well.

(U): By (U) of the projective topology.

(3): Let $\rho: E \to [0, \infty)$ be a seminorm that is bounded on bounded sets, then the topology induced by $\rho$ satisfies (1). By (U), $\rho$ is continuous with respect to $\mathcal{T}_{B}$, so $(E, \mathcal{T}_{B})$ is a bornological space.

(4): Let $\mathcal{S}$ be the inductive topology on $E$ induced by $\bracsn{\iota_B}_{B \in \mathcal{B}}$. Let $B \in \mathfrak{B}(E, \mathcal{T})$, $U \in \cn_{\mathcal{S}}(0)$, and $\rho: E \to [0, \infty)$ be the gauge of $U$. Since $\iota_{\ol B}\in L(E_{\ol B}; E, \mathcal{S})$, there exists $\lambda > 0$ such that $\rho \le \lambda \iota_{\ol B}\circ \rho_{\ol B}$. Thus $2\lambda U \supset B$ and $(E, \mathcal{S})$ satisfies (1). By (U), $\mathcal{T}_{B} \supset \mathcal{S}$.

On the other hand, since $\mathfrak{B}(E, \mathcal{T}_{B}) = \mathfrak{B}(E, \mathcal{T})$, for each $B \in \mathcal{B}$, $\iota_{B}: E_{B} \to (E, \mathcal{T}_{B})$ is continuous. Therefore by (U) of the inductive topology, $\mathcal{S}\supset \mathcal{T}_{B}$.$\square$