Proposition 2

— Let E and F be two locally convex spaces and let $\mathfrak{S}$ be a set of bounded subsets of E.

1) The $\mathfrak{S}$-topology on $\mathscr{L}(E; F)$ is identical with the $\tilde{\mathfrak{S}}$-topology, where $\tilde{\mathfrak{S}}$ denotes the smallest adapted, bornology (III, p. 3) on E which contains $\mathfrak{S}$.

2) Suppose that $\{0\}$ is not dense in F and let $\mathfrak{S}'$ be another set of bounded subsets of E. Then the $\mathfrak{S}'$-topology is coarser than the $\mathfrak{S}$-topology if and only if $\mathfrak{S}' \subset \tilde{\mathfrak{S}}$.

Let $u \in \mathscr{L}(E; F)$, $M \in \mathfrak{S}$ and let $p$ be a continuous semi-norm on F. Since $p \circ u$ is a continuous semi-norm on E, this is the same as saying that $p \circ u$ is bounded above by 1 on M or on the closed, convex balanced envelope $\tilde{M}$ of M; in other words, we have $p_M = p_{\tilde{M}}$. Moreover, it is clear that we have $p_{\lambda M} = \lambda p_M$ for all $\lambda > 0$ and $p_{M \cup M'} = \sup(p_M, p_{M'})$, from which the first assertion follows, since $\tilde{\mathfrak{S}}$ has the set of homothetics of the closed convex balanced envelopes of finite unions of sets of $\mathfrak{S}$ as a base.

We now prove the second assertion : first, if F is the base field, it follows from the definition that the $\tilde{\mathfrak{S}}$-topology on $E' = \mathscr{L}(E; F)$ has as a fundamental system of neighbourhoods of 0, the set of polars of the sets of $\tilde{\mathfrak{S}}$. Let A be a bounded subset of E, whose polar $A^\circ$ is a neighbourhood of 0 for the $\tilde{\mathfrak{S}}$-topology; then there exists a closed convex balanced set $B \in \tilde{\mathfrak{S}}$ such that $A^\circ \supset B^\circ$, and so $A \subset B^{\circ \circ}$; but by cor. 3 of II, p. 45, we have $B^{\circ \circ} = B$, and hence $A \subset B$ and $A \in \tilde{\mathfrak{S}}$. Therefore if $\mathfrak{S}'$ is a set of bounded subsets of E, the $\mathfrak{S}'$-topology is coarser than the $\mathfrak{S}$-topology on $E'$ if and only if $\mathfrak{S}' \subset \tilde{\mathfrak{S}}$. The general case follows immediately, since if $y \in F$ is not in the closure of 0, we can verify that the mapping which makes $f \in E'$ correspond to the mapping $x \mapsto f(x)\ y$ is an isomorphism of the locally convex spaces $E'_\mathfrak{S}$ onto its image in $\mathscr{L}_{\tilde{\mathfrak{S}}}(E; F)$.