Proposition 7

— Let M be a subset of $E'$. The following conditions are equivalent :

(i) $M$ is equicontinuous ; (ii) $M$ is contained in the polar of a neighbourhood of 0 in E ; (iii) the polar of M is a neighbourhood of 0 in E.

If M is equicontinuous, there exists a convex balanced neighbourhood V of 0 such that $|u(x)| \leq 1$ for all $x \in V$ and all $u \in M$; then we have that $M \subset V^\circ$ and (i) implies (ii). With the same notations, if $M \subset V^\circ$ then $V \subset V^{\circ\circ} \subset M^\circ$ and (ii) implies (iii). Finally, if $M^\circ$ contains a convex balanced neighbourhood V of 0, then $M \subset M^{\circ\circ} \subset V^\circ$ and the relations $x \in \varepsilon V,\ u \in M$ imply $|u(x)| \leq \varepsilon$ for all $\varepsilon > 0$, which proves that (iii) implies (i).

We remark that every $x \in E$ defines a mapping $j(x): u \mapsto u(x)$ from $E'$ into K. Hence we can talk of the $\mathscr{S}$-topology on E, where $\mathscr{S}$ is a family of subsets of $E'$: this is the inverse image under $j$ of the $\mathscr{S}$-topology on $K^{E'}$. We verify immediately that if $\mathscr{S}$ is a convex bornology on $E'$, then the polars of sets of $\mathscr{S}$ form a fundamental system of neighbourhoods of 0 for the $\mathscr{S}$-topology on E. This is so, in particular, when $\mathscr{S}$ is the family of equicontinuous subsets of $E'$ and prop. 7 implies :