Theorem 1 (Banach-Dieudonné)

— Let E be a locally convex metrizable space. The following topologies coincide on the dual $E'$ of E :

a) the topology $\mathscr{T}_\mathfrak{N}$ of $\mathfrak{N}$-convergence, where $\mathfrak{N}$ is the family of subsets of E each consisting of points of a sequence converging to 0 ; b) the topology $\mathscr{T}_c$ of uniform convergence on compact subsets of E ; c) the topology $\mathscr{T}_{pc}$ of uniform convergence on precompact subsets of E ; d) the topology $\mathscr{T}_f$ which is the finest topology inducing the same topology as $\sigma(E', E)$ on every equicontinuous subset of $E'$.

First observe that a subset A of $E'$ is closed for $\mathscr{T}_f$ if and only if $A \cap H$ is closed for $\sigma(E', E)$ for every subset H of $E'$ which is equicontinuous and closed for $\sigma(E', E)$. The weak topology $\sigma(E', E)$ and $\mathscr{T}_{pc}$ induce the same topology on every equicontinuous subset of $E'$ (III, p. 17, prop. 5). Consequently each of the topologies $\mathscr{T}_\mathfrak{N}$, $\mathscr{T}_c$, $\mathscr{T}_{pc}$, $\mathscr{T}_f$ is coarser than the one following it. It is therefore enough to prove that $\mathscr{T}_\mathfrak{N}$ is finer than $\mathscr{T}_f$. Moreover, every translation in $E'$ is a homeomorphism for $\mathscr{T}_f$. Hence it is enough to prove that, if F is a subset of $E'$ which is closed for $\mathscr{T}_f$, and does not contain 0, then there exists a set $S \in \mathfrak{N}$ such that $S^\circ \cap F = \varnothing$.

Let $(U_n)_{n \geq 0}$ be a decreasing sequence of neighbourhoods of 0 in E forming a fundamental system of neighbourhoods of 0. We shall construct, by induction on $n \geq 0$, finite sets $X_n$ such that we have

for every integer $n \geqslant 0$. Let $m \geqslant 0$ be an integer such that $X_n$ has been constructed for $0 \leqslant n < m$ and satisfies (4) and (5) for $0 \leqslant n < m$. For every $x \in U_m$, put

Formula (5) with $n = m - 1$ implies that $\bigcap_{x \in U_m} F_x = \varnothing$. Further, the set $U_{m+1}^{\circ}$ is equicontinuous, and compact for $\sigma(E', E)$. In view of the definition of $\mathscr{T}_f$, each of the sets $F_x$ is compact for $\sigma(E', E)$; therefore there exists a finite subset $X_m$ of $U_m$ such that $\bigcap_{x \in X_m} F_x = \varnothing$, i.e. relation (5) is satisfied for $n = m$.

Put $S = \bigcup_{n \geqslant 0} X_n$. We have $X_n \subset U_p$ for $n \geqslant p$, therefore $S$ is the set of points of a sequence which converges to 0 in $E$. From (5) we deduce that $S^{\circ} \cap U_{n+1}^{\circ} \cap F = \varnothing$, and since $E'$ is the union of the sequence of sets $U_{n+1}^{\circ}$, we get $S^{\circ} \cap F = \varnothing$.