Lemma 2

Let $\mathscr{T}$ be a locally convex topology on E such that every linear form on E which is continuous for $\mathscr{T}$ belongs to F. Then $\mathscr{T}$ is coarser than $\tau(E, F)$.

Let $\mathscr{U}$ be the set of convex, balanced neighbourhoods of 0 for $\mathscr{T}$. Let $\mathscr{S}$ be the set of polars in F of elements of $\mathscr{U}$. By cor. 2 of III, p. 17, we have $\mathscr{S} \subset \mathscr{S}_0$, and by cor. 1 of prop. 7 of III, p. 19, $\mathscr{T}$ is identical with the $\mathscr{S}'$-topology, where $\mathscr{S}'$ is the set of polars of sets of $\mathscr{U}$ in the dual $E'$ of E. But $E' \subset F$, by hypothesis, hence every set of $\mathscr{S}'$ is contained in a set of $\mathscr{S}$; and the lemma follows.

Let $\mathscr{T}$ be a topology on E compatible with the duality between E and F. Then $\mathscr{T}$ is coarser than $\tau(E, F)$ by lemma 2, and evidently $\mathscr{T}$ is finer than $\sigma(E, F)$. Conversely, F is the dual of E for the topology $\tau(E, F)$ (lemma 1) and for the topology $\sigma(E, F)$ (II, p. 43, prop. 3), hence also for every topology intermediate between $\tau(E, F)$ and $\sigma(E, F)$.