Lemma 1

The subspace F of E consists of all linear forms on E which are continuous for $\tau(E, F)$.

Every element of F is a continuous mapping for $\sigma(E, F)$, hence for $\tau(E, F)$.

Conversely, let $f \in E^*$ be continuous for $\tau(E, F)$. There exists a neighbourhood U of 0 in E (for $\tau(E, F)$), such that $|f| \leq 1$ on U; we can assume that there exists a set $A \in \mathscr{S}_0$ such that $U = A^\circ$. In other words, $f$ belongs to the bipolar $A^{\circ\circ}$ of A for the duality between $E^*$ and E. But the topology $\sigma(F, E)$ on F is induced by $\sigma(E^*, E)$; consequently A is convex, balanced and compact for $\sigma(E^*, E)$, and the theorem of bipolars (II, p. 44, th. 1) implies the equality $A = A^{\circ\circ}$. Therefore we have that $f \in F$, from which the lemma follows.