Definition 3

Let E and F be two vector spaces in duality, and $\mathscr{T}$ the family of subsets of F which are bounded for $\sigma(F, E)$. Then the $\mathscr{T}$-topology on F is denoted by $\beta(E, F)$.

Similarly, we define the topology $\beta(F, E)$ on F. It can be seen easily that the topology $\beta(E, F)$ is identical with $\beta(E, F/E^\circ)$, and we can reduce to the case when the duality between E and F is separating in F.

Remarks. — 3) Let $E_\sigma$ denote the space E endowed with the topology $\sigma(E, F)$. The barrels (III, p. 24) in $E_\sigma$ are the subsets of E which are convex, balanced closed and absorbent for $\sigma(E, F)$. These are none other than the polars of the subsets of F which are convex, balanced and bounded for $\sigma(F, E)$. Consequently, the family of all barrels in $E_\sigma$ is a fundamental system of neighbourhoods of 0 for the topology $\beta(E, F)$ in E. In other words, a semi-norm on E is continuous for $\beta(E, F)$ if and only if it is lower semi-continuous for $\sigma(E, F)$ (cf. III, p. 24, prop. 1).

4) Let $\mathscr{T}$ be a topology on E compatible with the duality between E and F. By prop. 1, (ii) of IV, p. 1, the topology $\beta(F, E)$ on F is none other than the strong topology on F, when F is identified with the dual of E (with the topology $\mathscr{T}$).

5) The topology $\beta(E, F)$ on E is finer than $\tau(E, F)$. It is not, in general compatible with the duality between E and F (cf. however § 2). In particular, a subset of E which is bounded for $\sigma(E, F)$ is not necessarily bounded for $\beta(E, F)$.