Lemma 1

Let A be a closed convex balanced subset of E and let u be a linear form on E whose restriction to A is continuous. Then for every \varepsilon > 0, there exists an x' \in E' such that

Let \varepsilon > 0. There exists a closed convex balanced neighbourhood U of 0 in E such that |u(x)| \leq \varepsilon for all x \in U \cap A. We know that the polar U^\circ of U in E* is contained in E' and is compact for the topology \sigma(E^*, E) (III, p. 17, cor. 2). Since the polar A^\circ of A in E* is closed for \sigma(E^*, E), it follows that A^\circ + U^\circ is a closed convex subset of E* (GT, III, § 4, No. 1, cor. 1).

Let C be a closed convex balanced subset of E. Then C is closed for \sigma(E, E') (II, p. 45, cor. 3), hence also for \sigma(E, E^*), and as a consequence, we have C = C^{\circ\circ} (for the duality between E and E*). As a result, we have

from which, we get

Since the linear form \varepsilon^{-1}u belongs to (A \cap U)^\circ, there exist v \in A^\circ and w \in U^\circ such that u = \varepsilon(v + w). Hence x' = \varepsilon w belongs to E' and u - x' = \varepsilon v is bounded above in absolute value by \varepsilon on A; hence the lemma.

Now let E be a locally convex Hausdorff space and \hat{E} its completion. Every continuous linear form f on E extends to \hat{E} by continuity; hence we have (\hat{E})' = E'

(III, p. 16) and every element of $\hat{E}$ defines a linear form on $E'$; that is, an element of the algebraic dual ${E'}^*$ of $E'$. In addition, the duality between $E$ (resp. $\hat{E}$) and $E'$ is separating (II, p. 24, cor. 1). Consequently $E$ and $\hat{E}$ can be identified with vector subspaces of ${E'}^*$.