Proposition 10

Let $G$ be a real vector space and $F = G^*$ its dual with the topology $\sigma(G^*, G)$.

(i) The mapping $V \mapsto V^\circ$ is a bijection of the set of vector subspaces of $G$ on the set of closed vector subspaces of $F$.

(ii) Every closed vector subspace of $F$ is a product of lines and has a topological complement.

By the bipolar theorem (II, p. 45, cor. 2) $V \mapsto V^\circ$ is a bijection of the set of vector subspaces $V$ of $G$, closed for $\sigma(G, G^*)$ on the set of closed vector subspaces of $F$. But, by definition, every linear form on $G$ is continuous for $\sigma(G, G^*)$, therefore every vector subspace in $G$ is closed, being defined by a system of equations $\langle y, y_\lambda^* \rangle = 0$ (where $y_\lambda^* \in G^*$); this proves (i).

Now let $W$ be a closed subspace of $F$; we have then $W = V^\circ$ with $V = W^\circ$ in $G$. Let $V'$ be a complement of $V$ in $G$. We know that $F = G^*$ can be canonically identified with $V^* \oplus {V'}^*$, and ${V'}^*$ identified with $V^\circ = W$ (A, II, § 2.6, cor. to prop. 10); further (II, p. 50, prop. 8) the topology $\sigma(G^*, G)$ can be identified with the product of the topologies $\sigma(V^*, V)$ and $\sigma({V'}^*, V')$; this proves assertion (ii).

Though, for the topology $\sigma(G, G^*)$, every vector subspace of $G$ is closed, we note that if $G$ is of infinite dimension then the topology $\sigma(G, G^*)$ is not the finest locally convex topology on $G$, every neighbourhood of 0 for $\sigma(G, G^*)$ containing a vector subspace of infinite dimension : it is however the finest of the weak topologies on $G$ (II, p. 43, cor. 3).