Proposition 4

(i) For any set M of F, the polar set $M^\circ$ is a convex set that contains 0 and is closed in G for the topology $\sigma(G, F)$.

(ii) If M is a cone of vertex 0, then $M^\circ$ is a cone of vertex 0 and it is also the set of $y \in G$ such that $\langle x, y \rangle \geq 0$ for all $x \in M$.

(iii) If M is a vector subspace of F, then $M^\circ$ is a vector subspace of G, and it is also the set of $y \in G$ such that $\langle x, y \rangle = 0$ for all $x \in M$.

(i) Since the linear forms $y \mapsto \langle x, y \rangle$ are continuous for $\sigma(G, F)$ the statement follows immediately from the definitions and the fact that a half-space determined by a hyperplane is convex.

(ii) If M is a cone with vertex 0 and if $x \in M, y \in M^\circ$, then as $\lambda x \in M$, for all $\lambda > 0$, we have $\langle \lambda x, y \rangle \geq -1$, i.e. $\lambda \langle x, y \rangle \geq -1$. Since this holds for all $\lambda > 0$, it follows that $\langle x, y \rangle \geq 0$, and (ii) is proved.

(iii) Similarly, if M is a vector subspace of F, the relations $x \in M, y \in M^\circ$ imply, this time, that $\lambda \langle x, y \rangle \geq -1$ for all real $\lambda$ which is possible only if $\langle x, y \rangle = 0$.

If M is a vector subspace of F we say that $M^\circ$ is the orthogonal of M in G; if $G \subset F^*$, then $M^\circ$ is the intersection of G, and of the subspace orthogonal to M in the algebraic dual $F^*$ of F (A, II, § 2.4, def. 4).

For a vector subspace M of F and a vector subspace N of G we say that M and N are orthogonal if $M \subset N^o$ (or, equivalently, if $N \subset M^\circ$).