Proposition 6

Suppose that the duality between $F$ and $G$ (resp. $F_1$ and $G_1$) is separating in $G$ (resp. $F_1$). Let $u$ be a linear mapping of $F$ in $F_1$ that is continuous for $\sigma(F, G)$ and $\sigma(F_1, G_1)$. Let $A$ be a set in $F$ and $A_1$ a set in $F_1$; then:

(i) We have $(u(A))^{\circ} = {}^t u^{-1}(A^{\circ})$.

(ii) We have ${}^t u(A_1^{\circ}) \subset (u^{-1}(A_1))^{\circ}$; further, if $A$ is closed, (for $\sigma(F_1, G_1)$) convex, and contains the origin, then we have ${}^t u(A_1^{\circ}) = (u^{-1}(A_1))^{\circ}$.

Let $z_1 \in G_1$, the relation $z_1 \in (u(A))^{\circ}$ is equivalent to $\langle u(y), z_1 \rangle \geq -1$ for all $y \in A$, and the relation ${}^t u(z_1) \in A^{\circ}$ is equivalent to $\langle y, {}^t u(z_1) \rangle \geq -1$ for all $y \in A$ and our assertion (i) follows using (1). Next interchanging $u$ and ${}^t u$ and applying (i) to the set $A_1^{\circ}$ of $G_1$ we get

from which, on taking polars

We have ${}^t u(A_1^{\circ}) \subset ({}^t u(A_1^{\circ}))^{\circ \circ}$ by the bipolar theorem (II, p. 44, th. 1); the final statement follows from (2) and the bipolar theorem since then $A_1^{\circ \circ} = A_1$ and ${}^t u(A_1^{\circ})$ is convex and contains the origin.