(E, F), (G, H): Weakly Continuous t from E to G, (t(A))° = t'-1(A°)

For Dual Pairs (E, F), (G, H), and a Weakly Continuous Linear Operator t from E to G, (t(A))° = t'-1(A°)

Proposition 1: Let $(E, F)$ and $(G, H)$ be dual pairs and let $t : E \to G$ be a linear operator. If $A \subseteq E$ and $t$ is weakly continuous then $(t(A))^{\circ} = t'^{-1}(A^{\circ})$.

• Proof: $(t(A))^{\circ}$ is the set of all points $w \in H$ such that:

(1)

\begin{align} \quad |\langle t(x), w \rangle| = |\langle x, t'(w) \rangle| \leq 1 \end{align}

• for all $x \in A$. Note that for each $w \in H$, since $t$ is weakly continuous, $t'(H) \subseteq F$ so that $t'(w) \in F$ and so that $|\langle x, t'(w) \rangle|$ is valid in the above expression.

• since $t : E \to G$ is weakly continuous, for each $w \in H$ there exists a $y_w \in F$ such that $y_w = t'(w)$ and then $|\langle x, y_w \rangle| \leq 1$ for all $x \in A$, so that $y_w \in A^{\circ}$.

• Hence $(t(A))^{\circ}$ is the set of all points $w \in H$ such that $w$ is the preimage under $t'$ of a point in $A^{\circ}$, that is:

(2)

\begin{align} \quad (t(A))^{\circ} = t'^{-1}(A^{\circ}) \quad \blacksquare \end{align}