Proposition 16.2.4 ([IV.1.4, SW99]).label Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$ and $A \subset E$, then
- (1)
$A^{\circ}$ is a $\sigma(F, E)$-closed convex subset of $F$ containing $0$.
- (2)
If $A$ is circled, then so is $A^{\circ}$.
- (3)
If $A$ is a subspace of $E$, then
\[A^{\circ} = A^{\perp} = \bracs{y \in F| \dpn{x, y}{E} = 0 \forall x \in A}\]and $A^{\circ}$ is a subspace of $F$.
Proof. (1): For each $x \in E$,
is the sublevel set of a continuous $\real$-linear functional, so it is $\sigma(F, E)$-closed and convex. Since $A^{\circ} = \bigcap_{x \in A}\bracs{x}^{\circ}$, $A$ is also $\sigma(F, E)$-closed and convex.
(2): If $A$ is circled, then by Proposition 16.2.3,
For any $y \in A^{\circ}$ and $\alpha \in K \setminus \bracs{0}$ with $|\alpha| \le 1$. Since $y \in \alpha^{-1}A^{\circ}$, $\alpha y \in A^{\circ}$, so $A^{\circ}$ is circled.$\square$