Proof, [IV.1.5, SW99]. By Proposition 16.2.4, $A^{\circ \circ}$ is a $\sigma(E, F)$-closed, convex set that contains $0$. Since $A^{\circ \circ}\supset A$, it is sufficient to show that $A^{\circ\circ}\subset \ol{\conv}(A \cup \bracs{0})$.

Let $x_{0} \in E \setminus \ol{\conv}(A \cup \bracs{0})$, then by the [Hahn-Banach Theorem]theorem:hahn-banach-geometric-2, there exists $\phi: E \to \real$ such that:

1. (1)

   $\phi$ is $\sigma(E, F)$-continuous.

2. (2)

   $\phi(\ol{\conv}(A \cup \bracs{0})) \subset (-\infty, 1)$ and $\phi(x_{0}) > 1$.

If $K = \real$, let $\Phi = \phi$. If $K = \complex$, then by Proposition 10.4.1, the mapping $\Phi(x) = \phi(x) - i\phi(ix)$ is a $\sigma(E; F)$-continuous $K$-linear map on $E$ such that $\text{Re}(\Phi) = \phi$. By Lemma 16.1.3, there exists $y \in F$ such that $\dpn{x, y}{\lambda}= \Phi(x)$ for all $x \in E$. In which case, for any $x \in E$,

Therefore $y \in A^{\circ}$, but $\text{Re}\dpn{x_0, y}{\lambda}> 1$, so $x_{0} \not\in A^{\circ\circ}$.$\square$