Theorem 17.1.6 (Goldstine).label Let $\dpn{E, F}{\lambda}$ be a duality over $K \in \RC$, $A \subset E$ be non-empty, convex, circled, and $\sigma(E, F)$-compact, and $B$ be the closed unit ball of $E_{A}^{*}$, then $B \cap F$ is $\sigma(E_{A}^{*}, E_{A})$-dense in $B$.

Proof. For any $S \subset E$ or $S \subset F$, denote $S^{\circ}$ as the polar of $S$ with respect to $\dpn{E, F}{\lambda}$. For any $S \subset E_{A}$ or $S \subset E_{A}^{*}$, denote $S^{\bullet}$ as the polar of $S$ with respect to $\dpn{E_A, E_A^*}{E_A}$.

Since $B_{E_A}(0, 1)$ is circled and $A = \ol{B_{E_A}(0, 1)}^{E_A}$ is compact in $E$,

\[B \cap F = \bracsn{\phi \in F| \text{Re}\dpn{x, \phi}{\lambda} \le 1 \forall x \in A}= A^{\circ}\]

is the polar of $A$ with respect to $\dpn{E, F}{\lambda}$. Now, as $A$ is convex, circled, and compact, the Bipolar Theorem implies that

\begin{align*}A^{\circ\bullet}&= \bracsn{x \in E_A|\text{Re}\dpn{x, \phi}{\lambda} \le 1 \forall \phi \in A^\circ}\\&= E_{A} \cap \bracsn{x \in E|\text{Re}\dpn{x, \phi}{\lambda} \le 1 \forall \phi \in A^\circ}= E_{A} \cap A^{\circ\circ}= A\end{align*}

Given that $B \cap F$ is a convex and circled subset of $E_{A}^{*}$,

\[\ol{B \cap F}^{\sigma(E_A^*, E_A)}= (B \cap F)^{\bullet\bullet}= A^{\circ\bullet\bullet}= A^{\bullet} = B\]

$\square$

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