The Polar of an Intersection of σ(E, F)-Closed Absolutely Convex Sets

# The Polar of an Intersection of σ(E, F)-Closed Absolutely Convex Sets

 Proposition 1: Let $(E, F)$ be a dual pair and let $\{ A_{\alpha} \}_{\alpha}$ be a collection of $\sigma(E, F)$-closed absolutely convex subsets of $E$. Then $\displaystyle{\left ( \bigcap_{\alpha} A_{\alpha} \right )^{\circ} = \overline{\mathrm{abs \: conv} \left ( \bigcup_{\alpha} A_{\alpha}^{\circ} \right )}^{\sigma(F, E)}}$, that is, $\displaystyle{\left (\bigcap_{\alpha} A_{\alpha} \right )^{\circ}}$ is the $\sigma(F, E)$-closed absolutely convex hull of $\displaystyle{\bigcup_{\alpha} A_{\alpha}^{\circ}}$.
• Proof: Since $(E, F)$ is a dual pair, so is $(F, E)$, and $E \subseteq E \subseteq F^*$.
• Taking the polar of $\displaystyle{\bigcup_{\alpha} A_{\alpha}^{\circ}}$ (in $E$) and we get that:
(1)
\begin{align} \quad \left ( \bigcup_{\alpha} A_{\alpha}^{\circ} \right )^{\circ} = \bigcap_{\alpha} A_{\alpha}^{\circ \circ} \end{align}
(2)
\begin{align} \quad \left ( \bigcup_{\alpha} A_{\alpha}^{\circ} \right )^{\circ} = \bigcap_{\alpha} A_{\alpha} \end{align}
• Taking the polar of both sides of the equation above yields:
(3)
\begin{align} \left ( \bigcup_{\alpha} A_{\alpha}^{\circ} \right )^{\circ \circ} = \left ( \bigcap_{\alpha} A_{\alpha} \right )^{\circ} \end{align}
• By the theorem referenced above, $\displaystyle{\left ( \bigcup_{\alpha} A_{\alpha}^{\circ} \right )^{\circ \circ}}$ is the $\sigma(F, E)$-closed absolutely convex hull of $\displaystyle{\bigcup_{\alpha} A_{\alpha}^{\circ}}$, and thus:
(4)
\begin{align} \left ( \bigcap_{\alpha} A_{\alpha} \right )^{\circ} = \overline{\mathrm{abs \: conv} \left ( \bigcup_{\alpha} A_{\alpha}^{\circ} \right )}^{\sigma(F, E)} \quad \blacksquare \end{align}