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:

\begin{align} \quad \left ( \bigcup_{\alpha} A_{\alpha}^{\circ} \right )^{\circ} = \bigcap_{\alpha} A_{\alpha}^{\circ \circ} \end{align}

- by one of the properties in a proposition on The Polar of a Set page. But each $A_{\alpha}^{\circ \circ}$ is the $\sigma(E, F)$-closed absolutely convex hull of $A_{\alpha}$ by the theorem on the If E ⊆ G ⊆ F* and (E, F) is a Dual Pair then A°° is the σ(G, F)-Closed Absolutely Convex Hull of A page. But each $A_{\alpha}$ is already assumed to be $\sigma(E, F)$-closed and absolutely convex, so $A_{\alpha}^{\circ \circ} = A_{\alpha}$, and thus:

\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:

\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:

\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}