A°° is the σ(G, F)-Closed Absolutely Convex Hull of A

# If E ⊆ G ⊆ F* and (E, F) is a Dual Pair then A°° is the σ(G, F)-Closed Absolutely Convex Hull of A

Theorem 1: Let $E$, $F$, and $G$ be vector spaces such that $E \subseteq G \subseteq F^*$ (as subspaces), and let $(E, F)$ be a dual pair. Let $A \subseteq E$. Then $\displaystyle{A^{\circ \circ} = \overline{\mathrm{abs \: conv} (A)}^{\sigma (G, F)}}$, that is, the bipolar $A^{\circ \circ}$ is the $\sigma(G, F)$-closed absolutely convex hull of $A$. |

**Proof:**Let:

\begin{align} B = A^{\circ \circ} = \overline{\mathrm{abs \: conv} (A)}^{\sigma (G, F)} \end{align}

- Our goal is to show that $A^{\circ \circ} = B$. First observe that since $E \subseteq G \subseteq F^*$ and since $(E, F)$ is a dual pair, by the result on the If E ⊆ G ⊆ F* and (E, F) is a Dual Pair, A ⊆ A°° page, we have that $A \subseteq A^{\circ \circ}$. Furthermore, $A^{\circ \circ}$ is absolutely convex and $\sigma(G, F)$-closed as it is the polar in $G$ of $A^{\circ}$ and by applying one of the propositions on The Polar of a Set page. Since $B$ is the smallest such $\sigma(G, F)$-closed and absolutely convex set containing $A$, we must have that:

\begin{align} \quad B \subseteq A^{\circ \circ} \end{align}

- For the reverse inclusion, we will show that $B^C \subseteq A^{\circ \circ C}$. Let $a \in B^C$. Then $\displaystyle{a \not \in B = \overline{B}^{\sigma(G, F)}}$. Since $G$ equipped with the $\sigma (G, F)$ topology is a locally convex topological vector space and since $B$ is absolutely convex with $a \not \in \overline{B}^{\sigma(G, F)}$, by one of the corollaries to The Hahn-Banach Separation Theorem, there exists a $\sigma(G, F)$-continuous linear form $f$ on $G$ such that $|\langle b, f \rangle| \leq 1$ for all $b \in B$ and such that $\langle a, f \rangle > 1$.

- Since $A \subseteq B$ (as again, $B$ is the smallest $\sigma(G, F)$-closed absolutely convex set containing $A$), we must have that $f \in A^{\circ}$ since:

\begin{align} \quad \sup \{ |\langle x, f \rangle| : x \in A \} \leq \sup \{ \underbrace{|\langle b, f \rangle |}_{\leq 1} : b \in B \} \leq 1 \end{align}

- However, since $E \subseteq G \subseteq F^*$ and $(E, F)$ is a dual, and:

\begin{align} \quad |\langle a, f \rangle| > 1 \geq \sup \{ |\langle x, f \rangle| : x \in A \} \end{align}

- we have by the proposition on the If E ⊆ G ⊆ F* and (E, F) is a Dual Pair, A ⊆ A°° page that $a \not \in A^{\circ \circ}$. So $a \in A^{\circ \circ C}$. Thus $B^C \subseteq A^{\circ \circ C}$ and consequentially:

\begin{align} \quad A^{\circ \circ} \subseteq B \end{align}

- Therefore $A^{\circ \circ} = B$. $\blacksquare$