# If E ⊆ G ⊆ F* and (E, F) is a Dual Pair, A ⊆ A°°

Recall from The Bipolar of a Set page that if $(E, F)$ and $(F, G)$ are dual pairs and if $A \subseteq E$ then the bipolar of $A$ in $G$, $A^{\circ \circ}$, is defined to be the polar (in $G$) of the polar of $A$ in $F$.

Let $E$, $F$, and $G$ be vector spaces such that $E \subseteq G \subseteq F^*$. Then observe that if $(E, F)$ is a dual pair, then so is $(F, G)$, with the duality pairing given for all $y \in F$ and for all $z \in G$ by:

(1)(Noting that $g \in F^*$ since $G \subseteq F^*$). The case when $(E, F)$ is a dual pair and $E \subseteq G \subseteq F^*$ will be of crucial importance to us.

If $E$ is a Hausdorff locally convex topological vector space then $(E, E')$ is a dual pair. Also, $(E', E'^*)$ is a dual pair and is such that:

(2)Furthermore, since $(E, E')$ is a dual pair, so is $(E', E)$, and is such that:

(3)Lemma 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 $z \in A^{\circ \circ}$ if and only if $\displaystyle{|\langle z, y \rangle| \leq \sup \{ |\langle x, y \rangle| : x \in A \}}$ for all $y \in F$. |

**Proof:**Since $E \subseteq G \subseteq F^*$ and since $(E, F)$ is a dual pair, so is $(F, G)$ from the remarks made earlier so that the bipolar $A^{\circ \circ}$ in $G$ is defined.

- Observe that $z \in A^{\circ \circ}$ if and only if:

- Or equivalently, $|\langle z, y \rangle| \leq 1$ for all $y \in A^{\circ}$. But $y \in A^{\circ}$ if and only if $\sup \{ |\langle x, y \rangle| : x \in A \} \leq 1$. Therefore $z \in A^{\circ \circ}$ if and only if:

- for all $y \in F$ whenever $\sup \{ |\langle x, y \rangle| : x \in A \} \leq 1$ for all $y \in F$. But $|\langle y, z \rangle|$ and $\displaystyle{\sup \{ |\langle x, y \rangle| : x \in A \}}$ are both seminorms on $F$, and so by one of the propositions on the Seminorms and Norms on Vector Spaces page we have that the above criterion is equivalent to:

- for all $y \in F$. $\blacksquare$

Proposition 2: 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 $A \subseteq A^{\circ \circ}$. |

**Proof:**Let $z \in A$. Since $A \subseteq E \subseteq G$ we have that $z \in G$. But then trivially:

- so that by Lemma 1 we have that $z \in A^{\circ \circ}$. Thus $A \subseteq A^{\circ \circ}$. $\blacksquare$