Polar Topologies

# Polar Topologies

 Definition: Let $(E, F)$ be a dual pair and let $\mathcal A$ be a collection of $\sigma(E, F)$-weakly bounded subsets of $E$. The Topology of $\mathcal A$-Convergence is the coarsest locally convex topology for which $\{ A^{\circ} : A \in \mathcal A \}$ are neighbourhoods.

It should be remarked that this topology is well-defined. Indeed, if $\mathcal A$ is a collection of $\sigma(E, F)$-weakly bounded subsets of $E$, then by the result on the Criteria for a Subset A to be σ(E, F)-Weakly Bounded, $A^{\circ}$ is absorbent for each $A \in \mathcal A$.

Furthermore, $A^{\circ}$ is absolutely convex for each $A \in \mathcal A$ as proven on The Polar of a Set page.

So $\{ A^{\circ} : A \in \mathcal A \}$ is a collection of absolutely convex and absorbent subsets of $F$. For each $A \in \mathfrak{A}$, let $p'_A : F \to \mathbb{F}$ be the seminorm defined for all $y \in F$ by:

(1)
\begin{align} \quad p'_A(y) := \sup \{ |\langle x, y \rangle| : x \in A \} \end{align}

Note that these are indeed seminorms since each $A \in \mathcal A$ is $\sigma(E, F)$-weakly bounded. Furthermore, note that $p'_A = p_{A^{\circ}}$ where $p_{A^{\circ}}$ is the gauge of $A^{\circ}$.

Then the topology of $\mathcal A$ convergence is the coarsest locally convex topology for which the collection of seminorms $\{ p'_A : A \in \mathcal A \}$ are made continuous, and a base of neighbourhoods for this topology is obtained from sets of the form:

(2)
\begin{align} \epsilon \bigcap_{i=1}^{n} A_i^{\circ} = \bigcap_{i=1}^{n} (\epsilon A_i^{\circ}) = \bigcap_{i=1}^{n} (\epsilon^{-1} A_i)^{\circ} = \left ( \bigcup_{i=1}^{n} \epsilon^{-1} A_i \right )^{\circ} = \left ( \epsilon^{-1} \bigcup_{i=1}^{n} A_i \right )^{\circ} \end{align}

where $\epsilon > 0$ and $A_1, A_2, ..., A_n \in \mathcal A$.

 Proposition 1: Let $(E, F)$ be a dual pair and let $\mathcal A$ be a collection of $\sigma(E, F)$-weakly bounded subsets of $E$. Consider the following properties (B1), (B2), and (B3): (B1) If $A, B \in \mathcal A$ then there exists a $C \in \mathcal A$ such that $A \cup B \subseteq C$. (B2) If $A \in \mathcal A$ and $\lambda \in \mathbf{F}$ then $\lambda A \in \mathcal A$. (B3) $\displaystyle{\bigcup_{A \in \mathcal A} A}$ spans $E$. Then: (1) If properties (B1) and (B2) hold, then the collection $\{ A^{\circ} : A \in \mathcal A \}$ is a base for the topology of $\mathcal A$-convergence on $F$. (2) If property (B3) holds, then the topology of $\mathcal A$-convergence is Hausdorff.
• Proof of (1): Suppose that properties (B1) and (B2) hold. Consider any element $V$ on the base of neighbourhoods for the topology of $\mathcal A$-convergence. Then $V$ has the form:
(3)
\begin{align} \quad V = \left ( \epsilon^{-1} \bigcup_{i=1}^{n} A_i \right )^{\circ} \end{align}
• where $\epsilon > 0$ and $A_1, A_2, ..., A_n \in \mathcal A$. From property (B1) used successively and property (B2), there exists an $A \in \mathcal A$ such that:
(4)
\begin{align} \quad \epsilon^{-1} \bigcup_{i=1}^{n} A_i \subseteq A \end{align}
• Taking the polars of both sides and using a result from the proposition on The Polar of a Set page we see that:
(5)
\begin{align} \quad A^{\circ} \subseteq \left ( \epsilon^{-1} \bigcup_{i=1}^{n} A_i \right )^{\circ} \end{align}
• Thus $\{ A^{\circ} : A \in \mathcal A \}$ is a base for the topology of $\mathcal A$-convergence. $\blacksquare$
 Definition: Let $(E, F)$ be a dual pair. A Polar Topology is a topology on $F$ that is the topology of $\mathcal A$-convergence, for some collection $\mathcal A$ of $\sigma(E, F)$-weakly bounded subsets of $E$.

If $\mathcal A$ is a collection of $\sigma(E, F)$-weakly bounded sets, then the topology of $\mathcal A$-convergence is a polar topology. If $\mathcal A'$ consists of all finite unions and scalar multiples of the sets in $\mathcal A$ then $\mathcal A'$ is also a collection of $\sigma(E, F)$-weakly bounded sets, and the topology of $\mathcal A'$-convergence is a polar topology that coincides with the topology of $\mathcal A$-convergence.

Furthermore, note that if $\mathrm{abs \: conv} (\overline{\mathcal A}^{\sigma(E, F)})$ consists of all the $\sigma(E, F)$-weakly closed absolute convex hulls of the sets in $\mathcal A$ then again, the topology of $\mathrm{abs \: conv} (\overline{\mathcal A}^{\sigma(E, F)})$ convergence coincides with the topology of $\mathcal A$-convergence.

Thus, from this point on, we will assume that $\mathcal A$ consists of $\sigma(E, F)$-weakly closed and absolutely convex sets that satisfy properties (B1) and (B2) (as well as (B3))