Equicontinuity Classification of Hausdorff Locally Compact Topologies

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Theorem 1: Let

$E$

be a Hausdorff locally convex topological vector space, let

\tau

denote its Hausdorff and locally convex topology and let

\mathcal A

be the collection of all

\tau

-equicontinuous sets of linear forms on

$E$

. Then

\tau

is the topology of

\mathcal A

-convergence.

Proof: Let $\mathcal A$ be the set of all $\tau$ -equicontinuous sets of linear forms on $$E$$ . We first need to show that $\mathcal A$ is a collection of $\sigma (E', E)$ -weakly bounded sets to even consider the topology $\mathcal A$ -convergence.

Since $\tau$ is a locally convex topology, it has a base of closed and absolutely convex neighbourhoods and absorbent of the origin as proven on the Every LCTVS Has a Base of Closed Absolutely Convex Absorbent Neighbourhoods of the Origin page. Let $\mathcal U$ denote this base.

Let $V \in \mathcal A$ . Since $\tau$ is Hausdorff and since $$V$$ is $\tau$ -equicontinuous, by The Polar Criterion for Equicontinuity of a Set of Linear Forms there exists a $U \in \mathcal U$ such that $V \subseteq U^{\circ}$ . Taking polars again and using one of the properties on The Polar of a Set page, and we see that $U^{\circ \circ} \subseteq V^{\circ}$ .

Since $\tau$ is Hausdorff and since each $U \in \mathcal U$ is closed and absolutely convex, by considering the pair of dual pairs $(E, E^{\tau'})$ and $(E^{\tau'}, E)$ , we have by the theorem on the If E is a Hausdorff LCTVS then A°° (in E) is the Closed Absolutely Convex Hull of A page that for each $U \in \mathcal U$ :

(1)

\begin{align} U^{\circ \circ} = \overline{\mathrm{abs \: conv} (U)} = U \end{align}

Therefore $U \subseteq V^{\circ}$ . But since $$U$$ is absorbent, so is $V^{\circ}$ . Indeed, if $x \in E$ then by the absorbency of $$U$$ there exists a $\lambda > 0$ such that if $\mu \in \mathbf{F}$ is such that $|\mu| \geq \lambda$ then $x \in \mu U$ . Since $U \subseteq V^{\circ}$ we see that if $\mu \in \mathbf{F}$ is such that $|\mu| \geq \lambda$ then $x \in \mu V^{\circ}$ . So $V^{\circ}$ is absorbent (in $$E$$ ). So by the Criteria for a Subset A to be σ(E, F)-Weakly Bounded, each $V \in \mathcal A$ is a $\sigma(E', E)$ -weakly bounded set.

All that remains to show is that $\tau$ is the topology of $\mathcal A$ -convergence.

First note that since $U = U^{\circ \circ}$ for all $U \in \mathcal U$ , we have that $\tau$ is the topology of uniform convergence of the sets $\{ U^{\circ} : U \in \mathcal U \}$ and that $\{ U^{\circ} : U \in \mathcal U \}$ is a base for $\tau$ . If $V \in \mathcal A$ then there exists a $U \in \mathcal U$ such that:

(2)

\begin{align} V \subseteq U^{\circ} \end{align}

However, $\{ V^{\circ} : V \in \mathcal A \}$ is a base of neighbourhoods for the topology of $\mathcal A$ -convergence, while $\{ U^{\circ \circ} : U \in \mathcal U \} = \mathcal U$ is a base of neighbourhoods for $\tau$ . Therefore the topology of $\mathcal A$ -convergence is coarser than $\tau$ .

(3)

\begin{align} U \subseteq V^{\circ} \end{align}

So every neighbourhood in $\mathcal U$ is contained in a neighbourhood of the topology of $\mathcal A$ -convergence. Therefore $\tau$ is coarser than the topology of $\mathcal A$ -convergence.

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