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Given $(X,s)$ a (real) barreled locally convex space (that is, every closed convex and absorbing set in $(X,s)$ is a neighborhood of the origin), is there a (strictly) finer, non-barreled linear locally convex topology $\tau$ on $X$ that is compatible with the duality $(X,X^*)$?

In other words $s\preceq\tau$, $(X,s)^*=(X,\tau)^*$, and $\tau$ is not barreled.

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2 Answers 2

A slightly different argument: Any barrelled space $X$ with continuous dual $X'$ carries the strong topology $\beta(X,X')$ as the latter is the finest polar topology on $X$ of the dual pair $(X,X')$ and - as the space is barrelled - every closed polar is a neighbourhood of $0$.

So you have $\beta(X,X')$ compatible with the duality, hence by Mackey-Arens $\beta(X,X') = \tau(X,X')$, the Mackey topology, which in turn is the finest topology compatible with the duality. So there is no finer topology compatible with the duality.

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Thank you for the new argument. –  Neutral Element Nov 16 '12 at 21:11
up vote 1 down vote accepted

The answer is negative. Every linear locally convex topology $\tau$ on $X$ that is compatible with the duality has the same closed convex sets as $s$. Hence, every $\tau-$closed convex and absorbing set $M$ is $s-$closed convex and absorbing, and since $s$ is barreled, $M$ is an $s-$neighborhood of $0$. But because $s\preceq\tau$, $M$ is also a $\tau-$neighborhood of $0$. Therefore every $\tau-$closed convex and absorbing set is a $\tau-$neighborhood of $0$, that is, $\tau$ is barreled.

Conclusion: -all topologies compatible with the duality and finer that a given barreled topology are barreled.

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