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Let $E$ be a locally convex space, let $E^{\prime}$ be its continuous dual space and let $F$ be a subspace of $E^{\prime}$ which is dense with respect to the strong topology on $E^{\prime}$ (i.e. the topology of uniform convergence on bounded subsets on $E$). Under what assumptions on $E$ is it possible to conclude that the weak topology induced by $F$ equals the one induced by $E^{\prime}$, i.e. when does $\sigma(E,F) = \sigma(E,E^{\prime})$ hold?

I think that the statement is false in general, but I don't know a counterexample at the moment.

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As a condition on $E$, you need that $E$ is finite-dimensional (or, if Hausdorffness is not assumed, that $E/\overline{\{0\}}$ is finite-dimensional).

Since $F$ is dense in $b(E',E)$, it separates points on $E$, and thus $\sigma(E,F)$ is a locally convex Hausdorff topology, and we have

$$\sigma(E,F)' = F; \quad \sigma(E,E')' = E',$$

so if $\sigma(E,F) = \sigma(E,E')$, it follows that $F = E'$.

That means to have $\overline{F} = E' \Rightarrow \sigma(E,F) = \sigma(E,E')$ you need a condition that implies $\overline{F} = E' \Rightarrow F = E'$. The only such condition, as far as I'm aware, is finite-dimensionality.

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  • $\begingroup$ I follow you to the point that $F = E'$. But how does this imply that $E$ is finite-dimensional? $\endgroup$ – Sebastian Feb 4 '14 at 15:47
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    $\begingroup$ That does not imply finite-dimensionality. It goes the other way, finite-dimensionality of $E$ guarantees that a dense subspace of $E'$ is all of $E'$. If $E$ is infinite-dimensional, $E'$ has - if I remember correctly, under all circumstances - proper dense subspaces, and such an $F$ induces a different topology than $\sigma(E,E')$. To conclude $\sigma(E,F) = \sigma(E,E')$ from the denseness of $F$, you need a condition that says $F = E'$, i.e. $E'$ has no proper dense subspaces (hence must be finite-dimensional). $\endgroup$ – Daniel Fischer Feb 4 '14 at 15:53

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