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Originally asked on MSE.

Let $E$ be a metrizable locally convex topological vector space and let $E^{*}$ be its dual space endowed with the strong topology = topology of uniform convergence on (closed convex balanced) bounded subsets of $E$. Let $F\subset E^{*}$ be a linear subspace.

Is it true that $\overline{F}=F_1$, where $F_1$ is the set of all linear functionals on $E$, whose restrictions on every (closed convex balanced) bounded subset of $E$ are continuous in the weak $\sigma(E,F)$ topology?

It seems that I can adapt the proof of Grothendieck's completion theorem to prove this. Indeed, every linear functional continuous on a closed set $B$ in $E$ can be uniformly approximated on $B$ by an element of $E^{*}$ (I guess this is Grothendieck's lemma). Hence, we only need to show that $F_1\subset E^{*}$. But this follows from the fact, that since $E$ is metrizable, $E^{*}$ is complete, and so it is a closed subset of $E'$ (the algebraic dual), andowed with the uniformity of uniform convergence on bounded sets.

If this is correct, I hope this result is contained in some textbook on locally convex spaces, and so a reference would be highly appreciated.

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  • $\begingroup$ Is $F$ intended to be a linear subspace? The definition of $F_1$ implies that it is a linear space. $\endgroup$ Apr 29 '18 at 0:40
  • $\begingroup$ @RobertFurber Yes, thank you, I'll add that. $\endgroup$
    – erz
    Apr 29 '18 at 6:32
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I don't know a reference, but I think you can prove this without using the algebraic dual: To see $F_1\subseteq E^*$ you can use the fact that metrizable spaces are bornological, i.e., a linear map on $E$ is continuous if all restrictions to bounded sets are continuous. This shows that $F_1$ is a closed subspace of $(E^*, \beta(E^*,E))$ which clearly contains $F$.

For the other inclusion, given $f\in F_1$ and a $0$-neighbourhood $U$ of $(E^*, \beta(E^*,E))$, there is a (closed absolutely convex) bounded set $B$ in $E$ with $B^\circ \subseteq U$. Since $f|_B$ is $\sigma(E,F)$-continuous there is a finite set $M\subseteq F$ with $f\in \left(B\cap M^\bullet\right)^\circ$ (where the $\bullet$ indicates the polar in $E$). The theorem of bipolars then gives $f\in B^\circ + M^{\bullet\circ} = B^\circ + \Gamma(M) \subseteq U+F$ (because the absolutely convex hull $\Gamma(M)$ of a finite set is compact and hence closed). This shows $f\in \overline{F}^{\beta(E^*,E)}$.

This proof uses metrizability only to obtain bornologicity. The statement thus holds e.g. for strong duals of distinguished Frechet spaces.

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  • $\begingroup$ Thank you for your answer, but I don't see major differences between your proof and mine. In fact, your second paragraph is very similar to the proof of Grothendieck's lemma. I was aware that metrizability was only used to get bornologicity, just thought (perhaps mistakingly), that it is better to state the question this way. $\endgroup$
    – erz
    Apr 29 '18 at 15:46

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