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

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

I tried to adapt the Grothendieck completion theorem to prove this, but $\sigma(F,E)$ closure of a bounded set can fail to be bounded.

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

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I think that this is true for any locally convex space $F$. For a subspace $E$ of the topological dual $F^*$ we want to prove $$ \overline{E}^{\beta(F^*,F)}= \{f\in F^*: f|_B \text{ $\sigma(F,E)$-continuous for all $F$-bounded sets $B\}$}$$ where $\beta(F^*,F)$ is the strong toplogy of uniform convergence on all bounded subsets of $B$. Let us denote the set on the right hand side by $\tilde E$ and first prove the simpler implication $\overline E\subseteq \tilde E$: Given $f\in \tilde E$, $B\subseteq F$ absolutely convex (so that continuity is the same as continuity at $0$) and bounded and $\varepsilon>0$ we need a $\sigma(F,E)$-neighbourhood $U$ of $0$ with $|f(x)|<\varepsilon$ for all $x\in U\cap B$. Since $f\in \overline E$ there is $g\in E$ with $|f-g|<\varepsilon/2$ on $B$ and thus $U=\{|g|<\varepsilon/2\}$ does the job.

The other inclusion is based on a lemma which I believe is somewhere (perhaps more or less explicit) in Grothendieck's book on topological vector spaces (sorry, that I don't have a precise reference).

Lemma. Let $B$ be an absolutely convex subset of a locally convex space $(F,\tau)$ which is closed in some finer locally convex topology on $F$ and $\psi:X\to\mathbb K$ be a linear map such that the restriction $\psi|_B$ is $\tau$-continuous. For every $\varepsilon>0$ there is $\varphi\in F^*$ with $|\psi-\varphi|\le\varepsilon$ on $B$.

The lemma is a consequence of the bipolar theorem. Indeed, from the continuity of the restriction we get a closed absolutely convex $0$-neighbourhood $U$ with $|\psi(x)|\le\varepsilon$ for all $x\in B\cap U$, i.e., $\psi\in \varepsilon(B\cap U)^\circ$ where the polar is taken in the algebraic dual $F^\#$ of $F$, i.e., the dual of $F$ endowed with the finest locally convex topology on $F$. Since $B$ and $U$ are closed also with respect to this finer topology the bipolar theorem (the bullet $\bullet$ denote the polar in the predual, and $\Gamma$ denotes absolutely convex hulls) implies $$ (B\cap U)^\circ =(B^{\circ\bullet}\cap U^{\circ\bullet})^\circ = (B^\circ\cup U^\circ)^{\bullet\circ} =\overline{\Gamma(B^\circ \cup U^\circ)}^{\sigma(F^\#,F)}\subseteq B^\circ + U^\circ $$ because this set is closed as the sum of a closed and a compact set (by Alaoglu's theorem, the polar of $U$ is compact and in weak$^*$-dual which is continuously included in $(F^\#,\sigma(F^\#,F)$). Moreover, every element of the polar $U^\circ$ in $F^\#$ is in fact $\tau$-continuous since $U$ is a $\tau$-neighbourhood of $0$. Therefore, there is $\varphi\in (X,\tau)^*$ with $\psi- \varphi\in \delta B^\circ$ which is the conclusion of the lemma.

Applying this lemma to $\tau=\sigma(F,E)$ such that $(F,\tau)^*=E$ and all $F$-closed absolutely convex bounded sets gives the inclusion $\tilde E\subseteq \overline E$.

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  • $\begingroup$ @Tomasz Kania: You might be interested in my try on that old question. $\endgroup$
    – Jochen
    Commented May 12, 2022 at 14:42

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