# Closure in the strong dual topology.

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.

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.
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 $$\tilde E\subseteq \overline E$$.