[Originally asked on MSE.](https://math.stackexchange.com/questions/2739990/closure-in-the-strong-dual-topology)


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.



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> 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?
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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.



Closure in the strong dual topology
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.


