[I have posted this question on MSE some time ago, but received no answer.]


It is known, that if two locally convex topologies on a vector space determine the same collection of continuous linear functionals, then the classes of closed convex sets are the same, as well as the classes of bounded sets. Consequently, the classes of closed bounded convex sets also have to coincide.



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> I am wondering about the converse of this statement.
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In particular I am interested if the following holds. Let $E$ and $F$ be vector spaces in duality, and let $H\subset F$ be such that the class of closed bounded absolutely convex sets with respect to $\sigma(E,H)$ is the same as with respect to $\sigma(E,F)$.



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> What can be said about $H$? Is it true that $H=F$?
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Topologies of pairs and closed bounded convex sets
No. Let $E=\bigoplus\_{\mathbb N} \mathbb R$ be the space of scalar sequences with only finitely many non-zero entries, $F=\mathbb R^{\mathbb N}$ with the duality $\langle x,y\rangle=\sum x\_ny\_n$, and $H$ be an algebraic complement in $F$ of the one-dimensional space generated by $(1,1,1,\ldots)$. Every $\sigma(E,H)$-bounded absolutely convex set $B\subseteq E$ is then finite dimensional (this should be a standard proof by contradiction like the one that $\sigma(E,F)$-bounded sets are finite-dimensional). 


Then $H$ and $F$ yield the same $\sigma$-bounded and $\sigma$-closed subsetes of $E$.


