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[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.

I am wondering about the converse of this statement.

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 $\sigma(E,F)$.

What can be said about $H$? Is it true that $H=F$?

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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 $E$.

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This is just to fill the details of Jochen Wengenroth's answer.

Every $\sigma(E,F)$- bounded set is finitely dimensional. Indeed, one can show that for any linearly independent sequence $A=\{a_n\}_{n=1}^{\infty}\subset E$ and $b=(b_1,b_2,...)\in \mathbb{R}^{\mathbb{N}}$ there is $x\in F=\mathbb{R}^{\mathbb{N}}$, such that $\left<a_k,x\right>=b_k$. In particular we can choose $b\not\in l^{\infty}$, which will show that $A$ is not $\sigma(E,F)$- bounded.

Let $H$ be a subspace of $F$ of finite co-dimension, which separates points of $E$. Then the set $\{(\left<a_n,x\right>)_{n=1}^{\infty},x\in H\}$ is a liner subspace of $\mathbb{R}^{\mathbb{N}}$ of finite co-dimension, and so cannot be contained in $l^{\infty}$, which shows that $A$ is not $\sigma(E,H)$- bounded. Thus, all $\sigma(E,H)$- bounded subsets of $E$ are also finitely dimensional.

Hence, any $\sigma(E,F)$-/$\sigma(E,H)$- [bounded & closed] set has to be a relatively closed set of any finite-dimensional subspace of $E$. But such sets are closed in any Hausdorff locally convex topology. Thus, these classes coincide.

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    $\begingroup$ There is a slight mistake in the argument of you and Jochen--you need $H$ to be weakly dense (which is the same as saying that $H$ separates points). Otherwise in e.g. Jochen's description $H$ could be the the sequences that are zero in the first coordinate. When $H$ is dense your explanation is OK except that you should add Hausdorff" before locally convex" in the penultimate sentence. $\endgroup$ – Bill Johnson Sep 1 '17 at 16:00
  • $\begingroup$ @BillJohnson You are right, thank you! $\endgroup$ – erz Sep 1 '17 at 19:46

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