Recall that a locally convex space $E$ is said to be barreled if every closed absolutely convex and absorbing set (i.e. every barrel) is a 0-neighborhood.
A countable base of bounded sets in $E$ is any sequence $(B_n)_n$ consisting of bounded sets such that every bounded set in $E$ is contained in some $B_n$.
The strong dual of $E$ is the topological dual $E'$ of $E$ endowed with the strong topology, i.e. the topology of uniform convergence on the bounded subsets of $E$.
The question is whether there is an example of a locally convex spaces with a countable base of bounded sets and non barreled strong dual.
Notice that the space $E$ cannot be bornological, since its strong dual would be Fréchet, then barreled.
Dear Lahbib Oubbi ,
It seems to me, I can give a relatively simple example of the kind you are looking for. Please, see the enclosed .pdf file.
Best, Vladimir
Thank you Professor Kadets.
I appreciate very much your interest in the question.
I have however a remark :
You assume that such a Z exists. Do not you need the space X to be separable or at least the quotient ?
Dear Lahbib Oubbi ,
No separability assumption is needed. By the definition, a sequence in Banach space is called a basic sequence if it forms Schauder basis of the closed linear span of that sequence. According to Masur's theorem (Theorem 1.a.5 of Lindenstrauss, Tzafriri "Classical Banach Spaces I. Sequence spaces."), every infinite-dimensional Banach space contains a basic sequence. So, if $X^{**}/X$ is infinite-dimensional, then it contains a basic sequence $(e_n)$. The linear span $Z$ of $(e_n)$ will be a non-barreled normed space. Indeed, an example of barrel $U$ in $Z$ with empty interior is the set of all vectors of the form $\sum_{n =1}^N a_n e_n/ ||e_n||$, where $N$ is arbitrarily large and $|a_n| \le 1/n$. The closedness of $U$ in $Z$ follows from continuity of coordinate functionals: that is why we need a basic sequence.
Best, Vladimir
Dear Professor Kadets,
I think one can contract a concrete example by considering, as you suggested, c_0 and a non barreled subspace S of $l^\infty$ containing c_0. Such an example is given in "S. Saxon , Non-Barreled Dense β-φ Subspaces, J. Math. Anal. App. 196 (1995), 428-441" (Example 2. 4). Then consider on l^1 the weak topology defined by S and continu as you did.
All the best
Dear Professor Lahbib Oubbi ,
In the case of $c_0$ one can also find a concrete example of corresponding basic sequence in $\ell_\infty/c_0$. Namely, take any pairwise disjoint sequence $(A_n)$ of infinite subsets of naturals (say, $A_n$ consists of all powers of n_th prime number). Denote $x_n = (x_{n,1}, x_{n,2}, \ldots) \in \ell_\infty$ the element, whose coordinates $x_{n,k}$ are equal to 1 when $k \in A_n$, and are equal to 0 when $k \in \mathbb N \setminus A_n$. Then equivalent classes of these $x_n$ form a basic sequence in $\ell_\infty/c_0$. The corresponding non-barreled subspace $S$ of $\ell_\infty$ containing $c_0$ will be just the linear span of $c_0$ and the sequence $x_1, x_2, \ldots$.
All the best, Vladimir
Dear Professor Kadets,
Thank you again for the fruitful conversation. The question is now settled.
Seeing forward to discussing a new question,
All the best
