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For a locally convex space $E$ let $E_\beta$ be the space $E$ endowed with the locally convex topology $\beta(E,E')$ whose neighborhood base at zero consists of barrels, i.e., closed absolutely convex absorbing sets.

Observe that a space $E$ is barrelled if and only if $E=E_\beta$.

Question 1. Is $(E_\beta)_\beta=E_\beta$ for any locally convex space? Equivalently, is the space $E_\beta$ always barrelled?

If not, then we can ask a more restricted version of Question 1.

Question 2. Let $E$ be a locally convex space such that the space $E_\beta$ is normable (and separable). Is $E_\beta$ barrelled?

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    $\begingroup$ @SergeiAkbarov The passage from $E$ to $E_\beta$ can change the dual. Just consider the space $E=C_p(K)$ of real-valued continuous functions on an infinite compact Hausdorff space $K$, endowed with the topology of pointwise convergence. The dual $E'$ is the space of finitely supported signed measures on $K$. On the other hand, $E_\beta$ coincides with the Banach space $C(K)$ and has much larger dual (consisting of all regular sign-measures). At least $C(K)'$ contain the space $\ell_1(K)$ of countably supported sign-measures. $\endgroup$
    – Taras Banakh
    Commented Jun 4, 2019 at 17:36
  • $\begingroup$ Taras, yes, excuse me, you are right. $\endgroup$
    – Sergei Akbarov
    Commented Jun 4, 2019 at 17:49
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    $\begingroup$ For my purposes it would be very desirable to have affirmative answer at least to Question 2. I cannot imagine how a counterexample can be constructed (if it exists at all). $\endgroup$
    – Taras Banakh
    Commented Jun 4, 2019 at 17:52

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The answer to question 1 is negative. There are Frechet spaces $X$ whose strong duals $(X',\beta(X',X))$ are not barrelled (equivalently, not bornological by a theorem of Grothendieck). The first examples of such non-distinguished Frechet spaces were constructed by Köthe and Grothendieck but there are also examples which are very easy to describe: According to Taskinen the Frechet space $C(\mathbb R) \cap L^1(\mathbb R)$ of continuous Lebesgue-intergrable functions (endowed with the seminorms $\int|f(x)|dx + \sup\{|f(x)|: x\in [-n,n]\}$) is not distinguished. This answers question 1 with $E=(X',\sigma(X',X))$.

EDIT: If $M$ is a barrel in $E$ then its polar $M^\circ$ is a $\sigma(X,X')$-bounded subset of $X$ and hence it is bounded in the Frechet topology so that $M=M^{\circ\circ}$ is a $\beta(X',X)$-neighbourhood of $0$. Conversely, the typical $0$-neighbourhood $B^\circ$ in $\beta(X',X)$ with a bounded subset $B$ of $X$ is a barrel in $(X',\sigma(X',X))$.

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  • $\begingroup$ Is it always true that for the dual space $E=(X',\sigma(X',X))$ of a Frechet space $X$ the topology generated by barrels in $E$ coincides with the topology $\beta(X',X)$? $\endgroup$
    – Taras Banakh
    Commented Jun 5, 2019 at 7:24
  • $\begingroup$ I have editet the post. Hopefully, this answers your question. $\endgroup$
    – Jochen Wengenroth
    Commented Jun 5, 2019 at 7:33
  • $\begingroup$ Yes, thank you. $\endgroup$
    – Taras Banakh
    Commented Jun 5, 2019 at 7:35
  • $\begingroup$ @JochenWengenroth: You mentioned Koethe. Is there another example with sequence spaces? $\endgroup$
    – Abdelmalek Abdesselam
    Commented Jun 5, 2019 at 12:47
  • $\begingroup$ Of course, the first examples were sequence spaces. You can see a characterization of distinguished Köthe space $\lambda^1(A)$ in terms of the matrix in proposition 27.17 and the classical example in 27.19 in the book Introduction to Functional Analysis of Meise and Vogt. $\endgroup$
    – Jochen Wengenroth
    Commented Jun 5, 2019 at 16:33
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Many examples of spaces $C_p(X)$ whose strong dual is not barrelled can be found in a recent paper "Examples of Nondistinguished Function Spaces $C_p(X)$", Journal of Convex Analysis, 2019.

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No. Let $E$ be the space $C[0,1]$ of continuous functions on $[0,1]$ with the norm $\|x\|=\int_0^1 |x(t)|dt$. Then $E=E_\beta$ is normable and separable, but it is not barrelled. The set $B=\{x\in E: \sup_{t\in [0,1]} |x(t)|\le1\}$ is a barrell in $E$, but it is not neighborhood at zero.

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    $\begingroup$ It seems to me that the space $E_\beta$ is the Banach $C[0,1]$ endowed with the sup-norm, so $E_\beta$ is barrelled and $E\ne E_\beta$. So your example does not work. $\endgroup$
    – Taras Banakh
    Commented Jun 5, 2019 at 7:16

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