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 $E=E\_\beta$.



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> **Question 1.** Is $(E\_\beta)\_\beta=E\_\beta$ for any locally convex space? Equivalently, is the space $E\_\beta$ always barrelled?
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If not, then we can ask a more restricted version of Question 1.



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> **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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Is the strong topology of a locally convex space always barrelled?
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 $(X',\sigma(X',X))$.


