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I have an infinite-dimensional (Hausdorff separated, non-metrizable) locally convex space $(X,\tau)$ with topological dual $X^*$.

My questions are:

Under what conditions is there a barreled topology on $X^*$ that is finer than the weak-star topology?

If $(X,\tau)$ is complete then is the answer to the previous question positive?

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  • $\begingroup$ can you plz define the barreled topology ? $\endgroup$ – Red shoes Jun 17 '17 at 3:47
  • $\begingroup$ @Ashkan see en.wikipedia.org/wiki/Barrelled_space $\endgroup$ – Henno Brandsma Jun 17 '17 at 8:07
  • $\begingroup$ @HennoBrandsma Thanks. So does question say, put a topology on $X^*$ and extract its barreled topology , and see whether it is finer than weak-star ? Or we should consider the barreled topology coming from weak-star topology on $X^*$ Can you plz explain it? $\endgroup$ – Red shoes Jun 17 '17 at 8:24
  • $\begingroup$ The question seems to be: we have the weak-$\ast$ topology on $X^\ast$, which is just $\sigma(X^\ast, X)$. Is there a TVS-topology $\tau'$ on $X^\ast$ such that $\tau'$ is barreled (so every barrel is a neighbourhood of $0$) and $\sigma(X^\ast,X) \subsetneq \tau'$ ? @Ashkan $\endgroup$ – Henno Brandsma Jun 17 '17 at 8:29
  • $\begingroup$ Your title does not match the question.... $\endgroup$ – Henno Brandsma Jun 17 '17 at 9:51
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Hint: For any vector space $Y$, the finest (the biggest) topology that makes $Y$ into a locally convex topological vector space is $\sigma(Y,Y') $ Where $Y'$ is the set of all linear functionals on $Y$! This topology is called core convex topology and indeed it is a barreled topology, You can verify this easily via looking at the Topological basis of this topology look at my paper in arXiv https://128.84.21.199/abs/1704.06932v1 .

So Now Put $Y = X^*$

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  • $\begingroup$ So you need $X$ to be nonreflexive. $\endgroup$ – Henno Brandsma Jun 17 '17 at 11:54
  • $\begingroup$ Why are you saying that? @HennoBrandsma Even let $X$ be a separable Hilbert space. $\endgroup$ – Red shoes Jun 17 '17 at 12:05
  • $\begingroup$ @Ashkan if $X$ is reflexive this core convex topology is just the weak-star topology and the OP wanted a finer one than that. $\endgroup$ – Henno Brandsma Jun 17 '17 at 12:11
  • $\begingroup$ Note that the Core convex topology has nothing to do with the topology on $X$ ! it is independently defined on $X$ (or $X^*$ in our case ) . @HennoBrandsma $\endgroup$ – Red shoes Jun 17 '17 at 12:38

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