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Q1. Does there exist a separable Banach space $X$ satisfying in the following property?

1- $X^*$ is non separable.

2- For every countable subset $F\subset X^*$ there exists $0\neq x_F\in X$ such that $f(x_F)=0$ for all $f\in F$.

Q2. If it is impossible, what about if we replace $X$ by a separable topological vector space?

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Suppose that $X$ is a separable Banach space. Then $X^*$ is weak*-separable, so take a countable weak*-dense set $F$ in $X^*$. Consequently, $F$ is total, hence the only element on which all elements of $F$ simultaneously vanish is 0.

This works as long as $X$ is a locally convex space and $X^*$ is weak*-separable. Note that, by Goldstine's theorem, this will work for every $X=Y^*$, where $Y$ is a separable Banach space.

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  • $\begingroup$ Thanks. Any example of separable locally convex space $X$ whose dual is not weak*-separable? Perhaps it serves an answer for Q2! $\endgroup$
    – ABB
    Commented May 11, 2018 at 14:48
  • $\begingroup$ $\ell_1$ is separable then its second dual $X=\ell_\infty^{*}$ is weak star separable. But its dual is not separable any more. Right? $\endgroup$
    – ABB
    Commented May 11, 2018 at 19:40
  • $\begingroup$ @AliBagheri, certainly $\ell_1^*$ is not norm-separable, however, it is weak*-separable (as is $\ell_1^{**}$). You may be interested in this paper: carma.newcastle.edu.au/brailey/Research_papers/… $\endgroup$
    – Tomasz Kania
    Commented May 12, 2018 at 9:09

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