2

Possible Duplicate:
Does separability follow from weak-* sequential separability of dual space?
$\omega^*$-separability of $l_\infty^*$.

Recently I read a Theorem stating, Let $X$ be a Banach space which is separable then every weakly compact subset is metrizable. I noticed that the separability is used in order to prove that $X^{*}$ is weakly* separable and in the rest of proof separability of $X$ is no longer used.

Motivated by that my question is, let $X$ be a Banach space if $X^*$ is $\mathrm{weakly}^*$ separable then does this imply that $X$ is separable, if not this would weaken the hypothesis.

Community
  • 1
clark
  • 15,655
  • Related: http://math.stackexchange.com/questions/43365/does-separability-follow-from-weak-sequential-separability-of-dual-space – Asaf Karagila Jul 03 '12 at 17:30
  • This is probably inspired by a proof of the Eberlein–Šmulian theorem, so you might be interested in the references given here (I recommend having a look at Whitley's paper mentioned in the accepted answer). Seeing that the thread linked to by @Asaf gives a negative answer to a weaker question I'm voting to close as a duplicate. – t.b. Jul 03 '12 at 17:54
  • Actually $(\ell^\infty)^\ast$ is weak$^\ast$-separable, as I argue here, so that's a much more elementary counterexample. – t.b. Jul 03 '12 at 18:30
  • @t.b. thanks that is much better!Also could this be a good counterexample because of the schur property of $l_1$ ? – clark Jul 03 '12 at 18:53
  • 1
    I'm not exactly sure what you have in mind. It's also a good example because $\ell^\infty$ is a Grothendieck space :) – t.b. Jul 04 '12 at 01:55

0 Answers0