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  1. When the Banach space $V^*$ is reflexive, we have the unit ball in $V^*$ is weak$^*$ sequentially compact.

  2. For a Banach space $V^*$ that might not be reflexive, we have to assume that $V$ is separable, then the unit-ball in $V^*$ is weak$^*$ weak$^*$ sequentially compact result.

Question:

When the space is reflexive, what is the intuitive reason that we do not require the pairing space to be separable for the sequential compactness result?

Proof of 1 (followed from book):

We prove in terms of weak convergence instead of weak$^*$ convergence since the space is reflexive.

(a) First if $V = V^{**}$ is separable, then the result is true from (2).

(b) Now assume $V = V^{**}$ is not separable, given $\{v_n^*\}$ in the unit ball, if we let $W^*$ to be the clousure of the subspace generated by $\{v_n^*\}$, then $W^*$ is separable, reflexive and $W^{**}$ is also separable, reflexive (there is a ref in my book for this). Using the result of (a), we know there exists a weakly convergent subsequence under $\sigma(W^*, W^{**})$, since $V^{**} \subset W^{**}$ the subsequence is also convergent under $\sigma(V^*, V^{**}) =\sigma(V^*, V)$.

To answer my question, i guess that the result $W^{**}$ is separable might not hold when the space $V^*$ is not reflexive?

Thank you very much!

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    $\begingroup$ $V$ is reflexive iff $V^\ast$ is. So if $V^\ast$ is not, $V^{\ast\ast}$ is too big, in a way. I believe when we start with $V = \ell_\infty$, the unit ball in $V^\ast$ is not weak* sequentially compact. $\endgroup$ – Henno Brandsma Nov 30 '14 at 9:35
  • $\begingroup$ Which book you followed from ? I mean the proof. $\endgroup$ – Yan kai Nov 30 '14 at 11:19
  • $\begingroup$ @Yankai Variational analysis for Sobolev Space and BV Space. In the book, the original prove is that that when $V$ is reflexive, then every bounded sequence has a weakly convergent subsequence. $\endgroup$ – Xiao Nov 30 '14 at 13:20

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