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.


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!

  • 1
    $\begingroup$ $V$ 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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