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A Banach space $X$ is weakly sequentially complete (WSC) if every weakly Cauchy sequence in $X$ is weakly convergent.

I will use the following classical result:

Rosenthal's $\ell_1$ theorem: Every bounded sequence in a Banach space $X$ either contains a weakly Cauchy subsequence, or a subsequence equivalent to the $\ell_1$ basis.

It follows from the fact that a closed subspace of a WSC space is WSC, that every subpace of a WSC space is either reflexive or contains $\ell_1$. This observation, if localized, actually gives us a characterization of WSC spaces, namely:

Proposition. For a Banach space $X$ TFAE:

  1. $X$ is WSC;
  2. Every bounded sequence in $X$ has either a weakly convergent subsequence, or a subsequence equivalent to the $\ell_1$ basis;
  3. Every weakly precompact (i.e. not containing a sequence equivalent to the $\ell_1$ basis) bounded set in $X$ is relatively weakly compact.

It would thus feel plausible that the "nonlocalized" version of this Proposition could be true as well. Namely, I would like to know if:

Question: Let $X$ be a Banach space such that every closed subspace $Y$ of $X$ is either reflexive or contains $\ell_1$, is $X$ then weakly sequentially complete?

As mentioned above, the converse implication is true. I have, however, not managet to prove/disprove the Question above. Any help will be appreciated.

KeeperOfSecrets
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1 Answers1

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The article by Azimi & Hagler provides a Banach space $X$ with a basis $(e_n)$ such that

  • $X$ is hereditarily $\ell^1$ : every infinite-dimensional closed subspace of $X$ contains a copy of $\ell^1$.
  • $(e_n)$ is weakly Cauchy but not weakly convergent.
Onur Oktay
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  • Thank you for the answer! I have some questions:
    1. How do we know that $T$ is weakly precompact?
    2. How do we know that $T$ is onto $Y$ (or at least that its range contains $Z$)?
    3. What is the need to define $V$, cannot we just take $V=\ell_1$ and proceed in the smae way?

    Thanks again!

     – KeeperOfSecrets Sep 20 '23 at 12:37
  • @KeeperOfSecrets You spotted a mistake in your (2.) Please see the edited answer. It turns out that there exist non-wsc hereditarily $\ell^1$ Banach spaces. – Onur Oktay Sep 20 '23 at 15:17
  • Thank you for the answer :) – KeeperOfSecrets Sep 20 '23 at 15:37
  • @KeeperOfSecrets cheers! – Onur Oktay Sep 20 '23 at 15:40