0
$\begingroup$

Recall that a Banach space $X$ is said to have the Schur property if any weakly null sequence in $X$ is norm null, equivalently, every weakly Cauchy sequence in $X$ is norm Cauchy. It follows from Rosenthal's $l_{1}$-theorem that if a Banach space $X$ has the Schur property and contains no isomorphic copy of $l_{1}$, then $X$ is finite dimensional.

Question. Is there an infinite-dimensional Banach space $X$ such that $X^{*}$ has the Schur property and $X^{**}$ contains no isomorphic copies of $l_{1}$?

Thank you!

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ According to what you said, $X^*$ contains $\ell_1$, which obviously implies $X^{**}$ contains $\ell_1$ as well. $\endgroup$
    – Narutaka OZAWA
    Commented Sep 23, 2021 at 9:12
  • $\begingroup$ I do not know $X^{**}$ would contain $l_{1}$ whenever $X^{*}$ contains $l_{1}$. Could you give a detailed proof ? $\endgroup$
    – Dongyang Chen
    Commented Sep 23, 2021 at 9:17
  • $\begingroup$ @DongyangChen If a Banach space $Y$ contains a copy of $\ell^1$, then $Y^{*}$ contains a copy of the space of measures $M([0,1]) = C([0,1]) ^{*}$, which contains a copy of $\ell^1([0,1])$. This was proven by Pełczyński, see Proposition 3.3 in https:/doi.org/10.4064/sm-30-2-231-246 $\endgroup$
    – Onur Oktay
    Commented Sep 23, 2021 at 10:18
  • $\begingroup$ A particular case of Pełczyński's result can also be found in Corollary A in arxiv.org/abs/2108.03057v1. $\endgroup$
    – Onur Oktay
    Commented Sep 23, 2021 at 10:50
  • $\begingroup$ Thanks, Narutaka and Onur. You are right. $\endgroup$
    – Dongyang Chen
    Commented Sep 23, 2021 at 11:01

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .