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A Banach space $E$ is called Grothendieck if every weak* convergent sequence in the dual space $E^*$ is weakly convergent. A typical example of a Grothendieck space is $\ell_\infty$. Diestel proved the following characterization of Grothendieck Banach spaces:

For every Banach space $E$, TFAE:

  1. $E$ is Grothendieck,
  2. for each separable Banach space $F$, every bounded operator $T\colon E\to F$ is weakly compact,
  3. every bounded operator $T\colon E\to c_0$ is weakly compact.

My question: Can we replace the space $c_0$ in item 3 by some/any other fixed separable space? In particular, can we use here $\ell_1$?

(By "other" I of course mean "non-isomorphic to $c_0$".)

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    $\begingroup$ Every bounded linear operator $c_0 \to \ell^1$ is compact (this is a version of Pitt's theorem), hence weakly compact. But $c_0$ is, of course, not Grothendieck. $\endgroup$
    – Jochen Glueck
    Commented Jun 1, 2023 at 10:05
  • $\begingroup$ @JochenGlueck, good point! Thanks! So, $\ell_1$ (and all $\ell_p$'s for $1<p<\infty$ as well) is eliminated. $\endgroup$
    – Damian Sobota
    Commented Jun 1, 2023 at 10:10
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    $\begingroup$ Yes - but please note that $\ell^p$ for $1 < p < \infty$ aren't suitable candidates anyway since those spaces are reflexive and every bounded linear operator from a Banach space into a reflexive space is weakly compact. $\endgroup$
    – Jochen Glueck
    Commented Jun 1, 2023 at 10:28
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    $\begingroup$ Obviously you can replace $c_0$ with any separable space that has a subspace isomorphic to $c_0$. Your question should have been "if you can replace $c_0$ by $X$ in item 3, must $X$ have a subspace that is isomorphic to $c_0$". $\endgroup$
    – Bill Johnson
    Commented Jun 2, 2023 at 15:01

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In the light of Professor Johnson's comment, we cannot replace $c_0$ with another $X$ unless $X$ contains a copy of $c_0$.

Indeed, let $X$ be a Banach space that contains no isomorphic copy of $c_0$. Consider a $C(K)$ space that is not a Grothendieck space (e.g., $C([0,1])$ ). Any bounded linear $T:C(K)\to X$ is unconditionally converging since it does not fix a copy of $c_0$, thus weakly compact since $C(K)$ has property $(V)$.

Generally, if $E$ (may not be a Grothendieck space) has property (V) & $X$ contains no copy of $c_0$, then every $T\in B(E,X)$ is weakly compact.

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  • $\begingroup$ An unconditionally convergent operator? $\endgroup$
    – Jochen Wengenroth
    Commented Jun 3, 2023 at 9:28
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    $\begingroup$ Professor @JochenWengenroth, an operator is said to be unconditionally converging (u.c.) if it maps w.u.c. series to unconditionally convergent series. These are characterized by the fact that an operator $T$ is u.c. iff $T$ doesn't fix a copy of $c_0$. $\endgroup$
    – Onur Oktay
    Commented Jun 3, 2023 at 10:18
  • $\begingroup$ Dear Onur, thanks for your answer! $\endgroup$
    – Damian Sobota
    Commented Jun 12, 2023 at 10:35

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