If $X$ is a Banach space with the Dunford Pettis Property (DPP), then no infinite reflexive subspace can be complemented. Suppose now that the Banach space has the property, that no infinite reflexive subspace is complemented. Is it true that $X$ has DPP?
$\begingroup$
$\endgroup$
4
-
$\begingroup$ Thank you Bill, do you know additional conditions on the space $X$, such that the question is true? $\endgroup$– user44155Commented Mar 15, 2014 at 21:34
-
$\begingroup$ Oh, you saw before I deleted--my answer was wrong. Tomek gave in his answer what is probably the simplest example. $\endgroup$– Bill JohnsonCommented Mar 15, 2014 at 21:51
-
$\begingroup$ I do not know a reasonable extra condition on the space that would make the answer affirmative. $\endgroup$– Bill JohnsonCommented Mar 15, 2014 at 21:54
-
$\begingroup$ The example with the Schreier space actually shows that a space failing DPP can have no reflexive subspaces at all. $\endgroup$– Tomasz KaniaCommented Mar 15, 2014 at 21:56
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
0
No. The Schreier space $S$ is $c_0$-saturated (every closed infinite-dimensional subspace of $S$ contains a copy of $c_0$) as it embeds into $C[0,\omega^\omega]$ which has this property, yet if fails the Dunford-Pettis Property (you will find an easy proof here (p. 168)).
answered Mar 15, 2014 at 21:33