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Recently, I have been asked if are there any sufficient conditions for a Banach space with Pelczynski property (V*) to have dual with property (V). Since it was a long standing open problem, I guess that many spaces with property (V*) have dual with (V). Thus, my question is:

  1. What additional properties of a space $E$ with property (V*) force $E^*$ to have property (V)?

  2. Assuming $E$ has (V), what can we assume on $E$ to ensure that $E^{**}$ has (V)?

Surely, these questions are more interesting for non-rexlefive spaces as they have both (V) and (V*).

EDIT: Definition.

Let us say that an operator is unconditionally converging if it does not fix a copy of $c_0$. A Banach space $E$ has property (V) if any unconditionally converging operator defined on $E$ with values in an arbitrary Banach space is weakly compact.

$C(K)$ spaces as well as C*-algebras have property (V).

The definition of (V*) is a bit more tricky and is given for example here

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    $\begingroup$ Can you please tell us what properties (V) and (V*) are? Thanks. $\endgroup$
    – Kevin Beanland
    Commented May 21, 2012 at 12:06
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    $\begingroup$ I also do not recall the definitions of these obscure properties. If neither Kevin nor I knows the definitions, probably no regular contributor to MO does. $\endgroup$
    – Bill Johnson
    Commented May 21, 2012 at 16:46
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    $\begingroup$ It is not clear what you are looking for. The Saab-Saab paper to which you link gives a space $E$ with $(V^*)$ whose dual fails $(V)$. This $E$ is a Banach lattice. An easier example, and one which has an unconditional basis, is $E=(\sum \ell_\infty^n))_1$, which clearly has $(V^*)$. It's second dual contains $\ell_\infty$, hence is not weakly sequentially complete, whence $E^*$ fails $(V)$. $\endgroup$
    – Bill Johnson
    Commented May 25, 2012 at 19:43

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