Radon-Nikodým property of $\ell^\infty$ I am wondering whether $\ell^\infty(\mathbb N)$ has the Radon-Nikodým property. Of course $\ell^1(\mathbb N)$ does, but I was unable to find out whether (e.g.) duals of spaces with the R-N property have the R-N property themselves.
UPDATE: A Banach space is Asplund if and only if its dual has the Radon-Nikodým property. On the other hand, a separable space is Asplund if and only if its dual is separable, too. This rules out the possibility that $\ell^\infty(\mathbb N)$ has the R-N property. But is there any other (more) elementary argument for this assertion?
 A: For dual spaces, there is an important  characterization: $X^*$ has the Radon-Nikodym property if and only if $X$ is Asplund (its separable subspaces have separable duals). Of course, $\ell_1$ is not Asplund.
A: Since a closed subspace of a space with RNP clearly also has RNP, in order to get the requested elementary example, it suffices to display a measure on the unit interval with values in $c_0$ which is absolutely continuous with respect to Lebesgue measure but whose 
derivative does not take its values there.  This can be done via  the usual trick with the sequence $\frac 1 n \cos n x$.  I guess this example is in the classical text by Diestel and Uhl which is probably still the best reference for the RNP.
A: See Proposition 1.2.9 in Arendt-Batty-Hieber-Neubrander. (I look at the first edition now) 
A: This post is concerned with "whether duals of spaces with the R-N property have the R-N property themselves".
Every reflexive Banach space $X$ and its dual $X^{*}$ have RNP. For a non-reflexive (but quasi-reflexive) example, James space $J$ and its dual $J^{*}$ (generally all higher duals) have RNP.
