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?



Radon-Nikodým property of $\ell^\infty$
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.


