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.

Thank you in advance.

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?


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.


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.

  • $\begingroup$ (I see the update just after posting) $\endgroup$ – Pietro Majer May 7 '13 at 17:52
  • $\begingroup$ Uh, thanks a lot. I found exactly the same answer simultaneously. I am not familiar with the theory of Asplund spaces, so why is $\ell^1$ "obviously" no Asplund space? For the reason I mention or is there some simpler explanation? $\endgroup$ – Delio Mugnolo May 7 '13 at 17:54
  • $\begingroup$ Separable dual spaces are ok (Dunford-Pettis theorem). $\endgroup$ – András Bátkai May 7 '13 at 18:00
  • $\begingroup$ András, I do not understand this comment, either. This shows that $\ell^1$ has the R-N property, of course. But why does anything about $\ell^\infty$, or weighted version thereof, follow? $\endgroup$ – Delio Mugnolo May 8 '13 at 4:34
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    $\begingroup$ $l^\infty$ has a subspace isometric to $l^1$. Indeed $l^\infty$ has a subspace isometric to any given separable space. $\endgroup$ – Gerald Edgar May 8 '13 at 12:42

See Proposition 1.2.9 in Arendt-Batty-Hieber-Neubrander. (I look at the first edition now)

  • $\begingroup$ András, I am not sure I understand. That Proposition is about $c_0$, right? $\endgroup$ – Delio Mugnolo May 8 '13 at 4:33
  • $\begingroup$ See the later answer by jbc. $\endgroup$ – András Bátkai May 8 '13 at 5:59

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