Some isometric preduals of $\ell_1$ are of the form $C_0(K)$ where $K$ is countable. I am wondering whether this is a general rule.

Question: Is there a measure $\mu$ and a (preferably separable) Banach space $X$ without a subspace isomorphic to $c_0$ which has $X^*=L_1(\mu)$ isometrically?

I apologise for three questions in a such short period of time. Now I'll take my time.

EDIT: Corrected according to Philip's remarks.

  • $\begingroup$ The first sentence of your post is incorrect; what is true is that when $C_0(K)^\ast$ is isometric to $\ell_1$ it is necessarily the case that $K$ is countable. There are isometric preduals of $\ell_1$ that are not isomorphic to a space $C_0(K)$; the first example is due to Benyamini and Lindenstrauss, A predual of $\ell_1$ which is not isomorphic to a $C(K)$ space, Israel J. Math. 13 (1972), 246-254. Other constructions have since been given, see Gasparis' preprint arxiv.org/pdf/1205.4317.pdf for a brief survey. Gasparis' paper contains a new approach to constructing an $\ell_$ $\endgroup$ – Philip Brooker Jun 24 '12 at 9:49
  • $\begingroup$ isometric predual of $\ell_1$. The difference between his space and earlier constructions is that his space does not contain a subspace isomorphic to $C_0([0,\omega^\omega])$. You can see a video of him presenting a talk on his paper at birs.ca/events/2012/5-day-workshops/12w5019/videos $\endgroup$ – Philip Brooker Jun 24 '12 at 9:52

Zippin proved that every isometric $L_1$ predual contains $c_0$ isometrically.

Zippin, M. On some subspaces of Banach spaces whose duals are L1 spaces. Proc. Amer. Math. Soc. 23 1969 378–385.

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  • $\begingroup$ Is anything known about containment of $c$ isometrically? $\endgroup$ – Jan Vardøen Jun 24 '12 at 12:00
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    $\begingroup$ @Jan: $c$ does not embed isometrically into $c_0$. $\endgroup$ – Philip Brooker Jul 4 '12 at 6:12

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