Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
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 $\ell_$ –  Philip Brooker Jun 24 '12 at 9:49
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 –  Philip Brooker Jun 24 '12 at 9:52

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
Is anything known about containment of $c$ isometrically? –  Jan Vardøen Jun 24 '12 at 12:00
@Jan: $c$ does not embed isometrically into $c_0$. –  Philip Brooker Jul 4 '12 at 6:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.