Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a nice characterization of the predual of $L^1$? So, what does the space $X$ look like, such that $X^*=L^1$, where the star denotes the dual of a space. How do you start to find such preduals in general? Thanks four your help.



share|cite|improve this question
Note also that the question what is "the" predual does not make sense in general. The space $\ell^1$ has many non-isomorphic preduals, for example $C(K)$ for $K$ countable and compact. – t.b. Apr 27 '12 at 18:03
@ t.b: Thanks for pointing this out! To be honest, since I didn't know predual of $L^1$ I didn't think there could be various. – math Apr 28 '12 at 15:50
I think $K$ also needs to be Hausdorff no? – Christian Bueno Mar 9 '15 at 16:21
up vote 12 down vote accepted

In fact, $L_1[0,1]$ has no pre-dual. More is true: $L_1$ cannot be embedded is a separable dual space. See, e.g., Theorem 6.3.7 in Kalton and Albiac's Topics in in Banach Space Theory.

share|cite|improve this answer
Can one show this for general $ L^1(X)$ ? I will check the book at the library and accept your answer afterwards. – math Apr 27 '12 at 15:25
@Math: $\ell^1=L^1(\mathbb{N})$ (with counting measure) famously has predual $c_0$, the subspace of $L^{\infty}(\mathbb{N})$ consisting of sequences converging to $0$. – Chris Eagle Apr 27 '12 at 15:38
@math I should point out that it is not hard to show that $L_1[0,1]$ is not isometric to a dual space using an extreme point argument and the Krein-Milman Theorem. – David Mitra Apr 27 '12 at 15:55
@math A result due to Johnson and Zippen: Let $X$ be a separable $L_1(\mu)$ predual. Then there is a subspace $Y$ of $C(\Delta)$, where $\Delta$ is the Cantor set, such that $X$ is isometric to $C(\Delta)/Y$. Here is the reference: W.B. Johnson and M. Zippen, Every separable predual of an $L_1$-space is a quotient of $C(\Delta)$. Israel J. Math 16 (1973), 198-202. See also here. – David Mitra Apr 27 '12 at 16:43
David's comment expanded. In a dual space, the unit ball has lots of extreme points (since, in the weak* topology, it is the closed convex hull of its extreme points: Krein-Milman theorem). But the unit ball of $L^1[0,1]$ has no extreme points at all! This shows $L^1[0,1]$ is not isometric to a dual space. The isomorphic theorem is harder. – GEdgar Apr 27 '12 at 17:44

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.