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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.



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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 at 16:21

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up vote 10 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.

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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

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