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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 Banach space. How do you start to find such preduals in general?

For some context, it is well known that given a measure space $(S, \Sigma, \mu)$, $L^p := L^p(S, \mu)$ is a Banach space for $p\in (1,\infty)$ and that $L^p \cong (L^q)^*$ where $q$ is the Holder conjugate of $p$, that is $\frac 1p + \frac 1q =1$. It is also known that $L^1$ is the predual of $L^\infty$. This leaves the above questions as the only remaining case.

When $S$ is (for example) finite of course the question is moot. If you like one can consider only very simple measure space, like $[0,1]$ with the Lebesgue measure.

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    $\begingroup$ 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. $\endgroup$
    – t.b.
    Commented Apr 27, 2012 at 18:03
  • $\begingroup$ @ 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. $\endgroup$
    – math
    Commented Apr 28, 2012 at 15:50
  • $\begingroup$ I think $K$ also needs to be Hausdorff no? $\endgroup$ Commented Mar 9, 2015 at 16:21

2 Answers 2

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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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    $\begingroup$ @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 $0$. $\endgroup$ Commented Apr 27, 2012 at 15:38
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    $\begingroup$ @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. $\endgroup$ Commented Apr 27, 2012 at 15:55
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    $\begingroup$ @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. $\endgroup$ Commented Apr 27, 2012 at 16:43
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    $\begingroup$ 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. $\endgroup$
    – GEdgar
    Commented Apr 27, 2012 at 17:44
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    $\begingroup$ @XanderHenderson I've added an answer for the Krein-Milman argument which should answer this question and the duplicate source: I'm afraid the more general argument in this answer is beyond me. $\endgroup$
    – B. Mehta
    Commented Jul 30, 2018 at 22:33
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Banach-Alaoglu-Bourbaki: For $X$ a Banach space, the closed unit ball of $X^*$ is weak*-compact.

Krein-Milman: For $X$ a locally convex topological vector space, and $K$ a compact convex subset, $K$ is the closed convex hull of its extreme points.

Suppose $X^* = L^1[0,1]$. By BAB, the closed unit ball of $L^1[0,1]$ is weak*-compact. In addition, weak* topologies are always locally convex (see, for instance, here) and $L^1[0,1]$ is clearly convex so Krein-Milman gives that it is the closed convex hull of its extreme points.

But the closed unit ball of $L^1[0,1]$ has no extreme points: Suppose $\|f\|_1 = \alpha \leq 1$, $f \neq 0$. Then $H(s) = \int_0^s |f(t)| \, dt$ is a continuous function, with $H(0) = 0$ and $H(1) = \alpha$. Thus there is some $s_0 \in (0,1)$ with $H(s) = \frac{\alpha}{2}$, so set $$g(t) = 2f \chi_{[0,s_0]} \quad h(t) = 2f \chi_{[s_0,1]}$$ which satisfy $\|g\|_1 = \|h\|_1 = \|f\|_1 = \alpha$ and $\frac{1}{2}(g + h) = f$, so $f$ is not an extreme point. Clearly $0$ is not an extreme point either, so the unit ball of $L^1[0,1]$ has no extreme points.

Thus we have a contradiction, and $L^1[0,1]$ has no pre-dual.

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