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Where could I find a direct proof showing that the dual ball of $L_1(\mu)$ is weak*-sequentially compact?

Since $(L_1(\mu))^*=L_\infty(\mu)$, I mean the unit ball $B_{L_\infty(\mu)}$ of $L_\infty(\mu)$ with the topology $\sigma(L_\infty(\mu),L_1(\mu))$.

(Eberlein-Smulian yields that $B_{L^\infty}$ is $\omega$-sequentially compact.)

Tomasz Kania
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  • $L^1$ as in $L^1(\Omega,\mu)$? What properties is the measure $\mu$ assumed to have? – Daniel Fischer Apr 14 '15 at 19:47
  • Ok. $\mu$ is a probability. –  Apr 14 '15 at 20:01
  • That implies $L^1$ is separable, iirc. If $X$ is separable, then the closed unit ball of $X^\ast$ is metrisable in the weak$^\ast$-topology, and compact metric spaces are sequentially compact. – Daniel Fischer Apr 14 '15 at 20:07
  • Thank you. Now I only have to figure out why L^1 is separable ... –  Apr 14 '15 at 20:27
  • ... but I'm afraid that L^1 is not always separable! Even when $\mu$ is a probability –  Apr 16 '15 at 19:34
  • Then it seems I didn't recall correctly. Have you an example of a finite measure where $L^1$ isn't separable? – Daniel Fischer Apr 16 '15 at 21:04
  • well, to be honest I don't have any counterexample off the top of my head. I need to search more. But I always find extraconditions in every theorem. For instance, if $\mu$ is $\sigma$-finite and the $\sigma$-algebra is countably generated, then $L^p$ ($1\leq p <\infty$) is separable. This made me think that we cannot guarantee $L^1$ to be sepable ... –  Apr 17 '15 at 22:12
  • Could be that finiteness of the measure is not sufficient for $L^1$ to be separable. In that case, I have at the moment no idea for a direct proof of sequential compactness. – Daniel Fischer Apr 17 '15 at 22:25
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    Daniel, $L_1({0,1}^{\omega_1})$ is non-separable even though the Haar measure on ${0,1}^{\omega_1}$ is finite. – Tomasz Kania Apr 22 '15 at 12:01
  • Zap, would you please revisit my answer? – Tomasz Kania May 23 '15 at 20:15

1 Answers1

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zap, the unit ball of $L_\infty(\mu)$ is not weakly compact (hence not weakly sequentially compact either) because $L_\infty(\mu)$ is not reflexive apart from the trivial cases.

The result you are interested in follows from two facts.

Theorem (Amir–Lindenstrauss). Let $X$ be a weakly compactly generated Banach space. Then weak*-compact subsets of $X^*$ are weak*-sequentially compact.

D. Amir and J. Lindenstrauss, The structure of weakly compact sets in Banach spaces, Ann. Math. 88 (1968), 35–46.

Fact. Let $\mu$ be a finite measure. Then $L_1(\mu)$ is weakly compactly generated. Indeed, the inclusion map from the reflexive space $L_2(\mu)$ to $L_1(\mu)$ has dense range (and is injective of course).

Observation. We can extend our conclusion to $\sigma$-finite measures $\mu$. Indeed, the Radon–Nikodym theorem yields an isometry from $L_1(\mu)$ to some $L_1(\nu)$ where $\nu$ is a finite measure.

hardmath
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Tomasz Kania
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  • Now, if I am not wrong, I see that it is not necessary the Amir-Lindenstrauss Theorem. Indeed, Fact I: Due to the Eberlein-Smulian theorem and since $L_2^=L_2$, we have that the dual unit ball of $L^2$ is weak-sequentially compact. Fact II: It is not difficult to show that the class of Banach spaces having weak-∗ sequentially compact dual ball is closed under the operation of taking dense continuous linear images. Fact II: It suffices to consider the injection $L_2\rightarrow L_1$, which has dense range. –  Apr 05 '16 at 15:47