When is the dual ball of $L_1(\mu)$ weak*-sequentially compact?

Question

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.)

Answer 1

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.