Exercise 24, § 2

Let $\mu$ be a positive measure on T, u a linear mapping of $L^1(\mu)$ into a Banach space F, transforming the unit ball of $L^1(\mu)$ into a subset of F relatively compact for $\sigma(F, F')$. Show that there exists a $\mu$-measurable mapping f of T into F such that $|f(t)| \leq \|u\|$ for all $t \in T$ and such that

for every function $g \in \overline{\mathscr{L}}^1(\mu)$ (Dunford–Pettis–Phillips theorem). (First reduce to the case that T is compact, with the help of Prop. 14 of Ch. IV, §5, No. 9. The unit ball B of $L^\infty(\mu) \subset L^1(\mu)$ is then relatively compact for $\sigma(L^1, L^\infty)$ (Ch. V, §5, Exer. 15); using Exer. 23, show that the closed linear subspace of F generated by $u(L^1)$ is separable. One can therefore reduce to the case that F is separable and $u(L^1) = F$. The closure A in F of the image under u of the unit ball of $L^1$ is then compact for $\sigma(F, F')$ (Ch. IV, §7, Exer. 10); show that it is metrizable for this topology (TVS, III, §3, No. 4, Cor. 2 of Prop. 6). Observe that A may then be identified with the unit ball of the dual of the normed space G obtained by equipping $F'$ with the norm equal to the gauge of $A^\circ$. Show that G is separable, and thus reduce to applying the Dunford–Pettis theorem (Cor. 2 of Th. 1).)