TITLE: Dual of Banach-valued $L^p$
QUESTION [6 upvotes]: Let $X$ be an infinite-dimensional  Banach space and let $p\in(1,+\infty)$. We may define $L^p(\mathbb R;X)$. Is it always true that the topological dual of $L^p(\mathbb R;X)$ is $L^{p'}(\mathbb R;X^*)$? Maybe some reflexivity or separability is needed for the Radon-Nikodym argument to work.
Is there a simple realization of  the dual of $L^1(\mathbb R;X)$?

REPLY [9 votes]: Diestel-Uhl, Vector measures, Section IV.1., Theorem 1:
Let $(\Omega,\mu)$ be a $\sigma$-finite measure space, $1\leq p<\infty$ and $\frac{1}{p}+\frac{1}{p'}=1$.
The dual of $L^p(\mu;X)$ is $L^{p'}(\mu,X^\ast)$ if and only if $X^\ast$ has the Radon-Nikodym property. 
This is especially the case if $X$ is reflexive.