TITLE: Proof that $L^2(0,T;X)^* = L^2(0,T;X^*)$
QUESTION [6 upvotes]: How is the proof that 
$$[L^2(0,T;X)]' = L^2(0,T;X')$$
looking like, where $X$ is a Hilbert space? I am asking for the proof that the dual space of $L^2(0,T;X)$ is the space $L^2(0,T;X^*)$. 
Is the proof much different to the $L^p(a,b)$ case?
I can't find any easy to understand proofs.

REPLY [6 votes]: To give you a reference: Diestel-Uhl, Vector measures, page 98, Chapter 4, Theorem 1:
$$L^p(\mu,X)^\ast = L^q(\mu,X^\ast)$$
if and only if $X^\ast$ has the Radon-Nikodym property with respect to $\mu$. 
Here $\mu$ is a finite measure and $p$ and $q$ as usual. Proof can be read there.
ADDED: It should be noted that $L^q(\mu,X^\ast) \subset L^p(\mu,X)^\ast $ always holds, without any condition on the Banach space $X$. The proof of this is quite the same as in the scalar case.