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.



Proof that $L^2(0,T;X)^* = L^2(0,T;X^*)$
To give you a reference: [Diestel-Uhl, Vector measures](http://books.google.hu/books?id=NCm4E2By8DQC&printsec=frontcover&hl=de#v=onepage&q&f=false), 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.


