Dual space of Bochner space Let $B$ be a refexive Banach space. I want to show that 
$$(L^2(0,T;B))^* = L^2(0,T;B^*)$$ and that 
the dual pairing is
$$\langle F,f \rangle_{L^2(0,T;B^*), L^2(0,T;B)} = \int_0^T \langle F(t), f(t) \rangle_{B^*,B}.$$
Can anyone help me with either part? Thanks.
 A: No proof in the generality that is desired is going to be very brief. But all the essential ideas come from the usual, scalar case. First we need the following:

Definition. A Banach space $V$ is said to have the Radon-Nikodym property with respect to a measure space $(X, \Sigma, \mu)$ if, for every $V$-valued vector measure of bounded variation $\nu$ which is absolutely continuous with respect to $\mu$, there exists a Bochner integrable $f \in L^1(X,V)$ such that $\nu(E) = \int_E f \, d\mu$ for every $E \in \Sigma$. 

In other words, spaces with the Radon-Nikodym property are precisely those in which the generalization of the Radon-Nikodym theorem for scalar functions holds. Then the key result that one needs to prove is due to Phillips:

Reflexive Banach spaces have the Radon-Nikodym property.

A proof of this requires a decent amount of machinery and a proof of this fact and other related facts is the subject of Chapter 2 of Vector Measures by Diestel and Uhl. Now, with this fact under our belt, we have the following result:

Theorem. Let $(X, \Sigma, \mu)$ be a finite measure space, and let $V$ be a Banach space such that $V'$ has the Radon-Nikodym property. Then, for $p \in [1, \infty)$, $L^p(X,V)' \cong L^q(X,V')$ isometrically, where $1/p + 1/q = 1$. 

Proof. Suppose $g \in L^q(X,V')$. Then, for $f \in L^p(X,V)$, the map $\Lambda_g(f) = \int_X \langle f, g \rangle \, d\mu$ is clearly a continuous linear functional, so $L^q(X,V') \subseteq L^p(X,V)'$. To prove the reverse inclusion, let $\Lambda \in L^p(X,V)'$. For $h \in L^{\infty}(X,V)$, $|\Lambda(h)| \le C \|h\|_p \le  C \|h\|_{\infty}$, and hence $\Lambda$ is also a continuous functional on $L^{\infty}(X,V)$. Now, $\Lambda$ induces a $V'$-valued vector measure absolutely continuous with respect with $\mu$ given by
$$ \langle \nu(E), x \rangle = \Lambda(x 1_E)$$
for each $x \in V$. Then, as $V'$ has the Radon-Nikodym property, then there exists $g \in L^1(X,V')$ such that $\nu(E) = \int_E g \, d\mu$. Then, by approximating each $f \in L^p(X,V)$ by simple functions, we have that $\Lambda(f) = \int_X \langle f, g\rangle \, d\mu$. Finally, it is straightforward to show that $g$ is in fact also in $L^q$. 
