I want to prove following theorem:
Let X be separable and reflexive Banach space, $1<p<\infty$ than $$ L^p((0,1),X)^* = L^q((0,1),X^*) $$ where $\frac1{p}+\frac1{q} = 1$, with representation $v\in L^q((0,1),X^*), u \in L^p((0,1),X)$ $$ \left<v,u\right> = \int_0^1 \left<v(t),u(t)\right> {\rm d}t. $$
I would like to know why I need the reflexivity and separability of $X$. I sketch my proof here, but it seams I do not need neither reflexivity nor separability.
Proof sketch:
The hard inclusion is $$ L^p((0,1),X)^* \subset L^q((0,1),X^*) $$
So let $v^* \in L^p((0,1),X)^*$ than we want to find $v \in L^q((0,1),X^*)$ such that $$ \left<v^*,u\right> = \int_0^1 \left<v(t),u(t)\right> {\rm d}t $$ for all $u \in L^p((0,1),X)$.
I will construct piece-wise constant approximation $v_n$ of $v^*$ in such a way that: \begin{align} v_n(t) &= \sum_{k=0}^{n-1} \chi_{(\frac{k}{n},\frac{k+1}{n})}(t) v_n^k \qquad &v_n^k \in X^* \\ \left< v^* , x \chi_{(\frac{k}{n},\frac{k+1}{n})} \right> &= \frac{1}{n} \left< v_n^k, x \right> \qquad &\forall x \in X \end{align} where $\chi_A$ is characteristic function of set $A$.
It should be fairly easy to prove that for any simple function $u\in L^p((0,1),X)$ $$ \lim_{n \rightarrow \infty}\left< v_n , u \right> = \left< v^* , u \right>, $$
Now comes the part where I might need reflexivity and separability. I need to show that $v_n$ actually converges to some function $v\in L^q((0,1),X^*)$. I guess there might be some problem with measurability, so the separability of $X$ might be handy but it is not clear to me.
If we know that $v_n$ converges to some function $v\in L^q((0,1),X^*)$ than we know that $$ \left<v^*,u\right> = \left<v,u\right> $$ for all simple functions $u$. Since $\left<v^*,\cdot\right>,\left<v,\cdot\right>$ are both continuous than $$ \left<v^*,u\right> = \left<v,u\right> $$ for all functions $u \in L^p((0,1),X)$
So the only problem that I see is how to show that $v_n$ converges to some function $v$.
Wild guess is that from reflexivity I can show that $v_n$ converges weakly to $v$ but that gives me only weak measurability of $v$, so I need separability of $X$ to get strong measurability of $v$.
The conclusion of the theorem you want to prove holds if and only if $X^*$ has the Radon–Nikodym property with respect to the Lebesgue measure on $(0,1)$. This is precisely what you need to recover Rademacher's theorem about differentiation of Lipschitz maps from $(0,1)$ to $X^*$. Historically, the first description of $(L_p(X))^*$ was given by Bochner and Taylor:
They have identified the condition for $X$ which is needed for the identification $(L_p(X))^*\cong L_{p^*}(X^*)$, and it is what we call these days the Radon–Nikodym property.
Separability in your proof is redundant but it makes it easier as you may then apply freely Pettis' measurability theorem. As I said, it is not essential though as all reflexive spaces have the Radon–Nikodym property.
See also this question.