Let $(\Omega, \mathscr F, \mu)$ be a $\sigma$-finite measure space and $X$ be a Banach space, and assume that $X^*$ has the Radon-Nikodym property with respect to $(\Omega, \mathscr F, \mu)$. I'm trying to prove the reflexivity of the Bochner space $L^p(\Omega;X)$ when $1< p <\infty$. The following is my proof:
Let $1/p + 1/q =1 $ and consider the map $\Phi_{p,X} : L^q(\Omega; X^*) \to L^p(\Omega; X)^* $ defined by $$\langle f, \Phi_{p,X}g \rangle := \int_\Omega \langle f,g\rangle \,d\mu, \quad f\in L^p(\Omega; X),\;g\in L^q(\Omega; X^*) .$$ Then by the Radon-Nikodym property, $\Phi_{p,X}$ is an isometric isomorphism. Note that, for all $f\in L^p(\Omega; X)$ and $g\in L^q(\Omega; X^*) $ we have $$ \langle f, \Phi_{p,X}g \rangle = \int_\Omega \langle f,g\rangle \,d\mu = \int_\Omega \langle g,j_Xf\rangle \,d\mu = \langle g, \Phi_{q,X^*}(j_Xf) \rangle, $$ where $j_X:X\to X^{**}$ is the canonical injection. Now, let $J:L^p(\Omega; X)\to L^p(\Omega; X)^{**} $ be the canonical injection. Then for all $\Lambda \in L^p(\Omega; X)^* $ and $f\in L^p(\Omega; X) $, we have $$ \langle \Lambda, Jf \rangle = \langle f,\Lambda \rangle = \langle f,\Phi_{p,X}\Phi_{p,X}^{-1}\Lambda \rangle = \langle \Phi_{p,X}^{-1}\Lambda,\Phi_{q,X^*}(j_Xf) \rangle = \langle \Lambda,(\Phi_{p,X}^{-1})^*\Phi_{q,X^*}(j_Xf) \rangle, $$ where $(\Phi_{p,X}^{-1})^*$ is the adjoint operator of $\Phi_{p,X}^{-1}$. Therefore $J$ is surjective. That is, $L^p(\Omega; X)$ is reflexive.
Is the above proof correct ?
Edited: I found that reflexivity of $X$ is also needed.