Let $X$ be a Banach space and $\ell^p(X)$ denote the space of sequences $x_i\in X$ for which the norm $\big(\sum_{i=1}^\infty\|x_i\|^p\big)^\frac1p$ is finite, when $X=\mathbb{R}$ we get the usual $\ell^p$. For $1<p<\infty$ and $\frac1p+\frac1q=1$ we have $(l^p)^*\simeq l^q$.
Is it always true that $\ell^p(X)^*\simeq \ell^q(X^*)$ for $1<p<\infty$ and $\frac1p+\frac1q=1$? This fails for big $L_p([0,1];X)$ without some condition on $X$, like reflexivity. Does it hold for any $L_p(\Omega,\mu;X)$ when $X$ is reflexive? References appriciated.
The duality $\ell^p(X)^*=\ell^q(X^*)$ for $1<p<\infty$ holds for every Banach space. Indeed, $c_{00}(X)$, the space of finitely supported sequences, is dense in $l^p(X)$. Therefore, every linear functional on $l^p(X)$ is determined by its values on sequences with one nonzero element. This identifies such a functional with an $X^*$-valued sequence $(x_n^*)$. Then one shows $\sum \|x_n^*\|^q<\infty$ just as in the proof of $(l^p)^*=l^q$.
The duality $L_p(X)^*=L^q(X^*)$ holds when $X$ is reflexive, and more generally, when $X$ has the Radon-Nikodym property. See here.