Let $(X,\mu)$ be a measure space and let $1<p<\infty$.
Question. Is the space $L^p(X,\ell^p)$,
$$
\lVert f\rVert_p=\Bigl(\int_X\sum_{i=1}^\infty \lvert f_i\rvert^p\, dx\Bigr)^{1/p},
\qquad
f=(f_i)_{i=1}^\infty,
$$
uniformly convex?
If the answer if "yes" I would appreciate a reference to the statement/proof.
I know that $L^p(X,\ell^p_M)$, where $\ell^p_M$ is finite dimensional $\ell^p$ space
$$
\lVert f\rVert_p=\Bigl(\int_X\sum_{i=1}^M \lvert f_i\rvert^p\, dx\Bigr)^{1/p},
\qquad
f=(f_1,\dotsc,f_M)
$$
is uniformly convex. This result was proved by Clarkson in Uniformly convex spaces where he introduced the notion; see also Uniformly convex Banach spaces. I haven't checked whether the argument applies to the case of values into $\ell^p$.
Best Answer
By Proposition 1.2.24 in T. Hytonen, J. Van Neerven, M. Veraar and L. Weis, Analyis in Banach spaces Vol I, Springer, the spaces $L^p(X; \ell^p)$ and $L^p(X\times \mathbb N)$, $X \times \mathbb N$ with the product measure, are isometric (at least when $\mu$ is $\sigma$-finite). Then the uniform convexity of $L^p(X; \ell^p)$ follows from that of $L^p$-spaces.