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](https://doi.org/10.2307/1989630) where he introduced the notion; see also [Uniformly convex Banach spaces](https://mathoverflow.net/q/420011/121665). I haven't checked whether the argument applies to the case of values into $\ell^p$.



Is $L^p(X,\ell^p)$, $1<p<\infty$, uniformly convex?
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.


