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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 $\ell^p$.

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    $\begingroup$ Isn't $L^p(X,\ell^p)$ isometric to $L^p(X\times \mathbb N)$, $X\times \mathbb N$ with the product measure? $\endgroup$ Apr 21 at 16:42
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    $\begingroup$ @GiorgioMetafune I think you are correct and I feel like a moron. If you post your comment as n answer, I will accept it. $\endgroup$ Apr 21 at 17:44

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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.

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