2
$\begingroup$

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 $\ell^p$.

$\endgroup$
2
  • 4
    $\begingroup$ 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, 2022 at 16:42
  • 2
    $\begingroup$ @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, 2022 at 17:44

1 Answer 1

8
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.