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Theorem. If $X$ and $Y$ are uniformly convex Banach spaces, then for $1<p<\infty$ the space $$ X\oplus_pY=X\times Y, \qquad \Vert(x,y)\Vert:=(\Vert x\Vert_X^p+\Vert y\Vert_Y^p)^{1/p} $$ is uniformly convex.

This is a consequence of a classical theorem of Clarkson [Theorem1, C] who introduced the notion of uniform convexity and proved a much more general result than the one quoted above. In addition to the original reference I would like to quote a more recent one, however, I was unable to find anything.

Question. Is there a more recent reference to the result that I quoted above?

While Clarkson's proof is easy to follow, his theorem is very general and it takes a while to check that the result I stated above is a consequence of Clarkson's theorem. I would like to know if there is a more straightforward proof of the above result.

[C] J. A. Clarkson, Uniformly convex spaces. Trans. Amer. Math. Soc. 40 (1936), no. 3, 396–414.

Edit. Based on the answers it seems this simple and important result did not make into any textbooks which I find rather surprising. I am still looking for a good reference if there is one.

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    $\begingroup$ Professor @Hajlasz, the proof in doi.org/10.1016/s0022-247x(02)00282-2 is a bit lengthy but self-involved & straightforward in a way that one may even consider to suggest it for advanced undergrad or novice grad students. I hope you find it useful. $\endgroup$
    – Onur Oktay
    Commented Apr 9, 2022 at 18:18
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    $\begingroup$ @OnurOktay Maybe post it as an answer? $\endgroup$
    – Willie Wong
    Commented Apr 10, 2022 at 15:50
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    $\begingroup$ If you have at hand the theory of ultra powers of Banach spaces, the result is very easy. It is trivial that the $p$-direct sum of two strictly convex spaces is strictly convex (here $1<p<\infty$) and it is equally easy that a Banach space is uniformly convex if and only if its ultra powers are strictly convex. The downside of this line of argument is that you do not get any estimate on the modulus of convexity of the direct sum if you know the modulus of convexity of the direct summands. $\endgroup$
    – Bill Johnson
    Commented Apr 10, 2022 at 22:17

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I'm copying my comment above in here following Willie Wong's suggestion.

The proof in https://doi.org/10.1016/s0022-247x(02)00282-2 is a bit lengthy but self-involved & straightforward in a way that one may even consider to suggest it for advanced undergrad or novice grad students.

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  • $\begingroup$ This is an interesting reference worth to know, however, since the result is quite general the proof is more difficult than that of Clarkson. I am quite surprised that the particular case of Clarkson's theorem that I stated in my post cannot be found in any textbook. $\endgroup$
    – Piotr Hajlasz
    Commented Apr 17, 2022 at 3:27

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