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Does the second Clarkson's inequality hold for any two vector in $\mathbb{R}^N$? That is, for any $p\in(1,2)$ and $z,w\in\mathbb{R}^N$, $$ \left|\frac{z+w}{2}\right|^q+\left|\frac{z-w}{2}\right|^q\leq\left[\frac{1}{2}\left(|z|^p+|w|^p\right)\right]^{\frac{1}{p-1}}. $$ Where $|\cdot|$ denotes the Euclidean norm in $\mathbb{R}^N$? $\left(\frac{1}{p}+\frac{1}{q}=1\right)$

Cameron Buie
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Taukhtn
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  • @user54504 I update my answers. – Elias Costa Dec 28 '12 at 17:51
  • Just a remark: any two vectors in $\mathbb R^N$ lie in a two-dimensional subspace (just consider their span). So it suffices to consider $N=2$, where complex arithmetics may help. –  Dec 28 '12 at 18:26

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See the article A Note On Clarkson’s Inequality In The Real Case by Hiroyasu Mizuguchi and Kichi-suke Saito in Journal of Mathematical Inequalities.

And see Theorem 1 of Some Uniformly Convex Spaces by R. P. Boas, Jr. This last article answers your question in a more general spaces $L^p$ and $l^p$ instead of $\mathbb{R}^N$. Note that $\mathbb{R}^N$ can be regarded as a subspace of $l^p$.

Elias Costa
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  • I update my answers. – Elias Costa Dec 28 '12 at 17:45
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    You can use boldface and italics instead of capital letters. It's more readable too. – Asaf Karagila Dec 28 '12 at 18:10
  • Thank Elias very much. However, I have problem with the vector case, the scalar case is ok. The book of Adam presents the proof for R^2 by using complex numbers. So I can not see an obvious way to generalize to $\mathbb{R}^N$. – Taukhtn Dec 28 '12 at 17:40
  • Thank Elias again for your help. Actually, I have considered $R^N$ as a $l^p$ space, but since we are using Euclidean norm, it turns out that $p=2$ and then I can not apply the origianl Clarkson' inequality in $L^p$ space. However, the generalized theorem you suggest is indeed useful. I think I can find the solution now. Thank you very much for your kindness. – Taukhtn Dec 29 '12 at 12:17