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Is it true that for $p\in (1,2)$ the following inequalities holds: $$ 2^{p-1} (|x|^p+|y|^p)\leq |x+y|^p+|x-y|^p \leq 2 (|x|^p+|y|^p)$$ for $x, y \in \mathbb{R}$ ?


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up vote 4 down vote accepted

We first show that $x_1^p+x_2^p\geq (x_1^2+x_2^2)^{p/2}$ for non-negative $x_1,x_2$. By homogeneity it's enough to show that $f(t):=x^p+1-(x^2+1)^{p/2}$ is non-negative. Its derivative is $f'(x)=px^{p-1}-\frac p22x(x^2+1)^{p/2-1}\geq px^{p-1}-pxx^{p-2}=0$ since the map $s\mapsto s^{p/2-1}$ is decreasing ($p<2$). Now we apply it to $x_1=\frac{x+y}2$ and $x_2=\frac{x-y}2$. We get $$\left|\frac{x+y}2\right|^p+\left|\frac{x-y}2\right|^p\geq\left(\left|\frac{x+y}2\right|^2+\left|\frac{x-y}2\right|^2\right)^{p/2}=\left(\frac{x^2}2+\frac{y^2}2\right)^{p/2}$$ and since the map $t\mapsto t^{p/2}$ is concave, we get the first inequality.

To get the second one, put $u=x+y$ and $v=x-y$ and apply the first inequality to $u$ and $v$. Since $x=\frac{u+v}2$ and $y=\frac{u-v}2$ we have $$2^{p-1}\left(\left|\frac{u+v}2\right|^p+\left|\frac{u-v}2\right|^p\right)\leq |u|^p+|v|^p$$ so $2^{p-1}(|x|^p+|y|^p)\leq 2^p(|x+y|+|x-y|)$ and we are done.

For $p\geq 2$, the first inequality is reversed.

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If I understood correctly, to get the second inequality, we use the first inequality $2^{p-1} [|u|^p+|v|^p] \leq |u+v|^p+|u-v|^p$ with $u=(x+y)/2$, $v=(x-y)/2$. – Richard Mar 9 '12 at 18:25
Yes, we have to use this substitution. – Davide Giraudo Mar 9 '12 at 18:26
Thanks a lot for answer. – Richard Mar 9 '12 at 18:28
You're welcome. – Davide Giraudo Mar 9 '12 at 18:29

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