Peetre’s inequality

Theorem  [Peetre’s inequality] [1, 2] If $t$ is a real number and $x,y$ are vectors in ${ℝ}^n$, then

$${\left(\frac{1+{|x|}^2}{1+{|y|}^2}\right)}^t\le 2^{|t|}{(1+{|x-y|}^2)}^{|t|}.$$

Proof. (Following [1].) Suppose $b$ and $c$ are vectors in ${ℝ}^n$. Then, from ${(|b|-|c|)}^2\ge 0$, we obtain

$$2|b|\cdot |c|\le {|b|}^2+{|c|}^2.$$

Using this inequality and the Cauchy-Schwarz inequality, we obtain

$1+{|b-c|}^2$ $=$ $1+{|b|}^2-2b\cdot c+{|c|}^2$

$\le$ $1+{|b|}^2+2|b||c|+{|c|}^2$

$\le$ $1+2{|b|}^2+2{|c|}^2$

$\le$ $2\left(1+{|b|}^2+{|c|}^2+{|b|}^2{|c|}^2\right)$

$=$ $2(1+{|b|}^2)(1+{|c|}^2)$

Let us define $a=b-c$. Then for any vectors $a$ and $b$, we have

${\displaystyle \frac{1+{|a|}^2}{1+{|b|}^2}}\le 2(1+{|a-b|}^2).$ (1)

Let us now return to the given inequality. If $t=0$, the claim is trivially true for all $x,y$ in ${ℝ}^n$. If $t>0$, then raising both sides in inequality 1 to the power of $t$, using $t=|t|$, and setting $a=x$, $b=y$ yields the result. On the other hand, if , then raising both sides in inequality 1 to the power to $-t$, using $-t=|t|$, and setting $a=y$, $b=x$ yields the result. $\mathrm{\square }$

Title	Peetre’s inequality
Canonical name	PeetresInequality
Date of creation	2013-03-22 13:55:26
Last modified on	2013-03-22 13:55:26
Owner	Koro (127)
Last modified by	Koro (127)
Numerical id	10
Author	Koro (127)
Entry type	Theorem
Classification	msc 15-00
Classification	msc 15A39