# Peetre’s inequality

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

 $\Big{(}\frac{1+|x|^{2}}{1+|y|^{2}}\Big{)}^{t}\leq 2^{|t|}(1+|x-y|^{2})^{|t|}.$

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

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

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

 $\displaystyle 1+|b-c|^{2}$ $\displaystyle=$ $\displaystyle 1+|b|^{2}-2b\cdot c+|c|^{2}$ $\displaystyle\leq$ $\displaystyle 1+|b|^{2}+2|b||c|+|c|^{2}$ $\displaystyle\leq$ $\displaystyle 1+2|b|^{2}+2|c|^{2}$ $\displaystyle\leq$ $\displaystyle 2\big{(}1+|b|^{2}+|c|^{2}+|b|^{2}|c|^{2}\big{)}$ $\displaystyle=$ $\displaystyle 2(1+|b|^{2})(1+|c|^{2})$

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

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

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

## References

• 1 J. Barros-Neta, An introduction to the theory of distributions, Marcel Dekker, Inc., 1973.
• 2 F. Treves, Introduction To Pseudodifferential and Fourier Integral Operators, Vol. I, Plenum Press, 1980.
Title Peetre’s inequality PeetresInequality 2013-03-22 13:55:26 2013-03-22 13:55:26 Koro (127) Koro (127) 10 Koro (127) Theorem msc 15-00 msc 15A39