Peetre’s inequality

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


Proof. (Following [1].) Suppose b and c are vectors in n. Then, from (|b|-|c|)20, we obtain


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

1+|b-c|2 = 1+|b|2-2bc+|c|2
= 2(1+|b|2)(1+|c|2)

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

1+|a|21+|b|22(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 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.


  • 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
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