Peetre’s inequality
Proof. (Following [1].) Suppose and are vectors in . Then, from , we obtain
Using this inequality and the Cauchy-Schwarz inequality, we obtain
Let us define . Then for any vectors and , we have
(1) |
Let us now return to the given inequality. If , the claim is trivially true for all in . If , then raising both sides in inequality 1 to the power of , using , and setting , yields the result. On the other hand, if , then raising both sides in inequality 1 to the power to , using , and setting , yields the result.
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 |
---|---|
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 |