proof of Clarkson inequality

Suppose 2p< and f,gLp.

f+g2pp+f-g2pp = |f+g2|p𝑑μ+|f-g2|p𝑑μ (1)
= 12p(|f+g|p𝑑μ+|f-g|p𝑑μ). (2)

By the triangle inequalityMathworldMathworldPlanetmath, we have the following two inequalitiesMathworldPlanetmath

|f+g|p|f|p+|g|p  and  |f-g|p|f|p+|g|p,

and summing the two inequalities we get


This means that expression (2) above is less than or equal to

12p-1(|f|p+|g|p)𝑑μ. (3)

Hence it follows that

f+g2pp+f-g2pp 12p-1(|f|p𝑑μ+|g|p𝑑μ)
= 12p-1(fpp+gpp),

which since p2 directly implies the desired inequality.

Title proof of Clarkson inequality
Canonical name ProofOfClarksonInequality
Date of creation 2013-03-22 16:24:46
Last modified on 2013-03-22 16:24:46
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 8
Author CWoo (3771)
Entry type Proof
Classification msc 28A25