another proof of Young inequality


Let

F(x)=0xϕ(t)𝑑t, and G(x)=0xϕ-1(t)𝑑t.

Since ϕ-1 is strictly increasingPlanetmathPlanetmath, G is strictly convex, hence lies above its supporting line, i.e. for every c and xc

G(b)>G(c)+G(c)(b-c)=G(c)+ϕ-1(c)(b-c).

In particular, for c=ϕ(a) we have

F(a)+G(b)>F(a)+G(ϕ(a))+a(b-ϕ(a))=ab,

because F(a)+G(ϕ(a))=aϕ(a).

Title another proof of Young inequality
Canonical name AnotherProofOfYoungInequality
Date of creation 2013-03-22 15:45:38
Last modified on 2013-03-22 15:45:38
Owner a4karo (12322)
Last modified by a4karo (12322)
Numerical id 10
Author a4karo (12322)
Entry type Proof
Classification msc 26D15