proof of Hölder inequality


First we prove the more general form (in measure spacesMathworldPlanetmath).

Let (X,μ) be a measure space and let fLp(X), gLq(X) where p,q[1,+] and 1p+1q=1.

The case p=1 and q= is obvious since

|f(x)g(x)|gL|f(x)|.

Also if f=0 or g=0 the result is obvious. Otherwise notice that (applying http://planetmath.org/node/YoungInequalityYoung inequalityMathworldPlanetmathPlanetmath) we have

fg1fpgq=X|f|fp|g|gq𝑑μ1pX(|f|fp)p𝑑μ+1qX(|g|gq)q𝑑μ=1p+1q=1

hence the desired inequalityMathworldPlanetmath holds

X|fg|=fg1fpgq=(X|f|p)1p(X|g|q)1q.

If x and y are vectors in n or vectors in p and q-spaces we can specialize the previous result by choosing μ to be the counting measure on .

In this case the proof can also be rewritten, without using measure theory, as follows. If we define

xp=(k|xk|p)1p

we have

|kxkyk|xpyqk|xk||yk|xpyq=k|xk|xp|yk|yq1pk|xk|pxpp+1qk|yk|qyqq=1p+1q=1.
Title proof of Hölder inequality
Canonical name ProofOfHolderInequality
Date of creation 2013-03-22 13:31:16
Last modified on 2013-03-22 13:31:16
Owner paolini (1187)
Last modified by paolini (1187)
Numerical id 10
Author paolini (1187)
Entry type Proof
Classification msc 15A60
Classification msc 46E30
Synonym proof of Hölder inequality
Synonym proof of Holder’s inequality