proof of Hölder inequality
First we prove the more general form (in measure spaces).
Let (X,μ) be a measure space and let f∈Lp(X), g∈Lq(X) where p,q∈[1,+∞] and 1p+1q=1.
The case p=1 and q=∞ is obvious since
|f(x)g(x)|≤∥g∥L∞|f(x)|. |
Also if f=0 or g=0 the result is obvious.
Otherwise notice that (applying http://planetmath.org/node/YoungInequalityYoung inequality) we have
∥fg∥1∥f∥p⋅∥g∥q=∫X|f|∥f∥p⋅|g|∥g∥q𝑑μ≤1p∫X(|f|∥f∥p)p𝑑μ+1q∫X(|g|∥g∥q)q𝑑μ=1p+1q=1 |
hence the desired inequality holds
∫X|fg|=∥fg∥1≤∥f∥p⋅∥g∥q=(∫X|f|p)1p(∫X|g|q)1q. |
If x and y are vectors in ℝn or vectors in ℓp and ℓq-spaces we can specialize the previous result by choosing μ to be the counting measure on ℕ.
In this case the proof can also be rewritten, without using measure theory, as follows. If we define
∥x∥p=(∑k|xk|p)1p |
we have
|∑kxkyk|∥x∥p⋅∥y∥q≤∑k|xk||yk|∥x∥p⋅∥y∥q=∑k|xk|∥x∥p|yk|∥y∥q≤1p∑k|xk|p∥x∥pp+1q∑k|yk|q∥y∥qq=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 |