proof of Lp-norm is dual to Lq


Let (X,𝔐,μ) be a σ-finite measure space and p,q be Hölder conjugates. Then, we show that a measurable functionMathworldPlanetmath f:X→ℝ has Lp-norm

∥f∥p=sup⁡{∥f⁢g∥1:g∈Lq,∥g∥q=1}. (1)

Furthermore, if either p<∞ and ∥f∥p<∞ or p=1 then μ is not required to be σ-finite.

If ∥f∥p=0 then f is zero almost everywhere, and both sides of equality (1) are zero. So, we only need to consider the case where ∥f∥p>0.

Let K be the right hand side of equality (1). For any g∈Lq with ∥g∥q=1, the Hölder inequalityMathworldPlanetmath gives ∥f⁢g∥1≤∥f∥p, so K≤∥f∥p. Only the reverse inequality remains to be shown.

If 1<p<∞ and ∥f∥p<∞ then, setting g=|f|p-1 gives

∥g∥q=(∫|f|p⁢𝑑μ)1q=∥f∥pp-1<∞.

Therefore, g∈Lq and,

K≥∥f⁢(g/∥g∥q)∥1=∥|f|p∥1/∥g∥q=∥f∥pp/∥f∥pp-1=∥f∥p.

On the other hand, if p=1 so that q=∞, then setting g=1 gives ∥g∥q=1 and

K≥∥f⁢g∥1=∥f∥1.

So, we have shown that K=∥f∥p when p<∞ and ∥f∥p<∞, and when p=1. From now on, it is assumed that the measureMathworldPlanetmath is σ-finite. Then there is a sequence An∈𝔐 increasing to the whole of X and such that μ⁢(An)<∞.

Now consider the case where 1<p<∞ and ∥f∥p=∞. Let fn be the sequence of functions

fn=1An⁢1|f|≤n⁢f

then, |fn|≤|f| and monotone convergence gives ∥fn∥p→∥f∥p=∞. Therefore,

K≥sup⁡{∥fn⁢g∥1:g∈Lq,∥g∥q=1}=∥fn∥p.

and letting n go to infinity gives K=∞.

We finally consider p=∞. Then, for any L<∥f∥p there exists a set A∈𝔐 with μ⁢(A)>0 such that |f|≥L on A. Also, monotone convergence gives μ⁢(A∩An)→μ⁢(A) and, therefore, μ⁢(A∩An)>0 eventually. Replacing A by A∩An if necessary, we may suppose that μ⁢(A)<∞. So, setting g=1A/μ⁢(A) gives ∥g∥1=1 and,

K≥∥f⁢g∥1=∫A|f|⁢𝑑μ/μ⁢(A)≥L.

Letting L increase to ∥f∥p gives K≥∥f∥p as required.

Title proof of Lp-norm is dual to Lq
Canonical name ProofOfLpnormIsDualToLq
Date of creation 2013-03-22 18:38:16
Last modified on 2013-03-22 18:38:16
Owner gel (22282)
Last modified by gel (22282)
Numerical id 4
Author gel (22282)
Entry type Proof
Classification msc 46E30
Classification msc 28A25