𝕃p vs 𝕃q

Let (X,𝒜,μ) be a measure spaceMathworldPlanetmath and 1p,q. Generally there is no connection between 𝕃p(μ) and 𝕃q(μ) as sets. However, for some special measures, there is an interesting relationship between them. A few examples:

  1. 1.

    If λn is the Lebesgue measureMathworldPlanetmath on n and pq, then 𝕃p(λn)𝕃q(λn) for all n+. Here is an example for n=1 and 1p<q<. Let




    This gives fqq=pq-p, gpp=qq-p and fp=gq=. So f𝕃q𝕃p and g𝕃p𝕃q. For the -norm, χ𝕃𝕃q, where χ is the characteristic functionMathworldPlanetmathPlanetmathPlanetmathPlanetmath, and also f𝕃.

  2. 2.

    If p<q then lplq. This is trivial if q=. Now let x=(x0,x1,)lp and q<. Then


    so xlq.

  3. 3.

    If μ(X) is finite and p<q, then 𝕃q𝕃p. This is easy if q=, because |f|f almost everywhere, so fpp=|f|p𝑑μfp𝑑μ=fpμ(X)<. Now let q<, thus

    fpp =|f|p𝑑μ

Finally, we prove an interesting property for p-norms: if X is a finite measure space, then for any measurable functionMathworldPlanetmath f on X the equality limpfp=f holds. We have already seen that fpfμ(X)1p. Now for any ε(0,f) define Aε:={xX:|f(x)|f-ε}, δε:=μ(Aε)>0 and g:=(f-ε)χAε. Since |g||f|, we have gp=(f-ε)δε1pfpfμ(X)1p. Now we take lim inf on the left and lim sup on the right side: f-εlim infpfplim suppfpf. Taking ε0 gives limpfp=f.

Title 𝕃p vs 𝕃q
Canonical name mathbbLpVsmathbbLq
Date of creation 2013-03-22 15:22:05
Last modified on 2013-03-22 15:22:05
Owner yark (2760)
Last modified by yark (2760)
Numerical id 11
Author yark (2760)
Entry type Topic
Classification msc 28A25