Tuesday, 9 September 2014

functional analysis - If all Lp norms are bounded then the Linfty is bounded



Suppose that ||f||pK for all 1p< for some K>0.

How to show that the essential supremum exists and bounded by K that i s||f||K?



I know how to prove that if fL then
limp||f||p=||f||
but this already assume that fL in this question we have to show that f has an essential supremum. To be more precise I don't think I can use a technique when I define
Aϵ={x| |f(x)|>||f||ϵ}




I feel like here we have to use some converges theorem.
Thanks for any help


Answer



Assume f=. Let M>0. Define AM={|f|M}. Then μ(AM)>0. Take p so large that μ(AM)1/p1/2, then fp(μ(AM)Mp)1/pM/2. Since M was arbitrary, f is not uniformly bounded in Lp, and your result follows by contraposition.



This is essentially the argument suggested by John Ma in the comments, but decoupling M and p.


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