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 . Let M>0. Define A_M=\{ |f| \geq M \}. Then \mu(A_M)>0. Take p so large that \mu(A_M)^{1/p} \geq 1/2, then \| f \|_p \geq (\mu(A_M) M^p)^{1/p} \geq M/2. Since M was arbitrary, f is not uniformly bounded in L^p, 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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