Suppose that ||f||p≤K for all 1≤p<∞ 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 f∈L∞ then
limp→∞||f||p=||f||∞
but this already assume that f∈L∞ 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/p≥1/2, then ‖f‖p≥(μ(AM)Mp)1/p≥M/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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