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 ‖. 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.
No comments:
Post a Comment