brian_m.
Nov20-11, 04:55 AM
Hi,
let g \in L^\infty(\mathbb R) with g(x+1)=g(x) in \mathbb R and \int_0^1 g(t)dt =\lambda.
Then for f_n(x)=g(nx) the sequence f_n converges weakly to \lambda in L^p(I) , whereas 1<p<\infty and I denotes an open, bounded intervall.
How to prove this?
Thanks in advance for your help!
Bye,
brian
let g \in L^\infty(\mathbb R) with g(x+1)=g(x) in \mathbb R and \int_0^1 g(t)dt =\lambda.
Then for f_n(x)=g(nx) the sequence f_n converges weakly to \lambda in L^p(I) , whereas 1<p<\infty and I denotes an open, bounded intervall.
How to prove this?
Thanks in advance for your help!
Bye,
brian