Does bounded sequence of functions have subsequence convergent a.e.?
There is a equivalent question :
Does weak convergent sequence of functions have subsequence convergent a.e.?
These two are equivalent because of Banach-Steinhaus theorem.
Before I ask here, I thought Bolzano-Weierstrass theorem which is every bounded sequence has convergent subsequence. And my question comes up from 'this theorem can be applied to sequence of functions?'.
If the space of functions should be special (like reflexive or compact or housdorff), please mention about it.
I didn't study functional analysis but attended courses : Partial differential equation and real analysis for graduate student.
Best Answer
There is no subsequence of $\sin (nx)$ that converges a.e. In fact, every subsequence $\sin (n_kx)$ diverges a.e. For a proof of this, see Pointwise almost everywhere convergent subsequence of $\{\sin (nx)\}$