Let {$f_n$} be a sequence of continuous functions on $\mathbb R$ which is pointwise bounded.
Does there exist subsequence {${f_n}_i$} such that
{${f_n}_i(x)$} converges for all $x\in \mathbb R$?
My attempt:
By Cantor diagonal argument I got subsequence coverging pointwise on rational numbers but I dont know how to proceed further. So now I suspect that there is some counter example. Any help is welcome!!!
I think I found a counterexample. Let $f_n(x)= \sin(nx)$ for $x \in [0,\pi]$ and $f_n(x) = 0$ otherwise. This sequence is continuous and uniformly bounded but it does not converge pointwise along any subsequence.
Indeed, note that if $n \neq m$, we have $$\int^\pi_{0} (f_n(x) - f_m(x))^2 dx = \int^\pi_0 [\sin^2(nx) - 2\sin(nx)\sin(mx) +\sin^2(mx) ]dx.$$ The middle term goes to zero by orthogonality and : $$\int^\pi_0 \sin^2(nx) dx = \frac{\pi}{2}$$ for any integer $n$. Thus $$\int^\pi_{0} (f_n(x) - f_m(x))^2 dx = \pi.$$ If $\{f_{n_i}\}$ converged at all points, then $$(f_{n_{i+1}} - f_{n_i})^2 \to 0$$ and then by the dominated convergence theorem we would have $$\int^\pi_0 (f_{n_{i+1}}( x) - f_{n_i}( x) )^2 dx \to 0$$ which is impossible.