Proof that the sequence of functions $f_n(x) = \sin(nx)$ doesn't have a convergent subsequence in $C[0,2 \pi]$.
I have came across this question in Rudin's PRINCIPLES OF
MATHEMATICAL ANALYSIS (Chapter 7) but the proof is using dominated convergence theorem. See below
Is there any proof which doesn't use Measure Theory?
Best Answer
I'd recommend to put the problem into the body of your question, it's not good that it's only in the title.
Not knowing that book, I can't imagine why anybody would need the dominated convergence theorem for a proof: convergence in $C[0,2 \pi]$ is uniform convergence, so integrals of such functions converge as well, without any measure theory.
Whatever: let $g_k(x)=\sin n_k x$. If that's a convergent sequence, you must have $\|g_k-g_l\|<\epsilon$ for $k,l>N_\epsilon$, and that implies $\int^{2\pi}_0(g_k(x)-g_l(x))^2\,dx<2\pi\epsilon^2$, while in reality $\int^{2\pi}_0(g_k(x)-g_l(x))^2\,dx=\int^{2\pi}_0g_k(x)^2\,dx+\int^{2\pi}_0g_l(x)^2\,dx=2\pi$, since $2\int^{2\pi}_0g_k(x)g_l(x)\,dx=\int^{2\pi}_0(\cos(n_k-n_l)x-\cos(n_k+n_l)x)\,dx=0$.