Suppose there exists a sequence $\{n_k\}$ such that $\{\sin n_k x\}$ converges for every $x \in [0,2\pi]$.
Calculate $\int_{0}^{2\pi}(\sin n_k x-\sin n_{k+1}x)^2dx$
I calculated this and got $0$ (I take $\sin n_k x = \sin n_{k+1}x = \sin nx$, since $\{\sin n_k x\}$ converges)
but the answer is $2\pi$.
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This is an example to show that
"Even if {$f_n$} is a uniformly bounded sequence of continuous functions on a compact set E, there need not exist a subsequence which converges pointwise on E."
Original example is here:
Let $f_n(x)=\sin nx$ ($0 \le x \le 2\pi$, n=1,2,3...)
Suppose there exists a sequence $\{n_k\}$ such that $\{\sin n_k x\}$ converges for every $x \in [0,2\pi]$.
In that case $\lim_{k \to \infty}(\sin n_k x-\sin n_{k+1}x)=0$ ($0 \le x \le 2\pi$)
hence, $\lim_{k \to \infty}(\sin n_k x-\sin n_{k+1}x)^2=0$ ($0 \le x \le 2\pi$).
By Lebesgue's theorem concerning integration of boundedly convergent sequences,
$\lim_{k \to \infty}\int_{0}^{2\pi}(\sin n_k x-\sin n_{k+1}x)^2dx=0$
but simple calculation shows that $\int_{0}^{2\pi}(\sin n_k x-\sin n_{k+1}x)^2dx=2\pi$.