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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 $\int_{0}^{2\pi}(\sin n_k x-\sin n_{k+1}x)^2dx=2\pi$.

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    By dominated convergence theorem, $\lim_{k \to\infty}\int_{0}^{2\pi}(\sin n_k x-\sin n_{k+1}x)^2dx=0$, but $\int_{0}^{2\pi}(\sin n_k x-\sin n_{k+1}x)^2dx=2\pi$. I don't know how to get the answer.2012-12-10

1 Answers 1

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For every nonzero integers $k\ne\ell$, $\displaystyle\int_0^{2\pi}(\sin(kx)-\sin(\ell x))^2\mathrm dx=2\pi$.

To prove this, expand the square, use the identities $ -2\sin(kx)\sin(\ell x)=\cos((k+\ell)x)-\cos((k-\ell)x),\quad \sin^2(nx)=\tfrac12-\tfrac12\cos(2nx), $ and the fact that for every integer $n\ne0$, $ \int_0^{2\pi}\cos(nx)\mathrm dx=0. $

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    Good: indeed every $n_k$ is an integer, and $n_k\lt n_{k+1}$. Hence my post fully answers the question.2012-12-10