The van der Corput Lemma states

Van der Corput Lemma: Let $(x_n)$ be a bounded sequence in a Hilbert space $H$. Define a sequence $(s_n)$ by $$s_h = \limsup_{N \to \infty} \left | \frac1N \sum_{n = 1}^N \langle x_{n + h}, x_h \rangle \right |.$$ If there now holds that $$\lim_{H \to \infty} \frac1H \sum_{h = 0}^{H - 1} s_h = 0,$$ then we have that $$\lim_{N \to \infty} \left \| \frac1N \sum_{n = 1}^N x_n \right \| = 0.$$

We should be able to prove using this lemma that ($\{x\}$ denotes the fractional part of $x$) $\{n^2 \alpha \}$ is equidistributed where $\alpha$ is irrational.

Does someone have a hint how to do this? If I solve it I will modify my question to give the full solution. I assume that I am missing something simple.

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    $\begingroup$ Don't forget to use Weyl's criterion. $\endgroup$ – Yuval Filmus May 29 '11 at 17:36

Hint: Take $\displaystyle x_n (t) = e^{2\pi i n^2 \alpha t}$ in $L^2(S^1)$.

  • $\begingroup$ Okay, I have fixed the typos. What is your Hilbert space? $L^2$? Then what is the variable because $\alpha$ is a fixed irrational number? $\endgroup$ – Jonas Teuwen May 29 '11 at 14:31
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    $\begingroup$ Typo-day today :) I think now the hint's fine. $\endgroup$ – t.b. May 29 '11 at 14:40
  • $\begingroup$ Okay, I have picked this up again. I'm not sure what norm you have on your space, for me $L^2(S^1)$ has the norm $\|f\| = \left (\int_0^1 f(e^{2 \pi i \theta}) \, d\theta \right )^\frac12$. This norm does not make any sense in this case and replacing $f(e^{2 \pi i t})$ with $f(t)$ does not seem to give me an interesting conclusion. I'm really missing something I think. Say it is the norm with $f(t)$, then I have $L^2$ convergence and from that I can conclude nothing pointwise except for a subsequence. Could you extend your hint a little bit? $\endgroup$ – Jonas Teuwen Jun 3 '11 at 22:58
  • $\begingroup$ @Jonas: I'll think about it again and post something. But please be a bit patient, I'm pretty busy these days. Sorry if my suggestion didn't seem to help. $\endgroup$ – t.b. Jun 4 '11 at 8:02
  • $\begingroup$ Okay, thanks :). I'm not in a hurry. $\endgroup$ – Jonas Teuwen Jun 4 '11 at 19:29

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