Use the van der Corput Lemma to prove the equidistribution of $\{\alpha n^2\}$

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