I am stuck at an exercise in Stein's book: Fourier Analysis, and it's exercise 8 in chapter 4.
Show that for any $a\ne0$, and $\sigma$ with $0<\sigma<1$, the sequence $an^\sigma$ is equidistributed in $[0,1)$.
There are two hints:
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$\sum_{n=1}^{N}e^{2\pi ibn^\sigma}-\int_1^Ne^{2\pi ibx^\sigma}dx=O(\sum_{n=1}^{N}n^{-1+\sigma})$
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$\sum_{n=1}^{N}e^{2\pi ibn^\sigma}=O(N^\sigma)+O(N^{1-\sigma})$
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Below is my attempt.
$$|\sum_{n=1}^{N}e^{2\pi ibn^\sigma}-\int_1^Ne^{2\pi ibx^\sigma}dx|$$
$$=|\sum_{n=1}^{N-1}\int_n^{n+1}(e^{2\pi ibn^\sigma}-e^{2\pi ibx^\sigma})dx+e^{2\pi ibN^\sigma}|$$
$$\leq\sum_{n=1}^{N-1}\int_n^{n+1}|e^{2\pi ibn^\sigma}-e^{2\pi ibx^\sigma}|dx+1$$
$$\leq2\pi b\sum_{n=1}^{N-1}\int_n^{n+1}|x^\sigma-n^\sigma|dx+1$$
$$\leq2\pi b\sum_{n=1}^{N-1}\int_n^{n+1}|(n+1)^\sigma-n^\sigma|dx+1$$
$$=2\pi bN^\sigma-2\pi b+1$$
$$=O(N^\sigma)$$
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However my attempt doesn't solve two hints and problem. Can someone prove it following the hints? Thank you.
Best Answer
To prove the theorem, you don't need to use the first hint.
Note by Weyl's criterion, it suffices to prove
$$\frac{1}{N}\sum_{n=1}^{N}e^{2\pi ibn^\sigma}\to 0$$ First we show $\int_1^Ne^{2\pi ibx^\sigma}dx=O(N^{1-\sigma})$ $$\begin{align} \int_1^Ne^{2\pi ibx^\sigma}dx&=\int_1^N\frac{x^{1-\sigma}}{2\pi ib\sigma} (2\pi ib\sigma)x^{\sigma-1}e^{2\pi ibx^\sigma}dx\\& =\int_1^N\frac{x^{1-\sigma}}{2\pi ib\sigma} de^{2\pi ibx^\sigma}\\ &=\frac{x^{1-\sigma}}{2\pi ib\sigma} e^{2\pi ibx^\sigma}\bigg|_1^N-\int_1^N\frac{(1-\sigma )x^{-\sigma}}{2\pi ib\sigma} e^{2\pi ibx^\sigma}dx\\ \end{align}$$ Hence $$\begin{align} \bigg|\int_1^Ne^{2\pi ibx^\sigma}dx\bigg|&\le\bigg|\frac{N^{1-\sigma}e^{2\pi ibN^\sigma}-e^{2\pi ib}}{2\pi ib\sigma} \bigg|+\int_1^N\frac{(1-\sigma )x^{-\sigma}}{|2\pi ib\sigma|} dx\\ &\leq \frac{N^{1-\sigma}+1}{2\pi |b|\sigma} +\frac{(1-\sigma )}{2\pi |b|\sigma}\int_1^Nx^{-\sigma}dx\\&=O(N^{1-\sigma}) \end{align}$$
Then from what you have done,
$$|\sum_{n=1}^{N}e^{2\pi ibn^\sigma}-\int_1^Ne^{2\pi ibx^\sigma}dx|=O(N^\sigma)$$
we have $\sum_{n=1}^{N}e^{2\pi ibn^\sigma}=O(N^\sigma)+O(N^{1-\sigma})$
Hence $\frac{1}{N}\sum_{n=1}^{N}e^{2\pi ibn^\sigma}=O(N^{\sigma-1})+O(N^{-\sigma})\to 0$ since $0<\sigma<1$.
Thus it's equidistributed.
I will follow your approach to prove hint1:
$$\begin{align} |\sum_{n=1}^{N}e^{2\pi ibn^\sigma}-\int_1^Ne^{2\pi ibx^\sigma}dx|&=|\sum_{n=1}^{N-1}\int_n^{n+1}(e^{2\pi ibn^\sigma}-e^{2\pi ibx^\sigma})dx+e^{2\pi ibN^\sigma}|\\ &\leq\sum_{n=1}^{N-1}\int_n^{n+1}|e^{2\pi ibn^\sigma}-e^{2\pi ibx^\sigma}|dx+1\\ &\leq2\pi b\sum_{n=1}^{N-1}\int_n^{n+1}|x^\sigma-n^\sigma|dx+1\\ &\leq2\pi b\sum_{n=1}^{N-1}\int_n^{n+1}|\sigma \xi_n^{\sigma-1}||x-n|dx+1\quad\xi_n\in(n,x)\quad \text{MVT}\\ &\leq2\pi b\sigma \sum_{n=1}^{N-1}n^{\sigma-1}\int_n^{n+1}(x-n)dx+1\\ &=O(\sum_{n=1}^{N}n^{\sigma-1}) \end{align}$$