Equidistribution of $an^\sigma$ for $\sigma\in(0,1)$

Question

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:

• $\sum_{n=1}^{N}e^{2\pi ibn^\sigma}-\int_1^Ne^{2\pi ibx^\sigma}dx=O(\sum_{n=1}^{N}n^{-1+\sigma})$

• $\sum_{n=1}^{N}e^{2\pi ibn^\sigma}=O(N^\sigma)+O(N^{1-\sigma})$

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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.

Answer 1

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.

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I will follow your approach to prove hint1:

$$\begin{align} |\sum_{n=1}^{N}e^{2\pi ibn^\sigma}-\int_1^Ne^{2\pi ibx^\sigma}dx|&=|\sum_{n=1}^{N-1}\int_n^{n+1}(e^{2\pi ibn^\sigma}-e^{2\pi ibx^\sigma})dx+e^{2\pi ibN^\sigma}|\\ &\leq\sum_{n=1}^{N-1}\int_n^{n+1}|e^{2\pi ibn^\sigma}-e^{2\pi ibx^\sigma}|dx+1\\ &\leq2\pi b\sum_{n=1}^{N-1}\int_n^{n+1}|x^\sigma-n^\sigma|dx+1\\ &\leq2\pi b\sum_{n=1}^{N-1}\int_n^{n+1}|\sigma \xi_n^{\sigma-1}||x-n|dx+1\quad\xi_n\in(n,x)\quad \text{MVT}\\ &\leq2\pi b\sigma \sum_{n=1}^{N-1}n^{\sigma-1}\int_n^{n+1}(x-n)dx+1\\ &=O(\sum_{n=1}^{N}n^{\sigma-1}) \end{align}$$