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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.

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  • $\begingroup$ Just curious why is $\vert e^{2{\pi}{i}n^{\sigma}}- e^{2{\pi}{i}x^{\sigma}} \vert \leq 2\pi{b} \vert x^{\sigma} - n^{\sigma} \vert?$ $\endgroup$
    – user135520
    May 19, 2017 at 21:42
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    $\begingroup$ Think of two points on the circle. The arclength is always bigger than the direct distance between the two points. $\endgroup$ Jan 12, 2018 at 22:13

1 Answer 1

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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:

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

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  • $\begingroup$ Thank you. I am also curious about how we can prove the hint 1, and can hint 1 deduces the hint 2 to some degree. $\endgroup$
    – gaoxinge
    Nov 7, 2014 at 13:19
  • $\begingroup$ @gaoxinge You just need to use MVT twice. $\endgroup$
    – John
    Nov 7, 2014 at 13:55
  • $\begingroup$ @JohnZHANG How is the use of the MVT in the second inequality for the proof of hint 1 justified? I thought for complex-valued functions we cannot use the MVT, as in $e^{xi}$ on $[0,2\pi]$. Are there specific instances we can use it? $\endgroup$ Sep 23, 2016 at 20:48

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