Let $\mu$ be a continuous measure on $[0,1]$ (i.e. each individual point has $0$ measure). As usual, denote by $\hat\mu(n)=\int_0^1e^{2\pi inx}d\mu(x)$ the Fourier transform of $\mu$, and let $\lfloor x\rfloor$ denote the floor of $x\in\mathbb R$. Is it true that $$\lim_{N\to\infty}sup_{M\in\mathbb N}\frac1N\sum_{n=M}^{M+N}\left|\hat\mu\Big(\big\lfloor n^{3/2}\big\rfloor\Big)\right|=0~?$$

Observe that if $\mu$ is absolutely continuous, then the answer is yes by the Riemann-Lebesgue lemma.

If instead of $\hat\mu(\lfloor n^{3/2}\rfloor)$ we consider $\hat\mu(n)$, then the answer is again yes (it follows from Wiener's theorem).

If we don't take a supremum over all shifts of the interval $\{1,\dots,N\}$, then the result is again well known because the sequences $n\mapsto\lfloor n^{3/2}\rfloor x$ are uniformly distributed for any irrational $x$.

The motivation for this question comes from the fact that $$\lim_{N\to\infty}sup_{M\in\mathbb N}\frac1N\sum_{n=M}^{M+N}\left|\hat\mu\big(n^2\big)\right|=0.$$ This follows easily from the fact that the sequences $n\mapsto n^2 x$ are well distributed for every irrational $x$, but this is not the case for the sequences $n\mapsto\lfloor n^{3/2}\rfloor x$.

  • $\begingroup$ The Fourier coefficients are complex numbers, so you need to take an absolute value somewhere before you can take the sup. Where exactly do you put the $|\cdot|$? $\endgroup$ Sep 14, 2016 at 20:32
  • $\begingroup$ Oops, you're right, I just added them $\endgroup$ Sep 14, 2016 at 20:37

1 Answer 1


Really, you have almost answered it yourself, just left the very final words out.

Take $M=4N^4$. Then $\lfloor (M+n)^{3/2}\rfloor=8N^6+3N^2n$ for $n=1,\dots,N$, so $\frac 1N\sum_{n=1}^N e^{-2\pi i \lfloor (M+n)^{3/2}\rfloor x}$ is $1$ when $x=q/N^2$ and nearly $1$ on a small open neighborhood $U_N$ of those points. Now take some very fast increasing sequence of $N$ so that $\cap_N U_N$ contains a Cantor-type set and put the measure on it. The exact formulae do not matter, of course. What mattered a bit was having long linear pieces of arbitrarily large slope, but even that wasn't crucial. As you noticed yourself, the (particular kind of) absence of good distribution was the key.

  • $\begingroup$ Thanks for the answer but I didn't understand something: it seems to me that the intervals $U_N$ could shrink so that the intersection will be a single point at most. How do we guarantee we can fit a Cantor set in there? $\endgroup$ Sep 16, 2016 at 1:22
  • $\begingroup$ Oh, never mind I see it now, each $U_N$ consists of $N^2$ intervals, not just one. $\endgroup$ Sep 16, 2016 at 1:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.