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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 $n\mapsto\lfloor n^{3/2}\rfloor x$.

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  • $\begingroup$ 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

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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.

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  • $\begingroup$ 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

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