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 $$\lim_{N\to\infty}sup_{M\in\mathbb N}\left|\frac1N\sum_{n=M}^{M+N}\hat\mu\Big(\big\lfloor n^{3/2}\big\rfloor\Big)\right|=0~?$$

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

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

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 $$\lim_{N\to\infty}sup_{M\in\mathbb N}\frac1N\sum_{n=M}^{M+N}\hat\mu\Big(\big\lfloor n^{3/2}\big\rfloor\Big)=0~?$$

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

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 $$\lim_{N\to\infty}sup_{M\in\mathbb N}\left|\frac1N\sum_{n=M}^{M+N}\hat\mu\Big(\big\lfloor n^{3/2}\big\rfloor\Big)\right|=0~?$$

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