Let $Z=\sqrt{2}\mathbb Z$, and consider the sequence on $\mathbb{Z}$ $$\xi(k)= 1_{Z\cap [k,k+1]\neq\emptyset}.$$(remark that the intersection is either empty or with one point.)

The question is: is this sequence periodic?

My mathematical intuition shouts me that "no", but I cannot find the error in the following argument. Would you agree with me or show how the argument is wrong? Remark that $\sqrt{2}$ can be replaced by any non-zero real number, but it is not very interesting for a rational number.

The basic idea is to consider $Z$ as a measure and write $\xi$ as the convolution $$\xi(k)=Z\star \psi(k)=\sum_{m\in \mathbb{Z}}\sum_{x\in Z}\psi(x-m)$$ where $\psi(y)=1_{[0,1]}(y)$.

One can prove in the right space that $\hat \xi=\hat Z\times \hat \psi$, and $\hat Z$ is purely atomic with Poisson summation formula. In particular, $\hat\xi$'s support is discrete, and this means with classical spectral theory (Helson, Szego, ...) that $\xi$ is periodic. I wrote all this down but I fear I have made a mistake, that I do not want to reproduce here not to influence you.

I will validate any answer that gives me another proof or ref that this is true or finds the problem in the argument.

Let $Z=\sqrt{2}\mathbb Z$, and consider the sequence on $\mathbb{Z}$ $$\xi(k)= 1_{Z\cap [k,k+1]\neq\emptyset}.$$(remark that the intersection is either empty or with one point.)

Thanks to the comments, it is clear that this sequence is not periodic (@wojowu: the density of such $k$ goes to $1/\sqrt{2}$).

The problem is that it contradicts the following argument. Could you help me find where the error is below?

The basic idea is to consider $Z$ as a measure and write $\xi$ as the convolution $$\xi(k)=Z\star \psi(k)=\sum_{m\in \mathbb{Z}}\sum_{x\in Z}\psi(x-m)$$ where $\psi(y)=1_{[0,1]}(y)$.

One can prove in the right space that $\hat \xi=\hat Z\times \hat \psi$, and $\hat Z$ is purely atomic with Poisson summation formula. In particular, $\hat\xi$'s support is discrete, and this means with classical spectral theory (Helson, Szego, ...) that $\xi$ is periodic. I wrote all this down but I do not want to reproduce it here so that my mistake does not influence you.

Let $Z=\sqrt{2}\mathbb Z$, and consider the sequence on $\mathbb{Z}$ $$\xi(k)= 1_{Z\cap [k,k+1]\neq\emptyset}.$$(remark that the intersection is either empty or with one point.)

The question is: is this sequence periodic?

My mathematical intuition shouts me that "no", but I cannot find the error in the following argument. Would you agree with me (or show how the argument)? Remark that $\sqrt{2}$ can be replaced by any non-zero real number, but it is not very interesting for a rational number.

The basic idea is to consider $Z$ as a measure and write $\xi$ as the convolution $$\xi(k)=Z\star \psi(k)=\sum_{m\in \mathbb{Z}}\sum_{x\in Z}\psi(x-m)$$ where $\psi(y)=1_{[0,1]}(y)$.

One can prove in the right space that $\hat \xi=\hat Z\times \hat \psi$, and $\hat Z$ is purely atomic with Poisson summation formula. In particular, $\hat\xi$'s support is discrete, and this means with classical spectral theory (Helson, Szego, ...) that $\xi$ is periodic. I wrote all this down but I fear I have made a mistake, that I do not want to reproduce here not to influence you.

I will validate any answer that gives me another proof or ref that this is true or finds the problem in the argument.

Let $Z=\sqrt{2}\mathbb Z$, and consider the sequence on $\mathbb{Z}$ $$\xi(k)= 1_{Z\cap [k,k+1]\neq\emptyset}.$$(remark that the intersection is either empty or with one point.)

The question is: is this sequence periodic?

My mathematical intuition shouts me that "no", but I cannot find the error in the following argument. Would you agree with me or show how the argument is wrong? Remark that $\sqrt{2}$ can be replaced by any non-zero real number, but it is not very interesting for a rational number.

The basic idea is to consider $Z$ as a measure and write $\xi$ as the convolution $$\xi(k)=Z\star \psi(k)=\sum_{m\in \mathbb{Z}}\sum_{x\in Z}\psi(x-m)$$ where $\psi(y)=1_{[0,1]}(y)$.

One can prove in the right space that $\hat \xi=\hat Z\times \hat \psi$, and $\hat Z$ is purely atomic with Poisson summation formula. In particular, $\hat\xi$'s support is discrete, and this means with classical spectral theory (Helson, Szego, ...) that $\xi$ is periodic. I wrote all this down but I fear I have made a mistake, that I do not want to reproduce here not to influence you.

I will validate any answer that gives me another proof or ref that this is true or finds the problem in the argument.

Let $Z=\sqrt{2}\mathbb Z$, and consider the sequence on $\mathbb{Z}$ $$\xi(k)= 1_{Z\cap [k,k+1]\neq\emptyset}.$$(remark that the intersection is either empty or with one point.)

The question is: is this sequence periodic?

My mathematical intuition shouts me that "no", but I cannot find the error in the following argument. Would you agree with me (or have a proof that this is not periodic)? Remark that $\sqrt{2}$ can be replaced by any non-zero real number, but it is not very interesting for a rational number.

The basic idea is to consider $Z$ as a measure and write $\xi$ as the convolution $$\xi(k)=Z\star \psi(k)=\sum_{m\in \mathbb{Z}}\sum_{x\in Z}\psi(x-m)$$ where $\psi(y)=1_{[0,1]}(y)$.

One can prove in the right space that $\hat \xi=\hat Z\times \hat \psi$, and $\hat Z$ is purely atomic with Poisson summation formula. In particular, $\hat\xi$'s support is discrete, and this means with classical spectral theory (Helson, Szego, ...) that $\xi$ is periodic. I wrote all this down but I fear I have made a mistake, that I do not want to reproduce here not to influence you.

Let $Z=\sqrt{2}\mathbb Z$, and consider the sequence on $\mathbb{Z}$ $$\xi(k)= 1_{Z\cap [k,k+1]\neq\emptyset}.$$(remark that the intersection is either empty or with one point.)

The question is: is this sequence periodic?

My mathematical intuition shouts me that "no", but I cannot find the error in the following argument. Would you agree with me (or show how the argument)? Remark that $\sqrt{2}$ can be replaced by any non-zero real number, but it is not very interesting for a rational number.

The basic idea is to consider $Z$ as a measure and write $\xi$ as the convolution $$\xi(k)=Z\star \psi(k)=\sum_{m\in \mathbb{Z}}\sum_{x\in Z}\psi(x-m)$$ where $\psi(y)=1_{[0,1]}(y)$.

One can prove in the right space that $\hat \xi=\hat Z\times \hat \psi$, and $\hat Z$ is purely atomic with Poisson summation formula. In particular, $\hat\xi$'s support is discrete, and this means with classical spectral theory (Helson, Szego, ...) that $\xi$ is periodic. I wrote all this down but I fear I have made a mistake, that I do not want to reproduce here not to influence you.

I will validate any answer that gives me another proof or ref that this is true or finds the problem in the argument.