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I am wondering if either implication is true:

$$(a_n) \text{ is equidistributed in } [0,1] \iff\text{ For each } x\in [0,1],\ \vert a_n-x\vert<\frac{1}{n}\text{ for infinitely many } n\in\mathbb{N}.$$

I'm not sure either way, but if yes, then the right-side statement would be an interesting equivalent definition of "equidistributed sequence".

Obviously for $\Rightarrow$ I thought of assuming the right side of the proposition was false and proving by contradiction, but didn't get anywhere. I also couldn't construct a counter-example for $\Rightarrow$ or $\Leftarrow$.

I don't see how anything like Weyl's Criterion could be useful here.

Adam Rubinson
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  • I think $\Rightarrow$ fails. Consider the sequence $0, 1/2, 0, 1/3, 2/3, 0, 1/4, 2/4, 3/4, \dots$. I haven't checked carefully but the criterion on the right appears to fail for $x=1$. If it doesn't fail, I think we can make it fail by padding our sequence at the beginning with a bunch of $0$s, which shouldn't affect equidistribution but should prevent the criterion on the right. – user806050 Sep 19 '24 at 11:53
  • You might be right, but I don't see how padding a bunch of $0'$s at the beginning makes the right side of my implication fail. – Adam Rubinson Sep 19 '24 at 12:12
  • Same idea as my deleted comment, but here's a sketch that $\Leftarrow$ fails. For each $N\in\mathbb{N}$, there is a finite covering of $[0,1]$ by $k$ balls of radius $1/N, 1/(N+2), ..., 1/(N+2k)$ respectively for some $k$. Let $y_1,\dots,y_k$ be the centers of the balls, and define $s(N) = (y_1, 0, y_2, 0, \dots, y_k)$ and $e(N)=k$. Now consider the sequence $(a_n)$ defined recursively by concatenating the sequences $s(1)$ and for $i > 1$, $s(e(i-1))$. Sorry that this is messy, but this should not be equidistributed. – user806050 Sep 19 '24 at 12:14
  • Looking at it again, you don't need to pad with $0$s to get the counterexample. I was thinking that padding could improve the counterexample because even if the sequence $(a_n)$ approaches $1$ arbitrarily closely to $1$ infinitely many times, it would not do so with the $1/n$ dependence once we add $0$s to the beginning of the sequence, because the terms of the sequence getting within $1/n$ of $1$ would be too "far out" in the sequence. – user806050 Sep 19 '24 at 12:20

1 Answers1

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The implication $\Rightarrow$ is false. Let $p_n$ be the sequence of prime numbers. Consider the sequence $${1\over p_1},{2\over p_1},{1\over p_2},{2\over p_2},{3\over p_2},\ldots ,{1\over p_k},{2\over p_k},\ldots , {p_k\over p_k},\ldots $$ and $x={1\over 2}.$ The sequence is equidistributed. Indeed, consider the part of the subsequence terminating at ${m\over p_{k+1}}.$ The first part of the subsequence starting at ${1\over p_1}$ and terminating at ${p_k\over p_k},$ consists of $s_k:=p_1+p_2+\ldots +p_k$ terms and is equally distributed. The remaining part starting at ${1\over p_{k+1}}$ and ending at ${m\over p_{k+1}}$ consists of at most $p_{k+1}$ terms. Therefore it does not affect the equidistribution, as $${p_{k+1}\over s_k}={p_{k+1}\over p_1+\ldots +p_k}\to 0$$ In particular there is $k_0$ such that $${p_{k+1}\over s_k}< {1\over 2}\quad k\ge k_0 $$ Observe that the term ${m\over p_{k+1}},$ where $1\le m\le p_{k+1},$ has the index equal $s_k+m.$ Next for $k\ge k_0$ we have $$|x-a_{s_k+m}|=\left |{1\over 2}-{m\over p_{k+1}}\right |={|p_{k+1}-2m|\over 2p_{k+1}}\\ \ge {1\over 2p_{k+1}}>{1\over s_k}>{1\over s_k+m}$$ Therefore $|1/2-a_n|<1/n$ for finitely many $n.$

The implication $\Leftarrow $ is false as well. We will construct the sequence $a_n$ violating the equidistribution property inductively. Assume that $N_j$ terms of $a_n$ are defined. Let the next $N_j^2-N_j+2^{N_j}$ terms be $${1\over N_j^2},{2\over N_j^2},{3\over N_j^2},\ldots, {N_j^2-N_j\over N_j^2},\underbrace{0,0,\ldots 0}_{2^{N_j}\ {\rm times}}$$ For any $0\le x\le {N_j^2-N_j\over N_j^2}$ there is $1\le k\le N_j^2-N_j$ such that $$0\le x-a_{N_j+k}=x-{k\over N_j^2}< {1\over N_j^2}\color{red}{\le }{1\over N_j+k}\quad (*)$$ Let $N_{j+1}$ be the quantity of the terms $a_n$ defined so far. This value can be determined explicitly $$N_{j+1}=N_j+(N_j^2-N_j)+2^{N_j}=N_j^2+2^{N_j}$$ As ${N_j^2-N_j\over N_j}\to 1$ the sequence $a_n$ satisfies the assumptions in view of $(*).$ However this sequence is not equidistributed as $${1\over N_{j+1}}\sum_{n=1}^{N_{j+1}}a_n= {1\over N_{j+1}}\sum_{n=1}^{N_j^2}a_n\le {1\over 2^{N_j}}\sum_{n=1}^{N_j^2}a_n\le {N_j^2\over 2^{N_j}}\to 0$$

Remark I am pretty sure that the construction of sequences can be made simpler.

Ryszard Szwarc
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