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The question is about the existence of a number $x$ for which we know the existence of $c>0$ such that for all $u>0,n\in\mathbb{N}^*$ that $$ \frac {1}{nu}\sum_{j=1}^{n}1_{d(jx,\mathbb Z)<u}<c $$ This holds for each $u>0$ for $x$ irrational (i.e. with $c$ depending on $u$, see references on "well distributed numbers"), but not uniformy in $u$.

Ideally it would be great to show that the series above is uniformly bounded from below by some $a>0$.

EDIT: the original statement was probably false, as noticed Anthony Quas, so I weakened it.

kaleidoscop
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  • Do you have a reference for this? –  Oct 26 '19 at 11:50
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    For the last sentence? See for instance the intro of "On the measure of well-distributed sequences" by Alan Zame, but it seems to come from a result of Koskma. It basically means that if $x$ is irrational, $(nx)$ is "well-distributed". – kaleidoscop Oct 26 '19 at 12:00
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    Could you say more clearly and separately what is known, and what you're asking? – YCor Oct 26 '19 at 12:11
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    It cannot hold uniformly in u. Rather for each irrational x, for all sufficiently large n, an inequality of this type holds. To see this, note that the left side is either 0 or greater than $1/(nu)$ so that for any a, b and n, fox sufficiently small u, this is impossible. – Anthony Quas Oct 26 '19 at 20:43
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    Previous questions on related topic by this user, for context: https://mathoverflow.net/questions/344202/well-distributed-sequence-uniformly-over-small-intervals and https://mathoverflow.net/questions/344293/very-badly-approximable-numbers – Gerry Myerson Oct 27 '19 at 00:44
  • Hi, thanks for your comments! So i hope the new statement is more clear. Following the remark of Anthony, it was probably false, so I weakened the statement. I hope someone has an idea! – kaleidoscop Oct 28 '19 at 14:58

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Yes, quadratic irrationalities satisfy such inequality. Denote $\|t\|=d(t,\mathbb{Z})$. We know that if $x$ is a fixed quadratic irrationality, say $x=\sqrt{2}$, then $\|mx\|\geqslant C/m$ for fixed $C$ depending only on $x$ and any positive integer $m$. Thus if $\|j_1x\|<u$ and $\|j_2x\|<u$ for two non-negative integers $j_1<j_2$, we get $2u>\|(j_2-j_1)x\|\geqslant C/(j_2-j_1)$, thus $j_2-j_1>C/(2u)$. Therefore if $0=j_0<j_1<j_2<\ldots<j_k\leqslant n$ are all numbers from 0 to $n$ for which $\|j_ix\|<u$, we get $n\geqslant \sum_{i=1}^k (j_i-j_{i-1})\geqslant Ck/(2u)$ and $k/(nu)\leqslant 2/C$, as desired.

Fedor Petrov
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