2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Do you have a reference for this? $\endgroup$ – Matt F. Oct 26 at 11:50
  • 1
    $\begingroup$ 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". $\endgroup$ – kaleidoscop Oct 26 at 12:00
  • 6
    $\begingroup$ Could you say more clearly and separately what is known, and what you're asking? $\endgroup$ – YCor Oct 26 at 12:11
  • 2
    $\begingroup$ 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. $\endgroup$ – Anthony Quas Oct 26 at 20:43
  • 1
    $\begingroup$ Previous questions on related topic by this user, for context: mathoverflow.net/questions/344202/… and mathoverflow.net/questions/344293/… $\endgroup$ – Gerry Myerson Oct 27 at 0:44
2
+100
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Concerning the final question, the series cannot be uniformly bounded from below, since for each $n$ you can find $u$ such that the sum if empty, i.e., has value 0. $\endgroup$ – Kurisuto Asutora Oct 30 at 9:12
  • $\begingroup$ @KurisutoAsutora ah, of course. Possibly OP had in mind something different. $\endgroup$ – Fedor Petrov Oct 30 at 9:17
  • $\begingroup$ Thanks! Your method actually helped me also with numbers with larger irrationality exponents $\endgroup$ – kaleidoscop Nov 1 at 13:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.