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.



Is there a number for which we know precisely the approximation by rationals?
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.


