Let $0 < a < 1$ be an irrational number. Is it true that


$$\liminf\_{n \in \mathbb N, n \to \infty} n \{na\} = 0?$$


*Note: Here $\{\cdot\}$ denotes the fractional part.*



The liminf of an expression involving an irrational rotation
The answer is no, because for some real $c>0$ and all integers $q>0$ we have
$$q\{q\sqrt2\}=q(q\sqrt2-\lfloor q\sqrt2\rfloor)
=q|q\sqrt2-\lfloor q\sqrt2\rfloor| \\
\ge q\,\inf\_{p\in\mathbb Z}|q\sqrt 2-p|
=q^2\,\inf\_{p\in\mathbb Z}|\sqrt 2-p/q|\ge c.$$
So, $\liminf\limits\_{q\to\infty} q\{q\sqrt2\}\ge c>0$.


More specifically, letting $p:=p\_q:=\lfloor q\sqrt2\rfloor$ and $q\to\infty$, we have $p<q\sqrt2$, $p\sim q\sqrt2$, and hence
$$q\{q\sqrt2\}=q(q\sqrt2-p)
=q\frac{2q^2-p^2}{q\sqrt2+p} 
\ge q\frac1{q\sqrt2+p}\sim\frac1{2\sqrt2}.$$
So,
$$\liminf\_{q\to\infty} q\{q\sqrt2\}\ge\frac1{2\sqrt2}>0.$$


