Extension of 1991 IMO question

Question

Problem 6 in the 1991 IMO asks for a bounded real sequence $x_n$ such that for $i\ne j$ we have

$|x_i-x_j|\cdot |i-j|^{1+\epsilon} \geq r$ for some $r>0$

The solution is to make $x_n$ the fractional part of $\sqrt{2}\cdot n$ which in fact allows $\epsilon=0$.

But what if we replaced $1+\epsilon$ with $1-\epsilon$ ? Does such a sequence exist? And has its existence something to do with Diophantine approximations?

Answer 1

Given $r >0$, $\varepsilon > 0$, and $C>0$, if $x_1,...x_n$ lie in $[-C,C]$, then for $n$ sufficiently large, for some distinct $i, j$, we have $|x_j-x_i| < \frac{r}{|j-i|^{1-\varepsilon}}$.

Indeed, assume that $x_1,...,x_n$ lie in $[-C,C]$. Then you can slice $[-C,C]$ in $n-1$ intervals of length at most $\frac{2C}{n-1}$, and apply pigeonhole principle. Thus, for some distinct $i,j$, $|x_j-x_i| \le \frac{2C}{n-1}$.

Now $|x_j-x_i| \cdot |j-i|^{1-\varepsilon} \le |x_j-x_i| \cdot n^{1-\varepsilon} \le \frac{2Cn}{n-1} \frac{1}{n^\varepsilon} \le \frac{4C}{n^{\varepsilon}}$, and we also have $\frac{4C}{n^{\varepsilon}} \underset{n \to +\infty}{\longrightarrow} 0$.

If $n$ is great enough, then for some distinct $i,j$, we have $|x_j-x_i| \cdot |j-i|^{1-\varepsilon} < r$.

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Hence, if we have $(x_n)$ a real sequence, $\varepsilon >0$ and $r > 0$ such that for all $i \neq j$, $|x_j-x_i| \ge \frac{r}{|j-i|^{1-\varepsilon}}$, then $(x_n)$ is unbounded.

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In this thread (Bounded sequence in the plane...) on Mathoverflow, they consider the same problem in the plane.