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I suppose this is une cause perdue, but it would be nice if the following held.

Let $\theta$ be an irrational, and let $c_m = {{2m}\choose m}$, the central binomial. For a real number $r$, let $d(r)$ denote the distance from $r$ to the nearest integer (often indicated by $\| r \|$). Is it true that for every irrational $\theta$, the Cesaro sums behave reasonably well: $$ (*) \quad \limsup_{N\to \infty} \frac 1N \sum_{m=1}^{N} d(c_m \theta) > 0 \,? $$ Equivalently, does there exist $c >0$---depending on $\theta$---such that $$ \limsup_{N\to \infty} \frac{\left| \left\{m \leq N\left.\right| d(c_m \theta) > c\right\}\right|}N > 0\,? $$

For almost all $\theta$, $d(c_m\theta)$ is uniformly distributed, but not for all irrational $\theta$ (since the ratio of consecutive coefficients exceeds $1$). I don't know much more about it. It's still possible (although seemingly remote) that (*) holds for all irrational $\theta$. We can also replace the central binomial terms by the Catalan numbers and the questions are probably equivalent. (An affirmative answer to either would be fine.)

A weaker version would be whether (for all irrational $\theta$) $$ \limsup_{N\to \infty} \frac{\left| \left\{m \leq N\left.\right| d(c_m \theta) > c\right\}\right|}{N^{1/2 + \delta}} = \infty $$ for some $\delta > 0$ (depending on $\theta$); an affirmative answer would be useful.

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  • $\begingroup$ May help the question if you wrote it in a way that it is immediate at first look what is known and what is to be proved -- just my "even prime number" cents :-) $\endgroup$ – Suvrit Aug 10 '15 at 14:42
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I don't have an answer, but I can show that it's not true if you replace $\limsup$ by $\liminf$.

Let $F_N(\theta) = \dfrac{1}{N} \sum_{m=1}^N d(c_m \theta)$ for $\theta \in \mathbb R$. These are continuous functions of $\theta$.

If $\theta$ is rational $a/\prod_{j=1}^k p_j^{d_j}$, $d(c_m \theta) = 0$ when the $p_j$-adic orders $\nu_{p_j}(c_m) \ge d_j$ for all $j$, and using Kummer's theorem it's easy to prove that the fraction of $m \in \{1,2,\ldots,N\}$ for which this is true goes to $1$ as $N \to \infty$. Thus $F_N(\theta) \to 0$ as $N \to \infty$ for rational $\theta$.

As a consequence of this and the fact that $\{\theta: F_N(\theta) < t\}$ is open, $\{\theta: \liminf_{N \to \infty} F_N(\theta) = 0\}$ is a dense $G_\delta$. On the other hand, since the mean of $F_N(\theta)$ over any interval goes to $1/4$ as $N \to \infty$, $\{\theta: \limsup_{N \to \infty} F_N(\theta) \ge 1/4 \}$ is also a dense $G_\delta$. So a generic $\theta$ (in the Baire category sense) has $\liminf_{N \to \infty} F_N(\theta) = 0$ and $\limsup_{N \to \infty} F_N(\theta) \ge 1/4$.

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