The Central Limit Theorem for Subsequences in Probabilistic Number Theory

http://dx.doi.org/10.4153/CJM-2011-074-1
Canad. J. Math. 64(2012), 1201-1221

  Published:2011-12-23
 Printed: Dec 2012

Let $(n_k)_{k \geq 1}$ be an increasing sequence of positive integers, and let $f(x)$ be a real function satisfying \begin{equation} \tag{1} f(x+1)=f(x), \qquad \int_0^1 f(x) ~dx=0,\qquad \operatorname{Var_{[0,1]}} f \lt \infty. \end{equation} If $\lim_{k \to \infty} \frac{n_{k+1}}{n_k} = \infty$ the distribution of \begin{equation} \tag{2} \frac{\sum_{k=1}^N f(n_k x)}{\sqrt{N}} \end{equation} converges to a Gaussian distribution. In the case $$ 1 \lt \liminf_{k \to \infty} \frac{n_{k+1}}{n_k}, \qquad \limsup_{k \to \infty} \frac{n_{k+1}}{n_k} \lt \infty $$ there is a complex interplay between the analytic properties of the function $f$, the number-theoretic properties of $(n_k)_{k \geq 1}$, and the limit distribution of (2). In this paper we prove that any sequence $(n_k)_{k \geq 1}$ satisfying $\limsup_{k \to \infty} \frac{n_{k+1}}{n_k} = 1$ contains a nontrivial subsequence $(m_k)_{k \geq 1}$ such that for any function satisfying (1) the distribution of $$ \frac{\sum_{k=1}^N f(m_k x)}{\sqrt{N}} $$ converges to a Gaussian distribution. This result is best possible: for any $\varepsilon\gt 0$ there exists a sequence $(n_k)_{k \geq 1}$ satisfying $\limsup_{k \to \infty} \frac{n_{k+1}}{n_k} \leq 1 + \varepsilon$ such that for every nontrivial subsequence $(m_k)_{k \geq 1}$ of $(n_k)_{k \geq 1}$ the distribution of (2) does not converge to a Gaussian distribution for some $f$. Our result can be viewed as a Ramsey type result: a sufficiently dense increasing integer sequence contains a subsequence having a certain requested number-theoretic property.