# Rate of convergence of the average of an equidistributed sequence

Let $f : \mathbb R\to\mathbb C$ be an $1$-periodic and sufficiently smooth function, which has zero average, and let $\alpha$ irrational. We know the following:

a. $\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n f(k\alpha)=0$.

b. The sequence $s_n=\sum_{k=1}^n f(k\alpha)$ is not necessarily bounded for every such $\alpha$. In fact, we are (to the best of my knowledge) certain that $s_n$ is bounded for every such $\alpha$ only if $f$ a trigonometric polynomial.

c. Combining a. & b. we obtain that $s_n=o(n)$.

Is it the best estimate we can have for $s_n$?

This is a follow-up question to my earlier question:

Does equidistribution of zero average, due to irrationality, imply boundedness?

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You are more or less asking about the Denjoy-Koksma inequality –  Asaf Dec 5 '13 at 11:06
And of course the related Koksma–Hlawka inequality –  Asaf Dec 5 '13 at 11:11
Does any of these inequalities (Denjoy-Koksma & Koksma-Hlawka) improve the estimate for the sequence of the partial sums? Or, does it suggest this estimate is the optimal one? –  smyrlis Dec 5 '13 at 11:53
Could you perhaps be more specific about the order of quantifiers? That is, do you want an estimate which works for fixed $f$ and for all irrational $\alpha$, or an estimate which works for fixed irrational $\alpha$ and all $f$ within a given smoothness class, or...? –  Ian Morris Dec 5 '13 at 12:45
If the irrational is fixed, then $f$ can be restricted according to the measure of irrationality of the irrational. I need an estimate which holds for all irrationals, for $f$ real analytic. Apparently, we do not expect to have a uniform bound, for all irrationals. I am asking for a possibly optimal (sublinear) order of growth of this sequence, which holds for all irrationals (with different constants). –  smyrlis Dec 5 '13 at 22:22

By Koksma's inequality, $s_n$ is bounded by $N$ times the variation of $f$, multiplied with the so-called discrepancy $D_N$ of the sequence $(\alpha, 2\alpha, \dots, N \alpha)$. The discrepancy of course depends on $\alpha$. More precisely, there is a close relation between the discrepancy of $(n \alpha)$ and the continued fraction expansion of $\alpha$.

In particular, for almost all $\alpha$ the discrepancy is of order $N^{-1}(\log N (\log \log N)^{1+\varepsilon})$ (Khintchine). Thus $s_n$ is for almost all $\alpha$ of order $\log N (\log \log N)^{1+\varepsilon}$.

As to whether $s_n$ may even be bounded, check the following references:

Kesten, Harry: On a conjecture of Erdős and Szüsz related to uniform distribution mod 1. Acta Arith. 12 1966/1967 193–212.

Roçadas, Luís; Schoißengeier, Johannes On the local discrepancy of (nα)-sequences. J. Number Theory 131 (2011), no. 8, 1492–1497

and a recent manuscript of Alan Hanyes: http://arxiv.org/abs/1311.7277

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