Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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
1  
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
1  
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
add comment

1 Answer

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.