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Let $f:\mathbb R\to\mathbb C$ be a sufficiently smooth and $1$-periodic function of average zero (i.e., $\int_0^{1}f(x)\,dx=0$), and let $\alpha\in(0,1)\smallsetminus\mathbb Q$. We know that

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

Let $s_n=\sum_{k=1}^n f(k\alpha)$. Is it true that the sequence $\{s_n\}_{n\in\mathbb N}$ is bounded?

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1 Answer 1

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I believe the answer depends on the irrationality measure of $\alpha$.

Per Fourier, Write $f= \sum_{m=-\infty}^\infty a_m e^{ 2 \pi i m x}$, then

$$s_n = \sum_{m=-\infty}^\infty a_m \frac{ e^{2\pi i m (n+1) \alpha} - 1}{e^{2 \pi i m \alpha} - 1}$$

so we have the bound:

$$ |s_n| < 2\sum_{m=-\infty}^\infty \frac{ |a_m|} { | e^{ 2 \pi i m \alpha} - 1 |} $$

If for some number $\mu$, the equation:

$$ \left| \alpha- \frac{p}{q} \right| < \frac{1}{q^\mu} $$

has finitely many solutions, then we have the bound $1/ | e^{ 2 \pi i m \alpha} - 1 | = O( m^{\mu-1} ) $, by setting $q=m$ and $p=$ the closest integer approximation to $m\alpha$. If $f$ is $\lfloor\mu\rfloor+1$ times continuously differentiable, we have the bound $|a_m|=o(m^{\lfloor\mu\rfloor+1})$. Multiplying these, we see that the terms of our sum are $o(1/m^{2+\lfloor \mu\rfloor - \mu})= o(1/m^{1+\epsilon})$, so the sum converges. Working more carefully, one could probably improve this to $\lfloor \mu \rfloor$ times differentiable using the fact that solutions $p,q$ are very rare.

However, for Liouville numbers, this bound does not hold for any $\mu$. Specifically, if $\alpha$ is the Liouville constant then for $m=10^{a!}$, $1/|e^{2\pi i m \alpha} - 1|$ is about $10^{(a+1)!}=m^{a+1}$, so even for smooth $f$ this bound is useless. In fact, for $\alpha$ the Liouville constant, $s_n$ cannot be bounded at all:

Take $f$ to be the smooth function

$$f(x) = \sum_{t=1}^\infty \frac{e^{ 2 \pi i 10^{t!}x} }{10^{ (t+1) !}}$$

Then by choosing $n = \sum_{b=1}^c \frac{ 10^{(b+1)! - b!}}{2} $, the first $c$ turns will all be around $(1 - e^{ 2\pi i/2})/i=-2i$ and the remaining terms will be vanishingly small, so the sum will go to $\infty$ as $c$ goes to $\infty$.


Suppose that for every $f$ that has $r$ derivatives with the last derivative $L^1$, $s_n(f)$ is bounded independent of $n$. Then the irrationality measure of $\alpha$ is at most $r+1$.

Let $X$ be the Banach space of functions that have $r$ derivatives with the last derivative $L^1$. Consider the sequence of linear functions from $X$ to $\mathbb R$ whose $n$th element sends $f \to s_n(f)$. If this sequence is bounded when applied to any element of $X$, then by the uniform boundedness principle it is uniformly bounded, hence there is an absolute constant $C$ such that $s_n( e^{2\pi i m x} ) < C m^r$ for all $n,m$. By ordinary equidstribution, the supremum over all $n$ of $s_n(e^{ 2 \pi i q x})$ is $2/(1 - e^{ 2 pi i q\alpha} = 1/ O (|q\alpha-p|)$ for intgers $p$, which gives us the inequality

$$|q\alpha-p| > C' q^{-r}$$

$$|\alpha-\frac{p}{q} | > C' q^{-r-1}$$

So the irrationality measure of $\alpha$ is at most $r+1$.

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Very good counterexample. I am wondering whether the following is true: Is it possible for every smooth 1-periodic and 0-average function (which is not a trigonometric polynomial) to find an irrational, such that the sequence in my question is not bounded? –  smyrlis Dec 5 '13 at 12:14
    
I believe so, by restricting to successive intervals. Suppose we have already chosen $\alpha$ to lie in some very small interval such that for some number of $m$, $e^{2 \pi i m \alpha} - 1$ is quite small, and for some $n$, $s_n$ is very large. Let $w$ be the width of the interval. Then for all $m> 1/w $, we can choose an appropriate $\alpha$ such that $e^{ 2\pi m \alpha}$ is any point on the unit circle we want. So fix some $m>1/w$ such that $a_m\neq 0$, and choose an $\alpha$ such that $e^{2 \pi m \alpha}$ is incredibly close to $1$, making some appropriate $s_n$ arbitrarily large. –  Will Sawin Dec 5 '13 at 15:31
    
Actually choose an interval of appropriate $\alpha$s. Then we can repeat the process arbitrarily often, making a bigger $s_n$ each time. The only difficulty is making sure that later $s_n$s don't mess up earlier $s_n$s. But we control the contribution from $m$ that come just after our $m$, because $e^{2 \pi i m\alpha]-1$ will be large for there, and we can control the contribution from far-away terms because the trigonometric sum will be at most $n$ and $a_m$ will be very small. So it should be possible to do this, but the calculation will probably be messy. –  Will Sawin Dec 5 '13 at 15:59

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