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

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?

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 $$r+1$$.

• 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? 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. 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. Dec 5 '13 at 15:59