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? 
 A: 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 $r+1$.
