Possible generalizations of $\sum_{k=1}^n \cos k$ being bounded

I claim that the sequence $s_n=\sum_{k=0}^{n-1} f(kq)$ is bounded, for a given $q$, if the following hypotheses are satisfied:

• $f$ is $p$-periodic, and its integral over an entire period vanishes,
• $q$ is an irrational multiple of $p$,
• the irrationality measure of $q/p$ is finite (say, $q/p$ has irrationality measure $\mu$),
• $f$ is $C^k$ for some $k>\mu$.

In particular, if $f$ is $C^3$, the sequence is bounded for almost all $q$, including all $q$ where $q/p$ is algebraic but irrational (by the Thue-Siegel-Roth theorem).

First of all, we may suppose without loss of generality that $f$ has period $2\pi$. As $\mu \geq 2$ for all irrational $q$, we need only consider those $f$ which are at least $C^3$. In particular, this means $f$ has an absolutely convergent Fourier series $f(x)=\sum_{-\infty}^\infty a_m e^{imx}$. Since $\int_0^{2\pi} f = 0$, $a_0=0$. Now, $$ s_n=\sum_{k=0}^{n-1} f(kq)= \sum_{k=0}^{n-1} \sum_{m=-\infty}^\infty a_m e^{imkq}=\sum_{m=-\infty}^\infty a_m \left(\sum_{k=0}^{n-1}e^{imkq}\right) $$ (where we've used absolute convergence to swap the order of summation).

As $a_0=0$ and $q/2\pi$ is irrational, the common ratio $e^{imq}$ in this last sum is never $1$, and so we can apply the general form of the geometric series formula to obtain $$ s_n=\sum_{m=-\infty}^\infty a_m \left(\frac{1-e^{imnq}}{1-e^{imq}}\right)=\sum_{m=-\infty}^\infty a_m e^{im(n-1)q/2} \frac{\sin \frac{mnq}{2}}{\sin \frac{mq}{2}} \, . $$

Now, let $\Phi_n(x)=\frac{\sin (nx/2)}{\sin (x/2)}$. It suffices to show that $\sum_{m=-\infty}^\infty |a_m \Phi_n(mq)|$ is bounded independently of $n$. Note that, for any $x$ with $0 < |x| < \pi$, we have $|\Phi_n(x)| \leq \left|\frac{1}{\sin (x/2)}\right| \leq \frac{\pi}{|x|}$.

Choose $\alpha$ with $\mu < \alpha < k$. Since $\alpha > \mu$, the inequality \begin{equation} \left| \frac{q}{2\pi} - \frac{\ell}{m}\right| > \frac{1}{|m|^\alpha} \end{equation} holds for all but finitely many choices of $(\ell, m) \in \Bbb{Z}^2$.

We will now divide and conquer: first we will show that the sum of all the terms in which the inequality holds for $m$ is bounded independent of $n$; then we will show that the sum of all the terms in which it does not hold is also so bounded.

Suppose the inequality holds for some $m$; choose $\ell \in \Bbb{Z}$ so that $|mq-2\pi\ell|<\pi$. Then by periodicity we have $ \Phi_n(mq)=\Phi_n(mq-2\pi\ell) $; by the bound we just derived on $\Phi_n$ it follows that $$ \left|\Phi_n(mq)\right|<\frac{\pi}{|mq-2\pi \ell|} = \frac{1}{2|m||q/2\pi - \ell/m|} < \frac{1}{2}|m|^{\alpha-1} $$ by our irrationality measure inequality. Since $f$ is $C^k$, its Fourier coefficients $a_m$ are all bounded by $\frac{2C}{|m|^k}$ for some uniform constant $C$. So we can bound the sum over all the $m$ such that the irrationality measure inequality holds by $$ \sum_{m=-\infty}^\infty \frac{C}{|m|^k} |m|^{\alpha-1}=\sum_{m=-\infty}^\infty C |m|^{\alpha-k-1} \, ; $$ since $\alpha < k$, this is a convergent sum (and is independent of $n$, since $C$ depends only on $f$).

On the other hand, there are only finitely many $m$ such that the inequality does not hold. Let $\{m_1,\dots\,m_j\}$ be those $m$; then it follows immediately from our inequality for $\Phi$ that the sum of these terms is bounded by $\sum_j a_{m_j} \frac{\pi}{|m_j q - 2\pi \ell_j|}$, where the $\ell_j$ are chosen so that $|m_j q - 2\pi \ell_j|\leq \pi$. Since this is a finite sum independent of $n$, we are done.