A friend gave me this problem as a "Christmas gift":
If $f$ is periodic with irrational period $r$ and $\int_0^r f(x) \mathrm{d}x = 0$, prove or disprove that $\left\{\sum\limits_{i=1}^n f(i)\right\}_{n \in \mathbb{N}}$ is bounded.
One construction that works is $f(x) = \begin{cases} 1 &\mbox{if } x \equiv k \pmod{r}, k \in \mathbb{Z} \\ 0 & \mbox{otherwise } \end{cases}$ (I guess this only works if we use the Lebesgue integral); another is $f(x) = \tan(x-\frac{\pi}{2})$.
However, he accidentally forgot to specify that $f$ must be continuous, which invalidates both of these constructions. Since $\mathbb{N} \pmod{r}$ is uniformly distributed in $[0, r)$, I'm pretty sure that we have $$\lim_{n \to \infty}\frac{1}{n}\sum_{i=1}^{n}f(i) = \int_{0}^{r}f(x)\mathrm{d}x = 0.$$ Unfortunately, this is weaker than the original statement.
My question is, is the original statement true, and if not, what's a counterexample? I'm not sure if this problem can be solved with elementary knowledge, or if it would require some more theory.
No this is not true for continuous functions. Take $r = 2\pi$. Continuous functions with vanishing integral form Banach space. Define linear operators from this space: $$T_n (f) = f(1) + f(2) + \dots +f(n)$$ If the claim were true then (by Banach-Steinhaus, a.k.a. uniform boundedenss principle) $||T_n|| \lt C$ where $C$ is a constant independent from $n$. In particular, taking for $f_k(t) = \exp(ikt)$ we would get (for any $k \neq 0$) $$|\exp(ik \cdot 1) + \exp(ik \cdot 2) + \dots + \exp(ik \cdot n)| \lt C$$ But the sum on the left is: $$\frac{|\exp(ik\cdot n) - 1|}{|\exp(ik) - 1|}$$ and this should be bounded by a constant independent of $k$ and $n$. Here is where irrationality of $r$ comes into play (that is uniform distribution of $r\cdot k$ and $r\cdot n\cdot k$). First take $k$ so that denominator is small and then ($k$ is now fixed and $r\cdot k$ is irrational) take $n$ so that nominator is not small.