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If $f:\mathbb{N}\to \{-1,+1\}$ then is it true that for every $C>0$ there exist some $d,m\geq 1$ such that \[\left\lvert \sum_{1\leq k\leq m}f(kd)\right\rvert > C?\]
The 'Erdős discrepancy problem'. This is true, and was proved by Tao [Ta16], who also proved the more general case when $f$ takes values on the unit sphere.
Find the smallest $h(d)$ such that the following holds. There exists a function $f:\mathbb{N}\to\{-1,1\}$ such that, for every $d\geq 1$, \[\max_{P_d}\left\lvert \sum_{n\in P_d}f(n)\right\rvert\leq h(d),\] where $P_d$ ranges over all finite arithmetic progressions with common difference $d$.
Cantor, Erdős, Schreiber, and Straus [Er66] proved that $h(d)\ll d!$ is possible. Van der Waerden's theorem implies that $h(d)\to \infty$. Beck [Be17] has shown that $h(d) \leq d^{8+\epsilon}$ is possible for every $\epsilon>0$. Roth's famous discrepancy lower bound [Ro64] implies that $h(d)\gg d^{1/2}$.
Let $A_1,A_2,\ldots$ be an infinite collection of infinite sets of integers, say $A_i=\{a_{i1}<a_{i2}<\cdots\}$. Does there exist some $f:\mathbb{N}\to\{-1,1\}$ such that \[\max_{m, 1\leq i\leq d} \left\lvert \sum_{1\leq j\leq m} f(a_{ij})\right\rvert \ll_d 1\] for all $d\geq 1$?
Erdős remarks 'it seems certain that the answer is affirmative'. This was solved by Beck [Be81]. Recently Beck [Be17] proved that one can replace $\ll_d 1$ with $\ll d^{4+\epsilon}$ for any $\epsilon>0$.
Let $z_1,z_2,\ldots \in [0,1]$ be an infinite sequence. Must there exist some interval $I\subseteq [0,1]$ such that \[\limsup_{N\to \infty}\lvert \#\{ n\leq N : z_n\in I\} - N\lvert I\rvert \rvert =\infty?\]