 2 solved out of 4 shown
$500 If$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 $\limsup_{N\to \infty}\lvert \#\{ n\leq N : z_n\in I\} - N\lvert I\rvert \rvert =\infty?$