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Erdős #524 · analysis / probability

For any t∈(0,1)t\in (0,1) let t=∑k=1∞ϵk(t)2−kt=\sum_{k=1}^\infty \epsilon_k(t)2^{-k} (where ϵk(t)∈{0,1}\epsilon_k(t)\in \{0,1\}). What is the correct order of magnitude (for almost all t∈(0,1)t\in(0,1)) for

M_n(t)=\max_{x\in [-1,1]}\left\lvert \sum…
Source reports OPENAs of 2026-09-08Source & evidence ↓
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Problem statement

For any t∈(0,1)t\in (0,1) let t=∑k=1∞ϵk(t)2−kt=\sum_{k=1}^\infty \epsilon_k(t)2^{-k} (where ϵk(t)∈{0,1}\epsilon_k(t)\in \{0,1\}). What is the correct order of magnitude (for almost all t∈(0,1)t\in(0,1)) for

Mn(t)=max⁡x∈[−1,1]∣∑k≤n(−1)ϵk(t)xk∣?M_n(t)=\max_{x\in [-1,1]}\left\lvert \sum_{k\leq n}(-1)^{\epsilon_k(t)}x^k\right\rvert?

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Background

A problem of Salem and Zygmund [SaZy54]. Chung showed that, for almost all tt, there exist infinitely many nn such that

Mn(t)≪(nlog⁡log⁡n)1/2.M_n(t) \ll \left(\frac{n}{\log\log n}\right)^{1/2}.

Erdős (unpublished) showed that for almost all tt and every ϵ>0\epsilon>0 we have lim⁡n→∞Mn(t)/n1/2−ϵ=∞\lim_{n\to \infty}M_n(t)/n^{1/2-\epsilon}=\infty.

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The source labels this entry OPEN in the 2026-09-08 snapshot. Editorial progress notes are reproduced as source commentary, not independently verified results. Prize amounts and Lean badges at the source are not local awards or verification.

References
  1. [1]Thomas Bloom (ed.), Erdős Problems #524 link

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