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Erdős #995 · analysis / discrepancy

Let n1<n2<⋯n_1<n_2<\cdots be a lacunary sequence of integers and f∈L2([0,1])f\in L^2([0,1]). Estimate the growth of, for almost all α\alpha,

∑1≤k≤Nf({αnk}).\sum_{1\leq k\leq N}f(\{ \alpha n_k\}).

For example, is it true that, for almost all…

Source reports OPENAs of 2026-09-08Source & evidence ↓
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Problem statement

Let n1<n2<⋯n_1<n_2<\cdots be a lacunary sequence of integers and f∈L2([0,1])f\in L^2([0,1]). Estimate the growth of, for almost all α\alpha,

∑1≤k≤Nf({αnk}).\sum_{1\leq k\leq N}f(\{ \alpha n_k\}).

For example, is it true that, for almost all α\alpha,

∑1≤k≤Nf({αnk})=o(Nlog⁡log⁡N)?\sum_{1\leq k\leq N}f(\{ \alpha n_k\})=o(N\sqrt{\log\log N})?

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Background

Erdős [Er49d] constructed a lacunary sequence and f∈L2([0,1])f\in L^2([0,1]) such that, for every ϵ>0\epsilon>0, for almost all α\alpha

lim sup⁡N→∞1N(log⁡log⁡N)12−ϵ∑1≤k≤Nf({αnk})=∞.\limsup_{N\to \infty}\frac{1}{N(\log\log N)^{\frac{1}{2}-\epsilon}}\sum_{1\leq k\leq N}f(\{\alpha n_k\})=\infty.

Erdős also proved that, for every lacunary sequence and f∈L2f\in L^2, for every ϵ>0\epsilon>0, for almost all α\alpha,

∑1≤k≤N∑1≤k≤Nf({αnk})=o(N(log⁡N)12+ϵ).\sum_{1\leq k\leq N}\sum_{1\leq k\leq N}f(\{\alpha n_k\})=o( N(\log N)^{\frac{1}{2}+\epsilon}).

Erdős [Er64b] thought that his lower bound was closer to the truth.

Progress & source assessment

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 #995 link

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