次缺项序列的模1密度：Peres-Schlag论证的应用
Density modulo 1 of sublacunary sequences: application of Peres-Schlag's arguments
摘要：
设实数序列$\{t_n\}_{n=1}^{\infty}$满足条件$ \frac{t_{n+1}}{t_n} \ge 1+ \fracγ{n^β},0\le β<1, γ>0. $，则集合$ \{α\in [0,1]: \exists \varkappa > 0 \forall n \in \mathbb{N} ||t_n α|| > \frac{\varkappa}{n^β\log (n+1)} \} $是不可数的.而且它的豪斯多夫维数等于1。考虑$2^n3^m$形式的自然数集合，让序列$ s_1=1, s_2=2, s_3=3, s_4=4, s_5=6, s_6 = 8,... $作为递增序列执行此集合。则集合$ \{α\in [0,1]: \exists \varkappa > 0 \forall n \in \mathbb{N} ||s_n α|| > \frac{\varkappa}{\sqrt{n}\log (n+1)} \} $也具有等于1的豪斯多夫维数。所得结果采用了Y. Peres和W.施拉格
英文摘要：
Let the sequence $\{t_n\}_{n=1}^{\infty}$ of reals satisfy the condition $ \frac{t_{n+1}}{t_n} \ge 1+ \fracγ{n^β},0\le β<1, γ>0. $ Then the set $ \{α\in [0,1]: \exists \varkappa > 0 \forall n \in \mathbb{N} ||t_n α|| > \frac{\varkappa}{n^β\log (n+1)} \} $ is uncountable. Moreover its Hausdorff dimension is equal to 1. Consider the set of naturals of the form $2^n3^m$ and let the sequence $ s_1=1, s_2=2, s_3=3, s_4=4, s_5=6, s_6 = 8,... $ performs this set as an increasing sequence. Then the set $ \{α\in [0,1]: \exists \varkappa > 0 \forall n \in \mathbb{N} ||s_n α|| > \frac{\varkappa}{\sqrt{n}\log (n+1)} \} $ also has Hausdorff dimension equal to 1. The results obtained use an original approach due to Y. Peres and W. Schlag.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
