A remark on exceptional sets in Erdös-Rényi limit theorem
A remark on exceptional sets in Erdös-Rényi limit theorem
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DOI：
10.1007/s00605-016-0974-1
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发表时间：
2016-09
期刊：
影响因子：
--
通讯作者：
Yu Sun;Jian Xu
中科院分区：
文献类型：
--
作者：
Yu Sun;Jian Xu
Let (xi)i=1+∞\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(x_i)_{i=1}^{+\infty }$$\end{document} be the digits sequence in the unique terminating dyadic expansion of x∈[0,1)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$x\in [0,1)$$\end{document}. The run-length function ln(x)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$l_n(x)$$\end{document} is defined by ln(x):=maxj:xi+1=xi+2=⋯=xi+j=1for some0≤i≤n-j.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} l_n(x):=\max \left\{ j:x_{i+1}=x_{i+2}=\cdots =x_{i+j}=1\ \text {for some}\ 0\le i\le n-j\right\} . \end{aligned}$$\end{document}Erdös and Rényi proved that limn→+∞ln(x)log2n=1,a.e.x∈[0,1).\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \lim _{n\rightarrow +\infty }\frac{l_n(x)}{\log _2{n}}=1, \text {a.e.}\ x\in [0,1). \end{aligned}$$\end{document}In this note, we show that for each pair of numbers α,β∈[0,+∞]\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha ,\beta \in [0,+\infty ]$$\end{document} with α≤β\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha \le \beta $$\end{document}, the following exceptional set Eα,β=x∈[0,1):lim infn→+∞ln(x)log2n=α,lim supn→+∞ln(x)log2n=β\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} E_{\alpha ,\beta }=\left\{ x\in [0,1):\liminf _{n\rightarrow +\infty }\frac{l_n(x)}{\log _2{n}}=\alpha ,\ \limsup _{n\rightarrow +\infty }\frac{l_n(x)}{\log _2{n}}=\beta \right\} \end{aligned}$$\end{document}has Hausdorff dimension one.
Let (xi)i=1+∞\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(x_i)_{i=1}^{+\infty }$$\end{document} be the digits sequence in the unique terminating dyadic expansion of x∈[0,1)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$x\in [0,1)$$\end{document}. The run-length function ln(x)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$l_n(x)$$\end{document} is defined by ln(x):=maxj:xi+1=xi+2=⋯=xi+j=1for some0≤i≤n-j.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} l_n(x):=\max \left\{ j:x_{i+1}=x_{i+2}=\cdots =x_{i+j}=1\ \text {for some}\ 0\le i\le n-j\right\} . \end{aligned}$$\end{document}Erdös and Rényi proved that limn→+∞ln(x)log2n=1,a.e.x∈[0,1).\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \lim _{n\rightarrow +\infty }\frac{l_n(x)}{\log _2{n}}=1, \text {a.e.}\ x\in [0,1). \end{aligned}$$\end{document}In this note, we show that for each pair of numbers α,β∈[0,+∞]\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha ,\beta \in [0,+\infty ]$$\end{document} with α≤β\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha \le \beta $$\end{document}, the following exceptional set Eα,β=x∈[0,1):lim infn→+∞ln(x)log2n=α,lim supn→+∞ln(x)log2n=β\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} E_{\alpha ,\beta }=\left\{ x\in [0,1):\liminf _{n\rightarrow +\infty }\frac{l_n(x)}{\log _2{n}}=\alpha ,\ \limsup _{n\rightarrow +\infty }\frac{l_n(x)}{\log _2{n}}=\beta \right\} \end{aligned}$$\end{document}has Hausdorff dimension one.