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Hausdorff dimension of Dirichlet non-improvable set versus well-approximable set

Published online by Cambridge University Press:  04 August 2022

BIXUAN LI*
Affiliation:
School of Mathematics and Statistics, Huazhong University of Science and Technology, Wuhan 430074, China (e-mail: bwei_wang@hust.edu.cn, xujian@hust.edu.cn)
BAOWEI WANG
Affiliation:
School of Mathematics and Statistics, Huazhong University of Science and Technology, Wuhan 430074, China (e-mail: bwei_wang@hust.edu.cn, xujian@hust.edu.cn)
JIAN XU
Affiliation:
School of Mathematics and Statistics, Huazhong University of Science and Technology, Wuhan 430074, China (e-mail: bwei_wang@hust.edu.cn, xujian@hust.edu.cn)

Abstract

Dirichlet’s theorem, including the uniform setting and asymptotic setting, is one of the most fundamental results in Diophantine approximation. The improvement of the asymptotic setting leads to the well-approximable set (in words of continued fractions)

$$ \begin{align*} \mathcal{K}(\Phi):=\{x:a_{n+1}(x)\ge\Phi(q_{n}(x))\ \textrm{for infinitely many }n\in \mathbb{N}\}; \end{align*} $$
the improvement of the uniform setting leads to the Dirichlet non-improvable set
$$ \begin{align*} \mathcal{G}(\Phi):=\{x:a_{n}(x)a_{n+1}(x)\ge\Phi(q_{n}(x))\ \textrm{for infinitely many }n\in \mathbb{N}\}. \end{align*} $$
Surprisingly, as a proper subset of Dirichlet non-improvable set, the well-approximable set has the same s-Hausdorff measure as the Dirichlet non-improvable set. Nevertheless, one can imagine that these two sets should be very different from each other. Therefore, this paper is aimed at a detailed analysis on how the growth speed of the product of two-termed partial quotients affects the Hausdorff dimension compared with that of single-termed partial quotients. More precisely, let $\Phi _{1},\Phi _{2}:[1,+\infty )\rightarrow \mathbb {R}^{+}$ be two non-decreasing positive functions. We focus on the Hausdorff dimension of the set $\mathcal {G}(\Phi _{1})\!\setminus\! \mathcal {K}(\Phi _{2})$ . It is known that the dimensions of $\mathcal {G}(\Phi )$ and $\mathcal {K}(\Phi )$ depend only on the growth exponent of $\Phi $ . However, rather different from the current knowledge, it will be seen in some cases that the dimension of $\mathcal {G}(\Phi _{1})\!\setminus\! \mathcal {K}(\Phi _{2})$ will change greatly even slightly modifying $\Phi _1$ by a constant.

Type
Original Article
Copyright
© The Author(s), 2022. Published by Cambridge University Press

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