Metrical properties of weighted products of consecutive Lüroth digits

Adam Brown-Sarre Adam Brown-Sarre, Department of Mathematical and Physical Sciences, La Trobe University, Bendigo 3552, Australia. 20356213@students.latrobe.edu.au , Gerardo González Robert Gerardo González Robert, Department of Mathematical and Physical Sciences, La Trobe University, Bendigo 3552, Australia. G.Robert@latrobe.edu.au and Mumtaz Hussain Mumtaz Hussain, Department of Mathematical and Physical Sciences, La Trobe University, Bendigo 3552, Australia. m.hussain@latrobe.edu.au

Abstract.

The Lüroth expansion of a real number $x\in(0,1]$ is the series

x=\frac{1}{d_{1}}+\frac{1}{d_{1}(d_{1}-1)d_{2}}+\frac{1}{d_{1}(d_{1}-1)d_{2}(d_{2}-1)d_{3}}+\cdots,

with $d_{j}\in\mathbb{N}_{\geq 2}$ for all $j\in\mathbb{N}$ . Given $m\in\mathbb{N}$ , $\mathbf{t}=(t_{0},\ldots,t_{m-1})\in\mathbb{R}_{>0}^{m-1}$ and any function $\Psi:\mathbb{N}\to(1,\infty)$ , define

\mathcal{E}_{\mathbf{t}}(\Psi)\colon=\left\{x\in(0,1]:d_{n}^{t_{0}}\cdots d_{n+m}^{t_{m-1}}\geq\Psi(n)\text{ for infinitely many}\ n\in\mathbb{N}\right\}.

We establish a Lebesgue measure dichotomy statement (a zero-one law) for $\mathcal{E}_{\mathbf{t}}(\Psi)$ under a natural non-removable condition $\liminf_{n\to\infty}\Psi(n)>\leavevmode\nobreak\ 1$ . Let $B$ be given by

\log B\colon=\liminf_{n\to\infty}\frac{\log(\Psi(n))}{n}.

For any $m\in\mathbb{N}$ , we compute the Hausdorff dimension of $\mathcal{E}_{\mathbf{t}}(\Psi)$ when either $B=1$ or $B=\infty$ . We also compute the Hausdorff dimension of $\mathcal{E}_{\mathbf{t}}(\Psi)$ when $1<B<\infty$ for $m=2$ .

1. Introduction

J. Lüroth showed in 1883 that every $x\in(0,1]$ can be uniquely expressed as a series of the form

x=\frac{1}{d_{1}(x)}+\frac{1}{d_{1}(x)(d_{1}(x)-1)d_{2}(x)}+\frac{1}{d_{1}(x)(d_{1}(x)-1)d_{2}(x)(d_{2}(x)-1)d_{3}(x)}+\cdots,

where each $d_{j}(x)$ is an integer at least $2$ . Analogous to regular continued fractions with the Gauss map, Lüroth series can be interpreted through dynamical means using the Lüroth map $\mathscr{L}$ (see Section 2 for its definition).

The metrical theory of Lüroth expansions has been thoroughly studied. In [9], H. Jager and C. de Vroedt showed that the Lebesgue measure $\lambda$ on $(0,1]$ is $\mathscr{L}$ -ergodic. They also noted that the digits in the Lüroth expansion are independent when regarded as random variables. Hence, the Lüroth series analogue of the Borel-Bernstein Theorem (for regular continued fractions) [11, Theorem 30] is an immediate consequence of the classical Borel-Cantelli Lemma (see, for example, [10, Theorem 4.18]) and the following observation:

\lambda\left(\left\{x\in(0,1]:d_{n}(x)\geq m\right\}\right)=\frac{1}{m}\quad\text{ for all }n\in\mathbb{N}\text{ and all }m\in\mathbb{N}_{\geq 2}.

Throughout this paper, $\Psi:\mathbb{N}\to\mathbb{R}_{>0}$ denotes a positive function. The Borel-Bernstein Theorem for Lüroth series provides the Lebesgue measure $\lambda$ of the set

\mathcal{E}(\Psi)\colon=\left\{x\in(0,1]:d_{n}(x)\geq\Psi(n)\ \text{\rm for infinitely many }n\in\mathbb{N}\right\}.

Theorem 1.1 ([9, Theorem 2.1]).

The Lebesgue measure of $\mathcal{E}(\Psi)$ is given by

\lambda\left(\mathcal{E}(\Psi)\right)=\begin{cases}0,&\text{\rm if }\quad\displaystyle\sum_{n=1}^{\infty}\Psi(n)^{-1}<\infty,\\[8.61108pt] 1,&\text{\rm if }\quad\displaystyle\sum_{n=1}^{\infty}\Psi(n)^{-1}=\infty.\end{cases}

Theorem 1.1 appeared first in [9], although $\Psi$ was unnecessarily assumed to be increasing.

It is well-known that to Hausdorff dimension is an appropriate notion to distinguish between the null sets (Lebesgue measure zero sets). L. Shen [15] calculated the Hausdorff dimension of the set $\mathcal{E}(\Psi)$ which is the Lüroth analogue of a celebrated theorem of B. Wang and J. Wu [19]. To analyse the Hausdorff dimension of $\mathcal{E}(\Psi)$ and related sets, define

(1)

\log B=\liminf_{n\to\infty}\frac{\log\Psi(n)}{n}\;\text{ and }\;\log b=\liminf_{n\to\infty}\frac{\log\log\Psi(n)}{n}.

Theorem 1.2 ([15, Theorem 4.2] ).

The Hausdorff dimension of $\mathcal{E}(\Psi)$ is as follows:

1.

If $B=1$ , then $\dim_{\operatorname{H}}\mathcal{E}(\Psi)=1$ .
2.

If $1<B<\infty$ , then $\dim_{\operatorname{H}}\mathcal{E}(\Psi)=s(B)$ , where $s=s(B)$ is the solution of

$\sum_{k=2}^{\infty}\left(\frac{1}{Bk(k-1)}\right)^{s}=1.$
3.
When $B=\infty$ , we have three cases:
1. (i)
  
  If $b=1$ , then $\dim_{\operatorname{H}}\mathcal{E}(\Psi)=\frac{1}{2}$ .
2. (ii)
  
  If $1<b<\infty$ , then $\dim_{\operatorname{H}}\mathcal{E}(\Psi)=\frac{1}{1+b}$ .
3. (iii)
  
  If $b=\infty$ , then $\dim_{\operatorname{H}}\mathcal{E}(\Psi)=0$ .

In [18], B. Tan and Q. Zhou extended Shen’s result by considering the product of two consecutive partial quotients. Define the set

\mathcal{E}_{(1,1)}(\Psi)\colon=\left\{x\in(0,1]:d_{n}(x)d_{n+1}(x)\geq\Psi(n)\ \text{\rm for infinitely many }n\in\mathbb{N}\right\}.

Theorem 1.3 ([18, Lemma 3.1]).

Let $t=t(B)$ be the unique solution of the equation

\sum_{d=2}^{\infty}\frac{1}{B^{2s}d^{s}(d-1)^{s}}=1.

If $B$ and $b$ are given by (1), then

\dim_{\operatorname{H}}\mathcal{E}_{(1,1)}(\phi)=\begin{cases}1,&\text{\rm if }\quad B=1,\\ t(B),&\text{\rm if }\quad 1<B<\infty,\\ \frac{1}{1+b},&\text{\rm if }\quad B=\infty.\end{cases}

In this paper, we consider the weighted product of consecutive partial quotients and establish the Lebesgue measure and Hausdorff dimension for the corresponding limsup set. Given $m\in\mathbb{N}$ and $\mathbf{t}=(t_{0},\ldots,t_{m-1})\in\mathbb{R}_{>0}^{m}$ , define the set

\mathcal{E}_{\mathbf{t}}(\Psi)\colon=\left\{x\in(0,1]:\prod_{i=0}^{m-1}d_{n+i}^{t_{i}}(x)\geq\Psi(n)\text{ for infinitely many }n\in\mathbb{N}\right\},

and the numbers

t\colon=\min\{t_{0},t_{1},\ldots,t_{m-1}\},\quad T\colon=\max\{t_{0},t_{1},\ldots,t_{m-1}\},\quad\ell(\mathbf{t})\colon=\#\{j\in\{0,\ldots,m\}:t_{j}=T\}.

Theorem 1.4.

Let $m\in\mathbb{N}$ and $\mathbf{t}\in\mathbb{R}_{>0}^{m}$ be arbitrary. If

(2)

\liminf_{n\to\infty}\Psi(n)>1,

then, for $B$ and $b$ given by (1), we have

(3)

\lambda\left(\mathcal{E}_{\mathbf{t}}(\Psi)\right)=\begin{cases}0,&\text{\rm if }\quad\displaystyle\sum_{n=1}^{\infty}\frac{\left(\log\Psi(n)\right)^{\ell(\mathbf{t})-1}}{\Psi(n)^{\frac{1}{T}}}<\infty,\\[8.61108pt] 1,&\text{\rm if }\quad\displaystyle\sum_{n=1}^{\infty}\frac{\left(\log\Psi(n)\right)^{\ell(\mathbf{t})-1}}{\Psi(n)^{\frac{1}{T}}}=\infty.\end{cases}

Assumption (2) essentially says that there exists some $n_{0}\in\mathbb{R}_{>1}$ such that $\Psi(n)\geq n_{0}$ for all $n\in\mathbb{N}$ . Theorem 1.4 might fail without this condition (see Section 3 for the justification).

Theorem 1.5.

Take any $m\in\mathbb{N}$ and $\mathbf{t}\in\mathbb{R}_{>0}^{m}$ and let $B$ and $b$ be given by (1). Then

\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(\Psi)=\begin{cases}1,&\text{\rm if }\quad B=1,\\ \frac{1}{b+1},&\text{\rm if }\quad B=\infty.\end{cases}

If $m=2$ , we can compute the $\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(\Psi)$ when $1<B<\infty$ .

Theorem 1.6.

Let $B$ and $b$ be given by (1) and assume that $1<B<\infty$ . For a given $\mathbf{t}=(t_{0},t_{1})\in\mathbb{R}_{>0}^{2}$ , define

f_{t_{0},t_{1}}(s)\colon=\frac{s^{2}}{t_{0}t_{1}\max\left\{\frac{s}{t_{1}}+\frac{1-s}{t_{0}},\frac{s}{t_{0}}\right\}}.

The Hausdorff dimension of $\mathcal{E}_{\mathbf{t}}(\Psi)$ is the unique solution of

\sum_{d=2}^{\infty}\frac{1}{d^{s}(d-1)^{s}B^{f_{t_{0},t_{1}}(s)}}=1.

For completeness, it is worth mentioning that there have been abundance of work regarding the metrical properties of product of consecutive partial quotients. The results of the paper provide full Lüroth analogues of very recent work of Bakhtawar-Hussain-Kleinbock-Wang [2]. Note that paper [2] was a generalisation of previous works [3, 7, 8, 12, 19].

The organization of the paper is as follows. In Section 2, we recall some basic facts on Lüroth series. In Section 3, we prove Theorem 1.4. Section 4 is dedicated to the proof of Theorem 1.5. In Section 5, we prove Theorem 1.6. Lastly, in Section 6, we give some final remarks and a conjecture.

Notation.

We adopt the Vinogradov symbol $\ll$ for asymptotic behavior. If $(x_{n})_{n\geq 1}$ and $(y_{n})_{n\geq 1}$ are two sequences of positive real numbers, we write $x_{n}\ll y_{n}$ if there exists a constant $C>0$ such that $x_{n}\leq Cy_{n}$ holds for all $n\in\mathbb{N}$ . When the constant $C$ depends on some parameter $m$ , we write $x_{n}\ll_{m}y_{n}$ . If we have $x_{n}\ll y_{n}$ and $y_{n}\ll x_{n}$ , we write $x_{n}\asymp y_{n}$ . If the implied constants depend of some parameter $m$ , we write $x_{n}\asymp_{m}y_{n}$ . We write $\mathscr{D}\colon=\mathbb{N}_{\geq 2}$ . If $\mathbf{a}=(a_{1},\ldots,a_{n})\in\mathscr{D}^{n}$ and $\mathbf{b}=(b_{1},\ldots,b_{m})\in\mathscr{D}^{m}$ , then $\mathbf{a}\mathbf{b}\in\mathscr{D}^{n+m}$ is $\mathbf{a}\mathbf{b}\colon=(a_{1},\ldots,a_{n},b_{1},\ldots,b_{m})$ . We denote by $\lambda$ the Lebesgue measure on $\mathbb{R}$ .

Acknowledgements The research of Mumtaz Hussain and Gerardo González Robert is supported by the Australian Research Council Discovery Project (200100994).

2. Elements of Lüroth series

Let $d_{1}:(0,1]\to\mathscr{D}\colon=\mathbb{N}_{\geq 2}$ be the function associating to each $x\in(0,1]$ the natural number $d_{1}(x)\geq 2$ determined by

\frac{1}{d_{1}(x)}<x\leq\frac{1}{d_{1}(x)-1}.

That is, if $\lfloor\,\cdot\,\rfloor$ represents the floor function, we define $d_{1}(x)\colon=\lfloor\frac{1}{x}\rfloor+1$ . The Lüroth map is the function $\mathscr{L}(x):[0,1]\to[0,1]$ given by

\mathscr{L}(x)=\begin{cases}d_{1}(x)(d_{1}(x)-1)x-(d_{1}(x)-1),\text{ if }x\in(0,1],\\ 0,\text{ if }x=0.\end{cases}

For any $x\in(0,1]$ and $n\geq 2$ , we define $d_{n}(x)\colon=d_{1}(\mathscr{L}^{n-1}(x))$ , the exponent denotes iteration. For any $\mathbf{d}:=(d_{1},\ldots,d_{n})\in\mathscr{D}^{n}$ , the cylinder of level $n$ based at $\mathbf{d}$ is the set

I_{n}(\mathbf{d})\colon=\left\{x\in(0,1]:d_{1}(x)=d_{1},\ldots,d_{n}(x)=d_{n}\right\}.

The cylinders are intervals of the form^a^aaSome authors, for example [4], define the Lüroth map in a slightly different manner. For any $x\in(0,1)$ they choose $n\in\mathbb{N}$ such that $n^{-1}\leq x<(n-1)^{-1}$ , they map $x$ to $n(n-1)x-(n-1)$ , and they leave $0$ as a fixed point. Under this definition, cylinders are intervals of the for $[\alpha,\beta)$ . Our definition follows [6, 9] among others. The difference between these two approaches is irrelevant for our purposes, because it only affects the countable set $\mathbb{Q}\cap[0,1]$ . $(\alpha,\beta]$ for $0<\alpha<\beta\leq 1$ .

Lüroth series induce a continuous map $\Lambda:\mathscr{D}^{\mathbb{N}}\to(0,1]$ given by

	$\displaystyle\Lambda(d_{1},d_{2},d_{3},\ldots)$	$\displaystyle\colon=\langle d_{1},d_{2},d_{3},\ldots\rangle$
		$\displaystyle\colon=\frac{1}{d_{1}}+\frac{1}{d_{1}(d_{1}-1)d_{2}}+\frac{1}{d_{1}(d_{1}-1)d_{2}(d_{2}-1)d_{3}}+\cdots.$

Denote by $\sigma$ the left shift on $\mathscr{D}^{\mathbb{N}}$ . Then, the dynamical system $(\mathscr{D}^{\mathbb{N}},\sigma)$ is an extension of $((0,1],\mathscr{L})$ in the sense that $\Lambda:\mathscr{D}^{\mathbb{N}}\to(0,1]$ is a continuous onto map satisfying $\Lambda\circ\sigma=\mathscr{L}\circ\Lambda$ . Clearly, these systems cannot be topologically conjugated, because $\mathscr{D}^{\mathbb{N}}$ is totally disconnected and $(0,1]$ is connected (see [1, Proposition 2.1]).

For each $(d_{1},\ldots,d_{n})\in\mathscr{D}^{n}$ , write

\langle d_{1},\ldots,d_{n}\rangle\colon=\frac{1}{d_{1}}+\frac{1}{d_{1}(d_{1}-1)d_{2}}+\cdots+\frac{1}{d_{1}(d_{1}-1)d_{2}(d_{2}-1)\cdots d_{n-1}(d_{n-1}-1)d_{n}}.

Proposition 2.1.

For every $n\in\mathbb{N}$ and every $\mathbf{c}=(c_{1},\ldots,c_{n})\in\mathscr{D}^{n}$ , we have

I_{n}(\mathbf{c})=\left(\langle c_{1},\ldots,c_{n}\rangle,\langle c_{1},\ldots,c_{n}-1\rangle\right);

and therefore,

|I_{n}(\mathbf{c})|=\prod_{j=1}^{n}\frac{1}{c_{j}(c_{j}-1)}.

Proof.

The proof follows from applying mathematical induction on $n$ . ∎

3. Proof of Theorem 1.4

In 1967, W. Philipp published a quantitative version of the Borel-Cantelli Lemma [14, Theorem 3]. As noted by D. Kleinbock and N. Wadleigh [12, Remark 3.2], Philipp’s theorem can be strengthened to obtain Lemma 3.1. In this lemma and afterwards, we denote the number of elements in a given set $Y$ by $\#Y$ .

Lemma 3.1.

Let $(X,\mathscr{B},\mu)$ be a probability space and let $(E_{n})_{n\geq 1}$ be a sequence of measurable sets. For each $N\in\mathbb{N}$ and each $t\in X$ , define

A(N,t)\colon=\#\left\{n\in\{1,\ldots,N\}\colon t\in E_{n}\right\}

and

\varphi(N)\colon=\sum_{n=1}^{N}\mu(E_{n}).

Suppose that there is a summable sequence of non-negative real numbers $(C_{j})_{j\geq 1}$ such that for any $k,m,n\in\mathbb{N}$ satisfying $n+k<m$ we have

\mu(E_{n}\cap E_{m})\leq\mu(E_{n})\mu(E_{m})+\mu(E_{m})C_{m-n}.

Then, $\mu$ -almost every $t\in X$ satisfies

A(N,t)=\varphi(N)+\mathscr{O}\left(\sqrt{\varphi(N)}\log^{\frac{3}{2}+\varepsilon}\varphi(N)\right)\quad\text{ for all }\varepsilon>0.

Lemma 3.2.

Fix $k\in\mathbb{N}$ . Let $(A_{n})_{n\geq 1}$ be a sequence of at most countable unions of cylinders of level $k$ . We have

\lambda\left(\limsup_{n\to\infty}\mathscr{L}^{-n}[A_{n}]\right)=\begin{cases}0,&\text{\rm if }\quad\displaystyle\sum_{n=1}^{\infty}\lambda(A_{n})<\infty,\\ 1,&\text{\rm if }\quad\displaystyle\sum_{n=1}^{\infty}\lambda(A_{n})=\infty.\end{cases}

Proof.

The convergence case follows from the convergence part of the Borel-Cantelli Lemma. Assume that

(4)

\sum_{n=1}^{\infty}\lambda\left(A_{n}\right)=\infty.

The $\mathscr{L}$ -invariance of $\lambda$ and (4) imply

\sum_{n=1}^{\infty}\lambda\left(\mathscr{L}^{-n}[A]\right)=\infty.

Let $\mathscr{B}$ be the Borel $\sigma$ -algebra of $(0,1]$ . It is shown in [9, Equation (3.4)] that any cylinder $I$ of level $k$ satisfies

(5)

\lambda\left(\mathscr{L}^{-u}[B]\cap I\right)=\lambda(B)\lambda(I)\;\text{ for all }B\in\mathscr{B}\text{ and }u\in\mathbb{N}_{\geq k}.

Pick an arbitrary $n\in\mathbb{N}$ and write $A_{n}$ as an at most countable and disjoint union of cylinders of level $k$ , say

A_{n}=\bigcup_{j}I_{j}^{n}.

By (5) and the $\mathscr{L}$ -invariance of $\lambda$ , for any $m\in\mathbb{N}_{>n+k}$ we have

	$\displaystyle\lambda\left(\mathscr{L}^{-n}[A_{n}]\cap\mathscr{L}^{-m}[A_{m}]\right)$	$\displaystyle=\lambda\left(A_{n}\cap\mathscr{L}^{-(m-n)}[A_{m}]\right)$
		$\displaystyle=\lambda\left(\bigcup_{j}I_{j}^{n}\cap\mathscr{L}^{-(m-n)}[A_{m}]\right)$
		$\displaystyle=\sum_{j}\lambda\left(I_{j}^{n}\cap\mathscr{L}^{-(m-n)}[A_{m}]\right)$
		$\displaystyle=\sum_{j=1}^{\infty}\lambda\left(I_{j}^{n}\right)\lambda(A_{m})=\lambda(A_{n})\lambda(A_{m}).$

The result now follows from Lemma 3.1. ∎

We start with an estimate in the spirit of A. Khinchin’s proof of the existence of the Lévy-Khinchin constant [11, Theorem 31].

Lemma 3.3.

If $m\in\mathbb{N}$ and $g\geq 2^{m}$ , then

\idotsint\limits_{\begin{subarray}{c}x_{1}\cdots x_{m}>g\\ x_{1},\ldots,x_{m}\geq 2\end{subarray}}\frac{\,\mathrm{d}x_{1}\cdots\,\mathrm{d}x_{m}}{x_{1}^{2}\cdots x_{m}^{2}}\asymp_{m}\frac{\log^{m-1}g}{g}.

Proof.

The proof is by induction on $m$ . The base $m=1$ is clear: if $g>2$ , then

\int\limits_{x_{1}>g}\frac{\,\mathrm{d}x_{1}}{x_{1}^{2}}=\frac{1}{g}.

Assume that for $m=M\in\mathbb{N}$ , we have

\idotsint\limits_{\begin{subarray}{c}x_{1}\cdots x_{M}>g\\ x_{1},\ldots,x_{M}\geq 2\end{subarray}}\frac{\,\mathrm{d}x_{1}\cdots\,\mathrm{d}x_{M}}{x_{1}^{2}\cdots x_{M}^{2}}\asymp_{M}\frac{\log^{M-1}g}{g}\;\text{ for all }g>2^{M}.

Let $g>2^{M+1}$ be a real number. Then,

	$\displaystyle\idotsint\limits_{\begin{subarray}{c}x_{1}\cdots x_{M}x_{M+1}>g\\ x_{1},\ldots,x_{M},x_{M+1}\geq 2\end{subarray}}\frac{\,\mathrm{d}x_{1}\cdots\,\mathrm{d}x_{M}\,\mathrm{d}x_{M+1}}{x_{1}^{2}\cdots x_{M}^{2}x_{M+1}^{2}}=$
(6)		$\displaystyle=\int_{2}^{\frac{g}{2^{M}}}\frac{1}{x_{M+1}^{2}}\idotsint\limits_{\begin{subarray}{c}x_{1}\cdots x_{M}>\frac{g}{x_{M+1}}\\ x_{1},\ldots,x_{M}\geq 2\end{subarray}}\frac{\,\mathrm{d}x_{1}\cdots\,\mathrm{d}x_{M}}{x_{1}^{2}\cdots x_{M}^{2}}\,\mathrm{d}x_{M+1}+\int_{\frac{g}{2^{M}}}^{\infty}\frac{1}{x_{M+1}^{2}}\idotsint\limits_{\begin{subarray}{c}x_{1}\cdots x_{M}>\frac{g}{x_{M+1}}\\ x_{1},\ldots,x_{M}\geq 2\end{subarray}}\frac{\,\mathrm{d}x_{1}\cdots\,\mathrm{d}x_{M}}{x_{1}^{2}\cdots x_{M}^{2}}\,\mathrm{d}x_{M+1}.$

When $\frac{g}{2^{M}}<x_{M+1}$ and $\min\{x_{1},\ldots,x_{M}\}\geq 2$ , we have $x_{1}\cdots x_{M}x_{M+1}\geq 2^{M}>\frac{g}{x_{M+1}}$ , so the second term in (6) is

(7)

\int_{\frac{g}{2^{M}}}^{\infty}\frac{1}{x_{M+1}^{2}}\idotsint\limits_{\begin{subarray}{c}x_{1}\cdots x_{M}>\frac{g}{x_{M+1}}\\ x_{1},\ldots,x_{M}\geq 2\end{subarray}}\frac{\,\mathrm{d}x_{1}\cdots\,\mathrm{d}x_{M}}{x_{1}^{2}\cdots x_{M}^{2}}\,\mathrm{d}x_{M+1}=\frac{1}{2^{M}}\int_{\frac{g}{2^{M}}}^{\infty}\frac{1}{x_{M+1}^{2}}\,\mathrm{d}x_{M+1}=\frac{1}{g}.

If $2\leq x_{M+1}\leq\frac{g}{2^{M}}$ , then $2^{M}<\frac{g}{x_{M+1}}$ and the induction hypothesis leads us to

	$\displaystyle\int_{2}^{\frac{g}{2^{M}}}\frac{1}{x_{M+1}^{2}}\idotsint\limits_{\begin{subarray}{c}x_{1}\cdots x_{M}>\frac{g}{x_{M+1}}\\ x_{1},\ldots,x_{M}\geq 2\end{subarray}}\frac{\,\mathrm{d}x_{1}\cdots\,\mathrm{d}x_{M}}{x_{1}^{2}\cdots x_{M}^{2}}\,\mathrm{d}x_{M+1}$	$\displaystyle\asymp_{M}\int_{2}^{\frac{g}{2^{M}}}\frac{1}{x_{M+1}^{2}}\frac{\log^{M-1}\left(\frac{g}{x_{M+1}}\right)}{\left(\frac{g}{x_{M+1}}\right)}\,\mathrm{d}x_{M+1}$
		$\displaystyle=\frac{1}{g}\int_{2}^{\frac{g}{2^{M}}}\frac{(\log g-\log x_{M+1})^{M-1}}{x_{M+1}}\,\mathrm{d}x_{M+1}$
		$\displaystyle=\frac{\log^{M-1}g}{g}\int_{2}^{\frac{g}{2^{M}}}\frac{1}{x_{M+1}}\left(1-\frac{\log x_{M+1}}{\log g}\right)^{M-1}\,\mathrm{d}x_{M+1}.$

The inequality $0<1-\frac{\log x_{M+1}}{\log g}<1$ holds when $2\leq x_{M+1}\leq\frac{g}{2^{M}}$ , then

\frac{\log^{M-1}g}{g}\int_{2}^{\frac{g}{2^{M}}}\frac{1}{x_{M+1}^{2}}\idotsint\limits_{\begin{subarray}{c}x_{1}\cdots x_{M}>\frac{g}{x_{M+1}}\\ x_{1},\ldots,x_{M}\geq 2\end{subarray}}\frac{\,\mathrm{d}x_{1}\cdots\,\mathrm{d}x_{M}}{x_{1}^{2}\cdots x_{M}^{2}}\,\mathrm{d}x_{M+1}\ll_{M+1}\frac{\log^{M}g}{\log g}.

Observe that

(8)

2\leq x_{M+1}\leq g^{\frac{1}{2}}\quad\text{ implies }\quad\frac{1}{2}\leq 1-\frac{\log x_{M+1}}{\log g}.

We consider two cases for the lower bound: $2^{2M}<g$ and $2^{2M}\geq g$ . If $2^{2M}\leq g$ , then $g^{\frac{1}{2}}\leq\frac{g}{2^{M}}$ , so

	$\displaystyle\frac{1}{g}\int_{2}^{\frac{g}{2^{M}}}\frac{(\log g-\log x_{M+1})^{M-1}}{x_{M+1}}\,\mathrm{d}x_{M+1}$	$\displaystyle\geq\frac{\log^{M-1}g}{g}\int_{2}^{\sqrt{g}}\frac{1}{x_{M+1}}\left(1-\frac{\log x_{M+1}}{\log g}\right)^{M-1}\,\mathrm{d}x_{M+1}$
		$\displaystyle\geq\frac{1}{2^{M-1}}\left(\frac{1}{2}\log g-\log 2\right)\frac{\log^{M-1}g}{g}$
		$\displaystyle=\frac{1}{2^{M-1}}\left(\frac{1}{2}-\frac{\log 2}{\log g}\right)\frac{\log^{M}g}{g}$
		$\displaystyle\geq\frac{1}{2^{M}}\left(1-\frac{1}{M}\right)\frac{\log^{M}g}{g}.$

We have used $2^{2M}\leq g$ in the last inequality.

For the second case, observe that $2^{M+1}<g\leq 2^{2M}$ yields $\frac{g}{2^{M}}\leq g^{\frac{1}{2}}$ . This means that $2\leq x_{M+1}\leq\frac{g}{2^{M}}$ implies $2\leq x_{M+1}\leq g^{\frac{1}{2}}$ and, by (8),

	$\displaystyle\frac{1}{g}\int_{2}^{\frac{g}{2^{M}}}\frac{(\log g-\log x_{M+1})^{M}}{x_{M+1}}\,\mathrm{d}x_{M+1}$	$\displaystyle\geq\frac{1}{2^{M}}\frac{\log^{M-1}g}{g}\int_{2}^{\frac{g}{2^{M}}}\,\frac{1}{x_{M+1}}\,\mathrm{d}x_{M+1}$
		$\displaystyle\geq\frac{1}{2^{M}}\frac{\log^{M}g}{g}\left(1-\frac{(M+1)\log 2}{\log g}\right)>0.$

These estimates along with (7) yield

	$\displaystyle\idotsint\limits_{\begin{subarray}{c}x_{1}\cdots x_{M+1}>g\\ x_{1},\ldots,x_{M+1}\geq 2\end{subarray}}\frac{\,\mathrm{d}x_{1}\cdots\,\mathrm{d}x_{M+1}}{x_{1}^{2}\cdots x_{M+1}^{2}}$	$\displaystyle\asymp_{M+1}\frac{\log^{M}g}{g}\left(\min\left\{1-\frac{(M+1)\log 2}{\log g},1-\frac{1}{M}\right\}+\frac{1}{g\log^{M}g}\right)$
		$\displaystyle\asymp_{M+1}\frac{\log^{M}g}{g}.$

∎

Lemma 3.4.

If $m\in\mathbb{N}$ , $\mathbf{t}\in\mathbb{R}_{>0}^{m}$ , and $g\geq 2^{mt}$ , then

\sum_{\begin{subarray}{c}d_{1}^{t_{0}}\cdots d_{m}^{t_{m-1}}\geq g\\ d_{1},\ldots,d_{m}\geq 2\end{subarray}}\;\prod_{j=1}^{m}\frac{1}{d_{j}(d_{j}-1)}\asymp_{m,\mathbf{t}}\frac{\log^{\ell-1}g}{g^{1/T}}.

Proof.

Without loss of generality, we may assume that $t=t_{m-1}\leq\ldots\leq t_{1}\leq t_{0}=T$ , then

g^{1/t_{0}}\leq\ldots\leq g^{1/t_{m-2}}\leq g^{1/t_{m-1}}.

Under this assumption on $\mathbf{t}$ , the function $\ell$ has the following properties:

i.

If $\ell(t_{0},\ldots,t_{m-2},t_{m-1})=m$ , then $\ell(t_{0},\ldots,t_{m-2})=m-1$ .
ii.

If $1\leq\ell(t_{0},\ldots,t_{m-2},t_{m-1})\leq m-1$ , then $\ell(t_{0},\ldots,t_{m-2},t_{m-1})=\ell(t_{0},\ldots,t_{m-2})$ .

For every $m$ , $\mathbf{t}$ , and $g$ as in the statement, define

S_{m}(\mathbf{t};g)\colon=\sum_{\begin{subarray}{c}d_{1}^{t_{0}}\cdots d_{m}^{t_{m-1}}\geq g\\ d_{1},\ldots,d_{m}\geq 2\end{subarray}}\;\prod_{j=1}^{m}\frac{1}{d_{j}(d_{j}-1)}.

As such, we aim to show that $S_{m}(\mathbf{t};g)\asymp_{m,\mathbf{t}}(\log^{\ell(\mathbf{t})-1}g)/g^{1/T}$ . The proof is by induction on $m$ . For $m=1$ and $g>2^{t_{0}}$ , we have

\frac{g^{1/t_{0}}}{2}\leq\lceil g^{1/t_{0}}\rceil-1<g^{1/t_{0}}

and so

S_{1}(t_{0};g)=\sum_{d_{1}^{t_{0}}>g}\frac{1}{d_{1}(d_{1}-1)}=\sum_{d_{1}>\lceil g^{1/t_{0}}\rceil}\frac{1}{d_{1}(d_{1}-1)}=\frac{1}{\lceil g^{1/t_{0}}\rceil-1}\asymp\frac{1}{g^{1/t_{0}}}

Assume that for some $m=M\in\mathbb{N}$ , every $\mathbf{t}\in\mathbb{R}_{>0}^{m}$ satisfies

(9)

S_{M}(\mathbf{t};g)\asymp_{\mathbf{t},M}\frac{\log^{\ell(\mathbf{t})-1}g}{g^{1/T}}\;\text{ for all }g\geq 2^{mt_{M-1}}.

If $\ell(\mathbf{t})=M+1$ , then $S_{M+1}(\mathbf{t};g)=S_{m}((1,\ldots,1);g^{1/t})$ and Lemma 3.3 gives the result. Suppose that $1\leq\ell(\mathbf{t})\leq M$ . Write

u(M+1,g)\colon=\begin{cases}\left\lfloor\frac{g^{1/t_{M}}}{2^{Mt_{M-1}/t_{M}}}\right\rfloor,\text{ if }\frac{g^{1/t_{M}}}{2^{Mt_{M-1}/t_{M}}}\not\in\mathbb{N},\\[8.61108pt] \frac{g^{1/t_{M}}}{2^{Mt_{M-1}/t_{M}}}-1,\text{ otherwise.}\end{cases}

On the one hand, when $d_{M+1}\geq u(M+1,g)+1$ , every $(d_{1},\ldots,d_{M})\in\mathscr{D}^{M}$ satisfies

d_{1}^{t_{0}}\cdots d_{M}^{t_{M-1}}\geq 2^{Mt_{M-1}}>\frac{g}{(u(M+1,g)+1)^{t_{M}}}\geq\frac{g}{d_{M+1}^{t_{M}}}.

As a consequence, we may express $S_{M+1}(\mathbf{t};g)$ as follows:

	$\displaystyle S_{M+1}(\mathbf{t};g)$	$\displaystyle=\sum_{d_{M+1}=2}^{u(M+1,g)}\frac{1}{d_{M+1}(d_{M+1}-1)}S_{M}\left((t_{0},\ldots,t_{M-1}),\frac{g}{d_{M}^{t_{M+1}}}\right)+$
(10)			$\displaystyle\quad+\sum_{d_{M+1}=u(M+1,g)+1}^{\infty}\frac{1}{d_{M+1}(d_{M+1}-1)}\sum_{d_{1},\ldots,d_{M}\geq 2}\prod_{j=1}^{M}\frac{1}{d_{j}(d_{j}-1)}.$

We apply the induction hypothesis (9) on the first term in (10) and use $\ell(\mathbf{t})=\ell(t_{0},\ldots,t_{M-1})$ to get

	$\displaystyle\sum_{d_{M+1}=2}^{u(M+1,g)}$	$\displaystyle\frac{1}{d_{M+1}(d_{M+1}-1)}S_{M}\left((t_{0},\ldots,t_{M-1}),\frac{g}{d_{M+1}^{t_{M}}}\right)\asymp_{\mathbf{t},M}$
		$\displaystyle\asymp_{\mathbf{t},M}\sum_{d_{M+1}=2}^{u(M+1,g)}\frac{\log^{\ell(\mathbf{t})-1}(g)\left(1-\frac{t_{M}\log d_{M+1}}{\log g}\right)^{\ell(\mathbf{t})-1}}{d_{M+1}(d_{M+1}-1)\left(\frac{g}{d_{M+1}^{t_{M}}}\right)^{1/T}}$
		$\displaystyle\asymp_{\mathbf{t},M}\frac{\log^{\ell(\mathbf{t})-1}(g)}{g^{1/T}}\sum_{d_{M+1}=2}^{u(M+1,g)}\frac{1}{d_{M+1}^{2-t_{M}/T}}\left(1-\frac{t_{M}\log d_{M+1}}{\log g}\right)^{\ell(\mathbf{t})-1}.$

Since $2-t_{M}/T>1$ , the last expression satisfies

	$\displaystyle\frac{\log^{\ell(\mathbf{t})-1}(g)}{g^{1/T}}\sum_{d_{M+1}=2}^{u(M+1,g)}\frac{1}{d_{M+1}^{2-t_{M}/T}}\left(1-\frac{t_{M}\log d_{M+1}}{\log g}\right)^{\ell(\mathbf{t})-1}$	$\displaystyle\leq\frac{\log^{\ell(\mathbf{t})-1}(g)}{g^{1/T}}\sum_{d_{M+1}=2}^{u(M+1,g)}\frac{1}{d_{M+1}^{2-t_{M}/T}}$
		$\displaystyle\asymp_{\mathbf{t},M}\frac{\log^{\ell(\mathbf{t})-1}(g)}{g^{1/T}}.$

We now obtain the lower estimate. If $g\leq 2^{2Mt_{M-1}+2t_{M}}$ , then

\log g\leq(2Mt_{M-1}+2t_{M})\log 2.

Therefore, since $d_{M+1}\leq u(M+1;g)$ , we have

\frac{t_{M}\log d_{M+1}}{\log g}<1-\frac{Mt_{M-1}\log 2}{\log g}\leq 1-\frac{Mt_{M-1}\log 2}{(2Mt_{M-1}+2t_{M})\log 2},

which means

\frac{Mt_{M-1}}{2Mt_{M-1}+2t_{M}}\leq 1-\frac{t_{M}\log d_{M+1}}{\log g}

and we conclude

\frac{\log^{\ell(\mathbf{t})-1}(g)}{g^{1/T}}\sum_{d_{M+1}=2}^{u(M+1,g)}\frac{1}{d_{M+1}^{2-t_{M}/T}}\left(1-\frac{t_{M}\log d_{M+1}}{\log g}\right)^{\ell(\mathbf{t})-1}\gg_{\mathbf{t},M+1}\frac{1}{4}\frac{\log^{\ell(\mathbf{t})-1}(g)}{g^{1/T}}.

If $g\geq 2^{2Mt_{M-1}+2t_{M}}$ , then $g^{1/(2t_{M})}\geq 2^{1+Mt_{M-1}/t_{M}}$ and $2<u(M+1,g^{1/2})<u(M+1,g)$ , so

	$\displaystyle\frac{\log^{\ell(\mathbf{t})-1}(g)}{g^{1/T}}\sum_{d_{M+1}=2}^{u(M+1,g)}\frac{1}{d_{M+1}^{2-t_{M}/T}}\left(1-\frac{t_{M}\log d_{M+1}}{\log g}\right)^{\ell(\mathbf{t})-1}$	$\displaystyle\geq\frac{1}{2^{\ell(\mathbf{t})-1}}\frac{\log^{\ell(\mathbf{t})-1}(g)}{g^{1/T}}\sum_{d_{M+1}=2}^{u(M+1,g^{1/2})}\frac{1}{d_{M+1}^{2-t_{M}/T}}$
		$\displaystyle\geq\frac{1}{2^{\ell(\mathbf{t})+1}}\frac{\log^{\ell(\mathbf{t})-1}(g)}{g^{1/T}}\quad(cfr.\eqref{EQ:LEM:KHIN:02}).$

∎

Proof of Theorem 1.4.

If there are infinitely many $n\in\mathbb{N}$ such that $1<\Psi(n)\leq 2^{mt}$ , we can pick a real number $x$ and a strictly increasing sequence of natural numbers $(n_{j})_{j\geq 1}$ such that

1<x<\Psi(n_{j})\leq 2^{mt}\;\text{ for all }j\in\mathbb{N}.

Then,

\log^{\ell-1}x<\log^{\ell(\mathbf{t})-1}\Psi(n_{j})\quad\text{ and }\quad 2^{-mt/T}<\Psi(n_{j})^{-1/T}\quad\text{ for all }j\in\mathbb{N}

and, therefore,

\sum_{n=1}^{\infty}\frac{\log^{\ell-1}\Psi(n)}{\Psi(n)^{1/T}}\geq\sum_{j=1}^{\infty}\frac{\log^{\ell-1}\Psi(n_{j})}{\Psi(n_{j})^{1/T}}=\infty\quad\text{ and }\quad\mathcal{E}_{\mathbf{t}}(\Psi)=(0,1].

Assume that $\Psi(n)>2^{mt}$ holds for all $n\in\mathbb{N}$ . For each $n\in\mathbb{N}$ , define

G_{n}^{\mathbf{t}}(\Psi)\colon=\left\{\mathbf{d}\in\mathscr{D}^{m}:d_{1}^{t_{0}}\cdots d_{m-1}^{t_{m}}\geq\Psi(n)\right\}\quad\text{ and }\quad A_{n}\colon=\bigcup_{\mathbf{d}\in G_{n}(\Psi)}I_{m}(\mathbf{d}).

In view of Proposition 2.1, we have

\lambda(A_{n})=\sum_{\mathbf{d}\in G_{n}(\Psi)}\lambda(I_{m}(\mathbf{d}))=\sum_{\mathbf{d}\in G_{n}(\Psi)}\prod_{j=1}^{m}\frac{1}{d_{j}(d_{j}-1)}

and, by Lemma 3.4,

(11)

\lambda(A_{n})\asymp_{m}\frac{\log\Psi^{\ell(\mathbf{t})-1}(n)}{\Psi(n)^{\frac{1}{T}}}.

The definitions of $\mathcal{E}_{\mathbf{t}}(\Psi)$ and $(A_{n})_{n=1}^{\infty}$ entail

	$\displaystyle\mathcal{E}_{\mathbf{t}}(\Psi)$	$\displaystyle=\left\{x\in(0,1]:\mathscr{L}^{n-1}(x)\in A_{n}\text{ for infinitely many }n\in\mathbb{N}\right\}$
		$\displaystyle=\limsup_{n\to\infty}\mathscr{L}^{-n}(A_{n}).$

We deduce (3) from the $\mathscr{L}$ -invariance of $\lambda$ , Lemma 3.2, and (11).

Finally, we show that Theorem 1.4 may fail if (2) is dropped. Let $t>0$ and $m\in\mathbb{N}$ be arbitrary and put $\mathbf{t}=(t,\ldots,t)\in\mathbb{R}_{>0}^{m}$ . Choose a real number $r$ such that

0<r<\min\left\{1,\frac{1}{t}\log 2\right\}.

Define $\Psi_{t}:\mathbb{N}\to\mathbb{R}_{\geq 1}$ by

\quad\Psi_{t}(n)\colon=\exp\left(r^{n}\right)\quad\text{ for all }n\in\mathbb{N}.

Then, $\Psi_{t}(n)>1$ for all $n\in\mathbb{N}$ , $\Psi_{t}(n)\to 1$ as $n\to\infty$ , and

\sum_{n=0}^{\infty}\frac{\log^{m-1}\Psi_{t}(n)}{\Psi_{t}(n)^{\frac{1}{t}}}=\sum_{n=0}^{\infty}\frac{r^{n(m-1)}}{\Psi_{t}(n)^{\frac{1}{t}}}<\infty.

However, the definition of $\Psi_{t}$ implies $\Psi_{t}(n)=\exp(r^{n})<2^{t}$ for all $n\in\mathbb{N}$ , so every $\mathbf{d}=(d_{n})_{n\geq 1}\in\mathscr{D}^{\mathbb{N}}$ satisfies

d_{n}^{t}d_{n+1}^{t}\cdots d_{n+m-1}^{t}\geq 2^{mt}>\Psi_{t}(n)\quad\text{ for all }n\in\mathbb{N}.

In other words, $\mathcal{E}_{t}(\Psi_{t})=(0,1]$ and $\lambda(\mathcal{E}_{\mathbf{t}}(\Psi_{t}))=1$ . ∎

4. Proof of Theorem 1.5

We recall two results on Lüroth series. The first one is an analogue of T. Łuczak’s Theorem on the Hausdorff dimension of sets of continued fractions with rapidly growing partial quotients [13]. The second result is the Lüroth analogue of a theorem by B. Wang and J. Wu [19, Theorem 3.1].

For every pair of real numbers $a,b$ strictly larger than $1$ , define the sets

	$\displaystyle E(a,b)$	$\displaystyle\colon=\left\{x\in(0,1]:d_{n}(x)\geq a^{b^{n}}\text{ for all }n\in\mathbb{N}\right\},$
	$\displaystyle\widetilde{E}(a,b),$	$\displaystyle\colon=\left\{x\in(0,1]:d_{n}(x)\geq a^{b^{n}}\text{ for infinitely many }n\in\mathbb{N}\right\}.$

Lemma 4.1 ([16, Theorem 3.1]).

For any $a>1$ and $b>1$ , we have

\dim_{\operatorname{H}}E(a,b)=\dim_{\operatorname{H}}\widetilde{E}(a,b)=\frac{1}{b+1}.

For every $B>1$ , define the number

s(B)\colon=\dim_{\operatorname{H}}\left\{x\in(0,1]:d_{n}(x)>B^{n}\text{ for infinitely many }n\in\mathbb{N}\right\}.

Lemma 4.2 ([15, Lemma 2.3]).

The function $s:\mathbb{R}_{>1}\to\mathbb{R}$ is continuous, $\displaystyle\lim_{B\to\infty}s(B)=\frac{1}{2}$ , and $s=s(B)$ is the only solution of

\sum_{k=2}^{\infty}\left(\frac{1}{Bk(k-1)}\right)^{s}=1.

Proof of Theorem 1.5.

First, assume that $B=1$ . From

\left\{x\in(0,1]:d_{n}(x)\geq\Psi(n)^{1/t_{0}}\text{ for infinitely many }n\in\mathbb{N}\right\}\subseteq\mathcal{E}_{\mathbf{t}}(\Psi),

$B=1$ , and Theorem 1.2, we conclude $\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(\Psi)=1$ .

Suppose that $B=\infty$ . Assume that $b>1$ . Every number $c\in(1,b)$ satisfies $\frac{\log\log\Psi(n)}{n}\geq\log c$ or, equivalently, $\Psi(n)\geq e^{c^{n}}$ for every large $n$ . Then,

	$\displaystyle\mathcal{E}_{\mathbf{t}}(\Psi)$	$\displaystyle\subseteq\left\{x\in(0,1]:\prod_{i=1}^{m}d_{n+i}^{t_{i}}(x)\geq e^{c^{n}}\text{ for infinitely many }n\right\}$
		$\displaystyle\subseteq\left\{x\in(0,1]:d_{n+i}^{\,t_{i}}(x)\geq e^{\frac{c^{n}}{m}}\text{ for some }i\in\{1,\ldots,m\}\text{ and infinitely many }n\right\}$
		$\displaystyle\subseteq\left\{x\in(0,1]:d_{n+i}(x)\geq(e^{\frac{1}{mT}})^{c^{n}}\text{ for some }i\in\{1,\ldots,m\}\text{ and infinitely many }n\right\}$

and, by Lemma 4.1,

\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(\Psi)\leq\frac{1}{1+c}.

The previous inequality give us two implications:

\text{ if }b<\infty,\text{ then }\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(\Psi)\leq\frac{1}{b+1},

\text{ if }b=\infty,\text{ then }\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(\Psi)=0.

Now we obtain the lower bound for $\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(\Psi)$ when $1<b<\infty$ . For all $c>b$ , we have

\left\{x\in(0,1]:d_{n}(x)^{t_{0}}\geq e^{c^{n}}\text{ for all }n\in\mathbb{N}\right\}=\left\{x\in(0,1]:d_{n}(x)\geq(e^{\frac{1}{t_{0}}})^{c^{n}}\text{ for all }n\in\mathbb{N}\right\}\subseteq\mathcal{E}_{\mathbf{t}}(\Psi).

Thus, applying Lemma 4.1,

\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(\Psi)\geq\frac{1}{1+c}\to\frac{1}{1+b}\quad\text{ when }\quad c\to b.

Lastly, assume that $b=1$ . Then, for any $\varepsilon>0$ , we have $\Psi(n)\leq e^{(1+\varepsilon)^{n}}$ infinitely often, which gives

\mathcal{E}_{\mathbf{t}}(\Psi)\supseteq\left\{x\in(0,1]:d_{n}^{t_{0}}(x)\geq e^{(1+\varepsilon)^{n}}\text{ for infinitely many }n\in\mathbb{N}\right\}

and, by Lemma 4.1,

\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(\Psi)\geq\frac{1}{2+\varepsilon}\to\frac{1}{2}\quad\text{ when }\quad\varepsilon\to 0.

For the upper bound, note that $B=\infty$ implies that for each $A>0$ there is some $N(A)\in\mathbb{N}$ such that $A^{n}<\Psi(n)$ whenever $n\geq N(A)$ . Hence, for all $x\in\mathcal{E}_{\mathbf{t}}(\Psi)$ we have

\prod_{i=0}^{m-1}d_{n+i}^{t_{i}}(x)\geq A^{n}\;\text{ for infinitely many }n\in\mathbb{N}_{\geq N(A)},

and, thus,

d_{n+1}^{t_{i}}\geq A^{\frac{1}{m}}\;\text{ for some }0\leq i\leq m-1\;\text{ and infinitely many }n\in\mathbb{N}_{\geq N(A)}.

As a consequence,

\mathcal{E}_{\mathbf{t}}(\Psi)\subseteq\left\{x\in(0,1]:d_{n}(x)\geq(A^{\frac{1}{Tm}})^{n}\text{ for infinitely many }n\in\mathbb{N}\right\},

and Lemma 4.2 guarantees

\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(\Psi)\leq s\left(A^{\frac{1}{Tm}}\right)\to\frac{1}{2}\;\text{ as }\;A\to\infty.

∎

5. Proof of Theorem 1.6

We obtain Theorem 1.6 from Theorem 5.1 below and its proof. In Theorem 5.1, we compute the Hausdorff dimension of $\mathcal{E}_{\mathbf{t}}(\Psi)$ for any $\mathbf{t}\in\mathbb{R}_{>0}^{2}$ and a particular choice of $\Psi$ . For each $B\in(1,\infty)$ , define the set

\mathcal{E}_{\mathbf{t}}(B)\colon=\left\{x\in(0,1]:d_{n}^{t_{0}}(x)d_{n+1}^{t_{1}}(x)\geq B^{n}\text{ for infinitely many }n\in\mathbb{N}\right\}.

Theorem 5.1.

Take any $B>1$ , $\mathbf{t}=(t_{0},t_{1})\in\mathbb{R}_{>0}^{2}$ , and let $f_{t_{0},t_{1}}$ be as in Theorem 1.6. The Hausdorff dimension of $\mathcal{E}_{\mathbf{t}}(B)$ is the unique solution $s=s_{0}(B)$ of the equation

\sum_{d=2}^{\infty}\frac{1}{d^{s}(d-1)^{s}B^{f_{t_{0},t_{1}}(s)}}=1.

Moreover, the map $B\mapsto s_{0}(B)$ is continuous.

5.1. Continuity of $B\mapsto s_{0}(B)$

Lemma 5.2.

For any $(t_{0},t_{1})\in\mathbb{R}_{>0}^{2}$ , the function $f_{t_{0},t_{1}}$ is strictly increasing.

Proof.

Let $(t_{0},t_{1})\in\mathbb{R}_{>0}^{2}$ be given. For any $s\geq 0$ we have

f_{t_{0},t_{1}}(s)=\min\left\{\frac{s^{2}}{t_{0}s+(1-s)t_{1}},\frac{s}{t_{1}}\right\}.

Thus, it suffices to show that the functions $s\mapsto\frac{s}{t_{1}}$ and $s\mapsto\frac{s^{2}}{t_{0}s+(1-s)t_{1}}$ are strictly increasing. This is obvious for $s\mapsto\frac{s}{t_{1}}$ . When $t_{0}\leq t_{1}$ , the function $s\mapsto t_{0}s+(1-s)t_{1}$ is non-increasing, so $s\mapsto\frac{s^{2}}{t_{0}s+(1-s)t_{1}}$ is strictly increasing. If $t_{1}<t_{0}$ , then the derivative of

s\mapsto\frac{s^{2}}{(t_{0}-t_{1})s+t_{1}}

is positive for $s>0$ and it is $0$ for $s=0$ . Therefore, the function $s\mapsto\frac{s^{2}}{(t_{0}-t_{1})s+t_{1}}$ is strictly increasing. ∎

For each $n\in\mathbb{N}$ , $\mathbf{t}=(t_{0},t_{1})\in\mathbb{R}_{>0}^{2}$ , and $B>1$ , define the map $g_{n}^{\mathbf{t}}(\,\cdot\,;B):\mathbb{R}_{\geq 0}\to\mathbb{R}_{>0}$ by

g_{n}^{\mathbf{t}}(\rho;B)\colon=\sum_{d=2}^{n}\frac{1}{d^{\rho}(d-1)^{\rho}B^{f_{t_{0},t_{1}}(\rho)}}\;\text{ for all }\rho\in\mathbb{R}_{>0}

and $g^{\mathbf{t}}(\,\cdot\,;B):\mathbb{R}_{\geq 0}\to\mathbb{R}_{>0}\cup\{\infty\}$ by

g(\rho;B)\colon=\sum_{d=2}^{\infty}\frac{1}{d^{\rho}(d-1)^{\rho}B^{f_{t_{0},t_{1}}(\rho)}}\;\text{ for all }\rho\in\mathbb{R}_{>0}.

Observe that for all $B>1$ we have

g^{\mathbf{t}}(1;B)=\sum_{d=2}^{\infty}\frac{1}{d(d-1)B^{1/t_{1}}}<\sum_{d=2}^{\infty}\frac{1}{d(d-1)}=1

and

g^{\mathbf{t}}\left(\frac{1}{2};B\right)=\sum_{d=2}^{\infty}\frac{1}{d^{1/2}(d-1)^{1/2}B^{f_{t_{0},t_{1}}(1/2)}}=\frac{1}{B^{f_{t_{0},t_{1}}(1/2)}}\sum_{d=2}^{\infty}\frac{1}{d^{1/2}(d-1)^{1/2}}=\infty,

hence $\frac{1}{2}<s_{0}(B)<1$ . Also, every $m,n\in\mathbb{N}$ with $m<n$ and every $\rho>0$ satisfy

g_{m}^{\mathbf{t}}(\rho;B)<g_{n}^{\mathbf{t}}(\rho;B)<g^{\mathbf{t}}(\rho;B).

As a consequence, the sequence $(s_{n}(B))_{n\geq 1}$ is strictly increasing and each of its terms is bounded above by $s_{0}(B)$ .

Lemma 5.3.

We have

\lim_{n\to\infty}s_{n}(B)=s_{0}(B).

Proof.

The discussion preceding the lemma implies

\lim_{n\to\infty}s_{n}(B)\leq s_{0}(B).

Pick any positive number $t<s_{0}(B)$ . Then, $g(t;B)>1$ and every large $n\in\mathbb{N}$ satisfies $g_{n}(t;B)>1$ , so $t<s_{n}(B)$ and

s_{0}(B)\leq\lim_{n\to\infty}s_{n}(B).

∎

Lemma 5.4.

The function $B\mapsto s_{0}(B)$ is continuous.

Proof.

Fix $B>1$ . Take $n\in\mathbb{N}$ . Let $\varepsilon>0$ be arbitrary and

0<\delta=\frac{1}{2}\min\left\{B-B^{\frac{f_{t_{0},t_{1}}(s_{n}(B))}{f_{t_{0},t_{1}}(s_{n}(B)+\varepsilon)}},B^{\frac{f_{t_{0},t_{1}}(s_{n}(B))}{f_{t_{0},t_{1}}(s_{n}(B)-\varepsilon)}}-B\right\}.

Then, we have

	$\displaystyle g_{n}(s_{n}(B)+\varepsilon;B-\delta)$	$\displaystyle=\sum_{d=2}^{n}\frac{1}{d^{s_{n}(B)+\varepsilon}(d-1)^{s_{n}(B)+\varepsilon}(B-\delta)^{f_{t_{0},t_{1}}(s_{n}(B)+\varepsilon)}}$
		$\displaystyle\leq\frac{B^{f_{t_{0},t_{1}}(s_{n}(B))}}{(B-\delta)^{f_{t_{0},t_{1}}(s_{n}(B)+\varepsilon)}}\sum_{d=2}^{n}\frac{1}{d^{s_{n}(B)}(d-1)^{s_{n}(B)}B^{f_{t_{0},t_{1}}(s_{n}(B))}}$
		$\displaystyle=\frac{B^{f_{t_{0},t_{1}}(s_{n}(B))}}{(B-\delta)^{f_{t_{0},t_{1}}(s_{n}(B)+\varepsilon)}}<1.$

Since $s_{n}$ is non-increasing on $B$ , we conclude

s_{n}(B)\leq s_{n}(B-\delta)<s_{n}(B)+\varepsilon.

Similarly, we can show $g_{n}(s_{n}(B)-\varepsilon;B+\delta)>1$ and, hence,

s_{n}(B+\delta)\leq s_{n}(B)<s_{n}(B+\delta)+\varepsilon.

As a consequence, $s_{n}$ is continuous.

Since $s_{n}(B)\to s_{0}(B)$ as $n\to\infty$ and since $f_{t_{0},t_{1}}$ is continuous and strictly increasing, we may pick $\delta>0$ such that every large $m\in\mathbb{N}$ verifies

0<\delta=\frac{1}{2}\min\left\{B-B^{\frac{f_{t_{0},t_{1}}(s_{m}(B))}{f_{t_{0},t_{1}}(s_{m}(B)+\varepsilon)}},B^{\frac{f_{t_{0},t_{1}}(s_{m}(B))}{f_{t_{0},t_{1}}(s_{m}(B)-\varepsilon)}}-B\right\}.

For such $m\in\mathbb{N}$ , every $B^{\prime}\in\mathbb{R}_{>1}$ with the property $|B-B^{\prime}|<\delta$ satisfies $|s_{m}(B)-s_{m}(B^{\prime})|<\varepsilon$ . Letting $m\to\infty$ , we conclude $|s_{0}(B)-s_{0}(B^{\prime})|\leq\varepsilon$ . Therefore, $s_{0}$ is continuous. ∎

Our argument actually proves the next result.

Theorem 5.5.

Let $f:\mathbb{R}_{>0}\to\mathbb{R}_{>0}$ be a strictly increasing continuous function. For each $n\in\mathbb{N}$ , let $s=s_{n}(B,f)$ be the unique solution $s$ of

\sum_{d=2}^{n}\frac{1}{d^{s}(d-1)^{s}B^{f(s)}}=1.

Call $s_{0}(B,f)$ the unique solution $s$ of

\sum_{d=2}^{\infty}\frac{1}{d^{s}(d-1)^{s}B^{f(s)}}=1.

Then, $s_{n}(B,f)\to s_{0}(B,f)$ as $n\to\infty$ .

5.2. Hausdorff dimension estimates

We split the proof of Theorem 5.1 into two parts. First, we use a particular family of coverings of $\mathcal{E}_{\mathbf{t}}(\Psi)$ to obtain an upper bound for $\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(\Psi)$ . The lower bound is proved considering two cases. In one of them, we use Lemma 4.2. In the other case, we apply the Mass Distribution Principle (Lemma 5.8).

The hypothesis $B>1$ implies $\liminf_{n}\Psi(n)=\infty$ . Then, without loss of generality, we may assume that

\Psi(n)\geq 2^{mt}\quad\text{ for all }n\in\mathbb{N}.

5.2.1. Upper bound

Proof of Lemma 5.1. Upper bound..

For each real number $A$ satisfying $1<A<B$ , define the sets

	$\displaystyle\mathcal{E}_{\mathbf{t}}^{\prime}(A)$	$\displaystyle\colon=\left\{x\in(0,1]:d_{n}^{t_{0}}(x)\leq A^{n}\text{ and }d_{n+1}^{t_{1}}(x)\geq\frac{B^{n}}{d_{n}^{t_{0}}(x)}\text{ for infinitely many }n\in\mathbb{N}\right\},$
	$\displaystyle\mathcal{E}_{\mathbf{t}}^{\prime\prime}(A)$	$\displaystyle\colon=\left\{x\in(0,1]:d_{n}^{t_{0}}(x)\geq A^{n}\text{ for infinitely many }n\in\mathbb{N}\right\}.$

Then, $\mathcal{E}_{\mathbf{t}}(B)\subseteq\mathcal{E}_{\mathbf{t}}^{\prime}(A)\cup\mathcal{E}_{\mathbf{t}}^{\prime\prime}(A)$ , so

\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(B)\leq\max\left\{\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}^{\prime}(A),\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}^{\prime\prime}(A)\right\}.

Lemma 4.2 implies

\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}^{\prime\prime}(A)=s\left(A^{1/t_{0}}\right).

We take advantage of $\mathcal{E}_{\mathbf{t}}^{\prime}(A)$ being a limsup-set to give an upper estimate for its dimension. For each $\mathbf{d}=(d_{1},\ldots,d_{n})\in\mathscr{D}^{n}$ , define

J_{n}(\mathbf{d})\colon=\bigcup\left\{I_{n+1}(\mathbf{d}d_{n+1}):d_{n+1}\geq\max\left\{2,\left(\frac{B^{n}}{d_{n}^{t_{0}}}\right)^{1/t_{1}}\right\}\right\}

Observe that $d_{n}^{t_{0}}\leq A^{n}$ and $d_{n+1}^{t_{1}}\geq B^{n}/d_{n}^{t_{0}}$ imply

d_{n+1}\geq\left(\frac{B}{A}\right)^{n/t_{1}},

and $(B/A)^{n/t_{1}}>2$ for every large $n$ (depending on $B$ , $A$ , and $t_{1}$ ). For such $n\in\mathbb{N}$ , we have

	$\displaystyle\|J_{n}(\mathbf{d})\|$	$\displaystyle=\left(\prod_{j=1}^{n}\frac{1}{d_{j}(d_{j}-1)}\right)\frac{1}{\left\lceil\frac{B^{n}}{d_{n}^{t_{0}/t_{1}}}\right\rceil-1}$
		$\displaystyle\asymp\left(\prod_{j=1}^{n}\frac{1}{d_{j}(d_{j}-1)}\right)\frac{d_{n}^{t_{0}/t_{1}}}{B^{n/t_{1}}}$
		$\displaystyle\asymp\frac{1}{d_{n}^{2-\frac{t_{0}}{t_{1}}}B^{\frac{n}{t_{1}}}}\prod_{j=1}^{n-1}\frac{1}{d_{j}(d_{j}-1)}=\frac{\|I_{n-1}(d_{1},\ldots,d_{n-1})\|}{d_{n}^{2-\frac{t_{0}}{t_{1}}}B^{\frac{n}{t_{1}}}}.$

Take any $s>0$ and let $\mathcal{H}^{s}$ be the $s$ -Hausdorff measure on $\mathbb{R}$ . From the inequality

\mathcal{E}^{\prime}_{\mathbf{t}}(A)=\bigcap_{N=1}^{\infty}\bigcup_{n=N}^{\infty}\bigcup_{\mathbf{d}\in\mathscr{D}^{n-1}}\bigcup_{d_{n}=2}^{\lfloor A^{n/t_{0}}\rfloor}J_{n}(\mathbf{d}\,d_{n})

we conclude that

	$\displaystyle\mathcal{H}^{s}(\mathcal{E}_{\mathbf{t}}^{\prime}(A))$	$\displaystyle\leq\liminf_{N\to\infty}\sum_{n\geq N}\sum_{\mathbf{d}\in\mathscr{D}^{n-1}}\sum_{d_{n}=2}^{\lfloor A^{n/t_{0}}\rfloor}\|J_{n}(\mathbf{d}\,d_{n})\|^{s}$
		$\displaystyle\asymp\liminf_{N\to\infty}\sum_{n\geq N}\sum_{\mathbf{d}\in\mathscr{D}^{n-1}}\sum_{d_{n}=2}^{\lfloor A^{n/t_{0}}\rfloor}\|I_{n-1}(\mathbf{d})\|^{s}d_{n}^{-s\left(2-\frac{t_{0}}{t_{1}}\right)}B^{-\frac{sn}{t_{1}}}$
		$\displaystyle=\liminf_{N\to\infty}\sum_{n\geq N}\sum_{\mathbf{d}\in\mathscr{D}^{n-1}}\frac{\|I_{n-1}(\mathbf{d})\|^{s}}{B^{\frac{sn}{t_{1}}}}\sum_{d_{n}=2}^{\lfloor A^{n/t_{0}}\rfloor}d_{n}^{-s\left(2-\frac{t_{0}}{t_{1}}\right)}$
		$\displaystyle=\liminf_{N\to\infty}\sum_{n\geq N}\left(\sum_{d\in\mathscr{D}}\left(B^{\frac{1}{t_{1}}}d(d-1)\right)^{-s}\right)^{n}\sum_{d_{n}=2}^{\lfloor A^{n/t_{0}}\rfloor}d_{n}^{-s\left(2-\frac{t_{0}}{t_{1}}\right)}$
		$\displaystyle\ll\liminf_{N\to\infty}\sum_{n\geq N}\left(\sum_{d\in\mathscr{D}}\left(B^{\frac{1}{t_{1}}}d(d-1)\right)^{-s}\right)^{n}\max\left\{1,A^{\frac{n}{t_{0}}\left(1-s\left(2-\frac{t_{0}}{t_{1}}\right)\right)}\right\}$
		$\displaystyle=\liminf_{N\to\infty}\sum_{n\geq N}\left(\sum_{d\in\mathscr{D}}\left(B^{\frac{1}{t_{1}}}d(d-1)\right)^{-s}\max\left\{1,A^{\frac{1}{t_{0}}\left(1-s\left(2-\frac{t_{0}}{t_{1}}\right)\right)}\right\}\right)^{n}.$

Therefore, if $S(A)$ is the solution of

\sum_{d=2}^{\infty}\left(B^{\frac{1}{t_{1}}}d(d-1)\right)^{-s}\max\left\{1,A^{\frac{1}{t_{0}}\left(1-s\left(2-\frac{t_{0}}{t_{1}}\right)\right)}\right\}=1,

we conclude that $\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(B)\leq S(A)$ . Let $A>1$ be such that $S(A)=s(A)$ , which occurs precisely when

\frac{1}{A^{s/t_{0}}}=\frac{\max\{1,A^{(1-s(2-t_{0}/t_{1}))/t_{0}}\}}{B^{s/t_{1}}},

or equivalently

-f_{t_{0}}(s)\log A=\max\left\{0,\frac{1-2s}{t_{0}}+\frac{s}{t_{1}}\right\}\log A-f_{t_{1}}(s)\log B,

which is

f_{t_{0},t_{1}}(s)\log B=f_{t_{0}}(s)\log A.

Using the exact same argument as in [2, Lemma 5.1], we conclude $1<A<B$ and, therefore, $\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(B)\leq s(A)=s_{0}(B)$ . ∎

5.2.2. Lower bound

We consider two cases:

\frac{s_{0}(B)}{t_{1}}-\frac{2s_{0}(B)-1}{t_{0}}\leq 0\quad\text{ and }\quad\frac{s_{0}(B)}{t_{1}}-\frac{2s_{0}(B)-1}{t_{0}}>0.

5.2.3. Lower bound: first case

Assume that $\frac{s_{0}(B)}{t_{1}}-\frac{2s_{0}(B)-1}{t_{0}}\leq 0$ .

Lemma 5.6.

If $\frac{s_{0}(B)}{t_{1}}-\frac{2s_{0}(B)-1}{t_{0}}\leq 0$ , then $s=s_{0}(B)$ is the unique solution of

\sum_{d=2}^{\infty}\left(\frac{1}{d(d-1)B^{1/t_{1}}}\right)^{s}=1.

Proof.

Since $f_{t_{0},t_{1}}(s)\leq\frac{s}{t_{1}}$ for all $s>0$ , we have

\sum_{d=2}^{\infty}\left(\frac{1}{d(d-1)B^{1/t_{1}}}\right)^{s}\leq\sum_{d=2}^{\infty}\frac{1}{d^{s}(d-1)^{s}B^{f_{t_{0},t_{1}}(s)}}.

Therefore, for all $\varepsilon>0$ , the inequality

\sum_{d=2}^{\infty}\left(\frac{1}{d(d-1)B^{1/t_{1}}}\right)^{s_{0}(B)+\varepsilon}<1,

holds and we conclude $s(B^{1/t_{1}})\leq s_{0}(B)$ .

We consider two further sub-cases. First, if $\frac{s_{0}(B)}{t_{1}}-\frac{2s_{0}(B)-1}{t_{0}}<0$ , then $f_{t_{0},t_{1}}(s)=\frac{s}{t_{1}}$ for every $s$ sufficiently close to $s_{0}$ and for any sufficiently small $\varepsilon>0$ we have

1<\sum_{d=2}^{\infty}\left(\frac{1}{d(d-1)B^{1/t_{1}}}\right)^{s_{0}(B)-\varepsilon}=\sum_{d=2}^{\infty}\frac{1}{d^{s}(d-1)^{s}B^{f_{t_{0},t_{1}}(s_{0}(B)-\varepsilon)}}.

This implies $s_{0}(B)\leq s(B^{1/t_{1}})+\varepsilon$ and, letting $\varepsilon\to 0$ , we get $s_{0}(B)\leq s(B^{1/t_{1}})$ . Second, assume that $\frac{s_{0}(B)}{t_{1}}-\frac{2s_{0}(B)-1}{t_{0}}=0$ . Let $\delta^{\prime}>0$ be arbitrary. Since $f_{t_{0},t_{1}}$ is continuous, for any $\varepsilon>0$ , there is $\delta^{\prime\prime}>0$ such that, for all $s\in\mathbb{R}$ ,

s_{0}(B)-\delta^{\prime\prime}<s<s_{0}(B)\quad\text{ implies }\quad f_{t_{0},t_{1}}(s)>\frac{s_{0}(B)}{t_{1}}-\varepsilon.

As a consequence, for $\delta=\min\{\delta^{\prime},\delta^{\prime\prime}\}$ and any $s$ such that $s_{0}(B)-\delta<s<s_{0}(B)$ , we have

1\leq\sum_{d=2}^{\infty}\frac{1}{d^{s}(d-1)^{s}B^{f_{t_{0},t_{1}}(s)}}\leq\sum_{d=2}^{\infty}\frac{1}{d^{s}(d-1)^{s}B^{\frac{s_{0}(B)}{t_{1}}-\varepsilon}}.

Hence, $s(B^{1/t_{1}})-\delta\geq s_{0}(B)$ . Letting $\delta^{\prime}\to 0$ , we have $\delta\to 0$ and, thus, $s_{0}(B)\leq s(B^{1/t_{1}})$ . ∎

Lemma 5.7.

If $\frac{s_{0}(B)}{t_{1}}-\frac{2s_{0}(B)-1}{t_{0}}\leq 0$ , then $\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(B)\geq s_{0}(B)$ .

Proof.

The definition of $\mathcal{E}_{\mathbf{t}}(B)$ yields

\left\{x\in(0,1]:d_{n+1}^{t_{1}}(x)\geq B^{n}\text{ for infinitely many }n\right\}\subseteq\mathcal{E}_{\mathbf{t}}(B).

Then, by Lemmas 4.2 and 5.6, we conclude $\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(B)\geq s(B^{1/t_{1}})=s_{0}(B)$ . ∎

5.2.4. Lower bound: second case

Assume that

(12)

\frac{s_{0}(B)}{t_{1}}-\frac{2s_{0}(B)-1}{t_{0}}>0.

In what follows, for all $x\in\mathbb{R}$ and all $r>0$ , we write $B(x;r)\colon=(x-r,x+r)$ .

Lemma 5.8 (Mass Distribution Principle).

Let $F\subseteq\mathbb{R}$ be a non-empty set and let $\mu$ be a finite measure satisfying $\mu(F)>0$ . If there are constants $c>0$ , $r_{0}>0$ , and $s\geq 0$ such that

\mu\left(B(x;r)\right)\leq cr^{s}\;\text{ for all }\;x\in F\;\text{ and }\;r\in(0,r_{0}),

then $\dim_{\operatorname{H}}F\geq s$ .

Proof.

see [5, Proposition 2.1]. ∎

Construction of the Cantor set.

By Lemma 5.3 and (12), we may take an $M\in\mathbb{N}_{\geq 3}$ such that $s\colon=s_{M}(B)$ is so close to $s_{0}(B)$ that

\frac{1}{2}<s<1\quad\text{ and }\quad f_{t_{0},t_{1}}(s)=\frac{sf_{t_{0}}(s)}{t_{1}\left(f_{t_{0}}(s)+\frac{s}{t_{1}}-\frac{2s-1}{t_{0}}\right)}.

Let $A$ be such that

f_{t_{0},t_{1}}(s)\log A=f_{t_{0}}(s)\log B;

as above, $1<A<B$ . Let $(\ell_{k})_{k\geq 1}$ be a sequence in $\mathbb{N}$ such that $\ell_{k}\gg e^{\ell_{1}+\ldots+\ell_{k-1}}$ for all $k\in\mathbb{N}$ . Write

\alpha_{0}\colon=A^{1/t_{0}}\quad\text{ and }\quad\alpha_{1}\colon=\left(\frac{B}{A}\right)^{1/t_{1}}

and let $N\in\mathbb{N}$ be such that $2<\alpha_{0}^{N}$ . Define the sequence $(n_{j})_{j\geq 1}$ by

n_{1}=\ell_{1}N+1\quad\text{ and }\quad n_{k+1}-n_{k}=\ell_{k+1}N+2\;\text{ for all }k\in\mathbb{N}.

We can take the sequence $(\ell_{k})_{k\geq 1}$ so sparse that $(n_{k})_{k\geq 1}$ satisfies

(13)

\frac{(2k-1)\log\alpha_{0}}{\log\alpha_{1}}<n_{1}+\ldots+n_{k}\quad\text{ for all }k\in\mathbb{N}_{\geq 2}.

Let $E$ be the set formed by the real numbers $x=\langle d_{1},d_{2},d_{3},\ldots\rangle$ satisfying the following conditions:

For every $k\in\mathbb{N}$ , we have

\alpha_{0}^{n_{k}}A^{n_{k}/t_{0}}\leq d_{n_{k}}\leq 2\alpha_{0}^{n_{k}}\;\text{ and }\;\alpha_{1}^{n_{k}}\leq d_{n_{k}+1}\leq 2\alpha_{1}^{n_{k}}.

ii.

If $n\in\mathbb{N}\setminus\{n_{k}:k\in\mathbb{N}\}$ , then $2\leq d_{n}\leq M$ .

Let us exhibit the Cantor structure of $E$ . Define $D\colon=\Lambda^{-1}[E]$ (see Section 2) and for all $n\in\mathbb{N}$ define

D_{n}\colon=\left\{(d_{1},\ldots,d_{n})\in\mathscr{D}^{n}:(d_{j})_{j\geq 1}\in D\right\}.

For each $\mathbf{d}\in D_{n}$ define the compact interval

J_{n}(\mathbf{d})\colon=\bigcup_{\begin{subarray}{c}d_{n+1}\in\mathscr{D}\\ \mathbf{d}d\in D_{n+1}\end{subarray}}\overline{I}_{n+1}(\mathbf{d}d_{n+1}).

We refer to the sets of the form $J_{n}(\mathbf{d})$ as fundamental intervals of order $n$ . Clearly,

E=\bigcap_{n\in\mathbb{N}}\bigcup_{\mathbf{d}\in D_{n}}J_{n}(\mathbf{d}).

A probability measure. Observe that for every $\mathbf{d}=(d_{j})_{j\geq 1}\in D$ there is a finite collection of words $\mathbf{w}_{1}^{j}$ , $\ldots$ , $\mathbf{w}_{\ell_{j}}^{j}$ in $\{2,\ldots,M\}^{N}$ for $j\in\mathbb{N}$ such that, writing $\mathbf{W}_{j}\colon=\mathbf{w}_{1}^{j}\cdots\mathbf{w}_{\ell_{j}}^{j}$ ,

	$\displaystyle\mathbf{d}$	$\displaystyle=\mathbf{w}_{1}^{1}\mathbf{w}_{2}^{1}\ldots\mathbf{w}_{\ell_{1}}^{1}d_{n_{1}}d_{n_{1}+1}\mathbf{w}_{1}^{2}\mathbf{w}_{2}^{2}\ldots\mathbf{w}_{\ell_{2}}^{2}d_{n_{2}}d_{n_{2}+1}\cdots\mathbf{w}_{1}^{k}\mathbf{w}_{2}^{k}\ldots\mathbf{w}_{\ell_{k}}^{k}d_{n_{k}}d_{n_{k}+1}\cdots$
		$\displaystyle=\mathbf{W}_{1}d_{n_{1}}d_{n_{1}+1}\mathbf{W}_{2}d_{n_{2}}d_{n_{2}+1}\ldots\mathbf{W}_{k}d_{n_{k}}d_{n_{k}+1}\ldots.$

Take $n\in\mathbb{N}$ and $\mathbf{d}\in D_{n}$ . First, assume that $n\leq n_{1}+1$ .

i.

If $n=N\ell$ for some $\ell\in\{1,\ldots,\ell_{1}\}$ , define

$\mu\left(J_{\ell N}(\mathbf{d})\right)\colon=\frac{|I_{N\ell}(\mathbf{d})|^{s}}{\alpha_{0}^{\ell Ns}}.$

ii.

If there is some $\ell\in\{1,\ldots,\ell_{1}\}$ such that $(\ell-1)N+1\leq n\leq\ell N-1$ , put

\mu\left(J_{n}(\mathbf{d})\right)\colon=\sum_{\begin{subarray}{c}\mathbf{b}\in\mathscr{D}^{\ell N-n}\\ \mathbf{d}\mathbf{b}\in D_{\ell N}\end{subarray}}\mu\left(J_{\ell N}(\mathbf{d}\mathbf{b})\right).

iii.

If $n=n_{1}$ , then

\mu\left(J_{n_{1}}(\mathbf{d})\right)\colon=\frac{\mu\left(J_{n_{1}-1}(d_{1},\ldots,d_{n_{1}-1})\right)}{\lfloor 2\alpha_{0}^{n_{1}}\rfloor-\lceil\alpha_{0}^{n_{1}}\rceil}.

iv.

If $n=n_{1}+1$ , define

\mu\left(J_{n_{1}+1}(\mathbf{d})\right)\colon=\frac{\mu\left(J_{n_{1}-1}(d_{1},\ldots,d_{n_{1}-1})\right)}{\left(\lfloor 2\alpha_{1}^{n_{1}}\rfloor-\lceil\alpha_{1}^{n-1}\rceil\right)\left(\lfloor 2\alpha_{0}^{n_{1}}\rfloor-\lceil\alpha_{0}^{n-1}\rceil\right)}

Assume that we have defined $\mu$ on the fundamental sets of level $1,\ldots,n_{k}+1$ for some $k\in\mathbb{N}$ . Suppose that $n\in\{n_{k}+2,\ldots,n_{k+1}+1\}$ .

If $n=n_{k}+1+N\ell$ for some $\ell\in\{1,\ldots,\ell_{k+1}\}$ , write

\mu\left(J_{n_{k}+1+N\ell}(\mathbf{d})\right)\colon=\mu\left(J_{n_{k}+1}(d_{1},\ldots,d_{n_{k}+1}))\right)\frac{\left|I_{N\ell}(\mathbf{w}_{1}^{k+1}\cdots\mathbf{w}_{\ell}^{k+1})\right|^{s}}{\alpha_{0}^{sN\ell}}.

ii.

When $n_{k}+1+(\ell-1)N<n<n_{k}j+1+\ell N$ for some $\ell\in\{1,\ldots,\ell_{k+1}\}$ , define

$\mu\left(J_{n}(\mathbf{d})\right)\colon=\sum_{\mathbf{b}}\mu\left(J_{n_{k}+1+\ell N}(\mathbf{d}\mathbf{b})\right),$

where the sum runs along those words $\mathbf{b}$ such that $\mathbf{d}\mathbf{b}\in D_{n_{k}+1+\ell N}$ .

iii.

When $n=n_{k+1}$ , put

\mu\left(J_{n_{k+1}}(\mathbf{d})\right)\colon=\frac{\mu\left(J_{n_{k+1}-1}(\mathbf{d})\right)}{\lfloor 2\alpha_{0}^{n_{k+1}}\rfloor-\lceil\alpha_{0}^{n_{k+1}}\rceil}.

iv.

Whenever $n=n_{k+1}+1$ , put

\mu\left(J_{n_{k+1}+1}(\mathbf{d})\right)\colon=\frac{\mu\left(J_{n_{k+1}-1}(d_{1},\ldots,d_{n_{k+1}-1})\right)}{\left(\lfloor 2\alpha_{1}^{n_{k+1}}\rfloor-\lceil\alpha_{1}^{n_{k+1}}\rceil\right)\left(\lfloor 2\alpha_{0}^{n_{k+1}}\rfloor-\lceil\alpha_{0}^{n_{k+1}}\rceil\right)}

The procedure defines a probability measure on the fundamental sets of a given level. The choice of $A$ and $B$ and the definition of $\mu$ ensure the consistency conditions. Hence, by the Daniell-Kolmogorov Consistency Theorem [10, Theorem 8.23], the function $\mu$ is indeed a probability measure on $E$ .

Gap estimates.

For $n\in\mathbb{N}$ and $\mathbf{d}\in D_{n}$ , let $G_{n}(\mathbf{d})$ be the distance between $J_{n}(\mathbf{d})$ and the fundamental interval of level $n$ closest to it; that is,

G_{n}(\mathbf{d})\colon=\inf\left\{d\left(J_{n}(\mathbf{d}),J_{n}(\mathbf{e})\right):\mathbf{e}\in D_{n},\,\mathbf{e}\neq\mathbf{d}\right\}.

Lemma 5.9.

For any $n\in\mathbb{N}$ and any $\mathbf{d}\in D_{n}$ , we have

G_{n}(\mathbf{d})\geq\frac{1}{M}|I_{n}(\mathbf{d})|.

Proof.

The proof is by induction on $n$ . Pick $d\in\{2,\ldots,M\}$ . If $2\leq d\leq M-1$ , then

\inf J_{1}(d)-\sup J_{1}(d+1)=\inf J_{1}(d)-\sup I_{1}(d+1)=\frac{\left|I_{1}(d)\right|}{M}.

If $3\leq d\leq M$ , then

\sup I_{1}(d)-\inf J_{1}(d-1)=\frac{1}{M}\left|I_{1}(d-1)\right|>\frac{1}{M}\left|I_{1}(d)\right|.

This shows the result for $n=1$ . Assume that the lemma holds for $n=\tilde{n}-1\in\mathbb{N}$ . Suppose that either

(14)

1<\tilde{n}\leq N\ell_{1}-1\;\text{ or }\;n_{k}+1\leq\tilde{n}\leq n_{k}+1+(\ell_{k}N-1)\;\text{ for some }k\in\mathbb{N}

and consider $\mathbf{d}=(d_{1},\ldots,d_{\tilde{n}})\in D_{\tilde{n}}$ . When $2\leq d_{\tilde{n}}\leq M-1$ , we have

	$\displaystyle\inf J_{\tilde{n}}(\mathbf{d})-\sup J_{\tilde{n}}\left(d_{1},\ldots,d_{\tilde{n}}+1\right)$	$\displaystyle=\inf J_{\tilde{n}}(\mathbf{d})-\sup I_{\tilde{n}}\left(d_{1},\ldots,d_{\tilde{n}}+1\right)$
		$\displaystyle=\inf J_{\tilde{n}}(\mathbf{d})-\inf I_{\tilde{n}}\left(\mathbf{d}\right)$
		$\displaystyle=\frac{1}{M}\left\|I_{\tilde{n}}\left(\mathbf{d}\right)\right\|.$

If $3\leq d_{\tilde{n}}\leq M$ , then

	$\displaystyle\inf J_{\tilde{n}}(d_{1},\ldots,d_{\tilde{n}}-1)-\sup J_{\tilde{n}}(\mathbf{d})$	$\displaystyle=\inf J_{\tilde{n}}(d_{1},\ldots,d_{\tilde{n}}-1)-\sup I_{\tilde{n}}(\mathbf{d})$
		$\displaystyle=\inf J_{\tilde{n}}(d_{1},\ldots,d_{\tilde{n}}-1)-\inf I_{\tilde{n}}(d_{1},\ldots,d_{\tilde{n}}-1)$
		$\displaystyle=\frac{1}{M}\left\|I_{\tilde{n}}(d_{1},\ldots,d_{\tilde{n}}-1)\right\|$
		$\displaystyle>\frac{1}{M}\left\|I_{\tilde{n}}(\mathbf{d})\right\|.$

We conclude that, provided $3\leq d_{\tilde{n}}\leq M-1$ or $\mathbf{d}=(2,2,\ldots,2)$ , we have $G_{\tilde{n}}(\mathbf{d})>M^{-1}|I_{\tilde{n}}(\mathbf{d})|$ . Assume that $d_{\tilde{n}}=2$ and let $j\in\{1,\ldots,\tilde{n}-1\}$ be the largest index such that $d_{j}\geq 3$ . Then, the neighbor to the right of $J_{\tilde{n}}(\mathbf{d})$ is $J_{\tilde{n}}(d_{1},\ldots,d_{j-1},M,\ldots,M)$ and, using the induction hypothesis on the second inequality, we have

	$\displaystyle\inf J_{\tilde{n}}(d_{1},\ldots,d_{j-1},M,\ldots,M)-\sup J_{\tilde{n}}(\mathbf{d})$	$\displaystyle>\inf J_{j}(d_{1},\ldots,d_{j-1})-\sup J_{j}(d_{1},\ldots,d_{j})$
		$\displaystyle>\frac{1}{M}\left\|I_{j}(d_{1},\ldots,d_{j})\right\|$
		$\displaystyle>\frac{1}{M}\left\|I_{\tilde{n}}(\mathbf{d})\right\|.$

A similar argument holds when $d_{\tilde{n}}=M$ . This proves the result for $\tilde{n}$ assuming (14).

Suppose that $\tilde{n}=n_{k}-1$ for some $k\in\mathbb{N}$ . If $2\leq d_{\tilde{n}}\leq M-1$ , then

	$\displaystyle\inf J_{\tilde{n}}(\mathbf{d})$	$\displaystyle-\sup J_{\tilde{n}}(d_{1},\ldots,d_{\tilde{n}}+1)=$
		$\displaystyle=\left(\inf J_{\tilde{n}}(\mathbf{d})-\inf I_{\tilde{n}}(\mathbf{d})\right)+\left(\sup I_{\tilde{n}}(d_{1},\ldots,d_{\tilde{n}}+1)-\sup J_{\tilde{n}}(d_{1},\ldots,d_{\tilde{n}}+1)\right)$
		$\displaystyle=\frac{1}{\lceil 2\alpha_{0}^{n_{j}}\rceil}\left\|I_{\tilde{n}}(\mathbf{d})\right\|+\left(1-\frac{1}{\lfloor\alpha_{0}^{n_{j}}\rfloor-1}\right)\left\|I_{\tilde{n}}(d_{1},\ldots,d_{\tilde{n}}+1)\right\|$
		$\displaystyle>\left(1-\frac{1}{\lfloor\alpha_{0}^{n_{j}}\rfloor-1}\right)\frac{d_{\tilde{n}}-1}{d_{\tilde{n}}+1}\left\|I_{\tilde{n}}(\mathbf{d})\right\|$
		$\displaystyle=\left(1-\frac{1}{\lfloor\alpha_{0}^{n_{j}}\rfloor-1}\right)\left(1-\frac{2}{d_{\tilde{n}}+1}\right)\left\|I_{\tilde{n}}(\mathbf{d})\right\|$
		$\displaystyle>\left(1-\frac{1}{\lfloor\alpha_{0}^{n_{j}}\rfloor-1}\right)\left(1-\frac{2}{M}\right)\left\|I_{\tilde{n}}(\mathbf{d})\right\|>\frac{1}{M}\left\|I_{\tilde{n}}(\mathbf{d})\right\|.$

The last inequality follows from $M>4$ . A similar argument shows that, when $3\leq d_{\tilde{n}}\leq M$ ,

\inf J_{\tilde{n}}(d_{1},\ldots,d_{\tilde{n}}-1)-\sup J_{\tilde{n}}(\mathbf{d})>\frac{1}{M}\left|I_{\tilde{n}}(d_{1},\ldots,d_{\tilde{n}}-1)\right|\\ >\frac{1}{M}\left|I_{\tilde{n}}(\mathbf{d})\right|.

A slight modification of this argument yields the result for $\tilde{n}=n_{k}$ . ∎

Length estimates.

For any $n\in\mathbb{N}$ , let $c_{n}$ and $C_{n}$ be

c_{n}\colon=\min\{d_{n}:(d_{j})_{j\geq 1}\in D\}\quad\text{ and }\quad C_{n}\colon=\max\{d_{n}:(d_{j})_{j\geq 1}\in D\}.

Hence, when $\mathbf{d}=(d_{1},\ldots,d_{n})\in D_{n}$ , we have

|J_{n}(\mathbf{d})|=\left(\frac{1}{c_{n+1}-1}-\frac{1}{C_{n+1}}\right)\prod_{j=1}^{n}\frac{1}{d_{j}(d_{j}-1)}=\left(\frac{1}{c_{n+1}-1}-\frac{1}{C_{n+1}}\right)|I_{n}(\mathbf{d})|

In particular, when $n\neq n_{k}-1$ and $n\neq n_{k}$ for all $k\in\mathbb{N}$ ,

|J_{n}(\mathbf{d})|=\left(1-\frac{1}{M}\right)\prod_{j=1}^{n}\frac{1}{d_{j}(d_{j}-1)}=\left(1-\frac{1}{M}\right)|I_{n}(\mathbf{d})|,

which gives

(15)

\frac{1}{2}|I_{n}(\mathbf{d})|\leq|J_{n}(\mathbf{d})|\leq\left(1-\frac{1}{M}\right)|I_{n}(\mathbf{d})|.

For $n=n_{k}-1$ , we have

|J_{n_{k}-1}(\mathbf{d})|=\left(\frac{1}{\lceil\alpha_{0}^{n_{k}}\rceil-1}-\frac{1}{\lfloor 2\alpha_{0}^{n_{k}}\rfloor}\right)\prod_{j=1}^{n_{k}-1}\frac{1}{d_{j}(d_{j}-1)}=\left(\frac{1}{\lceil\alpha_{0}^{n_{k}}\rceil-1}-\frac{1}{\lfloor 2\alpha_{0}^{n_{k}}\rfloor}\right)\left|I_{n_{k}-1}(\mathbf{d})\right|,

(16)

\frac{1}{2\alpha_{0}^{n_{k}}}|I_{n_{k}-1}(\mathbf{d})|\leq|J_{n_{k}-1}(\mathbf{d})|\leq\frac{1}{\alpha_{0}^{n_{k}}}|I_{n_{k}-1}(\mathbf{d})|.

We can replace the constant $1$ in the upper bound with an arbitrary constant strictly larger than $\frac{1}{2}$ , but we would have to consider larger values of $N$ .

Similarly, for $n=n_{k}$ ,

|J_{n_{k}}(\mathbf{d})|=\left(\frac{1}{\lceil\alpha_{1}^{n_{k}}\rceil-1}-\frac{1}{\lfloor 2\alpha_{1}^{n_{k}}\rfloor}\right)\prod_{j=1}^{n_{k}}\frac{1}{d_{j}(d_{j}-1)}=\left(\frac{1}{\lceil\alpha_{1}^{n_{k}}\rceil-1}-\frac{1}{\lfloor 2\alpha_{1}^{n_{k}}\rfloor}\right)\left|I_{n_{k}}(\mathbf{d})\right|

\frac{1}{2\alpha_{1}^{n_{k}}}|I_{n_{k}}(\mathbf{d})|\leq|J_{n_{k}}(\mathbf{d})|\leq\frac{1}{\alpha_{1}^{n_{k}}}|I_{n_{k}}(\mathbf{d})|.

Again, we can replace the constant $1$ in the upper bound with an arbitrary constant strictly larger than $\frac{1}{2}$ at the expense of a larger $N$ .

Lemma 5.10.

Let $k\in\mathbb{N}$ be arbitrary.

For any $\mathbf{d}=(d_{1},\ldots,d_{n_{k}})\in D_{n_{k}}$ , we have

\frac{1}{8\alpha_{0}^{n_{k}}\alpha_{1}^{n_{k}}}\left|J_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right|<\left|J_{n_{k}}(\mathbf{d})\right|.

For any $\mathbf{d}=(d_{1},\ldots,d_{n_{k}+1})\in D_{n_{k}+1}$ , we have

\frac{1}{2^{6}\alpha_{0}^{n_{k}}\alpha_{1}^{2n_{k}}}\left|J_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right|<\left|J_{n_{k}+1}(\mathbf{d})\right|.

Proof.

By our previous discussion,

$\displaystyle\left\|J_{n_{k}}(\mathbf{d})\right\|$	$\displaystyle=\left(\frac{1}{\lceil\alpha_{1}^{n_{k}}\rceil-1}-\frac{1}{\lfloor 2\alpha_{1}^{n_{k}}\rfloor}\right)\left\|I_{n_{k}}(\mathbf{d})\right\|$
	$\displaystyle=\left(\frac{1}{\lceil\alpha_{1}^{n_{k}}\rceil-1}-\frac{1}{\lfloor 2\alpha_{1}^{n_{k}}\rfloor}\right)\frac{\left\|I_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right\|}{d_{n_{k}}(d_{n_{k}}-1)}$
	$\displaystyle>\frac{1}{8\alpha_{1}^{n_{k}}\alpha_{0}^{n_{k}}\alpha_{0}^{n_{k}}}\left\|I_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right\|$
	$\displaystyle>\frac{1}{8\alpha_{1}^{n_{k}}\alpha_{0}^{n_{k}}}\left\|J_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right\|$	$\displaystyle\text{(by \eqref{Ec:LEST:02})}.$

Similarly,

$\displaystyle\left\|J_{n_{k}+1}(\mathbf{d})\right\|$	$\displaystyle=\left(1-\frac{1}{M}\right)\left\|I_{n_{k}+1}(\mathbf{d})\right\|$
	$\displaystyle=\left(1-\frac{1}{M}\right)\frac{\left\|I_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right\|}{d_{n_{k}}(d_{n_{k}}-1)d_{n_{k}+1}(d_{n_{k}+1}-1)}$
	$\displaystyle>\frac{1}{2^{6}}\frac{\left\|I_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right\|}{\alpha_{0}^{2n_{k}}\alpha_{1}^{2n_{k}}}$
	$\displaystyle>\frac{1}{2^{6}}\frac{\left\|J_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right\|}{\alpha_{0}^{n_{k}}\alpha_{1}^{2n_{k}}}$	$\displaystyle\text{(by \eqref{Ec:LEST:02})}.$

∎

Measure of the fundamental intervals.

We now compute upper estimates of $\mu(J_{n}(\mathbf{d}))$ for all $n\in\mathbb{N}$ and $\mathbf{d}\in D_{n}$ .

Lemma 5.11.

The following statements hold:

\frac{1}{\alpha_{0}}=\left(\frac{1}{\alpha_{0}\alpha_{1}}\right)^{s}\quad\text{ and }\quad\frac{1}{\alpha_{0}\alpha_{1}}\leq\left(\frac{1}{\alpha_{0}\alpha_{1}^{2}}\right)^{s}.

Proof.

See [2, Lemma 6.6]. ∎

For each $k\in\mathbb{N}$ , define the number $\gamma_{k}>0$ by

\gamma_{k}^{-1}\colon=\left(1-\frac{2}{\alpha_{0}^{n_{k}}}\right)\left(1-\frac{2}{\alpha_{1}^{n_{k}}}\right).

We will use the following obvious facts:

1.

Since $\alpha_{0}>1$ and $\alpha_{1}>1$ , the series $\sum_{k}\alpha_{0}^{-n_{k}}$ and $\sum_{k}\alpha_{0}^{-n_{k}}$ are convergent and, thus, so is the product $C^{\prime}\colon=\prod_{k\in\mathbb{N}}\gamma_{k}$ (see [17, Theorem 7.32]).

For each $k\in\mathbb{N}$ ,

\frac{1}{\left(\lfloor 2\alpha_{1}^{n_{k-1}}\rfloor-\lceil\alpha_{1}^{n_{k-1}}\rceil\right)\left(\lfloor 2\alpha_{0}^{n_{k-1}}\rfloor-\lceil\alpha_{0}^{n_{k-1}}\rceil\right)}<\frac{\gamma_{k}}{\alpha_{0}^{n_{k}}\alpha_{0}^{n_{k}}}.

Lemma 5.12.

For every $k\in\mathbb{N}_{\geq 2}$ and every $\mathbf{d}\in D_{n_{k}-1}$ , we have

\mu\left(J_{n_{k}-1}(\mathbf{d})\right)<\gamma_{k-1}\left(\frac{1}{\alpha_{0}^{N\ell_{k}}\alpha_{1}^{n_{k-1}}}\right)^{s}\left(\frac{\left|I_{N\ell_{k}}(\mathbf{W}_{k})\right|}{\alpha_{0}^{n_{k-1}}\alpha_{1}^{n_{k-1}}}\right)^{s}\mu\left(J_{n_{k-1}-1}(d_{1},\ldots,d_{n_{k-1}-1})\right).

Proof.

Take $k$ and $\mathbf{d}$ as in the statement. The lemma follows from the definition of $\mu$ and $n_{k}-1=n_{k-1}+N\ell_{k}+1$ :

	$\displaystyle\mu\left(J_{n_{k}-1}(\mathbf{d})\right)$	$\displaystyle=\frac{\left\|I_{N\ell_{k}}\left(\mathbf{W}_{k}\right)\right\|^{s}}{\alpha_{0}^{N\ell_{k}s}}\mu\left(J_{n_{k-1}+1}(d_{1},\ldots,d_{n_{k-1}+1})\right)$
		$\displaystyle=\frac{\left\|I_{N\ell_{k}}\left(\mathbf{W}_{k}\right)\right\|^{s}}{\alpha_{0}^{N\ell_{k}s}}\frac{\mu\left(J_{n_{k-1}-1}(d_{1},\ldots,d_{n_{k-1}-1})\right)}{\left(\lfloor 2\alpha_{1}^{n_{k-1}}\rfloor-\lceil\alpha_{1}^{n_{k-1}}\rceil\right)\left(\lfloor 2\alpha_{0}^{n_{k-1}}\rfloor-\lceil\alpha_{0}^{n_{k-1}}\rceil\right)}$
		$\displaystyle<\frac{\gamma_{k-1}}{\alpha_{0}^{N\ell_{k}s}}\frac{\left\|I_{N\ell_{k}}\left(\mathbf{W}_{k}\right)\right\|^{s}}{\alpha_{0}^{n_{k-1}}\alpha_{1}^{n_{k-1}}}\mu\left(J_{n_{k-1}-1}(d_{1},\ldots,d_{n_{k-1}-1})\right)$
		$\displaystyle<\frac{\gamma_{k-1}}{\alpha_{0}^{N\ell_{k}s}}\frac{\left\|I_{N\ell_{k}}\left(\mathbf{W}_{k}\right)\right\|^{s}}{\alpha_{0}^{sn_{k-1}}\alpha_{1}^{2sn_{k-1}}}\mu\left(J_{n_{k-1}-1}(d_{1},\ldots,d_{n_{k-1}-1})\right).$

∎

Lemma 5.13.

There exists a constant $C=C(B,M,N,s,\mathbf{t})>0$ such that

\mu\left(J_{n}(\mathbf{d})\right)\leq C|J_{n}(\mathbf{d})|^{s}\quad\text{ for all }n\in\mathbb{N}\text{ and all }\mathbf{d}\in D_{n}.

Proof.

Pick $n\in\mathbb{N}$ . First, we further assume that $1\leq n\leq n_{1}+1$ .

Suppose that $n=\ell N$ for some $\ell\in\{1,\ldots,\ell_{1}\}$ . When $1\leq\ell\leq\ell_{1}-1$ , using (15) we get

\mu\left(J_{\ell N}(\mathbf{d})\right)=\frac{1}{\alpha_{0}^{N\ell s}}\,|I_{\ell N}(\mathbf{d})|^{s}=\frac{2^{s}}{\alpha_{0}^{N\ell s}}\,\frac{|I_{\ell N}(\mathbf{d})|^{s}}{2^{s}}<\frac{2^{s}}{\alpha_{0}^{N\ell s}}\,|J_{\ell N}(\mathbf{d})|^{s}<|J_{\ell N}(\mathbf{d})|^{s}.

When $\ell=\ell_{1}$ , we have $n=\ell_{1}N=n_{1}-1$ and, by (16),

\mu(J_{\ell_{1}N}(\mathbf{d}))=2^{s}\alpha_{0}^{s}\,\frac{|I_{n_{1}-1}(\mathbf{d})|^{s}}{2^{s}\alpha_{0}^{sn_{1}}}<2^{s}\alpha_{0}^{s}\,|J_{\ell_{1}N}(\mathbf{d})|^{s}<2\alpha_{0}\,|J_{\ell_{1}N}(\mathbf{d})|^{s}.

ii.

If $(\ell-1)N+1\leq n\leq\ell N-1$ for some $\ell\in\{1,\ldots,\ell_{1}\}$ , then

	$\displaystyle\mu\left(J_{n}(\mathbf{d})\right)$	$\displaystyle=\frac{1}{\alpha_{0}^{N\ell s}}\,\sum_{\begin{subarray}{c}\mathbf{b}\in\mathscr{D}^{\ell N-n}\\ \mathbf{d}\mathbf{b}\in D_{\ell N}\end{subarray}}\left\|I_{\ell N}(\mathbf{d}\mathbf{b})\right\|^{s}$
		$\displaystyle=\frac{\|I_{n}(\mathbf{d})\|^{s}}{\alpha_{0}^{ns}}\,\sum_{\begin{subarray}{c}\mathbf{b}\in\mathscr{D}^{\ell N-n}\\ \mathbf{d}\mathbf{b}\in D_{\ell N}\end{subarray}}\frac{\left\|I_{\ell N-n}(\mathbf{b})\right\|^{s}}{\alpha_{0}^{(N\ell-n)s}}$
		$\displaystyle=\frac{\|I_{n}(\mathbf{d})\|^{s}}{\alpha_{0}^{ns}}\,\left(\sum_{d=2}^{M}\frac{1}{d^{s}(d-1)^{s}\alpha_{0}^{s}}\right)^{\ell N-n}$
		$\displaystyle=\frac{\|I_{n}(\mathbf{d})\|^{s}}{\alpha_{0}^{ns}}\,=\frac{2^{s}}{\alpha_{0}^{ns}}\,\frac{\|I_{n}(\mathbf{d})\|^{s}}{2^{s}}<\frac{2^{s}}{\alpha_{0}^{ns}}\,\|J_{n}(\mathbf{d})\|^{s}<\|J_{n}(\mathbf{d})\|^{s}.$

iii.

If $n=n_{1}$ , then

	$\displaystyle\mu\left(J_{n_{1}}(\mathbf{d})\right)$	$\displaystyle=\frac{\mu\left(J_{n_{1}-1}(d_{1},\ldots,d_{n_{1}-1})\right)}{\lfloor 2\alpha_{0}^{n_{1}}\rfloor-\lceil\alpha_{0}^{n_{1}}\rceil}$
		$\displaystyle<\frac{2}{\alpha_{0}^{n_{1}}}\mu\left(J_{n_{1}-1}(d_{1},\ldots,d_{n_{1}-1})\right)$
		$\displaystyle<\frac{2}{\alpha_{0}^{n_{1}}}\,2\alpha_{0}\|J_{n_{1}-1}(d_{1},\ldots,d_{n_{1}-1})\|^{s}$
		$\displaystyle=2^{2}\,\alpha_{0}8\frac{\|J_{n_{1}-1}(d_{1},\ldots,d_{n_{1}-1})\|^{s}}{8\left(\alpha_{0}^{n_{1}}\alpha_{1}^{n_{1}}\right)^{s}}<2^{5}\alpha_{0}\|J_{n_{1}}(\mathbf{d})\|^{s}.$

We used Lemma 5.11 in the last inequality.

iv.

When $n=n_{1}+1$ , we have

	$\displaystyle\mu\left(J_{n_{1}+1}(\mathbf{d})\right)$	$\displaystyle<2^{2}\frac{\mu\left(J_{n_{1}+1}(d_{1},\ldots,d_{n_{k}-1})\right)}{\alpha_{0}^{n_{1}}\alpha_{1}^{n_{1}}}$
		$\displaystyle\leq 2^{2}\frac{\mu\left(J_{n_{1}+1}(d_{1},\ldots,d_{n_{k}-1})\right)}{\alpha_{0}^{n_{1}s}\alpha_{1}^{2n_{1}s}}$
		$\displaystyle<2^{3}\alpha_{0}\,\frac{\left\|J_{n_{1}+1}(d_{1},\ldots,d_{n_{k}-1})\right\|^{s}}{\alpha_{0}^{n_{1}s}\alpha_{1}^{2n_{1}s}}<2^{9}\alpha_{0}\,\left\|J_{n_{1}+1}(\mathbf{d})\right\|^{s}.$

Assume now that $n_{1}+1<n$ and pick $k\in\mathbb{N}_{\geq 2}$ such that $n_{k}+1<n<n_{k+1}+1$ .

If $n=n_{k}-1$ , we apply Lemma 5.12 repeatedly to obtain

$\displaystyle\mu\left(J_{n_{k}-1}(\mathbf{d})\right)$	$\displaystyle<\gamma_{k-1}\left(\frac{1}{\alpha_{0}^{N\ell_{k}}\alpha_{1}^{n_{k-1}}}\right)^{s}\left(\frac{\left\|I_{N\ell_{k}}(\mathbf{W}_{k})\right\|}{\alpha_{0}^{n_{k-1}}\alpha_{1}^{n_{k-1}}}\right)^{s}\mu\left(J_{n_{k-1}-1}(d_{1},\ldots,d_{n_{k-1}-1})\right)$
	$\displaystyle<\gamma_{1}\cdots\gamma_{k-1}\left(\frac{1}{\alpha_{0}^{N\left(\ell_{1}+\ldots+\ell_{k}\right)}\alpha_{1}^{n_{1}+\ldots+n_{k-1}}}\right)^{s}\left(\frac{\left\|I_{N\ell_{1}}(\mathbf{W}_{1})\right\|\cdots\left\|I_{N\ell_{k}}(\mathbf{W}_{k})\right\|}{\alpha_{0}^{n_{1}}\alpha_{1}^{n_{1}}\cdots\alpha_{0}^{n_{k-1}}\alpha_{1}^{n_{k-1}}}\right)^{s}$
	$\displaystyle<C^{\prime}\left(\frac{1}{\alpha_{0}^{N\left(\ell_{1}+\ldots+\ell_{k}\right)}\alpha_{1}^{n_{1}+\ldots+n_{k-1}}}\right)^{s}\left\|I_{n_{k}-1}(\mathbf{d})\right\|^{s}$
	$\displaystyle\leq C^{\prime}\left(\frac{2\alpha_{0}^{n_{k}}}{\alpha_{0}^{N\left(\ell_{1}+\ldots+\ell_{k}\right)}\alpha_{1}^{n_{1}+\ldots+n_{k-1}}}\right)^{s}\left\|J_{n_{k}-1}(\mathbf{d})\right\|^{s}$
	$\displaystyle<2C^{\prime}\left(\frac{1}{\alpha_{0}^{N\left(\ell_{1}+\ldots+\ell_{k}\right)-n_{k}}\alpha_{1}^{n_{1}+\ldots+n_{k-1}}}\right)^{s}\left\|J_{n_{k}-1}(\mathbf{d})\right\|^{s}$
	$\displaystyle=2C^{\prime}\left(\frac{\alpha_{0}^{2k-1}}{\alpha_{1}^{n_{1}+\ldots+n_{k-1}}}\right)^{s}\left\|J_{n_{k}-1}(\mathbf{d})\right\|^{s}$
	$\displaystyle<C^{\prime}\left\|J_{n_{k}-1}(\mathbf{d})\right\|^{s}$	$\displaystyle\text{ (by \eqref{EQ:LWEtB:00}) }.$

ii.

Suppose we had $n=n_{k}$ , then

$\displaystyle\mu\left(J_{n_{k}}(\mathbf{d})\right)$	$\displaystyle<\frac{2}{\alpha_{0}^{n_{k}}}\mu\left(J_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right)$
	$\displaystyle<\frac{2}{\alpha_{0}^{n_{k}}}C^{\prime}\left\|J_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right\|^{s}$
	$\displaystyle<2C^{\prime}\frac{\left\|J_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right\|^{s}}{\alpha_{0}^{2n_{k}s}\alpha_{1}^{2n_{k}s}}$	(by Lemma 5.11)
	$\displaystyle<2^{4}C^{\prime}\left\|J_{n_{k}}(\mathbf{d})\right\|^{s}$	(by Lemma 5.10).

iii.

If $n=n_{k}+1$ , then

$\displaystyle\mu\left(J_{n_{k}+1}(\mathbf{d})\right)$	$\displaystyle<\frac{2^{2}}{\alpha_{0}^{n_{k}}\alpha_{1}^{n_{k}}}\mu\left(J_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right)$
	$\displaystyle<\frac{2^{2}}{\alpha_{0}^{n_{k}}\alpha_{1}^{n_{k}}}C^{\prime}\left\|J_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right\|^{s}$
	$\displaystyle<2^{2}C^{\prime}\left(\frac{\left\|J_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right\|}{\alpha_{0}^{n_{k}}\alpha_{1}^{2n_{k}}}\right)^{s}$
	$\displaystyle<2^{2}\,2^{6s}\,C^{\prime}\left\|J_{n_{k}+1}(\mathbf{d})\right\|^{s}$	(by Lemma 5.11)
	$\displaystyle<2^{8}\,C^{\prime}\left\|J_{n_{k}+1}(\mathbf{d})\right\|^{s}.$

iv.

If $n=n_{k}+1+\ell N$ for some $1\leq\ell<\ell_{k+1}$ , then

	$\displaystyle\mu\left(J_{n_{k}+1+\ell N}(\mathbf{d})\right)$	$\displaystyle=\frac{\left\|I_{N\ell}(\mathbf{w}_{1}^{k+1}\cdots\mathbf{w}_{\ell}^{k+1})\right\|^{s}}{\alpha_{0}^{N\ell s}}\mu\left(J_{n_{k}+1}(d_{1},\ldots,d_{n_{k}+1})\right)$
		$\displaystyle=\frac{2^{8}\,C^{\prime}}{\alpha_{0}^{N\ell s}}\left\|I_{N\ell}(\mathbf{w}_{1}^{k+1}\cdots\mathbf{w}_{\ell}^{k+1})\right\|^{s}\left\|J_{n_{k}+1}(d_{1},\ldots,d_{n_{k}+1})\right\|^{s}$
		$\displaystyle<2^{8}\,C^{\prime}\left\|I_{n_{k}+1}(d_{1},\ldots,d_{n_{k}+1})\right\|^{s}\frac{\left\|I_{N\ell}(\mathbf{w}_{1}^{k+1}\cdots\mathbf{w}_{\ell}^{k+1})\right\|^{s}}{\alpha_{0}^{N\ell s}}$
		$\displaystyle<2^{8}\,C^{\prime}\left\|J_{n_{k}+1+N\ell}(\mathbf{d})\right\|^{s}.$

The last inequality is shown as in the case $k=1$ .

Assume that $n_{k}+2+(\ell-1)N\leq n<n_{k}+1+\ell N$ with $1\leq\ell\leq\ell_{k}$ , then

	$\displaystyle\mu(J_{n}(\mathbf{d}))$	$\displaystyle<\mu\left(J_{n_{k}+1+(\ell-1)N}(d_{1},\ldots,d_{n_{k}+1+(\ell-1)N})\right)$
		$\displaystyle<2^{8}\,C^{\prime}\left\|J_{n_{k}+1+(\ell-1)N}(d_{1},\ldots,d_{n_{k}+1+(\ell-1)N})\right\|^{s}.$

The discussion preceding Lemma 5.10 tells us that

	$\displaystyle\left\|J_{n_{k}+1+(\ell-1)N}(d_{1},\ldots,d_{n_{k}+1+(\ell-1)N})\right\|$	$\displaystyle\leq\left(1-\frac{1}{M}\right)\left\|I_{n_{k}+1+(\ell-1)N}(d_{1},\ldots,d_{n_{k}+1+(\ell-1)N})\right\|$
		$\displaystyle=\left(1-\frac{1}{M}\right)\left\|I_{n}(\mathbf{d})\right\|\prod_{j=n_{k}+2+(\ell-1)N}^{n}d_{j}(d_{j}-1)$
		$\displaystyle<\left(1-\frac{1}{M}\right)M^{2N}\left\|I_{n}(\mathbf{d})\right\|$
		$\displaystyle<2\left(1-\frac{1}{M}\right)M^{2N}\left\|J_{n}(\mathbf{d})\right\|.$

As a consequence, we have

\mu(J_{n}(\mathbf{a}))<2^{9}C^{\prime}\left(1-\frac{1}{M}\right)M^{2N}\left|J_{n}(\mathbf{d})\right|^{s}.

∎

Measure of balls. We now estimate the measure of balls with center on $E$ and small radius. Define

r_{0}\colon=\min\left\{G_{1}(d):2\leq d\leq M\right\}.

Lemma 5.14.

There is a constant $C^{\prime\prime}>0$ such that

\mu\left(B(x;r)\right)\leq C^{\prime\prime}r\quad\text{ for all }x\in E\text{ and all }r\in(0,r_{0}).

Take any $x=\langle d_{1},d_{2},\ldots\rangle$ and any $r\in(0,r_{0})$ . Pick $n\in\mathbb{N}$ such that

G_{n+1}(d_{1},\ldots,d_{n+1})\leq r<G_{n}(d_{1},\ldots,d_{n}).

By definition of $G_{n}$ , the ball $B(x;r)$ intersects exactly one fundamental interval of order $n$ , namely $J_{n}(d_{1},\ldots,d_{n})$ .

Let us further assume that $n_{k}+1\leq n\leq n_{k+1}-1$ for some $k\in\mathbb{N}$ . Taking $C$ as in Lemma 5.13,

$\displaystyle\mu\left(B(x;r)\right)$	$\displaystyle\leq\mu\left(J_{n}(d_{1},\ldots,d_{n})\right)$
	$\displaystyle\leq C\left\|J_{n}(d_{1},\ldots,d_{n})\right\|^{s}$
	$\displaystyle=C\left\|I_{n}(d_{1},\ldots,d_{n})\right\|^{s}\left(\frac{M-1}{M}\right)^{s}$
	$\displaystyle=C\left\|I_{n+1}(d_{1},\ldots,d_{n},d_{n+1})\right\|^{s}\left(\frac{M-1}{M}\right)^{s}d_{n+1}^{s}(d_{n+1}-1)^{s}$
	$\displaystyle<CM^{3}G_{n+1}(d_{1},\ldots,d_{n+1})^{s}$	(by lemma (5.9))
	$\displaystyle<CM^{3}r^{s}.$

Suppose now that $n=n_{k}-1$ . We consider two cases:

(17)

r<\frac{\left|I_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right|}{\alpha_{0}^{2n_{k}}}\quad\text{ and }\quad r\geq\frac{\left|I_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right|}{\alpha_{0}^{2n_{k}}}.

In the first case, $B(x;r)$ intersects at most three fundamentals intervals of level $n_{k}$ . As a consequence, we have

	$\displaystyle\mu\left(B(x;r)\right)$	$\displaystyle\leq 3\mu\left(J_{n_{k}}(d_{1},\ldots,d_{n_{k}})\right)$
		$\displaystyle\leq 3C\left\|J_{n_{k}}(d_{1},\ldots,d_{n_{k}})\right\|^{s}\leq 3CMG_{n_{k}}(d_{1},\ldots,d_{n_{k}})^{s}\leq 3CMr^{s}.$

Assume the second inequality in (17). All the cylinders of level $n_{k}$ contained in $J_{n_{j}-1}(d_{1},\ldots,d_{n_{k}-1})$ are of the form

I_{n_{k}}(d_{1},\ldots,d_{n_{k}-1},a)\quad\text{ with }\quad a\in\{\lceil\alpha_{0}^{n_{k}}\rceil,\ldots,\lfloor 2\alpha_{0}^{n_{k}}\rfloor\},

\left|I_{n_{k}}(d_{1},\ldots,d_{n_{k}-1},a)\right|=\frac{\left|I_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right|}{a(a-1)}\geq\frac{\left|I_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right|}{\alpha_{0}^{2n_{k}}};

If $T$ is the total amount of cylinders of level $n_{k}$ contained in $B(x;r)$ , then

T\leq\frac{\alpha_{0}^{2n_{k}}}{\left|I_{n_{k-1}}(d_{1},\ldots,d_{n_{k-1}})\right|}\,2r

and the total amount of cylinders of level $n_{k}$ intersecting $B(x;r)$ is at most

\frac{2r\alpha_{0}^{2n_{k}}}{\left|I_{n_{k-1}}(d_{1},\ldots,d_{n_{k-1}})\right|}+2<\frac{4r\alpha_{0}^{2n_{k}}}{\left|I_{n_{k-1}}(d_{1},\ldots,d_{n_{k-1}})\right|}.

Since each cylinder of level $n_{k}$ contains at most one fundamental interval of level $n_{k}$ , we have

	$\displaystyle\mu\left(B(x;r)\right)$	$\displaystyle<\frac{4r\alpha_{0}^{2n_{k}}}{\left\|I_{n_{k-1}}(d_{1},\ldots,d_{n_{k-1}})\right\|}\mu\left(J_{n_{k}}(d_{1},\ldots,d_{n_{k}-1},d_{n_{k}})\right)$
		$\displaystyle<\frac{8r\alpha_{0}^{2n_{k}}}{\left\|I_{n_{k-1}}(d_{1},\ldots,d_{n_{k-1}})\right\|}\mu\left(J_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right).$

The second inequality follows from the definition of $\mu\left(J_{n_{k}}(d_{1},\ldots,d_{n_{k}-1},d_{n_{k}})\right)$ . Take $C$ as in Lemma 5.13. Then, since $\min\{a,b\}\leq a^{1-s}b^{s}$ for all positive $a,b$ ,

	$\displaystyle\mu\left(B(x;r)\right)$	$\displaystyle\leq\min\left\{\mu(J_{n_{k-1}}(d_{1},\ldots,d_{n_{k-1}})),8r\alpha_{0}^{2n_{k}}\,\frac{\mu\left(J_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right)}{\left\|I_{n_{k-1}}(d_{1},\ldots,d_{n_{k-1}})\right\|}\right\}$
		$\displaystyle=\mu(J_{n_{k-1}}(d_{1},\ldots,d_{n_{k-1}}))\min\left\{1,\frac{8r\alpha_{0}^{2n_{k}}}{\left\|I_{n_{k-1}}(d_{1},\ldots,d_{n_{k-1}})\right\|}\right\}$
		$\displaystyle<8^{s}r^{s}\alpha_{0}^{2sn_{k}}\,\frac{\mu(J_{n_{k-1}}(d_{1},\ldots,d_{n_{k-1}}))}{\left\|I_{n_{k-1}}(d_{1},\ldots,d_{n_{k-1}})\right\|^{s}}$
		$\displaystyle<8Cr^{s}.$

In the last inequality, we have used (16). A similar argument holds for $n=n_{k}$ , since we have distributed uniformly the mass $\mu(J_{n_{k-1}}(d_{1},\ldots,d_{n_{k}-1}))$ among the $\lfloor 2\alpha_{1}^{n_{k}}\rfloor-\lceil\alpha_{1}^{n_{k}}\rceil$ fundamental intervals of level $n_{k}$ contained in $J_{n_{k-1}}(d_{1},\ldots,d_{n_{k}-1})$ when defining $\mu$ .

Proof of Theorem 1.6. Lower bound.

The Mass Distribution Principle tells us that $\dim_{\operatorname{H}}E\geq s$ , so $\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(B)\geq s$ . Letting $s\to s_{0}(B)$ , we conclude $\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(B)\geq s_{0}(B)$ . ∎

5.3. Proof of Theorem 1.6

Proof.

The upper bound follows from Theorem 5.1. Certainly, for any $\varepsilon>0$ , every large $n\in\mathbb{N}$ satisfies $\Psi(n)\geq(B-\varepsilon)^{n}$ , so $\mathcal{E}_{\mathbf{t}}(\Psi)\subseteq\mathcal{E}_{\mathbf{t}}(B-\varepsilon)$ and

\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(\Psi)\leq\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(B-\varepsilon)\to\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(B)\;\text{ as }\;\varepsilon\to 0.

The lower bound is obtained in essentially the same way as in Theorem 5.1. The case $\frac{s_{0}(B)}{t_{1}}-\frac{2s_{0}(B)-1}{t_{0}}\leq 0$ is solved without significant modifications. Assume that

(18)

\frac{s_{0}(B)}{t_{1}}-\frac{2s_{0}(B)-1}{t_{0}}>0.

We shall only define a useful Cantor set contained in $\mathcal{E}_{\mathbf{t}}(\Psi)$ and a probability measure supported on it. Let $\widetilde{B}>B$ be so close to $B$ that (18) still holds for $s_{0}(\widetilde{B})$ . Let $\widetilde{A}$ be such that

f_{t_{0}}\log\widetilde{A}=f_{t_{0},t_{1}}(s)\log\widetilde{B}.

Consider a strictly increasing sequence $(n_{j})_{j\geq 1}$ in $\mathbb{N}$ such that

\Psi(n_{k})\leq\widetilde{B}^{n_{k}}\quad\text{ for all }k\in\mathbb{N}.

Write

\beta_{0}\colon=\widetilde{A}^{1/t_{0}}\quad\text{ and }\quad\beta_{1}\colon=\widetilde{B}^{1/t_{0}}.

Now, let $(\ell_{k})_{k\geq 1}$ and $(i_{k})_{k\geq 1}$ be the sequences of integers determined by

(n_{k}-1)-(n_{k-1}-1)=\ell_{k}N+i_{k}\text{ with }0\leq i_{k}<N.

Call $\widetilde{E}$ be the subset of $\mathcal{E}_{\mathbf{t}}(\Psi)$ whose elements $x=\langle d_{1},d_{2},\ldots\rangle$ satisfy:

i.

For each $k\in\mathbb{N}$ , we have

$\beta_{0}^{n_{k}}\leq d_{n_{k}}\leq 2\beta_{0}^{n_{k}}\quad\text{ and }\quad\beta_{1}^{n_{k}}\leq d_{n_{k}+1}<2\beta_{1}^{n_{k}}.$
ii.

We have $d_{1}=\ldots=d_{i_{1}}=2$ and $d_{n_{k}+2}=\ldots=d_{n_{k}+1+i_{k}}=2$ for $k\in\mathbb{N}$ .
iii.

For every other natural number $n$ , we have $2\leq d_{n}\leq M$ .

Hence, if $x=\langle d_{1},d_{2},\ldots,\rangle\in\widetilde{E}$ , there are $\ell_{k}$ words $\mathbf{w}_{1}^{k}$ , $\ldots$ , $\mathbf{w}_{\ell_{k}}^{k}$ in $\{2,\ldots,m\}^{N}$ for $k\in\mathbb{N}$ such that

\mathbf{d}=\underbrace{2\cdots 2}_{i_{1}\text{ times }}\mathbf{w}_{1}^{1}\cdots\mathbf{w}_{\ell_{1}}^{1}d_{n_{1}}d_{n_{1}+1}\underbrace{2\cdots 2}_{i_{2}\text{ times }}\mathbf{w}_{1}^{2}\cdots\mathbf{w}_{\ell_{2}}^{2}d_{n_{2}}d_{n_{2}+1}\ldots\underbrace{2\cdots 2}_{i_{k}\text{ times }}\mathbf{w}_{1}^{k}\cdots\mathbf{w}_{\ell_{k}}^{k}d_{n_{k}}d_{n_{k}+1}\ldots.

Define $\widetilde{D}\colon=\Lambda^{-1}\left[\widetilde{E}\right]$ . For each $n\in\mathbb{N}$ consider $\widetilde{D}_{n}\colon=\left\{(d_{1},\ldots,d_{n}):(d_{j})_{j\geq 1}\in\widetilde{D}\right\}$ and define the set $\widetilde{J}_{n}$ by adapting the definition of $J_{n}$ into our current context. Take any $1\leq n\leq n_{1}+1$ and $\mathbf{d}\in\widetilde{D}_{n}$ .

i.

If $n\in\{1,\ldots,i_{1}\}$ , put $\widetilde{\mu}(\widetilde{J}_{n}(2,\ldots,2))\colon=1$ .

ii.

If $n=i_{1}+1+N\ell$ with $1\leq\ell\leq\ell_{1}$ , we write

\widetilde{\mu}\left(J_{n}(\mathbf{d})\right)\colon=\frac{1}{\beta_{0}^{sN\ell}}\left|I_{N\ell}(\mathbf{w}_{1}^{1}\cdots\mathbf{w}_{\ell}^{1})\right|^{s}.

iii.

If $i_{1}+1+N(\ell-1)+1\leq n\leq i_{1}+1+N(\ell-1)-1$ , where $1\leq\ell\leq\ell_{1}$ , write

\widetilde{\mu}\left(J_{n}(\mathbf{d})\right)\colon=\sum_{\mathbf{b}}\widetilde{\mu}\left(J_{i_{1}+1+N\ell}(\mathbf{d}\mathbf{b})\right),

where the sum runs along all those words $\mathbf{b}$ such that $\mathbf{d}\mathbf{b}$ belongs to $\widetilde{D}_{i_{1}+1+N\ell}$ .

iv.

When $n=n_{1}$ , write

\widetilde{\mu}(J_{n_{1}}(\mathbf{d}))\colon=\frac{\widetilde{\mu}\left(J_{n_{1}-1}(d_{1},\ldots,d_{n_{1}-1})\right)}{\lfloor 2\beta_{0}^{n_{1}}\rfloor-\lceil\beta_{0}^{n_{1}}\rceil}.

When $n=n_{1}+1$ , write

\widetilde{\mu}(J_{n_{1}+1}(\mathbf{d})\colon=\frac{\widetilde{\mu}\left(J_{n_{1}-1}(d_{1},\ldots,d_{n_{1}-1})\right)}{\left(\lfloor 2\beta_{0}^{n_{1}}\rfloor-\lceil\beta_{0}^{n_{1}}\rceil\right)\left(\lfloor 2\beta_{1}^{n_{1}}\rfloor-\lceil\beta_{1}^{n_{1}}\rceil\right)}.

Assume that we have already defined $\widetilde{\mu}$ for the fundamental intervals of order up to $n_{k}+1$ for some $k\in\mathbb{N}$ . Take $n\in\mathbb{N}$ such that $n_{k}+2\leq n\leq n_{k+1}+1$ .

i.

If $n_{k}+2\leq n\leq n_{k}+1+i_{k}$ , write

$\widetilde{\mu}\left(J_{n}(\mathbf{d})\right)=\widetilde{\mu}\left(J_{n_{k}+1}(d_{1},\ldots,d_{n_{k}+1})\right).$

ii.

If $n=n_{k}+i_{k+1}+1+N\ell$ for some $1\leq\ell\leq\ell_{k}$ , define

\widetilde{\mu}\left(J_{n}(d_{1},\ldots,d_{n})\right)\colon=\frac{1}{2^{N\ell s}\beta_{0}^{sN\ell}}\,\widetilde{\mu}\left(J_{n_{k}+i_{k+1}+1}(d_{1},\ldots,d_{n_{k}+1+i_{k+1}})\right).

iii.

If $n_{k}+1+i_{k+1}+N(\ell-1)+1\leq n<n_{k}+1+i_{k+1}+N\ell$ , then

\widetilde{\mu}(J_{n}(\mathbf{d}))\colon=\sum_{\mathbf{b}}\widetilde{\mu}\left(J_{n_{k}+1+i_{k+1}+N\ell}(\mathbf{d}\mathbf{b})\right),

where the sum runs along the words $\mathbf{b}$ such that $\mathbf{d}\mathbf{b}\in D_{n_{k}+1+i_{k+1}+N\ell}$ .

iv.

If $n=n_{k}$ , write

\widetilde{\mu}(J_{n_{1}}(\mathbf{d}))\colon=\frac{\widetilde{\mu}\left(J_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right)}{\lfloor 2\beta_{0}^{n_{k}}\rfloor-\lceil\beta_{0}^{n_{k}}\rceil}.

If $n=n_{k}+1$ , write

\widetilde{\mu}(J_{n_{1}+1}(\mathbf{d}))\colon=\frac{\widetilde{\mu}\left(J_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right)}{\left(\lfloor 2\beta_{0}^{n_{k}}\rfloor-\lceil\beta_{0}^{n_{k}}\rceil\right)\left(\lfloor 2\beta_{1}^{n_{k}}\rfloor-\lceil\beta_{1}^{n_{k}}\rceil\right)}.

∎

6. Final remarks

Our investigations give rise to a natural question: what happens when $1<B<\infty$ and $m\geq 3$ ? Unfortunately, our argument is not strong enough to solve this problem. However, based on [2], we state a conjecture on the Hausdorff dimension of $\mathcal{E}_{\mathbf{t}}(\Psi)$ . For any $m\in\mathbb{N}$ and $\mathbf{t}=(t_{0},\ldots,t_{m-1})\in\mathbb{R}_{>0}^{m}$ , define the functions $f_{t_{0}},f_{t_{0},t_{1}},\ldots,f_{t_{0},t_{1},\ldots,t_{m-1}}$ as follows: $f_{t_{0}}(s)=\frac{s}{t_{0}}$ and

f_{t_{0},\ldots,t_{j}}(s)=\frac{sf_{t_{0},\ldots,t_{j-1}}(s)}{t_{j}f_{t_{0},\ldots,t_{j-1}}(s)+\max\left\{0,s-\frac{2s-1}{\max\{t_{0},\ldots,t_{j-1}\}}\right\}}

for all $j\in\{2,\ldots,m\}$ .

Conjecture 6.1.

Let $m\in\mathbb{N}_{\geq 3}$ be arbitrary and let $B$ be as in (1). If $1<B<\infty$ , then $\dim_{\operatorname{H}}\mathcal{E}_{\mathbf{t}}(\Psi)$ is the unique solution $s$ of

\sum_{d=2}^{\infty}\frac{1}{d^{s}(d-1)^{s}B^{f_{t_{0},\ldots,t_{m-1}}(s)}}=1.

References

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[7] L. Huang, J. Wu, and J. Xu. Metric properties of the product of consecutive partial quotients in continued fractions. Israel J. Math., 238(2):901–943, 2020.
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[14] W. Philipp. Some metrical theorems in number theory. Pacific J. Math., 20:109–127, 1967.
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	$\displaystyle\inf J_{\tilde{n}}(\mathbf{d})$	$\displaystyle-\sup J_{\tilde{n}}(d_{1},\ldots,d_{\tilde{n}}+1)=$
		$\displaystyle=\left(\inf J_{\tilde{n}}(\mathbf{d})-\inf I_{\tilde{n}}(\mathbf{d})\right)+\left(\sup I_{\tilde{n}}(d_{1},\ldots,d_{\tilde{n}}+1)-\sup J_{\tilde{n}}(d_{1},\ldots,d_{\tilde{n}}+1)\right)$
		$\displaystyle=\frac{1}{\lceil 2\alpha_{0}^{n_{j}}\rceil}\left\|I_{\tilde{n}}(\mathbf{d})\right\|+\left(1-\frac{1}{\lfloor\alpha_{0}^{n_{j}}\rfloor-1}\right)\left\|I_{\tilde{n}}(d_{1},\ldots,d_{\tilde{n}}+1)\right\|$
		$\displaystyle>\left(1-\frac{1}{\lfloor\alpha_{0}^{n_{j}}\rfloor-1}\right)\frac{d_{\tilde{n}}-1}{d_{\tilde{n}}+1}\left\|I_{\tilde{n}}(\mathbf{d})\right\|$
		$\displaystyle=\left(1-\frac{1}{\lfloor\alpha_{0}^{n_{j}}\rfloor-1}\right)\left(1-\frac{2}{d_{\tilde{n}}+1}\right)\left\|I_{\tilde{n}}(\mathbf{d})\right\|$
		$\displaystyle>\left(1-\frac{1}{\lfloor\alpha_{0}^{n_{j}}\rfloor-1}\right)\left(1-\frac{2}{M}\right)\left\|I_{\tilde{n}}(\mathbf{d})\right\|>\frac{1}{M}\left\|I_{\tilde{n}}(\mathbf{d})\right\|.$

$\displaystyle\left\|J_{n_{k}}(\mathbf{d})\right\|$	$\displaystyle=\left(\frac{1}{\lceil\alpha_{1}^{n_{k}}\rceil-1}-\frac{1}{\lfloor 2\alpha_{1}^{n_{k}}\rfloor}\right)\left\|I_{n_{k}}(\mathbf{d})\right\|$
	$\displaystyle=\left(\frac{1}{\lceil\alpha_{1}^{n_{k}}\rceil-1}-\frac{1}{\lfloor 2\alpha_{1}^{n_{k}}\rfloor}\right)\frac{\left\|I_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right\|}{d_{n_{k}}(d_{n_{k}}-1)}$
	$\displaystyle>\frac{1}{8\alpha_{1}^{n_{k}}\alpha_{0}^{n_{k}}\alpha_{0}^{n_{k}}}\left\|I_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right\|$
	$\displaystyle>\frac{1}{8\alpha_{1}^{n_{k}}\alpha_{0}^{n_{k}}}\left\|J_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right\|$	$\displaystyle\text{(by \eqref{Ec:LEST:02})}.$

$\displaystyle\left\|J_{n_{k}+1}(\mathbf{d})\right\|$	$\displaystyle=\left(1-\frac{1}{M}\right)\left\|I_{n_{k}+1}(\mathbf{d})\right\|$
	$\displaystyle=\left(1-\frac{1}{M}\right)\frac{\left\|I_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right\|}{d_{n_{k}}(d_{n_{k}}-1)d_{n_{k}+1}(d_{n_{k}+1}-1)}$
	$\displaystyle>\frac{1}{2^{6}}\frac{\left\|I_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right\|}{\alpha_{0}^{2n_{k}}\alpha_{1}^{2n_{k}}}$
	$\displaystyle>\frac{1}{2^{6}}\frac{\left\|J_{n_{k}-1}(d_{1},\ldots,d_{n_{k}-1})\right\|}{\alpha_{0}^{n_{k}}\alpha_{1}^{2n_{k}}}$	$\displaystyle\text{(by \eqref{Ec:LEST:02})}.$

	$\displaystyle\mu\left(J_{n_{k}-1}(\mathbf{d})\right)$	$\displaystyle=\frac{\left\|I_{N\ell_{k}}\left(\mathbf{W}_{k}\right)\right\|^{s}}{\alpha_{0}^{N\ell_{k}s}}\mu\left(J_{n_{k-1}+1}(d_{1},\ldots,d_{n_{k-1}+1})\right)$
		$\displaystyle=\frac{\left\|I_{N\ell_{k}}\left(\mathbf{W}_{k}\right)\right\|^{s}}{\alpha_{0}^{N\ell_{k}s}}\frac{\mu\left(J_{n_{k-1}-1}(d_{1},\ldots,d_{n_{k-1}-1})\right)}{\left(\lfloor 2\alpha_{1}^{n_{k-1}}\rfloor-\lceil\alpha_{1}^{n_{k-1}}\rceil\right)\left(\lfloor 2\alpha_{0}^{n_{k-1}}\rfloor-\lceil\alpha_{0}^{n_{k-1}}\rceil\right)}$
		$\displaystyle<\frac{\gamma_{k-1}}{\alpha_{0}^{N\ell_{k}s}}\frac{\left\|I_{N\ell_{k}}\left(\mathbf{W}_{k}\right)\right\|^{s}}{\alpha_{0}^{n_{k-1}}\alpha_{1}^{n_{k-1}}}\mu\left(J_{n_{k-1}-1}(d_{1},\ldots,d_{n_{k-1}-1})\right)$
		$\displaystyle<\frac{\gamma_{k-1}}{\alpha_{0}^{N\ell_{k}s}}\frac{\left\|I_{N\ell_{k}}\left(\mathbf{W}_{k}\right)\right\|^{s}}{\alpha_{0}^{sn_{k-1}}\alpha_{1}^{2sn_{k-1}}}\mu\left(J_{n_{k-1}-1}(d_{1},\ldots,d_{n_{k-1}-1})\right).$

	$\displaystyle\mu\left(J_{n}(\mathbf{d})\right)$	$\displaystyle=\frac{1}{\alpha_{0}^{N\ell s}}\,\sum_{\begin{subarray}{c}\mathbf{b}\in\mathscr{D}^{\ell N-n}\\ \mathbf{d}\mathbf{b}\in D_{\ell N}\end{subarray}}\left\|I_{\ell N}(\mathbf{d}\mathbf{b})\right\|^{s}$
		$\displaystyle=\frac{\|I_{n}(\mathbf{d})\|^{s}}{\alpha_{0}^{ns}}\,\sum_{\begin{subarray}{c}\mathbf{b}\in\mathscr{D}^{\ell N-n}\\ \mathbf{d}\mathbf{b}\in D_{\ell N}\end{subarray}}\frac{\left\|I_{\ell N-n}(\mathbf{b})\right\|^{s}}{\alpha_{0}^{(N\ell-n)s}}$
		$\displaystyle=\frac{\|I_{n}(\mathbf{d})\|^{s}}{\alpha_{0}^{ns}}\,\left(\sum_{d=2}^{M}\frac{1}{d^{s}(d-1)^{s}\alpha_{0}^{s}}\right)^{\ell N-n}$
		$\displaystyle=\frac{\|I_{n}(\mathbf{d})\|^{s}}{\alpha_{0}^{ns}}\,=\frac{2^{s}}{\alpha_{0}^{ns}}\,\frac{\|I_{n}(\mathbf{d})\|^{s}}{2^{s}}<\frac{2^{s}}{\alpha_{0}^{ns}}\,\|J_{n}(\mathbf{d})\|^{s}<\|J_{n}(\mathbf{d})\|^{s}.$

Metrical properties of weighted products of consecutive Lüroth digits

Abstract.

1. Introduction

Theorem 1.1 ([9, Theorem 2.1]).

Theorem 1.2 ([15, Theorem 4.2] ).

Theorem 1.3 ([18, Lemma 3.1]).

Theorem 1.4.

Theorem 1.5.

Theorem 1.6.

Notation.

2. Elements of Lüroth series

Proposition 2.1.

Proof.

3. Proof of Theorem 1.4

Lemma 3.1.

Lemma 3.2.

Proof.

Lemma 3.3.

Proof.

Lemma 3.4.

Proof.

Proof of Theorem 1.4.

4. Proof of Theorem 1.5

Lemma 4.1 ([16, Theorem 3.1]).

Lemma 4.2 ([15, Lemma 2.3]).

Proof of Theorem 1.5.

5. Proof of Theorem 1.6

Theorem 5.1.

5.1. Continuity of B↦s0​(B)B\mapsto s_{0}(B)

Lemma 5.2.

Proof.

Lemma 5.3.

Proof.

Lemma 5.4.

Proof.

Theorem 5.5.

5.2. Hausdorff dimension estimates

5.2.1. Upper bound

Proof of Lemma 5.1. Upper bound..

5.2.2. Lower bound

5.2.3. Lower bound: first case

Lemma 5.6.

Proof.

Lemma 5.7.

Proof.

5.2.4. Lower bound: second case

Lemma 5.8 (Mass Distribution Principle).

Proof.

Construction of the Cantor set.

Gap estimates.

Lemma 5.9.

Proof.

Length estimates.

Lemma 5.10.

Proof.

Measure of the fundamental intervals.

Lemma 5.11.

Proof.

Lemma 5.12.

Proof.

Lemma 5.13.

Proof.

Lemma 5.14.

Proof of Theorem 1.6. Lower bound.

5.3. Proof of Theorem 1.6

Proof.

6. Final remarks

Conjecture 6.1.

References

5.1. Continuity of $B\mapsto s_{0}(B)$