$HD(M\setminus L)>0.353$

TL;DR

This paper investigates the structure and Hausdorff dimension of the complement of the Lagrange spectrum in the Markov spectrum, providing new lower bounds and explicit elements, including the largest known member of this set.

Contribution

It describes the local structure of $M\setminus L$ near a non-isolated point and constructs a Cantor set with matching Hausdorff dimension, establishing a lower bound of 0.353 for the Hausdorff dimension of $M\setminus L$.

Findings

Hausdorff dimension of $M\setminus L$ exceeds 0.353

Constructed a Cantor set with dimension matching $M\setminus L$ near a special point

Explicitly identified new elements of $M\setminus L$, including its largest known member

Abstract

The complement $M\setminus L$ of the Lagrange spectrum $L$ in the Markov spectrum $M$ was studied by many authors (including Freiman, Berstein, Cusick and Flahive). After their works, we disposed of a countable collection of points in $M\setminus L$. In this article, we describe the structure of $M\setminus L$ near a non-isolated point $\alpha_{\infty}$ found by Freiman in 1973, and we use this description to exhibit a concrete Cantor set $X$ whose Hausdorff dimension coincides with the Hausdorff dimension of $M\setminus L$ near $\alpha_{\infty}$. A consequence of our results is the lower bound $HD(M\setminus L)>0.353$ on the Hausdorff dimension $HD(M\setminus L)$ of $M\setminus L$. Another by-product of our analysis is the explicit construction of new elements of $M\setminus L$, including its largest known member $c\in M\setminus L$ (surpassing the former largest known number $\alpha_4\in M\setminus L$ obtained by Cusick and Flahive in 1989).

Equations

$$L:=\left\{\underset{\begin{array}{c}p,q\to \infty \\ p,q\in \mathbb{Z}\end{array}}{\mathrm{limsup}}\frac{1}{|q(q\alpha -p)|}<\infty :\alpha \in \mathbb{R}-\mathbb{Q}\right\}$$

$$M:=\left\{\frac{1}{\underset{\begin{array}{c}(x,y)\in {\mathbb{Z}}^2\\ (x,y)\ne (0,0)\end{array}}{\inf }|q(x,y)|}<\infty :q(x,y)=a{x}^2+bxy+c{y}^2\text{ real indefinite, }{b}^2-4ac=1\right\}.$$

$$L\cap (-\infty ,3)=M\cap (-\infty ,3)=\left\{\sqrt{5}<\sqrt{8}<\frac{\sqrt{221}}{5}<\dots \right\}$$

$$c_F:=\frac{2221564096+283748\sqrt{462}}{491993569}\simeq 4.5278\dots .$$

$$0.353<HD(M\setminus L)$$

$$0.353<HD(X)=HD((M\setminus L)\cap (b_{\infty },B_{\infty })),$$

$$c=\frac{77+\sqrt{18229}}{82}+\frac{17633692-\sqrt{151905}}{24923467}=3.29304447990138\dots .$$

$$\alpha =[a_0;a_1,a_2,\dots ]=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{\ddots }}}$$

$$[a_0;a_1,\dots ,a_n]:=a_0+\frac{1}{a_1+\frac{1}{\ddots +\frac{1}{a_n}}}:=[a_0;a_1,\dots ,a_n,\infty ,\dots ]$$

$${\lambda }_i(A):=[a_i;a_{i+1},a_{i+2},\dots ]+[0;a_{i-1},a_{i-2},\dots ].$$

$$\ell (A)=\underset{i\to \infty }{\mathrm{limsup}}{\lambda }_i(A)\text{and}m(A)=\underset{i\in \mathbb{Z}}{\sup }{\lambda }_i(A).$$

$$L=\{\ell (A)<\infty :A\in ({\mathbb{N}}^{\ast }{)}^{\mathbb{Z}}\}\text{and}M=\{m(A)<\infty :A\in ({\mathbb{N}}^{\ast }{)}^{\mathbb{Z}}\}.$$

$$K(B):=\{[0;{\gamma }_1,{\gamma }_2,\dots ]:{\gamma }_i\in B\forall i⩾1\}\subset [0,1]$$

$${\alpha }_{\infty }:={\lambda }_0(A_{\infty }):=[2;\overline{1_2,2_3,1,2}]+[0;1,2_3,1_2,2,1,\overline{2}]\in M\setminus L.$$

$${\alpha }_n:={\lambda }_0(A_n):=[2;\overline{1_2,2_3,1,2}]+[0;1,2_3,1_2,2,1,2_n,\overline{1,2,1_2,2_3}]\in M\setminus L$$

$$X:=\{[0;\gamma ]:\gamma \in \{1,2{\}}^{\mathbb{N}}\text{ not containing the subwords in }P\}$$

$$P:=\{21212,2121_3,1_{3}212,12121_2,1_{2}2121,2_{3}121_2{2}_{2}1,12_2{1}_{2}212_3,12_{3}121_2{2}_2,2_2{1}_{2}212_{3}1\}$$

$$b_{\infty }:=[2;\overline{1_2,2_3,1,2}]+[0;\overline{1,2_3,1_2,2}]=3.2930442439\dots$$

$$B_{\infty }$$

$${\lambda }_j(B)=[2;1,2,\dots ]+[0;1,2,\dots ]>[2;1,2]+[0;1,2]=\frac{10}{3}>{\alpha }_{\infty }+10^{-2}.$$

$${\lambda }_j(B)=[2;1_3,\dots ]+[0;1,2,\dots ]>[2;1_4]+[0;1,2_2,1]=\frac{33}{10}>{\alpha }_{\infty }+10^{-3}.$$

$${\lambda }_j(B)$$

Full text

In this article, we describe the structure of $M\setminus L$ near a non-isolated point $\alpha_{\infty}$ found by Freiman in 1973, and we use this description to exhibit a concrete Cantor set $X$ whose Hausdorff dimension coincides with the Hausdorff dimension of $M\setminus L$ near $\alpha_{\infty}$.

A consequence of our results is the lower bound $HD(M\setminus L)>0.353$ on the Hausdorff dimension $HD(M\setminus L)$ of $M\setminus L$. Another by-product of our analysis is the explicit construction of new elements of $M\setminus L$, including its largest known member $c\in M\setminus L$ (surpassing the former largest known number $\alpha_{4}\in M\setminus L$ obtained by Cusick and Flahive in 1989).

1. Introduction

1.1. Statement of the main results

The Lagrange and Markov spectra are subsets of the real line related to classical Diophantine approximation problems. More precisely, the Lagrange spectrum is

[TABLE]

and the Markov spectrum is

[TABLE]

Markov proved in 1879 that

[TABLE]

consists of an explicit increasing sequence of quadratic surds accumulating only at $3$.

Hall proved in 1947 that $L\cap[c,\infty)=[c,\infty)$ for some constant $c>3$. For this reason, a half-line $[c,\infty)$ contained in the Lagrange spectrum is called a Hall ray.

Freiman determined in 1975 the biggest half-line $[c_{F},\infty)$ contained in the Lagrange spectrum, namely,

[TABLE]

The constant $c_{F}$ is called Freiman’s constant.

In general, it is known that $L\subset M$ are closed subsets of $\mathbb{R}$. The results of Markov, Hall and Freiman mentioned above imply that the Lagrange and Markov spectra coincide below $3$ and above $c_{F}$. Nevertheless, Freiman showed in 1968 that $M\setminus L\neq\emptyset$ by exhibiting a number $\sigma\simeq 3.1181\dots\in M\setminus L$. On the other hand, some authors believe that the Lagrange and Markov spectra coincide111Added in proof: This was conjectured by Cusick in 1975: see [3, p. 516]. As it turns out, after this article was completed, we discovered that Cusick’s conjecture is false, namely $M\setminus L$ contains some numbers near $3.7096\dots$: see [8] for more details. above $\sqrt{12}\simeq 3.4641\dots$.

The reader is invited to consult the excellent book [4] of Cusick-Flahive for a review of the literature on the Lagrange and Markov spectrum until the mid-eighties.

The main theorem of this paper concerns the Hausdorff dimension of $M\setminus L$:

Theorem 1.1**.**

The Hausdorff dimension $HD(M\setminus L)$ of $M\setminus L$ satisfies:

[TABLE]

The proof of Theorem 1.1 is based on a refinement of the analysis in Chapter 3 of Cusick-Flahive book [4] of a sequence $\alpha_{n}\in M\setminus L$, $n\geqslant 4$, converging to a number $\alpha_{\infty}\simeq 3.293\dots\in M\setminus L$ in order to exhibit a Cantor set $X$ such that

[TABLE]

where $(b_{\infty},B_{\infty})$ is the largest interval disjoint from $L$ containing $\alpha_{\infty}$.

Remark 1.2**.**

The Cantor set $X$ is described in (3.1) below: it is a Cantor set defined in terms of explicit restrictions on continued fraction expansions. In particular, one can use the “thermodynamical arguments” of Bumby [2], Hensley [7], Jenkinson-Pollicott [11] and Falk-Nussbaum [5] to compute $HD(X)$.

In this direction, we implemented the algorithm of Jenkinson-Pollicott and we obtained the heuristic approximation $HD(X)=0.4816\cdots$.

In principle, this heuristic approximation can be made rigorous, but we have not pursued this direction. Instead, we exhibit a Cantor set $K(\{1,2_{2}\})\subset X$ whose Hausdorff dimension can be easily (and rigorously) estimated as $0.353<HD(K(\{1,2_{2}\}))<0.35792$ via some classical arguments explained in Palis-Takens book [12]: see Section 4 below.

As it turns out, the first term $\alpha_{4}=3.29304427\dots$ of the sequence $(\alpha_{n})_{n\geqslant 4}$ mentioned above was the largest known element of $M\setminus L$ since 1989 (see page 35 of [4]). By exploiting the arguments establishing Theorem 1.1, we are able to exhibit new numbers in $M\setminus L$, including a constant $c\in M\setminus L$ with $c>\alpha_{4}$:

Proposition 1.3**.**

The largest element of $(M\setminus L)\cap(b_{\infty},B_{\infty})$ is

[TABLE]

In particular, $c$ is the largest known element of $M\setminus L$.

Remark 1.4**.**

One has $\frac{c-\alpha_{\infty}}{\alpha_{4}-\alpha_{\infty}}=32.58\dots$. In other words, if the coordinates are centered at $\alpha_{\infty}$, then $c$ is more than $32$ times larger than $\alpha_{4}$.

1.2. Organization of the article

In Section 2, we recall some classical facts about continued fractions and Perron’s characterization of $L$ and $M$. In Section 3, we show that $HD((M\setminus L)\cap(b_{\infty},B_{\infty}))=HD(X)$, where $X$ is a Cantor set of real numbers in $[0,1]$ whose continued fraction expansions correspond to the elements of $\{1,2\}^{\mathbb{N}}$ not containing nine explicit finite words. In particular, this reduces the proof of Theorem 1.1 to the computation of lower bounds on $HD(X)$. In Section 4, we complete the proof of Theorem 1.1 by showing that $HD(K(\{1,2_{2}\}))>0.353$, where $K(\{1,2_{2}\})\subset X$ is the Cantor set of real numbers in $[0,1]$ whose continued fraction expansions associated to elements of $\{1,2\}^{\mathbb{N}}$ given by concatenations of the finite words $1$ and $2,2$. In Section 5, we pursue the arguments in Section 3 in order to establish Proposition 1.3. In Appendix A, we show that $(b_{\infty},B_{\infty})$ is the largest interval disjoint from $L$ containing $\alpha_{\infty}$: in particular, we correct some claims made by Berstein in Theorem 1 at page 47 of [1] concerning $(b_{\infty},B_{\infty})$. Finally, in Appendix B, we show that the largest element $\alpha_{2}$ of the sequence $(\alpha_{n})_{n\in\mathbb{N}}$ constructed by Cusick-Flahive in Chapter 3 of [4] belongs to the Lagrange spectrum.

Acknowledgements

We are thankful to Thomas Cusick, Dmitry Gayfulin and Nikolay Moshchevitin for their immense help in giving us access to the references [1] and [6].

2. Some preliminaries

2.1. Continued fractions

Given an irrational number $\alpha$, we denote by

[TABLE]

its continued fraction expansion, and we let

[TABLE]

be its $n$th convergent.

A standard comparison tool for continued fractions is the following lemma222Compare with Lemmas 1 and 2 in Chapter 1 of Cusick-Flahive book [4].:

Lemma 2.1**.**

Let $\alpha=[a_{0};a_{1},\dots,a_{n},a_{n+1},\dots]$ and $\beta=[a_{0};a_{1},\dots,a_{n},b_{n+1},\dots]$ with $a_{n+1}\neq b_{n+1}$. Then:

• •

$\alpha>\beta$* if and only if $(-1)^{n+1}(a_{n+1}-b_{n+1})>0$;*

• •

$|\alpha-\beta|<1/2^{n-1}$.

Remark 2.2**.**

For later use, note that Lemma 2.1 implies that if $a_{0}\in\mathbb{Z}$ and $a_{i}\in\mathbb{N}^{*}$ for all $i\geqslant 1$, then $[a_{0};a_{1},\dots,a_{n},\dots]<[a_{0};a_{1},\dots,a_{n}]$ when $n\geqslant 1$ is odd, and $[a_{0};a_{1},\dots,a_{n},\dots]>[a_{0};a_{1},\dots,a_{n}]$ when $n\geqslant 0$ is even.

2.2. Perron’s description of the Lagrange and Markov spectra

Given a bi-infinite sequence $A=(a_{n})_{n\in\mathbb{Z}}\in(\mathbb{N}^{*})^{\mathbb{Z}}$ and $i\in\mathbb{Z}$, let

[TABLE]

Define the quantities

[TABLE]

In 1921, Perron showed that

[TABLE]

In the sequel, we will work exclusively with these characterizations of $L$ and $M$.

2.3. Gauss-Cantor sets

Given a finite alphabet $B=\{\beta_{1},\dots,\beta_{m}\}$, $m\geqslant 2$, consisting of finite words $\beta_{j}\in(\mathbb{N}^{*})^{r_{j}}$, $1\leqslant j\leqslant m$, such that $\beta_{i}$ does not begin by $\beta_{j}$ for all $i\neq j$, we denote by

[TABLE]

the Gauss-Cantor set associated to $B$.

2.4. Some notations

Given a finite word $\beta=(b_{1},\dots,b_{r})\in(\mathbb{N}^{*})^{r}$, we denote by $\beta^{T}:=(b_{r},\dots,b_{1})$ the transpose of $\beta$.

Also, we abreviate periodic continued fractions and bi-infinite sequences which are periodic in one or both sides by putting a bar over the period: for instance, $[\overline{2,1,1}]=[2;1,1,2,1,1,2,1,1,\dots]$ and $\overline{1},2,1,2,\overline{1,2}=\dots,1,1,1,2,1,2,1,2,1,2,1,2,\dots$.

Moreover, we shall use subscripts to indicate the multiplicity of a digit in a sequence: for example, $[2;1_{2},2_{3},1,2,\dots]=[2;1,1,2,2,2,1,2,\dots]$.

3. $HD(M\setminus L)>0$

In 1973, Freiman [6] showed that

[TABLE]

In a similar vein, Theorem 4 in Chapter 3 of Cusick-Flahive book [4] asserts that

[TABLE]

for all $n\geqslant 4$. In particular, $\alpha_{\infty}$ is not isolated in $M\setminus L$.

In what follows, we shall revisit Freiman’s arguments as described in Chapter 3 of Cusick-Flahive book [4] in order to prove the following result. Let $X$ be the Cantor set

[TABLE]

where

[TABLE]

Also, let

[TABLE]

and

[TABLE]

The remainder of this section is devoted to the proof of the following result:

Theorem 3.1**.**

$HD((M\setminus L)\cap(b_{\infty},B_{\infty}))=HD(X)$* (where $X$ is the Cantor set in (3.1)).*

3.1. Description of $M\setminus L$ near $\alpha_{\infty}$

Our description of $(M\setminus L)\cap(b_{\infty},B_{\infty})$ needs the following improvements of Lemma 1 in [4, Chapter 3]:

Lemma 3.2**.**

If $B\in\{1,2\}^{\mathbb{Z}}$ contains any of the subsequences:

• (i)

$212^{*}12$**

• (ii)

$212^{*}1_{3}$**

• (iii)

$1212^{*}1_{2}$**

• (iv)

$2_{3}12^{*}1_{2}2_{2}1$**

• (v)

$212_{3}12^{*}1_{2}2_{3}$**

• (vi)

$1_{2}2_{3}12^{*}1_{2}2_{4}$**

• (vii)

$1_{2}2_{3}12^{*}1_{2}2_{3}1_{2}$**

• (viii)

$1_{3}2_{3}12^{*}1_{2}2_{3}12$**

• (ix)

$21_{2}2_{3}12^{*}1_{2}2_{3}12_{2}$**

• (x)

$2_{2}1_{2}2_{3}12^{*}1_{2}2_{3}121$**

• (xi)

$1_{2}21_{2}2_{3}12^{*}1_{2}2_{3}121_{2}2$**

then $\lambda_{j}(B)>\alpha_{\infty}+10^{-6}$ where $j$ indicates the position in asterisk.

Proof.

If (i) occurs, then Remark 2.2 implies that

[TABLE]

If (ii) occurs, then Remark 2.2 says that

[TABLE]

If (iii) occurs, then Remark 2.2 implies that

[TABLE]

If (iv) occurs, then Remark 2.2 says that

[TABLE]

If (v) occurs, then Remark 2.2 implies that

[TABLE]

If (vi) occurs, then Remark 2.2 says that

[TABLE]

If (vii) occurs, then Remark 2.2 says that

[TABLE]

If (viii) occurs, then Remark 2.2 implies that

[TABLE]

If (ix) occurs, then Remark 2.2 says that

[TABLE]

If (x) occurs, then Remark 2.2 implies that

[TABLE]

If (xi) occurs, then Remark 2.2 says that

[TABLE]

∎

Lemma 3.3**.**

Let $B\in\{1,2\}^{\mathbb{Z}}$.

• (xii’)

If $B$ contains $1212_{2}121_{2}2_{3}12^{*}1_{2}2_{3}121_{2}2_{3}121_{2}2$, then $\lambda_{j}(B)>B_{\infty}+6\times 10^{-9}$ where $j$ indicates the position in asterisk.

• (xii”)

If $B$ contains $2_{2}12_{2}121_{2}2_{3}12^{*}1_{2}2_{3}121_{2}2_{3}121_{2}2$, then $\lambda_{j}(B)<B_{\infty}-10^{-9}$ where $j$ indicates the position in asterisk.

Proof.

If (xii’) occurs, then Lemma 2.1 says that

[TABLE]

If (xii”) occurs, then Lemma 2.1 says that

[TABLE]

∎

We will also need the following result (extracted from Lemma 2 in Chapter 3 of [4]):

Lemma 3.4**.**

If $B\in\{1,2\}^{\mathbb{Z}}$ contains any of the subsequences

• (a)

$1^{*}$**

• (b)

$22^{*}$**

• (c)

$1_{2}2^{*}1_{2}$**

• (d)

$2_{2}12^{*}1_{2}21$**

• (e)

$12_{2}12^{*}1_{2}2$**

• (f)

$2_{4}12^{*}1_{2}2_{3}$**

then $\lambda_{j}(B)<\alpha_{\infty}-10^{-5}$ where $j$ indicates the position in asterisk.

Proof.

If (a) occurs, then $\lambda_{j}(B)=1+[0;\dots]+[0;\dots]<3<\alpha_{\infty}-10^{-1}$.

If (b) occurs, then Remark 2.2 implies that

[TABLE]

If (c) occurs, then Remark 2.2 says that

[TABLE]

If (d) occurs, then Remark 2.2 implies that

[TABLE]

If (e) occurs, then Remark 2.2 says that

[TABLE]

If (f) occurs, then Remark 2.2 implies that

[TABLE]

∎

By putting together Lemma 3.2 and Lemma 3.4, we obtain:

Lemma 3.5**.**

Let $B=(B_{m})_{m\in\mathbb{Z}}\in\{1,2\}^{\mathbb{Z}}$ be a bi-infinite sequence. Suppose that $\lambda_{n}(B)\leqslant\alpha_{\infty}+10^{-6}$ at a certain position $n\in\mathbb{Z}$. Then, the sole possible situations are:

• •

$B_{n}=1$* and $\lambda_{n}(B)<\alpha_{\infty}-10^{-5}$;*

• •

$B_{n-1}B_{n}=22$* and $\lambda_{n}(B)<\alpha_{\infty}-10^{-5}$;*

• •

$B_{n}B_{n+1}=22$* and $\lambda_{n}(B)<\alpha_{\infty}-10^{-5}$;*

• •

$B_{n-2}B_{n-1}B_{n}B_{n+1}B_{n+2}=11211$* and $\lambda_{n}(B)<\alpha_{\infty}-10^{-5}$;*

• •

$B_{n-3}\dots B_{n+4}\in\{21121221,22121121\}$* and $\lambda_{n}(B)<\alpha_{\infty}-10^{-5}$;*

• •

$B_{n-4}\dots B_{n+3}\in\{12112122,12212112\}$* and $\lambda_{n}(B)<\alpha_{\infty}-10^{-5}$;*

• •

$B_{n-5}\dots B_{n+5}\in\{22211212222,22221211222\}$* and $\lambda_{n}(B)<\alpha_{\infty}-10^{-5}$;*

• •

$B_{n-5}\dots B_{n+4}=1222121122$;

• •

$B_{n-4}\dots B_{n+5}=2211212221$.

In particular, the subwords $212^{*}12$, $212^{*}1_{3}$, $1_{3}2^{*}12$, $1212^{*}1_{2}$, $1_{2}2^{*}121$, $2_{3}12^{*}1_{2}2_{2}1$ and $12_{2}1_{2}2^{*}12_{3}$ are forbidden (where the asterisk indicates the $n$th position).

Proof.

By items (a) and (b) of Lemma 3.4, if $B_{n}=1$, $B_{n-1}B_{n}=22$ or $B_{n}B_{n+1}=22$, then $\lambda_{n}(B)<\alpha_{\infty}-10^{-5}$.

By items (i), (ii) and (iii) of Lemma 3.2, our assumption $\lambda_{n}(B)\leqslant\alpha_{\infty}+10^{-6}$ implies that the subwords $212^{*}12$, $212^{*}1_{3}$, $1_{3}2^{*}12$, $1212^{*}1_{2}$ and $1_{2}2^{*}121$ are forbidden for $B_{n}=2^{*}$. So, if $B_{n-1}B_{n}B_{n+1}=121$, then one has just three possibilities:

• •

$B_{n-2}\dots B_{n+2}=11211$ and, by item (c) of Lemma 2, $\lambda_{n}(B)<\alpha_{\infty}-10^{-5}$;

• •

$B_{n-3}\dots B_{n+3}\in\{2112122,2212112\}$.

Suppose that $B_{n-3}\dots B_{n+3}\in\{2112122,2212112\}$. By items (d) and (e) of Lemma 3.4, if $B_{n+4}=1$ or $B_{n-4}=1$, i.e., if

[TABLE]

then $\lambda_{n}(B)<\alpha_{\infty}-10^{-5}$.

Assume that $B_{n-4}\dots B_{n+4}\in\{221121222,222121122\}$. By item (f) of Lemma 3.4, if $(B_{n-5},B_{n+5})=(2,2)$, i.e.,

[TABLE]

then $\lambda_{n}(B)<\alpha_{\infty}-10^{-5}$.

Consider the case $B_{n-4}\dots B_{n+4}\in\{221121222,222121122\}$ and $(B_{n-5},B_{n+5})\neq(2,2)$. Our assumption $\lambda_{n}(B)\leqslant\alpha_{\infty}+10^{-6}$ and the item (iv) of Lemma 3.2 say that the subwords $22212^{*}11221$ and $122112^{*}1222$ are forbidden for $B_{n}=2^{*}$. Therefore, we have just two possibilities in this situation:

[TABLE]

This proves the desired lemma. ∎

By further exploiting Lemma 3.2, we also get the following results:

Lemma 3.6**.**

Let $B=(B_{m})_{m\in\mathbb{Z}}\in\{1,2\}^{\mathbb{Z}}$ be a bi-infinite sequence. Suppose that $\lambda_{n}(B)\leqslant\alpha_{\infty}+10^{-6}$ for some $n\in\mathbb{Z}$.

• •

If $B_{n-5}\dots B_{n+4}=1222121122$, then $B_{n-8}\dots B_{n+8}=12112221211222121$;

• •

If $B_{n-4}\dots B_{n+5}=2211212221$, then $B_{n-8}\dots B_{n+8}=12122211212221121$.

Proof.

Since $2211212221=(1222121122)^{T}$, it suffices to show the lemma in the first case $B_{n-5}\dots B_{n+4}=1222121122$.

By succesively using items (iv), (v), (vi), (vii), (viii), (ix) and (x) of Lemma 3.2 together with our assumption $\lambda_{n}(B)\leqslant\alpha_{\infty}+10^{-6}$, we see that, in our setting, the only possible way to extend $B_{n-5}\dots B_{n+4}=1222121122$ is $B_{n-8}\dots B_{n+8}=1211222121122$. ∎

Lemma 3.7**.**

Let $B=(B_{m})_{m\in\mathbb{Z}}\in\{1,2\}^{\mathbb{Z}}$ be a bi-infinite sequence. Suppose that $\lambda_{n-7}(B),\lambda_{n}(B),\lambda_{n+7}(B)\leqslant\alpha_{\infty}+10^{-6}$ for some $n\in\mathbb{Z}$.

• •

If $B_{n-5}\dots B_{n+4}=12_{3}121_{2}2_{2}$, then:

• –

either $B_{n-10}\dots B_{n+11}=2_{2}121_{2}2_{3}121_{2}2_{3}121_{2}21$,

• –

or $B_{n-10}\dots B_{n+11}=2_{2}121_{2}2_{3}121_{2}2_{3}121_{2}2_{2}$ and, in particular, the vicinity of $B_{n+7}=2$ is $B_{n+2}\dots B_{n+11}=12_{3}121_{2}2_{2}$.

• •

If $B_{n-4}\dots B_{n+5}=2_{2}1_{2}212_{3}1$, then:

• –

either $B_{n-11}\dots B_{n+10}=121_{2}212_{3}1_{2}212_{3}1_{2}212_{2}$,

• –

or $B_{n-11}\dots B_{n+10}=2_{2}1_{2}212_{3}1_{2}212_{3}1_{2}212_{2}$ and, in particular, the vicinity of $B_{n-7}=2$ is $B_{n-11}\dots B_{n-2}=2_{2}1_{2}212_{3}1$.

Proof.

Since $2211212221=(1222121122)^{T}$, it suffices to show the lemma in the first case $B_{n-5}\dots B_{n+4}=1222121122$.

By Lemma 3.6, we have from our hypothesis $\lambda_{n}(B)\leqslant\alpha_{\infty}+10^{-6}$ that $B_{n-8}\dots B_{n+8}=121_{2}2_{3}121_{2}2_{3}121$.

From our assumption $\lambda_{n+7}(B)\leqslant\alpha_{\infty}+10^{-6}$ and the items (i), (ii) of Lemma 3.2, the only way to extend $B_{n-8}\dots B_{n+8}$ is

[TABLE]

From our assumption $\lambda_{n}(B)\leqslant\alpha_{\infty}+10^{-6}$ and the item (xi) of Lemma 3.2, the only way to extend $B_{n-8}\dots B_{n+10}$ is

[TABLE]

From our assumption $\lambda_{n-7}(B)\leqslant\alpha_{\infty}+10^{-6}$ and the item (iii) of Lemma 3.2, the only way to extend $B_{n-9}\dots B_{n+10}$ is

[TABLE]

Thus, $B_{n-10}\dots B_{n+10}$ extends as

[TABLE]

∎

Next, we employ Lemmas 3.2 and 3.3 to get the following statement:

Lemma 3.8**.**

Let $A\in\{1,2\}^{\mathbb{Z}}$ be a bi-infinite sequence. Suppose that, for some $n\in\mathbb{Z}$ and $a\in\mathbb{N}$, one has $\lambda_{n\pm 7}(A)\leqslant B_{\infty}+6\times 10^{-9}$, $\lambda_{n\pm(17+6k)}(A)\leqslant\alpha_{\infty}+10^{-6}$ for each $k=1,\dots,2a$, and $\lambda_{n\pm(7+6j)}(A)\leqslant\alpha_{\infty}+10^{-6}$ for each $j=1,\dots,2a$.

If $A_{n-10}\dots A_{n+11}$ or $(A_{n-11}\dots A_{n+10})^{T}$ equals $2_{2}121_{2}2_{3}121_{2}2_{3}121_{2}21$, then

[TABLE]

In particular, the subsequence $2_{2}121_{2}2_{3}121_{2}2_{3}121_{2}21$ or its transpose $121_{2}212_{3}1_{2}212_{3}1_{2}212_{2}$ is not contained in a bi-infinite sequence $A\in\{1,2\}^{\mathbb{Z}}$ with $m(A)<B_{\infty}$.

Proof.

We can assume that $A_{n-10}\dots A_{n+11}=2_{2}121_{2}2_{3}121_{2}2_{3}121_{2}21$: indeed, the other case $(A_{n-11}\dots A_{n+10})^{T}=2_{2}121_{2}2_{3}121_{2}2_{3}121_{2}21$ is completely similar.

By Lemma 2.1, if $A_{n-10}\dots A_{n+11}=2_{2}121_{2}2_{3}121_{2}2_{3}121_{2}21$, then

[TABLE]

From our assumption $\lambda_{n-7}(A)\leqslant B_{\infty}+6\times 10^{-9}<\alpha_{\infty}+10^{-6}$, we deduce from the items (v), (viii), (x) and (xi) of Lemma 3.2 that

[TABLE]

By Lemma 2.1, one has

[TABLE]

It follows from our assumption $\lambda_{n-7}(A)\leqslant B_{\infty}+6\times 10^{-9}$ and Lemma 3.3 that

[TABLE]

By Lemma 2.1, we get that

[TABLE]

We proceed now by induction. On one hand, by recursively using

• •

the item (iii) of Lemma 3.2 and our assumption $\lambda_{n+1+12j}(A)\leqslant\alpha_{\infty}+10^{-6}$ for $j=1,\dots,a$,

• •

Lemma 2.1, and

• •

the items (i), (ii), (iv) and (v) of Lemma 3.2 and our assumption $\lambda_{n+7+12j}(A)\leqslant\alpha_{\infty}+10^{-6}$ for $j=1,\dots,a$,

we derive that $\lambda_{n}(A)$ is minimized when $A_{n+4+12j}\dots A_{n+15+12j}=2_{2}121_{2}21_{2}212$ for $j=1,\dots,a$. On the other hand, by recursively using

• •

the items (i), (ii), (iv) and (v) of Lemma 3.2 our assumption $\lambda_{n-11-12k}(A)\leqslant\alpha_{\infty}+10^{-6}$ for $k=1,\dots,a$, and

• •

Lemma 2.1,

• •

the item (iii) of Lemma 3.2 and our assumption $\lambda_{n-17-12k}(A)\leqslant\alpha_{\infty}+10^{-6}$ for $k=1,\dots,a$,

we derive that $\lambda_{n}(A)$ is minimized when $A_{n-13-12k}\dots A_{n-24-12k}=121_{2}212_{3}121$ for $k=1,\dots,a$. Therefore,

[TABLE]

Finally, suppose that $A\in\{1,2\}^{\mathbb{Z}}$ is a bi-infinite sequence with $m(A)<B_{\infty}$ containing $2_{2}121_{2}2_{3}121_{2}2_{3}121_{2}21$ or its transpose, say $A_{l-10}\dots A_{l+11}$ or $(A_{l-11}\dots A_{l+10})^{T}$ equals $2_{2}121_{2}2_{3}121_{2}2_{3}121_{2}21$ for some $l\in\mathbb{Z}$. The previous discussion would then imply that

[TABLE]

a contradiction. This completes the proof of the lemma. ∎

At this point, we are ready to describe $M\cap(b_{\infty},B_{\infty})$.

Proposition 3.9**.**

If $\alpha\in M\cap(b_{\infty},B_{\infty})$, then $\alpha\notin L$.

Proof.

Our argument is inspired by the proof of Theorem 4 in Chapter 3 of Cusick-Flahive book [4].

Suppose that $\alpha\in L\cap(b_{\infty},B_{\infty})$. Let $B\in\{1,2\}^{\mathbb{Z}}$ be a bi-infinite sequence such that $\ell(B):=\limsup\limits_{i\to\infty}\lambda_{i}(B)=\alpha$.

Since $\alpha_{\infty}-10^{-5}<b_{\infty}<\alpha<B_{\infty}$, we can fix $N\in\mathbb{N}$ large enough such that

[TABLE]

for all $|n|\geqslant N$, and we can select a monotone sequence $\{n_{k}\}_{k\in\mathbb{N}}$ such that $|n_{k}|\geqslant N$ and $\lambda_{n_{k}}(B)\geqslant\alpha_{\infty}-10^{-5}$ for all $k\in\mathbb{Z}$. Moreover, by reversing $B$ if necessary, we can assume that $n_{k}\to+\infty$ as $k\to\infty$ and $\limsup\limits_{n\to+\infty}\lambda_{n}(B)=\alpha$.

We have two possibilities:

• •

either the sequence $B_{n}B_{n+1}\dots$ contains the subsequence $2_{2}121_{2}2_{3}121_{2}2_{3}121_{2}21$ or its transpose $121_{2}212_{3}1_{2}212_{3}1_{2}212_{2}$ for all $n\geqslant N$,

• •

or there exists $R\geqslant N$ such that $B_{R}B_{R+1}\dots$ does not contain the subsequence $2_{2}121_{2}2_{3}121_{2}2_{3}121_{2}21$ or its transpose $121_{2}212_{3}1_{2}212_{3}1_{2}212_{2}$.

In the first scenario, let $\{m_{k}\}_{k\in\mathbb{N}}$ be a monotone sequence such that $B_{m_{k}-10}\dots B_{m_{k}+11}$ or $(B_{m_{k}-11}\dots B_{m_{k}+10})^{T}$ equals $2_{2}121_{2}2_{3}121_{2}2_{3}121_{2}21$ for all $k\in\mathbb{N}$ and $m_{k}\to+\infty$ as $k\to\infty$. By Lemma 3.8, the fact that $\lambda_{n}(B)<B_{\infty}$ for all $n\geqslant N$ would imply that

[TABLE]

where $a_{k}=\lfloor\frac{m_{k}-17-N}{6}\rfloor$. Since $a_{k}\to\infty$ as $k\to\infty$, it would follow that

[TABLE]

a contradiction.

In the second scenario, we note that, by Lemma 3.5, for each $k\in\mathbb{N}$, we have

• •

either $B_{n_{k}-5}\dots B_{n_{k}+4}=12_{3}121_{2}2_{2}$,

• •

or $B_{n_{k}-4}\dots B_{n_{k}+5}=2_{2}1_{2}212_{3}1$.

If the first possibility occurs for some $k_{0}\in\mathbb{N}$ with $n_{k_{0}}\geqslant R+10$, then the facts that $\lambda_{n}(B)<B_{\infty}<\alpha_{\infty}+10^{-6}$ for all $n\geqslant N$ and the sequence $B_{R}B_{R+1}\dots$ does not contain the subsequence $2_{2}121_{2}2_{3}121_{2}2_{3}121_{2}21$ allow to repeatedly apply Lemma 3.7 at the positions $n_{k_{0}}+7a$, $a\in\mathbb{N}$, to deduce that the sequence $B$ has the form

[TABLE]

If the second possibility occurs for all $n_{k}>R+10$, then the facts that $\lambda_{n}(B)<B_{\infty}<\alpha_{\infty}+10^{-6}$ for all $n\geqslant N$ and the sequence $B_{R}B_{R+1}\dots$ does not contain the subsequence $121_{2}212_{3}1_{2}212_{3}1_{2}212_{2}$ allow to apply $d_{k}:=\lfloor\frac{n_{k}-4-R}{7}\rfloor$ times Lemma 3.7 at the positions $n_{k}-7(j-1)$, $j=1,\dots,d_{k}$, to deduce that the sequence $B$ has the form

[TABLE]

Because $R-4\leqslant n_{k}-7d_{k}\leqslant R+11$ and $n_{k}\to+\infty$, we deduce that $B$ has the form $\dots\overline{212_{3}1_{2}}$.

In any case, the second scenario would imply that

[TABLE]

a contradiction.

In summary, the existence of $\alpha\in L\cap(b_{\infty},B_{\infty})$ leads to a contradiction in any scenario. This proves the proposition. ∎

Remark 3.10**.**

As it was first observed in Theorem 1, pages 47 to 49 of Berstein’s article [1], one can improve Proposition 3.9 by showing that $(b_{\infty},B_{\infty})$ is the largest interval disjoint from $L$ containing $\alpha_{\infty}$.

Actually, it does not take much more work to get this improved version of Proposition 3.9: in fact, since this proposition ensures that $L\cap(b_{\infty},B_{\infty})=\emptyset$, and we have that $b_{\infty}=\ell(\overline{1_{2}2_{3}12})\in L$, it suffices to prove that $B_{\infty}\in L$. For the sake of completeness (and also to correct some mistakes in [1]), we show that $B_{\infty}\in L$ in Appendix A below.

Proposition 3.11**.**

Let $m\in M\cap(b_{\infty},B_{\infty})$. Then, $m=m(B)=\lambda_{0}(B)$ for a sequence $B\in\{1,2\}^{\mathbb{Z}}$ with the following properties:

• •

$B_{-10}\dots B_{0}B_{1}\dots B_{7}\dots=2_{2}121_{2}2_{3}12\overline{1_{2}2_{3}12}$;

• •

there exists $N\geqslant 11$ such that $\dots B_{-N-1}B_{-N}$ is a word on $1$ and $2$ satisfying:

• –

it does not contain the subwords $21212$, $2121_{3}$, $1_{3}212$, $12121_{2}$, $1_{2}2121$, $2_{3}121_{2}2_{2}1$, $12_{2}1_{2}212_{3}$ and $12_{3}121_{2}2_{2}$,

• –

if it contains the subword $2_{2}1_{2}212_{3}1=B_{n-4}\dots B_{n+5}$, then

[TABLE]

Proof.

Let $B\in\{1,2\}^{\mathbb{Z}}$ be a bi-infinite sequence such that $m=m(B)$. Since $m<B_{\infty}$, Proposition 3.9 implies that $\limsup\limits_{i\to\infty}\lambda_{i}(B)=\ell(B)\leqslant b_{\infty}<m$.

Therefore, we can select $N_{0}$ large enough such that $\lambda_{n}(B)<\frac{b_{\infty}+m}{2}<m$ for all $|n|\geqslant N_{0}$. In particular, $m=m(B)=\lambda_{n_{0}}(B)$ for some $|n_{0}|<N_{0}$.

It follows that we can shift $B$ in order to obtain a sequence – still denoted by $B$ – such that $\lambda_{0}(B)=m(B)=m$. Since $m>b_{\infty}>\alpha_{\infty}-10^{-5}$, Lemma 3.5 says that

[TABLE]

Thus, by reversing $B$ if necessary, we obtain a bi-infinite sequence $B\in\{1,2\}^{\mathbb{Z}}$ such that $m=m(B)=\lambda_{0}(B)$ and $B_{-5}\dots B_{4}=1222121122$.

Because $\lambda_{n}(B)\leqslant m<B_{\infty}$ for all $n\in\mathbb{Z}$, we know from Lemma 3.8 that $B$ does not contain the subsequence $2_{2}121_{2}2_{3}121_{2}2_{3}121_{2}21$, and, thus, we can successively apply Lemma 3.7 at the positions $7k$, $k\in\mathbb{N}$, to get that

[TABLE]

Moreover, Lemma 3.5 implies that the word $\dots B_{-11}$ does not contain the subwords $21212$, $2121_{3}$, $1_{3}212$, $12121_{2}$, $1_{2}2121$, $2_{3}121_{2}2_{2}1$ and $12_{2}1_{2}212_{3}$.

Furthermore, the subword $12_{3}121_{2}2_{2}$ can not appear in $\dots B_{n}$ for all $n\leqslant-11$. Indeed, if this happens, since $m(B)=m<B_{\infty}$, it would follow from Lemma 3.8 that $B$ does not contain the subsequence $2_{2}121_{2}2_{3}121_{2}2_{3}121_{2}21$ and, hence, one could repeatedly apply Lemma 3.7 to deduce that $B=\overline{1_{2}2_{3}12}$, a contradiction because this would mean that $b_{\infty}<m=m(B)=m(\overline{1_{2}2_{3}12})=b_{\infty}$.

In summary, we showed that there exists $N\geqslant 11$ such that the word $\dots B_{-N}$ does not contain the subwords $21212$, $2121_{3}$, $1_{3}212$, $12121_{2}$, $1_{2}2121$, $2_{3}121_{2}2_{2}1$, $12_{2}1_{2}212_{3}$ and $12_{3}121_{2}2_{2}$.

Finally, if the word $\dots B_{-11}$ contains the subword $2_{2}1_{2}212_{3}1=B_{n-4}\dots B_{n+5}$, since $B$ does not contain the subsequence $121_{2}212_{3}1_{2}212_{3}1_{2}212_{2}$ (thanks to Lemma 3.8 and the fact that $\lambda_{n}(B)<B_{\infty}$ for all $n\in\mathbb{Z}$), then one can apply Lemma 3.7 at the positions $n-7k$ for all $k\in\mathbb{N}$ to get that

[TABLE]

This completes the proof of the proposition. ∎

Remark 3.12**.**

We use Proposition 3.11 to detect new numbers in $M\setminus L$: see Appendix 5.

3.2. Comparison between $M\setminus L$ near $\alpha_{\infty}$ and the Cantor set $X$

The description of $(M\setminus L)\cap(b_{\infty},B_{\infty})=M\cap(b_{\infty},B_{\infty})$ provided by Propositions 3.9 and 3.11 allows us to compare this piece of $M\setminus L$ with the Cantor set

[TABLE]

where

[TABLE]

introduced in (3.1) above.

Proposition 3.13**.**

$(M\setminus L)\cap(\alpha_{\infty}-10^{-8},\alpha_{\infty}+10^{-8})$* contains the set*

[TABLE]

Proof.

Consider the sequence

[TABLE]

where $2_{3}\gamma\in\{1,2\}^{\mathbb{N}}$ does not contain subwords in $P$ and ; serves to indicate the zeroth position.

On one hand, Remark 2.2 implies that

[TABLE]

and

[TABLE]

and items (a), (b) and (f) of Lemma 3.4 imply that

[TABLE]

for all positions $n\geqslant-12$ except possibly for $n=7k$ with $k\geqslant 1$.

On the other hand,

[TABLE]

so that $\lambda_{0}(B)-\lambda_{7k}(B)>[0;1,2_{3},1_{2},2,1,2_{4}]-[0;1,2_{3},1_{2},2,1,2_{3},1_{2},2,1]>10^{-9}$ for all $k\geqslant 1$.

Moreover, since $2_{3}\gamma$ does not contain subwords in $P$, it follows from (the proof of) Lemma 3.5 that $\lambda_{n}(B)<\alpha_{\infty}-10^{-5}$ for all $n\leqslant-13$.

This shows that $m(B)=\lambda_{0}(B)=[2;\overline{1_{2},2_{3},1,2}]+[0;1,2_{3},1_{2},2,1,2_{4},\gamma]$ belongs to $(M\setminus L)\cap(\alpha_{\infty}-10^{-8},\alpha_{\infty}+10^{-8})$. ∎

Proposition 3.14**.**

$(M\setminus L)\cap(b_{\infty},B_{\infty})$* is contained in the union of*

[TABLE]

and the sets

[TABLE]

where $\delta$ is a finite word in $1$ and $2$.

Proof.

By Proposition 3.11, if $m\in(M\setminus L)\cap(b_{\infty},B_{\infty})$, then $m=m(B)=\lambda_{0}(B)$ with

[TABLE]

where the asterisk indicates the zeroth position, $\delta$ is a finite word in $1$ and $2$, and the infinite word $\gamma$ satisfies:

• •

$\gamma^{T}$ does not contain the subwords $21212$, $2121_{3}$, $1_{3}212$, $12121_{2}$, $1_{2}2121$, $2_{3}121_{2}2_{2}1$, $12_{2}1_{2}212_{3}$ and $12_{3}121_{2}2_{2}$,

• •

if $\gamma^{T}$ contains the subword $2_{2}1_{2}212_{3}1$, then $\gamma^{T}=\overline{212_{3}1_{2}}\mu^{T}$ with $\mu$ a finite word in $1$ and $2$.

It follows that:

• •

if $\gamma^{T}$ contains $2_{2}1_{2}212_{3}1$, then

[TABLE]

where $\theta=\delta\mu$ is a finite word in $1$ and $2$, i.e., $m(B)\in\mathcal{C}$;

• •

otherwise,

[TABLE]

where $\gamma$ does not contain the subwords $21212$, $2121_{3}$, $1_{3}212$, $12121_{2}$, $1_{2}2121$, $2_{3}121_{2}2_{2}1$, $12_{2}1_{2}212_{3}$, $12_{3}121_{2}2_{2}$ and $2_{2}1_{2}212_{3}1$, i.e., $m(B)\in\mathcal{D}(\delta)$.

This completes the proof of the proposition. ∎

3.3. Proof of Theorem 3.1

By putting together Propositions 3.13 and 3.14, we can derive Theorem 3.1.

Indeed, by Proposition 3.13, $(M\setminus L)\cap(b_{\infty},B_{\infty})$ contains a set diffeomorphic to $X$ and, hence,

[TABLE]

By Proposition 3.14, $(M\setminus L)\cap(b_{\infty},B_{\infty})$ is contained in

[TABLE]

Since $\mathcal{C}$ is a countable set and $\{\mathcal{D}(\delta):\delta\in\{1,2\}^{n},n\in\mathbb{N}\}$ is a countable family of subsets diffeomorphic to $X$, it follows that

[TABLE]

This proves Theorem 3.1.

3.4. Lower bounds on $HD(M\setminus L)$

Note that the definition of $X$ in (3.1) implies that $X$ contains the Gauss-Cantor set $K(\{1,2_{2}\})$. Thus, Theorem 3.1 implies that:

Corollary 3.15**.**

One has $HD(M\setminus L)\geqslant HD(X)\geqslant HD(K(\{1,2_{2}\}))>0$.

In Section 4 below, we complete the proof of Theorem 1.1 by employing some classical bounds on Hausdorff dimensions of dynamical Cantor sets discussed in [12, pp. 68–70] to obtain the following refinement of the previous corollary:

Proposition 3.16**.**

One has $HD(M\setminus L)\geqslant HD(K(\{1,2_{2}\}))>0.353$.

Remark 3.17**.**

Of course, this estimate can be improved by computing the value $HD(X)$ using one of the several methods in the literature (e.g., [2], [7], [12], [10], [11], [5]).

4. $0.353<HD(K(\{1,2_{2}\}))<0.35792$

In this section, we revisit pages 68, 69 and 70 of Palis-Takens book [12] to give some bounds on the Hausdorff dimension of the Gauss-Cantor set $K(\{1,2_{2}\})$.

By Lemma 2.1, the convex hull of $K(\{1,2_{2}\})$ is the interval $I$ with extremities $[0;\overline{2}]$ and $[0;1,\overline{2}]$. The images $I_{1}:=\phi_{1}(I)$ and $I_{22}:=\phi_{22}(I)$ of $I$ under the inverse branches

[TABLE]

of the first two iterates of the Gauss map $G(x):=\{1/x\}$ provide the first step of the construction of the Cantor set $K(\{1,2_{2}\})$. In general, given $n\in\mathbb{N}$, the collection $\mathcal{R}^{n}$ of intervals of the $n$th step of the construction of $K(\{1,2_{2}\})$ is given by

[TABLE]

By definition, $K(\{1,2_{2}\})$ is a dynamically defined Cantor set associated to the expanding map $\Psi:I_{1}\cup I_{22}\to I$ with $\Psi|_{I_{1}}=G$, $\Psi|_{I_{22}}=G^{2}$. Following [12, pp. 68–69], given $R\in\mathcal{R}^{n}$, let

[TABLE]

and define $\alpha_{n}\in[0,1]$, $\beta_{n}\in[0,1]$ by

[TABLE]

It is shown in [12, pp. 69–70] that $\alpha_{n}\leqslant HD(K(\{1,2_{2}\}))\leqslant\beta_{n}$ for all $n\in\mathbb{N}$.

Therefore, we can estimate on $K(\{1,2_{2}\})$ by computing $\alpha_{n}$ and $\beta_{n}$ for some particular values of $n\in\mathbb{N}$.

In this direction, let us notice that the quantities $\lambda_{n,R}$ and $\Lambda_{n,R}$ can be calculated along the following lines.

Since:

• •

$G^{\prime}(x)=-1/x^{2}$;

• •

the interval $R=\psi_{x_{1}}\circ\dots\circ\psi_{x_{n}}(I)\in\mathcal{R}^{n}$ associated to a string $(x_{1},\dots,x_{n})\in\{0,1\}^{n}$ has extremities $[0;x_{1},\dots,x_{n},\overline{2}]$ and $[0;x_{1},\dots,x_{n},1,\overline{2}]$, and

• •

$(\Psi^{n})^{\prime}|_{R}$ is monotone333Because $(\Psi^{n})|_{R}$ is a Möbius transformation induced by an integral matrix of determinant $\pm 1$. on each $R\in\mathcal{R}^{n}$,

we have that

[TABLE]

and

[TABLE]

Hence, $\alpha_{n}$ and $\beta_{n}$ are the solutions of

[TABLE]

and

[TABLE]

A computer search444See the Mathematica routine available at ‘www.impa.br/$\sim$cmateus/files/G(1,22)vPT.nb’. for the values of $\alpha_{12}$ and $\beta_{12}$ reveals that

[TABLE]

In particular, $0.353<\alpha_{12}\leqslant HD(K(\{1,2_{2}\}))\leqslant\beta_{12}<0.35792$, so that the proof of Proposition 3.16 and, a fortiori, Theorem 1.1 is now complete.

Remark 4.1**.**

In general, the approximations $\alpha_{n}$ and $\beta_{n}$ given in [12, pp.68–70] converge slowly to the actual value of the Hausdorff dimension: indeed, as it is explained in [12, pp.70], one has $\beta_{n}-\alpha_{n}=O(1/n)$. Hence, it is unlikely that further computations with $\alpha_{n}$ and $\beta_{n}$ will lead to the determination of the first ten decimal digits of $HD(K(\{1,2_{2}\}))$.

On the other hand, a quick implementation555See the Mathematica routine available at ‘www.impa.br/$\sim$cmateus/files/G(1,22)vJP.nb’. of the “thermodynamical” algorithm described in Jenkinson-Pollicott [10] provided the heuristic approximations

[TABLE]

for $HD(K(\{1,2_{2}\}))$. In particular, the super-exponential convergence666I.e., $|s_{n}-HD(K(\{1,2_{2}\}))|=O(\theta^{n^{2}})$ for some $0<\theta<1$. of this algorithm indicates that $HD(K(\{1,2_{2}\}))=0.355\dots$. In principle, this heuristics can be made rigorous along the lines of the recent paper [11], but we have not pursued this direction.

5. New numbers in $M\setminus L$

Consider the sequences $g,G\in\{1,2\}^{\mathbb{Z}}$ given by

[TABLE]

and

[TABLE]

where the asterisks serve to indicate the zeroth position.

In this section, we show that

[TABLE]

and

[TABLE]

are the largest and smallest elements of $(M\setminus L)\cap(b_{\infty},B_{\infty})$.

5.1. The largest element of $(M\setminus L)\cap(b_{\infty},B_{\infty})$

We start the discussions by showing that $c\in M$:

Lemma 5.1**.**

One has $c=\lambda_{0}(G)=m(G)\in M$.

Proof.

By items (a) and (b) of Lemma 3.4, we have $\lambda_{j}(G)<\alpha_{\infty}-10^{-5}<c=\lambda_{0}(G)$ for all $j\in\mathbb{Z}\setminus\{0\}$ except possibly for

• •

$j=-22-12k$, $k\geqslant 0$,

• •

$j=-19-12k$, $k\geqslant 0$,

• •

$j=-16-12k$, $k\geqslant 0$,

• •

$j=-7$,

• •

$j=7k$, $k\geqslant 1$.

By item (c) of Lemma 3.4, we have $\lambda_{-19-12k}(G)<\alpha_{\infty}-10^{-5}<c$ for all $k\geqslant 0$. By item (d) of Lemma 3.4, we also have $\lambda_{-22-12k}(G),\lambda_{-16-12k}(G)<\alpha_{\infty}-10^{-5}<c$ for all $k\geqslant 0$. By item (e) of Lemma 3.4, we get $\lambda_{-7}(G)<\alpha_{\infty}-10^{-5}<c$.

Moreover, by Lemma 2.1, we have that

[TABLE]

for all $k\geqslant 1$.

In summary, we proved that $\lambda_{j}(G)<\lambda_{0}(G)$ for all $j\neq 0$, and, hence, $c=\lambda_{0}(G)=m(G)\in M$. ∎

Let us now prove that $m\leqslant c$ whenever $m\in M\cap(b_{\infty},B_{\infty})$:

Lemma 5.2**.**

If $m\in M\cap(b_{\infty},B_{\infty})$, then $m\leqslant c$.

Proof.

By Proposition 3.14, an element $m\in M\cap(b_{\infty},B_{\infty})$ has the form

[TABLE]

By Lemma 2.1, we have

[TABLE]

Since $\lambda_{0}(B)=m<B_{\infty}$, it follows from Lemma 3.8 that

[TABLE]

By Lemma 2.1, we deduce that

[TABLE]

Since $\lambda_{-16}(B)\leqslant m<B_{\infty}<\alpha_{\infty}+10^{-6}$, it follows from items (i), (ii), (iv) and (v) of Lemma 3.2 that

[TABLE]

By Lemma 2.1, we have

[TABLE]

Since $\lambda_{-22}(B)\leqslant m<B_{\infty}<\alpha_{\infty}+10^{-6}$, it follows from item (iii) of Lemma 3.2 that

[TABLE]

At this point, we proceed by induction: if we apply repeatedly Lemma 2.1, items (i), (ii), (iv) and (v) of Lemma 3.2 at the positions $-16-12k$ for $k\geqslant 1$, and item (iii) of Lemma 3.2 at the positions $-22-12k$ for $k\geqslant 1$, then we obtain

[TABLE]

This completes the proof. ∎

At this point, Proposition 1.3 is an immediate consequence of Lemmas 5.1 and 5.2.

5.2. The smallest element of $(M\setminus L)\cap(b_{\infty},B_{\infty})$

Similarly to the previous subsection, we begin our discussion by showing that $\gamma\in M$:

Lemma 5.3**.**

One has $\gamma=\lambda_{0}(g)=m(g)\in M$.

Proof.

From items (a) and (b) of Lemma 3.4, it follows that $\lambda_{j}(g)<\alpha_{\infty}-10^{-5}<\gamma$ for all $j\in\mathbb{Z}\setminus\{0\}$ except possibly for

• •

$j=-15-8k$, $k\geqslant 0$,

• •

$j=-7$

• •

$j=7k$, $k\geqslant 1$

By item (f) of Lemma 3.4, $\lambda_{-15-8k}(g),\lambda_{-7}(g)<\alpha_{\infty}-10^{-5}<\gamma$ (for $k\geqslant 0$). Also, by Lemma 2.1, we have

[TABLE]

for each $k\geqslant 1$.

In other terms, we showed that $\lambda_{j}(g)<\lambda_{0}(g)$ for all $j\neq 0$, and, a fortiori, $\gamma=\lambda_{0}(g)=m(g)\in M$. ∎

Let us now establish the fact $m\geqslant\gamma$ for all $m\in M\cap(b_{\infty},B_{\infty})$:

Lemma 5.4**.**

If $m\in M\cap(b_{\infty},B_{\infty})$, then $m\geqslant\gamma$.

Proof.

By Proposition 3.13, any $m\in M\cap(b_{\infty},B_{\infty})$ has the form:

[TABLE]

We claim there exists a smallest integer $k_{0}\in\mathbb{N}$ such that $B_{-11-7k_{0}},B_{-12-7k_{0}}\neq 2,1$: otherwise, since $m(B)<B_{\infty}<\alpha_{\infty}+10^{-6}$, we could recursively apply Lemma 3.7 at the positions $n=-7k$ to deduce that $B=\overline{2,1_{2},2_{3},1}$, and, hence $b_{\infty}=m(\overline{2,1_{2},2_{3},1})=m(B)$, a contradiction with our assumption $m(B)>b_{\infty}$.

Note that the definition of $k_{0}$ and Lemma 3.7 imply that

[TABLE]

with $B_{-11-7k_{0}},B_{-12-7k_{0}}\neq 2,1$.

If $B_{-11-7k_{0}}=1$, then we are done because Lemma 2.1 says that

[TABLE]

If $B_{-11-7k_{0}}=2$, then $B_{-11-7k_{0}},B_{-12-7k_{0}}\neq 2,1$ forces $B_{-12-7k_{0}}=2$, and, thus,

[TABLE]

By Lemma 2.1, it follows that

[TABLE]

At this point, we recursively apply items (i), (ii) and (iv) of Lemma 3.2 at the positions $j=-15-8k-7k_{0}$, $k\geqslant 0$ together with Lemma 2.1 to obtain that

[TABLE]

In any case, we proved that $m\geqslant\gamma$, as desired. ∎

Appendix A Berstein’s interval around $\alpha_{\infty}$

In this appendix, we prove that $(b_{\infty},B_{\infty})$ is the largest interval disjoint from $L$ containing $\alpha_{\infty}$.

Remark A.1**.**

The first attempt to describe the largest interval $(b_{\infty},B_{\infty})$ disjoint from $L$ containing $\alpha_{\infty}$ was made by Berstein [1] in 1973: for this reason, we refer to $(b_{\infty},B_{\infty})$ as Berstein’s interval around $\alpha_{\infty}$. As it turns out, his description of $b_{\infty}$ and $B_{\infty}$ in Theorem 1, page 47 of [1] is slightly different from ours (perhaps due to some typographical errors). More precisely:

• •

our value of $b_{\infty}=\ell(\overline{2,1_{2},2_{3},1})=3.2930442439\dots$ is slightly smaller than the value $[2;\overline{1_{2},2_{3},1,2}]+[0;1,2_{3},1_{2},2,1,2_{5},\overline{1,2,1_{2},2_{4}}]=3.2930442642\dots$ proposed by Berstein777Actually, this value proposed by Berstein coincides with the smallest element $\gamma$ of $M\cap(b_{\infty},B_{\infty})$: see Appendix 5.;

• •

our value of $B_{\infty}=3.2930444814\dots$ coincides with the numerical value proposed by Berstein, but curiously enough Berstein also claims that $3.2930444814\dots$ equals888We guess that Berstein wanted to write $[2;1,\overline{1,2_{3},1,2,1_{2},2,1_{2},2}]+[0;\overline{1,2_{3},1_{2},2}]=3.293044481451\dots$ here, but this quantity is slightly larger than $B_{\infty}=3.293044481438\dots$ anyway. $[2;1,\overline{1,2_{3},1,2,1_{2},2,1_{2},2}]+[0;1,\overline{2_{3},1_{2}}]$, which is certainly not true (as this last number is $3.29306183\dots$).

As we pointed out in Remark 3.10, since Proposition 3.9 ensures that $(b_{\infty},B_{\infty})\cap L=\emptyset$ and $b_{\infty}=\ell(\overline{2,1_{2},2_{3},1})\in L$, our task is reduced to the following lemma:

Lemma A.2**.**

One has $B_{\infty}\in L$.

Proof.

Since $L$ is a closed subset of the real line, it suffices to find a sequence $(P_{a})_{a\in\mathbb{N}}$ of finite words in $1$ and $2$ such that

[TABLE]

We claim that the finite words

[TABLE]

given by concatenation of the blocks

[TABLE]

[TABLE]

and

[TABLE]

satisfy $\lim\limits_{a\to\infty}\ell(\overline{P_{a}})=B_{\infty}$.

Indeed, we start by noticing that Lemma 2.1 implies that $B_{\infty}+\frac{1}{2^{12a-1}}>\lambda_{j}(\overline{P_{a}})>B_{\infty}$ whenever the $j$th position of $\overline{P_{a}}$ corresponds to $2^{*}$ in a copy of the block $R$: for the sake of convenience, we denote by $\mathcal{C}_{a}$ the set of such positions. Next, we observe that items (a) and (b) of Lemma 3.4 imply that $\lambda_{j}(\overline{P_{a}})<\alpha_{\infty}-10^{-5}$ except possibly when the $j$th position of $\overline{P_{a}}$ corresponds to $2$ in a copy of $Q_{a}$, $R$ or $S_{a}$ whose immediate neighborhood is $1,2,1$. By inspecting the blocks $Q_{a}$, $R$, $S_{a}$, we see that if the $j$th position of $\overline{P_{a}}$ corresponds to $2$ in a copy of $Q_{a}$, $R$ or $S_{a}$ whose immediate neighborhood is $1,2,1$, then:

• •

either $j\in\mathcal{C}_{a}$ corresponds to $2^{*}$;

• •

or $j+7\in\mathcal{C}_{a}$ and its neighborhood in $\overline{P_{a}}$ is $2_{2},1,2_{2},1,2,1_{2},2_{3},1,2,1_{2},2_{3},1,2,1_{2},2$;

• •

or the neighborhood of the $j$th position is $1_{2},\widetilde{2},1_{2}$ or $1,2,1_{2},\widetilde{2},1,2_{2}$ or $2_{2},1,\widetilde{2},1_{2},2,1$ or $1,2_{2},1,\widetilde{2},1_{2},2$ (where $\widetilde{2}$ indicates the $j$th position).

In the second case, Lemma 3.3 implies that $\lambda_{j}(\overline{P_{a}})<B_{\infty}-10^{-9}$. In the third case, the items (c), (d) and (e) of Lemma 3.4 says that $\lambda_{j}(\overline{P_{a}})<\alpha_{\infty}-10^{-5}$.

It follows from this discussion that

[TABLE]

where $j\in\mathcal{C}_{a}$ corresponds to $2^{*}$ in a copy of the block $R$. This proves the claim. ∎

Appendix B On Cusick-Flahive sequence $(\alpha_{n})_{n\in\mathbb{N}}$

Recall that Theorem 4 in Chapter 3 of Cusick-Flahive book [4] proves that

[TABLE]

for all $n\geqslant 4$.

In this appendix, we show that the largest element $\alpha_{2}$ of the sequence $(\alpha_{n})_{n\in\mathbb{N}}$ belongs to the Lagrange spectrum:

Proposition B.1**.**

One has $\alpha_{2}=3.2930444886\dots\in L$. In particular, $\alpha_{4}$ is the largest element of the sequence $(\alpha_{n})_{n\in\mathbb{N}}$ belonging to $M\setminus L$.

During the proof of this proposition, we will need the following two lemmas:

Lemma B.2**.**

Let $B\in(\mathbb{N}^{*})^{\mathbb{Z}}$ be a bi-infinite sequence. Suppose that $B$ contains the subsequence $2_{3}121_{2}2_{3}12^{*}1_{2}2_{3}121_{2}2_{2}$. Then,

[TABLE]

where $j$ is the position indicated by the asterisk.

Proof.

By Remark 2.2, if $B$ contains $2_{3}121_{2}2_{3}12^{*}1_{2}2_{3}121_{2}2_{2}$, then

[TABLE]

∎

Lemma B.3**.**

Let $B\in(\mathbb{N}^{*})^{\mathbb{Z}}$ be a bi-infinite sequence. Suppose that $B$ contains the subsequence $121_{2}2_{3}121_{2}2_{3}12^{*}1_{2}2_{3}121_{2}21_{2}212_{3}12$. Then,

[TABLE]

where $j$ is the position indicated by the asterisk.

Proof.

By Remark 2.2, if $B$ contains $121_{2}2_{3}121_{2}2_{3}12^{*}1_{2}2_{3}121_{2}21_{2}212_{3}12$, then

[TABLE]

∎

After these preliminaries, we are ready to show Proposition B.1:

Proof of Proposition B.1.

Since $L$ is a closed subset of the real line, it suffices to find a sequence $(T_{a})_{a\in\mathbb{N}}$ of finite words in $1$ and $2$ such that

[TABLE]

We affirm that the finite words

[TABLE]

where

[TABLE]

[TABLE]

and

[TABLE]

satisfy $\lim\limits_{a\to\infty}\ell(\overline{T_{a}})=\alpha_{2}$.

Indeed, we start by observing that Lemma 2.1 implies that $|\lambda_{j}(\overline{T_{a}})-\alpha_{2}|<\frac{1}{2^{7a}}$ whenever the $j$th position of $\overline{T_{a}}$ corresponds to $2^{**}$ or $2^{*}$ in a copy of the block $V$: for the sake of convenience, we denote by $\mathcal{D}_{a}$ the set of such positions. Now, we note that items (a) and (b) of Lemma 3.4 imply that $\lambda_{j}(\overline{T_{a}})<\alpha_{\infty}-10^{-5}$ except possibly when the $j$th position of $\overline{P_{a}}$ corresponds to $2$ in a copy of $U_{a}$, $V$ or $W_{a}$ whose immediate neighborhood is $1,2,1$. By inspecting the blocks $U_{a}$, $V$, $W_{a}$, we see that if the $j$th position of $\overline{P_{a}}$ corresponds to $2$ in a copy of $U_{a}$, $V$ or $W_{a}$ whose immediate neighborhood is $1,2,1$, then:

• •

either $j\in\mathcal{D}_{a}$ corresponds to $2^{**}$ or $2^{*}$;

• •

or $j\notin\mathcal{D}_{a}$ corresponds to $2$ in a copy of $V$ and its neighborhood is $2,1_{2},2,1,2_{2},1$ or $1,2_{2},1,2,1_{2},2$;

• •

or $j$ corresponds to $2$ in a copy of $U_{a}$ or $W_{a}$ and its neighborhood is

• –

$2_{2},1,\widetilde{2},1_{2},2,1$ or $1_{2},\widetilde{2},1_{2}$ or $1,2,1_{2},\widetilde{2},1,2_{2}$,

• –

or $2,1,2_{3},1,2,1_{2},2,1_{2},2,1,2_{3},1_{2},\widetilde{2},1,2_{3},1_{2},2,1,2_{3},1_{2},2,1$,

• –

or $1,2,1_{2},2_{3},1,2,1_{2},2_{3},1,\widetilde{2},1_{2},2_{3},1,2,1_{2},2,1_{2},2,1,2_{3},1,2$,

• –

or $2_{2},1_{2},2,1,2_{3},1_{2},\widetilde{2},1,2_{3},1_{2},2,1,2_{3}$,

• –

or $2_{3},1,2,1_{2},2_{3},1,\widetilde{2},1_{2},2_{3},1,2,1_{2},2_{2}$,

where $\widetilde{2}$ indicates the $j$th position.

In the second and third cases, it follows from items (c), (d), (e) of Lemma 3.4 and Lemmas B.2 and B.3 that $\lambda_{j}(\overline{T_{a}})<\alpha_{2}-6\times 10^{-9}$.

Since $1/2^{7a}<6/10^{9}$ when $a\geqslant 4$, our discussion so far implies that

[TABLE]

for all $a\geqslant 4$. This concludes the argument. ∎