Posted by: matheuscmss | March 14, 2017

## HD(M-L) > 0.353

My friend Gugu and I have just uploaded to arXiv our paper ${0.353 < HD(M\setminus L)}$.

In this article, we study the complement ${M\setminus L}$ of the Lagrange spectrum ${L}$ in the Markov spectrum ${M}$ near a non-isolated point ${\alpha_{\infty}}$ found by Freiman, and, as a by-product, we prove that its Hausdorff dimension ${HD(M\setminus L)}$ is

$\displaystyle HD(M\setminus L) > 0.353$

Remark 1 Currently, this paper deals exclusively with lower bounds on ${HD(M\setminus L)}$. In its next version, Gugu and I will include upper bounds on ${HD(M\setminus L)}$.

In what follows, we present a streamlined version of our proof of ${HD(M\setminus L) > 0.353}$ based on the construction of an explicit Cantor set ${K\subset M\setminus L}$ with ${HD(K)>0.353}$.

Remark 2 W e refer to our paper for more refined informations about the structure of ${M\setminus L}$ near ${\alpha_{\infty}}$.

1. Perron’s characterization of the classical spectra

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

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

Here,

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

is the usual continued fraction expansion, and

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

is the ${n}$th convergent.

In 1921, Perron showed that the classical Lagrange and Markov spectra ${L}$ and ${M}$ are the sets

$\displaystyle L=\{\ell(A)<\infty: A\in(\mathbb{N}^*)^{\mathbb{Z}}\} \quad \textrm{and} \quad M=\{m(A)<\infty: A\in(\mathbb{N}^*)^{\mathbb{Z}} \}$

where

$\displaystyle \ell(A)=\limsup\limits_{i\rightarrow\infty}\lambda_i(A) \quad \textrm{and} \quad m(A) = \sup\limits_{i\in\mathbb{Z}} \lambda_i(A)$

2. Freiman’s number ${\alpha_{\infty}}$

In 1973, Freiman showed that

$\displaystyle \alpha_{\infty}:=\lambda_0(A_{\infty}):=[2; \overline{1_2, 2_3, 1, 2}] + [0; 1, 2_3, 1_2, 2, 1, \overline{2}]\in M \setminus L$

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

$\displaystyle \alpha_n:=\lambda_0(A_n):= [2; \overline{1_2, 2_3, 1, 2}] + [0; 1, 2_3, 1_2, 2, 1, 2_n, \overline{1, 2, 1_2, 2_3}]\in M\setminus L$

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

Remark 3 As it turns out, ${\alpha_4}$ is the largest known number in ${M\setminus L}$: see page 35 of Cusick-Flahive book.

In what follows, we shall revisit Freiman’s arguments as described in Chapter 3 of Cusick-Flahive book in order to prove the following result:

Theorem 1 Consider the alphabet ${B=\{\beta_1, \beta_2\}}$ consisting of the words ${\beta_1 = 1\in\mathbb{N}^*}$ and ${\beta_2 = 2_2 = (2,2)\in(\mathbb{N}^*)^2}$. Then,

$\displaystyle K:=\{[2; \overline{1_2, 2_3, 1, 2}] + [0; 1, 2_3, 1_2, 2, 1, 2_4,\gamma_1,\gamma_2,\dots]: \gamma_i\in B \,\,\,\,\forall\,i\geq 1\}\subset M\setminus L$

3. A standard comparison tool

In the sequel, we use the following standard comparison tool for continued fractions is the following lemma (cf. Lemmas 1 and 2 in Chapter 1 of Cusick-Flahive book):

Lemma 2 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 4 For later use, note that Lemma 2 implies that if ${a_0\in\mathbb{Z}}$ and ${a_i\in\mathbb{N}^*}$ for all ${i\geq 1}$, then ${[a_0; a_1,\dots, a_n,\dots]<[a_0; a_1,\dots, a_n,\infty, ...]:=[a_0; a_1,\dots, a_n]}$ when ${n\geq 1}$ is odd, and ${[a_0; a_1,\dots, a_n,\dots]>[a_0; a_1,\dots, a_n]}$ when ${n\geq 0}$ is even.

4. Proof of Theorem 1

Similarly to the discussions in Cusick-Flahive book, we shall use the next lemma (extracted from Lemma 2 in Chapter 3 of this book):

Lemma 3 If ${B\in\{1,2\}^{\mathbb{Z}}}$ contains any of the subsequences

• (a) ${1^*}$
• (b) ${22^*}$
• (f) ${2_4 1 2^* 1_2 2_3}$

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 4 implies that

$\displaystyle \lambda_j(B) = [2;2,\dots]+[0;\dots]<[2; 1, 2, 1] + [0; 2, 2, 1] = \frac{89}{28} < \alpha_{\infty}-10^{-1}$

If (f) occurs, then Remark 4 implies that

$\displaystyle \begin{array}{rcl} \lambda_j(B) &=& [2; 1_2, 2_3, \dots]+[0; 1, 2_4, \dots] \\ &<& [2; 1_2, 2_4, 1] + [0; 1, 2_5, 1] = \frac{45641}{13860} < \alpha_{\infty} - 10^{-5} \end{array}$

$\Box$

We shall also need the following fact:

Lemma 4 If ${A\in(\mathbb{N}^*)^{\mathbb{Z}}}$ is a bi-infinite sequence such that

$\displaystyle \alpha_{\infty}-10^{-8} < m(A) < \alpha_{\infty} + 10^{-8}$

then ${m(A)\in M\setminus L}$.

Proof: See the proof of Theorem 4 in Chapter 3 of Cusick-Flahive book (especially the last paragraph at page 40). $\Box$

These lemmas allow us to conclude the proof of Theorem 1 along the following lines.

Proposition 5 Given a bi-infinite sequence

$\displaystyle B=\dots,\gamma_2^T,\gamma_1^T, 2_4, 1, 2, 1_2, 2_3, 1, 2; \overline{1_2, 2_3, 1, 2}$

where ${\gamma_i\in \{1, 2_2\}}$ for all ${i\geq 1}$ and ${;}$ serves to indicate the zeroth position, then

$\displaystyle m(B) = [2; \overline{1_2, 2_3, 1, 2}] + [0; 1, 2_3, 1_2, 2, 1, 2_4,\gamma_1,\gamma_2,\dots]\in [\alpha_{\infty}-10^{-8}, \alpha_{\infty}+10^{-8}]$

Proof: On one hand, Remark 4 implies that

$\displaystyle \lambda_0(B)\leq [2; \overline{1_2, 2_3, 1, 2}] + [0; 1, 2_3, 1_2, 2, 1, 2_4, 1, 2, 1] < \alpha_{\infty}+10^{-8}$

and

$\displaystyle \lambda_0(B)\geq [2; \overline{1_2, 2_3, 1, 2}] + [0; 1, 2_3, 1_2, 2, 1, 2_4, 2, 1] > \alpha_{\infty} - 10^{-8},$

and items (a), (b) and (f) of Lemma 3 imply that

$\displaystyle \lambda_i(B)<\alpha_{\infty}-10^{-5}$

for all positions ${i}$ except possibly for ${i=7k}$ with ${k\geq 1}$.

On the other hand,

$\displaystyle \begin{array}{rcl} \lambda_{7k}(B) &=& [2; \overline{1_2, 2_3, 1, 2}] + [0;\underbrace{1, 2_3, 1_2, 2, \dots, 1, 2_3, 1_2, 2}_{k \textrm{ times }}, 1, 2_3, 1_2, 2, 1, 2_4,\dots] \\ &<& [2; \overline{1_2, 2_3, 1, 2}] + [0; 1, 2_3, 1_2, 2, 1, 2_3, 1_2, 2, 1] \\ &<& [2; \overline{1_2, 2_3, 1, 2}] + [0; 1, 2_3, 1_2, 2, 1, 2_4] \\ &<& [2; \overline{1_2, 2_3, 1, 2}] + [0; 1, 2_3, 1_2, 2, 1, 2_4,\dots] = \lambda_0(B), \end{array}$

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 ${k\geq 1}$. This proves the proposition. $\Box$

At this point, the proof of Theorem 1 is complete: in fact, Proposition 5 and Lemma 4 together imply that

$\displaystyle K:=\{[2; \overline{1_2, 2_3, 1, 2}] + [0; 1, 2_3, 1_2, 2, 1, 2_4,\gamma_1,\gamma_2,\dots]: \gamma_i\in B \,\,\,\,\forall\,i\geq 1\}$

is contained in ${M\setminus L}$.

5. Lower bounds on ${HD(M\setminus L)}$

The Gauss map ${G:(0,1)\rightarrow[0,1]}$, ${G(x):=\{1/x\}}$ (where ${\{y\}}$ is the fractional part of ${y}$) acts on continued fractions as a shift operator:

$\displaystyle G([0;a_1, a_2, a_3, \dots]) = [0; a_2, a_3, \dots]$

Therefore, we can use the iterates of the Gauss map ${G}$ to build a bi-Lipschitz map between the Cantor set ${K}$ introduced above and the dynamical Cantor set

$\displaystyle K(\{1, 2_2\}):=\{[0;\gamma_1,\gamma_2,\dots]: \gamma_i\in B \,\,\,\,\forall\,i\geq 1\}$

Since the Hausdorff dimension is preserved by bi-Lipschitz maps, an immediate corollary of Theorem 1 is:

Corollary 6 One has ${HD(M\setminus L)\geq HD(K) = HD(K(\{1, 2_2\}))}$.

On the other hand, the Hausdorff dimension ${HD(K(\{1, 2_2\}))}$ was estimated in Subsection 2.2 of this previous post here. In particular, it was shown that:

Proposition 7 One has ${HD(K(\{1, 2_2\})) > 0.353}$.

By putting Corollary 6 and Proposition 7, we conclude the desired estimate

$\displaystyle HD(M\setminus L)>0.353$

in the title of this post.