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.

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