Intersections of middle- $\alpha$ Cantor sets with a fixed translation

Yan Huang College of Mathematics and Statistics, Chongqing University, 401331, Chongqing, P.R.China [email protected] and Derong Kong College of Mathematics and Statistics, Chongqing University, 401331, Chongqing, P.R.China [email protected]

Abstract.

For $\lambda\in(0,1/3]$ let $C_{\lambda}$ be the middle- $(1-2\lambda)$ Cantor set in $\mathbb{R}$ . Given $t\in[-1,1]$ , excluding the trivial case we show that

\Lambda(t):=\left\{\lambda\in(0,1/3]:C_{\lambda}\cap(C_{\lambda}+t)\neq\emptyset\right\}

is a topological Cantor set with zero Lebesgue measure and full Hausdorff dimension. In particular, we calculate the local dimension of $\Lambda(t)$ , which reveals a dimensional variation principle. Furthermore, for any $\beta\in[0,1]$ we show that the level set

\Lambda_{\beta}(t):=\left\{\lambda\in\Lambda(t):\dim_{H}(C_{\lambda}\cap(C_{\lambda}+t))=\dim_{P}(C_{\lambda}\cap(C_{\lambda}+t))=\beta\frac{\log 2}{-\log\lambda}\right\}

has equal Hausdorff and packing dimension $(-\beta\log\beta-(1-\beta)\log\frac{1-\beta}{2})/\log 3$ . We also show that the set of $\lambda\in\Lambda(t)$ for which $\dim_{H}(C_{\lambda}\cap(C_{\lambda}+t))\neq\dim_{P}(C_{\lambda}\cap(C_{\lambda}+t))$ has full Hausdorff dimension.

Key words and phrases:

Intersection; Cantor set; Hausdorff dimension; Packing dimension; Level set; Digit frequency

2010 Mathematics Subject Classification:

Primary:28A78, 37A10; Secondary: 28D10, 28A80, 37B10

1. Introduction

Intersections of Cantor sets on the real line appear in the setting of homoclinic bifurcations in dynamical systems (cf. [4]). Moreira and Yoccoz [5] studied the stable intersections of regular Cantor sets, and gave an affirmative answer to Palis’ conjecture. Intersections of Cantor sets also appear in number theory. Hall [9] proved that any real number can be written as the sum of two numbers whose continued fractional coefficients are at most $4$ , and from this it follows that the Lagrange spectrum contains a whole half-line. Note that the middle- $\alpha$ Cantor set is an affine Cantor set, which minimizes the Hausdorff dimension with a given thickness (cf. [12, 16]). Motivated by the above works there is a great interest in the study of intersections of middle- $\alpha$ Cantor set with its translations. Kraft [13] gave a complete description when the intersection of middle- $\alpha$ Cantor set with its translation is a single point. Li and Xiao [14] calculated the Hausdorff and packing dimensions of the intersection of middle- $\alpha$ Cantor set with its translation for $\alpha\geq 1/3$ . When $\alpha\in(0,1/3)$ , Zou, Lu and Li [19] determined when the intersection of middle- $\alpha$ Cantor set with its translation is a self-similar set under the condition that the translation has a unique coding. Recently, Baker and the second author [2] proved that for $\alpha\in(0,1/3)$ it is possible that the intersection of middle- $\alpha$ Cantor set with its translation contains only Liouville numbers.

For $\lambda\in(0,{1/2})$ let $C_{\lambda}$ be the middle- $\alpha$ Cantor set with $\alpha=1-2\lambda$ . Then $C_{\lambda}$ is a self-similar set generated by the iterated function system (IFS): $\left\{g_{i}(x)=\lambda x+i(1-\lambda):i=0,1\right\}.$ In other words, $C_{\lambda}$ is the unique nonempty compact set satisfying $C_{\lambda}=g_{0}(C_{\lambda})\cup g_{1}(C_{\lambda})$ . So,

(1.1)

C_{\lambda}=\left\{(1-\lambda)\sum_{n=1}^{\infty}i_{n}\lambda^{n-1}:i_{n}\in\{0,1\}\leavevmode\nobreak\ \forall n\geq 1\right\}.

Observe that $C_{\lambda}\cap(C_{\lambda}+t)\neq\emptyset$ if and only if $t\in C_{\lambda}-C_{\lambda}$ . By (1.1) it follows that

(1.2)

E_{\lambda}:=C_{\lambda}-C_{\lambda}=\left\{(1-\lambda)\sum_{n=1}^{\infty}{i_{n}}\lambda^{n-1}:{i_{n}}\in\{-1,0,1\}\leavevmode\nobreak\ \forall n\geq 1\right\}.

Clearly, if $\lambda\in(0,1/3)$ , then $E_{\lambda}$ is a Cantor set having zero Lebesgue measure. And each $t\in E_{\lambda}$ has a unique coding ${(i_{n})}\in\left\{-1,0,1\right\}^{\mathbb{N}}$ such that $t=(1-\lambda)\sum_{n=1}^{\infty}{i_{n}}\lambda^{n-1}.$ Li and Xiao [14, Theorem 3.4] gave the Hausdorff and packing dimensions of $C_{\lambda}\cap(C_{\lambda}+t)$ :

(1.3)

\begin{split}&\dim_{H}(C_{\lambda}\cap(C_{\lambda}+t))=\frac{\log 2}{-\log\lambda}\,\liminf_{n\to\infty}\frac{\#\{1\leq j\leq n:i_{j}=0\}}{n},\\ &\dim_{P}(C_{\lambda}\cap(C_{\lambda}+t))=\frac{\log 2}{-\log\lambda}\,\limsup_{n\to\infty}\frac{\#\{1\leq j\leq n:i_{j}=0\}}{n},\end{split}

where $\#A$ denotes the cardinality of a set $A$ . If $\lambda=1/3$ , then except for a countable set of points in $E_{\lambda}$ having two different codings, all other $t$ s in $E_{\lambda}$ have a unique coding; and the dimension formulae (1.3) still hold. If $\lambda\in(1/3,1/2)$ , then $E_{\lambda}=[-1,1]$ , and it is well known that Lebesgue almost every $t\in[-1,1]$ has a continuum of codings (cf. [17]). In this case the dimension formulae (1.3) fail for typical $t\in[-1,1]$ , but we still have the dimension formulae (1.3) if $t$ has a unique coding (see [11]).

Refer to caption — Figure 1. The fourth level approximation of the master set $\Gamma$ which consists of all vectors $(\lambda,t)\in(0,1/3]\times[0,1]$ such that $C_{\lambda}\cap(C_{\lambda}+t)\neq\emptyset$ . Each curve corresponds to a unique coding $(i_{n})\in\left\{-1,0,1\right\}^{\mathbb{N}}$ via the equation $t=(1-\lambda)\sum_{n=1}^{\infty}i_{n}\lambda^{n-1}$ . For any $(\lambda,t)\in\Gamma$ the vertical fiber is $E_{\lambda}\cap[0,1]$ , and the horizontal fiber is $\Lambda(t)$ .

Let $\tilde{\Gamma}:=\left\{(\lambda,t)\in(0,1/2)\times[-1,1]:C_{\lambda}\cap(C_{\lambda}+t)\neq\emptyset\right\}$ be the master set defined by the intersections of Cantor sets. Since for any $\lambda\in(1/3,1/2)$ and $t\in[-1,1]$ the intersection $C_{\lambda}\cap(C_{\lambda}+t)$ is always non-empty, and $C_{\lambda}\cap(C_{\lambda}+t)\neq\emptyset$ if and only if $C_{\lambda}\cap(C_{\lambda}-t)\neq\emptyset$ , it is then interesting to consider the subset (see Figure 1)

\Gamma:=\left\{(\lambda,t)\in(0,\frac{1}{3}]\times[0,1]:C_{\lambda}\cap(C_{\lambda}+t)\neq\emptyset\right\}=\left\{(\lambda,t){\in(0,\frac{1}{3}]\times[0,1]}:t\in E_{\lambda}\right\},

where the second equality holds because $C_{\lambda}\cap(C_{\lambda}+t)\neq\emptyset$ if and only $t\in C_{\lambda}-C_{\lambda}=E_{\lambda}$ . Observe that for $\lambda\in(0,1/3]$ the vertical fiber $\Gamma(\lambda)=\left\{t\in[0,1]:t\in E_{\lambda}\right\}=E_{\lambda}\cap[0,1]$ is a Cantor set having Hausdorff dimension $-\log 3/\log\lambda$ . Since $E_{\lambda}$ is a self-similar set, a lot is known about this vertical fiber $\Gamma(\lambda)$ (cf. [10]). On the other hand, for any $t\in[0,1]$ let

(1.4)

\Lambda(t):=\left\{\lambda\in(0,\frac{1}{3}]:C_{\lambda}\cap(C_{\lambda}+t)\neq\emptyset\right\}=\left\{\lambda\in(0,\frac{1}{3}]:t\in E_{\lambda}\right\}

be a horizontal fiber of $\Gamma$ . By (1.2) it follows that $\Lambda(0)=\Lambda(1)=(0,1/3]$ , and $\Lambda(1/3)=\{1/3\}$ . So it is interesting to study $\Lambda(t)$ for $t\in(0,1)\setminus\{1/3\}$ .

Our first result shows that $\Lambda(t)$ is a topological Cantor set, which is a non-empty compact set having neither interior nor isolated points.

Theorem 1.1.

Let $t\in(0,1)\setminus\{1/3\}$ .

(i)

The set $\Lambda(t)$ is a topological Cantor set with ${\min}\Lambda(t)=\min\{t,\frac{1-t}{2}\}$ and ${\max}\Lambda(t)=1/3$ ;
(ii)

$\Lambda(t)$ is a Lebesgue null set having full Hausdorff dimension;
(iii)

For any $\lambda\in\Lambda(t)$ we have

$\lim_{\delta\to 0^{+}}\dim_{H}(\Lambda(t)\cap(\lambda-\delta,\lambda+\delta))=\frac{\log 3}{-\log\lambda}.$

Remark 1.2.

Note that $E_{\lambda}$ is a self-similar set having Hausdorff dimension $-\log 3/\log\lambda$ . Then Theorem 1.1 (iii) shows that the local dimension of $\Lambda(t)$ at $\lambda$ is the same as the local dimension of $E_{\lambda}$ at $t$ . In view of Figure 1 it follows that the local dimensions of $\Gamma$ at any point $(\lambda,t){\in\Gamma}$ through the horizontal and the vertical fibers are the same; this can be viewed as a dimensional ‘variation principle’ for the set $\Gamma$ .

Note by Theorem 1.1 (i) that $\Lambda(t)$ is a topological Cantor set. Then it can be obtained by successively removing a sequence of open intervals from the closed interval $[\min\Lambda(t),1/3]$ . A geometrical construction of $\Lambda(1/2)$ is plotted in Figure 2 (left). By Theorem 1.1 (iii) it follows that

\dim_{H}(\Lambda(t)\cap(0,\lambda])=\sup_{\gamma\in\Lambda(t)\cap(0,\lambda)}\frac{\log 3}{-\log\gamma}\qquad\forall\lambda\in(0,1/3].

Therefore, the dimension function $\psi_{t}:\lambda\mapsto\dim_{H}(\Lambda(t)\cap(0,\lambda])$ , which describes the distribution of $\Lambda(t)$ , is a Cádlág function (see Figure 2, right). It is locally constant almost everywhere, left continuous with right-hand limits everywhere, and thus it has countably infinitely many discontinuities; and it has no downward jumps.

Figure 2. Left: the geometrical construction of

\Lambda(1/2)

. Right: the graph of

\psi_{1/2}:\lambda\mapsto\dim_{H}(\Lambda(1/2)\cap(0,\lambda])

for

\lambda\in(\min\Lambda(1/2),1/3]=(1/4,1/3]

Observe by (1.3) that the dimension of $C_{\lambda}\cap(C_{\lambda}+t)$ is determined by the frequency of digit zero in the coding of $t$ in base $\lambda$ . Then for $\lambda\in(0,1/3]$ the vertical fiber $\Gamma(\lambda)$ can be partitioned as

\Gamma(\lambda)=\Gamma_{not}(\lambda)\cup\bigcup_{\beta\in[0,1]}\Gamma_{\beta}(\lambda),

where

	$\displaystyle\Gamma_{not}(\lambda)$	$\displaystyle:=\left\{t\in[0,1]:\dim_{H}(C_{\lambda}\cap(C_{\lambda}+t))\neq\dim_{P}(C_{\lambda}\cap(C_{\lambda}+t))\right\},$
	$\displaystyle\Gamma_{\beta}(\lambda)$	$\displaystyle:=\left\{t\in[0,1]:\dim_{H}(C_{\lambda}\cap(C_{\lambda}+t))=\dim_{P}(C_{\lambda}\cap(C_{\lambda}+t))=\beta\frac{\log 2}{-\log\lambda}\right\}.$

Li and Xiao [14, Theorem 4.3] showed that

\dim_{H}\Gamma_{\beta}(\lambda)=\dim_{P}\Gamma_{\beta}(\lambda)=\frac{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}{-\log\lambda},

where for a probability vector $(p_{1},p_{2},p_{3})$

(1.5)

{h(p_{1},p_{2},p_{3}):=-\sum_{i=1}^{3}p_{i}\log p_{i}}.

Here we adopt the convention $0\log 0=0$ . Furthermore, an application of [3] gives that $\dim_{H}\Gamma_{not}(\lambda)=-\log 3/\log\lambda$ .

Inspired by the works of [3] and [14] we consider the level sets of the horizontal fiber $\Lambda(t)$ . Given $t\in(0,1)\setminus\left\{1/3\right\}$ , for $\beta\in[0,1]$ let

\Lambda_{\beta}(t):=\left\{\lambda\in\Lambda(t):\dim_{H}\big{(}C_{\lambda}\cap(C_{\lambda}+t)\big{)}=\dim_{P}\big{(}C_{\lambda}\cap(C_{\lambda}+t)\big{)}=\beta\;\frac{\log 2}{-\log\lambda}\right\}.

Then the horizontal fiber $\Lambda(t)$ can be partitioned as

\Lambda(t)=\Lambda_{not}(t)\cup\bigcup_{\beta\in[0,1]}\Lambda_{\beta}(t),

where

\displaystyle\Lambda_{not}(t)

\displaystyle:=\left\{\lambda\in\Lambda(t):\dim_{H}\big{(}C_{\lambda}\cap(C_{\lambda}+t)\big{)}\neq\dim_{P}\big{(}C_{\lambda}\cap(C_{\lambda}+t)\big{)}\right\}.

Theorem 1.3.

For any $t\in(0,1)\setminus\left\{1/3\right\}$ and $\beta\in[0,1]$ the sets $\Lambda_{not}(t)$ and $\Lambda_{\beta}(t)$ are both dense in $\Lambda(t)$ . Furthermore,

\dim_{H}\Lambda_{not}(t)=1\quad\textrm{and}\quad\dim_{H}\Lambda_{\beta}(t)=\dim_{P}\Lambda_{\beta}(t)=\frac{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}{\log 3}.

Remark 1.4.

Our proof of Theorem 1.3 can be adapted to showing that for any $t\in(0,1)\setminus\left\{1/3\right\}$

\Lambda_{fin}(t)={\left\{\lambda\in\Lambda(t):\#C_{\lambda}\cap(C_{\lambda}+t)<+\infty\right\}}

has the same Hausdorff dimension as $\Lambda_{0}(t)$ , that is $\dim_{H}\Lambda_{fin}(t)=\log 2/\log 3$ .

The rest of the paper is organized as follows. In the next section we define the coding map $\Phi_{t}$ which maps each $\lambda\in\Lambda(t)$ to its coding of $t$ in base $\lambda$ , and show that $\Phi_{t}$ is continuous and piecewise monotonic in $\Lambda(t)$ . Based on this map $\Phi_{t}$ we show in Section 3 that $\Lambda(t)$ is a topological Cantor set, and it has full Hausdorff dimension. In Section 4 we calculate the local dimension of $\Lambda(t)$ , and prove Theorem 1.1. Motivated by the works of non-normal numbers we show in Section 5 that $\Lambda_{not}(t)$ is a dense subset of $\Lambda(t)$ with full Hausdorff dimension. Finally, in Section 6 we calculate the dimension of $\Lambda_{\beta}(t)$ and prove Theorem 1.3.

2. Preliminaries

In this section we will define a map $\Phi_{t}$ which maps $\Lambda(t)$ to the symbolic space $\{-1,0,1\}^{\mathbb{N}}$ . First we recall some terminology from symbolic dynamics (cf. [15]). Let $\left\{-1,0,1\right\}^{\mathbb{N}}$ be the set of all infinite sequences over the alphabet $\left\{-1,0,1\right\}$ . By a word we mean a finite string of digits over $\left\{-1,0,1\right\}$ . Denote by $\left\{-1,0,1\right\}^{*}$ the set of all finite words including the empty word $\epsilon$ . For two words $\mathbf{c}=c_{1}\ldots c_{m}$ and $\mathbf{d}=d_{1}\ldots d_{n}$ we write $\mathbf{cd}=c_{1}\ldots c_{m}d_{1}\ldots d_{n}$ for their concatenation. In particular, for any $k\in\mathbb{N}$ we denote by $\mathbf{c}^{k}$ the $k$ -fold concatenation of $\mathbf{c}$ with itself, and by $\mathbf{c}^{\infty}$ the periodic sequence which is obtained by the infinite concatenation of $\mathbf{c}$ with itself. For a word $\mathbf{c}=c_{1}\ldots c_{m}\in\left\{-1,0,1\right\}^{*}$ with $m\geq 1$ , if $c_{m}<1$ we write $\mathbf{c}^{+}:=c_{1}\ldots c_{m-1}(c_{m}+1)$ ; and if $c_{m}>-1$ , we write $\mathbf{c}^{-}:=c_{1}\ldots c_{m-1}(c_{m}-1)$ . Therefore, $\mathbf{c}^{+}$ and $\mathbf{c}^{-}$ are both words over the alphabet $\left\{-1,0,1\right\}$ . Throughout the paper we will use lexicographical order $\prec,\preccurlyeq,\succ$ or $\succcurlyeq$ between sequences in $\left\{-1,0,1\right\}^{\mathbb{N}}$ . For example, we say $(i_{n})\succ(j_{n})$ if $i_{1}>j_{1}$ , or there exists $n\in\mathbb{N}$ such that $i_{1}\ldots i_{n}=j_{1}\ldots j_{n}$ and $i_{n+1}>j_{n+1}$ . And we write $(i_{n})\succcurlyeq(j_{n})$ if $(i_{n})\succ(j_{n})$ or $(i_{n})=(j_{n})$ . Similarly, we say $(i_{n})\prec(j_{n})$ if $(j_{n})\succ(i_{n})$ , and say $(i_{n})\preccurlyeq(j_{n})$ if $(j_{n})\succcurlyeq(i_{n})$ . For two infinite sequences ${\mathbf{c}}=(c_{n}),{\mathbf{d}}=(d_{n})\in\left\{-1,0,1\right\}^{\mathbb{N}}$ with ${\mathbf{c}}\prec{\mathbf{d}}$ we write

({\mathbf{c}},{\mathbf{d}}):=\left\{(i_{n})\in\left\{-1,0,1\right\}^{\mathbb{N}}:{\mathbf{c}}\prec(i_{n})\prec{\mathbf{d}}\right\},\quad[{\mathbf{c}},{\mathbf{d}}]:=\left\{(i_{n})\in\left\{-1,0,1\right\}^{\mathbb{N}}:{\mathbf{c}}\preccurlyeq(i_{n})\preccurlyeq{\mathbf{d}}\right\}.

Similarly, we set

({\mathbf{c}},{\mathbf{d}}]:=\left\{(i_{n})\in\left\{-1,0,1\right\}^{\mathbb{N}}:{\mathbf{c}}\prec(i_{n})\preccurlyeq{\mathbf{d}}\right\},\quad[{\mathbf{c}},{\mathbf{d}}):=\left\{(i_{n})\in\left\{-1,0.1\right\}^{\mathbb{N}}:{\mathbf{c}}\preccurlyeq(i_{n})\prec{\mathbf{d}}\right\}.

For $\lambda\in(0,1/3]$ we note by (1.2) that $E_{\lambda}$ is a self-similar set generated by the IFS $\left\{g_{i}(x)=\lambda x+i(1-\lambda):i=-1,0,1\right\}$ . This induces a map $\pi_{\lambda}$ defined by

\pi_{\lambda}:\left\{-1,0,1\right\}^{\mathbb{N}}\to E_{\lambda};\quad(i_{n})\mapsto(1-\lambda)\sum\limits_{n=1}^{\infty}i_{n}\lambda^{n-1}.

It is clear that the map $\pi_{\lambda}$ is bijective if $\lambda\in(0,1/3)$ ; and the map $\pi_{\lambda}$ is bijective up to a countable set if $\lambda=1/3$ . Now, based on $\pi_{\lambda}$ we define the master map $\Pi$ by

(2.1)

\Pi:\{-1,0,1\}^{\mathbb{N}}\times(0,1/3]\to[-1,1];\quad((i_{n}),\lambda)\mapsto\pi_{\lambda}((i_{n}))=(1-\lambda)\sum_{n=1}^{\infty}i_{n}\lambda^{n-1}.

Note that the symbolic space $\left\{-1,0,1\right\}^{\mathbb{N}}$ becomes a compact metric space under the metric $\rho$ defined by

\rho((i_{n}),(j_{n}))=3^{1-\inf\left\{n\geq 1:i_{n}\neq j_{n}\right\}}.

First we show that $\Pi$ is continuous under the product topology induced by the metric $\rho$ on $\left\{-1,0,1\right\}^{\mathbb{N}}$ and the Euclidean metric $|\cdot|$ on $\mathbb{R}$ .

Lemma 2.1.

The map $\Pi:\left\{-1,0,1\right\}^{\mathbb{N}}\times(0,1/3]\to[-1,1]$ is continuous and onto.

Proof.

Note that $\Pi(\left\{-1,0,1\right\}^{\mathbb{N}}\times\left\{1/3\right\})=[-1,1]$ . It suffices to prove the continuity of $\Pi$ , which follows from the following observation: for any two $((i_{n}),\lambda_{1}),((j_{n}),\lambda_{2})\in\left\{-1,0,1\right\}^{\mathbb{N}}\times(0,1/3]$ with $k=\inf\left\{n\geq 1:i_{n}\neq j_{n}\right\}$ , by (2.1) it follows that

	$\displaystyle\|\Pi((i_{n}),\lambda_{1})-\Pi((j_{n}),\lambda_{2})\|$	$\displaystyle\leq\|\Pi((i_{n}),\lambda_{1})-\Pi((i_{n}),\lambda_{2})\|+\|\Pi((i_{n}),\lambda_{2})-\Pi((j_{n}),\lambda_{2})\|$
		$\displaystyle\leq\sup_{\lambda\in(0,1/3]}\|\Pi_{2}((i_{n}),\lambda)\|\cdot\|\lambda_{1}-\lambda_{2}\|+\left\|(1-\lambda_{2})\sum_{n=1}^{\infty}(i_{n}-j_{n})\lambda_{2}^{n-1}\right\|$
		$\displaystyle\leq 2\|\lambda_{1}-\lambda_{2}\|+\left\|(1-\lambda_{2})\sum_{n=k}^{\infty}2\lambda_{2}^{n-1}\right\|$
		$\displaystyle\leq 2\|\lambda_{1}-\lambda_{2}\|+2\rho((i_{n}),(j_{n})),$

where the second inequality follows by the mean value theorem that

(2.2)

\Pi_{2}((i_{n}),\lambda):=\frac{\partial\Pi((i_{n}),\lambda)}{\partial\lambda}=\sum_{n=1}^{\infty}i_{n}n\left(\frac{n-1}{n}-\lambda\right)\lambda^{n-2},

and the third inequality follows by $\left|\Pi_{2}((i_{n}),\lambda)\right|\leq 1+\sum_{n=2}^{\infty}n\left(\frac{n-1}{n}-\lambda\right)\lambda^{n-2}=2.$ ∎

Next we show that $\Pi$ is monotonic in its first variable.

Lemma 2.2.

Given $\lambda\in(0,1/3]$ , the function $\Pi(\cdot,\lambda)$ is increasing in $\left\{-1,0,1\right\}^{\mathbb{N}}$ with respect to the lexicographical order. In particular, if $\lambda\in(0,1/3)$ , the function $\Pi(\cdot,\lambda)$ is strictly increasing.

Proof.

Take $(i_{n}),(j_{n})\in\{-1,0,1\}^{\mathbb{N}}$ with $(i_{n}){\prec}(j_{n})$ . Then there exists $m\in\mathbb{N}$ such that $i_{n}=j_{n}$ for all $n<m$ , and $i_{m}<j_{m}$ . By (2.1) this implies

	$\displaystyle\Pi((j_{n}),\lambda)-\Pi((i_{n}),\lambda)$	$\displaystyle=(1-\lambda)\sum_{n=m}^{\infty}(j_{n}-i_{n})\lambda^{n-1}$
		$\displaystyle\geq(1-\lambda)\lambda^{m-1}-(1-\lambda)\sum_{n=m+1}^{\infty}2\lambda^{n-1}$
		$\displaystyle=(1-\lambda)\lambda^{m-1}-2\lambda^{m}\geq 0,$

where the last inequality follows by $\lambda\in(0,1/3]$ ; and this inequality is strict if $\lambda\in(0,1/3)$ . ∎

However, $\Pi$ is not always monotonic with respect to its second variable (see Figure 3, left), which complicates our proofs in many places of the paper. Note that the symbolic space $\left\{-1,0,1\right\}^{\mathbb{N}}$ is symmetric, and $\Pi((i_{n}),\cdot)$ and $\Pi((-i_{n}),\cdot)$ have opposite monotonicity. Moreover, $\Pi(0^{\infty},\lambda)\equiv 0$ and $\Pi(1^{\infty},\lambda)\equiv 1$ . So it suffices to consider the piecewise monotonicity of $\Pi((i_{n}),\cdot)$ for $(i_{n})\in(0^{\infty},1^{\infty})$ , which will be divided into four subintervals (see Figure 3, right):

(0^{\infty},001^{\infty}],\quad(001^{\infty},01(-1)0^{\infty}),\quad[01(-1)0^{\infty},01^{\infty}]\quad\textrm{{and}}\quad(01^{\infty},1^{\infty}).

Lemma 2.3.

Let $(i_{n})\in(0^{\infty},1^{\infty})$ .

(i)

If $(i_{n})\in(0^{\infty},001^{\infty}]$ , then $\Pi((i_{n}),\cdot)$ is strictly increasing in $(0,1/3]$ ;
(ii)

If $(i_{n})\in(001^{\infty},01(-1)0^{\infty})$ , then $\Pi((i_{n}),\cdot)$ is strictly concave in $(0,1/3]$ . Furthermore, there exists a unique $\lambda_{(i_{n})}\in[{1}/{4},{1}/{3})$ such that $\Pi((i_{n}),\cdot)$ is strictly increasing in $(0,\lambda_{(i_{n})}]$ and strictly decreasing in $[\lambda_{(i_{n})},{1}/{3}]$ ;
(iii)

If $(i_{n})\in[01(-1)0^{\infty},01^{\infty}]$ , then $\Pi((i_{n}),\cdot)$ is strictly increasing in $(0,1/3]$ ;
(iv)

If $(i_{n})\in(01^{\infty},1^{\infty})$ , then $\Pi((i_{n}),\cdot)$ is strictly decreasing in $(0,1/3]$ .

Proof.

For (i) we take $(i_{n})\in(0^{\infty},001^{\infty}]$ . Then there exists $k\geq 3$ such that $i_{1}\ldots i_{k}=0^{k-1}1$ . By (2.2) it follows that

	$\displaystyle\Pi_{2}((i_{n}),\lambda)$	$\displaystyle\geq k\left(\frac{k-1}{k}-\lambda\right)\lambda^{k-2}-\sum_{n=k+1}^{\infty}n\left(\frac{n-1}{n}-\lambda\right)\lambda^{n-2}$
		$\displaystyle=\lambda^{k-2}(k-1-2k\lambda)>0,$

where the last inequality follows by $\lambda\in(0,1/3)$ and $k\geq 3$ . This proves (i).

Next we consider (ii). Take $(i_{n})\in(001^{\infty},01(-1)0^{\infty})$ . Then $01(-1)^{\infty}\preccurlyeq(i_{n})\prec 01(-1)0^{\infty}$ , and thus $i_{1}i_{2}i_{3}=01(-1)$ and $i_{4}i_{5}\ldots\prec 0^{\infty}$ . By (2.2) it follows that

(2.3)

\begin{split}\Pi_{2}((i_{n}),\lambda)&=2\left(\frac{1}{2}-\lambda\right)-3\left(\frac{2}{3}-\lambda\right)\lambda+\sum_{n=4}^{\infty}i_{n}n\left(\frac{n-1}{n}-\lambda\right)\lambda^{n-2}\\ &=1-4\lambda+3\lambda^{2}+\sum_{n=4}^{\infty}i_{n}n\left(\frac{n-1}{n}-\lambda\right)\lambda^{n-2},\end{split}

and therefore

(2.4)

\frac{\partial\Pi_{2}((i_{n}),\lambda)}{\partial\lambda}=6\lambda-4+\sum_{n=4}^{\infty}i_{n}(n-1)[(n-2)-n\lambda]\lambda^{n-3}<6\lambda-4<0.

So, $\Pi((i_{n}),\cdot)$ is strictly concave in $(0,1/3]$ . Observe by (2.3) that $\Pi_{2}((i_{n}),\lambda)\geq 1-4\lambda>0$ for any $\lambda\in(0,1/4)$ , and $\Pi_{2}((i_{n}),1/3)<0$ since $i_{4}i_{5}\cdots\prec 0^{\infty}$ . So by (2.4) there exists a unique $\lambda_{(i_{n})}\in[1/4,1/3)$ such that $\Pi((i_{n}),\cdot)$ is strictly increasing in $(0,\lambda_{(i_{n})}]$ and strictly decreasing in $[\lambda_{(i_{n})},1/3]$ .

For (iii) we take $(i_{n})\in[01(-1)0^{\infty},01^{\infty}]$ . Then $i_{1}i_{2}=01$ . If $i_{3}=-1$ , then $i_{4}i_{5}\ldots\succeq 0^{\infty}$ . By (2.2) we obtain that $\Pi_{2}((i_{n}),\lambda){\geq}1-4\lambda+3\lambda^{2}>0$ for any $\lambda\in(0,1/3)$ ; and thus $\Pi((i_{n}),\cdot)$ is strictly increasing in $(0,1/3]$ . If $i_{3}\in\left\{0,1\right\}$ , then by (2.2) it follows that

\displaystyle\Pi_{2}((i_{n}),\lambda)\geq{\Pi(010(-1)^{\infty},\lambda)}=2\left(\frac{1}{2}-\lambda\right)-\sum_{n=4}^{\infty}n\left(\frac{n-1}{n}-\lambda\right)\lambda^{n-2}=1-2\lambda-3\lambda^{2}>0

for all $\lambda\in(0,1/3)$ . This proves (iii).

Finally we consider (iv). Take $(i_{n})\in(01^{\infty},1^{\infty})$ . Then $1(-1)^{\infty}\preccurlyeq(i_{n})\prec 1^{\infty}$ , which gives $i_{1}=1$ . By (2.2) it follows that

\displaystyle\Pi_{2}((i_{n}),\lambda)

\displaystyle=-1+\sum_{n=2}^{\infty}i_{n}n\left(\frac{n-1}{n}-\lambda\right)\lambda^{n-2}<-1+\sum_{n=2}^{\infty}n\left(\frac{n-1}{n}-\lambda\right)\lambda^{n-2}=0,

which proves (iv). ∎

Remark 2.4.

From the proof of Lemma 2.3 (ii) it follows that the critical value $\lambda_{(i_{n})}$ is the unique zero in $[1/4,1/3)$ of (2.3). This implies that the map $(i_{n})\mapsto\lambda_{(i_{n})}$ is continuous and strictly increasing in $(001^{\infty},01(-1)0^{\infty}]=[01(-1)^{\infty},01(-1)0^{\infty}]$ with $\lambda_{01(-1)^{\infty}}=1/4$ and $\lambda_{01(-1)0^{\infty}}=1/3$ .

Given $t\in(0,1)\setminus\left\{1/3\right\}$ , by (1.4) it follows that

\Lambda(t)=\left\{\lambda\in(0,1/3]:\Pi((i_{n}),\lambda)=t\textrm{ for some }(i_{n})\in(0^{\infty},1^{\infty})\right\}.

Note by Lemma 2.2 that each $\lambda\in\Lambda(t)\setminus\left\{1/3\right\}$ corresponds to a unique coding $(i_{n})$ satisfying $\Pi((i_{n}),\lambda)=t$ , while for $\lambda=1/3$ there might be two codings $(i_{n})$ satisfying $\Pi((i_{n}),\lambda)=t$ , one ends with $(-1)^{\infty}$ and the other ends with $1^{\infty}$ . Now we define the map

(2.5)

\Phi_{t}:\Lambda(t)\to(0^{\infty},1^{\infty});\quad{\lambda\mapsto(i_{n})\quad\textrm{with}\quad\Pi((i_{n}),\lambda)=t,}

where for $\lambda=1/3$ we set

\left\{\begin{array}[]{lll}\Phi_{t}(1/3)\textrm{ does not end with }1^{\infty}&\textrm{ if }&t\in(0,1/9]\cup[4/27,1/3),\\ \Phi_{t}(1/3)\textrm{ does not end with }(-1)^{\infty}&\textrm{ if }&t\in(1/9,4/27)\cup(1/3,1).\end{array}\right.

So, if $t\in(0,1/9]\cup[4/27,1/3)$ , then $\Phi_{t}(1/3)$ is the greedy triadic coding of $t$ ; and if $t\in(1/9,4/27)\cup(1/3,1)$ , then $\Phi_{t}(1/3)$ is the lazy triadic coding of $t$ . Note that the definition of $\Phi_{t}(1/3)$ depends on the monotonicity of $\Pi(\Phi_{t}(1/3),\cdot)$ (see Figure 3, right).

In the remaining part of this section we will show that $\Phi_{t}$ is continuous and piecewise monotonic, which will be vital in our study of $\Lambda(t)$ and its level sets.

Lemma 2.5.

For any $t\in(0,1)\setminus\left\{1/3\right\}$ the map $\Phi_{t}$ is continuous in $\Lambda(t)$ .

Proof.

The proof is similar to that of [18, Lemma 2.2]. For completeness we sketch its main idea. Take $\lambda_{*}\in\Lambda(t)$ . Suppose on the contrary that $\Phi_{t}$ is not continuous at $\lambda_{*}$ . Then there exist a large $N\in\mathbb{N}$ and a sequence $(\lambda_{k})\subset\Lambda(t)$ such that

(2.6)

\lim_{k\to\infty}\lambda_{k}=\lambda_{*}\quad\textrm{and}\quad{{\rho(\Phi_{t}(\lambda_{k}),\Phi_{t}(\lambda_{*}))}\geq 3^{-N+1}\quad\forall k\geq 1}.

Write $\Phi_{t}(\lambda_{k})=(i_{n}^{(k)})$ and $\Phi_{t}(\lambda_{*})=(i_{n}^{*})$ . By (2.6) it follows that $i_{1}^{(k)}\ldots i_{N}^{(k)}\neq i_{1}^{*}\ldots i_{N}^{*}$ for all $k\geq 1$ . Since $(\left\{-1,0,1\right\}^{\mathbb{N}},\rho)$ is a compact metric space, there exists a subsequence $(k_{j})\subset\mathbb{N}$ such that $\lim_{j\to\infty}(i_{n}^{(k_{j})})$ exists, say $(i_{n}^{\prime})\in\left\{-1,0,1\right\}^{\mathbb{N}}$ . Then

(2.7)

i_{1}^{\prime}\ldots i_{N}^{\prime}\neq i_{1}^{*}\ldots i_{N}^{*}.

Note that

(2.8)

\Pi((i_{n}^{(k_{j})}),\lambda_{k_{j}})=t=\Pi((i_{n}^{*}),\lambda_{*})\quad\forall j\geq 1.

Letting $j\to\infty$ in (2.8), by using (2.6) and Lemma 2.1 it follows that

(2.9)

\Pi((i_{n}^{\prime}),\lambda_{*})=t=\Pi((i_{n}^{*}),\lambda_{*}).

If $\lambda_{*}\in(0,1/3)$ , then by (2.9) and Lemma 2.2 it follows that $(i_{n}^{\prime})=(i_{n}^{*})$ , leading to a contradiction with (2.7). If $\lambda_{*}=1/3$ , then we consider the following two cases.

Case (I). $t\in(0,1/9]\cup[4/27,1/3)$ . Then $(i_{n}^{*})=\Phi_{t}(1/3)\in(0^{\infty},001^{\infty}]\cup[01(-1)0^{\infty},01^{\infty})$ . By Lemma 2.3 it follows that $\Pi((i_{n}^{*}),\cdot)$ is strictly increasing in $(0,1/3]$ . So by (2.8) and Lemma 2.2 it follows that $(i_{n}^{(k_{j})})\succcurlyeq(i_{n}^{*})$ for all $j\geq 1,$ and thus $(i_{n}^{\prime})\succcurlyeq(i_{n}^{*})$ . Since $(i_{n}^{*})=\Phi_{t}(1/3)$ is the greedy triadic coding of $t$ , we must have $(i_{n}^{\prime})=(i_{n}^{*})$ , contradicting to (2.7).

Case (II). $t\in(1/9,4/27)\cup(1/3,1)$ . Then $(i_{n}^{*})=\Phi_{t}(1/3)\in(001^{\infty},01(-1)0^{\infty})\cup(01^{\infty},1^{\infty})$ . Then by Lemma 2.3 there exists $\delta>0$ such that $\Pi((i_{n}^{*}),\cdot)$ is strictly decreasing in $(1/3-\delta,1/3]$ . So by (2.8) and Lemma 2.2 it follows that $(i_{n}^{(k_{j})})\preccurlyeq(i_{n}^{*})$ for all $j\geq 1,$ and therefore $(i_{n}^{\prime})\preccurlyeq(i_{n}^{*})$ . Since $(i_{n}^{*})$ is the lazy triadic coding of $t$ , we must have $(i_{n}^{\prime})=(i_{n}^{*})$ , again leading to a contradiction with (2.7).

Hence, $\Phi_{t}$ is continuous at $\lambda_{*}$ . Since $\lambda_{*}\in\Lambda(t)$ was arbitrary, $\Phi(t)$ is continuous in $\Lambda(t)$ . ∎

We will end this section by showing that $\Phi_{t}$ is piecewise monotonic.

Proposition 2.6 (Key proposition).

Let $t\in(0,1)\setminus\left\{1/3\right\}$ .

(i)

If $t\in(0,{1/9}]$ , then $\Phi_{t}$ is strictly decreasing in $\Lambda(t)$ ;
(ii)

If $t\in(1/9,4/27)$ , then there exist a unique $\tau{=\tau(t)}\in[1/4,1/3)$ such that $\Phi_{t}$ is strictly decreasing in $(0,\tau]\cap\Lambda(t)$ and strictly increasing in $[\tau,1/3]\cap\Lambda(t)$ ;
(iii)

If $t\in[{4}/{27},{1}/{3})$ , then $\Phi_{t}$ is strictly decreasing in $\Lambda(t)$ ;
(iv)

If $t\in({1}/{3},1)$ , then $\Phi_{t}$ is strictly increasing in $\Lambda(t)$ .

The proof of Proposition 2.6 will be split into several lemmas. First we consider (iii) and (iv).

Lemma 2.7.

(i)

If $t\in[{4}/{27},{1}/{3})$ , then $\Phi_{t}$ is strictly decreasing in $\Lambda(t)$ ;
(ii)

If $t\in({1}/{3},1)$ , then $\Phi_{t}$ is strictly increasing in $\Lambda(t)$ .

Proof.

In view of Lemma 2.3, the proof of (ii) is very similar to (i). Here we only prove (i). Take $t\in[4/27,1/3)$ . Then $\Phi_{t}(1/3)\in[01(-1)0^{\infty},01^{\infty})$ . By Lemma 2.3 (iii) it follows that for any $\lambda\in\Lambda(t)\setminus\left\{1/3\right\}$

\Pi(\Phi_{t}(\lambda),\lambda)=t=\Pi(\Phi_{t}(1/3),1/3)>\Pi(\Phi_{t}(1/3),\lambda),

which, together with Lemma 2.2, implies $\Phi_{t}(\lambda)\succ\Phi_{t}(1/3)$ . Furthermore, note that $\Phi_{t}(\lambda)\preccurlyeq 01^{\infty}$ for any $\lambda\in\Lambda(t)\setminus\left\{1/3\right\}$ , since otherwise $t=\Pi(\Phi_{t}(\lambda),\lambda)\geq\Pi(1(-1)^{\infty},\lambda)=1-2\lambda>1/3$ , a contradiction. So, $\Phi_{t}(\lambda)\in[01(-1)0^{\infty},01^{\infty}{]}$ for all $\lambda\in\Lambda(t)$ . By Lemma 2.3 (iii) it follows that for any $\lambda_{1},\lambda_{2}\in\Lambda(t)$ with $\lambda_{1}<\lambda_{2}$

\Pi(\Phi_{t}(\lambda_{1}),\lambda_{1})=t=\Pi(\Phi_{t}(\lambda_{2}),\lambda_{2})>\Pi(\Phi_{t}(\lambda_{2}),\lambda_{1}),

which implies $\Phi_{t}(\lambda_{1})\succ\Phi_{t}(\lambda_{2})$ . This completes the proof. ∎

In the following we consider $t\in(0,4/27)$ . Note by Lemma 2.3 (ii) that $\Pi((i_{n}),\cdot)$ is not globally monotonic in $(0,1/3]$ for $(i_{n})\in[01(-1)^{\infty},01(-1)0^{\infty})$ , which complicates our proofs of Proposition 2.6 (i) and (ii).

Lemma 2.8.

Let $t\in(0,4/27)$ , and let $\lambda_{\diamond}=\lambda_{\diamond}(t)$ be the unique root in $(0,1/3)$ of the equation $\lambda_{\diamond}(1-\lambda_{\diamond})^{2}=t$ . Then $\Phi_{t}$ is strictly decreasing in $(0,\lambda_{\diamond}]\cap\Lambda(t)$ .

Proof.

The proof is similar to Lemma 2.7. Since $t=\lambda_{\diamond}(1-\lambda_{\diamond})^{2}=\Pi(01(-1)0^{\infty},\lambda_{\diamond})$ , by Lemma 2.3 (iii) it follows that for any $\lambda\in(0,\lambda_{\diamond}]\cap\Lambda(t)$

\Pi(\Phi_{t}(\lambda),\lambda)=t=\Pi(01(-1)0^{\infty},\lambda_{\diamond})\geq\Pi(01(-1)0^{\infty},\lambda),

which gives $\Phi_{t}(\lambda)\succcurlyeq 01(-1)0^{\infty}$ . Furthermore, $\Phi_{t}(\lambda)\preccurlyeq 01^{\infty}$ since $t<1/3$ . Therefore, $\Phi_{t}(\lambda)\in[01(-1)0^{\infty},01^{\infty}]$ for any $\lambda\in(0,\lambda_{\diamond}]\cap\Lambda(t)$ . By the same argument as in the proof of Lemma 2.7 (i) one can prove that $\Phi_{t}$ is strictly decreasing in $(0,\lambda_{\diamond}]\cap\Lambda(t)$ . ∎

Now we are ready to prove Proposition 2.6 (i). Note by Lemma 2.3 (ii) that for any $(i_{n})\in(001^{\infty},01(-1)0^{\infty})$ there exists a unique $\lambda_{(i_{n})}\in[1/4,1/3)$ such that $\Pi((i_{n}),\cdot)$ is strictly increasing in $(0,\lambda_{(i_{n})}]$ and strictly decreasing in $[\lambda_{(i_{n})},1/3]$ .

Lemma 2.9.

Let $t\in(0,1/9]$ . Then $\Phi_{t}$ is strictly decreasing in $\Lambda(t)$ .

Proof.

First we show that $\Phi_{t}$ is strictly decreasing in $[\sqrt{t},1/3]\cap\Lambda(t)$ . Observe that $t=(\sqrt{t})^{2}=\Pi(001^{\infty},\sqrt{t})$ . By the same argument as in the proof of Lemma 2.8 we can prove that $\Phi_{t}(\lambda)\in(0^{\infty},001^{\infty}]$ for any $\lambda\in[\sqrt{t},1/3]\cap\Lambda(t)$ , and then by Lemma 2.3 (i) it follows that $\Phi_{t}$ is strictly decreasing in $[\sqrt{t},1/3]\cap\Lambda(t)$ . Note by Lemma 2.2 that $\Pi(01(-1)0^{\infty},\lambda_{\diamond})=t=\Pi(001^{\infty},\sqrt{t})<\Pi(01(-1)0^{\infty},\sqrt{t}).$ Then by Lemma 2.3 (iii) it follows that $\lambda_{\diamond}<\sqrt{t}$ . So, by Lemma 2.8 it suffices to prove that $\Phi_{t}$ is strictly decreasing in $(\lambda_{\diamond},\sqrt{t})\cap\Lambda(t)$ . Note that $\Pi(01(-1)0^{\infty},\lambda_{\diamond})=t=\Pi(001^{\infty},\sqrt{t})$ . Then by Lemma 2.3 it follows that for any $\lambda\in(\lambda_{\diamond},\sqrt{t})\cap\Lambda(t)$

\Pi(001^{\infty},\lambda)<\Pi(001^{\infty},\sqrt{t})=t=\Pi(\Phi_{t}(\lambda),\lambda)=\Pi(01(-1)0^{\infty},\lambda_{\diamond})<\Pi(01(-1)0^{\infty},\lambda),

which implies

(2.10)

\Phi_{t}(\lambda)\in(001^{\infty},01(-1)0^{\infty}).

Now we claim that $\lambda<\lambda_{\Phi_{t}(\lambda)}$ for all $\lambda\in(\lambda_{\diamond},\sqrt{t})\cap\Lambda(t)$ . Suppose on the contrary that $\lambda\geq\lambda_{\Phi_{t}(\lambda)}$ . Then by the definition of $\lambda_{\Phi_{t}(\lambda)}$ it follows that for any $\lambda\in(\lambda_{\diamond},\sqrt{t})\cap\Lambda(t)$

\Pi(\Phi_{t}(\lambda),\sqrt{t})<\Pi(\Phi_{t}(\lambda),\lambda)=t=\Pi(001^{\infty},\sqrt{t}),

which yields $\Phi_{t}(\lambda)\prec 001^{\infty}$ , leading to a contradiction with (2.10). This proves the claim.

Therefore, for any $\lambda_{1},\lambda_{2}\in(\lambda_{\diamond},\sqrt{t})\cap\Lambda(t)$ with $\lambda_{1}<\lambda_{2}$

\Pi(\Phi_{t}(\lambda_{1}),\lambda_{1})=t=\Pi(\Phi_{t}(\lambda_{2}),\lambda_{2})>\Pi(\Phi_{t}(\lambda_{2}),\lambda_{1}),

where the inequality follows by $\lambda_{1}<\lambda_{2}<\lambda_{\Phi_{t}({\lambda_{2}})}$ . This gives $\Phi_{t}(\lambda_{1})\succ\Phi_{t}(\lambda_{2})$ , completing the proof. ∎

In the following it remains to prove Proposition 2.6 (ii). First we consider $t\in(1/9,1/8]$ .

Lemma 2.10.

Let $t\in(1/9,1/8]$ . Then $\Phi_{t}$ is strictly decreasing in $(0,1/4]\cap\Lambda(t)$ and strictly increasing in $[1/4,1/3]\cap\Lambda(t)$ .

Proof.

Note that $\Pi(01(-1)0^{\infty},\lambda_{\diamond})=t\leq{1}/{8}<{9}/{64}=\Pi(01(-1)0^{\infty},{1}/{4}).$ Then by Lemma 2.3 (iii) it follows that $\lambda_{\diamond}=\lambda_{\diamond}(t)<1/4$ for all $t\in(1/9,1/8]$ . So, by Lemma 2.8 it suffices to prove that $\Phi_{t}$ is strictly decreasing in $(\lambda_{\diamond},1/4]\cap\Lambda(t)$ and strictly increasing in $[1/4,1/3]\cap\Lambda(t)$ . Note that $\Pi(01(-1)0^{\infty},\lambda_{\diamond})=t>1/9=\Pi(001^{\infty},1/3)$ . Then by Lemma 2.3 it follows that for any $\lambda\in(\lambda_{\diamond},1/3)\cap\Lambda(t)$

\Pi(001^{\infty},\lambda)\leq\Pi(001^{\infty},1/3)<\Pi(\Phi_{t}(\lambda),\lambda)=t=\Pi(01(-1)0^{\infty},\lambda_{\diamond})<\Pi(01(-1)0^{\infty},\lambda).

This implies that

(2.11)

\Phi_{t}(\lambda)\in(001^{\infty},01(-1)0^{\infty})=[01(-1)^{\infty},01(-1)0^{\infty}).

So, by Lemma 2.3 (ii) it follows that for any $\lambda_{1},\lambda_{2}\in\Lambda(t)\cap(\lambda_{\diamond},1/4]$ with $\lambda_{1}<\lambda_{2}$

\Pi(\Phi_{t}(\lambda_{1}),\lambda_{1})=t=\Pi(\Phi_{t}(\lambda_{2}),\lambda_{2})>\Pi(\Phi_{t}(\lambda_{2}),\lambda_{1}),

which gives $\Phi_{t}(\lambda_{1})\succ\Phi_{t}(\lambda_{2})$ .

Next we show that $\Phi_{t}$ is strictly increasing in $[1/4,1/3]\cap\Lambda(t)$ . We claim that $\lambda>\lambda_{\Phi_{t}(\lambda)}$ in this case. Suppose on the contrary that $\lambda\leq\lambda_{\Phi_{t}(\lambda)}$ for some $\lambda\in(1/4,1/3)\cap\Lambda(t)$ . Then

\Pi(\Phi_{t}(\lambda),1/4)<\Pi(\Phi_{t}(\lambda),\lambda)=t\leq 1/8=\Pi(01(-1)^{\infty},1/4),

which implies $\Phi_{t}(\lambda)\prec 01(-1)^{\infty}$ , leading to a contradiction with (2.11). This proves the claim, and then it follows that for any $\lambda_{1},\lambda_{2}\in(1/4,1/3)\cap\Lambda(t)$ with $\lambda_{1}<\lambda_{2}$

\Pi(\Phi_{t}(\lambda_{2}),\lambda_{2})=t=\Pi(\Phi_{t}(\lambda_{1}),\lambda_{1})>\Pi(\Phi_{t}(\lambda_{1}),\lambda_{2}),

where the inequality follows by $\lambda_{2}>\lambda_{1}>\lambda_{\Phi_{t}(\lambda_{1})}$ . This gives $\Phi_{t}(\lambda_{2})\succ\Phi_{t}(\lambda_{1})$ . ∎

To prove Proposition 2.6 (ii) for $t\in(1/8,4/27)$ we need two lemmas on the critical value $\lambda_{(i_{n})}$ .

Lemma 2.11.

The map $\phi:(i_{n})\mapsto\Pi((i_{n}),\lambda_{(i_{n})})$ is strictly increasing and continuous in the interval $[01(-1)^{\infty},01(-1)0^{\infty})$ .

Proof.

Note by Remark 2.4 that $(i_{n})\mapsto\lambda_{(i_{n})}$ is continuous. Then the continuity of $\phi$ follows by Lemma 2.1. And the monotonicity of $\phi$ follows by Lemma 2.2 that for any $(i_{n}),(j_{n})\in[01(-1)^{\infty},01(-1)0^{\infty})$ with $(i_{n})\prec(j_{n})$

\Pi((i_{n}),\lambda_{(i_{n})})<\Pi((j_{n}),\lambda_{(i_{n})})\leq\Pi((j_{n}),\lambda_{(j_{n})}),

where the last inequality follows by Lemma 2.3 (ii) that $\Pi((j_{n}),\cdot)$ attains its maximum value at $\lambda_{(j_{n})}\in[1/4,1/3)$ . ∎

Lemma 2.12.

Let $t\in(1/8,4/27)$ . Then there exists a unique $\tau=\tau(t)\in(1/4,1/3)\cap\Lambda(t)$ such that $\lambda_{\Phi_{t}(\tau)}=\tau$ . Furthermore, the map $t\mapsto\tau(t)$ is strictly increasing.

Proof.

Note by Remark 2.4 that $\lambda_{01(-1)^{\infty}}=1/4$ and $\lambda_{01(-1)0^{\infty}}=1/3$ . Then

\phi(01(-1)^{\infty})=\Pi\left(01(-1)^{\infty},\frac{1}{4}\right)=\frac{1}{8},\quad\phi(01(-1)0^{\infty})=\Pi\left(01(-1)0^{\infty},\frac{1}{3}\right)=\frac{4}{27}.

So, by Lemma 2.11 it follows that $\phi$ bijectively maps $(01(-1)^{\infty},01(-1)0^{\infty})$ to $(1/8,4/27)$ . Since $t\in(1/8,4/27)$ , by Lemma 2.11 there exists a unique $(j_{n})\in(01(-1)^{\infty},01(-1)0^{\infty})$ such that

(2.12)

t=\Pi((j_{n}),\lambda_{(j_{n})}).

Observe by Remark 2.4 that the map $(i_{n})\mapsto\lambda_{(i_{n})}$ is strictly increasing, $\lambda_{01(-1)^{\infty}}=1/4$ and $\lambda_{01(-1)0^{\infty}}=1/3$ . Then by (2.12) it follows that $\tau:=\lambda_{(j_{n})}\in(1/4,1/3)\cap\Lambda(t)$ and $(j_{n})=\Phi_{t}(\tau)$ . Thus, $\lambda_{\Phi_{t}(\tau)}=\lambda_{(j_{n})}=\tau$ .

Next we prove that the map $t\mapsto\tau(t)$ is strictly increasing. Take $t_{1},t_{2}\in(1/8,4/27)$ with $t_{1}<t_{2}$ . Write $\tau_{i}=\tau(t_{i})$ for $i=1,2$ . Since $\Pi(\Phi_{t_{1}}(\tau_{1}),\tau_{1})=t_{1}<t_{2}=\Pi(\Phi_{t_{2}}(\tau_{2}),\tau_{2}),$ by Lemma 2.11 we have $\Phi_{t_{1}}(\tau_{1})\prec\Phi_{t_{2}}(\tau_{2})$ . Hence, by Remark 2.4 it follows that

\tau_{1}=\lambda_{\Phi_{t_{1}}(\tau_{1})}<\lambda_{\Phi_{t_{2}}(\tau_{2})}=\tau_{2}

as desired. ∎

Lemma 2.13.

Let $t\in(1/8,4/27)$ . Then $\Phi_{t}$ is strictly decreasing in $(0,\tau]\cap\Lambda(t)$ and strictly increasing in $[\tau,1/3]\cap\Lambda(t)$ , where $\tau=\tau(t)$ is defined as in Lemma 2.12.

Proof.

Note by the proof of Lemma 2.12 that $\Phi_{t}(\tau)\in(01(-1)^{\infty},01(-1)0^{\infty})$ for any $t\in(1/8,4/27)$ . Then by Lemma 2.2 it follows that $\Pi(01(-1)0^{\infty},\lambda_{\diamond})=t=\Pi(\Phi_{t}(\tau),\tau)<\Pi(01(-1)0^{\infty},\tau),$ which implies $\lambda_{\diamond}<\tau$ by Lemma 2.3 (iii). So, by Lemma 2.8 it suffices to prove that $\Phi_{t}$ is strictly decreasing in $(\lambda_{\diamond},\tau]\cap\Lambda(t)$ and strictly increasing in $[\tau,1/3]\cap\Lambda(t)$ . First we claim that $\lambda\leq\lambda_{\Phi_{t}(\lambda)}$ for all $\lambda\in(\lambda_{\diamond},\tau]\cap\Lambda(t)$ . Note that for $\lambda\in(\lambda_{\diamond},\tau]\cap\Lambda(t)$

\Pi(\Phi_{t}(\lambda),\lambda)=t=\Pi(\Phi_{t}(\tau),\tau)\geq\Pi(\Phi_{t}(\tau),\lambda),

which gives $\Phi_{t}(\lambda)\succcurlyeq\Phi_{t}(\tau)$ . Since the map $(i_{n})\mapsto\lambda_{(i_{n})}$ is strictly increasing, by Lemma 2.12 we conclude that $\lambda_{\Phi_{t}(\lambda)}\geq\lambda_{\Phi_{t}(\tau)}=\tau\geq\lambda$ , proving the claim. Therefore, for any $\lambda_{1},\lambda_{2}\in(\lambda_{\diamond},\tau]\cap\Lambda(t)$ with $\lambda_{1}<\lambda_{2}$

\Pi(\Phi_{t}(\lambda_{1}),\lambda_{1})=t=\Pi(\Phi_{t}(\lambda_{2}),\lambda_{2})>\Pi(\Phi_{t}(\lambda_{2}),\lambda_{1}),

which yields $\Phi_{t}(\lambda_{1})\succ\Phi_{t}(\lambda_{2})$ .

Next we prove that $\Phi_{t}$ is strictly increasing in $[\tau,1/3]\cap\Lambda(t)$ . We claim that $\lambda\geq\lambda_{\Phi_{t}(\lambda)}$ in this case. Suppose on the contrary that $\lambda<\lambda_{\Phi_{t}(\lambda)}$ for some $\lambda\in[\tau,1/3]\cap\Lambda(t)$ . Then

\Pi(\Phi_{t}(\tau),\tau)=t=\Pi(\Phi_{t}(\lambda),\lambda)\geq\Pi(\Phi_{t}(\lambda),\tau),

which implies that $\Phi_{t}(\tau)\succcurlyeq\Phi_{t}(\lambda)$ . Since the map $(i_{n})\mapsto\lambda_{(i_{n})}$ is strictly increasing, by Lemma 2.12 it follows that $\lambda_{\Phi_{t}(\lambda)}\leq\lambda_{\Phi_{t}(\tau)}=\tau\leq\lambda$ , leading to a contradiction. This proves the claim, and then it follows that for any $\lambda_{1},\lambda_{2}\in[\tau,1/3]\cap\Lambda(t)$ with $\lambda_{1}<\lambda_{2}$

\Pi(\Phi_{t}(\lambda_{2}),\lambda_{2})=t=\Pi(\Phi_{t}(\lambda_{1}),\lambda_{1})>\Pi(\Phi_{t}(\lambda_{1}),\lambda_{2}),

which gives $\Phi_{t}(\lambda_{2})\succ\Phi_{t}(\lambda_{1})$ . This completes the proof. ∎

Proof of Proposition 2.6.

The proposition follows by Lemmas 2.7, 2.9, 2.10 and 2.13. ∎

By the proof of Proposition 2.6 (i) and (ii) it follows that

(2.13)

t<\lambda_{\diamond}(t)<\sqrt{t}\leq\frac{1}{3}\leavevmode\nobreak\ \textrm{ if }0<t\leq\frac{1}{9},\qquad t<\lambda_{\diamond}(t)<\tau(t)<\frac{1}{3}\leavevmode\nobreak\ \textrm{ if }\frac{1}{9}<t<\frac{4}{27}.

Here $\tau(t)=1/4$ for any $t\in(1/9,1/8]$ , and $\tau(t)$ is defined as in Lemma 2.12 for $t\in(1/8,4/27)$ .

3. Topology and Hausdorff dimension of $\Lambda(t)$

In this section we will show that $\Lambda(t)$ is a topological Cantor set (Proposition 3.1) and it has full Hausdorff dimension (Corollary 3.7).

3.1. Topology of $\Lambda(t)$

Recall that a topological Cantor set in $\mathbb{R}$ is a non-empty perfect set with no interior points.

Proposition 3.1.

For any $t\in(0,1)\setminus\left\{1/3\right\}$ the set $\Lambda(t)$ is a topological Cantor set with

\min\Lambda(t)=\min\left\{t,\frac{1-t}{2}\right\}\quad\textrm{and}\quad\max\Lambda(t)=\frac{1}{3}.

First we determine the extreme points of $\Lambda(t)$ .

Lemma 3.2.

For any $t\in(0,1)\setminus\{1/3\}$ we have $\min\Lambda(t)=\min\{t,\frac{1-t}{2}\}$ and $\max(\Lambda(t))=1/3$ .

Proof.

Note that $E_{\lambda}=[-1,1]$ for $\lambda=1/3$ . This implies that $\max\Lambda(t)=1/3$ for any $t\in[-1,1]$ . For the minimum value of $\Lambda(t)$ we consider two cases: (I) $t\in(0,1/3)$ ; (II) $t\in(1/3,1)$ .

(I) $t\in(0,1/3)$ . Then $\Phi_{t}(\lambda)\in(0^{\infty},01^{\infty}]$ for any $\lambda\in(0,1/3]\cap\Lambda(t)$ . By Lemmas 2.7 (i) and 2.8 it follows that the minimum value $\lambda_{*}=\min\Lambda(t)$ satisfies $\Pi(01^{\infty},\lambda_{*})=t.$ So, $\min\Lambda(t)=\lambda_{*}=t$ .

(II) $t\in(1/3,1)$ . Then $\Phi_{t}(\lambda)\in[1(-1)^{\infty},1^{\infty}]$ for any $\lambda\in\Lambda(t)$ . Then by Lemma 2.7 (ii) it follows that $\lambda_{*}=\min\Lambda(t)$ satisfies $\Pi(1(-1)^{\infty},\lambda_{*})=t$ , which implies that $t=1-2\lambda_{*}$ . Hence, $\min\Lambda(t)=\lambda_{*}=\frac{1-t}{2}$ . ∎

Next we show that $\Lambda(t)$ is a topological Cantor set.

Lemma 3.3.

For any $t\in(0,1)\setminus\left\{1/3\right\}$ , $\Lambda(t)$ is a non-empty perfect set.

Proof.

By Lemma 3.2 it suffices to prove that $\Lambda(t)$ is closed and has no isolated points. First we prove the closeness of $\Lambda(t)$ . Let $(\lambda_{j})\subset\Lambda(t)$ with $\lim_{j\to\infty}\lambda_{j}=\lambda_{0}$ . Suppose on the contrary that $\lambda_{0}\notin\Lambda(t)$ , i.e., $t\notin E_{\lambda_{0}}$ . Since $E_{\lambda_{0}}$ is compact, we have $dist(t,E_{\lambda_{0}}):=\inf\left\{|x-t|:x\in E_{\lambda_{0}}\right\}>0.$ Note that $E_{\lambda_{0}}=\bigcap_{n=1}^{\infty}E_{\lambda_{0}}(n)$ , where for $\lambda\in(0,1/3]$

(3.1)

E_{\lambda}(n):=\bigcup_{i_{1}\ldots i_{n}\in\left\{-1,0,1\right\}^{n}}g_{i_{1}}\circ\cdots\circ g_{i_{n}}([-1,1])

with $g_{i}(x)=\lambda x+i(1-\lambda)$ for $i=-1,0,1$ . Since $E_{\lambda_{0}}(n)\supset E_{\lambda_{0}}(n+1)$ for all $n\geq 1$ , there exists $N\in\mathbb{N}$ such that

(3.2)

d_{0}:=dist(t,E_{\lambda_{0}}(N))>0.

Note that $\lim_{j\to\infty}\lambda_{j}=\lambda_{0}$ , and by (3.1) each $E_{\lambda}(N)$ is the union of $3^{N}$ pairwise disjoint intervals of equal length $2\lambda^{N}$ . Then $E_{\lambda_{j}}(N)$ converges to $E_{\lambda_{0}}(N)$ under the Hausdorff metric $d_{H}$ as $j\to\infty$ . So there exists a large $j\in\mathbb{N}$ such that

dist(t,E_{\lambda_{0}}(N))\leq d_{H}(E_{\lambda_{j}}(N),E_{\lambda_{0}}(N))\leq\frac{d_{0}}{2},

where the first inequality follows by $t\in E_{\lambda_{j}}\subset E_{\lambda_{j}}(N)$ . This leads to a contradiction with (3.2). So, $\lambda_{0}\in\Lambda(t)$ , and thus $\Lambda(t)$ is closed.

Next we prove that $\Lambda(t)$ has no isolated points. Take $\lambda\in\Lambda(t)$ and let $(i_{n})=\Phi_{t}(\lambda)$ . Then there exists a subsequence $(n_{j})\subset\mathbb{N}$ such that $i_{n_{j}}\in\left\{-1,0\right\}$ for all $j\geq 1$ , or $i_{n_{j}}\in\left\{0,1\right\}$ for all $j\geq 1$ . Without lose of generality we assume that $n_{1}$ is sufficiently large, and $i_{n_{j}}\in\left\{-1,0\right\}$ for all $j\geq 1$ . For any $j\geq 1$ we define $\lambda_{j}\in(0,1/3]\cap(\lambda-\delta,\lambda+\delta)$ for some small $\delta>0$ , such that $\Phi_{t}(\lambda_{j})=i_{1}\ldots i_{n_{j}-1}(i_{n_{j}}+1)0^{\infty}.$ Then it is clear that $(\lambda_{j})\subset\Lambda(t)$ and $\lambda_{j}\to\lambda$ as $j\to\infty$ . Hence, $\lambda$ is not isolated in $\Lambda(t)$ . ∎

Proof of Proposition 3.1.

By Lemmas 3.2 and 3.3 it suffices to prove that $\Lambda(t)$ has no interior points. Take $\lambda_{1},\lambda_{2}\in\Lambda(t)$ with $\lambda_{1}<\lambda_{2}$ . It suffices to prove $(\lambda_{1},\lambda_{2})\setminus\Lambda(t)\neq\emptyset$ . By Proposition 2.6 there exists $\hat{\lambda}\in(\lambda_{1},\lambda_{2})$ such that $\Phi_{t}$ is strictly monotonic in $(\hat{\lambda},\lambda_{2})$ . Without lose of generality we may assume that $\Phi_{t}$ is strictly increasing in $(\hat{\lambda},\lambda_{2})$ . Write $(i_{n})=\Phi_{t}(\hat{\lambda})$ and $(j_{n})=\Phi_{t}(\lambda_{2})$ . Then $(i_{n})\prec(j_{n})$ . So there exists $N\in\mathbb{N}$ such that $i_{1}\ldots i_{N-1}=j_{1}\ldots j_{N-1}$ and $i_{N}<j_{N}$ . Suppose $j_{N}-i_{N}=1$ , since otherwise we can choose a larger $\hat{\lambda}$ . Let

{\mathbf{c}}=i_{1}\ldots i_{N}1^{\infty}\quad\textrm{and}\quad{\mathbf{d}}=j_{1}\ldots j_{N}(-1)^{\infty}.

Then by Lemma 2.2 it follows that

(3.3)

\Pi({\mathbf{c}},\lambda_{2})<\Pi({\mathbf{d}},\lambda_{2})\leq\Pi((j_{n}),\lambda_{2})=t=\Pi((i_{n}),\hat{\lambda})\leq\Pi({\mathbf{c}},\hat{\lambda})<\Pi({\mathbf{d}},\hat{\lambda}).

Denote the open interval by $I_{\lambda}:=(\Pi({\mathbf{c}},\lambda),\Pi({\mathbf{d}},\lambda))$ . Observe that the map $\lambda\mapsto cl(I_{\lambda})$ is continuous with respect to the Hausdorff metric $d_{H}$ . Then (3.3) implies that $t\in I_{\lambda}$ for some $\lambda\in(\hat{\lambda},\lambda_{2})\subset(\lambda_{1},\lambda_{2})$ . Since $I_{\lambda}\cap E_{\lambda}=\emptyset$ , we have $t\notin E_{\lambda}$ , i.e., $\lambda\notin\Lambda(t)$ . So, $(\lambda_{1},\lambda_{2})\setminus\Lambda(t)\neq\emptyset$ , completing the proof. ∎

3.2. Hausdorff dimension of $\Lambda(t)$

Now we turn to prove that $\Lambda(t)$ has full Hausdorff dimension, which can be deduced from the following result.

Proposition 3.4.

For any $t\in(0,1)\setminus\{1/3\}$ we have

{\lim_{\delta\to 0^{+}}\dim_{H}\big{(}\Lambda(t)\cap(1/3-\delta,1/3+\delta)\big{)}=1.}

Our strategy to prove Proposition 3.4 is to construct a sequence of subsets of $\Lambda(t)\cap(1/3-\delta,1/3+\delta)$ whose Hausdorff dimension can be arbitrarily close to one. In view of $\Phi_{t}(1/3)$ defined in (2.5), we consider two cases: $t\in(0,1/9]\cup[4/27,1/3)$ and $t\in(1/9,4/27)\cup(1/3,1)$ .

First we consider $t\in(0,1/9]\cup[4/27,1/3)$ . Then $\Phi_{t}(1/3)\in(0^{\infty},001^{\infty}]\cup[01(-1)0^{\infty},01^{\infty})$ , and $\Phi_{t}(1/3)$ does not end with $1^{\infty}$ . So there exists a subsequence $\{n_{j}\}\subset\mathbb{N}_{\geq 4}$ such that $x_{n_{j}}<1$ for all $j\geq 1$ . Note by Proposition 2.6 (i) and (iii) that $\Phi_{t}$ is strictly decreasing in $\Lambda(t)$ . Then for each $j\geq 1$ there exists a unique $\eta_{j}\in(0,1/3)$ such that

t=\Pi(x_{1}\ldots x_{n_{j}}^{+}1^{\infty},\eta_{j}).

Accordingly, for $k\geq 1$ we let

\Delta^{+}_{k,j}(t):=\{\lambda\in\Lambda(t):\Phi_{t}(\lambda)=x_{1}\dots x_{n_{j}}^{+}d_{1}d_{2}\dots\textrm{ with }d_{ik}\neq-1\leavevmode\nobreak\ \forall i\geq 1\}.

Then by Proposition 2.6 (i) and (iii) it follows that

(3.4)

\Delta^{+}_{k,j}(t)\subset\Lambda(t)\cap(\eta_{j},1/3)\leavevmode\nobreak\ \forall k\geq 1;\quad\textrm{ and }\quad\eta_{j}\nearrow 1/3\leavevmode\nobreak\ \textrm{ {as} }j\to\infty.

Lemma 3.5.

Let $t\in(0,1/9]\cup[4/27,1/3)$ . Then for any $k,j\in\mathbb{N}$ there exists a constant $C>0$ such that

|\pi_{\eta_{j}}(\Phi_{t}(\lambda_{1}))-\pi_{\eta_{j}}(\Phi_{t}(\lambda_{2}))|\leq C|\lambda_{1}-\lambda_{2}|

for any $\lambda_{1},\lambda_{2}\in\Delta^{+}_{k,j}(t)$ .

Proof.

Let $\lambda_{1},\lambda_{2}\in\Delta^{+}_{k,j}(t)$ with $\lambda_{1}<\lambda_{2}$ . Write $(a_{n})=\Phi_{t}(\lambda_{1})$ and $(b_{n})=\Phi_{t}(\lambda_{2})$ . Since $\Phi_{t}$ is strictly decreasing in $[\eta_{j},1/3]\cap\Lambda(t)$ , we have $(a_{n})\succ(b_{n})$ ; and then there exists a $N>n_{j}$ such that $a_{1}\cdots a_{N-1}=b_{1}\cdots b_{N-1}$ and $a_{N}>b_{N}$ . So,

(3.5)

\begin{split}|\pi_{\eta_{j}}(\Phi_{t}(\lambda_{1}))-\pi_{\eta_{j}}(\Phi_{t}(\lambda_{2}))|&=\left|(1-\eta_{j})\left(\sum_{n=N}^{\infty}a_{n}\eta_{j}^{n-1}-\sum_{n=N}^{\infty}b_{n}\eta_{j}^{n-1}\right)\right|\\ &\leq(1-\eta_{j})\sum_{n=N}^{\infty}2\eta_{j}^{n-1}=2\eta_{j}^{N-1}.\end{split}

On the other hand, by Lemma 2.2 it follows that

(1-\lambda_{1})\sum_{n=1}^{\infty}a_{n}\lambda_{1}^{n-1}=t=(1-\lambda_{2})\sum_{n=1}^{\infty}b_{n}\lambda_{2}^{n-1}\leq(1-\lambda_{2})\left(\sum_{n=1}^{N}a_{n}\lambda_{2}^{n-1}-\sum_{n=N+1}^{\infty}\lambda_{2}^{n-1}\right),

which implies

(3.6)

\begin{split}(1-\lambda_{1})\sum_{n=N+1}^{\infty}(a_{n}+1)\lambda_{1}^{n-1}&\leq(1-\lambda_{2})\left(\sum_{n=1}^{N}a_{n}\lambda_{2}^{n-1}-\sum_{n=N+1}^{\infty}\lambda_{2}^{n-1}\right)\\ &\qquad-(1-\lambda_{1})\left(\sum_{n=1}^{N}a_{n}\lambda_{1}^{n-1}-\sum_{n=N+1}^{\infty}\lambda_{1}^{n-1}\right)\\ &=\Pi(a_{1}\ldots a_{N}(-1)^{\infty},\lambda_{2})-\Pi(a_{1}\ldots a_{N}(-1)^{\infty},\lambda_{1})\\ &\leq 2(\lambda_{2}-\lambda_{1}),\end{split}

where the last inequality follows by (2.2) that $\left|\Pi_{2}((i_{n}),\lambda)\right|\leq 2$ for any $(i_{n})\in(0^{\infty},1^{\infty})$ and $\lambda\in(0,1/3]$ . Note by the definition of $\Delta_{k,j}^{+}(t)$ that $a_{N+1}a_{N+2}\cdots\succcurlyeq((-1)^{k-1}0)^{\infty}$ . Then $(a_{N+1}+1)(a_{N+2}+1)\cdots\succcurlyeq(0^{k-1}1)^{\infty}$ . So, by (3.6) and using $\lambda_{1}\in(\eta_{j},1/3)$ it follows that

2(\lambda_{2}-\lambda_{1})>(1-\lambda_{1})\lambda_{1}^{N+k-1}>\frac{2}{3}\eta_{j}^{N+k-1}.

This, together with (3.5), implies that

{|\pi_{\eta_{j}}(\Phi_{t}(\lambda_{1}))-\pi_{\eta_{j}}(\Phi_{t}(\lambda_{2}))|}\leq\frac{6}{\eta_{j}^{k}}|\lambda_{1}-\lambda_{2}|

as required. ∎

Next we consider $t\in(1/9,4/27)\cup(1/3,1)$ . Then $\Phi_{t}(1/3)\in(001^{\infty},01(-1)0^{\infty})\cup(01^{\infty},1^{\infty})$ , and it does not end with $(-1)^{\infty}$ . So there exists a subsequence $\{n_{j}\}\subset\mathbb{N}$ such that $x_{n_{j}}>-1$ for any $j\geq 1$ . By Proposition 2.6 (ii) and (iv) there exists $\delta>0$ such that $\Phi_{t}$ is strictly increasing in $\Lambda(t)\cap(1/3-\delta,1/3]$ . So, without loss of generality, by deleting the first finitely many terms of $\left\{n_{j}\right\}$ , we can assume that for any $j\geq 1$ the equation

t=\Pi(x_{1}\ldots x_{n_{j}}^{-}(-1)^{\infty},\gamma_{j})

determines a unique $\gamma_{j}\in(1/3-\delta,1/3)$ . Note by Lemma 2.3 (ii) that if $x_{1}\ldots x_{n_{j}}^{-}(-1)^{\infty}\in(001^{\infty},01(-1)0^{\infty})$ , the above equation may determine two different $\gamma_{j}$ s in $(0,1/3)$ , but only one is in $(1/3-\delta,1/3)$ . Accordingly, for $k\geq 1$ we set

\Delta^{-}_{k,j}(t):=\{\lambda\in\Lambda(t)\cap(1/3-\delta,1/3):\Phi_{t}(\lambda)=x_{1}\dots x_{n_{j}}^{-}d_{1}d_{2}\dots\textrm{ with }d_{ik}\neq-1\leavevmode\nobreak\ \forall i\geq 1\}.

Then by Proposition 2.6 (ii) and (iv) it follows that

(3.7)

\Delta_{k,j}^{-}(t)\subset\Lambda(t)\cap(\gamma_{j},1/3)\leavevmode\nobreak\ \forall k\geq 1;\quad\textrm{ and }\quad\gamma_{j}\nearrow 1/3\leavevmode\nobreak\ \textrm{ as }j\to\infty.

Lemma 3.6.

Let $t\in(1/9,4/27)\cup(1/3,1)$ . Then for any $k,j\in\mathbb{N}$ there exists a constant $C>0$ such that

{|\pi_{\gamma_{j}}(\Phi_{t}(\lambda_{1}))-\pi_{\gamma_{j}}(\Phi_{t}(\lambda_{2}))|\leq C|\lambda_{1}-\lambda_{2}|}

for any $\lambda_{1},\lambda_{2}\in\Delta^{-}_{k,j}(t)$ .

Proof.

Note that $\Phi_{t}$ is strictly increasing in $[\gamma_{j},1/3]\cap\Lambda(t)$ . Then by a similar argument as in the proof of Lemma 3.5 one can verify that $|\pi_{\gamma_{j}}(\Phi_{t}(\lambda_{1}))-\pi_{\gamma_{j}}(\Phi_{t}(\lambda_{2}))|\leq\frac{6}{\gamma_{j}^{k}}|\lambda_{1}-\lambda_{2}|$ for any $\lambda_{1},\lambda_{2}\in\Delta_{k,j}^{-}(t).$ ∎

Proof of Proposition 3.4.

Take $t\in(0,1)\setminus\left\{1/3\right\}$ . Since the proof for $t\in(1/9,4/27)\cup(1/3,1)$ is similar, we only consider $t\in(0,1/9]\cup[4/27,1/3)$ . By (3.4) and Lemma 3.5 it follows that

(3.8)

\dim_{H}(\Lambda(t)\cap(\eta_{j},1/3))\geq\dim_{H}\Delta_{k,j}^{+}(t)\geq\dim_{H}\pi_{\eta_{j}}(\Phi_{t}(\Delta_{k,j}^{+}(t)))

for all $k,j\in\mathbb{N}$ . Note by the definition of $\Delta_{k,j}^{+}(t)$ that $\Phi_{t}(\Delta_{k,j}^{+}(t))$ consists of all sequences $(i_{n})\in\left\{-1,0,1\right\}^{\mathbb{N}}$ with a prefix $i_{1}\ldots i_{n_{j}}=x_{1}\ldots x_{n_{j}}^{+}$ and $i_{n_{j}+mk}\neq-1$ for all $m\geq 1$ . So, by using $\eta_{j}\in(0,1/3)$ and (3.8) we obtain that

(3.9)

\dim_{H}(\Lambda(t)\cap(\eta_{j},1/3))\geq\frac{(k-1)\log 3+\log 2}{-k\log\eta_{j}}\to\frac{\log 3}{-\log\eta_{j}}{\quad\textrm{as }k\to\infty.}

Hence, by (3.4) and (3.9) it follows that

\lim_{\delta\to 0^{+}}\dim_{H}(\Lambda(t)\cap(1/3-\delta,1/3+\delta))\geq\lim_{j\to\infty}\dim_{H}(\Lambda(t)\cap(\eta_{j},1/3))\geq\lim_{j\to\infty}\frac{\log 3}{-\log\eta_{j}}=1.

The reverse inequality is obvious, since $\dim_{H}(\Lambda(t)\cap(1/3-\delta,1/3+\delta))\leq 1$ for any $\delta>0$ . ∎

The following corollary follows directly from Proposition 3.4.

Corollary 3.7.

For any $t\in(0,1)\setminus\left\{1/3\right\}$ we have $\dim_{H}\Lambda(t)=1$ .

4. Proof of Theorem 1.1

In this section we will determine the local dimension of $\Lambda(t)$ , and prove Theorem 1.1.

Proposition 4.1.

Let $t\in(0,1)\setminus\left\{1/3\right\}$ . Then for any $\lambda\in\Lambda(t)$

(4.1)

\lim_{\delta\to 0^{+}}\dim_{H}\big{(}\Lambda(t)\cap(\lambda-\delta,\lambda+\delta)\big{)}=\frac{\log 3}{-\log\lambda}.

Note by Proposition 3.4 that (4.1) holds for $\lambda=1/3$ . So, in the following we only need to prove (4.1) for $\lambda\in\Lambda(t)\setminus\left\{1/3\right\}$ , which will be split into two subsections.

4.1. A lower bound on the local dimension of $\Lambda(t)$

Take $\lambda_{*}\in\Lambda(t)\setminus\{1/3\}$ and let $(x_{i})=\Phi_{t}(\lambda_{*})$ be the unique coding of $t$ defined as in (2.5), i.e., $t=\Pi((x_{i}),\lambda_{*})$ . We will prove in this subsection that

\lim_{\delta\to 0^{+}}\dim_{H}\big{(}\Lambda(t)\cap(\lambda_{*}-\delta,\lambda_{*}+\delta)\big{)}\geq\frac{\log 3}{-\log\lambda_{*}}.

For this we consider two cases: (I) $(x_{i})$ does not end with $1^{\infty}$ ; (II) $(x_{i})$ ends with $1^{\infty}$ .

Case (I). $(x_{i})=\Phi_{t}(\lambda_{*})$ does not end with $1^{\infty}$ . Then there exists a subsequence $(n_{j})\subset{\mathbb{N}}$ such that $x_{n_{j}}<1$ for all $j\geq 1$ . By Proposition 2.6 there exists a $\delta>0$ such that $(\lambda_{*}-\delta,\lambda_{*}+\delta)\subset(0,1/3)$ , and $\Phi_{t}$ is monotonic in $(\lambda_{*}-\delta,\lambda_{*})\cap\Lambda(t)$ and $(\lambda_{*},\lambda_{*}+\delta)\cap\Lambda(t)$ , respectively. Furthermore, if $\Phi_{t}$ is strictly decreasing in $(\lambda_{*},\lambda_{*}+\delta)\cap\Lambda(t)$ , then so is $\Phi_{t}$ in $(\lambda_{*}-\delta,\lambda_{*}+\delta)\cap\Lambda(t)$ . By deleting the first finitely many terms from $(n_{j})$ we can assume that for any $j\geq 1$ the equation

(4.2)

t=\Pi(x_{1}\ldots x_{n_{j}}^{+}1^{\infty},\eta_{j})

determines a unique root $\eta_{j}\in(\lambda_{*}-\delta,\lambda_{*})$ . Accordingly, for $k\geq 1$ we set

\Delta_{k,j}^{+}(\lambda_{*},t):=\left\{\lambda\in\Lambda(t)\cap(\lambda_{*}-\delta,\lambda_{*}):\Phi_{t}(\lambda)=x_{1}\ldots x_{n_{j}}^{+}d_{1}d_{2}\ldots\textrm{ with }d_{ik}\neq{-1}\leavevmode\nobreak\ \forall i\geq 1\right\}.

Then by Proposition 2.6 it follows that

(4.3)

\Delta_{k,j}^{+}(\lambda_{*},t)\subset(\eta_{j},\lambda_{*})\cap\Lambda(t)\leavevmode\nobreak\ \forall k\geq 1;\quad\textrm{and}\quad\eta_{j}\nearrow\lambda_{*}\leavevmode\nobreak\ \textrm{ {as} }j\to\infty.

If $\Phi_{t}$ is strictly increasing in $(\lambda_{*},\lambda_{*}+\delta)\cap\Lambda(t)$ , then we can assume that for any $j\geq 1$ (4.2) determines a unique $\tilde{\eta}_{j}\in(\lambda_{*},\lambda_{*}+\delta)$ . Accordingly, for $k\geq 1$ we let

\tilde{\Delta}_{k,j}^{+}(\lambda_{*},t):=\left\{\lambda\in\Lambda(t)\cap(\lambda_{*},\lambda_{*}+\delta):\Phi_{t}(\lambda)=x_{1}\ldots x_{n_{j}}^{+}d_{1}d_{2}\ldots\textrm{ with }d_{ik}\neq{-1}\leavevmode\nobreak\ \forall i\geq 1\right\}.

Therefore,

(4.4)

\tilde{\Delta}_{k,j}^{+}(\lambda_{*},t)\subset(\lambda_{*},\tilde{\eta}_{j})\cap\Lambda(t)\leavevmode\nobreak\ \forall k\geq 1;\quad\textrm{and}\quad\tilde{\eta}_{j}\searrow\lambda_{*}\leavevmode\nobreak\ \textrm{ {as} }j\to\infty.

Lemma 4.2.

Let $t\in(0,1)\setminus\left\{1/3\right\}$ and $\lambda_{*}\in\Lambda(t)\setminus\left\{1/3\right\}$ . Suppose $\Phi_{t}(\lambda_{*})$ does not end with $1^{\infty}$ .

(i)

If $\Phi_{t}$ is strictly decreasing in $(\lambda_{*},\lambda_{*}+\delta)\cap\Lambda(t)$ , then for any $k,j\in\mathbb{N}$ there exists $C_{1}>0$ such that

$|\pi_{\eta_{j}}(\Phi_{t}(\lambda_{1}))-\pi_{\eta_{j}}(\Phi_{t}(\lambda_{2}))|\leq C_{1}|\lambda_{1}-\lambda_{2}|$

for any $\lambda_{1},\lambda_{2}\in\Delta_{k,j}^{+}(\lambda_{*},t)$ .
(ii)

If $\Phi_{t}$ is strictly increasing in $(\lambda_{*},\lambda_{*}+\delta)\cap\Lambda(t)$ , then for any $k,j\in\mathbb{N}$ there exists $C_{2}>0$ such that

$|\pi_{\lambda_{*}}(\Phi_{t}(\lambda_{1}))-\pi_{\lambda_{*}}(\Phi_{t}(\lambda_{2}))|\leq C_{2}|\lambda_{1}-\lambda_{2}|$

for any $\lambda_{1},\lambda_{2}\in\tilde{\Delta}_{k,j}^{+}(\lambda_{*},t)$ .

Proof.

Since the proofs of (i) and (ii) are similar, we only prove (i). Suppose $\Phi_{t}$ is strictly decreasing in $(\lambda_{*},\lambda_{*}+\delta)\cap\Lambda(t)$ . Then so is $\Phi_{t}$ in $[\eta_{j},\lambda_{*}]\cap\Lambda(t)$ for any $j\geq 1$ . By a similar argument as in the proof of Lemma 3.5 one can verify that for any $\lambda_{1},\lambda_{2}\in\Delta_{k,j}^{+}(\lambda_{*},t)$

|\pi_{\eta_{j}}(\Phi_{t}(\lambda_{1}))-\pi_{\eta_{j}}(\Phi_{t}(\lambda_{2}))|\leq\frac{6}{\eta_{j}^{k}}|\lambda_{1}-\lambda_{2}|,

proving (i). ∎

Case (II). $(x_{i})=\Phi_{t}(\lambda_{*})$ ends with $1^{\infty}$ . Then there exists a subsequence $(n_{j})\subset\mathbb{N}$ such that $x_{n_{j}}>-1$ for all $j\geq 1$ . The proof is similar to Case (I). By Proposition 2.6 there exists a $\delta>0$ such that $(\lambda_{*}-\delta,\lambda_{*}+\delta)\subset(0,1/3)$ , and $\Phi_{t}$ is monotonic in $(\lambda_{*}-\delta,\lambda_{*})\cap\Lambda(t)$ and $(\lambda_{*},\lambda_{*}+\delta)\cap\Lambda(t)$ , respectively. Furthermore, if $\Phi_{t}$ is strictly decreasing in $(\lambda_{*},\lambda_{*}+\delta)\cap\Lambda(t)$ , then $\Phi_{t}$ is also strictly decreasing in $(\lambda_{*}-\delta,\lambda_{*}+\delta)\cap\Lambda(t)$ . So, by deleting the first finitely many terms from $(n_{j})$ we can assume that for any $j\geq 1$ the equation

(4.5)

t=\Pi(x_{1}\ldots x_{n_{j}}^{-}(-1)^{\infty},\gamma_{j})

determines a unique root $\gamma_{j}\in(\lambda_{*},\lambda_{*}+\delta)$ . Accordingly, for $k\geq 1$ let

\Delta_{k,j}^{-}(\lambda_{*},t):=\left\{\lambda\in\Lambda(t)\cap(\lambda_{*},\lambda_{*}+\delta):\Phi_{t}(\lambda)=x_{1}\ldots x_{n_{j}}^{-}d_{1}d_{2}\ldots\textrm{ with }d_{ik}\neq{-1}\leavevmode\nobreak\ \forall i\geq 1\right\}.

Then by Proposition 2.6 it follows that

(4.6)

{\Delta_{k,j}^{-}(\lambda_{*},t)}\subset(\lambda_{*},\gamma_{j})\cap\Lambda(t)\leavevmode\nobreak\ \forall k\geq 1;\quad\textrm{and}\quad\gamma_{j}\searrow\lambda_{*}\leavevmode\nobreak\ \textrm{ {as} }j\to\infty.

If $\Phi_{t}$ is strictly increasing in $(\lambda_{*},\lambda_{*}+\delta)\cap\Lambda(t)$ , then we can assume that for any $j\geq 1$ (4.5) determines a unique $\tilde{\gamma}_{j}\in(\lambda_{*}-\delta,\lambda_{*})$ . Accordingly, for $k\geq 1$ we set

\tilde{\Delta}_{k,j}^{-}(\lambda_{*},t):=\left\{\lambda\in\Lambda(t)\cap(\lambda_{*}-\delta,\lambda_{*}):\Phi_{t}(\lambda)=x_{1}\ldots x_{n_{j}}^{-}d_{1}d_{2}\ldots\textrm{ with }d_{ik}\neq{-1}\leavevmode\nobreak\ \forall i\geq 1\right\}.

Therefore,

(4.7)

{\tilde{\Delta}_{k,j}^{-}(\lambda_{*},t)\subset(\tilde{\gamma}_{j},\lambda_{*})\cap\Lambda(t)\leavevmode\nobreak\ \forall k\geq 1;\quad\textrm{and}\quad\tilde{\gamma}_{j}\nearrow\lambda_{*}\leavevmode\nobreak\ \textrm{ {as} }j\to\infty.}

By a similar augment as in the proof of Lemma 3.5 we have the following Lipschitz property.

Lemma 4.3.

Let $t\in(0,1)\setminus\left\{1/3\right\}$ and $\lambda_{*}\in\Lambda(t)\setminus\left\{1/3\right\}$ . Suppose $\Phi_{t}(\lambda_{*})$ ends with $1^{\infty}$ .

(i)

If $\Phi_{t}$ is strictly decreasing in $(\lambda_{*},\lambda_{*}+\delta)\cap\Lambda(t)$ , then for any $k,j\in\mathbb{N}$ there exists $C_{1}>0$ such that

$|\pi_{\lambda_{*}}(\Phi_{t}(\lambda_{1}))-\pi_{\lambda_{*}}(\Phi_{t}(\lambda_{2}))|\leq C_{1}|\lambda_{1}-\lambda_{2}|$

for any $\lambda_{1},\lambda_{2}\in\Delta_{k,j}^{-}(\lambda_{*},t)$ .

(ii)

If $\Phi_{t}$ is strictly increasing in $(\lambda_{*},\lambda_{*}+\delta)\cap\Lambda(t)$ , then for any $k,j\in\mathbb{N}$ there exists $C_{2}>0$ such that

|\pi_{\tilde{\gamma}_{j}}(\Phi_{t}(\lambda_{1}))-\pi_{\tilde{\gamma}_{j}}(\Phi_{t}(\lambda_{2}))|\leq C_{2}|\lambda_{1}-\lambda_{2}|

for any $\lambda_{1},\lambda_{2}\in\tilde{\Delta}_{k,j}^{-}(\lambda_{*},t)$ .

Proof of Proposition 4.1 (a lower bound).

Let $t\in(0,1)\setminus\left\{1/3\right\}$ and $\lambda_{*}\in\Lambda(t)\setminus\left\{1/3\right\}$ . If $\Phi_{t}(\lambda_{*})$ does not end with $1^{\infty}$ and $\Phi_{t}$ is strictly decreasing in $(\lambda_{*},\lambda_{*}+\delta)\cap\Lambda(t)$ for some $\delta>0$ , then by (4.3) and Lemma 4.2 (i) it follows that for any $j\geq 1$ ,

	$\displaystyle\dim_{H}(\Lambda(t)\cap(\lambda_{}-\delta,\lambda_{}+\delta))$	$\displaystyle\geq\dim_{H}\pi_{\eta_{j}}(\Phi_{t}(\Delta_{k,j}^{+}(\lambda_{*},t)))$
		$\displaystyle\geq\frac{(k-1)\log 3+\log 2}{-k\log\eta_{j}}\to\frac{\log 3}{-\log\eta_{j}}{\quad\textrm{as }k\to\infty}.$

Since $\eta_{j}\nearrow\lambda_{*}$ as $j\to\infty$ by (4.3), we conclude that

(4.8)

\lim_{\delta\to 0^{+}}\dim_{H}(\Lambda(t)\cap(\lambda_{*}-\delta,\lambda_{*}+\delta))\geq\lim_{j\to\infty}\frac{\log 3}{{-\log\eta_{j}}}=\frac{\log 3}{-\log\lambda_{*}}.

Similarly, if $\Phi_{t}(\lambda_{*})$ does not end with $1^{\infty}$ and $\Phi_{t}$ is strictly increasing in $(\lambda_{*},\lambda_{*}+\delta)\cap\Lambda(t)$ , then by (4.4), Lemma 4.2 (ii) and the same argument as above we can prove (4.8). Moreover, if $\Phi_{t}(\lambda_{*})$ ends with $1^{\infty}$ , then (4.8) can be proved by a similar argument as above together with (4.6), (4.7) and Lemma 4.3. ∎

4.2. An upper bound on the local dimension of $\Lambda(t)$

Now we turn to prove that

\lim_{\delta\to 0^{+}}\dim_{H}(\Lambda(t)\cap(\lambda_{*}-\delta,\lambda_{*}+\delta))\leq\frac{\log 3}{-\log\lambda_{*}}\quad\forall\lambda_{*}\in\Lambda(t)\setminus\left\{1/3\right\}.

In view of our proof of Proposition 2.6 we consider $t\in(0,1/9]$ , $t\in(1/9,4/27)$ and $t\in[4/27,1/3)\cup(1/3,1)$ , separately.

Lemma 4.4.

Let $t\in[4/27,1/3)\cup(1/3,1)$ . Then for any $\theta\in(\min\Lambda(t),1/3)$ there exists $C>0$ such that

|\pi_{\theta}(\Phi_{t}(\lambda_{1}))-\pi_{\theta}(\Phi_{t}(\lambda_{2}))|\geq C|\lambda_{2}-\lambda_{1}|

for any $\lambda_{1},\lambda_{2}\in\Lambda(t)\cap(0,\theta]$ .

Proof.

First we consider $t\in[4/27,1/3)$ . Take $\theta\in(\min\Lambda(t),1/3)$ . By the proof of Lemma 2.7 it follows that $\Phi_{t}(\lambda)\in[01(-1)0^{\infty},01^{\infty}]$ for any $\lambda\in\Lambda(t)\cap(0,\theta]$ . Furthermore, by the proof of Lemma 2.3 (iii) we obtain that

(4.9)

\begin{split}\Pi_{2}((i_{n}),\lambda)&\geq\min\left\{1-4\lambda+3\lambda^{2},1-2\lambda-3\lambda^{2}\right\}\geq 1-4\theta+3\theta^{2}=:C_{1}>0\end{split}

for any $(i_{n})\in[01(-1)0^{\infty},01^{\infty}]$ and $\lambda\in\Lambda(t)\cap(0,\theta]$ .

Now take $\lambda_{1},\lambda_{2}\in\Lambda(t)\cap(0,\theta]$ with $\lambda_{1}<\lambda_{2}$ . Then $(i_{n}):=\Phi_{t}(\lambda_{1})\succ\Phi_{t}(\lambda_{2})=:(j_{n})$ by Lemma 2.7. So there exists $N\in\mathbb{N}_{\geq 3}$ such that $i_{1}\ldots i_{N-1}=j_{1}\ldots j_{N-1}$ and $i_{N}>j_{N}$ . Thus,

(4.10)

\begin{split}|\pi_{\theta}(\Phi_{t}(\lambda_{1}))-\pi_{\theta}(\Phi_{t}(\lambda_{2}))|&=(1-\theta)\sum_{n=N}^{\infty}(i_{n}-j_{n})\theta^{n-1}\\ &\geq(1-\theta)\theta^{N-1}-(1-\theta)\sum_{n=N+1}^{\infty}2\theta^{n-1}=(1-3\theta)\theta^{N-1}.\end{split}

On the other hand, note that $(1-\lambda_{1})\sum_{n=1}^{\infty}i_{n}\lambda_{1}^{n-1}=t=(1-\lambda_{2})\sum_{n=1}^{\infty}j_{n}\lambda_{2}^{n-1}.$ Then by (4.9) it follows that

	$\displaystyle\quad(1-\lambda_{1})\sum_{n=N}^{\infty}(i_{n}-1)\lambda_{1}^{n-1}-(1-\lambda_{2})\sum_{n=N}^{\infty}(j_{n}-1)\lambda_{2}^{n-1}$
	$\displaystyle=\left((1-\lambda_{2})\sum_{n=1}^{N-1}j_{n}\lambda_{2}^{n-1}+(1-\lambda_{2})\sum_{n=N}^{\infty}\lambda_{2}^{n-1}\right)$
	$\displaystyle\qquad-\left((1-\lambda_{1})\sum_{n=1}^{N-1}j_{n}\lambda_{1}^{n-1}+(1-\lambda_{1})\sum_{n=N}^{\infty}\lambda_{1}^{n-1}\right)$
	$\displaystyle=\Pi(j_{1}\ldots j_{N-1}1^{\infty},\lambda_{2})-\Pi(j_{1}\ldots j_{N-1}1^{\infty},\lambda_{1})$
	$\displaystyle=\Pi_{2}(j_{1}\ldots j_{N-1}1^{\infty},\lambda_{*})(\lambda_{2}-\lambda_{1})\geq C_{1}(\lambda_{2}-\lambda_{1})$

for some $\lambda_{*}\in[\lambda_{1},\lambda_{2}]$ , where the third equality follows by the mean value theorem. So,

(4.11)

\begin{split}\lambda_{2}-\lambda_{1}&\leq\frac{1}{C_{1}}\left[(1-\lambda_{1})\sum_{n=N}^{\infty}2\lambda_{1}^{n-1}+(1-\lambda_{2})\sum_{n=N}^{\infty}2\lambda_{2}^{n-1}\right]\\ &\leq\frac{2}{C_{1}}(\lambda_{1}^{N-1}+\lambda_{2}^{N-1})\leq\frac{4}{C_{1}}\theta^{N-1}.\end{split}

Hence, by (4.10) and (4.11) we conclude that

|\pi_{\theta}(\Phi_{t}(\lambda_{1}))-\pi_{\theta}(\Phi_{t}(\lambda_{2}))|\geq\frac{C_{1}(1-3\theta)}{4}{|\lambda_{1}-\lambda_{2}|}

as required.

Next we consider $t\in(1/3,1)$ . Then by the proof of Lemma 2.7 there exists $k\in\mathbb{N}$ such that $\Phi_{t}(\lambda)\in[1(-1)^{\infty},1^{k}(-1)^{\infty}]$ for any $\lambda\in\Lambda(t)\cap(0,\theta]$ . Since $\min\Lambda(t)=(1-t)/2$ , by (2.2) it follows that

|\Pi_{2}((i_{n}),\lambda)|\geq|\Pi_{2}(1^{k}(-1)^{\infty},\lambda)|=2k\lambda^{k-1}\geq 2k\left(\frac{1-t}{2}\right)^{k-1}=:C_{2}>0.

By a similar argument as above one can prove that for any $\lambda_{1},\lambda_{2}\in\Lambda(t)\cap(0,\theta]$

|\pi_{\theta}(\Phi_{t}(\lambda_{1}))-\pi_{\theta}(\Phi_{t}(\lambda_{2}))|\geq\frac{C_{2}(1-3\theta)}{4}|\lambda_{1}-\lambda_{2}|,

completing the proof. ∎

Recall from Lemma 2.8 that $\lambda_{\diamond}=\lambda_{\diamond}(t)\in(0,1/3)$ satisfies $\lambda_{\diamond}(1-\lambda_{\diamond})^{2}=t$ . Furthermore, by (2.13) we have $t<\lambda_{\diamond}(t)<\sqrt{t}\leq 1/3$ for all $t\in(0,1/9]$ .

Lemma 4.5.

Let $t\in(0,1/9]$ .

(i)

For any $\theta\in[t,\lambda_{\diamond}]$ there exists $C>0$ such that for any $\lambda_{1},\lambda_{2}\in\Lambda(t)\cap(0,\lambda_{\diamond}]$

$|\pi_{\theta}(\Phi_{t}(\lambda_{1}))-\pi_{\theta}(\Phi_{t}(\lambda_{2}))|\geq C|\lambda_{2}-\lambda_{1}|;$
(ii)

For any $\theta\in(\lambda_{\diamond},\sqrt{t})$ there exists $C^{\prime}>0$ such that for any $\lambda_{1},\lambda_{2}\in\Lambda(t)\cap(\lambda_{\diamond},\sqrt{t})$

$|\pi_{\theta}(\Phi_{t}(\lambda_{1}))-\pi_{\theta}(\Phi_{t}(\lambda_{2}))|\geq C^{\prime}|\lambda_{1}-\lambda_{2}|;$
(iii)

For any $\theta\in[\sqrt{t},1/3)$ there exists $C^{\prime\prime}>0$ such that for any $\lambda_{1},\lambda_{2}\in\Lambda(t)\cap[\sqrt{t},\theta]$

$|\pi_{\theta}(\Phi_{t}(\lambda_{1}))-\pi_{\theta}(\Phi_{t}(\lambda_{2}))|\geq C^{\prime\prime}|\lambda_{2}-\lambda_{1}|.$

Proof.

For (i), note by the proof of Lemma 2.8 that $\Phi_{t}(\lambda)\in[01(-1)0^{\infty},01^{\infty}]$ for any $\lambda\in\Lambda(t)\cap(0,\lambda_{\diamond}]$ . Then by (4.9) it follows that $\Pi_{2}((i_{n}),\lambda)\geq 1-4\lambda_{\diamond}+3\lambda_{\diamond}^{2}>0$ for all $(i_{n})\in[01(-1)0^{\infty},01^{\infty}]$ and $\lambda\in(0,\lambda_{\diamond}]$ . So, (i) follows by the same argument as in the proof of Lemma 4.4.

For (ii), note by (2.10) that for any $\lambda\in\Lambda(t)\cap(\lambda_{\diamond},\sqrt{t})$ we have $\Phi_{t}(\lambda)\in(001^{\infty},01(-1)0^{\infty})$ , and the proof of Lemma 2.9 yields that $\lambda<\lambda_{\Phi_{t}(\lambda)}$ . Then $\Pi_{2}(\Phi_{t}(\lambda),\lambda)>0$ for all $\lambda\in\Lambda(t)\cap(\lambda_{\diamond},\sqrt{t})$ . Furthermore, observe by Lemma 2.3 that $\Pi_{2}(\Phi_{t}(\lambda_{\diamond}),\lambda_{\diamond})=\Pi_{2}(01(-1)0^{\infty},\lambda_{\diamond})>0$ and $\Pi_{2}(\Phi_{t}(\sqrt{t}),\sqrt{t})=\Pi_{2}(001^{\infty},\sqrt{t})>0$ . Since the map $t\mapsto\Pi_{2}(\Phi_{t}(\lambda),\lambda)$ is continuous by Lemma 2.5 and $\Lambda(t)$ is compact by Proposition 3.1, it follows that

C_{1}:=\inf_{\lambda\in[\lambda_{\diamond},\sqrt{t}]\cap\Lambda(t)}\Pi_{2}(\Phi_{t}(\lambda),\lambda)=\min_{\lambda\in[\lambda_{\diamond},\sqrt{t}]\cap\Lambda(t)}\Pi_{2}(\Phi_{t}(\lambda),\lambda)>0.

Now take $\lambda_{1},\lambda_{2}\in\Lambda(t)\cap(\lambda_{\diamond},\sqrt{t})$ with $\lambda_{1}<\lambda_{2}$ . Then by Lemma 2.9 we have $(i_{n}):=\Phi_{t}(\lambda_{1})\succ\Phi_{t}(\lambda_{2})=(j_{n})$ . So, there exists $N\in\mathbb{N}_{\geq 4}$ such that $i_{1}\ldots i_{N-1}=j_{1}\ldots j_{N-1}$ and $i_{N}>j_{N}$ . By the same argument as in the proof of Lemma 4.4 one can show that

|\pi_{\theta}(\Phi_{t}(\lambda_{1}))-\pi_{\theta}(\Phi_{t}(\lambda_{2}))|\geq(1-3\theta)\theta^{N-1},

and there exists $\lambda_{*}\in[\lambda_{1},\lambda_{2}]$ such that

		$\displaystyle(1-\lambda_{1})\sum_{n=N}^{\infty}(i_{n}-1)\lambda_{1}^{n-1}-(1-\lambda_{2})\sum_{n=N}^{\infty}(j_{n}-1)\lambda_{2}^{n-1}$
	$\displaystyle=$	$\displaystyle\Pi(j_{1}\ldots j_{N-1}1^{\infty},\lambda_{2})-\Pi(j_{1}\ldots j_{N-1}1^{\infty},\lambda_{1})$
	$\displaystyle=$	$\displaystyle\Pi_{2}(j_{1}\ldots j_{N-1}1^{\infty},\lambda_{*})(\lambda_{2}-\lambda_{1})$
	$\displaystyle\geq$	$\displaystyle\Pi_{2}(j_{1}\ldots j_{N-1}1^{\infty},\lambda_{2})(\lambda_{2}-\lambda_{1})$
	$\displaystyle\geq$	$\displaystyle\Pi_{2}(\Phi_{t}(\lambda_{2}),\lambda_{2})(\lambda_{2}-\lambda_{1})\geq C_{1}(\lambda_{2}-\lambda_{1}),$

where the first inequality follows by the concavity of $\Pi(j_{1}\ldots j_{N-1}1^{\infty},\cdot)$ and the second inequality follows by Lemma 2.2. Therefore, one can deduce from this that

|\pi_{\theta}(\Phi_{t}(\lambda_{1}))-\pi_{\theta}(\Phi_{t}(\lambda_{2}))|\geq\frac{C_{1}(1-3\theta)}{4}|\lambda_{1}-\lambda_{2}|,

establishing (ii).

For (iii), by the proof of Lemma 2.9 it follows that $\Phi_{t}(\lambda)\in[\Phi_{t}(1/3),\Phi_{t}(\sqrt{t})]\subset(0^{\infty},001^{\infty}]$ for all $\lambda\in\Lambda(t)\cap[\sqrt{t},1/3].$ So, by the proof of Lemma 2.3 (i) there exists $k\in\mathbb{N}_{\geq 3}$ such that

\displaystyle\Pi_{2}((i_{n}),\lambda)

\displaystyle\geq\Pi_{2}(\Phi_{t}(1/3),\lambda)\geq\lambda^{k-2}(k-1-2k\lambda)\geq\min_{\lambda\in[\sqrt{t},\theta]}\lambda^{k-1}(k-1-2k\lambda){=:C_{2}>0}.

for all $(i_{n})\in[\Phi_{t}(1/3),\Phi_{t}(\sqrt{t})]$ and $\lambda\in[\sqrt{t},\theta]$ . Thus (iii) follows by the same argument as in the proof of Lemma 4.4.∎

Note by the proof of Proposition 2.6 (ii) that $\tau=\tau(t)\in[1/4,1/3)$ for $t\in(1/9,4/27)$ . Furthermore, by (2.13) we have $t<\lambda_{\diamond}(t)<\tau(t)<1/3$ for all $t\in(1/9,4/27)$ .

Lemma 4.6.

Let $t\in(1/9,4/27)$ .

(i)

For any $\theta\in[t,\lambda_{\diamond}]$ there exists $C>0$ such that for any $\lambda_{1},\lambda_{2}\in\Lambda(t)\cap(0,\lambda_{\diamond}]$

$|\pi_{\theta}(\Phi_{t}(\lambda_{1}))-\pi_{\theta}(\Phi_{t}(\lambda_{2}))|\geq C|\lambda_{2}-\lambda_{1}|;$
(ii)

For any $\theta\in(\lambda_{\diamond},\tau)$ there exists $C^{\prime}>0$ such that for any $\lambda_{1},\lambda_{2}\in\Lambda(t)\cap[\lambda_{\diamond},\theta]$

$|\pi_{\theta}(\Phi_{t}(\lambda_{1}))-\pi_{\theta}(\Phi_{t}(\lambda_{2}))|\geq C^{\prime}|\lambda_{2}-\lambda_{1}|;$
(iii)

For any $\theta\in(\tau,1/3)$ there exists $C^{\prime\prime}>0$ such that for any $\lambda_{1},\lambda_{2}\in\Lambda(t)\cap[\theta,1/3]$

$|\pi_{\theta}(\Phi_{t}(\lambda_{1}))-\pi_{\theta}(\Phi_{t}(\lambda_{2}))|\geq C^{\prime\prime}|\lambda_{2}-\lambda_{1}|.$

Proof.

The proof is similar to that for Lemma 4.5. More precisely, (i) follows by the same argument as in the proof of Lemma 4.5 (i). In view of the proofs of Lemmas 2.10 and 2.13, (ii) and (iii) follow by a similar argument as in the proof of Lemma 4.5 (ii). ∎

Proof of Proposition 4.1 (an upper bound).

Let $\lambda_{*}\in\Lambda(t)\setminus\left\{1/3\right\}$ . We consider the following two cases.

Case I. $t\in[4/27,1/3)\cup(1/3,1)$ . Then by Lemma 4.4 it follows that for any $\delta\in(0,1/3-\lambda_{*})$ ,

\dim_{H}(\Lambda(t)\cap(\lambda_{*}-\delta,\lambda_{*}+\delta))\leq\dim_{H}\pi_{\lambda_{*}+\delta}(\Phi_{t}(\Lambda(t)\cap(\lambda_{*}-\delta,\lambda_{*}+\delta)))\leq\frac{\log 3}{-\log(\lambda_{*}+\delta)},

which implies

(4.12)

\lim_{\delta\to 0^{+}}\dim_{H}(\Lambda(t)\cap(\lambda_{*}-\delta,\lambda_{*}+\delta))\leq\frac{\log 3}{-\log\lambda_{*}}

as desired.

Case II. $t\in(0,4/27)$ . Then by Lemmas 4.5 and 4.6 it follows that for any $\delta\in(0,1/3-\lambda_{*})$

\dim_{H}(\Lambda(t)\cap(\lambda_{*}-\delta,\lambda_{*}))\leq\frac{\log 3}{-\log\lambda_{*}}\quad\textrm{and}\quad\dim_{H}(\Lambda(t)\cap(\lambda_{*},\lambda_{*}+\delta))\leq\frac{\log 3}{-\log(\lambda_{*}+\delta)}.

Letting $\delta\to 0^{+}$ we also obtain (4.12). ∎

Proof of Theorem 1.1.

By Proposition 3.1, Corollary 3.7 and Proposition 4.1 it suffices to prove that $\Lambda(t)$ has zero Lebesgue measure for all $t\in(0,1)\setminus\left\{1/3\right\}$ . Note by Proposition 4.1 that for each $\lambda\in\Lambda(t)\setminus\left\{1/3\right\}$ there exists $\delta_{\lambda}>0$ such that $\dim_{H}(\Lambda(t)\cap(\lambda-\delta_{\lambda},\lambda+\delta_{\lambda}))<1$ , and thus $\Lambda(t)\cap(\lambda-\delta_{\lambda},\lambda+\delta_{\lambda})$ has zero Lebesgue measure. Since by Proposition 3.1 that $\Lambda(t)$ is compact, for any $n\in\mathbb{N}_{\geq 4}$ the segment $\Lambda(t)\cap(0,1/3-1/n]$ can be covered by finitely many open intervals, say $\left\{(\lambda_{i}-\delta_{\lambda_{i}},\lambda_{i}+\delta_{\lambda_{i}}):i=1,\ldots,N\right\}$ . This implies that $\Lambda(t)\cap(0,1/3-1/n]$ has zero Lebesgue measure for all $n\geq 4$ . Therefore, $\Lambda(t)=\left\{1/3\right\}\cup\bigcup_{n=4}^{\infty}(\Lambda(t)\cap(0,1/3-1/n])$ is a Lebesgue null set. ∎

5. Intersections with different Hausdorff and packing dimensions

In this section we consider the set

\Lambda_{not}(t):=\{\lambda\in\Lambda(t):\dim_{H}(C_{\lambda}\cap(C_{\lambda}+t))\neq\dim_{P}(C_{\lambda}\cap(C_{\lambda}+t))\},

and show that it has full Hausdorff dimension. Our proof is motivated by the work of [1] to construct subsets of $\Lambda_{not}(t)$ with Hausdorff dimension arbitrarily close to $1$ . Recall by (1.3) that

\dim_{H}(C_{\lambda}\cap(C_{\lambda}+t))=\frac{\log 2}{-\log\lambda}\underline{freq}_{0}(\Phi_{t}(\lambda)),\quad\dim_{P}(C_{\lambda}\cap(C_{\lambda}+t))=\frac{\log 2}{-\log\lambda}\overline{freq}_{0}(\Phi_{t}(\lambda)),

where for $(i_{n})\in\left\{-1,0,1\right\}^{\mathbb{N}}$ we denote

\underline{freq}_{0}((i_{n})):=\liminf_{n\to\infty}\frac{\#\{1\leq k\leq n:i_{k}=0\}}{n},\quad\overline{freq}_{0}((i_{n})):=\limsup_{n\to\infty}\frac{\#\{1\leq k\leq n:i_{k}=0\}}{n}.

Then $\Lambda_{not}(t)$ can be rewritten as

(5.1)

{\Lambda_{not}(t)=\left\{\lambda\in\Lambda(t):\underline{freq}_{0}(\Phi_{t}(\lambda))\neq\overline{freq}_{0}(\Phi_{t}(\lambda))\right\}.}

Proposition 5.1.

For any $t\in(0,1)\setminus\{1/3\}$ we have $\dim_{H}\Lambda_{not}(t)=1$ .

Similar to the proof of Proposition 3.4, we consider two cases: $t\in(0,1/9]\cup[4/27,1/3)$ and $t\in(1/9,4/27)\cup(1/3,1)$ . Note by Proposition 2.6 that the map $\Phi_{t}$ is strictly monotonic for $t\in(0,1/9]\cup[4/27,1/3)$ but piecewise monotonic for $t\in(1/9,4/27)\cup(1/3,1)$ . Since the proofs of the two cases are similar, in the following we only consider $t\in(1/9,4/27)\cup(1/3,1)$ . By Proposition 2.6 it follows that there exists a small $\delta>0$ such that $\Phi_{t}$ is strictly increasing in $\Lambda(t)\cap[1/3-\delta,1/3]$ . Let $(x_{i})=\Phi_{t}(1/3)$ . Then $(x_{i})$ does not end with $(-1)^{\infty}$ . So there exists a subsequence $\{n_{j}\}\subset\mathbb{N}$ such that $x_{n_{j}}>-1$ for all $j\geq 1$ . By deleting the first finitely many terms from $(n_{j})$ we may assume that for each $j\geq 1$ the equation

t=\Pi(x_{1}\ldots x_{n_{j}}^{-}(-1)^{\infty},\gamma_{j})

determines a unique $\gamma_{j}\in(1/3-\delta,1/3)$ . Clearly, $\gamma_{j}\nearrow 1/3$ as $j\to\infty$ .

Now, for $q\in\mathbb{N}$ let $\Sigma_{q,j}(t)$ consists of all $\lambda\in\Lambda(t)\cap(1/3-\delta,1/3)$ such that

(5.2)

\begin{split}\Phi_{t}(\lambda)=\leavevmode\nobreak\ &x_{1}\ldots x_{n_{j}}^{-}d_{1}d_{2}\ldots\\ =\leavevmode\nobreak\ &x_{1}\ldots x_{n_{j}}^{-}\\ &d_{r_{0}+1}d_{r_{0}+2}\ldots d_{r_{0}+3q}110\\ &d_{r_{1}+1}d_{r_{1}+2}\ldots d_{r_{1}+2\cdot 3q}111100\\ &\cdots\\ &d_{r_{m-1}+1}d_{r_{m-1}+2}\ldots d_{r_{m-1}+2^{m-1}\cdot 3q}1^{2^{m}}0^{2^{m-1}}\\ &d_{r_{m}+1}d_{r_{m}+2}\ldots d_{r_{m}+2^{m}\cdot 3q}1^{2^{m+1}}0^{2^{m}}\\ &\cdots,\end{split}

where

r_{m}:=3(q+1)(1+2+\cdots+2^{m-1})=3(q+1)(2^{m}-1)\quad\textrm{and}\quad d_{r_{m}+kq}\neq-1

for all $k\in\left\{1,\ldots,{2^{m}\cdot 3}\right\}$ and $m\geq 0$ . Here we point out that $r_{0}=0$ . Then by Proposition 2.6 it follows that $\Sigma_{q,j}(t)\subset\Lambda(t)\cap(\gamma_{j},1/3)$ for all $q\geq 1.$ In the following we show that $\Sigma_{q,j}(t)$ is a subset of $\Lambda_{not}(t)$ .

Lemma 5.2.

Let $t\in(1/9,4/27)\cup(1/3,1)$ . Then

\Sigma_{q,j}(t)\subset\Lambda_{not}(t)\cap(\gamma_{j},1/3)\quad\forall q,j\in\mathbb{N}.

Proof.

Take $\lambda\in\Sigma_{q,j}(t)$ . Then $\Phi_{t}(\lambda)=x_{1}\ldots x_{n_{j}}^{-}d_{1}d_{2}\ldots$ is defined as in (5.2). By (5.1) it suffices to prove that the frequency of digit zero in the sequence $(d_{i})$ does not exist. For $n\in\mathbb{N}$ let $N_{0}((d_{i}),n)$ be the number of zeros in the word $d_{1}\ldots d_{n}$ . Then

{N_{0}((d_{i}),r_{m})=\xi_{m}((d_{i}))+(1+2+\cdots+2^{m-1})=\xi_{m}((d_{i}))+2^{m}-1,}

where $\xi_{m}((d_{i}))$ denotes the number of zeros in the word

{d_{r_{0}+1}d_{r_{0}+2}\ldots d_{r_{0}+3q}\;d_{r_{1}+1}d_{r_{1}+2}\ldots d_{r_{1}+2\cdot 3q}\;\cdots\;d_{r_{m-1}+1}d_{r_{m-1}+2}\ldots d_{r_{m-1}+2^{m-1}\cdot 3q}.}

This implies that

(5.3)

\lim_{m\to\infty}\frac{N_{0}((d_{i}),r_{m})}{r_{m}}=\lim_{m\to\infty}\frac{\xi_{m}((d_{i}))+2^{m}-1}{3(q+1)(2^{m}-1)}=\frac{1}{3(q+1)}\left(1+\lim_{m\to\infty}\frac{\xi_{m}((d_{i}))}{2^{m}}\right).

Similarly, for $\ell_{m}:=r_{m}-2^{m-1}$ we have

N_{0}((d_{i}),\ell_{m})=\xi_{m}((d_{i}))+(1+2+\cdots+2^{m-2})=\xi_{m}((d_{i}))+2^{m-1}-1,

and thus

(5.4)

\begin{split}\lim_{m\to\infty}\frac{N_{0}((d_{i}),\ell_{m})}{\ell_{m}}&=\lim_{m\to\infty}\frac{\xi_{m}((d_{i}))+2^{m-1}-1}{3(q+1)(2^{m}-1)-2^{m-1}}\\ &=\frac{1}{6q+5}\left(1+2\lim_{m\to\infty}\frac{\xi((d_{i}),m)}{2^{m}}\right).\end{split}

Combining (5.3) with (5.4), if the limit $\lim_{m\to\infty}\frac{\xi_{m}((d_{i}))}{2^{m}}$ does not exist, then both limits $\lim_{m\to\infty}\frac{N_{0}((d_{i}),r_{m})}{r_{m}}$ and $\lim_{m\to\infty}\frac{N_{0}((d_{i}),\ell_{m})}{\ell_{m}}$ do not exist, and thus the frequency of digit zero in $(d_{i})$ does not exist. If the limit $\lim_{m\to\infty}\frac{\xi_{m}((d_{i}))}{2^{m}}$ exists, call this limit $\zeta$ , then since $\xi_{m}((d_{i}))\leq 3q(1+2+\ldots+2^{m-1})=3q(2^{m}-1)$ , we have $\zeta\leq 3q$ . Therefore, by (5.3) and (5.4) it follows that

\lim_{m\to\infty}\frac{N_{0}((d_{i}),r_{m})}{r_{m}}=\frac{\zeta+1}{3(q+1)}>\frac{2\zeta+1}{6q+5}=\lim_{m\to\infty}\frac{N_{0}((d_{i}),\ell_{m})}{\ell_{m}},

which again implies that the frequency of digit zero in $(d_{i})$ does not exist. So, $\lambda\in\Lambda_{not}(t)$ , completing the proof. ∎

Next we give a lower bound for the Huasdorff dimension of $\Sigma_{q,j}(t)$ .

Lemma 5.3.

Let $t\in(1/9,4/27)\cup(1/3,1)$ . Then for any $q,j\in\mathbb{N}$ we have

dim_{H}\Sigma_{q,j}(t)\geq\frac{(q-1)\log 3+\log 2}{-(q+1)\log\gamma_{j}}.

Proof.

Note by Lemma 5.2 that $\Sigma_{q,j}(t)\subset\Lambda_{not}(t)\cap(\gamma_{j},1/3)$ . Then for any $\lambda\in\Sigma_{q,j}(t)$ the sequence $\Phi_{t}(\lambda)$ begins with $x_{1}\ldots x_{n_{j}}^{-}$ and the tail sequence does not contain $q$ consecutive $(-1)$ s. Then by the same argument as in the proof of Lemma 3.6 there exists $C>0$ such that

|\pi_{\gamma_{j}}(\Phi_{t}(\lambda_{1}))-\pi_{\gamma_{j}}(\Phi_{t}(\lambda_{2}))|\leq C|\lambda_{1}-\lambda_{2}|

for any $\lambda_{1},\lambda_{2}\in\Sigma_{q,j}(t)$ . This implies that $\dim_{H}\Sigma_{q,j}(t)\geq\dim_{H}\pi_{\gamma_{j}}(\Phi_{t}(\Sigma_{q,j}(t)))$ . So it suffices to prove

(5.5)

\dim_{H}\pi_{\gamma_{j}}(\Phi_{t}(\Sigma_{q,j}(t)))\geq\frac{(q-1)\log 3+\log 2}{-(q+1)\log\gamma_{j}}.

Observe by (5.2) that

(5.6)

\dim_{H}\pi_{\gamma_{j}}(\Phi_{t}(\Sigma_{q,j}(t)))=\dim_{H}\pi_{\gamma_{j}}{\left(\prod_{m=0}^{\infty}E_{m}(q)\right)},

where

{E_{m}(q)}:=\left(\left\{-1,0,1\right\}^{q-1}\times\left\{0,1\right\}\right)^{3\cdot 2^{m}}\times\left\{1^{2^{m+1}}0^{2^{m}}\right\}.

Note that each word in $E_{m}(q)$ has length $3(q+1)2^{m}$ , and $\#E_{m}(q)=(3^{q-1}\cdot 2)^{3\cdot 2^{m}}$ . Furthermore, $\prod_{m=0}^{\infty}E_{m}(q)$ is the set of infinite sequences by concatenating words from each $E_{m}(q)$ . So $\pi_{\gamma_{j}}(\prod_{m=0}^{\infty}E_{m}(q))$ is a homogeneous Moran set satisfying the strong separation condition. Hence, by [8, Theorem 2.1] it follows that

	$\displaystyle\dim_{H}\pi_{\gamma_{j}}{\left({\prod_{m=0}^{\infty}E_{m}(q)}\right)}$	$\displaystyle\geq\liminf_{m\to\infty}\frac{\log\prod_{\ell=0}^{m-1}(3^{q-1}\cdot 2)^{3\cdot 2^{\ell}}}{-{\sum_{l=0}^{m-1}3(q+1)2^{l}}\log\gamma_{j}}$
		$\displaystyle=\liminf_{m\to\infty}\frac{\sum_{\ell=0}^{m-1}3\cdot 2^{\ell}\log(3^{q-1}\cdot 2)}{-3(q+1)(2^{m}-1)\log\gamma_{j}}$
		$\displaystyle=\liminf_{m\to\infty}\frac{3(2^{m}-1)\log(3^{q-1}\cdot 2)}{-3(q+1)(2^{m}-1)\log\gamma_{j}}=\frac{(q-1)\log 3+\log 2}{-(q+1)\log\gamma_{j}}.$

This, together with (5.6), proves (5.5). ∎

Proof of Proposition 5.1.

Take $t\in(0,1)\setminus\left\{1/3\right\}$ . Since the proof for $t\in(0,1/9]\cup[4/27,1/3)$ is analogous, we only consider for $t\in(1/9,4/27)\cup(1/3,1)$ . By Lemmas 5.2 and 5.3 it follows that for any $q\in{\mathbb{N}}$

\dim_{H}\Lambda_{not}(t)\geq\dim_{H}\Sigma_{q,j}(t)\geq\frac{(q-1)\log 3+\log 2}{-(q+1)\log\gamma_{j}}\to\frac{(q-1)\log 3+\log 2}{(q+1)\log 3}\quad\textrm{as }j\to\infty.

Letting $q\to\infty$ we obtain $\dim_{H}\Lambda_{not}(t)\geq 1$ . Since the reverse inequality is obvious, this proves $\dim_{H}\Lambda_{not}(t)=1$ . ∎

6. Proof of Theorem 1.3

In this section we consider the level set

\Lambda_{\beta}(t):=\left\{\lambda\in\Lambda(t):\dim_{H}\big{(}C_{\lambda}\cap(C_{\lambda}+t)\big{)}=\dim_{P}\big{(}C_{\lambda}\cap(C_{\lambda}+t)\big{)}=\beta\frac{\log 2}{-\log\lambda}\right\}

for $\beta\in[0,1]$ , and prove Theorem 1.3. Recall from (1.5) the entropy function $h(p_{1},p_{2},p_{3})=-\sum_{i=1}^{3}p_{i}\log p_{i}$ for a probability vector $(p_{1},p_{2},p_{3})$ .

Proposition 6.1.

Let $t\in(0,1)\setminus\left\{1/3\right\}$ . Then for any $\beta\in[0,1]$

\dim_{H}\Lambda_{\beta}(t)=\dim_{P}\Lambda_{\beta}(t)=\frac{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}{\log 3}.

It is worth mentioning that the dimension of $\Lambda_{\beta}(t)$ is independent of $t$ . Since the proof for $t\in(0,1/9]\cup(4/27,1/3)$ is similar, we only consider $t\in(1/9,4/27)\cup(1/3,1)$ . Then $(x_{i})=\Phi_{t}(1/3)$ does not end with $(-1)^{\infty}$ . So there exists a subsequence $(n_{j})\subset\mathbb{N}$ such that $x_{n_{j}}>-1$ for all $j\geq 1$ . By Proposition 2.6 there exists $\delta>0$ such that $\Phi_{t}$ is strictly increasing in $\Lambda(t)\cap(1/3-\delta,1/3)$ . By deleting the first finitely many terms from $(n_{j})$ we may assume that for each $j\geq 1$ the equation

t=\pi_{\gamma_{j}}(x_{1}\ldots x_{n_{j}}^{-}(-1)^{\infty})

determines a unique $\gamma_{j}\in(1/3-\delta,1/3)$ . Then by Lemma 2.5 it follows that $\gamma_{j}\nearrow 1/3$ as $j\to\infty$ .

Given $\beta\in[0,1]$ , we first prove the lower bound

\dim_{H}\Lambda_{\beta}(t)\geq\frac{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}{\log 3}.

For $k,j\in{\mathbb{N}}$ let $\mathcal{F}_{k,j}^{-}(t,\beta)$ consist of all $\lambda\in\Lambda(t)\cap(1/3-\delta,1/3)$ such that

\Phi_{t}(\lambda)=x_{1}\ldots x_{n_{j}}^{-}d_{1}d_{2}\ldots

satisfying

(6.1)

d_{nk+1}\ldots d_{nk+k}\neq(-1)^{k}\;\;{\forall n\geq 0}\quad\textrm{and}\quad{freq_{0}((d_{i})):=\underline{freq}_{0}((d_{i}))=\overline{freq}_{0}((d_{i}))=\beta}.

Then by Proposition 2.6 it follows that for each $j\in{\mathbb{N}}$ ,

(6.2)

\mathcal{F}_{k,j}^{-}(t,\beta)\subset\Lambda_{\beta}(t)\cap(\gamma_{j},1/3)\quad\forall k\geq 1.

Note that for any sequence $(c_{i})\in\Phi_{t}(\mathcal{F}_{k,j}^{-}(t,\beta))$ the tail sequence $c_{n_{j}+1}c_{n_{j}+2}\ldots$ does not contain $2k-1$ consecutive $(-1)$ s. Then by the same argument as in the proof of Lemma 3.5 it follows that

(6.3)

\dim_{H}\mathcal{F}_{k,j}^{-}(t,\beta)\geq\dim_{H}\pi_{\gamma_{j}}(\Phi_{t}(\mathcal{F}_{k,j}^{-}(t,\beta))).

So it is necessary to consider a lower bound of $\dim_{H}\pi_{\gamma_{j}}(\Phi_{t}(\mathcal{F}_{k,j}^{-}(t,\beta)))$ .

To do this, for each $k,j\in\mathbb{N}$ we construct a measure $\hat{\mu}_{k,j}$ on the tree

{\mathbf{T}_{k,j}(t):=\left\{x_{1}\ldots x_{n_{j}}^{-}d_{1}d_{2}\ldots:d_{nk+1}\ldots d_{nk+k}\neq(-1)^{k}\leavevmode\nobreak\ \forall n\geq 0\right\}}.

Let $(p_{-1},p_{0},p_{1})=(\theta_{k},1-2\theta_{k},\theta_{k})$ be a probability vector with $\theta_{k}\in[0,1/2]$ satisfying

(6.4)

\frac{1-2\theta_{k}}{1-(\theta_{k})^{k}}=\beta.

Then $\theta_{k}\to(1-\beta)/2$ as $k\to\infty$ . For each cylinder set of the tree $\mathbf{T}_{k,j}(t)$ we set $\hat{\mu}_{0}([x_{1}\ldots x_{n_{j}}^{-}])=1,$ and for $n\geq 1$ , we let

\hat{\mu}_{n}([x_{1}\ldots x_{n_{j}}^{-}d_{1}\ldots d_{nk}])=\prod_{i=0}^{n-1}\frac{p_{d_{ik+1}}p_{d_{ik+2}}\cdots p_{d_{ik+k}}}{1-p_{-1}^{k}}.

By Kolmogorov’s extension theorem (cf. [6]) there exists a unique probability measure $\hat{\mu}=\hat{\mu}_{k,j}$ on the tree $\mathbf{T}_{k,j}(t)$ satisfying $\hat{\mu}([x_{1}\ldots x_{n_{j}}^{-}d_{1}\ldots d_{nk}])=\hat{\mu}_{n}([x_{1}\ldots x_{n_{j}}^{-}d_{1}\ldots d_{nk}])$ for any $n\geq 0.$

Lemma 6.2.

$\hat{\mu}(\Phi_{t}(\mathcal{F}_{k,j}^{-}(t,\beta)))=1.$

Proof.

Note that $\Phi_{t}(\mathcal{F}_{k,j}^{-}(t,\beta))\subset\mathbf{T}_{k,j}(t)$ . By (6.1) it suffices to prove that for $\hat{\mu}$ -a.e. $(c_{i})\in\mathbf{T}_{k,j}(t)$ we have $freq_{0}((c_{i}))=\beta$ . Note that $c_{1}\ldots c_{n_{j}}=x_{1}\ldots x_{n_{j}}^{-}$ is fixed, and the choice of each block $c_{n_{j}+nk+1}\ldots c_{n_{j}+nk+k}$ is independent and identical distributed for all $n\geq 0$ according to our definition of $\hat{\mu}$ . So, by the law of large numbers it follows that for $\hat{\mu}$ -a.e. $(c_{i})\in\mathbf{T}_{k,j}(t)$

	$\displaystyle{freq_{0}((c_{i}))}$	$\displaystyle=\lim_{n\to\infty}\frac{\#\left\{i\in[1,n]:c_{i}=0\right\}}{n}$
		$\displaystyle=\lim_{n\to\infty}\frac{\#\left\{i\in[n_{j}+1,n_{j}+nk]:c_{i}=0\right\}}{nk}$
		$\displaystyle=\sum_{d_{1}\ldots d_{k}\neq(-1)^{k}}\frac{\#\left\{i\in[1,k]:d_{i}=0\right\}}{k}\hat{\mu}([x_{1}\ldots x_{n_{j}}d_{1}\ldots d_{k}])$
		$\displaystyle=\sum_{\ell=1}^{k}\frac{\ell}{k}\binom{k}{\ell}2^{k-\ell}\frac{(\theta_{k})^{k-\ell}(1-2\theta_{k})^{\ell}}{1-(\theta_{k})^{k}},$

where the last equality follows since the number of blocks $d_{1}\ldots d_{k}\in\left\{-1,0,1\right\}^{k}\setminus\left\{(-1)^{k}\right\}$ with precisely $\ell(\geq 1)$ zeros is $\binom{k}{\ell}2^{k-\ell}$ . Rearranging the above summation and by (6.4) we conclude that

{freq_{0}((c_{i}))}=\frac{1}{1-(\theta_{k})^{k}}\sum_{\ell=1}^{k}\frac{(k-1)!}{(\ell-1)!(k-\ell)!}(2\theta_{k})^{k-\ell}(1-2\theta_{k})^{\ell}=\frac{1-2\theta_{k}}{1-(\theta_{k})^{k}}=\beta,

completing the proof. ∎

By the same argument as in the proof of Lemma 6.2 one can verify that for $\hat{\mu}$ -a.e. $(c_{i})\in\mathbf{T}_{k,j}(t)$ the frequencies of digits $1$ and $-1$ in $(c_{i})$ are given respectively by

(6.5)

freq_{1}((c_{i}))=\frac{\theta_{k}}{1-(\theta_{k})^{k}}\to\frac{1-\beta}{2}\quad\textrm{and}\quad freq_{-1}((c_{i}))=\frac{\theta_{k}-(\theta_{k})^{k}}{1-(\theta_{k})^{k}}\to\frac{1-\beta}{2}

as $k\to\infty$ , where the limits follow by $\lim_{k\to\infty}\theta_{k}=(1-\beta)/2$ .

Proof of Proposition 6.1 (a lower bound).

Note by Lemma 6.2 the set $\pi_{\gamma_{j}}(\Phi_{t}(\mathcal{F}_{k,j}^{-}(t,\beta)))$ has full $\mu:=\hat{\mu}\circ\pi_{\gamma_{j}}^{-1}$ measure. Then for $\mu$ -a.e. $y=\pi_{\gamma_{j}}(x_{1}\ldots x_{n_{j}}^{-}d_{1}d_{2}\ldots)\in\pi_{\gamma_{j}}(\Phi_{t}(\mathcal{F}_{k,j}^{-}(t,\beta)))$

	$\displaystyle\liminf_{r\to 0^{+}}\frac{\log\mu(B(y,r))}{\log r}$	$\displaystyle=\liminf_{n\to\infty}\frac{\log\hat{\mu}([x_{1}\ldots x_{n_{j}}^{-}d_{1}\ldots d_{nk}])}{\log\gamma_{j}^{n_{j}+nk}}$
		$\displaystyle=\liminf_{n\to\infty}\frac{\log\prod_{i=1}^{nk}p_{d_{i}}-\log(1-p_{-1}^{k})^{n}}{(n_{j}+nk)\log\gamma_{j}}$
		$\displaystyle=\liminf_{n\to\infty}\frac{\sum_{s=-1,0,1}N_{s}(nk)\log{p_{s}}-n\log(1-p_{-1}^{k})}{nk\log\gamma_{j}},$

where $N_{s}(nk)$ denotes the number of digit $s$ in the block $d_{1}\ldots d_{nk}$ . Therefore, by Lemma 6.2 and (6.5) it follows that for $\mu$ -a.e. $y\in\pi_{\gamma_{j}}(\Phi_{t}(\mathcal{F}_{k,j}^{-}(t,\beta)))$

	$\displaystyle\liminf_{r\to 0^{+}}\frac{\log\mu(B(y,r))}{\log r}$	$\displaystyle=\frac{\frac{\theta_{k}-(\theta_{k})^{k}}{1-(\theta_{k})^{k}}\log\theta_{k}+\beta\log(1-2\theta_{k})+\frac{\theta_{k}}{1-(\theta_{k})^{k}}\log\theta_{k}}{\log\gamma_{j}}{-\frac{\log(1-(\theta_{k})^{k})}{k\log\gamma_{j}}}$
		$\displaystyle\to\quad\frac{\frac{1-\beta}{2}\log\frac{1-\beta}{2}+\beta\log\beta+\frac{1-\beta}{2}\log\frac{1-\beta}{2}}{\log\gamma_{j}}=\frac{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}{-\log\gamma_{j}}$

as $k\to\infty$ . Then by Billingsley’s Lemma (cf. [7]) it follows that

\dim_{H}\pi_{\gamma_{j}}(\Phi_{t}(\mathcal{F}_{k,j}^{-}(t,\beta)))\geq\frac{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}{-\log\gamma_{j}}\quad{\forall j\geq 1.}

This, together with (6.2) and (6.3), implies

\dim_{H}(\Lambda_{\beta}(t)\cap(\gamma_{j},1/3))\geq\frac{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}{-\log\gamma_{j}}\quad\forall j\geq 1.

Since $\gamma_{j}\nearrow 1/3$ as $j\to\infty$ , we conclude that $\dim_{H}\Lambda_{\beta}(t)\geq{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}/{\log 3}.$ ∎

Next we consider the upper bound.

Lemma 6.3.

Let $t\in[4/27,1/3)\cup(1/3,1)$ and $\beta\in[0,1]$ . Then

\dim_{P}\Lambda_{\beta}(t)\leq\frac{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}{\log 3}.

Proof.

By the same argument as in the proof of Lemma 4.4 it follows that for any $\theta\in(t,1/3)$ there exists $C>0$ , such that $|\pi_{\theta}(\Phi_{t}(\lambda_{1}))-\pi_{\theta}(\Phi_{t}(\lambda_{2}))|\geq C|\lambda_{1}-\lambda_{2}|$ for any $\lambda_{1},\lambda_{2}\in\Lambda_{\beta}(t)\cap(0,\theta]$ . This implies that

\dim_{P}(\Lambda_{\beta}(t)\cap(0,\theta])\leq\dim_{P}{\pi_{\theta}}(\Phi_{\theta}(\Lambda_{\beta}(t)))\leq\dim_{P}\pi_{1/3}(\Phi_{t}(\Lambda_{\beta}(t))).

By the countable stability of packing dimension we obtain

(6.6)

\dim_{P}\Lambda_{\beta}(t)\leq\dim_{P}\pi_{1/3}(\Phi_{t}(\Lambda_{\beta}(t))).

Define a Bernoulli measure $\hat{\nu}$ on the symbolic space $\{-1,0,1\}^{\mathbb{N}}$ such that

\hat{\nu}([d_{1}\ldots d_{n}])=\prod_{i=1}^{n}p_{d_{i}}\quad\forall n\geq 0,

where $p_{-1}=p_{1}=(1-\beta)/2$ and $p_{0}=\beta$ . Let $\nu=\hat{\nu}\circ\pi_{1/3}^{-1}$ . Note that for each $(d_{i})\in\Phi_{t}(\Lambda_{\beta}(t))$ we have $freq_{0}((d_{i}))=\beta$ . Then for any $y\in\pi_{1/3}(\Phi_{t}(\Lambda_{\beta}(t)))$ with $(c_{i})=\Phi_{t}(1/3)$

	$\displaystyle\limsup_{r\to 0^{+}}\frac{\log\nu(B(y,r))}{\log r}$	$\displaystyle=\limsup_{n\to\infty}\frac{\log\prod_{i=1}^{n}p_{c_{i}}}{-\log 3^{N+n}}=\frac{1}{-\log 3}\limsup_{n\to\infty}\frac{\sum_{i=1}^{n}\log p_{c_{i}}}{n}$
		$\displaystyle=\frac{1}{-\log 3}\limsup_{n\to\infty}\frac{1}{n}\left(N_{0}(n)\log\beta+(n-N_{0}(n))\log\frac{1-\beta}{2}\right)$
		$\displaystyle=\frac{1}{-\log 3}\left(\beta\log\beta+(1-\beta)\log\frac{1-\beta}{2}\right)=\frac{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}{\log 3},$

where $N_{0}(n)$ denotes the number of digit zero in the block $c_{1}\ldots c_{n}$ . So, by [7, Proposition 2.3] and (6.6) it follows that

\dim_{P}\Lambda_{\beta}(t)\leq\dim_{P}\pi_{1/3}(\Phi_{t}(\Lambda_{\beta}(t)))\leq\frac{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}{\log 3},

completing the proof. ∎

Proof of Proposition 6.1 (an upper bound).

By Lemma 6.3 it suffices to consider $t\in(0,4/27)$ . If $t\in(0,1/9]$ , then by Lemma 4.5 and the same argument as in the proof of Lemma 6.3 it follows that

\begin{split}\dim_{P}(\Lambda_{\beta}(t)\cap(0,\lambda_{\diamond}])&\leq\frac{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}{-\log\lambda_{\diamond}}\leq\frac{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}{\log 3},\\ \dim_{P}(\Lambda_{\beta}(t)\cap(\lambda_{\diamond},\sqrt{t}))&\leq\frac{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}{-\log\sqrt{t}}\leq\frac{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}{\log 3}\\ \dim_{P}(\Lambda_{\beta}(t)\cap[\sqrt{t},1/3])&\leq\frac{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}{\log 3}.\end{split}

This implies $\dim_{P}\Lambda_{\beta}(t)\leq{h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})}/{\log 3}$ as desired.

Similarly, if $t\in(1/9,4/27)$ , then by Lemma 4.6 and the same argument as in the proof of Lemma 6.3 we can also prove $\dim_{P}\Lambda_{\beta}(t)\leq h(\frac{1-\beta}{2},\beta,\frac{1-\beta}{2})/\log 3.$ ∎

Proof of Theorem 1.3.

By Propositions 5.1 and 6.1 it suffices to prove that for any $\beta\in[0,1]$ both $\Lambda_{not}(t)$ and $\Lambda_{\beta}(t)$ are dense in $\Lambda(t)$ . Since the proofs are similar, we only prove it for $\Lambda_{\beta}(t)$ . Take $\lambda_{*}\in\Lambda(t)$ and $\delta>0$ . Then we can always find $\lambda\in\Lambda(t)$ such that $\Phi_{t}(\lambda)$ has a long common prefix with $\Phi_{t}(\lambda_{*})$ , and the tail sequence of $\Phi_{t}(\lambda)$ has digit zero frequency equaling $\beta$ . In other words, $\lambda\in\Lambda_{\beta}(t)\cap(\lambda_{*}-\delta,\lambda_{*}+\delta)$ . So, $\Lambda_{\beta}(t)$ is dense in $\Lambda(t)$ . ∎

Acknowledgements

The second author was supported by NSFC No. 11971079.

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	$\displaystyle\|\Pi((i_{n}),\lambda_{1})-\Pi((j_{n}),\lambda_{2})\|$	$\displaystyle\leq\|\Pi((i_{n}),\lambda_{1})-\Pi((i_{n}),\lambda_{2})\|+\|\Pi((i_{n}),\lambda_{2})-\Pi((j_{n}),\lambda_{2})\|$
		$\displaystyle\leq\sup_{\lambda\in(0,1/3]}\|\Pi_{2}((i_{n}),\lambda)\|\cdot\|\lambda_{1}-\lambda_{2}\|+\left\|(1-\lambda_{2})\sum_{n=1}^{\infty}(i_{n}-j_{n})\lambda_{2}^{n-1}\right\|$
		$\displaystyle\leq 2\|\lambda_{1}-\lambda_{2}\|+\left\|(1-\lambda_{2})\sum_{n=k}^{\infty}2\lambda_{2}^{n-1}\right\|$
		$\displaystyle\leq 2\|\lambda_{1}-\lambda_{2}\|+2\rho((i_{n}),(j_{n})),$

Intersections of middle-α\alpha Cantor sets with a fixed translation

Abstract.

Key words and phrases:

2010 Mathematics Subject Classification:

1. Introduction

Theorem 1.1.

Remark 1.2.

Theorem 1.3.

Remark 1.4.

2. Preliminaries

Lemma 2.1.

Proof.

Lemma 2.2.

Proof.

Lemma 2.3.

Proof.

Remark 2.4.

Lemma 2.5.

Proof.

Proposition 2.6 (Key proposition).

Lemma 2.7.

Proof.

Lemma 2.8.

Proof.

Lemma 2.9.

Proof.

Lemma 2.10.

Proof.

Lemma 2.11.

Proof.

Lemma 2.12.

Proof.

Lemma 2.13.

Proof.

Proof of Proposition 2.6.

3. Topology and Hausdorff dimension of Λ​(t)\Lambda(t)

3.1. Topology of Λ​(t)\Lambda(t)

Proposition 3.1.

Lemma 3.2.

Proof.

Lemma 3.3.

Proof.

Proof of Proposition 3.1.

3.2. Hausdorff dimension of Λ​(t)\Lambda(t)

Proposition 3.4.

Lemma 3.5.

Proof.

Lemma 3.6.

Proof.

Proof of Proposition 3.4.

Corollary 3.7.

4. Proof of Theorem 1.1

Proposition 4.1.

4.1. A lower bound on the local dimension of Λ​(t)\Lambda(t)

Lemma 4.2.

Proof.

Lemma 4.3.

Proof of Proposition 4.1 (a lower bound).

4.2. An upper bound on the local dimension of Λ​(t)\Lambda(t)

Lemma 4.4.

Proof.

Lemma 4.5.

Proof.

Lemma 4.6.

Proof.

Proof of Proposition 4.1 (an upper bound).

Proof of Theorem 1.1.

5. Intersections with different Hausdorff and packing dimensions

Proposition 5.1.

Lemma 5.2.

Proof.

Lemma 5.3.

Proof.

Proof of Proposition 5.1.

6. Proof of Theorem 1.3

Proposition 6.1.

Lemma 6.2.

Proof.

Proof of Proposition 6.1 (a lower bound).

Lemma 6.3.

Intersections of middle- $\alpha$ Cantor sets with a fixed translation

3. Topology and Hausdorff dimension of $\Lambda(t)$

3.1. Topology of $\Lambda(t)$

3.2. Hausdorff dimension of $\Lambda(t)$

4.1. A lower bound on the local dimension of $\Lambda(t)$

4.2. An upper bound on the local dimension of $\Lambda(t)$