Thue-Morse-Sturmian words and critical bases for ternary alphabets

TL;DR

This paper investigates the structure of unique beta-expansions over three-letter alphabets using S-adic words, extending known thresholds for binary alphabets to more complex ternary cases.

Contribution

It introduces a new generalization of the Komornik-Loreti constant for three-letter alphabets using Thue-Morse and Sturmian words.

Findings

Determined the value of the generalized Komornik-Loreti constant for ternary alphabets.

Connected S-adic words with thresholds for unique expansions.

Extended binary alphabet results to ternary cases.

Abstract

The set of unique $\beta$-expansions over the alphabet $\{0,1\}$ is trivial for $\beta$ below the golden ratio and uncountable above the Komornik-Loreti constant. Generalisations of these thresholds for three-letter alphabets were studied by Komornik, Lai and Pedicini (2011, 2017). We use S-adic words including the Thue-Morse word (which defines the Komornik-Loreti constant) and Sturmian words (which characterise generalised golden ratios) to determine the value of a certain generalisation of the Komornik-Loreti constant to three-letter alphabets.

Equations

$${\pi }_{\beta }(u_1{u}_2\cdots )=\sum _{k=1}^{\infty }\frac{u_k}{{\beta }^k}.$$

$$U_{\beta }(A)=\{\mathbf{u}\in A^{\infty }:{\pi }_{\beta }(\mathbf{u})\ne {\pi }_{\beta }(\mathbf{v})\text{mbox}forall\mathbf{v}\in A^{\infty }\setminus \{\mathbf{u}\}\}.$$

$$\mathcal{G}(A)=\inf \{\beta >1:|U_{\beta }(A)|>2\}$$

$$\mathcal{K}(A)=\inf \{\beta >1:U_{\beta }(A)\text{mbox}isuncountable\}$$

$$U_{\beta }(m)=U_{\beta }(\{0,1,m\}),\mathcal{G}(m)=\mathcal{G}(\{0,1,m\}),\mathcal{K}(m)=\mathcal{K}(\{0,1,m\}).$$

$$\mathcal{L}(m)=\inf \{\beta >1:U_{\beta }(m)\cap \{0,1{\}}^{\infty }\text{mbox}isuncountable\},$$

$$\displaystyle L:\$$

$$1\mapsto 01,$$

$$f_{\mathbf{u}}(m){\pi }_{f_{\mathbf{u}}(m)}(\sup O(\mathbf{u}))=m\text{mbox}and(g_{\mathbf{u}}(m)-1)(1+{\pi }_{g_{\mathbf{u}}(m)}(\inf O(\mathbf{u})))=m$$

$$f_{\mathbf{u}}({\mu }_{\mathbf{u}})=g_{\mathbf{u}}({\mu }_{\mathbf{u}}),$$

$$\mathcal{G}(m)=\{\begin{array}{ll}{f}_{\sigma (\overline{0})}(m)& \text{mbox}ifm\in [{\mu }_{\sigma (1\overline{0})},{\mu }_{\sigma (\overline{0})}],\sigma \in \{L,R{\}}^{\ast }M,\\ g_{\sigma (\overline{0})}(m)& \text{mbox}ifm\in [{\mu }_{\sigma (\overline{0})},{\mu }_{\sigma (0\overline{1})}],\sigma \in \{L,R{\}}^{\ast }M,\\ f_{\overline{1}}(m)& \text{mbox}ifm\in [{\mu }_{0\overline{1}},2],\\ 1+\sqrt{m}& \text{mbox}ifm={\mu }_{\mathbf{u}},\mathbf{u}\in {\mathcal{S}}_{\{L,R\}};\end{array}$$

$$\mathcal{L}(m)=\{\begin{array}{ll}{g}_{\sigma (1\overline{0})}(m)& \text{mbox}ifm\in [{\mu }_{\sigma (1\overline{0})},{\mu }_{\sigma (01\overline{0})}],\sigma \in \{L,M,R{\}}^{\ast }M,\\ f_{\sigma (0\overline{1})}(m)& \text{mbox}ifm\in [{\mu }_{\sigma (10\overline{1})},{\mu }_{\sigma (0\overline{1})}],\sigma \in \{L,M,R{\}}^{\ast }M,\\ g_{0\overline{1}}(m)& \text{mbox}ifm\in [{\mu }_{0\overline{1}},2],\\ f_{\mathbf{u}}(m)& \text{mbox}ifm={\mu }_{\mathbf{u}},\mathbf{u}\in {\mathcal{S}}_{\{L,M,R\}}.\end{array}$$

$$\mathcal{L}(m)=\left\{\begin{array}[]{ll}g_{0\overline{01}}(m)&\mbox{if}\ m\in[\mu_{0\overline{01}},\mu_{110\overline{01}}]\approx[1.281972,1.46811],\\ f_{1\overline{10}}(m)&\mbox{if}\ m\in[\mu_{001\overline{10}},\mu_{1\overline{10}}]\approx[1.516574,1.55496].\end{array}\right.$$

$$\mathcal{L}(m)=\left\{\begin{array}[]{ll}g_{001\overline{0110}}(m)&\mbox{if}\ m\in[\mu_{001\overline{0110}},\mu_{1101001\overline{0110}}]\approx[1.47571,1.503114],\\ f_{110\overline{1001}}(m)&\mbox{if}\ m\in[\mu_{0010110\overline{1001}},\mu_{110\overline{1001}}]\approx[1.504152,1.509304].\end{array}\right.$$

$$2\le \mathcal{G}(m)\le 1+\sqrt{m}\le \mathcal{K}(m)\le \mathcal{L}(m)\le g_{1\overline{0}}(m)=1+m,$$

$$T:\,[0,\tfrac{m}{\beta-1}]\to[0,\tfrac{m}{\beta-1}],\ x\mapsto\beta x-d(x),\ d(x)=\left\{\begin{array}[]{cl}0&\mbox{if}\ x<\frac{1}{\beta},\\ 0\ \mbox{or}\ 1&\mbox{if}\ \frac{1}{\beta}\leq x\leq\frac{m}{\beta(\beta-1)},\\ 1&\mbox{if}\ \frac{m}{\beta(\beta-1)}<x<\frac{m}{\beta},\\ 1\ \mbox{or}\ m&\mbox{if}\ \frac{m}{\beta}\leq x\leq\frac{1}{\beta}+\frac{m}{\beta(\beta-1)},\\ m&\mbox{if}\ x>\frac{1}{\beta}+\frac{m}{\beta(\beta-1)},\end{array}\right.$$

$$\mathbf{u}\in U_{\beta }(m)⟺{\pi }_{\beta }(u_k{u}_{k+1}\cdots )\notin [\frac{1}{\beta },\frac{m}{\beta (\beta -1}]\cup [\frac{m}{\beta },\frac{1}{\beta }+\frac{m}{\beta (\beta -1)}]\text{mbox}forallk\ge 1.$$

$$\pi_{\beta}(u_{k}u_{k+1}\cdots)<\tfrac{m}{\beta},\ \pi_{\beta}(u_{k+1}u_{k+2}\cdots)>\tfrac{m}{\beta-1}-1\ \mbox{for all $k\geq 1$ such that $u_{k}=1$},$$

$$\beta {\pi }_{\beta }(\sup (\mathbf{u}))\le m\le (\beta -1)(1+{\pi }_{\beta }({\inf }_1(\mathbf{u}))),$$

$${\inf }_1(u_1{u}_2\cdots )=\inf \{u_{k+1}{u}_{k+2}\cdots :k\ge 1,u_k=1\},$$

$$\max (f_{\mathbf{u}}(m),g_{\mathbf{u}}(m))\ge 2,$$

$$\beta {\pi }_{\beta }(\sup (\mathbf{u}))<m\text{mbox}forall\beta >f_{\mathbf{u}}(m),$$

$$(\beta -1)(1+{\pi }_{\beta }(\inf (\mathbf{u}))>m\text{mbox}forall\beta >g_{\mathbf{u}}(m),$$

$$f_{\mathbf{u}}(m)>f_{\mathbf{u}}(m^{\prime })\text{mbox}and{g}_{\mathbf{u}}(m)<g_{\mathbf{u}}(m^{\prime })\text{mbox}ifm<m^{\prime },$$

$$f_{\mathbf{u}}(m)<f_{{\mathbf{u}}^{\prime }}(m)\text{mbox}if\sup (\mathbf{u})<\sup ({\mathbf{u}}^{\prime })\text{mbox}and{f}_{\mathbf{u}}(m)\ge 2,$$

$$g_{\mathbf{u}}(m)>g_{{\mathbf{u}}^{\prime }}(m)\text{mbox}if\inf (\mathbf{u})<\inf ({\mathbf{u}}^{\prime })\text{mbox}and{g}_{{\mathbf{u}}^{\prime }}(m)\ge 2.$$

$$\beta\pi_{\beta}(\sup(\mathbf{u}))\geq 1+\sum_{i=1}^{k-1}\frac{v_{i}}{\beta^{i}}+\frac{1}{\beta^{k}},\ (\beta-1)(1+\pi_{\beta}(\inf(\mathbf{u})))\leq(\beta-1)\bigg{(}1+\sum_{i=1}^{k-1}\frac{v_{i}}{\beta^{i}}\bigg{)}+\frac{1}{\beta^{k}},$$

$${\sup }_0(u_1{u}_2\cdots )=\sup \{u_{k+1}{u}_{k+2}\cdots :k\ge 1,u_k=0\},$$

$$\inf (L(\mathbf{u}))=L(\inf (\mathbf{u})),\inf (R(\mathbf{u}))=R(\inf (\mathbf{u})),0\sup (L(\mathbf{u}))=L(\sup (\mathbf{u})).$$

$$\sup (R(\mathbf{u}))=1R(\sup (\mathbf{u})),\sup (M(\mathbf{u}))=1M(\sup (\mathbf{u})).$$

$$\mathcal{L}(m)\le \{\begin{array}{ll}{g}_{\sigma (1\overline{0})}(m)& \text{mbox}ifm\ge {\mu }_{\sigma (1\overline{0})},\sigma \in \{L,M,R{\}}^{\ast }M,\\ f_{\sigma (0\overline{1})}(m)& \text{mbox}ifm\le {\mu }_{\sigma (0\overline{1})},\sigma \in \{L,M,R{\}}^{\ast }M,\\ g_{0\overline{1}}(m)& \text{mbox}ifm\ge {\mu }_{0\overline{1}},\\ g_{\mathbf{u}}(m)& \text{mbox}ifm\ge {\mu }_{\mathbf{u}},\mathbf{u}\in {\mathcal{S}}_{\{L,M,R\}},\\ f_{\mathbf{u}}(m)& \text{mbox}ifm\le {\mu }_{\mathbf{u}},\mathbf{u}\in {\mathcal{S}}_{\{L,M,R\}}.\end{array}$$

$$\inf (\sigma (\mathbf{v}))\ge \inf (\sigma (\overline{10(01{)}^{h-1}0}))\text{mbox}and\sup (\sigma (\mathbf{v}))\le \sup (\sigma (\overline{(01{)}^{h+1}0}))$$

$$\inf (\sigma (\overline{10(01{)}^{h-1}0}))\to \inf (\sigma M(1\overline{0})),\sup (\sigma (\overline{(01{)}^{h+1}0}))\to \sup (\sigma M(\overline{0}))(h\to \infty ).$$

$${\sup }_0(\mathbf{u})={\sup }_0(01{\mathbf{u}}^{\prime })=\sup (01{\mathbf{u}}^{\prime }),{\inf }_1(\mathbf{u})={\inf }_1(01{\mathbf{u}}^{\prime })={\inf }_1(1{\mathbf{u}}^{\prime })=\inf (1{\mathbf{u}}^{\prime }).$$

Full text

This work was supported by the Agence Nationale de la Recherche through the project CODYS (ANR-18-CE40-0007).

1. Introduction and main results

For a base $\beta>1$ and a sequence of digits $u_{1}u_{2}\cdots\in A^{\infty}$, with $A\subset\mathbb{R}$, let

[TABLE]

We say that $u_{1}u_{2}\cdots$ is a $\beta$-expansion of this number. This paper deals with unique $\beta$-expansions over $A$, that is with

[TABLE]

We know from [DK93] that $U_{\beta}(\{0,1\})$ is trivial if and only if $\beta\leq\frac{1+\sqrt{5}}{2}$, where trivial means that $U_{\beta}(\{0,1\})=\{\overline{0},\overline{1}\}$, $\overline{a}$ being the infinite repetition of $a$. Therefore,

[TABLE]

is called generalised golden ratio of $A$. By [GS01], the set $U_{\beta}(\{0,1\})$ is uncountable if and only if $\beta$ is larger than the Komornik–Loreti constant $\beta_{\mathrm{KL}}\approx 1.787$; we call

[TABLE]

generalised Komornik–Loreti constant of $A$. (We can replace uncountable throughout the paper by has the cardinality of the continuum.) The precise structure of $U_{\beta}(\{0,1\})$ was described in [KKL17]. For integers $M\geq 2$, $\mathcal{G}(\{0,1,\ldots,M\})$ was determined by [Bak14], and $U_{\beta}(\{0,1,\ldots,M\})$ was described in [KLLdV17, ABBK19].

For $x,y\in\mathbb{R}$, $x\neq 0$, we have $(xu_{1}+y_{1})(xu_{2}+y_{2})\cdots\in U_{\beta}(xA+y)$ if and only if $u_{1}u_{2}\cdots\in U_{\beta}(A)$, thus $\mathcal{G}(xA+y)=\mathcal{G}(A)$ and $\mathcal{K}(xA+y)=\mathcal{K}(A)$. Hence, the only two-letter alphabet to consider is $\{0,1\}$, and we can restrict to $\{0,1,m\}$, $m\in(1,2]$, for three-letter alphabets; another possibility is $m\geq 2$ as in [KLP11]. We write

[TABLE]

It was established in [KLP11, Lai11, BS17] that the generalised golden ratio $\mathcal{G}(m)$ is given by mechanical words, i.e., Sturmian words and their periodic counterparts; in particular, we can restrict to sequences $\mathbf{u}\in\{0,1\}^{\infty}$. Calculating $\mathcal{K}(m)$ seems to be much harder since this restriction is not possible. Therefore, we study

[TABLE]

following [KP17], where this quantity was determined for certain intervals. We give a complete characterisation in Theorem 1 below.

To this end, we use the substitutions (or morphisms)

[TABLE]

which act on finite and infinite words by $\sigma(u_{1}u_{2}\cdots)=\sigma(u_{1})\sigma(u_{2})\cdots$. The monoid generated by a set of substitutions $S$ (with the usual product of substitutions) is denoted by $S^{*}$. An infinite word $\mathbf{u}$ is a limit word of a sequence of substitutions $(\sigma_{n})_{n\geq 1}$ (or an $S$-adic word if $\sigma_{n}\in S$ for all $n\geq 1$) if there is a sequence of words $(\mathbf{u}^{(n)})_{n\geq 1}$ with $\mathbf{u}^{(1)}=\mathbf{u}$, $\mathbf{u}^{(n)}=\sigma_{n}(\mathbf{u}^{(n+1)})$ for all $n\geq 1$. The sequence $(\sigma_{n})_{n\geq 1}$ is primitive if for each $k\geq 1$ there is an $n\geq k$ such that both words $\sigma_{k}\sigma_{k+1}\cdots\sigma_{n}(0)$ and $\sigma_{k}\sigma_{k+1}\cdots\sigma_{n}(1)$ contain both letters [math] and $1$. For $S=\{L,M,R\}$, this means that there is no $k\geq 1$ such that $\sigma_{n}=L$ for all $n\geq k$ or $\sigma_{n}=R$ for all $n\geq k$. Let $\mathcal{S}_{S}$ be the set of limit words of primitive sequences of substitutions in $S^{\infty}$. Then $\mathcal{S}_{\{L,R\}}$ consists of Sturmian words, and $\mathcal{S}_{\{M\}}$ consists of the Thue-Morse word $0\mathbf{u}=0110100110010110\cdots$, which defines the Komornik–Loreti constant by $\pi_{\beta_{\mathrm{KL}}}(\mathbf{u})=1$, and its reflection by $0\leftrightarrow 1$. We call the elements of $\mathcal{S}_{\{L,M,R\}}$, which to our knowledge have not been studied yet, Thue–Morse–Sturmian words. For details on $S$-adic and other words, we refer to [Lot02, BD14].

For $\mathbf{u}\in\{0,1\}^{\infty}$ and $m\in(1,2]$, define $f_{\mathbf{u}}(m)$ (if $\mathbf{u}$ contains at least two ones) and $g_{\mathbf{u}}(m)$ as the unique positive solutions of

[TABLE]

respectively, where $O(u_{1}u_{2}\cdots)=\{u_{k}u_{k+1}\cdots:\,k\geq 1\}$ denotes the shift orbit and infinite words are ordered by the lexicographic order. For the existence and monotonicity properties of $f_{\mathbf{u}}(m)$ and $g_{\mathbf{u}}(m)$, see Lemma 2 below. We define $\mu_{\mathbf{u}}$ by

[TABLE]

i.e., $f_{\mathbf{u}}(\mu_{\mathbf{u}})=g_{\mathbf{u}}(\mu_{\mathbf{u}})=\beta$ with $\beta\,\pi_{\beta}(\sup O(\mathbf{u}))=(\beta-1)(1+\pi_{\beta}(\inf O(\mathbf{u})))$.

The main result of [KLP11] can be written as

[TABLE]

cf. [BS17, Proposition 3.18], where substitutions $\tau_{h}=L^{h}R$ are used and $f,g,\mu,\mathcal{S}$ are defined slightly differently. Our main theorem looks similar, but we need $\{L,M,R\}$ instead of $\{L,R\}$, and the roles of $f$ and $g$ are exchanged.

Theorem 1**.**

The function $\mathcal{L}(m)$, $1<m\leq 2$, is given by

[TABLE]

The Hausdorff dimension of $\pi_{\beta}(U_{\beta}(m))$ is positive for all $\beta>\mathcal{L}(m)$.

The graphs of $\mathcal{G}(m)$ and $\mathcal{L}(m)$ are drawn in Figure 1. For example, $\sigma=M$ gives

[TABLE]

Taking $\sigma=M^{2}$, we have $\sigma(0)=0110$, $\sigma(1)=1001$, and

[TABLE]

Subintervals of the first three intervals were also given by [KP17].

By [KLP11, KP17], we have, for all $m\in(1,2]$,

[TABLE]

with $\mathcal{G}(m)=\mathcal{L}(m)$ if and only if $m\in\{\mu_{\sigma(1\overline{0})},\mu_{\sigma(0\overline{1})}\}$, $\sigma\in\{L,R\}^{*}M$, or $m=\mu_{\mathbf{u}}$, $\mathbf{u}\in\mathcal{S}_{\{L,R\}}$. Besides those $m$, the value of $\mathcal{K}(m)$ is known only for $m=2$ from [dVK09, AF09, KP17], with $\mathcal{K}(2)\approx 2.536<\frac{3+\sqrt{5}}{2}=\mathcal{L}(2)$. The functions $\mathcal{G}(m)$, $\mathcal{K}(m)$ and $\mathcal{L}(m)$ are continuous for $m>1$ by [KLP11, KP17]; at least for the generalised golden ratio, this also holds for larger alphabets by [BS17].

2. Proof of the main theorem

We first establish relations between $f_{\mathbf{u}}(m)$, $g_{\mathbf{u}}(m)$ and $\mathbf{u}\in U_{\beta}(m)$. For convenience, we write $\inf(\mathbf{u})$ for $\inf O(\mathbf{u})$ and $\sup(\mathbf{u})$ for $\sup O(\mathbf{u})$ in the following.

Lemma 1**.**

Let $m\in(1,2]$, $\beta\in(1,1+m]$. For $\mathbf{u}\in\{0,1\}^{\infty}$, we have $\mathbf{u}\in U_{\beta}(m)$ if and only if $0\mathbf{u}\in U_{\beta}(m)$. For $\mathbf{u}\in 1\{0,1\}^{\infty}\setminus\{1\overline{0}\}$, $\mathbf{u}\in U_{\beta}(m)$ implies that $\beta\geq\max(f_{\mathbf{u}}(m),g_{\mathbf{u}}(m))$, and $\beta>\max(f_{\mathbf{u}}(m),g_{\mathbf{u}}(m))$ implies that $\mathbf{u}\in U_{\beta}(m)$.

Proof.

For $\beta\in(1,1+m]$, $\mathbf{u}=u_{1}u_{2}\cdots\in\{0,1,m\}^{\infty}$, $x\in[0,\frac{m}{\beta-1}]$, we have $\pi_{\beta}(\mathbf{u})=x$ if and only if $u_{k}=d(T^{k-1}(x))$ for all $k\geq 1$, with the branching $\beta$-transformation

[TABLE]

see Figure 2. We have thus

[TABLE]

For $\mathbf{u}\in\{0,1\}^{\infty}\setminus\{\overline{0}\}$, this means that $\beta>2$ and

[TABLE]

see [BS17, Lemma 3.9], i.e.,

[TABLE]

with strict equalities if the supremum and infimum are attained. In particular, we have $\mathbf{u}\in U_{\beta}(m)$ if and only if $0\mathbf{u}\in U_{\beta}(m)$. If $\mathbf{u}$ starts with $1$, then $\inf_{1}(\mathbf{u})=\inf(\mathbf{u})$, and the first lines of Lemma 2 below conclude the proof of the lemma. ∎

Lemma 2**.**

Let $m\in(1,2]$, $\mathbf{u},\mathbf{u}^{\prime}\in\{0,1\}^{\infty}$. Then $g_{\mathbf{u}}(m)$ is well defined. If $\mathbf{u}$ contains at least two ones, then $f_{\mathbf{u}}(m)$ and $\mu_{\mathbf{u}}$ are well defined, and we have

[TABLE]

Proof.

Let $\sup(\mathbf{u})=\mathbf{v}$ and set $h_{\mathbf{v}}(x,m)=x\pi_{x}(\mathbf{v})-m$. Then $h_{\mathbf{v}}(x,m)$ is strictly decreasing in $x$ and $m$ (for $x>1$). If $\mathbf{u}$ contains at least two ones, then this also holds for $\mathbf{v}$, thus $\lim_{x\to 1}h_{\mathbf{v}}(x,m)\geq 2-m$ and $\lim_{x\to\infty}h_{\mathbf{v}}(x,m)=1-m$. Therefore, there is, for each $m\in(1,2]$, a unique $x_{m,\mathbf{v}}\geq 1$ such that $h_{\mathbf{v}}(x_{m,\mathbf{v}},m)=0$, i.e., $f_{\mathbf{u}}(m)=x_{m,\mathbf{v}}$. If $m<m^{\prime}$, then we have $x_{m,\mathbf{v}}>x_{m^{\prime},\mathbf{v}}$. If $\mathbf{v}<\mathbf{v}^{\prime}$ and $x\geq 2$, then we have $h_{\mathbf{v}}(x,m)<h_{\mathbf{v}^{\prime}}(x,m)$, thus $x_{m,\mathbf{v}}<x_{m,\mathbf{v^{\prime}}}$ if $x_{m,\mathbf{v}}\geq 2$.

Let now $\inf(\mathbf{u})=\mathbf{v}$ and set $h_{\mathbf{v}}(x,m)=\frac{m}{x-1}-\pi_{x}(\mathbf{v})-1$. Since $\frac{m}{x-1}=\pi_{x}(\overline{m})$, $h_{\mathbf{v}}(x,m)$ is strictly decreasing in $x$ (for $x>1$) and strictly increasing in $m$. Again, there is, for each $m\in(1,2]$, a unique $x_{m,\mathbf{v}}>1$ such that $h_{\mathbf{v}}(x_{m,\mathbf{v}},m)=0$, i.e., $g_{\mathbf{u}}(m)=x_{m,\mathbf{v}}$. We have $h_{\mathbf{v}}(x,m)<0$ if $x>x_{m,\mathbf{v}}$, $x_{m,\mathbf{v}}<x_{m^{\prime},\mathbf{v}}$ if $m<m^{\prime}$, and $h_{\mathbf{v}}(x,m)>h_{\mathbf{v}^{\prime}}(x,m)$ if $\mathbf{v}<\mathbf{v}^{\prime}$, $x\geq 2$, thus $x_{m,\mathbf{v}}>x_{m,\mathbf{v}^{\prime}}$ if $x_{m,\mathbf{v}^{\prime}}\geq 2$.

Since $f_{\mathbf{u}}(m)$ is strictly decreasing, $g_{\mathbf{u}}(m)$ is strictly increasing, $\lim_{m\to 1}f_{\mathbf{u}}(m)=\infty$, $f_{\mathbf{u}}(2)\leq 2$ and $g_{\mathbf{u}}(2)\geq 2$, we have $f_{\mathbf{u}}(m)=g_{\mathbf{u}}(m)$ for a unique $m\in(1,2]$.

Let $\beta=f_{\mathbf{u}}(\mu_{\mathbf{u}})=g_{\mathbf{u}}(\mu_{\mathbf{u}})$, i.e., $\beta\,\pi_{\beta}(\sup(\mathbf{u}))=(\beta-1)(1+\pi_{\beta}(\inf(\mathbf{u})))$. We have $\sup(\mathbf{u})\geq 1\inf(\mathbf{u})$. If equality holds, then $\beta=2$. Otherwise, $\sup(\mathbf{u})$ starts with $1v_{1}\cdots v_{k-1}1$ and $\inf(\mathbf{u})$ starts with $v_{1}\cdots v_{k-1}0$ for some $v_{1}\cdots v_{k-1}$, $k\geq 1$. Then

[TABLE]

thus $\beta\geq 2$. By the monotonicity properties that are proved above, this implies that $\max(f_{\mathbf{u}}(m),g_{\mathbf{u}}(m))\geq 2$ for all $m\in(1,2]$. ∎

Therefore, it is crucial to determine $\inf(\mathbf{u})$ and $\sup(\mathbf{u})$. We set

[TABLE]

similarly to $\inf_{1}(u_{1}u_{2}\cdots)=\inf\{u_{k+1}u_{k+2}\cdots:\,k\geq 1,u_{k}=1\}$.

Lemma 3**.**

For all $\mathbf{u}\in\{0,1\}^{\infty}$, we have

[TABLE]

If $\inf(\mathbf{u})=\inf_{1}(\mathbf{u})$, then $\inf(M(\mathbf{u}))=0M(\inf(\mathbf{u}))$. If $\sup(\mathbf{u})=\sup_{0}(\mathbf{u})$, then

[TABLE]

For each $\sigma\in\{L,M,R\}^{*}$, there is a suffix $w$ of $\sigma(1)$ such that $\inf_{1}(\sigma(\mathbf{u}))=\inf(\sigma(\mathbf{u}))=w\sigma(\inf(\mathbf{u}))$ for all $\mathbf{u}\in\{0,1\}^{\infty}$ with $\inf(\mathbf{u})=\inf_{1}(\mathbf{u})$.

For each $\sigma\in\{L,M,R\}^{*}M\cup\{L,M,R\}^{*}R$, there is a suffix $w$ of $\sigma(0)$ such that $\sup_{0}(\sigma(\mathbf{u}))=\sup(\sigma(\mathbf{u}))=w\sigma(\sup(\mathbf{u}))$ for all $\mathbf{u}\in\{0,1\}^{\infty}$ with $\sup(\mathbf{u})=\sup_{0}(\mathbf{u})$.

For each $\sigma\in\{L,M,R\}^{*}L$, there is a prefix $w$ of $\sigma(\overline{0})$ such that $w\sup_{0}(\sigma(\mathbf{u}))=w\sup(\sigma(\mathbf{u}))=\sigma(\sup(\mathbf{u}))$ for all $\mathbf{u}\in\{0,1\}^{\infty}$ with $\sup(\mathbf{u})=\sup_{0}(\mathbf{u})$.

Proof.

The first statements follow from the facts that $L,M,R$ are order-preserving on infinite words and that $\inf(\mathbf{u})=\inf_{1}(\mathbf{u})$, $\sup(\mathbf{u})=\sup_{0}(\mathbf{u})$ mean that $1\inf(\mathbf{u})$, $0\sup(\mathbf{u})$ are in the closure of $O(\mathbf{u})$.

We claim that, for each $\sigma\in\{L,M,R\}^{*}$, there is a suffix $1w$ of $\sigma(1)$ such that $\inf_{1}(\sigma(\mathbf{u}))=\inf(\sigma(\mathbf{u}))=w\sigma(\inf(\mathbf{u}))$ for all $\mathbf{u}\in\{0,1\}^{\infty}$ with $\inf(\mathbf{u})=\inf_{1}(\mathbf{u})$. If $1w$ is a suffix of $\sigma(1)$, then $1L(w)$, $10M(w)$ and $1R(w)$ are suffixes of $L\sigma(1)$, $M\sigma(1)$ and $R\sigma(1)$ respectively. Therefore, this claim holds for $L\sigma$, $M\sigma$ and $R\sigma$ when it holds for $\sigma$. Since it holds for $\sigma=\mathrm{id}$, it holds for all $\sigma\in\{L,M,R\}^{*}$.

Next we claim that, for each $\sigma\in\{L,M,R\}^{*}\{M,R\}$, there is a suffix $01w$ of $\sigma(0)$ such that $\sup_{0}(\sigma(\mathbf{u}))=\sup(\sigma(\mathbf{u}))=1w\sigma(\sup(\mathbf{u}))$ for all $\mathbf{u}\in\{0,1\}^{\infty}$ with $\sup(\mathbf{u})=\sup_{0}(\mathbf{u})$. This holds for $\sigma\in\{M,R\}$. If $01w$ is a suffix of $\sigma(0)$, then $01L(w)$, $01M(1w)$ and $01R(1w)$ are suffixes of $L\sigma(0)$, $M\sigma(0)$ and $R\sigma(0)$ respectively. Therefore, this claim holds for all $\sigma\in\{L,M,R\}^{*}\{M,R\}$.

Finally we claim that, for each $\sigma\in\{L,M,R\}^{*}L$, there is a prefix $w0$ of $\sigma(\overline{0})$ such that $w0\sup_{0}(\sigma(\mathbf{u}))=w0\sup(\sigma(\mathbf{u}))=\sigma(\sup(\mathbf{u}))$ for all $\mathbf{u}\in\{0,1\}^{\infty}$ with $\sup(\mathbf{u})=\sup_{0}(\mathbf{u})$. This holds for $\sigma=L$ . If $w0$ is a prefix of $\sigma(\overline{0})$, then $L(w0)0$, $M(w)0$ and $R(w)0$ are prefixes of $L\sigma(\overline{0})$, $M\sigma(\overline{0})$ and $R\sigma(\overline{0})$ respectively. Therefore, this claim holds for all $\sigma\in\{L,M,R\}^{*}L$. ∎

Now we can prove that Theorem 1 gives an upper bound for $\mathcal{L}(m)$, cf. Figure 3.

Proposition 1**.**

Let $m\in(1,2]$. We have

[TABLE]

If $\beta$ is above this bound, then the Hausdorff dimension of $\pi_{\beta}(U_{\beta}(m))$ is positive.

Proof.

Let $\sigma\in\{L,M,R\}^{*}$. For all $h\geq 1$, $\mathbf{v}\in 1\{0(01)^{h},0(01)^{h+1}\}^{\infty}$, we have

[TABLE]

by Lemma 3, with

[TABLE]

Therefore, we have for each $\beta>\max(f_{\sigma M(\overline{0})}(m),g_{\sigma M(1\overline{0})}(m))$ some $h\geq 1$ such that $\sigma(\{0(01)^{h},0(01)^{h+1}\}^{\infty})\subseteq U_{\beta}(m)$. If $m\geq\mu_{\sigma M(1\overline{0})}$, then $f_{\sigma M(\overline{0})}(m)=f_{\sigma M(1\overline{0})}(m)\leq g_{\sigma M(1\overline{0})}(m)$, thus $U_{\beta}(m)\cap\{0,1\}^{\infty}$ is uncountable (and has the cardinality of the continuum) for all $\beta>g_{\sigma M(1\overline{0})}(m)$, i.e., $\mathcal{L}(m)\leq g_{\sigma M(1\overline{0})}(m)$. By symmetry, sequences in $\sigma(\{1(10)^{h},1(10)^{h+1}\}^{\infty})$ give that $\mathcal{L}(m)\leq f_{\sigma M(0\overline{1})}(m)$ for $m\leq\mu_{\sigma M(0\overline{1})}$. Similarly, sequences in $1\{01^{h},01^{h+1}\}^{\infty}$ give that $\mathcal{L}(m)\leq g_{0\overline{1}}(m)$ for $m\geq\mu_{0\overline{1}}$.

Let now $\mathbf{u}$ be a limit word of a primitive sequence $(\sigma_{n})_{n\geq 1}\in\{L,M,R\}^{\infty}$, and set $\sigma^{\prime}_{n}=\sigma_{1}\sigma_{2}\cdots\sigma_{n}$. Then $\inf(\sigma^{\prime}_{n}(1\overline{0}))\leq\inf(\mathbf{u})\leq\inf(\sigma^{\prime}_{n}(10\overline{1}))$ for all $n\geq 1$, thus $\inf(\sigma^{\prime}_{n}(1\overline{0}))\to\inf(\mathbf{u})$ and (by symmety) $\sup(\sigma^{\prime}_{n}(0\overline{1}))\to\sup(\mathbf{u})$ as $n\to\infty$. Therefore, for each $\beta>\max(f_{\mathbf{u}}(m),g_{\mathbf{u}}(m))$ there is $n\geq 1$ such that $\sigma^{\prime}_{n}(\mathbf{v})\in U_{\beta}(m)$ for all $\mathbf{v}\in\{0,1\}^{\infty}\setminus\{\overline{0},\overline{1}\}$, hence $\mathcal{L}(m)\leq g_{\mathbf{u}}(m)$ for $m\geq\mu_{\mathbf{u}}$ and $\mathcal{L}(m)\leq f_{\mathbf{u}}(m)$ for $m\leq\mu_{\mathbf{u}}$.

If $\{v,w\}^{\infty}\subseteq U_{\beta}(m)$, then by [Hut81] we have $\dim_{H}(\pi_{\beta}(U_{\beta}(m)))\geq r$, with $r>0$ such that $\beta^{-|v|r}+\beta^{-|w|r}=1$, where $|v|$ and $|w|$ denote the lengths of $v$ and $w$. ∎

For the lower bound, we use Lemma 5 below, which tells us that, if the orbit of a sequence satisfies inequalities that hold for all non-trivial images of $\sigma\in\{L,M,R\}^{*}$, then it is eventually in the image of $\sigma$. In particular, with $\sigma=M^{n}$, $n\geq 0$, this yields that $U_{\beta}(\{0,1\})$ is countable for all $\beta$ less than the Komornik–Loreti constant; cf. [GS01]. First we show that the conditions of Lemma 3 are satisfied for a suffix.

Lemma 4**.**

Let $\mathbf{u}\in\{0,1\}^{\infty}$ with $\mathbf{u}\neq 0^{k}\overline{1}$ and $\mathbf{u}\neq 1^{k}\overline{0}$ for all $k\geq 0$. There is a suffix $\mathbf{v}$ of $\mathbf{u}$ such that $\inf(\mathbf{v})=\inf_{1}(\mathbf{v})=\inf_{1}(\mathbf{u})$ and $\sup(\mathbf{v})=\sup_{0}(\mathbf{v})=\sup_{0}(\mathbf{u})$.

Proof.

If $\inf(\mathbf{u})=\inf_{1}(\mathbf{u})$ and $\sup(\mathbf{u})=\sup_{0}(\mathbf{u})$, then we can take $\mathbf{v}=\mathbf{u}$. Otherwise, assume that $\inf(\mathbf{u})\neq\inf_{1}(\mathbf{u})$, the case $\sup(\mathbf{u})\neq\sup_{0}(\mathbf{u})$ being symmetric. Then we have $\inf(\mathbf{u})=\mathbf{u}=0^{k}01\mathbf{u}^{\prime}$ for some $k\geq 0$, $\mathbf{u}^{\prime}\in\{0,1\}^{\infty}\setminus\{\overline{1}\}$,

[TABLE]

If $\inf_{1}(01\mathbf{u}^{\prime})\neq\inf(01\mathbf{u}^{\prime})$, then $\mathbf{u}^{\prime}=1^{n}01\mathbf{u}^{\prime\prime}$ with $n\geq 0$, $\mathbf{u}^{\prime\prime}>\mathbf{u}^{\prime}$, which implies that $\sup_{0}(\mathbf{u})=\sup_{0}(1\mathbf{u}^{\prime})=\sup(1\mathbf{u}^{\prime})$. Hence, we can take $\mathbf{v}=01\mathbf{u}^{\prime}$ or $\mathbf{v}=1\mathbf{u}^{\prime}$. ∎

Lemma 5**.**

Let $\mathbf{u}\in\{0,1\}^{\infty}\!\!$, $\sigma\in\{L,M,R\}^{*}$, with $\inf(\mathbf{u})\geq\inf(\sigma(1\overline{0}))$, $\sup(\mathbf{u})\leq\sup(\sigma(0\overline{1}))$. Then $\mathbf{u}$ ends with $\sigma(\mathbf{v})$ for some $\mathbf{v}\in\{0,1\}^{\infty}$ or with $\sigma^{\prime}(\overline{0})$, $\sigma^{\prime}\in\{L,M,R\}^{*}M$, $\sigma\in\sigma^{\prime}\{L,M,R\}^{*}$.

Proof.

The statement is trivially true when $\sigma$ is the identity. Suppose that it holds for some $\sigma\in\{L,M,R\}^{*}$, let $\varphi\in\{L,M,R\}$ and $\mathbf{u}\in\{0,1\}^{\infty}$ with $\inf(\mathbf{u})\geq\inf(\varphi\sigma(1\overline{0}))$, $\sup(\mathbf{u})\leq\sup(\varphi\sigma(0\overline{1}))$.

If $\varphi=L$, then $\sup(\mathbf{u})\leq\overline{10}$, thus every $1$ in $\mathbf{u}$ is followed by a [math], hence $\mathbf{u}=L(\mathbf{v})$ or $\mathbf{u}=1L(\mathbf{v})$ for some $\mathbf{v}\in\{0,1\}^{\infty}$. Similary, if $\varphi=R$, then $\inf(\mathbf{u})\geq\overline{01}$, hence $\mathbf{u}=R(\mathbf{v})$ or $\mathbf{u}=0R(\mathbf{v})$ for some $\mathbf{v}\in\{0,1\}^{\infty}$. If $\varphi=M$, then $\inf(\mathbf{u})\geq 0\overline{01}$ and $\sup(\mathbf{u})\leq 1\overline{10}$. Hence, for all $k\geq 1$, $0(01)^{k}$ as well as $1(10)^{k}$ is always followed in $\mathbf{u}$ by $01$ or $10$. Since $\mathbf{u}$ contains $001$ or $110$ if $\mathbf{u}\notin\{M(\overline{0}),M(\overline{1})\}$, we obtain that $\mathbf{u}$ ends with $M(\mathbf{v})$ for some $\mathbf{v}\in\{0,1\}^{\infty}$.

We can assume that $\mathbf{v}\in\{\overline{0},\overline{1}\}$ or $\inf_{1}(\mathbf{v})=\inf(\mathbf{v})$ and $\sup_{0}(\mathbf{v})=\sup(\mathbf{v})$, by Lemma 4. If $\mathbf{v}\neq\overline{0}$, then we cannot have $\inf(\mathbf{v})<\inf(\sigma(1\overline{0}))$ because this would imply that $\inf(\varphi(\mathbf{v}))<\inf(\varphi\sigma(1\overline{0}))$ by Lemma 3. Similarly, we obtain that $\sup(\mathbf{v})\leq\sup(\sigma(1\overline{0}))$ if $\mathbf{v}\neq\overline{1}$. If $\mathbf{v}=\overline{0}$, $\varphi\in\{L,R\}$, then $\inf(\varphi(\overline{0}))\geq\inf(\varphi\sigma(1\overline{0}))$ implies that $\inf(\sigma(1\overline{0}))=\overline{0}$, thus $\mathbf{v}=\sigma(\overline{0})$. Similarly, if $\mathbf{v}=\overline{1}$ and $\varphi\in\{L,R\}$, then $\sup(\varphi(\overline{1}))\leq\sup(\varphi\sigma(0\overline{1}))$ implies that $\sup(\sigma(0\overline{1}))=\overline{1}$, thus $\mathbf{v}=\sigma(\overline{1})$. If $\mathbf{v}\in\{\overline{0},\overline{1}\}$, $\varphi=M$, then $\mathbf{u}$ ends with $M(\overline{0})$ since $M(\overline{1})=1M(\overline{0})$. Therefore, $\mathbf{u}$ ends with $\varphi\sigma(\mathbf{v})$ or with $\sigma^{\prime}(\overline{0})$, $\sigma^{\prime}\in\{L,M,R\}^{*}M$, $\varphi\sigma\in\sigma^{\prime}\{L,M,R\}^{*}$. ∎

We obtain the following lower bound for $\mathcal{L}(m)$, cf. Figure 3.

Proposition 2**.**

Let $m\in(1,2]$. We have $\mathcal{L}(m)\geq g_{0\overline{1}}(m)$ and

[TABLE]

Proof.

For all $\mathbf{v}\in 1\{0,1\}^{\infty}\setminus\{\overline{1}\}$, we have $\inf(\mathbf{v})\leq 0\overline{1}$. Then $\mathbf{v}\in U_{\beta}(m)$ implies that $\beta\geq g_{0\overline{1}}(m)$ by Lemma 1, hence $\mathcal{L}(m)\geq g_{0\overline{1}}(m)$.

Suppose that $U_{\beta}(m)\cap\{0,1\}^{\infty}$ is uncountable for $\beta<g_{\sigma(1\overline{0})}(m)$, $m\leq\mu_{\sigma(01\overline{0})}$, $\sigma\in\{L,M,R\}^{*}M$, thus $\beta<g_{\sigma(01\overline{0})}(m)\leq f_{\sigma(01\overline{0})}(m)$. Then $U_{\beta}(m)$ contains an aperiodic sequence $\mathbf{v}\in 1\{0,1\}^{\infty}$, with $f_{\mathbf{v}}(m)<f_{\sigma(01\overline{0})}(m)$ and $g_{\mathbf{v}}(m)<g_{\sigma(1\overline{0})}(m)$ by Lemma 1, thus $\inf(\mathbf{v})>\inf(\sigma(1\overline{0}))$ and $\sup(\mathbf{v})<\sup(\sigma(01\overline{0}))$ by Lemma 2. By Lemma 5, $\mathbf{v}$ ends with $\sigma(\mathbf{v}^{\prime})$ for some (aperiodic) $\mathbf{v}^{\prime}\in\{0,1\}^{\infty}$, contradicting that $\sup(\mathbf{v})<\sup(\sigma(01\overline{0}))$. Symetrically, we get that $\mathcal{L}(m)\geq f_{\sigma(0\overline{1})}(m)$ for $m\geq\mu_{\sigma(10\overline{1})}$.

If $\mathbf{u}$ is a limit word of a primitive sequence $(\sigma_{n})_{n\geq 1}\in\{L,M,R\}^{\infty}$, then we have $\mu_{\sigma^{\prime}_{n}(01\overline{0})}\to\mu_{\mathbf{u}}$ for $\sigma^{\prime}_{n}=\sigma_{1}\sigma_{2}\cdots\sigma_{n}$ as $n\to\infty$, thus $\beta<g_{\mathbf{u}}(m)$, $m\leq\mu_{\mathbf{u}}$ implies that $\beta<\min(g_{\sigma^{\prime}_{n}(01\overline{0})}(m),f_{\sigma^{\prime}_{n}(01\overline{0})}(m))$ for some $n\geq 1$, and we obtain as in the previous paragraph that $U_{\beta}(m)\cap\{0,1\}^{\infty}$ is at most countable. Therefore, we have $\mathcal{L}(m)\geq g_{\mathbf{u}}(m)$ and, similarly, $\mathcal{L}(m)\geq f_{\mathbf{u}}(m)$ for $m\geq\mu_{\mathbf{u}}$. ∎

Propositions 1 and 2 prove the formula for $\mathcal{L}(m)$ in Theorem 1. It remains to show that this covers all $m\in(1,2]$.

For the characterisation of $\mathcal{G}(m)$, in [BS17, Proposition 3.3] the partition

[TABLE]

for intervals of sequences in $\{0,1\}^{\infty}$ is used, which is a consequence of the partition

[TABLE]

We have to refine these partitions. For $\boldsymbol{\sigma}=(\sigma_{n})_{n\geq 1}\in\{L,M,R\}^{\infty}$, set

[TABLE]

[TABLE]

Note that, for a primitive sequence $\boldsymbol{\sigma}$, $\inf(\mathbf{u})$ as well as $\sup(\mathbf{u})$ does not depend on the limit word $\mathbf{u}$. We order sequences in $\{L,M,R\}^{\infty}$ lexicographically.

Lemma 6**.**

In $\{0,1\}^{\infty}$, we have

[TABLE]

If $\boldsymbol{\sigma}<\boldsymbol{\sigma}^{\prime}$, then $\mathbf{v}<\mathbf{v}^{\prime}$ for all $\mathbf{v}\in I_{\boldsymbol{\sigma}}$, $\mathbf{v}^{\prime}\in I_{\boldsymbol{\sigma}^{\prime}}$, and for all $\mathbf{v}\in J_{\boldsymbol{\sigma}}$, $\mathbf{v}^{\prime}\in J_{\boldsymbol{\sigma}^{\prime}}$.

Proof.

We clearly have $I_{\boldsymbol{\sigma}}\subset(\overline{0},0\overline{1})$ for all $\boldsymbol{\sigma}\in\{L,M,R\}^{\infty}$. For all $\sigma\in\{L,M,R\}^{*}$, Lemma 3 gives that $\inf(\sigma(1\overline{0}))=\inf(\sigma L(1\overline{0}))$, $\inf(\sigma L(10\overline{1}))=\inf(\sigma M(1\overline{0}))$, and we have $M(\overline{1})=R(1\overline{0})$, $R(10\overline{1})=10\overline{1}$, thus

[TABLE]

(in this order). Inductively, we obtain that the sets $I_{\boldsymbol{\sigma}}$ are ordered by the lexicographical order on $\{L,M,R\}^{\infty}$. Moreover, the union of sets $I_{\boldsymbol{\sigma}}$ with $\boldsymbol{\sigma}$ ending in $M\overline{L}$ or $M\overline{R}$ covers $(\inf(1\overline{0}),\inf(10\overline{1}))=(\overline{0},0\overline{1})$, except for points lying in the intersection of nested intervals $\bigcap_{n\geq 1}(\inf(\sigma_{1}\cdots\sigma_{n}(1\overline{0})),\inf(\sigma_{1}\cdots\sigma_{n}(10\overline{1})))$ for some $\boldsymbol{\sigma}=(\sigma_{n})_{n\geq 1}\in\{L,M,R\}^{\infty}$. Since $\sigma_{1}\cdots\sigma_{n}(\overline{0})$ is close to $\sigma_{1}\cdots\sigma_{n}(0\overline{1})$ for large $n$, these intervals tend to some $\mathbf{v}\in\{0,1\}^{\infty}$. If $\boldsymbol{\sigma}$ is primitive, then $I_{\boldsymbol{\sigma}}=\{\mathbf{v}\}$. If $\sigma_{n+1}\sigma_{n+2}\cdots$ is $\overline{L}$ or $\overline{R}$, then we have $\mathbf{v}=\inf(\sigma_{1}\cdots\sigma_{n}(1\overline{0}))$ or $\mathbf{v}=\inf(\sigma_{1}\cdots\sigma_{n}(10\overline{1}))$, which are not in the intersection.

The proof for $(1\overline{0},\overline{1})=\bigcup_{\boldsymbol{\sigma}\in\{L,M,R\}^{\infty}}J_{\boldsymbol{\sigma}}$ is similar, with

[TABLE]

Hence, the $J_{\boldsymbol{\sigma}}$ are also ordered by the lexicographical order on $\{L,M,R\}^{\infty}$. ∎

Proposition 3**.**

We have the partition

[TABLE]

Proof.

For $m\in(1,\mu_{0\overline{1}})$, $\boldsymbol{\sigma}\in\{L,M,R\}^{\infty}$, let

[TABLE]

[TABLE]

By Lemmas 2 and 6, we have

[TABLE]

(Note that $f_{\mathbf{u}}(m)$ is close to $f_{\mathbf{u}^{\prime}}(m)$ if $\sup(\mathbf{u})$ is close to $\sup(\mathbf{u}^{\prime})$, $g_{\mathbf{u}}(m)$ is close to $g_{\mathbf{u}^{\prime}}(m)$ if $\inf(\mathbf{u})$ is close to $\inf(\mathbf{u}^{\prime})$.) If $\boldsymbol{\sigma}<\boldsymbol{\sigma}^{\prime}$, then we have $\beta>\beta^{\prime}$ if $\beta\in I^{\prime}_{\boldsymbol{\sigma}}(m)$, $2\leq\beta^{\prime}\in I^{\prime}_{\boldsymbol{\sigma}^{\prime}}(m)$, and $\beta<\beta^{\prime}$ if $2\leq\beta\in J^{\prime}_{\boldsymbol{\sigma}}(m)$, $\beta^{\prime}\in J^{\prime}_{\boldsymbol{\sigma}^{\prime}}(m)$, by Lemmas 2 and 6. Since $\max(f_{\mathbf{u}}(m),g_{\mathbf{u}}(m))\geq 2$ for all $\mathbf{u}\in\{0,1\}^{\infty}$ and $\inf(\sigma M(1\overline{0}))\leq\inf(\sigma M(\overline{0}))$, $\sup(\sigma M(0\overline{1}))\geq\sup(\sigma M(\overline{1}))$ for all $\sigma\in\{L,M,R\}^{*}$, we have $I^{\prime}_{\boldsymbol{\sigma}}(m)\subset[2,\infty)$ or $J^{\prime}_{\boldsymbol{\sigma}}(m)\subset[2,\infty)$ for all $\boldsymbol{\sigma}\in\{L,M,R\}^{\infty}$. Therefore, we have $I^{\prime}_{\boldsymbol{\sigma}}(m)\cap J^{\prime}_{\boldsymbol{\sigma}}(m)\neq\emptyset$ for some $\boldsymbol{\sigma}\in\{L,M,R\}^{\infty}$. If $\boldsymbol{\sigma}$ is primitive, this means that $m=\mu_{\mathbf{u}}$. If $\sigma_{n}\sigma_{n+1}\cdots=M\overline{L}$, then we have $g_{\sigma_{1}\cdots\sigma_{n}(1\overline{0})}(m)\in[f_{\sigma_{1}\cdots\sigma_{n}(\overline{0})}(m),f_{\sigma_{1}\cdots\sigma_{n}(01\overline{0})}(m)]$, which means that $m\in[\mu_{\sigma_{1}\cdots\sigma_{n}(1\overline{0})},\mu_{\sigma_{1}\cdots\sigma_{n}(01\overline{0})}]$, see Figure 3. Similarly, if $\sigma_{n}\sigma_{n+1}\cdots=M\overline{R}$, then we have that $m\in[\mu_{\sigma_{1}\cdots\sigma_{n}(10\overline{1})},\mu_{\sigma_{1}\cdots\sigma_{n}(0\overline{1})}]$. ∎

Proof of Theorem 1.

This is a direct consequence of Propositions 1, 2 and 3. ∎

3. Final remarks and open questions

By [KLP11, BS17, Kwo18], there are simple formulas for $\mu_{\sigma(1\overline{0})}$, $\mu_{\sigma(\overline{0})}$ and $\mu_{\sigma(0\overline{1})}$, $\sigma\in\{L,R\}^{*}M$, and for $\mu_{\mathbf{u}}$, $\mathbf{u}\in\mathcal{S}_{L,R}$. This is because, for $\mathbf{u}\in\{\sigma(1\overline{0}),\sigma(0\overline{1})\}$, $\sigma\in\{L,R\}^{*}M$, or $\mathbf{u}\in\mathcal{S}_{L,R}$, we have $\inf(\mathbf{u})=0\mathbf{v}$, $\sup(\mathbf{u})=1\mathbf{v}$ for some $\mathbf{v}$, thus $(\beta-1)(1+\pi_{\beta}(0\mathbf{v}))=(\beta-1)^{2}=\beta\pi_{\beta}(1\mathbf{v})$, where $\beta>1$ is defined by $\pi_{\beta}(20\mathbf{v})=1$, which gives that $\mu_{\mathbf{u}}=(\beta-1)^{2}$. For $\mathbf{u}=\sigma(\overline{0})$, we have $\inf(\mathbf{u})=\overline{0w1}$, $\sup(\mathbf{u})=\overline{1w0}$, with $\sigma(0)=0w1$, and

[TABLE]

where $\beta>1$ is defined by $\pi_{\beta}(20w\overline{0})=1$ and $|\sigma(0)|$ is the length of $\sigma(0)$, hence $\mu_{\sigma(\overline{0})}=(\beta-1)^{2}\beta^{|\sigma(0)|}/(\beta^{|\sigma(0)|}-1)$. Are there similar formulas for $\sigma\in\{L,M,R\}^{*}M$?

In [BS17, Kwo18], it was proved that the Hausdorff dimension of $\{\mu_{\mathbf{u}}:\,\mathbf{u}\in\mathcal{S}_{L,R}\}$ is [math], using that the number of balanced words grows polynomially. What is the complexity of $\mathcal{S}_{L,M,R}$?

As mentioned in the Introduction, we know the generalised Komornik–Loreti constant $\mathcal{K}(m)$ only for $m=2$ and when $\mathcal{G}(m)=1+\sqrt{m}=\mathcal{K}(m)=\mathcal{L}(m)$. This is due to the fact that it is usually difficult to study maps with two holes; see Figure 2. (For $m=2$, we can use the symmetry of the map $T$, and for $\mathcal{L}(m)=1+\sqrt{m}$, we can restrict to sequences in $\{0,1\}^{\infty}$. ) New ideas are needed for the general case.

Finally, Sturmian holes are key ingredients in [Sid14], where supercritical holes for the doubling map are studied. Do our Thue–Morse–Sturmian sequences also play a role in this context?