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# Quantization dimension for self-similar measures of overlapping construction

MRINAL KANTI ROYCHOWDHURY  Mrinal Kanti Roychowdhury, University of Texas Rio
Grande Valley, 1201 West University Drive, Edinburg, TX 78539-2999, USA
<EMAIL_ADDRESS>and  Károly Simon  Károly Simon, Budapest
University of Technology and Economics, MTA-BME Stochastics Research Group,
P.O. Box 91, 1521 Budapest, Hungary<EMAIL_ADDRESS>
###### Abstract.

Quantization dimension has been computed for many invariant measures of
dynamically defined fractals having well separated cylinders (that is the
cases when the so-called Open Set Condition, OSC holds). To attack the same
problem in case of heavy overlaps between the cylinders, we consider a family
of self-similar measures, for which the underlying Iterated Function System
satisfies the so-called Weak Separation Property (WSP) but does not satisfy
the OSC since complete overlaps occur in between the cylinders. Our work also
shows that the quantization dimension determined for the set of overlap self-
similar construction satisfying the WSP has a relationship with the
temperature function of the thermodynamic formalism.

###### Key words and phrases:

Quantization dimension, self-similar measure, Weak Separation Property.

###### 2000 Mathematics Subject Classification:

Primary 28A80; Secondary 37A50, 94A15, 60D05

##  1\. Introduction

The basic goal of quantization for probability distribution is to reduce the
number of values, which is typically uncountable, describing a probability
distribution to some finite set and thus approximation of a continuous
probability distribution by a discrete distribution. It has broad applications
in signal processing, telecommunications, data compression, image processing
and cluster analysis. Over the years, many useful theorems have been proved
and numerous other results and algorithms have been obtained in quantization.
For a detailed survey on the subject and comprehensive lists of references to
the literature one is referred to [B, BW, G, GG, GKL, GL2, GL1, GL4, GN, Z1,
Z2]. Rigorous mathematical treatment of the fundamentals of the quantization
theory is provided in Graf-Luschgy’s book (see [GL1]). In general, these
theorems almost exclusively involve absolutely continuous probability measures
on $\mathbb{R}^{d}$. Two main goals of the theory are: $(1)$ finding the exact
configuration of a so-called ‘$n$-optimal set’ which corresponds to the
support of the quantized version of the distribution, and $(2)$ estimating the
rate at which some specified measure of the error goes to zero as $n$ goes to
infinity. This paper deals with the second kind of problem.

Given a Borel probability measure $\mu$ on $\mathbb{R}^{d}$, a number
$r\in(0,+\infty)$ and a natural number $n\in\mathbb{N}$, the $n$th
quantization error of order $r$ for $\mu$, is defined by

$V_{n,r}:=V_{n,r}(\mu)=\text{inf}\left\\{\int
d(x,\alpha)^{r}d\mu(x):\alpha\subset\mathbb{R}^{d},\,1\leq\text{card}(\alpha)\leq
n\right\\},$

where $d(x,\alpha)$ denotes the distance from the point $x$ to the set
$\alpha$ with respect to a given norm $\|\cdot\|$ on $\mathbb{R}^{d}$. If
$\alpha$ is a finite set, the error $\int d(x,\alpha)^{r}d\mu(x)$ is often
referred to as the cost, or distortion error of order $r$ for $\mu$ and
$\alpha$. It is known that for a Borel probability measure $\mu$, if its
support contains infinitely many elements and $\int\|x\|^{r}d\mu(x)$ is
finite, then an optimal set of $n$-means always has exactly $n$-elements (see
[AW, GKL, GL, GL1]). This set $\alpha$ can then be used to give a best
approximation of $\mu$ by a discrete probability supported on a set with no
more than $n$ points. Such a set $\alpha$ for which the infimum occurs and
contains no more than $n$ points is called an optimal set of $n$-means, or
optimal set of $n$-quantizers (of order $r$). Under suitable conditions this
can be done by giving each point $a\in\alpha$ a mass corresponding to
$\mu(A_{a})$, where $A_{a}$ is the set of points $x\in\mathbb{R}^{d}$ such
that $d(x,\alpha)=d(x,a)$. So, $\\{A_{a}:a\in\alpha\\}$ is the ‘Voronoi’
partition of $\mathbb{R}^{d}$ induced by $\alpha$. Of course, the idea of
‘best approximation’ is, in general, dependent on the choice of $r$. For some
recent work in this direction of optimal sets of $n$-means, one can see [CR,
DR1, DR2, GL5, R, R1, R2, R3, R4, R5, R6, RR1]. The set of all optimal sets of
$n$-means for a probability measure $\mu$ is denoted by
$\mathcal{C}_{n,r}(\mu)$. i.e.,

(1.1)
$\mathcal{C}_{n,r}(\mu):=\\{\alpha\subset\mathbb{R}^{d}:1\leq\text{card}(\alpha)\leq
n\text{ and }V_{n,r}(\mu)=\int d(x,\alpha)^{r}d\mu(x)\\}.$

Write $e_{n,r}(\mu):=V_{n,r}^{\frac{1}{r}}(\mu)$. The numbers

(1.2) $\underline{D}_{r}(\mu):=\liminf_{n\to\infty}\frac{\log n}{-\log
e_{n,r}(\mu)},\text{ and
}\overline{D}_{r}(\mu):=\limsup_{n\to\infty}\frac{\log n}{-\log
e_{n,r}(\mu)},$

are called the lower and the upper quantization dimensions of $\mu$ of order
$r$, respectively. If $\underline{D}_{r}(\mu)=\overline{D}_{r}(\mu)$, the
common value is called the quantization dimension of $\mu$ of order $r$ and is
denoted by $D_{r}(\mu)$. Quantization dimension measures the speed at which
the specified measure of the error goes to zero as $n$ tends to infinity. For
any $\kappa>0$, the numbers $\liminf_{n}n^{\frac{r}{\kappa}}V_{n,r}(\mu)$ and
$\limsup_{n}n^{\frac{r}{\kappa}}V_{n,r}(\mu)$ are called the
$\kappa$-dimensional lower and upper quantization coefficients for $\mu$,
respectively. The quantization coefficients provide us with more accurate
information about the asymptotics of the quantization error than the
quantization dimension. Compared to the calculation of quantization dimension,
it is usually much more difficult to determine whether the lower and the upper
quantization coefficients are finite and positive. It follows from [GL1,
Proposition 11.3] that if

(1.3) $0\leq t<\underline{D}_{r}<s,\text{ then
}\lim_{n\to\infty}ne^{t}_{n,r}=+\infty\text{ and
}\liminf_{n\to\infty}ne^{s}_{n,r}=0.$

and if

(1.4) $0\leq t<\overline{D}_{r}<s,\text{ then
}\limsup_{n\to\infty}ne^{t}_{n,r}=+\infty\text{ and
}\lim_{n\to\infty}ne^{s}_{n,r}=0.$

For probabilities with non-vanishing absolutely continuous parts the numbers
$D_{r}$ are all equal to the dimension $d$ of the underlying space, but for
singular probabilities the family $\\{D_{r}\\}_{r>0}$ gives an interesting
description of their geometric (multifractal) structures. A detailed account
of this theory can be found in [GL1]. There the quantization dimension is
introduced as a new type of fractal dimension spectrum and a formula for its
determination is derived in the case of self-similar probabilities with the
strong separation property. In [GL2], Graf and Luschgy extended the above
result and determined the quantization dimension $D_{r}$ of self-similar
probabilities with the weaker open set condition, but there it remained open
whether the $D_{r}$-dimensional lower quantization coefficient is positive.
Later they answered it in [GL3]. Under the open set condition, Lindsay and
Mauldin (see [LM]) determined the quantization dimension function $D_{r}$,
where $r\in(0,+\infty)$, of an $F$-conformal measure $m$ associated with a
conformal iterated function system determined by finitely many conformal
mappings. Subsequently, quantization dimension has been computed for many
invariant measures of dynamically defined fractals having well separated
cylinders (that is the cases when the so-called Open Set Condition holds), for
example, one can see [MR, R7, R8, R9, R10, R11, R12, R13, R14, R15]. From all
the known results it can be seen that if the quantization dimension function
$D_{r}(\mu)$ for $r>0$ of a fractal probability measure $\mu$ exists, it has a
relationship with the temperature function of the thermodynamic formalism that
arises in multifractal analysis of $\mu$ (see Figure 1).

Figure 1. To determine $D_{r}$ first find the point of intersection of
$y=\beta(q)$ and the line $y=rq$. Then $D_{r}$ is the $y$-intercept of the
line through this point and the point (1, 0).

In a very recent paper, among many other interesting applications, S. Zhu
[Zhu] has solved the problem of the computation of quantization dimension for
the complete overlapping case in the following very special situation: Let
$\mathcal{S}=\\{S_{k}\\}_{k\leq m}$ be a self-similar IFS with the following
properties: $(1)$ there are distinct $i$ and $j$ with $S_{i}=S_{j}$. $(2)$ The
self-similar IFS $\mathcal{S}=\\{S_{k}\\}_{k\leq m,k\neq j}$ satisfies the so-
called strong separation property. That is, for any two distinct
$u,v\in\\{1,\dots,m\\}\setminus\\{j\\}$, $S_{u}(\Lambda)\cap
S_{v}(\Lambda)=\emptyset$, where $\Lambda$ is the attractor of the IFSs
$\mathcal{S}$. In general, if we are given a self-similar IFS which satisfies
the so-called Weak Separation Property (WSP), but does not satisfy the OSC we
cannot get rid of the overlapping feature of the system just by throwing out
some of the mappings of the IFS. Even if we get rid of one of the mappings
which causes the total overlapping feature, we still have overlaps in the
remaining system (which are not complete overlaps).

In this paper we make a step towards our goal to determine the quantization
dimension for self-similar measures on the line in the case when the
underlying self-similar system satisfies the WSP (for the definition and basic
properties see [Zer]). Namely, we solve this problem for a special family
which has the above mentioned properties. Our work also shows that the
quantization dimension determined for a set of overlap self-similar
construction satisfying the WSP has a relationship with the temperature
function of the thermodynamic formalism.

After the paper was written we have just realized that a more general result
was recently proved in the paper [KNZ] by Kessebohmer et al. But, their method
is completely different than the method that we have given in this paper.

##  2\. An overlapping self-similar IFS on the line

We consider the following self-similar IFS on $\mathbb{R}$

(2.1)
$\mathcal{S}=\left\\{S_{i}(x)=\frac{1}{3}x+i\right\\}_{i\in\left\\{0,1,3\right\\}}.$

Figure 2. $I_{i_{1}\dots i_{n}}:=S_{i_{1}\dots i_{n}}(I)$,
$(i_{1},\dots,i_{n})\in\mathcal{A}^{n}$. $I_{10}=I_{03}$

Naturally the alphabet corresponds to this IFS is
$\mathcal{A}:=\left\\{0,1,3\right\\}$. We write $\Sigma$ ($\Sigma^{*}$) for
the set of infinite (finite) words, respectively, over the alphabet
$\mathcal{A}$. As usual we write $\sigma$ for the left shift on
$\Sigma\cup\Sigma^{*}$. We write

(2.2) $\mathbf{i}^{-}:=(i_{1},\dots,i_{n-1})\quad\text{ for an
}\quad\mathbf{i}=(i_{1},\dots,i_{n})\in\mathcal{A}^{n}.$

We say that $\Gamma\subset\Sigma^{*}$ is a maximal finite antichain if for
every $\mathbf{i}\in\Sigma$ there exists a unique $n$ such that
$\mathbf{i}|_{n}\in\Gamma$.

Let $\Lambda$ be the attractor of $\mathcal{S}$. That is, $\Lambda$ is the
unique non-empty compact set satisfying
$\Lambda=\bigcup\limits_{i\in\mathcal{A}}S_{i}(\Lambda)$. The smallest
interval that contains $\Lambda$ is $I=\left[0,\frac{9}{2}\right]$. Put
$I_{i_{1}\dots i_{n}}:=S_{i_{1}\dots i_{n}}(I)$, where we use the shorthand
notation $S_{i_{1}\dots i_{n}}=S_{i_{1}}\circ\cdots\circ S_{i_{n}}$.

The natural projection $\Pi:\Sigma\cup\Sigma^{*}\to\Lambda$ is defined by

(2.3) $\Pi(\mathbf{i}):=\sum_{k=1}^{|\mathbf{i}|}i_{k}3^{-(k-1)},$

where $|\mathbf{i}|=n$ if $\mathbf{i}\in\mathcal{A}^{n}$ and
$|\mathbf{i}|=\infty$ if $\mathbf{i}\in\Sigma$. For a finite word
$\mathbf{i}\in\Sigma^{*}$ the projection $\Pi(\mathbf{i})$ is the left end
point of the interval $I_{\mathbf{i}}$. That is

(2.4) $\text{ for }\mathbf{i},\mathbf{j}\in\Sigma^{*}\text{ with
}|\mathbf{i}|=|\mathbf{j}|\text{ we have
}\Pi(\mathbf{i})=\Pi(\mathbf{j})\Longleftrightarrow
I_{\mathbf{i}}=I_{\mathbf{j}}.$

We are given a probability vector $\mathbf{p}:=(p_{0},p_{1},p_{3})$. That is,
$p_{i}>0$ and $\sum_{i\in\mathcal{A}}p_{i}=1$.

We form the corresponding product measure $\mu:=\mathbf{p}^{\mathbb{N}}$ and
define its push forward measure $\nu:=\Pi_{*}\mu$. Then, $\nu$ is the self-
similar measure corresponding to the probability vector $\mathbf{p}$. That is,
for every maximal finite antichain $\Gamma\subset\Sigma^{*}$ we have

(2.5) $\nu=\sum_{\mathbf{i}\in\Gamma}p_{\mathbf{i}}\cdot\nu\circ
S_{\mathbf{i}}^{-1}.$

The peculiarity of this IFS $\mathcal{S}$ is that we have complete overlap:
$I_{0,3}=I_{1,0}.$ Moreover, an easy case analysis yields:

######  Fact 2.1.

Assume that $\mathbf{i},\mathbf{j}\in\mathcal{A}^{n}$, with $i_{1}<j_{1}$ such
that

$\Pi(\mathbf{i})=\Pi(\mathbf{j})\quad\text{\emph{but}}\quad\Pi(\mathbf{i}|_{k})\neq\Pi(\mathbf{j}|_{k}),\quad\forall
k<n.$

Then, $\mathbf{i}=(\underbrace{1,\dots,1}_{n-1},0)$ and
$\mathbf{j}=(0,\underbrace{3,\dots,3}_{n-1})$.

######  Definition 2.2.

Let

(2.6) $A:=\left(\begin{array}[]{ccc}1&1&0\\\ 1&1&1\\\ 1&1&1\\\
\end{array}\right)\qquad B:=\left(\begin{array}[]{ccc}1&1&1\\\ 0&1&1\\\
1&1&1\\\ \end{array}\right),$

where we index the rows and columns with ${0,1,3}$ in increasing order. We
introduce the subshifts of finite types

$\Sigma_{A}:=\left\\{\mathbf{i}\in\Sigma:(i_{k},i_{k+1})\neq(0,3),\forall
k\right\\},\
\Sigma_{B}:=\left\\{\mathbf{i}\in\Sigma:(i_{k},i_{k+1})\neq(1,0),\forall
k\right\\}.$

Analogously, we define

$\mathcal{T}_{n}:=\left\\{\mathbf{i}\in\mathcal{A}^{n}:(i_{k},i_{k+1})\neq(0,3),\forall
k<n\right\\},$

and
$\mathcal{U}_{n}:=\left\\{\mathbf{i}\in\mathcal{A}^{n}:(i_{k},i_{k+1})\neq(1,0),\forall
k<n\right\\}.$ Set

$\Sigma_{A}^{*}:=\bigcup_{n=1}^{\infty}\mathcal{T}_{n}\cup\flat\text{ and
}\Sigma_{B}^{*}:=\bigcup_{n=1}^{\infty}\mathcal{U}_{n}\cup\flat,$

where $\flat$ is the empty word.

An obvious case analysis shows (see [BSS, Fact 4.2.18]) that the following
fact holds:

######  Fact 2.3.

  1. (a)

Assume that for the distinct $\mathbf{i},\mathbf{j}\in\mathcal{T}_{n}$ we have
$I_{\mathbf{i}}\cap I_{\mathbf{j}}\neq\emptyset$. Then
$|\mathbf{i}\wedge\mathbf{j}|=n-1$ and
$\left\\{i_{n},j_{n}\right\\}=\left\\{0,1\right\\}$, where
$\mathbf{i}\wedge\mathbf{j}$ is the common prefix of the words $\mathbf{i}$
and $\mathbf{j}$.

  2. (b)

Assume that for the distinct $\mathbf{i},\mathbf{j}\in\mathcal{U}_{n}$ we have
$I_{\mathbf{i}}\cap I_{\mathbf{j}}\neq\emptyset$. Then, there exists a $k\leq
n-2$ and an $\omega\in\mathcal{T}_{k}$ such that $\omega_{k}\neq 1$,
$\mathbf{i}=\omega 0\overline{3}^{n-k-1}$ and
$\mathbf{j}=\omega\overline{1}^{n-k}$.

For an $\mathbf{i}\in\mathcal{T}_{n}$ there can be exponentially many
$\mathbf{j}\in\mathcal{A}^{n}$ with $I_{\mathbf{i}}\cap
I_{\mathbf{j}}\neq\emptyset$. However, the previous Fact implies the following
corollary.

######  Corollary 2.4.

If $\mathbf{i}\in\mathcal{T}_{n}$, then there is at most one
$\mathbf{j}\in\mathcal{T}_{n}\setminus\left\\{\mathbf{i}\right\\}$ such that
$I_{\mathbf{i}}\cap I_{\mathbf{j}}\neq\emptyset$. The same remains valid if we
replace $\mathcal{T}_{n}$ with $\mathcal{U}_{n}$.

If $p_{1}\geq p_{3}$, then we should work with $\Sigma_{A}$ and
$\mathcal{T}_{n}$, $n\in\mathbb{N}$. On the other hand, if $p_{3}>p_{1}$, then
we should work with $\Sigma_{B}$, and $\mathcal{U}_{n}$, $n\in\mathbb{N}$.

######  Principal Assumption 2.5.

We always assume in this note that

(2.7) $p_{3}\leq p_{1}.$

For the symmetry pointed out in Corollary 2.4, we may assume without any loss
of generality for the rest of the paper that (2.7) holds.

It is immediate from part (a) of Fact 2.3 that

(2.8)
$\mathbf{i},\mathbf{j}\in\mathcal{T}_{n},\qquad\mathbf{i}=\mathbf{j}\Longleftrightarrow
S_{\mathbf{i}}=S_{\mathbf{j}}\Longleftrightarrow\Pi(\mathbf{i})=\Pi(\mathbf{j}).$

##  3\. The main result

######  Definition 3.1 (Pressure of a potential).

We call a continuous function $f:\Sigma^{*}_{A}\to[0,\infty)$ a potential. The
pressure of the potential $f$ is defined by

(3.1)
$P(f):=\lim\limits_{n\to\infty}\frac{1}{n}\log\sum_{\mathbf{i}\in\mathcal{T}_{n}}f(\mathbf{i}),$

if the limit exists, otherwise we replace the limit with limsup.

###  3.1. The main result

As we mentioned above we assume that $p_{1}\geq p_{3}$. If $p_{1}\leq p_{3}$,
then all $\mathcal{T}_{n}$ below should be replaced by $\mathcal{U}_{n}$ and
all the results remain unchanged. We define

(3.2)
$\mathcal{I}_{\mathbf{i}}:=\left\\{\boldsymbol{\eta}\in\mathcal{A}^{n}:S_{\boldsymbol{\eta}}=S_{\mathbf{i}}\right\\}\quad\text{
and
}\quad\psi(\mathbf{i}):=\sum_{\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{i}}}p_{\boldsymbol{\eta}},\quad\text{
for every }\mathbf{i}\in\Sigma^{*}_{A},$

and we define

(3.3) $\psi(\flat):=1,\text{ where }\flat\text{ is the empty word. }$

We will prove that the limit in the following definition exists:

(3.4)
$p(t):=P(\psi^{t})=\lim\limits_{n\to\infty}\frac{1}{n}\log\sum_{\mathbf{i}\in\mathcal{T}_{n}}\psi^{t}(\mathbf{i}),\quad
t\geq 0.$

Moreover, we will prove in Fact 4.3 that there exists a unique $t_{0}\in(0,1)$
such that

(3.5) $p(t_{0}):=rt_{0}\log 3.$

Then, we define $\chi_{r}$ such that

(3.6) $t_{0}=\frac{\chi_{r}}{r+\chi_{r}},\text{ that is,
}\quad\chi_{r}=\frac{t_{0}r}{1-t_{0}}.$

######  Theorem 3.2.

The quantization dimension of the measure $\nu$ is $D_{r}(\nu)=\chi_{r}$.

###  3.2. The main Proposition

To prove our main result we need our Main Proposition (Proposition 7.19)
below. To state it we need some further notation.

####  3.2.1.  A projection
$\Phi:\Sigma\cup\Sigma^{*}\to\Sigma_{A}\cup\Sigma_{A}^{*}$

Recall that by definition $\Sigma\cup\Sigma^{*}$ is the collection of finite
or infinite words over the alphabet $\mathcal{A}$. Similarly,
$\Sigma_{A}\cup\Sigma_{A}^{*}\subset\Sigma\cup\Sigma^{*}$ is the collection of
those elements of $\Sigma\cup\Sigma^{*}$, which do not contain the sequence
$(0,3)$. First we define such a mapping
$\Phi:\Sigma\cup\Sigma^{*}\to\Sigma_{A}\cup\Sigma_{A}^{*}$ which has the
following properties: For every $\boldsymbol{\eta}\in\Sigma\cup\Sigma^{*}$

  1. (a)

$\Phi$ preserves the length of every word:
$|\boldsymbol{\eta}|=|\Phi(\boldsymbol{\eta})|$,

  2. (b)

$\Pi(\boldsymbol{\eta})=\Pi(\Phi(\boldsymbol{\eta}))$.

In the rest of the paper we frequently use the following notation: For a digit
$a$ and $n\in\mathbb{N}\cup\\{\infty\\}$,

$\overline{a}^{n}:=\underbrace{a,a,\dots,a}_{n}.$

######  Definition 3.3.

We define $\Phi:\Sigma\cup\Sigma^{*}\to\Sigma_{A}\cup\Sigma_{A}^{*}$ as
follows: Let $\boldsymbol{\eta}\in\Sigma$. We obtain
$\mathbf{i}:=(i_{1},i_{2},\dots):=\Phi(\boldsymbol{\eta})\in\Sigma_{A}$ from
$\boldsymbol{\eta}\in\Sigma$ by successive substitutions as follows

  1. (a)

For every $1\leq k<\ell<\infty$ such that
$(\eta_{k},\dots,\eta_{\ell})=(0,\overline{3}^{\ell-k})$ and
$\eta_{\ell+1}\neq 3$, we define
$(i_{k},\dots,i_{\ell}):=(\overline{1}^{\ell-k},0)$.

  2. (b)

If there exists a $1\leq k<\infty$ such that
$(\eta_{k},\eta_{k+1},\eta_{k+1},\dots)=(0,\overline{3}^{\infty})$, then we
define $(i_{k},i_{k+1},\dots):=(\overline{1}^{\infty})$.

Then, we got rid of all the $(\eta_{k},\eta_{k+1})=(0,3)$ in
$\boldsymbol{\eta}$, so
$\Phi(\boldsymbol{\eta})\in\Sigma_{A}\cup\Sigma_{A}^{*}$ and by Fact 2.1:

(3.7)
$\Pi(\Phi(\boldsymbol{\eta}))=\Pi(\boldsymbol{\eta}),\qquad\forall\boldsymbol{\eta}\in\Sigma\cup\Sigma^{*}.$

However, considerable technical difficulties are caused by the fact that

(3.8) $\Phi\circ\sigma\neq\sigma\circ\Phi.$

Namely, $\sigma\Phi(0,3)=\sigma 10=0\neq 3=\Phi(3)=\Phi(\sigma 03)$.

Moreover,

(3.9) $\exists\,\boldsymbol{\eta},\text{ and
}\exists\,m<|\boldsymbol{\eta}|,\qquad\Phi(\boldsymbol{\eta}|_{m})\neq\Phi(\boldsymbol{\eta})|_{m}.$

For example,

$\Phi(0,0,3)=(0,1,0),\text{ but
}\Phi((0,0,3)|_{2})=\Phi(0,0)=(0,0)\neq(0,1)=\Phi(0,0,3)|_{2}.$

D.J. Feng [F] introduced a very important family of potentials. This family
was termed quasi-multiplicative potentials by A. Käenmäki and H.W. Reeve
[KR]).

####  3.2.2. Weak quasi-multiplicative potentials

######  Definition 3.4.

We say that a function $\phi:\Sigma_{A}^{*}\to[0,\infty)$ is a weak quasi-
multiplicative potential on $\Sigma^{*}_{A}$ if the following two conditions
hold:

  1. (a)

There is an $\boldsymbol{\ell}\in\Sigma_{A}^{*}$ which is not the empty word
such that $\phi(\boldsymbol{\ell})>0$. Moreover, there exist $C_{1},C_{2}>0$
such that

(3.10) $\phi(\mathbf{i}\mathbf{j})\leq
C_{1}\phi(\mathbf{i})\phi(\mathbf{j}),\qquad\mathbf{i}\mathbf{j}\in\Sigma_{A}^{*}.$

  2. (b)

There exists a $z\in\mathbb{N}$ such that

(3.11)
$\forall\mathbf{i},\mathbf{j}\in\Sigma_{A}^{*},\exists\mathbf{k}\in\bigcup_{\ell=1}^{z}\mathcal{T}_{\ell}\cup\flat\text{
such that }\mathbf{i}\mathbf{k}\mathbf{j}\in\Sigma_{A}^{*}\text{ and
}\phi(\mathbf{i})\phi(\mathbf{j})\leq
C_{2}\phi(\mathbf{i}\mathbf{k}\mathbf{j}).$

First we introduce a potential $\widehat{\psi}:\Sigma^{*}_{A}\to(0,1]$ as
follows:

(3.12)
$\widehat{\psi}(\mathbf{i}):=\left\\{\begin{array}[]{ll}\max\left\\{\psi(\mathbf{i}),\psi(\mathbf{i}^{-}0)\right\\},&\hbox{if
$\mathbf{i}_{|\mathbf{i}|}=1$;}\\\ \psi(\mathbf{i}),&\hbox{if
$\mathbf{i}_{|\mathbf{i}|}\neq 1$,}\end{array}\right.\quad\text{ for
}\mathbf{i}\in\Sigma^{*}_{A}.$

####  3.2.3. The statement of the Main Proposition

######  Proposition 3.5.

The following properties hold:

  1. Property-1

$\widehat{\psi}$ is a weak quasi-multiplicative potential on $\Sigma^{*}_{A}$.

  2. Property-2

There exists a $C_{3}>0$ such that for every $n\geq 1$

(3.13) $1\leq\frac{\widehat{\psi}(\mathbf{i})}{\psi(\mathbf{i})}\leq
C_{3}\cdot n,\qquad\forall\mathbf{i}\in\mathcal{T}_{n}.$

  3. Property-3

For every $n\geq 1$

(3.14)
$\sum_{\mathbf{i}\in\mathcal{T}_{n}}\mathbbm{1}_{\mathcal{I}_{\mathbf{i}}}(x)\leq
2.$

  4. Property-4

For every $n\in\mathbb{N}\cup\\{\infty\\}$,
$\boldsymbol{\eta}\in\mathcal{A}^{n}$, $1\leq z<n$,
$\mathbf{i}:=\Phi(\boldsymbol{\eta})|_{z}$

(3.15)
$\boldsymbol{\eta}|_{n}\subset\mathcal{I}_{\mathbf{i}}\cup\mathcal{I}_{\mathbf{i}^{-}0}.$

Property-2 is an immediate consequence of Fact 7.14,

Property-3 is an immediate consequence of Part (a) of Fact 2.3,

Property-4 is proved in Part (c) of Fact 7.5.

The organization of the rest of the paper

  1. (a)

In Section 4 we introduce further pressure functions and finite maximal
antichains. Moreover, we prove some of their properties, assuming Proposition
7.19.

  2. (b)

In Sections 5 and 6 we prove Theorem 3.2 using only those properties of
$\widehat{\psi}(\mathbf{i})$ which are listed in Proposition 7.19.

  3. (c)

In Section 7 we prove that $\widehat{\psi}(\mathbf{i})$ satisfies the
Properties-1,2 listed in Proposition 7.19.

##  4\. Pressure functions and finite maximal antichans

We will use the following theorem due to D.J. Feng [F, Theorem 5.5].

######  Theorem 4.1 (Feng).

Let $\phi$ be a weak quasi-multiplicative potential on $\Sigma^{*}_{A}$. Then,
there exists a unique invariant ergodic measure $\mathfrak{m}$ on $\Sigma_{A}$
with the following property

(4.1)
$\mathfrak{m}(\mathbf{i})\thickapprox\frac{\phi(\mathbf{i})}{\sum_{\mathbf{j}\in\mathcal{T}_{n}}\phi(\mathbf{j})}\thickapprox\phi(\mathbf{i})\exp\left(-nP(\phi)\right),$

where $a(\mathbf{i})\thickapprox b(\mathbf{i})$ if there exists a $c>0$ such
that $\frac{1}{c}b(\mathbf{i})\leq a(\mathbf{i})\leq cb(\mathbf{i})$ for all
$\mathbf{i}\in\Sigma_{A}^{*}$ and

(4.2)
$P(\phi)=\lim\limits_{n\to\infty}\log\sum\limits_{\mathbf{i}\in\mathcal{T}_{n}}\phi(\mathbf{i}).$

###  4.1. Various pressure functions

By Theorem 4.1, we obtain that

(4.3) $\text{ the limit
}\quad\lim\limits_{n\to\infty}\frac{1}{n}\log\sum_{\mathbf{i}\in\mathcal{T}_{n}}\left(\widehat{\psi}(\mathbf{i})\right)^{t}\quad\text{exists.}$

Recall that the pressure function $p(t)=P(\psi^{t})$ was defined in (3.4) with
the comment that the existence of the limit in (3.4) would be proved later.
Using Property-2 and (4.3) we get that the second equation below holds:

(4.4)
$p(t)=\lim\limits_{n\to\infty}\frac{1}{n}\log\sum_{\mathbf{i}\in\mathcal{T}_{n}}\left(\psi(\mathbf{i})\right)^{t}=\lim\limits_{n\to\infty}\frac{1}{n}\log\sum_{\mathbf{i}\in\mathcal{T}_{n}}\left(\widehat{\psi}(\mathbf{i})\right)^{t}.$

In particular, the first limit (which was defined as $p(t)$ in (3.4) exists.

For every $t>0$ we define the potentials

(4.5) $\phi_{t}(\mathbf{i}):=\left(\psi(\mathbf{i})\cdot
3^{-|\mathbf{i}|r}\right)^{t}\quad\text{ and
}\quad\widehat{\phi}_{t}(\mathbf{i}):=\left(\widehat{\psi}(\mathbf{i})\cdot
3^{-|\mathbf{i}|r}\right)^{t}.$

We obtain from Property-1 that

######  Corollary 4.2.

For every $t>0$ the potential
$\mathbf{i}\mapsto\widehat{\phi}_{t}(\mathbf{i})$ is also quasi-
multiplicative.

We cannot say the same about the potential $\phi_{t}$ but by (4.4) the
pressure functions of $\phi_{t}$ and $\widehat{\phi}_{t}$ are the same:

(4.6)
$P(t):=\lim\limits_{n\to\infty}\frac{1}{n}\log\sum_{\mathbf{j}\in\mathcal{T}_{n}}\phi_{t}(\mathbf{j})=\lim\limits_{n\to\infty}\frac{1}{n}\log\sum_{\mathbf{j}\in\mathcal{T}_{n}}\widehat{\phi}_{t}(\mathbf{j})=p(t)-rt\log
3.$

A standard argument shows that the function $t\mapsto p(t)$ is strictly
decreasing and continuous, $p(0)=\dim_{\rm H}\Lambda=0.876036.\dots$ and
$p(1)=0$. Hence, $t\mapsto P(t)$ is also strictly decreasing,
$P(0)=p(0)=0.876036.\dots$ and $P(1)=-r\log 3$. So, we obtain that

######  Fact 4.3.

There is a unique $t_{0}\in(0,1)$ such that

(4.7) $p(t_{0})=rt_{0}\log 3,\quad P(t_{0})=0,\quad P(t)>0\text{ if
}t\in[0,t_{0})\text{ and }P(t)<0\text{ if }t>t_{0}.$

Finally, we introduce the potential
$\widehat{\phi}:\Sigma^{*}_{A}\to[0,\infty)$,

(4.8)
$\widehat{\phi}(\mathbf{i}):=\widehat{\phi}_{t_{0}}(\mathbf{i})=\left(\widehat{\psi}(\mathbf{i})\cdot
3^{-|\mathbf{i}|r}\right)^{t_{0}}.$

Then by definition, the pressure of $\widehat{\phi}$ is equal to $0$:

$P(\widehat{\phi})=\lim\limits_{n\to\infty}\log\sum\limits_{\mathbf{i}\in\mathcal{T}_{n}}\widehat{\phi}(\mathbf{i})=0.$

Moreover, as a corollary of Feng Theorem (Theorem 4.1) and Corollary 4.2 we
obtain:

######  Proposition 4.4.

There is a $C_{4}>1$ and a unique invariant ergodic measure $\mathfrak{m}$ on
$\Sigma_{A}$ such that

(4.9)
$C_{4}^{-1}<\frac{\mathfrak{m}([\mathbf{i}])}{\widehat{\phi}(\mathbf{i})}<C_{4},\quad\text{
for all }\quad\mathbf{i}\in\Sigma_{A}^{*}.$

###  4.2. Finite maximal antichains of $\Sigma_{A}$ and $\Sigma$

Let $X$ be either $\Sigma_{A}$ or $\Sigma$. A finite collection $\Gamma$ of
finite words $\mathbf{i}$ of $X$ is a finite maximal antichain of $X$ if for
every $\boldsymbol{\omega}\in X$ we can find a unique $\mathbf{i}\in\Gamma$
such that $\boldsymbol{\omega}\in[\mathbf{i}]$.

######  Definition 4.5.

Let $\widetilde{q}_{i}:=\left(p_{i}3^{-r}\right)^{t_{0}}$ and
$\varepsilon_{0}:=\min\limits_{i\in\mathcal{A}}\widetilde{q}_{i}$. For an
$0<\varepsilon<\varepsilon_{0}$ we define

(4.10)
$\widehat{\Gamma}(\varepsilon)\\!:\\!=\Big{\\{}\mathbf{i}\in\Sigma^{*}_{A}:\widehat{\phi}(\mathbf{i})<\varepsilon,\
\forall p<|\mathbf{i}|,\
\widehat{\phi}(\mathbf{i}|_{p})\geq\varepsilon\Big{\\}}\text{ and
}\widehat{\Gamma}^{-}(\varepsilon):=\Big{\\{}\mathbf{i}^{-}\\!:\\!\mathbf{i}\in\widehat{\Gamma}(\varepsilon)\Big{\\}}.$

It is clear that both $\widehat{\Gamma}(\varepsilon)$ and
$\widehat{\Gamma}^{-}(\varepsilon)$ are maximal antichains of $\Sigma_{A}$.
From this and from (4.9), we get that

(4.11)
$\sum_{\mathbf{i}\in\widehat{\Gamma}(\varepsilon)}\widehat{\phi}(\mathbf{i})\leq
C_{4}.$

######  Lemma 4.6.

There exists a $\gamma^{\prime}>0$ such that for all
$0<\varepsilon<\varepsilon_{0}$, we have

(4.12)
$\frac{\sum\limits_{\mathbf{i}\in\widehat{\Gamma}(\varepsilon)}\widehat{\phi}(\mathbf{i})}{\sum\limits_{\mathbf{i}\in\widehat{\Gamma}(\varepsilon)}\widehat{\phi}(\mathbf{i}^{-})}>\gamma^{\prime}.$

###### Proof.

Observe that

(4.13)
$\sum\limits_{\mathbf{i}\in\widehat{\Gamma}(\varepsilon)}\widehat{\phi}(\mathbf{i}^{-})\leq
3\sum\limits_{\mathbf{j}\in\widehat{\Gamma}^{-}(\varepsilon)}\widehat{\phi}(\mathbf{j})\leq
3C_{4}\sum\limits_{\mathbf{j}\in\widehat{\Gamma}^{-}(\varepsilon)}\mathfrak{m}([\mathbf{i}])\leq
3C_{4}.$

Namely, the first inequality follows from the fact that for every
$\mathbf{j}\in\widehat{\Gamma}^{-}(\varepsilon)$ there are at most three
$\mathbf{i}\in\widehat{\Gamma}(\varepsilon)$ such that
$\mathbf{j}=\mathbf{i}^{-}$. The second inequality is immediate from (4.9).
The third inequality is a consequence of the fact that
$\widehat{\Gamma}^{-}(\varepsilon)$ is a maximal antichain of $\Sigma_{A}$. On
the other hand,

(4.14)
$\sum\limits_{\mathbf{i}\in\widehat{\Gamma}(\varepsilon)}\widehat{\phi}(\mathbf{i})\geq
C_{4}^{-1}\sum\limits_{\mathbf{i}\in\widehat{\Gamma}(\varepsilon)}\mathfrak{m}(\mathbf{i})=C_{4}^{-1}.$

Putting together (4.13) and (4.14) we obtain that (4.12) holds with the choice
of $\gamma^{\prime}:=\frac{1}{3C_{4}^{2}}$. ∎

######  Corollary 4.7.

Let $0<\varepsilon<\varepsilon_{0}$. Then, we have

(4.15) $\\#\widehat{\Gamma}(\varepsilon)\leq\frac{3C_{4}^{3}}{\varepsilon}.$

###### Proof.

It follows from (4.9) that
$C_{4}>\sum\limits_{\mathbf{i}\in\widehat{\Gamma}(\varepsilon)}\widehat{\phi}(\mathbf{i})$.
Putting together this, (4.12), and the fact that
$\widehat{\phi}(\mathbf{i}^{-})\geq\varepsilon$, we obtain

$C_{4}>\gamma^{\prime}\sum\limits_{\mathbf{i}\in\widehat{\Gamma}(\varepsilon)}\widehat{\phi}(\mathbf{i}^{-})>\gamma^{\prime}\varepsilon\\#\widehat{\Gamma}(\varepsilon).$

This implies that (4.15) holds since $\gamma^{\prime}=\frac{1}{3C_{4}^{2}}$. ∎

Although $\widehat{\Gamma}(\varepsilon)$ is a finite maximal antichain the
same is not true (in general) for
$\bigcup\limits_{\mathbf{i}\in\widehat{\Gamma}(\varepsilon)}\mathcal{I}_{\mathbf{i}}$,
where we defined $\mathcal{I}_{\mathbf{i}}$ in (3.2). So, we need to introduce
one more step:

######  Definition 4.8.

(4.16)
$\Theta(\varepsilon):=\left\\{\mathbf{i}^{-}0:\mathbf{i}\in\widehat{\Gamma}(\varepsilon)\text{
and }i_{|\mathbf{i}|}=1\right\\}\subset\Sigma_{A}^{*}.$

Moreover, let

(4.17)
$\widehat{\Gamma}_{E}(\varepsilon):=\widehat{\Gamma}(\varepsilon)\cup\Theta(\varepsilon)\quad\text{
and
}\quad\widehat{\Gamma}_{\Sigma}(\varepsilon):=\bigcup\limits_{\mathbf{j}\in\widehat{\Gamma}_{E}(\varepsilon)}\mathcal{I}_{\mathbf{j}}.$

The following claim states a simple but important property.

######  Claim 4.9.

For all $\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)$ we have
$\widehat{\phi}(\mathbf{i})<\varepsilon$.

###### Proof.

We need to check this only for $\mathbf{i}\in\Theta(\varepsilon)$ since by
definition $\widehat{\phi}(\mathbf{i})<\varepsilon$ holds for all
$\mathbf{i}\in\widehat{\Gamma}(\varepsilon)$. If
$\mathbf{i}\in\Theta(\varepsilon)$, then $\mathbf{i}=\mathbf{i}^{-}0$ and
$\mathbf{i}^{-}1\in\widehat{\Gamma}(\varepsilon)$. By (3.12),

$\widehat{\psi}(\mathbf{i})=\psi(\mathbf{i})=\psi(\mathbf{i}^{-}0)\leq\max\\{\psi(\mathbf{i}^{-}0),\psi(\mathbf{i}^{-}1)\\}=\widehat{\psi}(\mathbf{i}^{-}1).$

So, we get that

(4.18)
$\forall\mathbf{i}\in\Theta(\varepsilon),\quad\widehat{\phi}(\mathbf{i})=\left(\widehat{\psi}(\mathbf{i})\cdot
3^{-|\mathbf{i}|r}\right)^{t_{0}}\leq\left(\widehat{\psi}(\mathbf{i}^{-}1)\cdot
3^{-|\mathbf{i}|r}\right)^{t_{0}}=\widehat{\phi}(\mathbf{i}^{-}1)<\varepsilon,$

where in the last step we used that
$\mathbf{i}^{-}1\in\widehat{\Gamma}(\varepsilon)$. ∎

We will use the following immediate consequence of Claim 4.9:

(4.19)
$\left(\psi(\mathbf{i})3^{-|\mathbf{i}|r}\right)^{t_{0}}\leq\left(\widehat{\psi}(\mathbf{i})3^{-|\mathbf{i}|r}\right)^{t_{0}}=\widehat{\phi}(\mathbf{i})<\varepsilon\text{
holds for all }\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon).$

######  Lemma 4.10.

For every $0<\varepsilon<\varepsilon_{0}$, we have

(4.20)
$\Sigma=\bigcup\limits_{\boldsymbol{\tau}\in\widehat{\Gamma}_{\Sigma}(\varepsilon)}[\boldsymbol{\tau}].$

###### Proof.

Let $\boldsymbol{\eta}\in\Sigma$. Then, there is a unique
$\mathbf{i}\in\widehat{\Gamma}(\varepsilon)$ such that
$\Phi(\boldsymbol{\eta})\in[\mathbf{i}]$. Let $n:=|\mathbf{i}|$. Then,
$\mathbf{i}=\Phi(\boldsymbol{\eta})|_{n}$. It follows from Property-4 that
$\boldsymbol{\eta}|_{n}\in\mathcal{I}_{\mathbf{i}}\cup\mathcal{I}_{\mathbf{i}^{-}0}$.
∎

For $\mathbf{i},\mathbf{j}\in\Sigma_{A}^{*}$ we say that $\mathbf{i}$ is a
proper prefix of $\mathbf{j}$ if $|\mathbf{i}|<|\mathbf{j}|$ and
$\mathbf{j}|_{|\mathbf{i}|}=\mathbf{i}$. In this case we write
$\mathbf{i}\precneqq\mathbf{j}$. Let

(4.21)
$\widetilde{\Gamma}_{\Sigma}(\varepsilon):=\left\\{\mathbf{j}\in\widehat{\Gamma}_{\Sigma}(\varepsilon):\nexists\
\mathbf{i}\in\widehat{\Gamma}_{\Sigma}(\varepsilon)\text{ such that
}\mathbf{i}\precneqq\mathbf{j}\right\\}.$

The elements of $\widetilde{\Gamma}_{\Sigma}(\varepsilon)$ are incomparable
and by (4.20) we have
$\Sigma=\bigcup\limits_{\boldsymbol{\tau}\in\widetilde{\Gamma}_{\Sigma}(\varepsilon)}[\boldsymbol{\tau}]$.
In this way we have proved that

######  Fact 4.11.

The collection of cylinders $\widetilde{\Gamma}_{\Sigma}(\varepsilon)$ is a
finite maximal antichain for $\Sigma$. That is, for every
$\boldsymbol{\eta}\in\Sigma$ there is a unique
$\mathbf{i}\in\widetilde{\Gamma}_{\Sigma}(\varepsilon)$ such that
$\boldsymbol{\eta}\in[\mathbf{i}]$.

######  Lemma 4.12.

For an $0<\varepsilon<\varepsilon_{0}$, we have

(4.22)
$\sum_{\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)}\widehat{\phi}(\mathbf{i})\leq
2C_{4}.$

###### Proof.

Observe that

(4.23)
$\sum\limits_{\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)}\widehat{\phi}(\mathbf{i})\leq\underbrace{\sum\limits_{\mathbf{i}\in\widehat{\Gamma}(\varepsilon)}\widehat{\phi}(\mathbf{i})}_{S_{1}}+\underbrace{\sum\limits_{\mathbf{i}\in\widehat{\Gamma}(\varepsilon):i|_{|\mathbf{i}|=1}}\widehat{\phi}(\mathbf{i}^{-}0)}_{S_{2}}.$

Putting together formulas (4.5) and (3.12) we get that $S_{2}\leq S_{1}$. This
and (4.11) together imply that (4.22) holds. ∎

######  Definition 4.13.

Using the notation introduced in Definition 4.5 we choose $m,n\in\mathbb{N}$
satisfying

(4.24) $\frac{m}{n}<\varepsilon_{0}^{2}\quad\text{ and
}\quad\varepsilon:=6\frac{m}{n}C_{4}^{3}.$

Fix such an $m,n$ and $\varepsilon$ for the rest of this section.

Then, we have from Lemma 4.12 and Corollary 4.7 that

(4.25) $\\#\widehat{\Gamma}_{E}(\varepsilon)\leq\frac{n}{m}.$

Namely, Corollary 4.7 we have $\\#\widehat{\Gamma}_{E}(\varepsilon)\leq
2\\#\widehat{\Gamma}(\varepsilon)\leq\frac{6C_{4}^{3}}{\varepsilon}=\frac{n}{m}$
by the choice of $\varepsilon$.

##  5\. The upper estimate for the quantization domension of the measure $\nu$

From now on we follow [GL1] and [R3]. Recall that $\nu=\Pi_{*}\mu$, where
$\mu=\mathbf{p}^{\mathbb{N}}$ is the infinite product measure on
$\mathcal{A}^{\mathbb{N}}$. Also recall that

(5.1) $V_{n,r}(\nu):=\inf\left\\{\int
d(x,\alpha)^{r}d\nu(x):\alpha\subset\mathbb{R}^{d},\operatorname{card}(\alpha)\leq
n\right\\},$

where $d(x,\alpha)$ denotes the distance from the point $x$ to the set
$\alpha$. Recall that we introduced $C_{n,r}(\mu)$ in (1.1). It was proved in
[GL1] that $C_{n,r}(\mu)\neq\emptyset$.

In the argument below we will use frequently the potential
$\widehat{\phi}(\mathbf{i})$ which was defined in (4.8), the set
$\mathcal{I}_{\mathbf{i}}$ which was defined in (3.2), the constant $C_{4}$
defined in (4.9).

###  5.1. An upper estimate for $V_{n,r}(\nu)$

First we need to prove a Fact which is a slight modification of [GL1, Lemma
4.14].

######  Fact 5.1.

Let $\mu,\mu_{i}$ for $i=1,\dots,L$ be Borel probability measures on
$\mathbb{R}$ such that $\int\|x\|^{r}d\mu_{i}(x)<\infty$ for all
$i=1,\dots,L$. We are given $q_{1},\dots,q_{L}$ positive numbers such that

(5.2) $\mu\leq\sum_{m=1}^{L}q_{m}\mu_{m}.$

Moreover, we are given natural numbers $\left\\{n_{i}\right\\}_{i=1}^{L}$ such
that $n_{i}\geq 1$ and $\sum_{i=1}^{L}n_{i}\leq n$. Then,

(5.3) $V_{n,r}(\mu)\leq\sum_{i=1}^{L}q_{i}V_{n_{i},r}(\mu_{i}).$

###### Proof.

Let $\alpha_{i}\in C_{n_{i},r}(\mu_{i})$ for $i=1,\dots,L$ and
$\alpha:=\cup_{i=1}^{L}\alpha_{i}$.

(5.4) $\displaystyle V_{n,r}(\mu)$ $\displaystyle\leq$
$\displaystyle\int\min_{a\in\alpha}\|x-a\|^{r}d\mu(x)\leq\sum_{i=1}^{L}q_{i}\int\min_{a\in\alpha}\|x-a\|^{r}d\mu_{i}(x)$
(5.5) $\displaystyle\leq$
$\displaystyle\sum_{i=1}^{L}q_{i}\int\min_{a\in\alpha_{i}}\|x-a\|^{r}d\mu_{i}(x)=\sum_{i=1}^{L}q_{i}V_{n_{i},r}(\mu_{i}).$

∎

To apply this Fact we prove that the condition (5.2) holds in our case:

######  Fact 5.2.

(5.6)
$\nu\leq\sum_{\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)}\psi(\mathbf{i})\nu_{\mathbf{i}},$

where we write

(5.7) $\nu_{\boldsymbol{\eta}}:=(S_{\boldsymbol{\eta}})_{*}\nu,\text{ for an
}\boldsymbol{\eta}\in\mathcal{A}^{*}.$

Note that if $\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{i}}$ then,
$S_{\boldsymbol{\eta}}=S_{\mathbf{i}}$. Consequently,

(5.8) $\nu_{\boldsymbol{\eta}}=\nu_{\mathbf{i}},\text{ for all
}\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{i}}.$

###### Proof.

Using that
$\widetilde{\Gamma}_{\Sigma}(\varepsilon)\subset\widehat{\Gamma}_{E}(\varepsilon)$
is a maximal finite antichain

$\displaystyle\nu$ $\displaystyle=$
$\displaystyle\sum_{\boldsymbol{\eta}\in\widetilde{\Gamma}_{\Sigma}(\varepsilon)}p_{\boldsymbol{\eta}}\nu_{\boldsymbol{\eta}}\leq\sum_{\boldsymbol{\eta}\in\widehat{\Gamma}_{\Sigma}(\varepsilon)}p_{\boldsymbol{\eta}}\nu_{\boldsymbol{\eta}}=\sum_{\mathbf{j}\in\widehat{\Gamma}_{E}(\varepsilon)}\sum_{\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{j}}}p_{\boldsymbol{\eta}}\nu_{\boldsymbol{\eta}}$
$\displaystyle=$
$\displaystyle\sum_{\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)}\left(\sum_{\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{i}}}p_{\boldsymbol{\eta}}\right)\nu_{\mathbf{i}}=\sum_{\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)}\psi(\mathbf{i})\nu_{\mathbf{i}}.$

∎

######  Fact 5.3.

Let us fix an $n_{\mathbf{i}}\in\mathbb{N}$ for every
$\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)$ such that $n_{\mathbf{i}}\geq
1$ and
$\sum\limits_{\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)}n_{\mathbf{i}}\leq
n$. Then,

(5.9)
$V_{n,r}(\nu)\leq\sum_{\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)}\psi(\mathbf{i})V_{n_{\mathbf{i}},r}(\nu_{\mathbf{i}}).$

###### Proof.

This follows from the combination of Fact 5.1 and Fact 5.2. ∎

######  Lemma 5.4.

(5.10)
$V_{n,r}(\nu)\leq\inf\left\\{\sum_{\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)}3^{-|\mathbf{i}|r}\psi(\mathbf{i})\cdot
V_{n_{\mathbf{i}},r}(\nu):1\leq n_{\mathbf{i}},\
\sum_{\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)}n_{\mathbf{i}}\leq
n\right\\}.$

###### Proof.

Let $\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)$, $n_{\mathbf{i}}\geq 1$,
and $\alpha_{\mathbf{i}}\in C_{n_{\mathbf{i}},r}(\nu)$. Below we prove that

(5.11) $V_{n_{\mathbf{i}},r}(\nu_{\mathbf{i}})\leq
3^{-|\mathbf{i}|r}V_{n_{\mathbf{i}},r}(\nu).$

We obtain the assertion of the lemma from the combination of Fact 5.3 and
(5.11). Now we prove (5.11).

(5.12) $\displaystyle V_{n_{\mathbf{i}},r}(\nu_{\mathbf{i}})$
$\displaystyle\leq$ $\displaystyle\int
d(x,S_{\mathbf{i}}(\alpha_{\mathbf{i}}))^{r}d\nu_{\mathbf{i}}(x)=\int
d(x,S_{\mathbf{i}}(\alpha_{\mathbf{i}}))^{r}d(\nu\circ
S_{\mathbf{i}}^{-1})(x)$ $\displaystyle=$ $\displaystyle\int
d(S_{\mathbf{i}}(x),S_{\mathbf{i}}(\alpha_{\mathbf{i}}))^{r}d\nu(x)=3^{-|\mathbf{i}|r}\int
d(x,\alpha_{\mathbf{i}})^{r}d\nu(x)$ $\displaystyle=$ $\displaystyle
3^{-|\mathbf{i}|r}V_{n_{\mathbf{i}},r}(\nu).$

∎

######  Proposition 5.5.

There exists a constant $C_{9}>0$ such that

(5.13) $\limsup\limits_{n\to\infty}n\cdot e_{n,r}^{\chi_{r}}(\nu)\leq
C_{9}m\cdot e_{m,r}^{\chi_{r}}(\nu)<\infty.$

###### Proof.

Recall the definition of $n,m,\varepsilon$ from Definition 4.13. For every
$\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)$ we define
$n_{\mathbf{i}}:=m$. Then, by (4.25) we have
$\sum_{\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)}n_{\mathbf{i}}=\\#\widehat{\Gamma}_{E}(\varepsilon)m\leq
n$. We apply Lemma 5.4 in the first step below:

$\displaystyle V_{n,r}(\nu)$ $\displaystyle\leq$
$\displaystyle\sum_{\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)}3^{-|\mathbf{i}|r}\psi(\mathbf{i})V_{m,r}(\nu)$
$\displaystyle=$
$\displaystyle\sum_{\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)}\underbrace{\left(3^{-|\mathbf{i}|r}\psi(\mathbf{i})\right)^{t_{0}}}_{\leq\widehat{\phi}(\mathbf{i})}\underbrace{\left(3^{-|\mathbf{i}|r}\psi(\mathbf{i})\right)^{1-t_{0}}}_{\leq\varepsilon^{\frac{1-t_{0}}{t_{0}}}}V_{m,r}(\nu)$
$\displaystyle\leq$
$\displaystyle\underbrace{\varepsilon^{\frac{1-t_{0}}{t_{0}}}}_{\varepsilon^{\frac{r}{\chi_{r}}}}V_{m,r}(\nu)\underbrace{\sum_{\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon)}\widehat{\phi}(\mathbf{i})}_{\leq
2C_{4}}$

The reasoning for the $4$ underbraces above are as follows:

  1. (1)

This is immediate from the definition $\widehat{\phi}$ and from the fact that
$\psi(\mathbf{i})\leq\widehat{\psi}(\mathbf{i})$ for all $\mathbf{i}$.

  2. (2)

If $\mathbf{i}\in\widehat{\Gamma}_{E}(\varepsilon),$ then
$\left(3^{-|\mathbf{i}r|}\psi(\mathbf{i})\right)^{t_{0}}<\varepsilon$ (see
(4.19)).

  3. (3)

This follows from definition:
$t_{0}=\frac{\chi_{r}}{r+\chi_{r}}\Longrightarrow\frac{1-t_{0}}{t_{0}}=\frac{r}{\chi_{r}}$.

  4. (4)

This follows from Lemma 4.12.

So, we have proved that

(5.14)
$V_{n,r}(\nu)\leq\varepsilon^{\frac{r}{\chi_{r}}}V_{m,r}(\nu)=\left(\frac{m}{n}\right)^{\frac{r}{\chi_{r}}}C_{9}^{\frac{r}{\chi_{r}}}V_{m,r}(\nu),$

where $C_{9}:=6C_{4}^{3}$. Hence,

$nV_{n,r}^{\frac{\chi_{r}}{r}}(\nu)\leq
C_{9}mV_{m,r}^{\frac{\chi_{r}}{r}}(\nu).$

Letting $n$ to infinity we obtain that (5.13) holds. ∎

Putting together Proposition (5.5) and (1.4) we obtain that

(5.15) $\overline{D}_{r}(\nu)\leq\chi_{r}.$

##  6\. The lower estimate

First we prove a Fact similar to Fact 5.2.

######  Fact 6.1.

For every $n$ we have

(6.1)
$\nu=\sum_{\mathbf{i}\in\mathcal{T}_{n}}\psi(\mathbf{i})\nu_{\mathbf{i}}.$

###### Proof.

Fix an $n$. Then, $\mathcal{A}^{n}$ is a finite maximal antichain for
$\Sigma$. Hence,

(6.2)
$\nu=\sum_{\boldsymbol{\eta}\in\mathcal{A}^{n}}p_{\boldsymbol{\eta}}\cdot\nu_{\boldsymbol{\eta}},$

where $\nu_{\boldsymbol{\eta}}:=\nu\circ S_{\boldsymbol{\eta}}^{-1}$.
Moreover, let $f:\Sigma\to\mathbb{R}$ be a continuous function. Using that
$S_{\boldsymbol{\eta}}=S_{\mathbf{i}}$ if
$\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{i}}$ we get

(6.3) $\displaystyle\int f(x)d\nu(x)$ $\displaystyle=$
$\displaystyle\sum_{\boldsymbol{\eta}\in\mathcal{A}^{n}}p_{\boldsymbol{\eta}}\int
f(x)d\nu_{\boldsymbol{\eta}}(x)=\sum_{\mathbf{i}\in\mathcal{T}_{n}}\sum_{\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{i}}}p_{\boldsymbol{\eta}}\int
f(x)d(\nu\circ S_{\boldsymbol{\eta}}^{-1})(x).$ $\displaystyle=$
$\displaystyle\sum_{\mathbf{i}\in\mathcal{T}_{n}}\underbrace{\sum_{\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{i}}}p_{\boldsymbol{\eta}}}_{\psi(\mathbf{i})}\int
f(x)d(\nu\circ
S_{\mathbf{i}}^{-1})(x)=\sum_{\mathbf{i}\in\mathcal{T}_{n}}\psi(\mathbf{i})\int
f(x)d\nu_{\mathbf{i}}(x).$

∎

Let $U:=\text{int}(I)=\left(0,\frac{9}{2}\right)$. Recall that
$I=\overline{U}\supset\Lambda$. For an $\boldsymbol{\eta}\in\Sigma^{*}$ we
write $U_{\boldsymbol{\eta}}:=S_{\boldsymbol{\eta}}(U)\subset U$. Following
Graf and Luschgy [GL3] we introduce

(6.4) $u_{n,r}(\nu):=\inf\left\\{\int d(x,\alpha\cup
U^{c})^{r}d\nu(x):\alpha\subset\mathbb{R}^{d},\operatorname{card}(\alpha)\leq
n\right\\}.$

As an analogue of [GL3, Lemma 4.4] we need the following assertion:

######  Lemma 6.2.

There exists a set $\alpha_{n}\subset\mathbb{R}^{n}$ such that
$\\#\alpha_{n}\leq n$ and

(6.5) $u_{n,r}(\nu)=\int d\left(x,\alpha_{n}\cup U^{c}\right)^{r}d\nu(x).$

###### Proof.

Define $f:I^{n}\to\mathbb{R}$ by $f(x_{1},\dots,x_{n}):=\int
d(x,{x_{1},\dots,x_{n},0,9/2})^{r}d\nu(x).$ It is easy to see that this
function is continuous. Hence, it attains its infimum on $I^{n}$. Any place
where the infimum is attained can be chosen as $\alpha_{n}$. ∎

######  Lemma 6.3.

Fix an $m$. Then, there exists an $n_{0}$ such that for all $n\geq n_{0}$
there exists a sequence
$\left\\{n_{\mathbf{i}}\right\\}_{\mathbf{i}\in\mathcal{T}_{m}}$ such that

(6.6) $1\leq
n_{\mathbf{i}}<n,\quad\sum_{\mathbf{i}\in\mathcal{T}_{m}}n_{\mathbf{i}}\leq
2n,\quad
u_{n,r}(\nu)\geq\sum_{\mathbf{i}\in\mathcal{T}_{m}}\psi(\mathbf{i})3^{-|\mathbf{i}|r}u_{n_{\mathbf{i}},r}(\nu).$

###### Proof.

Fix an $m$ and to shorten the notation we write $\Gamma:=\mathcal{T}_{m}$. We
consider any $\boldsymbol{\tau}\in\Sigma^{*}$ such that
$S_{\boldsymbol{\tau}}(I)\subset U$. For example $\boldsymbol{\tau}=(1)$ will
do since
$I_{1}=S_{1}(I)=\left[1,\frac{5}{2}\right]\subset\left(0,\frac{9}{2}\right)$.
Let $\varepsilon:=\text{dist}(S_{\tau}(I),U^{c})$. If we choose $\tau=1$, then
$\varepsilon=1$. Let $\delta:=3^{-m}$. Then,

(6.7) $d(x,U^{c})\geq
d(x,S_{\mathbf{i}}(U^{c}))\geq\delta\varepsilon,\quad\text{ if }x\in
I_{\mathbf{i}\boldsymbol{\tau}}\text{ and }\mathbf{i}\in\mathcal{T}_{m}.$

For each $n$ let $\alpha_{n}$ be the optimal set for $u_{n,r}(\nu)$. This
exists according to Lemma 6.2. We define

(6.8) $\delta_{n}:=\max\left\\{d(x,\alpha_{n}\cup
U^{c}):x\in\Lambda\right\\}.$

Since $\delta_{n}\to 0$ we can choose an $n_{0}$ such that
$\delta_{n}<\delta\cdot\varepsilon$ if $n\geq n_{0}$. Let
$x\in\Lambda_{\mathbf{i}\boldsymbol{\tau}}$. Then, $x\in
S_{\mathbf{i}}(U)\subset U$, for all $\mathbf{i}\in\Gamma$. By compactness,
there exists an $a\in\alpha_{n}\cup U^{c}$ such that

$d(x,\alpha_{n}\cup U^{c})=d(x,a)\leq\delta_{n}<\delta\varepsilon,$

where the one but last inequality holds since $x\in\Lambda$. So, by (6.7) we
have $a\not\in S_{\mathbf{i}}(U^{c})$. That is, $a\in S_{\mathbf{i}}(U)\subset
U$ but $a\in\alpha_{n}\cup U^{c}$, hence $a\in\alpha_{n}$. Let

$n_{\mathbf{i}}:=\\#\alpha_{n}\cap
S_{\mathbf{i}}(U),\quad\alpha_{n_{\mathbf{i}}}:=\alpha_{n}\cap
S_{\mathbf{i}}(U).$

Then, we have just proved that $a\in\alpha_{\mathbf{n}_{\mathbf{i}}}$. That
is,

(6.9) $n_{\mathbf{i}}\geq 1,\quad\forall\mathbf{i}\in\Gamma.$

It follows from the optimal property of $\alpha_{n}$ that
$\alpha_{n}\not\subset\alpha_{n_{\mathbf{i}}}$. Hence, $n_{\mathbf{i}}<n$ for
all $\mathbf{i}\in\Gamma$.

It follows from Property-3 that

(6.10) $\\#\left\\{\mathbf{i}\in\Gamma:x\in S_{n_{\mathbf{i}}}(I)\right\\}\leq
2,\quad\forall x\in I.$

Thus,

(6.11) $\sum_{\mathbf{i}\in\Gamma}n_{\mathbf{i}}\leq 2n.$

Finally we prove that

(6.12)
$u_{n,r}(\nu)\geq\sum_{\mathbf{i}\in\Gamma}\psi(\mathbf{i})3^{-mr}u_{n_{\mathbf{i}},r}(\nu).$

To verify this we will use the following trivial observation

(6.13) $\alpha_{n}\cup
S_{\mathbf{i}}(U^{c})=\alpha_{n}\cup(S_{\mathbf{i}}(U))^{c}=\left(\alpha_{n}\cap
S_{\mathbf{i}}(U)\right)\cup\left(S_{\mathbf{i}}(U)\right)^{c}.$

Using this we get

(6.14) $\int\\!d\left(S_{\mathbf{i}}(x),\alpha_{n}\cup
S_{\mathbf{i}}(U^{c})\right)^{r}d\nu(x)\\!\\!=\\!\\!\int\\!d\left(S_{\mathbf{i}}(x),(\alpha_{n}\cap
S_{\mathbf{i}}(U))\cup S_{\mathbf{i}}(U)^{c}\right)^{r}d\nu(x)\\\ =3^{-mr}\int
d\left(x,S_{\mathbf{i}}^{-1}\bigg{[}\underbrace{\big{(}\alpha_{n}\cap
S_{\mathbf{i}}(U)\big{)}}_{\alpha_{n_{\mathbf{i}}}}\cup
S_{\mathbf{i}}(U^{c})\bigg{]}\right)^{r}d\nu(x)\\\ =3^{-mr}\int
d\left(x,S_{\mathbf{i}}^{-1}(\alpha_{n_{\mathbf{i}}})\cup
U^{c}\right)^{r}d\nu(x)\geq 3^{-mr}u_{n_{\mathbf{i}},r}(\nu).$

Now we put all of these together:

(6.15) $\displaystyle u_{n,r}(\nu)$ $\displaystyle=$ $\displaystyle\int
d(x,\alpha_{n}\cup
U^{c})^{r}d\nu(x)=\sum_{\mathbf{i}\in\Gamma}\psi(\mathbf{i})\int
d\left(S_{\mathbf{i}}(x),\alpha_{n}\cup U^{c}\right)^{r}d\nu(x)$
$\displaystyle\geq$
$\displaystyle\sum_{\mathbf{i}\in\Gamma}\psi(\mathbf{i})\int
d\left(S_{\mathbf{i}}(x),\alpha_{n}\cup
S_{\mathbf{i}}(U^{c})\right)^{r}d\nu(x)\geq\sum_{\mathbf{i}\in\Gamma}\psi(\mathbf{i})3^{-mr}u_{n_{\mathbf{i}},r}(\nu),$

where at the second step we used (6.3), at the third step we used that
$S_{\mathbf{i}}(U^{c})\supset U^{c}$. At the fourth step we used the previous
inequality. ∎

######  Proposition 6.4.

Let $0<\ell<\chi_{r}$. Then,

(6.16)
$\liminf\limits_{n\to\infty}n\cdot\left(u_{n,r}(\nu)\right)^{\frac{\ell}{r}}>0.$

Our proof follows the line [GL3, Lemma 4.4] and [R3, Proposition 3.12].

###### Proof.

Fix an $\ell\in\left(0,\chi_{r}\right)$. Then, $\frac{\ell}{r+\ell}<t_{0}$.
Recall the definition of the pressure function $P(t)$ from (4.6). It follows
from (4.7) that

(6.17)
$P\left(\frac{\ell}{r+\ell}\right)=\lim\limits_{m\to\infty}\frac{1}{m}\log\sum_{\mathbf{i}\in\mathcal{T}_{m}}\left(\psi(\mathbf{i})3^{-mr}\right)^{\frac{\ell}{r+\ell}}>0.$

Fix an $m$ such that

(6.18)
$\sum_{\mathbf{i}\in\mathcal{T}_{m}}\left(\psi(\mathbf{i})3^{-mr}\right)^{\frac{\ell}{r+\ell}}>2.$

For this $m$ we choose $n_{0}$ as in Lemma 6.3. Let

(6.19) $C:=\min\left\\{q^{\frac{r}{\ell}}u_{q,r}(\nu):q\leq n_{0}\right\\}.$

Clearly, $u_{n,r}(\nu)>0$. Hence, $C>0$. Choose an $n$ such that

(6.20) $n\geq n_{0}\quad\&\ \quad k<n\Longrightarrow
k^{\frac{r}{\ell}}u_{k,r}(\nu)\geq C.$

Below we prove that

(6.21) $n^{\frac{r}{\ell}}u_{n,r}(\nu)\geq C$

For this $m$ and $n$ we choose $n_{\mathbf{i}}$ for every
$\mathbf{i}\in\mathcal{T}_{m}$, as in Lemma 6.3, such that the inequalities of
(6.6) hold.

(6.22) $\displaystyle n^{\frac{r}{\ell}}u_{n,r}(\nu)$ $\displaystyle\geq$
$\displaystyle
n^{\frac{r}{\ell}}\sum_{\mathbf{i}\in\Gamma}\psi(\mathbf{i})3^{-mr}u_{n_{\mathbf{i}},r}(\nu)$
$\displaystyle=$ $\displaystyle
n^{\frac{r}{\ell}}\sum_{\mathbf{i}\in\Gamma}\psi(\mathbf{i})3^{-mr}u_{n_{\mathbf{i}},r}(\nu)(n_{\mathbf{i}})^{-\frac{r}{\ell}}\underbrace{(n_{\mathbf{i}})^{\frac{r}{\ell}}u_{n_{\mathbf{i}},r}(\nu)}_{\geq
C}$ (6.23) $\displaystyle\geq$ $\displaystyle
C\underbrace{\sum_{\mathbf{i}\in\Gamma}\psi(\mathbf{i})3^{-mr}\left(\frac{n_{\mathbf{i}}}{n}\right)^{-\frac{r}{\ell}}}_{a_{n}},$

where at first step we used (6.15), and at the last step we used that
$n_{\mathbf{i}}<n$ so by (6.20) we have $u_{n_{\mathbf{i}},r}(\nu)\geq C$. Set
$a_{n}:=\sum\limits_{\mathbf{i}\in\Gamma}\psi(\mathbf{i})3^{-mr}\left(\frac{n_{\mathbf{i}}}{n}\right)^{-\frac{r}{\ell}}$.
To verify (6.21), it is enough to prove that

(6.24) $a_{n}\geq 1.$

To see this we use the so-called reversed Hölder inequality: Let
$\left\\{x_{k}\right\\}_{k=1}^{M}$ and $\left\\{y_{k}\right\\}_{k=1}^{M}$ be
finite sequences of positive numbers and let $p\in(1,\infty)$. Then

(6.25)
$\sum_{k=1}^{M}x_{k}y_{k}\geq\left(\sum_{k=1}^{M}x_{k}^{\frac{1}{p}}\right)^{p}\left(\sum_{k=1}^{M}y_{k}^{-\frac{1}{p-1}}\right)^{-(p-1)}.$

In our case the summation is taken for $\mathbf{i}\in\Gamma$,
$x_{\mathbf{i}}=\psi(\mathbf{i})3^{-mr}$ and
$y_{\mathbf{i}}=\left(\frac{n_{\mathbf{i}}}{n}\right)^{-\frac{r}{\ell}}$.
Finally, $p:=1+\frac{r}{\ell}$. That is, $\frac{1}{p}=\frac{\ell}{r+\ell}$,
$\frac{1}{p-1}=\frac{\ell}{r}$ and $-(p-1)=-\frac{r}{\ell}$. Hence, from the
reversed Hölder inequality we get

(6.26)
$a_{n}\geq\left(\sum_{\mathbf{i}\in\Gamma}\left(\psi(\mathbf{i})3^{-mr}\right)^{\frac{\ell}{r+\ell}}\right)^{1+\frac{r}{\ell}}\left(\sum_{\mathbf{i}\in\Gamma}\left(\frac{n_{\mathbf{i}}}{n}\right)^{\left(-\frac{r}{\ell}\right)\left(-\frac{\ell}{r}\right)}\right)^{-\frac{r}{\ell}}.$

Putting together this, (6.11) and (6.18) we get that

$a_{n}\geq 2^{1+\frac{r}{\ell}}\cdot 2^{-\frac{r}{\ell}}=2.$

In this way we have proved that (6.24) holds which implies that
$n^{\frac{r}{\ell}}u_{n,r}(\nu)>C$ for all $n$. Thus, (6.16) holds. ∎

Clearly, $u_{n,r}(\nu)\leq V_{n,r}(\nu)$. Hence, we get from Proposition 6.4
that

(6.27) $\liminf\limits_{n\to\infty}n\cdot e_{n,r}^{\ell}>0.$

Combining this with (1.3) we obtain that

(6.28) $\chi_{r}\leq\underline{D}_{r}(\nu).$

###### Proof of Theorem 3.2.

Putting together (5.15) and (6.28) we obtain that

(6.29) $D_{r}(\nu)=\chi_{r}.$

∎

##  7\. Checking Properties 1-4

###  7.1. Good blocks and bad blocks

Let $n\in\mathbb{N}\cup\\{\infty\\}$. For arbitrary $1\leq k<\ell\leq n$ and
we use the shorthand notation

$[n]:=\left\\{1,\dots,n\right\\},\quad[k,\ell]:=\left\\{k,\dots,\ell\right\\}.$

######  Definition 7.1.

Bad blocks, Good blocks
Let $n\in\mathbb{N}\cup\\{\infty\\}$ and
$\boldsymbol{\eta}=(\eta_{1},\dots,\eta_{n})$. For a $k<\ell\leq n$ we say
that $[k,\ell]$ is a bad block of $[n]$ with respect to an
$\boldsymbol{\eta}=(\eta_{1},\dots,\eta_{n})\in\mathcal{A}^{n}$ if
$(\eta_{k},\dots,\eta_{\ell})=(0,3,3,\dots,3)$. We say that $[k,\ell]$ is a
maximal bad block if either $\ell=\infty$ or $\eta_{\ell+1}\neq 3$. Similarly,
$[k,\ell]$ is a good block of $[n]$ with respect to $\boldsymbol{\eta}$ if
$(\eta_{k},\dots,\eta_{\ell})=(1,1,1,\dots,1,0)$. We say that $[k,\ell]$ is a
maximal good block if either $k=1$ or $\eta_{k-1}\neq 1$. We write
$B(\boldsymbol{\eta})$, ($G(\boldsymbol{\eta})$) for the collection of maximal
bad (good) blocks with respect to $\boldsymbol{\eta}$, respectively. That is,

(7.1) $B(\boldsymbol{\eta}):=\left\\{[k,\ell]:1\leq k<\ell<\infty,\
\eta_{k}=0,\ \eta_{k+1}=\cdots=\eta_{\ell}=3,\eta_{\ell+1}\neq 3\right\\}\\\
\bigcup\left\\{[k,\infty]:1\leq k,\ \eta_{k}=0,\
\eta_{k+1}=\eta_{k+2}=\eta_{k+3}=\cdots=3\right\\}.$

Similarly,

(7.2) $G(\boldsymbol{\eta}):=\left\\{[k,\ell]:1\leq k<\ell<\infty,\
\eta_{k}=\cdots=\eta_{\ell-1}=1,\ \eta_{\ell}=0,\eta_{k-1}\neq 1\right\\}\\\
\bigcup\left\\{[k,\infty]:1\leq k,\
\eta_{k}=\eta_{k+1}=\eta_{k+2}=\eta_{k+3}=\cdots=1\right\\}.$

######  Definition 7.2.

A "bad" and a "good" partition of $[n]$
Given an $n\in\mathbb{N}\cup\\{\infty\\}$ and a
$\boldsymbol{\eta}\in\mathcal{A}^{n}$, the following definitions are meant to
be with respect to $\boldsymbol{\eta}$.

  1. (1)

$A_{\text{Good}}(\boldsymbol{\eta}):=[n]\setminus\bigcup\limits_{[k,\ell]\in
G(\boldsymbol{\eta})}[k,\ell],\text{ and
}A_{\text{Bad}}(\boldsymbol{\eta}):=[n]\setminus\bigcup\limits_{[k,\ell]\in
B(\boldsymbol{\eta})}[k,\ell],$

  2. (2)

$B_{\text{Good}}(\boldsymbol{\eta}):=\bigcup\limits_{[k,\ell]\in
G(\boldsymbol{\eta})}\left\\{\ell\right\\},\text{ and
}B_{\text{Bad}}(\boldsymbol{\eta}):=\bigcup\limits_{[k,\ell]\in
B(\boldsymbol{\eta})}\left\\{\ell\right\\},$

  3. (3)

$C_{\text{Good}}(\boldsymbol{\eta}):=\bigcup\limits_{[k,\ell]\in
G(\boldsymbol{\eta})}[k,\ell-1],\text{ and
}C_{\text{Bad}}(\boldsymbol{\eta}):=\bigcup\limits_{[k,\ell]\in
B(\boldsymbol{\eta})}[k,\ell-1]$,

  4. (4)

$D_{\text{Good}}(\boldsymbol{\eta}):=A_{\text{Good}}(\boldsymbol{\eta})\cup
B_{\text{Good}}(\boldsymbol{\eta})$,  and
$D_{\text{Bad}}(\boldsymbol{\eta}):=A_{\text{Bad}}(\boldsymbol{\eta})\cup
B_{\text{Bad}}(\boldsymbol{\eta})$.

We use most frequently $C_{\text{Bad}}(\boldsymbol{\eta})$ and
$D_{\text{Bad}}(\boldsymbol{\eta})$, so we also explain their meaning in
words:

$C_{\text{Good}}(\boldsymbol{\eta})$ is the collection of indices which are in
a good box of $[n]$ with respect to $\boldsymbol{\eta}$ but not as a right
endpoint of a maximal good box (with respect to $\boldsymbol{\eta}$).

$D_{\text{Good}}(\boldsymbol{\eta})$ is the collection of indices which are
either the right endpoint of a maximal good box of $[n]$ or not contained in
any good boxes of $[n]$ (with respect to $\boldsymbol{\eta}$).

In this way we partition $[n]$ into $C_{\text{Good}}(\boldsymbol{\eta})\cup
D_{\text{Good}}(\boldsymbol{\eta})$. The indices in
$D_{\text{Good}}(\boldsymbol{\eta})$ cause less problem than the ones in
$C_{\text{Good}}(\boldsymbol{\eta})$. This is indicated by the following fact.
Before stating it recall that $\mathcal{I}_{\mathbf{i}}$ was defined in (3.2).
Using this definition we get that for an $\mathbf{i}\in\Sigma_{A}^{*}$

(7.3)
$\mathcal{I}_{\mathbf{i}}=\left\\{\boldsymbol{\eta}\in\mathcal{A}^{|\mathbf{i}|}:\Pi(\boldsymbol{\eta})=\Pi(\mathbf{i})\right\\}.$

######  Fact 7.3.

Let $\mathbf{i}\in\mathcal{T}_{n}$ and let $z\in D_{\text{Good}}(\mathbf{i})$.
Then, for every $\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{i}}$ we have

(7.4)
$\boldsymbol{\eta}|_{z}\in\mathcal{I}_{\mathbf{i}|_{z}},\quad\sigma^{z}\boldsymbol{\eta}\in\mathcal{I}_{\sigma^{z}\mathbf{i}}$

where $\boldsymbol{\tau}|_{z}:=(\tau_{1},\dots,\tau_{z})$ if
$\boldsymbol{\tau}=(\tau_{1},\dots,\tau_{n})\in\mathcal{A}_{n}$ and $z\leq n$.

Note that for all
$\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}\in\Sigma_{A}^{*}$

(7.5) $\Pi(\mathbf{a})=\Pi(\mathbf{b})\ \&\
\Pi(\mathbf{c})=\Pi(\mathbf{d})\Longrightarrow\Pi(\mathbf{a},0,\underbrace{3,\dots,3}_{\ell-1},\mathbf{c})=\Pi(\mathbf{b},\underbrace{1,\dots,1}_{\ell-1},0,\mathbf{d}).$

This is immediate from the definition (2.3) of the natural projection $\Pi$
since

(7.6) $\Pi(0,\underbrace{3,\dots,3}_{\ell-1})=0\cdot
3^{-1}+\sum_{k=2}^{\ell}3\cdot 3^{-(k-1)}=\sum_{k=1}^{\ell-1}1\cdot
3^{-(k-1)}+0\cdot 3^{-(\ell-1)}=\Pi(\underbrace{1,\dots,1}_{\ell-1},0).$

Now we summarize some important properties of the mapping $\Phi$ introduced in
Definition 3.3:

######  Fact 7.4.

  1. (a)

For every $\boldsymbol{\eta}\in\mathcal{A}^{n}$ we have
$\Phi(\boldsymbol{\eta})\in\mathcal{T}_{n}$,

  2. (b)

$\Pi(\boldsymbol{\eta})=\Pi(\Phi(\boldsymbol{\eta}))$ if
$\boldsymbol{\eta}\in\mathcal{A}^{n}$,

  3. (c)

$\mathcal{I}_{\mathbf{i}}=\left\\{\boldsymbol{\eta}\in\mathcal{A}^{n}:\Phi(\boldsymbol{\eta})=\mathbf{i}\right\\}$,
for any $\mathbf{i}\in\mathcal{T}_{n}$.

###### Proof.

Part (a) is obvious from Definition 3.3 since we kill all bad blocks of
$\boldsymbol{\eta}$.

To prove part (b) we apply (7.5) with $\mathbf{a}=\mathbf{c}$ and
$\mathbf{b}=\mathbf{d}$ in every step of the construction of
$\Phi(\boldsymbol{\eta})$ in Definition 3.3.

To prove part (c), observe that the inclusion "$\supset$" follows from part
(b) and (7.3). In order to verify the inclusion "$\subset$" in (c), we assume
the opposite to get a contradiction. That is, we assume that there exists an
$\boldsymbol{\eta}\in\mathcal{A}^{|\mathbf{i}|}$ such that
$\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{i}}$ but
$\Phi(\boldsymbol{\eta})\neq\mathbf{i}$. But as we saw in part (b),
$\Pi(\boldsymbol{\eta})=\Pi(\Phi(\boldsymbol{\eta}))$. In this way the
distinct $\mathbf{i},\Phi(\boldsymbol{\eta})\in\mathcal{T}_{n}$ satisfy
$\Pi(\Phi(\boldsymbol{\eta}))=\Pi(\mathbf{i})$. This is impossible by part (a)
of Fact 2.3. ∎

Part (c) of Fact 7.6 and the definition of $\Phi$ (Definition 3.3) imply that

(7.7) $\mathbf{i}\in\mathcal{T}_{n}\quad\&\quad
G(\mathbf{i})=\emptyset\quad\Longrightarrow\quad\mathcal{I}_{\mathbf{i}}=\left\\{\mathbf{i}\right\\}.$

We partition $\mathcal{A}^{n}$ as follows

(7.8)
$\mathcal{A}^{n}=\bigcup_{\mathbf{i}\in\mathcal{T}_{n}}\mathcal{I}_{\mathbf{i}},\text{
where
}\mathcal{I}_{\mathbf{i}}:=\left\\{\boldsymbol{\eta}\in\mathcal{A}^{n}:\Pi(\boldsymbol{\eta})=\Pi(\mathbf{i})\right\\}=\left\\{\boldsymbol{\eta}\in\mathcal{A}^{n}:\Phi(\boldsymbol{\eta})=\mathbf{i}\right\\}.$

Using Fact 7.4 and (2.8) this is a partition in deed.

###### Proof of Fact 7.3.

Let $\boldsymbol{\eta}\in\mathcal{A}^{n}$ for an $n\geq 2$. By part (c) of
Fact 7.4 we have $\Phi(\boldsymbol{\eta})=\mathbf{i}$. Then, by the definition
of the mapping $\Phi$ we get $\mathbf{i}$ from $\boldsymbol{\eta}$ by
replacing all maximal bad blocks of $\boldsymbol{\eta}$ by the corresponding
good blocks. That is, if $[k,\ell]\in B(\boldsymbol{\eta})$, then we define
$i_{k}=\cdots=i_{\ell-1}=1$ and $i_{\ell}=0$. Every bad block in
$\boldsymbol{\eta}$ is a good block in $\mathbf{i}$. If we stop at an index
$z\in D_{\text{Good}}(\mathbf{i})$, then the collection of maximal good blocks
of $\mathbf{i}|_{z}$ are the same as the collection of those maximal good
blocks of $\mathbf{i}$ which intersect $[z]=\left\\{1,\dots,z\right\\}$. So,
if we apply the definition of $\Phi$ to $\boldsymbol{\eta}|_{z}$ we get that
$\Phi(\boldsymbol{\eta}|_{z})=\mathbf{i}|_{z}$. The second part follows from
the first part and the definition of $\Pi$. ∎

Using a little modification of this argument we can also prove that

######  Fact 7.5.

Let $\boldsymbol{\eta}\in\mathcal{A}^{n}$, where
$n\in\mathbb{N}\cup\\{\infty\\}$. Let $z<n$ and
$\mathbf{i}:=\Phi(\boldsymbol{\eta})|_{z}$

  1. (a)

If $z\in D_{Bad}(\boldsymbol{\eta})$, then

(7.9) $\Phi(\boldsymbol{\eta}|_{z})=\Phi(\boldsymbol{\eta})|_{z}\quad\text{
that is }\quad\boldsymbol{\eta}|_{z}\in\mathcal{I}_{\mathbf{i}}.$

  2. (b)

If $z\in C_{Bad}(\boldsymbol{\eta})$, then the z-th coordinate of
$\Phi(\boldsymbol{\eta})$ is equal to $1$ and

(7.10)
$\Phi(\boldsymbol{\eta}|_{z})=\Phi(\boldsymbol{\eta})|_{z-1}0\neq\Phi(\boldsymbol{\eta})|_{z}\quad\text{
that is }\quad\boldsymbol{\eta}|_{z}\in\mathcal{I}_{\mathbf{i}^{-}0}.$

  3. (c)

Consequently, we get that for all $z<n$ we have

(7.11)
$\boldsymbol{\eta}|_{z}\in\mathcal{I}_{\mathbf{i}}\cup\mathcal{I}_{\mathbf{i}^{-}0}.$

######  Example 7.6.

Let $\boldsymbol{\eta}=(0,\underbrace{3,\dots,3}_{n-1})$,
$\mathbf{i}=(\underbrace{1,\dots,1}_{n-1},0)$ and let $k=n-1$. Moreover, let
$\boldsymbol{\tau}^{\ell}:=(\underbrace{1,\dots
1}_{\ell},0,\underbrace{3,\dots,3}_{n-1-\ell})$, where $0\leq\ell\leq n-1$.
Then,

$\mathcal{I}_{\mathbf{i}}=\left\\{\boldsymbol{\tau}^{0},\dots,\boldsymbol{\tau}^{n-1}\right\\}\quad\text{
and
}\quad\mathcal{I}_{\mathbf{i}^{-}}=\big{\\{}(\underbrace{1,\dots,1}_{n-1})\big{\\}},$

where $\mathbf{i}^{-}$ was defined in (2.2).

That is, there exists $n-1$ elements
$\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{i}}$ such that
$\Phi(\boldsymbol{\eta}|_{n-1})\not\in\mathcal{I}_{\mathbf{i}|_{n-1}}$.

######  Fact 7.7.

Let $\mathbf{i}\in\mathcal{T}_{n}$,
$\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{i}}$. Let $[k,\ell]\in
G(\mathbf{\mathbf{i}})$ and $[k^{\prime},\ell^{\prime}]\in
B(\boldsymbol{\eta})$ such that
$[k,\ell]\cap[k^{\prime},\ell^{\prime}]\neq\emptyset$. Then,
$\ell=\ell^{\prime}$.

Namely, $\ell^{\prime}>\ell$ is not possible since this would follow that
$i_{\ell}=0,i_{\ell+1}=3$. $\ell^{\prime}<\ell$ is not possible since this
would follow $k\leq\ell^{\prime}<\ell$ and $i_{\ell^{\prime}}=0$ which is not
possible since by definition $i_{\ell^{\prime}}=1$.

###  7.2. The cases when $\psi(\mathbf{i})$ is multiplicative

For the rest of the paper we fix some notation. Let

(7.12)
$\chi:=\chi(\mathbf{i}):=\left\\{\begin{array}[]{ll}\max\left\\{\ell:[k,\ell]\in
G(\mathbf{i})\right\\},&\hbox{if $G(\mathbf{i})\neq\emptyset$;}\\\
0,&\hbox{otherwise.}\end{array}\right.$

Moreover, let

(7.13) $\xi:=\xi(\mathbf{j}):=\min\left\\{k:\forall\ell>k,\
j_{\ell}=1\right\\}.$

Clearly,

(7.14) $G(\mathbf{i})\neq\emptyset\Longrightarrow(i_{\chi-1},i_{\chi})=(1,0).$

We also get by definition that

(7.15)
$\chi(\boldsymbol{\ell})\leq\xi(\boldsymbol{\ell}),\quad\forall\boldsymbol{\ell}\in\mathcal{T}_{p},\quad\forall
p.$

######  Fact 7.8.

For $\ell\geq\chi$

(7.16) $\psi(\mathbf{i})=\psi(\mathbf{i}|_{\ell})\cdot p_{i_{\ell+1}\dots
i_{n}}.$

###### Proof.

If $G(\mathbf{i})\neq\emptyset$, then the proof is immediate from Definition
3.3. In the case when $G(\mathbf{i})=\emptyset$, then $\chi=0$,
$\psi(\mathbf{i}|_{0})=\psi(\flat)=1$. Then, the fact follows from (7.7). ∎

######  Lemma 7.9.

Let $n\geq 2$, $\mathbf{i}\in\mathcal{T}_{n}$ and $z\in
D_{\text{Good}}(\mathbf{i})$. Then,

(7.17)
$\psi(\mathbf{i})=\psi(\mathbf{i}|_{z})\cdot\psi(\sigma^{z}\mathbf{i}).$

###### Proof.

$\psi(\mathbf{i})=\sum_{\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{i}}}p_{\boldsymbol{\eta}}=\sum_{\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{i}}}p_{\boldsymbol{\eta}|_{z}}\cdot
p_{\sigma^{z}\boldsymbol{\eta}}\leq\sum_{\boldsymbol{\omega}\in\mathcal{I}_{\mathbf{i}|_{z}}}p_{\boldsymbol{\omega}}\cdot\sum_{\boldsymbol{\tau}\in\mathcal{I}_{\sigma^{z}\mathbf{i}}}p_{\boldsymbol{\tau}}=\psi(\mathbf{i}|_{z})\cdot\psi(\sigma^{z}\mathbf{i}),$

where in the third step we used that by Fact 7.3 for an
$\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{i}}$ we have
$\boldsymbol{\eta}|_{z}\in\mathcal{I}_{\mathbf{i}|_{z}}$ and
$\sigma^{z}\boldsymbol{\eta}\in\mathcal{I}_{\sigma^{z}\mathbf{i}}$ since $z\in
D_{\text{Good}}(\mathbf{i})$. On the other hand,

$\psi(\mathbf{i}|_{z})\cdot\psi(\sigma^{z}\mathbf{i})=\sum_{\boldsymbol{\omega}\in\mathcal{I}_{\mathbf{i}|_{z}}}p_{\boldsymbol{\omega}}\cdot\sum_{\boldsymbol{\tau}\in\mathcal{I}_{\sigma^{z}\mathbf{i}}}p_{\boldsymbol{\tau}}=\sum_{\boldsymbol{\omega}\in\mathcal{I}_{\mathbf{i}|_{z}}\atop\boldsymbol{\tau}\in\mathcal{I}_{\sigma^{z}\mathbf{i}}}p_{\boldsymbol{\omega}\boldsymbol{\tau}}\leq\sum_{\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{i}}}p_{\boldsymbol{\eta}}=\psi(\mathbf{i}),$

where in the one but last step we used that $(i_{z},i_{z+1})\neq(0,3)$ since
$\mathbf{i}\in\mathcal{T}_{n}$. Moreover, we also used that by (2.3),

(7.18)
$\Pi(\boldsymbol{\eta})=\Pi(\boldsymbol{\eta}|_{z})+3^{-z}\Pi(\sigma^{z}\boldsymbol{\eta}),\quad\forall\boldsymbol{\eta}\in\mathcal{A}^{n}.$

Hence, $\boldsymbol{\omega}\in\mathcal{I}_{\mathbf{i}|_{z}}$ and
$\boldsymbol{\tau}\in\mathcal{I}_{\sigma^{z}\mathbf{i}}$ implies that
$\boldsymbol{\eta}=\boldsymbol{\omega}\boldsymbol{\tau}\in\mathcal{I}_{\mathbf{i}}$.
∎

###  7.3. The properties of the sequence $\left\\{a_{w}\right\\}_{w}$

Observe that for a $w\geq 1$

(7.19)
$\mathcal{I}_{\overline{1}^{w}}=\left\\{\overline{1}^{w}\right\\}\quad\text{
and
}\quad\mathcal{I}_{\overline{1}^{w-1}0}=\left\\{\overline{1}^{w-1-\ell}0\overline{3}^{\ell}\right\\}_{\ell=0}^{w-1}.$

Hence,

(7.20) $\psi(\overline{1}^{w})=p_{1}^{w}\quad\text{ and
}\quad\psi\left(\overline{1}^{w-1}0\right)=a_{w}:=\frac{p_{0}}{p_{1}}p_{1}^{w}\sum_{\ell=0}^{w-1}\left(\frac{p_{3}}{p_{1}}\right)^{\ell}.$

So,

(7.21)
$\psi\left(\overline{1}^{w-1}0\right)>\psi(\overline{1}^{w})\Longleftrightarrow\sum\limits_{\ell=0}^{w-1}\left(\frac{p_{3}}{p_{1}}\right)^{\ell}>\frac{p_{1}}{p_{0}}$

This motivates the following definition. For an $n\geq 1$ and
$\mathbf{i}\in\mathcal{T}_{n}$ we define

(7.22)
$\mathbf{i}^{*}:=\left\\{\begin{array}[]{ll}\mathbf{i}^{-}0,&\hbox{if\qquad$n-\xi(\mathbf{i})\geq
1$ and
$\sum\limits_{\ell=0}^{n-\xi(\mathbf{i})-1}\left(\frac{p_{3}}{p_{1}}\right)^{\ell}>\frac{p_{1}}{p_{0}}$;}\\\
\mathbf{i},&\hbox{otherwise.}\end{array}\right.$

We introduce

(7.23) $\mathfrak{A}:=\left\\{q\in\mathbb{N}:q\geq 1\ \&\
\sum\limits_{\ell=0}^{q-1}\left(\frac{p_{3}}{p_{1}}\right)^{\ell}>\frac{p_{1}}{p_{0}}\right\\}.$

Observe that

(7.24)
$\mathbf{i}^{*}\neq\mathbf{i}\Longleftrightarrow|\mathbf{i}|-\xi(\mathbf{i})\in\mathfrak{A}.$

Recall that in (2.7) we assumed that $p_{3}\leq p_{1}$. A simple calculation
shows that

(7.25) $\mathfrak{A}\neq\emptyset\Longleftrightarrow p_{1}-p_{3}<p_{0}$

Putting together (7.21) and (7.23) we get that

(7.26)
$q\in\mathfrak{A}\Longleftrightarrow\widehat{\psi}(\overline{1}^{q})>\psi(\overline{1}^{q}),$

We define

(7.27) $q_{0}:=\left\\{\begin{array}[]{ll}\min\mathfrak{A},&\hbox{if
$p_{1}-p_{3}<p_{0}$;}\\\ \infty,&\hbox{otherwise.}\end{array}\right.$

Now we prove the sub-multiplicative property of the sequence
$\left\\{a_{w}\right\\}_{w=1}^{\infty}$.

######  Fact 7.10.

Let $u,v\geq 1$. Then, there are constants $C_{6},C_{7}>0$ such that

(7.28) $a_{u+v}<C_{6}a_{u}a_{v},$

and

(7.29)
$\sum_{\ell=0}^{u+v}\left(\frac{p_{3}}{p_{1}}\right)^{\ell}<C_{7}\cdot\sum_{\ell=0}^{u-1}\left(\frac{p_{3}}{p_{1}}\right)^{\ell}\cdot\sum_{\ell=0}^{v-1}\left(\frac{p_{3}}{p_{1}}\right)^{\ell}$

###### Proof.

First we prove (7.29). Recall that we assumed that $0<p_{3}\leq p_{1}$. If
$p_{1}=p_{3}$, then we can clearly choose a constant $C_{7}>0$ such that
$u+v+1<C_{7}\cdot u\cdot v$. If $p_{3}<p_{1}$, then (7.29) holds with the
choice of $C_{7}=\frac{1}{1-\frac{p_{3}}{p_{1}}}$. It is immediate that (7.28)
follows from (7.29). ∎

The following fact is an important but trivial consequence of the definitions.

######  Fact 7.11.

Given a $\mathbf{k}\in\mathcal{T}_{n}$ with $k_{n}\neq 1$ and a natural number
$w\geq 1$, we have

(7.30)
$\psi\left(\left(\mathbf{k}\overline{1}^{w}\right)^{*}\right)=\psi(\mathbf{k})\psi\left((\overline{1}^{w})^{*}\right).$

###### Proof.

By definition, $\xi(\mathbf{k}\overline{1}^{w})=n$. Therefore,

$(\mathbf{k}\overline{1}^{w})^{*}=\mathbf{k}\overline{1}^{w-1}0\Longleftrightarrow
w\geq q_{0}\Longleftrightarrow(\overline{1}^{w})^{*}=\overline{1}^{w-1}0.$

Hence, $(\mathbf{k}\overline{1}^{w})^{*}=\mathbf{k}(\overline{1}^{w})^{*}$.
Observe that by $k_{n}\neq 1$ we have $n\in
D_{\text{Good}}(\mathbf{k}(\overline{1}^{w})^{*})$. Using this and Fact 7.9 we
get

$\psi((\mathbf{k}\overline{1}^{w})^{*})=\psi(\mathbf{k}(\overline{1}^{w})^{*})=\psi(\mathbf{k})\psi((\overline{1}^{w})^{*}).$

∎

The second condition in first line of (7.22) is to guarantee that
$\widehat{\psi}(\mathbf{i})>\psi(\mathbf{i})$.

We know by (3.12) that
$\widehat{\psi}(\mathbf{i})=\max\\{\psi(\mathbf{i}),\psi(\mathbf{i}^{-}0)\\}$.

######  Fact 7.12.

For every $n\geq 1$ and $\mathbf{i}\in\mathcal{T}_{n}$

(7.31) $\mathbf{i}^{*}=\left\\{\begin{array}[]{ll}\mathbf{i}^{-}0,&\hbox{if
$\widehat{\psi}(\mathbf{i})>\psi(\mathbf{i})$;}\\\ \mathbf{i},&\hbox{if
$\widehat{\psi}(\mathbf{i})=\psi(\mathbf{i})$.}\end{array}\right.$

###### Proof.

If $i_{n}\neq 1$ then by (3.12) we have
$\widehat{\psi}(\mathbf{i})=\psi(\mathbf{i})$. In this case
$n-\xi(\mathbf{i})=0$. Hence, it follows from (7.22) that
$\mathbf{i}^{*}=\mathbf{i}$. So, we may assume that $i_{n}=1$. That is,
$n-\xi(\mathbf{i})\geq 1$. Hence,
$\mathbf{i}=\mathbf{i}|_{\xi(\mathbf{i})}\overline{1}^{n-\xi(\mathbf{i})}$.
Using Fact 7.11, we get

(7.32)
$\psi(\mathbf{i}^{*})=\psi(\mathbf{i}|_{\xi(\mathbf{i})})\cdot\psi\left(\left(\overline{1}^{n-\xi(\mathbf{i})}\right)^{*}\right).$

Using that $\xi(\mathbf{\ell})\in D_{\text{Good}}(\mathbf{i})$ we obtain from
Lemma 7.9 that

(7.33)
$\psi(\mathbf{i})=\psi(\mathbf{i}|_{\xi(\mathbf{i})})\cdot\psi\left(\overline{1}^{n-\xi(\mathbf{i})}\right).$

Putting together the last two displayed formulas with (7.24) and (7.26) we get
the assertion of Fact. ∎

Observe that in virtue of Fact 7.12 we have

(7.34) $\widehat{\psi}(\mathbf{i})=\psi(\mathbf{i}^{*}).$

###  7.4. The Proof of Property-2

######  Lemma 7.13.

There exists a constant $C_{6}>1$ such that the following holds. Let $q\geq 1$
and let $\mathbf{j}\in\mathcal{T}_{q}$ be arbitrary. Let $u,v\in\mathcal{A}$
such that $\mathbf{j}u,\mathbf{j}v\in\mathcal{T}_{q+1}$. Then,

(7.35)
$C_{6}^{-1}(|\mathbf{j}|+1)^{-1}<\frac{\psi(\mathbf{j}v)}{\psi(\mathbf{j}u)}<C_{6}(|\mathbf{j}|+1).$

###### Proof.

We may assume that $u\neq v$ and

(7.36) $\text{ if }0\in\left\\{u,v\right\\},\text{ then }v=0.$

Using that any good block ends with a $0$, this implies that
$\chi(\mathbf{j}v)\geq\chi(\mathbf{j}u)$. More precisely,

(7.37) $\left(G(\mathbf{j}v)\neq\emptyset\ \&\
\chi(\mathbf{j}u)\neq\chi(\mathbf{j}v)\right)\Longrightarrow\chi(\mathbf{j}u)<\chi(\mathbf{j}v)=q+1.$

Recall from (7.13) that

$\xi:=\xi(\mathbf{j})=\min\left\\{k\in[0,q]:\forall\ell>k,\
j_{\ell}=1\right\\},$

where $\xi=q$ means that $\mathbf{j}$ does not end with a block of $1$s. If
$\xi<q$, then we can write

(7.38) $\mathbf{j}=\mathbf{j}|_{\xi}\overline{1}^{q-\xi}.$

By definition, we get that

(7.39) $\xi\in\left\\{0\right\\}\cup D_{\text{Good}}(\mathbf{j}).$

By Fact 7.3 we get that

(7.40) $\boldsymbol{\omega}\in\mathcal{I}_{\mathbf{j}u},\
\boldsymbol{\tau}\in\mathcal{I}_{\mathbf{j}v}\Longrightarrow\Pi(\boldsymbol{\omega}|_{\xi})=\Pi(\boldsymbol{\tau}|_{\xi})=\Pi(\mathbf{j}|_{\xi})=\Pi(\mathbf{j}u|_{\xi})=\Pi(\mathbf{j}v|_{\xi}).$

Moreover,

(7.41)
$\mathcal{I}_{\mathbf{j}u}=\left\\{\boldsymbol{\tau}\overline{1}^{q-\xi}u:\boldsymbol{\tau}\in\mathcal{I}_{\mathbf{i}|_{\xi}}\right\\},\text{
if
}u\in\left\\{1,3\right\\},\quad\mathcal{I}_{\mathbf{j}0}=\bigcup_{\ell=0}^{q-\xi}\left\\{\boldsymbol{\tau}\overline{1}^{q-\xi-\ell}0\overline{3}^{\ell}:\boldsymbol{\tau}\in\mathcal{I}_{\mathbf{i}|_{\xi}}\right\\}.$

Using that for $u\in\left\\{1,3\right\\}$, $\chi(\mathbf{j}u)\leq\xi$ it
follows from Fact 7.8 that

(7.42) $\psi(\mathbf{j}u)=\psi(\mathbf{j}|_{\xi})p_{1}^{q-\xi}p_{u}.$

Using the second part of (7.41) we get

(7.43)
$\psi(\mathbf{j}0)=\psi(\mathbf{j}|_{\xi})\cdot\sum_{\ell=0}^{q-\xi}p_{1}^{q-\xi-\ell}p_{3}^{\ell}p_{0}.$

Hence,

(7.44)
$\frac{\psi(\mathbf{j}v)}{\psi(\mathbf{j}u)}=\left\\{\begin{array}[]{ll}\frac{\sum_{\ell=0}^{q-\xi}p_{1}^{q-\xi-\ell}p_{3}^{\ell}p_{0}}{p_{1}^{q-\xi}p_{u}}=\frac{p_{0}}{p_{u}}\sum_{\ell=0}^{q-\xi}\left(\frac{p_{3}}{p_{1}}\right)^{\ell},&\hbox{
if $v=0$;}\\\ \frac{p_{v}}{p_{u}},&\hbox{if $v\neq 0$.}\end{array}\right.$

That is, by (2.7) we obtain

(7.45)
$\frac{p_{\min}}{p_{\max}}\leq\frac{\psi(\mathbf{j}v)}{\psi(\mathbf{j}u)}\leq
C_{14}(|\mathbf{j}|+1-\xi),\quad u\in\left\\{1,3\right\\}\text{ and
}v\in\left\\{0,1,3\right\\}\setminus\\{u\\},$

where
$C_{14}:=\left\\{\begin{array}[]{ll}\frac{p_{0}p_{1}}{p_{3}(p_{1}-p_{3})},&\hbox{if
$p_{1}>p_{3}$;}\\\ \frac{p_{0}}{p_{3}},&\hbox{if
$p_{0}=p_{3}$.}\end{array}\right.$ ∎

We obtain from (7.43) that

(7.46)
$\xi(\mathbf{i})\leq|\mathbf{i}|-1\Longrightarrow\psi(\mathbf{i}^{-}0)=\psi(\mathbf{i}|_{\xi(\mathbf{i})})\cdot\frac{p_{0}}{p_{1}}p_{1}^{|\mathbf{i}|-\xi(\mathbf{i})}\sum_{\ell=0}^{|\mathbf{i}|-\xi(\mathbf{i})-1}\left(\frac{p_{3}}{p_{1}}\right)^{\ell}=\psi(\mathbf{i}|_{\xi(\mathbf{i})})\cdot
a_{|\mathbf{i}|-\xi(\mathbf{i})},$

where the sequence $\left\\{a_{q}\right\\}_{q=1}^{\infty}$ was defined in
(7.20). Hence, by Fact 7.12

(7.47)
$\mathbf{i}^{*}\neq\mathbf{i}\Longrightarrow\widehat{\psi}(\mathbf{i})=\psi(\mathbf{i}^{-}0)=\psi(\mathbf{i}|_{\xi(\mathbf{i})})\cdot
a_{|\mathbf{i}|-\xi(\mathbf{i})}.$

######  Fact 7.14.

Let $\mathbf{i}\in\mathcal{T}_{n}$. If $\xi(\mathbf{i})=n$, then $i_{n}\neq
1$. So, by definition $\psi(\mathbf{i})=\widehat{\psi}(\mathbf{i})$. There
exists a $C_{3}>0$ such that whenever $\xi(\mathbf{i})\leq n-1$, then we have

(7.48) $\psi(\mathbf{i})\leq\widehat{\psi}(\mathbf{i})\leq
C_{3}\cdot(n-\xi(\mathbf{i}))\cdot\psi(\mathbf{i}).$

Moreover,

(7.49)
$\mathbf{i}\neq\mathbf{i}^{*}\Longleftrightarrow\psi(\mathbf{i})<\widehat{\psi}(\mathbf{i})\Longleftrightarrow\max\left\\{1,q_{0}\right\\}\leq
n-\xi(\mathbf{i}).$

Observe that Fact 7.14 implies that Property-2 holds.

###### Proof.

Let $\mathbf{i}\in\mathcal{T}_{n}$ and $\mathbf{j}:=\mathbf{i}^{-}$. It is
assumed that $\xi(\mathbf{i})\leq n-1$. This implies that
$\mathbf{i}=\mathbf{j}1$. If $\xi(\mathbf{i})=n-1$, then
$\widehat{\psi}(\mathbf{i})=\psi(\mathbf{i})$ by definition. We may assume
that $\xi(\mathbf{i})\leq n-2$. Then, either
$\widehat{\psi}(\mathbf{i})=\psi(\mathbf{i})$ or if not then by Fact 7.12
$\mathbf{i}^{*}=\mathbf{j}0$ and by (7.22) and (7.44)

$1\leq\frac{\widehat{\psi}(\mathbf{i})}{\psi(\mathbf{i})}\leq\frac{p_{0}}{p_{1}}(n-\xi).$

This completes the proof of the first part with the choice of
$C_{3}:=\frac{p_{0}}{p_{1}}$.

To verify the second part we first observe that if $\xi(\mathbf{i})\geq n-1$,
then $\mathbf{i}=\mathbf{i}^{*}$ so
$\widehat{\psi}(\mathbf{i})=\psi(\mathbf{i})$. So, we may assume that $2\leq
n-\xi(\mathbf{i})$. In this case
$\widehat{\psi}(\mathbf{i})=\psi(\mathbf{j}0)$ and
$\psi(\mathbf{i})=\psi(\mathbf{j}1)$. We write $\xi:=\xi(\mathbf{j})$. Then,
by (7.42) and (7.43) we get that

(7.50) $\psi(\mathbf{i})<\widehat{\psi}(\mathbf{i})\Longleftrightarrow
p_{1}^{n-\xi}<\sum_{\ell=0}^{n-1-\xi}p_{1}^{n-1-\xi-\ell}p_{3}^{\ell}p_{0}\Longleftrightarrow
n-\xi\geq q_{0}.$

∎

###  7.5. The Proof of Property-1

######  Lemma 7.15.

There exists a $C_{10}>1$ such that for any $n,m\geq 1$,
$\mathbf{k}\in\mathcal{T}_{n+m}$

(7.51) $\widehat{\psi}(\mathbf{k})\leq
C_{10}\widehat{\psi}(\mathbf{k}|_{n})\widehat{\psi}(\sigma^{n}\mathbf{k}).$

###### Proof.

Fix a $\mathbf{k}\in\mathcal{T}_{n+m}$. Let

$\mathbf{i}=(i_{1},\dots,i_{n}):=\mathbf{k}|_{n}\in\mathcal{T}_{n}\quad\text{
and
}\quad\mathbf{j}=(j_{1},\dots,j_{m}):=\sigma^{n}\mathbf{k}\in\mathcal{T}_{m}.$

We distinguish two cases:

Case 1 $\xi(\mathbf{k})\leq n-1$. Then, $\xi(\mathbf{k})=\xi(\mathbf{i})$ and
$1=i_{n}=j_{1}=\cdots=j_{m}$. That is, $\mathbf{j}=\overline{1}^{m}$. We
distinguish three cases again:

  1. (a)

$\mathbf{i}^{*}\neq\mathbf{i}$ (this implies that
$\mathbf{k}^{*}\neq\mathbf{k}$). Then, $\xi(\mathbf{i})=\xi(\mathbf{k})$.
Hence,

(7.52)
$\frac{\psi(\mathbf{k}^{*})}{\psi(\mathbf{i}^{*})}=\frac{\psi(\mathbf{i}|_{\xi(\mathbf{i})})\cdot
a_{n+m-\xi(\mathbf{i})}}{\psi(\mathbf{i}|_{\xi(\mathbf{i})})\cdot
a_{n-\xi(\mathbf{i})}}=p_{1}^{m}\cdot\underbrace{\frac{\sum\limits_{\ell=0}^{n+m-\xi(\mathbf{i})-1}\left(\frac{p_{3}}{p_{1}}\right)^{\ell}}{\sum\limits_{\ell=0}^{n-\xi(\mathbf{i})-1}\left(\frac{p_{3}}{p_{1}}\right)^{\ell}}}_{r(\mathbf{i},\mathbf{j})}.$

    1. (I)

If $m\leq q_{0}$, then by definition $\mathbf{j}^{*}=\mathbf{j}$ and then, by
Fact 7.12, $\widehat{\psi}(\mathbf{j})=\psi(\mathbf{j})=p_{1}^{m}$ and
$r(\mathbf{i},\mathbf{j})\leq C_{5}\cdot q_{0}$ for a $C_{5}>0$. So, in this
case (7.51) holds since in this situation,
$\widehat{\psi}(\mathbf{i})=\psi(\mathbf{i}^{-}0)$ and
$\widehat{\psi}(\mathbf{k})=\psi(\mathbf{k}^{-}0)$.

    2. (II)

Assume that $m>q_{0}$. Then, as we have said, $\mathbf{j}=\overline{1}^{m}$
and

(7.53)
$\widehat{\psi}(\mathbf{j})=\psi(\mathbf{j}^{-}0)=a_{m}=\frac{p_{0}}{p_{1}}p_{1}^{m}\sum_{\ell=0}^{m-1}\left(\frac{p_{3}}{p_{1}}\right)^{\ell}.$

So, in order to verify (7.51) for this case, we have to show that

(7.54) $r(\mathbf{i},\mathbf{j})\leq\text{ Const
}\cdot\sum_{\ell=0}^{m-1}\left(\frac{p_{3}}{p_{1}}\right)^{\ell}.$

This follows from (7.29).

  2. (b)

$\mathbf{i}^{*}=\mathbf{i}$ but $\mathbf{k}^{*}\neq\mathbf{k}$. Then,

(7.55) $n-\xi(\mathbf{i})<q_{0}\leq n+m-\xi(\mathbf{i}).$

We write $\xi:=\xi(\mathbf{i})$. Then,
$\widehat{\psi}(\mathbf{k})=\psi(\mathbf{k}^{-}0)$ and
$\widehat{\psi}(\mathbf{i})=\psi(\mathbf{i})$,

(7.56)
$\frac{\psi(\mathbf{k}^{*})}{\psi(\mathbf{i}^{*})}=\frac{\psi(\mathbf{i}|_{\xi})\frac{p_{0}}{p_{1}}p_{1}^{n-\xi+m}\sum\limits_{\ell=0}^{n-\xi+m-1}\left(\frac{p_{1}}{p_{3}}\right)^{\ell}}{\psi(\mathbf{i}|_{\xi})p_{1}^{n-\xi}}=\frac{p_{0}}{p_{1}}p_{1}^{m}\sum\limits_{\ell=0}^{n-\xi+m-1}\left(\frac{p_{1}}{p_{3}}\right)^{\ell}.$

    1. (I)

If $m\leq q_{0}$, then $\widehat{\psi}(\mathbf{j})=p_{1}^{m}$. Then
$n-\xi+m<2q_{0}$. Hence, (7.51) holds in this case.

    2. (II)

If $m>q_{0}$, then
$\widehat{\psi}(\mathbf{j})=\widehat{\psi}(\overline{1}^{m})=a_{m}=\frac{p_{0}}{p_{1}}p_{1}^{m}\sum_{\ell=0}^{m-1}\left(\frac{p_{3}}{p_{1}}\right)^{\ell}$.
Using this, (7.56) and the fact that $n-\xi<q_{0}$ (which holds since
$\mathbf{i}^{*}=\mathbf{i}$) we obtain that (7.51) holds also in this case.

  3. (c)

$\mathbf{i}^{*}=\mathbf{i}$ and $\mathbf{k}^{*}=\mathbf{k}$ (this implies that
$\mathbf{j}^{*}=\mathbf{j}$ since $\xi(\mathbf{k})\leq n-1$). Then,
$\widehat{\psi}(\mathbf{i})=\psi(\mathbf{i})$,
$\widehat{\psi}(\mathbf{j})=\psi(\mathbf{j})$ and
$\widehat{\psi}(\mathbf{k})=\psi(\mathbf{k})$. Hence,

$\frac{\widehat{\psi}(\mathbf{k})}{\widehat{\psi}(\mathbf{i})}=\frac{\psi(\mathbf{i}|_{\xi})p_{1}^{n+m-\xi}}{\psi(\mathbf{i}|_{\xi})p_{1}^{n-\xi}}=p_{1}^{m}=\widehat{\psi}(\mathbf{j}).$

This verifies (7.51) for this case.

Case 2 $\xi(\mathbf{k})\geq n$. In this case
$m-\xi(\mathbf{j})=n+m-\xi(\mathbf{k})$. This means that
$\mathbf{k}^{*}\neq\mathbf{k}$ if and only if $\mathbf{j}^{*}\neq\mathbf{j}$.
In particular, $\xi(\mathbf{k})\geq n$ implies that

(7.57) $\mathbf{k}^{*}=\mathbf{i}\mathbf{j}^{*}.$

Now we distinguish two cases:

  1. (a)

$n\in D_{\text{Good}}(\mathbf{k}^{*})$. Then, it follows from Lemma 7.9 that
$\psi(\mathbf{k}^{*})=\psi(\mathbf{i})\psi(\mathbf{j}^{*})$. Here we used that
by (7.57), we have $\sigma^{n}(\mathbf{k}^{*})=(\sigma^{n}\mathbf{k})^{*}$.
Using that $\psi(\mathbf{i})\leq\widehat{\psi}(\mathbf{i})$ we obtain that
(7.51) holds in this case.

  2. (b)

$n\in C_{\text{Good}}(\mathbf{k}^{*})$. This means that there is a good block
$[u,v+1]\in\mathcal{G}(\mathbf{k}^{*})$ such that

(7.58) $1\leq u\leq n\leq v\leq n+m-1,\quad\text{and}\quad k_{v+1}=0.$

Using that either $\xi(\mathbf{k}^{*})=n+m$ or $k_{\xi(\mathbf{k}^{*})}\neq
1$, we get that

(7.59) $\xi(\mathbf{k}^{*})\in D_{\text{Good}}(\mathbf{k}^{*}).$

Moreover,

(7.60) $u-1<v+1\leq\xi(\mathbf{k}^{*}),\quad\text{ and }\quad u-1,v+1\in
D_{\text{Good}}(\mathbf{k}^{*}).$

Hence, by Lemma 7.9 we have

(7.61)
$\psi(\mathbf{k}^{*})=\psi(\mathbf{i}|_{u-1})\psi(\overline{1}^{v-u+1}0)\psi(\sigma^{v+1}\mathbf{k}|_{\xi(\mathbf{k})})\psi\left(\left(\overline{1}^{n+m-\xi(\mathbf{k})}\right)^{*}\right),$

where we remark that the first and last words in (7.61), $\mathbf{i}|_{u-1}$
and $(\overline{1}^{n+m-\xi(\mathbf{k})})^{*}$, respectively, can be the empty
words $\flat$. In this case we recall that $\psi(\flat)=1$. Using Fact 7.10 in
the third step we get

(7.62) $\displaystyle\psi(\overline{1}^{v-u+1}0)$ $\displaystyle=$
$\displaystyle\psi(\overline{1}^{n-u+1}\overline{1}^{v-n}0)=a_{(n-u+1)+(v-n+1)}\leq
C_{6}a_{n-u+1}a_{v-n+1}.$ $\displaystyle=$ $\displaystyle
C_{6}\psi(\overline{1}^{n-u}0)\psi(\overline{1}^{v-n}0)\leq
C_{6}\psi\left(\left(\overline{1}^{n-u+1}\right)^{*}\right)\psi(\overline{1}^{v-n}0).$

Now we substitute this into (7.61) and get

(7.63) $\psi(\mathbf{k}^{*})\leq
C_{6}\underbrace{\psi(\mathbf{i}|_{u-1})\psi\left(\left(\overline{1}^{n-u+1}\right)^{*}\right)}_{\psi(\mathbf{i}^{*})}\underbrace{\psi(\overline{1}^{v-n}0)\psi(\sigma^{v+1}\mathbf{k}|_{\xi(\mathbf{k})})\psi\left(\left(\overline{1}^{n+m-\xi(\mathbf{k})}\right)^{*}\right)}_{\psi(\mathbf{j}^{*})},$

where we used Fact 7.11 and in the last step we used that
$n+m-\xi(\mathbf{k})=m-\xi(\mathbf{j})$. That is, (7.51) holds also in this
last possible case.

∎

We will need the following Claim:

######  Claim 7.16.

For an arbitrary $\mathbf{i}\in\mathcal{T}_{n}$ we have

(7.64) $\psi(\mathbf{i}0)\geq p_{0}\cdot\widehat{\psi}(\mathbf{i}).$

###### Proof.

Fix an $\mathbf{i}\in\mathcal{T}_{n}$.

  1. (a)

Assume that $i_{n}\neq 1$. Then, $\mathbf{i}^{*}=\mathbf{i}$ and $n\in
D_{\text{Good}}(\mathbf{i}0)$. Then, by Lemma 7.9, we have
$\psi(\mathbf{i}0)=\psi(\mathbf{i})\psi(0)=p_{0}\widehat{\psi}(\mathbf{i})$.
So, (7.64) holds in this case.

  2. (b)

Assume that $i_{n}=1$. Then, $\xi:=\xi(\mathbf{i})\leq n-1$ and $\xi\in
D_{\text{Good}}(\mathbf{i}0)$. Then, by Lemma 7.9 we have

(7.65)
$\psi(\mathbf{i}0)=\psi(\mathbf{i}|_{\xi})\psi\left(\sigma^{\xi}(\mathbf{i}0)\right)=\psi(\mathbf{i}|_{\xi})\psi\left(\overline{1}^{n-\xi}0\right)=\psi(\mathbf{i}|_{\xi})\cdot
a_{n-\xi+1}.$

On the other hand, by Fact 7.11

(7.66)
$\widehat{\psi}(\mathbf{i})=\left\\{\begin{array}[]{ll}\psi(\mathbf{i}|_{\xi})p_{1}^{n-\xi},&\hbox{if
$n-\xi<q_{0}$;}\\\ \psi(\mathbf{i}|_{\xi})a_{n-\xi},&\hbox{if $n-\xi\geq
q_{0}$.}\end{array}\right.$

Putting together (7.65) and (7.66) we get that (7.64) holds.

∎

######  Claim 7.17.

Let $\mathbf{j}\in\mathcal{T}_{m}$ for an $m\geq 1$. Then, we have

(7.67) $\widehat{\psi}(1\mathbf{j})\geq p_{1}\widehat{\psi}(\mathbf{j}).$

###### Proof.

First we consider the case when $\mathbf{j}=\overline{1}^{m}$. This case can
be subdivided into the three cases when $m<q_{0}-1$, $m=q_{0}-1$ and $m\geq
q_{0}$. Using formula (7.20), one can easily point out in each of these three
cases that (7.67) holds. So, from now on we may assume that there exists a
$p\in\left\\{1,\dots,m\right\\}$ such that $j_{p}\neq 1$. Using this, one can
easily see in the same way as in Case 2 of the proof of Lemma 7.15 that

(7.68) $\left(1\mathbf{j}\right)^{*}=1\mathbf{j}^{*}.$

Observe that

(7.69) $\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{j}^{*}}\Longrightarrow
1\boldsymbol{\eta}\in\mathcal{I}_{1\mathbf{j}^{*}}.$

Namely,

(7.70)
$\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{j}^{*}}\Longleftrightarrow\Pi(\boldsymbol{\eta})=\Pi(\mathbf{j}^{*})\Longrightarrow
S_{1}(\Pi(\boldsymbol{\eta}))=S_{1}(\Pi(\mathbf{j}^{*}))\\\
\Longrightarrow\Pi(1\boldsymbol{\eta})=\Pi(1\mathbf{j}^{*})\Longrightarrow
1\boldsymbol{\eta}\in\mathcal{I}_{1\mathbf{j}^{*}}.$

By definition

(7.71)
$\psi(1\mathbf{j}^{*})=\sum_{\boldsymbol{\omega}\in\mathcal{I}_{1\mathbf{j}^{*}}}p_{\boldsymbol{\omega}}\geq\sum_{\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{j}^{*}}}p_{1\boldsymbol{\eta}}=p_{1}\sum_{\boldsymbol{\eta}\in\mathcal{I}_{\mathbf{j}^{*}}}p_{1\boldsymbol{\eta}}=p_{1}\psi(\mathbf{j}^{*}).$

∎

######  Lemma 7.18.

There exist a $C_{11}>0$ such that

(7.72) $\widehat{\psi}(\mathbf{i}01\mathbf{j})\geq
C_{11}\widehat{\psi}(\mathbf{i})\widehat{\psi}(\mathbf{j}),\quad\forall\mathbf{i},\mathbf{j}\in\Sigma_{A}^{*}.$

###### Proof.

Given $\mathbf{i}\in\mathcal{T}_{n}$ and $\mathbf{j}\in\mathcal{T}_{m}$.
Clearly,
$\widetilde{\mathbf{k}}:=\mathbf{i}01\mathbf{j}\in\mathcal{T}_{n+m+2}$. Using
that $\xi(\widetilde{\mathbf{k}})\leq n+1$. Hence, by the same argument that
we used in Case 2, part of the proof of Lemma 7.15, we obtain that

(7.73) $\widetilde{\mathbf{k}}^{*}=\mathbf{i}0(1\mathbf{j})^{*}.$

Observe that $n+1\in D_{\text{Good}}(\widetilde{\mathbf{k}}^{*})$. Hence,
first using Lemma 7.9 and then using Claims (7.16) and (7.17) we get that

(7.74)
$\widehat{\psi}(\widetilde{\mathbf{k}})=\psi(\widetilde{\mathbf{k}}^{*})=\psi(\mathbf{i}0)\psi\left(\left(1\mathbf{j}\right)^{*}\right)\geq
p_{0}\widehat{\psi}(\mathbf{i})p_{1}\widehat{\psi}(\mathbf{j}).$

∎

Putting together Lemmas 7.15 and 7.18 we obtain that

######  Corollary 7.19.

For every $t>0$, $\widehat{\psi}^{t}$ is a quasi-multiplicative potential in
the sense of Definition 3.4.

This proves that Property-1 holds.

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