# Ahlfors regular conformal dimension and Gromov-Hausdorff convergence

Nicola Cavallucci

###### Abstract

We prove that the Ahlfors regular conformal dimension is upper semicontinuous
with respect to Gromov-Hausdorff convergence when restricted to the class of
uniformly perfect, uniformly quasi-selfsimilar metric spaces. Moreover we show
the continuity of the Ahlfors regular conformal dimension in case of limit
sets of discrete, quasiconvex-cocompact group of isometries of uniformly
bounded codiameter of $\delta$-hyperbolic metric spaces under equivariant
pointed Gromov-Hausdorff convergence of the spaces.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Ahlfors regular spaces
    2. 2.2 Quasi-selfsimilar spaces
  3. 3 Combinatorial modulus
  4. 4 Combinatorial modulus on quasi-selfsimilar spaces
  5. 5 Upper semicontinuity of the conformal dimension
  6. 6 Gromov-hyperbolic spaces
    1. 6.1 The proof of Theorem B

##  1 Introduction

The Ahlfors regular conformal gauge of a metric space $(X,d)$ is the set
$\mathcal{J}(X,d)$ of all metrics on $X$ that are quasisymmetric equivalent to
$d$. By definition a homeomorphism $F\colon(X,d_{X})\to(Y,d_{Y})$ is a
quasisymmetric equivalence if there exists a strictly increasing map
$\eta\colon[0,+\infty)\to[0,+\infty)$ with $\eta(0)=0$ such that

$\frac{d_{Y}(F(x),F(x^{\prime}))}{d_{Y}(F(x),F(x^{\prime\prime}))}\leq\eta\left(\frac{d_{X}(x,x^{\prime})}{d_{X}(x,x^{\prime\prime})}\right)$

for every $x,x^{\prime},x^{\prime\prime}\in X$ with
$d_{X}(x,x^{\prime\prime})>0$. The notion of quasisymmetric maps was
introduced in [TV80] and it has played an important role in the study of
quasiconformal structure on metric spaces. The _Ahlfors regular conformal
dimension_ of a metric space $(X,d)$ is defined as

$\text{CD}(X,d):=\inf\\{\text{HD}(X,d^{\prime})\text{ s.t.
}d^{\prime}\in\mathcal{J}(X,d)\\},$

where HD denotes the Hausdorff dimension. In general $\mathcal{J}(X,d)$ can be
empty, implying CD$(X,d)=+\infty$. On the other hand the conformal dimension
of any doubling, uniformly perfect metric space is always finite by [Hei01,
Corollary 14.5]. There is a special class of metric spaces that are doubling
and uniformly perfect: the class of perfect quasi-selfsimilar metric spaces
(Proposition 2.2).

######  Definition 1.

Let $\rho_{0}>0$ and $L_{0}\geq 1$. A compact metric space $(X,d)$ is said
$(L_{0},\rho_{0})$-quasi-selfsimilar (shortly $(L_{0},\rho_{0})$-q.s.s.) if
for every open ball $B(x,\rho)$ in $X$ with $0<\rho\leq\rho_{0}$ there is a
map $\Phi\colon\left(B(x,\rho),\frac{\rho_{0}}{\rho}\cdot d\right)\to X$ which
is $L_{0}$-biLipschitz and such that $\Phi(B(x,\rho))\supseteq
B(\Phi(x),\frac{\rho_{0}}{L_{0}})$.

The notation $\frac{\rho_{0}}{\rho}\cdot d$ means the metric obtained by
multiplying the original metric $d$ by the positive number
$\frac{\rho_{0}}{\rho}$. In other words a metric space is
$(L_{0},\rho_{0})$-quasi-selfsimilar if every ball of radius smaller than
$\rho_{0}$ is, up to rescaling it to the right size, biLipschitz comparable to
a ball of radius exactly $\rho_{0}$ of the same space.

This notion arises naturally in the study of limit sets of Gromov-hyperbolic
groups and semi-hyperbolic rational fractions, see for instance [Sul82],
[Haï07], [BK13] and [BM17]. Examples of spaces that are quasi-selfsimilar
include Lipschitz manifolds, simplicial complexes with a metric of fixed
constant curvature on each simplex, self-similar fractals, boundaries of
cocompact Gromov-hyperbolic spaces and Julia sets of semi-hyperbolic rational
fractions. Of course if one puts natural geometric constraints to each of the
above classes then it is possible to quantify the quasi-selfsimilarity
constants in terms of the constraints.

The purpose of this paper is to study the behaviour of the Ahlfors regular
conformal dimension on quasi-selfsimilar metric spaces under Gromov-Hausdorff
convergence.

######  Theorem A.

Let $(X_{n},d_{n})$ be a sequence of compact, $a_{0}$-uniformly perfect,
$(L_{0},\rho_{0})$-q.s.s. spaces. Suppose it converges in the Gromov-Hausdorff
sense to $(X_{\infty},d_{\infty})$. Then
$\textup{CD}(X_{\infty},d_{\infty})\geq\limsup_{n\to+\infty}\textup{CD}(X_{n},d_{n})$.

For quasi-selfsimilar spaces uniform perfectness is quantitatively equivalent
to a uniform lower bound of the diameter of the balls
$B(\Phi(x),\frac{\rho_{0}}{L_{0}})$ appearing in Definition 1, see Proposition
2.2.
Theorem A is false if the spaces are not $(L_{0},\rho_{0})$-q.s.s.: the
sequence $X_{n}=[0,1/n]\subseteq\mathbb{R}$ converges in the Gromov-Hausdorff
sense to $X_{\infty}=\\{0\\}$, but CD$(X_{n},d_{E})=1$ for every $n$ while
CD$(X_{\infty},d_{E})=0$. Here $d_{E}$ is the standard Euclidean metric.
Moreover the upper semicontinuity in Theorem A cannot be improved to
continuity in general.

######  Example 1.

Let $X_{n}$ be the set built in this way: we start with $[0,1]$ and we remove
the central segment of length $\frac{1}{2n+1}$. We do the same for each of the
two remaining parts. We continue this procedure infinitely many times and we
call $X_{n}$ the resulting metric space endowed with the Euclidean metric
$d_{E}$. For instance $X_{1}$ is the standard Cantor set. The sequence $X_{n}$
converges to $X_{\infty}=[0,1]$ in the Gromov-Hausdorff sense. However
CD$(X_{n},d_{E})=0$ for every $n$ by [DS97, Proposition 15.11] (see also
[Pia11, Theorem 2.16]), while CD$(X_{\infty},d_{E})=1$.

On the other hand we have continuity in a particular setting. In [Cav21] the
author studied the class $\mathcal{M}(\delta,D)$ of triples $(X,x,\Gamma)$,
where $X$ is a proper $\delta$-hyperbolic metric space, $\Gamma$ is a
discrete, non-elementary, quasiconvex-cocompact, torsion-free group of
isometries of $X$ with codiameter bounded above by $D$ and $x$ belongs to the
quasiconvex hull of the limit set $\Lambda(\Gamma)$. We refer to Section 6 for
more details about these terms. One of the main results of [Cav21] is the
closeness of $\mathcal{M}(\delta,D)$ under equivariant pointed Gromov-
Hausdorff convergence. Under this convergence it is possible to prove that the
limit sets $\Lambda(\Gamma_{n})$ and $\Lambda(\Gamma_{\infty})$ are
quasisymmetric equivalent for $n$ big enough (see Section 6.1). As a
consequence we get the following.

######  Theorem B.

Let $(X_{n},x_{n},\Gamma_{n})\subseteq\mathcal{M}(\delta,D)$ be a sequence of
triples converging in the equivariant pointed Gromov-Hausdorff sense to
$(X_{\infty},x_{\infty},\Gamma_{\infty})$. Then, for suitable metrics, the
sequence $\Lambda(\Gamma_{n})$ is uniformly perfect and uniformly q.s.s. and
converges in the Gromov-Hausdorff sense to $\Lambda(\Gamma_{\infty})$.
Moreover
$\textup{CD}(\Lambda(\Gamma_{\infty}))=\lim_{n\to+\infty}\textup{CD}(\Lambda(\Gamma_{n})).$

Here the Ahlfors regular conformal dimension is computed with respect to any
visual metric on the limit sets, see Proposition 6.3 and the discussion below.
Motivated by Theorem B, we propose the following question.

######  Question 1.

Are there conditions on the metric spaces $X_{n}$ that ensures continuity of
the conformal dimension under Gromov-Hausdorff convergence?

It is useful to consider the following example (the author thanks M.Murugan
for bringing the reference [Tys00] to his attention).

######  Example 2.

Let us repeat the construction of Example 1 in dimension $2$. For every $n$ we
define $X_{n}$ in the following way: we start with $[0,1]^{2}$, we divide it
into $(\frac{1}{2n+1})^{2}$ squares and we delete the central one. Now we do
the same for every remaining squares. We repeat the procedure infinitely many
times. We endow $X_{n}$ with the Euclidean metric $d_{E}$. For instance the
space $X_{1}$ is the standard Sierpinski carpet. The sequence $(X_{n},d_{E})$
converges in the Gromov-Hausdorff sense to $X_{\infty}=[0,1]^{2}$, whose
Ahlfors regular conformal dimension is $2$. In this case we have
$\lim_{n\to+\infty}\textup{CD}(X_{n},d_{E})=2$ by a well-known argument (see
for instance [Tys00, Theorem 3.4 and Example 3.2]).

We will briefly discuss Question 1 and the example above at the end of Section
5.

##  2 Preliminaries

We denote a metric space by $(X,d)$. The open (resp. closed) ball of center
$x\in X$ and radius $\rho>0$ is denoted by $B(x,\rho)$ (resp.
$\overline{B}(x,\rho)$).
Given $r>0$ and $Y\subseteq X$ we say that a subset $S$ of $Y$ is
$r$-separated if $d(x,y)>r$ for all $x,y\in S$, while a subset $N$ of $Y$ is a
$r$-net if for all $y\in Y$ there exists $x\in N$ such that $d(x,y)\leq r$. It
is straightforward from the definitions that a maximal $r$-separated subset of
$Y$ is a $r$-net.
A metric space $(X,d)$ is said $D$-doubling if the cardinality of any
$\frac{\rho}{2}$-separated subset inside any ball of radius $\rho$ is at most
$D$.
A metric space $(X,d)$ is perfect if it has no isolated points, while it is
said $a$-uniformly perfect, $0<a<1$, if $\overline{B}(x,\rho)\setminus
B(x,a\cdot\rho)\neq\emptyset$ for all $x\in X$ and $0\leq\rho<\text{Diam}(X)$,
where Diam denotes the diameter. Clearly every uniformly perfect metric space
is perfect.
Gromov-Hausdorff convergence will be always considered in the class of compact
metric spaces. With the notation
$(X_{n},d_{n})\underset{\text{GH}}{\longrightarrow}(X_{\infty},d_{\infty})$ we
mean that the sequence of compact metric spaces $(X_{n},d_{n})$ converges in
the Gromov-Hausdorff sense to the compact metric space
$(X_{\infty},d_{\infty})$.

###  2.1 Ahlfors regular spaces

A metric space $(X,d)$ is said $(A,s)$-Ahlfors regular, for given $A,s\geq 0$,
if there is a measure $\mu$ on $X$ satisfying

$\frac{1}{A}\cdot\rho^{s}\leq\mu(B(x,\rho))\leq A\cdot\rho^{s}$

for all $x\in X$ and all $0<\rho\leq\text{Diam}(X)$. The following lemma is
classical.

######  Lemma 2.1.

Let $A_{0}\geq 1$ and $s_{0}>0$. Then there exists
$0<a_{0}=a_{0}(A_{0},s_{0})<1$ such that every $(A,s)$-Ahlfors regular metric
space $(X,d)$ with $A\leq A_{0}$ and $s\geq s_{0}$ is $a_{0}$-uniformly
perfect.

###### Proof.

We claim that $(X,d)$ is $a$-uniformly perfect for all $a<A^{-2/s}$. Indeed
for any such $a$, for every $x\in X$ and every $0<\rho\leq\text{Diam}(x)$ we
have

$\mu(\overline{B}(x,\rho))\geq\frac{1}{A}\rho^{s},\qquad\mu(B(x,a\cdot\rho))\leq
A\cdot a^{s}\rho^{s}<\frac{1}{A}\cdot\rho^{s}.$

Hence $\mu(\overline{B}(x,\rho)\setminus B(x,a\cdot\rho))>0$ and in particular
this set is not empty. It is then clear we can choose any
$a_{0}<A_{0}^{-2/s_{0}}$. ∎

###  2.2 Quasi-selfsimilar spaces

We collect now some basic properties of quasi-selfsimilar metric spaces, see
Definition 1. We say a quasi-selfimilar metric space $(X,d)$ has diameters
bounded below by some $c_{0}>0$ if the ball
$B(\Phi(x),\frac{\rho_{0}}{L_{0}})$ that appears in Definition 1 has diameter
$\geq c_{0}$ for every $x\in X$ and every $0<\rho\leq\rho_{0}$.

######  Proposition 2.2 (Compare with [Pia11, Lemma 2.2 and Proposition 2.3]).

Let $(X,d)$ be a quasi-selfsimilar metric space as in Definition 1. Then:

  * (i)

it is doubling;

  * (ii)

if it is perfect then it is uniformly perfect;

  * (iii)

it is uniformly perfect if and only if it has diameters bounded below,
quantitatively in terms of the relative constants and the diameter of $X$.

###### Proof.

If (i) is not true then for every $n\in\mathbb{N}$ there exist $x_{n}\in X$
and $\rho_{n}>0$ such that there is a $\frac{\rho_{n}}{2}$-separated set
inside $B(x_{n},\rho_{n})$ of cardinality $\geq n$. Up to pass to a
subsequence we can suppose
$\lim_{n\to+\infty}\rho_{n}=\rho_{\infty}\in[0,+\infty)$ and that $x_{n}$
converges to $x_{\infty}\in X$. If $\rho_{\infty}>0$ then $X$ is not totally
bounded, a contradiction. If $\rho_{\infty}=0$ we can find $L_{0}$-biLipschitz
maps $\Phi_{n}\colon\left(B(x_{n},\rho_{n}),\frac{\rho_{0}}{\rho_{n}}\cdot
d\right)\to X$ such that $\Phi_{n}(B(x_{n},\rho_{n}))\supseteq
B(\Phi_{n}(x_{n}),\frac{\rho_{0}}{L_{0}})$ for every $n$ big enough. Hence
there exists a $(\rho_{0}/2L_{0})$-separated set inside
$B(\Phi_{n}(x_{n}),\frac{\rho_{0}}{L_{0}})$ with cardinality $\geq n$. Once
again this contradicts the compactness of $X$.

Now we show (ii). Since $X$ is perfect and compact we have the following
property, as in [Pia11, Lemma 2.2]: for all $\rho>0$ there exists $d(\rho)>0$
such that Diam$(B(x,\rho))\geq d(\rho)$ for every $x\in X$. Suppose now $X$ is
not uniformly perfect: then for all $n\in\mathbb{N}$ there exist $x_{n}\in X$
and $0<\rho_{n}\leq\text{Diam}(X)$ such that
$\overline{B}(x_{n},\rho_{n})\setminus B(x_{n},\rho_{n}/n)=\emptyset$. Up to
take a subsequence we can suppose that $x_{n}$ converges to $x_{\infty}$ and
$\rho_{n}$ converges to $\rho_{\infty}$. Suppose first $\rho_{\infty}>0$. Let
$y$ be a point inside $B(x_{\infty},\rho_{\infty})$. It also belongs to
$B(x_{n},\rho_{n})$ for $n$ big enough, and so it belongs to
$B(x_{n},\rho_{n}/n)$. In other words $d(x_{\infty},y)\leq
d(x_{\infty},x_{n})+\frac{\rho_{n}}{n}$ for every $n$ big enough, i.e.
$d(x_{\infty},y)=0$ and $x_{\infty}$ is an isolated point. This shows that $X$
is not perfect, a contradiction.
Suppose now $\rho_{\infty}=0$. For all $n$ big enough we take the map
$\Phi_{n}\colon B(x_{n},\rho_{n})\to X$ given by Definition 1. Therefore

$B\left(\Phi_{n}(x_{n}),\frac{r_{0}}{L_{0}}\right)\subseteq\Phi_{n}(B(x_{n},\rho_{n}))=\Phi_{n}\left(B\left(x_{n},\frac{\rho_{n}}{n}\right)\right).$

From one side we have
Diam$\left(B\left(\Phi(x_{n}),\frac{r_{0}}{L_{0}}\right)\right)\geq
d(\frac{r_{0}}{L_{0}})>0$ for every $n$. On the other hand

$\text{Diam}\left(\Phi_{n}\left(B\left(x_{n},\frac{\rho_{n}}{n}\right)\right)\right)\leq
L_{0}\cdot\frac{\rho_{0}}{\rho_{n}}\cdot\frac{2\rho_{n}}{n}\underset{n\to+\infty}{\longrightarrow}0,$

which is a contradiction.

Finally we prove (iii). Let us suppose $X$ has diameters bounded below by
$c_{0}>0$. We fix $x\in X$ and $0<\rho\leq\rho_{0}$. We take the map
$\Phi\colon B(x,\rho)\to X$ given by the definition of quasi-selfsimilarity.
Since $\Phi(B(x,\rho))$ contains a set with diameter $\geq c_{0}$, then there
exists $y\in B(x,\rho)$ such that $d(\Phi(x),\Phi(y))\geq\frac{c_{0}}{2}$.
Therefore
$d(x,y)\geq\frac{1}{L_{0}}\cdot\frac{\rho}{\rho_{0}}\cdot\frac{c_{0}}{2}=a(L_{0},\rho_{0},c_{0})\cdot\rho$,
with $0<a(L_{0},\rho_{0},c_{0})=:a<1$. If $\rho$ is bigger than $\rho_{0}$
then we apply what said above to $\rho_{0}$ finding $B(x,\rho_{0})\setminus
B(x,a\cdot\rho_{0})\neq\emptyset$, so $B(x,\rho)\setminus
B(x,\frac{a\cdot\rho_{0}}{\rho}\cdot\rho)\neq\emptyset$. Since $X$ is compact
we have
$\frac{a\cdot\rho_{0}}{\rho}\geq\frac{a\cdot\rho_{0}}{\text{Diam}(X)}=:a_{0}>0$,
showing that $X$ is $a_{0}$-uniformly perfect and $a_{0}$ depends only on
$L_{0},\rho_{0},c_{0}$ and the diameter of $X$.
Viceversa if $X$ is $a_{0}$-uniformly perfect then
$\overline{B}(\Phi(x),\frac{\rho_{0}}{L_{0}})\setminus
B(\Phi(x),a_{0}\cdot\frac{\rho_{0}}{L_{0}})\neq\emptyset$ for all $x\in X$ and
$\Phi$ as in Definition 1. Therefore
Diam$(B(\Phi(x),\frac{\rho_{0}}{L_{0}}))\geq
a_{0}\cdot\frac{\rho_{0}}{L_{0}}=:c_{0}$. ∎

We remark that the diameter bounded below condition is part of the definition
of quasi-selfsimilar spaces in [Kle06] and in [Pia11]. When an upper bound on
the diameter of the metric space is fixed this condition is equivalent to
bounded uniform perfectness of the metric space by Proposition 2.2. For
instance in the context of Theorem A there is a uniform upper bound on the
diameter of the spaces $X_{n}$, so the spaces $X_{n}$ have all diameter
bounded below by some $c_{0}>0$ if and only if they are all $a_{0}$-uniformly
perfect for some $0<a_{0}<1$.

##  3 Combinatorial modulus

It is known that the conformal dimension of a metric space is closely related
to the combinatorial modulus, see for instance [BK13], [Pia12], [MT10] and the
references therein. In this section we recall the definition of combinatorial
modulus and we prove some technical lemma.

From now on we fix a $D$-doubling metric space $(X,d)$. For every
$k\in\mathbb{N}$ we choose a $10^{-k}$-net $X_{k}$ of $X$. To simplify the
notation, given a real number $\lambda>0$ and $k\in\mathbb{N}$, we will denote
by $B_{\lambda,k}(x)$ the open ball of center $x$ and radius $\lambda\cdot
10^{-k}$, namely $B(x,\lambda\cdot 10^{-k})$. The same convention holds for
closed balls.

A $(\lambda,k)$-path is a finite collection $\gamma=\\{q_{j}\\}_{j=0}^{M}$ of
elements of $X_{k}$ satisfying
$\overline{B}_{\lambda,k}(q_{j})\cap\overline{B}_{\lambda,k}(q_{j+1})\neq\emptyset$
for all $j=0,\ldots,M-1$. The points $q_{0}$ and $q_{M}$ are called
respectively the starting and the ending point of the path.

Given two subsets $E,F\subseteq X$ we denote by $P_{\lambda,k}(E,F)$ the set
of $(\lambda,k)$-paths with starting point in $E$ and ending point in $F$. We
denote by $\mathcal{A}_{\lambda,k}(E,F)$ the set of admissible functions, i.e.
functions $f\colon X_{k}\to[0,+\infty)$ such that $\sum_{i=0}^{M}f(q_{i})\geq
1$ for every $\\{q_{i}\\}_{i=0}^{M}\in P_{\lambda,k}(E,F)$.

Given a real number $p\geq 0$ we define

$p\text{-Mod}_{\lambda,k}(E,F)=\inf_{f\in\mathcal{A}_{\lambda,k}(E,F)}\sum_{q\in
X_{k}}f(q)^{p},$

and we call it the $p$-modulus of the couple $(E,F)$ at level $(\lambda,k)$.
The infimum is actually realized: any admissible function realizing the
minimum is said optimal. If there are no $(\lambda,k)$-paths joining $E$ and
$F$ we set $p\text{-Mod}_{\lambda,k}(E,F)=0$.

######  Lemma 3.1.

If $E^{\prime}\subseteq E$ and $F^{\prime}\subseteq F$ then
$p\textup{-Mod}_{\lambda,k}(E^{\prime},F^{\prime})\leq
p\textup{-Mod}_{\lambda,k}(E,F)$.

###### Proof.

If $P_{\lambda,k}(E^{\prime},F^{\prime})=\emptyset$ then the result is trivial
by definition. Otherwise every path in $P_{\lambda,k}(E^{\prime},F^{\prime})$
belongs to $P_{\lambda,k}(E,F)$. This implies that
$\mathcal{A}_{\lambda,k}(E,F)\subseteq\mathcal{A}_{\lambda,k}(E^{\prime},F^{\prime})$,
and the result follows from the definition. ∎

Let $1\leq L_{1}<L_{2}$ be two real numbers. For every $i\in\mathbb{N}$ and
every point $y\in X_{i}$ we set

$p\text{-Mod}_{\lambda,k}^{L_{1},L_{2}}(y):=p\text{-Mod}_{\lambda,i+k}(\overline{B}_{L_{1},i}(y),X\setminus
B_{L_{2},i}(y)).$

We remark that this is a modulus at level $(\lambda,i+k)$. Finally we define

$p\text{-Mod}_{\lambda,k}^{L_{1},L_{2}}(X)=\sup_{i\in\mathbb{N}}\sup_{y\in
X_{i}}p\text{-Mod}_{\lambda,k}^{L_{1},L_{2}}(y).$

We want to control how this quantity changes when $L_{1},L_{2}$ and $\lambda$
change. We recall that $D$ denotes the doubling constant of $X$.

######  Lemma 3.2.

(cp. [Pia11], Lemma 4.4) Let $k,\lambda,p$ be fixed quantities as above. Let
$1\leq L_{1}^{\prime}\leq L_{1}<L_{2}\leq L_{2}^{\prime}$. Then there exist
$\ell\in\mathbb{N}$ and $C>0$, depending only on
$L_{1},L_{1}^{\prime},L_{2},L_{2}^{\prime}$ and $D$, such that

$p\textup{-Mod}_{\lambda,k}^{L_{1}^{\prime},L_{2}^{\prime}}(X)\leq
p\textup{-Mod}_{\lambda,k}^{L_{1},L_{2}}(X)$

and

$p\textup{-Mod}_{\lambda,k+\ell}^{L_{1},L_{2}}(X)\leq C\cdot
p\textup{-Mod}_{\lambda,k}^{L_{1}^{\prime},L_{2}^{\prime}}(X).$

###### Proof.

For every $y\in X_{i}$, $i\in\mathbb{N}$, we have
$\overline{B}_{L_{1}^{\prime},i}(y)\subseteq\overline{B}_{L_{1},i}(y)$ and
$X\setminus B_{L_{2}^{\prime},i}(y)\subseteq X\setminus B_{L_{2},i}(y)$, so
the first inequality follows by Lemma 3.1.
In order to prove the second inequality we define $\ell$ as the minimum
integer satisfying
$10^{-\ell}\leq\frac{L_{2}-L_{1}}{L_{2}^{\prime}+L_{1}^{\prime}}$. We fix
$y\in X_{i}$ for some $i\in\mathbb{N}$ and we consider the set
$X_{i+\ell}(y)=\\{z\in X_{i+\ell}\text{ s.t.
}B_{L_{1}^{\prime},i+\ell}(z)\cap\overline{B}_{L_{1},i}(y)\neq\emptyset\\}$.
We fix any $(\lambda,i+\ell+k)$-path $\gamma=\\{q_{j}\\}_{j=0}^{M}$ joining
$\overline{B}_{L_{1},i}(y)$ and $X\setminus B_{L_{2},i}(y)$. This means in
particular that $d(y,q_{0})\leq L_{1}\cdot 10^{-i}$ and $d(y,q_{M})\geq
L_{2}\cdot 10^{-i}$. We can find $z\in X_{i+\ell}$ such that $d(z,q_{0})\leq
10^{-i-\ell}$, so by definition $z\in X_{i+\ell}(y)$. We claim that the
$(\lambda,i+\ell+k)$-path $\gamma$ joins
$\overline{B}_{L_{1}^{\prime},i+\ell}(z)$ and $X\setminus
B_{L_{2}^{\prime},i+\ell}(z)$. Indeed we know that $d(z,q_{0})\leq
10^{-i-\ell}\leq L_{1}^{\prime}\cdot 10^{-i-\ell}$. Moreover $d(z,y)\leq
L_{1}\cdot 10^{-i}+L_{1}^{\prime}\cdot 10^{-i-\ell}$. Therefore

$d(z,q_{N})\geq L_{2}\cdot 10^{-i}-L_{1}\cdot 10^{-i}-L_{1}^{\prime}\cdot
10^{-i-\ell}\geq L_{2}^{\prime}\cdot 10^{-i-\ell}$

by the choice of $\ell$. This means that any path $\gamma\in
P_{\lambda,i+\ell+k}(\overline{B}_{L_{1},i}(y),X\setminus B_{L_{2},i}(y))$
belongs to
$P_{\lambda,i+\ell+k}(\overline{B}_{L_{1}^{\prime},i+\ell}(z),X\setminus
B_{L_{2}^{\prime},i+\ell}(z))$ for some $z\in X_{i+\ell}(y)$.
For each $z\in X_{i+\ell}(y)$ we take optimal functions
$f_{z}\in\mathcal{A}_{\lambda,i+\ell+k}(\overline{B}_{L_{1}^{\prime},i+\ell}(z),X\setminus
B_{L_{2}^{\prime},i+\ell}(z))$ and we define the function $f\colon
X_{i+\ell+k}\to[0,+\infty)$ as

$f(q)=\max_{z\in X_{i+\ell}(y)}f_{z}(q).$

We claim
$f\in\mathcal{A}_{\lambda,i+\ell+k}(\overline{B}_{L_{1},i}(y),X\setminus
B_{L_{2},i}(y)).$ Indeed every path $\\{q_{j}\\}_{j=0}^{M}\in
P_{\lambda,i+\ell+k}(\overline{B}_{L_{1},i}(y),X\setminus B_{L_{2},i}(y))$
belongs to $P_{\lambda,i+\ell+k}(B_{L_{1}^{\prime},i+\ell}(z),X\setminus
B_{L_{2}^{\prime},i+\ell}(z))$ for some $z\in X_{i+\ell}(y)$, therefore

$\sum_{j=0}^{M}f(q_{j})\geq\sum_{j=0}^{M}f_{z}(q_{j})\geq 1.$

Finally we have:

$\displaystyle\sum_{q\in X_{i+\ell+k}}f(q)^{p}$ $\displaystyle=\sum_{q\in
X_{i+\ell+k}}\max_{z\in X_{i+\ell}(y)}f_{z}(q)^{p}\leq\sum_{z\in
X_{i+\ell}(y)}\sum_{q\in X_{i+\ell+k}}f_{z}(q)^{p}$ $\displaystyle=\sum_{z\in
X_{i+\ell}(y)}p\text{-Mod}^{L_{1}^{\prime},L_{2}^{\prime}}_{\lambda,k}(z)\leq
C\cdot p\text{-Mod}^{L_{1}^{\prime},L_{2}^{\prime}}_{\lambda,k}(X),$

where $C$ is a constant depending only on the doubling constant $D$, on $\ell$
and on $L_{1}^{\prime}$. This shows that

$p\text{-Mod}^{L_{1},L_{2}}_{\lambda,k+\ell}(y)\leq C\cdot
p\text{-Mod}^{L_{1}^{\prime},L_{2}^{\prime}}_{\lambda,k}(X).$

Since this is true for every $y\in X_{i}$ and for every $i$ we get

$p\text{-Mod}^{L_{1},L_{2}}_{\lambda,k+\ell}(X)\leq C\cdot
p\text{-Mod}^{L_{1}^{\prime},L_{2}^{\prime}}_{\lambda,k}(X).$

∎

######  Lemma 3.3.

Let $k\in\mathbb{N}$, $p\geq 0$, $1\leq L_{1}<L_{2}$ and
$2<\lambda\leq\lambda^{\prime}$. Then there exist $\ell\in\mathbb{N}$ and
$C>0$ depending only on $\lambda,\lambda^{\prime},D$ such that

$p\textup{-Mod}^{L_{1},L_{2}}_{\lambda,k}(X)\leq
p\textup{-Mod}^{L_{1},L_{2}}_{\lambda^{\prime},k}(X)$

and

$p\textup{-Mod}^{L_{1},L_{2}}_{\lambda^{\prime},k+\ell}(X)\leq C\cdot
p\textup{-Mod}^{L_{1},L_{2}}_{\lambda,k}(X)$

for all $k>k_{0}=\log_{10}\left(\frac{2}{L_{2}-L_{1}}\right)$.

###### Proof.

For every $y\in X_{i}$, $i\in\mathbb{N}$, we have

$P_{\lambda,k}(\overline{B}_{L_{1},i}(y),X\setminus B_{L_{2},i}(y))\subseteq
P_{\lambda^{\prime},k}(\overline{B}_{L_{1},i}(y),X\setminus B_{L_{2},i}(y)).$

Therefore arguing as in the proof of Lemma 3.1 we get

$p\text{-Mod}_{\lambda,k}^{L_{1},L_{2}}(y)\leq
p\text{-Mod}_{\lambda^{\prime},k}^{L_{1},L_{2}}(y).$

Taking the supremum on $i\in\mathbb{N}$ and $y\in X_{i}$ we obtain the first
inequality.
In order to show the second inequality we define $\ell$ as the smallest
integer such that $\lambda^{\prime}\cdot 10^{-\ell}\leq\frac{\lambda}{2}-1$.
It is well defined since $\lambda>2$. We restrict the attention to the
integers $k$ bigger than $k_{0}$, so that $10^{-k}<\frac{L_{2}-L_{1}}{2}$.
We fix $y\in X_{i}$, $i\in\mathbb{N}$, and a
$(\lambda^{\prime},i+k+\ell)$-path $\gamma=\\{q_{j}\\}_{j=0}^{M}$ joining
$\overline{B}_{L_{1},i}(y)$ to $X\setminus B_{L_{2},i}(y)$. For every
$j=0,\ldots,M$ we take a point $\tilde{q}_{j}\in X_{i+k}$ such that
$d(q_{j},\tilde{q}_{j})\leq 10^{-i-k}$. We claim
$\tilde{\gamma}=\\{\tilde{q}_{j}\\}_{j=0}^{M}$ is a $(\lambda,i+k)$-path
joining $\overline{B}_{L_{1}^{\prime},i}(y)$ to $X\setminus
B_{L_{2}^{\prime},i}(y)$, where $L_{1}^{\prime}=L_{1}+10^{-k}$ and
$L_{2}^{\prime}=L_{2}-10^{-k}$. Indeed we have:

$d(y,\tilde{q}_{0})\leq d(y,q_{0})+d(q_{0},\tilde{q}_{0})\leq L_{1}\cdot
10^{-i}+10^{-i-k}=L_{1}^{\prime}\cdot 10^{-i},$ $d(y,\tilde{q}_{M})\geq
d(y,q_{M})-d(q_{M},\tilde{q}_{M})\geq L_{2}\cdot
10^{-i}-10^{-i-k}=L_{2}^{\prime}\cdot 10^{-i}$

and

$\displaystyle d(\tilde{q}_{j},\tilde{q}_{j+1})$ $\displaystyle\leq
d(\tilde{q}_{j},q_{j})+d(q_{j},q_{j+1})+d(q_{j+1},\tilde{q}_{j+1})$
$\displaystyle\leq 2\cdot 10^{-i-k}+2\lambda^{\prime}\cdot 10^{-i-k-\ell}$
$\displaystyle\leq 2\cdot 10^{-i-k}+2\left(\frac{\lambda}{2}-1\right)\cdot
10^{-i-k}$ $\displaystyle=\lambda\cdot 10^{-i-k}$

for every $j=0,\ldots,M-1$. Observe that the condition on $k$ implies
$L_{1}^{\prime}<L_{2}^{\prime}$. We are ready to compare the combinatorial
moduli. We take an optimal function
$\tilde{f}\in\mathcal{A}_{\lambda,i+k}(\overline{B}_{L_{1}^{\prime},i}(y),X\setminus
B_{L_{2}^{\prime},i}(y))$ and we define the function $f\colon
X_{i+k+\ell}\to[0,+\infty)$ by

$f(q):=\max\\{\tilde{f}(\tilde{q})\text{ s.t. }\tilde{q}\in X_{i+k}\text{ and
}d(q,\tilde{q})\leq 10^{-i-k}\\}.$

First of all we show that
$f\in\mathcal{A}_{\lambda^{\prime},i+k+\ell}(\overline{B}_{L_{1},i}(y),X\setminus
B_{L_{2},i}(y))$. Indeed we have seen that given any
$(\lambda^{\prime},i+k+\ell)$-path $\\{q_{j}\\}_{j=0}^{M}$ joining
$\overline{B}_{L_{1},i}(y)$ to $X\setminus B_{L_{2},i}(y)$ there is an
associated $(\lambda,i+k)$-path $\\{\tilde{q}_{j}\\}_{j=0}^{M}$ joining
$\overline{B}_{L_{1}^{\prime},i}(y)$ to $X\setminus B_{L_{2}^{\prime},i}(y)$
such that $d(q_{j},\tilde{q}_{j})\leq 10^{-i-k}$ for every $j=0,\ldots,M$.
Therefore by definition of $f$ we have

$\sum_{j=0}^{M}f(q_{j})\geq\sum_{j=0}^{M}\tilde{f}(\tilde{q}_{j})\geq 1.$

Finally we observe that

$\displaystyle p\text{-Mod}^{L_{1},L_{2}}_{\lambda^{\prime},k+\ell}(y)$
$\displaystyle\leq\sum_{q\in X_{i+k+\ell}}f^{p}(q)\leq
C^{\prime}\cdot\sum_{\tilde{q}\in X_{i+k}}\tilde{f}^{p}(\tilde{q})$
$\displaystyle=C^{\prime}\cdot
p\text{-Mod}^{L_{1}^{\prime},L_{2}^{\prime}}_{\lambda,k}(y)\leq
C^{\prime}\cdot p\text{-Mod}^{L_{1}^{\prime},L_{2}^{\prime}}_{\lambda,k}(X),$

where $C^{\prime}$ is a constant depending only on $D$. Since this is true for
every $y\in X_{i}$ and for every $i\in\mathbb{N}$ we conclude that

$p\text{-Mod}^{L_{1},L_{2}}_{\lambda^{\prime},k+\ell}(X)\leq C^{\prime}\cdot
p\text{-Mod}^{L_{1}^{\prime},L_{2}^{\prime}}_{\lambda,k}(X).$

This inequality is true for all $k\geq k_{0}$. Choosing
$L_{1}^{\prime\prime}=L_{1}+10^{-k_{0}}$ and
$L_{2}^{\prime\prime}=L_{2}-10^{-k_{0}}$ one concludes, using the easy
inequality in Lemma 3.2, that

$p\text{-Mod}^{L_{1},L_{2}}_{\lambda^{\prime},k+\ell}(X)\leq C^{\prime}\cdot
p\text{-Mod}^{L_{1}^{\prime\prime},L_{2}^{\prime\prime}}_{\lambda,k}(X)$

for all $k\geq k_{0}$. An application of the non-trivial inequality of Lemma
3.2 concludes the proof. ∎

In order to normalize the notation from now on we choose for technical reasons
$\lambda=10$, $L_{1}=3$, $L_{2}=4$ and we set

$p\text{-Mod}_{k}(y):=p\text{-Mod}_{10,i+k}(\overline{B}_{3,i}(y),X\setminus
B_{4,i}(y))$

for every $i\in\mathbb{N}$ and every point $y\in X_{i}$. In the same way we
put

$p\text{-Mod}_{k}(X)=\sup_{i\in\mathbb{N}}\sup_{y\in
X_{i}}p\text{-Mod}_{k}(y).$

##  4 Combinatorial modulus on quasi-selfsimilar spaces

In this section we consider the class of quasi-selfsimilar metric spaces as
defined in Definition 1. On these spaces the computation of the combinatorial
moduli is easier. Before that we need an easy result.

######  Lemma 4.1.

If $X$ is $(L_{0},\rho_{0})$-q.s.s. then it is $(L_{0},\rho_{1})$-q.s.s. for
every $0<\rho_{1}\leq\rho_{0}$.

###### Proof.

We fix $x\in X$ and $0<\rho\leq\rho_{1}$. We apply the definition of
$(L_{0},\rho_{0})$-quasi-selfsimilarity to the ball
$B(x,\frac{\rho_{0}}{\rho_{1}}\rho)$: we can find a $L_{0}$-biLipschitz map
$\Phi\colon\left(B(x,\frac{\rho_{0}}{\rho_{1}}\rho),\frac{\rho_{1}}{\rho}\cdot
d\right)\to X$ such that $\Phi(B(x,\frac{\rho_{0}}{\rho_{1}}\rho))\supseteq
B(\Phi(x),\frac{\rho_{0}}{L_{0}})$. Then it is straightforward to see that the
restriction of $\Phi$ to $\left(B(x,\rho),\frac{\rho_{1}}{\rho}\cdot d\right)$
is still $L_{0}$-biLipschitz. We now take any point $z\in
B(\Phi(x),\frac{\rho_{1}}{L_{0}})$. We know there exists a point $y\in
B(x,\frac{\rho_{0}}{\rho_{1}}\rho)$ such that $z=\Phi(y)$. By the property of
$\Phi$ we get $d(x,y)<\rho$. This shows that $\Phi(B(x,\rho))\supseteq
B(\Phi(x),\frac{\rho_{1}}{L_{0}})$. ∎

Let $X$ be a $(L_{0},\rho_{0})$-q.s.s. space. We denote by $i_{0}$ the
smallest integer such that $2(L_{0}+5)^{2}\cdot 10^{-i_{0}}\leq\rho_{0}$. We
define $I_{0}=\\{i\in\mathbb{N}\text{ s.t. }(L_{0}+5)\cdot 10^{-i}\geq
10^{-i_{0}}\\}$. Observe that the set $I_{0}$ is of the form
$\\{1,\ldots,n_{0}\\}$ where $n_{0}$ depends only on $L_{0}$ and $\rho_{0}$.
We set

$p\text{-Mod}_{k}(X,n_{0}):=\sup_{i\leq n_{0}}\sup_{y\in
X_{i}}p\text{-Mod}_{k}(y).$

The following is the main result of the section: it allows to use _the fixed
sizes up to $n_{0}$_ to estimate the combinatorial modulus. Since the explicit
doubling constant of our metric space plays an important role we sometimes add
it in the definition: we say a metric space is $(L_{0},\rho_{0},D_{0})$-q.s.s.
if it is $(L_{0},\rho_{0})$-q.s.s. and $D_{0}$-doubling.

######  Proposition 4.2.

Let $X$ be $(L_{0},\rho_{0},D_{0})$-q.s.s. Then there exist a constant
$C_{0}\geq 1$ and an integer $\ell_{0}$, both depending only on $L_{0}$ and
$D_{0}$, such that

$p\textup{-Mod}_{k+\ell_{0}}(X,n_{0})\leq p\textup{-Mod}_{k+\ell_{0}}(X)\leq
C_{0}\cdot p\textup{-Mod}_{k}(X,n_{0})$

for every $k\in\mathbb{N}$.

###### Proof.

The first inequality is trivial since we are doing a supremum among less
elements. In order to show the second inequality we fix $i\in\mathbb{N}$ and
$y\in X_{i}$. Clearly we can suppose $i>n_{0}$. By Lemma 4.1 we know that $X$
is also $(L_{0},2(L_{0}+5)^{2}\cdot 10^{-i_{0}})$-q.s.s.. Since
$2(L_{0}+5)^{3}\cdot 10^{-i}<2(L_{0}+5)^{2}\cdot 10^{-i_{0}}$, then there is a
$L_{0}$-biLipschitz map

$\Phi\colon\left(B(y,2(L_{0}+5)^{3}\cdot
10^{-i}),\frac{10^{-i_{0}}}{(L_{0}+5)\cdot 10^{-i}}\cdot d\right)\to X.$

We choose a point $x\in X_{i_{0}}$ such that $d(x,\Phi(y))\leq 10^{-i_{0}}$.
We consider any $(10,i+k)$-path $\\{q_{j}\\}_{j=0}^{M}$ joining
$\overline{B}_{1,i}(y)$ to $X\setminus B_{(L_{0}+5)^{3},i}(y)$. This means

  * -

$q_{0}\in\overline{B}(y,10^{-i})$ and $q_{M}\notin B(y,(L_{0}+5)^{3}\cdot
10^{-i})$;

  * -

$\overline{B}(q_{j},10\cdot 10^{-i-k})\cap\overline{B}(q_{j+1},10\cdot
10^{-i-k})\neq\emptyset$ for every $j=0,\ldots,M-1$.

Suppose that $q_{j}\in B_{2(L_{0}+5)^{3},i}(y)$ for every $j=0,\ldots,M$. Then
we can choose a point $\tilde{q}_{j}\in X_{i_{0}+k-1}$ such that
$d(\tilde{q}_{j},\Phi(q_{j}))\leq 10^{-i_{0}-k+1}$ for every $j=0,\ldots,M$.
By the property of $\Phi$ we get
$d(\Phi(y),\Phi(q_{0}))\leq\frac{L_{0}}{L_{0}+5}\cdot 10^{-i_{0}}$ and
$d(\Phi(y),\Phi(q_{M}))\geq\frac{(L_{0}+5)^{2}}{L_{0}}\cdot 10^{-i_{0}}$.
Therefore we have:

$d(x,\tilde{q}_{0})\leq
d(x,\Phi(y))+d(\Phi(y),\Phi(q_{0}))+d(\Phi(q_{0}),\tilde{q}_{0})\leq 3\cdot
10^{-i_{0}},$ $d(x,\tilde{q}_{M})\geq
d(\Phi(y),\Phi(q_{M}))-d(x,\Phi(y))-d(\Phi(q_{M}),\tilde{q}_{M})\geq(L_{0}+3)10^{-i_{0}}\geq
4\cdot 10^{-i_{0}}.$

Moreover we know that $d(q_{j},q_{j+1})\leq 20\cdot 10^{-i-k}$ for every
$j=0,\ldots,M-1$. Therefore we get

$\displaystyle d(\tilde{q}_{j},\tilde{q}_{j+1})$ $\displaystyle\leq
d(\tilde{q}_{j},\Phi(q_{j}))+d(\Phi(q_{j}),\Phi(q_{j+1})+d(\Phi(q_{j+1}),\tilde{q}_{j+1})$
$\displaystyle\leq 10^{-i_{0}-k+1}+20\cdot 10^{-i_{0}-k}+10^{-i_{0}-k+1}$
$\displaystyle\leq 10\cdot 10^{-i_{0}-k+1}.$

In other words $\\{\tilde{q}_{j}\\}_{j=0}^{M}$ is a $(10,i_{0}+k-1)$-path
joining $\overline{B}_{3,i_{0}}(x)$ to $X\setminus B_{4,i_{0}}(x)$.
We take an optimal map
$\tilde{f}\in\mathcal{A}_{10,i_{0}+k-1}(\overline{B}_{3,i_{0}}(x),X\setminus
B_{4,i_{0}}(x))$. We define the map $f\colon X_{i+k}\to[0,+\infty)$ by

$f(q):=\max\\{\tilde{f}(\tilde{q}):\tilde{q}\in
X_{i_{0}+k-1}\cap\overline{B}(\Phi(q),10^{-i_{0}-k+1})\\}$

if $q\in\overline{B}(y,2(L_{0}+5)^{3}\cdot 10^{-i})$ and $0$ otherwise. We
want to show that $f\in\mathcal{A}_{i+k}(\overline{B}_{1,i}(y),X\setminus
B_{(L_{0}+5)^{3},i}(y))$. We consider any path $\\{q_{j}\\}_{j=0}^{M}\in
P_{10,i+k}(\overline{B}_{1,i}(y),X\setminus B_{(L_{0}+5)^{3},i}(y))$. First of
all we can extract the minimal subpath $\\{q_{j}\\}_{j=0}^{M^{\prime}}$ such
that $q_{M^{\prime}}\notin B_{(L_{0}+5)^{3},i}(y)$. Clearly if
$\sum_{j=0}^{M^{\prime}}f(q_{j})\geq 1$ then also $\sum_{j=0}^{M}f(q_{j})\geq
1$, so it is enough to check the admissibility condition on this minimal
subpath. For such a minimal subpath we can construct the path
$\\{\tilde{q}_{j}\\}_{j=0}^{M^{\prime}}$ as in the first part of the proof
since $q_{j}\in B_{2(L_{0}+5)^{3},i}(y)$ for every $j=0,\ldots,M^{\prime}$. By
definition it holds:

$\sum_{j=0}^{M^{\prime}}f(q_{j})\geq\sum_{j=0}^{M^{\prime}}\tilde{f}(\tilde{q}_{j})\geq
1.$

Moreover we have

$\sum_{q\in X_{i+k}}f(q)^{p}\leq D_{0}\cdot\sum_{\tilde{q}\in
X_{i_{0}+k-1}}\tilde{f}(\tilde{q})^{p}\leq D_{0}\cdot
p\text{-Mod}_{k-1}(x)\leq D_{0}\cdot p\text{-Mod}_{k-1}(X,n_{0}),$

since $x\in X_{i_{0}}$ and $1\leq i_{0}\leq n_{0}$ by definition. By the
arbitrariness of $i\in\mathbb{N}$ and $y\in X_{i}$ we conclude

$p\text{-Mod}_{10,k}^{1,(L_{0}+5)^{3}}(X)\leq D_{0}\cdot
p\text{-Mod}_{k-1}(X,n_{0})$

for every $k\geq 1$. Using Lemma 3.2 we obtain the second inequality, indeed:

$\displaystyle p\text{-Mod}_{k}(X)=p\text{-Mod}_{10,k}^{3,4}(X)$
$\displaystyle\leq C\cdot p\text{-Mod}_{10,k-\ell}^{1,(L_{0}+5)^{3}}(X)$
$\displaystyle\leq C\cdot D_{0}\cdot p\text{-Mod}_{k-\ell-1}(X,n_{0}),$

where $C$ and $\ell$ are constants depending only on $L_{0}$ and $D_{0}$. The
thesis follows choosing $C_{0}=C\cdot D_{0}$ and $\ell_{0}=\ell+1$.

∎

The Ahlfors regular conformal dimension of a compact, doubling, uniformly
perfect metric space $(X,d)$ coincides with the critical exponent of the
combinatorial modulus.

######  Theorem 4.3 ([Pia11, Theorem 4.5]).

Let $(X,d)$ be a compact, doubling, uniformly perfect metric space. Then

$\textup{CD}(X,d)=\inf\\{p\geq 0\textup{ s.t.
}\liminf_{k\to+\infty}p\textup{-Mod}_{k}(X)=0\\}.$

By Lemma 3.2 and Lemma 3.3, the right hand quantity does not depend on our
specific choices of $\lambda=10$, $L_{1}=3$ and $L_{2}=4$ in the definition of
$p\text{-Mod}_{k}(X)$: the critical exponent associated to any other
admissible choice of $\lambda,L_{1},L_{2}$ equals the Ahlfors regular
conformal dimension of $(X,d)$. Moreover, following again [Pia11] and [BK13],
in the quasi-selfsimilar setting it is possible to find a uniform estimate
which will be the key ingredient of the proof of Theorem A.

######  Proposition 4.4.

Let $(X,d)$ be a perfect $(L_{0},\rho_{0},D_{0})$-q.s.s. metric space and let
$p<\textup{CD}(X,d)$. Then there exists a constant $\lambda_{0}$ depending
only on $D_{0}$, $L_{0}$ and $p$ such that

$p\textup{-Mod}_{k}(X,n_{0})\geq\lambda_{0}>0$

for all $k>0$.

###### Proof.

$(X,d)$ is uniformly perfect and doubling by Proposition 2.2, so the Ahlfors
regular conformal dimension of $(X,d)$ can be computed as in Theorem 4.3. The
result follows by a submultiplicative estimate. [Pia11, Lemma 4.9] proves

$p\text{-Mod}_{10,k+h}^{1,4}(X)\leq C\cdot
p\text{-Mod}_{10,k}^{\frac{11}{10},\frac{39}{10}}(X)\cdot
p\text{-Mod}_{10,h}^{1,4}(X)$

for all $k,h\geq 0$. Here $C$ is a constant depending only on $p$ and $D_{0}$.
Applying Lemma 3.2 we get

$p\text{-Mod}_{k+h}(X)\leq C^{\prime}\cdot p\text{-Mod}_{k-\ell}(X)\cdot
p\text{-Mod}_{h}(X)$

for all $k\geq\ell$ and $h\geq 0$, where $C^{\prime}$ is a constant depending
only on $p$ and $D_{0}$ and $\ell$ is a universal constant. Let us denote by
$a_{k}$ the quantity $p\text{-Mod}_{k}(X)$. The inequality above is
$a_{k+h}\leq C^{\prime}\cdot a_{k-\ell}\cdot a_{h}$. By Theorem 4.3
$\liminf_{k\to+\infty}a_{k}>0$ since $p<\text{CD}(X,d)$. This implies that
$a_{k}\geq\frac{1}{C^{\prime}}$ for all $k>0$. Indeed, if there exists $k>0$
such that $C^{\prime}\cdot a_{k}<(1-\varepsilon)$ for some $\varepsilon>0$
then

$a_{n(k+\ell)}\leq C^{\prime}\cdot a_{k}\cdot
a_{(n-1)(k+\ell)}\leq\cdots\leq(1-\varepsilon)^{n}$

for all $n\in\mathbb{N}$. Therefore the subsequence
$\\{a_{n(k+\ell)}\\}_{n\in\mathbb{N}}$ would converge to $0$, which is a
contradiction. Hence we have found a constant $\lambda>0$ depending only on
$p$ and $D_{0}$ such that $a_{k}\geq\lambda$ for all $k>0$. An application of
Proposition 4.2 gives the thesis. ∎

##  5 Upper semicontinuity of the conformal dimension

Our scope is to study the behaviour of the Ahlfors regular conformal dimension
under Gromov-Hausdorff convergence. For technical reasons it is often useful
to study ultralimits instead of Gromov-Hausdorff limits. It essentially avoids
to extract converging subsequences. For more detailed notions on ultralimits
we refer to [DK18] and [CS21]. A non-principal ultrafilter $\omega$ is a
finitely additive measure on $\mathbb{N}$ such that $\omega(A)\in\\{0,1\\}$
for every $A\subseteq\mathbb{N}$ and $\omega(A)=0$ for every finite subset of
$\mathbb{N}$. Accordingly we write $\omega$-a.s. and for $\omega$-a.e.$(n)$ in
the usual measure theoretic sense.
Given a bounded sequence $(a_{n})$ of real numbers and a non-principal
ultrafilter $\omega$ there exists a unique $a\in\mathbb{R}$ such that for
every $\varepsilon>0$ the set $\\{n\in\mathbb{N}\text{ s.t.
}|a_{n}-a|<\varepsilon\\}$ has $\omega$-measure $1$, see for instance [DK18,
Lemma 10.25]. The real number $a$ is called the ultralimit of the sequence
$a_{n}$ and it is denoted by $\omega$-$\lim a_{n}$.
If $(X_{n},d_{n},x_{n})$ is a sequence of pointed metric spaces, we denote by
$(X_{\omega},d_{\omega},x_{\omega})$ the ultralimit pointed metric space. It
is the set of sequences $(y_{n})$, where $y_{n}\in X_{n}$ for every $n$, such
that $\omega$-$\lim d(x_{n},y_{n})<+\infty$ for $\omega$-a.e.$(n)$, modulo the
relation $(y_{n})\sim(y_{n}^{\prime})$ if and only if $\omega$-$\lim
d(y_{n},y_{n}^{\prime})=0$. The point of $X_{\omega}$ defined by the class of
the sequence $(y_{n})$ is denoted by $y_{\omega}=\omega$-$\lim y_{n}$. The
formula $d_{\omega}(\omega$-$\lim y_{n},\omega$-$\lim
y_{n}^{\prime})=\omega$-$\lim d(y_{n},y_{n}^{\prime})$ defines a metric on
$X_{\omega}$ which is called the ultralimit distance on $X_{\omega}$.
The relation between Gromov-Hausdorff convergence and ultralimits is
summarized here.

######  Proposition 5.1 ([Jan17], Proposition 3.11; [Cav21], Proposition
3.13).

Let $(X_{n},d_{n},x_{n})$ be a sequence of pointed, compact metric spaces and
let $\omega$ be a non-principal ultrafilter.

  * (i)

If
$(X_{n},d_{n})\underset{\textup{GH}}{\longrightarrow}(X_{\infty},d_{\infty})$
then $(X_{\omega},d_{\omega})$ is isometric to $(X_{\infty},d_{\infty})$. In
particular the ultralimit does not depend on the choice of the basepoints.

  * (ii)

If $(X_{\omega},d_{\omega},x_{\omega})$ is compact then
$(X_{n_{k}},d_{n_{k}})\underset{\textup{GH}}{\longrightarrow}(X_{\omega},d_{\omega})$
for some subsequence $\\{{n_{k}}\\}$.

Let $(X_{n},d_{X_{n}},x_{n}),(Y_{n},d_{Y_{n}},y_{n})$ be two sequences of
pointed metric spaces and let $\omega$ be a non-principal ultrafilter. A
sequence of maps $f_{n}\colon X_{n}\to Y_{n}$ is said admissible if
$\omega$-$\lim d_{Y_{n}}(f_{n}(x_{n}),y_{n})<+\infty$. A sequence of
admissible $L$-Lipschitz maps $f_{n}$ defines a $L$-Lipschitz map
$f_{\omega}=\omega$-$\lim
f_{n}\colon(X_{\omega},x_{\omega})\to(Y_{\omega},y_{\omega})$ by
$f_{\omega}(\omega$-$\lim x_{n})=\omega$-$\lim f_{n}(x_{n})$.
The class of uniformly perfect $(L_{0},\rho_{0})$-quasi-selfsimilar metric
spaces is closed under Gromov-Hausdorff convergence.

######  Proposition 5.2.

Let $(X_{n},d_{n})$ be a sequence of compact, $a_{0}$-uniformly perfect,
$(L_{0},\rho_{0})$-q.s.s. metric spaces. Suppose it converges in the Gromov-
Hausdorff sense to a metric space $(X_{\infty},d_{\infty})$. Then
$(X_{\infty},d_{\infty})$ is a compact, $a_{0}$-uniformly perfect,
$(L_{0},\rho_{0})$-q.s.s. metric space.

###### Proof.

$X_{\infty}$ is compact by definition of Gromov-Hausdorff convergence. We fix
a non-principal ultrafilter $\omega$ and we call $X_{\omega}$ the ultralimit
space: it does not depend on the basepoints and it is isometric to
$X_{\infty}$ by Proposition 5.1. We fix a point $x_{\omega}=\omega$-$\lim
x_{n}\in X_{\omega}$ and a positive real number $\rho\leq\rho_{0}$. For every
$n$ there exists a $L_{0}$-biLipschitz map
$\Phi_{n}\colon\left(B(x_{n},\rho),\frac{\rho_{0}}{\rho}\cdot d_{n}\right)\to
X_{n}$ with $\Phi_{n}(B(x_{n},\rho))\supseteq
B(\Phi_{n}(x_{n}),\frac{\rho_{0}}{L_{0}})$. The sequence of maps $\Phi_{n}$ is
clearly admissible, so it defines a ultralimit $L_{0}$-biLipschitz map
$\Phi_{\omega}$, which is defined on the ultralimit space of the sequence
$\left(B(x_{n},\rho),\frac{\rho_{0}}{\rho}\cdot d_{n}\right)$. We observe that
this ultralimit space contains $B(x_{\omega},\rho)$. Indeed if
$y_{\omega}=\omega$-$\lim y_{n}\in B(x_{\omega},\rho)$, then
$d_{n}(y_{n},x_{n})<\rho$ $\omega$-a.s. Moreover the ultralimit metric of the
metrics $\frac{\rho_{0}}{\rho}\cdot d_{n}$ is $\frac{\rho_{0}}{\rho}\cdot
d_{\omega}$. So we can restrict $\Phi_{\omega}$ to a $L_{0}$-biLipschitz map
from $\left(B(x_{\omega},\rho),\frac{\rho_{0}}{\rho}\cdot d_{\omega}\right)\to
X_{\omega}$. We need to show that $\Phi_{\omega}(B(x_{\omega},\rho))\supseteq
B(\Phi_{\omega}(x_{\omega}),\frac{\rho_{0}}{L_{0}})$. We take
$y_{\omega}=\omega$-$\lim y_{n}$ such that
$d_{\omega}(y_{\omega},\Phi_{\omega}(x_{\omega}))\leq(1-2\varepsilon)\cdot\frac{\rho_{0}}{L_{0}}$,
with $\varepsilon>0$. By definition
$d_{n}(y_{n},\Phi_{n}(x_{n}))\leq(1-\varepsilon)\cdot\frac{\rho_{0}}{L_{0}}$
for $\omega$-a.e.$(n)$. By assumption we can find points $z_{n}\in
B(x_{n},\rho)$ such that $\Phi_{n}(z_{n})=y_{n}$, $\omega$-a.s. These points
satisfy

$\frac{\rho_{0}}{\rho}\cdot d_{n}(z_{n},x_{n})\leq L_{0}\cdot
d_{n}(y_{n},\Phi_{n}(x_{n}))\leq(1-\varepsilon)\cdot\rho_{0},$

so $d_{n}(x_{n},z_{n})\leq(1-\varepsilon)\cdot\rho$. Clearly the point
$z_{\omega}=\omega$-$\lim z_{n}$ belongs to $B(x_{\omega},\rho)$ and satisfies
$\Phi_{\omega}(z_{\omega})=y_{\omega}$.
It remains only to prove that $X_{\omega}$ is $a_{0}$-uniformly perfect. We
fix $x_{\omega}=\omega$-$\lim x_{n}\in X_{\omega}$ and
$0<\rho\leq\text{Diam}(X_{\omega})$. For every $\varepsilon>0$ we have
$(1-\varepsilon)\rho\leq\text{Diam}(X_{n})$ for $\omega$-a.e.$(n)$, therefore
there exists a point $y_{n}^{\varepsilon}\in X_{n}$ with
$d(x_{n},y_{n}^{\varepsilon})\leq(1-\varepsilon)\rho$ and
$d(x_{n},y_{n}^{\varepsilon})\geq a_{0}(1-\varepsilon)\rho$ for
$\omega$-a.e.$(n)$. We consider the ultralimit point
$y_{\omega}^{\varepsilon}=\omega$-$\lim y_{n}^{\varepsilon}\in X_{\omega}$. It
satisfies $d(x_{\omega},y_{\omega}^{\varepsilon})\leq(1-\varepsilon)\rho$ and
$d(x_{\omega},y_{\omega}^{\varepsilon})\geq a_{0}(1-\varepsilon)\rho$. Since
this is true for every $\varepsilon>0$ and since $X_{\omega}$ is compact we
can find a point $y_{\omega}\in X_{\omega}$ such that
$d(x_{\omega},y_{\omega})\leq\rho$ and $d(x_{\omega},y_{\omega})\geq
a_{0}\rho$, showing that $X_{\omega}$ is $a_{0}$-uniformly perfect. ∎

######  Remark 5.3.

This proposition, together with Proposition 2.2, implies that the Gromov-
Hausdorff limit of a sequence of compact $(L_{0},\rho_{0})$-q.s.s. metric
spaces with diameters bounded below by $c_{0}>0$, as considered by [Kle06] and
[Pia11], is still uniformly perfect.

We can now give the

###### Proof of Theorem A.

We notice that since $X_{\infty}$ is compact then the diameters of $X_{n}$ are
uniformly bounded above by some $\Delta_{0}\geq 0$. We proceed in several
steps.
Step 1: _there exists $D_{0}\geq 0$ such that $X_{n}$ is $D_{0}$-doubling for
every $n$._
Suppose it is not true: then for every $j\in\mathbb{N}$ there exist $n_{j}$,
$x_{n_{j}}\in X_{n_{j}}$ and $\rho_{n_{j}}>0$ such that there is a
$\frac{\rho_{n_{j}}}{2}$-separated set inside $B(x_{n_{j}},\rho_{n_{j}})$ of
cardinality $\geq j$. Up to pass to a subsequence we can suppose
$\lim\rho_{n_{j}}=\rho_{\infty}\in[0,+\infty)$. Clearly $X_{\infty}$ is not
totally bounded when $\rho_{\infty}>0$, and this is impossible since
$X_{\infty}$ is compact. If $\rho_{\infty}=0$ we use the quasi-selfsimilarity
to get $L_{0}$-biLipschitz maps
$\Phi_{j}\colon\left(B(x_{n_{j}},\rho_{n_{j}}),\frac{\rho_{0}}{\rho_{n_{j}}}\cdot
d_{n_{j}}\right)\to X_{n_{j}}$ such that
$\Phi_{j}(B(x_{n_{j}},\rho_{n_{j}}))\supseteq
B(\Phi_{j}(x_{n_{j}}),\frac{\rho_{0}}{L_{0}})$ for every $j$ for which
$\rho_{n_{j}}\leq\rho_{0}$. Hence we can find a
$\frac{\rho_{0}}{2L_{0}}$-separated set inside
$B(\Phi_{j}(x_{n_{j}}),\frac{\rho_{0}}{L_{0}})$ with cardinality $\geq j$.
Once again this contradicts the compactness of $X_{\infty}$.

In order to simplify the notations we fix a non-principal ultrafilter $\omega$
and we call $X_{\omega}$ the ultralimit space, which is isometric to
$X_{\infty}$ by Proposition 5.1.

Step 2: _Let $k\in\mathbb{N}$. We fix a maximal $10^{-k}$-separated subset
$X_{k,n}$ of $X_{n}$. Then:_

  * (i)

_the cardinality of_ $X_{k,n}$ is uniformly bounded from above, and each
$X_{k,n}$ is a $10^{-k}$-net of $X_{n}$;

  * (ii)

_the set_ $X_{k,\omega}:=\\{\omega\textup{-}\lim q_{n}\textup{ s.t. }q_{n}\in
X_{k,n}\\}$ is a $10^{-k}$-net of $X_{\omega}$;

  * (iii)

_there exists_ $A_{k}\subseteq\mathbb{N}$, $\omega(A_{k})=1$, such that the
function $\pi_{n}\colon X_{k,\omega}\to X_{k,n}$,
$\pi_{n}(\omega\textup{-}\lim q_{n}):=q_{n}$ is well defined and bijective for
all $n\in A_{k}$.

By Step 1 we know that each $X_{n}$ is $D_{0}$-doubling, therefore the
cardinality of $X_{k,n}$ is uniformly bounded above in terms of $D_{0}$ and
$k$. The second statement of (i) has been explained in Section 2.
We take two points $\omega$-$\lim q_{n}$, $\omega$-$\lim q_{n}^{\prime}\in
X_{k,\omega}$. If $d_{\omega}(\omega\text{-}\lim q_{n},\omega\text{-}\lim
q_{n}^{\prime})<10^{-k}$ then $d_{n}(q_{n},q_{n}^{\prime})<10^{-k}$
$\omega$-a.s. and by definition $q_{n}=q_{n}^{\prime}$ $\omega$-a.s., implying
$\omega\text{-}\lim q_{n}=\omega\text{-}\lim q_{n}^{\prime}$. Since
$X_{\omega}$ is compact we conclude that the set $X_{k,\omega}$ is finite and
that $\pi_{n}$ is well defined $\omega$-a.s.
We suppose $X_{k,\omega}$ is not a $10^{-k}$-net of $X_{\omega}$. Therefore we
can find $y_{\omega}=\omega$-$\lim y_{n}\in X_{\omega}$ such that
$d_{\omega}(y_{\omega},q_{\omega})>10^{-k}$ for all $q_{\omega}\in
X_{k,\omega}$. Since $X_{k,\omega}$ is finite we know that
$d_{n}(y_{n},q_{n})>10^{-k}$ for all $q_{n}\in X_{k,n}$, $\omega$-a.s. This
contradicts the fact that $X_{k,n}$ is a $10^{-k}$-net for every $n$, so also
$X_{k,\omega}$ is a $10^{-k}$-net of $X_{\omega}$.
Since

$|d_{\omega}(q_{\omega},q_{\omega}^{\prime})-d_{n}(q_{n},q_{n}^{\prime})|<\frac{10^{-k}}{2}$

for all $q_{\omega}=\omega\text{-}\lim
q_{n},q_{\omega}^{\prime}=\omega\text{-}\lim q_{n}^{\prime}\in X_{k,\omega}$
and for $\omega$-a.e.$(n)$ we conclude that $\pi_{n}$ is injective
$\omega$-a.s. Finally suppose $\pi_{n}$ is not surjective $\omega$-a.s. Then
it is possible to find a set $A\in\omega$ such that for every $n\in A$ there
exists $q_{n}\in X_{k,n}$ which is not in the image of $\pi_{n}$. In this case
we consider the point $\omega$-$\lim q_{n}$ that belongs to $X_{k,\omega}$,
finding a contradiction. This ends the proof of (iii).

Since we have fixed $10^{-k}$-nets $X_{k,n}$ and $X_{k,\omega}$ of $X_{n}$ and
$X_{\omega}$ respectively, every path will be intended with respect to these
sets.

Step 3: _Let $k\in\mathbb{N}$. There exists a subset
$B_{k}\subseteq\mathbb{N}$ of $\omega$-measure $1$ such that_

  * (i)

_the map_ $\pi_{n}\colon X_{k,\omega}\to X_{k,n}$ from Step 2 is well defined
and bijective for every $n\in B_{k}$;

  * (ii)

_for every_ $n\in B_{k}$, for every $(10,k)$-path
$\gamma_{n}=\\{q_{j}^{n}\\}_{j=0}^{M}$ of $X_{n}$ the associated path
$\gamma_{\omega}=\\{\pi_{n}^{-1}(q_{j}^{n})\\}_{j=0}^{M}$ is a $(30,k)$-path
of $X_{\omega}$.

Since $X_{k,\omega}$ is finite we can find a subset $B_{k}$ of $A_{k}$ with
$\omega$-measure $1$ such that

$|d_{\omega}(q_{\omega},q_{\omega}^{\prime})-d_{n}(\pi_{n}(q_{\omega}),\pi_{n}(q_{\omega}^{\prime}))|\leq
10\cdot 10^{-k}$

for all $q_{\omega},q_{\omega}^{\prime}\in X_{k,\omega}$ and for all $n\in
B_{k}$. Let us take a $(10,k)$-path $\gamma_{n}=\\{q_{j}^{n}\\}_{j=0}^{M}$ of
$X_{n}$, $n\in B_{k}$. This means $d_{n}(q_{j}^{n},q_{j+1}^{n})\leq 20\cdot
10^{-k}$ for all $j=0,\ldots,M-1$. Therefore
$d_{\omega}(\pi_{n}^{-1}(q_{j}^{n}),\pi_{n}^{-1}(q_{j+1}^{n}))\leq 30\cdot
10^{-k}$, i.e. the thesis.

Step 4: _Let $i,k\in\mathbb{N}$ and $p\geq 0$. Then
$p\textup{-Mod}_{30,k}^{\frac{13}{4},\frac{15}{4}}(y_{\omega})\geq\omega\textup{-}\lim
p\textup{-Mod}_{k}(y_{n})$ for every $y_{\omega}=\omega$-$\lim y_{n}\in
X_{i,\omega}$._

We apply Step 3 to the integer $i+k$ finding $B_{i+k}\subseteq\mathbb{N}$,
$\omega(B_{i+k})=1$, and bijective maps $\pi_{n}\colon X_{i+k,\omega}\to
X_{i+k,n}$ for all $n\in B_{i+k}$. We take an optimal function
$f_{\omega}\in\mathcal{A}_{30,i+k}(\overline{B}_{13/4,i}(y_{\omega}),X_{\omega}\setminus
B_{15/4,i}(y_{\omega}))$. By definition $f_{\omega}$ maps points of
$X_{i+k,\omega}$ to $[0,+\infty)$. For all $n\in B_{i+k}$ we define the
functions $f_{n}\colon X_{i+k,n}\to[0,+\infty)$ by
$f_{n}(q)=f_{\omega}(\pi_{n}^{-1}(q))$. We find another subset
$C_{i+k,y_{\omega}}\subseteq B_{i+k}$ of $\omega$-measure $1$ such that

$|d_{\omega}(q_{\omega},y_{\omega})-d_{n}(\pi_{n}(q_{n}),y_{n})|\leq\frac{1}{4}\cdot
10^{-i}$

for all $q_{\omega}\in X_{i+k,\omega}$ and for all $n\in C_{i+k,y_{\omega}}$.
We want to check that
$f_{n}\in\mathcal{A}_{10,i+k}(\overline{B}_{3,i}(y_{n}),X_{n}\setminus
B_{4,i}(y_{n}))$ for all $n\in C_{i+k,y_{\omega}}$. We fix $n\in
C_{i+k,y_{\omega}}$ and we take a $(10,i+k)$-path
$\gamma_{n}=\\{q_{j}^{n}\\}_{j=0}^{M}$ such that $d_{n}(y_{n},q_{0}^{n})\leq
3\cdot 10^{-i}$ and $d_{n}(y_{n},q_{M}^{n})>4\cdot 10^{-i}$. We denote by
$\gamma_{\omega}=\\{\pi_{n}^{-1}(q_{j}^{n})\\}$ the $(30,i+k)$-path given by
Step 3. We observe that
$d_{\omega}(y_{\omega},\pi_{n}^{-1}(q_{0}^{n}))\leq\frac{13}{4}\cdot 10^{-i}$
and $d_{\omega}(y_{\omega},q_{M}^{\omega})>\frac{15}{4}\cdot 10^{-i}$, i.e.
the $(30,i+k)$-path $\gamma_{\omega}$ joins
$\overline{B}_{13/4,i}(y_{\omega})$ and $X_{\omega}\setminus
B_{15/4,i}(y_{\omega})$. By definition of $f_{n}$ we get

$\sum_{j=0}^{M}f_{n}(q_{j}^{n})=\sum_{j=0}^{M}f_{\omega}(\pi_{n}^{-1}(q_{j}^{n}))\geq
1.$

Moreover it holds

$p\text{-Mod}_{k}(y_{n})\leq\sum_{q\in X_{i+k,n}}f_{n}^{p}(q)=\sum_{q\in
X_{i+k,\omega}}f_{\omega}^{p}(\pi^{-1}_{n}(q))=p\text{-Mod}_{30,k}^{\frac{13}{4},\frac{15}{4}}(y_{\omega}).$

Since this is true for all $n\in C_{i+k,y_{\omega}}$ we get

$p\textup{-Mod}_{30,k}^{\frac{13}{4},\frac{15}{4}}(y_{\omega})\geq\omega\textup{-}\lim
p\textup{-Mod}_{k}(y_{n}).$

Step 5: _Conclusion._
We fix $k\in\mathbb{N}$ and $0\leq p<\omega$-$\lim\text{CD}(X_{n},d_{n})$. By
Proposition 4.4 we find a constant $\lambda_{0}>0$ depending only on $D_{0}$,
$L_{0}$ and $p$ such that

$\sup_{i\leq n_{0}}\sup_{y\in X_{i,n}}p\text{-Mod}_{k}(y)\geq\lambda_{0}$
(5.1)

for $\omega$-a.e.$(n)$. For all these $n$’s we take a point $y_{n}\in
X_{i_{n},n}$, $1\leq i_{n}\leq n_{0}$, realizing the supremum in (5.1). The
sequence $i_{n}$ is $\omega$-a.s. equal to some
$i_{*}\in\\{1,\ldots,n_{0}\\}$. So the limit point $y_{\omega}$ belongs to
$X_{i_{*},\omega}$. By Step 4 we get

$p\textup{-Mod}_{30,k}^{\frac{13}{4},\frac{15}{4}}(X_{\omega})\geq
p\textup{-Mod}_{30,k}^{\frac{13}{4},\frac{15}{4}}(y_{\omega})\geq\omega\textup{-}\lim
p\textup{-Mod}_{k}(y_{n})\geq\lambda_{0}.$

Applying the easy inequalities of Lemma 3.2 and Lemma 3.3 we conclude that
$p\textup{-Mod}_{k}(X_{\omega})\geq\lambda_{0}$ for every $k$. This shows
$\liminf_{k\to+\infty}p\textup{-Mod}_{k}(X_{\omega})>0$. $X_{\omega}$ is a
compact, doubling and uniformly perfect metric space by Proposition 5.2 and
Proposition 2.2, so we can apply Theorem 4.3 to get
$p\leq\text{CD}(X_{\omega},d_{\omega})$. In conclusion we proved

$\omega\text{-}\lim\text{CD}(X_{n},d_{n})\leq\text{CD}(X_{\omega},d_{\omega}).$

This inequality is true for every non-principal ultrafilter $\omega$, so by
Lemma 6.3 of [Cav21] we have
$\limsup_{n\to+\infty}\text{CD}(X_{n},d_{n})\leq\text{CD}(X_{\infty},d_{\infty}).$
∎

The main tools used in this proof are: the reduction of the computation of the
combinatorial modulus to a _finite_ set of scales and the uniform lower bound
on the combinatorial modulus, independent of $k$, given by Proposition 4.4.
The approach to the lower semicontinuity problem is more difficult because
from one side it is still possible to reduce the computation to a finite set
of scales, but from the other side there is no more any control on the
behaviour of the $p$-modulus, independent of $k$. If we take some
$p>\text{CD}(X_{n},d_{n})$ for every $n$ then by Theorem 4.3 it holds
$\liminf_{k\to+\infty}p\text{-Mod}_{k}(X_{n})=0$. But a priori it is not
possible to conclude that
$\liminf_{k\to+\infty}p\text{-Mod}_{k}(X_{\infty})=0$. Indeed for given
$\varepsilon>0$ we cannot control the threshold $k_{\varepsilon}$ such that
$p\text{-Mod}_{k}(X_{n})<\varepsilon$ for $k\geq k_{\varepsilon}$. Clearly if
we have this kind of uniform control on the spaces $X_{n}$ then the Ahlfors
regular conformal dimension of $X_{\infty}$ is equal to the limit of the
Ahlfors regular conformal dimensions of $X_{n}$.
Then Question 1 can be rephrased in the following way: are there (interesting)
geometric conditions on a quasi-selfsimilar space that gives a uniform control
on the thresholds $k_{\varepsilon}$ defined above?
This question seems to be related to (uniform) weak super-multiplicative
properties of the sequence $p$-Mod${}_{k}(X)$, as studied in relation with the
combinatorial Lowner property in [BK13, §4, §8], and in the special case of
the Sierpinski carpet in [Kwa20, Theorem 1.3]. This weak super-multiplicative
property seems to hold true only for spaces in which curves are uniformly
distributed in some sense, as suggested by the arguments used again in [BK13,
Lemma 4.3 and Lemma 8.1]. This observation gives a possible approach to the
question presented in the introduction in case of spaces satisfying a uniform
combinatorial Lowner property.

##  6 Gromov-hyperbolic spaces

In this second part of the paper we prove Theorem B. We briefly recall the
definition of Gromov-hyperbolic metric spaces. Good references are for
instance [BH13] and [CDP90]. Let $X$ be a metric space. Given three points
$x,y,z\in X$, the Gromov product of $y$ and $z$ with respect to $x$ is

$(y,z)_{x}=\frac{1}{2}\big{(}d(x,y)+d(x,z)-d(y,z)\big{)}.$

The space $X$ is said $\delta$-hyperbolic, $\delta\geq 0$, if for every four
points $x,y,z,w\in X$ the following 4-points condition hold:

$(x,z)_{w}\geq\min\\{(x,y)_{w},(y,z)_{w}\\}-\delta,\vspace{-2mm}$ (6.1)

or, equivalently,

$d(x,y)+d(z,w)\leq\max\\{d(x,z)+d(y,w),d(x,w)+d(y,z)\\}+2\delta.\vspace{-2mm}$
(6.2)

The space $X$ is Gromov hyperbolic if it is $\delta$-hyperbolic for some
$\delta\geq 0$.

Let $X$ be a $\delta$-hyperbolic metric space and $x$ be a point of $X$. The
Gromov boundary of $X$ is defined as the quotient

$\partial X=\\{(z_{n})_{n\in\mathbb{N}}\subseteq X\hskip 2.84526pt|\hskip
2.84526pt\lim_{n,m\to+\infty}(z_{n},z_{m})_{x}=+\infty\\}\hskip
2.84526pt/_{\approx},\vspace{-3mm}$

where $(z_{n})_{n\in\mathbb{N}}$ is a sequence of points in $X$ and $\approx$
is the equivalence relation defined by
$(z_{n})_{n\in\mathbb{N}}\approx(z_{n}^{\prime})_{n\in\mathbb{N}}$ if and only
if $\lim_{n,m\to+\infty}(z_{n},z_{m}^{\prime})_{x}=+\infty$. We will write
$z=[(z_{n})]\in\partial X$ for short, and we say that $(z_{n})$ converges to
$z$. This definition does not depend on the basepoint $x$. There is a natural
topology on $X\cup\partial X$ that extends the metric topology of $X$. The
Gromov product can be extended to points $z,z^{\prime}\in\partial X$ by

$(z,z^{\prime})_{x}=\sup_{(z_{n}),(z_{n}^{\prime})}\liminf_{n,m\to+\infty}(z_{n},z_{m}^{\prime})_{x}\vspace{-3mm}$

where the supremum is taken among all sequences such that $(z_{n})\in z$ and
$(z_{n}^{\prime})\in z^{\prime}$. For every
$z,z^{\prime},z^{\prime\prime}\in\partial X$ it continues to hold

$(z,z^{\prime})_{x}\geq\min\\{(z,z^{\prime\prime})_{x},(z^{\prime},z^{\prime\prime})_{x}\\}-\delta.\vspace{-3mm}$
(6.3)

Moreover for all sequences $(z_{n}),(z_{n}^{\prime})$ converging to
$z,z^{\prime}$ respectively it holds

$(z,z^{\prime})_{x}-\delta\leq\liminf_{n,m\to+\infty}(z_{n},z_{m}^{\prime})_{x}\leq(z,z^{\prime})_{x}.\vspace{-3mm}$
(6.4)

The Gromov product between a point $y\in X$ and a point $z\in\partial X$ is
defined in a similar way and it satisfies a condition analogue of (6.4).
The boundary of a $\delta$-hyperbolic metric space is metrizable. A metric
$D_{x,a}$ on $\partial X$ is called a visual metric of center $x\in X$ and
parameter $a\in\left(0,\frac{1}{2\delta\cdot\log_{2}e}\right)$ if there exists
$V>0$ such that for all $z,z^{\prime}\in\partial X$ it holds

$\frac{1}{V}e^{-a(z,z^{\prime})_{x}}\leq D_{x,a}(z,z^{\prime})\leq
Ve^{-a(z,z^{\prime})_{x}}.$ (6.5)

A visual metric is said standard if for all $z,z^{\prime}\in\partial X$ it
holds

$(3-2e^{a\delta})e^{-a(z,z^{\prime})_{x}}\leq D_{x,a}(z,z^{\prime})\leq
e^{-a(z,z^{\prime})_{x}}.$ (6.6)

For all $a$ as before and $x\in X$ there exists always a standard visual
metric of center $x$ and parameter $a$ (cp. [Pau96], [BS11]). Any two
different visual metrics are quasisymmetric equivalent, and the quasisymmetric
homeomorphism is the identity ([BS11, Lemma 6.1]). This defines a well defined
quasisymmetric gauge on $\partial X$, that we denote by $\mathcal{J}(\partial
X)$. If $C$ is a subset of $\partial X$ then the restriction of two visual
metrics on $C$ define again two quasisymmetric distances, so it is well
defined the quasisymmetric gauge $\mathcal{J}(C)$.

We will deal with proper metric spaces, i.e. spaces in which every closed ball
is compact. A metric space $X$ is $K$-almost geodesic if for all $x,y\in X$,
for all $t\in[0,d(x,y)]$ there exists $z\in X$ such that $|d(x,z)-t|\leq K$
and $|d(y,z)-(d(x,y)-t)|\leq K$. If we do not need to specify the value of $K$
we simply say that $X$ is almost geodesic. A metric space is geodesic if it is
$0$-almost geodesic. Let $X$ be a proper, geodesic, Gromov-hyperbolic metric
space. Every geodesic ray $\xi$ defines a point
$\xi^{+}=[(\xi(n))_{n\in\mathbb{N}}]$ of the Gromov boundary $\partial X$.
Moreover for every $z\in\partial X$ and every $x\in X$ it is possible to find
a geodesic ray $\xi_{x,z}$ such that $\xi(0)=x$ and $\xi^{+}=z$. Analogously,
given different points $z=[(z_{n})],z^{\prime}=[(z^{\prime}_{n})]\in\partial
X$ there exists a geodesic line $\gamma$ joining $z$ to $z^{\prime}$, i.e.
such that $\gamma|_{[0,+\infty)}$ and $\gamma|_{(-\infty,0]}$ join $\gamma(0)$
to $z,z^{\prime}$ respectively. We call $z$ and $z^{\prime}$ the positive and
negative endpoints of $\gamma$, respectively, denoted $\gamma^{\pm}$.
The quasiconvex hull of a subset $C$ of $\partial X$ is the union of all the
geodesic lines joining two points of $C$ and it is denoted by QC-Hull$(C)$. If
$X$ is proper and geodesic and $C$ has more than one point then QC-Hull$(C)$
is non-empty by the discussion above. We can say more.

######  Proposition 6.1.

Let $X$ be a proper, geodesic, $\delta$-hyperbolic metric space and let
$C\subseteq\partial X$ be a closed subset with at least two points. Then
$\textup{QC-Hull}(C)$ is proper, $36\delta$-almost geodesic and
$\delta$-hyperbolic. Moreover $\mathcal{J}(\partial\textup{QC-
Hull}(C))\cong\mathcal{J}(C)$, in the sense that there exists a homeomorphism
$F\colon C\to\partial\textup{QC-Hull}(C)$ which is a quasisymmetric
equivalence when we equip $C$ and $\partial\textup{QC-Hull}(C)$ with every
metrics in the gauges $\mathcal{J}(C)$ and $\mathcal{J}(\partial\textup{QC-
Hull}(C))$, respectively.

We need the following approximation result.

######  Lemma 6.2 ([Cav21], Lemma 4.6).

Let $X$ be a proper, geodesic, $\delta$-hyperbolic metric space. Let
$C\subseteq\partial X$ be a subset with at least two points and
$x\in\textup{QC-Hull}(C)$. Then $d(\xi_{x,z},\textup{QC-Hull}(C))\leq
14\delta$ for all $z\in C$.

###### Proof of Proposition 6.1.

QC-Hull$(C)$ is closed and $36\delta$-quasiconvex ([Cav21, Lemma 4.5]), i.e.
every point of every geodesic segment joining every two points $y,y^{\prime}$
of $\text{QC-Hull}(C)$ is at distance at most $36\delta$ from $\text{QC-
Hull}(C)$. This implies that $\text{QC-Hull}(C)$ is $36\delta$-almost geodesic
and proper. Condition (6.1) involves only the distance function, so $\text{QC-
Hull}(C)$ is $\delta$-hyperbolic. We define a map $F\colon
C\to\partial\textup{QC-Hull}(C)$ in the following way. We fix $x\in\textup{QC-
Hull}(C)$. For every $z\in C$ we take a sequence $(z_{n})\in\textup{QC-
Hull}(C)$ such that $d(\xi_{x,z}(n),z_{n})\leq 14\delta$, as provided by Lemma
6.2. The sequence $(z_{n})$ defines a point $\hat{z}\in\partial\textup{QC-
Hull}(C)$, since $\lim_{n,m\to+\infty}(z_{n},z_{m})_{x}=+\infty$. We set
$F(z):=\hat{z}$. It is straightforward to check that $F$ is well defined, i.e.
it does not depend on the choice of the sequence $(z_{n})$. The Gromov
products on $C$ and $\partial\textup{QC-Hull}(C)$ are comparable by (6.4),
namely

$(z,z^{\prime})_{x}-\delta\leq(F(z),F(z^{\prime}))_{x}\leq(z,z^{\prime})_{x}.$
(6.7)

Fix visual distances $D_{x,a}$ and $\hat{D}_{x,a}$ on $\partial X$ and
$\partial\textup{QC-Hull}(C)$. By (6.7) and (6.5) we have that $F$ is
injective. If moreover it is surjective then it is a quasisymmetric
homemorphism from $(C,D_{x,a})$ to $(\textup{QC-Hull}(C),\hat{D}_{x,a})$,
which is the thesis. So fix a point $\hat{z}\in\partial\textup{QC-Hull}(C)$.
By definition it is represented by a sequence $(z_{n})\in\textup{QC-Hull}(C)$
such that $\lim_{n,m\to+\infty}(z_{n},z_{m})_{x}=+\infty$. Let $\gamma_{n}$ be
a geodesic line of $X$ such that $\gamma_{n}^{\pm}\in C$ and
$z_{n}\in\gamma_{n}$. We claim that, up to change the orientation of
$\gamma_{n}$, it holds $\lim_{n\to+\infty}(\gamma_{n}^{+},z_{n})_{x}=+\infty$,
so that $\gamma_{n}^{+}$ converges to $\hat{z}\in\partial X$ as $n$ goes to
$+\infty$. Since $C$ is closed we deduce that $\hat{z}\in C$. Let us prove the
claim is false, so both $(z_{n},\gamma_{n}^{\pm})\leq M$ for every $n$, for
some $M$. By [CS20, Lemma 3.2] applied to both the segments
$[z_{n},\gamma_{n}^{\pm}]$ we get $d(x,[z_{n},\gamma_{n}^{\pm}])\leq
M+4\delta$. Let us call $p_{n}^{\pm}$ points on the rays
$[z_{n},\gamma_{n}^{\pm}]$ realizing the distance from $x$. The 4-point
condition (6.2) gives

$d(x,z_{n})+d(p_{n}^{+},p_{n}^{-})\leq\max\\{d(x,p_{n}^{+})+d(z_{n},p_{n}^{-}),d(x,p_{n}^{-})+d(z_{n},p_{n}^{+})\\}+2\delta.$

Since $d(p_{n}^{+},p_{n}^{-})=d(z_{n},p_{n}^{-})+d(z_{n},p_{n}^{+})$ the
inequality above implies

$d(x,z_{n})\leq\max\\{d(x,p_{n}^{+}),d(x,p_{n}^{-})\\}+2\delta\leq
2M+10\delta.$

But this is impossible since $\lim_{n\to+\infty}d(x,z_{n})=+\infty$. ∎

The quasisymmetric gauge of an almost geodesic Gromov-hyperbolic space is
preserved by quasi-isometries. Recall that a quasi-isometry is a map $f\colon
X\to Y$ between metric spaces for which there exist $K\geq 0$ and $\lambda\geq
1$ such that:

  * (i)

$f(X)$ is $K$-dense in $Y$;

  * (ii)

$\frac{1}{\lambda}d(x,x^{\prime})-K\leq d(f(x),f(x^{\prime}))\leq\lambda
d(x,x^{\prime})+K$ for all $x,x^{\prime}\in X$.

######  Proposition 6.3 ([BS11, Theorem 6.5]).

Let $X,Y$ be two almost geodesic, Gromov-hyperbolic metric spaces and let
$f\colon X\to Y$ be a quasi-isometry. Then $f$ induces a quasisymmetric
homeomorphism $\partial f\colon\partial X\to\partial Y$.

The statement means that for one (hence every) choice of metrics on
$\mathcal{J}(\partial X)$ and $\mathcal{J}(\partial Y)$, the map $\partial f$
is a quasisymmetric homeomorphism. In this case we write $\mathcal{J}(\partial
X)\cong\mathcal{J}(\partial Y)$ as in Proposition 6.1.

###  6.1 The proof of Theorem B

We recall the definition of the class $\mathcal{M}(\delta,D)$ appearing in
Theorem B. Let $X$ be a proper, geodesic, $\delta$-hyperbolic metric space.
Every isometry of $X$ acts naturally on $\partial X$ and the resulting map on
$X\cup\partial X$ is a homeomorphism. A group of isometries $\Gamma$ of $X$ is
said discrete if it is discrete in the compact-open topology. The limit set
$\Lambda(\Gamma)$ of a discrete group of isometries $\Gamma$ is the set of
accumulation points of the orbit $\Gamma x$ on $\partial X$, where $x$ is any
point of $X$. The group $\Gamma$ is called elementary if
$\\#\Lambda(\Gamma)\leq 2$. The set $\Lambda(\Gamma)$ is closed and
$\Gamma$-invariant so it is its quasiconvex hull. A discrete group of
isometries $\Gamma$ is quasiconvex-cocompact if its action on QC-
Hull$(\Lambda(\Gamma))$ is cocompact, i.e. if there exists $D\geq 0$ such that
for all $x,y\in\text{QC-Hull}(\Lambda(\Gamma))$ it holds $d(gx,y)\leq D$ for
some $g\in\Gamma$. The smallest $D$ satisfying this property is called the
codiameter of $\Gamma$.

_Given two real numbers $\delta\geq 0$ and $D>0$ we define
$\mathcal{M}(\delta,D)$ to be the class of triples $(X,x,\Gamma)$, where $X$
is a proper, geodesic, $\delta$-hyperbolic metric space, $\Gamma$ is a
discrete, non-elementary, torsion-free, quasiconvex-cocompact group of
isometries with codiameter $\leq D$ and $x\in\textup{QC-
Hull}(\Lambda(\Gamma))$._

Let $\Gamma$ be a finitely generated. Given a finite generating set $\Sigma$
of $\Gamma$ one can construct the Cayley graph $\text{Cay}(\Gamma,\Sigma)$ of
$\Gamma$ relative to $\Sigma$. Any two Cayley graphs, made with respect to
different generating sets, are quasi-isometric. $\Gamma$ is said to be Gromov-
hyperbolic if one (and hence all) of its Cayley graphs is Gromov-hyperbolic.
If it is the case the Gromov boundaries of every two Cayley graphs are
quasisymmetric equivalent, by Proposition 6.3. We denote the corresponding
quasisymmetric gauge by $\mathcal{J}(\partial\Gamma)$. A straightforward
modification of the classical proof of the Svarc-Milnor lemma (along the same
lines of [Cav21, Lemma 5.1]) says that every Cayley graph of $\Gamma$ is
quasi-isometric to $\text{QC-Hull}(\Lambda(\Gamma))$, if
$(X,x,\Gamma)\in\mathcal{M}(\delta,D)$. So both these spaces are Gromov-
hyperbolic and almost geodesic. By Proposition 6.3 the gauges
$\mathcal{J}(\partial\Gamma)$ and $\mathcal{J}(\partial\textup{QC-
Hull}(\Lambda(\Gamma)))\cong\mathcal{J}(\Lambda(\Gamma))$ are quasisymmetric
equivalent. The last equality is Proposition 6.1. We can already prove the
last part of Theorem B.

######  Proposition 6.4.

If $(X_{n},x_{n},\Gamma_{n})\underset{\textup{eq-
pGH}}{\longrightarrow}(X_{\infty},x_{\infty},\Gamma_{\infty})$, with
$(X_{n},x_{n},\Gamma_{n})\in\mathcal{M}(\delta,D)$, then
$\lim_{n\to+\infty}\textup{CD}(\Lambda(\Gamma_{n}))=\textup{CD}(\Lambda(\Gamma_{\infty}))$.

Here $(X_{n},x_{n},\Gamma_{n})\underset{\textup{eq-
pGH}}{\longrightarrow}(X_{\infty},x_{\infty},\Gamma_{\infty})$ means that the
triples $(X_{n},x_{n},\Gamma_{n})$ converge in the equivariant pointed Gromov-
Hausdorff sense to the triple $(X_{\infty},x_{\infty},\Gamma_{\infty})$. We
will not recall the definition, details can be found for instance in [Cav21].
If all the triples $(X_{n},x_{n},\Gamma_{n})$ belong to
$\mathcal{M}(\delta,D)$ then also
$(X_{\infty},x_{\infty},\Gamma_{\infty})\in\mathcal{M}(\delta,D)$, by [Cav21,
Theorem A]. In particular it is meaningful to talk about
$\Lambda(\Gamma_{\infty})$. Recall that the conformal dimensions are the
conformal dimensions of the quasisymmetic gauges
$\mathcal{J}(\Lambda(\Gamma_{n}))$, $n\in\mathbb{N}\cup\\{\infty\\}$.

###### Proof.

By [Cav21, Theorem A] the triple $(X_{\infty},x_{\infty},\Gamma_{\infty})$
belongs to $\mathcal{M}(\delta,D)$. Moreover [Cav21, Corollary 7.7] implies
that $\Gamma_{n}$ is isomorphic to $\Gamma_{\infty}$ for $n$ big enough. Using
Proposition 6.3 we conclude that
$\mathcal{J}(\partial\Gamma_{n})\cong\mathcal{J}(\partial\Gamma_{\infty})$.
The discussion above says that
$\mathcal{J}(\Lambda(\Gamma_{n}))\cong\mathcal{J}(\Lambda(\Gamma_{\infty}))$.
Therefore, by definition,
$\textup{CD}(\Lambda(\Gamma_{n}))=\textup{CD}(\Lambda(\Gamma_{\infty}))$ for
$n$ big enough. ∎

The next step is to show that, under the assumptions of Theorem B, the spaces
$\Lambda(\Gamma_{n})$ are uniformly perfect and uniformly quasi-selfsimilar,
when equipped with suitable visual metrics. Let
$(X,x,\Gamma)\in\mathcal{M}(\delta,D)$. We always consider a standard visual
metric $D_{x}$ centered at $x$ and with parameter
$a_{\delta}=\frac{1}{4\delta\log_{2}e}$. All the estimates will be done with
respect to this metric $D_{x}$.

The following three results are essentially known, see for instance [Kle06].
We provide quantified version of them. The critical exponent of $\Gamma$ is
$h_{\Gamma}=\lim_{T\to+\infty}\frac{1}{T}\log\\#\Gamma
x\cap\overline{B}(x,T).$

######  Proposition 6.5.

Let $\delta,D,H\geq 0$. There exists $A=A(\delta,D,H)>0$ such that for all
$(X,x,\Gamma)\in\mathcal{M}(\delta,D)$ with $h_{\Gamma}\leq H$ the limit set
$\Lambda(\Gamma)$ is $(A,\frac{h_{\Gamma}}{a_{\delta}})$-Ahlfors regular.

###### Proof.

It follows from [Cav21, Theorem 6.1 and Lemma 4.9]. ∎

######  Corollary 6.6.

Let $\delta,D,H\geq 0$. There exists $a_{0}=a_{0}(\delta,D,H)$ such that for
all $(X,x,\Gamma)\in\mathcal{M}(\delta,D)$ with $h_{\Gamma}\leq H$ the limit
set $\Lambda(\Gamma)$ is $a_{0}$-uniformly perfect.

###### Proof.

[Cav21, Proposition 5.2] says that $h_{\Gamma}\geq\frac{\log
2}{99\delta+10D}$. The conclusion follows by Proposition 6.5 and Lemma 2.1. ∎

######  Proposition 6.7.

Let $\delta,D\geq 0$. There are $L_{0}=L_{0}(\delta,D)$ and
$\rho_{0}=\rho_{0}(\delta,D)$ such that for all
$(X,x,\Gamma)\in\mathcal{M}(\delta,D)$ the set $\Lambda(\Gamma)$ is
$(L_{0},\rho_{0})$-q.s.s.

Before the proof of this last property we need a bit of preparation.

######  Lemma 6.8 ([Cav21, Lemma 4.2]).

Let $X$ be a proper, geodesic, $\delta$-hyperbolic metric space,
$z,z^{\prime}\in\partial X$ and $x\in X$.

  * (i)

If $(z,z^{\prime})_{x}\geq T$ then
$d(\xi_{x,z}(T-\delta),\xi_{x,z^{\prime}}(T-\delta))\leq 4\delta$.

  * (ii)

If $d(\xi_{x,z}(T),\xi_{x,z^{\prime}}(T))<2b$ then $(z,z^{\prime})_{x}>T-b$,
for all $b>0$.

######  Lemma 6.9 ([Cav21, Lemma 4.4]).

Let $X$ be a proper, geodesic, $\delta$-hyperbolic metric space. Then every
two geodesic rays $\xi,\xi^{\prime}$ with same endpoints at infinity are at
distance at most $8\delta$, more precisely there exist $t_{1},t_{2}\geq 0$
such that $t_{1}+t_{2}=d(\xi(0),\xi^{\prime}(0))$ and
$d(\xi(t+t_{1}),\xi^{\prime}(t+t_{2}))\leq 8\delta$ for all $t\geq 0$.

Recall that on $\partial X$ we always consider a visual metrics of parameter
$a_{\delta}$.

######  Corollary 6.10.

Let $(X,x,\Gamma)\in\mathcal{M}(\delta,D)$ and let $z,z^{\prime}\in\partial
X$. Let $\rho>0$ and $R$ be such that $e^{-a_{\delta}R}=\rho$. If
$D_{x}(z,z^{\prime})\leq\rho$ then $d(\xi_{x,z}(R),\xi_{x,z^{\prime}}(R))\leq
14\delta$.

###### Proof.

With this choice of $a_{\delta}$ we have
$\frac{1}{2}e^{-a_{\delta}(z,z^{\prime})_{x}}\leq D_{x}(z,z^{\prime})\leq
e^{-a_{\delta}(z,z^{\prime})_{x}}$. So, if $D_{x}(z,z^{\prime})\leq\rho$ then
$(z,z^{\prime})_{x}\geq R-\frac{\log(2)}{a_{\delta}}=R-4\delta$. By Lemma 6.8
we get $d(\xi_{x,z}(R-5\delta),\xi_{x,z^{\prime}}(R-5\delta))\leq 4\delta$, so
by triangle inequality $d(\xi_{x,z}(R),\xi_{x,z^{\prime}}(R))\leq 14\delta$. ∎

We can finally give the

###### Proof of Proposition 6.7.

We claim that $\rho_{0}=e^{-a_{\delta}\cdot D}$ works. We fix
$0<\rho\leq\rho_{0}$ and we call $R\geq 0$ the real number such that
$\rho=e^{-a_{\delta}\cdot R}$. Let $z\in\text{QC-Hull}(\Lambda(\Gamma))$ and
$\xi_{x,z}$ be a geodesic ray joining $x$ to $z$. By Lemma 6.2 there is a
point $y\in\text{QC-Hull}(\Lambda(\Gamma))$ such that $d(\xi_{x,z}(R),y)\leq
14\delta$. Moreover by definition of quasiconvex-cocompactness there exists
$g\in\Gamma$ such that $d(y,gx)\leq D$, so $d(\xi_{x,z}(R),gx)\leq
14\delta+D$. Observe that $d(x,gx)\leq R+14\delta+D$. We call $\Phi$ the map
induced by $g$ on $\Lambda(\Gamma)$, which is well defined since
$\Lambda(\Gamma)$ is $\Gamma$-invariant. We claim it satisfies the properties
required by Definition 1.
Let $w,w^{\prime}$ be two points of $B(z,\rho)\cap\Lambda(\Gamma)$, so
$D_{x}(w,z)$, $D_{x}(w^{\prime},z)<\rho$. Let $T\geq 0$ be the real number
such that $e^{-a_{\delta}\cdot T}=D_{x}(w,w^{\prime})$. Since
$D_{x}(w,w^{\prime})<2\rho$ we get $e^{-a_{\delta}\cdot
T}<2e^{-a_{\delta}\cdot R}$ and, by definition of $a_{\delta}$, $T>R-4\delta$.
We apply three times Corollary 6.10 to obtain

$\displaystyle d(\xi_{x,z}(R),\xi_{x,w}(R))\leq 14\delta,$ $\displaystyle
d(\xi_{x,z}(R),\xi_{x,w^{\prime}}(R))\leq 14\delta,$ $\displaystyle
d(\xi_{x,w}(T),\xi_{x,w^{\prime}}(T))\leq 14\delta.$

By triangle inequality we have

$\displaystyle|T-R|-D-28\delta\leq d(x,g\xi_{x,w}(T))\leq|T-R|+D+28\delta.$

Similar estimates hold for $d(x,g\xi_{x,w^{\prime}}(T))$. We want to estimate
$d(\xi_{x,gw}(|T-R|),g\xi_{x,w}(T))$. The two rays $\xi_{x,gw}$ and
$g\xi_{x,w}$ define the same point $gw$ of $\partial X$ and $d(x,gx)\leq
R+14\delta+D$, so by Lemma 6.9 there exist $t_{1},t_{2}\geq 0$ with
$t_{1}+t_{2}\leq R+14\delta+D$ such that
$d(\xi_{x,gw}(t+t_{1}),g\xi_{x,w}(t+t_{2}))\leq 8\delta$ for all $t\geq 0$. We
apply this property to $t=T-t_{2}+18\delta+D$, which is non-negative since
$T\geq R-4\delta$, finding

$d(\xi_{x,gw}(T-t_{2}+t_{1}+18\delta+D),g\xi_{x,w}(T+18\delta+D))\leq
8\delta.$

By this inequality and the estimates on $d(x,g\xi_{x,w}(T))$ we get

$|T-R|-2D-54\delta\leq
d(x,\xi_{x,gw}(T-t_{2}+t_{1}+18\delta+D))\leq|T-R|+2D+54\delta,$

so

$|T-R|-2D-54\delta\leq T-t_{2}+t_{1}+18\delta+D\leq|T-R|+2D+54\delta.$

Therefore by triangle inequality

$d(\xi_{x,gw}(|T-R|),g\xi_{x,w}(T))\leq 80\delta+3D.$

Analogously we get $d(\xi_{x,gw^{\prime}}(|T-R|),g\xi_{x,w^{\prime}}(T))\leq
80\delta+3D.$ Combining these two estimates we conclude
$d(\xi_{x,gw}(|T-R|),\xi_{x,gw^{\prime}}(|T-R|))\leq 160\delta+6D$. By Lemma
6.8 we have $(gw,gw^{\prime})_{x}>|T-R|-80\delta-3D$, so

$\displaystyle D_{x}(gw,gw^{\prime})$ $\displaystyle\leq
e^{-a_{\delta}(|T-R|-80\delta-3D)}$ $\displaystyle\leq
e^{a_{\delta}(80\delta+4D)}\cdot\frac{e^{-a_{\delta}D}}{e^{-a_{\delta}R}}\cdot
e^{-a_{\delta}T}$
$\displaystyle=e^{a_{\delta}(80\delta+4D)}\cdot\frac{\rho_{0}}{\rho}\cdot
D_{x}(w,w^{\prime}),$

where we used the definition of standard visual metric, $|T-R|\leq T-R$ and
$\rho_{0}=e^{-a_{\delta}D}$. We prove now the other inequality. We have
$D_{x}(w,w^{\prime})=e^{a_{\delta}T}\leq e^{a_{\delta}(w,w^{\prime})_{x}}$, so
$(w,w^{\prime})_{x}\leq T$. We set $b=47\delta+D$ and $T^{\prime}=T+b$. By
Lemma 6.8.(ii) we know that
$d(\xi_{x,w}(T^{\prime}),\xi_{x,w^{\prime}}(T^{\prime}))\geq 2b$. We can argue
in the same way as before with $t=T^{\prime}-t_{2}$ finding

$d(\xi_{x,gw}(T^{\prime}-R),g\xi_{x,w}(T^{\prime}))\leq 44\delta+D,$

and the analogous estimate for $w^{\prime}$. Therefore

$d(\xi_{x,gw}(T^{\prime}-R),\xi_{x,gw^{\prime}}(T^{\prime}-R))\geq
2b-88\delta-2D>4\delta.$

By Lemma 6.8.(i) we have
$(gw,gw^{\prime})_{x}<T^{\prime}-R+\delta=T-R+48\delta+D$. Therefore

$D_{x}(gw,gw^{\prime})\geq\frac{1}{2}e^{-a_{\delta}(T-R+48\delta+D)}=\frac{1}{2}e^{-48\cdot
a_{\delta}\cdot\delta}\cdot\frac{\rho_{0}}{\rho}\cdot D_{x}(w,w^{\prime}).$

Thus $\Phi$ is $L_{0}$-biLipschitz with
$L_{0}=L_{0}(\delta,D)=\max\\{e^{a_{\delta}(80\delta+4D)},2e^{48\cdot
a_{\delta}\cdot\delta}\\}$ from
$(B(z,\rho)\cap\Lambda(\Gamma),\frac{\rho_{0}}{\rho}\cdot
D_{x})\to\Lambda(\Gamma)$. We need to show that
$\Phi(B(z,\rho)\cap\Lambda(\Gamma))\supseteq
B(\Phi(z),\frac{\rho_{0}}{L_{0}})\cap\Lambda(\Gamma)$. The map $\Phi$ is a
well defined self-homeomorphism of $\Lambda(\Gamma)$, so every $w\in
B(\Phi(z),\frac{\rho_{0}}{L_{0}})\cap\Lambda(\Gamma)$ is of the form
$\Phi(w^{\prime})$ for some $w^{\prime}\in\Lambda(\Gamma)$. We know that
$D_{x}(\Phi(w^{\prime}),\Phi(z))\leq\frac{\rho_{0}}{L_{0}}$, then
$D_{x}(w^{\prime},z)\leq L_{0}\cdot\frac{\rho}{\rho_{0}}\cdot
D_{x}(\Phi(w^{\prime}),\Phi(z))\leq\rho$, i.e. $w^{\prime}\in
B(z,\rho)\cap\Lambda(\Gamma)$. This concludes the proof. ∎

######  Corollary 6.11.

Let $(X_{n},x_{n},\Gamma_{n})\underset{\textup{eq-
pGH}}{\longrightarrow}(X_{\infty},x_{\infty},\Gamma_{\infty})$, with
$(X_{n},x_{n},\Gamma_{n})\in\mathcal{M}(\delta,D)$. Then the spaces
$\Lambda(\Gamma_{n})$ are uniformly q.s.s. and uniformly perfect.

###### Proof.

By Proposition 6.7 all the spaces $\Lambda(\Gamma_{n})$ are compact and
$(L_{0},\rho_{0})$-q.s.s. By [Cav21, Corollary 5.9] there exists $H\geq 0$
such that $h_{\Gamma_{n}}\leq H$ for all $n\in\mathbb{N}$. Then
$\Lambda(\Gamma_{n})$ is $a_{0}$-uniformly perfect for the same $0<a_{0}<1$,
by Corollary 6.6. ∎

The last step we need is the following.

######  Proposition 6.12.

If $(X_{n},x_{n},\Gamma_{n})\underset{\textup{eq-
pGH}}{\longrightarrow}(X_{\infty},x_{\infty},\Gamma_{\infty})$, with
$(X_{n},x_{n},\Gamma_{n})\in\mathcal{M}(\delta,D)$, then there exist visual
metrics $D_{n}\in\mathcal{J}(\Lambda(\Gamma_{n}))$ for
$n\in\mathbb{N}\cup\\{\infty\\}$ such that
$(\Lambda(\Gamma_{n}),D_{n})\underset{\textup{GH}}{\longrightarrow}(\Lambda(\Gamma_{\infty}),D_{\infty})$,
up to a subsequence.

###### Proof.

We fix a non-principal ultrafilter $\omega$. We denote by
$(X_{\omega},x_{\omega},\Gamma_{\omega})$ the ultralimit triple of the
sequence $(X_{n},x_{n},\Gamma_{n})$ (cp. [Cav21]): it is equivariantly
isometric to $(X_{\infty},x_{\infty},\Gamma_{\infty})$ by [Cav21, Proposition
3.13]. We equip each $\Lambda(\Gamma_{n})$ with a standard visual metric
$D_{n}$ of center $x_{n}$ and parameter $a_{\delta}$. Every point of the space
$\omega$-$\lim(\Lambda(\Gamma_{n}),D_{n})$ is an equivalence class of
sequences $(z_{n})$ with $z_{n}\in\Lambda(\Gamma_{n})$. Associated to this
sequence there is a sequence of geodesic rays $\xi_{x_{n},z_{n}}$ of $X_{n}$.
It is classical (cp [CS21], Lemma A.7) that this sequence of geodesic rays
define a limit geodesic ray $\xi_{x_{\omega},z_{\omega}}$, with
$z_{\omega}\in\partial X_{\omega}$. The map
$\Psi\colon\omega$-$\lim(\Lambda(\Gamma_{n}),D_{n})\to\partial X_{\omega}$
defined by $\Psi((z_{n}))=z_{\omega}$ is a well defined homeomorphism by
[Cav21, Proposition 5.11]. Moreover the proof of [Cav21, Theorem A.(i)] shows
that the image of $\Psi$ is exactly $\Lambda(\Gamma_{\omega})$. We denote by
$D_{\omega}$ the distance induced by $\Psi$ on $\Lambda(\Gamma_{\omega})$. By
definition the two spaces $\omega$-$\lim(\Lambda(\Gamma_{n}),D_{n})$ and
$(\Lambda(\Gamma_{\omega}),D_{\omega})$ are isometric. Proposition 5.1 implies
that, up to a subsequence,
$(\Lambda(\Gamma_{n}),D_{n})\underset{\textup{GH}}{\longrightarrow}(\Lambda(\Gamma_{\omega}),D_{\omega})$.
We claim that $D_{\omega}$ is a visual metric on $\Lambda(\Gamma_{\omega})$.
This would imply, since $(X_{\infty},x_{\infty},\Gamma_{\infty})$ and
$(X_{\omega},x_{\omega},\Gamma_{\omega})$ are equivariantly isometric, that
$(\Lambda(\Gamma_{n}),D_{n})\underset{\textup{GH}}{\longrightarrow}(\Lambda(\Gamma_{\infty}),D_{\infty})$
for a visual metric $D_{\infty}$ on $\Lambda(\Gamma_{\infty})$.
Let us prove the claim. We take two points
$\Psi(z),\Psi(z^{\prime})\in\Lambda(\Gamma_{\omega})$, with $z=\omega$-$\lim
z_{n}$, $z^{\prime}=\omega$-$\lim
z_{n}^{\prime}\in\omega$-$\lim(\Lambda(\Gamma_{n}),D_{n})$. By definition
$D(\Psi(z),\Psi(z^{\prime}))=\omega$-$\lim
D_{n}(z_{n},z_{n}^{\prime})=e^{-a_{\delta}R}$ for some $R\geq 0$. From (6.6)
we have $\omega$-$\lim(z_{n},z_{n}^{\prime})_{x_{n}}\geq R-4\delta$, then
$\omega$-$\lim
d(\xi_{x_{n},z_{n}}(R-5\delta),\xi_{x_{n},z_{n}^{\prime}}(R-5\delta))\leq
4\delta$ by Lemma 6.8.(i). This implies
$d(\xi_{x_{\omega},z_{\omega}}(R-5\delta),\xi_{x_{\omega},z_{\omega}^{\prime}}(R-5\delta))\leq
4\delta$ and so $(\Psi(z),\Psi(z^{\prime}))_{x_{\omega}}>R-8\delta$ by Lemma
6.8.(ii). This means

$D_{\omega}(\Psi(z),\Psi(z^{\prime}))=e^{-a_{\delta}R}\geq e^{-a_{\delta}\cdot
8\delta}\cdot
e^{-a_{\delta}(\Psi(z),\Psi(z^{\prime}))_{x_{\omega}}}=\frac{1}{4}\cdot
e^{-a_{\delta}(\Psi(z),\Psi(z^{\prime}))_{x_{\omega}}}.$

Analogously, with the same notation as above, we have
$\omega$-$\lim(z_{n},z_{n}^{\prime})_{x_{n}}\leq R$ and so $\omega$-$\lim
d(\xi_{x_{n},z_{n}}(R+3\delta),\xi_{x_{n},z_{n}^{\prime}}(R+3\delta))\geq
6\delta$ by Lemma 6.8.(ii). By definition
$d(\xi_{x_{\omega},z_{\omega}}(R+3\delta),\xi_{x_{\omega},z_{\omega}^{\prime}}(R+3\delta))\geq
6\delta$, so $(\Phi(z),\Phi(z^{\prime}))_{x_{\omega}}\leq R+4\delta$ by Lemma
6.8.(i). This means

$D_{\omega}(\Psi(z),\Psi(z^{\prime}))=e^{-a_{\delta}R}\leq e^{a_{\delta}\cdot
4\delta}\cdot e^{-a_{\delta}(\Psi(z),\Psi(z^{\prime}))_{x_{\omega}}}=2\cdot
e^{-a_{\delta}(\Psi(z),\Psi(z^{\prime}))_{x_{\omega}}}.$

This shows that $D_{\omega}$ is a visual metric on $\Lambda(\Gamma_{\omega})$
and concludes the proof. ∎

Theorem B follows from Corollary 6.11, Proposition 6.12 and Proposition 6.4.

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