text:
\section{Introduction}
Given a metric space $(X,d)$ its Ahlfors regular conformal gauge 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
$$\frac{d_Y(F(x), F(x'))}{d_Y(F(x), F(x''))} \leq \eta\left( \frac{d_X(x,x')}{d_X(x,x'')} \right)$$
for every $x,x',x'' \in X$ with $d_X(x,x'')> 0$. The notion of quasisymmetric maps was introduced in \cite{TV80} and it has played an important role in the study of quasiconformal structure on metric spaces.\\
The \emph{Ahlfors regular conformal dimension} of a metric space $(X,d)$ is defined as
$$\text{CD}(X,d) := \inf\lbrace \text{HD}(X,d') \text{ s.t. } d'\in \mathcal{J}(X,d)\rbrace,$$
where  $\text{HD}$ denotes the Hausdorff dimension.
\\
In general $\mathcal{J}(X,d)$ can be empty. In that case clearly CD$(X,d)=+\infty$. On the other hand the conformal dimension of any doubling, uniformly perfect metric space is always finite by \cite{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 \ref{prop-perfect-implies-uniform}).
\begin{defin}
	\label{defin-qss}
	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 $\Phi(B(x,\rho)) \supseteq B(\Phi(x),\frac{\rho_0}{L_0})$.
\end{defin}
\noindent 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.
\vspace{2mm}

\noindent This notion arises naturally in the study of limit sets of Gromov-hyperbolic groups and semi-hyperbolic rational fractions, see for instance \cite{Sul82}, \cite{Hai07}, \cite{BK13} and \cite{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.
\vspace{2mm}

\noindent 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.
\begin{theorem}
	\label{theorem-main-intro}
	Let $(X_n,d_n)$ be a sequence of compact, $a_0$-uniformly perfect, $(L_0,\rho_0)$-q.s.s. Suppose it converges in the Gromov-Hausdorff sense to a compact metric space $(X_\infty, d_\infty)$. Then
	$\textup{CD}(X_\infty, d_\infty)\geq \limsup_{n\to +\infty} \textup{CD}(X_n, d_n)$.
\end{theorem}
The notion of uniform perfectness will be recalled in Section \ref{sec-basic}. For quasi-selfsimilar spaces it is quantitatively equivalent to require that the diameter of the balls $B(\Phi(x),\frac{\rho_0}{L_0})$ appearing in Definition \ref{defin-qss} is uniformly bounded below, see Proposition \ref{prop-perfect-implies-uniform}.\\
We remark that Theorem \ref{theorem-main-intro} is false if the spaces are not all $(L_0,\rho_0)$-q.s.s.: the sequence $X_n = [0,\frac{1}{n}] \subseteq \mathbb{R}$ converges in the Gromov-Hausdorff sense to $X_\infty = \lbrace 0 \rbrace$, 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 \ref{theorem-main-intro} cannot be improved to continuity in general.
\begin{example-no}
	\label{example-cantor-intro}
	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. It is easy to show that the sequence $X_n$ converges to $X_\infty = [0,1]$ in the Gromov-Hausdorff sense. However \textup{CD}$(X_n, d_E)=0$ for every $n$ by Proposition 15.11 of \cite{DS97} (see also Theorem 2.16 of \cite{Pia11}), while \textup{CD}$(X_\infty, d_E)=1$.
\end{example-no}

We present now an application of Theorem \ref{theorem-main-intro}. In \cite{Cav21ter} 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 \ref{sec-Gromov-application} for more details about these terms. One of the main results of \cite{Cav21ter} is the closeness of $\mathcal{M}(\delta,D)$ under equivariant pointed Gromov-Hausdorff convergence. Under this convergence there is also convergence in the Gromov-Hausdorff sense of the limit sets $\Lambda(\Gamma_n)$ to $\Lambda(\Gamma_\infty)$ up to quasisymmetric equivalence, as pointed out in Proposition \ref{prop-convergence}. This fact will allow us to apply Theorem \ref{theorem-main-intro} to this particular setting.

\begin{theorem}
	\label{theorem-main-introduction}
	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
	$\textup{CD}(\Lambda(\Gamma_\infty)) \geq \limsup_{n\to +\infty}\textup{CD}(\Lambda(\Gamma_n)).$
\end{theorem}
Here the Ahlfors regular conformal dimension is computed with respect to any visual metric on the limit sets, see Proposition \ref{prop-isometry-quasisymmetry} and the discussion below.
\vspace{2mm}

\noindent The following two questions remain open:
\begin{question}
	Do we have continuity of the conformal dimension in the setting of Theorem \ref{theorem-main-introduction}?
\end{question}
\begin{question}
	\label{question-continuity-general}
	Are there conditions on the metric spaces $X_n$ that ensures continuity of the conformal dimension under Gromov-Hausdorff convergence?
\end{question}
Related to the second question we consider the following example (the author thanks M.Murugan for bringing the reference \cite{Tys00} to his attention).
\begin{example-no}
	\label{conjecture}
	Let us repeat the construction of Example \ref{example-cantor-intro} 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. It is easy to show that $(X_n, d_E)$ 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 \cite{Tys00}, Theorem 3.4 and Example 3.2).
\end{example-no}
We will briefly discuss Question \ref{question-continuity-general} and the example above at the end of Section \ref{sec-upper}.

\section{Preliminaries}
\label{sec-basic}
A metric space is a couple $(X,d)$, where $d$ is a distance function on $X$. The open (resp. closed) ball of center some point $x\in X$ 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 $(X_\infty, d_\infty)$.

\subsection{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 quite classical.
\begin{lemma}
	\label{lemma-Ahlfors-perfect}
	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.
\end{lemma}
\begin{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 $a_0 < A_0^{-2/s_0}$.
\end{proof}

\subsection{Quasi-selfsimilar spaces}

We collect now some basic properties of quasi-selfsimilar metric spaces, see Definition \ref{defin-qss}.
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 \ref{defin-qss} has diameter $\geq c_0$ for every $x\in X$ and every $0<\rho\leq \rho_0$.

\begin{prop}[Compare with \cite{Pia11}, Lemma 2.2 and Proposition 2.3]
	\label{prop-perfect-implies-uniform}
	Let $(X,d)$ be a quasi-selfsimilar metric space as in Definition \ref{defin-qss}. Then:
	\begin{itemize}
		\item[(i)] it is doubling;
		\item[(ii)] if it is perfect then it is uniformly perfect;
		\item[(iii)] it is uniformly perfect if and only if it has diameters bounded below, quantitatively.
	\end{itemize}
\end{prop}
\begin{proof}
	If (i) is not true then for every $n\in \mathbb{N}$ there are $x_n \in X$ 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 $x_\infty \in X$. $X$ is not totally bounded if $\rho_\infty > 0$, a contradiction. If $\rho_\infty = 0$ we can find $L_0$-biLipschitz maps $\Phi_n\colon \left(B(x_{n}, \rho_{n}), \frac{\rho_0}{\rho_{n}}\cdot d_{n}\right) \to X_{n}$ 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. Clearly we can find a $\frac{\rho_0}{2L_0}$-separated set inside $B(\Phi_n(x_{n}), \frac{\rho_0}{L_0})$ with cardinality $\geq n$. Once again this contradicts the compactness of $X$.

	\noindent Now we show (ii). Since $X$ is perfect and compact we have the following property, as in Lemma 2.2 of \cite{Pia11}: 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 are $x_n\in X$ 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 $\rho_\infty$. If $\rho_\infty > 0$ then we show that $X$ is not perfect, a contradiction. Indeed let $y$ be any point inside $B(x_\infty, \rho_\infty)$. It 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.\\
	Suppose $\rho_\infty = 0$: then for all $n$ big enough we take the map $\Phi_n \colon B(x_n,\rho_n) \to X_n$ given by the definition of quasi-selfsimilarity. Therefore we get
	$$B\left(\Phi(x_n), \frac{r_0}{L_0}\right) \subseteq \Phi_n(B(x_n,\rho_n)) = \Phi_n\left(B\left(x_n,\frac{\rho_n}{n}\right)\right).$$
	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.

	\noindent 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 \ref{defin-qss}. Therefore Diam$(B(\Phi(x), \frac{\rho_0}{L_0})) \geq a_0\cdot\frac{\rho_0}{L_0} =: c_0$.
\end{proof}
We remark that the diameter bounded below condition is part of the definition of quasi-selfsimilar spaces in \cite{Kle06} and in \cite{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 \ref{prop-perfect-implies-uniform}. For instance in the context of Theorem \ref{theorem-main-intro} 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 $0 < a_0 < 1$.

\section{Combinatorial modulus}
It is known that the conformal dimension of a metric space is closely related to the combinatorial modulus, see for instance \cite{BK13}, \cite{Pia12}, \cite{MT10} and the references therein. In this section we recall the definition of combinatorial modulus and we prove some technical lemma.
\vspace{1mm}

\noindent 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.
\vspace{1mm}

\noindent A $(\lambda,k)$-path is a finite collection $\gamma = \lbrace q_j \rbrace_{j=0}^M$ 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.
\vspace{1mm}

\noindent 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 $\lbrace q_i\rbrace_{i=0}^M \in P_{\lambda,k}(E,F)$.
\vspace{1mm}

\noindent 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 $(\lambda,k)$. It is easy to see that 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 $p\text{-Mod}_{\lambda,k}(E,F) = 0$.

\begin{lemma}
	\label{lemma-base}
	If $E'\subseteq E$ and $F'\subseteq F$ then $p\textup{-Mod}_{\lambda,k}(E',F') \leq p\textup{-Mod}_{\lambda,k}(E,F)$.
\end{lemma}
\begin{proof}
	If $P_{\lambda,k}(E',F') = \emptyset$ then the result is trivial by definition. Otherwise every path in $P_{\lambda,k}(E',F')$ belongs to $P_{\lambda,k}(E,F)$. This implies that $\mathcal{A}_{\lambda,k}(E,F)\subseteq \mathcal{A}_{\lambda,k}(E',F')$, and the inequality follows directly from the definition.
\end{proof}

\noindent 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 $X$.
\begin{lemma}(cp. \cite{Pia11}, Lemma 4.4)
	\label{lemma-rescaling}
	Let $k,\lambda,p$ be fixed quantities as above.
	Let $1\leq L_1' \leq L_1 < L_2 \leq L_2'$. Then there exist $\ell\in\mathbb{N}$ and $C>0$, depending only on $L_1, L_1', L_2, L_2'$ and $D$, such that
	$$p\textup{-Mod}_{\lambda,k}^{L_1',L_2'}(X) \leq p\textup{-Mod}_{\lambda,k}^{L_1,L_2}(X)$$
	and
	$$p\textup{-Mod}_{\lambda,k + \ell}^{L_1,L_2}(X) \leq C\cdot p\textup{-Mod}_{\lambda,k}^{L_1',L_2'}(X).$$
\end{lemma}
\begin{proof}
	The first inequality is trivial. Indeed for every $y\in X_i$, $i\in \mathbb{N}$, we have $\overline{B}_{L_1',i}(y) \subseteq \overline{B}_{L_1,i}(y)$ and $X\setminus B_{L_2',i}(y) \subseteq X\setminus B_{L_2,i}(y)$, so the conclusion follows by Lemma \ref{lemma-base}.\\
	In order to prove the second inequality we define $\ell$ as the minimum integer satisfying $10^{-\ell}\leq \frac{L_2 - L_1}{L_2' + L_1'}$. We fix $y\in X_i$ for some $i\in \mathbb{N}$ and we consider the set $X_{i+\ell}(y) = \lbrace z\in X_{i+\ell} \text{ s.t. } B_{L_1',i+\ell}(z) \cap \overline{B}_{L_1,i}(y) \neq \emptyset \rbrace$.
	We fix any $(\lambda,i+\ell+k)$-path $\gamma = \lbrace q_j \rbrace_{j=0}^M$ 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 $\overline{B}_{L_1',i+\ell}(z)$ and $X\setminus B_{L_2',i+\ell}(z)$. Indeed we know that $d(z, q_0) \leq 10^{-i-\ell} \leq L_1' \cdot 10^{-i-\ell}$. Moreover $d(z,y)\leq L_1\cdot 10^{-i} + L_1'\cdot 10^{-i-\ell}$. Therefore
	$$d(z,q_N) \geq L_2\cdot 10^{-i} - L_1\cdot 10^{-i} - L_1'\cdot 10^{-i-\ell} \geq L_2'\cdot 10^{-i-\ell}$$
	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 $P_{\lambda, i+\ell+k}(\overline{B}_{L_1',i+\ell}(z), X\setminus B_{L_2',i+\ell}(z))$
	for some $z\in X_{i+\ell}(y)$.\\
	For each $z\in X_{i+\ell}(y)$ we take optimal functions $f_z \in \mathcal{A}_{\lambda,i+\ell+k}(\overline{B}_{L_1',i+\ell}(z), X\setminus B_{L_2',i+\ell}(z))$ 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 $\lbrace q_j \rbrace_{j=0}^M \in P_{\lambda, i+\ell+k}(\overline{B}_{L_1,i}(y), X\setminus B_{L_2,i}(y))$ belongs to $P_{\lambda, i+\ell+k}(B_{L_1',i+\ell}(z), X\setminus B_{L_2',i+\ell}(z))$ for some $z\in X_{i+\ell}(y)$, therefore
	$$\sum_{j=0}^Mf(q_j) \geq \sum_{j=0}^Mf_z(q_j) \geq 1.$$
	Finally we have:
	\begin{equation*}
		\begin{aligned}
			\sum_{q\in X_{i+\ell+k}}f(q)^p &= \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 \\
			&= \sum_{z\in  X_{i+\ell}(y)} p\text{-Mod}^{L_1',L_2'}_{\lambda,k}(z) \leq C\cdot p\text{-Mod}^{L_1',L_2'}_{\lambda,k}(X),
		\end{aligned}
	\end{equation*}
	where $C$ is a constant depending only on the doubling constant $D$, on $\ell$ and on $L_1'$. This shows that
	$$p\text{-Mod}^{L_1,L_2}_{\lambda, k+\ell}(y) \leq C\cdot p\text{-Mod}^{L_1',L_2'}_{\lambda,k}(X).$$
	Since this is true for every $y\in X_i$ and for every $i$ we get
	$$p\text{-Mod}^{L_1,L_2}_{\lambda,k+\ell}(X) \leq C\cdot p\text{-Mod}^{L_1',L_2'}_{\lambda,k}(X).$$
\end{proof}

\begin{lemma}
	\label{lemma-lambda-rescaling}
	Let $k\in\mathbb{N}$, $p\geq 0$, $1\leq L_1 < L_2$ and $2<\lambda \leq \lambda'$. Then there exist $\ell\in \mathbb{N}$ and $C>0$ depending only on $\lambda,\lambda',D$ such that
	$$p\textup{-Mod}^{L_1,L_2}_{\lambda,k}(X) \leq p\textup{-Mod}^{L_1,L_2}_{\lambda',k}(X)$$
	and
	$$p\textup{-Mod}^{L_1,L_2}_{\lambda',k + \ell}(X) \leq C\cdot p\textup{-Mod}^{L_1,L_2}_{\lambda,k}(X)$$
	for all $k> k_0 = \log_{10}\left(\frac{2}{L_2-L_1}\right)$.
\end{lemma}
\begin{proof}
	For every $y\in X_i$, $i\in \mathbb{N}$, we have $$P_{\lambda,k}(\overline{B}_{L_1,i}(y), X\setminus B_{L_2,i}(y)) \subseteq P_{\lambda',k}(\overline{B}_{L_1,i}(y), X\setminus B_{L_2,i}(y)).$$
	Therefore arguing as in the proof of Lemma \ref{lemma-base} we get
	$$p\text{-Mod}_{\lambda,k}^{L_1,L_2}(y) \leq p\text{-Mod}_{\lambda',k}^{L_1,L_2}(y).$$
	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 $\lambda'\cdot 10^{-\ell} \leq \frac{\lambda}{2} -1$. 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 $(\lambda', i+k + \ell)$-path $\gamma = \lbrace q_j \rbrace_{j=0}^M$ 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 $\tilde{\gamma} = \lbrace \tilde{q}_j\rbrace_{j=0}^M$ is a $(\lambda, i+k)$-path joining $\overline{B}_{L_1',i}(y)$ to $X\setminus B_{L_2',i}(y)$, where $L_1' = L_1 + 10^{-k}$ and $L_2' = L_2 - 10^{-k}$. Indeed we have:
	$$d(y,\tilde{q}_0) \leq d(y,q_0) + d(q_0,\tilde{q}_0) \leq L_1\cdot 10^{-i} + 10^{-i-k} = L_1'\cdot 10^{-i},$$
	$$d(y,\tilde{q}_M) \geq d(y,q_M) - d(q_M,\tilde{q}_M) \geq L_2\cdot 10^{-i} - 10^{-i-k} = L_2'\cdot 10^{-i}$$
	and
	\begin{equation*}
		\begin{aligned}
			d(\tilde{q}_j, \tilde{q}_{j+1}) &\leq d(\tilde{q}_j,q_j) + d(q_j,q_{j+1}) + d(q_{j+1}, \tilde{q}_{j+1}) \\
			&\leq 2\cdot 10^{-i-k} + 2\lambda'\cdot 10^{-i-k-\ell} \\
			&\leq 2\cdot 10^{-i-k} + 2\left(\frac{\lambda}{2} -1\right)\cdot 10^{-i-k} \\
			& = \lambda \cdot 10^{-i-k}
		\end{aligned}
	\end{equation*}
	for every $j=0,\ldots,M-1$.	Observe that the condition on $k$ implies $L_1'<L_2'$.
	We are ready to compare the combinatorial moduli. We take an optimal function $\tilde{f} \in \mathcal{A}_{\lambda,i+k}(\overline{B}_{L_1',i}(y), X\setminus B_{L_2',i}(y))$ and we define the function $f\colon X_{i+k+\ell} \to [0,+\infty)$ by
	$$f(q) := \max\lbrace \tilde{f}(\tilde{q}) \text{ s.t. } \tilde{q}\in X_{i+k} \text{ and } d(q,\tilde{q}) \leq 10^{-i-k} \rbrace.$$
	First of all we show that $f\in \mathcal{A}_{\lambda',i+k+\ell}(\overline{B}_{L_1,i}(y), X\setminus B_{L_2,i}(y))$. Indeed we have seen that given any $(\lambda',i+k+\ell)$-path $\lbrace q_j \rbrace_{j=0}^M$ joining $\overline{B}_{L_1,i}(y)$ to $X\setminus B_{L_2,i}(y)$ there is an associated $(\lambda, i+k)$-path $\lbrace \tilde{q}_j \rbrace_{j=0}^M$ joining $\overline{B}_{L_1',i}(y)$ to $X\setminus B_{L_2',i}(y)$ 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
	\begin{equation*}
		\begin{aligned}
			p\text{-Mod}^{L_1,L_2}_{\lambda',k+\ell}(y) &\leq \sum_{q\in X_{i+k+\ell}}f^p(q) \leq C' \cdot \sum_{\tilde{q}\in X_{i+k}}\tilde{f}^p(\tilde{q}) \\
			&= C'\cdot p\text{-Mod}^{L_1',L_2'}_{\lambda,k}(y) \leq C' \cdot p\text{-Mod}^{L_1',L_2'}_{\lambda,k}(X),
		\end{aligned}
	\end{equation*}
	where $C'$ 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
	$$p\text{-Mod}^{L_1,L_2}_{\lambda',k+\ell}(X) \leq C' \cdot p\text{-Mod}^{L_1',L_2'}_{\lambda,k}(X).$$
	This inequality is true for all $k\geq k_0$. Choosing $L_1'' = L_1 + 10^{-k_0}$ and $L_2'' = L_2 - 10^{-k_0}$ one concludes, using the easy inequality in Lemma \ref{lemma-rescaling}, that
	$$p\text{-Mod}^{L_1,L_2}_{\lambda',k+\ell}(X) \leq C' \cdot p\text{-Mod}^{L_1'',L_2''}_{\lambda,k}(X)$$
	for all $k\geq k_0$.
	An application of the non-trivial inequality of Lemma \ref{lemma-rescaling} concludes the proof.
\end{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
$$p\text{-Mod}_k(X) = \sup_{i\in\mathbb{N}}\sup_{y\in X_i}p\text{-Mod}_k(y).$$

\section{Combinatorial modulus on quasi-selfsimilar spaces}
In this section we consider the class of quasi-selfsimilar metric spaces as defined in Definition \ref{defin-qss}: on these spaces the computation of the combinatorial moduli is easier. Since the explicit doubling constant of our metric space plays an important role we give another 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 it is $D_0$-doubling. We start with a lemma.
\begin{lemma}
	\label{lemma-qss-propagation}
	If $X$ is $(L_0,\rho_0)$-q.s.s. then it is $(L_0,\rho_1)$-q.s.s. for every $0<\rho_1\leq \rho_0$.
\end{lemma}
\begin{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 it 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 $\Phi(B(x,\rho)) \supseteq B(\Phi(x),\frac{\rho_1}{L_0})$.
\end{proof}

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 $I_0 = \lbrace i \in \mathbb{N} \text{ s.t. } (L_0 + 5)\cdot 10^{-i} \geq 10^{-i_0}\rbrace$. Observe that the set $I_0$ is of the form $\lbrace 1,\ldots,n_0\rbrace$ 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 allow us to use \emph{the fixed sizes up to $n_0$} to estimate the combinatorial modulus.
\begin{prop}
	\label{prop-key}
	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 $k\in\mathbb{N}$.
\end{prop}
\begin{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 \ref{lemma-qss-propagation} 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 $\lbrace q_j \rbrace_{j=0}^M$ joining $\overline{B}_{1,i}(y)$ to $X\setminus B_{(L_0 + 5)^3,i}(y)$. This means
	\begin{itemize}
		\item[-] $q_0 \in \overline{B}(y,10^{-i})$ and $q_M \notin B(y,(L_0 + 5)^3\cdot 10^{-i})$;
		\item[-] $\overline{B}(q_j, 10\cdot 10^{-i-k}) \cap \overline{B}(q_{j+1}, 10\cdot 10^{-i-k}) \neq \emptyset$ for every $j = 0,\ldots,M-1$.
	\end{itemize}
	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
	\begin{equation*}
		\begin{aligned}
			d(\tilde{q}_j, \tilde{q}_{j+1}) &\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}) \\
			&\leq 10^{-i_0-k+1} + 20\cdot 10^{-i_0-k} + 10^{-i_0-k+1} \\
			&\leq 10 \cdot 10^{-i_0 - k + 1}.
		\end{aligned}
	\end{equation*}
	In other words $\lbrace \tilde{q}_j\rbrace_{j=0}^M$ 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
	$$f(q) := \max\lbrace \tilde{f}(\tilde{q}) : \tilde{q}\in X_{i_0+k-1}\cap \overline{B}(\Phi(q), 10^{-i_0-k+1})\rbrace$$
	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  $\lbrace q_j \rbrace_{j=0}^M \in P_{i+k}(\overline{B}_{1,i}(y), X\setminus B_{(L_0 + 5)^3,i}(y))$. First of all we can extract the minimal subpath $\lbrace q_j \rbrace_{j=0}^{M'}$ such that $q_{M'} \notin B_{(L_0 + 5)^3,i}(y)$. Clearly if $\sum_{j=0}^{M'} f(q_j) \geq 1$ 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 $\lbrace \tilde{q}_j \rbrace_{j=0}^{M'}$ as in the first part of the proof since  $q_j\in B_{2(L_0 + 5)^3,i}(y)$ for every $j=0,\ldots,M'$. By definition it holds:
	$$\sum_{j=0}^{M'} f(q_j) \geq \sum_{j=0}^{M'} \tilde{f}(\tilde{q}_j) \geq 1.$$
	Moreover we have
	\begin{equation*}
		\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),
	\end{equation*}
	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 \ref{lemma-rescaling} we obtain the second inequality, indeed:
	\begin{equation*}
		\begin{aligned}
			p\text{-Mod}_k(X) = p\text{-Mod}_{10,k}^{3,4}(X) &\leq C \cdot p\text{-Mod}_{10,k - \ell}^{1, (L_0+5)^3}(X) \\
			&\leq C\cdot D_0 \cdot p\text{-Mod}_{k- \ell -1}(X,n_0),
		\end{aligned}
	\end{equation*}
	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 $\ell_0 = \ell + 1$.

\end{proof}

The Ahlfors regular conformal dimension of a compact, doubling, uniformly perfect metric space $(X,d)$ coincides with the critical exponent of the combinatorial modulus.
\begin{theo}[\cite{Pia11}, Theorem 4.5]
	\label{theo-CD-modulus}
	Let $(X,d)$ be a compact, doubling, uniformly perfect metric space. Then
		$$\textup{CD}(X,d) = \inf\lbrace p\geq 0 \textup{ s.t. } \liminf_{k\to+\infty} p\textup{-Mod}_k(X) = 0 \rbrace.$$
\end{theo}

Clearly by Lemma \ref{lemma-rescaling} and Lemma \ref{lemma-lambda-rescaling} 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 \cite{Pia11} and \cite{BK13}, we can find in the quasi-selfsimilar setting a uniform estimate which will be the key ingredient of the proof of Theorem \ref{theorem-main-intro}.
\begin{prop}
	\label{prop-uniform-lower-bound-p-modulus}
	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 $k> 0$.
\end{prop}
\begin{proof}
	$(X,d)$ is uniformly perfect and doubling by Proposition \ref{prop-perfect-implies-uniform}, so the Ahlfors regular conformal dimension of $(X,d)$ can be computed as in Theorem \ref{theo-CD-modulus}. The result follows by a submultiplicative estimate. Lemma 4.9 of \cite{Pia11} 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 \ref{lemma-rescaling} we get
	$$p\text{-Mod}_{k+h}(X) \leq C'\cdot p\text{-Mod}_{k-\ell}(X) \cdot p\text{-Mod}_{h}(X)$$
	for all $k\geq \ell$ and $h\geq 0$, where $C'$ 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 simply $a_{k+h}\leq C'\cdot a_{k-\ell}\cdot a_h$. By Theorem \ref{theo-CD-modulus} $\liminf_{k\to+\infty}a_k > 0$ since $p< \text{CD}(X,d)$. This implies that $a_k \geq \frac{1}{C'}$ for all $k > 0$. Indeed, if there exists $k > 0$ such that $C'\cdot a_k < (1-\varepsilon)$ for some $\varepsilon > 0$ then
	$$a_{n(k+\ell)} \leq C'\cdot a_{k}\cdot a_{(n-1)(k+\ell)} \leq \cdots \leq (1-\varepsilon)^n$$
	for all $n\in\mathbb{N}$. Therefore the subsequence $\lbrace a_{n(k+\ell)}\rbrace_{n\in\mathbb{N}}$ 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 \ref{prop-key} gives the thesis.
\end{proof}

\section{Upper semicontinuity of the conformal dimension}
\label{sec-upper}
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 \cite{DK18} and \cite{CavS20}. A non-principal ultrafilter $\omega$ is a finitely additive measure on $\mathbb{N}$ such that $\omega(A) \in \lbrace 0,1 \rbrace$ 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 $\lbrace n \in \mathbb{N} \text{ s.t. } \vert a_n - a \vert < \varepsilon\rbrace$ has $\omega$-measure $1$, see for instance \cite{DK18}, Lemma 10.25. The real number $a$ is called the ultralimit of the sequence $a_n$ and it is denoted by $\omega$-$\lim a_n$.\\
Given a sequence of pointed metric spaces $(X_n, d_n, x_n)$ 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 $n$, for which there exists $M\in\mathbb{R}$ such that $d(x_n,y_n)\leq M$ for $\omega$-a.e.$(n)$, modulo the relation $(y_n)\sim (y_n')$ if and only if $\omega$-$\lim d(y_n,y_n') = 0$. 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 $d_\omega(\omega$-$\lim y_n, \omega$-$\lim y_n') = \omega$-$\lim d(y_n,y_n')$ 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 in the following
\begin{prop}[\cite{Jan17}, Proposition 3.11; \cite{Cav21ter}, Proposition 3.13]
	\label{prop-GH-ultralimit}
	Let $(X_n, d_n, x_n)$ be a sequence of pointed, compact metric spaces and let $\omega$ be a non-principal ultrafilter.
	\begin{itemize}
		\item[(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.
		\item[(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 $\lbrace{n_k}\rbrace$.
	\end{itemize}
\end{prop}

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 $f_\omega(\omega$-$\lim x_n) = \omega$-$\lim f_n(x_n)$.\\
First of all we show that the class of uniformly perfect $(L_0,\rho_0)$-quasi-selfsimilar metric spaces is closed under Gromov-Hausdorff convergence.
\begin{prop}
	\label{prop-closeness-GH}
	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.
\end{prop}
\begin{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 \ref{prop-GH-ultralimit}. We fix a generic point $x_\omega = \omega$-$\lim x_n \in X_\omega$ and $0<\rho\leq \rho_0$. For every $n$ there is a $L_0$-biLipschitz maps $\Phi_n\colon \left(B(x_n,\rho), \frac{\rho_0}{\rho}\cdot d_n \right) \to X_n$ 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.
\end{proof}
\begin{obs}
	This proposition, together with Proposition \ref{prop-perfect-implies-uniform}, 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 \cite{Kle06} and \cite{Pia11}, is still uniformly perfect.
\end{obs}

We can now give the
\begin{proof}[Proof of Theorem \ref{theorem-main-intro}]
	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. 	\\
	\noindent \textbf{Step 1:} \emph{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 are $n_j$, $x_{n_j} \in X_{n_j}$ 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 can find $L_0$-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 $\rho_{n_j} \leq \rho_0$. Clearly 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 $X_\infty$.
	\vspace{2mm}

	\noindent 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 \ref{prop-GH-ultralimit}.
	\vspace{1mm}

	\noindent \textbf{Step 2:} \emph{Let $k\in \mathbb{N}$. We fix a maximal $10^{-k}$-separated subset $X_{k,n}$ of $X_n$. Then:\vspace{-3mm}
		\begin{itemize}
			\item[(i)] the cardinality of $X_{k,n}$ is uniformly bounded from above, and each $X_{k,n}$ is a $10^{-k}$-net of $X_n$;
			\item[(ii)] the set $X_{k,\omega} := \lbrace \omega\textup{-}\lim q_n \textup{ s.t. } q_n \in X_{k,n}\rbrace$ is a $10^{-k}$-net of $X_\omega$;
			\item[(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 $n\in A_k$.
	\end{itemize} }
	\noindent 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 \ref{sec-basic}.\\
	We take two points $\omega$-$\lim q_n$, $\omega$-$\lim q_n' \in X_{k,\omega}$. If $d(\omega\text{-}\lim q_n,\omega\text{-}\lim q_n') < 10^{-k}$ then $d(q_n,q_n')<10^{-k}$ $\omega$-a.s. and by definition $q_n = q_n'$ $\omega$-a.s., implying $\omega\text{-}\lim q_n = \omega\text{-}\lim q_n'$. 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 $d(y_\omega, q_\omega) > 10^{-k}$ for all $q_\omega \in X_{k,\omega}$. Since $X_{k,\omega}$ is finite we know that $d(y_n, q_n) > 10^{-k}$ 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
	$$\vert d(q_\omega, q_\omega') - d(q_n, q_n')\vert < \frac{10^{-k}}{2}$$
	for all $q_\omega = \omega\text{-}\lim q_n, q_\omega' = \omega\text{-}\lim q_n' \in X_{k,\omega}$ 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 $n\in A$ it 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).
	\vspace{2mm}

	\noindent 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.
	\vspace{1mm}

	\noindent \textbf{Step 3:} \emph{Let $k\in\mathbb{N}$. There exists a subset $B_k\subseteq \mathbb{N}$ of $\omega$-measure $1$ such that
	\begin{itemize}
		\item[(i)] the map $\pi_n\colon X_{k,\omega} \to X_{k,n}$ from Step 2 is well defined and bijective for every $n\in B_k$;
		\item[(ii)] for every $n\in B_k$, for every $(10,k)$-path $\gamma_n = \lbrace q_j^n\rbrace_{j=0}^M$ of $X_n$ the associated path $\gamma_\omega = \lbrace \pi_n^{-1}(q_j^n)\rbrace_{j=0}^M$ is a $(30,k)$-path of $X_\omega$.
	\end{itemize}}
	\noindent Since $X_{k,\omega}$ is finite we can find a subset $B_k$ of $A_k$ with $\omega$-measure $1$ such that
	\begin{equation*}
		\vert d(q_\omega, q_\omega') - d(\pi_n(q_\omega), \pi_n(q_\omega')) \vert\leq 10\cdot 10^{-k}
	\end{equation*}
	for all $q_\omega,q_\omega' \in X_{k,\omega}$ and for all $n\in B_k$.
	Let us take a $(10,k)$-path $\gamma_n = \lbrace q_j^n\rbrace_{j=0}^M$ of $X_n$, $n\in B_k$. This means $d(q_j^n, q_{j+1}^n) \leq 20\cdot 10^{-k}$ for all $j=0,\ldots,M-1$.
	Therefore $d(\pi_n^{-1}(q_j^n), \pi_n^{-1}(q_{j+1}^n)) \leq 30\cdot 10^{-k}$, i.e. the thesis.
	\vspace{2mm}

	\noindent \textbf{Step 4:} \emph{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 $y_\omega = \omega$-$\lim y_n \in X_{i,\omega}$.}
	\vspace{1mm}

	\noindent 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
	\begin{equation*}
		\vert d(q_\omega, y_\omega) - d(\pi_n(q_n), y_n) \vert\leq \frac{1}{4}\cdot 10^{-i}
	\end{equation*}
	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 $\gamma_n = \lbrace q_j^n\rbrace_{j=0}^M$ such that $d(y_n, q_0^n) \leq 3\cdot 10^{-i}$ and $d(y_n,q_M^n)> 4\cdot 10^{-i}$. We denote by $\gamma_\omega = \lbrace \pi_n^{-1}(q_j^n)\rbrace$ the $(30,i+k)$-path given by Step 3.
	We observe that $d(y_\omega, \pi_n^{-1}(q_0^n)) \leq \frac{13}{4}\cdot 10^{-i}$ and $d(y_\omega,q_M^\omega)> \frac{15}{4}\cdot 10^{-i}$, 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
	$$p\textup{-Mod}_{30,k}^{\frac{13}{4},\frac{15}{4}}(y_\omega) \geq \omega\textup{-}\lim p\textup{-Mod}_k(y_n).$$
	\vspace{1mm}

	\noindent \textbf{Step 5:} \emph{Conclusion.}\\
	We fix $k\in \mathbb{N}$ and $0\leq p < \omega$-$\lim \text{CD}(X_n,d_n)$.

	By Proposition \ref{prop-uniform-lower-bound-p-modulus} we find a constant $\lambda_0 > 0$ depending only on $D_0$, $L_0$ and $p$ such that
	\begin{equation}
		\label{eq-uniform}
		\sup_{i\leq n_0}\sup_{y\in X_{i,n}}p\text{-Mod}_k(y) \geq \lambda_0
	\end{equation}
	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 \eqref{eq-uniform}. The sequence $i_n$ is $\omega$-a.s. equal to some $i_* \in \lbrace 1,\ldots,n_0\rbrace$. 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 \ref{lemma-rescaling} and Lemma \ref{lemma-lambda-rescaling} 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 \ref{prop-closeness-GH} and Proposition \ref{prop-perfect-implies-uniform}, so we can apply Theorem \ref{theo-CD-modulus} 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 \cite{Cav21ter} we have
	$\limsup_{n\to +\infty} \text{CD}(X_n,d_n) \leq \text{CD}(X_\infty,d_\infty).$
\end{proof}

The main tools used in this proof are: the reduction of the computation of the combinatorial modulus to a \emph{finite} set of scales and the uniform lower bound on the combinatorial modulus, independent of $k$, given by Proposition \ref{prop-uniform-lower-bound-p-modulus}. 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 \ref{theo-CD-modulus} 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 \ref{question-continuity-general} 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 strictly related to (uniform) weak super-multiplicative properties of the sequence $p$-Mod$_k(X)$, as studied in relation with the combinatorial Lowner property in \cite{BK13}, sections 4 and 8, and in the special case of the Sierpinski carpet in \cite{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 \cite{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.

\section{Application to Gromov-hyperbolic spaces}
\label{sec-Gromov-application}
We want to apply Theorem \ref{theorem-main-intro} to the case of limit sets of quasiconvex-cocompact groups of isometries of Gromov-hyperbolic spaces.\\
We recall briefly the definition and some properties of Gromov-hyperbolic metric spaces. Good references are for instance \cite{BH09} and \cite{CDP90}.

\noindent Let $X$ be a geodesic metric space. Given  three points  $x,y,z \in X$,  the {\em Gromov product} of $y$ and $z$ with respect to $x$  is defined as
\vspace{-3mm}

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

\noindent The space $X$ is said {\em $\delta$-hyperbolic}, $\delta \geq 0$, if   for every four points $x,y,z,w \in X$   the following {\em 4-points condition} hold:
\begin{equation}\label{hyperbolicity}
	(x,z)_w \geq \min\lbrace (x,y)_w, (y,z)_w \rbrace -  \delta.
\end{equation}

\noindent The space $X$ is   {\em Gromov hyperbolic} if it is $\delta$-hyperbolic for some $\delta \geq 0$.
\vspace{1mm}

\noindent Let $X$ be a proper, $\delta$-hyperbolic metric space and $x$ be a point of $X$. \\
The {\em Gromov boundary} of $X$ is defined as the quotient
$$\partial X = \lbrace (z_n)_{n \in \mathbb{N}} \subseteq X \hspace{1mm} | \hspace{1mm}   \lim_{n,m \to +\infty} (z_n,z_m)_{x} = + \infty \rbrace \hspace{1mm} /_\approx,$$
where $(z_n)_{n \in \mathbb{N}}$ is a sequence of points in $X$ and $\approx$ is the equivalence relation defined by $(z_n)_{n \in \mathbb{N}} \approx (z_n')_{n \in \mathbb{N}}$ if and only if $\lim_{n,m \to +\infty} (z_n,z_m')_{x} = + \infty$.  \linebreak
We will write $ z = [(z_n)] \in \partial X$ for short, and we say that $(z_n)$ {\em 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 $z,z'  \in \partial X$ by
$$(z,z')_{x} = \sup_{(z_n) , (z_n') } \liminf_{n,m \to + \infty} (z_n, z_m')_{x}$$
where the supremum is taken among all sequences such that $(z_n) \in z$ and $(z_n')\in z'$.
For every $z,z',z'' \in \partial X$ it continues to hold
\begin{equation}
	\label{hyperbolicity-boundary}
	(z,z')_{x} \geq \min\lbrace (z,z'')_{x}, (z',z'')_{x} \rbrace - \delta.
\end{equation}
Moreover for all sequences $(z_n),(z_n')$ converging to  $z,z'$ respectively it holds
\begin{equation}
	\label{product-boundary-property}
	(z,z')_{x} -\delta \leq \liminf_{n,m \to + \infty} (z_n,z_m')_{x} \leq (z,z')_{x}.
\end{equation}
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  \eqref{product-boundary-property}.

\noindent Every geodesic ray $\xi$ defines a point  $\xi^+=[(\xi(n))_{n \in \mathbb{N}}]$  of the Gromov boundary $ \partial X$: we  say that $\xi$ {\em joins} $\xi(0) = y$ {\em to} $\xi^+ = z$, and we denote it by  $[y, z]$. Moreover for every $z\in \partial X$ and every $x\in X$ it is possible to find a geodesic ray $\xi$ such that $\xi(0)=x$ and $\xi^+ = z$. Any such geodesic ray is denoted as $\xi_{x,z} = [x,z]$ even if it is possibly not unique.
Analogously, given different points $z = [(z_n)], z' = [(z'_n)] \in \partial X$ there always exists  a geodesic line $\gamma$ joining $z'$ to $z$, i.e. such that  $\gamma|_{[0, +\infty)}$ and $\gamma|_{(-\infty,0]}$ join   $\gamma(0)$ to $z,z'$ respectively (just  consider the limit $\gamma$ of the segments $[z_n,z'_n]$; notice that  all these segments intersect a ball   of fixed radius centered at $x_0$, since $(z_n,z'_m)_{x_0}$ is uniformly bounded above). We call $z$ and $z'$ the  {\em positive} and {\em negative endpoints} of $\gamma$, respectively,  denoted  $\gamma^\pm$.
The relation between Gromov product and geodesic ray is highlighted in the following well known lemma.
\begin{lemma}[\cite{Cav21ter}, Lemma 4.2]
	\label{product-rays}
	Let $X$ be a proper, $\delta$-hyperbolic metric space, $z,z'\in \partial X$ and $x\in X$.
	\begin{itemize}
		\item[(i)] If $(z,z')_{x} \geq T$ then $d(\xi_{x,z}(T - \delta),\xi_{x,z'}(T - \delta)) \leq 4\delta$.
		\item[(ii)] If $d(\xi_{x,z}(T),\xi_{x,z'}(T)) < 2b$ then $(z,z')_{x} > T - b$, for all $b>0$.
	\end{itemize}
\end{lemma}

\noindent The following is a standard computation, see for instance \cite{BCGS}.
\begin{lemma}
	\label{parallel-geodesics}
	Let $X$ be a proper, $\delta$-hyperbolic metric space. Then every two geodesic rays $\xi, \xi'$ with same endpoints at infinity are at distance at most $8\delta$, more precisely there exist $t_1,t_2\geq 0$ such that $t_1+t_2=d(\xi(0),\xi'(0))$ and  $d(\xi(t + t_1),\xi'(t+t_2)) \leq 8\delta$ for all $t\geq 0$.
\end{lemma}

The {\em 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$(C)$. It is $36\delta$-quasiconvex (\cite{Cav21ter}, Lemma 4.5), i.e. every point of every geodesic segment joining every two points $y,y'$ of $\text{QC-Hull}(C)$ is at distance at most $36\delta$ from $\text{QC-Hull}(C)$. We need the following approximation result.
\begin{lemma}[\cite{Cav21ter}, Lemma 4.6]
	\label{approximation-ray-line}
	Let $X$ be a proper, $\delta$-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 $z\in C$.
\end{lemma}

When $X$ is a proper, $\delta$-hyperbolic metric space it is known that the boundary $\partial X$ is metrizable. A metric $D_{x,a}$ on $\partial X$ is called a {\em visual metric} of center $x \in X$ and parameter $a\in\left(0,\frac{1}{2\delta\cdot\log_2e}\right)$ if there exists $V> 0$ such that for all $z,z' \in \partial X$ it holds
\begin{equation}
	\label{visual-metric}
	\frac{1}{V}e^{-a(z,z')_{x}}\leq D_{x,a}(z,z')\leq V e^{-a(z,z')_{x}}.
\end{equation}
A visual metric is said {\em standard} if for all $z,z'\in \partial X$ it holds
$$(3-2e^{a\delta})e^{-a(z,z')_{x}}\leq D_{x,a}(z,z')\leq e^{-a(z,z')_{x}}.$$
For all $a$ as before and $x\in X$ there exists always a standard visual metric of center $x$ and parameter $a$, see \cite{Pau96}.
The identity map $\text{id}\colon(\partial X, D_{x,a}) \to (\partial X, D_{x',a'})$ is a quasisymmetric equivalence for all possible choices of $x,x'\in X$ and $a,a'$. Indeed we have
\begin{prop}[\cite{Hai07}, Theorem 3.1]
	\label{prop-isometry-quasisymmetry}
	Let $X,Y$ be two proper, Gromov-hyperbolic metric spaces and let $f\colon X \to Y$ be an isometry. Then $f$ induces a quasisymmetric homeomorphism from $\partial X$ to $\partial Y$, for any choice of visual metrics $D_{x,a}$ and $D_{y,b}$ on $\partial X$ and $\partial Y$.
\end{prop}
As a consequence it is well defined the Ahlfors regulaar conformal dimension of a subset $C$ of $\partial X$ as $\text{CD}(C) := \text{CD}(C, D_{x,a}),$ where $D_{x,a}$ is any visual metric.

\subsection{The class $\mathcal{M}(\delta,D)$}
Let $X$ be a proper, $\delta$-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 {\em 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 $X$.

\noindent The group $\Gamma$ is called {\em elementary} if $\# \Lambda(\Gamma) \leq 2$.

\noindent The set $\Lambda(\Gamma)$ is $\Gamma$-invariant so it is its quasiconvex hull. A discrete group of isometries $\Gamma$ is {\em 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 {\em codiameter} of $\Gamma$.\vspace{2mm}

\noindent\emph{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 $x\in \textup{QC-Hull}(\Lambda(\Gamma))$. We set $a_\delta = \frac{1}{4\delta\log_2e}$ and we fix a standard visual metric $D_x$ of center $x$ and parameter $a_\delta$. We will always consider $\partial X$ endowed with this fixed metric.}
\vspace{2mm}

\begin{cor}
	\label{cor-product-rays}
	Let $(X,x,\Gamma)\in \mathcal{M}(\delta,D)$ and let $z,z'\in \partial X$. Let $\rho > 0$ and $R$ be such that $e^{-a_\delta R} =\rho$. If $D_x(z,z') \leq \rho$ then $d(\xi_{x,z}(R), \xi_{x,z'}(R)) \leq 14\delta$.
\end{cor}
\begin{proof}
	With this choice of $a_\delta$ we have
	\begin{equation}
		\label{eq-new-visual}
		\frac{1}{2}e^{-a_\delta(z,z')_{x}}\leq D_x(z,z')\leq e^{-a_\delta(z,z')_{x}}.
	\end{equation}
	So, if $D_x(z,z') \leq \rho$ then $(z,z')_x \geq R - \frac{\log(2)}{a_\delta} = R - 4\delta$. By Lemma \ref{product-rays} we get $d(\xi_{x,z}(R - 5\delta), \xi_{x,z'}(R-5\delta)) \leq 4\delta$, so by triangle inequality $d(\xi_{x,z}(R), \xi_{x,z'}(R)) \leq 14\delta$.
\end{proof}

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

\begin{prop}[\cite{Cav21ter}, Theorem 6.1]
	\label{prop-Ahlfors-regularity}
	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.
\end{prop}
\begin{proof}
	It follows applying Theorem 6.1 and Lemma 4.9 of \cite{Cav21ter}.
\end{proof}
\begin{cor}
	\label{cor-hyp-uniform-perfect}
	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.
\end{cor}
\begin{proof}
	By Proposition 5.2 of \cite{Cav21ter} we know that $h_\Gamma \geq \frac{\log 2}{99\delta + 10D}$. The conclusion follows by Proposition \ref{prop-Ahlfors-regularity} and Lemma \ref{lemma-Ahlfors-perfect}.
\end{proof}

\begin{prop}
	\label{prop-hyp-qss}
	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.
\end{prop}
\begin{proof}
	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 \ref{approximation-ray-line} 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 \ref{defin-qss}.\\
	Let $w,w'$ be two points of $B(z,\rho) \cap \Lambda(\Gamma)$, so $D_x(w,z)$, $D_x(w',z)< \rho$. Let $T \geq 0$ be the real number such that $e^{-a_\delta \cdot T} = D_x(w,w')$. Since $D_x(w,w') < 2\rho$ 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 \ref{cor-product-rays} to obtain
	\begin{equation*}
		\begin{aligned}
			&d(\xi_{x,z}(R),\xi_{x,w}(R)) \leq 14\delta,\\ &d(\xi_{x,z}(R),\xi_{x,w'}(R)) \leq 14\delta,\\ &d(\xi_{x,w}(T),\xi_{x,w'}(T)) \leq 14\delta.
		\end{aligned}
	\end{equation*}
	By triangle inequality we have
	\begin{equation*}
		\begin{aligned}
			d(x,g\xi_{x,w}(T)) &\leq d(x,g\xi_{x,z}(R)) + d(g\xi_{x,z}(R), g\xi_{x,w}(R)) + d(g\xi_{x,w}(R),g\xi_{x,w}(T))\\
			&\leq \vert T-R \vert +D+28\delta.
		\end{aligned}
	\end{equation*}
	Again by the triangle inequality it holds $d(x,g\xi_{x,w}(T)) \geq \vert T-R \vert -D-28\delta$. Similar estimates hold for $d(x,g\xi_{x,w'}(T))$. We want to estimate $d(\xi_{x,gw}(\vert T-R \vert), g\xi_{x,w}(T))$.
	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 \ref{parallel-geodesics} 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 $T\ge R-4\delta$, 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
	$$\vert T - R \vert - 2D - 54\delta \leq d(x,\xi_{x,gw}(T-t_2+t_1+18\delta+D)) \leq \vert T - R \vert + 2D + 54\delta,$$
	so
	$$\vert T - R \vert - 2D - 54\delta \leq T-t_2+t_1+18\delta+D \leq \vert T - R \vert + 2D + 54\delta.$$
	Therefore by triangle inequality
	$$d(\xi_{x,gw}(\vert T-R \vert), g\xi_{x,w}(T)) \leq 80\delta + 3D.$$
	Analogously we get $d(\xi_{x,gw'}(\vert T-R \vert), g\xi_{x,w'}(T)) \leq 80\delta + 3D.$
	Combining these two estimates we conclude $d(\xi_{x,gw}(\vert T-R \vert), \xi_{x,gw'}(\vert T-R \vert))\leq 160\delta + 6D$. By Lemma \ref{product-rays} we have $(gw,gw')_x > \vert T - R \vert - 80 \delta - 3D$, so
	\begin{equation*}
		\begin{aligned}
			D_x(gw,gw') &\leq e^{-a_\delta(\vert T - R \vert - 80\delta - 3D)} \\
			&\leq e^{a_\delta(80\delta + 4D)}\cdot \frac{e^{-a_\delta D}}{e^{-a_\delta R}}\cdot e^{-a_\delta T} \\
			&= e^{a_\delta(80\delta + 4D)}\cdot \frac{\rho_0}{\rho}\cdot D_x(w,w')   ,
		\end{aligned}
	\end{equation*}
	where we used the definition of standard visual metric, $\vert T - R \vert \leq T-R$ and $\rho_0 = e^{-a_\delta D}$. We prove now the other inequality. We have $D_x(w,w') = e^{a_\delta T} \leq e^{a_\delta (w,w')_x}$, so $(w,w')_x \leq T$. We set $b=47\delta + D$ and $T' = T + b$. By Lemma \ref{product-rays}.(ii) we know that $d(\xi_{x,w}(T'), \xi_{x,w'}(T')) \geq 2b$.
	We can argue in the same way as before with $t=T'-t_2$ finding
	$$d(\xi_{x,gw}(T'-R), g\xi_{x,w}(T')) \leq 44\delta + D,$$
	and the analogous estimate for $w'$. Therefore
	$$d(\xi_{x,gw}(T'-R), \xi_{x,gw'}(T'-R)) \geq 2b - 88\delta - 2D > 4\delta.$$
	By Lemma \ref{product-rays}.(i) we have $(gw,gw')_x < T'-R + \delta = T - R + 48\delta + D$.
	Therefore
	$$D_x(gw,gw') \geq \frac{1}{2}e^{-a_\delta(T - R + 48\delta + D)} = \frac{1}{2}e^{-48\cdot a_\delta\cdot\delta}\cdot \frac{\rho_0}{\rho}\cdot D_x(w,w').$$
	Thus $\Phi$ is $L_0$-biLipschitz with $L_0 = L_0(\delta,D) = \max\lbrace e^{a_\delta(80\delta + 4D)}, 2e^{48\cdot a_\delta\cdot\delta}\rbrace$ 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 $\Phi(w')$ for some $w'\in \Lambda(\Gamma)$. We know that $D_x(\Phi(w'), \Phi(z)) \leq \frac{\rho_0}{L_0}$, then $D_x(w',z) \leq L_0\cdot \frac{\rho}{\rho_0}\cdot D_x(\Phi(w'), \Phi(z)) \leq \rho$, i.e. $w'\in B(z,\rho) \cap \Lambda(\Gamma)$. This concludes the proof.
\end{proof}

\subsection{Semicontinuity of conformal dimension on $\mathcal{M}(\delta,D)$}
\label{subsec-convergence}
We consider triples $(X,x,\Gamma)$ where $(X,x)$ is a pointed, proper metric space and $\Gamma< \text{Isom}(X)$ is a closed subgroup.
The notion of equivariant pointed Gromov-Hausdorff convergence of such triples were introduced in \cite{Fuk86}. We do not repeat here the definition and we refer to \cite{Fuk86} and to \cite{Cav21ter} for details.
If a sequence of triples $(X_n,x_n,\Gamma_n)$ converge in this sense to a triple $(X_\infty,x_\infty,\Gamma_\infty)$ we write $(X_n,x_n,\Gamma_n)\underset{\text{eq-pGH}}{\longrightarrow} (X_\infty,x_\infty,\Gamma_\infty)$.
Since the metric spaces are proper and the groups are closed then the limit is unique up to equivariant isometry, if it exists (\cite{Fuk86}, Proposition 1.5).

\noindent In \cite{Cav21ter} and \cite{CavS20bis} the author introduced the notion of ultralimit group.
Given a sequence of triples $(X_n,x_n,\Gamma_n)$ as above and a non-principal ultrafilter $\omega$ we denote by $(X_\omega, x_\omega)$ the ultralimit space and by $\Gamma_\omega$ the set
$$\Gamma_\omega = \lbrace \omega\text{-}\lim g_n \text{ s.t. } g_n \in \Gamma_n \text{ for } \omega\text{-a.e.}(n)\rbrace.$$
In particular the elements of $\Gamma_\omega$ are ultralimits of admissible sequences. $\Gamma_\omega$ is a closed group of isometries of $X_\omega$, Proposition 3.8 of \cite{Cav21ter}.
The relation between the two type of convergence is similar to the one we saw in Proposition \ref{prop-GH-ultralimit}.
\begin{prop}[\cite{Cav21ter}, Proposition 3.13]
	\label{prop-GH-equivariant-ultralimit}
	Let $(X_n, x_n, \Gamma_n)$ be a sequence of triples and let $\omega$ be a non-principal ultrafilter.
	\begin{itemize}
		\item[(i)] If $(X_n,x_n,\Gamma_n) \underset{\textup{eq-pGH}}{\longrightarrow} (X,x,\Gamma)$ then $(X_\omega, x_\omega, \Gamma_\omega)$ is equivariantly isometric to $(X_\infty,x_\infty,\Gamma_\infty)$.
		\item[(ii)] If $X_\omega$ is proper then $(X_{n_k},x_{n_k},\Gamma_{n_k}) \underset{\textup{eq-pGH}}{\longrightarrow} (X_\omega,x_\omega,\Gamma_\omega)$ for some subsequence $\lbrace{n_k}\rbrace$.
	\end{itemize}
\end{prop}

If we focus the attention to the class $\mathcal{M}(\delta,D)$ we get
\begin{prop}
	\label{prop-convergence}
	Let $(X_n,x_n,\Gamma_n)\subseteq \mathcal{M}(\delta,D)$ and suppose $(X_n,x_n,\Gamma_n) \underset{\textup{eq-pGH}}{\longrightarrow} (X_\infty,x_\infty,\Gamma_\infty)$. Then
	\begin{itemize}
		\item[(i)] $(X_\infty,x_\infty,\Gamma_\infty) \in \mathcal{M}(\delta,D)$ (\cite{Cav21ter}, Theorem A);
		\item[(ii)] $\sup_{n\in\mathbb{N}}h_{\Gamma_n} < + \infty$ (\cite{Cav21ter}, Corollary 5.9);
		\item[(iii)] for every non-principal ultrafilter $\omega$ the ultralimit space $\omega$-$\lim(\Lambda(\Gamma_n), D_{x_n})$ is quasisymmetric equivalent to $(\Lambda(\Gamma_\infty), D_{x_\infty})$.
	\end{itemize}
\end{prop}
\begin{proof}
	We only need to show (iii). We denote by $(X_\omega, x_\omega, \Gamma_\omega)$ the ultralimit triple of the sequence $(X_n,x_n, \Gamma_n)$: it is equivariantly isometric to $(X_\infty,x_\infty,\Gamma_\infty)$ by Proposition \ref{prop-GH-equivariant-ultralimit}. Every point of the space $\omega$-$\lim(\Lambda(\Gamma_n), D_{x_n})$ 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 \cite{CavS20}, Lemma A.7) that this sequence of geodesic rays define a limit geodesic ray $\xi_{x_\omega, z_\omega}$, with $z_\omega \in \partial X_\omega$. Proposition 5.11 of \cite{Cav21ter} showed that the map $\Psi \colon \omega$-$\lim(\Lambda(\Gamma_n), D_{x_n}) \to \partial X_\omega$ defined by $\Psi((z_n)) = z_\omega$ is a well defined homeomorphism. Moreover by the proof of Theorem A.(i) of \cite{Cav21ter} we know that the image of $\Psi$ is exactly $\Lambda(\Gamma_\omega)$. We denote by $D$ the distance induced by $\Psi$ on $\Lambda(\Gamma_\omega)$. Our claim is that $D$ is a visual metric on $\Lambda(\Gamma_\omega)$.
	We take two points $\Psi(z),\Psi(z') \in \Lambda(\Gamma_\omega)$, with $z=\omega$-$\lim z_n$, $z' = \omega$-$\lim z_n' \in \omega$-$\lim(\Lambda(\Gamma_n), D_{x_n})$. By definition $D(\Psi(z),\Psi(z')) = \omega$-$\lim D_{x_n}(z_n,z_n') = e^{-a_\delta R}$ for some $R\geq 0$. From \eqref{eq-new-visual} we have $\omega$-$\lim (z_n,z_n')_{x_n} \geq R - 4\delta$, then $\omega$-$\lim d(\xi_{x_n,z_n}(R-5\delta), \xi_{x_n,z_n'}(R-5\delta)) \leq 4\delta$ by Lemma \ref{product-rays}.(i). This implies $d(\xi_{x_\omega,z_\omega}(R-5\delta), \xi_{x_\omega,z_\omega'}(R-5\delta)) \leq 4\delta$ and so $(\Psi(z), \Psi(z'))_{x_\omega} > R-8\delta$ by Lemma \ref{product-rays}.(ii). This means
	$$D(\Psi(z),\Psi(z')) = e^{-a_\delta R} \geq e^{-a_\delta \cdot 8\delta}\cdot e^{-a_\delta(\Psi(z), \Psi(z'))_{x_\omega}} = \frac{1}{4}\cdot e^{-a_\delta(\Psi(z), \Psi(z'))_{x_\omega}}.$$
	Analogously, with the same notation as above, we have $\omega$-$\lim (z_n,z_n')_{x_n} \leq R$ and so $\omega$-$\lim d(\xi_{x_n,z_n}(R+3\delta), \xi_{x_n,z_n'}(R+3\delta)) \geq 6\delta$ by Lemma \ref{product-rays}.(ii). By definition $d(\xi_{x_\omega,z_\omega}(R+3\delta), \xi_{x_\omega, z_\omega'}(R+3\delta)) \geq 6\delta$, so $(\Phi(z), \Phi(z'))_{x_\omega} \leq R+4\delta$ by Lemma \ref{product-rays}.(i). This means
	$$D(\Psi(z),\Psi(z')) = e^{-a_\delta R} \leq e^{a_\delta \cdot 4\delta}\cdot e^{-a_\delta(\Psi(z), \Psi(z'))_{x_\omega}} = 2\cdot e^{-a_\delta(\Psi(z), \Psi(z'))_{x_\omega}}.$$
	This shows that $D$ is a visual metric on $\Lambda(\Gamma_\omega)$, so it is quasisymmetric equivalent to any other visual metric on this set.\\
	Moreover any visual metric on $\Lambda(\Gamma_\omega)$ is quasisymmetric equivalent to any visual metric on $\Lambda(\Gamma_\infty)$ since $(X_\omega,x_\omega, \Gamma_\omega)$ and $(X_\infty,x_\infty,\Gamma_\infty)$ are equivariantly isometric and isometries between Gromov-hyperbolic spaces induce quasisymmetries between boundaries by Proposition \ref{prop-isometry-quasisymmetry}.
\end{proof}

Finally we can give the
\begin{proof}[Proof of Theorem \ref{theorem-main-introduction}]
	By Proposition \ref{prop-convergence} we know that $(X_\infty, x_\infty, \Gamma_\infty)$ belongs to $\mathcal{M}(\delta,D)$, so the conformal dimension of $\Lambda(\Gamma_\infty)$ is well defined. By the same proposition we know it exists $H\geq 0$ such that $h_{\Gamma_n} \leq H$ for all $n\in \mathbb{N}$. By Proposition \ref{prop-hyp-qss} we can find $L_0$ and $\rho_0$ so that all the spaces $\Lambda(\Gamma_n)$ are compact and $(L_0,\rho_0)$-q.s.s. By Corollary \ref{cor-hyp-uniform-perfect} they are also all $a_0$-uniformly perfect for the same $0<a_0<1$.
	We fix a non-principal ultrafilter $\omega$. Proposition \ref{prop-convergence}.(iii), together with Proposition \ref{prop-GH-ultralimit}, tells us that there is a subsequence $(\Lambda(\Gamma_{n_k}), D_{x_{n_k}})$ that converges in the Gromov-Hausdorff sense to $(\Lambda(\Gamma_\infty), D)$, where $(\Lambda(\Gamma_\infty), D)$ is quasisymmetric equivalent to $(\Lambda(\Gamma_\infty), D_{x_\infty})$. Applying directly Theorem \ref{theorem-main-intro} we get
	$$\text{CD}(\Lambda(\Gamma_\infty), D) \geq \limsup_{n\to +\infty} \text{CD}(\Lambda(\Gamma_n)).$$
	Since $D$ is quasisymmetric equivalent to $D_x$ we conclude that $\text{CD}(\Lambda(\Gamma_\infty)) = \text{CD}(\Lambda(\Gamma_\infty), D) \geq \limsup_{n\to +\infty} \text{CD}(\Lambda(\Gamma_n))$.
\end{proof}

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Home » News » Legal GRC Platform exterro announces acquisition of e-discovery provider zapproved
Legal GRC Platform exterro announces acquisition of e-discovery provider zapproved
Exterro, a leading provider of Legal Governance, Risk and Compliance (GRC) software, has acquired Zapproved, a provider of e-discovery software known for its strong customer outcomes. This acquisition is the latest step in furthering Exterro's vision to empower customers to proactively and defensibly manage their legal governance, risk and compliance obligations. The financial terms of the transaction were not disclosed.
Today, Exterro offers a comprehensive solution on a single unified platform that addresses all phases of the e-discovery and forensic investigative process. Converging market forces in privacy, digital forensics & incident response, including e-discovery have created a need for a comprehensive and unified solution that can identify, understand and manage data in a contextual manner within a consistent, defensible process that meticulously ensures and maintains data integrity. Modern enterprises must address critical business challenges within these disciplines as quickly and cost-effectively as possible, which is one of the driving forces behind Exterro's acquisition of Zapproved.
"I have known Monica Enand, President & CEO of Zapproved, since the early days of both of our companies' origins in Portland, Oregon. Since inception, we have shared a common vision for providing market-leading software and support for our clients and partners, and the combination of Exterro and Zapproved will continue that vision by delivering enhanced value to our constituents," said Bobby Balachandran, CEO & President of Exterro.
"Zapproved's mission has always been to make our customers ridiculously successful," said Monica Enand, President & CEO, Zapproved. "We are confident that combining Exterro's innovation engine with Zapproved's world-class customer experience brings the best of the industry together in one powerful and easy-to-use solution that comprehensively addresses the critical business challenges facing our customers today."
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India – Sri Lanka: The Movers And Shakers Of Policy And Protocol – Analysis
June 28, 2012 September 1, 2017 Salma Yusuf 2 Comments
By Salma Yusuf
A fact accorded scant credence in our times is that the personal and professional can be inextricably linked in the sphere of international and foreign relations. This was the perception distilled from an event held last week to mark the second death anniversary of veteran Sri Lankan diplomat Vernon L.B. Mendis.
The personal memoir rendered by retired civil servant M.D.D Pieris in memory of his colleague was heart-warming, to say the least. What came through very strongly is the foundation of personal goodwill, trust and affection between two of Sri Lanka's distinguished public servants, which undoubtedly played a key role in the conduct of their official functions. Pieris noted instances where mutual trust that was cultivated in their personal relationship became a key driver of efficiency and effectiveness when carrying out national duties: he reflected on instances where matters of state importance were discussed over a telephone call, the individual minute-taking at either end would suffice for documentation and decision – making purposes, with no further need for exchange of notes or letters of confirmation. This and the keynote address delivered by Ambassador Nihal Rodrigo, one time Foreign Secretary, and Secretary General of SAARC, were resonant with a key message for the public at large: foreign policy and international relations is as much about actors, personalities and relationships as it is about substantive policy and protocol.
MANAGING POLARIZED IDEOLOGIES
The seventies was a time fraught with complexity for Sri Lanka's external relations. Yet, a lesson to be learnt is that it is indeed possible to maintain productive and beneficial relations with somewhat opposing and polarized forces and ideologies: Prime Minister Sirimavo Bandaranaike was known for her close personal relationships with leaders from India, Pakistan and China which greatly contributed to Sri Lanka's strong regional positioning. The dividends of such close personal relationships with leaders such as Zulfikar Ali Bhutto are felt even today through the warm hospitality received when an ordinary Sri Lankan citizen visits Pakistan. At the international level, we had excellent relationships with the Middle East, the Western World, South Asia and China. Despite India, Pakistan and China having divergent foreign policy orientations, Sri Lanka managed its relationships with all three countries commendably.
INDIA IN THE WORLD
The potential role played by actors in the context of international relations, in relation to Sri Lanka specifically, could be illustrated even today: Shiv Shankar Menon and Nirupama Rao, both former Indian High Commissioners to Sri Lanka and having served in China, are currently occupying positions that can have a significant bearing on Sri Lanka's international and regional compasses. Shiv Shanker Menon is currently India's National Security Advisor and Nirupama Roa is India's Ambassador to the United States. The ability of actors who have been associated with Sri Lanka in the height of conflict accords the ability to provide a graphical and realistic representation of the intricacies, nuances and complexities in situational contextual analyses. India as our immediate neighbour, and hence crucial both geographically and strategically for Sri Lanka's international relations, coupled with India's growing position in international affairs needs to be considered seriously.
Furthermore, living in times where the India – China relationship is set to emerge as the most important relationship in Asia and the parallel emergence of the phenomenon of 'The Asian century' makes it imperative that we begin to acknowledge the role that actors and personalities can play in shaping international and foreign policy. The two meetings between Shiv Shankar Menon set for later this week with President Mahinda Rajapaksa when he visits Sri Lanka will once again provide an opportunity to clear misconceptions, outline challenges and note progress while reinforcing commitments : the fact that the only meetings scheduled for the visit are with the President underscores the important role that can be played by actors in the foreign policy discourse.
RELATIONSHIPS WITH THE INTERNATIONAL HUMAN RIGHTS COMMUNITY
As my previous article in these pages highlighting Sri Lanka's foreign policy challenges notes, most of the Sri Lanka's current bilateral and multilateral engagements, both regional and international, are haunted by a spectre of reconciliation and human rights concerns. A related though somewhat distinct challenge that has been highlighted is the lack of direct contact with the international human rights community and groups. There was a time, not so long ago, for instance in the eighties, when international human rights organisations such as Amnesty International and the government of the day would, through informal channels, clarify matters especially in relation to facts and figures that would be quoted in human rights reports, and which were in turn used to draw analyses and conclusions and used for purposes of advocacy.
The benefit of such direct contact and engagement is mutually beneficial for both parties – on one hand, maintaining credibility in respect of the findings of reports of the international organizations; and giving the state party an opportunity to clear the flawed or erroneous perceptions that are usually the cause of hostile relationships. The damage caused after misconceptions and misperceptions reach the public domain cannot adequately be rebutted through a reactionary right of reply. The need for proactive engagement on issues of genuine national interest and concern will contribute greatly to improve foreign positioning.
Furthermore, a relationship of hostility and non – engagement between the government of the day and the international human rights community will only lead to further mistrust and will jeopardize the ultimate goal of achieving human rights protection and promotion. In this regard, the role of the diaspora in maintaining relationships with the international human rights community must also be accorded serious consideration, given the ability it has to influence the perceptions of the international human rights community in its analyses and conclusions.
THE TAMIL EELAM SUPPORTERS CONFERENCE
A final example for consideration is the upcoming conference that is being planned by the Dravida Munnetra Kazhagam (DMK) in India for holding the Tamil Eelam Supporters' Conference (TESO) in Villupuram on August 5 2012. The Hindu reported this week the DMK President M. Karunanidhi, who is also the Chairman of TESO, saying that the venue was being prepared for the meet. The reception committee headed by DMK's treasurer M.K. Stalin comprises five secretaries – former DMK minister K. Ponmudi, Dravidar Kazhagam general secretary Kali Poonguntran, Viduthalai Chiruthaikal Katchi general secretary Ravikumar and DMK functionaries K.S. Radhakrishnan and advocate H.M. Jinnah. "We want to hold the conference on the lines of the one held in Madurai last time in 1986. Besides inviting national leaders, we are planning to bring human rights activists from across the world and international leaders who are supporting the Tamil Eelam cause," said one of the secretaries, as reported in the news item.
The role of actors, once again, becomes important at such events which will be well within the cynosure of international attention. Relationships with international human rights organisations and leaders who are specifically mentioned as invitees will have a further bearing on the presentations and projections tabled. Moreover, the personalities that represent the country's standing on matters of national importance and interest such as reconciliation and human rights protection will be critical to the outcome at such exchanges.
Ultimately what Sri Lanka requires is a robust foreign policy grounded in strong interest – driven national positions: the effectiveness of the policy will in turn largely depend on the actors involved in its implementation.
This article appeared in The Daily Mirror and is reprinted with permission.
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Salma Yusuf
Salma Yusuf is a Visiting Lecturer, Masters in Human Rights, University of Colombo and University of Sydney; Visiting Lecturer, Bachelor of Laws, University of Northumbria – Regional Campus for Sri Lanka & Maldives; LL.M, Queen Mary, University of London; Queen Mary Scholar 2008-2009; LL.B (Hons), University of London. She provides legal and policy advisory services on both national and international programmes in the fields of human rights law, transitional justice, comparative social justice, and peace-building. She has authored publications for the Sri Lanka Journal of International Law; the Seattle Journal for Social Justice; the Complutense University of Madrid; the Institue of Human Rights; and the School of Oriental and African Studies, University of London. Email: [email protected]
2 thoughts on "India – Sri Lanka: The Movers And Shakers Of Policy And Protocol – Analysis"
Dr. A.S.W. Tammita-Delgoda
A considered and thought provoking piece.
SRTD
Tamils are eagerly waiting for the regime change in New Delhi, India to take a firm stand against the Rajapakse regime's criminal activities. it is well known that the current UPA regime and the Malayalee mafias in Delhi alleged to have collaborated with the Sri Lankan regime in committing human rights abuses against the Tamils.
Obama has proved that he is a preacher and not an achiever and recently Mitt Romney has said that "Talk is Cheap". Obama was at the White House when Tamils were brutalized and massacred and he failed in every respect to save mankind and humans despite Obama has said during his election campaign that he will tear down rogue regimes.
Although George Bush was bad, when another leader fails to listen to him, he take bold and strong action against the nation. I believe George Bush would have attacked Sri Lanka if he was in power when the Sri Lankan regime was committing crimes against humanity.
Despite Selvi Jeyalalithaa, the Tamil Nadu Chief Minister who demanded the Delhi regime to declare Rajapakse as a war criminal and take him to the ICC to stand trial for the crimes against humanity, the Delhi regime failed to do it as it may expose their own alleged crimes against humanity to the world.
Tamils are confident that the Congress will be wiped out in Tamil Nadu in the next election. It is great that the Tamils in Tamil Nadu are more knowledgeable about the events in Eelam, participating and demanding for action against the Sri Lankan regime. The struggle for Eelam will be alive with the help of the Tamil Diaspora which is spread all over the world. Tamil Nadu Tamils will start a spring for independence as the Delhi regime fails in democracy, Rule of Law, Justice, Transparency, R2P, International Law and respect the Tamils' legitimate aspirations.
The Sinhala Buddhist Apartheid Rajapakse regime's terror is temporary and it should end sooner than later as it may save more lives and sufferings if ends early!


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