\begin{document}

\title{The relative volume function and the capacity of sphere on asymptotically hyperbolic manifolds}
\author{Xiaoshang Jin}
\date{}
\maketitle
\begin{abstract}
Following the work of Li-Shi-Qing, we propose the definition of the relative volume function for an AH manifold. It is not a constant function in general and then we study the regularity of this function. We also use this function to give an accurate characterization of the height of the geodesic defining function for the AH manifold with a given boundary metric. As an application, we provide a new simple method of proving the important relative volume inequality, i.e. theorem 1.3 in \cite{li2017gap} of Li-Shi-Qing. At last, we use this function to research  the capacity of balls in AH manifold and provide some limit results.
\end{abstract}
\section{Introduction}  \label{sect1}
Let $\overline{X}$ be an $n+1$ dimensional smooth manifold with boundary. We use $X$ and $\partial X$ to denote the interior and the boundary of $\overline{X}.$ A complete noncompact metric $g^+$ on $X$ is called conformally compact if there exists a smooth defining function $\rho$ on  $\overline{X}$ such that the conformal metric $g=\rho^2g^+$ can be continuously extended to $\overline{X}.$
The  defining function satisfies that
\begin{equation}\label{1.1}
  \rho>0\ \ in \ \ X,\ \ \ \ \rho=0\ \ on\ \ \partial X,\ \ \ \ d\rho\neq 0\ \ on \ \ \partial X.
\end{equation}
We say that $(X,g^+)$ is $C^{m,\alpha}\ (W^{k,p},$ smoothly) conformally compact if $g=\rho^2g^+$ is $C^{m,\alpha}\ (W^{k,p},$ smooth) on $\overline{X}.$ Here $m,k\in \mathbb{N},\alpha\in[0,1)$ and $p\geq 1.$  The restriction of $g$ on the boundary $\hat{g}=g|_{\partial X}$ is called the boundary metric. It is well known that $g^+$ induces a conformal class $(\partial X,[\hat{g}])$ on the boundary, called the conformal infinity of $g^+.$ If $g^+$ is $C^2$ conformally compact, then a straightforward calculation yields that
the curvature of $(X,g^+)$ is of the following form \cite{graham2000volume}:
\begin{equation}\label{1.2}
  R_{ijkl}[g^+]=|d\rho|^2_{g}(g^+_{ik}g^+_{jl}-g^+_{il}g^+_{jk})+O(\rho^{-3})
\end{equation}
near $\partial X.$ Here $O(\rho^{-3})$ denotes the (0,4) tensors and take norm with respect to metric $g.$  Readers can see \cite{besse2007einstein} for the conformal transformation
law of curvatures. Hence the sectional curvature $K[g^+]=-|d\rho|^2_{g}+O(\rho)$ is uniformly approaching to $-|d\rho|^2_{g}$ (see \cite{mazzeo1988hodge}).
\par Thus if $g^+$ is at least $C^2$ conformally compact and $|d\rho|^2_{g}|_{\partial X}=1,$ we say $(X,g^+)$ is an asymptotically hyperbolic manifold or AH manifold for short. Given an  AH manifold $(X,g^+)$ and a boundary representative $\hat{g}\in[\hat{g}],$ Graham and Lee
proved that there exists a unique defining function $x$ such that $|dx|^2_{x^2g^+}=1$ in a neighborhood of the boundary $\partial X\times[0,\delta)$
for some small $\delta$ in \cite{graham1991einstein}. We call $x$ the geodesic defining function of $(X,g^+)$ associated with $\hat{g}.$
However, the new geodesic compactification may not be as smooth as the initial one. In fact, if $g=\rho^2g^+$ is a $C^{m,\alpha}$ compactification with $m \geq 2$ and $\alpha\geq 0,$ then the geodesic compactification $\bar{g}=x^2g^+$ with the same boundary metric is, at least $C^{m-1,\alpha},$ see \cite{lee1994spectrum} or \cite{chrusciel2005boundary}.
\par Let $\delta>0$ be the supremum where the geodesic defining function is well defined as above. One interesting problem is: how big (small) is $\delta?$ it seems that $\delta$ is dependent on $g^+$ and $\hat{g},$  can we give a mathematical expression between them? We give an answer
to the problems in this paper.
\\ \par If in addition, the AH manifold $(X,g^+)$ also satisfies the Einstein equation,
$$ Ric[g^+]=-ng^+$$
 then we call $g^+$ a conformally compact Einstein metric or Poincar\'e-Einstein metric.  In recent years it has become the main theme in the study of conformal geometry as it plays an important role in the proposal of AdS/CFT correspondence \cite{maldacena1999large} in theoretic physics.
\par In this paper, we mainly research the AH manifold of order $2,$ i.e. the sectional curvature
satisfies that
\begin{equation}\label{1.3}
  K[g^+]+1=O(\rho^2)
\end{equation}
We emphasize that a conformally compact Einstein metric $g^+$ satisfies the curvature condition (\ref{1.3}), (one can see section 2 in \cite{jin2021finite} or lemma 1.4 in \cite{li2017gap}, or lemma 3.1 in \cite{chrusciel2005boundary} for more information). Therefore we also call $g^+$
an asmyptotically hyperbolic Einstein (or AHE for short) metric.
\\ \par The motivation of this paper is mainly from the work of Li-Shi-Qing when they study the gap phenomena and curvature estimates for AH manifolds. They prove that:
\begin{lemma}\cite{li2017gap}
  Let $(X^{n+1},g^+)$ be an $n+1-$ dimensional AH manifold with a $C^2$ conformally compactification $g=\rho^2g^+,$ if the sectional curvature satisfies (\ref{1.3}), then for any $p\in X,\frac{Vol(\partial B_p^{g^+}(t),g^+)}{\sinh^nt}$ is convergent as $t$ tends to infinity.
\end{lemma}
Here we illustrate that the original thought is from Shi and Tian (lemma 2.2 in \cite{shi2005rigidity}) where they studied the rigidity of
ALH manifold with an intrinsic geometric condition. By analysing the geodesic sphere of radius $\rho$ and centered at a fixed pole $o,$ they
showed that $e^{-2\rho}g^+$ is uniformly equivalent to the flat metric $\delta_{ij}$ and hence $\frac{Vol(\partial
B_o^{g^+}(\rho),g^+)}{\sinh^n\rho}$ is uniformly bounded. Furthermore, the curvature  and diameter of $(\partial
B_o^{g^+}(\rho),\sinh^{-2}\rho g^+)$ are also bounded. Then they proved that $(\partial B_o^{g^+}(\rho),\sinh^{-2}\rho g^+)$ converges in
the weakly $W^{2,p}$-topology to a manifold which is conformally equivalent to a sphere and finally get a rigidity result. Later the research was developed by Dutta and Javaheri in \cite{dutta2010rigidity} and they assumed that the conformal infinity of the AH manifold is
the standard round sphere metric. They compared the distance functions $t$ and $r=-\ln\frac{x}{2}$ of $g_+$ and showed that $t-r$ has a
continuous extension to the boundary $\mathbb{S}^n$ and $(\Sigma_r=\{y:r(y)=r\},4e^{-2t}g^+)$ converges in the weak $W^{1,2}$ topology to some metric on the sphere. Finally, by computing the volume of the hypersurface they obtained the rigidity result. A more general case was studied by Li-Shi-Qing in \cite{li2017gap} and they get a relative volume inequality for AHE manifolds. This inequality also provides a new curvature pinching estimate for AHE manifolds with conformal infinities having large Yamabe invariant.
\par It is nature to ask: whether the limit in lemma 1.1 is dependent on the choice of the base point $p?$ As it is showed in example \ref{ex4.2}, the answer is yes , even in hyperbolic manifold. Then it is meaningful to introduce the following definition:
\begin{definition}
Suppose that $(X,g^+)$ is an $n+1-$ dimensional AH manifold of order 2, then for any $p\in X,$ we define
\begin{equation}\label{1.4}
  \mathcal{A}(p)=2^n\cdot\lim\limits_{t\rightarrow+\infty}e^{-nt}\cdot Vol(\partial B_p^{g^+}(t),g^+)
\end{equation}
we call $\mathcal{A}$  the relative volume function on $X.$
\end{definition}
Here we remark that the condition (\ref{1.3})  is a sufficient and unnecessary condition for the existence of the limit. In fact, if $Ric[g^+]\geq ng^+,$ then we can still define the relative volume function because of the volume comparison theorem.
\\ \par In the first part of this paper we will study the property of $\mathcal{A}.$ Firstly, we use the triangle inequality to prove that $\ln\mathcal{A}$ is a Lipschitz continuous function. Then we show that there is a relationship between $\mathcal{A}$ and the 'height' of the geodesic defining function.
More concretely,
\begin{theorem}
  Let $(X^{n+1},g^+)$ be an $n+1-$ dimensional AH manifold of order 2, then we have the followings:\\
  (1) The function$\mathcal{A}$ is positive on $X$ and $\ln\mathcal{A}$ is Lipschitz continuous on X in the sense that
\begin{equation}\label{1.5}
  |d \ln\mathcal{A}|_{g^+}\leq n\ \ i.e.\ \ |d \mathcal{A}|_{g^+}\leq n\mathcal{A}.
\end{equation}
  (2) Let $g=x^2g^+$ be the $C^2$ geodesic compactification of $g^+$ and the boundary metric of $g$ is $\hat{g}=g|_{\partial X}.$ If $|dx|^2_g=1$ and $x$ is smooth on $X_\delta=\partial X\times(0,\delta),$ then
  for any $p\in X\setminus X_\delta,$
\begin{equation}\label{1.6}
  \mathcal{A}(p)\leq(\frac{2}{\delta})^n\cdot Vol(\partial X,\hat{g})\leq e^{n\cdot diam(X\setminus X_\delta,g^+)}\mathcal{A}(p).
\end{equation}
 Here $diam(X\setminus X_\delta,g^+)$ denotes the diameter of $X\setminus X_\delta$ in $(X,g^+).$
Furthermore, if $Ric[g^+]\geq -ng^+,$ then we have the following estimate:
\begin{equation}\label{1.7}
 \delta\cdot e^{diam(X\setminus X_\delta, g^+)}\geq 2\cdot(\frac{Vol(\partial X,\hat{g})}{\mathcal{A}(p)})^{\frac{1}{n}}\geq 2\cdot(\frac{Vol(\partial X,\hat{g})}{Vol(\mathbb{S}^n,g_{\mathbb{S}})})^{\frac{1}{n}}
\end{equation}
\end{theorem}
\par (\ref{1.7}) provides a lower bound of $\delta$ in some sense. It is easy to see that, if the complement set of $X_\delta$  has a
uniformly bounded diameter, then the height of the geodesic function $\delta$ is bounded from below. This is very useful when we study the
compactness problems of AHE metrics.
\par Here we refer the work of Cai-Galloway in \cite{cai1999boundaries}. They obtained an estimate of the
diameter if the mean curvature is bounded from below by $n.$ Combining it with (\ref{1.7}), we could provide a new proof of the classical result  of Li-Qing-Shi. That is:
\begin{theorem}[Li-Qing-Shi,(2017),theorem 1.3 in \cite{li2017gap}]
  Let $(X^{n+1}, g^+)$ be an AH manifold of $C^3$ regularity whose
conformal infinity $(\partial X,[\hat{g}])$ is of positive Yamabe type. Let  $p\in X^{n+1}$ be a fixed point. Assume
further that
\begin{equation}\label{1.8}
   Ric[g^+] \geq -ng^+ \ \ and \ \ R[g^+] + n(n +1) = o(e^{-2t})
\end{equation}
for the distance function t from p. Then
\begin{equation}\label{1.9}
(\frac{Y(\partial X,[\hat{g}])}{Y(\mathbb{S}^n,[g_{\mathbb{S}}])})^{\frac{n}{2}}\leq\frac{Vol(\partial B^{g^+}_p(t))}{Vol(\partial B^\mathbb{H}_o(t))}\leq\frac{Vol(B^{g^+}_p(t))}{Vol(B^\mathbb{H}_o(t))}\leq 1.
\end{equation}
where  $B^{g^+}_p(t)$ and $B^\mathbb{H}_o(t)$ are geodesic balls.
\end{theorem}
The above inequality not only leads to the rigidity of conformally compact Einstein manifolds
when the conformal infinity is exactly the round sphere in any dimension, but also gives an interesting curvature
pinching estimate which only relies on the Yamabe constant of the conformal infinity. Hence the relative volume inequality plays an important role in the studying of the compactness and uniqueness problems of AHE metrics, at least nearby hyperbolic space case, see \cite{chang2020compactness} and \cite{chang2021compactness}.
\par The idea of estimating of the curvature and volume bounds on the level sets was firstly introduced in \cite{shi2005rigidity} and was later developed in \cite{dutta2010rigidity}. Both of the two papers deal with the rigidity problem when the conformal infinity is the standard sphere. Finally, the authors in \cite{li2017gap} showed the results in a more general case, i.e. how the Yamabe constant of the
conformal infinity is used to control the relative volume growth of geodesic balls. They also provided a complete and rigorous proof. One of the difficulties in proving the relative volume inequality is that the equation (4.8) may not hold in \cite{li2017gap} because the function $\Psi$ is only Lipschitz continuous. The authors used the deep understanding of the structure of cut locus to overcome the challenge. Another difficulty in their proof is that one cannot obtain the upper bound for the scalar curvature (4.9) without (4.1). By careful analysis of the Riccati equations, the authors derive the upper bounds and lower bounds of
the principal curvature, then they showed that the part of the level set $\Sigma_r$ where estimate (4.1)
does not holds has arbitrarily small volume. Then (4.10) was obtained in\cite{li2017gap}.
\par In this paper, we do not use the limit of the yamabe functional of the level set $\Sigma_r$ ((4.10) in\cite{li2017gap}) to calculate the Yamabe constant of the boundary $(\partial X,[\hat{g}]).$ Instead, by analysing the property of relative volume function, i.e.(\ref{1.6}) or(\ref{1.7}), and making an estimate of the diameter of $(X\setminus X_\delta,g^+)$ with mean curvature, we could directly get a lower bound of the
 relative volume function $\mathcal{A},$ then (\ref{1.9}) is obtained and the proof is simpler.
\\ \par  We notice that (\ref{1.6}) could also provide the upper bound of such $\delta.$ It depends on the n-dimensional volume of the boundary manifold $(\partial X,\hat{g})$ and the relative volume function on $(X,g^+).$ As an immediate corollary of theorem 1.3 and theorem 1.4, we get the following estimate:
\begin{corollary}
  Let $(X^{n+1}, g^+)$ be an AH manifold satisfies conditions in theorem 1.4. Suppose that $g=x^2g^+$ is a $C^2$ geodesic comactification with boundary metric $\hat{g}$. Let $\delta$  be the supremum of $x$ where $|dx|_g^2=1$ and $x$ is smooth on $\partial X\times(0,\delta).$
 Then we have the upper bound estimate of $\delta:$
\begin{equation}\label{1.10}
  \delta\leq 2\cdot (\frac{Vol(\partial X,\hat{g})}{Vol(\mathbb{S}^n,g_{\mathbb{S}})})^{\frac{1}{n}}\cdot(\frac{Y(\mathbb{S}^n,[g_{\mathbb{S}}])}{Y(\partial X,[\hat{g}])})^{\frac{1}{2}}
\end{equation}
\end{corollary}

The inequality (\ref{1.7}) and (\ref{1.10}) is sharp. In fact, if we consider the hyperbolic space: (Poincar\'e ball model)
$$(\mathbb{B}^{n+1},\ g_\mathbb{H}=(\frac{2}{1-|x|^2})^2|dx|^2),$$
where $\rho=\frac{1-|x|^2}{2}$ is defining function.  Let $s=2\frac{1-|x|}{1+|x|},$ then
$$g_\mathbb{H}=s^{-2}(ds^2+\frac{(4-s^2)^2}{16}g_{\mathbb{S}^n}),$$
where $s$ is the geodesic defining function. Hence the conformal infinity is the standard conformal round sphere $(\mathbb{S}^n,[g_{\mathbb{S}^n}]).$
We find that $\delta=2$ and the equality in (\ref{1.7}) and (\ref{1.10}) holds.
\\ \par In the second part of this paper. We do some calculations on some AH manifolds. Firstly, we show that the relative volume function on hyperbolic space is exactly the volume of the standard sphere $\omega_n$ and the inverse conclusion also holds provided the Ricci curvature is bounded from below. More concretely, we have the following rigidity result:
\begin{theorem}
  Let $(X,g^+)$ be an $n+1-$ dimensional AH metric and the Ricci curvature satisfies $Ric[g^+]\geq -ng^+$, then $\mathcal{A}\leq \omega_n$ and the equality holds at one point if and only if $X$ is isometric to a hyperbolic space.
\end{theorem}
We also calculate the relative volume function on hyperbolic manifolds and on Ads-Schwarzchild space, and obtain that such functions are not constant on these spaces. See example \ref{ex4.2} and \ref{ex4.7}.
\\
\par In the third part of paper, we provide an application of the relative volume function. Recall the work in \cite{bray2008capacity}, where the authors derive an upper bound for the capacity of an asymptotically flat 3-manifold with nonnegative scalar curvature and also showed a rigidity result, i.e. the upper bound is achieved if and only if $(M^3,g)$ is isometric to a spatial Schwarzschild manifold.
\par In this paper, we study the capacity of an AH manifold. Since the pivotal of an AH manifold is that it has a good structure at infinity, i.e. a compactification, then we will study the behavior of the capacity at infinity. Using the relative volume function, we give an upper bound of the capacity of balls in AH manifold when its radius goes to infinity. More concretely, it is easy to calculate the capacity of balls in hyperbolic space, it has the formula (\cite{grigor1999isoperimetric}):
\begin{equation}\label{1.11}
 \lim\limits_{t\rightarrow+\infty}\frac{cap_p(B^\mathbb{H}_o(t))}{e^{nt}}=\frac{1}{2^n}(\frac{n}{p-1})^{p-1}\omega_n.
\end{equation}
Noticing that $\omega_n$ is the value of the relative volume function on hyperbolic space.  We prove that the right side of (\ref{1.11}) is the upper bound of an AH manifold with bounded Ricci curvature from below and we also get a rigidity result. Besides, thanks to the work in \cite{hijazi2020cheeger}, if the AH manifold satisfies the conditions in theorem 1.4, we could give the accurate expression of the capacity at infinity. That is,
\begin{theorem}
  Let $(X,g^+)$ be an be an $n+1-$ dimensional AH manifold of $C^{3,\alpha}$ regularity ($\alpha\in (0,1)).$ If the Ricci curvature satisfies $Ric[g^+]+ng^+\geq 0,$ then for any $p>1$ and $q\in X,$
\begin{equation}\label{1.12}
\limsup\limits_{t\rightarrow+\infty}\frac{cap_p(B^{g^+}_q(t))}{e^{nt}}\leq\frac{1}{2^n}(\frac{n}{p-1})^{p-1}\omega_n
\end{equation}
The equality holds if and only if  $(X,g^+)$ is isometric to a hyperbolic space.
\\
   If in addition, the conformal infinity is of nonnegative Yamabe constant and the scalar curvature satisfies $R[g^+]+n(n-1)=o(x^2)$ where $x$ is a defining function. Then
\begin{equation}\label{1.13}
  \lim\limits_{t\rightarrow+\infty}\frac{cap_p(B^{g^+}_q(t))}{e^{nt}}=\frac{1}{2^n}(\frac{n}{p-1})^{p-1}\mathcal{A}(q).
\end{equation}
   for any $p>1$ and $q\in X.$
\end{theorem}
Theorem 1.7 is exactly an extension of (\ref{1.11}) to an AH manifold.
\par In the end, we give an estimate of the capacity of balls in a general AH manifold without any other curvature condition. i.e.
\begin{theorem}
  Let $(X,g^+)$ be an $n+1-$ dimensional AH manifold of $C^2$ regularity, then for any $q\in X,p>1,$
\begin{equation}\label{1.14}
cap_p(B^{g^+}_q(t))=O(e^{nt}),\ \ t\rightarrow+\infty.
\end{equation}
\end{theorem}

The paper is organized as follows. In section 2, we firstly review the key steps in \cite{dutta2010rigidity} and \cite{li2017gap} proving lemma 1.1. and give an method to compute the function $\mathcal{A}$ in (\ref{2.4}). Then with the help of triangle inequality, we show that
 $\ln\mathcal{A}$ is Lipschitz continuous.
\par In order to prove theorem 1.3 (2), we introduce the concept of the relative volume function acting on a compact set in section 3. Then study the relationship between the two functions which we use to obtain the upper and lower estimate of $\delta.$ In the end, we offer a new proof of theorem 1.4.
\par In section 4 we compute the relative volume functions on some AH manifolds. Firstly, we prove that $\mathcal{A} \equiv \omega_n$ on $n+1-$dimensional hyperbolic space. Afterwards we do some calculations on hyperbolic manifolds and get the expression of $\mathcal{A}.$ In special, the relative volume function $\mathcal{A}$ of hyperbolic manifolds  $\mathcal{A}<\omega_n.$  And then a rigidity result is proved when the Ricci curvature is greater than or equal to $-n.$ We also prove that the relative volume functions $\mathcal{A}$ is not a constant function on hyperbolic manifold with small $\lambda$ and on Ads-Schwarzchild space when the mass is sufficiently small.
\par In section 5 and 6, we prove theorem 1.7 and 1.8. The main techniques are classic and are from  Maz’ya's work.
\section{The Lipschitz property of the relative volume function}
We will firstly give a short proof of the existence of the relative volume fnction $\mathcal{A}(q),$ where more details can be found in \cite{dutta2010rigidity} and \cite{li2017gap}.
\par Let $p\in X$ and $B(t)=B^{g^+}_p(t)$ be the geodesic sphere in $(X,g^+)$ centred at $p$ and set

 \begin{equation}\label{2.1}
   s_p(\cdot)=dist_{g^+}(p,\cdot)
 \end{equation}
 be the distance function from $p.$ We state that the symbol $\nabla$ and $|.|$ are the Levi-Civita connection and the norm with respect to the metric $g^+.$ Let $g=x^2g^+$  be the geodesic compactification on $X_\delta=\partial X\times(0,\delta)$ for some small $\delta.$
 We set $r=-\ln x,$ then $|\nabla r|=|\nabla^gx|_g=1$ on $X_\delta,$ which means that $r$ is a distance function on $(X_\delta,g^+).$ By triangle inequality we obtain that $s_p-r$ is bounded on $X_\delta.$ Moreover, if the curvature satisfies the condition \ref{1.3} ,then $|d(r-s_p)|_g$ is also bounded when $r$ is large enough and $s_p$ is smooth. (Lemma 4.1 in \cite{li2017gap} and lemma 3.1 in \cite{dutta2010rigidity}).
\par  Consider the conformal metric

\begin{equation}\label{2.2}
  g^p=e^{-2s_p}g^+=e^{2(r-s_p)}g
\end{equation}
on $X_\delta,$ then $g^p$ is Lipschitz continuous to the boundary. Let $\hat{g}^p=g^p|_{\partial X}$ be the boundary metric of $g^p.$ If we regard $\partial B(t)$ as a Lipschitz graph over $\partial X$ in $(X,g^p),$ we could obtain that
\begin{equation}\label{2.3}
\lim\limits_{t\rightarrow+\infty}Vol(\partial B(t),g^p|_{\partial B(t)})=Vol(\partial X,\hat{g}^p)
\end{equation}

 By the definition,
\begin{equation}\label{2.4}
\begin{aligned}
\mathcal{A}(p)&=2^n\cdot\lim\limits_{t\rightarrow+\infty}e^{-nt}Vol(\partial B_p^{g^+}(t),g^+)\\
&=2^n\cdot\lim\limits_{t\rightarrow+\infty}Vol(\partial B(t),g^p|_{\partial B(t)})
\\&=2^n\cdot Vol(\partial X,\hat{g}^p)
\end{aligned}
\end{equation}
Hence $\mathcal{A}(p)>0$ for any $p\in X.$
\par For any $p,q\in X,$ we use $s_p$ and $s_q$ to denote the distance function from $p$ and $q$ respectively in $(X,g^+).$
 Set
\begin{equation}\label{2.5}
  g^p=e^{-2s_p}g^+\ \ and\ \ g^q=e^{-2s_q}g^+
\end{equation}
 Suppose  $d=dist_{g^+}(p,q),$ then $|s_p(\cdot)-s_q(\cdot)|\leq d.$
We could use (\ref{2.4}) to get that
\begin{equation}\label{2.6}
\begin{aligned}
\mathcal{A}(p)&=2^n\cdot Vol(\partial X,\hat{g}^p)=2^n\int_{\partial X}dV_{\hat{g}^p}\\
&=2^n\int_{\partial X}e^{-n(s_p-s_q)}dV_{\hat{g}^q}
\end{aligned}
\end{equation}
Hence
\begin{equation}\label{2.7}
e^{-nd}\mathcal{A}(q)\leq\mathcal{A}(p)\leq e^{nd}\mathcal{A}(q)
\end{equation}
It means that
\begin{equation}\label{2.8}
  \frac{|\ln\mathcal{A}(p)-\ln\mathcal{A}(q)|}{dist_{g^+}(p,q)}\leq n.
\end{equation}
Hence $\ln\mathcal{A}$ is Lipschitz continuous.
\\ \par The proof above uses the property that $s_p-s_q$ has a continuous extension to the boundaryly because the curvature satisfies the condition $(\ref{1.3}).$ In the following, we provider another method to show that $\ln\mathcal{A}$ is Lipschitz continuous without $(\ref{1.3})$ as long as $\mathcal{A}(p)$ is well defined. The Lipschitz constant is $n+1$ in this case.
\par Firstly, in $X,$ we have that
\begin{equation}\label{2.9}
  \det g^p=e^{-2(n+1)s_p}\det g^+=e^{-2(n+1)(s_p-s_q)}\det g^q\leq e^{2(n+1)d}\det g^q
\end{equation}
 Let $B(t)=B^{g^+}_p(t)$ and $D(t)=B^{g^+}_q(t),$ then for any $t>0$ big enough,
\begin{equation}\label{2.10}
  B(t)\subseteq D(t+d)\ \ \ \ D(\frac{t}{2})\subseteq B(\frac{t}{2}+d)
\end{equation}
 Hence
\begin{equation}\label{2.11}
  \begin{aligned}
Vol(B(t)\setminus B(\frac{t}{2}+d),g^p)&=\int_{B(t)\setminus B(\frac{t}{2}+d)}dV_{g^p}\leq e^{(n+1)d}\int_{B(t)\setminus B(\frac{t}{2}+d)}dV_{g^q}\\
&=e^{(n+1)d}\cdot Vol(B(t)\setminus B(\frac{t}{2}+d),g^q)
\\ &\leq e^{(n+1)d}\cdot Vol(D(t+d)\setminus D(\frac{t}{2}),g^q).
\end{aligned}
\end{equation}
Which implies
\begin{equation}\label{2.12}
  \int_{\frac{t}{2}+d}^t Vol(\partial B(u),g^p)du\leq e^{(n+1)d} \int_{\frac{t}{2}}^{t+d} Vol(\partial D(u),g^q)du.
\end{equation}
Integral mean value theorem tells us that $\exists\xi_t\in[\frac{t}{2}+d,t]$ and $\exists\eta_t\in[\frac{t}{2},t+d]$
such that
\begin{equation}\label{2.13}
 (\frac{t}{2}-d)\cdot Vol(\partial B(\xi_t),g^p)\leq  e^{(n+1)d}\cdot(\frac{t}{2}+d)\cdot Vol(\partial D(\eta_t),g^q)
\end{equation}
Multiply both sides by $\frac{2}{t}$ and let $t$ tends to infinity, we deduce that
\begin{equation}\label{2.14}
  \mathcal{A}(p)\leq e^{(n+1)d}\cdot\mathcal{A}(q)
\end{equation}
We could also prove that $\mathcal{A}(q)\leq e^{(n+1)d}\cdot\mathcal{A}(p)$ in the same method. Hence  $\ln\mathcal{A}$ is Lipschitz continuous and the Lipschitz constant is $n+1.$

\section{Estimates of $\delta:$ the height of the geodesic defining function}
In order to get the upper bounder of $\delta,$ we need to extend the definition of the function $\mathcal{A}.$
Let $E\subset X$ be a compact subset, we use $s_E$ to denote the distance function of $E,$ i.e.
\begin{equation}\label{3.1}
  \forall q\in X,\ \ s_E(q)=dist_{g^+}(E,q)=\inf\limits_{y\in X}dist_{g^+}(y,q).
\end{equation}
Recall the geodesic compactification $g=x^2g^+$ with boundary metric $\hat{g}$ and let $r=-\ln x$ near boundary. Similar to the beginning of section 2, we set the conformal change $g^E=e^{-2s_E}g^+$ on $X$  and could also prove that $|d(s_E-r)|_{g}$ is bounded  for sufficiently large $r$ almost everywhere. Hence $g^E$ is a Lipschitz continuous metric on $\overline{X}.$ Then we can also define the relative volume function of $E,$
\begin{equation}\label{3.2}
\begin{aligned}
  \mathcal{A}(E)&=2^n\cdot\lim\limits_{t\rightarrow+\infty}e^{-nt}Vol(\partial B_E^{g^+}(t),g^+)\\
&=2^n\cdot\lim\limits_{t\rightarrow+\infty}Vol(\partial B_E^{g^+}(t),g^E|_{\partial B_E(t)})\\
&=2^n\cdot Vol(\partial X,\hat{g}^E)
\end{aligned}
\end{equation}
 where $\hat{g}^E=g^E|_{\partial X}.$
\par We observe that the relationship between $\mathcal{A}(p)$ and $\mathcal{A}(E)$ is :
\begin{lemma}
If $E$ is a compact subset of an AH manifold  $(X^{n+1},g^+),$ then
\begin{equation}\label{3.3}
  \forall p\in E,\mathcal{A}(p)\leq \mathcal{A}(E)\leq e^{n\cdot diam(E,g^+)}\mathcal{A}(p).
\end{equation}
Here $diam(E,g^+)=\sup\limits_{p,q\in E}dist_{g^+}(p,q)$ denotes the diameter of $E.$
\end{lemma}
\begin{proof}
  By the definition of distance function, $\forall q\in X,s_p(q)\geq s_E(q).$ As $s_p-s_E$ is  Lipschitz continuous to the boundary,
  $(s_p-s_E)|_{\partial X}\geq 0.$ Noticing that
\begin{equation}\label{3.4}
  \hat{g}^p=e^{-2s_p}g^+=e^{-2(s_p-s_E)}\hat{g}^E,
\end{equation}
we obtain
\begin{equation}\label{3.5}
\begin{aligned}
  \mathcal{A}(p)&=2^n\int_{\partial X}dV_{\hat{g}^p}=2^n\int_{\partial X}e^{-n(s_p-s_E)}dV_{\hat{g}^E}\\
&\leq2^n\int_{\partial X}dV_{\hat{g}^E}=\mathcal{A}(E).
\end{aligned}
\end{equation}
The second inequality could be proved in the same way with the triangle inequality
\begin{equation}\label{3.6}
  s_p(q)\leq s_E(q)+diam(E,g^+),\ \ \ \forall q\in X.
\end{equation}
\end{proof}

Suppose that $g=x^2g^+$ is the $C^2$ geodesic compactification with boundary metric $\hat{g}$ and $|\nabla^{g} x|_g= 1$ on $X_\delta=\partial X\times[0,\delta),$ we need to explain that $\partial X\times\{\delta\}$ may not be homeomorphism to $\partial X$ as it may degenerate into a lower dimension manifold. However we always have that $\partial X\times\{\delta'\}$ is homeomorphism to $\partial X$ for all $\delta'<\delta.$ We are now going to prove that the estimate (\ref{1.6}) holds for all $\delta'<\delta.$
\par Let $E'=X\setminus X_{\delta'}$ be a compact subset. Hence
\begin{equation}\label{3.7}
  \forall p\in X\setminus X_{\delta}\subseteq E',\ \ \ \ \ \mathcal{A}(p)\leq \mathcal{A}(E')\leq e^{n\cdot diam(E',g^+)}\mathcal{A}(p)
\end{equation}

We define function $r:$ $r(\cdot)=\ln\frac{\delta'}{x(\cdot)}$ on $X_{\delta'}$ and $r=0$ on $E'.$
Then $r$ is continuous on $X$ and smooth on $X_{\delta'}.$
Since $E'$ is compact, for any $q\in X_{\delta'},$ there exists a point $q'\in \partial E'=\partial X\times\{\delta'\}$ such that
\begin{equation}\label{3.8}
  dist_{g^+}(q,E')=dist_{g^+}(q,q')
\end{equation}
For any smooth curve $\sigma:[0,l]\rightarrow X_{\delta'}$ with  $\sigma(0)=q$ and $\sigma(l)=q',$
\begin{equation}\label{3.9}
  \begin{aligned}
Length(\sigma,g^+)&=\int_0^l|\dot{\sigma}(t)|dt=\int_0^l|\nabla r|\cdot|\dot{\sigma}(t)|dt
\\ &\geq\int_0^l|g^+(\nabla r,\dot{\sigma})|dt=\int_0^l|(r\circ\sigma)'(t)|dt
\\ &\geq|\int_0^l (r\circ\sigma)'(t)dt|=|r(q')-r(q)|
  \end{aligned}
\end{equation}
(One can see Chapter 5 in \cite{petersen2006riemannian} for more details.) If we choose $\sigma$ as an integral curve for $\nabla r,$ then $Length(\sigma,g^+)=|r(q')-r(q)|.$
Hence
\begin{equation}\label{3.10}
  dist_{g^+}(q,q')=|r(q')-r(q)|=r(q)
\end{equation}
Then $dist_{g^+}(q,E')=r(q),$ which means
\begin{equation}\label{3.11}
  r(\cdot)=s_{E'}(\cdot) \ \ on\ \ X.
\end{equation}
\par With the preparations above, we could compute $\mathcal{A}(E').$ We notice that near  the boundary,
\begin{equation}\label{3.12}
  g^{E'}=e^{-2s_{E'}}g^+=e^{-2r}g^+=\frac{x^2}{(\delta')^2}g^+=\frac{1}{(\delta')^2}g
\end{equation}
Therefore, $\hat{g}^{E'}=\frac{1}{(\delta')^2}\hat{g}$ and
\begin{equation}\label{3.13}
  \mathcal{A}(E')=2^n\cdot Vol(\partial X,\hat{g}^E)=(\frac{2}{\delta'})^n\cdot Vol(\partial X,\hat{g})
\end{equation}
Let $\delta'\rightarrow\delta$ and combine it with (\ref{3.7}), we get the estimate of $\delta$ in (\ref{1.6}). At last,
(\ref{1.7}) holds because of the rigidity result, i.e. theorem 1.6 (or theorem 4.5.)
\\
\par At the end of this section, we use (\ref{1.7}) to prove theorem 1.4 in a new way. By the Bishop-Gromov relative volume
comparison theorem, we only need to prove that
\begin{equation}\label{3.14}
(\frac{Y(\partial X,[\hat{g}])}{Y(\mathbb{S}^n,[g_{\mathbb{S}}])})^{\frac{n}{2}}\leq\lim\limits_{t\rightarrow+\infty}\frac{Vol(\partial B^{g^+}_p(t))}{Vol(\partial B^\mathbb{H}_o(t))}=\frac{\mathcal{A}(p)}{\omega_n}.
\end{equation}
 The only thing we need to do is to calculate the mean curvature of $\partial X_\delta=\partial X\times\{\delta\}$ in $(X\setminus X_{\delta},g^+)$ for sufficiently small $\delta.$ In fact, the mean curvature $H$ is
\begin{equation}\label{3.15}
  H=\Delta_{g^+} r=\Delta_{g^+}(-\ln x)=n-x\Delta_gx.
\end{equation}
On the other hand, as it is showed of (3-3) in \cite{jin2019boundary}, the scalar curvature of g can be expressed as
\begin{equation}\label{3.16}
  S=-2n\frac{\Delta_gx}{x}+tr_g(Ric[g^+]+ng^+)=-2n\frac{\Delta_gx}{x}+o(1).
\end{equation}
Lemma 4.1 in \cite{jin2019boundary} (or lemma 1.2 in \cite{anderson2003boundary})tells us that when restricted on the boundary,
\begin{equation}\label{3.17}
  S|_{\partial X}=\frac{n}{n-1}S_{\hat{g}}
\end{equation}
If $(\partial X,[\hat{g}])$ is of the positive Yamabe constant, then $S_{\hat{g}}>0,$ hence $S$ is also positive, at least in a neighbourhood of $\partial X.$ For $\delta>0$ is small, let
\begin{equation}\label{3.18}
a(\delta)=\frac{1}{n}\min\{-x\Delta_gx|_{\partial X_\delta}\}>0.
\end{equation}
Then $H\geq n(1+a(\delta))$ on $\partial X_\delta.$  As it is showed in proposition 2 in \cite{cai1999boundaries}, we have the  diameter estimate:
\begin{equation}\label{3.19}
diam(X\setminus X_\delta,g^+)\leq \coth^{-1}(1+a(\delta))
\end{equation}
Then by (\ref{1.7}), we deduce that
\begin{equation}\label{3.20}
2\cdot(\frac{Vol(\partial X,\hat{g})}{\mathcal{A}(p)})^{\frac{1}{n}}\leq \delta e^{\coth^{-1}(1+a(\delta))}
=\delta e^{\frac{1}{2}\ln\frac{a(\delta)+2}{a(\delta)}}=\delta(1+\frac{2}{a(\delta)})^{\frac{1}{2}}
\end{equation}
(\ref{3.16}) and (\ref{3.17}) imply that
\begin{equation}\label{3.21}
\lim\limits_{\delta\rightarrow 0}\delta(1+\frac{2}{a(\delta)})^{\frac{1}{2}}=\lim\limits_{\delta\rightarrow 0}(\frac{2\delta^2}{a(\delta)})^{\frac{1}{2}}=(\frac{4n(n-1)}{\min S_{\hat{g}}})^{\frac{1}{2}}
\end{equation}
Combining (\ref{3.20}) and (\ref{3.21}), we finally get that
\begin{equation}\label{3.22}
(Vol(\partial X,\hat{g}))^{\frac{2}{n}}\cdot \min S_{\hat{g}}\leq n(n-1)(\mathcal{A}(p))^{\frac{2}{n}}.
\end{equation}
If we choose the Yamabe metric $\hat{g}$ on $\partial X,$ then (\ref{3.14}) is obtained by (\ref{3.22}).

\section{Examples and a rigidity result}
\begin{example}
The hyperbolic space:
\begin{equation}\label{4.1}
(\mathbb{H}^{n+1},g_\mathbb{H}=dt^2+\sinh^2tg_{\mathbb{S}^n})
\end{equation}
 where $g_{\mathbb{S}^n}$ is the standard metric on the $n-$sphere. Let $o$ be the centre point, then the distance function from o, $s_o=t.$
 So
\begin{equation}\label{4.2}
\mathcal{A}(o)=2^n\cdot\lim\limits_{t\rightarrow+\infty}e^{-nt}\omega_n\sinh^nt=\omega_n.
\end{equation}
Here $\omega_n$ denotes the volume of the standard unit sphere $\mathbb{S}^n.$
Since $\mathcal{A}$ is an invariant under the isometry transformation and ${H}^{n+1}$ is a homogeneous space, we derive that
$\mathcal{A}$ is a constant function on $\mathbb{H}^{n+1}$ and $\mathcal{A}\equiv \omega_n.$
\end{example}

\begin{example}\label{ex4.2}
Hyperbolic manifold:
\begin{equation}\label{4.3}
  (\mathbb{R}^n\times \mathbb{S}^1(\lambda),g^+=dt^2+\sinh^2tg_{\mathbb{S}^{n-1}}+\cosh^2tg_{\mathbb{S}^1(\lambda)}).
\end{equation}
Here $n\geq 2,$ and we use the polar coordinate system $\mathbb{R}^n=[0,+\infty)\times\mathbb{S}^{n-1}.$
The conformal infinity of $g^+$ is
$$(\mathbb{S}^{n-1}\times\mathbb{S}^1,[g_{\mathbb{S}^{n-1}}+g_{\mathbb{S}^1(\lambda)}]).$$
Then we have
\begin{equation}\label{4.4}
  \mathcal{A}(t_0,\cdot,\cdot)=\int_{\mathbb{S}^{n-1}\times\mathbb{S}^1(\lambda)}e^{nb_{\gamma_+}(t_0,w,\theta)}d\Theta_{n-1}d\Theta_\lambda.
\end{equation}
Here $d\Theta_{n-1}$ and $d\Theta_\lambda$  are the standard metric on the $n-1$ sphere $\mathbb{S}^{n-1}$ and $\mathbb{S}^1(\lambda).$
$\gamma_+$ is a geodesic ray starting at $(0_n,\theta_0)$ and $b_{\gamma_+}$ is the buseman function with respect to $\gamma_+.$
\begin{proof}
Let $E=\{0_n\}\times\mathbb{S}^1(\lambda)$ be a compact set where $0_n$ is the centre of $\mathbb{R}^n.$ We know that $t$ is the distance function from $E,$ i.e. $t=s_E.$ Let $g^E=e^{-2t}g^+,$ then
\begin{equation}\label{4.5}
  \hat{g}^E=g^E|_{\partial (\mathbb{R}^n\times \mathbb{S}^1(\lambda))}=\frac{1}{4}(g_{\mathbb{S}^{n-1}}+g_{\mathbb{S}^1(\lambda)}).
\end{equation}

Now we are going to calculate $\mathcal{A}(p_0)$ for some $p_0=(t_0,w_0,\theta_0)\in \mathbb{R}^n\times \mathbb{S}^1(\lambda).$
Firstly we need to study the distance function from $p_0.$ In fact, let $y=(t,w,\theta)\in \mathbb{R}^n\times \mathbb{S}^1(\lambda)$
where $t\geq t_0.$ If we use $d$ to denote the distance function $dist_{g^+}$,  then
\begin{equation}\label{4.6}
  d(p_0,y)=d((t_0,w,\theta),(t,w_0,\theta_0))
\end{equation}
We still use $s_{p_0}(\cdot)$ and $s_E(\cdot)$ to denote the distance function from $p_0$ and $E.$ Hence
\begin{equation}\label{4.7}
\lim\limits_{t\rightarrow+\infty}[s_E(y)-s_{p_0}(y)]=\lim\limits_{t\rightarrow\infty}
[t-d((t_0,w,\theta),(t,w_0,\theta_0))]=b_{\gamma_+}(t_0,w,\theta)
\end{equation}
where $\gamma_+(t)=(t,w_0,\theta_0),t\geq 0$ is a geodesic ray.
Then
\begin{equation}\label{4.8}
\begin{aligned}
  \mathcal{A}(p_0)&=2^n\int_{\partial X}dV_{\hat{g}^{p_0}}=2^n\int_{\partial X}e^{-n(s_{p_0}-s_E)|_{\partial X}}dV_{\hat{g}^E}\\
  &=\int_{\mathbb{S}^{n-1}\times\mathbb{S}^1(\lambda)}e^{nb_{\gamma_+}(t_0,w,\theta)}d\Theta_{n-1}d\Theta_\lambda.
  \end{aligned}
\end{equation}
By an orthogonal transformation, we find that $\mathcal{A}(p_0)$ is not depending on $(w_0,\theta_0),$ and hence it is a function of $t_0.$
\end{proof}
\begin{lemma}
If $t_0=0,$ then
\begin{equation}\label{4.9}
  \mathcal{A}(p_0)=2\omega_{n-1}\int_0^{\lambda\pi}\frac{1}{\cosh^n y}dy.
\end{equation}
\end{lemma}

\begin{proof}
Let $p_0=(0_0,\theta_0)$ and $y=(t,w,\theta), t\geq 0.$ Choose $q=(0_n,\theta).$ Then the geodesic line $\overline{p_0q}$ is  perpendicular to $\overline{qy}$ at $q.$ We also know that
\begin{equation}\label{4.10}
dist_{g^+}(y,q)=t,\ \ \ \
dist_{g^+}(p_0,q)=dist_{\mathbb{S}^1(\lambda)}(\theta,\theta_0)=\lambda|\theta-\theta_0|.
\end{equation}
Recall Pythagorean theorem in hyperbolic manifold, i.e.
\begin{equation}\label{4.11}
\cosh t\cdot\cosh (\lambda|\theta-\theta_0|)=\cosh (dist_{g^+}(p_0,y)).
\end{equation}
Hence
\begin{equation}\label{4.12}
\begin{aligned}
  b_{\gamma_+}(0_n,\theta)&=\lim\limits_{t\rightarrow\infty}[s_E(y)-s_{p_0}(y)]
  \\&=\lim\limits_{t\rightarrow\infty}[t-\cosh^{-1}(\cosh t\cdot\cosh(\lambda|\theta-\theta_0|))]  \\
  &= -\ln\cosh (\lambda|\theta-\theta_0|).
\end{aligned}
\end{equation}
Then
\begin{equation}\label{4.13}
  \mathcal{A}(p_0)=\int_{\mathbb{S}^{n-1}\times\mathbb{S}^1(\lambda)}\frac{1}{\cosh^n (\lambda|\theta-\theta_0|)}d\Theta_{n-1}d\Theta_\lambda.
\end{equation}
Then (\ref{4.9}) is achieved by the variable substitution of multiple integrals.

\end{proof}
Notice that (\ref{4.9}) can be obtained by partial integration and recursion. Furthermore, we have the following estimate
\begin{equation}\label{4.14}
  0<\mathcal{A}(0_n,\theta_0)<2\omega_{n-1}\int_0^{+\infty}\frac{1}{\cosh^n y}dy=\omega_n.
\end{equation}
We observe that $\mathcal{A}((0_n,\theta_0))$ can take all values in $(0,\omega_n)$ for $\lambda\in(0,+\infty).$
\begin{lemma}
If $\lambda$ is small enough, then $\mathcal{A}$ is not a constant function.
\end{lemma}

\begin{proof}
For any $p_0=(t_0,w_0,\theta_0)\in [0,+\infty)\times\mathbb{S}^{n-1}\times\mathbb{S}^1(\lambda)$ and $y=(t,w,\theta)$ where $t>t_0$
let $d=dist_{g^+}$ be the distance function. We have that

 \begin{equation}\label{4.15}
 \begin{aligned}
d(p_0,y)&\leq d(y,(t_0,w,\theta))+d((t_0,w,\theta),(t_0,w_0,\theta))+d((t_0,w_0,\theta),p_0)\\
&\leq t-t_0+\sinh t_0|w-w_0|+\lambda\cosh t_0|\theta-\theta_0|
\end{aligned}
\end{equation}
Therefore
 \begin{equation}\label{4.16}
 \begin{aligned}
 \mathcal{A}(p_0)&=2^n\int_{\partial X}dV_{\hat{g}^{p_0}}=2^n\int_{\partial X}e^{-n(s_{p_0}-s_E)|_{\partial X}}dV_{\hat{g}^E}\\
 &\geq \int_{\mathbb{S}^{n-1}\times\mathbb{S}^1(\lambda)}e^{n(t_0-\sinh t_0|w-w_0|-\lambda\cosh t_0|\theta-\theta_0|)}d\Theta_{n-1}d\Theta_\lambda
\\&=e^{nt_0}\int_0^{\lambda\pi} 2e^{-n\cosh t_0\cdot x}dx\int_0^\pi \omega_{n-2}\sin^{n-2}y\cdot e^{-n\sinh t_0\cdot y}dy
\\&=2\omega_{n-2} e^{nt_0}\frac{1-e^{-n\lambda\pi \cosh t_0}}{n\cosh t_0}\int_0^\pi\sin^{n-2}y\cdot e^{-n\sinh t_0\cdot y}dy
\end{aligned}
\end{equation}
If $n=2,$ then we get
 \begin{equation}\label{4.17}
 \mathcal{A}(p_0)\geq 4 e^{2t_0}\frac{1-e^{-2\lambda\pi \cosh t_0}}{2\cosh t_0}\frac{1-e^{-2\pi \sinh t_0}}{2\sinh t_0}
\end{equation}
which imlies $\liminf\limits_{t_0\rightarrow+\infty}\mathcal{A}(p_0)\geq 4.$
If $n=3,$ then we get
 \begin{equation}\label{4.18}
 \mathcal{A}(p_0)\geq 4\pi e^{3t_0}\frac{1-e^{-3\lambda\pi \cosh t_0}}{3\cosh t_0}\cdot\frac{1+e^{-3\pi\sinh t_0}}{1+9\sinh^2t_0}
\end{equation}
which implies that $\liminf\limits_{t_0\rightarrow+\infty}\mathcal{A}(p_0)\geq \frac{32}{39}\pi.$ Similarly, there exists a constant $C=C(n)>0,$ only depending on the dimension, such that
\begin{equation}\label{4.19}
\liminf\limits_{t_0\rightarrow+\infty}\mathcal{A}(p_0)\geq C.
\end{equation}
It means that the function $\mathcal{A}$ is uniformly bounded from below at infinity. On the other hand, (\ref{4.9}) tells us that
$\mathcal{A}(0_n,\cdot)$ tends to 0 as $\lambda$ goes to 0. Hence $\mathcal{A}$ is not constant as long as $\lambda$ is sufficiently small.

\end{proof}

\end{example}
From (\ref{4.14})we deduce that $\mathcal{A}(0_n,\theta_0) \nearrow\omega_n$ as $\lambda$ tends to infinity when
the hyperbolic manifold $(\mathbb{R}^n\times \mathbb{S}^1(\lambda),g^+)$ is convergent to the hyperbolic space in pointed Cheeger-Gromov topology. In fact, we have the following rigidity result:
\begin{theorem}[theorem 1.6]
  Let $(X,g^+)$ be an $n+1-$ dimensional AH metric and the Ricci curvature satisfies $Ric[g^+]\geq -ng^+$, then $\mathcal{A}\leq \omega_n$ and the equality holds at one point if and only if $X $is isometric to a hyperbolic space.
\end{theorem}
\begin{proof}
    Noticing that when the Ricci curvature is bounded from below, we can use the Bishop-Gromov volume  comparison theorem, i.e.
  $\frac{Vol(\partial B^{g^+}_p(t))}{Vol(\partial B^\mathbb{H}_o(t))}$ is monotonically decreasing on $t$ and
\begin{equation}\label{4.20}
  \frac{Vol(\partial B^{g^+}_p(t))}{Vol(\partial B^\mathbb{H}_o(t))}\leq 1, \ \ \forall t>0
\end{equation}
 The monotonicity guarantees the existence of limit when $t\rightarrow+\infty$ and we derive that $\mathcal{A}(p)\leq \omega_n.$
  \\ If $\mathcal{A}(p)=\omega_n$ for some $p\in X,$ then we can use the property of monotonically decreasing to get
\begin{equation}\label{4.21}
1=\lim\limits_{t\rightarrow +\infty}\frac{Vol(\partial B^{g^+}_p(t))}{Vol(\partial B^\mathbb{H}_o(t))}\leq\lim\limits_{t\rightarrow 0^+}\frac{Vol(\partial B^{g^+}_p(t))}{Vol(\partial B^\mathbb{H}_o(t))}=1
\end{equation}
  Hence $\frac{Vol(\partial B^{g^+}_p(t))}{Vol(\partial B^\mathbb{H}_o(t))}=1$ for any $t>0,$ as a consequence, $X$ is isometric to a hyperbolic space.
\end{proof}

As the end of this section, we give an estimate of the relative volume function on Ads-Schwarzchild space, which was firstly introduced in \cite{hawking1983thermodynamics}.
\begin{example}\label{ex4.7}

\begin{equation}\label{4.24}
 (X=\mathbb{R}^2\times\mathbb{S}^2,\ \ g^m=\lambda^2Vg_{\mathbb{S}^1}+V^{-1}dr^2+r^2g_{\mathbb{S}^2})
\end{equation}
where $V=V(r)=1+r^2-\frac{2m}{r}$ and $m>0$ denotes the mass. The parameter r runs over the interval $[r_m,+\infty)$ and $r_m$ is the positive root for $V(r)=0.$ Consider the change of coordinates $t=t(r)$ such that
\begin{equation}\label{4.25}
V^{-1}dr^2=dt^2,\ \ i.e.\ \ t=\int_{r_m}^r[V(\tau)]^{-1/2}d\tau.
\end{equation}
We get that
\begin{equation}\label{4.26}
g^m=dt^2+\lambda^2Vg_{\mathbb{S}^1}+r^2g_{\mathbb{S}^2}
\end{equation}
and we regard $V$ and $r$ as the functions of $t$ by the inverse function theorem. The metric is smooth at the centre of $R^2$ if and only if
\begin{equation}\label{4.27}
\lambda\sqrt{V}|_{t=0}=0,\ \ \frac{d}{dt}|_{t=0}(\lambda\sqrt{V})=1
\end{equation}
Then we get that
\begin{equation}\label{4.28}
\lambda=\frac{2r_m}{3r_m^2+1}.
\end{equation}
A direct calculations show hat
\begin{equation}\label{4.29}
-1-\frac{m}{r^3}\leq K[g^m]\leq -1+\frac{2m}{r^3},\ \ \ Ric[g^m]=-3g^m.
\end{equation}
Hence $(X,g^m)$ is an AHE manifold, although the sectional curvature could be arbitrarily large provided $m$ is small.
\par About the relative volume function, we have the following estimate and omit the proof as it is nearly the same as that in lemma 4.3 and
lemma 4.4.
\begin{lemma}
  If the mass $m$ is small enough, then the relative volume function $\mathcal{A}$ on X is not a constant function. Moreover, for any $m>0,$
\begin{equation}\label{4.30}
 \liminf\limits_{t_0\rightarrow+\infty} \mathcal{A}(t_0,\cdot,\cdot)\geq\frac{32}{27}\pi.
\end{equation}
and
\begin{equation}\label{4.31}
 \lim\limits_{m\rightarrow 0^+} \mathcal{A}(r_m,\cdot,\cdot)=0.
\end{equation}
\end{lemma}
\end{example}

\section{The capacity on AH manifolds}
The concept of capacity plays an important role in researching the isoperimetric inequalities and Sobolev inequalities in  Euclidean space. Many results are extended to some special manifold. Let's start with the notion on manifold.
\begin{definition}
Let $(M,g)$ be a Riemannian manifold. Suppose that $F\subseteq \Omega\subseteq M$ where $F$ is compact and $\Omega$ is open in $M.$ For each $p\geq 1,$ the p-Capacity is defined by
\begin{equation}\label{5.1}
  cap_p(F,\Omega)=\inf\limits_{u}\int_M|\nabla u|^p dV_g
\end{equation}
where the infimum is taken over all Lipschitz functions $u$ satisfying $u = 1$ on
$F$ and $u$ has compact support in $\Omega.$  If $\Omega=M,$ we write $cap_p(F,M)$ as $cap_p(F).$
\end{definition}
We study the capacity of balls in manifold, the formulas of which in simply connected space forms are already known. One can see section
2.2.4 in \cite{maz2013sobolev} for the Euclidean case.  We can directly get the capacity of balls in hyperbolic space $\mathbb{H}^{n+1},$
that is \cite{grigor1999isoperimetric}:
\begin{equation}\label{5.2}
  cap_p(B_r,B_R)=\omega_n[\int_r^R (\sinh t)^\frac{n}{1-p} dt]^{1-p}
\end{equation}
for $p>1.$ Let $R$ tends to infinity, we get that
\begin{equation}\label{5.3}
  cap_p(B_t)=\frac{1}{2^n}(\frac{n}{p-1})^{p-1}\omega_n e^{nt}+o(e^{nt}),\ \ t\rightarrow+\infty.
\end{equation}

\par Let's firstly recall some notions in the previous section. Suppose $(X,g^+)$ is an AH manifold and $q\in X.$  We denote the geodesic sphere by $B(t)=B^{g^+}_q(t)$ and claim that the symbol $\nabla ,|.|$ are with respect to $g^+.$
We also use $A(s)=Vol(\partial B(s),g^+)$ to denote the n-dimensional volume of the geodesic sphere.
\par Now we use the classical methods from \cite{polya1951isoperimetric} to get the upper bound of capacity via flux. Let $t$ be a fixed big number and define a fuction
\begin{equation}\label{5.4}
  f:[t,+\infty)\rightarrow [0,1], \ f(t)=1,f(+\infty)=\lim\limits_{s\rightarrow+\infty}f(s)=0
\end{equation}
Define $u=1$ on $B(t)$ and $u=f(s_q)$ on $X\setminus B(t)$ where $s_q=dist_{g^+}(q,\cdot),$ then
\begin{equation}\label{5.5}
  \int_X|\nabla u|^pdV_{g^+}=\int_{X\setminus B(t)}[f'(s)]^p|\nabla s|^pdV_{g^+}=\int_t^{+\infty}[f'(s)]^p A(s)ds
\end{equation}
On the other hand,
\begin{equation}\label{5.6}
\begin{aligned}
1&=-\int_t^{+\infty}f'(s)ds=\int_t^{+\infty}-f'(s)[A(s)]^{\frac{1}{p}}[A(s)]^{-\frac{1}{p}}ds\\
&\leq(\int_t^{+\infty}[f'(s)]^pA(s)ds)^{\frac{1}{p}}(\int_t^{+\infty}[A(s)]^{\frac{1}{1-p}}ds)^{1-\frac{1}{p}}
\end{aligned}
\end{equation}
Thus when
\begin{equation}\label{5.7}
  f(s)=-(\int_t^{+\infty}[A(\tau)]^{\frac{1}{1-p}}d\tau)^{-1}\cdot\int_t^s([A(\tau)]^{\frac{1}{1-p}}d\tau+1
\end{equation}
we have the following equality:
\begin{equation}\label{5.8}
  \int_t^{+\infty}[f'(s)]^pA(s)ds=(\int_t^{+\infty}[A(s)]^{\frac{1}{1-p}}ds)^{1-p}
\end{equation}
\par With the above preparations, we could prove theorem 1.7.
\par We firstly study the capacity of AH manifold in the case the Ricci curvature is bounded from below by $-n,$ i.e. $Ric[g^+]\geq-ng^+.$ By the volume comparison theorem, $\frac{A(s)}{\sinh^ns}$ is monotonically decreasing to $\mathcal{A}(q)$ as $s\rightarrow +\infty.$ Hence the function $\mathcal{A}$ is well-defined.
$$\forall\varepsilon>0,\exists T<0,\forall t>T,\frac{A(t)}{(e^t/2)^n}<\mathcal{A}(q)+\varepsilon.$$
Thus
\begin{equation}\label{5.9}
\begin{aligned}
(\int_t^{+\infty}[A(s)]^{\frac{1}{1-p}}ds)^{1-p}&\leq\int_t^{+\infty}[(\mathcal{A}(q)+\varepsilon)(\frac{e^s}{2})^n]^{\frac{1}{1-p}}ds)^{1-p}
\\ &=(\mathcal{A}(q)+\varepsilon)\frac{1}{2^n}(\int_t^{+\infty} e^{\frac{ns}{1-p}}ds)^{1-p}
\\ &=\frac{1}{2^n}(\frac{n}{p-1})^{p-1}(\mathcal{A}(q)+\varepsilon)e^{nt}
\end{aligned}
\end{equation}
 Let $\varepsilon\rightarrow 0,$ we deduce that

\begin{equation}\label{5.10}
\limsup\limits_{t\rightarrow+\infty}\frac{cap_p(B^{g^+}_q(t))}{e^{nt}}\leq\frac{1}{2^n}(\frac{n}{p-1})^{p-1}\mathcal{A}(q)
\end{equation}
 Hence the upper bound is achieved. Then the first part of theorem 1.7 holds because of (\ref{5.10}) and theorem 1.6.
\\
 \par It is well known that the lower estimates for $p-$capacity could be described via the isoperimetric function. That is,
 an isoperimetric function $I(\tau)=$ the infimum of $Vol(\partial\Omega)$
for all precompact open set $\Omega\subseteq X$ such that $Vol(\Omega)\geq \tau.$
We now recall Maz’ya's isocapacitary inequalitiese of (2.2.8) in \cite{maz2013sobolev},
\begin{equation}\label{5.11}
  cap_p(F,\Omega)\geq (\int_{\mu(F)}^{\mu(\Omega)}[I(\tau)]^{\frac{p}{1-p}}d\tau)^{1-p}
\end{equation}
for $p>1.$  In our paper, $\mu(\Omega)$ is the $n+1$ dimensional Riemannian volume of $\Omega.$
\par Recall the Cheeger isoperimetric constant of $X,$
\begin{equation}\label{5.12}
\mathcal{C}h(X,g^+)=\inf\limits_{\Omega}\frac{Vol(\partial\Omega)}{Vol(\Omega)}
\end{equation}
 where the infimum is taken over all the compact (smooth) domains in $X.$
 If the Ricci curvature and  the scalar curvature of $X$ satisfy (\ref{1.8}) , then it is showed
 in \cite{hijazi2020cheeger} that the Cheeger isoperimetric constant
\begin{equation}\label{5.13}
\mathcal{C}h(X,g^+)= n\Leftrightarrow Y(\partial X,[\hat{g}])\geq 0.
\end{equation}
 Hence under the conditions of theorem 1.7, we derive that the isoperimetric function of $X$ is $I(\tau)=n\tau$ for any $\tau>0.$  Therefore,
\begin{equation}\label{5.14}
cap_p(B^{g^+}_q(t))\geq (\int_{Vol(B^{g^+}_q(t))}^{+\infty}(n\tau)^{\frac{p}{1-p}}d\tau)^{1-p}=\frac{n^p}{(p-1)^{p-1}}Vol(B^{g^+}_q(t))
\end{equation}
Then
\begin{equation}\label{5.15}
\begin{aligned}
\liminf\limits_{t\rightarrow+\infty}\frac{cap_p(B^{g^+}_q(t))}{e^{nt}}&
\geq\liminf\limits_{t\rightarrow+\infty}\frac{n^p}{(p-1)^{p-1}}\frac{Vol(B^{g^+}_q(t))}{e^{nt}}
\\&=\frac{n^p}{(p-1)^{p-1}}\lim\limits_{t\rightarrow+\infty}\frac{\frac{d}{dt}[Vol(B^{g^+}_q(t))]}{ne^{nt}}
\\&=\frac{1}{2^n}(\frac{n}{p-1})^{p-1}\mathcal{A}(q)
\end{aligned}
\end{equation}
 Finally, (\ref{5.10}) and (\ref{5.15}) imply that
\begin{equation}\label{5.16}
\lim\limits_{t\rightarrow+\infty}\frac{cap_p(B^{g^+}_q(t))}{e^{nt}}=\frac{1}{2^n}(\frac{n}{p-1})^{p-1}\mathcal{A}(q)
\end{equation}

 \section{The capacity of  a general AH manifold}
 In this section, we study the capacity of an AH manifold $(X,g^+).$ The curvature conditions (\ref{1.3}) and (\ref{1.8}) may  not hold, thus the relative volume function $\mathcal{A}$ may not exist. However, we can still estimate the capacity of balls in $X$ with the property of asymptotically hyperbolic.
 \par Let $g=x^2g^+$  be the $C^1$ geodesic compactification on $X_\delta=\partial X\times(0,\delta)$ for some small $\delta<1$ and set $r=-\ln x.$ We know that $r$  is a distance function on $(X_\delta,g^+).$  For any $q\in X,$ let $s_q(\cdot)$ be the distance of function from $q.$ We still have the property that $s_q-r$  is bounded on $X_\delta,$  although $|d(s-r)|_g$ may be unbounded. Assume that $|s_q-r|<D$ on $X_\delta.$
 \par Recall that $B(t)=B^{g^+}_q(t).$
For any $t$ large enough, we have that
$$B(t)\subseteq X_{\varepsilon(t)}\subseteq B(t+2D)$$
and here $\varepsilon(t)=e^{-(t+D)}.$ Then
\begin{equation}\label{6.1}
  cap_p(B(t))\leq cap_p(X_{\varepsilon(t)})\leq cap_p(B(t+2D)).
\end{equation}
Hence
\begin{equation}\label{6.2}
  cap_p(B(t))=O(e^{nt})\Leftrightarrow cap_p(X_{\varepsilon(t)})=O(e^{n\varepsilon(t)}),\ t\rightarrow+\infty.
\end{equation}
\par We will prove that $cap_p(X_{\varepsilon})=O(\varepsilon^{-n}),\ \  \varepsilon\rightarrow 0,$ which  would imply that theorem 1.8 holds.
\par We will use the same method as that in section 5 to prove it. For a small $\varepsilon>0,$ define $f:[0,\varepsilon)\rightarrow [0,1], f(0)=0, \ f(\varepsilon)=1$
and consider the function $u(\cdot)=f(x(\cdot))$ on $X_\varepsilon$ and $u(\cdot)=1$ on $X\setminus X_\varepsilon.$ Then for any $p>1,$

\begin{equation}\label{6.3}
\begin{aligned}
\int_X|\nabla^{g^+} u|_{g^+}^pdV_{g^+}&=\int_{X\varepsilon}x^{p-n-1}|\nabla^gu|_g^pdV_g
\\&=\int_{X\varepsilon}x^{p-n-1}|f'(x)|^p\cdot|\nabla^gx|_g^pdV_g
\\&=\int_0^\varepsilon x^{p-n-1}|f'(x)|^p\cdot Vol(\Sigma_x)dx
\end{aligned}
\end{equation}
where $Vol(\Sigma_x)$ denote the $n-$dimensional volume of the level set $\Sigma_x$ of the geodesic defining function.
Now let $T(x)=x^{p-n-1}\cdot Vol(\Sigma_x).$

\begin{equation}\label{6.4}
\begin{aligned}
1&=\int_0^\varepsilon f'(x)dx=\int_0^\varepsilon f'(x)T^{\frac{1}{p}}(x)T^{-\frac{1}{p}}(x)dx\\
&\leq (\int_0^\varepsilon [f'(x)]^pT(x)dx)^{\frac{1}{p}}(\int_0^\varepsilon T^{\frac{1}{1-p}}(x)dx)^{1-\frac{1}{p}}
\end{aligned}
\end{equation}
The equality holds if and only if $$f(x)=(\int_0^\varepsilon T^{\frac{1}{1-p}}(\tau)d\tau)^{-1}\int_0^xT^{\frac{1}{1-p}}(\tau)d\tau$$ for $x\in[0,\varepsilon].$ Then we infer that

\begin{equation}\label{6.5}
\begin{aligned}
cap_p(X_\varepsilon)&\leq \int_X|\nabla^{g^+} u|_{g^+}^pdV_{g^+}=(\int_0^\varepsilon T^{\frac{1}{1-p}}(x)dx)^{1-p}
\\&=(\int_0^\varepsilon  x^{\frac{p-n-1}{1-p}}\cdot [Vol(\Sigma_x)]^{\frac{1}{1-p}}dx)^{1-p}
\end{aligned}
\end{equation}

$g$ is a $C^1$ geodesic compactification, let $\hat{g}$ be its boundary metric, hence by Gauss lemma,
\begin{equation}\label{6.6}
g=dx^2+g_x=dx^2+\hat{g}+O(x)
\end{equation}
holds near boundary. Then
\begin{equation}\label{6.7}
\frac{\det g_x}{\det \hat{g}}=\det(\hat{g}^{-1}g_x)=\det(I_n+O(x))=1+O(x)
\end{equation}

Therefore
\begin{equation}\label{6.8}
Vol(\Sigma_x)=\int_{\Sigma_x}dV_{g_x}=\int_{\partial X}\sqrt{\frac{\det g_x}{\det \hat{g}}}dV_{\hat{g}}=Vol(\partial X,\hat{g})+O(x)
\end{equation}
Back to (\ref{6.5}), we obtain
\begin{equation}
\begin{aligned}
cap_p(X_\varepsilon)&\leq (\int_0^\varepsilon  x^{\frac{p-n-1}{1-p}}\cdot [Vol(\partial X,\hat{g})+O(x))]^{\frac{1}{1-p}}dx)^{1-p}\\
&=Vol(\partial X,\hat{g})(\int_0^\varepsilon  x^{\frac{p-n-1}{1-p}}(1+O(x))dx)^{1-p}\\
&=Vol(\partial X,\hat{g})(\frac{n}{p-1})^{p-1}\varepsilon^{-n}+O(\varepsilon^{-n+1-p})
\end{aligned}
\end{equation}

Hence $cap_p(X_\varepsilon)=O(\varepsilon^{-n}),$ and it is easy to get the upper bound
\begin{equation}
\limsup\limits_{\varepsilon\rightarrow 0}\varepsilon^n\cdot cap_p(X_\varepsilon)\leq Vol(\partial X,\hat{g})(\frac{n}{p-1})^{p-1}.
\end{equation}

\noindent{Xiaoshang Jin}\\
  School of mathematics and statistics, Huazhong University of science and technology, Wuhan, P.R. China. 430074
 \\Email address: jinxs@hust.edu.cn

\end{document}