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\title{A simple  generalization of a Willmore-type inequality}

\alttitle{Une généralisation simple d'une inégalité de type Willmore}

\author{\firstname{Seong-Hun} \lastname{Paeng}}

\address{Department of Mathematics, Konkuk University,  120 Neungdong-ro, Gwangjin-gu, Seoul 05029, Korea}

\email{shpaeng@konkuk.ac.kr}

\thanks{This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT)(No. RS-2023-00242367).} 

 \CDRGrant[NRF]{RS-2023-00242367}

\keywords{\kwd{Willmore-type inequality} \kwd{Ricci curvature}}

\altkeywords{\kwd{Inégalité de type Willmore} \kwd{courbure de Ricci}}


\subjclass{53C20, 53C42}

\begin{abstract} 
   We give a simple generalization of the Willmore-type inequality in \tralicstex{[1]}{\cite{AFM}}  without assuming that  Ricci curvature is nonnegative everywhere.
\end{abstract}

\begin{altabstract} 
Nous donnons une généralisation simple de l'inégalité de type Willmore dans \tralicstex{[1]}{\cite{AFM}} sans supposer que la courbure de Ricci est partout non négative.
\end{altabstract}


\begin{document}


\maketitle

\section{Introduction}

The classical Willmore inequality for a bounded domain $\Omega$ of $\mathbb R^3$ with smooth boundary says that 
\[
\int_{\partial \Omega}H^2\,\dd \vol_{\partial\Omega}\ge16\pi,
\]
where $H$ is the mean curvature of $\partial\Omega$ and $\,\dd \vol_{\partial\Omega}$ is the volume form of $\partial\Omega$. 
In  \cite{AFM}, Agostiniani, Fogagnolo, and Mazzieri obtained the following Willmore-type inequality for an $n$-dimensional complete noncompact Riemannian manifold $(M,g)$ with the nonnegative Ricci curvature and Euclidean volume growth:
\begin{equation}\label{Wineq}
\int_{\partial \Omega}\Big\vert\frac{H}{n-1}\Big\vert^{n-1}\,\dd \vol_{\partial\Omega}\ge\AVR (g)|S^{n-1}|,
\end{equation}
where $|S^{n-1}|$ is the volume of the standard sphere and $\AVR (g)$ is  the asymptotic volume ratio, i.e. 
\[
\AVR (g)=\lim_{r\to\infty}\frac{n\vol(B(p,r))}{r^n|S^{n-1}|},
\]
where $B(p,r)$ is the $r$-ball  centered at $p\in M$.
They used the monotonicity-rigidity properties of the function $U_\beta(t)$ which is defined as follows:
\[
U_\beta(t)=t^{-\beta(\frac{n-1}{n-2})}\int_{\{u=t\}}|\nabla u|^{\beta+1},
\]
where $u$ is  the harmonic function which vanishes at infinity and  $u=1$ on $\partial \Omega$.

In this paper, we will  generalize the inequality without  assuming $\Ric \ge0$ everywhere. The condition $\Ric \ge0$ is essential  in \cite{AFM} to apply results on harmonic functions.

 Let $\rho(q)=\max\{ (-\Ric (v,v))_+\mid |v|=1, ~v\in T_qM\}$, where $f_+=\max\{0,f\}$.
We define  integral norms ${\mathcal R}_{n-1}$ as follows:
\[
\begin{aligned}
{\mathcal R}_{n-1}&=\int_{M}\rho^{n-1} \,\dd V,
\end{aligned}
\]
where $\dd V$ is the volume form of $M$. 

\begin{theorem}\label{thm1} 
Let $(M,g)$ be an $n$-dimensional complete noncompact Riemannian manifold with Euclidean volume growth and $\Ric \ge0$ outside $B(p,R_0)$. If $\Omega\subset M$ is a bounded and open subset with smooth boundary, then 
\begin{equation}
\begin{aligned}
\left(\int_{\partial \Omega}\left\vert\,\frac{H}{n-1}\right\vert^{n-1}\, \dd \vol_{\partial\Omega}\right)^{\frac{1}{n-1}}
&\ge\left({\AVR(g)|S^{n-1}|}\right)^\frac{1}{n-1}-\frac{(2R_0)^\frac{n-1}{n-2}(\mathcal R_{n-1})^{\frac{1}{n-1}}}{n-1}.
\end{aligned}
\end{equation}
\end{theorem}

If $\Ric \ge0$, then $\mathcal R_{n-1}=0$, so we obtain \eqref{Wineq}.

\section{Proof of Theorem \ref{thm1}}


We will use the following notations. Let
\[
\Omega_t=\{x\in M\mid d(x,\Omega)\le t\}
\]
\[
\partial \Omega_t=\{x\in M\mid d(x,\Omega)=t\}.
\]
 Let  $\gamma_q$ be the outward normal geodesic such that
$\gamma_q(0)=q$ and $\gamma_q'(0)$ is perpendicular to $\partial \Omega$ for
$q\in \partial \Omega$. Let
\[
t_q=\max\{t\mid d(\gamma_q(t),\partial \Omega)=t\}.
\]
Then we have 
\[
M\setminus\Omega=\bigcup_{q\in\partial \Omega}\{\gamma_q(t)\mid t\le t_q\}.
\]

 We
denote by $g_t$ the induced metric of $\partial \Omega_t$ from the metric  $g$ of $M$. Let
$\dd \vol_t$ be the volume form of $\partial \Omega_t$ induced from $g_t$.
Then the volume form $\dd \vol_{\partial \Omega}$ of $\partial \Omega$ is $\dd \vol_0$ and the volume form of $M$ satisfies   $\dd V=\dd t\wedge
\dd \vol_t$. By identifying $\gamma_q(t)\in \partial \Omega_t$  with $q\in\mathbb \partial \Omega$ for $t\le t_q$, we define $\omega(t,q)$ and $h(t,q)$  as follows:
\begin{align}\label{dvol00}
\dd \vol_t&=\omega(t,\cdot)\,\dd \vol_{\partial \Omega}
\\
\label{domega00}
 \frac{\partial}{\partial t}\omega(t,q)&=h(t,q)\omega(t,q),
 \end{align}
where $h(t,\cdot)$ is the mean curvature of $\partial \Omega_t$. So $H=h(0,\cdot)$. We abbreviate $\omega(t,q)$ and $h(t,q)$ to $\omega(t)$ and $h(t)$,
respectively.
For $t< t_q$,  $h$ satisfies 
\begin{equation}\label{riccati}
h'+\frac{h^2}{n-1}\le-\Ric \left(\frac{\partial}{\partial t},\frac{\partial}{\partial t}\right),
\end{equation}
where $\frac{\partial}{\partial t}=\gamma_q'(t)$ and $|\frac{\partial}{\partial t}|=1$.
For $t> t_q$, we let $\omega(t)=0$. Then $\omega(t)$ is defined for all $t>0$ and $\vol(\Omega_t\setminus \Omega)=\int_0^t\int_{\partial\Omega}\omega(s,q)\,\dd \vol_{\partial\Omega} \, \dd s$. Furthermore, $\omega'(t)=h(t)\omega(t)$ for $t<t_q$ and $\omega'(t)=0$ for $t>t_q$.
We denote  left and right limits $\lim_{\delta\to0-}f(t+\delta)$ and  $\lim_{\delta\to0+}f(t+\delta)$ by $f(t-)$ and $f(t+)$, respectively.

Let
\begin{equation}\label{psi00000}
\psi(t)=\begin{cases}
h_+(t)&\text{~if~}t< t_q\\
h_+(t-)&\text{~if~}t=t_q\\
0 &\text{~if~} t>t_q,
\end{cases}
\end{equation}
where $h_+(t)=\max\{h(t),0\}$.
Then we have
\begin{equation}\label{omegader}
\omega'(t)\le\psi(t)\omega(t)
\end{equation} 
for $t<t_q$. Also $\psi'=h'$ on $\{h(t)>0\}$ and $\psi'=0$ for $t>t_q$.

Although both  $\omega$ and $\psi$ are continuous along $\gamma_q(t)$ for $t\ne t_q$, they are left continuous for all $t>0$, i.e. $\omega(t-)=\omega(t)$ and $\psi(t-)=\psi(t)$ for all $t>0$. 
%(We also denote $f(t+)=\lim_{\delta\to0+}f(t+\delta)$.)
Even if $\omega$ is  not continuous at $t=t_q$, the left derivative $\lim_{\delta\to0-}\frac{\omega(t+\delta)-\omega(t)}{\delta}$ is well defined for any $t>0$.
% where the left derivative $f'(t-)=\lim_{\delta\to0-}\frac{f(t+\delta)-f(t)}{\delta}$. 
%Note that   $f'(t-)$ used in this paper is different  $\lim_{\delta\to0-}f'(t+\delta)$. 
On the other hand,  the right derivative $\lim_{\delta\to0+}\frac{\omega(t+\delta)-\omega(t)}{\delta}$ is not well defined, i.e. it can be $-\infty$.
For a while, we will abbreviate the left derivative $\lim_{\delta\to0-}\frac{f(t+\delta)-f(t)}{\delta}$ to $f'(t)$ for simplicity.

The left derivative $\omega'(t)$ is bounded on $[0,T]\times\partial\Omega$ for $T>0$ by the Jacobi equation $J''=R(\frac{\partial}{\partial t},J)\frac{\partial}{\partial t}$, where $R$ is the curvature tensor  and $J$ is the Jacobi field. Also $(\psi^{n-1})'(t)$ is bounded on $[0,T]\times\partial\Omega$ since $h'(t)\ne-\infty$ (i.e. $\gamma_q(t)$ is not a focal point) at $t$ when $h(t)\ge0$.

Since $\rho=\max\{(-\Ric (v,v))_+~|~|v|=1,~v\in T_qM\}$, \eqref{riccati} becomes
\begin{equation}\label{psi}
\psi'+\frac{\psi^2}{n-1}\le\rho,
\end{equation}
where we also let $\rho(\gamma_q(t))=0$ if $t>t_q$.
We let 
\[
\mathcal H_{n-1}(t)=\int_{\partial\Omega_t}\psi^{n-1}\dd \vol_t=\int_{\partial\Omega}\psi^{n-1}\omega.
\]
Since $(\psi^{n-1}\omega)(t-)=(\psi^{n-1}\omega)(t)$ for every $t$,  $\mathcal H_{n-1}(t)$ is left continuous, i.e. 
\begin{equation}\label{lc}
\mathcal H_{n-1}(t-)=\lim_{\delta\to0-}\int_{\partial\Omega}(\psi^{n-1}\omega)(t+\delta)=\int_{\partial\Omega}(\psi^{n-1}\omega)(t)=\mathcal H_{n-1}(t)\end{equation}
by   the dominated convergence theorem. 


% from the Jacobi equation $J''=R(\frac{\partial}{\partial t},J)\frac{\partial}{\partial t}$, where $R$ is the curvature tensor  and $J$ is the Jacobi field. 
%Let $f(t,q)$ be a nonnegative function  with a bounded left derivative $\frac{\partial f}{\partial t}$ on $[0,r]\times\partial\Omega$ for any $r>0$ and $f(t,q)=0$ for $t>t_q$. Also $f$ is continuous for $t<t_q$. Similarly as $\omega$, we abbreviate $f(t,q)$ to $f(t)$. Then $f$ is left continuous, i.e. $f(t)=f(t-)=\lim_{\delta\to 0-}f(t+\delta)$ even if $f$ is not necessarily continuous. Also we denote  $\lim_{\delta\to 0+}f(t+\delta)$ by $f(t+)$. 
%Let $W_t=\{q\in\partial\Omega~|~t\ne t_q\}$ be the set of discontinuous points for $\omega$ and $f$. 
  Since  $\frac{(\psi^{n-1}\omega)(t+\delta)-(\psi^{n-1}\omega)(t)}{\delta}\le0$ for $t>t_q$ and $\delta<0$, we obtain that
\begin{multline}\label{intder}
\int_{\partial \Omega}\frac{\psi^{n-1}(t+\delta)\omega( t+\delta)-\psi^{n-1}(t)\omega(t)}{\delta}\\
\le\int_{\{q\in\partial\Omega~|~t\le t_q\}}\frac{(\psi^{n-1}(t+\delta)-\psi^{n-1}(t))\omega( t+\delta)+\psi^{n-1}(t)(\omega(t+\delta)-\omega(t))}{\delta}\\
\to\int_{\{q\in\partial\Omega~|~t\le t_q\}}(\psi^{n-1})'\omega+\psi^{n-1}\omega'=\int_{\partial\Omega}(\psi^{n-1})'\omega+\psi^{n-1}\omega'
\end{multline}
as $\delta\to0-$ by the dominated convergence theorem for $t\le t_q$ and $\omega(t)=\psi(t)=0$ for $t>t_q$.

From now on, we redefine $f'(t)=\frac{\dd }{\dd t}f(t)$ as follows:
\[
f'(t):=\limsup_{\delta\to0-}\frac{f(t+\delta)-f(t)}{\delta}.
\]
 We call $f'(t)$ the left derivative of $f$ for simplicity. Then the left derivative $\mathcal H_{n-1}'(t)$ satisfies
\begin{equation}\label{intder1}
\begin{aligned}
\frac{\dd }{\dd t}\int_{\partial\Omega_t}\psi^{n-1}\dd \vol_t&=\frac{\dd }{\dd t}\int_{\partial\Omega}\psi^{n-1}\omega\\
&=\limsup_{\delta\to0-}\left(\int_{\partial\Omega}\frac{\psi^{n-1}(t+\delta)\omega( t+\delta)-\psi^{n-1}(t)\omega(t)}{\delta}\right)\\
%&\le\limsup_{\delta\to0}\int_{W_t^\delta}\frac{f(t+\delta)\omega( t+\delta)-f(t)\omega(t)}{\delta}\\
%&=\int_{\partial\Omega} (f\omega)'\\
&\le\int_{\partial\Omega}(\psi^{n-1})'\omega+\psi^{n-1}\omega'\\
&\le\int_{\partial \Omega}\psi^{n-2}((n-1)\psi'+\psi^{2})\omega\\
%&=\int_{\partial\Omega_t}(f'+f\psi)\dd \vol_t\\
%&\le\int_{\partial \Omega_t}\psi^{n-2}((n-1)\psi'+\psi^{2})\omega\\
&\le(n-1)\int_{\partial \Omega_t}\psi^{n-2}\rho \dd \vol_t\\
&\le(n-1)\left(\int_{\partial \Omega_t}\psi^{n-1}\dd \vol_t\right)^{\frac{n-2}{n-1}}
\left(\int_{\partial \Omega_t}\rho^{n-1}\dd \vol_t\right)^{\frac{1}{n-1}}
\end{aligned}
\end{equation}
by \eqref{psi}, \eqref{intder}.


\begin{proposition}\label{lem1}For sufficiently large $t$,  we have 
\begin{equation}
\left(\int_{\partial \Omega}|H|^{n-1}\, \dd \vol_{\partial\Omega}\right)^{\frac{1}{n-1}}\ge\mathcal H_{n-1}(t)^{\frac{1}{n-1}}-(2R_0)^\frac{n-1}{n-2}(\mathcal R_{n-1})^{\frac{1}{n-1}}.
\end{equation}
\end{proposition}
\begin{proof} Since $\int_{\partial \Omega}|H|^{n-1}\ge \mathcal H_{n-1}(0)$, we only need to show that 
\[
\mathcal H_{n-1}(t)^\frac{1}{n-1}-\mathcal H_{n-1}(0)^\frac{1}{n-1}\le (2R_0)^\frac{n-1}{n-2}(\mathcal R_{n-1})^{\frac{1}{n-1}}.
\]
By \eqref{intder1},
\begin{equation}\label{13}
\frac{\mathcal H_{n-1}'(t)}{(n-1)\mathcal H_{n-1}(t)^{\frac{n-2}{n-1}}}\le\left(\int_{\partial \Omega_t}\rho^{n-1}\dd \vol_t\right)^{\frac{1}{n-1}}
\end{equation}
if $\mathcal H_{n-1}(t)\ne0$.


If $g(x)=x^{\frac{1}{n-1}}$ and $f$ is left continuous, then the left derivative $g(f)'$ satisfies that
%$g'(x)$ is decreasing and $g'(x)\ge \frac{g(y)-g(x)}{y-x}\ge g'(y)$ for $x<y$. So we obtain that
\begin{equation}\label{14}
\begin{aligned}
g(f)'(a)&=\limsup_{\delta\to0-}\left(\frac{g(f(a+\delta))-g(f(a))}{f(a+\delta)-f(a)}\right)\left(\frac{f(a+\delta)-f(a)}{\delta}\right)\\
& =g'(f(a))f'(a)
\end{aligned}
\end{equation}
if $f(a)\ne0$.
By \eqref{13} and \eqref{14}, the left derivative $\frac{\dd }{\dd t}{\mathcal H_{n-1}(t)}^{\frac{1}{n-1}}$ satisfies that
\begin{equation}\label{15}
\frac{\dd }{\dd t}\Big({\mathcal H_{n-1}(t)}^{\frac{1}{n-1}}\Big)\le\left(\int_{\partial \Omega_t}\rho^{n-1}\,\dd \vol_t\right)^{\frac{1}{n-1}}
\end{equation}
if $\mathcal H_{n-1}(t)\ne0$. If $\mathcal H_{n-1}(t)=0$, 
then the left derivative $(\mathcal H_{n-1}(t))^{\frac{1}{n-1}})'\le0$ 
since $\mathcal H_{n-1}(s)^{\frac{1}{n-1}}\ge0$ for any $s$. Thus \eqref{15} holds for any $t$.

\begin{lemma}\label{lem3}
Let $f$  and $g\ge0$ be left continuous  on $[a,b]$ and $f(t)\ge f(t+)$ for all $t\in[a,b]$.  If the left derivative $f'$ satisfies that $f'(t)\le g(t)$ for any $t$ and $g$ is bounded, then  
\[
f(b)-f(a)\le \int_a^bg.
\]
\end{lemma}
\begin{proof}
For $\eta>0$, let $I=\{x\in[a,b]~|~f(b)-f(x)\le \int_x^b(g(t)+\eta)dt\}$ and $A=\inf I$. Then it is clear that $b\in I$. First, we show that $A\in I$. If not, there is a sequence $a_n\in I$ such that $a_n\to A$. Since $f(A)\ge f(A+)$,  we have
\[
f(b)-f(A)\le f(b)-f(A+)\le\lim_{n\to\infty}\int_{a_n}^b(g(t)+\eta)\, \dd t=\int_{A}^b(g(t)+\eta)\, \dd t.
\]
Hence we obtain that $A\in I$.

Now we show that $A=a$. If not, for $s$ satisfying $a\le  s<A$,
\[
f(b)-f(s)>\int_s^b(g(t)+\eta)\, \dd t.
\]
Since $f(b)-f(A)\le\int_A^b(g(t)+\eta)\, \dd t$, we have
\[
f(A)-f(s)>\int_s^A(g(t)+\eta)\, \dd t.
\]
Then
\[
f'(A)=\limsup_{s\to A-}\frac{f(A)-f(s)}{A-s}\ge\limsup_{s\to A-}\frac{1}{A-s}\int_s^A(g(t)+\eta)\, \dd t=g(A)+\eta
\]
since $g$ is left continuous.
But this is a contradiction to our assumption. So we obtain that $A=a$, i.e. $f(b)-f(a)\le\int_a^b(g(t)+\eta)\dd t$. Since $\eta$ is arbitrarily chosen, letting $\eta\to0$, we obtain Lemma~\ref{lem3}.
\end{proof}




By the same reason that ${\mathcal H_{n-1}}$ is left continuous, $\big(\int_{\partial \Omega_t}\rho^{n-1}\,\dd \vol_t\big)^{\frac{1}{n-1}}$ is left continuous.
Hence, by \eqref{15} and Lemma \ref{lem3} with $f={\mathcal H_{n-1}}$ and $g=\big(\int_{\partial \Omega_t}\rho^{n-1}\,\dd \vol_t\big)^{\frac{1}{n-1}}$,
\begin{equation}
\mathcal H_{n-1}(t)^{\frac{1}{n-1}}-\mathcal H_{n-1}(0)^{\frac{1}{n-1}}
\le\int_0^t\left(\int_{\partial \Omega_s}\rho^{n-1}\dd \vol_s\right)^{\frac{1}{n-1}} \, \dd s.
\end{equation}
Since the diameter of $B(p,R_0)$ is not larger than $2R_0$, $B(p,R_0)\setminus\Omega\subset \Omega_{t_0+2R_0}\setminus \Omega_{t_0}$ for some $t_0\ge0$.  
For $t>t_0+2R_0$,
\begin{equation}
\begin{aligned}
\mathcal H_{n-1}(t)^{\frac{1}{n-1}}-\mathcal H_{n-1}(0)^{\frac{1}{n-1}}
&\le\int_{t_0}^{t_0+2R_0}\left(\int_{\partial \Omega_s}\rho^{n-1}\, \dd \vol_s\right)^{\frac{1}{n-1}} \, \dd s\\
&\le\left(\int_{M}\rho^{n-1} \, \dd V\right)^{\frac{1}{n-1}}(2R_0)^\frac{n-2}{n-1},
\end{aligned}
\end{equation}
so we have Proposition \ref{lem1}.
\end{proof}

Now we calculate $\mathcal H_{n-1}(t)$ with $\vol(\partial\Omega_t)$.
Let $A(t)=\vol_t(\partial\Omega_t)=\int_{\partial\Omega}\omega(t,\cdot)$. 
Then $A(t)$ is left continuous similarly as in \eqref{lc}. Similarly as in \eqref{intder1},
\begin{equation}\label{Vder}
\begin{aligned}
A'(t)&\le\int_{\partial\Omega}\omega'\\
&\le\int_{\partial\Omega_t}\psi \dd \vol_{t}\\
&\le\left(\int_{\partial\Omega_t}\psi^{n-1}\,\dd \vol_t\right)^{\frac{1}{n-1}}
\left(\int_{\partial\Omega_t}\,\dd \vol_t\right)^{\frac{n-2}{n-1}}\\
&\le\mathcal H_{n-1}(t)^{\frac{1}{n-1}}A(t)^{\frac{n-2}{n-1}}.
\end{aligned}
\end{equation} 
by \eqref{omegader}.
By \eqref{14}, \eqref{Vder}, and the left continuity of $A(t)$, 
\begin{equation}\label{Ader1}
(n-1)\big(A^{\frac{1}{n-1}}(t)\big)'\le\mathcal H_{n-1}(t)^\frac{1}{n-1}.
\end{equation}
From Proposition \ref{lem1}, we obtain that
\begin{equation}\label{willpre}
\left(\int_{\partial \Omega}|H|^{n-1}\, \dd \vol_{\partial\Omega}\right)^{\frac{1}{n-1}}\ge(n-1)\big(A^{\frac{1}{n-1}}(t)\big)'-(2R_0)^\frac{n-1}{n-2}(\mathcal R_{n-1})^{\frac{1}{n-1}}.
\end{equation}

Now we calculate  the left derivative $\big(A^{\frac{1}{n-1}}(t)\big)'$ with $\AVR (g)$.
\begin{lemma}\label{lem4}
Let $f:[0,\infty)\to\mathbb R$ have left derivative and satisfy $f(t+)\le f(t)$ for all $t$.
If $\lim\sup_{t\to\infty}\frac{f(t)}{t}=a>0$, then the left derivative $f'(t)$ satisfies 
\begin{equation}\label{fp1}
\limsup_{t\to\infty}f'(t)\ge a.
\end{equation}
\end{lemma}

\begin{proof} 
Assume $\limsup_{t\to\infty}f'(t)< a-2\epsilon<a$. Then there exists $r_0>0$ such that $f'(t)<a-\epsilon$ for $t\ge r_0$. 
Take a large $R>r_0$.
By Lemma \ref{lem3},
\begin{equation}
f(R)- f(r_0)\le\int_{r_0}^R(a-\epsilon)=(a-\epsilon)(R-r_0),\end{equation}
which implies that
\[
\lim\sup_{t\to\infty}\frac{f(t)}{t}\le a-\epsilon.
\]
It is a contradiction to our condition.  Hence, 
 \begin{equation}\label{fp}
 \limsup_{t\to\infty}f'(t)\ge a,
 \end{equation}
which completes the proof of Lemma \ref{lem4}.
\end{proof}






We let $f(t)=\vol(\Omega_t\setminus\Omega)^\frac{1}{n}$, where $\vol(\Omega_t\setminus\Omega)=\int_0^t\int_{\partial\Omega}\omega(s,q)\,\dd \vol_{\partial\Omega} \, \dd s$. 
Although $\vol(\Omega_t\setminus\Omega)$ may not be differentiable since $\omega(t)$ is not continuous at $t_q$, $f$ is continuous.  We can obtain  the left derivative as follows:
\begin{equation}
\frac{\dd }{\dd t}\vol(\Omega_t\setminus\Omega)=\int_{\partial\Omega}\omega(t,q)\, \dd \vol_{\partial\Omega}=A(t).
\end{equation}
Since $\vol(\Omega_t\setminus\Omega)$ is continuous, the left derivative satisfies
\[
f'(t)=\frac{1}{n}\frac{A(t)}{\vol(\Omega_{t}\setminus\Omega)^\frac{n-1}{n}}
\]
by  \eqref{14}.


Since
$\frac{\vol(\Omega_R\setminus\Omega)}{\vol(B(p,R))}\to1$ as $R\to\infty$, we have 
\begin{equation}\label{fquo}
\lim_{t\to\infty}\frac{f(t)}{t}=\left(\frac{\AVR (g)|S^{n-1}|}{n}\right)^\frac{1}{n}.
\end{equation} 
From $f'(t)=\frac{1}{n}\frac{A(t)}{\vol(\Omega_{t}\setminus\Omega)^\frac{n-1}{n}}$ and Lemma \ref{lem4}, we obtain that
\begin{equation}\label{fratio}
\begin{aligned}
\left(\frac{\AVR (g)|S^{n-1}|}{n}\right)^\frac{1}{n}&\le\limsup_{t\to\infty} f'(t)=\limsup_{t\to\infty}\frac{1}{n}\frac{A(t)}{\vol(\Omega_{t}\setminus\Omega)^\frac{n-1}{n}}\\
&=\limsup_{t\to\infty}\frac{1}{n}\frac{A(t)}{\vol(B(p,t))^\frac{n-1}{n}}.
\end{aligned}
\end{equation}

Although $A^{\frac{1}{n-1}}(t)$ may  not be continuous, we have  $A^{\frac{1}{n-1}}(t+)\le A^{\frac{1}{n-1}}(t)$ since $\omega(t)\ge\omega(t+)$. By Lemma \ref{lem4} and \eqref{fquo}, \eqref{fratio},
 \[
\begin{aligned}
\limsup_{t\to\infty}(A^{\frac{1}{n-1}}(t))'&\ge\limsup_{t\to\infty} \frac{A^{\frac{1}{n-1}}(t)}{t}\\
&\ge n^\frac{1}{n-1} \lim_{t\to\infty}\left(\frac{\AVR (g)|S^{n-1}|}{n}\right)^\frac{1}{n(n-1)}\frac{\vol(B(p,t))^\frac{1}{n}}{t}\\
&=\Big({\AVR (g)|S^{n-1}|}\Big)^\frac{1}{n-1}.\end{aligned}
\]
Consequently, from \eqref{willpre},
\[
\left(\int_{\partial\Omega}\left(\frac{|H|}{n-1}\right)^{n-1}\right)^\frac{1}{n-1}\ge\Big({\AVR (g)|S^{n-1}|}\Big)^\frac{1}{n-1}-\frac{1}{n-1}(2R_0)^\frac{n-1}{n-2}(\mathcal R_{n-1})^{\frac{1}{n-1}},
\]
which completes the proof of Theorem \ref{thm1}.

\section*{Acknowledgements} 

The author would like to express his gratitude to the referee for their helpful comments, which improve this paper very much.

\section*{Declaration of interests}
The authors do not work for, advise, own shares in, or receive funds from
any organization that could benefit from this article, and have declared no
affiliations other than their research organizations.

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