# A Minkowski type inequality for manifolds with positive spectrum

Ovidiu Munteanu and Jiaping Wang  Department of Mathematics, University of
Connecticut, Storrs, CT 06268, USA<EMAIL_ADDRESS>School of
Mathematics, University of Minnesota, Minneapolis, MN 55455, USA
<EMAIL_ADDRESS>
###### Abstract.

The classical Minkowski inequality implies that the volume of a bounded convex
domain is controlled from above by the integral of the mean curvature of its
boundary. In this note, we establish an analogous inequality without the
convexity assumption for all bounded smooth domains in a complete manifold
with its bottom spectrum being suitably large relative to its Ricci curvature
lower bound. An immediate implication is the nonexistence of embedded compact
minimal hypersurfaces in such manifolds. This nonexistence issue is also
considered for steady and expanding Ricci solitons.

##  1\. Introduction

On a complete Riemannian manifold $(M,g),$ the Laplacian $\Delta$ is a self-
adjoint operator according to [15]. So the spectrum $\sigma(M)$ of $M,$
defined as the spectrum $\sigma(-\Delta)$ of $-\Delta,$ is a closed subset of
$[0,\infty).$ The bottom spectrum is given by

$\lambda_{1}(M):=\min\\{\lambda\in\sigma(M)\\}.$

Alternatively, it is characterized as the best constant for the Poincaré
inequality

$\lambda_{1}\,\int_{M}\phi^{2}\leq\int_{M}\left|\nabla\phi\right|^{2}$

for all compactly supported smooth functions $\phi$ on $M.$

A result by McKean [26] says that $\lambda_{1}(M)\geq\frac{(n-1)^{2}}{4}$ for
an $n$-dimensional, simply connected, complete manifold $M^{n}$ with sectional
curvature $K\leq-1$.

The famous Sullivan-Patterson theory [31, 33] computes the bottom spectrum for
the quotient space $\mathbb{H}^{n}/\Gamma$ of the $n$-dimensional real
hyperbolic space $\mathbb{H}^{n},$ where $\Gamma$ is a discrete, finitely
generated, group of isometries of $\mathbb{H}^{n}.$ Namely,
$\lambda_{1}(\mathbb{H}^{n}/\Gamma)=\frac{(n-1)^{2}}{4}$ if
$d_{\Gamma}\leq\frac{n-1}{2}$ and
$\lambda_{1}(\mathbb{H}^{n}/\Gamma)=d_{\Gamma}\,(n-1-d_{\Gamma})$ when
$d_{\Gamma}\geq\frac{n-1}{2},$ where $d_{\Gamma}$ is the Hausdorff dimension
of the limit set of $\Gamma,$ that is, those points $\theta$ in the ideal
boundary at infinity $S_{\infty}(\mathbb{H}^{n})$ of $\mathbb{H}^{n}$ such
that $\theta=\lim_{i\to\infty}\gamma_{i}(x)$ for some $x\in\mathbb{H}^{n}$ and
a sequence of $\gamma_{i}\in\Gamma$.

Another notable result, due to Brooks [4], is that $\lambda_{1}(M)>0$ for a
covering space $M$ of a compact manifold $N$ if and only if the covering group
is nonamenable.

Finally, we mention a result of Lee [21]. Recall that a Riemannian manifold
$(M,g)$ is conformally compact if topologically it is the interior of a
compact manifold $\overline{M}$ with boundary $N$ and its metric
$g=\rho^{-2}\,g_{\overline{M}}$ for some metric $g_{\overline{M}}$ on
$\overline{M}$ and smooth function $\rho$ on $\overline{M}$ with $\rho=0$ on
$N$ and $d\rho\neq 0$ on $N.$ Note that different pairs of $\rho$ and
$g_{\overline{M}}$ induce the same conformal class on $N.$

######  Theorem 1.1 (Lee).

Let $(M^{n},g)$ be a conformally compact Einstein manifold with its Ricci
curvature normalized to be $-(n-1).$ If its boundary $N$ with the induced
conformal metric has nonnegative scalar curvature, then
$\lambda_{1}(M)=\frac{(n-1)^{2}}{4}.$

A different proof of the result was given by X. Wang [34].

Concerning the upper bound of the bottom spectrum, we have the following
classical result due to Cheng [9].

######  Theorem 1.2 (Cheng).

Let $M^{n}$ be a complete Riemannian manifold with
$\mathrm{Ric}\geq-(n-1)\kappa$ for some nonnegative constant $\kappa.$ Then

$\lambda_{1}(M)\leq\lambda_{1}\left(\mathbb{H}^{n}(-\kappa)\right)=\frac{(n-1)^{2}}{4}\,\kappa.$

The rigidity issue has been studied by Li and the second author in [23, 24].

######  Theorem 1.3 (Li-Wang).

Suppose $(M^{n},g)$ is complete, $n\geq 3,$ with
$\lambda_{1}\geq\frac{(n-1)^{2}}{4}\,\kappa$ and
$\mathrm{Ric}\geq-(n-1)\kappa.$ Then either $M$ is connected at infinity or
$M^{n}=\mathbb{R}\times N^{n-1}$ for some compact $N$ with
$g=dt^{2}+e^{2\sqrt{\kappa}\,t}\,g_{N}$ for $n\geq 3$ or
$g=dt^{2}+\cosh^{2}(\sqrt{\kappa}\,t)\,g_{N}$ when $n=3.$

Note that as $\kappa$ goes to $0,$ the result recovers a weak version of the
famous Cheeger-Gromoll [5] splitting theorem for complete manifolds with
nonnegative Ricci curvature.

Our main purpose here is to establish the following Minkowski type inequality
for complete manifolds with positive bottom spectrum.

######  Theorem 1.4.

Let $\left(M^{n},g\right)$ be a complete Riemannian manifold of dimension
$n\geq 5$ with $\mathrm{Ric}\geq-\left(n-1\right)$ and

$\lambda_{1}\left(M\right)\geq\left(\frac{n-2}{n-1}\right)^{2}\left(2n-3\right).$

Then for any compact smooth domain $\Omega\subset M,$

$\frac{2}{3}\sqrt{n}\
\lambda_{1}\left(M\right)\mathrm{Vol}\left(\Omega\right)\leq\int_{\partial\Omega}\left|H\right|^{\frac{2n-3}{n-1}},$

where $H$ is the mean curvature of $\partial\Omega.$

The result seems to be new even for the hyperbolic space $\mathbb{H}^{n}.$ We
remark that it is necessary to assume $\lambda_{1}(M)>n-2.$ Indeed, for
$M^{n}=\mathbb{R}\times N^{n-1}$ with $g=dt^{2}+\cosh^{2}(t)\,g_{N},$
$\lambda_{1}(M)=n-2$ and $\mathrm{Ric}\geq-(n-1)$ when
$\mathrm{Ric_{N}}\geq-(n-2).$ Yet, the domain $\Omega$ given by
$\\{0<t<\varepsilon\\}$ violates the inequality when $\varepsilon$ is small.

Certainly, this example also shows that the result can not hold for $n=3.$
However, it remains unclear what to expect for $n=4.$ One may wish to compare
the result to the classical Minkowski inequality [27] for the Euclidean space
$\mathbb{R}^{n}$ and that for the hyperbolic space $\mathbb{H}^{n}$ [16]. The
advantage here is that no convexity is assumed for the domains.

######  Theorem 1.5 (Minkowski).

If $\Omega\subset\mathbb{R}^{n},$ $n\geq 3,$ is a convex domain with smooth
boundary $\Sigma$ and $\mathrm{H}$ is the mean curvature of $\Sigma$ with
respect to the outward unit normal, then there exists a sharp constant $c(n)$
so that

$\mathrm{Vol}\left(\Omega\right)\leq
c(n)\,\left(\int_{\Sigma}H\right)^{\frac{n}{n-2}}.$

Equality holds if and only if $\Omega$ is a ball.

The convexity can be relaxed to mean convex and star shaped by the work of
Guan-Li [17, 18], where they produced a different proof using a new mean
curvature flow. In fact, their proof yields the more general Alexandrov-
Fenchel inequalities of quermassintegrals and extends to other space forms as
well. For more related results, we refer to [2, 3, 7, 13, 19].

An immediate consequence of our result is the nonexistence of compact minimal
hypersurfaces.

######  Corollary 1.6.

Let $\left(M^{n},g\right)$ be a complete Riemannian manifold of dimension
$n\geq 5$ with $\mathrm{Ric}\geq-\left(n-1\right)$ and

$\lambda_{1}\left(M\right)=\frac{\left(n-1\right)^{2}}{4}.$

Then $M$ has no embedded compact minimal hypersurface. In particular, this
holds for a conformally compact Einstein manifold with its boundary having
nonnegative scalar curvature.

Note that the result is not true for $n=3$. Indeed, for
$M^{3}=\mathbb{R}\times N^{2}$ with $g=dt^{2}+\cosh^{2}(t)\,g_{N},$
$\lambda_{1}(M)=1$ and $\mathrm{Ric}\geq-2$ when $\mathrm{Ric_{N}}\geq-1.$
Yet, the hypersurface given by $\\{t=0\\}$ is totally geodesic.

The corollary follows from Theorem 1.4 by verifying that $\Sigma$ must enclose
a bounded domain $\Omega$ in $M.$ Indeed, observe that $M$ must be connected
at infinity as otherwise by Theorem 1.3,

$M=\mathbb{R}\times N,\ \ ds_{M}^{2}=dt^{2}+e^{2t}\,ds_{N}.$

Since $f(t,y)=t$ on $M$ is convex, by maximum principle, $\Sigma$ must be one
of the level sets of $f.$ However, each level set has constant mean curvature
$n-1$, which is a contradiction. The same argument shows that every double
cover of $M$ is connected at infinity as well. One then concludes from a
result by Carron and Pedon [6] that the integral homology
$H_{n-1}(M,\mathbb{Z})=0.$ It then follows that $\Sigma$ must enclose a
bounded domain $\Omega$ in $M.$

We now quickly sketch the proof of Theorem 1.4. First, there exists $v>0$ such
that $\Delta v=-\lambda_{1}(M)\,v$. Consider the function $h=\ln v$, for which
$\Delta h=-\lambda_{1}(M)-|\nabla h|^{2}$. Then

$\displaystyle\lambda_{1}\left(M\right)\mathrm{Vol}\left(\Omega\right)$
$\displaystyle\leq$
$\displaystyle\int_{\Omega}\left(\lambda_{1}\left(M\right)+\left|\nabla
h\right|^{2}\right)$ $\displaystyle=$ $\displaystyle-\int_{\Omega}\Delta h$
$\displaystyle=$ $\displaystyle\int_{\partial\Omega}h_{\nu},$

where $\nu$ is the inward unit normal to $\partial\Omega$.

The proof is then reduced to estimating $\int_{\partial\Omega}h_{\nu}.$ To do
so we consider the harmonic function $u$ on $M\setminus\Omega$ obtained as
$u=\lim_{R\to\infty}u_{R},$ where $\Delta u_{R}=0$ on
$\left(M\setminus\Omega\right)\cap B_{p}(R)$ with $u_{R}=1$ on
$\partial\Omega$ and $u_{R}=0$ on $\partial B_{p}(R).$ The upshot is to show

$c(n)\,\int_{\partial\Omega}h_{\nu}\leq\int_{\partial\Omega}\left(\left|\nabla
u\right|^{\alpha}\right)_{\nu}-\int_{\partial\Omega}\left(u^{\beta}\right)_{\nu}\left|\nabla
u\right|^{\alpha},$

where $\alpha=\frac{n-2}{n-1}$ and $\beta=\frac{n-2}{3n-5}.$ For that, we drew
inspiration from the monotonicity formulas for the Green’s function on
manifolds with nonnegative Ricci curvature [11, 12, 1], as well as on
$3$-dimensional manifolds with scalar curvature bounded below [28]. In the
process, the following generalized Poincaré inequality also comes into play.

######  Proposition 1.7.

Let $\left(M,g\right)$ be a complete manifold with
$\mathrm{Ric}\geq-\left(n-1\right)$ and $\lambda_{1}\left(M\right)>0.$ Let
$K\subset M$ be an open subset with (possibly noncompact) boundary $\partial
K.$ Then the Poincaré inequality

$\lambda_{1}\left(M\right)\int_{K}\phi^{2}\leq\int_{K}\left|\nabla\phi\right|^{2}-\int_{\partial
K}h_{\nu}\,\phi^{2}$

holds for any Lipschitz function $\phi$ with compact support in
$\overline{K},$ where $\nu$ is the outward unit normal to $\partial K.$

Concerning the nonexistence of compact minimal hypersurfaces, we also extend
our consideration to Ricci solitons.

######  Theorem 1.8.

Let $\left(M^{n},g,f\right)$ be a steady Ricci soliton. If there exists a
smooth compact embedded minimal hypersurface $\Sigma$ in $M,$ then
$\left(M,g\right)$ splits isometrically as a direct product
$\mathbb{R}\times\Sigma.$

A similar result is also established for expanding Ricci solitons. Recall that
a gradient Ricci soliton is a manifold $\left(M,g\right)$ such that there
exists a smooth function $f$ satisfying

$\mathrm{Ric}_{f}=\mathrm{Ric}+\mathrm{Hess}\left(f\right)=\lambda g$

for some constant $\lambda\in\mathbb{R}.$ Solitons are classified as
shrinking, steady or expanding, according to $\lambda>0,\ \lambda=0$ or
$\lambda<0,$ respectively. The function $f$ is called the potential.
Customarily, the constant $\lambda$ is assumed to be $1/2,$ $0,$ or $-1/2$ by
scaling. Obviously, Ricci solitons are natural generalizations of Einstein
manifolds. More significantly, they are the self similar solutions to the
Ricci flows, and play a crucial role in the study of singularities of the
flows [10].

As pointed out in [29], an important feature of a steady Ricci soliton to us
is that its bottom spectrum with respect to the weighted Laplacian
$\Delta_{f},$ defined by $\Delta_{f}u=\Delta u-\langle\nabla u,\nabla
f\rangle,$ always achieves the maximum value $\frac{1}{4}$ among all weighted
manifolds $(M,g,e^{-f}dv)$ with $\mathrm{Ric}_{f}\geq 0$ and $\left|\nabla
f\right|\leq 1.$

The paper is arranged as follows. In Section 2, we prove our main result
Theorem 1.4. In Section 3, we consider Ricci solitons and prove Theorem 1.8.

##  2\. Minkowski inequality

In this section, we prove Theorem 1.4. First, we make some general
consideration. Throughout this section, unless otherwise noted,
$\left(M^{n},g\right)$ is assumed to be an $n$-dimensional complete manifold
with positive spectrum $\lambda_{1}\left(M\right)>0$ and its Ricci curvature
$\mathrm{Ric}\geq-\left(n-1\right).$

It is well-known [14] that there exists $v>0$ such that

(2.1) $\Delta v=-\lambda_{1}\left(M\right)v.$

Hence,

(2.2) $h=\ln v$

satisfies

(2.3) $\Delta h=-\lambda_{1}\left(M\right)-\left|\nabla h\right|^{2}.$

Also, by [35] (or see Chapter 6 in [22]), positive solutions of (2.1) satisfy
the gradient estimate

(2.4) $\left|\nabla
h\right|\leq\frac{n-1}{2}+\sqrt{\frac{\left(n-1\right)^{2}}{4}-\lambda_{1}\left(M\right)}.$

The following generalized Poincaré inequality will be of use later.

######  Proposition 2.1.

Let $\left(M,g\right)$ be a complete manifold with
$\mathrm{Ric}\geq-\left(n-1\right)$ and $\lambda_{1}\left(M\right)>0.$ Let
$K\subset M$ be an open subset with boundary $\partial K$. Then

$\lambda_{1}\left(M\right)\int_{K}\phi^{2}\leq\int_{K}\left|\nabla\phi\right|^{2}-\int_{\partial
K}\left\langle\nabla h,\nu\right\rangle\phi^{2}$

holds for any Lipschitz function $\phi$ with compact support in
$\overline{K},$ where $\nu$ is the outward unit normal to the boundary
$\partial K.$ In particular,

$\lambda_{1}\left(M\right)\int_{K}\phi^{2}\leq\int_{K}\left|\nabla\phi\right|^{2}+A\int_{\partial
K}\phi^{2},$

where

$A=\frac{n-1}{2}+\sqrt{\frac{\left(n-1\right)^{2}}{4}-\lambda_{1}\left(M\right)}.$

###### Proof.

According to (2.3), for any Lipschitz function $\phi$ with compact support in
$\overline{K}$ we have

$\displaystyle\lambda_{1}\left(M\right)\int_{K}\phi^{2}$ $\displaystyle=$
$\displaystyle\int_{K}\left(-\Delta h-\left|\nabla
h\right|^{2}\right)\phi^{2}$ $\displaystyle=$
$\displaystyle\int_{K}\left(\left\langle\nabla
h,\nabla\phi^{2}\right\rangle-\left|\nabla h\right|^{2}\phi^{2}\right)$
$\displaystyle-\int_{\partial K}h_{\nu}\phi^{2}.$

Observe that

$2\phi\left\langle\nabla h,\nabla\phi\right\rangle\leq\left|\nabla
h\right|^{2}\phi^{2}+\left|\nabla\phi\right|^{2}.$

Therefore,

$\lambda_{1}\left(M\right)\int_{K}\phi^{2}\leq\int_{K}\left|\nabla\phi\right|^{2}-\int_{\partial
K}\left\langle\nabla h,\nu\right\rangle\phi^{2}.$

This proves the result. ∎

We will apply this Poincaré inequality on the sublevel sets of the harmonic
function $u$ constructed below. Given a compact domain $\Omega\subset M,$
according to [24], an unbounded component of $M\setminus\Omega$ is parabolic
if it has finite volume, and nonparabolic if it has infinite volume. Let
$E_{1},..,E_{k}$ be all the infinite volume connected components of
$M\setminus\Omega.$ Denote with $E=E_{1}\cup\cdots\cup E_{k}$ and

(2.5) $\displaystyle D$ $\displaystyle=$ $\displaystyle M\setminus E$
$\displaystyle\Sigma$ $\displaystyle=$ $\displaystyle\partial D=\partial E.$

Alternatively, $D$ is the union of $\Omega$ with all the finite volume
components of $M\setminus\Omega.$

Consider the following function $u_{i}$ with respect to a sequence
$R_{i}\to\infty.$

(2.6) $\displaystyle\Delta u_{i}$ $\displaystyle=$ $\displaystyle 0\text{ \ on
}B_{p}\left(R_{i}\right)\setminus D$ $\displaystyle u_{i}$ $\displaystyle=$
$\displaystyle 1\text{ \ on }\partial D$ $\displaystyle u_{i}$
$\displaystyle=$ $\displaystyle 0\text{ on }\partial
B_{p}\left(R_{i}\right)\cap E.$

As $\lambda_{1}\left(M\right)>0,$ from [23], the sequence
$\left\\{u_{i}\right\\}_{i=1}^{\infty}$ converges to a positive nonconstant
harmonic function $u:M\setminus D\rightarrow\left[0,1\right]$ such that $u=1$
on $\partial D.$ The strong maximum principle implies that $\left|\nabla
u\right|>0$ on $\Sigma=\partial D$. Moreover, by [24]

(2.7) $\int_{M\setminus\left(D\cup B_{p}\left(R\right)\right)}u^{2}\leq
C\,e^{-2\sqrt{\lambda_{1}\left(M\right)}R}$

for all $R>0$ large enough. As $D$ is the union of $\Omega$ together with all
the finite volume components of $M\setminus\Omega,$ by [24] the following
volume estimate holds.

(2.8) $\mathrm{Vol}\left(D\setminus B_{p}\left(R\right)\right)\leq
C\,e^{-2\sqrt{\lambda_{1}\left(M\right)}R}.$

We denote with

$\displaystyle L\left(\alpha,\beta\right)$ $\displaystyle=$
$\displaystyle\left\\{x\in M\setminus D:\alpha<u\left(x\right)<\beta\right\\}$
$\displaystyle\ell\left(t\right)$ $\displaystyle=$ $\displaystyle\left\\{x\in
M\setminus D:u\left(x\right)=t\right\\}.$

Note that these sets may be noncompact in general. However, (2.7) implies that

(2.9)
$\mathrm{Vol}\left(L\left(\alpha,\beta\right)\right)\leq\frac{1}{\alpha^{2}}\int_{M\setminus
D}u^{2}<\infty.$

According to [25],

(2.10) $\zeta=\int_{\ell\left(t\right)}\left|\nabla u\right|$

is a constant independent of $t\in\left[0,1\right]$. Hence, for any function
$F,$ by the co-area formula,

(2.11) $\int_{L\left(\alpha,\beta\right)}\left|\nabla
u\right|^{2}F\left(u\right)=\zeta\int_{\alpha}^{\beta}F\left(t\right)dt.$

The gradient estimate for positive harmonic functions states that

(2.12) $\left|\nabla u\right|\leq C\,u\text{ \ on }M\setminus D,$

where the constant $C$ depends only on the dimension $n$ and the maximum of
the mean curvature $\max_{\Sigma}\left|H_{\Sigma}\right|.$

Recall the Bochner formula

(2.13) $\frac{1}{2}\Delta\left|\nabla
u\right|^{2}\geq\left|\nabla^{2}u\right|^{2}-\left(n-1\right)\left|\nabla
u\right|^{2}$

and the improved Kato inequality

(2.14) $\left|\nabla^{2}u\right|^{2}\geq\frac{n}{n-1}\left|\nabla\left|\nabla
u\right|\right|^{2}.$

We begin with the following preliminary estimates.

######  Lemma 2.2.

Let $\left(M^{n},g\right)$ be a complete manifold with
$\mathrm{Ric}\geq-\left(n-1\right)$ and $\lambda_{1}\left(M\right)>0.$ There
exists a constant $C>0$ such that for all $0<t<1,$

(2.15) $\int_{L\left(t,1\right)}\left(u+\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-1}\right)\leq C\left(1-\ln
t\right)$

and

(2.16) $\int_{L\left(\frac{1}{2}t,t\right)}\left(u+\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-1}\right)\leq C.$

###### Proof.

We first prove (2.15). Let $\psi$ and $\chi$ be the cut-off functions

$\psi\left(x\right)=\left\\{\begin{array}[]{c}1\\\ R+1-r\left(x\right)\\\
0\end{array}\right.\begin{array}[]{l}\text{on }B_{p}\left(R\right)\\\ \text{on
}B_{p}\left(R+1\right)\setminus B_{p}\left(R\right)\\\ \text{on }M\setminus
B_{p}\left(R+1\right)\end{array}$

and

$\chi\left(x\right)=\left\\{\begin{array}[]{c}1\\\ \frac{\ln
u\left(x\right)-\ln\left(\frac{1}{2}t\right)}{\ln 2}\\\
0\end{array}\right.\begin{array}[]{l}\text{on }L\left(t,1\right)\\\ \text{on
}L\left(\frac{1}{2}t,t\right)\\\ \text{otherwise}\end{array}.$

We extend $u=1$ on $D$, and let $\phi=u^{\frac{1}{2}}\chi\psi$ in

$\lambda_{1}\left(M\right)\int_{M}\phi^{2}\leq\int_{M}\left|\nabla\phi\right|^{2}$

to obtain

$\displaystyle\lambda_{1}\left(M\right)\int_{M}u\chi^{2}\psi^{2}$
$\displaystyle\leq$ $\displaystyle 2\int_{M}\left|\nabla
u^{\frac{1}{2}}\right|^{2}\chi^{2}\psi^{2}+2\int_{M}u\left|\nabla\left(\chi\psi\right)\right|^{2}$
$\displaystyle\leq$ $\displaystyle\frac{1}{2}\int_{M}\left|\nabla
u\right|^{2}u^{-1}\chi^{2}\psi^{2}+4\int_{M}u\left|\nabla\chi\right|^{2}\psi^{2}$
$\displaystyle+4\int_{M}u\left|\nabla\psi\right|^{2}\chi^{2}.$

By (2.11) we immediately see that

$\int_{M}\left|\nabla
u\right|^{2}u^{-1}\chi^{2}\psi^{2}\leq\int_{L\left(\frac{1}{2}t,1\right)}\left|\nabla
u\right|^{2}u^{-1}=\zeta\ln\frac{2}{t}$

and that

(2.17) $\int_{M}u\left|\nabla\chi\right|^{2}\psi^{2}\leq\frac{1}{\left(\ln
2\right)^{2}}\int_{L\left(\frac{1}{2}t,t\right)}\left|\nabla
u\right|^{2}u^{-1}=\frac{\zeta}{\ln 2}.$

Finally, by (2.7) and (2.8) we have

(2.18)
$\int_{M}u\left|\nabla\psi\right|^{2}\chi^{2}\leq\frac{2}{t}\int_{M\setminus
B_{p}\left(R\right)}u^{2}\leq\frac{C}{t}e^{-2\sqrt{\lambda_{1}\left(M\right)}R}.$

This proves that

$\int_{L\left(t,1\right)\cap B_{p}\left(R\right)}u\leq C\left(1-\ln
t\right)+\frac{C}{t}e^{-2\sqrt{\lambda_{1}\left(M\right)}R}$

for all $R\geq 1.$ Making $R\rightarrow\infty$ implies that

(2.19) $\int_{L\left(t,1\right)}u\leq C\left(1-\ln t\right)$

for all $0<t<1.$

By (2.13) and (2.14) we have that

$\Delta\left|\nabla u\right|\geq\frac{1}{n-1}\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-1}-\left(n-1\right)\left|\nabla
u\right|$

on $M\setminus D$. It then follows that

$\displaystyle\frac{1}{n-1}\int_{M\setminus D}\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-1}\chi^{2}\psi^{2}$
$\displaystyle\leq$ $\displaystyle\int_{M\setminus
D}\chi^{2}\psi^{2}\Delta\left|\nabla u\right|$
$\displaystyle+\left(n-1\right)\int_{M\setminus D}\left|\nabla
u\right|\chi^{2}\psi^{2}.$

Note that by the gradient estimate (2.12) and (2.19) we have

$\displaystyle\int_{M\setminus D}\left|\nabla u\right|\chi^{2}\psi^{2}$
$\displaystyle\leq$ $\displaystyle C\int_{L\left(\frac{1}{2}t,1\right)}u$
$\displaystyle\leq$ $\displaystyle C\left(1-\ln t\right).$

Moreover, integrating by parts implies that

$\displaystyle\int_{M\setminus D}\chi^{2}\psi^{2}\Delta\left|\nabla u\right|$
$\displaystyle=$ $\displaystyle-\int_{M\setminus
D}\left\langle\nabla\left(\chi^{2}\psi^{2}\right),\nabla\left|\nabla
u\right|\right\rangle+\int_{\ell\left(1\right)}\left|\nabla u\right|_{\nu}$
$\displaystyle\leq$ $\displaystyle\frac{1}{2\left(n-1\right)}\int_{M\setminus
D}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-1}\chi^{2}\psi^{2}$
$\displaystyle+2\left(n-1\right)\int_{M\setminus D}\left|\nabla
u\right|\left|\nabla\left(\chi\psi\right)\right|^{2}+\int_{\ell\left(1\right)}\left|\nabla
u\right|_{\nu}$ $\displaystyle\leq$
$\displaystyle\frac{1}{2\left(n-1\right)}\int_{M\setminus
D}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-1}\chi^{2}\psi^{2}$ $\displaystyle+C\left(1-\ln
t\right)+\frac{C}{t}e^{-2\sqrt{\lambda_{1}\left(M\right)}R},$

where we have used (2.12), (2.17) and (2.18) to obtain the last line. Plugging
this estimate in (2) and making $R\rightarrow\infty$ we obtain that

$\int_{L\left(t,1\right)}\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-1}\leq C\left(1-\ln t\right)$

as claimed.

The second estimate (2.16) follows verbatim from the preceding argument by
modifying the function $\chi$ to

$\chi\left(x\right)=\left\\{\begin{array}[]{c}1\\\ \frac{\ln
u\left(x\right)-\ln\left(\frac{1}{4}t\right)}{\ln 2}\\\
\frac{\ln\left(2t\right)-\ln u}{\ln 2}\\\
0\end{array}\right.\begin{array}[]{l}\text{on }L\left(\frac{1}{2}t,t\right)\\\
\text{on }L\left(\frac{1}{4}t,\frac{1}{2}t\right)\\\ \text{on
}L\left(t,2t\right)\\\ \text{otherwise}\end{array}.$

∎

We are ready to prove the main result of this section.

######  Theorem 2.3.

Let $\left(M^{n},g\right)$ be a complete Riemannian manifold of dimension
$n\geq 5$ with $\mathrm{Ric}\geq-\left(n-1\right)$ and

$\lambda_{1}\left(M\right)\geq\left(\frac{n-2}{n-1}\right)^{2}\left(2n-3\right).$

Then for any compact smooth domain $\Omega\subset M,$

$\frac{2}{3}\sqrt{n}\
\lambda_{1}\left(M\right)\mathrm{Vol}\left(\Omega\right)\leq\int_{\partial\Omega}\left|H\right|^{\frac{2n-3}{n-1}},$

where $H$ is the mean curvature of $\partial\Omega.$

###### Proof.

As in (2.5) we let $D$ be the union of $\Omega$ with all the finite volume
components of $M\setminus\Omega.$ Define the harmonic function $u$ on
$M\setminus D$ as the limit of a subsequence of
$\left\\{u_{i}\right\\}_{i=1}^{\infty}$ from (2.6).

Let $\psi$ and $\chi$ be the cut-off functions

$\psi\left(x\right)=\left\\{\begin{array}[]{c}1\\\ R+1-r\left(x\right)\\\
0\end{array}\right.\begin{array}[]{l}\text{on }B_{p}\left(R\right)\\\ \text{on
}B_{p}\left(R+1\right)\setminus B_{p}\left(R\right)\\\ \text{on }M\setminus
B_{p}\left(R+1\right)\end{array}$

and

$\chi\left(x\right)=\left\\{\begin{array}[]{c}1\\\ \frac{\ln
u\left(x\right)-\ln\left(\frac{1}{2}t\right)}{\ln 2}\\\
0\end{array}\right.\begin{array}[]{l}\text{on }L\left(t,1\right)\\\ \text{on
}L\left(\frac{1}{2}t,t\right)\\\ \text{otherwise}\end{array}.$

The Bochner formula (2.13) and the inequality (2.14) imply the inequality (cf.
[23])

(2.21) $\Delta\left|\nabla u\right|^{\alpha}\geq-\left(n-2\right)\left|\nabla
u\right|^{\alpha},$

where

(2.22) $\alpha=\frac{n-2}{n-1}.$

For

(2.23) $\beta>\frac{1}{n-1}$

to be specified later, we multiply (2.21) by $u^{\beta}\chi^{2}\psi^{2}$ and
integrate it over $M\setminus D$ to obtain that

(2.24) $-\int_{M\setminus D}\left(\Delta\left|\nabla
u\right|^{\alpha}\right)u^{\beta}\chi^{2}\psi^{2}\leq\left(n-2\right)\int_{M\setminus
D}\left|\nabla u\right|^{\alpha}u^{\beta}\chi^{2}\psi^{2}.$

Integrating by parts one sees that the left side of (2.24) becomes

$\displaystyle-\int_{M\setminus D}\left(\Delta\left|\nabla
u\right|^{\alpha}\right)u^{\beta}\chi^{2}\psi^{2}$ $\displaystyle=$
$\displaystyle\int_{M\setminus D}\left\langle\nabla\left|\nabla
u\right|^{\alpha},\nabla u^{\beta}\right\rangle\chi^{2}\psi^{2}$
$\displaystyle+\int_{M\setminus D}\left\langle\nabla\left|\nabla
u\right|^{\alpha},\nabla\chi^{2}\right\rangle u^{\beta}\psi^{2}$
$\displaystyle+\int_{M\setminus D}\left\langle\nabla\left|\nabla
u\right|^{\alpha},\nabla\psi^{2}\right\rangle u^{\beta}\chi^{2}$
$\displaystyle-\int_{\ell\left(1\right)}\left(\left|\nabla
u\right|^{\alpha}\right)_{\nu},$

where $\nu=\frac{\nabla u}{\left|\nabla u\right|}$ is the inward unit normal
to $\partial D=\ell\left(1\right).$

By Lemma 2.2, (2.12), and (2.7) we have that

$\displaystyle\left|\int_{M\setminus D}\left\langle\nabla\left|\nabla
u\right|^{\alpha},\nabla\psi^{2}\right\rangle u^{\beta}\chi^{2}\right|$
$\displaystyle\leq$ $\displaystyle
2\alpha\int_{L\left(\frac{1}{2}t,1\right)\setminus
B_{p}\left(R\right)}\left|\nabla\left|\nabla u\right|\right|\left|\nabla
u\right|^{\alpha-1}u^{\beta}$ $\displaystyle\leq$ $\displaystyle
Ce^{-\sqrt{\lambda_{1}\left(M\right)}R}\int_{L\left(\frac{1}{2}t,1\right)}\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-1}$
$\displaystyle+Ce^{\sqrt{\lambda_{1}\left(M\right)}R}\int_{L\left(\frac{1}{2}t,1\right)\setminus
B_{p}\left(R\right)}u^{2\left(\alpha+\beta\right)-1}$ $\displaystyle\leq$
$\displaystyle Ce^{-\sqrt{\lambda_{1}\left(M\right)}R}\left(1-\ln t\right)$
$\displaystyle+\frac{C}{t}e^{\sqrt{\lambda_{1}\left(M\right)}R}\int_{L\left(\frac{1}{2}t,1\right)\setminus
B_{p}\left(R\right)}u^{2}$ $\displaystyle\leq$
$\displaystyle\frac{C}{t}e^{-\sqrt{\lambda_{1}\left(M\right)}R}.$

Moreover, Lemma 2.2 also implies that

$\displaystyle\left|\int_{M\setminus D}\left\langle\nabla\left|\nabla
u\right|^{\alpha},\nabla\chi^{2}\right\rangle u^{\beta}\psi^{2}\right|$
$\displaystyle\leq$ $\displaystyle
C\int_{L\left(\frac{1}{2}t,t\right)}\left|\nabla\left|\nabla
u\right|\right|\left|\nabla u\right|^{\alpha}u^{\beta-1}$ $\displaystyle\leq$
$\displaystyle
Ct^{\beta-\frac{1}{n-1}}\int_{L\left(\frac{1}{2}t,t\right)}\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-1}$
$\displaystyle+\frac{C}{t^{\beta-\frac{1}{n-1}}}\int_{L\left(\frac{1}{2}t,t\right)}\left|\nabla
u\right|^{1+2\alpha}u^{2\beta-2}$ $\displaystyle\leq$ $\displaystyle
Ct^{\beta-\frac{1}{n-1}}+\frac{C}{t^{\beta-\frac{1}{n-1}}}\int_{L\left(\frac{1}{2}t,t\right)}\left|\nabla
u\right|^{2}u^{2\left(\alpha+\beta\right)-3},$

where in the last line we have applied (2.12). On the other hand, by (2.22),
(2.23), and (2.11) we get

$\displaystyle\int_{L\left(\frac{1}{2}t,t\right)}\left|\nabla
u\right|^{2}u^{2\left(\alpha+\beta\right)-3}$ $\displaystyle=$
$\displaystyle\zeta\int_{\frac{1}{2}t}^{t}r^{2\beta-\frac{n+1}{n-1}}dr$
$\displaystyle\leq$
$\displaystyle\frac{1}{2\left(\beta-\frac{1}{n-1}\right)}t^{2\left(\beta-\frac{1}{n-1}\right)}\zeta.$

In conclusion,

$\left|\int_{M\setminus D}\left\langle\nabla\left|\nabla
u\right|^{\alpha},\nabla\chi^{2}\right\rangle u^{\beta}\psi^{2}\right|\leq
Ct^{\beta-\frac{1}{n-1}}.$

Hence, this proves that

$\displaystyle-\int_{M\setminus D}\left(\Delta\left|\nabla
u\right|^{\alpha}\right)u^{\beta}\chi^{2}\psi^{2}$ $\displaystyle\geq$
$\displaystyle\int_{M\setminus D}\left\langle\nabla\left|\nabla
u\right|^{\alpha},\nabla u^{\beta}\right\rangle\chi^{2}\psi^{2}$
$\displaystyle-\int_{\ell\left(1\right)}\left(\left|\nabla
u\right|^{\alpha}\right)_{\nu}$
$\displaystyle-\frac{C}{t}e^{-\sqrt{\lambda_{1}\left(M\right)}\,R}-Ct^{\beta-\frac{1}{n-1}}.$

We now estimate the first term on the right hand side. Integration by parts
implies that

$\displaystyle\int_{M\setminus D}\left\langle\nabla\left|\nabla
u\right|^{\alpha},\nabla u^{\beta}\right\rangle\chi^{2}\psi^{2}$
$\displaystyle=$ $\displaystyle-\int_{M\setminus D}\left|\nabla
u\right|^{\alpha}\left(\Delta u^{\beta}\right)\chi^{2}\psi^{2}$
$\displaystyle-\int_{M\setminus D}\left|\nabla
u\right|^{\alpha}\left\langle\nabla
u^{\beta},\nabla\left(\chi^{2}\psi^{2}\right)\right\rangle$
$\displaystyle+\int_{\ell\left(1\right)}\left(u^{\beta}\right)_{\nu}\left|\nabla
u\right|^{\alpha}.$

By (2.7) and (2.12) we have that

$\displaystyle\int_{M\setminus D}\left|\nabla
u\right|^{\alpha}\left|\left\langle\nabla
u^{\beta},\nabla\psi^{2}\right\rangle\chi^{2}\right|$ $\displaystyle\leq$
$\displaystyle C\int_{L\left(\frac{1}{2}t,1\right)\setminus
B_{p}\left(R\right)}u^{\alpha+\beta}\chi^{2}$ $\displaystyle\leq$
$\displaystyle\frac{C}{t^{2-\left(\alpha+\beta\right)}}\int_{L\left(\frac{1}{2}t,1\right)\setminus
B_{p}\left(R\right)}u^{2}$ $\displaystyle\leq$
$\displaystyle\frac{C}{t}e^{-2\sqrt{\lambda_{1}\left(M\right)}R}.$

Moreover, (2.11) and (2.12) imply that

$\displaystyle\int_{M\setminus D}\left|\nabla
u\right|^{\alpha}\left\langle\nabla
u^{\beta},\nabla\chi^{2}\right\rangle\psi^{2}$ $\displaystyle\leq$
$\displaystyle C\int_{L\left(\frac{1}{2}t,t\right)}\left|\nabla
u\right|^{\alpha+2}u^{\beta-2}$ $\displaystyle\leq$ $\displaystyle
C\int_{L\left(\frac{1}{2}t,t\right)}\left|\nabla
u\right|^{2}u^{\beta-\frac{n}{n-1}}$ $\displaystyle\leq$
$\displaystyle\frac{C}{\beta-\frac{1}{n-1}}t^{\left(\beta-\frac{1}{n-1}\right)}\zeta.$

Plugging these estimates in (2) yields that

$\displaystyle-\int_{M\setminus D}\left(\Delta\left|\nabla
u\right|^{\alpha}\right)u^{\beta}\chi^{2}\psi^{2}$ $\displaystyle\geq$
$\displaystyle\beta\left(1-\beta\right)\int_{M\setminus D}\left|\nabla
u\right|^{\alpha+2}u^{\beta-2}\chi^{2}\psi^{2}$
$\displaystyle-\int_{\ell\left(1\right)}\left(\left|\nabla
u\right|^{\alpha}\right)_{\nu}+\int_{\ell\left(1\right)}\left(u^{\beta}\right)_{\nu}\left|\nabla
u\right|^{\alpha}$
$\displaystyle-\frac{C}{t}e^{-\sqrt{\lambda_{1}\left(M\right)}R}-Ct^{\beta-\frac{1}{n-1}}.$

Therefore (2.24) becomes that

$\displaystyle\left(n-2\right)\int_{M\setminus D}\left|\nabla
u\right|^{\alpha}u^{\beta}\chi^{2}\psi^{2}$ $\displaystyle\geq$
$\displaystyle\beta\left(1-\beta\right)\int_{M\setminus D}\left|\nabla
u\right|^{\alpha+2}u^{\beta-2}\chi^{2}\psi^{2}$
$\displaystyle-\int_{\ell\left(1\right)}\left(\left|\nabla
u\right|^{\alpha}\right)_{\nu}+\int_{\ell\left(1\right)}\left(u^{\beta}\right)_{\nu}\left|\nabla
u\right|^{\alpha}$
$\displaystyle-\frac{C}{t}e^{-\sqrt{\lambda_{1}\left(M\right)}R}-Ct^{\beta-\frac{1}{n-1}}.$

We now estimate the left hand side. By Young’s inequality we have that

$\left|\nabla u\right|^{\alpha}\leq\frac{\alpha}{\alpha+2}A^{-2}\left|\nabla
u\right|^{\alpha+2}u^{-2}+\frac{2}{\alpha+2}A^{\alpha}u^{\alpha},$

where $A>0$ is a constant to be specified later. Hence, we obtain

$\displaystyle\int_{M\setminus D}\left|\nabla
u\right|^{\alpha}u^{\beta}\chi^{2}\psi^{2}$ $\displaystyle\leq$
$\displaystyle\frac{\alpha}{\alpha+2}A^{-2}\int_{M\setminus D}\left|\nabla
u\right|^{\alpha+2}u^{\beta-2}\chi^{2}\psi^{2}$
$\displaystyle+\frac{2}{\alpha+2}A^{\alpha}\int_{M\setminus
D}u^{\alpha+\beta}\chi^{2}\psi^{2}.$

Plugging this into (2) yields

$\displaystyle\Lambda_{1}\int_{M\setminus D}\left|\nabla
u\right|^{\alpha+2}u^{\beta-2}\chi^{2}\psi^{2}$ $\displaystyle\leq$
$\displaystyle\Lambda_{2}\int_{M\setminus D}u^{\alpha+\beta}\chi^{2}\psi^{2}$
$\displaystyle+\int_{\ell\left(1\right)}\left(\left|\nabla
u\right|^{\alpha}\right)_{\nu}-\int_{\ell\left(1\right)}\left(u^{\beta}\right)_{\nu}\left|\nabla
u\right|^{\alpha}$
$\displaystyle+\frac{C}{t}e^{-\sqrt{\lambda_{1}\left(M\right)}R}+Ct^{\beta-\frac{1}{n-1}},$

where

$\displaystyle\Lambda_{1}$ $\displaystyle=$
$\displaystyle\beta\left(1-\beta\right)-\frac{\alpha\left(n-2\right)}{\alpha+2}A^{-2}$
$\displaystyle\Lambda_{2}$ $\displaystyle=$
$\displaystyle\frac{2\left(n-2\right)}{\alpha+2}A^{\alpha}.$

We apply Proposition 2.1 for $K=M\setminus D$ and
$\phi=u^{\frac{\alpha+\beta}{2}}\chi\psi$. As $\partial K=\ell(1)$, and
$\nu=\frac{\nabla u}{|\nabla u|}$ is the outward unit normal to $K$, we obtain
that

$\lambda_{1}\left(M\right)\int_{M\setminus
D}u^{\alpha+\beta}\chi^{2}\psi^{2}\leq\int_{M\setminus
D}\left|\nabla\left(u^{\frac{\alpha+\beta}{2}}\chi\psi\right)\right|^{2}-\int_{\ell\left(1\right)}h_{\nu}.$

Note that

$\displaystyle\int_{M\setminus
D}\left|\nabla\left(u^{\frac{\alpha+\beta}{2}}\chi\psi\right)\right|^{2}$
$\displaystyle=$
$\displaystyle\frac{\left(\alpha+\beta\right)^{2}}{4}\int_{M\setminus
D}\left|\nabla u\right|^{2}u^{\alpha+\beta-2}\chi^{2}\psi^{2}$
$\displaystyle+\int_{M\setminus
D}u^{\alpha+\beta}\left|\nabla\left(\chi\psi\right)\right|^{2}$
$\displaystyle+\frac{1}{2}\int_{M\setminus D}\left\langle\nabla
u^{\alpha+\beta},\nabla\left(\chi\psi\right)^{2}\right\rangle.$

By (2.7), (2.22), and (2.23) we conclude that

$\displaystyle\int_{M\setminus
D}u^{\alpha+\beta}\left|\nabla\psi\right|^{2}\chi^{2}$ $\displaystyle\leq$
$\displaystyle\int_{L\left(\frac{1}{2}t,1\right)\setminus
B_{p}\left(R\right)}u^{\alpha+\beta}\chi^{2}$ $\displaystyle\leq$
$\displaystyle\frac{2}{t}\int_{L\left(\frac{1}{2}t,1\right)\setminus
B_{p}\left(R\right)}u^{2}\chi^{2}$ $\displaystyle\leq$
$\displaystyle\frac{C}{t}e^{-2\sqrt{\lambda_{1}\left(M\right)}R}.$

Similarly, by additionally using (2.12) we get that

$\frac{1}{2}\int_{M\setminus D}\left\langle\nabla
u^{\alpha+\beta},\nabla\psi^{2}\right\rangle\chi^{2}\leq\frac{C}{t}e^{-2\sqrt{\lambda_{1}\left(M\right)}R}.$

By (2.11), (2.22), and (2.23) we have

$\displaystyle\int_{M}u^{\alpha+\beta}\left|\nabla\chi\right|^{2}\psi^{2}$
$\displaystyle\leq$ $\displaystyle
C\int_{L\left(\frac{1}{2}t,t\right)}\left|\nabla
u\right|^{2}u^{\beta-\frac{n}{n-1}}$ $\displaystyle\leq$
$\displaystyle\frac{C}{\beta-\frac{1}{n-1}}t^{\beta-\frac{1}{n-1}}.$

Similarly,

$\int_{M}\left\langle\nabla
u^{\alpha+\beta},\nabla\chi^{2}\right\rangle\psi^{2}\leq\frac{C}{\beta-\frac{1}{n-1}}t^{\beta-\frac{1}{n-1}}.$

Combining all these estimates we arrive at

$\displaystyle\lambda_{1}\left(M\right)\int_{M\setminus
D}u^{\alpha+\beta}\chi^{2}\psi^{2}$ $\displaystyle\leq$
$\displaystyle\frac{\left(\alpha+\beta\right)^{2}}{4}\int_{M\setminus
D}\left|\nabla u\right|^{2}u^{\alpha+\beta-2}\chi^{2}\psi^{2}$
$\displaystyle-\int_{\ell\left(1\right)}h_{\nu}+\frac{C}{t}e^{-2\sqrt{\lambda_{1}\left(M\right)}R}+Ct^{\beta-\frac{1}{n-1}}.$

By Young’s inequality,

$\left|\nabla u\right|^{2}\leq\frac{2}{\alpha+2}B^{-\alpha}\left|\nabla
u\right|^{\alpha+2}u^{-\alpha}+\frac{\alpha}{\alpha+2}B^{2}u^{2}$

for a constant $B>0$ to be specified later. This yields

$\displaystyle\int_{M\setminus D}\left|\nabla
u\right|^{2}u^{\alpha+\beta-2}\chi^{2}\psi^{2}$ $\displaystyle\leq$
$\displaystyle\frac{2}{\alpha+2}B^{-\alpha}\int_{M\setminus D}\left|\nabla
u\right|^{\alpha+2}u^{\beta-2}\chi^{2}\psi^{2}$
$\displaystyle+\frac{\alpha}{\alpha+2}B^{2}\int_{M\setminus
D}u^{\alpha+\beta}\chi^{2}\psi^{2}.$

Plugging this into (2), one concludes that

$\displaystyle\left(\lambda_{1}\left(M\right)-\frac{\alpha}{\alpha+2}\frac{\left(\alpha+\beta\right)^{2}}{4}B^{2}\right)\int_{M\setminus
D}u^{\alpha+\beta}\chi^{2}\psi^{2}$ $\displaystyle\leq$
$\displaystyle\frac{\left(\alpha+\beta\right)^{2}}{2\left(\alpha+2\right)}B^{-\alpha}\int_{M\setminus
D}\left|\nabla u\right|^{\alpha+2}u^{\beta-2}\chi^{2}\psi^{2}$
$\displaystyle-\int_{\ell\left(1\right)}h_{\nu}+\frac{C}{t}e^{-2\sqrt{\lambda_{1}\left(M\right)}R}+Ct^{\beta-\frac{1}{n-1}}.$

By the assumption,

$\lambda_{1}\left(M\right)\geq\frac{\left(n-1\right)^{2}\delta^{2}}{4}$

with

(2.30) $\delta=\frac{2\left(n-2\right)}{\left(n-1\right)^{2}}\sqrt{2n-3}.$

Therefore,

$\displaystyle\left(\frac{\left(n-1\right)^{2}\delta^{2}}{4}-\frac{\alpha}{\alpha+2}\frac{\left(\alpha+\beta\right)^{2}}{4}B^{2}\right)\int_{M\setminus
D}u^{\alpha+\beta}\chi^{2}\psi^{2}$ $\displaystyle\leq$
$\displaystyle\frac{\left(\alpha+\beta\right)^{2}}{2\left(\alpha+2\right)}B^{-\alpha}\int_{M\setminus
D}\left|\nabla u\right|^{\alpha+2}u^{\beta-2}\chi^{2}\psi^{2}$
$\displaystyle-\int_{\ell\left(1\right)}h_{\nu}+\frac{C}{t}e^{-2\sqrt{\lambda_{1}\left(M\right)}R}+Ct^{\beta-\frac{1}{n-1}}.$

We optimize this inequality by choosing

$B=\frac{\left(n-1\right)\delta}{\alpha+\beta}$

and obtain that

$\displaystyle\int_{M\setminus D}u^{\alpha+\beta}\chi^{2}\psi^{2}$
$\displaystyle\leq$
$\displaystyle\left(\frac{\alpha+\beta}{\left(n-1\right)\delta}\right)^{\alpha+2}\int_{M\setminus
D}\left|\nabla u\right|^{\alpha+2}u^{\beta-2}\chi^{2}\psi^{2}$
$\displaystyle-\frac{2\left(\alpha+2\right)}{\left(n-1\right)^{2}\delta^{2}}\int_{\ell\left(1\right)}h_{\nu}$
$\displaystyle+\frac{C}{t}e^{-2\sqrt{\lambda_{1}\left(M\right)}R}+Ct^{\beta-\frac{1}{n-1}}.$

Plugging this into (2), we conclude that

$\displaystyle\Lambda\int_{M\setminus D}\left|\nabla
u\right|^{\alpha+2}u^{\beta-2}\chi^{2}\psi^{2}$ $\displaystyle\leq$
$\displaystyle-\frac{2\left(\alpha+2\right)}{\left(n-1\right)^{2}\delta^{2}}\Lambda_{2}\int_{l\left(1\right)}h_{\nu}$
$\displaystyle+\int_{\ell\left(1\right)}\left(\left|\nabla
u\right|^{\alpha}\right)_{\nu}-\int_{\ell\left(1\right)}\left(u^{\beta}\right)_{\nu}\left|\nabla
u\right|^{\alpha}$
$\displaystyle+\frac{C}{t}e^{-\sqrt{\lambda_{1}\left(M\right)}R}+Ct^{\beta-\frac{1}{n-1}},$

where

$\Lambda_{2}=\frac{2\left(n-2\right)}{\alpha+2}A^{\alpha}$

and

$\displaystyle\Lambda$ $\displaystyle=$
$\displaystyle\Lambda_{1}-\left(\frac{\alpha+\beta}{\left(n-1\right)\delta}\right)^{\alpha+2}\Lambda_{2}$
$\displaystyle=$
$\displaystyle\beta\left(1-\beta\right)-\frac{\alpha\left(n-2\right)}{\alpha+2}A^{-2}$
$\displaystyle-\left(\frac{\alpha+\beta}{\left(n-1\right)\delta}\right)^{\alpha+2}\frac{2\left(n-2\right)}{\alpha+2}A^{\alpha}.$

We optimize $\Lambda$ by choosing

$A=\frac{\left(n-1\right)\delta}{\alpha+\beta}$

and obtain that

(2.32)
$\Lambda=\beta\left(1-\beta\right)-\frac{\left(n-2\right)\left(\alpha+\beta\right)^{2}}{\left(n-1\right)^{2}\delta^{2}}.$

Hence, (2) becomes

$\displaystyle\Lambda\int_{M\setminus D}\left|\nabla
u\right|^{\alpha+2}u^{\beta-2}\chi^{2}\psi^{2}$ $\displaystyle\leq$
$\displaystyle-\frac{4\left(n-2\right)}{\left(\alpha+\beta\right)^{\alpha}\left(\left(n-1\right)\delta\right)^{2-\alpha}}\int_{\ell\left(1\right)}h_{\nu}$
$\displaystyle+\int_{\ell\left(1\right)}\left(\left|\nabla
u\right|^{\alpha}\right)_{\nu}-\int_{\ell\left(1\right)}\left(u^{\beta}\right)_{\nu}\left|\nabla
u\right|^{\alpha}$
$\displaystyle+\frac{C}{t}e^{-\sqrt{\lambda_{1}\left(M\right)}R}+Ct^{\beta-\frac{1}{n-1}}.$

Recall that

$\displaystyle\alpha$ $\displaystyle=$ $\displaystyle\frac{n-2}{n-1}$
$\displaystyle\delta^{2}$ $\displaystyle=$
$\displaystyle\frac{4\left(n-2\right)^{2}}{\left(n-1\right)^{4}}\left(2n-3\right),$

as specified in (2.22) and (2.30). We let

(2.34) $\beta=\frac{n-2}{3n-5}.$

Note that for any $n\geq 5$ we have $\beta>\frac{1}{n-1}$, as required by
(2.23).

Furthermore, it follows that $\Lambda=0$ by direct calculation. Consequently,
(2) reduces to

$\displaystyle 0$ $\displaystyle\leq$
$\displaystyle-\frac{4\left(n-2\right)}{\left(\alpha+\beta\right)^{\alpha}\left(\left(n-1\right)\delta\right)^{2-\alpha}}\int_{\ell\left(1\right)}h_{\nu}$
$\displaystyle+\int_{\ell\left(1\right)}\left(\left|\nabla
u\right|^{\alpha}\right)_{\nu}-\int_{\ell\left(1\right)}\left(u^{\beta}\right)_{\nu}\left|\nabla
u\right|^{\alpha}$
$\displaystyle+\frac{C}{t}e^{-\sqrt{\lambda_{1}\left(M\right)}R}+Ct^{\beta-\frac{1}{n-1}}.$

However, by (2.3),

$\displaystyle\lambda_{1}\left(M\right)\mathrm{Vol}\left(D\cap
B_{p}\left(R\right)\right)$ $\displaystyle\leq$
$\displaystyle\lambda_{1}\left(M\right)\int_{D}\psi^{2}$ $\displaystyle=$
$\displaystyle-\int_{D}\left(\Delta h+\left|\nabla
h\right|^{2}\right)\psi^{2}$ $\displaystyle=$
$\displaystyle\int_{\ell\left(1\right)}h_{\nu}+\int_{D}\left\langle\nabla
h,\nabla\psi^{2}\right\rangle$ $\displaystyle-\int_{D}\left|\nabla
h\right|^{2}\psi^{2}$ $\displaystyle\leq$
$\displaystyle\int_{\ell\left(1\right)}h_{\nu}+\int_{D}\left|\nabla\psi\right|^{2},$

where $\nu=\frac{\nabla u}{\left|\nabla u\right|}$ is the inward unit normal
to $\partial D=\ell\left(1\right).$ According to (2.8) it follows that

$\lambda_{1}\left(M\right)\mathrm{Vol}\left(D\cap
B_{p}\left(R\right)\right)\leq\int_{\ell\left(1\right)}h_{\nu}+Ce^{-2\sqrt{\lambda_{1}\left(M\right)}R}.$

Making $R\rightarrow\infty$ yields

$\lambda_{1}\left(M\right)\mathrm{Vol}\left(D\right)\leq\int_{\ell\left(1\right)}h_{\nu}.$

As $\Omega\subset D,$ we conclude from (2) that

$\displaystyle\frac{4\left(n-2\right)\lambda_{1}\left(M\right)}{\left(\alpha+\beta\right)^{\alpha}\left(\left(n-1\right)\delta\right)^{2-\alpha}}\mathrm{Vol}\left(\Omega\right)$
$\displaystyle\leq$ $\displaystyle\int_{\ell\left(1\right)}\left(\left|\nabla
u\right|^{\alpha}\right)_{\nu}-\int_{\ell\left(1\right)}\left(u^{\beta}\right)_{\nu}\left|\nabla
u\right|^{\alpha}$
$\displaystyle+\frac{C}{t}e^{-\sqrt{\lambda_{1}\left(M\right)}R}+Ct^{\beta-\frac{1}{n-1}}.$

Letting $R\rightarrow\infty$ first and then $t\rightarrow 0$ in (2) we arrive
at

$\frac{4\left(n-2\right)\lambda_{1}\left(M\right)}{\left(\alpha+\beta\right)^{\alpha}\left(\left(n-1\right)\delta\right)^{2-\alpha}}\mathrm{Vol}\left(\Omega\right)\leq\int_{\ell\left(1\right)}\left(\left|\nabla
u\right|^{\alpha}\right)_{\nu}-\int_{\ell\left(1\right)}\left(u^{\beta}\right)_{\nu}\left|\nabla
u\right|^{\alpha}.$

Note that the mean curvature of

$\Sigma=\ell\left(1\right)=\partial\left(M\setminus D\right)$

satisfies

$H_{\Sigma}=-\frac{\left\langle\nabla\left|\nabla u\right|,\nabla
u\right\rangle}{\left|\nabla u\right|^{2}}.$

Hence,

$\left(\left|\nabla u\right|^{\alpha}\right)_{\nu}=-\alpha
H_{\Sigma}\left|\nabla u\right|^{\alpha}.$

This proves that

$\frac{4\left(n-2\right)\lambda_{1}\left(M\right)}{\left(\alpha+\beta\right)^{\alpha}\left(\left(n-1\right)\delta\right)^{2-\alpha}}\mathrm{Vol}\left(\Omega\right)\leq-\alpha\int_{\Sigma}H_{\Sigma}\left|\nabla
u\right|^{\alpha}-\beta\int_{\Sigma}\left|\nabla u\right|^{\alpha+1}.$

From Young’s inequality that

$\alpha\left|H_{\Sigma}\right|\left|\nabla
u\right|^{\alpha}\leq\beta\left|\nabla
u\right|^{\alpha+1}+\frac{\alpha}{\alpha+1}\left(\frac{\alpha^{2}}{\left(\alpha+1\right)\beta}\right)^{\alpha}\left|H_{\Sigma}\right|^{\alpha+1},$

we conclude

(2.37)
$\Gamma\,\mathrm{Vol}\left(\Omega\right)\leq\int_{\Sigma}\left|H_{\Sigma}\right|^{\alpha+1}\leq\int_{\partial\Omega}\left|H\right|^{\alpha+1},$

where

$\Gamma=\frac{4\left(n-2\right)}{\left(\alpha+\beta\right)^{\alpha}\left(\left(n-1\right)\delta\right)^{2-\alpha}}\frac{\alpha+1}{\alpha}\left(\frac{\left(\alpha+1\right)\beta}{\alpha^{2}}\right)^{\alpha}\lambda_{1}\left(M\right)$

and

$\displaystyle\alpha$ $\displaystyle=$ $\displaystyle\frac{n-2}{n-1},\ \ \ \
\beta=\frac{n-2}{3n-5},$ $\displaystyle\delta$ $\displaystyle=$
$\displaystyle\frac{2\left(n-2\right)}{\left(n-1\right)^{2}}\sqrt{2n-3}.$

For $n\geq 5$ we have that

$\displaystyle\delta$ $\displaystyle<$ $\displaystyle 1,$ $\displaystyle\ \
\alpha+\beta$ $\displaystyle<$ $\displaystyle\frac{4}{3},$
$\displaystyle\frac{\left(\alpha+1\right)\beta}{\alpha^{2}}$ $\displaystyle>$
$\displaystyle\frac{1}{2},$ $\displaystyle\ \ \left(n-1\right)^{2-\alpha}$
$\displaystyle<$ $\displaystyle\sqrt{2}\left(n-1\right).$

Therefore,

$\Gamma>\frac{2}{3}\sqrt{n}\ \lambda_{1}\left(M\right).$

In conclusion, by (2.37) we have

$\frac{2}{3}\sqrt{n}\
\lambda_{1}\left(M\right)\mathrm{Vol}\left(\Omega\right)\leq\int_{\partial\Omega}\left|H\right|^{\frac{2n-3}{n-1}}.$

∎

##  3\. Splitting of Ricci solitons

In this section, we address the issue of nonexistence of compact minimal
hypersurfaces in Ricci solitons. We begin with the case of steady solitons.
Let $\left(M,g,f\right)$ be a gradient steady Ricci soliton. Then the
potential $f$ satisfies the soliton equation

$\mathrm{Ric}+\mathrm{Hess}\left(f\right)=0.$

It is known [20] that $f$ may be normalized so that

$S+\left|\nabla f\right|^{2}=1,$

where $S$ is the scalar curvature. It is also known [8] that $S>0$ unless
$\left(M,g\right)$ is Ricci flat.

######  Theorem 3.1.

Let $\left(M^{n},g,f\right)$ be a steady Ricci soliton. Assume that there
exists a smooth compact embedded minimal hypersurface $\Sigma$ in $M.$ Then
$\left(M,g\right)$ splits isometrically as a direct product
$\mathbb{R}\times\Sigma.$

###### Proof.

By the splitting theorem in [29] we may assume that $M$ and its double covers
all have one end. Hence, according to Proposition 5.2 in [6], the integral
homology

$H_{n-1}\left(M,\mathbb{Z}\right)=\left\\{0\right\\}.$

In particular, $\Sigma$ bounds a compact domain $D$ in $M.$

In [29] it was proved that $\Delta_{f}$ has positive spectrum. Consequently,
$M$ is $f$-nonparabolic. This implies that there exists $w>0$ on $M\setminus
D$ such that

$\displaystyle\Delta_{f}w$ $\displaystyle=$ $\displaystyle 0\text{ \ on
}M\setminus D$ $\displaystyle w$ $\displaystyle=$ $\displaystyle 1\text{ on
}\Sigma$ $\displaystyle\inf_{M\setminus D}w$ $\displaystyle=$ $\displaystyle
0.$

Moreover,

(3.1) $\int_{M\setminus D}\left|\nabla w\right|^{2}e^{-f}<\infty.$

The Bochner formula implies that

(3.2) $\frac{1}{2}\Delta_{f}\left|\nabla
w\right|^{2}\geq\left|\nabla\left|\nabla w\right|\right|^{2}.$

We now prove, similar to Proposition 2.1, that

(3.3) $0\leq\int_{M\setminus
D}\left|\nabla\phi\right|^{2}e^{-f}-\int_{\Sigma}f_{\nu}\phi^{2}e^{-f}$

for any smooth function $\phi$ on $M\setminus D$ that vanishes at infinity,
where $\nu=\frac{\nabla w}{\left|\nabla w\right|}$ is the outward unit normal
to $\partial\left(M\setminus D\right)=\Sigma=\left\\{w=1\right\\}.$

Note that

$\displaystyle\Delta_{f}\left(f\right)$ $\displaystyle=$ $\displaystyle\Delta
f-\left|\nabla f\right|^{2}$ $\displaystyle=$ $\displaystyle-S-\left|\nabla
f\right|^{2}$ $\displaystyle=$ $\displaystyle-1.$

Therefore,

$\displaystyle\int_{M\setminus D}\phi^{2}e^{-f}$ $\displaystyle=$
$\displaystyle-\int_{M\setminus
D}\left(\Delta_{f}\left(f\right)\right)\phi^{2}e^{-f}$ $\displaystyle=$
$\displaystyle\int_{M\setminus D}\left\langle\nabla\phi^{2},\nabla
f\right\rangle e^{-f}-\int_{\Sigma}f_{\nu}\phi^{2}e^{-f}$ $\displaystyle\leq$
$\displaystyle\int_{M\setminus D}\phi^{2}e^{-f}+\int_{M\setminus
D}\left|\nabla\phi\right|^{2}e^{-f}$
$\displaystyle-\int_{\Sigma}f_{\nu}\phi^{2}e^{-f}.$

This proves (3.3).

Let $\psi$ be a smooth function on $M\setminus D$ with $\psi=1$ on $\Sigma$
and $\psi=0$ outside a sufficiently large ball $B_{p}(R).$ Setting
$\phi=\left|\nabla w\right|\psi$ in (3.3) we get that

$\displaystyle 0$ $\displaystyle\leq$ $\displaystyle\int_{M\setminus
D}\left|\nabla\left(\left|\nabla
w\right|\psi\right)\right|^{2}e^{-f}-\int_{\Sigma}f_{\nu}\left|\nabla
w\right|^{2}e^{-f}$ $\displaystyle=$ $\displaystyle\int_{M\setminus
D}\left|\nabla\left|\nabla
w\right|\right|^{2}\psi^{2}e^{-f}+\frac{1}{2}\int_{M\setminus
D}\left\langle\nabla\left|\nabla w\right|^{2},\nabla\psi^{2}\right\rangle
e^{-f}$ $\displaystyle+\int_{M\setminus
D}\left|\nabla\psi\right|^{2}\left|\nabla
w\right|^{2}e^{-f}-\int_{\Sigma}f_{\nu}\left|\nabla w\right|^{2}e^{-f}.$

By (3.2), this yields

$\displaystyle 0$ $\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{M\setminus D}\left(\Delta_{f}\left|\nabla
w\right|^{2}\right)\psi^{2}+\frac{1}{2}\int_{M\setminus
D}\left\langle\nabla\left|\nabla w\right|^{2},\nabla\psi^{2}\right\rangle
e^{-f}$ $\displaystyle+\int_{M\setminus
D}\left|\nabla\psi\right|^{2}\left|\nabla
w\right|^{2}e^{-f}-\int_{\Sigma}f_{\nu}\left|\nabla w\right|^{2}e^{-f}$
$\displaystyle=$ $\displaystyle\frac{1}{2}\int_{\Sigma}\left(\left|\nabla
w\right|^{2}\right)_{\nu}e^{-f}+\int_{M\setminus
D}\left|\nabla\psi\right|^{2}\left|\nabla w\right|^{2}e^{-f}$
$\displaystyle-\int_{\Sigma}f_{\nu}\left|\nabla w\right|^{2}e^{-f}.$

However, as $\nu=\frac{\nabla w}{\left|\nabla w\right|}$ and $H_{\Sigma}=0,$
we see that

$\displaystyle\frac{1}{2}\left(\left|\nabla w\right|^{2}\right)_{\nu}$
$\displaystyle=$ $\displaystyle\left\langle\nabla\left|\nabla w\right|,\nabla
w\right\rangle$ $\displaystyle=$ $\displaystyle\left(\Delta
w\right)\left|\nabla w\right|$ $\displaystyle=$
$\displaystyle\left\langle\nabla f,\nabla w\right\rangle\left|\nabla w\right|$
$\displaystyle=$ $\displaystyle f_{\nu}\left|\nabla w\right|^{2}.$

Hence, (3) becomes an equality by letting $R\to\infty,$ which in turn forces
(3.2) to be an equality. This implies the splitting of the manifold as a
direct product $\mathbb{R}\times\Sigma.$ We refer to [29] for the details. ∎

An analogous result for expanding Ricci solitons holds true as well. Recall
that an expanding Ricci soliton $\left(M,g,f\right)$ satisfies the equation

$\mathrm{Ric}+\mathrm{Hess}\left(f\right)=-\frac{1}{2}g.$

We may normalize $f$ (see [20]) such that

$S+\left|\nabla f\right|^{2}=-f.$

Moreover, the scalar curvature $S\geq-\frac{n}{2}$ on $M$ by [32].

######  Theorem 3.2.

Let $\left(M^{n},g,f\right)$ be an expanding gradient Ricci soliton with
$S\geq-\frac{n-1}{2}$ on $M.$ Assume that there exists an embedded compact
minimal hypersurface $\Sigma$ in $M.$ Then $M$ splits isometrically as a
direct product $\mathbb{R}\times\Sigma.$

###### Proof.

Recall that by [30] such an expanding Ricci soliton must have one end or it
splits as a direct product. Hence, by [6] we may assume as before that the
integral homology

$H_{n-1}\left(M,\mathbb{Z}\right)=\left\\{0\right\\}$

and that $\Sigma$ bounds a compact domain $D$ in $M.$

As $\Delta_{f}$ has positive spectrum [30], $M$ is $f$-nonparabolic. In
particular, there exists function $w>0$ on $M\setminus D$ such that

$\displaystyle\Delta_{f}w$ $\displaystyle=$ $\displaystyle 0\text{ \ on
}M\setminus D$ $\displaystyle w$ $\displaystyle=$ $\displaystyle 1\text{ on
}\Sigma$ $\displaystyle\inf_{M\setminus D}w$ $\displaystyle=$ $\displaystyle
0.$

Moreover,

$\int_{M\setminus D}\left|\nabla w\right|^{2}e^{-f}<\infty.$

We now prove, similar to Proposition 2.1, that

(3.5) $\frac{1}{2}\int_{M\setminus D}\phi^{2}e^{-f}\leq\int_{M\setminus
D}\left|\nabla\phi\right|^{2}e^{-f}-\int_{\Sigma}f_{\nu}\phi^{2}e^{-f}$

for any smooth function $\phi$ on $M\setminus D$ that vanishes near infinity,
where $\nu=\frac{\nabla w}{\left|\nabla w\right|}$ is the unit normal to
$\Sigma=\left\\{w=1\right\\}.$ Direct calculation gives

$\displaystyle\Delta_{f}\left(f\right)$ $\displaystyle=$ $\displaystyle\Delta
f-\left|\nabla f\right|^{2}$ $\displaystyle=$
$\displaystyle-\frac{n}{2}-S-\left|\nabla f\right|^{2}$ $\displaystyle\leq$
$\displaystyle-\frac{1}{2}-\left|\nabla f\right|^{2}.$

So we have

$\displaystyle\frac{1}{2}\int_{M\setminus D}\phi^{2}e^{-f}$
$\displaystyle\leq$ $\displaystyle-\int_{M\setminus
D}\left(\Delta_{f}\left(f\right)\right)\phi^{2}e^{-f}-\int_{M\setminus
D}\left|\nabla f\right|^{2}\phi^{2}e^{-f}$ $\displaystyle=$
$\displaystyle\int_{M\setminus D}\left\langle\nabla\phi^{2},\nabla
f\right\rangle e^{-f}-\int_{\Sigma}f_{\nu}\phi^{2}e^{-f}-\int_{M\setminus
D}\left|\nabla f\right|^{2}\phi^{2}e^{-f}$ $\displaystyle\leq$
$\displaystyle\int_{M\setminus
D}\left|\nabla\phi\right|^{2}e^{-f}-\int_{\Sigma}f_{\nu}\phi^{2}e^{-f}.$

This proves (3.5). We apply (3.5) to $\phi=\left|\nabla w\right|\psi,$ where
$\psi$ is a cut-off function as in Theorem 3.1 that $\psi=1$ on $\Sigma$ and
that $\psi=0$ outside the geodesic ball $B_{p}(R)$ when $R$ is large. It
follows that

$\displaystyle\frac{1}{2}\int_{M\setminus D}\left|\nabla
w\right|^{2}\psi^{2}e^{-f}$ $\displaystyle\leq$ $\displaystyle\int_{M\setminus
D}\left|\nabla\left|\nabla w\right|\right|^{2}\psi^{2}e^{-f}$
$\displaystyle+\int_{M\setminus D}\left|\nabla
w\right|^{2}\left|\nabla\psi\right|^{2}e^{-f}$
$\displaystyle+\frac{1}{2}\int_{M\setminus D}\left\langle\nabla\left|\nabla
w\right|^{2},\nabla\psi^{2}\right\rangle e^{-f}$
$\displaystyle-\int_{\Sigma}f_{\nu}|\nabla w|^{2}e^{-f}.$

Recall the Bochner formula

$\frac{1}{2}\Delta_{f}\left|\nabla w\right|^{2}\geq\left|\nabla\left|\nabla
w\right|\right|^{2}-\frac{1}{2}\left|\nabla w\right|^{2}\text{ \ on
}M\setminus D.$

Plugging into (3) yields that

$\displaystyle\frac{1}{2}\int_{M\setminus D}\left|\nabla
w\right|^{2}\psi^{2}e^{-f}$ $\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{M\setminus D}\left(\Delta_{f}\left|\nabla
w\right|^{2}\right)\psi^{2}e^{-f}+\frac{1}{2}\int_{M\setminus D}\left|\nabla
w\right|^{2}\psi^{2}e^{-f}$ $\displaystyle+\int_{M\setminus D}\left|\nabla
w\right|^{2}\left|\nabla\psi\right|^{2}e^{-f}+\frac{1}{2}\int_{M\setminus
D}\left\langle\nabla\left|\nabla w\right|^{2},\nabla\psi^{2}\right\rangle
e^{-f}$ $\displaystyle-\int_{\Sigma}f_{\nu}\left|\nabla w\right|^{2}e^{-f}$
$\displaystyle=$ $\displaystyle\frac{1}{2}\int_{\Sigma}\left(\left|\nabla
w\right|^{2}\right)_{\nu}e^{-f}+\frac{1}{2}\int_{M\setminus D}\left|\nabla
w\right|^{2}\psi^{2}e^{-f}$ $\displaystyle+\int_{M\setminus D}\left|\nabla
w\right|^{2}\left|\nabla\psi\right|^{2}e^{-f}-\int_{\Sigma}f_{\nu}\left|\nabla
w\right|^{2}e^{-f}.$

In conclusion,

$\displaystyle 0$ $\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{\Sigma}\left(\left|\nabla
w\right|^{2}\right)_{\nu}e^{-f}-\int_{\Sigma}f_{\nu}\left|\nabla
w\right|^{2}e^{-f}$ $\displaystyle+\int_{M\setminus D}\left|\nabla
w\right|^{2}\left|\nabla\psi\right|^{2}e^{-f}.$

Making $R\rightarrow\infty,$ we get

$0\leq\frac{1}{2}\int_{\Sigma}\left(\left|\nabla
w\right|^{2}\right)_{\nu}e^{-f}-\int_{\Sigma}f_{\nu}\left|\nabla
w\right|^{2}e^{-f}.$

Since $\nu=\frac{\nabla w}{\left|\nabla w\right|},$ it follows that

$0\leq\int_{\Sigma}\left\langle\nabla\left|\nabla w\right|,\nabla
w\right\rangle e^{-f}-\int_{\Sigma}\left\langle\nabla f,\nabla
w\right\rangle\left|\nabla w\right|e^{-f}.$

However, as $w$ is $f$-harmonic,

$\left\langle\nabla\left|\nabla w\right|,\nabla
w\right\rangle-\left\langle\nabla f,\nabla w\right\rangle\left|\nabla
w\right|=-H_{\Sigma}\left|\nabla w\right|^{2}.$

Since $\Sigma$ is minimal, this again shows the above inequality must be
equality, which in turn forces the Bochner formula itself is also an equality.
This suffices to conclude that $M=\mathbb{R}\times\Sigma.$ One may refer to
[30] for details. ∎

Acknowledgment: We wish to thank Pengfei Guan for his interest and comments.
The first author was partially supported by the NSF grant DMS-1811845 and by a
Simons Foundation grant.

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