# Comparison theorems for three-dimensional manifolds with scalar curvature
bound

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.

Two sharp comparison results are derived for three-dimensional complete
noncompact manifolds with scalar curvature bounded from below. The first one
concerns the Green’s function. When the scalar curvature is nonnegative, it
states that the rate of decay of an energy quantity over the level set is
strictly less than that of the Euclidean space unless the manifold itself is
isometric to the Euclidean space. The result is in turn converted into a sharp
area comparison for the level set of the Green’s function when in addition the
Ricci curvature of the manifold is assumed to be asymptotically nonnegative at
infinity. The second result provides a sharp upper bound of the bottom
spectrum in terms of the scalar curvature lower bound, in contrast to the
classical result of Cheng which involves a Ricci curvature lower bound.

##  1\. Introduction

The classical Laplacian comparison theorem [21, Chapter I] states that the
Laplacian of a geodesic distance function is at most that of the corresponding
space form under a Ricci curvature lower bound. It is obvious that such a
result is no longer true under a scalar curvature lower bound. The purpose of
this note is to search for suitable alternatives for complete three-
dimensional manifolds.

Our first result deals with the case of nonnegative scalar curvature. Recall
that a complete manifold is called nonparabolic if it admits a positive
Green’s function [11, Chapter 20]. It is well-known that in this case the
minimal positive Green’s function $G(x,y)$ may be obtained as the limit of the
Dirichlet Green’s function of a sequence of compact exhaustive domains of the
manifold. Then

$\Delta_{x}G\left(x,y\right)=-\delta\left(x,y\right),$
$G\left(x,y\right)=G\left(y,x\right)>0$

and

$\liminf_{y\to\infty}G(x,y)=0.$

Throughout the paper we fix $p\in M$ and let

$G\left(x\right)=G\left(p,x\right).$

We also use the following notations to denote the level and sublevel sets of
$G(x).$

$\displaystyle L\left(a,b\right)$ $\displaystyle=$ $\displaystyle\left\\{x\in
M:\ a<G\left(x\right)<b\right\\}$ $\displaystyle l\left(t\right)$
$\displaystyle=$ $\displaystyle\left\\{x\in M:\ G\left(x\right)=t\right\\}.$

We have the following sharp comparison theorem concerning the minimal positive
Green’s function.

######  Theorem 1.1.

Let $\left(M,g\right)$ be a complete noncompact three-dimensional manifold
with nonnegative scalar curvature. Assume that $M$ has one end and its first
Betti number $b_{1}\left(M\right)=0.$ If $M$ is nonparabolic and the minimal
positive Green’s function $G\left(x\right)=G\left(p,x\right)$ satisfies
$\lim_{x\rightarrow\infty}G(x)=0,$ then

$\frac{d}{dt}\left(\frac{1}{t}\int_{l\left(t\right)}\left|\nabla
G\right|^{2}-4\pi t\right)\leq 0$

for all $t>0.$ Moreover, equality holds for some $T>0$ if and only if the
super level set $\left\\{x\in M,G(x)>T\right\\}$ is isometric to a ball in the
Euclidean space $\mathbb{R}^{3}.$

Some remarks are in order. First, the conclusion may be restated as

$\frac{d}{dt}\left(\frac{1}{t}\int_{l\left(t\right)}\left|\nabla
G\right|^{2}\right)\leq\frac{d}{dt}\left(\frac{1}{t}\int_{\bar{l}\left(t\right)}\left|\nabla\bar{G}\right|^{2}\right)$

for all $t>0,$ where $\bar{G}(\bar{x})=\frac{1}{4\pi|\bar{x}|}$ is the Green’s
function of $\mathbb{R}^{3}$ and $\bar{l}(t)$ the level set of $\bar{G}.$ As
such, it may be viewed as a comparison of the decay rate concerning the energy
quantity $\frac{1}{t}\int_{l\left(t\right)}\left|\nabla G\right|^{2}$ of $M$
with that of $\mathbb{R}^{3}.$ Second, the fact that $\lim_{x\to\infty}G(x)=0$
together with the topological information of $M$ is to ensure that the level
set $l(t)$ is compact and connected. Without those assumptions, one may work
with the Dirichlet Green’s function of an arbitrary bounded domain instead.
The resulting conclusion now depends on the number of components of $l(t)$ as
well. Third, it is unclear to us if an analogous conclusion holds in the
higher dimension, though our proof is very dimension specific.

Under the additional assumption that the Ricci curvature is nonnegative at
infinity, the preceding result may be converted into an area comparison
theorem for the level sets of the Green’s function.

######  Corollary 1.2.

Let $\left(M,g\right)$ be a complete noncompact three-dimensional manifold
with nonnegative scalar curvature and asymptotically nonnegative Ricci
curvature, that is,

$\liminf_{x\rightarrow\infty}\mathrm{Ric}\left(x\right)\geq 0.$

Assume that $M$ has one end and its first Betti number
$b_{1}\left(M\right)=0.$ If $M$ is nonparabolic and the minimal Green’s
function $G\left(x\right)=G\left(p,x\right)$ satisfies
$\lim_{x\rightarrow\infty}G(x)=0,$ then

$\int_{l\left(t\right)}\left|\nabla G\right|^{2}\leq 4\pi t^{2}$

and

$\mathrm{Area}\left(l\left(t\right)\right)\geq\frac{1}{4\pi t^{2}}$

for all $t>0.$ Moreover, if equality holds for some $T>0,$ then the super
level set $\left\\{G>T\right\\}$ is isometric to a ball in $\mathbb{R}^{3}.$

Note that in the case that $M$ has nonnegative Ricci curvature, its minimal
positive Green’s function $G$ always satisfies $\lim_{x\to\infty}G(x)=0$ by
[16] and the number of ends is necessarily one due to the Cheeger-Gromoll
splitting theorem [3]. Therefore, Theorem 1.1 and Corollary 1.2 are both
applicable to the universal cover of $M.$

Theorem 1.1 is motivated by the work of Colding [8] and Colding-Minicozzi [6],
where monotonicity formulas for functionals of the form

$w_{p}\left(r\right)=\frac{1}{r^{n-1}}\int_{b=r}\left|\nabla b\right|^{p}$

are derived for $n$-dimensional manifolds with nonnegative Ricci curvature,
where the function $b=G^{-\frac{1}{n-2}}.$ So Theorem 1.1 concerns $w_{2}$ for
dimension $n=3,$ while the exponent $p=3$ in [8], and more generally
$p\geq\frac{2n-3}{n-1}$ in [6], for all dimension $n.$ These monotonicity
results have been applied to the study of uniqueness of the tangent cones for
Ricci flat manifolds with Euclidean volume growth [7]. We refer the readers to
[5] for an exposition on monotonicity formulas in geometric analysis, and [1]
for their applications to Willmore type inequalities.

Our second result concerns the bottom spectrum. Recall that the bottom
spectrum $\lambda_{1}\left(M\right)$ is characterized by

$\lambda_{1}\left(M\right)=\inf_{f\in
C_{0}^{\infty}\left(M\right)}\frac{\int_{M}\left|\nabla
f\right|^{2}}{\int_{M}f^{2}}.$

According to Cheng’s theorem [4], for an $n$-dimensional complete manifold
$M,$

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

if the Ricci curvature of $\left(M,g\right)$ satisfies
$\mathrm{Ric}\geq-\left(n-1\right)K$ for some nonnegative constant $K.$ By
considering the example of the form
$M=\mathbb{H}^{2}\times\mathbb{S}^{n-2}(r),$ where $\mathbb{H}^{2}$ is the
standard hyperbolic plane of sectional curvature $-1$ and
$\mathbb{S}^{n-2}(r)$ the sphere of radius $r$ in $\mathbb{R}^{n-1},$ one sees
that a direct extension of Cheng’s result to the scalar curvature lower bound
is not possible for dimension $n\geq 4.$ Indeed, by choosing $r$ accordingly,
the scalar curvature of $M$ can be made as large as one desires, yet
$\lambda_{1}(M)=\frac{1}{4}.$ However, for $n=3,$ one does have the following
theorem.

######  Theorem 1.3.

Let $\left(M,g\right)$ be a three-dimensional complete noncompact Riemannian
manifold with scalar curvature $S\geq-6K$ on $M$ for some nonnegative constant
$K.$ Suppose that $M$ has finitely many ends and its first Betti number
$b_{1}(M)<\infty.$ Moreover, the Ricci curvature of $M$ is bounded from below
and the volume $V_{x}(1)$ of unit ball $B_{x}(1)$ satisfies

$V_{x}(1)\geq C(\epsilon)\,\exp\left(-2\sqrt{K+\epsilon}\,r(x)\right)$

for every $\epsilon>0$ and all $x\in M,$ where $r(x)$ is the geodesic distance
from $x$ to a fixed point $p.$ Then the bottom spectrum of the Laplacian
satisfies

$\lambda_{1}\left(M\right)\leq K.$

Note that in the case that the Ricci curvature of $M$ is bounded by
$\mathrm{Ric}\geq-2K,$ the Gromov-Bishop volume comparison theorem [11,
Chapter 2] readily implies that volume lower bound for $V_{x}(1)$ holds. So,
by considering the universal cover of $M$ if necessary and modulo the
topological assumption of finitely many ends, Theorem 1.3 provides a faithful
generalization of Cheng’s result to three-dimensional manifolds with only
scalar curvature lower bound.

######  Corollary 1.4.

Let $\left(M,g\right)$ be a three-dimensional complete noncompact Riemannian
manifold with nonpositive sectional curvature. Assume that the scalar
curvature $S$ is bounded by

$S\geq-6K\text{ \ on }M$

for some nonnegative constant $K.$ Then

$\lambda_{1}\left(M\right)\leq K.$

This is because Theorem 1.3 is applicable to the universal cover $\tilde{M}$
of $M$ as $\tilde{M}$ is a Cartan-Hadamard manifold with bounded curvature.
Since the bottom spectrum satisfies
$\lambda_{1}(M)\leq\lambda_{1}(\tilde{M}),$ the corollary follows.

Both Theorem 1.1 and Theorem 1.3 are proved by working with the minimal
positive Green’s function $G$ of $M.$ The idea of using Green’s function to
bound the bottom spectrum was introduced by the first author in [17]. Roughly
speaking, one takes a test function $f=\left|\nabla
G\right|^{\frac{1}{2}}\phi$ with a carefully chosen cut-off function $\phi$
which in turn depends on $G.$ The proofs of both theorems hinge on
manipulating the Bochner formula for the Green’s function.

$\Delta\left|\nabla
G\right|=\left(\left|G_{ij}\right|^{2}-\left|\nabla\left|\nabla
G\right|\right|^{2}\right)\left|\nabla G\right|^{-1}+\mathrm{Ric}\left(\nabla
G,\nabla G\right)\left|\nabla G\right|^{-1}.$

A crucial point is to rewrite the Ricci curvature term by mimicking a trick
originated from the work of Schoen and Yau [18, 19, 20] on stable minimal
surfaces in three-dimensional manifolds.

$\mathrm{Ric}\left(\nabla G,\nabla G\right)\left|\nabla
G\right|^{-2}=\frac{1}{2}S-\frac{1}{2}S_{l\left(r\right)}+\frac{1}{\left|\nabla
G\right|^{2}}\left(\left|\nabla\left|\nabla
G\right|\right|^{2}-\frac{1}{2}\left|\nabla^{2}G\right|^{2}\right),$

where $S_{l\left(r\right)}$ denotes the scalar curvature of the level set
$l\left(r\right).$ One can then proceed by integrating the formula over the
level sets and applying the Gauss-Bonnet theorem. We note that this kind of
idea has been exploited recently in [23] and [2] as well.

In order to make the argument work, however, we need to ensure that the level
sets $l(t)$ of $G$ are compact with controlled number of components. This is
where all the extra assumptions are used to show the following.

(1.1) $\displaystyle\lim_{x\rightarrow\infty}G\left(x\right)$ $\displaystyle=$
$\displaystyle 0,$ (1.2)
$\displaystyle\\#\mathrm{Conn}\left(l\left(t\right)\right)$
$\displaystyle\leq$ $\displaystyle A$

for all $t,$ where $A$ is a fixed constant and
$\\#\mathrm{Conn}\left(l(t)\right)$ the number of connected components of the
level set $l(t).$

Ideally, one would like to prove Theorem 1.3 under the sole assumption that
the scalar curvature $S\geq-6K.$ Another natural question is what happens if
$\lambda_{1}(M)$ achieves its maximum value $K$ in Theorem 1.3. In the case of
the aforementioned Cheng’s theorem, there are rigidity results [12, 13].

The study of scalar curvature has a long history with many significant
results. The recent lecture notes [10] and the survey paper [22] are good
sources for the state of the affairs and references.

The structure of the paper is at follows. After collecting some preliminary
results in Section 2, we supply the proofs of Theorem 1.1 and Theorem 1.3 in
Section 3 and Section 4, respectively.

##  2\. Preliminaries

In this section we make some preparations for proving Theorem 1.1 and Theorem
1.3. Let us start with the following result. It relies on an idea from Schoen-
Yau’s work on minimal surfaces [18, 19, 20] and appears as Lemma 4.1 in [2].
We include details here for completeness.

######  Lemma 2.1.

Let $\left(M,g\right)$ be a three-dimensional complete noncompact Riemannian
manifold with scalar curvature $S$ and $u$ a harmonic function on $M.$ Then on
each regular level set $l\left(r\right)$ of $u,$

$\mathrm{Ric}\left(\nabla u,\nabla u\right)\left|\nabla
u\right|^{-2}=\frac{1}{2}S-\frac{1}{2}S_{l\left(r\right)}+\frac{1}{\left|\nabla
u\right|^{2}}\left(\left|\nabla\left|\nabla
u\right|\right|^{2}-\frac{1}{2}\left|\nabla^{2}u\right|^{2}\right),$

where $S_{l\left(r\right)}$ denotes the scalar curvature of $l\left(r\right).$

###### Proof.

On a regular level set $l\left(r\right)$ of $u,$ its unit normal vector is
given

$e_{1}=\frac{\nabla u}{\left|\nabla u\right|}.$

Choose $\left\\{e_{a}\right\\}_{a=2,3},$ unit vectors tangent to
$l\left(r\right),$ such that $\left\\{e_{1},e_{2},e_{3}\right\\}$ forms a
local orthonormal frame on $M.$ Since $u$ is harmonic, the second fundamental
form and the mean curvature of $l\left(r\right)$ are given by

$h_{ab}=\frac{u_{ab}}{\left|\nabla u\right|}\text{ \ and
}H=-\frac{u_{11}}{\left|\nabla u\right|},\text{ \ respectively. }$

By the Gauss curvature equation, we have

$S_{l\left(r\right)}=S-2R_{11}+H^{2}-h^{2}.$

Therefore,

$\displaystyle 2\mathrm{Ric}\left(\nabla u,\nabla u\right)\left|\nabla
u\right|^{-2}$ $\displaystyle=$ $\displaystyle 2R_{11}$ $\displaystyle=$
$\displaystyle S-S_{l\left(r\right)}+\frac{1}{\left|\nabla
u\right|^{2}}\left(\left|u_{11}\right|^{2}-\left|u_{ab}\right|^{2}\right)$
$\displaystyle=$ $\displaystyle S-S_{l\left(r\right)}+\frac{1}{\left|\nabla
u\right|^{2}}\left(2\left|\nabla\left|\nabla
u\right|\right|^{2}-\left|u_{ij}\right|^{2}\right),$

where we have used the fact that

$\left|\nabla\left|\nabla
u\right|\right|^{2}=\left|u_{11}\right|^{2}+\left|u_{1a}\right|^{2}$

and

$\left|u_{ij}\right|^{2}=\left|u_{11}\right|^{2}+2\left|u_{1a}\right|^{2}+\left|u_{ab}\right|^{2}.$

This proves the result. ∎

We will also use the following well known Kato inequality for harmonic
functions.

######  Lemma 2.2.

Let $\left(M,g\right)$ be a three-dimensional complete noncompact Riemannian
manifold and $u$ a harmonic function on $M.$ Then

$\left|\nabla^{2}u\right|^{2}\geq\frac{3}{2}\left|\nabla\left|\nabla
u\right|\right|^{2}\text{ \ on }M.$

###### Proof.

It suffices to prove this at points where $\left|\nabla u\right|\neq 0.$ Let

$e_{1}=\frac{\nabla u}{\left|\nabla u\right|}$

and choose $\left\\{e_{a}\right\\}_{a=2,3}$ such that
$\left\\{e_{1},e_{2},e_{3}\right\\}$ is a local orthonormal frame on $M.$ Then

$\displaystyle\left|u_{ij}\right|^{2}$ $\displaystyle=$
$\displaystyle\left|u_{11}\right|^{2}+2\left|u_{1a}\right|^{2}+\left|u_{ab}\right|^{2}$
$\displaystyle\geq$
$\displaystyle\left|u_{11}\right|^{2}+2\left|u_{1a}\right|^{2}+\frac{1}{2}\left|u_{22}+u_{33}\right|^{2}$
$\displaystyle=$
$\displaystyle\frac{3}{2}\left|u_{11}\right|^{2}+2\left|u_{1a}\right|^{2},$

where in the last line we have used the fact that $u$ is harmonic. Therefore,

$\displaystyle\left|u_{ij}\right|^{2}$ $\displaystyle\geq$
$\displaystyle\frac{3}{2}\left(\left|u_{11}\right|^{2}+\left|u_{1a}\right|^{2}\right)$
$\displaystyle=$ $\displaystyle\frac{3}{2}\left|\nabla\left|\nabla
u\right|\right|^{2}.$

∎

We also need the following topological lemma concerning the number of
components of the level sets of a proper Green’s function. The proof is
inspired by [15].

######  Lemma 2.3.

Let $\left(M,g\right)$ be a complete noncompact Riemannian manifold with $k$
ends and finite first Betti number $b_{1}(M).$ Assume that $\left(M,g\right)$
is nonparabolic and its minimal positive Green’s function $G$ satisfies
$\lim_{x\to\infty}G(x)=0.$ Then there exists $t_{0}>0$ such that the level set
$l(t)$ of $G$ has exactly $k$ components for all $t\leq t_{0}.$ In the case
that $M$ has only one end and the first Betti number $b_{1}\left(M\right)=0,$
the level set $l\left(t\right)$ is connected for all $t>0.$

###### Proof.

We first claim that

(2.1) $L\left(a,\infty\right)\text{ \ is connected}$

for all $a>0.$ Indeed, if this is not true, then there exists a connected
component $L_{0}$ of $L\left(a,\infty\right)$ which does not contain the pole
$p$ of $G.$ Moreover, $L_{0}$ is bounded by the fact that
$\lim_{x\to\infty}G(x)=0.$ Then the harmonic function $G$ on $L_{0}$ must
achieve its maximum in the interior of $L_{0},$ which is a contradiction.

We also note that $L(0,a)$ contains no bounded components for any $a>0.$
Otherwise, on such a bounded component the function $G$ would achieve its
minimum at an interior point.

Since the first Betti number of $M$ is finite, we may choose $t_{0}$
sufficiently small so that all the representatives of $H_{1}\left(M\right)$
lie in $L\left(t_{0},\infty\right).$ By arranging $t_{0}$ to be even smaller
if necessary, we may assume that $L(0,t)$ has exactly $k$ components for
$t\leq t_{0}$ by the fact that it contains no bounded components and that $M$
has $k$ ends.

For $t<t_{0}$ and $\rho>0$ such that $t+\rho<t_{0},$ since
$M=L\left(t,\infty\right)\cup L\left(0,t+\rho\right),$ we have the Mayer-
Vietoris sequence

$\displaystyle H_{1}\left(L\left(t,\infty\right)\right)\oplus
H_{1}\left(L\left(0,t+\rho\right)\right)\rightarrow
H_{1}\left(M\right)\rightarrow H_{0}\left(L\left(t,t+\rho\right)\right)$
$\displaystyle\rightarrow H_{0}\left(L\left(t,\infty\right)\right)\oplus
H_{0}\left(L\left(0,t+\rho\right)\right)\rightarrow H_{0}(M).$

Note that $H_{1}\left(L\left(0,t+\rho\right)\right)$ is trivial because all
the representatives of $H_{1}\left(M\right)$ lie inside
$L\left(t_{0},\infty\right).$ In view of (2.1), we therefore conclude that

$H_{0}\left(L\left(t,t+\rho\right)\right)=H_{0}\left(L\left(0,t+\rho\right)\right)=\mathbb{Z}\oplus..\oplus\mathbb{Z}$

with $k$ summands. In other words, $L\left(t,t+\rho\right)$ has $k$
components. Since $\rho>0$ can be arbitrarily small, this proves that

(2.2) $l\left(t\right)\text{ has at most }k\text{ components}$

for $t\leq t_{0}.$ It is obvious that $l(t)$ can not have fewer than $k$
components for any $t\leq t_{0}$ as $L(0,t_{0})$ has $k$ components.
Therefore, $l(t)$ has exactly $k$ components for $t\leq t_{0}.$

In the special case that $b_{1}\left(M\right)=0,$ the number $t_{0}$ can be
taken arbitrarily large. By (2.2), one concludes that $l\left(t\right)$ is
connected for all $t>0.$ ∎

##  3\. Nonnegative scalar curvature

In this section, we work with three-dimensional complete manifolds with
nonnegative scalar curvature and prove Theorem 1.1 which is restated below.

######  Theorem 3.1.

Let $\left(M,g\right)$ be a complete noncompact three-dimensional manifold
with nonnegative scalar curvature. Assume that $M$ has one end and its first
Betti number $b_{1}\left(M\right)=0.$ If $M$ is nonparabolic and the minimal
positive Green’s function $G\left(x\right)=G\left(p,x\right)$ satisfies
$\lim_{x\rightarrow\infty}G(x)=0,$ then

$\frac{d}{dt}\left(\frac{1}{t}\int_{l\left(t\right)}\left|\nabla
G\right|^{2}-4\pi t\right)\leq 0$

for all $t>0.$ Moreover, equality holds for some $T>0$ if and only if the
super level set $\left\\{x\in M,G(x)>T\right\\}$ is isometric to a ball in the
Euclidean space $\mathbb{R}^{3}.$

###### Proof.

Recall that

$\displaystyle l\left(t\right)$ $\displaystyle=$ $\displaystyle\left\\{x\in
M:G\left(x\right)=t\right\\}$ $\displaystyle L\left(a,b\right)$
$\displaystyle=$ $\displaystyle\left\\{x\in M:a<G\left(x\right)<b\right\\}.$

By the assumption that $\lim_{x\rightarrow\infty}G(x)=0,$ the level set
$l\left(t\right)$ is compact for every $t>0.$ Moreover, since $M$ is assumed
to have one end and $b_{1}\left(M\right)=0,$ Lemma 2.3 implies that
$l\left(t\right)$ is connected for all $t>0.$

Consider the function

$w\left(t\right)=\int_{l\left(t\right)}\left|\nabla G\right|^{2}.$

Whenever $l\left(t\right)$ is regular, its mean curvature $H$ is given by

$H_{l\left(t\right)}=\frac{\sum_{a}G_{aa}}{\left|\nabla
G\right|}=-\frac{G_{11}}{\left|\nabla
G\right|}=-\frac{\left\langle\nabla\left|\nabla G\right|,\nabla
G\right\rangle}{\left|\nabla G\right|^{2}},$

where $e_{1}=\frac{\nabla G}{\left|\nabla G\right|}$ and
$\left\\{e_{a}\right\\}_{a=2,3}$ are unit tangent vectors on $l\left(t\right)$
such that $\left\\{e_{1},e_{2},e_{3}\right\\}$ is a local orthonormal frame on
$M.$ It follows that

$\displaystyle\frac{dw}{dt}\left(t\right)$ $\displaystyle=$
$\displaystyle\int_{l\left(t\right)}\left(\frac{\left\langle\nabla\left|\nabla
G\right|^{2},\nabla G\right\rangle}{\left|\nabla
G\right|^{2}}+\frac{H_{l\left(t\right)}}{\left|\nabla G\right|}\left|\nabla
G\right|^{2}\right)$ $\displaystyle=$
$\displaystyle\int_{l\left(t\right)}\frac{\left\langle\nabla\left|\nabla
G\right|,\nabla G\right\rangle}{\left|\nabla G\right|}.$

Multiplying the equation by $t^{-2}$ we get

(3.1)
$t^{-2}\frac{dw}{dt}\left(t\right)=\int_{l\left(t\right)}\frac{\left\langle\nabla\left|\nabla
G\right|,\nabla G\right\rangle}{\left|\nabla G\right|}G^{-2}.$

On the other hand, by Green’s identity we have

$\displaystyle\int_{L\left(t,T\right)}\left(G^{-2}\Delta\left|\nabla
G\right|-\left|\nabla G\right|\Delta G^{-2}\right)$ $\displaystyle=$
$\displaystyle\int_{l\left(T\right)}\left(G^{-2}\frac{\left\langle\nabla\left|\nabla
G\right|,\nabla G\right\rangle}{\left|\nabla G\right|}-\left|\nabla
G\right|\frac{\left\langle\nabla G^{-2},\nabla G\right\rangle}{\left|\nabla
G\right|}\right)$
$\displaystyle-\int_{l\left(t\right)}\left(G^{-2}\frac{\left\langle\nabla\left|\nabla
G\right|,\nabla G\right\rangle}{\left|\nabla G\right|}-\left|\nabla
G\right|\frac{\left\langle\nabla G^{-2},\nabla G\right\rangle}{\left|\nabla
G\right|}\right).$

Since $G$ is harmonic on $L\left(t,T\right)$,

$\left|\nabla G\right|\Delta G^{-2}=6G^{-4}\left|\nabla G\right|^{3}.$

Note that as $x\rightarrow p,$

(3.2) $G\left(x\right)\simeq\frac{1}{4\pi}\frac{1}{r\left(x\right)}\text{ and
}\left|\nabla
G\right|\left(x\right)\simeq\frac{1}{4\pi}\frac{1}{r^{2}\left(x\right)}.$

Hence,

$\lim_{T\rightarrow\infty}\int_{l\left(T\right)}\left(G^{-2}\frac{\left\langle\nabla\left|\nabla
G\right|,\nabla G\right\rangle}{\left|\nabla G\right|}-\left|\nabla
G\right|\frac{\left\langle\nabla G^{-2},\nabla G\right\rangle}{\left|\nabla
G\right|}\right)=0$

and

$\int_{L\left(t,\infty\right)}\left|G^{-2}\Delta\left|\nabla
G\right|-\left|\nabla G\right|\Delta G^{-2}\right|<\infty.$

In conclusion, we obtain the following identity.

$\displaystyle\int_{l\left(t\right)}\left(G^{-2}\frac{\left\langle\nabla\left|\nabla
G\right|,\nabla G\right\rangle}{\left|\nabla G\right|}-\left|\nabla
G\right|\frac{\left\langle\nabla G^{-2},\nabla G\right\rangle}{\left|\nabla
G\right|}\right)$ $\displaystyle=$
$\displaystyle-\int_{L\left(t,\infty\right)}\left(G^{-2}\Delta\left|\nabla
G\right|-6G^{-4}\left|\nabla G\right|^{3}\right).$

Together with (3.1), we conclude that

$\displaystyle t^{-2}\frac{dw}{dt}\left(t\right)$ $\displaystyle=$
$\displaystyle\int_{l\left(t\right)}G^{-2}\frac{\left\langle\nabla\left|\nabla
G\right|,\nabla G\right\rangle}{\left|\nabla G\right|}$ $\displaystyle=$
$\displaystyle-2\int_{l\left(t\right)}G^{-3}\left|\nabla
G\right|^{2}-\int_{L\left(t,\infty\right)}G^{-2}\Delta\left|\nabla G\right|$
$\displaystyle+6\int_{L\left(t,\infty\right)}G^{-4}\left|\nabla G\right|^{3}.$

Note that by the co-area formula,

$\displaystyle\int_{L\left(t,\infty\right)}G^{-4}\left|\nabla G\right|^{3}$
$\displaystyle=$
$\displaystyle\int_{t}^{\infty}r^{-4}\int_{l\left(r\right)}\left|\nabla
G\right|^{2}$ $\displaystyle=$
$\displaystyle\int_{t}^{\infty}r^{-4}w\left(r\right)dr.$

Hence, (3) can be written as

$\displaystyle t^{-2}\frac{dw}{dt}\left(t\right)$ $\displaystyle=$
$\displaystyle-2t^{-3}w\left(t\right)+6\int_{t}^{\infty}r^{-4}w\left(r\right)dr$
$\displaystyle-\int_{L\left(t,\infty\right)}G^{-2}\Delta\left|\nabla
G\right|.$

We now estimate the last term. Using the Bochner formula

$\Delta\left|\nabla
G\right|=\left(\left|G_{ij}\right|^{2}-\left|\nabla\left|\nabla
G\right|\right|^{2}\right)\left|\nabla G\right|^{-1}+\mathrm{Ric}\left(\nabla
G,\nabla G\right)\left|\nabla G\right|^{-1},$

we have

$\displaystyle\int_{l\left(r\right)}\left|\nabla
G\right|^{-1}\Delta\left|\nabla G\right|$ $\displaystyle=$
$\displaystyle\int_{l\left(r\right)}\left(\left|G_{ij}\right|^{2}-\left|\nabla\left|\nabla
G\right|\right|^{2}\right)\left|\nabla G\right|^{-2}$
$\displaystyle+\int_{l\left(r\right)}\mathrm{Ric}\left(\nabla G,\nabla
G\right)\left|\nabla G\right|^{-2}.$

Applying Lemma 2.1 to $G$ gives

$\mathrm{Ric}\left(\nabla G,\nabla G\right)\left|\nabla
G\right|^{-2}=\frac{1}{2}S-\frac{1}{2}S_{l\left(r\right)}+\left(\left|\nabla\left|\nabla
G\right|\right|^{2}-\frac{1}{2}\left|G_{ij}\right|^{2}\right)\left|\nabla
G\right|^{-2},$

where $S_{l\left(r\right)}$ is the scalar curvature of $l\left(r\right).$ We
therefore conclude that

(3.5) $\int_{l\left(r\right)}\left|\nabla G\right|^{-1}\Delta\left|\nabla
G\right|=\frac{1}{2}\int_{l\left(r\right)}\left(\left|G_{ij}\right|^{2}\left|\nabla
G\right|^{-2}+S-S_{l\left(r\right)}\right).$

Note that by Lemma 2.2,

$\left|G_{ij}\right|^{2}\geq\frac{3}{2}\left|\nabla\left|\nabla
G\right|\right|^{2}.$

Also, since $l\left(r\right)$ is compact and connected for any $r>0,$ the
Gauss-Bonnet theorem implies that

$\int_{l\left(r\right)}S_{l\left(r\right)}=4\pi\chi\left(l\left(r\right)\right)\leq
8\pi$

whenever $r$ is a regular value of $G.$ Therefore, on any regular level set
$l(r),$ one obtains from (3.5) that

(3.6) $\int_{l\left(r\right)}\left|\nabla G\right|^{-1}\Delta\left|\nabla
G\right|\geq\frac{3}{4}\int_{l\left(r\right)}\left|\nabla\left|\nabla
G\right|\right|^{2}\left|\nabla G\right|^{-2}-4\pi.$

Observe from (3.1) that

$\displaystyle\left|w^{\prime}\left(r\right)\right|$ $\displaystyle=$
$\displaystyle\left|\int_{l\left(r\right)}\frac{\left\langle\nabla\left|\nabla
G\right|,\nabla G\right\rangle}{\left|\nabla G\right|}\right|$
$\displaystyle\leq$
$\displaystyle\int_{l\left(r\right)}\left|\nabla\left|\nabla G\right|\right|$
$\displaystyle\leq$
$\displaystyle\left(\int_{l\left(r\right)}\left|\nabla\left|\nabla
G\right|\right|^{2}\left|\nabla
G\right|^{-2}\right)^{\frac{1}{2}}\left(\int_{l\left(r\right)}\left|\nabla
G\right|^{2}\right)^{\frac{1}{2}},$

which says that

$\int_{l\left(r\right)}\left|\nabla\left|\nabla
G\right|\right|^{2}\left|\nabla
G\right|^{-2}\geq\frac{\left(w^{\prime}\right)^{2}}{w}\left(r\right).$

Combining with (3.6) we conclude that

(3.8) $\int_{l\left(r\right)}\left|\nabla G\right|^{-1}\Delta\left|\nabla
G\right|\geq\frac{3}{4}\frac{\left(w^{\prime}\right)^{2}}{w}\left(r\right)-4\pi.$

By (3.8) and the co-area formula it follows that

$\displaystyle-\int_{L\left(t,\infty\right)}G^{-2}\Delta\left|\nabla G\right|$
$\displaystyle=$
$\displaystyle-\int_{t}^{\infty}r^{-2}\int_{l\left(r\right)}\left|\nabla
G\right|^{-1}\Delta\left|\nabla G\right|$ $\displaystyle\leq$
$\displaystyle-\frac{3}{4}\int_{t}^{\infty}r^{-2}\frac{\left(w^{\prime}\right)^{2}}{w}\left(r\right)dr+\frac{4\pi}{t}.$

From the elementary inequality

$\displaystyle 0$ $\displaystyle\leq$ $\displaystyle
w\left(\frac{w^{\prime}}{w}-\frac{2}{r}\right)^{2}$ $\displaystyle=$
$\displaystyle\frac{\left(w^{\prime}\right)^{2}}{w}-\frac{4}{r}w^{\prime}+\frac{4}{r^{2}}w,$

one sees that

$-\frac{3}{4}\int_{t}^{\infty}r^{-2}\frac{\left(w^{\prime}\right)^{2}}{w}\left(r\right)dr\leq-3\int_{t}^{\infty}r^{-3}w^{\prime}\left(r\right)dr+3\int_{t}^{\infty}r^{-4}w\left(r\right)dr.$

Furthermore, integrating by parts implies that

$\displaystyle-3\int_{t}^{\infty}r^{-3}w^{\prime}\left(r\right)dr$
$\displaystyle=$
$\displaystyle-3r^{-3}w\left(r\right)|_{t}^{\infty}-9\int_{t}^{\infty}r^{-4}w\left(r\right)dr$
$\displaystyle=$ $\displaystyle
3t^{-3}w\left(t\right)-9\int_{t}^{\infty}r^{-4}w\left(r\right)dr.$

In conclusion,

(3.11)
$-\frac{3}{4}\int_{t}^{\infty}r^{-2}\frac{\left(w^{\prime}\right)^{2}}{w}\left(r\right)dr\leq
3t^{-3}w\left(t\right)-6\int_{t}^{\infty}r^{-4}w\left(r\right)dr.$

Plugging (3.11) into (3) we get

(3.12) $-\int_{L\left(t,\infty\right)}G^{-2}\Delta\left|\nabla G\right|\leq
3t^{-3}w\left(t\right)-6\int_{t}^{\infty}r^{-4}w\left(r\right)dr+\frac{4\pi}{t}.$

Hence, by (3.12) and (3) we obtain

$t^{-2}\frac{dw}{dt}\left(t\right)\leq t^{-3}w\left(t\right)+\frac{4\pi}{t},$

or equivalently,

(3.13) $\frac{d}{dt}\left(\frac{1}{t}w\left(t\right)-4\pi t\right)\leq 0.$

Finally, if

$\frac{d}{dt}\left(\frac{1}{t}w\left(t\right)-4\pi t\right)=0\text{ for }t=T,$

then all the inequalities in Lemma 2.2, (3) and (3) become equality. In
particular, the Hessian of $G$ on $L\left(T,\infty\right)$ must be of the form

$\displaystyle G_{11}$ $\displaystyle=$ $\displaystyle 32\pi^{2}G^{3}$
$\displaystyle G_{22}$ $\displaystyle=$ $\displaystyle-16\pi^{2}G^{3}$
$\displaystyle G_{33}$ $\displaystyle=$ $\displaystyle-16\pi^{2}G^{3}$
$\displaystyle G_{ij}$ $\displaystyle=$ $\displaystyle 0\;\;\text{ otherwise}$

and $\left|\nabla G\right|=4\pi G^{2}.$ Now consider the function
$f(x)=\frac{1}{4\pi\,G(x)}.$ Then $\left|\nabla f\right|=1$ and the Hessian of
the function $\frac{1}{2}\,f^{2}$ is the identity matrix. This immediately
implies that $L\left(T,\infty\right)$ is isometric to the ball
$B_{0}(\frac{1}{4\pi\,T})$ in $\mathbb{R}^{3}.$ ∎

We conclude this section with the following corollary.

######  Corollary 3.2.

Let $\left(M,g\right)$ be a complete noncompact three-dimensional manifold
with nonnegative scalar curvature and asymptotically nonnegative Ricci
curvature, that is,

(3.14) $\liminf_{x\rightarrow\infty}\mathrm{Ric}\left(x\right)\geq 0.$

Assume that $M$ has one end and its first Betti number
$b_{1}\left(M\right)=0.$ If $M$ is nonparabolic and the minimal Green’s
function $G\left(x\right)=G\left(p,x\right)$ satisfies
$\lim_{x\rightarrow\infty}G(x)=0,$ then

$\int_{l\left(t\right)}\left|\nabla G\right|^{2}\leq 4\pi t^{2}$

and

$\mathrm{Area}\left(l\left(t\right)\right)\geq\frac{1}{4\pi t^{2}}$

for all $t>0.$ Moreover, if equality holds for some $T>0,$ then the super
level set $\left\\{G>T\right\\}$ is isometric to a ball in $\mathbb{R}^{3}.$

###### Proof.

Let us note that by Theorem 3.1 we have

(3.15) $\frac{1}{t}w\left(t\right)-4\pi
t\leq\frac{1}{\delta}w\left(\delta\right)-4\pi\delta$

for all $0<\delta<t.$ Now the gradient estimate in [9] together with the
assumption (3.14) implies that for any $\varepsilon>0$ there exists
sufficiently small $\delta>0$ such that

(3.16) $\left|\nabla\ln G\right|\leq\varepsilon\text{ \ on
}L\left(0,\delta\right).$

Therefore,

$\frac{1}{\delta}w\left(\delta\right)=\frac{1}{\delta}\int_{l\left(\delta\right)}\left|\nabla
G\right|^{2}\leq\varepsilon,$

where we have used the fact that

(3.17) $\int_{l\left(\delta\right)}\left|\nabla G\right|=1.$

This shows that the right hand side of (3.15) goes to $0$ as $\delta\to 0.$
Hence, $w(t)\leq 4\pi t^{2},$ or

$\int_{l\left(t\right)}\left|\nabla G\right|^{2}\leq 4\pi t^{2}$

for all $t>0.$

We now derive a sharp area estimate for the level sets of the Green’s
function. Indeed,

$1=\int_{l\left(t\right)}\left|\nabla
G\right|\leq\left(\int_{l\left(t\right)}\left|\nabla
G\right|^{2}\right)^{\frac{1}{2}}\left(\mathrm{Area}\left(l\left(t\right)\right)\right)^{\frac{1}{2}}.$

It follows that

$\mathrm{Area}\left(l\left(t\right)\right)\geq\frac{1}{4\pi t^{2}}$

for all $t>0.$

Obviously, the equality case follows from that of Theorem 3.1. ∎

We note that under the hypothesis of Corollary 3.2 if

(3.18)
$\limsup_{t\rightarrow\infty}\left(\frac{1}{t}\int_{l\left(t\right)}\left|\nabla
G\right|^{2}-4\pi t\right)\geq 0,$

then $\left(M,g\right)$ is isometric to $\mathbb{R}^{3}.$ Indeed, by (3.15)
and (3.18) we have

$0\leq\frac{1}{\delta}\int_{l\left(\delta\right)}\left|\nabla
G\right|^{2}-4\pi\delta$

for any $\delta>0.$ Hence, equality must hold in Corollary 3.2 and
$\left(M,g\right)$ is isometric to the Euclidean space $\mathbb{R}^{3}.$

##  4\. Negative scalar lower bound

We now turn to the proof of Theorem 1.3. We start by establishing some lemmas
under the assumption that $\left(M,g\right)$ admits a positive Green’s
function $G$ satisfying

(4.1) $\displaystyle\lim_{x\rightarrow\infty}G\left(x\right)$ $\displaystyle=$
$\displaystyle 0,$ (4.2)
$\displaystyle\\#\mathrm{Conn}\left(l\left(t\right)\right)$
$\displaystyle\leq$ $\displaystyle A$

for all $t\leq t_{0},$ where $t_{0}$ and $A>0$ are fixed constants, and
$\\#\mathrm{Conn}\left(l(t)\right)$ denotes the number of connected components
of the level set $l(t)$ of $G.$

######  Lemma 4.1.

Let $\left(M,g\right)$ be a three-dimensional complete noncompact Riemannian
manifold satisfying (4.1) and (4.2). Assume that the Ricci curvature of $M$ is
bounded from below and the scalar curvature $S$ is bounded by

$S\geq-6K\text{ \ on }M.$

Then for any $\varepsilon>0,$

$\int_{L\left(\varepsilon,t_{0}\right)}\left|\nabla\left|\nabla
G\right|\right|^{2}\left|\nabla G\right|^{-1}\leq
4K\int_{L\left(\frac{1}{2}\varepsilon,2t_{0}\right)}\left|\nabla
G\right|+\frac{32\pi}{3}t_{0}A+C,$

where $C$ is a constant depending only on $t_{0}$ and the Ricci curvature
lower bound of $M,$ but not $\varepsilon.$

###### Proof.

According to the Bochner formula,

$\displaystyle\frac{1}{2}\Delta\left|\nabla G\right|^{2}$ $\displaystyle=$
$\displaystyle\left|G_{ij}\right|^{2}+\left\langle\nabla\Delta G,\nabla
G\right\rangle+\mathrm{Ric}\left(\nabla G,\nabla G\right)$ $\displaystyle=$
$\displaystyle\left|G_{ij}\right|^{2}+\mathrm{Ric}\left(\nabla G,\nabla
G\right)$

on $M\backslash\left\\{p\right\\}.$ Therefore,

(4.3) $\Delta\left|\nabla
G\right|=\left(\left|G_{ij}\right|^{2}-\left|\nabla\left|\nabla
G\right|\right|^{2}\right)\left|\nabla G\right|^{-1}+\mathrm{Ric}\left(\nabla
G,\nabla G\right)\left|\nabla G\right|^{-1}$

holds on $M\backslash\left\\{p\right\\}$ whenever $\left|\nabla G\right|\neq
0.$

Fix $0<\varepsilon<t_{0}<\infty$ and let $\phi$ be the Lipschitz function with
support in $L\left(\frac{1}{2}\varepsilon,2t_{0}\right)$ defined by

(4.4) $\phi=\left\\{\begin{array}[]{c}1\\\ \frac{\ln
G-\ln\left(\frac{1}{2}\varepsilon\right)}{\ln 2}\\\
\frac{\ln\left(2t_{0}\right)-\ln G}{\ln 2}\\\
0\end{array}\right.\begin{array}[]{c}\text{on }L\left(\varepsilon,1\right)\\\
\text{on }L\left(\frac{1}{2}\varepsilon,\varepsilon\right)\\\ \text{on
}L\left(t_{0},2t_{0}\right)\\\ \text{otherwise}\end{array}$

By the co-area formula, we have

$\displaystyle\int_{M}\left(\left|G_{ij}\right|^{2}-\left|\nabla\left|\nabla
G\right|\right|^{2}+\mathrm{Ric}\left(\nabla G,\nabla
G\right)\right)\left|\nabla G\right|^{-1}\phi^{2}$ $\displaystyle=$
$\displaystyle\int_{\frac{1}{2}\varepsilon}^{2t_{0}}\phi^{2}\left(r\right)\int_{l\left(r\right)}\left(\left|G_{ij}\right|^{2}-\left|\nabla\left|\nabla
G\right|\right|^{2}+\mathrm{Ric}\left(\nabla G,\nabla
G\right)\right)\left|\nabla G\right|^{-2}dr.$

However, Lemma 2.1 says that

$\displaystyle\left(\left|G_{ij}\right|^{2}-\left|\nabla\left|\nabla
G\right|\right|^{2}+\mathrm{Ric}\left(\nabla G,\nabla
G\right)\right)\left|\nabla G\right|^{-2}$ $\displaystyle=$
$\displaystyle\frac{1}{2}S-\frac{1}{2}S_{l\left(r\right)}+\frac{1}{2}\left|G_{ij}\right|^{2}\left|\nabla
G\right|^{-2}.$

Applying Lemma 2.2 for the last term, we conclude that

$\displaystyle\left(\left|G_{ij}\right|^{2}-\left|\nabla\left|\nabla
G\right|\right|^{2}+\mathrm{Ric}\left(\nabla G,\nabla
G\right)\right)\left|\nabla G\right|^{-2}$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}S-\frac{1}{2}S_{l\left(r\right)}+\frac{3}{4}\left|\nabla\left|\nabla
G\right|\right|^{2}\left|\nabla G\right|^{-2}.$

According to the Gauss-Bonnet theorem, on each regular connected component
$l_{k}\left(r\right)$ of $l\left(r\right),$

$\int_{l_{k}\left(r\right)}S_{l\left(r\right)}=4\pi\chi\left(l_{k}\left(r\right)\right)\leq
8\pi$

as $l_{k}\left(r\right)$ is compact by (4.1). Since by hypothesis (4.2) there
are at most $A$ connected components of $l\left(r\right),$ it follows that

$\int_{l\left(r\right)}S_{l\left(r\right)}\leq 8\pi A$

for all regular value $r$ with $r\leq t_{0}.$ Therefore, using that
$S\geq-6K,$ we conclude from (4) that

$\displaystyle\int_{l\left(r\right)}\left(\left|G_{ij}\right|^{2}-\left|\nabla\left|\nabla
G\right|\right|^{2}+\mathrm{Ric}\left(\nabla G,\nabla
G\right)\right)\left|\nabla G\right|^{-2}$ $\displaystyle\geq$
$\displaystyle\int_{l\left(r\right)}\left(\frac{1}{2}S-\frac{1}{2}S_{l\left(r\right)}+\frac{3}{4}\left|\nabla\left|\nabla
G\right|\right|^{2}\left|\nabla G\right|^{-2}\right)$ $\displaystyle\geq$
$\displaystyle-3K\mathrm{Area}\left(l\left(r\right)\right)+\frac{3}{4}\int_{l\left(r\right)}\left|\nabla\left|\nabla
G\right|\right|^{2}\left|\nabla G\right|^{-2}-4\pi A.$

Consequently, this implies that

$\displaystyle\int_{M}\left(\left|G_{ij}\right|^{2}-\left|\nabla\left|\nabla
G\right|\right|^{2}+\mathrm{Ric}\left(\nabla G,\nabla
G\right)\right)\left|\nabla G\right|^{-1}\phi^{2}$ $\displaystyle\geq$
$\displaystyle\int_{\frac{1}{2}\varepsilon}^{2t_{0}}\phi^{2}\left(r\right)\left(-3K\mathrm{Area}\left(l\left(r\right)\right)+\frac{3}{4}\int_{l\left(r\right)}\left|\nabla\left|\nabla
G\right|\right|^{2}\left|\nabla G\right|^{-2}\right)dr-8\pi t_{0}A$
$\displaystyle=$ $\displaystyle-3K\int_{M}\left|\nabla
G\right|\phi^{2}+\frac{3}{4}\int_{M}\left|\nabla\left|\nabla
G\right|\right|^{2}\left|\nabla G\right|^{-1}\phi^{2}-8\pi t_{0}A.$

Together with the Bochner formula (4.3), we arrive at

(4.6) $-3K\int_{M}\left|\nabla
G\right|\phi^{2}+\frac{3}{4}\int_{M}\left|\nabla\left|\nabla
G\right|\right|^{2}\left|\nabla
G\right|^{-1}\phi^{2}\leq\int_{M}\phi^{2}\Delta\left|\nabla G\right|+8\pi
t_{0}A.$

To estimate the right hand side, we make use of the gradient estimate for
positive harmonic functions. Choose $R_{0}>0$ so that
$L\left(0,2t_{0}\right)\subset M\backslash B\left(p,R_{0}\right),$ where $p$
is the pole of $G.$ Since the Ricci curvature is bounded from below on $M,$
applying Cheng-Yau’s gradient estimate [9] to the positive harmonic function
$G$ on $M\backslash B\left(p,\frac{R_{0}}{2}\right),$ one concludes that

(4.7) $\left|\nabla\ln G\right|\leq\Lambda\text{ \ on }M\backslash
B\left(p,R_{0}\right),$

where the constant $\Lambda$ depends only on $R_{0}$ and Ricci curvature lower
bound of $M.$

Integration by parts implies that

$\displaystyle\int_{M}\phi^{2}\Delta\left|\nabla G\right|$ $\displaystyle=$
$\displaystyle-\int_{M}\left\langle\nabla\phi^{2},\nabla\left|\nabla
G\right|\right\rangle$ $\displaystyle=$
$\displaystyle-\int_{L\left(\frac{1}{2}\varepsilon,\varepsilon\right)}\left\langle\nabla\phi^{2},\nabla\left|\nabla
G\right|\right\rangle$
$\displaystyle-\int_{L\left(t_{0},2t_{0}\right)}\left\langle\nabla\phi^{2},\nabla\left|\nabla
G\right|\right\rangle.$

Further integration by parts on each term leads to

$\displaystyle-\int_{L\left(\frac{1}{2}\varepsilon,\varepsilon\right)}\left\langle\nabla\phi^{2},\nabla\left|\nabla
G\right|\right\rangle$ $\displaystyle=$
$\displaystyle\int_{L\left(\frac{1}{2}\varepsilon,\varepsilon\right)}\left|\nabla
G\right|\Delta\phi^{2}-\int_{l\left(\varepsilon\right)}\left\langle\nabla\phi^{2},\nabla
G\right\rangle$
$\displaystyle-\int_{L\left(t_{0},2t_{0}\right)}\left\langle\nabla\phi^{2},\nabla\left|\nabla
G\right|\right\rangle$ $\displaystyle=$
$\displaystyle\int_{L\left(t_{0},2t_{0}\right)}\left|\nabla
G\right|\Delta\phi^{2}+\int_{l\left(t_{0}\right)}\left\langle\nabla\phi^{2},\nabla
G\right\rangle.$

Noting that $G$ is harmonic, we get that on
$L\left(\frac{1}{2}\varepsilon,\varepsilon\right)$

$\displaystyle\Delta\phi^{2}$ $\displaystyle=$ $\displaystyle
2\phi\Delta\phi+2\left|\nabla\phi\right|^{2}$ $\displaystyle=$ $\displaystyle
2\left(-\frac{1}{\ln 2}\phi+\frac{1}{\left(\ln
2\right)^{2}}\right)\left|\nabla\ln G\right|^{2}.$

Similarly, on $L\left(t_{0},2t_{0}\right),$

$\Delta\phi^{2}=2\left(\frac{1}{\ln 2}\phi+\frac{1}{\left(\ln
2\right)^{2}}\right)\left|\nabla\ln G\right|^{2}.$

In both cases, in view of (4.7), we have

$\displaystyle\left|\Delta\phi^{2}\right|$ $\displaystyle\leq$ $\displaystyle
c\left|\nabla\ln G\right|^{2}$ $\displaystyle\leq$ $\displaystyle
c\Lambda\left|\nabla\ln G\right|,$

where $c$ is a universal constant. Hence, by the co-area formula and the fact
that

(4.8) $\int_{l\left(r\right)}\left|\nabla G\right|=1,$

we get

$\displaystyle\left|\int_{L\left(\frac{1}{2}\varepsilon,\varepsilon\right)}\left|\nabla
G\right|\Delta\phi^{2}\right|$ $\displaystyle\leq$ $\displaystyle
c\Lambda\int_{L\left(\frac{1}{2}\varepsilon,\varepsilon\right)}\left|\nabla
G\right|^{2}G^{-1}$ $\displaystyle=$ $\displaystyle
c\Lambda\int_{\frac{1}{2}\varepsilon}^{\varepsilon}\frac{1}{r}dr$
$\displaystyle=$ $\displaystyle c\Lambda\ln 2.$

The other term is estimated as follows.

$\displaystyle\left|\int_{l\left(\varepsilon\right)}\left\langle\nabla\phi^{2},\nabla
G\right\rangle\right|$ $\displaystyle\leq$ $\displaystyle\frac{2}{\ln
2}\int_{l\left(\varepsilon\right)}\left|\nabla G\right|^{2}G^{-1}$
$\displaystyle\leq$ $\displaystyle c\Lambda.$

Similarly,

$\left|\int_{L\left(t_{0},2t_{0}\right)}\left|\nabla
G\right|\Delta\phi^{2}\right|+\left|\int_{l\left(t_{0}\right)}\left\langle\nabla\phi^{2},\nabla
G\right\rangle\right|\leq c\Lambda.$

In conclusion, we have shown that

(4.9) $\left|\int_{M}\phi^{2}\Delta\left|\nabla G\right|\right|\leq
c\Lambda\text{\ }\ \text{and }\int_{M}\left|\nabla\phi\right|^{2}\left|\nabla
G\right|\leq c\Lambda.$

Plugging into (4.6) implies that

$\frac{3}{4}\int_{M}\left|\nabla\left|\nabla G\right|\right|^{2}\left|\nabla
G\right|^{-1}\phi^{2}\leq 3K\int_{M}\left|\nabla
G\right|\phi^{2}+c\Lambda+8\pi t_{0}A,$

which is what to be proved. ∎

######  Lemma 4.2.

Let $\left(M,g\right)$ be a three-dimensional complete noncompact Riemannian
manifold satisfying (4.1) and (4.2). Assume that the Ricci curvature of $M$ is
bounded from below and the scalar curvature $S$ is bounded by

$S\geq-6K\text{ \ on }M.$

Then

$\lambda_{1}\left(M\right)\leq K.$

###### Proof.

According to Lemma 4.1,

(4.10) $\frac{3}{4}\int_{M}\left|\nabla\left|\nabla
G\right|\right|^{2}\left|\nabla G\right|^{-1}\phi^{2}\leq
3K\int_{M}\left|\nabla G\right|\phi^{2}+c\Lambda+8\pi t_{0}A,$

where $\phi$ is given in (4.4).

We now bound the left hand side from below using the bottom spectrum
$\lambda_{1}(M).$ Note that

$\displaystyle\lambda_{1}\left(M\right)\int_{M}\left|\nabla G\right|\phi^{2}$
$\displaystyle=$
$\displaystyle\lambda_{1}\left(M\right)\int_{M}\left(\left|\nabla
G\right|^{\frac{1}{2}}\phi\right)^{2}$ $\displaystyle\leq$
$\displaystyle\int_{M}\left|\nabla\left(\left|\nabla
G\right|^{\frac{1}{2}}\phi\right)\right|^{2}.$

Expanding the right side, we get

$\displaystyle\int_{M}\left|\nabla\left(\left|\nabla
G\right|^{\frac{1}{2}}\phi\right)\right|^{2}$ $\displaystyle=$
$\displaystyle\frac{1}{4}\int_{M}\left|\nabla\left|\nabla
G\right|\right|^{2}\left|\nabla G\right|^{-1}\phi^{2}+\int_{M}\left|\nabla
G\right|\left|\nabla\phi\right|^{2}$
$\displaystyle+\frac{1}{2}\int_{M}\left\langle\nabla\left|\nabla
G\right|,\nabla\phi^{2}\right\rangle.$

By (4.9) we have

$\int_{M}\left|\nabla
G\right|\left|\nabla\phi\right|^{2}+\frac{1}{2}\int_{M}\left\langle\nabla\left|\nabla
G\right|,\nabla\phi^{2}\right\rangle\leq c\Lambda.$

Therefore,

$\int_{M}\left|\nabla\left(\left|\nabla
G\right|^{\frac{1}{2}}\phi\right)\right|^{2}\leq\frac{1}{4}\int_{M}\left|\nabla\left|\nabla
G\right|\right|^{2}\left|\nabla G\right|^{-1}\phi^{2}+c\Lambda.$

In conclusion,

$\displaystyle\lambda_{1}\left(M\right)\int_{M}\left|\nabla G\right|\phi^{2}$
$\displaystyle\leq$ $\displaystyle\int_{M}\left|\nabla\left(\left|\nabla
G\right|^{\frac{1}{2}}\phi\right)\right|^{2}$ $\displaystyle\leq$
$\displaystyle\frac{1}{4}\int_{M}\left|\nabla\left|\nabla
G\right|\right|^{2}\left|\nabla G\right|^{-1}\phi^{2}+c\Lambda.$

Combining (4.10) and (4) we arrive at the following inequality.

(4.12) $\lambda_{1}\left(M\right)\int_{M}\left|\nabla G\right|\phi^{2}\leq
K\int_{M}\left|\nabla G\right|\phi^{2}+c\Lambda+\frac{8\pi}{3}t_{0}A,$

where $c$ is a universal constant. Assume by contradiction that

(4.13) $\lambda_{1}\left(M\right)>K.$

Then in view of the definition of $\phi$ in (4.4) and (4.12) we conclude that

(4.14)
$\left(\lambda_{1}\left(M\right)-K\right)\int_{L\left(\varepsilon,t_{0}\right)}\left|\nabla
G\right|\leq c\left(\Lambda+A\right).$

However, applying (4.8) and the co-area formula, we have

$\displaystyle\int_{L\left(\varepsilon,t_{0}\right)}\left|\nabla
G\right|^{2}G^{-1}$ $\displaystyle=$
$\displaystyle\int_{\varepsilon}^{t_{0}}\frac{1}{r}\left(\int_{l\left(r\right)}\left|\nabla
G\right|\right)dr$ $\displaystyle=$
$\displaystyle\ln\left(t_{0}\varepsilon^{-1}\right).$

On the other hand, Cheng-Yau’s gradient estimate shows that

$\int_{L\left(\varepsilon,t_{0}\right)}\left|\nabla
G\right|^{2}G^{-1}\leq\Lambda\int_{L\left(\varepsilon,t_{0}\right)}\left|\nabla
G\right|.$

We thus conclude that

(4.15) $\int_{L\left(\varepsilon,t_{0}\right)}\left|\nabla
G\right|\geq\Lambda^{-1}\ln\left(t_{0}\varepsilon^{-1}\right).$

Finally, we infer from (4.15) and (4.14) that

(4.16)
$\lambda_{1}\left(M\right)-K\leq\frac{c\Lambda\left(\Lambda+A\right)}{\ln\left(t_{0}\varepsilon^{-1}\right)}.$

Note that $c$ is a universal constant and that both $A$ and $\Lambda$ are
independent of $\varepsilon.$ Taking $\varepsilon\rightarrow 0$ leads to a
contradiction. Therefore, we have

$\lambda_{1}\left(M\right)\leq K.$

∎

We are now ready to prove Theorem 1.3. The only thing left to do is to verify
that both (4.1) and (4.2) hold.

######  Theorem 4.3.

Let $\left(M,g\right)$ be a three-dimensional complete noncompact Riemannian
manifold with scalar curvature $S\geq-6K$ on $M$ for some nonnegative constant
$K.$ Suppose that $M$ has finitely many ends and its first Betti number
$b_{1}(M)<\infty.$ Moreover, the Ricci curvature of $M$ is bounded below and
the volume $V_{x}(1)$ of unit ball $B_{x}(1)$ satisfies

$V_{x}(1)\geq C(\epsilon)\,\exp\left(-2\sqrt{K+\epsilon}\,r(x)\right)$

for every $\epsilon>0$ and all $x\in M,$ where $r(x)$ is the geodesic distance
from $x$ to a fixed point $p.$ Then the bottom spectrum of the Laplacian
satisfies

$\lambda_{1}\left(M\right)\leq K.$

###### Proof.

We prove the result by contradiction. Suppose that
$\lambda_{1}\left(M\right)>K.$ Then $M$ is necessarily nonparabolic as
$\lambda_{1}(M)>0$, see [11, Chapter 22]. Let $G$ be the minimal positive
Green’s function. We first show that $\lim_{x\rightarrow\infty}G(x)=0$ or
(4.1) holds.

According to a result of Li and the second author [14],

(4.17) $\int_{B_{p}\left(R+1\right)\backslash B_{p}\left(R-1\right)}G^{2}\leq
C\,e^{-2\sqrt{\lambda_{1}\,\left(M\right)}R}\int_{B_{p}\left(2\right)\backslash
B_{p}\left(1\right)}G^{2}$

for all $R>4,$ where $C>0$ is a constant depending only on
$\lambda_{1}\left(M\right).$ Since the Ricci curvature is bounded from below,
by (4.7) the mean value inequality holds for function $G,$ that is, for any
$x\in M\setminus B_{p}(2),$

(4.18)
$G^{2}\left(x\right)\leq\frac{C}{V_{x}\left(1\right)}\,\int_{B\left(x,1\right)}G^{2}$

for some constant $C$ only depending on the Ricci curvature lower bound. Now
fix $\epsilon>0$ such that $\lambda_{1}(M)\geq K+2\epsilon.$ Combining (4.17)
and (4.18), we conclude that

(4.19)
$G^{2}\left(x\right)\leq\frac{C}{V_{x}\left(1\right)}\,e^{-2\sqrt{K+2\epsilon}\,r(x)}\,\int_{B_{p}\left(2\right)\backslash
B_{p}\left(1\right)}G^{2},$

where $r(x)$ is the geodesic distance from $x$ to point $p.$ Using the
assumption that

$V_{x}(1)\geq C(\epsilon)\,\exp\left(-2\sqrt{K+\epsilon}\,r(x)\right),$

one sees immediately from (4.19) that $\lim_{x\to\infty}G(x)=0.$

The fact that $G$ satisfies (4.2) follows directly from Lemma 2.3. Therefore,
Lemma 4.2 is applicable and $\lambda_{1}(M)\leq K.$ ∎

We now draw a corollary.

######  Corollary 4.4.

Let $\left(M,g\right)$ be a three-dimensional complete noncompact Riemannian
manifold with nonpositive sectional curvature. Assume that the scalar
curvature $S$ is bounded by

$S\geq-6K\text{ \ on }M$

for some nonnegative constant $K.$ Then

$\lambda_{1}\left(M\right)\leq K.$

###### Proof.

This is because Theorem 4.3 is applicable to the universal cover $\tilde{M}$
of $M$ as $\tilde{M}$ is a Cartan-Hadamard manifold with bounded curvature.
Indeed, being diffeomorphic to $\mathbb{R}^{3},$ $\tilde{M}$ has first Betti
number $0$ and exactly one end. Also, the volume $V_{x}(1)$ of unit balls
$B_{x}(1)$ is at least that of the unit ball in $\mathbb{R}^{3}$ by the volume
comparison theorem. Therefore, $\lambda_{1}(\tilde{M})\leq K.$ However, the
bottom spectrum satisfies $\lambda_{1}(M)\leq\lambda_{1}(\tilde{M}),$ the
corollary follows. ∎

Acknowledgment. The first author was partially supported by NSF grant
DMS-1811845.

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