# Geometry of three-dimensional manifolds with scalar curvature lower 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.

The paper concerns three-dimensional complete manifolds with scalar curvature
bounded from below. One of the purposes is to establish a sharp comparison
theorem for the bottom spectrum in the spirit of the classical result of
Cheng. Another purpose is to derive volume and other geometric information in
terms of the scalar curvature when the Ricci curvature is asymptotically
nonnegative and the scalar curvature is positive. If the scalar curvature
decays no faster than linearly, then the manifold does not admit any positive
Green’s function. When the scalar curvature is bounded from below by a
positive constant, it is shown that the volume of unit balls must be bounded
from above by the lower bound of the scalar curvature at infinity. In
particular, in the case that the Ricci curvature is nonnegative, the volume of
the manifold must be of linear growth. This answers a question of Gromov in
the affirmative for dimension three. Volume estimates are also established for
the case when scalar curvature decays polynomially.

##  1\. Introduction

The paper concerns complete three dimensional manifolds with scalar curvature
lower bound. Our aim is to derive both analytic and geometric information of
such manifolds under suitable, yet necessary, topological assumptions. More
specifically, results concerning bottom spectrum, volume growth and existence
of positive Green’s functions are established. This is achieved through
working on the level sets of suitably chosen harmonic functions and by
employing a technique introduced by Schoen and Yau [34] and further exploited
in [19, 38, 4, 30].

Recall that the bottom spectrum $\lambda_{1}\left(M\right)$ of a complete
manifold $M$ is defined to be the smallest $\lambda\in\sigma(\Delta),$ the
spectrum of the Laplacian of $M.$ Alternatively, it is characterized as the
optimal constant of the Poincaré inequality

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

for all smooth function $\varphi$ with compact support.

A classical comparison result due to Cheng [7] provides a sharp upper bound
for the bottom spectrum $\lambda_{1}(M)$ of an $n$-dimensional complete
manifold $M$ with its Ricci curvature $\mathrm{Ric}\geq-\left(n-1\right)K$ for
some nonnegative constant $K.$ Namely,

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

Naturally, one may wonder if the result remains true under a scalar curvature
lower bound. Since the product manifold
$\mathbb{H}^{n-2}\times\mathbb{S}^{2}(r)$ of the hyperbolic space
$\mathbb{H}^{n-2}$ with the sphere $\mathbb{S}^{2}(r)$ of radius $r$ has
positive bottom spectrum, yet its scalar curvature can be made into an
arbitrary positive constant by choosing $r$ small, the answer is obviously
negative in the case $n\geq 4.$ It turns out the answer is negative for the
three dimensional case as well. Indeed, consider the universal cover $M$ of
the connected sum
$N=\left(\mathbb{S}^{2}\times\mathbb{S}^{1}\right)\\#\left(\mathbb{S}^{2}\times\mathbb{S}^{1}\right).$
Then $M$ carries a metric with positive scalar curvature as $N$ does so [17,
35]. Note that the bottom spectrum of $M$ is positive according to a result of
Brooks [5]. This is because the first fundamental group of $N$ is a free group
on two generators and nonamenable.

Nonetheless, we have the following result concerning $\lambda_{1}(M).$

######  Theorem 1.1.

Let $\left(M,g\right)$ be a three-dimensional complete 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
by a constant. Then the bottom spectrum of $M$ satisfies

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

The estimate is obviously sharp for the hyperbolic space $\mathbb{H}^{3}.$ The
preceding example clearly demonstrates the necessity of the topological
assumption that the number of ends of $M$ is finite.

Our second purpose is to derive geometric information for three-dimensional
manifolds with positive scalar curvature and asymptotically nonnegative Ricci
curvature. This is partly motivated by the following well-known question posed
by Yau [40].

######  Question 1.2.

Let $(M^{n},g)$ be a complete manifold with nonnegative Ricci curvature. Then
for any point $p\in M,$ do we have

$\limsup_{r\to\infty}r^{2-n}\,\int_{B_{p}(r)}S<\infty?$

A related version due to Gromov [16] asks whether the volume growth is at most
of order $n-2$ if the scalar curvature is bounded below by a positive
constant. As a consequence of our more general results, the answer to Gromov’s
problem is positive in the case $n=3.$ We refer to [31, 39, 41] for recent
progress on Yau’s problem.

We begin with a result on the nonexistence of positive Green’s functions.

Recall that a complete manifold $M$ is said to have asymptotically nonnegative
Ricci curvature if its Ricci curvature
$\mathrm{Ric}(x)\geq-k\left(r(x)\right)$ for a nonnegative nonincreasing
function $k(r)$ with $\int_{0}^{\infty}r\,k(r)\,dr<\infty,$ where $r(x)$ is
the geodesic distance from $x$ to a fixed point $p\in M$. We have the
following result.

######  Theorem 1.3.

Let $(M,g)$ be a complete three-dimensional manifold with finite first Betti
number and finitely many ends. Suppose that its Ricci curvature is
asymptotically nonnegative. If the scalar curvature $S$ is bounded below by

$\liminf_{x\rightarrow\infty}\,S\left(x\right)r\left(x\right)\geq C$

for some positive constant $C,$ then $M$ is parabolic, that is, it does not
admit any positive Green’s function.

In particular, the manifold is parabolic if the scalar curvature is bounded
from below by a positive constant. Our next result relates the lower bound of
the scalar curvature with the volume of unit balls.

######  Theorem 1.4.

Let $(M,g)$ be a complete three-dimensional manifold with finite first Betti
number and finitely many ends. Suppose that its Ricci curvature is
asymptotically nonnegative. Then there exists a constant $C$ depending on $k$
such that the scalar curvature $S$ of $M$ satisfies

$\liminf_{x\rightarrow\infty}S\left(x\right)\leq\frac{C}{\mathrm{V}_{p}\left(1\right)},$

where $\mathrm{V}_{p}\left(1\right)$ denotes the volume of the unit ball
$B_{p}\left(1\right).$

In fact, volume information is also obtained under the more general assumption
that $\liminf_{x\rightarrow\infty}S\left(x\right)r^{\alpha}\left(x\right)>0$
for some $\alpha>0.$ We refer to Theorem 5.6 for the precise statement. It
should be pointed out that it is necessary to assume that $M$ has
asymptotically nonnegative Ricci curvature for Theorem 1.4 to hold. Indeed,
consider complete three-dimensional manifold $\left(M,g\right)$ which is
isometric to $\left[2,\infty\right)\times\mathbb{S}^{2}$ with the warped
product metric $ds_{M}^{2}=dt^{2}+\frac{1}{\ln t}ds_{\mathbb{S}^{2}}^{2}$
outside a compact set. Then its scalar curvature satisfies
$\liminf_{x\rightarrow\infty}S\left(x\right)=\infty.$ Note that its Ricci
curvature $\mathrm{Ric}(x)\geq-k\left(r\left(x\right)\right)$ for
$k\left(r\right)=\frac{C}{r^{2}\ln r}$ when $r$ is sufficiently large.

As a special case of the above results, we have the following conclusion.
Recall by Cheeger-Gromoll splitting theorem [6], a complete manifold $M^{n}$
with nonnegative Ricci curvature has at most two ends. Also, by [29], the
first Betti number of such $M^{n}$ is finite and bounded by its dimension $n.$
Moreover, according to [27], $M^{n}$ is parabolic if and only if

$\int_{1}^{\infty}\frac{t}{\mathrm{V}_{p}(t)}\,dt=\infty,$

where $\mathrm{V}_{p}(t)$ is the volume of geodesic ball $B_{p}(t).$

######  Corollary 1.5.

Let $(M,g)$ be a complete three-dimensional manifold with non-negative Ricci
curvature.

  * •

If the scalar curvature is bounded below by

$\liminf_{x\rightarrow\infty}\,S\left(x\right)r\left(x\right)\geq C$

for some positive constant $C,$ where $r(x)$ is the distance function to a
fixed point $p\in M,$ then

$\int_{1}^{\infty}\frac{t}{\mathrm{V}_{p}(t)}\,dt=\infty.$

  * •

There exists a universal constant $C$ such that the scalar curvature $S$ of
$M$ satisfies

$\liminf_{x\rightarrow\infty}S\left(x\right)\leq\frac{C}{\mathrm{V}_{p}\left(1\right)}$

for every $p\in M.$

  * •

If the scalar curvature is bounded below by $S\geq 1$ on $M,$ then there
exists a universal constant $C>0$ such that

$\mathrm{V}_{p}\left(R\right)\leq C\,R$

for all $R>0$ and $p\in M.$

  * •

If the scalar curvature

$S\left(x\right)\geq\frac{C}{r^{\alpha}\left(x\right)+1}$

for some $\alpha\in\left[0,1\right],$ then

$\mathrm{V}_{p}\left(R_{i}\right)\leq C\,R_{i}^{\alpha+1}$

for a sequence $R_{i}\rightarrow\infty.$

We remark that the upper bound estimate of scalar curvature $S$ in terms of
the volume $\mathrm{V}_{p}(1)$ is sharp by considering the example of cylinder
$\mathbb{R}\times\mathbb{S}^{2}(r),$ where $\mathbb{S}^{2}(r)$ is the sphere
of radius $r$ in Euclidean space $\mathbb{R}^{3}$. As mentioned already, the
linear volume growth estimate in the corollary gives an affirmative answer in
dimension three to a question of Gromov [16].

We now indicate some of the ideas involved in the proofs. Theorem 1.1 was
previously proved in [30] under an additional assumption that the volume
$\mathrm{V}_{x}(1)$ of unit balls $B_{x}(1)$ satisfies

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

for every $\epsilon>0$ and all $x\in M,$ where $r(x)$ is the geodesic distance
from $x$ to a fixed point $p.$ This was used to ensure that the minimal
positive Green’s function $G(p,x)$ of $M$ with fixed $p$ goes to $0$ at
infinity in the case that $\lambda_{1}(M)>K.$ Together with the topological
assumptions, it was concluded that the regular level sets $l(t)$ of $G$ are
necessarily connected compact surfaces on each end. The proof then proceeded
by integrating the following Bochner formula over the level sets $l(t)$

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

and applying the Gauss-Bonnet theorem, where the Ricci curvature term is
rewritten into

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

with $S_{l\left(t\right)}$ being the scalar curvature of the level set
$l\left(t\right).$

The idea of rewriting the Ricci curvature term as (1.1) has origin in Schoen
and Yau [34], where they made the important observation that on a minimal
surface $N$ in a three-dimensional manifold $M,$

(1.2)
$\mathrm{Ric}\left(\nu,\nu\right)=\frac{1}{2}S-\frac{1}{2}S_{N}-\frac{1}{2}\left|A\right|^{2}$

with $\nu,$ $S_{N}$ and $A$ being the unit normal vector, the scalar curvature
and the second fundamental form of $N,$ respectively. This observation enabled
them to classify compact stable minimal surfaces in a three-dimensional
manifold with nonnegative scalar curvature. In fact, it can be used to reprove
the well-known classification by Fisher-Colbrie and Schoen [14] for complete
stable minimal surfaces as well (see [26]). More generally, an identity of the
nature (1.2) was derived for any surface $N,$ not necessarily minimal, by
Jezierski and Kijowski in [19]. It was applied to level sets of suitably
chosen functions to prove both positive energy and positive mass results in
[19, 18]. Recently, this identity was rediscovered by Stern [38] for the level
sets of harmonic functions. In [4] it has been subsequently used to reprove
the positive mass theorem of Schoen and Yau [36].

The lack of properness of the minimal positive Green’s function creates
serious difficulty for the above argument to work. To overcome this
difficulty, we modify the function $G$ by considering $u=G\,\psi,$ where
$\psi$ is a smooth cut-off function. While this guarantees that the positive
level sets of $u$ are compact, the price we pay is that the function $u$ is no
longer harmonic. This creates many new technical issues. For one, it becomes
unclear whether the intersection of level sets of $u$ with each end is
connected. To get around the issue, we consider only the component
$L(t,\infty)$ of the super level set $\\{u>t\\}$ with the fixed pole $p\in
L(t,\infty).$ It turns out that for each end $E$ of $M,$ the unbounded
component of $E\setminus L(t,\infty)$ has exactly one component of the
boundary of $L(t,\infty).$ This fact more or less suffices for our purpose,
though there could be other components of the boundary of $L(t,\infty).$
Another issue is that the Bochner formula for $u$ introduces many extra terms.
Fortunately, those terms can be controlled with the help of the exponential
decay estimate for Green’s function from [23] together with a judicious choice
of the cut-off function $\psi$ based on a result from Schoen and Yau [37].
Actually, instead of $G,$ we work with the so-called barrier function $f,$
namely, a harmonic function defined on $M\setminus D$ for an arbitrarily large
compact smooth domain $D$ such that $f=1$ on the boundary $\partial D$ of $D$
and $0<f<1.$ The advantage is that it leads to a slightly stronger conclusion,
that is, the smallest essential spectrum of $M$ is bounded from above by $K$
as well.

It remains to be seen whether Theorem 1.1 holds without assuming the Ricci
curvature is bounded from below. For the aforementioned Cheng’s estimate,
there are rigidity results when $\lambda_{1}(M)$ attains its maximum value
$\frac{\left(n-1\right)^{2}}{4}K$ (see [23, 24]). Analogously, one may ask
what happens if $\lambda_{1}(M)$ attains its upper bound $K$ in Theorem 1.1.

The proof of Theorem 1.3 is very much inspired by Theorem 1.1. However, the
technical details differ. Again, we work with a barrier function $f.$ Since
$f$ may not be proper, as above we consider $u=f\,\psi$ instead, where $\psi$
is a smooth cut-off function on $M.$ Unlike before, we no longer have the fact
that the function $f$ decays exponentially in the $L^{2}$ integral sense. To
compensate, we choose a better behaved cut-off function $\psi,$ which in turn
depends on the construction of a well controlled distance-like function. That
is where the stronger assumption on the Ricci curvature is used.

The proof of Theorem 1.4 relies on an elaboration of the following comparison
theorem in [30].

######  Theorem 1.6.

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 $\mathbb{R}^{3}.$

We point out that in [2] this type of result was applied to reprove the
positive mass theorem [36]. Recently, using a variant of the above theorem,
Chodosh and Li [9] have affirmed a conjecture of Schoen that a stable minimal
hypersurface in Euclidean space $\mathbb{R}^{4}$ must be flat. Historically,
more general versions of monotonicity formulas were established by Colding
[10] and Colding-Minicozzi [12] for $n$-dimensional manifolds with nonnegative
Ricci curvature and applied to the study of uniqueness of the tangent cones
for Ricci flat manifolds with Euclidean volume growth [13]. We refer the
readers to [11] for an exposition on monotonicity formulas in geometric
analysis, and [1] for their applications to Willmore type inequalities.

The observation we make is that Theorem 1.6 can be localized to each end of
$M$ by working with the barrier function of an end. Recall by Li and Tam [20],
each end $E$ of a complete manifold admits a positive harmonic function $u$
satisfying $u=0$ on the boundary of $E.$ Such $u$ is bounded if and only if
$E$ is nonparabolic. On the other hand, by a result of Nakai [32], $u$ can be
chosen to be proper in the case $E$ is parabolic. Applying the localized
version of Theorem 1.6 to such $u$ together with suitable estimates on $u$
then yields Theorem 1.4.

Finally, we mention that Theorem 1.6 or its localized version to an end may
also be used to show the nonexistence of proper positive Green’s functions
under a suitable assumption on scalar curvature. This is explicitly stated in
Theorem 5.6. By [27], since the minimal positive Green’s function must go to
$0$ at infinity for a complete nonparabolic manifold with nonnegative Ricci
curvature, this provides an alternative proof to Theorem 1.3 in this special
case.

The structure of the paper is as follows. After some preparations in Section
2, we complete the proof of Theorem 1.1 in Section 3. To address Theorem 1.3
and Theorem 1.4, we collect some preliminary facts in Section 4. Section 5 is
devoted to the proofs of these theorems.

Acknowledgment: We wish to thank Otis Chodosh for his interest and for sharing
with us the example that demonstrates the necessity of the topological
assumptions for Theorem 1.1. We would also like to thank Florian Johne for his
careful reading of the previous version and for his helpful comments. The
first author was partially supported by NSF grant DMS-1811845.

##  2\. Preliminaries when Ricci curvature bounded below

In this section we make some preparations for proving Theorem 1.1. First, we
follow Theorem 4.2 in [37] and construct a smooth distance-like function on an
arbitrary dimensional manifold $M.$ Denote by $r\left(x\right)$ the distance
to a fixed point $p\in M.$

######  Lemma 2.1.

Let $\left(M,g\right)$ be an $n$-dimensional complete Riemannian manifold with
Ricci curvature bounded from below by a constant. Then there exists a proper
function $\rho\in C^{\infty}\left(M\right)$ such that

(2.1) $\displaystyle\frac{1}{C}r$ $\displaystyle\leq$ $\displaystyle\rho\leq
Cr$ $\displaystyle\left|\nabla\rho\right|+\left|\Delta\rho\right|$
$\displaystyle\leq$ $\displaystyle C\ \ $

and

(2.2) $\left|\nabla\left(\Delta\rho\right)\right|\leq
C\left|\nabla^{2}\rho\right|$

for some constant $C>0.$

###### Proof.

Since the last estimate is not explicitly stated in Theorem 4.2 of [37], we
sketch the proof below. For $\lambda>0,$ to be determined later, we solve the
boundary value problem

$\displaystyle\Delta h_{R}$ $\displaystyle=$ $\displaystyle\lambda h_{R}\text{
\ in }B_{p}\left(R\right)\backslash B_{p}\left(1\right)$ $\displaystyle h_{R}$
$\displaystyle=$ $\displaystyle 1\text{ \ on }\partial B_{p}\left(1\right)$
$\displaystyle h_{R}$ $\displaystyle=$ $\displaystyle 0\text{ \ on }\partial
B_{p}\left(R\right).$

The maximum principle implies that $0<h_{R}<1$ and that $h_{R}$ is increasing
in $R$. Letting $R\rightarrow\infty,$ we obtain a positive function $h$ on
$M\backslash B_{p}\left(1\right)$ satisfying

$\displaystyle\Delta h$ $\displaystyle=$ $\displaystyle\lambda h\text{ \ in
}M\backslash B_{p}\left(1\right)$ $\displaystyle h$ $\displaystyle=$
$\displaystyle 1\text{ \ on }\partial B_{p}\left(1\right).$

By Cheng-Yau’s gradient estimate [8],

(2.3) $\left|\nabla\ln h\right|\leq C\left(\lambda\right)$

as the Ricci curvature of $M$ is bounded from below.

For any $\gamma>0,$ note that

$\displaystyle\lambda\int_{B_{p}\left(R\right)\backslash
B_{p}\left(1\right)}e^{\gamma r}h_{R}^{2}$ $\displaystyle=$
$\displaystyle\int_{B_{p}\left(R\right)\backslash
B_{p}\left(1\right)}e^{\gamma r}h_{R}\Delta h_{R}$ $\displaystyle=$
$\displaystyle-\int_{B_{p}\left(R\right)\backslash
B_{p}\left(1\right)}e^{\gamma r}\left|\nabla h_{R}\right|^{2}$
$\displaystyle-\gamma\int_{B_{p}\left(R\right)\backslash
B_{p}\left(1\right)}\left\langle\nabla h_{R},\nabla r\right\rangle
h_{R}e^{\gamma r}$ $\displaystyle-e^{\gamma}\int_{\partial
B_{p}\left(1\right)}h_{R}\frac{\partial h_{R}}{\partial\nu}$
$\displaystyle\leq$
$\displaystyle\frac{\gamma^{2}}{4}\int_{B_{p}\left(R\right)\backslash
B_{p}\left(1\right)}e^{\gamma r}h_{R}^{2}-e^{\gamma}\int_{\partial
B_{p}\left(1\right)}h_{R}\frac{\partial h_{R}}{\partial\nu}.$

Hence, setting $\gamma=\sqrt{\lambda}$ we conclude that

$\int_{B_{p}\left(R\right)\backslash
B_{p}\left(1\right)}e^{\sqrt{\lambda}r}h_{R}^{2}\leq C$

for all $R>1$. This implies that

$\int_{M}e^{\sqrt{\lambda}r}h^{2}\leq C.$

Together with (2.3) we get that

$\displaystyle h^{2}\left(x\right)$ $\displaystyle\leq$
$\displaystyle\frac{C}{\mathrm{V}_{x}\left(1\right)}\int_{B_{x}\left(1\right)}h^{2}$
$\displaystyle\leq$
$\displaystyle\frac{C}{\mathrm{V}_{x}\left(1\right)}e^{-\sqrt{\lambda}r\left(x\right)}.$

Since the Ricci curvature is bounded below,

$\frac{C}{\mathrm{V}_{x}\left(1\right)}\leq e^{Cr\left(x\right)}.$

Hence, by choosing $\lambda$ sufficiently large, we conclude that $h$ decays
exponentially and

(2.4) $h\left(x\right)\leq Ce^{-\frac{1}{4}\sqrt{\lambda}r\left(x\right)}.$

The function $\rho$ is essentially defined as $-\ln h$ with a slight
modification. More precisely,

$\rho\left(x\right)=\left(1-\eta\left(x\right)\right)\left(-\ln
h\left(x\right)\right)+\eta\left(x\right),$

where $\eta$ is a smooth cut-off function with $\eta=1$ on
$B_{p}\left(1\right)$ and $\eta=0$ on $M\backslash B_{p}\left(2\right).$

Now (2.1) follows from (2.3), (2.4) and the fact that $\Delta h=\lambda h.$
Indeed,

(2.5) $\Delta\rho=-\lambda+\left|\nabla\rho\right|^{2}\text{ \ on }M\backslash
B_{p}\left(2\right).$

Clearly, (2.5) implies (2.2). ∎

For given $R>0,$ let $\psi:\left(0,\infty\right)\rightarrow\mathbb{R}$ be a
smooth function such that $\psi=1$ on $\left(0,2R\right)$ and $\psi=0$ on
$\left(3R,\infty\right)$ satisfying

$\left|\psi^{\prime}\right|\leq\frac{C}{R},\ \
\left|\psi^{\prime\prime}\right|\leq\frac{C}{R^{2}}\text{ and
}\left|\psi^{\prime\prime\prime}\right|\leq\frac{C}{R^{3}}.$

Composing it with $\rho,$ we obtain a cut-off function

$\psi\left(x\right)=\psi\left(\rho\left(x\right)\right).$

Obviously,

$\displaystyle\psi$ $\displaystyle=$ $\displaystyle 1\text{ on
}D\left(2R\right)$ $\displaystyle\psi$ $\displaystyle=$ $\displaystyle 0\text{
on }M\backslash D\left(3R\right).$

Throughout this section,

(2.6) $D\left(t\right)=\left\\{x\in M:\rho\left(x\right)<t\right\\}.$

Recall that a complete manifold $\left(M,g\right)$ is nonparabolic if it
admits a positive symmetric Green’s function. Given a smooth connected bounded
domain $D\subset M,$ according to Li-Tam [20], there a exists a barrier
function $f>0$ on $M\backslash D$ satisfying

(2.7) $\displaystyle\Delta f=0\text{ \ on }M\backslash D$ $\displaystyle
f=1\text{ on }\partial D$
$\displaystyle\liminf_{x\rightarrow\infty}f\left(x\right)=0.$

Moreover, such $f$ can be obtained as a limit
$f=\lim_{R_{i}\rightarrow\infty}f_{i},$ where each $f_{i}$ is a harmonic
function on $D(R_{i})\setminus D$ with $f_{i}=1$ on $\partial D$ and $f_{i}=0$
on $\partial D\left(R_{i}\right).$ It is easy to see that

(2.8) $\int_{M\setminus D}\,\left|\nabla f\right|^{2}<\infty.$

In particular, by [25], for all $0<t\leq 1$ and $R\geq R_{0}$ with $D\subset
D(R_{0}),$

$-\int_{\partial D(R)}\,\frac{\partial
f}{\partial\nu}=\int_{\\{f=t\\}}\,\left|\nabla f\right|=C>0,$

where $\nu$ is the outward unit normal vector of $\partial D(R).$

The following gradient estimate follows from [8] together with the standard
partial differential equation theory.

######  Lemma 2.2.

Let $\left(M,g\right)$ be an $n$-dimensional complete noncompact Riemannian
manifold with Ricci curvature bounded below by a constant. Then

$\left|\nabla\ln f\right|\leq C\text{ \ on }M\backslash D.$

We now consider the function $u$ given by

(2.9) $u=f\psi.$

We summarize some properties of $u$ in the following result.

######  Lemma 2.3.

Let $\left(M,g\right)$ be an $n$-dimensional complete noncompact Riemannian
manifold with Ricci curvature bounded below by a constant. Then $u$ is
harmonic on $D\left(2R\right)\backslash D$ and

$\displaystyle\left|\nabla u\right|$ $\displaystyle\leq$ $\displaystyle
C\left(u+\frac{1}{R}\right)$ $\displaystyle\left|\Delta u\right|$
$\displaystyle\leq$ $\displaystyle\frac{C}{R}$
$\displaystyle\left|\nabla\left(\Delta u\right)\right|$ $\displaystyle\leq$
$\displaystyle\frac{C}{R}\left(\left|\nabla^{2}\rho\right|^{2}+\left|\nabla^{2}f\right|^{2}+1\right)$

on $M\backslash D.$

###### Proof.

By Lemma 2.1 we have that

(2.10) $\left|\nabla\psi\right|+\left|\Delta\psi\right|\leq\frac{C}{R}$

and

(2.11)
$\left|\nabla\left(\Delta\psi\right)\right|\leq\frac{C}{R}\left(\left|\nabla^{2}\rho\right|+1\right).$

Indeed, the last estimate follows from

$\displaystyle\left|\nabla\left(\Delta\psi\right)\right|$ $\displaystyle=$
$\displaystyle\left|\nabla\left(\psi^{\prime}\Delta\rho+\psi^{\prime\prime}\left|\nabla\rho\right|^{2}\right)\right|$
$\displaystyle\leq$
$\displaystyle\left|\psi^{\prime}\right|\left|\nabla\left(\Delta\rho\right)\right|+\left|\Delta\rho\right|\left|\nabla\psi^{\prime}\right|+\left|\psi^{\prime\prime}\right|\left|\nabla\left|\nabla\rho\right|^{2}\right|+\left|\nabla\rho\right|^{2}\left|\nabla\psi^{\prime\prime}\right|$
$\displaystyle\leq$
$\displaystyle\frac{C}{R}\left(\left|\nabla^{2}\rho\right|+1\right)$

by applying (2.1) and (2.2).

By (2.10) and Lemma 2.2 we immediately obtain

$\displaystyle\left|\nabla u\right|$ $\displaystyle\leq$
$\displaystyle\left|\nabla f\right|\psi+f\left|\nabla\psi\right|$
$\displaystyle\leq$ $\displaystyle Cu+\frac{C}{R}$

on $M\backslash D$, and

$\displaystyle\Delta u$ $\displaystyle=$ $\displaystyle 0\text{ \ on
}D\left(2R\right)\backslash D,$ $\displaystyle\left|\Delta u\right|$
$\displaystyle\leq$ $\displaystyle\frac{C}{R}\text{ \ on }M\backslash D.$

Moreover, from (2.11), we have

$\displaystyle\left|\nabla\left(\Delta u\right)\right|$ $\displaystyle\leq$
$\displaystyle\left|\nabla\left(f\Delta\psi\right)\right|+2\left|\nabla\left(\left\langle\nabla\psi,\nabla
f\right\rangle\right)\right|$ $\displaystyle\leq$
$\displaystyle\frac{C}{R}\left(\left|\nabla^{2}\rho\right|+\left|\nabla^{2}f\right|+1\right)$
$\displaystyle\leq$
$\displaystyle\frac{C}{R}\left(\left|\nabla^{2}\rho\right|^{2}+\left|\nabla^{2}f\right|^{2}+1\right).$

This proves the result. ∎

We extend both $f$ and $u$ to $M$ by setting $f=1$ and $u=1$ in $D.$ For
$0<t<1,$ denote by $L\left(t,\infty\right)$ the connected component of the
super level set $\left\\{u>t\right\\}$ that contains $D.$ Note that all
bounded components of $M\setminus D$ are contained in $L\left(t,\infty\right)$
as well. This is because $f=1$ there and each of the components intersects
with $D.$ Let

(2.12) $l\left(t\right)=\partial L\left(t,\infty\right).$

Since $u$ has compact support in $M,$ the closure
$\overline{L\left(t,\infty\right)}$ of $L\left(t,\infty\right)$ and its
boundary $l\left(t\right)$ are compact in $M.$

We now assume that $M$ has finite first Betti number and finitely many, say
$m,$ ends. Let the smooth connected bounded domain $D$ be chosen large enough
so that all representatives of the first homology group $H_{1}\left(M\right)$
are included in $D.$ Moreover, $M\backslash D$ has exactly $m$ unbounded
connected components. Then the same holds true for $M\setminus\tilde{D}$, for
any bounded domain $\tilde{D}$ with $D\subset\tilde{D}.$

We have the following estimate on the number of components of
$l\left(t\right).$ The argument in [30] needs to be modified as $u$ is no
longer harmonic on $M.$

######  Lemma 2.4.

Let $\left(M,g\right)$ be an $n$-dimensional complete Riemannian manifold with
$m$ ends and finite first Betti number $b_{1}(M).$ Assume that the Ricci
curvature is bounded from below by a constant. Then for any $0<t<1,$

(2.13) $l\left(t\right)\cap\overline{M_{t}}=\partial M_{t}\text{ has }m\text{
connected components}$

and

(2.14) $l\left(t\right)\cap\left(M\backslash\overline{M_{t}}\right)\subset
M\backslash D\left(2R\right),$

where $M_{t}$ is the union of all unbounded connected components of
$M\backslash\overline{L\left(t,\infty\right)}$ and $l\left(t\right)$ is
defined in (2.12).

###### Proof.

Recall that all representatives of the first homology $H_{1}\left(M\right)$
lie in $D$ and $M\setminus D$ has exactly $m$ unbounded components. Therefore,
as $D\subset L\left(t,\infty\right)$, $L\left(t,\infty\right)$ contains all
representatives of $H_{1}\left(M\right)$ and $M_{t}$ has $m$ connected
components for any $0<t<1.$ Note also $\partial M_{t}\subset l\left(t\right).$

For fixed $0<\delta<1$ such that $t+\delta<1,$ define $U$ to be the union of
$L\left(t,\infty\right)$ with all bounded components of
$M\backslash\overline{L\left(t+\delta,\infty\right)}$ and $V$ the union of all
unbounded components of $M\backslash\overline{L\left(t+\delta,\infty\right)}.$
Since $M=U\cup V,$ we have the following Mayer-Vietoris sequence

(2.15) $H_{1}\left(U\right)\oplus
H_{1}\left(V\right)\overset{j_{\ast}}{\rightarrow}H_{1}\left(M\right)\overset{\partial}{\rightarrow}H_{0}\left(U\cap
V\right)\overset{i_{\ast}}{\rightarrow}H_{0}\left(U\right)\oplus
H_{0}\left(V\right)\overset{j_{\ast}^{\prime}}{\rightarrow}H_{0}(M).$

The map $j_{\ast}$ is onto because all representatives of
$H_{1}\left(M\right)$ lie inside $U$. The map $j_{\ast}^{\prime}$ is also
onto. Note also that $V$ has $m$ components and $U$ is connected. The latter
is true because each component of
$M\backslash\overline{L\left(t+\delta,\infty\right)}$ intersects with
$L\left(t,\infty\right)$ as $\overline{L\left(t+\delta,\infty\right)}\subset
L\left(t,\infty\right).$ In view of (2.15), we therefore obtain the short
exact sequence

$0\rightarrow H_{0}\left(U\cap
V\right)\rightarrow\mathbb{Z}\oplus..\oplus\mathbb{Z\rightarrow
Z\rightarrow}0$

with $m+1$ summands. In conclusion,

$H_{0}\left(U\cap V\right)=\mathbb{Z}\oplus..\oplus\mathbb{Z}$

with $m$ summands. Since $\delta>0$ can be arbitrarily small, this proves that

$l\left(t\right)\cap\overline{M_{t}}\text{ has }m\text{ components}$

for $0<t<1.$ Hence, we have established (2.13).

For the second conclusion, write

$M\backslash\overline{L\left(t,\infty\right)}=M_{t}\cup\Omega,$

where $\Omega$ is the union of all bounded components of
$M\backslash\overline{L\left(t,\infty\right)}.$ Clearly,
$\partial\Omega\subset l\left(t\right)$ and
$l\left(t\right)\cap\left(M\backslash\overline{M_{t}}\right)\subset\partial\Omega.$

Note that $\partial M_{t}\subset l\left(t\right)$ and $u\leq f.$ It follows
that $f\geq t$ on $\partial
M_{t}=\partial\left(M\backslash\overline{M_{t}}\right).$ Since the function
$f$ is superharmonic on the bounded set $M\backslash\overline{M_{t}},$ the
strong maximum principle shows that $f>t$ in $M\backslash\overline{M_{t}}.$ In
particular, we have $f>t$ on $\overline{\Omega}\backslash\overline{M_{t}}.$

Now for $x_{1}\in\left(\partial\Omega\cap
D\left(2R\right)\right)\backslash\overline{M_{t}},$ we would have $u(x_{1})>t$
as $f>t$ on $\overline{\Omega}\backslash\overline{M_{t}}$ and $u=f$ on
$D\left(2R\right).$ This contradicts with $\partial\Omega\subset
l\left(t\right).$

Therefore, $\left(\partial\Omega\cap
D\left(2R\right)\right)\backslash\overline{M_{t}}=\emptyset.$ In conclusion,
we have

$l\left(t\right)\cap\left(M\backslash\overline{M_{t}}\right)\subset
M\backslash D\left(2R\right).$

This proves (2.14). ∎

Let us point out that $\left\\{u>t\right\\}$ may have other connected
components in addition to $L\left(t,\infty\right).$ We now estimate the size
of those components that lie in a fixed $L\left(\varepsilon,\infty\right).$
Hence, for fixed $0<\varepsilon<1,$ denote

(2.16)
$\widetilde{L}\left(t,\infty\right)=\left(\left\\{u>t\right\\}\backslash
L\left(t,\infty\right)\right)\cap L\left(\varepsilon,\infty\right)$

the union of all such connected components in
$L\left(\varepsilon,\infty\right)$ and its boundary

(2.17)
$\widetilde{l}\left(t\right)=\partial\widetilde{L}\left(t,\infty\right).$

The following result provides an area estimate for
$\widetilde{l}\left(t\right).$

######  Lemma 2.5.

Let $\left(M,g\right)$ be an $n$-dimensional complete Riemannian manifold with
Ricci curvature bounded from below by a constant. Then for any
$0<\varepsilon<1,$

$\int_{\varepsilon}^{1}\mathrm{Area}\left(\widetilde{l}\left(t\right)\right)dt\leq\frac{C}{\sqrt{R\varepsilon^{3}}}\int_{L\left(\varepsilon,1\right)}u^{2},$

where $C$ is a constant independent of $R$ and $\varepsilon$, and
$\widetilde{l}\left(t\right)$ is defined in (2.17).

###### Proof.

Applying the divergence theorem, for any regular set
$\widetilde{l}\left(t\right)$ we have

$\displaystyle\int_{\widetilde{l}\left(t\right)}\left|\nabla u\right|$
$\displaystyle=$
$\displaystyle-\int_{\widetilde{L}\left(t,\infty\right)}\Delta u$
$\displaystyle\leq$
$\displaystyle\int_{\widetilde{L}\left(t,\infty\right)}\left|\Delta u\right|$
$\displaystyle\leq$
$\displaystyle\frac{C}{R}\mathrm{Vol}\left(\widetilde{L}\left(t,\infty\right)\right),$

where the last line follows from Lemma 2.3. Hence, for any $\varepsilon<t<1,$

$\displaystyle\int_{\widetilde{l}\left(t\right)}\left|\nabla u\right|$
$\displaystyle\leq$
$\displaystyle\frac{C}{R\varepsilon^{2}}\int_{\widetilde{L}\left(t,\infty\right)}u^{2}$
$\displaystyle\leq$
$\displaystyle\frac{C}{R\varepsilon^{2}}\int_{L\left(\varepsilon,1\right)}u^{2}.$

Since

$1\leq\frac{1}{2}\sqrt{R\varepsilon^{3}}\frac{\left|\nabla
u\right|}{u^{2}}+\frac{1}{2\sqrt{R\varepsilon^{3}}}\frac{u^{2}}{\left|\nabla
u\right|},$

it follows that

(2.19)
$\mathrm{Area}\left(\widetilde{l}\left(t\right)\right)\leq\frac{1}{2}\sqrt{R\varepsilon^{3}}\int_{\widetilde{l}\left(t\right)}\frac{\left|\nabla
u\right|}{u^{2}}+\frac{1}{2\sqrt{R\varepsilon^{3}}}\int_{\widetilde{l}\left(t\right)}\frac{u^{2}}{\left|\nabla
u\right|}.$

By (2) we have that

$\displaystyle\int_{\widetilde{l}\left(t\right)}\frac{\left|\nabla
u\right|}{u^{2}}$ $\displaystyle=$
$\displaystyle\frac{1}{t^{2}}\int_{\widetilde{l}\left(t\right)}\left|\nabla
u\right|$ $\displaystyle\leq$
$\displaystyle\frac{C}{R\varepsilon^{2}}\left(\int_{L\left(\varepsilon,1\right)}u^{2}\right)\frac{1}{t^{2}}.$

After integrating from $t=\varepsilon$ to $t=1,$ we obtain

(2.20)
$\int_{\varepsilon}^{1}\int_{\widetilde{l}\left(t\right)}\frac{\left|\nabla
u\right|}{u^{2}}dt\leq\frac{C}{R\varepsilon^{3}}\left(\int_{L\left(\varepsilon,1\right)}u^{2}\right).$

Using the co-area formula we estimate

$\displaystyle\int_{\varepsilon}^{1}\int_{\widetilde{l}\left(t\right)}\frac{u^{2}}{\left|\nabla
u\right|}dt$ $\displaystyle\leq$
$\displaystyle\int_{\varepsilon}^{1}\int_{\left\\{u=t\right\\}\cap
L\left(\varepsilon,1\right)}\frac{u^{2}}{\left|\nabla u\right|}dt$
$\displaystyle=$ $\displaystyle\int_{L\left(\varepsilon,1\right)}u^{2}.$

Now integrating (2.19) in $t$ and applying (2.20) and (2) we arrive at

$\int_{\varepsilon}^{1}\mathrm{Area}\left(\widetilde{l}\left(t\right)\right)dt\leq\frac{C}{\sqrt{R\varepsilon^{3}}}\int_{L\left(\varepsilon,1\right)}u^{2}$

as claimed. ∎

We now restrict to three-dimensional manifolds. The following result relies on
an idea from [34] and follows as in [18] or Lemma 4.1 in [4]. For the sake of
completeness, we include details here.

######  Lemma 2.6.

Let $\left(M,g\right)$ be a three-dimensional complete Riemannian manifold
with scalar curvature $S$ and $u$ a smooth function on $M.$ Then on each
regular level set $\\{u=t\\}$ of $u,$

$\displaystyle\mathrm{Ric}\left(\nabla u,\nabla u\right)\left|\nabla
u\right|^{-2}$ $\displaystyle=$
$\displaystyle\frac{1}{2}S-\frac{1}{2}S_{t}+\frac{1}{2}\frac{1}{\left|\nabla
u\right|^{2}}\left(\left|\nabla\left|\nabla
u\right|\right|^{2}-\left|\nabla^{2}u\right|^{2}\right)$
$\displaystyle+\frac{1}{2}\frac{1}{\left|\nabla
u\right|^{2}}\left(\left(\Delta
u\right)^{2}-2\frac{\left\langle\nabla\left|\nabla u\right|,\nabla
u\right\rangle}{\left|\nabla u\right|}\Delta u+\left|\nabla\left|\nabla
u\right|\right|^{2}\right),$

where $S_{t}$ denotes the scalar curvature of the surface $\\{u=t\\}.$

###### Proof.

On a regular level set $\\{u=t\\}$ of $u,$ its unit normal vector is given by

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

Choose unit vectors $\left\\{e_{2},e_{3}\right\\}$ tangent to $\\{u=t\\}$ such
that $\left\\{e_{1},e_{2},e_{3}\right\\}$ forms a local orthonormal frame on
$M.$ The second fundamental form and the mean curvature of $\\{u=t\\}$ are
then given by

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

where indices $a$ and $b$ range from $2$ to $3.$ By the Gauss curvature
equation, we have

$S_{t}=S-2R_{11}+H^{2}-\left|h\right|^{2}.$

Therefore,

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

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. ∎

Finally, we recall the following well known Kato inequality for harmonic
functions (cf. [30]).

######  Lemma 2.7.

Let $\left(M,g\right)$ be a three-dimensional Riemannian manifold and $f$ a
harmonic function on $M\backslash D.$ Then

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

##  3\. Upper bound of bottom spectrum

We now turn to the proof of Theorem 1.1 which is restated below. We continue
to follow the notations from Section 2.

######  Theorem 3.1.

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.
Then the bottom spectrum of the Laplacian satisfies

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

###### Proof.

For given small $0<\varepsilon<1,$ let $R>R_{0}$ and

(3.1) $R>\frac{1}{\varepsilon^{4}},$

where $R_{0}$ is chosen to be large enough so that $D\subset D(R_{0})$ with
$D$ being a smooth connected bounded domain that contains all representatives
of $H_{1}\left(M\right)$ and that $M\backslash D$ has the maximal number $m$
of ends, while $D(t)$ is as defined in (2.6).

In the following, we use $C$ to denote a positive constant that is independent
of $R$ and $\varepsilon,$ while its value may change from line to line.

To prove the theorem, we may assume that $\lambda_{1}\left(M\right)>0.$ In
particular, $\left(M,g\right)$ is nonparabolic. Let $f$ be the Li-Tam barrier
function defined by (2.7). Set

$u=f\psi$

as in (2.9). Everywhere we denote with

$L\left(a,b\right)=L\left(a,\infty\right)\cap\left\\{a<u<b\right\\}.$

Recall that from [23],

$\int_{M\backslash B_{p}\left(r\right)}f^{2}\left(x\right)dx\leq
C\,e^{-2\sqrt{\lambda_{1}\left(M\right)}r}$

for all $r$ with $D\subset B_{p}(r).$ Therefore,

$\int_{M\backslash D}f^{2}\left(x\right)dx\leq C$

and by (2.1),

(3.2) $\int_{M\backslash D\left(r\right)}f^{2}\left(x\right)dx\leq
C\,e^{-\frac{1}{C}r}$

for all $r>R_{0}.$

Using (3.2) and that $f\geq u,$ we get

$\displaystyle\mathrm{Vol}\left(L\left(\varepsilon,1\right)\right)$
$\displaystyle\leq$
$\displaystyle\frac{1}{\varepsilon^{2}}\int_{L\left(\varepsilon,1\right)}u^{2}$
$\displaystyle\leq$ $\displaystyle\frac{1}{\varepsilon^{2}}\int_{M\backslash
D}f^{2}$ $\displaystyle\leq$ $\displaystyle\frac{C}{\varepsilon^{2}}.$

Similarly,

$\displaystyle\mathrm{Vol}\left(\left(M\backslash D\left(R\right)\right)\cap
L\left(\varepsilon,1\right)\right)$ $\displaystyle\leq$
$\displaystyle\frac{1}{\varepsilon^{2}}\int_{\left(M\backslash
D\left(R\right)\right)\cap L\left(\varepsilon,1\right)}u^{2}\left(x\right)dx$
$\displaystyle\leq$ $\displaystyle\frac{1}{\varepsilon^{2}}\int_{M\backslash
D\left(R\right)}f^{2}\left(x\right)dx$ $\displaystyle\leq$
$\displaystyle\frac{C}{\varepsilon^{2}}e^{-\frac{1}{C}R}$ $\displaystyle\leq$
$\displaystyle C,$

where the last line follows from (3.1).

By Lemma 2.3 and (3), for $\varepsilon<t<1,$

$\displaystyle\int_{L\left(t,1\right)}\left|\Delta u\right|$
$\displaystyle\leq$
$\displaystyle\frac{C}{R}\mathrm{Vol}\left(L\left(\varepsilon,1\right)\right)$
$\displaystyle\leq$ $\displaystyle\frac{C}{R\varepsilon^{2}}$
$\displaystyle\leq$ $\displaystyle C,$

where the last line is from (3.1). On the other hand, the divergence theorem
implies that

$\displaystyle\int_{L\left(t,1\right)}\Delta u$ $\displaystyle=$
$\displaystyle-\int_{\partial D}\frac{\partial
f}{\partial\nu}-\int_{l\left(t\right)}\left|\nabla u\right|$ $\displaystyle=$
$\displaystyle C-\int_{l\left(t\right)}\left|\nabla u\right|,$

where we have used that $u=f$ on $D\left(2R\right)\backslash D$ and $\nu$ is
the outward unit normal vector to $\partial D.$ In conclusion,

(3.5) $\int_{l\left(t\right)}\left|\nabla u\right|\leq C$

for all $\varepsilon<t<1.$

Let

$\phi(x)=\left\\{\begin{array}[]{c}\phi\left(u\left(x\right)\right)\\\
0\end{array}\right.\begin{array}[]{c}\text{on
}L\left(\varepsilon,\infty\right)\\\ \text{on }M\backslash
L\left(\varepsilon,\infty\right).\end{array}$

Here, the function $\phi\left(t\right)$ is smooth, $\phi\left(t\right)=1$ for
$2\varepsilon\leq t\leq 1$ and $\phi\left(t\right)=0$ for $t<\varepsilon.$
Moreover,

(3.6) $\left|\phi^{\prime}\right|\left(t\right)\leq\frac{C}{\varepsilon}\text{
\ and
}\left|\phi^{\prime\prime}\right|\left(t\right)\leq\frac{C}{\varepsilon^{2}}\text{
\ for }\varepsilon<t<2\varepsilon.$

Clearly, by the definition, the function $\phi\left(x\right)$ satisfies
$\phi=1$ on $L\left(2\varepsilon,\infty\right)$ and $\phi=0$ on $M\setminus
L\left(\varepsilon,\infty\right).$ Furthermore,

(3.7) $\left|\nabla\phi\right|\leq C.$

Indeed, $\nabla\phi=0$ except for $L\left(\varepsilon,2\varepsilon\right).$ On
$L\left(\varepsilon,2\varepsilon\right),$ according to Lemma 2.3 and (3.1),

$\displaystyle\left|\nabla u\right|$ $\displaystyle\leq$ $\displaystyle
C\,u+\frac{C}{R}$ $\displaystyle\leq$ $\displaystyle C\,\varepsilon$

and

$\left|\nabla\phi\right|\leq\frac{C}{\varepsilon}\left|\nabla u\right|\leq C.$

We first prove the following inequality.

$\text{{Claim 1: \ \ \ \ }}\frac{3}{4}\int_{D\left(R\right)\backslash
D}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-1}\phi^{2}\leq 3K\int_{D\left(\frac{1}{2}R\right)\backslash
D}\left|\nabla u\right|\phi^{2}+C.$

According to the Bochner formula,

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

on $M\backslash D.$ Note that

$\left\langle\nabla\Delta u,\nabla u\right\rangle\geq-\left|\nabla\left(\Delta
u\right)\right|\left|\nabla u\right|.$

Therefore,

(3.9) $\Delta\left|\nabla
u\right|\geq\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-1}+\mathrm{Ric}\left(\nabla
u,\nabla u\right)\left|\nabla u\right|^{-1}-\left|\nabla\left(\Delta
u\right)\right|$

holds on $M\backslash D$ whenever $\left|\nabla u\right|\neq 0.$

From (3.9), we have

$\displaystyle\int_{M\backslash D}\left(\Delta\left|\nabla
u\right|\right)\phi^{2}$ $\displaystyle\geq$ $\displaystyle\int_{M\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}$
$\displaystyle-\int_{M\backslash D}\left|\nabla\left(\Delta
u\right)\right|\phi^{2}.$

We estimate the last term. According to Lemma 2.3,

(3.11) $\int_{M\backslash D}\left|\nabla\left(\Delta
u\right)\right|\phi^{2}\leq\frac{C}{R}\int_{M\backslash
D}\left(\left|\nabla^{2}\rho\right|^{2}+\left|\nabla^{2}f\right|^{2}+1\right)\phi^{2}.$

Integrating by parts gives

$\displaystyle\int_{M\backslash D}\left|\nabla^{2}\rho\right|^{2}\phi^{2}$
$\displaystyle=$ $\displaystyle-\int_{M\backslash
D}\left\langle\nabla\left(\Delta\rho\right),\nabla\rho\right\rangle\phi^{2}-\int_{M\backslash
D}\mathrm{Ric}\left(\nabla\rho,\nabla\rho\right)\phi^{2}$
$\displaystyle-2\int_{M\backslash
D}\rho_{ij}\rho_{i}\phi_{j}\phi-\int_{\partial
D}\rho_{ij}\rho_{i}\nu_{j}\phi^{2}.$

By Lemma 2.1,

$\displaystyle-\int_{M\backslash
D}\left\langle\nabla\left(\Delta\rho\right),\nabla\rho\right\rangle\phi^{2}$
$\displaystyle\leq$ $\displaystyle C\int_{M\backslash
D}\left|\nabla^{2}\rho\right|\phi^{2}$ $\displaystyle\leq$
$\displaystyle\frac{1}{4}\int_{M\backslash
D}\left|\nabla^{2}\rho\right|^{2}\phi^{2}+C\,\int_{M\backslash D}\phi^{2}$
$\displaystyle\leq$ $\displaystyle\frac{1}{4}\int_{M\backslash
D}\left|\nabla^{2}\rho\right|^{2}\phi^{2}+\frac{C}{\varepsilon^{2}},$

where in the last line we have used (3). Since Ricci curvature is bounded from
below by a constant, we have

$\displaystyle-\int_{M\backslash
D}\mathrm{Ric}\left(\nabla\rho,\nabla\rho\right)\phi^{2}$ $\displaystyle\leq$
$\displaystyle C\int_{M\backslash D}\left|\nabla\rho\right|^{2}\phi^{2}$
$\displaystyle\leq$ $\displaystyle C\int_{M\backslash D}\phi^{2}$
$\displaystyle\leq$ $\displaystyle\frac{C}{\varepsilon^{2}}.$

Finally,

$\displaystyle-2\int_{M\backslash D}\rho_{ij}\rho_{i}\phi_{j}\phi$
$\displaystyle\leq$ $\displaystyle 2\int_{M\backslash
D}\left|\nabla^{2}\rho\right|\left|\nabla\phi\right|\phi$ $\displaystyle\leq$
$\displaystyle\frac{1}{4}\int_{M\backslash
D}\left|\nabla^{2}\rho\right|^{2}\phi^{2}+4\int_{L\left(\varepsilon,1\right)}\left|\nabla\phi\right|^{2}.$

Using (3) and (3.7) we have

$\int_{L\left(\varepsilon,1\right)}\left|\nabla\phi\right|^{2}\leq\frac{C}{\varepsilon^{2}}.$

This shows that

$-2\int_{M\backslash
D}\rho_{ij}\rho_{i}\phi_{j}\phi\leq\frac{1}{4}\int_{M\backslash
D}\left|\nabla^{2}\rho\right|^{2}\phi^{2}+\frac{C}{\varepsilon^{2}}.$

Certainly,

$\left|\int_{\partial D}\rho_{ij}\rho_{i}\nu_{j}\phi^{2}\right|\leq C.$

Plugging these estimates into (3) implies that

$\int_{M\backslash
D}\left|\nabla^{2}\rho\right|^{2}\phi^{2}\leq\frac{C}{\varepsilon^{2}}.$

Similarly, we see that

$\int_{M\backslash D}\left|\nabla^{2}f\right|^{2}\phi^{2}\leq C.$

Indeed, as $f$ is harmonic on $M\setminus D,$

$\displaystyle\int_{M\backslash D}\left|\nabla^{2}f\right|^{2}\phi^{2}$
$\displaystyle=$ $\displaystyle-\int_{M\backslash D}\mathrm{Ric}\left(\nabla
f,\nabla f\right)\phi^{2}$ $\displaystyle-2\int_{M\backslash
D}f_{ij}f_{i}\phi_{j}\phi-\int_{\partial D}f_{ij}f_{i}\nu_{j}\phi^{2}.$

Again, as the Ricci curvature is bounded from below by a constant, we have

$\displaystyle-\int_{M\backslash D}\mathrm{Ric}\left(\nabla f,\nabla
f\right)\phi^{2}$ $\displaystyle\leq$ $\displaystyle C\int_{M\backslash
D}\left|\nabla f\right|^{2}$ $\displaystyle\leq$ $\displaystyle C.$

Also,

$\displaystyle-2\int_{M\backslash D}f_{ij}f_{i}\phi_{j}\phi$
$\displaystyle\leq$ $\displaystyle 2\int_{M\backslash
D}\left|\nabla^{2}f\right|\left|\nabla f\right|\left|\nabla\phi\right|\phi$
$\displaystyle\leq$ $\displaystyle\frac{1}{4}\int_{M\backslash
D}\left|\nabla^{2}f\right|^{2}\phi^{2}+C\,\int_{M\backslash D}\left|\nabla
f\right|^{2}$ $\displaystyle\leq$ $\displaystyle\frac{1}{4}\int_{M\backslash
D}\left|\nabla^{2}f\right|^{2}\phi^{2}+C.$

This shows that

$\int_{M\backslash D}\left|\nabla^{2}f\right|^{2}\phi^{2}\leq C.$

We conclude from (3.11) and (3.1) that

$\displaystyle\int_{M\backslash D}\left|\nabla\left(\Delta
u\right)\right|\phi^{2}$ $\displaystyle\leq$
$\displaystyle\frac{C}{R\varepsilon^{2}}$ $\displaystyle\leq$ $\displaystyle
C.$

Thus, (3) becomes

$\displaystyle\int_{M\backslash D}\left(\Delta\left|\nabla
u\right|\right)\phi^{2}$ $\displaystyle\geq$ $\displaystyle\int_{M\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}-C.$

Using the co-area formula, we have

$\displaystyle\int_{M\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}$ $\displaystyle=$
$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{\left\\{u=t\right\\}}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle=$
$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt$
$\displaystyle+\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{\widetilde{l}\left(t\right)}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt,$

where $l(t)$ and $\widetilde{l}(t)$ were defined in (2.12) and (2.17),
respectively.

By Lemma 2.5 and Kato inequality we have

$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{\widetilde{l}\left(t\right)}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle\geq$
$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{\widetilde{l}\left(t\right)}\mathrm{Ric}\left(\nabla
u,\nabla u\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle\geq$
$\displaystyle-C\int_{\varepsilon}^{1}\mathrm{Area}\left(\widetilde{l}\left(t\right)\right)dt$
$\displaystyle\geq$
$\displaystyle-\frac{C}{\sqrt{R\varepsilon^{3}}}\int_{L\left(\varepsilon,1\right)}u^{2}.$

Hence, according to (3.2) and (3.1), it follows that

$\displaystyle\int_{M\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}$ $\displaystyle\geq$
$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle-C.$

Rewrite it into

$\displaystyle\int_{M\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}$ $\displaystyle\geq$
$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\overline{M_{t}}}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt$
$\displaystyle+\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\left(M\backslash\overline{M_{t}}\right)}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle-C.$

Recall that from Lemma 2.4,

$l\left(t\right)\cap\left(M\backslash\overline{M_{t}}\right)\subset
M\backslash D\left(R\right).$

Since

$\mathrm{Ric}\left(\nabla u,\nabla u\right)\geq-C\left|\nabla u\right|^{2},$

together with the Kato inequality
$\left|u_{ij}\right|^{2}\geq\left|\nabla\left|\nabla u\right|\right|^{2},$ it
follows that

$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\left(M\backslash\overline{M_{t}}\right)}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle\geq$
$\displaystyle-C\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\left(M\backslash\overline{M_{t}}\right)}dt$
$\displaystyle\geq$
$\displaystyle-C\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\left(M\backslash
D\left(R\right)\right)}dt$ $\displaystyle\geq$
$\displaystyle-C\int_{M\backslash D\left(R\right)}\left|\nabla
u\right|\phi^{2}$ $\displaystyle\geq$ $\displaystyle-C\int_{M\backslash
D\left(R\right)}\phi^{2}.$

By (3) we have that $\int_{M\backslash D\left(R\right)}\phi^{2}\leq C$. In
conclusion,

$\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\left(M\backslash\overline{M_{t}}\right)}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt\geq-C$

and (3) becomes

$\displaystyle\int_{M\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}$ $\displaystyle\geq$
$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\overline{M_{t}}}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt-C.$

By Lemma 2.6,

$\displaystyle\mathrm{Ric}\left(\nabla u,\nabla u\right)\left|\nabla
u\right|^{-2}$ $\displaystyle=$
$\displaystyle\frac{1}{2}S-\frac{1}{2}S_{t}+\frac{1}{2}\frac{1}{\left|\nabla
u\right|^{2}}\left(\left|\nabla\left|\nabla
u\right|\right|^{2}-\left|u_{ij}\right|^{2}\right)$
$\displaystyle+\frac{1}{2}\frac{1}{\left|\nabla
u\right|^{2}}\left(\left(\Delta
u\right)^{2}-2\frac{\left\langle\nabla\left|\nabla u\right|,\nabla
u\right\rangle}{\left|\nabla u\right|}\Delta u+\left|\nabla\left|\nabla
u\right|\right|^{2}\right)$

on any regular level set $l\left(t\right).$ It follows that

$\displaystyle\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}$ $\displaystyle=$
$\displaystyle\frac{1}{2}S-\frac{1}{2}S_{t}+\frac{1}{2}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-2}$
$\displaystyle+\frac{1}{2}\left(\left(\Delta
u\right)^{2}-2\frac{\left\langle\nabla\left|\nabla u\right|,\nabla
u\right\rangle}{\left|\nabla u\right|}\Delta u+\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-2}.$

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

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

Invoking Lemma 2.4 that $l\left(t\right)\cap\overline{M_{t}}=\partial M_{t}$
has $m$ connected components, one has

$\int_{l\left(t\right)\cap\overline{M_{t}}}S_{t}\leq 8\pi m\leq C$

for all regular value $t$ of $u$ with $0<t<1.$ Therefore, (3) together with
the assumption $S\geq-6K$ implies that

$\displaystyle\int_{l\left(t\right)\cap\overline{M_{t}}}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}\int_{l\left(t\right)\cap\overline{M_{t}}}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-2}$
$\displaystyle+\frac{1}{2}\int_{l\left(t\right)\cap\overline{M_{t}}}\left(\left(\Delta
u\right)^{2}-2\frac{\left\langle\nabla\left|\nabla u\right|,\nabla
u\right\rangle}{\left|\nabla u\right|}\Delta u+\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-2}$
$\displaystyle-3K\mathrm{Area}\left(l\left(t\right)\cap\overline{M_{t}}\right)-C.$

So (3) becomes

$\displaystyle\int_{M\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\overline{M_{t}}}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-2}dt$
$\displaystyle+\frac{1}{2}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\overline{M_{t}}}\left(\left(\Delta
u\right)^{2}-2\frac{\left\langle\nabla\left|\nabla u\right|,\nabla
u\right\rangle}{\left|\nabla u\right|}\Delta u+\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-2}dt$
$\displaystyle-3K\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\mathrm{Area}\left(l\left(t\right)\cap\overline{M_{t}}\right)dt-C\int_{\varepsilon}^{1}\phi^{2}\left(t\right)dt-C.$

Note that

$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\mathrm{Area}\left(l\left(t\right)\cap\overline{M_{t}}\right)dt$
$\displaystyle\leq$
$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\mathrm{Area}\left(l\left(t\right)\right)dt$
$\displaystyle=$ $\displaystyle\int_{M\backslash D}\left|\nabla
u\right|\phi^{2}$

and that $\int_{\varepsilon}^{1}\phi^{2}\left(t\right)dt=C.$ Moreover, by
Lemma 2.3 and (3),

$\displaystyle\int_{M\backslash D}\left|\nabla u\right|\phi^{2}$
$\displaystyle=$ $\displaystyle\int_{D\left(R\right)\backslash D}\left|\nabla
u\right|\phi^{2}+\int_{M\backslash D\left(R\right)}\left|\nabla
u\right|\phi^{2}$ $\displaystyle\leq$
$\displaystyle\int_{D\left(R\right)\backslash D}\left|\nabla
u\right|\phi^{2}+C.$

Plugging into (3) we get

$\displaystyle\int_{M\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\overline{M_{t}}}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-2}dt$
$\displaystyle+\frac{1}{2}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\overline{M_{t}}}\left(\left(\Delta
u\right)^{2}-2\frac{\left\langle\nabla\left|\nabla u\right|,\nabla
u\right\rangle}{\left|\nabla u\right|}\Delta u+\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-2}dt$
$\displaystyle-3K\int_{D\left(R\right)\backslash D}\left|\nabla
u\right|\phi^{2}-C.$

Using the Kato inequality

$\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla u\right|\right|^{2}\geq 0$

and Lemma 2.4 that

$l\left(t\right)\cap D\left(R\right)\subset\overline{M_{t}},$

the first term on the right hand side can be estimated as

$\displaystyle\frac{1}{2}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\overline{M_{t}}}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap
D\left(R\right)}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{D\left(R\right)\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-1}\phi^{2}.$

Since $u=f$ on $D\left(R\right),$

$\Delta u=0\text{ \ and
}\left|u_{ij}\right|^{2}\geq\frac{3}{2}\left|\nabla\left|\nabla
u\right|\right|^{2}\text{ \ in }D\left(R\right)\backslash D$

by Lemma 2.7. It follows from above that

$\displaystyle\frac{1}{2}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\overline{M_{t}}}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle\geq$
$\displaystyle\frac{1}{4}\int_{D\left(R\right)\backslash
D}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-1}\phi^{2}.$

Note that

$\left(\Delta u\right)^{2}-2\frac{\left\langle\nabla\left|\nabla
u\right|,\nabla u\right\rangle}{\left|\nabla u\right|}\Delta
u+\left|\nabla\left|\nabla u\right|\right|^{2}\geq 0.$

By Lemma 2.4, the second term is bounded by

$\displaystyle\frac{1}{2}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\overline{M_{t}}}\left(\left(\Delta
u\right)^{2}-2\frac{\left\langle\nabla\left|\nabla u\right|,\nabla
u\right\rangle}{\left|\nabla u\right|}\Delta u+\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap
D\left(R\right)}\left(\left(\Delta
u\right)^{2}-2\frac{\left\langle\nabla\left|\nabla u\right|,\nabla
u\right\rangle}{\left|\nabla u\right|}\Delta u+\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{D\left(R\right)\backslash D}\left(\left(\Delta
u\right)^{2}-2\frac{\left\langle\nabla\left|\nabla u\right|,\nabla
u\right\rangle}{\left|\nabla u\right|}\Delta u+\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-1}\phi^{2}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{D\left(R\right)\backslash
D}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-1}\phi^{2},$

where we have used that $\Delta u=0$ on $D\left(R\right)\backslash D$ in the
last line.

In view of (3) and (3), (3) becomes

$\displaystyle\int_{M\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}$ $\displaystyle\geq$
$\displaystyle\frac{3}{4}\int_{D\left(R\right)\backslash
D}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-1}\phi^{2}-3K\int_{D\left(R\right)\backslash D}\left|\nabla
u\right|\phi^{2}-C.$

Plugging into (3) yields

$\displaystyle\int_{M\backslash D}\left(\Delta\left|\nabla
u\right|\right)\phi^{2}$ $\displaystyle\geq$
$\displaystyle\frac{3}{4}\int_{D\left(R\right)\backslash
D}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-1}\phi^{2}$ $\displaystyle-3K\int_{D\left(R\right)\backslash
D}\left|\nabla u\right|\phi^{2}-C.$

To estimate the left hand side, we integrate by parts and get

(3.23) $\int_{M\backslash D}\left(\Delta\left|\nabla
u\right|\right)\phi^{2}=\int_{M\backslash D}\left|\nabla
u\right|\Delta\phi^{2}-\int_{\partial D}\phi^{2}\left|\nabla u\right|_{\nu}.$

Since $u=f$ on $D\left(2R\right)\backslash D,$ we have $\left|\nabla
u\right|_{\nu}=\left|\nabla f\right|_{\nu},$ where $\nu$ is the unit outward
normal to $\partial D.$ Hence,

$\int_{\partial D}\phi^{2}\left|\nabla u\right|_{\nu}=\int_{\partial
D}\phi^{2}\left|\nabla f\right|_{\nu}=C.$

To estimate the first term in (3.23), note that $\Delta\phi^{2}=0$ on
$M\setminus L\left(\varepsilon,2\varepsilon\right).$ On
$L\left(\varepsilon,2\varepsilon\right)$, by (3.6) and Lemma 2.3, we have

$\displaystyle\left|\Delta\phi^{2}\right|$ $\displaystyle\leq$
$\displaystyle\frac{C}{\varepsilon^{2}}\,\left(\left|\Delta
u\right|+\left|\nabla u\right|^{2}\right)$ $\displaystyle\leq$
$\displaystyle\frac{C}{R\varepsilon^{2}}+\frac{C}{\varepsilon^{2}}\left|\nabla
u\right|^{2}$ $\displaystyle\leq$ $\displaystyle
C\varepsilon^{2}+\frac{C}{\varepsilon}\left|\nabla u\right|,$

where in the last line we have used (3.1) and (3). In view of (3), we get

$\displaystyle\int_{M\backslash D}\left|\nabla
u\right|\left|\Delta\phi^{2}\right|$ $\displaystyle\leq$ $\displaystyle
C\varepsilon^{2}\mathrm{Vol}\left(L\left(\varepsilon,1\right)\right)+\frac{C}{\varepsilon}\int_{L\left(\varepsilon,2\varepsilon\right)}\left|\nabla
u\right|^{2}$ $\displaystyle\leq$ $\displaystyle
C+\frac{C}{\varepsilon}\int_{L\left(\varepsilon,2\varepsilon\right)}\left|\nabla
u\right|^{2}.$

The co-area formula and (3.5) imply that

$\displaystyle\frac{1}{\varepsilon}\int_{L\left(\varepsilon,2\varepsilon\right)}\left|\nabla
u\right|^{2}$ $\displaystyle=$
$\displaystyle\frac{1}{\varepsilon}\int_{\varepsilon}^{2\varepsilon}\int_{l\left(t\right)}\left|\nabla
u\right|dt$ $\displaystyle\leq$ $\displaystyle C.$

Hence,

(3.24) $\int_{M\backslash D}\left|\nabla
u\right|\left|\Delta\phi^{2}\right|\leq C$

and (3.23) becomes

$\int_{M\backslash D}\left(\Delta\left|\nabla u\right|\right)\phi^{2}\leq C.$

Note that the same argument also shows

(3.25) $\int_{M\backslash D}\left|\nabla
u\right|\left|\nabla\phi\right|^{2}\leq C.$

We therefore conclude from (3) that

(3.26) $\frac{3}{4}\int_{D\left(R\right)\backslash D}\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-1}\phi^{2}\leq
3K\int_{D\left(R\right)\backslash D}\left|\nabla u\right|\phi^{2}+C.$

To complete the proof of Claim 1, observe that

$\int_{D\left(R\right)\backslash D}\left|\nabla
u\right|\phi^{2}=\int_{D\left(\frac{1}{2}R\right)\backslash D}\left|\nabla
u\right|\phi^{2}+\int_{D\left(R\right)\backslash
D\left(\frac{1}{2}R\right)}\left|\nabla u\right|\phi^{2}$

and that by (3),

$\int_{D\left(R\right)\backslash D\left(\frac{1}{2}R\right)}\left|\nabla
u\right|\phi^{2}\leq C.$

Therefore,

$\int_{D\left(R\right)\backslash D}\left|\nabla
u\right|\phi^{2}\leq\int_{D\left(\frac{1}{2}R\right)\backslash D}\left|\nabla
u\right|\phi^{2}+C$

and (3.26) becomes

(3.27) $\frac{3}{4}\int_{D\left(R\right)\backslash D}\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-1}\phi^{2}\leq
3K\int_{D\left(\frac{1}{2}R\right)\backslash D}\left|\nabla
u\right|\phi^{2}+C.$

This verifies Claim 1.

We now establish the following claim.

$\text{{Claim 2:\ \
}}\lambda_{1}\left(M\right)\int_{D\left(\frac{1}{2}R\right)\backslash
D}\left|\nabla u\right|\phi^{2}\leq\frac{1}{4}\int_{D\left(R\right)\backslash
D}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-1}\phi^{2}+C.$

Let $\sigma:\left(0,\infty\right)\rightarrow\mathbb{R}$ be a smooth cut-off
function so that $\sigma\left(t\right)=1$ for $t\leq\frac{1}{2}R$ and
$\sigma\left(t\right)=0$ for $t\geq R.$ Moreover, it satisfies

$\left|\sigma^{\prime}\right|+\left|\sigma^{\prime\prime}\right|\leq\frac{C}{R}.$

Denote

$\sigma\left(x\right)=\sigma\left(\rho\left(x\right)\right).$

By Lemma 2.1 we have

(3.28) $\left|\nabla\sigma\right|+\left|\Delta\sigma\right|\leq\frac{C}{R}.$

Let $\xi$ be a Lipschitz cut-off such that $\xi=0$ on $D\left(R_{0}\right)$
and $\xi=1$ on $M\backslash D\left(R_{0}+1\right).$

Since the function $\left|\nabla u\right|\phi^{2}\left(\xi\sigma\right)^{2}$
has compact support in $M,$ we have

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

By (3.28), (3), and (3.25), the second last term is bounded by

$\displaystyle\int_{M}\left|\nabla
u\right|\left|\nabla\left(\phi\xi\sigma\right)\right|^{2}$ $\displaystyle\leq$
$\displaystyle\frac{C}{R^{2}}\int_{M\backslash D}\left|\nabla
u\right|\phi^{2}$ $\displaystyle+\int_{M\backslash D}\left|\nabla
u\right|\left|\nabla\phi\right|^{2}$ $\displaystyle+\int_{M\backslash
D}\left|\nabla u\right|\left|\nabla\xi\right|^{2}$ $\displaystyle\leq$
$\displaystyle\frac{C}{R^{2}}\mathrm{Vol}\left(L(\varepsilon,1)\right)+C$
$\displaystyle+\int_{D\left(R_{0}+1\right)\backslash
D\left(R_{0}\right)}\left|\nabla f\right|\left|\nabla\xi\right|^{2}$
$\displaystyle\leq$ $\displaystyle C.$

For the last term in (3), we first perform integration by parts and get

$\displaystyle\frac{1}{2}\int_{M}\left\langle\nabla\left|\nabla
u\right|,\nabla\left(\phi\xi\sigma\right)^{2}\right\rangle$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{M}\left\langle\nabla\left|\nabla
u\right|,\nabla\sigma^{2}\right\rangle\xi^{2}\phi^{2}$
$\displaystyle+\frac{1}{2}\int_{M}\left\langle\nabla\left|\nabla
u\right|,\nabla\phi^{2}\right\rangle\xi^{2}\sigma^{2}+C$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\int_{M}\left|\nabla
u\right|\left(\Delta\sigma^{2}\right)\xi^{2}\phi^{2}$
$\displaystyle-\frac{1}{2}\int_{M}\left|\nabla
u\right|\left(\Delta\phi^{2}\right)\xi^{2}\sigma^{2}$
$\displaystyle-\int_{M}\left|\nabla
u\right|\left\langle\nabla\sigma^{2},\nabla\phi^{2}\right\rangle\xi^{2}+C.$

By (3.24),

$\int_{M}\left|\nabla u\right|\left|\Delta\phi^{2}\right|\xi^{2}\sigma^{2}\leq
C.$

Moreover, by (3.28) and (3),

$\displaystyle-\frac{1}{2}\int_{M}\left|\nabla
u\right|\left(\Delta\sigma^{2}\right)\xi^{2}\phi^{2}$ $\displaystyle\leq$
$\displaystyle\int_{M}\left|\nabla
u\right|\left(\sigma\left|\Delta\sigma\right|+\left|\nabla\sigma\right|^{2}\right)\xi^{2}\phi^{2}$
$\displaystyle\leq$ $\displaystyle\frac{C}{R}\int_{M\backslash D}\left|\nabla
u\right|\phi^{2}$ $\displaystyle\leq$
$\displaystyle\frac{C}{R}\mathrm{Vol}\left(L(\varepsilon,1)\right)$
$\displaystyle\leq$ $\displaystyle C.$

From (3.7) it is easy to see that

$\displaystyle-\int_{M}\left|\nabla
u\right|\left\langle\nabla\sigma^{2},\nabla\phi^{2}\right\rangle\xi^{2}$
$\displaystyle\leq$ $\displaystyle\frac{C}{R}\int_{M\backslash D}\left|\nabla
u\right|\phi$ $\displaystyle\leq$
$\displaystyle\frac{C}{R}\mathrm{Vol}\left(L(\varepsilon,1)\right)$
$\displaystyle\leq$ $\displaystyle C.$

In conclusion,

$\frac{1}{2}\int_{M}\left\langle\nabla\left|\nabla
u\right|,\nabla\left(\phi\xi\sigma\right)^{2}\right\rangle\leq C.$

Plugging these estimates into (3) we arrive at

$\lambda_{1}\left(M\right)\int_{M}\left|\nabla
u\right|\phi^{2}\xi^{2}\sigma^{2}\leq\frac{1}{4}\int_{M}\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-1}\phi^{2}\xi^{2}\sigma^{2}+C.$

In view of the definition of $\sigma$ and $\xi,$ we have

(3.30) $\lambda_{1}\left(M\right)\int_{D\left(\frac{1}{2}R\right)\backslash
D\left(R_{0}+1\right)}\left|\nabla
u\right|\phi^{2}\leq\frac{1}{4}\int_{D\left(R\right)\backslash
D}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-1}\phi^{2}+C.$

Since $u=f$ on $D\left(R_{0}+1\right)\backslash D,$

$\int_{D\left(R_{0}+1\right)\backslash D}\left|\nabla
u\right|\phi^{2}\leq\int_{D\left(R_{0}+1\right)\backslash D}\left|\nabla
f\right|=C.$

Claim 2 now follows from (3.30).

Combining Claim 1 and Claim 2, or (3.30) and (3.27), we conclude that

(3.31) $\lambda_{1}\left(M\right)\int_{D\left(\frac{1}{2}R\right)\backslash
D}\left|\nabla u\right|\phi^{2}\leq
K\int_{D\left(\frac{1}{2}R\right)\backslash D}\left|\nabla
u\right|\phi^{2}+C.$

We now show $\lambda_{1}\left(M\right)\leq K.$ Assume by contradiction that

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

We then infer from (3.31) that

(3.33) $\int_{D\left(\frac{1}{2}R\right)\cap
L\left(2\varepsilon,1\right)}\left|\nabla u\right|\leq C$

for any $\varepsilon>0$ and $R>\max\\{R_{0},\frac{1}{\varepsilon^{4}}\\}$ with
the constant $C$ independent of $\varepsilon$ and $R.$

For arbitrary large $r$ such that $D\subset B_{p}(r),$ let

$t\left(r\right)=\min_{x\in\partial B_{p}\left(r\right)}f\left(x\right).$

The maximum principle implies that

$B_{p}\left(r\right)\subset\left\\{f>t\left(r\right)\right\\}.$

So for $\varepsilon<\frac{1}{2}t\left(r\right)$ and all $R$ sufficiently
large,

$B_{p}\left(r\right)\backslash D\subset D\left(\frac{1}{2}R\right)\cap
L\left(2\varepsilon,1\right).$

It follows from (3.33) that, as $u=f$ on $B_{p}\left(r\right)\backslash D,$

$\int_{B_{p}\left(r\right)\backslash D}\left|\nabla f\right|\leq C.$

Since $r$ is arbitrary, this proves

(3.34) $\int_{M\backslash D}\left|\nabla f\right|\leq C.$

According to [25],

(3.35) $\int_{\left\\{f=t\right\\}}\left|\nabla f\right|=C.$

Applying (3.35) and the co-area formula we get

$\displaystyle\int_{\left\\{\delta<f<1\right\\}}\left|\nabla
f\right|^{2}f^{-1}$ $\displaystyle=$ $\displaystyle
C\int_{\delta}^{1}\frac{1}{t}dt$ $\displaystyle=$ $\displaystyle-C\ln\delta.$

In view of Lemma 2.2 and (3.34), we conclude

$-C\ln\delta=\int_{\left\\{\delta<f<1\right\\}}\left|\nabla
f\right|^{2}f^{-1}\leq C\int_{M\backslash D}\left|\nabla f\right|\leq C$

for any $\delta>0.$ This is a contradiction. Therefore, (3.32) does not hold.
This proves the theorem. ∎

##  4\. Preliminaries for asymptotically nonnegative Ricci curvature

In this section, in anticipation of the applications in the next section, we
establish some preliminary results for manifolds with asymptotically
nonnegative Ricci curvature. Mainly, we show the existence of a distance-like
function that behaves better than the function $\rho$ constructed in Section
2. As before, we use $r\left(x\right)=d\left(p,x\right)$ to denote the
distance from $x$ to a fixed point $p\in M.$

######  Lemma 4.1.

Let $\left(M,g\right)$ be an $n$-dimensional complete Riemannian manifold with
Ricci curvature bounded below by

$\mathrm{Ric}\geq-\frac{C}{r^{2}+1}.$

Then there exists a smooth proper function $\eta$ such that

(4.1) $\displaystyle\frac{1}{C}\ln\left(r+2\right)$ $\displaystyle\leq$
$\displaystyle\eta\leq C\ln\left(r+2\right)$
$\displaystyle\left|\nabla\eta\right|$ $\displaystyle\leq$
$\displaystyle\frac{C}{r+1}$ $\displaystyle\left|\Delta\eta\right|$
$\displaystyle\leq$ $\displaystyle\frac{C}{\left(r+1\right)^{2}}\ \ $

and

(4.2)
$\left|\nabla\left(\Delta\eta\right)\right|\leq\frac{C}{\left(r+1\right)^{3}}+\frac{C}{r+1}\left|\nabla^{2}\eta\right|$

for some constant $C>0.$

###### Proof.

The argument is similar to that of Lemma 2.1 with the details spelled out in
[3] except for (4.2). For completeness, we include them here. For $\lambda>0$
to be determined later, we solve the boundary value problem.

$\displaystyle\Delta h_{R}$ $\displaystyle=$
$\displaystyle\frac{\lambda}{\rho^{2}+1}h_{R}\text{ \ in
}D\left(R\right)\backslash D\left(1\right)$ $\displaystyle h_{R}$
$\displaystyle=$ $\displaystyle 1\text{ \ on }\partial D\left(1\right)$
$\displaystyle h_{R}$ $\displaystyle=$ $\displaystyle 0\text{ \ on }\partial
D\left(R\right),$

where $\rho$ is defined in Lemma 2.1 and $D(R)$ in (2.6).

The maximum principle implies that $0<h_{R}<1$ and that $h_{R}$ is increasing
in $R.$ Letting $R\rightarrow\infty,$ we obtain a positive function $h$ on
$M\backslash D\left(1\right)$ satisfying

(4.3) $\displaystyle\Delta h$ $\displaystyle=$
$\displaystyle\frac{\lambda}{\rho^{2}+1}h\text{ \ in }M\backslash
D\left(1\right)$ $\displaystyle h$ $\displaystyle=$ $\displaystyle 1\text{ \
on }\partial D\left(1\right).$

By Cheng-Yau’s gradient estimate [8], it is easy to see that

(4.4) $\left|\nabla\ln h\right|\leq\frac{C\left(\lambda\right)}{\rho+1}.$

For any $m>0,$ note that

$\displaystyle\frac{\lambda}{2}\int_{D\left(R\right)\backslash
D\left(1\right)}\rho^{m-2}h_{R}^{2}$ $\displaystyle\leq$
$\displaystyle\int_{D\left(R\right)\backslash
D\left(1\right)}\frac{\lambda}{\rho^{2}+1}\rho^{m}h_{R}^{2}$ $\displaystyle=$
$\displaystyle\int_{D\left(R\right)\backslash
D\left(1\right)}\rho^{m}h_{R}\Delta h_{R}$ $\displaystyle=$
$\displaystyle-\int_{D\left(R\right)\backslash
D\left(1\right)}\rho^{m}\left|\nabla h_{R}\right|^{2}$
$\displaystyle-m\int_{D\left(R\right)\backslash
D\left(1\right)}\left\langle\nabla h_{R},\nabla\rho\right\rangle
h_{R}\rho^{m-1}$ $\displaystyle-$ $\displaystyle\int_{\partial
D\left(1\right)}h_{R}\frac{\partial h_{R}}{\partial\nu}.$

Since

$\displaystyle m\int_{D\left(R\right)\backslash
D\left(1\right)}\left|\left\langle\nabla
h_{R},\nabla\rho\right\rangle\right|h_{R}\rho^{m-1}$ $\displaystyle\leq$
$\displaystyle Cm\int_{D\left(R\right)\backslash D\left(1\right)}\left|\nabla
h_{R}\right|h_{R}\rho^{m-1}$ $\displaystyle\leq$
$\displaystyle\int_{D\left(R\right)\backslash
D\left(1\right)}\rho^{m}\left|\nabla h_{R}\right|^{2}$
$\displaystyle+Cm^{2}\int_{D\left(R\right)\backslash
D\left(1\right)}\rho^{m-2}h_{R}^{2},$

it follows that

$\int_{D\left(R\right)\backslash D\left(1\right)}\rho^{m-2}h_{R}^{2}\leq C$

by taking $\lambda>C\,m^{2}$ for some large $C.$ This shows that

$\int_{M}\rho^{m-2}h^{2}\leq C.$

Together with (4.4) and Lemma 2.1 we get that

$\displaystyle h^{2}\left(x\right)$ $\displaystyle\leq$
$\displaystyle\frac{C}{\mathrm{V}_{x}\left(1\right)}\int_{B_{x}\left(1\right)}h^{2}$
$\displaystyle\leq$
$\displaystyle\frac{C}{\mathrm{V}_{x}\left(1\right)}\frac{1}{R^{m-2}}$

for $x\in\partial B_{p}\left(R\right).$

The assumption that Ricci curvature is bounded below by
$\mathrm{Ric}\geq-\frac{C}{r^{2}+1}$ implies that (cf. [3])

$\frac{C}{\mathrm{V}_{x}\left(1\right)}\leq R^{C}.$

By arranging $m$ to be sufficiently large, we conclude that $h$ decays
polynomially and that

(4.5) $h\left(x\right)\leq\rho^{-\frac{m}{2}}\left(x\right)\text{ \ on
}M\backslash D\left(1\right).$

Now the desired function $\eta$ is defined as

$\eta\left(x\right)=\left(1-\xi\left(x\right)\right)\left(-\ln
h\left(x\right)\right)+\xi\left(x\right),$

where $\xi$ is a smooth cut-off function with $\xi=1$ on $D\left(1\right)$ and
$\xi=0$ on $M\backslash D\left(2\right).$

By (4.5) we have that

$\eta\geq\frac{1}{C}\ln\left(r+2\right).$

On the other hand, (4.4) implies that

$\eta\leq C\ln\left(r+2\right)$

and

$\left|\nabla\eta\right|\leq\frac{C}{r+1}.$

Furthermore, together with (4.3), we get

$\left|\Delta\eta\right|\leq\frac{C}{\left(r+1\right)^{2}}.$

Finally, using (4.3) again that

$\Delta\eta=-\frac{\lambda}{\rho^{2}+1}+\left|\nabla\eta\right|^{2}\text{ on
}M\backslash D\left(2\right),$

we conclude

$\left|\nabla\left(\Delta\eta\right)\right|\leq\frac{C}{\left(r+1\right)^{3}}+\frac{C}{r+1}\left|\nabla^{2}\eta\right|.$

This proves the result. ∎

For the remaining part of the paper $\eta$, rather than $\rho$, will be our
choice of smooth proper function. We henceforth denote with

(4.6) $D\left(t\right)=\left\\{x\in M:\eta\left(x\right)<t\right\\}.$

As in Section 2, we use $\eta$ to construct a cut-off function on $M.$ For
given $R>0,$ let $\psi:\left(0,\infty\right)\rightarrow\mathbb{R}$ be a smooth
function such that $\psi=1$ on $\left(0,\ln R\right)$ and $\psi=0$ on
$\left(2\ln R,\infty\right),$ and

$\left|\psi^{\prime}\right|\leq\frac{C}{\ln R},\ \
\left|\psi^{\prime\prime}\right|\leq\frac{C}{\ln^{2}R}\text{ and
}\left|\psi^{\prime\prime\prime}\right|\leq\frac{C}{\ln^{3}R}.$

Composing it with $\eta,$ we obtain a cut-off function

$\psi\left(x\right)=\psi\left(\eta\left(x\right)\right).$

Obviously,

$\displaystyle\psi$ $\displaystyle=$ $\displaystyle 1\text{ on }D\left(\ln
R\right)$ $\displaystyle\psi$ $\displaystyle=$ $\displaystyle 0\text{ on
}M\backslash D\left(2\ln R\right).$

In the following we assume that $\left(M,g\right)$ is nonparabolic and let $f$
be a barrier function on $M\setminus D$ as constructed in Section 2, where $D$
is a smooth connected bounded domain in $M.$ Again, we extend $f$ to $M$ by
setting $f=1$ on $D.$ The next result is well-known [8].

######  Lemma 4.2.

Let $\left(M,g\right)$ be an $n$-dimensional complete noncompact Riemannian
manifold with Ricci curvature bounded below by

$\mathrm{Ric}\geq-\frac{C}{r^{2}+1}.$

Then

$\left|\nabla\ln f\right|\leq\frac{C}{r}\text{ \ on }M\backslash D.$

We now consider the function $u$ given by

(4.7) $u=f\psi.$

######  Lemma 4.3.

Let $\left(M,g\right)$ be an $n$-dimensional complete noncompact Riemannian
manifold with Ricci curvature bounded below by

$\mathrm{Ric}\geq-\frac{C}{r^{2}+1}.$

Then $u$ is harmonic on $D\left(\ln R\right)\backslash D$ and on $M\backslash
D,$

$\displaystyle\left|\nabla u\right|$ $\displaystyle\leq$ $\displaystyle
u\,\left|\nabla\ln f\right|+\frac{C}{r}\frac{1}{\ln R}$
$\displaystyle\left|\Delta u\right|$ $\displaystyle\leq$
$\displaystyle\frac{C}{r^{2}\ln R}+\frac{C}{\ln R}\left|\nabla f\right|^{2}$
$\displaystyle\left|\nabla\left(\Delta u\right)\right|$ $\displaystyle\leq$
$\displaystyle\frac{Cr}{\ln
R}\left(\left|\nabla^{2}\eta\right|^{2}+\left|\nabla^{2}f\right|^{2}\right)+\frac{C}{r^{3}\ln
R}+\frac{C}{\ln R}\left|\nabla f\right|^{2}.$

###### Proof.

By Lemma 4.1,

$\displaystyle\left|\nabla\psi\right|$ $\displaystyle\leq$
$\displaystyle\frac{C}{r\ln R},$ $\displaystyle\left|\Delta\psi\right|$
$\displaystyle\leq$ $\displaystyle\frac{C}{r^{2}\ln R}$

and

$\displaystyle\left|\nabla\left(\Delta\psi\right)\right|$ $\displaystyle=$
$\displaystyle\left|\nabla\left(\psi^{\prime}\Delta\eta+\psi^{\prime\prime}\left|\nabla\eta\right|^{2}\right)\right|$
$\displaystyle\leq$
$\displaystyle\left|\psi^{\prime}\right|\left|\nabla\left(\Delta\eta\right)\right|+\left|\Delta\eta\right|\left|\nabla\psi^{\prime}\right|+\left|\psi^{\prime\prime}\right|\left|\nabla\left|\nabla\eta\right|^{2}\right|+\left|\nabla\eta\right|^{2}\left|\nabla\psi^{\prime\prime}\right|$
$\displaystyle\leq$ $\displaystyle\frac{C}{\ln
R}\left(\frac{1}{r}\left|\nabla^{2}\eta\right|+\frac{1}{r^{3}}\right).$

It follows that

$\displaystyle\left|\nabla u\right|$ $\displaystyle\leq$
$\displaystyle\left|\nabla f\right|\psi+f\left|\nabla\psi\right|$
$\displaystyle\leq$ $\displaystyle u\,\left|\nabla\ln f\right|+\frac{C}{r\ln
R}$

on $M\backslash D.$ Moreover, $\Delta u=0$ on $D\left(\ln R\right)\backslash
D$ as $u=f$ in $D\left(\ln R\right).$ Since

$\Delta u=f\Delta\psi+2\left\langle\nabla f,\nabla\psi\right\rangle,$

we get from above that

$\displaystyle\left|\Delta u\right|$ $\displaystyle\leq$
$\displaystyle\frac{C}{r^{2}\ln R}+\frac{C}{r\ln R}\left|\nabla f\right|$
$\displaystyle\leq$ $\displaystyle\frac{C}{r^{2}\ln R}+\frac{C}{\ln
R}\left|\nabla f\right|^{2}.$

Finally,

$\displaystyle\left|\nabla\left(\Delta u\right)\right|$ $\displaystyle\leq$
$\displaystyle\left|\nabla\left(f\Delta\psi\right)\right|+2\left|\nabla\left(\left\langle\nabla\psi,\nabla
f\right\rangle\right)\right|$ $\displaystyle\leq$ $\displaystyle\left|\nabla
f\right|\left|\Delta\psi\right|+f\left|\nabla\left(\Delta\psi\right)\right|+2\left|\nabla^{2}f\right|\left|\nabla\psi\right|$
$\displaystyle+2\left|\nabla^{2}\psi\right|\left|\nabla f\right|$
$\displaystyle\leq$ $\displaystyle\frac{C}{r\ln
R}\left(\left|\nabla^{2}\eta\right|+\left|\nabla^{2}f\right|+\frac{1}{r^{2}}\right)$
$\displaystyle+\frac{C}{r^{2}\ln R}\left|\nabla f\right|+\frac{C}{\ln
R}\left|\nabla f\right|\left|\nabla^{2}\eta\right|,$

where we have used that

$\displaystyle\left|\nabla^{2}\psi\right|$ $\displaystyle\leq$
$\displaystyle\left|\psi^{\prime}\right|\left|\nabla^{2}\eta\right|+\left|\psi^{\prime\prime}\right|\left|\nabla\eta\right|^{2}$
$\displaystyle\leq$ $\displaystyle\frac{C}{\ln
R}\left|\nabla^{2}\eta\right|+\frac{C}{r^{2}\ln^{2}R}.$

Consequently,

$\left|\nabla\left(\Delta u\right)\right|\leq\frac{Cr}{\ln
R}\left(\left|\nabla^{2}\eta\right|^{2}+\left|\nabla^{2}f\right|^{2}\right)+\frac{C}{r^{3}\ln
R}+\frac{C}{\ln R}\left|\nabla f\right|^{2}.$

This proves the lemma. ∎

In the case that $M$ has finite first Betti number and finitely many, say $m,$
ends, we choose $D$ to be large enough so that all representatives of the
first homology group $H_{1}\left(M\right)$ are included in $D.$ Moreover,
$M\backslash D$ has exactly $m$ unbounded connected components. Let
$L\left(t,\infty\right)$ be the connected component of the super-level set
$\left\\{u>t\right\\}$ that contains $D$ and

$l\left(t\right)=\partial L\left(t,\infty\right).$

The same argument as Lemma 2.4 gives the following.

######  Lemma 4.4.

Let $\left(M,g\right)$ be an $n$-dimensional complete Riemannian manifold with
$m$ ends and finite first Betti number $b_{1}(M).$ Assume that the Ricci
curvature is bounded below by

$\mathrm{Ric}\geq-\frac{C}{r^{2}+1}.$

Then for all $0<t<1,$

$l\left(t\right)\cap\overline{M_{t}}=\partial M_{t}\text{ has }m\text{
connected components}$

and

$l\left(t\right)\cap\left(M\backslash\overline{M_{t}}\right)\subset
M\backslash D\left(\ln R\right),$

where $M_{t}$ is the union of all unbounded connected components of
$M\backslash\overline{L\left(t,\infty\right)}.$

##  5\. Positive scalar curvature

In this section, we study the geometry of three-dimensional complete manifolds
with positive scalar curvature and prove Theorem 1.3 and Theorem 1.4. We
continue to follow the notations from Section 4.

Recall that an $n$-dimensional manifold $M$ has asymptotically nonnegative
Ricci curvature if its Ricci curvature is bounded by

$\mathrm{Ric}(x)\geq-k\left(r(x)\right)$

for a continuous nonincreasing function
$k:\left[0,\infty\right)\rightarrow\left[0,\infty\right)$ with

$\int_{0}^{\infty}rk\left(r\right)dr<\infty.$

It is well known [15] that $M$ has at most Euclidean area growth, that is, the
area $A_{p}(t)$ of geodesic sphere $\partial B_{p}(t)$ satisfies

(5.1) $\mathrm{A}_{p}\left(t\right)\leq C\,t^{n-1}$

for all $t>0.$ It is also clear that

$\mathrm{Ric}\geq-\frac{C}{r^{2}+1}.$

So the results from Section 4 are applicable to $M.$

In passing, we mention that Li and Tam [20] have shown that $M^{n}$ has
finitely many ends under the stronger assumption that
$\int_{0}^{\infty}r^{n-1}\,k\left(r\right)dr<\infty.$

We also note that according to Lemma 4.1,

(5.2) $D\left(2\ln R\right)\subset B_{p}\left(R^{C}\right)$

for some $C>0$, where $D(t)$ was defined in (4.6).

We are now ready to prove Theorem 1.3 which is restated below.

######  Theorem 5.1.

Let $\left(M,g\right)$ be a three-dimensional complete noncompact Riemannian
manifold with asymptotically nonnegative Ricci curvature. Suppose that its
scalar curvature satisfies

$\liminf_{x\rightarrow\infty}\,S\left(x\right)r\left(x\right)\geq C_{0}$

for some $C_{0}>0$ and that $M$ has finitely many ends and finite first Betti
number $b_{1}(M).$ Then $\left(M,g\right)$ is parabolic.

###### Proof.

Assume by contradiction that $\left(M,g\right)$ is nonparabolic. Let $f$ be a
barrier function on $M\backslash D,$ where the smooth connected compact domain
$D$ contains all representatives of $H_{1}\left(M\right)$ and $M\backslash D$
has $m$ ends. Note that the assumption on scalar curvature implies

(5.3) $S\geq\frac{1}{2}\frac{C_{0}}{r}\text{ \ on }M\backslash D.$

For given small $0<\varepsilon<1,$ consider $R$ large so that

(5.4) $\ln R>\frac{1}{\varepsilon^{2}}.$

In the following, constant $C>0$ is independent of $R$ and $\varepsilon.$ But
its value may change from line to line.

Define the function

$u=f\psi$

as in (4.7). Then by the Bochner formula and the inequality

$\left\langle\nabla\Delta u,\nabla u\right\rangle\geq-\left|\nabla\left(\Delta
u\right)\right|\left|\nabla u\right|,$

$\Delta\left|\nabla
u\right|\geq\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}\right)\left|\nabla u\right|^{-1}+\mathrm{Ric}\left(\nabla
u,\nabla u\right)\left|\nabla u\right|^{-1}-\left|\nabla\left(\Delta
u\right)\right|$

holds on $M\backslash D$ whenever $\left|\nabla u\right|\neq 0.$

Let $\phi\left(x\right)$ be the smooth cut-off function from the proof of
Theorem 3.1, that is,

$\phi(x)=\left\\{\begin{array}[]{c}\phi\left(u\left(x\right)\right)\\\
0\end{array}\right.\begin{array}[]{c}\text{on
}L\left(\varepsilon,\infty\right)\\\ \text{on }M\backslash
L\left(\varepsilon,\infty\right)\end{array}$

Here, the function $\phi\left(t\right)$ is smooth, $\phi\left(t\right)=1$ for
$2\varepsilon\leq t\leq 1$ and $\phi\left(t\right)=0$ for $t<\varepsilon.$
Moreover,

(5.5) $\left|\phi^{\prime}\right|\left(t\right)\leq\frac{C}{\varepsilon}\text{
\ and
}\left|\phi^{\prime\prime}\right|\left(t\right)\leq\frac{C}{\varepsilon^{2}}\text{
\ for }\varepsilon<t<2\varepsilon.$

Clearly, by definition, the function $\phi\left(x\right)$ satisfies $\phi=1$
on $L\left(2\varepsilon,\infty\right)$ and $\phi=0$ on $M\setminus
L\left(\varepsilon,\infty\right).$

It follows that

$\displaystyle\int_{M\backslash D}\left(\Delta\left|\nabla
u\right|\right)\phi^{2}$ $\displaystyle\geq$ $\displaystyle\int_{M\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}$
$\displaystyle-\int_{M\backslash D}\left|\nabla\left(\Delta
u\right)\right|\phi^{2}.$

To estimate the last term, we use Lemma 4.3 to obtain

$\displaystyle\int_{M\backslash D}\left|\nabla\left(\Delta
u\right)\right|\phi^{2}$ $\displaystyle\leq$ $\displaystyle\frac{C}{\ln
R}\int_{M\backslash D}\left|\nabla^{2}\eta\right|^{2}r\phi^{2}$
$\displaystyle+\frac{C}{\ln R}\int_{M\backslash
D}\left|\nabla^{2}f\right|^{2}r\phi^{2}$ $\displaystyle+\frac{C}{\ln
R}\int_{D\left(2\ln R\right)\backslash D}\frac{1}{r^{3}}$
$\displaystyle+\frac{C}{\ln R}\int_{M\backslash D}\left|\nabla f\right|^{2}.$

By (2.8), the last term is bounded. By (5.1) and (5.2) we have

$\displaystyle\frac{C}{\ln R}\int_{D\left(2\ln R\right)\setminus
D}\frac{1}{r^{3}}$ $\displaystyle\leq$ $\displaystyle\frac{C}{\ln
R}\int_{B_{p}\left(R^{C}\right)}\frac{1}{\left(r+1\right)^{3}}$
$\displaystyle\leq$ $\displaystyle\frac{C}{\ln
R}\int_{0}^{R^{C}}\frac{r^{2}}{\left(r+1\right)^{3}}\,dr$ $\displaystyle\leq$
$\displaystyle C.$

Now we deal with the first term in (5). For convenience, we extend $\phi$
everywhere on $M$ by setting $\phi(x)=1$ for $x\in D$. Integrating by parts
gives

$\displaystyle\int_{M}\left|\nabla^{2}\eta\right|^{2}\left(r+1\right)\phi^{2}$
$\displaystyle=$
$\displaystyle-\int_{M}\left\langle\nabla\left(\Delta\eta\right),\nabla\eta\right\rangle\left(r+1\right)\phi^{2}$
$\displaystyle-\int_{M}\mathrm{Ric}\left(\nabla\eta,\nabla\eta\right)\left(r+1\right)\phi^{2}$
$\displaystyle-\int_{M}\eta_{ij}\eta_{i}r_{j}\phi^{2}$
$\displaystyle-2\int_{M}\eta_{ij}\eta_{i}\phi_{j}(r+1)\phi.$

By Lemma 4.1

$\displaystyle-\int_{M}\left\langle\nabla\left(\Delta\eta\right),\nabla\eta\right\rangle\left(r+1\right)\phi^{2}$
$\displaystyle\leq$ $\displaystyle
C\int_{M}\frac{1}{r+1}\left|\nabla^{2}\eta\right|\phi^{2}+C\int_{D\left(2\ln
R\right)}\frac{1}{\left(r+1\right)^{3}}$ $\displaystyle\leq$
$\displaystyle\frac{1}{4}\int_{M}\left(r+1\right)\left|\nabla^{2}\eta\right|^{2}\phi^{2}+C\int_{D\left(2\ln
R\right)}\frac{1}{\left(r+1\right)^{3}}$ $\displaystyle\leq$
$\displaystyle\frac{1}{4}\int_{M}\left(r+1\right)\left|\nabla^{2}\eta\right|^{2}\phi^{2}+C\ln
R,$

where in the last line we have used (5). According to the Ricci lower bound,
we have

$\displaystyle-\int_{M}\mathrm{Ric}\left(\nabla\eta,\nabla\eta\right)\left(r+1\right)\phi^{2}$
$\displaystyle\leq$ $\displaystyle
C\int_{M}\frac{1}{r+1}\left|\nabla\eta\right|^{2}$ $\displaystyle\leq$
$\displaystyle C\int_{D\left(2\ln R\right)}\frac{1}{\left(r+1\right)^{3}}$
$\displaystyle\leq$ $\displaystyle C\ln R.$

Furthermore, we have

$\displaystyle-\int_{M}\eta_{ij}\eta_{i}r_{j}\phi^{2}$ $\displaystyle\leq$
$\displaystyle\frac{1}{4}\int_{M}\left(r+1\right)\left|\nabla^{2}\eta\right|^{2}\phi^{2}+\int_{D\left(2\ln
R\right)}\frac{1}{\left(r+1\right)^{3}}$ $\displaystyle\leq$
$\displaystyle\frac{1}{4}\int_{M}\left(r+1\right)\left|\nabla^{2}\eta\right|^{2}\phi^{2}+C\ln
R.$

Finally, we estimate the last term in (5) by

$\displaystyle-2\int_{M}\eta_{ij}\eta_{i}\phi_{j}\phi(r+1)$
$\displaystyle\leq$
$\displaystyle\frac{1}{4}\int_{M}\left(r+1\right)\left|\nabla^{2}\eta\right|^{2}\phi^{2}+4\int_{M}|\nabla\eta|^{2}|\nabla\phi|^{2}(r+1)$
$\displaystyle\leq$
$\displaystyle\frac{1}{4}\int_{M}\left(r+1\right)\left|\nabla^{2}\eta\right|^{2}\phi^{2}+C\int_{M}\frac{1}{r+1}|\nabla\phi|^{2}.$

Lemma 4.3 and Lemma 4.2 imply that on $L(\varepsilon,2\varepsilon)$,

$\displaystyle|\nabla u|$ $\displaystyle\leq$ $\displaystyle\frac{C}{r\ln
R}+2\varepsilon|\nabla\ln f|$ $\displaystyle\leq$ $\displaystyle\frac{C}{r\ln
R}+\frac{C\varepsilon}{r}$ $\displaystyle\leq$
$\displaystyle\frac{C\varepsilon}{r}.$

Using the definition of $\phi$ we see that

$\displaystyle\int_{M}\frac{1}{r+1}|\nabla\phi|^{2}$ $\displaystyle\leq$
$\displaystyle\frac{C}{\varepsilon^{2}}\int_{M\backslash D}\frac{1}{r}|\nabla
u|^{2}$ $\displaystyle\leq$ $\displaystyle\int_{D\left(2\ln
R\right)}\frac{1}{\left(r+1\right)^{3}}$ $\displaystyle\leq$ $\displaystyle
C\ln R.$

We have proved that

$-2\int_{M}\eta_{ij}\eta_{i}\phi_{j}\phi(r+1)\leq\frac{1}{4}\int_{M}\left(r+1\right)\left|\nabla^{2}\eta\right|^{2}\phi^{2}+C\ln
R.$

Plugging these estimates into (5) implies that

(5.11) $\frac{C}{\ln
R}\int_{M}\left|\nabla^{2}\eta\right|^{2}\left(r+1\right)\phi^{2}\leq C.$

Since $f$ is harmonic on $M\backslash D$, we similarly have

$\displaystyle\int_{M\backslash D}\left|\nabla^{2}f\right|^{2}r\phi^{2}$
$\displaystyle=$ $\displaystyle-\int_{M\backslash D}\mathrm{Ric}\left(\nabla
f,\nabla f\right)r\phi^{2}$ $\displaystyle-\int_{M\backslash
D}f_{ij}f_{i}r_{j}\phi^{2}-2\int_{M\backslash D}f_{ij}f_{i}\phi_{j}r\phi$
$\displaystyle-$ $\displaystyle\int_{\partial D}f_{ij}f_{i}\nu_{j}r\phi^{2}$
$\displaystyle\leq$ $\displaystyle\frac{1}{2}\int_{M\backslash
D}\left|\nabla^{2}f\right|^{2}r\phi^{2}+C\int_{M\backslash D}\left|\nabla
f\right|^{2}|\nabla\phi|^{2}r+C.$

However, by (5) and (2.8) we have that

$\displaystyle\int_{M\backslash D}\left|\nabla f\right|^{2}|\nabla\phi|^{2}r$
$\displaystyle\leq$ $\displaystyle C\int_{M\backslash
D}\frac{1}{r}\left|\nabla f\right|^{2}$ $\displaystyle\leq$ $\displaystyle C.$

In conclusion,

(5.12) $\frac{C}{\ln R}\int_{M\backslash
D}\left|\nabla^{2}f\right|^{2}\left(r+1\right)\phi^{2}\leq C.$

Plugging (5), (5.11), and (5.12) into (5) implies that

(5.13) $\int_{M\backslash B_{p}\left(R_{0}\right)}\left|\nabla\left(\Delta
u\right)\right|\phi^{2}\leq C.$

Consequently, (5) becomes

$\displaystyle\int_{M\backslash D}\left(\Delta\left|\nabla
u\right|\right)\phi^{2}$ $\displaystyle\geq$ $\displaystyle\int_{M\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}-C.$

By the co-area formula, we have

$\displaystyle\int_{M\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}$ $\displaystyle=$
$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{\left\\{u=t\right\\}}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle=$
$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt$
$\displaystyle+\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{\widetilde{l}\left(t\right)}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt,$

for $l(t)$ and $\widetilde{l}\left(t\right)$ defined in (2.12) and (2.17). By
the Kato inequality and Ricci curvature bound $\mathrm{Ric}\left(\nabla
u,\nabla u\right)\geq-\frac{C}{\left(r+1\right)^{2}}\left|\nabla u\right|^{2}$
we have

$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{\widetilde{l}\left(t\right)}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle\geq$
$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{\widetilde{l}\left(t\right)}\mathrm{Ric}\left(\nabla
u,\nabla u\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle\geq$
$\displaystyle-C\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{\left\\{u=t\right\\}}\frac{1}{\left(r+1\right)^{2}}dt$
$\displaystyle\geq$
$\displaystyle-C\int_{L\left(\varepsilon,1\right)}\frac{1}{\left(r+1\right)^{2}}\left|\nabla
u\right|,$

where the last line follows from the co-area formula. However, Lemma 4.3,
(2.8) and (5) imply that

$\displaystyle\int_{M\backslash D}\frac{1}{r^{2}}\left|\nabla u\right|$
$\displaystyle\leq$ $\displaystyle\int_{M\backslash
D}\frac{1}{\left(r+1\right)^{2}}\left|\nabla f\right|$
$\displaystyle+\frac{1}{\ln R}\int_{D(2\,\ln
R)}\frac{1}{\left(r+1\right)^{3}}$ $\displaystyle\leq$
$\displaystyle\int_{M\backslash
D}\frac{1}{\left(r+1\right)^{4}}+\int_{M\backslash D}\left|\nabla
f\right|^{2}+C$ $\displaystyle\leq$ $\displaystyle C.$

In conclusion,

$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{\widetilde{l}\left(t\right)}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle\geq$
$\displaystyle-C.$

Thus, the following holds

$\displaystyle\int_{M\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}$ $\displaystyle\geq$
$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle-C.$

Rewrite it into

$\displaystyle\int_{M\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}$ $\displaystyle=$
$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\overline{M_{t}}}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt$
$\displaystyle+\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\left(M\backslash\overline{M_{t}}\right)}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt.$

Recall that from Lemma 4.4,

$l\left(t\right)\cap\left(M\backslash\overline{M_{t}}\right)\subset
M\backslash D\left(\ln R\right).$

Since Ricci curvature is bounded below by

$\mathrm{Ric}\left(\nabla u,\nabla
u\right)\geq-\frac{C}{\left(r+1\right)^{2}}\left|\nabla u\right|^{2},$

together with the Kato inequality
$\left|u_{ij}\right|^{2}\geq\left|\nabla\left|\nabla u\right|\right|^{2},$ it
follows that

$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\left(M\backslash\overline{M_{t}}\right)}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt$ $\displaystyle\geq$
$\displaystyle-C\int_{\varepsilon}^{1}\int_{l\left(t\right)\cap\left(M\backslash\overline{M_{t}}\right)}\frac{1}{r^{2}}dt$
$\displaystyle\geq$
$\displaystyle-C\int_{\varepsilon}^{1}\int_{l\left(t\right)\cap\left(M\backslash
D\left(\ln R\right)\right)}\frac{1}{r^{2}}dt$ $\displaystyle\geq$
$\displaystyle-C\int_{M\backslash D\left(\ln
R\right)}\frac{1}{r^{2}}\left|\nabla u\right|$ $\displaystyle\geq$
$\displaystyle-C,$

where the last line follows from (5). In conclusion,

$\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\left(M\backslash\overline{M_{t}}\right)}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt\geq-C$

and (5) becomes

$\displaystyle\int_{M}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}$ $\displaystyle\geq$
$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\overline{M_{t}}}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}dt-C.$

By Lemma 2.6,

$\mathrm{Ric}\left(\nabla u,\nabla u\right)\left|\nabla
u\right|^{-2}\geq\frac{1}{2}S-\frac{1}{2}S_{t}+\frac{1}{2}\frac{1}{\left|\nabla
u\right|^{2}}\left(\left|\nabla\left|\nabla
u\right|\right|^{2}-\left|u_{ij}\right|^{2}\right)$

on any regular level set $l\left(t\right)$. It follows that

(5.18) $\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-2}\geq\frac{1}{2}S-\frac{1}{2}S_{t}.$

Note that by the Gauss-Bonnet theorem, one has

$\int_{l\left(t\right)\cap\overline{M_{t}}}S_{t}\leq 8\pi m\leq C$

for all regular value $t$ with $0<t<1$ as Lemma 4.4 says that
$l\left(t\right)\cap\overline{M_{t}}=\partial M_{t}$ has $m$ connected
components. Therefore, we conclude from (5.18) that

$\int_{l\left(t\right)\cap\overline{M_{t}}}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla
u\right|^{-2}\geq\frac{1}{2}\int_{l\left(t\right)\cap\overline{M_{t}}}S-C.$

Together with (5), this implies that

$\displaystyle\int_{M\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\overline{M_{t}}}Sdt-C.$

Recall that by Lemma 4.4,

$l\left(t\right)\cap D\left(\ln R\right)\subset\overline{M_{t}}.$

According to (5.3), $S>0$ on $M\backslash D.$ Hence,

$\displaystyle\frac{1}{2}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\overline{M_{t}}}Sdt$
$\displaystyle\geq$
$\displaystyle\frac{1}{2}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap
D\left(\ln R\right)}Sdt$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}\int_{D\left(\ln R\right)\cap
L(\varepsilon,1)}S\left|\nabla u\right|\phi^{2}.$

In conclusion,

$\displaystyle\int_{M\backslash
D}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}\int_{D\left(\ln R\right)\cap
L(\varepsilon,1)}S\left|\nabla u\right|\phi^{2}-C.$

Plugging into (5) yields

(5.20) $\int_{M\backslash D}\left(\Delta\left|\nabla
u\right|\right)\phi^{2}\geq\frac{1}{2}\int_{D\left(\ln R\right)\cap
L(\varepsilon,1)}S\left|\nabla u\right|\phi^{2}-C.$

To estimate the left hand side, we integrate by parts to get

$\displaystyle\int_{M\backslash D}\left(\Delta\left|\nabla
u\right|\right)\phi^{2}$ $\displaystyle=$ $\displaystyle\int_{M\backslash
D}\left|\nabla u\right|\Delta\phi^{2}-\int_{\partial D}\left|\nabla
u\right|_{\nu}$ $\displaystyle\leq$
$\displaystyle\frac{C}{\varepsilon}\int_{L\left(\varepsilon,2\varepsilon\right)}\left|\nabla
u\right|\,\left|\Delta
u\right|+\frac{C}{\varepsilon^{2}}\int_{L\left(\varepsilon,2\varepsilon\right)}\left|\nabla
u\right|^{3}+C.$

First, by (5), (5), and Lemma 4.3 we have

$\displaystyle\frac{C}{\varepsilon}\int_{L\left(\varepsilon,2\varepsilon\right)}\left|\nabla
u\right|\,\left|\Delta u\right|$ $\displaystyle\leq$
$\displaystyle\frac{C}{\ln R}\int_{D\left(2\ln R\right)\backslash
D}\left(\frac{1}{r^{3}}+\left|\nabla f\right|^{2}\right)$ $\displaystyle\leq$
$\displaystyle C.$

Similarly, we get

$\displaystyle\frac{C}{\varepsilon^{2}}\int_{L\left(\varepsilon,2\varepsilon\right)}\left|\nabla
u\right|^{3}$ $\displaystyle\leq$ $\displaystyle C\int_{D\left(2\ln
R\right)\backslash D}\frac{1}{r^{2}}\left|\nabla u\right|$ $\displaystyle\leq$
$\displaystyle\frac{C}{\ln R}\int_{D\left(2\ln R\right)\backslash
D}\frac{1}{r^{3}}+C\int_{D\left(2\ln R\right)\backslash
D}\frac{1}{r^{2}}\left|\nabla f\right|$ $\displaystyle\leq$ $\displaystyle
C+\int_{M\backslash D}\left|\nabla f\right|^{2}$ $\displaystyle\leq$
$\displaystyle C,$

where we have applied (5) in the first line, and used Lemma 4.3 and that
$u\left|\nabla\ln f\right|\leq\left|\nabla f\right|$ in the second line. In
conclusion,

$\int_{M\backslash D}\left(\Delta\left|\nabla u\right|\right)\phi^{2}\leq C.$

Plugging this into (5.20) yields

$\int_{\left(D\left(\ln R\right)\backslash D\right)\cap
L\left(2\varepsilon,1\right)}S\left|\nabla u\right|\leq C$

for any $R$ such that $\ln R>\frac{1}{\varepsilon^{2}}.$ As in the proof of
Theorem 3.1, for arbitrary large $r$ such that $D\subset B_{p}(r),$ let

$t\left(r\right)=\min_{x\in\partial B_{p}\left(r\right)}f\left(x\right).$

Then for $\varepsilon<\frac{1}{2}t\left(r\right)$ and all $R$ sufficiently
large,

$B_{p}\left(r\right)\backslash D\subset D\left(\ln R\right)\cap
L\left(2\varepsilon,1\right).$

It follows that

$\int_{B_{p}\left(r\right)\backslash D}S\,\left|\nabla f\right|\leq C.$

Since $r$ is arbitrary, this proves

(5.21) $\int_{M\backslash D}S\,\left|\nabla f\right|\leq C.$

On the other hand, since $\Delta f=0$ on $B_{p}\left(r\right)\backslash
B_{p}\left(R_{0}\right)$ for $R_{0}<r$ with $D\subset B_{p}(R_{0}),$

$\int_{\partial B_{p}\left(r\right)}\frac{\partial f}{\partial
r}=\int_{\partial B_{p}\left(R_{0}\right)}\frac{\partial f}{\partial
r}=-C_{1}<0.$

Therefore, using (5.3) we have

$\displaystyle\int_{B_{p}(T)\backslash B_{p}\left(R_{0}\right)}S|\nabla f|$
$\displaystyle=$ $\displaystyle\int_{R_{0}}^{T}\int_{\partial
B_{p}(r)}S|\nabla f|dr$ $\displaystyle\geq$
$\displaystyle\frac{C_{0}}{2}\int_{R_{0}}^{T}\frac{1}{r}\left(\int_{\partial
B_{p}(r)}|\nabla f|\right)dr$ $\displaystyle\geq$
$\displaystyle\frac{C_{0}}{2}\int_{R_{0}}^{T}\frac{1}{r}\left|\int_{\partial
B_{p}(r)}\frac{\partial f}{\partial r}\right|dr$ $\displaystyle=$
$\displaystyle\frac{1}{2}C_{0}C_{1}\ln\frac{T}{R_{0}}.$

Together with (5.21) we have

$\frac{1}{2}C_{0}C_{1}\ln\frac{T}{R_{0}}\leq C$

for any $T>R_{0}.$ This contradiction shows that $\left(M,g\right)$ is
parabolic. ∎

The proof in fact shows that Theorem 5.1 holds true provided the scalar
curvature $S(x)\geq\chi(r(x))$ for some nonnegative function $\chi(r)$ with
$\int_{0}^{\infty}\chi(r)\,dr=\infty.$

We now turn to the proof of Theorem 1.4. Assume from now on that $E$ is a
parabolic end of $\left(M,g\right)$ with respect to a smooth connected bounded
domain $D.$ A result of Nakai [32] says that the barrier function may be
chosen to be proper on $E,$ that is, there exists positive harmonic function
$u$ with

$u=1\text{ on }\partial E$

and

(5.22) $\lim_{x\rightarrow E\left(\infty\right)}u\left(x\right)=\infty.$

Denote by

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

When $\left(M,g\right)$ has finitely many ends and finite first Betti number,
assume that $D$ is chosen such that all representatives of
$H_{1}\left(M\right)$ lie in $D$ and $M\setminus D$ has maximal number of
ends. The following result follows as Lemma 2.4, and is essentially contained
in [30, 21].

######  Lemma 5.2.

Let $\left(M,g\right)$ be a parabolic complete manifold with finite first
Betti number and finite number of ends. Then $l\left(t\right)$ is connected
for all $t\geq 1.$

We first establish a localized version of Theorem 1.6 on the end $E.$ For any
regular level set $l\left(t\right)$ of $u,$ let

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

For $\alpha\in\mathbb{R},$ define

$\displaystyle H_{\alpha}\left(t\right)$ $\displaystyle=$ $\displaystyle
t^{\alpha}\frac{dw}{dt}\left(t\right)-\left(\alpha+3\right)t^{\alpha-1}w\left(t\right)+\frac{4\pi}{\alpha+1}t^{\alpha+1}$
$\displaystyle=$ $\displaystyle
t^{\alpha}\int_{l\left(t\right)}\frac{\left\langle\nabla\left|\nabla
u\right|,\nabla u\right\rangle}{\left|\nabla
u\right|}-\left(\alpha+3\right)t^{\alpha-1}\int_{l\left(t\right)}\left|\nabla
u\right|^{2}$ $\displaystyle+\frac{4\pi}{\alpha+1}t^{\alpha+1},$

where the last term should be replaced by $4\pi\ln t$ when $\alpha=-1.$

######  Lemma 5.3.

Let $\left(M,g\right)$ be a parabolic complete three-dimensional manifold with
finite first Betti number and finite number of ends. Then

$H_{\alpha}\left(T\right)\geq
H_{\alpha}\left(s\right)-\left(\alpha^{2}+2\alpha\right)\int_{s}^{T}w\left(t\right)t^{\alpha-2}dt+\frac{1}{2}\int_{L(s,T)}u^{\alpha}S|\nabla
u|$

for all $1\leq s<T<\infty.$

###### Proof.

For regular values $1\leq s<T<\infty$ we have

$\displaystyle\left(T^{\alpha}\frac{dw}{dT}\left(T\right)-\alpha
T^{\alpha-1}w\left(T\right)\right)-\left(s^{\alpha}\frac{dw}{ds}(s)-\alpha
s^{\alpha-1}w\left(s\right)\right)$ $\displaystyle=$
$\displaystyle\left(\int_{l\left(T\right)}u^{\alpha}\frac{\left\langle\nabla\left|\nabla
u\right|,\nabla u\right\rangle}{\left|\nabla
u\right|}-\int_{l\left(s\right)}u^{\alpha}\frac{\left\langle\nabla\left|\nabla
u\right|,\nabla u\right\rangle}{\left|\nabla u\right|}\right)$
$\displaystyle-\left(\int_{l\left(T\right)}\left|\nabla
u\right|\frac{\left\langle\nabla u^{\alpha},\nabla
u\right\rangle}{\left|\nabla u\right|}-\int_{l\left(s\right)}\left|\nabla
u\right|\frac{\left\langle\nabla u^{\alpha},\nabla
u\right\rangle}{\left|\nabla u\right|}\right)$ $\displaystyle=$
$\displaystyle\int_{L\left(s,T\right)}\left(u^{\alpha}\Delta\left|\nabla
u\right|-\left|\nabla u\right|\Delta u^{\alpha}\right)$ $\displaystyle=$
$\displaystyle\int_{L\left(s,T\right)}\frac{u^{\alpha}}{\left|\nabla
u\right|}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla u\right)\right)$
$\displaystyle-\alpha\left(\alpha-1\right)\int_{L\left(s,T\right)}\left|\nabla
u\right|^{3}u^{\alpha-2}.$

Note that the term $\frac{1}{\left|\nabla
u\right|}\left|\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right|\leq\left|\mathrm{Ric}\right|\left|\nabla u\right|$ is
integrable even if $u$ has critical points in $L(s,T).$ The same can be
concluded for $\frac{1}{\left|\nabla
u\right|}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}\right)$ via a classical regularization procedure [28] from
the above identity by noticing that it is nonnegative due to the Kato
inequality.

Using the co-area formula to rewrite the last term

$\int_{L\left(s,T\right)}\left|\nabla
u\right|^{3}u^{\alpha-2}=\int_{s}^{T}t^{\alpha-2}w\left(t\right)dt,$

we conclude

$\displaystyle\left(T^{\alpha}\frac{dw}{dT}\left(T\right)-\alpha
T^{\alpha-1}w\left(T\right)\right)-\left(s^{\alpha}\frac{dw}{ds}(s)-\alpha
s^{\alpha-1}w\left(s\right)\right)$ $\displaystyle=$
$\displaystyle\int_{L\left(s,T\right)}\frac{u^{\alpha}}{\left|\nabla
u\right|}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla u\right)\right)$
$\displaystyle-\alpha\left(\alpha-1\right)\int_{s}^{T}t^{\alpha-2}w\left(t\right)dt.$

On the other hand, by Lemma 2.6,

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

on any regular level set $l\left(t\right),$ where $S_{t}$ is the scalar
curvature of $l\left(t\right).$ Recall that by Lemma 5.2 $l\left(t\right)$ is
connected for all $t\geq 1.$ So the Gauss-Bonnet theorem and the Kato
inequality Lemma 2.7 imply that

(5.27) $\frac{1}{2}\int_{l\left(t\right)}\left(S-S_{t}+\frac{1}{\left|\nabla
u\right|^{2}}\left|u_{ij}\right|^{2}\right)\geq\frac{1}{2}\int_{l(t)}S-4\pi+\frac{3}{4}\int_{l\left(t\right)}\frac{1}{\left|\nabla
u\right|^{2}}\left|\nabla\left|\nabla u\right|\right|^{2}.$

Observe that

$\left|w^{\prime}\left(t\right)\right|\leq\int_{l\left(t\right)}\left|\nabla\left|\nabla
u\right|\right|\leq\left(\int_{l\left(t\right)}\frac{1}{\left|\nabla
u\right|^{2}}\left|\nabla\left|\nabla
u\right|\right|^{2}\right)^{\frac{1}{2}}\left(\int_{l\left(t\right)}\left|\nabla
u\right|^{2}\right)^{\frac{1}{2}},$

which implies

$\int_{l\left(t\right)}\frac{1}{\left|\nabla
u\right|^{2}}\left|\nabla\left|\nabla
u\right|\right|^{2}\geq\frac{\left(w^{\prime}\right)^{2}\left(t\right)}{w\left(t\right)}.$

Together with the elementary inequality

$\frac{\left(w^{\prime}\right)^{2}}{w}\left(t\right)\geq\frac{4}{t}w^{\prime}\left(t\right)-\frac{4}{t^{2}}w\left(t\right),$

(5.27) becomes

$\frac{1}{2}\int_{l\left(t\right)}\left(S-S_{t}+\frac{1}{\left|\nabla
u\right|^{2}}\left|u_{ij}\right|^{2}\right)\geq\frac{1}{2}\int_{l(t)}S-4\pi+\frac{3}{t}w^{\prime}\left(t\right)-\frac{3}{t^{2}}w\left(t\right).$

Hence, by (5) we get

$\displaystyle\int_{l\left(t\right)}\frac{1}{\left|\nabla
u\right|^{2}}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla u\right)\right)$
$\displaystyle\geq$
$\displaystyle\frac{1}{2}\int_{l(t)}S-4\pi+\frac{3}{t}w^{\prime}\left(t\right)-\frac{3}{t^{2}}w\left(t\right)$

for any regular level set $l\left(t\right).$ We now use it and the co-area
formula to conclude

$\displaystyle\int_{L\left(s,T\right)}\frac{u^{\alpha}}{\left|\nabla
u\right|}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla u\right)\right)$
$\displaystyle\geq$ $\displaystyle\frac{1}{2}\int_{L(s,T)}u^{\alpha}S|\nabla
u|+\int_{s}^{T}\left(-4\pi+\frac{3}{t}w^{\prime}\left(t\right)-\frac{3}{t^{2}}w\left(t\right)\right)t^{\alpha}dt.$

Integrating by parts and noting that $w$ is Lipschitz, we obtain

$\displaystyle\int_{L\left(s,T\right)}\frac{u^{\alpha}}{\left|\nabla
u\right|}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla u\right)\right)$
$\displaystyle\geq$ $\displaystyle\frac{1}{2}\int_{L(s,T)}u^{\alpha}S|\nabla
u|-\frac{4\pi}{\alpha+1}\left(T^{\alpha+1}-s^{\alpha+1}\right)$
$\displaystyle+$ $\displaystyle
3T^{\alpha-1}w\left(T\right)-3s^{\alpha-1}w\left(s\right)-3\alpha\int_{s}^{T}t^{\alpha-2}w\left(t\right)dt.$

Plugging this into (5) yields

$\displaystyle\left(T^{\alpha}\frac{dw}{dT}\left(T\right)-\alpha
T^{\alpha-1}w\left(T\right)\right)-\left(s^{\alpha}\frac{dw}{ds}(s)-\alpha
s^{\alpha-1}w\left(s\right)\right)$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}\int_{L(s,T)}u^{\alpha}S|\nabla
u|-\frac{4\pi}{\alpha+1}\left(T^{\alpha+1}-s^{\alpha+1}\right)+3T^{\alpha-1}w\left(T\right)-3s^{\alpha-1}w\left(s\right)$
$\displaystyle-\alpha\left(\alpha+2\right)\int_{s}^{T}t^{\alpha-2}w\left(t\right)dt$

for any $1\leq s<T<\infty.$ This proves the result. ∎

We use the monotonicity in Lemma 5.3 to prove the following integral estimate.

######  Corollary 5.4.

Let $\left(M,g\right)$ be a parabolic complete three-dimensional manifold with
finite first Betti number and finite number of ends. Assume the barrier $u$
defined on the end $E$ has bounded gradient,

$\sup_{E}\left|\nabla u\right|<\infty.$

Then there exists a constant $\Upsilon>0$ such that

$\int_{L(1,t)}S|\nabla u|\leq 8\pi t+\Upsilon,$

for all $t>1$.

###### Proof.

Applying Lemma 5.3 with $\alpha=0$ to $u$ we get

(5.28) $\frac{1}{2}\int_{L(1,t)}S|\nabla
u|\leq\frac{dw}{dt}-3\frac{w\left(t\right)}{t}+4\pi t+\Upsilon_{0}$

for all $t>1,$ where

$\Upsilon_{0}=-\left(\frac{dw}{dt}|_{t=1}-3w\left(1\right)+4\pi\right).$

Since $u$ is harmonic, we know that $\int_{l\left(t\right)}\left|\nabla
u\right|$ is constant in $t$ . Together with the assumption that $u$ has
bounded gradient, we obtain

$\displaystyle w\left(t\right)$ $\displaystyle=$
$\displaystyle\int_{l\left(t\right)}\left|\nabla u\right|^{2}$
$\displaystyle\leq$
$\displaystyle\Upsilon_{1}\int_{l\left(t\right)}\left|\nabla u\right|$
$\displaystyle=$ $\displaystyle\Upsilon_{2}.$

So, for every $t>1$ there exists $t_{i}\in\left(t,t+1\right)$

$\frac{dw}{dt}\left(t_{i}\right)\leq\Upsilon_{2}.$

Then (5.28) implies

$\displaystyle\int_{L(1,t)}S|\nabla u|$ $\displaystyle\leq$
$\displaystyle\int_{L(1,t_{i})}S|\nabla u|$ $\displaystyle\leq$ $\displaystyle
8\pi t+\Upsilon,$

where

$\Upsilon=2\Upsilon_{0}+2\Upsilon_{2}+8\pi.$

∎

Recall that, as $u$ is harmonic, $\int_{l\left(t\right)}\left|\nabla u\right|$
is constant in $t.$ Let us normalize $u$ from here on such that

(5.30) $\int_{l\left(t\right)}\left|\nabla u\right|=1$

for all $t\geq 1.$

######  Lemma 5.5.

Let $\left(M,g\right)$ be a parabolic complete three-dimensional Riemannian
manifold. Assume that the Ricci curvature of $M$ is bounded from below by

$\mathrm{Ric}\geq-k\left(r\left(x\right)\right)$

for a continuous nonincreasing function $k\left(r\right)$ satisfying
$\int_{0}^{\infty}rk\left(r\right)dr<\infty.$ Normalize the barrier $u$ as in
(5.30). Then

$M\left(R\right)=\sup_{B_{p}\left(2R\right)\backslash
B_{p}\left(R\right)}\left|\nabla u\right|$

satisfies

$M^{2}\left(R\right)\leq CR\left(\inf_{x\in B_{p}\left(2R\right)\backslash
B_{p}\left(R\right)}\mathrm{V}_{x}\left(\frac{R}{2}\right)\right)^{-1}\left(M\left(\frac{R}{2}\right)+M\left(R\right)+M\left(2R\right)\right)$

for a constant $C$ depending only on $k$ and all $R>2R_{0}$ large enough.

###### Proof.

We follow an argument in [33]. Pick $x\in\partial B_{p}\left(t\right)$, where
$R\leq t\leq 2R$, such that

$M\left(R\right)=\left|\nabla u\right|\left(x\right).$

On $B_{x}\left(\frac{R}{2}\right)$ we have

$\mathrm{Ric}\geq-\frac{C}{R^{2}}$

and

$\Delta\left|\nabla u\right|\geq-\frac{C}{R^{2}}\left|\nabla u\right|.$

Applying the mean value inequality from [22] we get

$\left|\nabla
u\right|^{2}\left(x\right)\leq\frac{C}{\mathrm{V}_{x}\left(\frac{R}{2}\right)}\int_{B_{x}\left(\frac{R}{2}\right)}\left|\nabla
u\right|^{2}.$

This shows that

(5.31)
$M^{2}\left(R\right)\leq\frac{C}{\mathrm{V}_{x}\left(\frac{R}{2}\right)}\int_{L\left(\alpha_{x},\beta_{x}\right)}\left|\nabla
u\right|^{2},$

where

$\alpha_{x}=\min_{B_{x}\left(\frac{R}{2}\right)}u\text{ and
}\beta_{x}=\max_{B_{x}\left(\frac{R}{2}\right)}u.$

By the co-area formula and (5.30) we have

$\displaystyle\int_{L\left(\alpha_{x},\beta_{x}\right)}\left|\nabla
u\right|^{2}$ $\displaystyle=$
$\displaystyle\int_{\alpha_{x}}^{\beta_{x}}\int_{l\left(t\right)}\left|\nabla
u\right|dt$ $\displaystyle=$ $\displaystyle\beta_{x}-\alpha_{x}$
$\displaystyle\leq$ $\displaystyle
R\sup_{B_{x}\left(\frac{R}{2}\right)}\left|\nabla u\right|.$

Since $x\in B_{p}\left(2R\right)\backslash B_{p}\left(R\right),$ it follows
that

$\int_{L\left(\alpha_{x},\beta_{x}\right)}\left|\nabla u\right|^{2}\leq
R\left(M\left(\frac{R}{2}\right)+M\left(R\right)+M\left(2R\right)\right).$

Together with (5.31) we conclude that

$M^{2}\left(R\right)\leq\frac{CR}{\mathrm{V}_{x}\left(\frac{R}{2}\right)}\left(M\left(\frac{R}{2}\right)+M\left(R\right)+M\left(2R\right)\right)$

for some $x\in B_{p}\left(2R\right)\backslash B_{p}\left(R\right)$ and all
$R>2R_{0}.$ This proves the result. ∎

We are now ready to prove Theorem 1.4. A manifold $\left(M,g\right)$ is said
to satisfy the volume doubling property if there exists a constant $\nu>0$
such that

(5.32)
$\frac{\mathrm{V}_{x}\left(2r\right)}{\mathrm{V}_{x}\left(r\right)}\leq\nu$

for any $x\in M$ and any $r>0.$

######  Theorem 5.6.

Let $\left(M,g\right)$ be a three-dimensional complete noncompact Riemannian
manifold with finitely many ends and finite first Betti number
$b_{1}(M)<\infty.$ Assume that the Ricci curvature of $M$ is bounded from
below by

$\mathrm{Ric}\left(x\right)\geq-k\left(r\left(x\right)\right)$

for a continuous nonincreasing function $k\left(r\right)$ satisfying
$\int_{0}^{\infty}rk\left(r\right)dr<\infty.$ Then

(a) there exists a constant $C_{0}>0,$ depending only on $k,$ such that

$\liminf_{x\rightarrow\infty}S\left(x\right)\leq\frac{C_{0}}{\mathrm{V}_{p}\left(1\right)},$

where $\mathrm{V}_{p}\left(1\right)$ is the volume of the geodesic ball
$B_{p}\left(1\right).$

(b) there exists a constant $\bar{C}_{0}>0,$ depending only on $k$ and $\nu,$
such that

$\liminf_{x\rightarrow\infty}\left(S\left(x\right)r^{\alpha}\left(x\right)\right)\leq\bar{C}_{0}\limsup_{R\rightarrow\infty}\frac{R^{\alpha+1}}{\mathrm{V}_{p}\left(R\right)}$

for any $0\leq\alpha\leq 1$ provided that $\left(M,g\right)$ satisfies the
volume doubling property (5.32).

###### Proof.

We first prove (a). Let us assume by contradiction that

(5.33)
$\liminf_{x\rightarrow\infty}S\left(x\right)>\frac{\Gamma}{\mathrm{V}_{p}\left(1\right)},$

where $\Gamma>0$ is a large enough constant to be specified later. In
particular,

(5.34) $S\geq\frac{1}{2}\frac{\Gamma}{\mathrm{V}_{p}\left(1\right)}\text{ \ on
}M\backslash B_{p}\left(R_{0}\right)$

for some large enough $R_{0}.$ Below, $C$ denotes a constant that depends only
on $k.$

By Theorem 5.1 we know that $M$ is parabolic. By localizing a barrier function
to an end as above, we may assume without loss of generality that $M$ has only
one end. According to Nakai [32], there exists a proper harmonic function
$u:M\backslash D\rightarrow\left(0,\infty\right)$ with $u=1$ on $\partial D.$
We normalize $u$ as in (5.30).

Since $M$ has asymptotically nonnegative Ricci curvature, a volume comparison
result in [21] implies

$\mathrm{V}_{x}\left(\frac{R}{2}\right)\geq\frac{1}{C}R\mathrm{V}_{p}\left(1\right)$

for $x\in M$ with $r(x)=R.$ Choosing $R_{0}>0$ sufficiently large enough with
$D\subset B_{p}(R_{0})$ and applying Lemma 5.5 we get

(5.35)
$M^{2}\left(R\right)\leq\frac{C_{1}}{\mathrm{V}_{p}\left(1\right)}\left(M\left(\frac{R}{2}\right)+M\left(R\right)+M\left(2R\right)\right)$

for all $R>2R_{0}$. On the other hand, by Lemma 4.2,

$\left|\nabla\ln u\right|\leq\frac{C}{r}\text{ \ on }M\backslash
B_{p}\left(2R_{0}\right).$

It follows that

$u\leq r^{C}\sup_{\partial B_{p}\left(2R_{0}\right)}u\text{ \ \ on
}M\backslash B_{p}\left(2R_{0}\right)$

and

$\left|\nabla u\right|\leq r^{C}\sup_{\partial
B_{p}\left(2R_{0}\right)}u\text{ \ on }M\backslash B_{p}\left(2R_{0}\right).$

In particular, there exist constant $C_{2}$ with

$C_{2}>2C_{1}$

and $R_{1}>2R_{0}$ sufficiently large such that

(5.36)
$M\left(R\right)\leq\frac{1}{\mathrm{V}_{p}\left(1\right)}R^{\frac{C_{2}}{2}}$

for all $R\geq R_{1}.$

We now claim that for $m\geq 1,$

(5.37)
$M\left(R\right)\leq\frac{2^{C_{2}}}{\mathrm{V}_{p}\left(1\right)}R^{\frac{C_{2}}{2^{m}}}$

for any $R\geq 2^{m}R_{1}$.

By (5.36) we see that (5.37) holds for $m=1.$ We assume by induction that it
holds for $m$ and prove it for $m+1.$ In other words, we aim to prove that

(5.38)
$M\left(R\right)\leq\frac{2^{C_{2}}}{\mathrm{V}_{p}\left(1\right)}R^{\frac{C_{2}}{2^{m+1}}}$

for $R\geq 2^{m+1}R_{1}.$ By the induction hypothesis (5.37) we get that

$\displaystyle M\left(\frac{R}{2}\right)$ $\displaystyle\leq$
$\displaystyle\frac{2^{C_{2}}}{\mathrm{V}_{p}\left(1\right)}\left(\frac{R}{2}\right)^{\frac{C_{2}}{2^{m}}},$
$\displaystyle M\left(R\right)$ $\displaystyle\leq$
$\displaystyle\frac{2^{C_{2}}}{\mathrm{V}_{p}\left(1\right)}R^{\frac{C_{2}}{2^{m}}},$
$\displaystyle M\left(2R\right)$ $\displaystyle\leq$
$\displaystyle\frac{2^{C_{2}}}{\mathrm{V}_{p}\left(1\right)}\left(2R\right)^{\frac{C_{2}}{2^{m}}}.$

Using (5.35) we therefore obtain

$\displaystyle M^{2}\left(R\right)$ $\displaystyle\leq$
$\displaystyle\frac{2^{C_{2}}C_{1}}{\left(\mathrm{V}_{p}\left(1\right)\right)^{2}}\left(\left(\frac{1}{2}\right)^{\frac{C_{2}}{2^{m}}}+1+2^{\frac{C_{2}}{2^{m}}}\right)R^{\frac{C_{2}}{2^{m}}}$
$\displaystyle\leq$
$\displaystyle\frac{2^{2C_{2}}}{\left(\mathrm{V}_{p}\left(1\right)\right)^{2}}R^{\frac{C_{2}}{2^{m}}},$

where the second line follows from $C_{2}>2C_{1}.$ In conclusion,

$M\left(R\right)\leq\frac{2^{C_{2}}}{\mathrm{V}_{p}\left(1\right)}R^{\frac{C_{2}}{2^{m+1}}}$

and (5.38) is proved. Hence, (5.37) holds for any $m\geq 1$ and $R\geq
2^{m}R_{1}.$ By taking $2^{m}=\left[\ln R\right],$ this readily implies that

$M\left(R\right)\leq\frac{C}{\mathrm{V}_{p}\left(1\right)}$

for all $R\geq R_{1}^{2}.$ In conclusion,

(5.39) $\sup_{M\backslash B_{p}\left(R_{2}\right)}\left|\nabla
u\right|\leq\frac{C}{\mathrm{V}_{p}\left(1\right)}$

for sufficiently large $R_{2}>R_{1}.$ In particular,

(5.40)
$u\left(x\right)\leq\frac{C}{\mathrm{V}_{p}\left(1\right)}r\left(x\right)\text{
\ on }M\backslash B_{p}\left(R_{2}\right),$

by assuming $R_{2}$ is large enough. By (5.40) we conclude that

$B_{p}\left(R\right)\backslash B_{p}\left(R_{2}\right)\subset
L\left(1,\frac{C}{\mathrm{V}_{p}\left(1\right)}R\right).$

Since $u$ has bounded gradient by (5.39), according to Corollary 5.4 we have

$\displaystyle\int_{B_{p}\left(R\right)\backslash
B_{p}\left(R_{2}\right)}S\left|\nabla u\right|$ $\displaystyle\leq$
$\displaystyle\int_{L\left(1,\frac{C}{\mathrm{V}_{p}\left(1\right)}R\right)}S\left|\nabla
u\right|$ $\displaystyle\leq$
$\displaystyle\frac{C}{\mathrm{V}_{p}\left(1\right)}R+\Upsilon$

for all $R>R_{2}$, where $\Upsilon$ is a constant.

Now by (5.34),

$S\geq\frac{1}{2}\frac{\Gamma}{\mathrm{V}_{p}\left(1\right)}\text{ \ on
}B_{p}\left(R\right)\backslash B_{p}\left(R_{2}\right)$

and by (5.30),

(5.42) $\int_{\partial B_{p}\left(r\right)}\frac{\partial u}{\partial r}=1$

for all $\ r>R_{3}.$ The co-area formula then implies

$\displaystyle\int_{B_{p}\left(R\right)\backslash
B_{p}\left(R_{2}\right)}S\left|\nabla u\right|$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}\frac{\Gamma}{\mathrm{V}_{p}\left(1\right)}\int_{B_{p}\left(R\right)\backslash
B_{p}\left(R_{2}\right)}\left|\nabla u\right|$ $\displaystyle\geq$
$\displaystyle\frac{1}{2}\frac{\Gamma}{\mathrm{V}_{p}\left(1\right)}\int_{R_{2}}^{R}\left(\int_{\partial
B_{p}\left(r\right)}\frac{\partial u}{\partial r}\right)dr$
$\displaystyle\geq$
$\displaystyle\frac{1}{2}\frac{\Gamma}{\mathrm{V}_{p}\left(1\right)}\left(R-R_{2}\right).$

From (5) we get

$\frac{\Gamma-C}{\mathrm{V}_{p}\left(1\right)}R\leq\Upsilon_{1},$

for some constant $\Upsilon_{1}>0$. Since $R>R_{3}$ is arbitrary, this forces
$\Gamma\leq C$ and part (a) of the theorem is proved.

We now prove part (b). Define

(5.43)
$A=\liminf_{x\rightarrow\infty}\left(S\left(x\right)r^{\alpha}\left(x\right)\right)$

if the right hand side is finite. Otherwise, take $A>0$ to be arbitrary. Since
there is nothing to prove if $A=0,$ we assume $A>0.$ In particular, by Theorem
5.1, $\left(M,g\right)$ is parabolic. As before, we may assume that $M$ has
one end and that $u:M\backslash D\rightarrow\left(0,\infty\right)$ with $u=1$
on $\partial D$ is a proper barrier function satisfying (5.22).

We normalize $u$ to satisfy (5.30) and assume by contradiction that

(5.44)
$A>\Gamma\limsup_{R\rightarrow\infty}\frac{R^{\alpha+1}}{\mathrm{V}_{p}\left(R\right)},$

for some large constant $\Gamma$ to be specified later.

Recall Lemma 5.5 that

(5.45)
$M^{2}\left(R\right)\leq\frac{C_{1}R}{\mathrm{V}_{x}\left(\frac{R}{2}\right)}\left(M\left(\frac{R}{2}\right)+M\left(R\right)+M\left(2R\right)\right)$

for some $x\in B_{p}\left(2R\right)\backslash B_{p}\left(R\right)$ and all
$R>2R_{0}$ with the constant $C_{1}$ depending only on $k.$

Denote

(5.46) $\delta=\frac{2C_{1}A\nu^{3}}{\Gamma}.$

Note that (5.44) implies

$\frac{R^{\alpha+1}}{\mathrm{V}_{p}\left(R\right)}\leq\frac{\delta}{C_{1}\nu^{3}}$

for all $R>2R_{0}$ large enough. By (5.32) we have

$\frac{\mathrm{V}_{p}\left(R\right)}{\mathrm{V}_{x}\left(\frac{R}{2}\right)}\leq\frac{\mathrm{V}_{x}\left(3R\right)}{\mathrm{V}_{x}\left(\frac{R}{2}\right)}\leq\nu^{3}.$

Therefore,

$\frac{R^{\alpha+1}}{\mathrm{V}_{x}\left(\frac{R}{2}\right)}\leq\frac{\delta}{C_{1}}$

for all $R>2R_{0}$ and (5.45) becomes

(5.47)
$M^{2}\left(R\right)\leq\frac{\delta}{R^{\alpha}}\left(M\left(\frac{R}{2}\right)+M\left(R\right)+M\left(2R\right)\right).$

By Lemma 4.2, there exists a constant $C_{2}>0$ depending only on $k$ so that

$M\left(R\right)\leq\delta R^{\frac{C_{2}}{2}-\alpha}$

for all $R\geq R_{1}$ large enough. A similar induction argument as above then
shows that

$M\left(R\right)\leq 2^{C_{2}}\delta R^{\frac{C_{2}}{2^{m}}-\alpha}$

for all $R\geq 2^{m}R_{1}.$ Assuming $R\geq R_{2}$ is even larger if
necessary, we may take $m=\left[\ln R\right]$ and conclude that

$M\left(R\right)\leq\frac{C\delta}{R^{\alpha}}.$

In other words,