# Bottom spectrum 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.

A classical result of Cheng states that the bottom spectrum of complete
manifolds of fixed dimension and Ricci curvature lower bound achieves its
maximal value on the corresponding hyperbolic space. The paper establishes an
analogous result for three-dimensional complete manifolds with scalar
curvature lower bound subject to some necessary topological assumptions. The
rigidity issue is also addressed and a splitting theorem is obtained for such
manifolds with the maximal bottom spectrum.

##  1\. Introduction

The spectrum of the Laplacian $\Delta$ of a complete manifold $M$ is a closed
subset of the half line $[0,\infty).$ Its smallest value is called the bottom
spectrum of $M$ and denoted by $\lambda_{1}(M).$ 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 functions $\varphi$ with compact support.

A classical comparison result due to Cheng [6] 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. The answer is obviously negative in the case $n\geq 4.$ Indeed,
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. We are thus led to
consider the case of dimension three.

######  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 constant $K\geq 0.$ Assume
that the Ricci curvature of $M$ is bounded from below by a constant. Then the
bottom spectrum of $M$ satisfies $\lambda_{1}\left(M\right)\leq K$ provided
that either of the topological assumptions (A) or (B) is satisfied.
(A) The second homology group $H_{2}\left(M,\mathbb{Z}\right)$ of $M$ contains
no spherical classes;
(B) $M$ has finitely many ends and its first Betti number $b_{1}(M)<\infty.$

Recall that an end of $M$ with respect to a bounded (smooth) domain $\Omega$
is simply an unbounded component of $M\setminus\Omega,$ and $M$ is said to
have finitely many ends if the number of ends is always bounded by a fixed
constant regardless of $\Omega.$ We point out that the theorem has been proved
by the authors in [17] for the case (B) under an additional assumption that
the volume of unit balls does not decay too fast at infinity. The following
example indicates the necessity of the topological assumptions.

######  Example 1.2.

Consider the connected sum
$N=\left(\mathbb{S}^{2}\times\mathbb{S}^{1}\right)\\#\left(\mathbb{S}^{2}\times\mathbb{S}^{1}\right)$.
According to [8, 21], $N$ carries a metric with positive scalar curvature. So
the scalar curvature of the pull-back metric on its universal cover $M$ is
positive as well. Note that the first fundamental group of $N$ is a free group
on two generators. In particular, it is non-amenable. By a result of Brooks
[4], the bottom spectrum of $M$ equipped with the pull-back metric satisfies
$\lambda_{1}(M)>0.$

Clearly, the estimate of Theorem 1.1 fails for $M.$ This failure is due to the
fact that both topological conditions (A) and (B) are violated. Indeed, for
condition (A), while the first Betti number of $M$ is $0,$ the number of ends
of $M$ must be infinite as this is the case for the free group on two
generators. For condition (B), obviously, $H_{2}(M,\mathbb{Z})$ contains
spherical classes.

Note that the estimate in Theorem 1.1 is sharp for the hyperbolic space
$\mathbb{H}^{3}.$ As such, we consider the corresponding rigidity issue and
prove the following result.

######  Theorem 1.3.

Let $\left(M,g\right)$ be a complete three-dimensional manifold with scalar
curvature bounded by $S\geq-6K$ and bottom of spectrum
$\lambda_{1}\left(M\right)=K$ for some constant $K>0.$ Then either $M$ has no
finite volume end or $M=\mathbb{R}\times\mathbb{T}^{2}$, the warped product of
a line and the flat torus with

$ds_{M}^{2}=dt^{2}+e^{2\sqrt{K}\,t}ds_{\mathbb{T}^{2}}^{2}$

provided that $H_{2}\left(M,\mathbb{Z}\right)$ contains no spherical classes
and that the Ricci curvature of $M$ is bounded from below by a constant.

Theorem 1.3 is very much motivated by the work of Li and the second author
[13, 14], where they have addressed the similar rigidity question for the
aforementioned result of Cheng.

######  Theorem 1.4 (Li-Wang).

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

The result says that $M$ must split off a line in an explicit manner if it is
not connected at infinity. It may also be viewed as a suitable generalization
of the famous Cheeger-Gromoll splitting theorem [5] for manifolds with non-
negative Ricci curvature. In comparison, Theorem 1.3 is considerably weaker as
it fails to say anything when all ends are of infinite volume. Surprisingly,
this happens for good reasons. In fact, the following example shows that even
within the class of warped product manifolds the metrics with $S\geq-6$ and
$\lambda_{1}\left(M\right)=1$ are not unique. Consequently, a strict analogue
of Theorem 1.4 is not to be expected.

######  Example 1.5.

Let $M=\mathbb{R}\times\mathbb{T}^{2}$ with
$ds_{M}^{2}=dt^{2}+\cosh^{\frac{2}{a}}\left(at\right)ds_{\mathbb{T}^{2}}^{2}$,
where $1\leq a\leq\frac{3}{2}.$ Then the scalar curvature of $M$ satisfies

$S=-6+\left(6-4a\right)\cosh^{-2}\left(at\right)\geq-6$

and $\lambda_{1}\left(M\right)=1$ as the positive function
$u\left(t\right)=\cosh^{-\frac{1}{a}}\left(at\right)$ satisfies

$\Delta
u=\left(a-1\right)\sinh^{2}\left(at\right)\cosh^{-\left(2+\frac{1}{a}\right)}\left(at\right)-a\cosh^{-\frac{1}{a}}\left(at\right)\leq-u.$

Among this family of manifolds, only the one corresponding to $a=1$ appears in
rigidity result Theorem 1.4 as all others violate the Ricci lower bound. Note
also that $M$ has constant scalar curvature $S=-6$ for $a=\frac{3}{2}.$

We now indicate some of the ideas involved in the proof of Theorem 1.1. To do
so, let us first quickly recall the argument in [17]. Assume by contradiction
that $\lambda_{1}(M)>K.$ Then $M$ is necessarily nonparabolic and admits
positive Green’s functions. The assumption on the volume of unit balls is used
to show that the minimal positive Green’s function $G(p,x)$ of $M$ with fixed
$p$ goes to zero at infinity. Moreover, the topological assumption (B) ensures
the regular level sets $l(t)$ of $G$ are necessarily connected compact
surfaces on each end. The proof proceeds as in [16] by taking
$\varphi=\psi\,\left|\nabla G\right|^{1/2}$ as a test function in the Poincaré
inequality with the cut-off function $\psi$ chosen to depend on the function
$G$ as well. One is then led to integrate the following Bochner formula over
the level sets $l(t).$

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

The Ricci curvature term is handled by first rewriting it into

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

and then applying the Gauss-Bonnet theorem, where $S_{l\left(t\right)}$ is 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 [20], where it was realized 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. An identity of the
nature (1.2) was derived subsequently for any surface $N,$ not necessarily
minimal, by Jezierski and Kijowski in [11]. Recently, this kind of identity
was discovered by Stern [23] as well for the level sets of harmonic functions.
Level set methods have been recently applied in various settings in [1, 2, 3,
9].

Now for the proof of Theorem 1.1, without the volume lower bound assumption on
the unit balls, the properness of the minimal positive Green’s function is no
longer guaranteed. As a result, one runs into the difficulty that the Gauss-
Bonnet formula can not be directly applied for the level sets $l(t)$ as they
may be non-compact. 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 $p\in L(t,\infty).$ The Gauss-Bonnet formula is then
applied to the compact connected surfaces obtained from the intersection of
the boundary of $L(t,\infty)$ with each unbounded component of $M\setminus
L(t,\infty).$ 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 the Green’s function from [13] together
with a judicious choice of the cut-off function $\psi$ based on a result from
Schoen and Yau [22]. 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.

The proof of Theorem 1.3 follows more or less from that of Theorem 1.1 with a
careful tracking of the constants. The advantage here is that we work with a
globally defined positive harmonic function. So no interior boundary terms
arise from the estimates. However, at one point, we do need to derive precise
estimates to balance the boundary terms at the infinity of two ends.

Finally, we mention a splitting result for three dimensional manifolds with
non-negative scalar curvature. It may be viewed as the limiting case of
Theorem 1.3 by letting $K$ go to zero.

######  Theorem 1.6.

Let $\left(M,g\right)$ be a complete three-dimensional parabolic manifold with
its scalar curvature $S\geq 0$ and Ricci curvature bounded below by a
constant. Assume that $H_{2}\left(M,\mathbb{Z}\right)$ contains no spherical
classes. Then either $M$ is connected at infinity or it splits isometrically
as a direct product $M=\mathbb{R}\times\mathbb{T}^{2}.$

The proof relies on a monotonicity formula for harmonic functions originated
in [17] and refined in [18]. In fact, a similar argument works for the case
$M$ is nonparabolic as well provided that each nonparabolic end carries a
proper barrier function.

The paper is organized as follows. In Section 2 we prove the bottom spectrum
estimate Theorem 1.1. In Section 3 we present the proof of Theorem 1.3.
Section 4 concerns splitting results for three-manifolds with non-negative
scalar curvature.

Acknowledgment: We wish to thank Otis Chodosh for his interest in our work and
for sharing Example 1.2 with us. 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 and by a Simons
Foundation grant.

##  2\. Spectral estimate

This section is devoted to the proof of Theorem 1.1. We first make some
preparations without making any dimension restriction on manifold $M.$ Denote
by $r\left(x\right)$ the distance to a fixed point $p\in M.$ From Theorem 4.2
in [22], we have a smooth distance-like function on an arbitrary dimensional
manifold $M.$

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

and

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

for some constant $C>0.$

Although (2.2) is not explicitly stated in [22], it follows immediately from
the construction of $\rho$. For given $R>0,$ let
$\psi:\left(0,\infty\right)\rightarrow\mathbb{R}$ be a smooth function such
that $\psi=1$ on $\left(0,R\right)$ and $\psi=0$ on $\left(2R,\infty\right)$
satisfying

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

Composing it with $\rho,$ we obtain a cut-off function
$\psi\left(x\right)=\psi\left(\rho\left(x\right)\right).$ According to Lemma
2.1 it satisfies

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

Throughout this section,

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

Obviously, $\psi=1$ on $D\left(R\right)$ and $\psi=0$ on $M\setminus
D\left(2R\right)$.

We now assume that $\left(M,g\right)$ has positive spectrum, i.e.,

$\lambda_{1}\left(M\right)>0.$

In particular, $\left(M,g\right)$ is nonparabolic, that is, it admits a
positive symmetric Green’s function. Then, for a given smooth connected
bounded domain $D\subset M,$ by Li-Tam [12], there a exists a barrier function
$f>0$ on $M\setminus D$ satisfying

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

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 $\int_{M\setminus
D}\,\left|\nabla f\right|^{2}<\infty.$ In particular, by [15], for all
$0<t\leq 1$ and $r\geq R_{0}$ with $D\subset D(R_{0}),$

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

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

According to Cheng-Yau gradient estimate [7], when the Ricci curvature of $M$
is bounded below by a constant,

(2.8) $\left|\nabla\ln f\right|\leq C\text{ \ on }M\setminus D.$

We now consider the function $u$ given by

(2.9) $u=f\psi.$

It is straightforward to check the following lemma by (2.8) and (2.4).

######  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 $u$ is
harmonic on $D\left(R\right)\setminus D$ 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 $M\setminus D.$

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 $D.$ We furthermore
denote with

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

Note that all bounded components of $M\setminus D$ are contained in
$L\left(t,\infty\right)$ as well. This is because each of them shares the
boundary with $D$ and $f=1$ there. Let

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

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 $M$.

######  Lemma 2.3.

Let $\left(M,g\right)$ be an $n$-dimensional complete Riemannian manifold, and
let $M_{t}$ be the union of all unbounded connected components of
$M\setminus\overline{L\left(t,\infty\right)}$. Then

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

###### Proof.

Write

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

where $\Omega_{t}$ is the union of all bounded components of
$M\setminus\overline{L\left(t,\infty\right)}.$ Clearly,

$l\left(t\right)\cap\left(M\setminus\overline{M_{t}}\right)\subset\partial\Omega_{t}\subset
l\left(t\right).$

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

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

Therefore, $\left(\partial\Omega_{t}\cap
D\left(R\right)\right)\setminus\overline{M_{t}}=\varnothing.$ In conclusion,
we have

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

∎

We need the following results concerning the connected components of
$l\left(t\right)\cap\overline{M_{t}}$ under the topological assumption (A) or
(B) in Theorem 1.1.

######  Lemma 2.4.

Let $\left(M,g\right)$ be an $n$-dimensional manifold such that its homology
$H_{n-1}\left(M,\mathbb{Z}\right)$ contains no spherical classes. Let $M_{t}$
be the union of all unbounded connected components of
$M\setminus\overline{L\left(t,\infty\right)}.$ Then either no component of
$l\left(t\right)\cap\overline{M_{t}}$ is homeomorphic to $\mathbb{S}^{n-1},$
or

$l\left(t\right)\cap\overline{M_{t}}=\partial M_{t}\text{ \ is connected}.$

###### Proof.

Note that

$M\setminus l\left(t\right)=L\left(t,\infty\right)\cup M_{t}\cup\Omega_{t},$

where $\Omega_{t}$ is the union of all bounded connected components of
$M\setminus\overline{L\left(t,\infty\right)}$. Assume that
$l_{1}\left(t\right)\subset l\left(t\right)\cap\overline{M_{t}}$ is spherical.
As $H_{n-1}\left(M,\mathbb{Z}\right)$ contains no spherical classes, it
follows that $M\setminus l_{1}\left(t\right)$ has a bounded connected
component. However, since $M_{t}$ is unbounded, and as
$l_{1}\left(t\right)\subset\overline{M_{t}}$, we conclude that this bounded
component must be $L\left(t,\infty\right)\cup\overline{\Omega_{t}}$. In
particular,

$l\left(t\right)\cap\overline{M_{t}}=l_{1}\left(t\right)$

is connected. ∎

In the case that $M$ has finite first Betti number and finitely many, say $m,$
ends, the smooth connected bounded domain $D$ may be chosen large enough to
contain all representatives of the first homology group $H_{1}\left(M\right).$
Moreover, $M\setminus D$ 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 $l\left(t\right)$, cf. [17].

######  Lemma 2.5.

Let $\left(M,g\right)$ be an $n$-dimensional complete Riemannian manifold with
$m$ ends and finite first Betti number $b_{1}(M).$ Then for any $0<t<1,$

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

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

###### 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 $D\subset L\left(t,\infty\right)$, then $L\left(t,\infty\right)$ 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
$M\setminus\overline{L\left(t+\delta,\infty\right)}$ and $V$ the union of all
unbounded components of $M\setminus\overline{L\left(t+\delta,\infty\right)}.$
Since $M=U\cup V,$ we have the following Mayer-Vietoris sequence

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

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
$M\setminus\overline{L\left(t+\delta,\infty\right)}$ intersects with
$L\left(t,\infty\right)$ as $\overline{L\left(t+\delta,\infty\right)}\subset
L\left(t,\infty\right).$ 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 $0<t<1.$ ∎

Recall that from [13],

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

for all $r$ with $D\subset B_{p}(r).$ Therefore, by (2.1) and (2.5),

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

for all $r>R_{0}$. In particular, since $f\geq u,$ we get

(2.12)
$\mathrm{Vol}\left(L\left(\varepsilon,1\right)\right)\leq\frac{1}{\varepsilon^{2}}\int_{L\left(\varepsilon,1\right)}u^{2}\leq\frac{C}{\varepsilon^{2}}.$

Similarly,

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

Let us point out that $\left\\{u>t\right\\}$ may have other connected
components in addition to $L\left(t,\infty\right)$. Denote with

(2.14) $\widetilde{L}\left(t,\infty\right)=\left(\left\\{u>t\right\\}\setminus
L\left(t,\infty\right)\right)\cap L\left(\varepsilon,1\right)$

the union of all such connected components contained in
$L\left(\varepsilon,1\right)$, and its boundary

(2.15)
$\widetilde{l}\left(t\right)=\partial\widetilde{L}\left(t,\infty\right).$

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

######  Lemma 2.6.

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

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

where $C$ is a constant independent of $R$ and $\varepsilon$.

###### Proof.

Applying the divergence theorem, and using Lemma 2.2, for any regular set
$\widetilde{l}\left(t\right)$ we have

(2.16) $\int_{\widetilde{l}\left(t\right)}\left|\nabla
u\right|=-\int_{\widetilde{L}\left(t,\infty\right)}\Delta
u\leq\frac{C}{R}\mathrm{Vol}\left(L\left(\varepsilon,\infty\right)\right)\leq\frac{C}{R\varepsilon^{2}}.$

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

(2.17)
$\mathrm{Area}\left(\widetilde{l}\left(t\right)\right)\leq\frac{1}{2}\sqrt{R\varepsilon^{3}}\int_{\widetilde{l}\left(t\right)}\frac{\left|\nabla
u\right|}{u^{2}}+\frac{1}{2\sqrt{R\varepsilon^{3}}}\int_{\widetilde{l}\left(t\right)}\frac{u^{2}}{\left|\nabla
u\right|}.$

Integrating (2.16) from $t=\varepsilon$ to $t=1,$ we obtain

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

Using the co-area formula we estimate

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

Now integrating (2.17) in $t$ we arrive at

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

which proves the claim. ∎

We also need the following result.

######  Lemma 2.7.

Let $\left(M,g\right)$ be an $n$-dimensional complete Riemannian manifold with
positive spectrum $\lambda_{1}\left(M\right)>0$, and with Ricci curvature
bounded from below. For any $\varepsilon>0$ and any $a,b>\varepsilon$, we have

$\left|\int_{L\left(a,b\right)}\left|\nabla
u\right|^{2}\varphi\left(u\right)-C_{0}\int_{a}^{b}\varphi\left(t\right)dt\right|\leq\frac{C}{\sqrt{R}\varepsilon^{2}}\sup_{\left(a,b\right)}\left|\varphi\right|,$

for any integrable function $\varphi$ on $\left(a,b\right)$ with the constant
$C$ independent of $R$ and $\varepsilon,$ where $C_{0}$ is the constant in
(2.7).

###### Proof.

For $t<1$ we have by (2.7) that

$\displaystyle\int_{L\left(t,1\right)}\Delta u$ $\displaystyle=$
$\displaystyle-\int_{\partial D}\frac{\partial
u}{\partial\nu}-\int_{l\left(t\right)}\left|\nabla u\right|$ $\displaystyle=$
$\displaystyle C_{0}-\int_{l\left(t\right)}\left|\nabla u\right|.$

However, by (2.12)

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

So we conclude

(2.18) $\left|\int_{l\left(t\right)}\left|\nabla
u\right|-C_{0}\right|\leq\frac{C}{R\varepsilon^{2}}\text{ \ for all
}t>\varepsilon.$

According to the co-area formula,

$\displaystyle\int_{L\left(a,b\right)}\left|\nabla
u\right|^{2}\varphi\left(u\right)$ $\displaystyle=$
$\displaystyle\int_{a}^{b}\varphi\left(t\right)\left(\int_{l\left(t\right)}\left|\nabla
u\right|\right)dt$
$\displaystyle+\int_{a}^{b}\varphi\left(t\right)\left(\int_{\widetilde{l}\left(t\right)\cap
L\left(a,b\right)}\left|\nabla u\right|\right)dt.$

Finally, Lemma 2.2 and Lemma 2.6 imply that

$\int_{a}^{b}\left|\varphi\left(t\right)\right|\left(\int_{\widetilde{l}\left(t\right)\cap
L\left(a,b\right)}\left|\nabla
u\right|\right)dt\leq\frac{C}{\sqrt{R\varepsilon^{3}}}\sup_{\left(a,b\right)}\left|\varphi\right|.$

The desired result then follows immediately. ∎

From now on, we restrict to the case that $M$ is of dimension $3.$ First, we
have the following corollary of Lemma 2.4 and Lemma 2.5.

######  Corollary 2.8.

Let $\left(M,g\right)$ be a three-dimensional complete Riemannian manifold.
Then on any regular level $l\left(t\right)$ as defined in (2.10),

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

for a constant $C$ independent of $t\in\left(0,1\right),$ where $S_{t}$
denotes the scalar curvature of $l\left(t\right),$ provided that either (A) or
(B) holds.
(A) $H_{2}\left(M,\mathbb{Z}\right)$ contains no spherical classes.
(B) $M$ has finitely many ends and finite first Betti number $b_{1}(M).$

###### Proof.

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

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

If $H_{2}\left(M,\mathbb{Z}\right)$ contains no spherical classes, then Lemma
2.4 implies that

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

If $M$ has $m$ ends and finite first Betti number, then Lemma 2.5 says that

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

In either case, the constant is independent of $t\in\left(0,1\right).$ ∎

The following result relies on an idea from [20] and follows as in [10] or
Lemma 4.1 in [2]. For the sake of completeness, we include details here.

######  Lemma 2.9.

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,

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

This proves the result. ∎

Finally, we recall the following well known Kato inequality for harmonic
functions (cf. [17]).

(2.19) $\left|\nabla^{2}f\right|^{2}\geq\frac{3}{2}\left|\nabla\left|\nabla
f\right|\right|^{2}\text{ \ on }M\setminus D.$

We are now ready to prove Theorem 1.1 which is restated below.

######  Theorem 2.10.

Let $\left(M,g\right)$ be a three-dimensional complete Riemannian manifold
with scalar curvature $S\geq-6K$ on $M$ for some nonnegative constant $K.$
Assume that the Ricci curvature of $M$ is bounded from below by a constant.
Then the bottom spectrum of $M$ satisfies

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

provided that either (A) or (B) holds.
(A) $H_{2}\left(M,\mathbb{Z}\right)$ contains no spherical classes.
(B) $M$ has finitely many ends and finite first Betti number.

###### Proof.

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

(2.20) $R>\frac{1}{\varepsilon^{6}},$

where $R_{0}$ is chosen to be large enough so that $D\subset D(R_{0})$. 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 (2.6) and set $u=f\psi$ as
in (2.9). Let

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

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,

(2.21)
$\left|\phi^{\prime}\right|\left(t\right)\leq\frac{C}{\varepsilon}\text{ \ and
}\left|\phi^{\prime\prime}\right|\left(t\right)\leq\frac{C}{\varepsilon^{2}}\text{
\ for }\varepsilon<t<2\varepsilon.$

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

We first prove the following inequality.

$\text{{Claim 1: \ \ }}\int_{M\setminus D}\left|\nabla
u\right|^{3}u^{-2}\phi^{2}\leq 8K^{\frac{3}{2}}\int_{M\setminus
D}u\phi^{2}+C.$

Applying the Bochner formula and the inequality $\left\langle\nabla\Delta
u,\nabla u\right\rangle\geq-\left|\nabla\left(\Delta
u\right)\right|\left|\nabla u\right|$, we have

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

on $M\setminus D$ and whenever $\left|\nabla u\right|\neq 0.$ Hence, it
follows that

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

According to Lemma 2.2,

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

Integrating by parts gives

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

By Lemma 2.1 and (2.12) we get

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

Since Ricci curvature is bounded from below by a constant, we similarly have

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

Finally, we have

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

where in the last line we have used (2.21) and Lemma 2.7. In conclusion, this
proves that

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

Similarly, we see that

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

We conclude from above and (2.22) that

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

Thus, we have proved that

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

Using the co-area formula and noting that $\phi=0$ outside
$L(\varepsilon,\infty),$ we have

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

where $l(t)$ and $\widetilde{l}(t)$ are defined in (2.10) and (2.15),
respectively. Since $\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}\geq 0$ on $\widetilde{l}\left(t\right)\setminus
D\left(R\right)$ and $u$ is harmonic on $D\left(R\right)$, by Lemma 2.6 and
(2.19) the last term is estimated as

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

Rewrite the second last term in (2) as

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

Applying the inequality $\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}\geq 0,$ together with the fact that the Ricci curvature is
bounded from below, we conclude

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

where in the second line we have used Lemma 2.3 and in the last line (2).
Hence, (2) becomes

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

On any regular level set $l\left(t\right)$, from Lemma 2.9, Corollary 2.8, and
the assumption $S\geq-6K$, we have

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

We note that

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

Accordingly, (2) becomes

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

Using the inequality $\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}\geq 0$ and Lemma 2.3, the first term on the right hand
side can be estimated as

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

where the last line is by (2.19) and the fact that $u=f$ on $D\left(R\right).$
Similarly, the second term on the right hand side of (2) is bounded by

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

Consequently, (2) becomes

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

On the other hand, the left side of (2) satisfies

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

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

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

Note furthermore that $\Delta\phi^{2}=0$ on $M\setminus
L\left(\varepsilon,2\varepsilon\right).$ On
$L\left(\varepsilon,2\varepsilon\right)$, we have by Lemma 2.2 and (2.21)

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

Hence, Lemma 2.7 and (2.12) imply that

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

This proves that

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

We therefore conclude from (2) and (2) that

$\displaystyle\frac{3}{4}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap
D\left(R\right)}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-2}dt$
$\displaystyle+\frac{1}{2}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{\widetilde{l}\left(t\right)\cap
D\left(R\right)}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-2}dt$ $\displaystyle\leq$ $\displaystyle 3K\int_{M\setminus
D}\left|\nabla u\right|\phi^{2}+C.$

Integrating by parts yields

$\displaystyle\int_{M\setminus D}\left\langle\nabla\left|\nabla
u\right|,\nabla u\right\rangle u^{-1}\phi^{2}$ $\displaystyle=$
$\displaystyle-\int_{M\setminus D}\left|\nabla u\right|u^{-1}\left(\Delta
u\right)\phi^{2}+\int_{M\setminus D}\left|\nabla u\right|^{3}u^{-2}\phi^{2}$
$\displaystyle-\int_{M\setminus D}\left|\nabla
u\right|u^{-1}\left\langle\nabla u,\nabla\phi^{2}\right\rangle-\int_{\partial
D}\left|\nabla u\right|u^{-1}u_{\nu}\phi^{2}$ $\displaystyle\geq$
$\displaystyle\int_{M\setminus D}\left|\nabla u\right|^{3}u^{-2}\phi^{2}-C.$

In the last line we have used Lemma 2.2 and Lemma 2.7 to conclude that

$\left|\int_{M\setminus D}\left|\nabla u\right|u^{-1}\left\langle\nabla
u,\nabla\phi^{2}\right\rangle\right|\leq\frac{C}{\varepsilon}\int_{L\left(\varepsilon,2\varepsilon\right)}\left|\nabla
u\right|^{3}u^{-1}\leq C$

as well as

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

by Lemma 2.2 and (2.12). Moreover, in view of (2),

$\int_{M\setminus D\left(R\right)}\left\langle\nabla\left|\nabla
u\right|,\nabla u\right\rangle u^{-1}\phi^{2}\leq C\int_{M\setminus
D\left(R\right)}\left|\nabla\left|\nabla
u\right|\right|^{2}\phi^{2}+C\varepsilon.$

Let $\sigma$ be a cut-off function with support in $M\setminus
D\left(\frac{R}{2}\right)$ so that $\sigma=1$ on $M\setminus D\left(R\right)$
and $\left|\nabla\sigma\right|\leq\frac{C}{R}$. According to the Bochner
formula, and using the Ricci curvature lower bound, we have

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

Hence, (2) implies that

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

In conclusion, we have proved that

(2.30) $\int_{M\setminus D}\left|\nabla
u\right|^{3}u^{-2}\phi^{2}\leq\int_{D\left(R\right)\setminus
D}\left\langle\nabla\left|\nabla u\right|,\nabla u\right\rangle
u^{-1}\phi^{2}+C.$

By the co-area formula,

$\displaystyle\int_{D\left(R\right)\setminus D}\left\langle\nabla\left|\nabla
u\right|,\nabla u\right\rangle u^{-1}\phi^{2}$ $\displaystyle=$
$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap
D\left(R\right)}\left\langle\nabla\left|\nabla u\right|,\nabla u\right\rangle
u^{-1}\,\left|\nabla u\right|^{-1}\,dt$
$\displaystyle+\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{\widetilde{l}\left(t\right)\cap
D\left(R\right)}\left\langle\nabla\left|\nabla u\right|,\nabla u\right\rangle
u^{-1}\,\left|\nabla u\right|^{-1}\,dt.$

The first term on the right side is estimated by

$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap
D\left(R\right)}\left\langle\nabla\left|\nabla u\right|,\nabla u\right\rangle
u^{-1}\,\left|\nabla u\right|^{-1}\,dt$ $\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap
D\left(R\right)}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-2}dt$ $\displaystyle+\frac{1}{2}\int_{D\left(R\right)\setminus
D}\left|\nabla u\right|^{3}u^{-2}\phi^{2},$

and the second term by

$\displaystyle\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{\widetilde{l}\left(t\right)\cap
D\left(R\right)}\left\langle\nabla\left|\nabla u\right|,\nabla u\right\rangle
u^{-1}\,\left|\nabla u\right|^{-1}\,dt$ $\displaystyle\leq$
$\displaystyle\frac{1}{4}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{\widetilde{l}\left(t\right)\cap
D\left(R\right)}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-2}dt$
$\displaystyle+\int_{\varepsilon}^{1}\phi^{2}(t)\int_{\widetilde{l}\left(t\right)\cap
D\left(R\right)}\left|\nabla u\right|^{2}u^{-2}dt.$

However, by Lemma 2.2 and Lemma 2.6,

$\int_{\varepsilon}^{1}\phi^{2}(t)\int_{\widetilde{l}\left(t\right)\cap
D\left(R\right)}\left|\nabla u\right|^{2}u^{-2}dt\leq
C\int_{\varepsilon}^{1}\mathrm{Area}\left(\widetilde{l}\left(t\right)\right)dt\leq
C\varepsilon.$

Hence, by (2),

$\displaystyle\int_{D\left(R\right)\setminus D}\left\langle\nabla\left|\nabla
u\right|,\nabla u\right\rangle u^{-1}\phi^{2}$ $\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap
D\left(R\right)}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-2}dt$
$\displaystyle+\frac{1}{4}\int_{\varepsilon}^{1}\phi^{2}\left(t\right)\int_{\widetilde{l}\left(t\right)\cap
D\left(R\right)}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-2}dt$ $\displaystyle+\frac{1}{2}\int_{D\left(R\right)\setminus
D}\left|\nabla u\right|^{3}u^{-2}\phi^{2}+C$ $\displaystyle\leq$
$\displaystyle 2K\int_{M\setminus D}\left|\nabla
u\right|\phi^{2}+\frac{1}{2}\int_{D\left(R\right)\setminus D}\left|\nabla
u\right|^{3}u^{-2}\phi^{2}+C.$

Combining this with (2.30), we conclude that

$\int_{M\setminus D}\left|\nabla u\right|^{3}u^{-2}\phi^{2}\leq
4K\int_{M\setminus D}\left|\nabla u\right|\phi^{2}+C.$

Now together with

$4K\int_{M\setminus D}\left|\nabla
u\right|\phi^{2}\leq\frac{1}{3}\int_{M\setminus D}\left|\nabla
u\right|^{3}u^{-2}\phi^{2}+\frac{16}{3}K^{\frac{3}{2}}\int_{M\setminus
D}u\phi^{2},$

which follows from the elementary inequality
$ab\leq\frac{a^{3}}{3}+\frac{2b^{\frac{3}{2}}}{3}$ with $a=\left|\nabla
u\right|u^{-\frac{2}{3}}$ and $b=4Ku^{\frac{2}{3}}$, we arrive at

(2.31) $\int_{M\setminus D}\left|\nabla u\right|^{3}u^{-2}\phi^{2}\leq
8K^{\frac{3}{2}}\int_{M\setminus D}u\phi^{2}+C.$

This verifies Claim 1.

We now turn to the following claim.

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

Let $\xi$ be a Lipschitz cut-off such that $\xi=0$ on $D\left(R_{0}\right)$
and $\xi=1$ on $M\setminus D\left(R_{0}+1\right)$. Since the function
$u\phi^{2}\xi^{2}$ has compact support in $M,$ we have

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

According to Lemma 2.7 we have $\int_{M\setminus
D}u\left|\nabla\phi\right|^{2}\leq C$. Furthermore, we perform integration by
parts and get from Lemma 2.2 and (2) that

$\int_{M}\left\langle\nabla
u,\nabla\left(\phi^{2}\xi^{2}\right)\right\rangle=-\int_{M}\left(\Delta
u\right)\phi^{2}\xi^{2}\leq C\varepsilon.$

In conclusion, we have proved that

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

Claim 2 now follows.

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

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

Combining Claim 1 and Claim 2 we conclude that

$\displaystyle\lambda_{1}\left(M\right)\int_{M\setminus D}u\phi^{2}$
$\displaystyle\leq$ $\displaystyle\frac{1}{4}\int_{M\setminus D}\left|\nabla
u\right|^{2}u^{-1}\phi^{2}+C$ $\displaystyle\leq$
$\displaystyle\frac{1}{4}\left(\int_{M\setminus D}\left|\nabla
u\right|^{3}u^{-2}\phi^{2}\right)^{\frac{2}{3}}\left(\int_{M\setminus
D}u\,\phi^{2}\right)^{\frac{1}{3}}+C$ $\displaystyle\leq$ $\displaystyle
K\int_{M\setminus D}u\phi^{2}+C\left(\int_{M\setminus
D}u\,\phi^{2}\right)^{\frac{1}{3}}+C.$

By (2.32), it implies that $\int_{M\setminus D}u\phi^{2}\leq C.$

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
$R>\max\\{R_{0},\frac{1}{\varepsilon^{4}}\\}$ large,

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

As $u=f$ on $D\left(R\right)\cap L\left(2\varepsilon,1\right),$ it follows
that

(2.33) $\int_{B_{p}\left(r\right)\setminus D}f\leq\int_{M\setminus
D}u\phi^{2}\leq C$

with the constant $C$ independent of $r.$ On the other hand, the co-area
formula and (2.8), (2.7) imply

$\displaystyle\int_{B_{p}\left(r\right)\setminus D}f$ $\displaystyle\geq$
$\displaystyle C\,\int_{B_{p}\left(r\right)\setminus D}\left|\nabla f\right|$
$\displaystyle\geq$ $\displaystyle C\,\int_{R_{0}}^{r}dt\int_{\partial
B_{p}(t)}\left|\nabla f\right|$ $\displaystyle\geq$ $\displaystyle
C\,C_{0}\,(r-R_{0})$

for any $r>R_{0}.$ This contradicts (2.33). Therefore, (2.32) does not hold.
The theorem is proved. ∎

##  3\. Manifolds with maximal bottom spectrum

In this section we prove Theorem 1.3. We start by making some general
discussions without restriction on the dimension $n$ of $M.$ Note that for a
manifold $M$ with a finite volume end and positive bottom spectrum
$\lambda_{1}(M)>0,$ since its volume must be infinite, it has an infinite
volume end as well. By Nakai [19] and Li-Tam [12] there exists a positive
harmonic function $f$ on $M$ such that $f$ is proper and goes to infinity on a
given finite volume end $F$ and it is bounded away from $F.$ Therefore,
without loss of generality, we may assume that $F=\\{f\geq t_{0}\\}$ for some
large positive number $t_{0}$ and $F$ is connected. We view $E=M\setminus F$
as an infinite volume end and denote $E\left(R\right)=E\cap D\left(R\right),$
where $D(R)=\\{\rho(x)<R\\}$ as in Section 2 and $\rho$ be the function from
Lemma 2.1. By Li-Wang [13], such $f$ may be chosen to satisfy

(3.1) $\int_{E\setminus E\left(R\right)}f^{2}\leq C\,e^{-\frac{1}{C}R}$

for some constant $C>0$ depending on $\lambda_{1}(M).$ In particular,

(3.2) $\inf_{E}f=0\text{ and }\int_{E}f^{2}<\infty.\\\ $

According to [15], for such $f,$ $\int_{\left\\{f=t\right\\}}\left|\nabla
f\right|$ is independent of $t.$ So we normalize $f$ such that

(3.3) $\int_{\left\\{f=t\right\\}}\left|\nabla f\right|=1.$

For given $R>0,$ define cut-off function $\psi$ on $M$ as in Section 2 by
$\psi\left(x\right)=\psi\left(\rho\left(x\right)\right)$ with $\psi$ given in
(2.3). Define the function

(3.4) $u\left(x\right)=\left\\{\begin{array}[]{c}\psi f\\\
f\end{array}\right.\begin{array}[]{c}\text{on}\\\
\text{on}\end{array}\begin{array}[]{c}E\\\ F.\end{array}$

Note that $u$ is smooth on $M$ and satisfies

$u=f\text{ \ on }F\cup E\left(R\right)\text{ and }u=0\text{\ on }E\setminus
E\left(2R\right).$

Moreover, the level set $\\{u=t\\}$ of $u$ is compact for each $t>0.$ Lemma
2.2 implies that on the end $E$

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

On the end $F,$ for $M$ with Ricci curvature lower bound, by (2.8),

(3.6) $\left|\nabla u\right|\leq Cu\text{ \ and }\Delta u=0.$

Similar to the proof of Theorem 1.1, define $L\left(t,\infty\right)$ to be the
connected component of the set of $\\{u>t\\}$ that contains $F.$ Also,

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

Parallel to Lemma 2.3 and Lemma 2.4, we have the following.

######  Lemma 3.1.

Let $M_{t}$ be the union of all unbounded connected components of
$M\setminus\overline{L\left(t,\infty\right)},$ and $l_{0}\left(t\right)$ the
union of all connected components of $l\left(t\right)$ that are homeomorphic
to $\mathbb{S}^{n-1}.$ If the homology group $H_{n-1}(M,\mathbb{Z})$ contains
no spherical classes, then

$l_{0}\left(t\right)\subset
l\left(t\right)\cap\left(M\setminus\overline{M_{t}}\right)\subset E\setminus
D\left(R\right).$

###### Proof.

For the first inclusion, note that as $H_{n-1}\left(M,\mathbb{Z}\right)$
contains no spherical classes, any component of $l_{0}\left(t\right)$, being
spherical, must bound a compact component of
$M\setminus\overline{L\left(t,\infty\right)}.$ Obviously, such a component is
disjoint from $M_{t}.$

For the second inclusion, the same argument as in Lemma 2.3 implies that

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

Since $u$ is harmonic on $F,$ the second inclusion of the lemma follows by the
maximum principle. ∎

As in (2) we have that

(3.8) $\mathrm{Vol}\left(\left(E\setminus D\left(R\right)\right)\cap
L\left(\delta,\infty\right)\right)\leq\frac{C}{\delta^{2}}e^{-\frac{1}{C}R}$

for any $\delta>0,$ where $C$ is independent of $R$ and $\delta.$

Denote

$\widetilde{L}\left(t,\infty\right)=\left(\left\\{u>t\right\\}\setminus
L\left(t,\infty\right)\right)\cap\left(L\left(\delta,\infty\right)\right)$

and

(3.9)
$\widetilde{l}\left(t\right)=\partial\widetilde{L}\left(t,\infty\right).$

Since $u=f$ on $F=L\left(t_{0},\infty\right)$ and $u$ has compact support on
$E,$ it follows that $\widetilde{L}\left(t,\infty\right)$ is bounded.
Moreover,

(3.10) $L\left(t,\infty\right)\subset F\text{ \ and \
}\widetilde{L}\left(t,\infty\right)=\varnothing\text{ \ for }t\geq t_{0}.$

Note that Lemma 2.6 and (3.2) imply that

(3.11)
$\int_{\delta}^{\infty}\mathrm{Area}\left(\widetilde{l}\left(t\right)\right)dt\leq\frac{C}{\sqrt{R\delta^{3}}}$

for a constant $C$ independent of $R$ and $\delta,$ and Lemma 2.7 also holds
with $C_{0}=1$ in view of the normalization (3.3).

######  Lemma 3.2.

For any $\delta>0$ and any $a,b>\delta$, we have

$\left|\int_{L\left(a,b\right)}\left|\nabla
u\right|^{2}\varphi\left(u\right)-\int_{a}^{b}\varphi\left(t\right)dt\right|\leq\frac{C}{\sqrt{R}\delta^{2}}\sup_{\left(a^{\prime},b^{\prime}\right)}\left|\varphi\right|$

for any integrable function $\varphi$ on $\left(a,b\right),$ where
$a^{\prime}=\min\left\\{t_{0},a\right\\}$ and
$b^{\prime}=\min\left\\{t_{0},b\right\\},$ and the constant $C$ is independent
of $R$ and $\delta.$

We also note the following.

######  Lemma 3.3.

For any $\delta>0$ and any Lipschitz function $0\leq\chi\leq 1$ with compact
support in $\left(\delta,\infty\right),$

$\displaystyle\lambda_{1}\left(M\right)\int_{L\left(\delta,\infty\right)}u\chi^{2}\left(u\right)$
$\displaystyle\leq$
$\displaystyle\frac{1}{4}\int_{\delta}^{\infty}t^{-1}\chi^{2}\left(t\right)dt+\left(1+\frac{C}{\sqrt{R}\delta^{2}}\right)\int_{\delta}^{\infty}t\left(\chi^{\prime}\right)^{2}\left(t\right)dt$
$\displaystyle+\frac{C}{\sqrt{R}\delta^{3}},$

where $C$ is independent of $R$ and $\delta.$

###### Proof.

Since $\chi(u)$ has compact support in $L\left(\delta,\infty\right),$

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

By Lemma 3.2,

$\int_{L\left(\delta,\infty\right)}u^{-1}\left|\nabla
u\right|^{2}\chi^{2}\leq\int_{\delta}^{\infty}t^{-1}\chi^{2}\left(t\right)dt+\frac{C}{\sqrt{R}\delta^{3}}$

and

$\int_{L\left(\delta,\infty\right)}u\left|\nabla\chi\right|^{2}\leq\left(1+\frac{C}{\sqrt{R}\delta^{2}}\right)\int_{\delta}^{\infty}t\left(\chi^{\prime}\right)^{2}\left(t\right)dt.$

Also, by (3.5) and (3.8),

$\int_{L\left(\delta,\infty\right)}\left\langle\nabla
u,\nabla\chi^{2}\right\rangle=-\int_{L\left(\delta,\infty\right)}\chi^{2}\Delta
u\leq\frac{C}{R\delta^{2}}.$

The desired result follows by combining these estimates. ∎

We are now ready to prove Theorem 1.3. By scaling, we may assume $K=1.$

######  Theorem 3.4.

Let $\left(M,g\right)$ be a complete three-dimensional manifold with scalar
curvature bounded by $S\geq-6$ and bottom spectrum
$\lambda_{1}\left(M\right)=1.$ Assume that $H_{2}\left(M,\mathbb{Z}\right)$
has no spherical classes and that the Ricci curvature of $M$ is bounded from
below by a constant. Then either all the ends of $M$ have infinite volume or
$M=\mathbb{R}\times\mathbb{T}^{2}$ equipped with the warped product metric

$ds_{M}^{2}=dt^{2}+e^{2t}ds_{\mathbb{T}^{2}}^{2}.$

###### Proof.

We apply the previous discussions to $\delta=\varepsilon^{2}$ for any given
$\varepsilon>0.$ In the following, $R$ is arbitrarily large and satisfies

(3.12) $R>\frac{1}{\varepsilon^{20}}.$

Denote with

(3.13) $\Omega\left(R\right)=F\cup E\left(R\right).$

Let $\phi=\phi\left(t\right)$ be a Lipschitz function such that $0\leq\phi\leq
1$ and

(3.14) $\displaystyle\text{{supp}}\left(\phi\right)$ $\displaystyle\subset$
$\displaystyle\left(\varepsilon^{2},\frac{1}{\varepsilon^{2}}\right)$
$\displaystyle\left|\phi^{\prime}\right|$ $\displaystyle\leq$
$\displaystyle\frac{C}{\varepsilon^{2}}.$

Define $\phi\left(x\right)=\phi\left(u\left(x\right)\right)$ on
$L\left(\varepsilon^{2},\varepsilon^{-2}\right).$ From here on, we will denote
by $C$ a constant independent of $\varepsilon$ and $R$, whose value may change
from line to line.

Again, starting from the Bochner formula, we have

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

Using (2.23) that

(3.16) $\int_{M}\left|\nabla\left(\Delta
u\right)\right|\phi^{2}\leq\frac{C}{R\varepsilon^{4}}\leq C\varepsilon$

and that

(3.17) $\int_{l\left(t\right)\cap\overline{M_{t}}}S_{t}\leq 0$

on each regular level $l\left(t\right)$ by Lemma 3.1 and the Gauss-Bonnet
theorem, after manipulating as in the proof of Theorem 1.1, we arrive at

$\displaystyle\int_{M}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla
u\right)\right)\left|\nabla u\right|^{-1}\phi^{2}$ $\displaystyle\geq$
$\displaystyle\frac{3}{4}\int_{\varepsilon^{2}}^{\varepsilon^{-2}}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\Omega\left(R\right)}\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-2}dt-3\int_{M}\left|\nabla
u\right|\phi^{2}$
$\displaystyle+\frac{1}{2}\int_{\varepsilon^{2}}^{\varepsilon^{-2}}\phi^{2}\left(t\right)\int_{\widetilde{l}\left(t\right)\cap\Omega\left(R\right)}\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-2}dt-C\varepsilon+\Gamma,$

where

(3.19)
$\Gamma=\frac{1}{2}\int_{\Omega(R)}\left(\left|f_{ij}\right|^{2}-\frac{3}{2}\left|\nabla\left|\nabla
f\right|\right|^{2}\right)\phi^{2}+\frac{1}{2}\int_{\varepsilon^{2}}^{\varepsilon^{-2}}\phi^{2}(t)\int_{l(t)\cap\Omega(R)}(S+6)dt$

collects the relevant nonnegative terms discarded there.

Plugging (3) into (3) yields

$\displaystyle-\int_{M}\left\langle\nabla\left|\nabla
u\right|,\nabla\phi^{2}\right\rangle$ $\displaystyle\geq$
$\displaystyle\frac{3}{4}\int_{\varepsilon^{2}}^{\varepsilon^{-2}}\phi^{2}\left(t\right)\int_{l\left(t\right)\cap\Omega\left(R\right)}\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-2}dt$
$\displaystyle+\frac{1}{2}\int_{\varepsilon^{2}}^{\varepsilon^{-2}}\phi^{2}\left(t\right)\int_{\widetilde{l}\left(t\right)\cap\Omega\left(R\right)}\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-2}dt$
$\displaystyle-3\int_{M}\left|\nabla u\right|\phi^{2}-C\varepsilon+\Gamma.$

We will make use of this inequality with different choices of $\phi$ below.
First, let $\phi=\chi\left(u\right)$ be a cut-off supported in $E,$ where

$\chi\left(t\right)=\left\\{\begin{array}[]{c}\frac{\ln
t-2\ln\varepsilon}{\ln\left(\varepsilon^{-1}\right)}\\\
\frac{\ln\left(2\varepsilon\right)-\ln t}{\ln
2}\end{array}\right.\begin{array}[]{c}\text{for}\\\
\text{for}\end{array}\begin{array}[]{c}\varepsilon^{2}\leq t\leq\varepsilon\\\
\varepsilon\leq t\leq 2\varepsilon.\end{array}$

It follows that

$\displaystyle\frac{3}{4}\int_{\varepsilon^{2}}^{2\varepsilon}\chi^{2}\left(t\right)\int_{l\left(t\right)\cap
E\left(R\right)}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-2}dt$
$\displaystyle+\frac{1}{2}\int_{\varepsilon^{2}}^{2\varepsilon}\chi^{2}\left(t\right)\int_{\widetilde{l}\left(t\right)\cap
E\left(R\right)}\left|\nabla\left|\nabla u\right|\right|^{2}\left|\nabla
u\right|^{-2}dt$ $\displaystyle\leq$ $\displaystyle 3\int_{M}\left|\nabla
u\right|\chi^{2}-\int_{M}\left\langle\nabla\left|\nabla
u\right|,\nabla\chi^{2}\right\rangle+C\varepsilon.$

Following (2) to (2.31) we obtain

$\int_{M}\left|\nabla u\right|^{3}u^{-2}\chi^{2}\leq 8\int_{M}u\chi^{2}+C.$

By Lemma 3.3 we have

$\displaystyle\int_{M}u\chi^{2}$ $\displaystyle\leq$
$\displaystyle\frac{1}{4}\int_{\varepsilon^{2}}^{2\varepsilon}t^{-1}\chi^{2}\left(t\right)dt+C\int_{\varepsilon^{2}}^{2\varepsilon}t\left(\chi^{\prime}\right)^{2}\left(t\right)dt+C$
$\displaystyle\leq$
$\displaystyle\frac{1}{12}\ln\left(\varepsilon^{-1}\right)+C.$

Hence, this proves

(3.23) $\int_{L\left(\varepsilon^{2},\varepsilon\right)}\left|\nabla
u\right|^{3}u^{-2}\chi^{2}\leq\frac{2}{3}\ln\left(\varepsilon^{-1}\right)+C.$

Conversely, applying $ab\leq\frac{a^{3}}{3}+\frac{2b^{\frac{3}{2}}}{3}$ for
$a=2u^{\frac{1}{3}}$ and $b=u^{-\frac{4}{3}}\left|\nabla u\right|^{2}$ yields

(3.24) $2\int_{M}u^{-1}\left|\nabla
u\right|^{2}\chi^{2}\leq\frac{8}{3}\int_{M}u\chi^{2}+\frac{2}{3}\int_{M}u^{-2}\left|\nabla
u\right|^{3}\chi^{2}.$

Lemma 3.2 implies that

$2\int_{M}u^{-1}\left|\nabla
u\right|^{2}\chi^{2}\geq\frac{2}{3}\ln\left(\varepsilon^{-1}\right)-C.$

Plugging this and (3) into (3.24) implies

$\int_{M}u^{-2}\left|\nabla
u\right|^{3}\chi^{2}\geq\frac{2}{3}\ln\left(\varepsilon^{-1}\right)-C.$

Since by Lemma 3.2 and (3.5)

$\int_{L\left(\varepsilon,2\varepsilon\right)}u^{-2}\left|\nabla
u\right|^{3}\chi^{2}\leq
C\int_{L\left(\varepsilon,2\varepsilon\right)}u^{-1}\left|\nabla
u\right|^{2}\leq C,$

we conclude

$\int_{L\left(\varepsilon^{2},\varepsilon\right)}u^{-2}\left|\nabla
u\right|^{3}\chi^{2}\geq\frac{2}{3}\ln\left(\varepsilon^{-1}\right)-C.$

Together with (3.23), this proves

(3.25)
$\left|\int_{L\left(\varepsilon^{2},\varepsilon\right)}u^{-2}\left|\nabla
u\right|^{3}\chi^{2}-\frac{2}{3}\ln\left(\varepsilon^{-1}\right)\right|\leq
C.$

A similar argument on the end $F$ implies

(3.26)
$\left|\int_{L\left(\varepsilon^{-1},\varepsilon^{-2}\right)}u^{-2}\left|\nabla
u\right|^{3}\eta^{2}-\frac{2}{3}\ln\left(\varepsilon^{-1}\right)\right|\leq
C,$

where

$\eta\left(t\right)=\left\\{\begin{array}[]{c}\frac{\ln
t-\ln\left(2\varepsilon\right)^{-1}}{\ln 2}\\\
\frac{\ln\left(\varepsilon^{-2}\right)-\ln
t}{\ln\left(\varepsilon^{-1}\right)}\end{array}\right.\begin{array}[]{c}\text{for}\\\
\text{for}\end{array}\begin{array}[]{c}\frac{1}{2}\varepsilon^{-1}\leq
t\leq\varepsilon^{-1}\\\ \varepsilon^{-1}\leq
t\leq\varepsilon^{-2}.\end{array}$

We now return to (3) and replace $\phi$ by $\phi^{\frac{3}{2}}$ to obtain

$\displaystyle-\int_{M}\left\langle\nabla\left|\nabla
u\right|,\nabla\phi^{3}\right\rangle$ $\displaystyle\geq$
$\displaystyle\frac{3}{4}\int_{\varepsilon^{2}}^{\varepsilon^{-2}}\phi^{3}\left(t\right)\int_{l\left(t\right)\cap\Omega\left(R\right)}\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-2}dt$
$\displaystyle+\frac{1}{2}\int_{\varepsilon^{2}}^{\varepsilon^{-2}}\phi^{3}\left(t\right)\int_{\widetilde{l}\left(t\right)\cap\Omega\left(R\right)}\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-2}dt$
$\displaystyle-3\int_{M}\left|\nabla u\right|\phi^{3}-C\varepsilon+\Gamma.$

Let us now set

$\phi\left(t\right)=\left\\{\begin{array}[]{c}\frac{\ln
t-2\ln\varepsilon}{\ln\left(\varepsilon^{-1}\right)}\\\ 1\\\
\frac{2\ln\left(\varepsilon^{-1}\right)-\ln
t}{\ln\left(\varepsilon^{-1}\right)}\end{array}\right.\begin{array}[]{c}\text{for}\\\
\text{for}\\\ \text{for}\end{array}\begin{array}[]{c}\varepsilon^{2}\leq
t\leq\varepsilon\\\ \varepsilon\leq t\leq\varepsilon^{-1}\\\
\varepsilon^{-1}\leq t\leq\varepsilon^{-2}.\end{array}$

We claim that

(3.28) $-\int_{M}\left\langle\nabla\left|\nabla
u\right|,\nabla\phi^{3}\right\rangle\leq\frac{C}{\ln\left(\varepsilon^{-1}\right)}.$

Indeed, we have

$\displaystyle-\int_{M}\left\langle\nabla\left|\nabla
u\right|,\nabla\phi^{3}\right\rangle$ $\displaystyle=$
$\displaystyle-\frac{3}{\ln\left(\varepsilon^{-1}\right)}\int_{L\left(\varepsilon^{2},\varepsilon\right)}\left\langle\nabla\left|\nabla
u\right|,\nabla u\right\rangle u^{-1}\phi^{2}$
$\displaystyle+\frac{3}{\ln\left(\varepsilon^{-1}\right)}\int_{L\left(\varepsilon^{-1},\varepsilon^{-2}\right)}\left\langle\nabla\left|\nabla
u\right|,\nabla u\right\rangle u^{-1}\phi^{2}.$

On the one hand, integration by parts gives

$\displaystyle-\int_{L\left(\varepsilon^{2},\varepsilon\right)}\left\langle\nabla\left|\nabla
u\right|,\nabla u\right\rangle u^{-1}\phi^{2}$ $\displaystyle=$
$\displaystyle\int_{L\left(\varepsilon^{2},\varepsilon\right)}u^{-1}\left|\nabla
u\right|\left(\Delta
u\right)\phi^{2}-\int_{L\left(\varepsilon^{2},\varepsilon\right)}u^{-2}\left|\nabla
u\right|^{3}\phi^{2}$
$\displaystyle+\int_{L\left(\varepsilon^{2},\varepsilon\right)}\left|\nabla
u\right|\left\langle\nabla u,\nabla\phi^{2}\right\rangle
u^{-1}-\int_{l\left(\varepsilon\right)}\left|\nabla u\right|^{2}u^{-1}$
$\displaystyle\leq$
$\displaystyle-\frac{2}{3}\ln\left(\varepsilon^{-1}\right)+C,$

where in the last line we have used (3.25) and Lemma 3.2. On the other hand,
integration by parts as above together with (3.26) implies

$\int_{L\left(\varepsilon^{-1},\varepsilon^{-2}\right)}\left\langle\nabla\left|\nabla
u\right|,\nabla u\right\rangle
u^{-1}\phi^{2}\leq\frac{2}{3}\ln\left(\varepsilon^{-1}\right)+C.$

Plugging these estimates into (3) we obtain (3.28). Hence, by (3), we have
established that

$\displaystyle\Gamma+\frac{3}{4}\int_{\varepsilon^{2}}^{\varepsilon^{-2}}\phi^{3}\left(t\right)\int_{l\left(t\right)\cap\Omega\left(R\right)}\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-2}dt$
$\displaystyle+\frac{1}{2}\int_{\varepsilon^{2}}^{\varepsilon^{-2}}\phi^{3}\left(t\right)\int_{\widetilde{l}\left(t\right)\cap\Omega\left(R\right)}\left|\nabla\left|\nabla
u\right|\right|^{2}\left|\nabla u\right|^{-2}dt$ $\displaystyle\leq$
$\displaystyle 3\int_{M}\left|\nabla
u\right|\phi^{3}+\frac{C}{\ln\left(\varepsilon^{-1}\right)}.$

Proceeding as in the proof of (2.30) we note that

$\displaystyle\int_{M}\left\langle\nabla\left|\nabla u\right|,\nabla
u\right\rangle u^{-1}\phi^{3}$ $\displaystyle=$
$\displaystyle-\int_{M}\left|\nabla u\right|u^{-1}\left(\Delta
u\right)\phi^{3}+\int_{M}\left|\nabla u\right|^{3}u^{-2}\phi^{3}$
$\displaystyle-\int_{M}\left|\nabla u\right|u^{-1}\left\langle\nabla
u,\nabla\phi^{3}\right\rangle.$

However, (3.25) and (3.26) imply that

$\left|\int_{M}\left|\nabla u\right|u^{-1}\left\langle\nabla
u,\nabla\phi^{3}\right\rangle\right|\leq\frac{C}{\ln\left(\varepsilon^{-1}\right)}.$

Also,

$\left|-\int_{M}\left|\nabla u\right|u^{-1}\left(\Delta
u\right)\phi^{3}\right|\leq C\,\varepsilon$

by (3.5) and (3.8). Therefore,

$\int_{M}\left|\nabla
u\right|^{3}u^{-2}\phi^{3}\leq\int_{\Omega\left(R\right)}\left\langle\nabla\left|\nabla
u\right|,\nabla u\right\rangle
u^{-1}\phi^{3}+\frac{C}{\ln\left(\varepsilon^{-1}\right)}.$

Now arguing as from (2.30) to (2.31), we conclude from (3) and the above
inequality that

(3.31) $\,\Gamma+\int_{M}\left|\nabla u\right|^{3}u^{-2}\phi^{3}\leq
8\int_{M}u\phi^{3}+\frac{C}{\ln\left(\varepsilon^{-1}\right)}.$

On the other hand, as $\lambda_{1}(M)=1,$ Lemma 3.3 implies that

(3.32) $\int_{M}u\phi^{3}\leq\frac{1}{4}\int_{M}\left|\nabla
u\right|^{2}u^{-1}\phi^{3}+\frac{C}{\ln\left(\varepsilon^{-1}\right)}.$

However, as in (3.24),

(3.33) $2u^{-1}\left|\nabla u\right|^{2}\leq\frac{2}{3}u^{-2}\left|\nabla
u\right|^{3}+\frac{8}{3}u.$

Applying (3.31) and (3.32) it follows that

$\displaystyle 0$ $\displaystyle\leq$
$\displaystyle\int_{M}\left(\frac{2}{3}u^{-2}\left|\nabla
u\right|^{3}+\frac{8}{3}u-2u^{-1}\left|\nabla u\right|^{2}\right)\phi^{3}$
$\displaystyle\leq$ $\displaystyle
8\int_{M}u\phi^{3}-2\int_{M}u^{-1}\left|\nabla
u\right|^{2}\phi^{3}-\frac{2}{3}\Gamma+\frac{C}{\ln\left(\varepsilon^{-1}\right)}$
$\displaystyle\leq$
$\displaystyle-\frac{2}{3}\Gamma+\frac{C}{\ln\left(\varepsilon^{-1}\right)}.$

In conclusion, this proves that

$\displaystyle\int_{L\left(\varepsilon,\varepsilon^{-1}\right)\cap\Omega(R)}\left(\frac{2}{3}f^{-2}\left|\nabla
f\right|^{3}+\frac{8}{3}f-2f^{-1}\left|\nabla f\right|^{2}\right)$
$\displaystyle+\frac{1}{3}\int_{L\left(\varepsilon,\varepsilon^{-1}\right)\cap\Omega(R)}\left(\left|f_{ij}\right|^{2}-\frac{3}{2}\left|\nabla\left|\nabla
f\right|\right|^{2}\right)$
$\displaystyle+\frac{1}{3}\int_{\varepsilon}^{\varepsilon^{-1}}\int_{l(t)\cap\Omega(R)}(S+6)dt$
$\displaystyle\leq$ $\displaystyle\frac{C}{\ln\left(\varepsilon^{-1}\right)}.$

Since $R$ and $\varepsilon$ are arbitrary, we conclude from here that
$\left|f_{ij}\right|^{2}=\frac{3}{2}\left|\nabla\left|\nabla
f\right|\right|^{2}$ and $\left|\nabla f\right|=2f$. As in [15], this proves
that $\left(M,g\right)$ splits as a warped product $M=\mathbb{R}\times N$ with
$ds_{M}^{2}=dt^{2}+e^{2h\left(t\right)}ds_{N}^{2}$, and
$f\left(t\right)=Ce^{-2t}$ if we choose $F=\left(-\infty,0\right)\times N$ to
be the parabolic end. Since $f$ is harmonic, it follows $h\left(t\right)=t$.
Moroever, as $S=-6$ by (3), it follows that $N$ is flat. The theorem is
proved. ∎

##  4\. Splitting for nonnegative scalar curvature

In this section we study complete three-dimensional manifolds of non-negative
scalar curvature with more than one end. We first prove Theorem 1.6.

######  Theorem 4.1.

Let $\left(M,g\right)$ be a complete three-dimensional parabolic manifold with
its scalar curvature $S\geq 0$ and Ricci curvature bounded below by a
constant. Assume that $H_{2}\left(M,\mathbb{Z}\right)$ contains no spherical
classes. Then either $M$ is connected at infinity or it splits isometrically
as a direct product $M=\mathbb{R}\times\mathbb{T}^{2}.$

###### Proof.

Let us assume that $M$ has more than one end. As $M$ is parabolic, so is each
end. Then by [19] and [12] there exists a harmonic function
$u:M\rightarrow\mathbb{R}$ such that

$\lim_{x\rightarrow\infty,\ x\in E}u\left(x\right)=\infty\text{ \ and
}\lim_{x\rightarrow\infty,\ x\in F}u\left(x\right)=-\infty,$

where $E$ is an end of $M$ and $F=M\setminus E.$ Denote with

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

where $l\left(t\right)=\left\\{u=t\right\\}.$ Denote also with
$L(s,t)=\\{s<u<t\\}.$

Fix $s\in\mathbb{R}$ and let $\phi=\phi\left(t\right)$ be a $C^{2}$ function
on $\left(s,\infty\right)$. Following [17], we note that for $t>s,$

$\displaystyle\left(\phi\left(t\right)w^{\prime}\left(t\right)-\phi^{\prime}\left(t\right)w\left(t\right)\right)-\left(\phi\left(s\right)w^{\prime}(s)-\phi^{\prime}\left(s\right)w\left(s\right)\right)$
$\displaystyle=$
$\displaystyle\int_{L\left(s,t\right)}\left(\phi\left(u\right)\Delta\left|\nabla
u\right|-\left|\nabla u\right|\Delta\phi\left(u\right)\right)$
$\displaystyle=$
$\displaystyle\int_{L\left(s,t\right)}\frac{\phi\left(u\right)}{\left|\nabla
u\right|}\left(\left|u_{ij}\right|^{2}-\left|\nabla\left|\nabla
u\right|\right|^{2}+\mathrm{Ric}\left(\nabla u,\nabla u\right)\right)$
$\displaystyle-\int_{s}^{t}\phi^{\prime\prime}\left(r\right)w\left(r\right)dr.$

Since $H_{2}\left(M,\mathbb{Z}\right)$ contains no spherical classes, the
Gauss-Bonnet formula implies

$\int_{l\left(r\right)}S_{r}\leq 0.$

By Lemma 2.9 and the fact that $S\geq 0,$ we get

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

where the last line follows from

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

Therefore, by (4) we conclude that

$\displaystyle\left(\phi\left(t\right)w^{\prime}\left(t\right)-\phi^{\prime}\left(t\right)w\left(t\right)\right)-\left(\phi\left(s\right)w^{\prime}(s)-\phi^{\prime}\left(s\right)w\left(s\right)\right)$
$\displaystyle\geq$
$\displaystyle-\int_{s}^{t}\phi^{\prime\prime}\left(r\right)w\left(r\right)dr+\frac{3}{4}\int_{s}^{t}\phi\left(r\right)\frac{\left(w^{\prime}\right)^{2}}{w}\left(r\right)dr$

for any $C^{2}$ function $\phi\geq 0$ on $\left(s,\infty\right).$ So

(4.3) $w^{\prime}(t)\geq w^{\prime}(s),$

for all $s<t$ by taking $\phi\left(r\right)=1.$ Letting $\phi(r)=r^{-\alpha}$
for $\alpha>0$ and $r\geq 1,$ we obtain that

$\displaystyle\left(t^{-\alpha}w^{\prime}\left(t\right)+\alpha
t^{-\alpha-1}w\left(t\right)\right)-\left(s^{-\alpha}w^{\prime}(s)+\alpha
s^{-\alpha-1}w\left(s\right)\right)$ $\displaystyle\geq$
$\displaystyle-\alpha(\alpha+1)\int_{s}^{t}r^{-\alpha-2}w\left(r\right)dr+\frac{3}{4}\int_{s}^{t}r^{-\alpha}\frac{\left(w^{\prime}\right)^{2}}{w}\left(r\right)dr$

for all $1\leq s\leq t.$

We rewrite the elementary inequality

$\frac{2\varepsilon}{r}w^{\prime}(r)\leq\frac{\left(w^{\prime}\right)^{2}}{w}\left(r\right)+\frac{\varepsilon^{2}}{r^{2}}w(r)$

into

$\frac{3}{4}\int_{s}^{t}r^{-\alpha}\frac{\left(w^{\prime}\right)^{2}}{w}\left(r\right)dr\geq\frac{3\varepsilon}{2}\int_{s}^{t}r^{-\alpha-1}w^{\prime}(r)dr-\frac{3\varepsilon^{2}}{4}\int_{s}^{t}r^{-\alpha-2}w(r)dr.$

After integrating by parts the first term on the right side we get

$\displaystyle\frac{3}{4}\int_{s}^{t}r^{-\alpha}\frac{\left(w^{\prime}\right)^{2}}{w}\left(r\right)dr$
$\displaystyle\geq$
$\displaystyle\frac{3\varepsilon}{2}\left(t^{-\alpha-1}w(t)-s^{-\alpha-1}w(s)\right)$
$\displaystyle+$
$\displaystyle\left(\frac{3\varepsilon}{2}(\alpha+1)-\frac{3\varepsilon^{2}}{4}\right)\int_{s}^{t}r^{-\alpha-2}w(r)dr.$

Plugging this inequality into (4) implies that

$\displaystyle\left(t^{-\alpha}w^{\prime}\left(t\right)-\left(\frac{3\varepsilon}{2}-\alpha\right)t^{-\alpha-1}w\left(t\right)\right)-\left(s^{-\alpha}w^{\prime}(s)-\left(\frac{3\varepsilon}{2}-\alpha\right)s^{-\alpha-1}w\left(s\right)\right)$
$\displaystyle\geq$
$\displaystyle\left(\frac{3\varepsilon}{2}(\alpha+1)-\frac{3\varepsilon^{2}}{4}-\alpha(\alpha+1)\right)\int_{s}^{t}r^{-\alpha-2}w\left(r\right)dr$

for any $1\leq s\leq t.$ For $\alpha=\varepsilon=\frac{1}{4}$ it follows that

(4.5)
$t^{-\frac{1}{4}}w^{\prime}\left(t\right)-{\frac{1}{8}}t^{-\frac{5}{4}}w\left(t\right)\geq
s^{-\frac{1}{4}}w^{\prime}(s)-{\frac{1}{8}}s^{-\frac{5}{4}}w\left(s\right).$

As the Ricci curvature of $M$ is bounded from below by a constant, the
gradient estimate (2.8) implies

$w(t)=\int_{l(t)}|\nabla u|^{2}\leq Ct\int_{l(t)}|\nabla u|=Ct$

for all $t\geq 1.$ In particular, there exists $t_{n}\rightarrow\infty$ such
that $w^{\prime}(t_{n})\leq C.$ Letting $t=t_{n}$ in (4.5) and making
$n\rightarrow\infty$ we conclude that

$s^{-\frac{1}{4}}w^{\prime}(s)\leq{\frac{1}{8}}s^{-\frac{5}{4}}w\left(s\right)$

for all $s\geq 1.$ Integrating this differential inequality from $s=1$ to
$s=r\geq 1$ we conclude that

(4.6) $w(r)\leq Cr^{\frac{1}{8}}$

for all $r\geq 1.$ A similar argument implies that

$w(r)\leq C|r|^{\frac{1}{8}}$

for all $r\leq-1$ as well. Since $w$ is convex by (4.3), this immediately
implies that $w$ is a constant function. Therefore, (4.3) becomes an equality.
Now tracing through the derivation of (4.3), one sees that $u_{ij}=0$ and
$S=0$. This proves the splitting. ∎

Recall that a nonparabolic manifold is called regular if the Green’s function
$G\left(p,x\right)$ satisfies

$\lim_{x\rightarrow\infty}G\left(p,x\right)=0.$

Similarly, a nonparabolic end $E$ of $M$ is called regular if $G$ is regular
on $E.$

######  Theorem 4.2.

Let $\left(M,g\right)$ be a nonparabolic complete three-dimensional manifold
with $S\geq 0$ and Ricci curvature bounded below. Assume that
$H_{2}\left(M,\mathbb{Z}\right)$ contains no spherical classes. If all
nonparabolic ends of $M$ are regular, then $M$ has one end.

###### Proof.

Suppose first that $M$ has at least one parabolic end. We let $F$ be the union
of all parabolic ends and $E=M\setminus F$. Then $E$ is nonparabolic. Again,
by [19, 12], there exists a harmonic function
$u:M\rightarrow\left(0,\infty\right)$ such that

$\displaystyle\lim_{x\rightarrow\infty,x\in E}u\left(x\right)$
$\displaystyle=$ $\displaystyle 0$ $\displaystyle\lim_{x\rightarrow\infty,x\in
F}u\left(x\right)$ $\displaystyle=$ $\displaystyle\infty.$

According to (4.3) we have $w^{\prime}\left(r\right)\geq
w^{\prime}\left(s\right)$ for all $r>s>0,$ whereas (4.6) implies
$w^{\prime}\left(r_{n}\right)\rightarrow 0$ for some sequence
$r_{n}\rightarrow\infty.$ Hence, $w^{\prime}\left(s\right)\leq 0$ for all
$s>0.$ Therefore, $w\left(r\right)\leq w\left(\varepsilon\right)$ for
$r>\varepsilon.$ Letting $\varepsilon\rightarrow 0$ implies $w=0.$ This
contradiction shows that all ends of $M$ must be nonparabolic.

Assume now that $M$ has a nonparabolic end $E$ and let $F=M\setminus E.$ Since
both $E$ and $F$ are regular, there exists a harmonic function
$u:M\rightarrow\left(0,1\right)$ such that

$\lim_{x\rightarrow\infty,x\in E}u\left(x\right)=0\text{ \ and
}\lim_{x\rightarrow\infty,x\in F}u\left(x\right)=1.$

For $\phi\left(r\right)=r$ in (4) we obtain

(4.7) $tw^{\prime}\left(t\right)-w\left(t\right)\geq
sw^{\prime}\left(s\right)-w\left(s\right).$

However, the gradient estimate implies that $0<w\left(s\right)\leq Cs.$ In
particular, there exists a sequence $s_{n}\rightarrow 0$ such that

$\liminf_{n\rightarrow\infty}w^{\prime}\left(s_{n}\right)\geq 0.$

Hence, (4.7) implies that $tw^{\prime}\left(t\right)-w\left(t\right)\geq 0$
for all $t\in\left(0,1\right).$ Integrating in $t$ we get

(4.8) $\frac{w\left(t\right)}{t}\geq\frac{w\left(s\right)}{s}$

for all $0<s<t<1.$ However, the gradient estimate implies that
$w\left(t\right)\leq C\left(1-t\right)$ for $t<1.$ After letting $t\rightarrow
1$ in (4.8) we conclude that $w=0.$ This contradiction shows that $M$ has only
one end. ∎

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