# Heat kernel estimate for the Laplace-Beltrami operator under Bakry-Émery
Ricci curvature condition and applications

Xingyu Song, Ling Wu, and Meng Zhu  School of Mathematical Sciences and
Shanghai Key Laboratory of PMMP, East China Normal University, Shanghai
200241, China<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We establish a Gaussian upper bound of the heat kernel for the Laplace-
Beltrami operator on complete Riemannian manifolds with Bakry-Émery Ricci
curvature bounded below. As applications, we first prove an $L^{1}$-Liouville
property for non-negative subharmonic functions when the potential function of
the Bakry-Émery Ricci curvature tensor is of at most quadratic growth. Then we
derive lower bounds of the eigenvalues of the Laplace-Beltrami operator on
closed manifolds. An upper bound of the bottom spectrum is also obtained.

##  1\. Introduction

Let $(M^{n},g)$ be an $n$-dimensional Riemmanian manifold. The Bakry-Émery
Ricci curvature tensor of $M$ ([1]) is defined as

$\operatorname{Ric}_{f}:=\operatorname{Ric}+\operatorname{Hess}f,$ (1.1)

where $f$ is a smooth function on $M$ (called the potential function), and
$\operatorname{Ric}$ and $\operatorname{Hess}f$ denote the Ricci curvature
tensor and the hessian of $f$, respectively. It is clear that when $f$ is a
constant, $\operatorname{Ric}_{f}$ reduces to the Ricci curvature tensor.
Also, manifolds with lower Bakry-Émery Ricci curvature bound are closely
related to the singularity analysis of the Ricci flow, Ricci limit spaces, and
stationary black holes (see e.g., [17, 32, 36, 37, 28, 12]). Therefore, many
efforts have been made in extending the results under Ricci curvature
condition to Bakry-Émery Ricci curvature condition.

Since $\operatorname{Ric}_{f}$ appears in a Bochner type formula for the
weighted Laplace operator $\Delta_{f}:=\Delta-\left<\nabla f,\nabla\right>$
(see [1]) which is self-adjoint with respect to the weighted measure
$e^{-f}dv$ on $M$, the Bakry-Émery Ricci curvature can be considered as the
”Ricci curvature” for the smooth metric measure space $(M^{n},g,e^{-f}dv)$.
Here $\Delta$ is the Laplace-Beltrami operator, $\nabla f$ is the gradient of
$f$, and $dv$ is the volume form of $g$. Thus, while doing analysis under
Bakry-Émery Ricci curvature condition, one would naturally consider replacing
$\Delta$ and $dv$ with the weighted Laplace operator $\Delta_{f}$ and volume
form $e^{-f}dv$ (see e.g., [1, 27, 40, 11, 30, 26, 39, 42, 44, 45]).

Let us mention that there are many similar studies under assumptions on the
$m$-Bakry-Émery Ricci curvature,
$\operatorname{Ric}^{m}_{f}=\operatorname{Ric}+\operatorname{Hess}f-\frac{\nabla
f\otimes\nabla f}{m-n}$, where $m\in(n,+\infty)$ is a constant (see e.g., [2,
23, 25, 38, 41]). The Bakry-Émery Ricci curvature corresponds to the case
$m=+\infty$, and hence it is also called the $\infty$-Bakry-Émery Ricci
curvature.

In this paper, we treat the Bakry-Émery Ricci curvature as a generalization of
the Ricci curvature on $(M^{n},g)$ with the original volume measure $dv$, and
study the properties of elliptic and the heat equations related to the
Laplace-Beltrami operator. This type of study has previously been conducted by
Q. Zhang and the third author, and was applied in the extension of the
Cheeger-Colding-Naber theory and the proof of compactness theorems for
gradient Ricci solitons (see [49], [50], [16], [18]).

First, we investigate estimates of the heat kernel. On manifolds with Ricci
curvature bounded below, there have already been a long history and many
classical results about the heat kernel estimates, the readers may refer to
[10, 13, 35] and the references therein. For manifolds with Bakry-Émery Ricci
curvature, or $m$-Bakry-Émery Ricci curvature bounded below, the heat kernel
estimates for the weighted Laplace operator $\Delta_{f}$ have been studied in
many literatures recently (see [34, 23, 24, 43, 44, 45]).

Our first main result is a Gaussian upper bound of the heat kernel of $\Delta$
depending on the lower bound of the Bakry-Émery Ricci curvature and the bound
of the potential function $f$.

######  Theorem 1.1.

(Gaussian upper bound of the heat kernel) Let $(M^{n},g)$ be a complete
Riemannian manifold and $H(x,y,t)$ the heat kernel of $\Delta$ on $M\times
M\times(0,+\infty)$. Pick a fixed point $o\in M$ and $R>0$. Suppose that
$\operatorname{Ric}_{f}\geq-Kg$ on $B_{3R}(o)$ for some constant $K\geq 0$.
Then for any $\epsilon>0$, there exist constants $C_{1}(n,\epsilon)$ and
$C_{2}(n)$, such that

$H(x,y,t)\leq\frac{C_{1}(n,\epsilon)e^{C_{2}(n)(Kt+L(R))}}{\operatorname{Vol}(B_{\sqrt{t}}(x))^{\frac{1}{2}}\operatorname{Vol}(B_{\sqrt{t}}(y))^{\frac{1}{2}}}e^{\left(-\frac{d^{2}(x,y)}{(4+\epsilon)t}\right)}$
(1.2)

for all $x,y\in B_{\frac{1}{2}R}(o)$ and $0<t<R^{2}/4$, where
$L(R)=\sup\limits_{B_{3R}(o)}|f|$ and $\lim\limits_{\epsilon\to
0}C_{1}(n,\epsilon)=\infty$.

######  Remark 1.2.

Note that in general the heat kernel $H(x,y,t)$ of the Laplace-Beltrami
operator $\Delta$ and the heat kernel $H_{f}(x,y,t)$ of the weighted Laplace
operator $\Delta_{f}$ are not equivalent. Hence, one cannot expect to derive
(1.2) directly from the estimates of $H_{f}(x,y,t)$. Indeed, on the Euclidean
space $(\mathbb{R}^{n},g_{0})$ with $f=x_{1}$, we have (see e.g. [44])

$H_{f}(x,y,t)=\frac{e^{\frac{2(x_{1}+y_{1})-t}{4}}}{(4\pi
t)^{-\frac{n}{2}}}e^{-\frac{|x-y|^{2}}{4t}}=e^{\frac{2(x_{1}+y_{1})-t}{4}}H(x,y,t).$

The idea of the proof is standard. A key ingredient is a relative volume
comparison theorem (Theorem 2.1 below) under the assumptions in the theorem
above. Using the volume comparison, we can derive a Sobolev inequality, from
which a parabolic mean value property follows. Then the mean value property
and Davies’ double integral estimate will imply the heat kernel upper bound.

With the help of the heat kernel upper bound, one can prove an
$L^{1}$-Liouville property for non-negative subharmonic functions. On
manifolds with Ricci curvature bounded below, this was first proved by Li-
Schoen [21] and Li [19] , where the lower bound of the Ricci curvature is
allowed to have certain decay. We show that the same result still holds on
manifolds with Bakry-Émery Ricci curvature bounded below.

######  Theorem 1.3.

Let $(M^{n},g)$ be a complete noncompact Riemannian manifold with
$\operatorname{Ric}_{f}\geq-Kg$ for some constant $K\geq 0$. Assume that there
exist non-negative constants $a$ and $b$ such that

$|f(x)|\leq ar^{2}(x)+b\ for\ all\ x\in M,$ (1.3)

where $r(x)=d(x,o)$ is the geodesic distance function to a fixed point $o\in
M$. Then any non-negative $L^{1}$-integrable subharmonic function on $M$ must
be identically constant. In particular, any $L^{1}$-integrable harmonic
function must be identically constant.

######  Remark 1.4.

In [47], Yau obtained the $L^{\infty}$ Liouville property of harmonic
functions on manifolds with nonnegative Ricci curvature. Later, Yau [48]
showed that for $1<p<+\infty$ the $L^{p}$ Liouville property for subharmonic
functions actually holds without any curvature assumption.

The $L^{1}$ Liouville property for subharmonic functions with respect to the
weighted Laplace operator $\Delta_{f}$ on manifolds with Bakry-Émery Ricci
curvature (or $m$-Bakry-Émery Ricci curvature, resp.) bounded below was proved
by Wu-Wu [45] (X. Li [23], resp.).

The conditions in the above theorem are especially satisfied on so called
gradient Ricci solitons. A gradient Ricci soliton is a Riemannian manifold
$(M^{n},g)$ with constant Bakry-Émery Ricci curvature, namely,

$\operatorname{Ric}+\operatorname{Hess}f=\lambda g$ (1.4)

for some constant $\lambda$. It is called a shrinking, steady, or expanding
Ricci soliton when $\lambda>0$, $=0$, or $<0$, respectively.

For gradient Ricci solitons, it is well known that

$S+|\nabla f|^{2}=2\lambda f+C,$ (1.5)

where $S$ is the scalar curvature of $M$. Then from the results in [8] and [5]
, we know that $|f|$ is at most of quadratic and linear growth on gradient
shrinking and steady Ricci solitons, respectively. This implies that (1.3)
holds.

For gradient expanding solitons, it is showed in [33] that $S\geq n\lambda$,
then (1.5) implies that
$\left|\nabla\sqrt{-f-\frac{C-n\lambda}{2\lambda}}\right|\leq\sqrt{-\frac{\lambda}{2}}$.
Hence

$\sqrt{-f(x)-\frac{C-n\lambda}{2\lambda}}\leq\sqrt{-\frac{\lambda}{2}}r(x)+\sqrt{-f(o)-\frac{C-n\lambda}{2\lambda}},$

and the conditions in Theorem 1.3 continue to hold with $f$ replaced by
$\tilde{f}=f(x)+\frac{C-n\lambda}{2\lambda}$.

Therefore, Theorem 1.3 implies the following Liouville theorem for complete
gradient Ricci solitons.

######  Theorem 1.5.

Let $(M^{n},g)$ be a complete gradient Ricci soliton. Then any non-negative
$L^{1}$-integrable subharmonic function on $M$ must be identically constant.
In particular, any $L^{1}$-integrable harmonic function on $M$ must be
identically constant.

######  Remark 1.6.

In [29], Munteanu-Sesum proved that on a gradient shrinking Kähler-Ricci
soliton or a gradient steady Ricci soliton, harmonic functions with finite
energy must be constant.

By Theorem 1.3, following the arguments of Li in [19] we can prove a
uniqueness theorem for $L^{1}$-solutions of the heat equation.

######  Theorem 1.7.

Let $(M^{n},g)$ be a complete noncompact Riemannian manifold with
$\operatorname{Ric}_{f}\geq-Kg$ for some constant $K\geq 0$. Assume that there
exist non-negative constants $a$ and $b$ such that

$|f|(x)\leq ar^{2}(x)+b\ for\ all\ x\in M.$

If $u(x,t)$ is a non-negative function defined on $M\times[0,+\infty)$
satisfying

$\left(\Delta-\partial_{t}\right)u(x,t)\geq 0,\ \int_{M}u(x,t)dv<+\infty$

for all $t>0$, and $\lim\limits_{t\rightarrow 0}\int_{M}u(x,t)=0$, then
$u(x,t)\equiv 0$.

In particular, any $L^{1}$-solution of the heat equation is uniquely
determined by its initial data in $L^{1}$.

Another application of the heat kernel estimate is that one can get Li-Yau
type [22] lower bound estimates for the eigenvalues of $\Delta$ on closed
manifolds. More precisely, we show that

######  Theorem 1.8.

Let $(M^{n},g)$ be a closed Riemannian manifold with
$\operatorname{Ric}_{f}\geq-Kg$ and $|f|\leq L$. Let
$0=\lambda_{0}<\lambda_{1}\leq\lambda_{2}\leq\cdots\leq\lambda_{k}\leq\cdots$
be the eigenvalues of the Laplace-Beltrami operator. Then there exist
constants $C_{3}(n)$ and $C_{4}(n)$ such that

$\lambda_{k}\geq\frac{C_{3}(n)(k+1)^{\frac{2}{n}}}{D^{2}}e^{-C_{4}(n)(KD^{2}+L)}$
(1.6)

for all $k\geq 1$, where $D$ is an upper bound of the diameter of $M$.

Finally, to get an upper bound of $\lambda_{1}$, we may relax the requirements
on the manifold and $f$. Indeed, we can obtain a Cheng type upper bound of the
bottom spectrum $\mu_{1}(\Delta)$ of $\Delta$ on complete manifolds with $f$
being of at most linear growth.

######  Theorem 1.9.

Let $(M^{n},g)$ be a complete Riemannian manifold with
$\operatorname{Ric}_{f}\geq-(n-1)Kg$ for some constant $K\geq 0$. Assume that
there exist non-negative constants $\tilde{a}$ and $\tilde{b}$ such that

$|f|(x)\leq\tilde{a}r(x,o)+\tilde{b}\ for\ all\ x\in M.$

Then we have

$\mu_{1}(\Delta)\leq\frac{1}{4}\left(2\tilde{a}+(n-1)\sqrt{K}\right)^{2}.$

In particular, if $f$ is of sublinear growth, then the bottom spectrum of the
Laplacian has the following sharp upper bound:

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

When $f$ is constant, i.e, $\operatorname{Ric}\geq-(n-1)Kg$, Theorem 1.9
reduces to Cheng’s theorem (see Theorem 4.2 in [4]). This incidentally
indicates that our estimate is sharp.

The rest of the paper is organized as follows. In section 2, we derive a
Laplacian and volume comparison theorem for complete Riemannian manifolds with
Bakry-Émery Ricci curvature bounded below. Using the comparison theorem, we
get Poincaré and Sobolev inequalities. Section 3 is devoted to the proofs of a
mean value inequality for non-negative subsolutions of the heat equation and
the Gaussian upper bound (1.2) of the heat kernel. The $L^{1}$-Liouville
property for non-negative subharmonic functions and the uniqueness theorem for
$L^{1}$-solutions of the heat equation will be obtained in section 4. Then for
the purpose of comparison, we take a detour in section 5 to discuss an
$L^{\infty}$-Liouville property for harmonic function with polynomial growth.
Finally, lower bounds of the eigenvalues and an upper bound of the bottom
spectrum of the Beltrami-Laplace operator are shown in section 6.

##  2\. Poincaré and Sobolev inequalities

In this section, we derive local Poincaré and Sobolev inequalities which are
instrumental in proving the main results of the paper. A crucial step is to
show the following volume comparison theorem, the idea of whose proof is
similar to [46], [30] and [49], but our result is slightly more general.

For a fixed point $o\in M$ and $R>0$, we define

$L(R)=\sup\limits_{x\in B_{3R}(o)}|f(x)|,$ (2.1)

where $B_{3R}(o)$ is the geodesic ball centered at $o\in M$ with radius $3R$.

######  Theorem 2.1.

Let $(M^{n},g)$ be a complete Riemannian manifold with
$\operatorname{Ric}_{f}\geq-Kg$ for some constant $K\geq 0$. Then the
following conclusions are true.
(a)(Laplacian comparison) Let $r=d(y,p)$ be the distance from any point $y$ to
some fixed point $p\in B_{R}(o)$ with $0<r<R$. Then for $0<r_{1}<r_{2}<R$, we
have

$\int_{r_{1}}^{r_{2}}(\Delta
r-\frac{n-1}{r})dr\leq\frac{K}{6}(r_{2}^{2}-r_{1}^{2})+6L(R).$ (2.2)

(b)(Volume element comparison)Take any point $p\in B_{R}(o)$ and denote the
volume form in geodesic polar coordinates centered at $p$ with
$J(r,\theta,p)drd\theta$, where $r>0$ and $\theta\in S_{p}(M)$, a unit tangent
vector at $p$. Then for $0<r_{1}<r_{2}<R$, we have

$\frac{J(r_{2},\theta,p)}{J(r_{1},\theta,p)}\leq\left(\frac{r_{2}}{r_{1}}\right)^{n-1}e^{\frac{K}{6}(r_{2}^{2}-r_{1}^{2})+6L(R)}.$
(2.3)

(c)(Volume comparison)For any $p\in B_{R}(o),\ 0<r_{1}<r_{2}<R$, we have

$\frac{\operatorname{Vol}(B_{r_{2}}(p))}{\operatorname{Vol}(B_{r_{1}}(p))}\leq\left(\frac{r_{2}}{r_{1}}\right)^{n}e^{\frac{K}{6}(r_{2}^{2}-r_{1}^{2})+6L(R)},$
(2.4)

where $\operatorname{Vol}(.)$ denotes the volume of a region.

Proof of part (a) Let $r=d(y,p)$ be the distance from any point $y$ to some
fixed point $p\in B_{R}(o)$ with $0<r<R$ and $\gamma:[0,r]\to M$ a normal
minimal geodesic with $\gamma(0)=p$ and $\gamma(r)=y$. Then we know
$\gamma(t)\subset B_{3R}(o)$ and $y\in B_{3R}(o)$.
From the Bochner formula, we have

$0=\frac{1}{2}\Delta|\nabla
r|^{2}=|\operatorname{Hess}r|^{2}+\left<\nabla\Delta r,\nabla
r\right>+\operatorname{Ric}(\partial r,\partial r).$

By using Cauchy-Schwarz inequality
$|\operatorname{Hess}r|^{2}\geq\frac{(\Delta r)^{2}}{n-1}$, it yields

$\frac{\partial}{\partial r}(\Delta r)+\frac{(\Delta
r)^{2}}{n-1}\leq-\operatorname{Ric}(\partial r,\partial r),$

which is equivalent to

$\frac{1}{r^{2}}\frac{\partial}{\partial r}(r^{2}\Delta
r)+\frac{1}{n-1}\left(\Delta
r-\frac{n-1}{r}\right)^{2}\leq\frac{n-1}{r^{2}}-\operatorname{Ric}(\partial
r,\partial r).$ (2.5)

Multiplying both sides of (2.5) by $r^{2}$ and integrating from $0$ to $r$, we
get

$\Delta
r\leq\frac{n-1}{r}-\frac{1}{r^{2}}\int_{0}^{r}t^{2}\operatorname{Ric}(\gamma^{\prime}(t),\gamma^{\prime}(t))dt.$
(2.6)

We observe that

$\displaystyle\operatorname{Ric}(\gamma^{\prime}(t),\gamma^{\prime}(t))$
$\displaystyle\geq-K-\operatorname{Hess}f(\gamma^{\prime}(t),\gamma^{\prime}(t))$
$\displaystyle=-K-\left<\nabla_{\gamma^{\prime}(t)}\nabla
f,\gamma^{\prime}(t)\right>$ $\displaystyle=-K-\gamma^{\prime}(t)\left<\nabla
f,\gamma^{\prime}(t)\right>+\left<\nabla
f,\nabla_{\gamma^{\prime}(t)}\gamma^{\prime}(t)\right>$
$\displaystyle=-K-f^{\prime\prime}(t),$

where $f(t):=f(\gamma(t))$.
Hence, (2.6) becomes

$\displaystyle\Delta r-\frac{n-1}{r}$
$\displaystyle\leq\frac{1}{r^{2}}\int_{0}^{r}\left[Kt^{2}+t^{2}f^{\prime\prime}(t)\right]dt$
(2.7)
$\displaystyle=\frac{1}{r^{2}}\left[\frac{K}{3}r^{3}+r^{2}f^{\prime}(r)-2\int_{0}^{r}f^{\prime}(t)tdt\right]$
$\displaystyle=\frac{K}{3}r+f^{\prime}(r)-\frac{2}{r}f(r)+\frac{2}{r^{2}}\int_{0}^{r}f(t)dt.$

For $0<r_{1}<r_{2}<R$, integrating (2.7) from $r_{1}$ to $r_{2}$ yields

$\displaystyle\int_{r_{1}}^{r_{2}}\left(\Delta r-\frac{n-1}{r}\right)dr$
$\displaystyle\leq\frac{K}{6}(r_{2}^{2}-r_{1}^{2})+\int_{r_{1}}^{r_{2}}\left(f^{\prime}(r)-\frac{2}{r}f(r)\right)dr-2\int_{r_{1}}^{r_{2}}\int_{0}^{r}f(t)dtd\frac{1}{r}$
(2.8)
$\displaystyle=\frac{K}{6}(r_{2}^{2}-r_{1}^{2})+f(r_{2})-f(r_{1})-2\int_{r_{1}}^{r_{2}}\frac{f(r)}{r}dr$
$\displaystyle\ \ \
-2\left[\left.\frac{1}{r}\int_{0}^{r}f(t)dt\right|^{r_{2}}_{r_{1}}-\int_{r_{1}}^{r_{2}}\frac{f(r)}{r}dr\right]$
$\displaystyle=\frac{K}{6}(r_{2}^{2}-r_{1}^{2})+f(r_{2})-f(r_{1})-\frac{2\int_{0}^{r_{2}}f(t)dt}{r_{2}}+\frac{2\int_{0}^{r_{1}}f(t)dt}{r_{1}}$
$\displaystyle\leq\frac{K}{6}(r_{2}^{2}-r_{1}^{2})+6L(R).$

Proof of part (b) By the first variation of the area

$\Delta r=\frac{J^{\prime}(r,\theta,p)}{J(r,\theta,p)},$

where $r(y)=d(y,p)$. For $0<r_{1}<r_{2}<R$, integrating this from $r_{1}$ to
$r_{2}$ we get

$\int_{r_{1}}^{r_{2}}\frac{\partial}{\partial r}\ln
J(r,\theta,p)dr\leq\int_{r_{1}}^{r_{2}}\frac{n-1}{r}dr+\frac{K}{6}(r_{2}^{2}-r_{1}^{2})+6L(R).$

Then we have

$\frac{J(r_{2},\theta,p)}{J(r_{1},\theta,p)}\leq\left(\frac{r_{2}}{r_{1}}\right)^{n-1}e^{\frac{K}{6}(r_{2}^{2}-r_{1}^{2})+6L(R)}.$

Proof of part (c) For $0<r_{1}<r_{2}<R$, we have

$\frac{\operatorname{Vol}(B_{r_{2}}(p))}{\operatorname{Vol}(B_{r_{1}}(p))}=\frac{\int_{0}^{r_{2}}\int_{S^{n-1}}J(r,\theta,p)d\theta
dr}{\int_{0}^{r_{1}}\int_{S^{n-1}}J(r,\theta,p)d\theta
dr}=\frac{\frac{r_{2}}{r_{1}}\int_{0}^{r_{1}}\int_{S^{n-1}}J(\frac{r_{2}}{r_{1}}t,\theta,p)d\theta
dt}{\int_{0}^{r_{1}}\int_{S^{n-1}}J(t,\theta,p)d\theta dt},$

where $S^{n-1}$ denotes the unit sphere in $\mathbb{R}^{n}$ and $d\theta$ is
its volume element.
By volume element comparison, for $0<t<r_{1}$, we have

$\displaystyle J\left(\frac{r_{2}}{r_{1}}t,\theta,p\right)$ $\displaystyle\leq
J(t,\theta,p)\left(\frac{r_{2}}{r_{1}}\right)^{n-1}e^{\frac{K}{6}\left(\frac{r_{2}^{2}}{r_{1}^{2}}t^{2}-t^{2}\right)+6L(R)}$
$\displaystyle\leq
J(t,\theta,p)\left(\frac{r_{2}}{r_{1}}\right)^{n-1}e^{\frac{K}{6}(r_{2}^{2}-r_{1}^{2})+6L(R)}.$

Integrating this gives that

$\frac{\operatorname{Vol}(B_{r_{2}}(p))}{\operatorname{Vol}(B_{r_{1}}(p))}\leq\left(\frac{r_{2}}{r_{1}}\right)^{n}e^{\frac{K}{6}(r_{2}^{2}-r_{1}^{2})+6L(R)}.$

∎

By Theorem 2.1, following Buser’s proof [3] or Saloff-Coste’s alternative
proof (Theorem 5.6.5 in [35]), we can get a local Neumann Poincaré inequality,
see also Lemma 3.1 in [30].

######  Lemma 2.2.

Let $(M^{n},g)$ be a complete Riemannian manifold with
$\operatorname{Ric}_{f}\geq-Kg$ for some constant $K\geq 0$. Then for any
$p\in B_{R}(o)$, there exist constants $c_{1}$ and $c_{2}$ depending only on
$n$ such that

$\int_{B_{r}(p)}|u-u_{B_{r}(p)}|^{2}\leq
c_{1}e^{c_{2}(Kr^{2}+L(R))}r^{2}\int_{B_{r}(p)}|\nabla u|^{2}$

for all $0<r<R$, where $u\in C^{\infty}\left(B_{r}(p)\right)$ and
$u_{B_{r}(p)}=\frac{\int_{B_{r}(p)}u}{\operatorname{Vol}(B_{r}(p))}.$

Combining Theorem 2.1 and Lemma 2.2, and using a similar argument as in the
proof of Lemma 3.2 in [30], we obtain a local Neumann Sobolev inequality.

######  Theorem 2.3.

Let $(M^{n},g)$ be a complete Riemannian manifold with
$\operatorname{Ric}_{f}\geq-Kg$ for some constant $K\geq 0$. Then there exist
constants $\mu=4n-2>2$, $c_{3}$ and $c_{4}$ depending only on $n$ such that

$\left(\int_{B_{r}(o)}|u-u_{B_{r}(o)}|^{\frac{2\mu}{\mu-2}}\right)^{\frac{\mu-2}{\mu}}\leq\frac{c_{3}e^{c_{4}(Kr^{2}+L(R))}}{\operatorname{Vol}(B_{r}(o))^{\frac{2}{\mu}}}r^{2}\int_{B_{r}(o)}|\nabla
u|^{2}$ (2.9)

for all $0<r<R$, where $u\in C^{\infty}\left(B_{r}(o)\right)$ and
$u_{B_{r}(o)}=\frac{\int_{B_{r}(o)}u}{\operatorname{Vol}(B_{r}(o))}.$

By the Minkowski inequality and applying (2.9), it is well known that one can
get the following Sobolev inequality.

######  Theorem 2.4.

Let $(M^{n},g)$ be a complete Riemannian manifold with
$\operatorname{Ric}_{f}\geq-Kg$ for some constant $K\geq 0$. Then there exist
constants $\mu=4n-2>2$, $c_{5}$ and $c_{6}$, all depending only on $n$ such
that

$\left(\int_{B_{r}(o)}|u|^{\frac{2\mu}{\mu-2}}\right)^{\frac{\mu-2}{\mu}}\leq\frac{c_{5}e^{c_{6}(Kr^{2}+L(R))}}{\operatorname{Vol}(B_{r}(o))^{\frac{2}{\mu}}}r^{2}\int_{B_{r}(o)}(|\nabla
u|^{2}+r^{-2}u^{2})$ (2.10)

for all $0<r<R$, where $u\in C^{\infty}\left(B_{r}(o)\right)$.

##  3\.  Mean value inequality and Gaussian upper bounds of the heat kernel

In this section, we apply Theorem 2.4 to prove a mean value inequality for the
non-negative subsolution of the heat equation and the Gaussian upper bound
(1.2) of the heat kernel on complete Riemannian manifolds with Bakry-Émery
Ricci curvature bounded below.

In the following context, for function $u$ on $M$, the $L^{q}$ norm on a
domain $\Omega\subset M$ is denoted by

$||u||_{q,\Omega}={\left(\int_{\Omega}|f|^{q}\right)^{\frac{1}{q}}}.$

$||u||_{q}$ denotes the $L^{q}$ norm of $u$ on $M$.

######  Proposition 3.1.

(Mean value inequality) Let $(M^{n},g)$ be a complete Riemannian manifold with
$\operatorname{Ric}_{f}\geq-Kg$ for some constant $K\geq 0$. For any real
number $s$ and any $0<\delta<\delta^{\prime}\leq 1$, let $u$ be a smooth non-
negative subsolution of the heat equation in the cylinder
$Q=B_{r}(o)\times(s-r^{2},s)$, $0<r<R$.

For $2\leq p<\infty$, there exist constants $\tilde{c}_{1}(n)$ and
$\tilde{c}_{2}(n)$ such that

$\sup\limits_{Q_{\delta}}u^{p}\leq\frac{\tilde{c}_{1}(n)e^{\tilde{c}_{2}(n)(Kr^{2}+L(R))}}{(\delta^{\prime}-\delta)^{4n}r^{2}\operatorname{Vol}(B_{r}(o))}\cdot\int_{Q_{\delta^{\prime}}}u^{p}dvdt.$
(3.1)

For $0<p<2$, there exist constants $\tilde{c}_{3}(n,p)$ and $\tilde{c}_{4}(n)$
such that

$\sup\limits_{Q_{\delta}}u^{p}\leq\frac{\tilde{c}_{3}(n,p)e^{\tilde{c}_{4}(n)(Kr^{2}+L(R))}}{(\delta^{\prime}-\delta)^{4n}r^{2}\operatorname{Vol}(B_{r}(o))}\cdot\int_{Q_{\delta^{\prime}}}u^{p}dvdt.$
(3.2)

Here $Q_{\delta}=B_{\delta r}(o)\times(s-\delta r^{2},s)$,
$Q_{\delta^{\prime}}=B_{\delta^{\prime}r}(o)\times(s-\delta^{\prime}r^{2},s)$,
$L(R)=\sup\limits_{B_{3R}(o)}|f|$.

###### Proof.

The proof is similar to Theorem 5.2.9 in [35]. We need to carefully examine
the explicit coefficients of the mean value inequality in terms of the Sobolev
constants in (2.10).

Without loss of generality we may assume $\delta^{\prime}=1.$ Denote
$B=B_{r}(o)$ for simplicity. For any non-negative function $\phi\in
C_{0}^{\infty}(B),$ we have

$\int_{B}(\phi u_{t}+\left<\nabla u,\nabla\phi\right>)dv\leq 0.$

In particular, when $\phi=\Phi^{2}u,\Phi\in C_{0}^{\infty}(B)$, we obtain

$\displaystyle\int_{B}(\Phi^{2}uu_{t}+\Phi^{2}|\nabla u|^{2})dv$
$\displaystyle\leq 2\left|\int_{B}u\Phi\left<\nabla
u,\nabla\Phi\right>dv\right|$ $\displaystyle\leq
3\int_{B}|\nabla\Phi|^{2}u^{2}dv+\frac{1}{3}\int_{B}\Phi^{2}|\nabla u|^{2}dv.$

It then implies that

$\displaystyle\int_{B}(2\Phi^{2}uu_{t}+|\nabla(\Phi u)|^{2})dv$
$\displaystyle\leq
2\int_{B}\Phi^{2}uu_{t}dv+\frac{4}{3}\int_{B}\Phi^{2}|\nabla
u|^{2}dv+4\int_{B}|\nabla\Phi|^{2}u^{2}dv$ $\displaystyle\leq
10||\nabla\Phi||_{\infty}^{2}\int_{\operatorname{supp}(\Phi)}u^{2}dv.$

For any smooth non-negative function $\lambda(t)$ of the time variable $t$,
which will be chosen later, we get

$\displaystyle\frac{\partial}{\partial t}$
$\displaystyle\left(\int_{B}(\lambda\Phi
u)^{2}dv\right)+\lambda^{2}\int_{B}|\nabla(\Phi u)|^{2}dv$ $\displaystyle\leq
2\lambda|\lambda^{\prime}|\sup\Phi^{2}\int_{\operatorname{supp}(\Phi)}u^{2}dv+\lambda^{2}\left(2\int_{B}\Phi^{2}uu_{t}dv+\int_{B}|\nabla(\Phi
u)|^{2}dv\right)$ $\displaystyle\leq
C\lambda(\lambda||\nabla\Phi||_{\infty}^{2}+|\lambda^{\prime}|\sup\Phi^{2})\int_{\operatorname{supp}(\Phi)}u^{2}dv,$

where $C$ is a constant which will change from line to line in the following.

Now we choose $\Phi$ and $\lambda(t)$ such that, for any
$0<\sigma^{\prime}<\sigma<1,\ w=\sigma-\sigma^{\prime},$

(1)$0\leq\Phi\leq 1$, supp($\Phi)\subset B_{\sigma r}(o)$, $\Phi=1$ in
$B_{\sigma^{\prime}r}(o)$ and $|\nabla\Phi|\leq 2(wr)^{-1}$;
(2)$0\leq\lambda\leq 1$, $\lambda=0$ in $(-\infty,s-\sigma r^{2})$,
$\lambda=1$ in $(s-\sigma^{\prime}r^{2},+\infty)$, and
$|\lambda^{\prime}(t)|\leq 2(wr)^{-2}$.

Let $I_{\sigma}=(s-\sigma r^{2},s)$ and
$I_{\sigma^{\prime}}=(s-\sigma^{\prime}r^{2},s)$. For any $t\in
I_{\sigma^{\prime}}$, integrating the above inequality over $(s-\sigma
r^{2},t)$, we obtain

$\sup\limits_{I_{\sigma^{\prime}}}\left(\int_{B}\Phi^{2}u^{2}dv\right)\leq
C(wr)^{-2}\int_{Q_{\sigma}}u^{2}dvdt,$ (3.3)

and

$\int_{B\times I_{\sigma^{\prime}}}|\nabla(\Phi u)|^{2}dvdt\leq
C(wr)^{-2}\int_{Q_{\sigma}}u^{2}dvdt.$ (3.4)

On the other hand, by the Hölder inequality and the Sobolev inequality in
Theorem 2.4, for some constant $\mu=4n-2$, we have

$\displaystyle\int_{B}g^{2(1+\frac{2}{\mu})}dv$
$\displaystyle\leq\left(\int_{B}|g|^{\frac{2\mu}{\mu-2}}dv\right)^{\frac{\mu-2}{\mu}}\left(\int_{B}|g|^{2}dv\right)^{\frac{2}{\mu}}$
(3.5)
$\displaystyle\leq\left(\int_{B}|g|^{2}dv\right)^{\frac{2}{\mu}}\left(E(B)\int_{B}(|\nabla
g|^{2}+r^{-2}g^{2})dv\right)$

for all $g\in C^{\infty}(B)$, where
$E(B)=\frac{\tilde{c}_{5}(n)e^{\tilde{c}_{6}(n)(Kr^{2}+L(R))}}{\operatorname{Vol}(B_{r}(o))^{\frac{2}{\mu}}}r^{2}$.

Setting $g=\Phi u$, $\theta=1+\frac{2}{\mu}$, (3.5) becomes

$\int_{B}(\Phi u)^{2\theta}dv\leq\left(\int_{B}(\Phi
u)^{2}dv\right)^{\frac{2}{\mu}}\left(E(B)\int_{B}(|\nabla(\Phi
u)|^{2}+r^{-2}(\Phi u)^{2})dv\right).$

Combining (3.3), (3.4) and integrating the above inequality over
$(s-\sigma^{\prime}r^{2},s)$, we obtain

$\int_{s-\sigma^{\prime}r^{2}}^{s}\int_{B}(\Phi
u)^{2\theta}dvdt\geq\int_{Q_{\sigma^{\prime}}}u^{2\theta}dvdt,$

$\int_{B}(\Phi u)^{2}dv\leq\sup\limits_{I_{\sigma^{\prime}}}\int_{B}(\Phi
u)^{2}dv\leq C(wr)^{-2}\int_{Q_{\sigma}}u^{2}dv,$

and

$\displaystyle\int_{s-\sigma^{\prime}r^{2}}^{s}\int_{B}\left(|\nabla(\Phi
u)|^{2}+r^{-2}(\Phi u)^{2}\right)dvdt$ $\displaystyle=\int_{B\times
I_{\sigma^{\prime}}}|\nabla(\Phi
u)|^{2}dvdt+\int_{s-\sigma^{\prime}r^{2}}^{s}\int_{B}r^{-2}(\Phi u)^{2}dvdt$
$\displaystyle\leq\int_{B\times I_{\sigma^{\prime}}}|\nabla(\Phi
u)|^{2}dvdt+\sigma^{\prime}r^{2}\times
r^{-2}\sup\limits_{I_{\sigma^{\prime}}}\int_{B}(\Phi u)^{2}dv$
$\displaystyle\leq C(wr)^{-2}\int_{Q_{\sigma}}u^{2}dvdt,$

which implies

$\int_{Q_{\sigma^{\prime}}}u^{2\theta}dvdt\leq
E(B)\left(C(wr)^{-2}\int_{Q_{\sigma}}u^{2}dvdt\right)^{\theta}.$

For any $m\geq 1$, $u^{m}$ is also a smooth non-negative subsolution of the
heat equation. Hence the above inequality indeed implies

$\int_{Q_{\sigma^{\prime}}}u^{2m\theta}dvdt\leq
E(B)\left(C(wr)^{-2}\int_{Q_{\sigma}}u^{2m}dvdt\right)^{\theta}$ (3.6)

for $m\geq 1$.

Let $w_{i}=(1-\delta)2^{-i}$, which satisfies
$\sum_{1}^{\infty}w_{i}=1-\delta$. Let $\sigma_{0}=1$,
$\sigma_{i+1}=\sigma_{i}-w_{i+1}\\\ =1-\sum_{j=1}^{i+1}w_{j}$. Applying (3.6)
for $m=\theta^{i}$, $\sigma=\sigma_{i}$, $\sigma^{\prime}=\sigma_{i+1}$, we
have

$\int_{Q_{\sigma_{i+1}}}u^{2\theta^{i+1}}dvdt\leq
E(B)\left[C^{i+1}((1-\delta)r)^{-2}\int_{Q_{\sigma_{i}}}u^{2\theta^{i}}dvdt\right]^{\theta},$

i.e.,

$\left(\int_{Q_{\sigma_{i+1}}}u^{2\theta^{i+1}}dvdt\right)^{\theta^{-(i+1)}}\leq
E(B)^{\theta^{-(i+1)}}C^{(i+1)\theta^{-i}}((1-\delta)r)^{-2\theta^{-i}}\left(\int_{Q_{\sigma_{i}}}u^{2\theta^{i}}dvdt\right)^{\theta^{-i}}.$

Iterating from $i=0$ to $\infty$, we obtain

$\sup\limits_{Q_{\delta}}u^{2}\leq
E(B)^{\sum_{i=0}^{\infty}\theta^{-(i+1)}}C^{\sum_{i=0}^{\infty}(i+1)\theta^{-i}}((1-\delta)r)^{-2\sum_{i=0}^{\infty}\theta^{-i}}\int_{Q}u^{2}dvdt.$

Therefore

$\sup\limits_{Q_{\delta}}u^{2}\leq
C(n)E(B)^{\frac{\mu}{2}}((1-\delta)r)^{-(\mu+2)}||u||_{2,Q}^{2},$

i.e.,

$\sup\limits_{Q_{\delta}}u^{2}\leq\frac{\tilde{c}_{7}(n)e^{\tilde{c}_{8}(n)(Kr^{2}+L(R))}}{(1-\delta)^{4n}r^{2}\operatorname{Vol}(B_{r}(o))}\int_{Q}u^{2}dvdt.$
(3.7)

Formula (3.7) is in fact an $L^{2}$-mean value inequality. The case $p\geq 2$
immediately follows, since for any smooth non-negative subsolution $u$ of the
heat equation, $u^{\frac{p}{2}}$, $p\geq 2$ is also a smooth non-negative
subsolution of the heat equation.

Next, for $0<p<2$, we will apply (3.7) to prove (3.2) by a different iterative
argument. Let $\sigma\in(0,1)$ and $\eta=\sigma+(1-\sigma)/4$. Then (3.7)
implies

$\sup\limits_{Q_{\sigma}}u\leq
F(B)(1-\sigma)^{(-1-\frac{\mu}{2})}\left(\int_{Q_{\eta}}u^{2}dvdt\right)^{\frac{1}{2}},$

where
$F(B)=\frac{\tilde{c}_{9}(n)e^{\tilde{c}_{10}(n)(Kr^{2}+L(R))}}{r\operatorname{Vol}(B_{r}(o))^{\frac{1}{2}}}$.

Since
$\left(\int_{Q_{\eta}}u^{2}dvdt\right)^{\frac{1}{2}}=\left(\int_{Q_{\eta}}u^{p}u^{2-p}dvdt\right)^{\frac{1}{2}}\leq\sup\limits_{Q_{\eta}}u^{1-\frac{p}{2}}\left(\int_{Q}u^{p}dvdt\right)^{\frac{1}{2}}$,
we have

$||u||_{\infty,Q_{\sigma}}=\sup\limits_{Q_{\sigma}}u\leq
F(B)(1-\sigma)^{(-1-\frac{\mu}{2})}||u||_{p,Q}^{\frac{p}{2}}||u||_{\infty,Q_{\eta}}^{1-\frac{p}{2}}.$
(3.8)

Now fix $\delta\in(0,1)$ and let $\sigma_{0}=\delta$,
$\sigma_{i+1}=\sigma_{i}+(1-\sigma_{i})/4$, which satisfy
$1-\sigma_{i}=(\frac{3}{4})^{i}(1-\delta)$. Applying (3.8) to
$\sigma=\sigma_{i}$, and $\eta=\sigma_{i+1}$, we have

$||u||_{\infty,Q_{\sigma_{i}}}\leq(\frac{4}{3})^{(1+\frac{\mu}{2})i}F(B)||u||_{p,Q}^{\frac{p}{2}}(1-\delta)^{(-1-\frac{\mu}{2})}||u||_{\infty,Q_{\sigma_{i+1}}}^{1-\frac{p}{2}}.$

Therefore, for any $i$,

$||u||_{\infty,Q_{\delta}}\leq(\frac{4}{3})^{(1+\frac{\mu}{2})\sum_{j}j(1-\frac{p}{2})^{j}}\times[F(B)||u||_{p,Q}^{\frac{p}{2}}(1-\delta)^{(-1-\frac{\mu}{2})}]^{\sum_{j}(1-\frac{p}{2})^{j}}||u||_{\infty,Q_{\sigma_{i}}}^{(1-\frac{p}{2})^{i}},$

where $\sum$ denotes the summations from $0$ to $i-1$. Letting $i\to\infty$ we
get

$||u||_{\infty,Q_{\delta}}\leq\left(\frac{4}{3}\right)^{\tilde{c}(n,p)}\times[F(B)||u||_{p,Q}^{\frac{p}{2}}(1-\delta)^{(-1-\frac{\mu}{2})}]^{\frac{2}{p}},$

that is

$\sup\limits_{Q_{\delta}}u^{p}\leq\frac{\tilde{c}_{11}(n,p)e^{\tilde{c}_{12}(n)(Kr^{2}+L(R))}}{(1-\delta)^{4n}r^{2}\operatorname{Vol}(B_{r}(o))}\int_{Q}u^{p}dvdt,\
0<p<2.$

Then the proposition follows.∎

To get the Gaussian upper bound of the heat kernel, let us first recall
Davies’ double integral estimate [10].

######  Lemma 3.2.

(Davies [10]) Let $(M^{n},g)$ be a complete Riemannian manifold and $H(x,y,t)$
the heat kernel. Let $\mu_{1}(M)\geq 0$ be the greatest lower bound for the
$L^{2}$-spectrum of the Laplacian $\Delta$ on $M$. Assume that $B_{1}$ and
$B_{2}$ are bounded subsets of $M$. Then

$\int_{B_{1}}\int_{B_{2}}H(x,y,t)dydx\leq\operatorname{Vol}(B_{1})^{\frac{1}{2}}\operatorname{Vol}(B_{2})^{\frac{1}{2}}e^{\left(-\frac{d^{2}(B_{1},B_{2})}{4t}-\mu_{1}(M)t\right)},$

where $d(B_{1},B_{2})$ denotes the distance between $B_{1}$ and $B_{2}$.

Now we prove the Gaussian upper bound of the heat kernel by applying the mean
value inequality in Proposition 3.1 and Lemma 3.2.

Proof of Theorem 1.1. Fixing $x\in B_{\frac{1}{2}R}(o)$ and applying
Proposition 3.1 to the heat kernel $H(x,y,t)$ with
$Q=B_{\sqrt{t}}(y)\times(t-(\sqrt{t})^{2},t),\delta=\frac{1}{8}$ and
$\delta^{\prime}=\frac{1}{4}$, we have

$\displaystyle H(x,y,t)\leq\sup\limits_{(z,s)\in Q_{\delta}}H(x,z,s)$
$\displaystyle\leq\frac{\overline{c}_{1}(n)e^{\overline{c}_{2}(n)(Kt+\sup\limits_{B_{3\sqrt{t}}(y)}|f|)}}{\left(\frac{1}{8}\right)^{4n}t\operatorname{Vol}(B_{\sqrt{t}}(y))}\int_{t-\frac{1}{4}t}^{t}\int_{B_{\frac{1}{4}\sqrt{t}}(y)}H(x,z,s)dzds$
(3.9)
$\displaystyle\leq\frac{\overline{c}_{1}(n)e^{\overline{c}_{2}(n)(Kt+L(R))}}{\left(\frac{1}{8}\right)^{4n}t\operatorname{Vol}(B_{\sqrt{t}}(y))}\int_{t-\frac{1}{4}t}^{t}\int_{B_{\frac{1}{4}\sqrt{t}}(y)}H(x,z,s)dzds$
$\displaystyle\leq\frac{\overline{c}_{1}(n)e^{\overline{c}_{2}(n)(Kt+L(R))}}{\left(\frac{1}{8}\right)^{4n}t\operatorname{Vol}(B_{\sqrt{t}}(y))}\int_{\frac{3}{4}t}^{t}\int_{B_{\sqrt{t}}(y)}H(x,z,s)dzds$
$\displaystyle=\frac{\overline{c}_{3}(n)e^{\overline{c}_{2}(n)(Kt+L(R))}}{\operatorname{Vol}(B_{\sqrt{t}}(y))}\int_{B_{\sqrt{t}}(y)}H(x,z,s^{\prime})dz$

for some $s^{\prime}\in\left(\frac{3}{4}t,t\right)$, where
$Q_{\delta}=B_{\frac{1}{8}\sqrt{t}}(y)\times(t-\frac{1}{8}(\sqrt{t})^{2},t)$
and $B_{\sqrt{t}}(y)\subset B_{R}(o)$ for any $y\in B_{\frac{1}{2}R}(o)$ and
$0<t<\frac{R^{2}}{4}$.

Fixing $z\in B_{\sqrt{t}}(y)$ and applying Proposition 3.1 to the heat kernel
$H(x,z,s^{\prime})$ with $Q=B_{\sqrt{t}}(x)\times(t-(\sqrt{t})^{2},t)$,
$\delta=\frac{1}{4}$ and $\delta^{\prime}=\frac{1}{2}$, we have

$\displaystyle
H(x,z,s^{\prime})\leq\sup\limits_{(x^{\prime},s^{\prime\prime})\in
Q_{\delta}}H(x^{\prime},z,s^{\prime\prime})$
$\displaystyle\leq\frac{\overline{c}_{4}(n)e^{\overline{c}_{5}(n)(Kt+L(R))}}{\left(\frac{1}{4}\right)^{4n}t\operatorname{Vol}(B_{\sqrt{t}}(x))}\int_{t-\frac{1}{2}t}^{t}\int_{B_{\frac{1}{2}\sqrt{t}}(x)}H(w,z,\tau)dwd\tau$
(3.10)
$\displaystyle\leq\frac{\overline{c}_{4}(n)e^{\overline{c}_{5}(n)(Kt+L(R))}}{\left(\frac{1}{4}\right)^{4n}t\operatorname{Vol}(B_{\sqrt{t}}(x))}\int_{\frac{1}{2}t}^{t}\int_{B_{\sqrt{t}}(x)}H(w,z,\tau)dwd\tau$
$\displaystyle=\frac{\overline{c}_{6}(n)e^{\overline{c}_{5}(n)(Kt+L(R))}}{\operatorname{Vol}(B_{\sqrt{t}}(x))}\int_{B_{\sqrt{t}}(x)}H(w,z,\tau^{\prime})dw$

for some $\tau^{\prime}\in\left(\frac{1}{2}t,t\right)$, where
$Q_{\delta}=B_{\frac{1}{4}\sqrt{t}}(x)\times(t-\frac{1}{4}(\sqrt{t})^{2},t)$
and $B_{\sqrt{t}}(x)\subset B_{R}(o)$ for any $x\in B_{\frac{1}{2}R}(o)$ and
$0<t<\frac{R^{2}}{4}$.
Combining (3.9) and (3.10), the heat kernel satisfies

$H(x,y,t)\leq\frac{\overline{c}_{7}(n)e^{\overline{c}_{8}(n)(Kt+L(R))}}{\operatorname{Vol}(B_{\sqrt{t}}(x))\operatorname{Vol}(B_{\sqrt{t}}(y))}\int_{B_{\sqrt{t}}(y)}\int_{B_{\sqrt{t}}(x)}H(w,z,\tau^{\prime})dwdz$
(3.11)

for any $x,\ y\in B_{\frac{1}{2}R}(o)$ and $0<t<\frac{R^{2}}{4}$.
Using Lemma 3.2 and noticing that
$\tau^{\prime}\in\left(\frac{1}{2}t,t\right)$, (3.11) becomes

$H(x,y,t)\leq\frac{\overline{c}_{7}(n)e^{\overline{c}_{8}(n)(Kt+L(R))}}{\operatorname{Vol}(B_{\sqrt{t}}(x))^{{}^{\frac{1}{2}}}\operatorname{Vol}(B_{\sqrt{t}}(y))^{{}^{\frac{1}{2}}}}e^{\left(-\frac{d^{2}(B_{\sqrt{t}}(x),B_{\sqrt{t}}(y))}{4t}\right)}$
(3.12)

for all $x,\ y\in B_{\frac{1}{2}R}(o)$ and $0<t<\frac{R^{2}}{4}$.
Notice that if $d(x,y)\leq 2\sqrt{t}$, then
$d(B_{\sqrt{t}}(x),B_{\sqrt{t}}(y))=0$, and hence

$-\frac{d^{2}(B_{\sqrt{t}}(x),B_{\sqrt{t}}(y))}{4t}=0\leq-\frac{d^{2}(x,y)}{4t},$

and if $d(x,y)>2\sqrt{t}$, then
$d(B_{\sqrt{t}}(x),B_{\sqrt{t}}(y))=d(x,y)-2\sqrt{t}$, and hence

$-\frac{d^{2}(B_{\sqrt{t}}(x),B_{\sqrt{t}}(y))}{4t}=-\frac{(d(x,y)-2\sqrt{t})^{2}}{4t}\leq
1-\frac{d^{2}(x,y)}{4(1+\epsilon)t}+\frac{1}{\epsilon}$

for any $\epsilon>0$. Combining the above two conditions, gives that

$e^{\left(-\frac{d^{2}(B_{\sqrt{t}}(x),B_{\sqrt{t}}(y))}{4t}\right)}\leq
e^{\left(-\frac{d^{2}(x,y)}{4(1+\epsilon)t}+1+\frac{1}{\epsilon}\right)}.$

Therefore, by (3.12) we have

$H(x,y,t)\leq\frac{\overline{c}_{9}(n,\epsilon)e^{\overline{c}_{8}(n)(Kt+L(R))}}{\operatorname{Vol}(B_{\sqrt{t}}(x))^{\frac{1}{2}}\operatorname{Vol}(B_{\sqrt{t}}(y))^{\frac{1}{2}}}e^{\left(-\frac{d^{2}(x,y)}{(4+\epsilon)t}\right)}$
(3.13)

for all $x,y\in B_{\frac{1}{2}R}(o)$ and $0<t<\frac{R^{2}}{4}$, where
$\lim\limits_{\epsilon\to 0}\overline{c}_{9}(n,\epsilon)=\infty$. (LABEL:4.9)
follows by letting $R\to\infty$.∎

Notice that Theorem 2.1 implies

$\displaystyle\operatorname{Vol}(B_{\sqrt{t}}(y))$
$\displaystyle\leq\operatorname{Vol}(B_{\sqrt{t}+d(x,y)}(x))$
$\displaystyle\leq\left(\frac{\sqrt{t}+d(x,y)}{\sqrt{t}}\right)^{n}e^{\frac{K}{6}(d^{2}(x,y)+2d(x,y)\sqrt{t})+6L(R)}\operatorname{Vol}(B_{\sqrt{t}}(x))$

for all $x,y\in B_{\frac{1}{4}R}(o)$ and $B_{\sqrt{t}+d(x,y)}(x)\subset
B_{R}(o)$ with $0<t<\frac{R^{2}}{16}$. Therefore, the upper bound in Theorem
1.1 can be rewritten as follows.

######  Corollary 3.3.

Under the same assumptions as Theorem 1.1, for any $\epsilon>0$, there exist
constants $C_{5}(n,\epsilon)$ and $C_{6}(n)$, such that

$H(x,y,t)\leq\frac{C_{5}(n,\epsilon)e^{C_{6}(n)(KR^{2}+L(R)+Kd^{2}(x,y))}}{\operatorname{Vol}(B_{\sqrt{t}}(y))}\left(1+\frac{d(x,y)}{\sqrt{t}}\right)^{\frac{n}{2}}e^{\left(-\frac{d^{2}(x,y)}{(4+\epsilon)t}\right)}$
(3.14)

for all $x,y\in B_{\frac{1}{4}R}(o)$ and $0<t<R^{2}/16$, where
$L(R)=\sup\limits_{B_{3R}(o)}|f|$ and $\lim\limits_{\epsilon\to
0}C_{5}(n,\epsilon)=\infty$.

##  4\.  $L^{1}$-Liouville theorem and uniqueness of $L^{1}$ solutions of the
heat equation

In this section, inspired by the work of Li [19], we prove the
$L^{1}$-Liouville theorem for non-negative subharmonic functions and the
uniqueness theorem for $L^{1}$-solutions of the heat equation on complete
noncompact Riemannian manifolds with Bakry-Émery Ricci curvature bounded below
and the potential function of at most quadratic growth.

We start from a useful lemma.

######  Lemma 4.1.

(Theorem 11.8 in [13]) Let $(M^{n},g)$ be a complete Riemannian manifold. If,
for some point $x_{0}\in M$,

$\int_{1}^{+\infty}\frac{R\,dR}{\ln(\operatorname{Vol}(B_{R}(x_{0})))}=\infty,$

then $(M^{n},g)$ is stochastically complete, i.e.,

$\int_{M}H(x,y,t)dy=1.$

Under our assumptions mentioned earlier, it is easy to check the stochastic
completeness of the manifold due to Theorem 2.1.

######  Proposition 4.2.

Let $(M^{n},g)$ be a complete non-compact Riemannian manifold with
$\operatorname{Ric}_{f}\geq-Kg$ for some constant $K\geq 0$. Assume there
exist non-negative constants $a$ and $b$ such that

$|f|(x)\leq ar^{2}(x)+b\ for\ all\ x\in M,$

where $r(x)=d(x,o)$ is the geodesic distance function to a fixed point $o\in
M$. Then $(M^{n},g)$ is stochastically complete.

###### Proof.

In (2.4), letting $r_{1}\to 0,\ r_{2}=R$, and $p=o\in M$ yields

$\operatorname{Vol}(B_{R}(o))\leq c(n,b)R^{n}e^{c(K,a)R^{2}}$

for all $R>1$. Hence

$\int_{1}^{+\infty}\frac{R\,dR}{\ln(\operatorname{Vol}(B_{R}(o)))}=\infty.$

By Lemma 4.1, this implies that $(M^{n},g)$ is stochastically complete.∎

Now, we are ready to check the integration by parts formula by using the upper
bound of the heat kernel in Corollary 3.3 and the mean value inequality in
Proposition 3.1.

######  Proposition 4.3.

Under the same assumptions as Proposition 4.2, for any non-negative
$L^{1}$-integrable subharmonic function $u$, we have

$\int_{M}\Delta_{y}H(x,y,t)u(y)dy=\int_{M}H(x,y,t)\Delta_{y}u(y)dy$

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

###### Proof.

Applying integration by parts on $B_{R}(o)$ gives

$\displaystyle\left|\int_{B_{R}(o)}\Delta_{y}H(x,y,t)u(y)dy-\int_{B_{R}(o)}H(x,y,t)\Delta_{y}u(y)dy\right|$
(4.1) $\displaystyle=\left|\int_{\partial B_{R}(o)}\frac{\partial}{\partial
r}H(x,y,t)u(y)dS-\int_{\partial B_{R}(o)}H(x,y,t)\frac{\partial}{\partial
r}u(y)dS\right|$ $\displaystyle\leq\int_{\partial
B_{R}(o)}|\nabla_{y}H|(x,y,t)u(y)dS+\int_{\partial
B_{R}(o)}H(x,y,t)|\nabla_{y}u|(y)dS,$

where $dS$ denotes the area measure on $\partial B_{R}(o)$. We shall show that
the above two boundary integrals vanish as $R\to\infty$. Without loss of
generality, we assume $R>1$ and $x\in B_{\frac{1}{8}R}(o)$.

Step 1. Since $|f|(x)\leq ar^{2}(x)+b$, by Proposition 3.1, we get

$\sup\limits_{B_{R}(o)}u\leq\frac{\tilde{C}_{1}(n)e^{\tilde{C}_{2}(n,K,a,b)R^{2}}}{\operatorname{Vol}(B_{2R}(o))}\int_{B_{2R}(o)}u\leq\frac{\tilde{C}_{1}(n)e^{\tilde{C}_{2}(n,K,a,b)R^{2}}}{\operatorname{Vol}(B_{2R}(o))}||u||_{1}.$
(4.2)

Let $\phi(y)=\phi(d(y,o))$ be a non-negative cut-off function satisfying
$0\leq\phi\leq 1$, $|\nabla\phi|\leq\sqrt{3}$, $\phi(y)=1$ on
$B_{R+1}(o)\backslash B_{R}(o)$, and $\phi(y)=0$ on
$B_{R-1}(o)\cup(M\backslash B_{R+2}(o))$. Since $u$ is subharmonic, by the
Cauchy-Schwarz inequality, we have

$\displaystyle 0\leq\int_{M}\phi^{2}u\Delta u$
$\displaystyle=-\int_{M}\left<\nabla(\phi^{2}u),\nabla u\right>$
$\displaystyle=-2\int_{M}\phi u\left<\nabla\phi,\nabla
u\right>-\int_{M}\phi^{2}|\nabla u|^{2}$ $\displaystyle\leq
2\int_{M}|\nabla\phi|^{2}u^{2}-\frac{1}{2}\int_{M}\phi^{2}|\nabla u|^{2},$

i.e.,

$\int_{M}\phi^{2}|\nabla u|^{2}\leq 4\int_{M}|\nabla\phi|^{2}u^{2}.$

It then follows from (4.2) that

$\displaystyle\int_{B_{R+1}(o)\backslash B_{R}(o)}|\nabla
u|^{2}\leq\int_{M}\phi^{2}|\nabla u|^{2}$ $\displaystyle\leq
4\int_{M}|\nabla\phi|^{2}u^{2}$ $\displaystyle\leq 12\int_{B_{R+2}(o)}u^{2}$
$\displaystyle\leq 12\sup\limits_{B_{R+2}(o)}u\times||u||_{1}$
$\displaystyle\leq\frac{\tilde{C}_{3}(n)e^{\tilde{C}_{4}(n,K,a,b)(R+2)^{2}}}{\operatorname{Vol}(B_{2R+4}(o))}||u||_{1}^{2}.$

On the other hand, the Cauchy-Schwarz inequality implies that

$\int_{B_{R+1}(o)\backslash B_{R}(o)}|\nabla
u|\leq\left(\int_{B_{R+1}(o)\backslash B_{R}(o)}|\nabla
u|^{2}\right)^{\frac{1}{2}}\cdot\left[\operatorname{Vol}(B_{R+1}(o))-\operatorname{Vol}(B_{R}(o))\right]^{\frac{1}{2}}.$

Combining the above two inequalities, we have

$\int_{B_{R+1}(o)\backslash B_{R}(o)}|\nabla
u|\leq\tilde{C}_{5}(n)e^{\tilde{C}_{6}(n,K,a,b)(R+2)^{2}}||u||_{1}.$ (4.3)

Step 2. By letting $\epsilon=1$ in Corollary 3.3, the heat kernel $H(x,y,t)$
satisfies

$H(x,y,t)\leq\frac{\tilde{C}_{7}(n)e^{\tilde{C}_{8}(n,K,a,b)[R^{2}+d^{2}(x,y)]}}{\operatorname{Vol}(B_{\sqrt{t}}(x))}\left(1+\frac{d(x,y)}{\sqrt{t}}\right)^{\frac{n}{2}}e^{\left(-\frac{d^{2}(x,y)}{5t}\right)}$
(4.4)

for all $x,y\in B_{\frac{1}{4}R}(o)$ and $0<t<\frac{R^{2}}{16}$.
Together with (4.3) we get

$\displaystyle J_{1}:$ $\displaystyle=\int_{B_{R+1}(o)\backslash
B_{R}(o)}H(x,y,t)|\nabla u|(y)dy$ $\displaystyle\leq\sup\limits_{y\in
B_{R+1}(o)\backslash B_{R}(o)}H(x,y,t)\cdot\int_{B_{R+1}(o)\backslash
B_{R}(o)}|\nabla u|(y)dy$
$\displaystyle\leq\frac{\tilde{C}_{9}(n)e^{\tilde{C}_{10}(n,K,a,b)[(R+2)^{2}+(d(x,o)+R+1)^{2}]-\frac{(R-d(x,o))^{2}}{5t}}||u||_{1}}{\operatorname{Vol}(B_{\sqrt{t}}(x))}\left(1+\frac{d(x,o)+R+1}{\sqrt{t}}\right)^{\frac{n}{2}}.$

Thus, for $T$ sufficiently small, all $t\in(0,T)$, and
$d(x,o)\leq\frac{R}{8}$, there exists a fixed constant $\beta>0$ such that

$\displaystyle
J_{1}\leq\frac{\hat{C}_{1}(n,K,a,b)||u||_{1}}{\operatorname{Vol}(B_{\sqrt{t}}(x))}\left(1+\frac{2R+1}{\sqrt{t}}\right)^{\frac{n}{2}}e^{\big{[}-\beta
R^{2}+\hat{C}_{2}(n,K,a,b)R\big{]}}.$

Hence for all $t\in(0,T)$ and all $x\in M$, $J_{1}\rightarrow 0$ as
$R\rightarrow\infty$.

Step 3. We show that $\displaystyle\int_{B_{R+1}(o)\backslash
B_{R}(o)}|\nabla_{y}H|(x,y,t)u(y)dy\to 0$ as $R\to\infty$. Frist, consider the
integral

$\displaystyle\int_{M}\phi^{2}(y)|\nabla_{y}H|^{2}(x,y,t)dy$
$\displaystyle=-2\int_{M}\left<H(x,y,t)\nabla\phi(y),\phi(y)\nabla_{y}H(x,y,t)\right>dy$
$\displaystyle\ \ \ \ -\int_{M}\phi^{2}(y)H(x,y,t)\Delta_{y}H(x,y,t)dy$
$\displaystyle\leq
2\int_{M}|\nabla\phi|^{2}(y)H^{2}(x,y,t)dy+\frac{1}{2}\int_{M}\phi^{2}(y)|\nabla_{y}H|^{2}(x,y,t)dy$
$\displaystyle\ \ \ -\int_{M}\phi^{2}(y)H(x,y,t)\Delta_{y}H(x,y,t)dy.$

This implies

$\displaystyle\int_{B_{R+1}(o)\backslash
B_{R}(o)}|\nabla_{y}H|^{2}(x,y,t)dy\leq\int_{M}\phi^{2}(y)|\nabla_{y}H|^{2}(x,y,t)dy$
(4.5) $\displaystyle\leq
4\int_{M}|\nabla\phi|^{2}(y)H^{2}(x,y,t)dy-2\int_{M}\phi^{2}(y)H(x,y,t)\Delta_{y}H(x,y,t)dy$
$\displaystyle\leq 12\int_{B_{R+2}(o)\backslash
B_{R-1}(o)}H^{2}(x,y,t)dy+2\int_{B_{R+2}(o)\backslash
B_{R-1}(o)}H(x,y,t)|\Delta_{y}H|(x,y,t)dy$ $\displaystyle\leq
12\int_{B_{R+2}(o)\backslash
B_{R-1}(o)}H^{2}(x,y,t)dy+2\left(\int_{B_{R+2}(o)\backslash
B_{R-1}(o)}H^{2}\right)^{\frac{1}{2}}\left(\int_{M}|\Delta_{y}H|^{2}(x,y,t)\right)^{\frac{1}{2}}.$

By Proposition 4.2, we have

$\int_{M}H(x,y,t)dy=1$

for all $x\in M$ and $t>0$. Combining this with (4.4) gives

$\displaystyle\int_{B_{R+2}(o)\backslash
B_{R-1}(o)}H^{2}(x,y,t)dy\leq\sup\limits_{y\in B_{R+2}(o)\backslash
B_{R-1}(o)}H(x,y,t)$ (4.6)
$\displaystyle\leq\frac{\tilde{C}_{11}(n)e^{\tilde{C}_{12}(n,K,a,b)[(R+2)^{2}+(d(x,o)+R+2)^{2}]-\frac{(R-1-d(o,x))^{2}}{5t}}}{\operatorname{Vol}(B_{\sqrt{t}}(x))}\left(1+\frac{d(x,o)+R+2}{\sqrt{t}}\right)^{\frac{n}{2}}.$

Form (29.15) in [20], we know that

$\int_{M}(\Delta_{y}H)^{2}(x,y,t)dy\leq\frac{\tilde{C}}{t^{2}}H(x,x,t).$ (4.7)

Combining (4.5), (4.6) and (4.7), we obtain

$\displaystyle\int_{B_{R+1}(o)\backslash
B_{R}(o)}|\nabla_{y}H|^{2}(x,y,t)dy\leq\tilde{C}_{13}(n)e^{\tilde{C}_{14}(n,K,a,b)[(R+2)^{2}+(d(x,o)+R+2)^{2}]-\frac{(R-1-d(x,o))^{2}}{10t}}$
$\displaystyle\times\left[\operatorname{Vol}(B_{\sqrt{t}}(x))^{-1}+\operatorname{Vol}(B_{\sqrt{t}}(x))^{-\frac{1}{2}}t^{-1}H^{\frac{1}{2}}(x,x,t)\right]\left(1+\frac{d(x,o)+R+2}{\sqrt{t}}\right)^{\frac{n}{2}}.$

By the Cauchy-Schwarz inequality, we get

$\displaystyle\int_{B_{R+1}(o)\backslash B_{R}(o)}|\nabla_{y}H|(x,y,t)dy$
(4.8)
$\displaystyle\leq\left[\operatorname{Vol}(B_{R+1}(o))-\operatorname{Vol}(B_{R}(o))\right]^{\frac{1}{2}}\cdot\left(\int_{B_{R+1}(o)\backslash
B_{R}(o)}|\nabla_{y}H|^{2}(x,y,t)dy\right)^{\frac{1}{2}}$
$\displaystyle\leq\operatorname{Vol}(B_{R+1}(o))^{\frac{1}{2}}\tilde{C}_{16}(n)e^{\tilde{C}_{15}(n,K,a,b)[(R+2)^{2}+(d(x,o)+R+2)^{2}]-\frac{(R-1-d(x,o))^{2}}{20t}}$
$\displaystyle\times\left[\operatorname{Vol}(B_{\sqrt{t}}(x))^{-1}+\operatorname{Vol}(B_{\sqrt{t}}(x))^{-\frac{1}{2}}t^{-1}H^{\frac{1}{2}}(x,x,t)\right]^{\frac{1}{2}}\left(1+\frac{d(x,o)+R+2}{\sqrt{t}}\right)^{\frac{n}{4}}.$

Therefore, by (4.2) and (4.8), we have

$\displaystyle J_{2}:$ $\displaystyle=\int_{B_{R+1}(o)\backslash
B_{R}(o)}|\nabla_{y}H|(x,y,t)u(y)dy$ $\displaystyle\leq\sup\limits_{y\in
B_{R+1}(o)\backslash B_{R}(o)}u(y)\cdot\int_{B_{R+1}(o)\backslash
B_{R}(o)}|\nabla_{y}H|(x,y,t)dy$
$\displaystyle\leq\frac{\tilde{C}_{18}(n)e^{\tilde{C}_{17}(n,K,a,b)(R+1)^{2}}}{\operatorname{Vol}(B_{2R+2}(o))^{\frac{1}{2}}}||u||_{1}e^{\tilde{C}_{15}(n,K,a,b)[(R+2)^{2}+(d(x,o)+R+2)^{2}]-\frac{(R-1-d(x,o))^{2}}{20t}}$
$\displaystyle\times\left[\operatorname{Vol}(B_{\sqrt{t}}(x))^{-1}+\operatorname{Vol}(B_{\sqrt{t}}(x))^{-\frac{1}{2}}t^{-1}H^{\frac{1}{2}}(x,x,t)\right]^{\frac{1}{2}}\left(1+\frac{d(x,o)+R+2}{\sqrt{t}}\right)^{\frac{n}{4}}$
$\displaystyle\leq\frac{\tilde{C}_{18}(n)e^{\tilde{C}_{19}(n,K,a,b)[(R+2)^{2}+(d(x,o)+R+2)^{2}]-\frac{(R-1-d(x,o))^{2}}{20t}}||u||_{1}}{\operatorname{Vol}(B_{2}(o))^{\frac{1}{2}}}$
$\displaystyle\times\left[\operatorname{Vol}(B_{\sqrt{t}}(x))^{-1}+\operatorname{Vol}(B_{\sqrt{t}}(x))^{-\frac{1}{2}}t^{-1}H^{\frac{1}{2}}(x,x,t)\right]^{\frac{1}{2}}\left(1+\frac{d(x,o)+R+2}{\sqrt{t}}\right)^{\frac{n}{4}}.$

Similar to the case of $J_{1}$, by choosing $T$ sufficiently small, for all
$t\in(0,T)$ and all $x\in M$, $J_{2}$ also tends to zero when $R$ tends to
infinity.

Step 4. By the mean value theorem, for all $R>0$ there exists
$\bar{R}\in(R,R+1)$ such that

$\displaystyle J:$ $\displaystyle=\int_{\partial
B_{\bar{R}}(o)}|\nabla_{y}H|(x,y,t)u(y)dS+\int_{\partial
B_{\bar{R}}(o)}H(x,y,t)|\nabla_{y}u|(y)dS$
$\displaystyle=\int_{B_{R+1}(o)\backslash
B_{R}(o)}|\nabla_{y}H|(x,y,t)u(y)dy+\int_{B_{R+1}(o)\backslash
B_{R}(o)}H(x,y,t)|\nabla_{y}u|(y)dy$ $\displaystyle=J_{2}+J_{1}.$

From step 2 and step 3, we know that by choosing $T$ sufficiently small, for
all $t\in(0,T)$ and all $x\in M$, $J$ tends to zero along a sequence of radii
tending to infinity. Since $H(x,y,t)\Delta u\geq 0$ and
$\Delta_{y}H(x,y,t)=\frac{\partial}{\partial t}H(x,y,t)$ is bounded by a
result of Grigor’yan [14], one can see from (4.1) that $J\rightarrow 0$ for
all $R\rightarrow\infty$. Therefore, Proposition 4.3 is true for $t$
sufficiently small.

Step 5. For all $t\in(0,T)$ and $s\in(0,+\infty)$, using the semigroup
property of the heat kernel, we have

$\displaystyle\int_{M}\Delta_{y}H(x,y,t+s)u(y)dy$
$\displaystyle=\int_{M}\int_{M}H(x,z,s)\Delta_{y}H(z,y,t)dzu(y)dy$
$\displaystyle=\int_{M}\left(\int_{M}\Delta_{y}H(z,y,t)u(y)dy\right)H(x,z,s)dz$
$\displaystyle=\int_{M}\left(\int_{M}H(z,y,t)\Delta_{y}u(y)dy\right)H(x,z,s)dz$
$\displaystyle=\int_{M}H(x,y,t+s)\Delta_{y}u(y)dy.$

This completes the proof of Proposition 4.3 for all time $t>0$.∎

Applying the regularity theory of harmonic functions, combining Proposition
4.2 and Propositon 4.3, we can obtain the $L^{1}$-Liouville property.

Proof of Theorem 1.3. Let $u(x)$ be a non-negative $L^{1}$-integrable
subharmonic function on $M$. We define a space-time function

$u(x,t)=\int_{M}H(x,y,t)u(y)dy$

with initial data $u(x,0)=u(x)$. From Proposition 4.3, we conclude that

$\displaystyle\frac{\partial}{\partial t}u(x,t)$
$\displaystyle=\int_{M}\frac{\partial}{\partial t}H(x,y,t)u(y)dy$
$\displaystyle=\int_{M}\Delta_{y}H(x,y,t)u(y)dy$
$\displaystyle=\int_{M}H(x,y,t)\Delta_{y}u(y)dy\geq 0,$

that is, $u(x,t)$ is increasing in $t$. By Proposition 4.2,

$\int_{M}H(x,y,t)dy=1$

for all $x\in M$ and $t>0$. So we have

$\int_{M}u(x,t)dx=\int_{M}\int_{M}H(x,y,t)u(y)dydx=\int_{M}u(y)dy.$

Since $u(x,t)$ is increasing in $t$, so $u(x,t)=u(x)$ and hence $u(x)$ is a
non-negative harmonic function, i.e., $\Delta u(x)=0$.

On the other hand, for any positive constant $\epsilon$, let us define a new
function $h(x)=\min\\{u(x),\epsilon\\}$. Then $h$ satisfies

$0\leq h(x)\leq u(x)$, $|\nabla h|\leq|\nabla u|$ and $\Delta h(x)\leq 0$.

So $h$ has the integration by parts formula as $u$. Similarly we define
$h(x,t)=\int_{M}H(x,y,t)h(y)dy$ and

$\displaystyle\frac{\partial}{\partial t}h(x,t)$
$\displaystyle=\int_{M}\frac{\partial}{\partial t}H(x,y,t)h(y)dy$
$\displaystyle=\int_{M}H(x,y,t)\Delta_{y}h(y)dy\leq 0.$

By the same argument, we have that $\Delta h(x)=0$.

By the regularity theory of harmonic functions, that is impossible unless
$h=u$ or $h=\epsilon$. Since $\epsilon$ is arbitrary and $u$ is non-negative,
so $u$ must be identically constant. The theorem then follows from the fact
that the absolute value of a harmonic function is a non-negative subharmonic
function.∎

With the $L^{1}$-Liouville property, one can prove the uniqueness of $L^{1}$
solution of the heat equation.

Proof of Theorem 1.7. Let $u(x,t)\in L^{1}$ be a non-negative function
satisfying the assumptions in Theorem 1.7. For $\epsilon>0$, we define a
space-time function

$u_{\epsilon}(x,t)=\int_{M}H(x,y,t)u(y,\epsilon)dy$ (4.9)

and

$F_{\epsilon}(x,t)=\max\\{0,u(x,t+\epsilon)-u_{\epsilon}(x,t)\\}.$

Then $F_{\epsilon}(x,t)$ is non-negative and satisfies

$\lim\limits_{t\rightarrow 0}F_{\epsilon}(x,t)=0,\
\left(\Delta-\frac{\partial}{\partial t}\right)F_{\epsilon}(x,t)\geq 0.$

Let $T>0$ be fixed. Let $h(x)=\int_{0}^{T}F_{\epsilon}(x,t)dt$, which implies

$\Delta h(x)=\int_{0}^{T}\Delta
F_{\epsilon}(x,t)dt\geq\int_{0}^{T}\partial_{t}F_{\epsilon}(x,t)dt=F_{\epsilon}(x,T)\geq
0,$ (4.10)

and

$\displaystyle\int_{M}h(x)dx$
$\displaystyle=\int_{0}^{T}\int_{M}F_{\epsilon}(x,t)dxdt\leq\int_{0}^{T}\int_{M}|u(x,t+\epsilon)-u_{\epsilon}(x,t)|dxdt$
$\displaystyle\leq\int_{0}^{T}\int_{M}u(x,t+\epsilon)dxdt+\int_{0}^{T}\int_{M}u_{\epsilon}(x,t)dxdt<\infty,$

where the first term on the right hand is finite from our assumption, and the
second term is finite because the heat semigroup is contractive on $L^{1}$.
Therefore, $h(x)$ is a non-negative $L^{1}$-integrable subharmonic function.
By Theorem 1.3, $h(x)$ must be constant. Combining with (4.10) we have
$F_{\epsilon}(x,t)=0$ for all $x\in M$ and $t>0$, which implies

$u_{\epsilon}(x,t)\geq u(x,t+\epsilon).$ (4.11)

Next we estimate the function $u_{\epsilon}(x,t)$ in (4.9). Applying the upper
bound estimate (3.14) of the heat kernel $H(x,y,t)$ and letting $\epsilon=1$,
$R=2d(x,y)+1$, we have

$u_{\epsilon}(x,t)\leq\frac{c}{\operatorname{Vol}(B_{\sqrt{t}}(x))}\int_{M}\left[e^{\tilde{c}d^{2}(x,y)-\frac{d^{2}(x,y)}{5t}}\left(1+\frac{d(x,y)}{\sqrt{t}}\right)^{\frac{n}{2}}\right]u(y,\epsilon)dy.$

For sufficiently small values of $t>0$, the right-hand side can be estimated
by

$\frac{C}{\operatorname{Vol}(B_{\sqrt{t}}(x))}\int_{M}u(y,\epsilon)dy.$

Hence as $\epsilon\rightarrow 0$, $u_{\epsilon}(x,t)\rightarrow 0$ since
$\int_{M}u(y,\epsilon)dy\rightarrow 0$. However, by the semigroup property,
$u_{\epsilon}(x,t)\rightarrow 0$ for all $x\in M$ and $t>0$. Combining with
(4.11) we get $u(x,t)\leq 0$. Therefore $u(x,t)\equiv 0$.

To prove that any $L^{1}$-solution of the heat equation is uniquely determined
by its initial data in $L^{1}$, we suppose that $u_{1}(x,t),\ u_{2}(x,t)$ are
two $L^{1}$-integrable solutions of the heat equation
$(\Delta-\partial_{t})u=0$ with the initial data $u(x,0)\in L^{1}$. Applying
this above result to $v(x,t)=|u_{1}(x,t)-u_{2}(x,t)|$, we see that
$v(x,t)\equiv 0$. The proof of Theorem 1.7 is finished.
∎

##  5\.  An $L^{\infty}$ Liouville Property for harmonic functions with
polynomial growth

In this section, we take a detour to prove an $L^{\infty}$ Liouville theorem
for harmonic functions with polynomial growth when the Bakry-Émery Ricci
curvature is non-negative and the potential function is bounded by using a
similar idea in [6]. This is not the main result of the paper, but can be
compared with the results in the last section and may be of independent
interest.

First, we give a gradient estimate in the integral sense.

######  Lemma 5.1.

Let $(M^{n},g)$ be a complete non-compact Riemannian manifold with
$\operatorname{Ric}_{f}\geq 0$ and $|f|\leq L$ on $M$ for some constant $L\geq
0$. For any point $o\in M$ and $R>0$, let $u$ be a harmonic function on
$B_{2R}(o)$, and $diam\ \partial B_{r}(o):=\sup\limits_{x,y\in\partial
B_{r}(o)}d(x,y)\leq\epsilon r$ with $\epsilon\in(0,\frac{1}{12})$ for all
$r\in[R,2R]$. Then we have

$\int_{B_{R}(o)}|\nabla
u|^{2}\leq\delta^{\frac{1}{\epsilon}-6}\int_{B_{2R}(o)}|\nabla u|^{2},$ (5.1)

where
$\delta=\left(\frac{9c_{1}e^{c_{2}L}}{1+9c_{1}e^{c_{2}L}}\right)^{\frac{1}{6}}$
and $c_{1}$, $c_{2}$ are constants depending only on $n$ from Lemma 2.2.

###### Proof.

Choosing $r\in[R+3\epsilon R,2R-3\epsilon R]$, where
$\epsilon\in(0,\frac{1}{12})$, the diameter hypothesis in the Lemma implies
that there exists some $x\in\partial B_{r}(o)$ such that

$B_{r+\epsilon R}(o)\backslash B_{r}(o)\subset B_{\epsilon R+\epsilon r}(x).$

Let $\eta(y)$ be a cut-off function with support in $B_{r+\epsilon R}(o)$ such
that

$\eta(y)=\begin{cases}1&\text{ $y\in B_{r}(o)$ },\\\ \frac{r+\epsilon
R-d(y,o)}{\epsilon R}&\text{ $y\in B_{r+\epsilon R}(o)\backslash B_{r}(o)$
},\\\ 0&\text{ $y\in M\backslash B_{r+\epsilon R}(o)$ }.\end{cases}$

We easily observe that

$\int_{M}|\nabla(\eta(u-c))|^{2}=\int_{M}\eta^{2}|\nabla(u-c)|^{2}+2\eta(u-c)\left<\nabla\eta,\nabla(u-c)\right>+(u-c)^{2}|\nabla\eta|^{2},$

where $c$ is a real number to be chosen later. Notice that

$\displaystyle\int_{M}\eta^{2}|\nabla(u-c)|^{2}+2\eta(u-c)\left<\nabla\eta,\nabla(u-c)\right>$
$\displaystyle=\int_{M}\left<\nabla((u-c)\eta^{2}),\nabla(u-c)\right>$
$\displaystyle=-\int_{M}(u-c)\eta^{2}\Delta(u-c)=0.$

Therefore,

$\int_{M}|\nabla(\eta(u-c))|^{2}=\int_{B_{r+\epsilon
R}(o)}(u-c)^{2}|\nabla\eta|^{2}.$

According to the definition of $\eta(y)$, by the above equality, we get that

$\displaystyle\int_{B_{r}(o)}|\nabla u|^{2}$
$\displaystyle\leq\int_{B_{r+\epsilon R}(o)}|\nabla(\eta(u-c))|^{2}$
$\displaystyle=\int_{B_{r+\epsilon R}(o)}(u-c)^{2}|\nabla\eta|^{2}$
$\displaystyle\leq\frac{1}{\epsilon^{2}R^{2}}\int_{B_{r+\epsilon
R}(o)\backslash B_{r}(o)}(u-c)^{2}$
$\displaystyle\leq\frac{1}{\epsilon^{2}R^{2}}\int_{B_{\epsilon R+\epsilon
r}(x)}(u-c)^{2}.$

For the right-hand side of the above inequality, if we choose
$c=u_{B_{\epsilon R+\epsilon r}(x)}$, then using the Poincaré inequality in
Lemma 2.2 and the fact that $r+R\leq 3R$, we have

$\int_{B_{r}(o)}|\nabla u|^{2}\leq 9c_{1}e^{c_{2}L}\int_{B_{3\epsilon
R}(x)}|\nabla u|^{2}.$

This implies that

$\int_{B_{r-3\epsilon R}(o)}|\nabla u|^{2}\leq\int_{B_{r}(o)}|\nabla
u|^{2}\leq 9c_{1}e^{c_{2}L}\int_{B_{r+3\epsilon R}(o)\backslash B_{r-3\epsilon
R}(o)}|\nabla u|^{2},$

that is,

$\int_{B_{r-3\epsilon R}(o)}|\nabla
u|^{2}\leq\frac{9c_{1}e^{c_{2}L}}{1+9c_{1}e^{c_{2}L}}\int_{B_{r+3\epsilon
R}(o)}|\nabla u|^{2},\ r\in[R+3\epsilon R,2R-3\epsilon R].$

Set $r=R+3\epsilon R$, then

$\int_{B_{R}(o)}|\nabla
u|^{2}\leq\frac{9c_{1}e^{c_{2}L}}{1+9c_{1}e^{c_{2}L}}\int_{B_{R+6\epsilon
R}(o)}|\nabla u|^{2}.$

Iterating this inequality $N$ times, we finally get

$\int_{B_{R}(o)}|\nabla
u|^{2}\leq\left(\frac{9c_{1}e^{c_{2}L}}{1+9c_{1}e^{c_{2}L}}\right)^{N}\int_{B_{2R}(o)}|\nabla
u|^{2}$

provided that $N6\epsilon R\leq R$. Thus, we can choose
$N=\left[\frac{1}{6\epsilon}\right]\geq\frac{1}{6\epsilon}-1>0$ for
$\epsilon\in\left(0,\frac{1}{12}\right)$.
Then we have

$\int_{B_{R}(o)}|\nabla
u|^{2}\leq\left(\frac{9c_{1}e^{c_{2}L}}{1+9c_{1}e^{c_{2}L}}\right)^{\frac{1}{6\epsilon}-1}\int_{B_{2R}(o)}|\nabla
u|^{2}.$

The desired result follows by choosing
$\delta=\left(\frac{9c_{1}e^{c_{2}L}}{1+9c_{1}e^{c_{2}L}}\right)^{\frac{1}{6}}$.
∎

Now we are ready to prove the Liouville property by using Lemma 5.1.

######  Theorem 5.2.

Let $(M^{n},g)$ be a complete noncompact Riemannian manifold with
$\operatorname{Ric}_{f}\geq 0$ and $|f|\leq L$. For a base point $o\in M$, if
the diameter of the geodesic sphere $\partial B_{R}(o)$ has a sublinear
growth, i.e., $diam\ \partial B_{R}(o)=o(R),\ R\rightarrow\infty$, then any
harmonic function with polynomial growth is constant.

###### Proof.

Let $u:M\rightarrow\mathbb{R}$ be a harmonic function with polynomial growth
of order $\nu$, namely

$|u(x)|\leq c(1+d(x,o))^{\nu}.$

For $R>>1$, we define

$I_{R}:=\int_{B_{R}(o)}|\nabla u|^{2}$

and

$\epsilon(r):=\sup\limits_{t\geq r}\frac{\rho(t)}{t},$

where $\rho(t):=\sup\limits_{x,y\in\partial B_{t}(o)}d(x,y).$

To estimate $I_{R}$, we introduce a cut-off function $\xi(x)$ such that

$\xi(x)=\begin{cases}1&\text{ $x\in B_{R}(o)$ },\\\ \frac{2R-d(x,o)}{R}&\text{
$x\in B_{2R}(o)\backslash B_{R}(o)$ },\\\ 0&\text{ $x\in M\backslash
B_{2R}(o)$ }.\end{cases}$

Then we have

$\displaystyle I_{R}$ $\displaystyle\leq\int_{B_{2R}(o)}|\nabla(\xi
u)|^{2}=\int_{B_{2R}(o)}|u|^{2}|\nabla\xi|^{2}+\xi^{2}|\nabla
u|^{2}+2u\xi\left<\nabla u,\nabla\xi\right>$
$\displaystyle=\int_{B_{2R}(o)}|u|^{2}|\nabla\xi|^{2}\leq\int_{B_{2R}(o)}c(1+2R)^{2\nu}\frac{1}{R^{2}}\leq\int_{B_{2R}(o)}c(3R)^{2\nu}\frac{1}{R^{2}}$
$\displaystyle=cR^{2\nu-2}\operatorname{Vol}(B_{2R}(o)).$

The volume comparison Theorem 2.1 shows that

$\frac{\operatorname{Vol}(B_{2R}(o))}{\operatorname{Vol}(B_{1}(o))}\leq
e^{6L}(2R)^{n}.$

Hence

$I_{R}\leq C(n,L,\operatorname{Vol}(B_{1}(o)))R^{2\nu+n-2}.$ (5.2)

On the other hand, if we iterate the inequality (5.1) proved in Lemma 5.1 $l$
times, we can show that, for all sufficiently large $R$ such that
$\epsilon(R)<\frac{1}{12}$,

$I_{R}\leq\delta^{-6l+\sum_{j=0}^{l-1}\frac{1}{\epsilon(2^{j}R)}}I_{2^{l}R},$

where
$\delta=\left(\frac{9c_{1}e^{c_{2}L}}{1+9c_{1}e^{c_{2}L}}\right)^{\frac{1}{6}}$.
Applying (5.2) to the right-hand side of the above inequality yields

$I_{R}\leq
C(n,L,\operatorname{Vol}(B_{1}(o)))e^{l\left[\left(\frac{\sum_{j=0}^{l-1}\frac{1}{\epsilon(2^{j}R)}}{l}-6\right)\ln\delta+(2\nu+n-2)\left(\ln
2+\frac{\ln R}{l}\right)\right]}.$ (5.3)

For all sufficiently large $R$,

$\lim\limits_{l\rightarrow\infty}\frac{\sum_{j=0}^{l-1}\frac{1}{\epsilon(2^{j}R)}}{l}=\lim\limits_{j\rightarrow\infty}\frac{1}{\epsilon(2^{j}R)}=+\infty.$

Meanwhile, we know that $0<\delta<1$ for any $R$. Therefore, letting
$l\rightarrow\infty$ in (5.3), we conclude that $I_{R}=0$ for all sufficiently
large $R$. Therefore $u$ is constant.

∎

##  6\.  Eigenvalue estimates

In this section, we derive lower bound estimations of eigenvalues of the
Laplace-Beltrami operator $\Delta$ on closed Riemannian manifolds with Bakry-
Émery Ricci curvature bounded below and bounded the potential function.

Denote the eigenvalues of $\Delta$ by
$0=\lambda_{0}<\lambda_{1}\leq\lambda_{2}\leq\cdots\leq\lambda_{k}\leq\cdots$.
First, we bound $\lambda_{1}$ from below. According to [7], it suffices to
bound Cheeger’s isoperimetric constant from below. Let us recall the
definitions of isoperimetric constants. We adapt the notations and definitions
in [20].

######  Definition 6.1.

Let $(M^{n},g)$ be a compact Riemannian manifold (with or without boundary).
For $\alpha>0$, The Neumann $\alpha$-isoperimetric constant of M is defined by

$IN_{\alpha}(M)=\inf_{\begin{subarray}{c}\partial\Omega_{1}=H=\partial\Omega_{2}\\\
M=\Omega_{1}\cup
H\cup\Omega_{2}\end{subarray}}\frac{\operatorname{Vol}(H)}{\min\\{{\operatorname{Vol}(\Omega_{1}),\operatorname{Vol}(\Omega_{2})}\\}^{\frac{1}{\alpha}}},$

where the infimum is taken over all hypersurfaces $H$ dividing $M$ into two
parts, denoted by $\Omega_{1}$ and $\Omega_{2}$.

In [7], Cheeger showed that

$\lambda_{1}\geq\frac{IN_{1}(M)^{2}}{4}$ (6.1)

on closed manifolds. Thus, one can get a lower bound of $\lambda_{1}$ by
bounding $IN_{1}(M)$ from below. We will accomplish this goal by following the
method of Dai-Wei-Zhang in [9] and using Theorem 2.1.

######  Theorem 6.2.

Let $(M^{n},g)$ be a complete Riemannian manifold with
$\operatorname{Ric}_{f}\geq-Kg$ and $|f|\leq L$ on $M$, where $K,\ L$ are non-
negative constants. Let $\Omega$ be a bounded convex domain in $M$. Then for
$1\leq\alpha\leq\frac{n}{n-1},$ we have

$IN_{\alpha}(\Omega)\geq
d^{-1}2^{-2n-1}5^{-n}e^{-(18-\frac{6}{\alpha})L-(17\frac{1}{6}-\frac{1}{6\alpha})Kd^{2}}\operatorname{Vol}(\Omega)^{1-\frac{1}{\alpha}},$
(6.2)

and for $0<\alpha<1,$ we have

$IN_{\alpha}(\Omega)\geq
d^{-1}2^{-2n-1}5^{-n}e^{-12L-17Kd^{2}}\operatorname{Vol}(\Omega)^{1-\frac{1}{\alpha}},$
(6.3)

where $d=diam(\Omega)$, the diameter of $\Omega$. In particular, if $M$ is
closed, then

$IN_{1}(M)\geq D^{-1}2^{-2n-1}5^{-n}e^{-12L-17KD^{2}},$ (6.4)

where $D$ is an upper bound of the diameter of $M$.

Before starting the proof of Theorem 6.2, we need the following lemma by
Gromov.

######  Lemma 6.3 ([15]).

Let $(M^{n},g)$ be a complete Riemannian manifold. Let $\Omega$ be a convex
domain in $M$ and $H$ a hypersurface dividing $\Omega$ into two parts
$\Omega_{1},\Omega_{2}$. For any Borel subsets $W_{i}\subset\Omega_{i},i=1,2$,
there exists $x_{1}$ in one of $W_{i}$, say $W_{1}$, and a subset $W$ in the
other part $W_{2}$, such that

$\operatorname{Vol}(W)\geq\frac{1}{2}\operatorname{Vol}(W_{2}),$ (6.5)

and for any $x_{2}\in W$, there is a unique minimal geodesic between $x_{1}$
and $x_{2}$ which intersects $H$ at some $z$ with

$d(x_{1},z)\geq d(x_{2},z),$ (6.6)

where $d(x_{1},z)$ denotes the distance between $x_{1}$ and $z$.

Combining Theorem 2.1 and Lemma 6.3, we get

######  Lemma 6.4.

Let $H,W$ and $x_{1}$ be as in Lemma 6.3. Then

$\operatorname{Vol}(W)\leq
D_{1}2^{n-1}e^{6L+\frac{K}{2}D_{1}^{2}}\operatorname{Vol}(H^{{}^{\prime}}),$
(6.7)

where $D_{1}=\sup_{x\in W}d(x_{1},x)$, and $H^{{}^{\prime}}$ is the set of
intersection points with $H$ of geodesics $\gamma_{x_{1},x}$ for all $x\in W$.

###### Proof.

Let $\Gamma\subset S_{x_{1}}(M)$ be the subset of unit vectors $\theta$ such
that $\gamma_{\theta}=\gamma_{x_{1},x_{2}}$ for some $x_{2}\in W$. Set polar
coordinates at $x_{1}$. The volume element of the metric $g$ is written as
$dv=J(t,\theta,x_{1})dtd\theta$ in the polar coordinates $(\theta,t)\in
S_{x_{1}}(M)\times\mathbb{R^{+}}$. For any $\theta\in\Gamma$, let $r(\theta)$
be the radius such that $exp_{x_{1}}(r(\theta)\theta)\in H$. Then form Lemma
6.3 $W\subset\\{exp_{x_{1}}(r\theta)|\theta\in\Gamma,r(\theta)\leq r\leq
2r(\theta)\\}$. We conclude

$\operatorname{Vol}(W)\leq\int_{\Gamma}\int_{r(\theta)}^{2r(\theta)}J(t,\theta,x_{1})dtd\theta.$

For $r(\theta)\leq t\leq 2r(\theta)\leq 2D_{1}$, by Theorem 2.1, it implies

$\frac{J(t,\theta,x_{1})}{t^{n-1}}\leq
e^{\frac{K}{6}(t^{2}-r(\theta)^{2})+6L}\frac{J(r(\theta),\theta,x_{1})}{r(\theta)^{n-1}},$

so

$J(t,\theta,x_{1})\leq
e^{\frac{K}{2}D_{1}^{2}+6L}2^{n-1}J(r(\theta),\theta,x_{1}).$

It gives

$\operatorname{Vol}(W)\leq
e^{\frac{K}{2}D_{1}^{2}+6L}2^{n-1}\int_{\Gamma}r(\theta)J(r(\theta),\theta,x_{1})d\theta\leq
D_{1}2^{n-1}e^{\frac{K}{2}D_{1}^{2}+6L}\operatorname{Vol}(H^{{}^{\prime}}).$

∎
Form Lemmas 6.3 and 6.4, we immediately have

######  Corollary 6.5.

Let $H$ be any hypersurface dividing a convex domain $\Omega$ into two parts
$\Omega_{1},\Omega_{2}$. For any ball $B=B_{r}(x)$ in $M$, we have

$\min(\operatorname{Vol}(B\cap\Omega_{1}),\operatorname{Vol}(B\cap\Omega_{2}))\leq
2^{n+1}re^{\frac{K}{2}d^{2}+6L}\operatorname{Vol}(H\cap(B_{2r}(x))),$ (6.8)

where $d=diam(\Omega)$. In particular, if $B\cap\Omega$ is divided equally by
$H$, then

$\operatorname{Vol}(B\cap\Omega)\leq
2^{n+2}re^{\frac{K}{2}d^{2}+6L}\operatorname{Vol}(H\cap(B_{2r}(x))).$ (6.9)

###### Proof.

Put $W_{i}=B\cap\Omega_{i}$ in the above lemma and use $D_{1}\leq 2r$ and
$H^{{}^{\prime}}\subset H\cap B_{2r}(x).$ ∎

Now we are ready to prove Theorem 6.2.

Proof of Theorem 6.2. Let $H$ be any hypersurface dividing $\Omega$ into two
parts, $\Omega_{1}$ and $\Omega_{2}$. We may assume that
$\operatorname{Vol}(\Omega_{1})\leq\operatorname{Vol}(\Omega_{2})$. For any
$x\in\Omega_{1}$, let $r_{x}$ be the smallest radius such that

$\operatorname{Vol}(B_{r_{x}}(x)\cap\Omega_{1})=\operatorname{Vol}(B_{r_{x}}(x)\cap\Omega_{2})=\frac{1}{2}\operatorname{Vol}(B_{r_{x}}(x)\cap\Omega).$

Let $d=diam(\Omega)$. By (6.9) we have,

$\operatorname{Vol}(B_{r_{x}}(x)\cap\Omega)\leq
2^{n+2}r_{x}e^{\frac{K}{2}d^{2}+6L}\operatorname{Vol}(H\cap(B_{2r_{x}}(x))).$
(6.10)

The domain $\Omega_{1}$ has a covering

$\Omega_{1}\subset\cup_{x\in\Omega_{1}}B_{2r_{x}}(x).$

By Vitali Covering Lemma, we can choose a countable family of disjoint balls
$B_{i}=B_{2r_{x_{i}}}(x_{i})$ such that
$\cup_{i}B_{10r_{x_{i}}}(x_{i})\supset\Omega_{1}.$ So

$\operatorname{Vol}(\Omega_{1})\leq\sum_{i}\operatorname{Vol}(B_{10r_{x_{i}}}(x_{i})\cap\Omega_{1}).$

Applying the volume comparison Theorem 2.1 in $\Omega_{1}$ gives

$\frac{\operatorname{Vol}(B_{10r_{x_{i}}}(x_{i})\cap\Omega_{1})}{(10r_{x_{i}})^{n}}\leq
e^{\frac{33}{2}Kr_{x_{i}}^{2}+6L}\frac{\operatorname{Vol}(B_{r_{x_{i}}}(x_{i})\cap\Omega_{1})}{(r_{x_{i}})^{n}}.$

On the other hand, since
$\operatorname{Vol}(\Omega_{1})\leq\operatorname{Vol}(\Omega_{2})$, we have
$r_{x}\leq d$ for any $x\in\Omega_{1}$. Thus,

$\displaystyle\operatorname{Vol}(B_{10r_{x_{i}}}(x_{i})\cap\Omega_{1})$
$\displaystyle\leq
10^{n}e^{\frac{33}{2}Kd^{2}+6L}\operatorname{Vol}(B_{r_{x_{i}}}(x_{i})\cap\Omega_{1})$
$\displaystyle=2^{-1}10^{n}e^{\frac{33}{2}Kd^{2}+6L}\operatorname{Vol}(B_{r_{x_{i}}}(x_{i})\cap\Omega).$

Therefore,

$\operatorname{Vol}(\Omega_{1})\leq
2^{-1}10^{n}e^{\frac{33}{2}Kd^{2}+6L}\sum_{i}\operatorname{Vol}(B_{r_{x_{i}}}(x_{i})\cap\Omega).$
(6.11)

Moreover, since the balls $B_{i}$ are disjoint, (6.10) gives

$\operatorname{Vol}(H)\geq\sum_{i}\operatorname{Vol}(B_{i}\cap H)\geq
2^{-n-2}e^{-\frac{K}{2}d^{2}-6L}\sum_{i}r_{x_{i}}^{-1}\operatorname{Vol}(B_{r_{x_{i}}}(x_{i})\cap\Omega).$
(6.12)

Firstly, for $1\leq\alpha\leq\frac{n}{n-1}$, it follows from (6.11) and (6.12)
that

$\begin{split}\frac{\operatorname{Vol}(H)}{\operatorname{Vol}(\Omega_{1})^{\frac{1}{\alpha}}}&\geq\frac{2^{-n-2}e^{-\frac{K}{2}d^{2}-6L}}{(2^{-1}10^{n}e^{\frac{33}{2}Kd^{2}+6L})^{\frac{1}{\alpha}}}\frac{\sum_{i}r_{x_{i}}^{-1}\operatorname{Vol}(B_{r_{x_{i}}}(x_{i})\cap\Omega)}{(\sum_{i}\operatorname{Vol}(B_{r_{x_{i}}}(x_{i})\cap\Omega))^{\frac{1}{\alpha}}}\\\
&\geq\frac{2^{-n-2}e^{-\frac{K}{2}d^{2}-6L}}{2^{-1}10^{n}e^{\frac{33}{2}Kd^{2}+6L}}\frac{\sum_{i}r_{x_{i}}^{-1}\operatorname{Vol}(B_{r_{x_{i}}}(x_{i})\cap\Omega)}{\sum_{i}\operatorname{Vol}(B_{r_{x_{i}}}(x_{i})\cap\Omega)^{\frac{1}{\alpha}}}\\\
&\geq
2^{-2n-1}5^{-n}e^{-12L-17Kd^{2}}\inf_{i}\frac{r_{x_{i}}^{-1}\operatorname{Vol}(B_{r_{x_{i}}}(x_{i})\cap\Omega)}{\operatorname{Vol}(B_{r_{x_{i}}}(x_{i})\cap\Omega)^{\frac{1}{\alpha}}}\\\
&=2^{-2n-1}5^{-n}e^{-12L-17Kd^{2}}\inf_{i}r_{x_{i}}^{-1}\operatorname{Vol}(B_{r_{x_{i}}}(x_{i})\cap\Omega)^{1-\frac{1}{\alpha}}.\end{split}$

We apply the volume comparison Theorem 2.1 in $\Omega$, then

$\frac{\operatorname{Vol}(B_{d}(x_{i})\cap\Omega)}{d^{n}}\leq
e^{\frac{K}{6}(d^{2}-r_{x_{i}}^{2})+6L}\frac{\operatorname{Vol}(B_{r_{x_{i}}}(x_{i})\cap\Omega)}{r_{x_{i}}^{n}}.$

By $1-\frac{1}{\alpha}\geq 0$, and $n(1-\frac{1}{\alpha})-1\leq 0$, we can
derive

$\inf_{i}r_{x_{i}}^{-1}\operatorname{Vol}(B_{r_{x_{i}}}(x_{i})\cap\Omega)^{1-\frac{1}{\alpha}}\geq
d^{-1}e^{-(\frac{K}{6}d^{2}+6L)(1-\frac{1}{\alpha})}\operatorname{Vol}(\Omega)^{1-\frac{1}{\alpha}}.$

Taking infimum over $H$, we conclude the following

$IN_{\alpha}(\Omega)\geq
d^{-1}2^{-2n-1}5^{-n}e^{-(18-\frac{6}{\alpha})L-(17\frac{1}{6}-\frac{1}{6\alpha})Kd^{2}}\operatorname{Vol}(\Omega)^{1-\frac{1}{\alpha}}.$

On the other hand, for $0<\alpha<1$, we have

$\begin{split}\frac{\operatorname{Vol}(H)}{\operatorname{Vol}(\Omega_{1})^{\frac{1}{\alpha}}}&=\frac{\operatorname{Vol}(H)}{\operatorname{Vol}(\Omega_{1})\operatorname{Vol}(\Omega_{1})^{\frac{1}{\alpha}-1}}\geq\frac{\operatorname{Vol}(H)}{\operatorname{Vol}(\Omega_{1})\operatorname{Vol}(\Omega)^{\frac{1}{\alpha}-1}}\\\
&\geq\frac{2^{-n-2}e^{-\frac{K}{2}d^{2}-6L}}{2^{-1}10^{n}e^{\frac{33}{2}Kd^{2}+6L}}\frac{\sum_{i}r_{x_{i}}^{-1}\operatorname{Vol}(B_{r_{x_{i}}}(x_{i})\cap\Omega)}{\sum_{i}\operatorname{Vol}(B_{r_{x_{i}}}(x_{i})\cap\Omega)}\operatorname{Vol}(\Omega)^{1-\frac{1}{\alpha}}\\\
&\geq
d^{-1}2^{-2n-1}5^{-n}e^{-12L-17Kd^{2}}\operatorname{Vol}(\Omega)^{1-\frac{1}{\alpha}}.\end{split}$

Taking infimum over $H$ finishes the proof.
∎

From (6.1) and Theorem 6.2, we immediately have the estimate of the first
eigenvalue.

######  Theorem 6.6.

Let $(M^{n},g)$ be a closed Riemannian manifold with
$\operatorname{Ric}_{f}\geq-Kg$ and $|f|\leq L$ on $M$, where $K,\ L$ are non-
negative constants. Then

$\lambda_{1}\geq D^{-2}2^{-4n-4}5^{-2n}e^{-24L-34KD^{2}}:=\alpha_{0},$ (6.13)

where $D$ is an upper bound of the diameter of $M$.

Next, we derive lower bounds for $\lambda_{k}\ (k\geq 2)$ by using the upper
bound estimate of the heat kernel in Theorem 1.1 and an argument of Li-Yau
[22].

Proof of Theorem 1.8. In Theorem 1.1, letting $\epsilon=1$, we have

$H(x,x,t)\leq\frac{\overline{C}_{1}(n)e^{\overline{C}_{2}(n)(Kt+L)}}{\operatorname{Vol}(B_{\sqrt{t}}(x))}$

for all $x\in M$ and $t>0$. Note that the heat kernel can be written as

$H(x,y,t)=\sum_{i=0}^{\infty}e^{-\lambda_{i}t}\phi_{i}(x)\phi_{i}(y),$

where $\phi_{i}$ is the eigenfunction of $\Delta$ corresponding to
$\lambda_{i}$ and $\\{\phi_{i}\\}_{i=0}^{\infty}$ form an orthonormal basis
with respect to the $L^{2}$-norm. So we have

$\int_{M}H(x,x,t)dx=\sum_{i=0}^{\infty}e^{-\lambda_{i}t}\leq\overline{C}_{1}(n)e^{\overline{C}_{2}(n)(Kt+L)}\int_{M}\operatorname{Vol}(B_{\sqrt{t}}(x))^{-1}dx.$

When $\sqrt{t}\geq D$,
$\operatorname{Vol}(B_{\sqrt{t}}(x))=\operatorname{Vol}(M)$. On the other
hand, when $\sqrt{t}\leq D$, since $|f|\leq L$ on $M$, by the volume
comparison theorem in (2.4), we get

$\frac{\operatorname{Vol}(B_{D}(x))}{\operatorname{Vol}(B_{\sqrt{t}}(x))}\leq
e^{\left(\frac{K}{6}D^{2}+6L\right)}\left(\frac{D}{\sqrt{t}}\right)^{n}$

for all $x\in M$. Then we conclude that

$\displaystyle\sum_{i=0}^{\infty}e^{-\lambda_{i}t}$
$\displaystyle\leq\overline{C}_{1}(n)e^{\overline{C}_{2}(n)(Kt+L)}\begin{cases}e^{\left(\frac{K}{6}D^{2}+6L\right)}\left(\frac{D}{\sqrt{t}}\right)^{n}&\text{
$t\leq D^{2}$ },\\\ 1&\text{ $t\geq D^{2}$},\end{cases}$
$\displaystyle\leq\begin{cases}\overline{C}_{3}(n)e^{\overline{C}_{4}(n)(KD^{2}+L)}\left(\frac{D}{\sqrt{t}}\right)^{n}&\text{
$t\leq D^{2}$ },\\\ \overline{C}_{3}(n)e^{\overline{C}_{4}(n)(Kt+L)}&\text{
$t\geq D^{2}$}.\end{cases}:=q(t)$

Fixing $k\geq 1$, and taking the first $(k+1)$ terms, we get

$(k+1)e^{-\lambda_{k}t}\leq q(t),$

that is, $k+1\leq q(t)e^{\lambda_{k}t}$, for any $t>0$. It is easy to see that
$q(t)e^{\lambda_{k}t}$ is continuous for $t>0$ and

$\inf\limits_{t>0}\left(\frac{D}{\sqrt{t}}\right)^{n}e^{\lambda_{k}t}=\left(\frac{D}{\sqrt{t}}\right)^{n}e^{\lambda_{k}t}\bigg{|}_{t=\frac{n}{2\lambda_{k}}}=\left(\frac{2e}{n}\right)^{\frac{n}{2}}\cdot\left(D\sqrt{\lambda_{k}}\right)^{n}.$

When $\frac{n}{2\lambda_{k}}\leq D^{2}$, i.e.,
$\lambda_{k}\geq\frac{n}{2D^{2}}$, we have

$\inf\limits_{t>0}q(t)e^{\lambda_{k}t}=q(t)e^{\lambda_{k}t}\bigg{|}_{t=\frac{n}{2\lambda_{k}}}=\overline{C}_{3}(n)e^{\overline{C}_{4}(n)(KD^{2}+L)}\left(\frac{2e}{n}\right)^{\frac{n}{2}}\cdot\left(D\sqrt{\lambda_{k}}\right)^{n}.$

Hence

$\overline{C}_{3}(n)e^{\overline{C}_{4}(n)(KD^{2}+L)}\left(\frac{2e}{n}\right)^{\frac{n}{2}}\cdot\left(D\sqrt{\lambda_{k}}\right)^{n}\geq
k+1,$

that is,

$\lambda_{k}\geq\frac{\overline{C}_{5}(n)(k+1)^{\frac{2}{n}}}{D^{2}}e^{-\overline{C}_{6}(n)(KD^{2}+L)}.$
(6.14)

When $\frac{n}{2\lambda_{k}}\geq D^{2}$, i.e.,
$\lambda_{k}D^{2}\leq\frac{n}{2}$, we have

$\inf\limits_{t>0}q(t)e^{\lambda_{k}t}=q(t)e^{\lambda_{k}t}\bigg{|}_{t=D^{2}}=\overline{C}_{3}(n)e^{\overline{C}_{4}(n)(KD^{2}+L)}e^{\lambda_{k}D^{2}}.$

Hence

$\alpha_{1}:=\overline{C}_{3}(n)e^{\overline{C}_{4}(n)(KD^{2}+L)}e^{\frac{n}{2}}\geq
k+1,$ (6.15)

which shows that there are only finitely many $\lambda_{k}$ in this case (the
total number $k$ only depends on $n,K,L,D$). Combining (6.15) and the lower
bound of $\lambda_{1}$ in (6.13), for these finitely many $\lambda_{k}$’s, one
can choose $\overline{C}_{7}(n)$ and $\overline{C}_{8}(n)$ such that

$\frac{\lambda_{k}D^{2}}{(k+1)^{\frac{2}{n}}}\geq\frac{\lambda_{1}D^{2}}{(k+1)^{\frac{2}{n}}}\geq\frac{\alpha_{0}D^{2}}{\alpha_{1}^{\frac{2}{n}}}\geq\overline{C}_{7}(n)e^{-\overline{C}_{8}(n)(KD^{2}+L)}.$
(6.16)

Combining (6.14) and (6.16) finishes the proof. ∎

Finally, we study the bottom spectrum of the Beltrami Laplacian $\Delta$ on
complete Riemannian manifolds with Bakry-Émery Ricci curvature bounded below.
Recall the definition of $\mu_{1}(\Delta):=\inf Spec(\Delta)$, the bottom
spectrum of $\Delta$. By the variational principle, we have

$\mu_{1}(\Delta)=\inf\limits_{\phi\in
T\backslash\\{0\\}}\frac{\int_{M}|\nabla\phi|^{2}}{\int_{M}\phi^{2}},$ (6.17)

where $T$ is any class of functions such that

$C_{0}^{\infty}(M)\subset T\subset W_{0}^{1}(M).$

Here $W_{0}^{1}(M)$ is the closure of $C_{0}^{\infty}(M)$ in $W^{1}(M)$, the
$L^{2}$ Sobolev space.

Following Munteanu and Wang’s method [31], the volume growth estimate and the
variational principle immediately imply the upper bound of $\mu_{1}(\Delta)$.
However, it is required that the volume should not exceed the exponential
linear growth, so we cannot use Theorem 2.1 to obtain the growth estimate of
the volume, where the calculated volume is the exponential square growth.
Therefore we adopt the method of Lemma 2.1 in [31] to obtain the volume growth
estimate.

######  Lemma 6.7.

Let $(M^{n},g)$ be a complete Riemannian manifold with
$\operatorname{Ric}_{f}\geq-(n-1)Kg$ for some constant $K\geq 0$. Assume that
there exist non-negative constants $\tilde{a}$ and $\tilde{b}$ such that

$|f|(x)\leq\tilde{a}r(x,o)+\tilde{b}\ for\ all\ x\in M.$

Then for any $\epsilon>0$, there exists a constant $\tilde{A}(\epsilon)>0$
such that the volume upper bound

$\operatorname{Vol}(B_{R}(o))\leq\tilde{A}(\epsilon)e^{(2\tilde{a}+(n-1)(\sqrt{K}+\epsilon))R}$

holds for all $R>0$. Here, $o$ is a fixed point on $M$, and $r(x,o)$ denotes
the distance from $x$ to $o$.

###### Proof.

By the Bochner formula, we have for $r(x)=r(x,o)$

$\displaystyle 0=\frac{1}{2}\Delta|\nabla r|^{2}$
$\displaystyle=|\operatorname{Hess}r|^{2}+\left<\nabla\Delta r,\nabla
r\right>+\operatorname{Ric}(\partial r,\partial r)$
$\displaystyle\geq\frac{(\Delta r)^{2}}{n-1}+\partial_{r}(\Delta
r)+\operatorname{Ric}(\partial_{r},\partial_{r})$
$\displaystyle\geq\frac{(\Delta r)^{2}}{n-1}+\partial_{r}(\Delta
r)-f^{\prime\prime}(r)-(n-1)K.$

Integrating this inequality from $1$ to $r$, we get

$\frac{1}{n-1}\int_{1}^{r}(\Delta t)^{2}dt+\Delta
r-f^{\prime}(r)\leq(n-1)Kr+\tilde{b}_{0}$

for some constant $\tilde{b}_{0}>0$ independent of $r$.
Then for any $r\geq 1$,

$\Delta_{f}r+\frac{1}{n-1}\int_{1}^{r}\left(\Delta_{f}t+f^{\prime}(t)\right)^{2}dt\leq(n-1)Kr+\tilde{b}_{0},$
(6.18)

The Cauchy-Schwarz inequality implies that

$\int_{1}^{r}\left(\Delta_{f}t+f^{\prime}(t)\right)^{2}dt\geq\frac{1}{r-1}\left(\int_{1}^{r}\left(\Delta_{f}t+f^{\prime}(t)\right)dt\right)^{2}.$

Therefore, from (6.18) we obtain

$\Delta_{f}r+\frac{1}{(n-1)r}\left(f(r)-f(1)+\int_{1}^{r}\left(\Delta_{f}t\right)dt\right)^{2}\leq(n-1)Kr+\tilde{b}_{0}.$
(6.19)

We now claim that for any $r\geq 1$ and any $\epsilon>0$,

$\int_{1}^{r}\left(\Delta_{f}t\right)dt\leq(\tilde{a}+(n-1)(\sqrt{K}+\epsilon))r+\tilde{a}+2\tilde{b}+\frac{\tilde{b}_{0}}{\sqrt{K}+\epsilon}.$
(6.20)

To prove this, define

$v(r):=(\tilde{a}+(n-1)(\sqrt{K}+\epsilon))r+\tilde{a}+2\tilde{b}+\frac{\tilde{b}_{0}}{\sqrt{K}+\epsilon}-\int_{1}^{r}\left(\Delta_{f}t\right)dt.$

We show instead that $v(r)>0$ for all $r\geq 1$. Clearly, $v(1)>0$.
Suppose that $v$ does not remain positive for all $r\geq 1$ and let $R>1$ be
the first number such that $v(R)=0$. Then $v^{\prime}(R)\leq 0$ and

$\int_{1}^{R}\left(\Delta_{f}t\right)dt=(\tilde{a}+(n-1)(\sqrt{K}+\epsilon))R+\tilde{a}+2\tilde{b}+\frac{\tilde{b}_{0}}{\sqrt{K}+\epsilon}.$

In other words,

$\displaystyle\frac{1}{(n-1)R}\left(f(R)-f(1)+\int_{1}^{R}\left(\Delta_{f}t\right)dt\right)^{2}$
$\displaystyle=\frac{1}{(n-1)R}\left(f(R)-f(1)+(\tilde{a}+(n-1)(\sqrt{K}+\epsilon))R+\tilde{a}+2\tilde{b}+\frac{\tilde{b}_{0}}{\sqrt{K}+\epsilon}\right)^{2}$
$\displaystyle\geq\frac{1}{(n-1)R}\left((n-1)(\sqrt{K}+\epsilon)R+\frac{\tilde{b}_{0}}{\sqrt{K}+\epsilon}\right)^{2}$
$\displaystyle\geq(n-1)(\sqrt{K}+\epsilon)^{2}R+2\tilde{b}_{0}.$

Plugging this into (6.19), we conclude that $\Delta_{f}R\leq-\tilde{b}_{0}<0$,
so $v^{\prime}(R)=\tilde{a}+(n-1)(\sqrt{K}+\epsilon)-(\Delta_{f}R)>0$, which
is a contradiction.
We have thus proved (6.20) is true for any $r\geq 1$ and $\epsilon>0$, so

$\ln J(r)-\ln
J(1)\leq(2\tilde{a}+(n-1)(\sqrt{K}+\epsilon))r+2\tilde{a}+4\tilde{b}+\frac{\tilde{b}_{0}}{\sqrt{K}+\epsilon}.$

In particular, for $R\geq 1$, any $\epsilon>0$ we have the volume bound of the
form

$\operatorname{Vol}(B_{R}(o))\leq\tilde{b}_{1}e^{(2\tilde{a}+(n-1)(\sqrt{K}+\epsilon))R},$

where the constant $\tilde{b}_{1}$ depends on
$\tilde{a},\tilde{b},\operatorname{Vol}(B_{1}(o)),\epsilon,K$ and $n$.
∎

Proof of Theorem 1.9. Let $R>1$ and $\psi$ a cut-off function on $B_{R}(o)$
such that $\psi=1$ on $B_{R-1}(o)$ and $|\nabla\psi|\leq 2$. Set
$\phi(y):=e^{-\frac{(2\tilde{a}+(n-1)(\sqrt{K}+\epsilon)+\delta)}{2}r(y,o)}\psi(y)$
as a test function in the variational principle (6.17) for $\mu_{1}(\Delta)$,
where $\delta>0$ and $\epsilon>0$ are arbitrary positive constants. Then by
Lemma 6.7, we obtain

$\mu_{1}(\Delta)\leq\frac{1}{4}\left(2\tilde{a}+(n-1)(\sqrt{K}+\epsilon)+\delta\right)^{2}.$

Since $\epsilon$ and $\delta$ are arbitrary, then
$\mu_{1}(\Delta)\leq\frac{1}{4}\left(2\tilde{a}+(n-1)\sqrt{K}\right)^{2}.$

In the case that $f$ is of sublinear growth, we can take $\tilde{a}=0$.
Therefore, $\mu_{1}(\Delta)\leq\frac{1}{4}(n-1)^{2}K$ and the theorem is
proved.
∎

## Data Availability

No data was used for the research described in the article.

## Acknowledgements

Research is partially supported by NSFC Grant No. 11971168, Shanghai Science
and Technology Innovation Program Basic Research Project STCSM 20JC1412900,
and Science and Technology Commission of Shanghai Municipality (STCSM) No.
22DZ2229014.

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