On the essential spectrum of the Laplacian and vague convergence of the curvature at infinity
Hironori Kumura
We shall prove that under some volume growth condition, the essential spectrum of the Laplacian contains the interval $[(n-1)^2K/4, \infty)$ if an $n$-dimensional Riemannian manifold has an end and the average of the part of the Ricci curvature on the end which lies below a nonpositive constant $(n-1)K$ converges to zero at infinity.
math0505523.tex
\documentclass[a4paper,10pt]{article}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{latexsym}
\usepackage{enumerate}

\pagestyle{plain}

\newtheorem{thm}{Theorem}[section]
\newtheorem{prop}{Proposition}[section]
\newtheorem{lem}{Lemma}[section]
\newtheorem{cor}{Corollary}[section]
\newtheorem{rem}{Remark}[section]
\newtheorem{defi}{Definition}
\def\bint{\displaystyle -\!\!\!\!\!\!\int}
\renewcommand{\theenumi}{\roman{enumi}}
\renewcommand{\labelenumi}{$(\theenumi)$}


\title{On the essential spectrum of the 
Laplacian and vague convergence of the curvature at infinity}

\author{Hironori Kumura}
\date{Department of Mathematics, Shizuoka University\\
Ohya, Shizuoka 422-8529, Japan\\
\textit{E-mail address}: smhkumu@ipc.shizuoka.ac.jp}


%%%%%%    TEXT START    %%%%%%
\begin{document}

\maketitle

\begin{abstract}
We shall prove that under some volume growth condition, 
the essential spectrum of the Laplacian contains the interval 
$[(n-1)^2K/4, \infty)$ 
if an $n$-dimensional Riemannian manifold has an end and the average 
of the part of the Ricci curvature on the end which lies below 
a nonpositive constant $(n-1)K$ 
converges to zero at infinity. 
\end{abstract}

%%%%%%    INTRODUCTION    %%%%%%

\section{Introduction}

The Laplace-Beltrami operator $\Delta$ 
on a noncompact complete Riemannian manifold $M$ is 
essentially self-adjoint on $C^{\infty}_0(M)$ 
and its self-adjoint extension to $L^2(M)$ has 
been studied by several authors from various points of view. 
For instance, Donnelly proved the following:
% 
%%%%%%    �藝 1.1    %%%%%%
%
\begin{thm}[\cite{d}]
Let $M$ be an $n$-dimensional Hadamard manifold, 
$x_0$ a fixed point of $M$, and $K$ a nonpositive constant. 
For a two plane $\pi$ in $T_xM$, 
we denote $K(x,\pi )$ to be   
the sectional curvature of $\pi$. 
Suppose that 
$$
\lim_{s\to \infty} \sup\{ |K(x,\pi )-K|;\mathrm{dist}(x,x_0)\ge s\}=0.
$$
Then the essential spectrum 
$\sigma _{\mathrm{ess}}(-\Delta)$ of $-\Delta$ 
is equal to $[-(n-1)^2K/4, \infty)$. 
\end{thm}

In this paper, we shall consider the case that the part of the 
Ricci curvature on an end of $M$ which lies below a nonpositive 
constant converges to zero in some average sense at infinity and 
some volume condition holds. And we 
study the essential spectrum of the Laplacian. 
A modification of Petersen-Wei's Lemma $2.2$ in 
\cite{p-w} is of crucial importance in our argument 
(see Lemma $2.3$ in Section $2$). 

Let us state our results more precisely. 
Let $M$ be an $n$-dimensional complete Riemannian manifold. 
Suppose that there exists an open subset $U$ of $M$ with compact smooth 
boundary $\partial U$ such that the outward pointing normal exponential 
map $\exp _{\partial U}:N^{+}(\partial U)\to M-U$ induces a diffeomorphism. 
Denote $r(*)=\mathrm{dist}(\partial U,*)$ on $M-U$ and let 
$(r,\theta )\in [0,\infty )\times \partial U$ be 
the geodesic coordinates induced from this diffeomorphism. 
We shall write the volume form as $dv_M=\sqrt{g}(r,\theta )drd\theta $ 
on the end $M-U$, where 
$d\theta$ stands for the Riemannian measure on $\partial U$ 
induced from $dv_M$. 
For latter use, 
we shall say that a plane $\pi \subset T_xM$ $(x\in M-U)$ is 
a radial plane if $\pi$ contains $\nabla r$, and by radial curvature 
we mean the restriction of the sectional curvature to all the radial 
planes. 

We introduce a new measure on the end $M-U$, which will be important tool 
in our arguments. 
For a given $C^2$-function $f:[0,\infty)\to \mathbf{R}$ satisfying 
$$\lim_{t\to \infty}f'(t)= \lim_{t\to \infty}f''(t)=0,$$ 
we introduce a measure 
$\mu_f$ on $M-U$ by 
$$\mu _f(A)=\int_A\exp (-cr(x)+f(r(x)))dv_M(x),$$
where $A$ is a measurable subset of $M-U$, $c=(n-1)\sqrt{-K}$ is a 
constant, and $K$ is a nonpositive constant. 
We also write $d\mu_f=\exp (-(n-1)\sqrt{-K}r+f(r))dv_M$. 
Moreover, we shall use the convenient notation 
$$
\frac{1}{\mu_f(A)}\int_Ah(x)d\mu_f(x)=\bint_Ah(x)d\mu_f(x),
$$
for the mean value of $h$ in $A$, when $\mu_f(A)>0$. 
This measure $d\mu_f$ is natural to consider the essential spectrum 
of the Laplacian, because  
the function $\exp (-(n-1)\sqrt{-K}r+f(r(x))$ cancels out the growth part 
of test functions and reduces our situation to subexponential growth case. 

Typical examples of the function 
$f$ are $f(t)=t^{\theta }$, $t/\log t$, $t/\log (\log t)$, and so on, 
where $\theta \in (0,1)$ is a constant. 
In this case, we should note that the weight function decomposes 
into two parts: the exponential decay part 
$\exp(-cr(x))$ and subexponential growth part $\exp(f(r))$.
The subexponential perturbation $\exp(f(r))$ will
have no effect on the essential spectrum of the Laplacian. 
 
In the sequel we will use the following notation:
$$
      A(a,b)=\{y\in M-U~|~a<\mathrm{dist}(\partial U,y)<b\}
      ~~\mathrm{for}~ 0\le a<b. 
$$
Moreover we write $h_+=\max\{h,0\}$ and 
$h_-=\max\{-h,0\}$ for a function $h$. 
In section $2$, we shall prove the following:
%
%%%%%%%    Theorem 1.2     %%%%%%%
%
\begin{thm}
Let $M$ be an $n$-dimensional complete Riemannian manifold. 
Suppose that there exists an open subset $U$ of $M$ with compact smooth 
boundary $\partial U$ such that the outward pointing normal exponential 
map $\exp _{\partial U}:N^{+}(\partial U)\to M-U$ induces a diffeomorphism. 
Denote $r(*)=\mathrm{dist}(\partial U,*)$ on $M-U$. 
We assume that the following three conditions hold:
\begin{enumerate}
	\item The measure of the end $M-U$ is infinite: 
$\displaystyle \lim_{t\to \infty}\mu_f(A(0,t))=+\infty ;$
\item there exists a constant $p>n/2$ such that \\
\hspace{20mm}
$
     \displaystyle \lim_{t\to \infty}
     \bint_{A(0,t)}\{(n-1)K-
     \mathrm{Ric}(\partial _r,\partial _r)\}_{+}^{p}d\mu_f=0;
$

\item there exists a nonnegative constant $c_1$ 
such that $\Delta r\ge -c_1$ on $M-U$. 
\end{enumerate}
Then the essential spectrum $\sigma_{\mathrm{ess}}(-\Delta )$ of $-\Delta$ 
contains the interval $[(n-1)^2K,\infty)$.
\end{thm}

We explain our assumptions in Theorem $1.2$. 
The assumption $(i)$ implies that the volume growth is greater than or 
equal to exponential except for the subexponential growth part. 
When the curvature of $M-U$ converges to a constant $K\le 0$ from {\it above} 
at infinity, $\mu_0(M-U)<\infty ~(f\equiv 0)$ may occurs. 
But choosing $f$ appropriately, 
we will be able to get the situation $\mu_f(M-U)=\infty$ in many cases. 
The assumption $(i)$ and $(ii)$ mean that the average of the part of the 
Ricci curvature on the end $M-U$ which lies below 
a nonpositive constant $(n-1)K$ converges to zero {\it at infinity}. 

Next we should add some comments about the condition $(iii)~\Delta r\ge -c_1$. 
If $M$ satisfies the assumptions of Theorem $1.1$, 
$\Delta r$ approaches to $(n-1)\sqrt{-K}$ at infinity 
and hence  
the condition $(iii)$ $\Delta r \ge -c_1$ in Theorem $1.2$ is 
indeed satisfied. 
We also note the following:
%%
%%%%%   Proposition   %%%%%
%%
\begin{prop}
Let $M$ be as in Theorem $1.2$. 
\begin{enumerate}
\item If the mean curvature of $\partial U$ is nonpositive, i.e., 
$\Delta r\ge 0$ on $\partial U$ and radial curvature on $M-U$ is 
nonpositive, then $\Delta r\ge 0$ on $M-U$.
\item If there exists a positive constant $A(>K)$ such that 
$\mathrm{Ric}(\partial r,\partial r)\ge -(n-1)A$ on $M-U$, 
then $\Delta r\ge -(n-1)\sqrt{A}$ on $M-U$.
\end{enumerate}
\end{prop} 
Thus, if either of the conditions in Proposition $1.1$ holds, then 
the condition $(iii)$ $\Delta r \ge -c_1$ in Theorem $1.2$ is satisfied. 
Proposition $1.1$ is proved by using the 
comparison theorem in Riemannian geometry in section $2$. 

The final remark is that 
in Theorem $1.2$ we do not assume that the average of 
the part of the curvature 
which lies {\it above} $K$ converges to zero at infinity, 
but we only assume that 
$(i)~\displaystyle \lim_{t\to \infty}\mu_f(A(0,t))=+\infty $. 
This condition $(i)$ bounds the curvature from the above very mildly,  
which seems to be worthy of note. 
%
%%%%%%    SECTION 2    %%%%%%
%
\section{Proof}
%
In this section, we shall prove Theorem $1.2$ by transplantation method 
and Proposition $1.1$ by the comparison theorem in Riemannian geometry. 

Theorem $1.2$ readily follows from the following five lemmas. 
In the following, we simply denote $c=(n-1)\sqrt{-K}$. 
First, we note the following obvious lemma:
%%%%%%    LEMMA 2.1    %%%%%%
%
\begin{lem}
Let $M$ be an $n$-dimensional complete Riemannian manifold. 
Suppose that there exists an open subset $U$ of $M$ with compact smooth 
boundary $\partial U$ such that the outward pointing normal exponential 
map $\exp _{\partial U}:N^{+}(\partial U)\to M-U$ induces a diffeomorphism. 
For any constants $\delta \in (0,1)$ 
and monotone increasing sequence $\{b_i\}$ 
with $\displaystyle \lim_{i\to \infty}b_i=\infty$, 
define the sequence $\{a_i\}$ by 
$\mu_f(A(0,a_i))=\delta\cdot \mu_f(A(0,b_i))$. 
Then the following two conditions are equivalent:\vspace{1mm}\\
\hspace{2mm}$(i)$~$\displaystyle\lim_{i\to \infty}a_i=\infty$.\vspace{1mm}\\
\hspace{2mm}$(i)$~$\displaystyle\lim_{t\to \infty}\mu_f(A(0,t))=\infty.$
\end{lem}
%
%%%%%%    LEMMA 2.2    %%%%%%
%
\begin{lem}
Let $M$ be as above. 
Suppose that for any constant $\delta \in (0,1)$ 
and monotone increasing sequence $\{d_i\}$ 
with $\displaystyle\lim_{i\to \infty}d_i=\infty$, 
there exists a sequence $\{a_i\}$ such that\vspace{1mm}\\
\hspace{2mm}$(i)$~$\mu_f(A(0,a_i))=\delta\cdot \mu_f(A(0,d_i))$;\vspace{1mm}\\
\hspace{2mm}$(ii)$~$\displaystyle\lim_{i\to \infty}a_i
=\lim_{i\to \infty}(d_i-a_i)=\infty$;\vspace{1mm}\\
\hspace{2mm}$(iii)$~$\displaystyle\lim_{i\to \infty}\bint_{A(a_i,d_i)}
|\Delta r-c|^2d\mu_f=0$.\vspace{1mm}\\
Then the essential spectrum of $-\Delta$ contains the 
interval $[c^2/4,\infty)$. 
\end{lem}
\begin{proof}
For an arbitrary increasing sequence $\{d_i\}$ 
with $\displaystyle \lim_{i\to \infty}d_i=\infty$, we define 
sequences $0<a_i<b_i<c_i<d_i$ $(i=1,2,\dots )$ 
by the following equation:
$$
   \mu_f(A(0,a_i))=\mu_f(A(a_i,b_i))
   =\mu_f(A(b_i,c_i))=\mu_f(A(c_i,d_i)).
$$
Then, our assumptions imply
\begin{align}
    &\lim_{i\to \infty}a_i
   =\lim_{i\to \infty}(b_i-a_i)
   =\lim_{i\to \infty}(c_i-b_i)
   =\lim_{i\to \infty}(d_i-c_i)
  =\infty ,\\
  &\lim_{i\to \infty}\bint_{A(a_i,d_i)}
  |\Delta r-c|^2d\mu_f=0.
\end{align}
Indeed, first, we see that $(i)$ and $(ii)$ imply that 
$\displaystyle\lim_{i\to \infty}a_i=\lim_{i\to \infty}b_i
=\lim_{i\to \infty}c_i=\lim_{i\to \infty}(d_i-c_i)=\infty$. 
Then, since $\frac{1}{2}\mu_f(A(0,b_i))=\mu_f(A(0,a_i))$ and 
$\displaystyle\lim_{i\to \infty}b_i=\infty$, 
$(i)$ and $(ii)$ with $d_i$ replaced by $b_i$ 
imply $\displaystyle\lim_{i\to \infty}(b_i-a_i)=\infty$. 
Similarly, we see that $\displaystyle\lim_{i\to \infty}(c_i-b_i)=\infty$. 

In view of (1), we can choose a sequence of functions 
$\varphi_i\in C^2_0(\mathbf{R})$ such that 
\begin{align*}
   &\varphi_i(t)=1 \quad (b_i\le t \le c_i); 
   ~~\varphi_i(t)=0 \quad (t\le a_i,~~d_i\le t);\\ 
   &0\le \varphi_i(t)\le 1~~\mathrm{for ~all}~ t\in \mathbf{R};\\
   &\sup |\varphi '|\le A_i,
   ~~\sup |\varphi ''|\le A_i,~~\lim_{i\to \infty}A_i=0.
\end{align*}
For any $\lambda >c^2/4$, we define functions 
$$
   \phi _i(x):=\sqrt{v_i}\cdot \varphi_i(r(x))\cdot 
   \exp\left(ar(x)+\frac{f(r(x))}{2}\right)
    \in C^2_0(M-U),
$$
where we set $v_i=\mu_f(A(a_i,d_i))^{-1}$ and 
$\displaystyle a=-\frac{c}{2}+\sqrt{-\left(\lambda-\frac{c^2}{4}\right)}
\in \mathbf{C}$. 
Then, a direct computation shows that 
\begin{align*}
   (\Delta \phi_i+\lambda \phi_i)
   =&\sqrt{v_i}\varepsilon(r)
     \exp \left(ar+\frac{f(r)}{2}\right)\\
   &  +\sqrt{v_i}\{\varphi_i'(r)+\varphi_i(r)f'(r)/2\}
     \cdot \Delta r\cdot \exp \left(ar+\frac{f(r)}{2}\right)\\
   &+\sqrt{v_i} a\varphi_i(r) (\Delta r-c) 
      \cdot \exp \left(ar+\frac{f(r)}{2}\right),
\end{align*}
where we set 
\begin{align*}
     \varepsilon(r)=&\varphi_i''(r)+2a \varphi_i'(r)+\varphi_i'(r)f'(r)
                      +\varphi_i(r)f''(r)/2\\
      &+a\varphi_i(r)f'(r)+\varphi_i(r)(f'(r))^2.
\end{align*}
Hence, we have 
\begin{align*}
   \int_{M-U}
   |\Delta \phi_i+\lambda \phi_i|^2dv_M
   \le &c_2\left(A_i^2+A_i^4+B_i^2+B_i^4\right)
   \,\bint_{A(a_i,d_i)}d\mu_f\\
   &+2\left(A_i^2+B_i^2\right)
   \,\bint_{A(a_i,d_i)}|\Delta r|^2 d\mu_f\\
   &+2|a|^2
   \,\bint_{A(a_i,d_i)}|\Delta r-c|^2d\mu_f,
\end{align*}
where $\displaystyle B_i=\max\{\sup_{t\ge a_i} |f'(t)|,\sup_{t\ge a_i} |f''(t)|\}$ 
and $c_2$ is a constant which depends only on $a$. 
Since 
$$
    A_i^2\bint_{A(a_i,d_i)}|\Delta r|^2d\mu_c 
    \le 2A_i^2\left(|c|^2+
    \bint_{A(a_i,d_i)}|\Delta r-c|^2d\mu_f\right),
$$
we see from $(2)$ and $\displaystyle\lim_{i\to \infty}B_i=0$ that 
\begin{equation}
     \lim_{i\to \infty }\int_M|\Delta \phi_i+\lambda \phi_i|^2dv_M=0.
\end{equation}
On the other hand, 
\begin{equation}
   1\ge \int_M|\phi_i|^2dv_M
   \ge v_i\cdot \mu_f(A(b_i,c_i))=\frac{1}{3}.
\end{equation}
In view of our construction of $\{\phi_i\}$, $(3)$ and $(4)$ imply that 
$\lambda \in \sigma_{\mathrm{ess}}(-\Delta)$. 
Since 
$\lambda > c^2/4$ is arbitrary, 
we obtain $[c^2/4,\infty)\subset \sigma_{\mathrm{ess}}(-\Delta)$. 
\end{proof}

A modification of the argument of the proof of 
Lemma $2.2$ in Petersen-Wei \cite{p-w} 
yields the following lemma, which is of crucial importance in our argument.
%
%%%%%%    LEMMA 2.3    %%%%%%
%
\begin{lem}
Let $M$ be an $n$-dimensional complete Riemannian manifold. 
Suppose that there exists an open subset $U$ of $M$ with compact smooth 
boundary $\partial U$ such that the outward pointing normal exponential 
map $\exp _{\partial U}:N^{+}(\partial U)\to M-U$ induces a diffeomorphism. 
Let $K \le 0$ be a constant 
and set $c=(n-1)\sqrt{-K}$. 
As a technical condition, we assume that 
$c\ge f'(t)$ for $t\ge 0$. 
Then if $p>n/2$, we have 
\begin{align*}
   &\int^r_0\{\Delta r(r,\theta)
   -h_K(r)\}_{+}^{2p}
      \exp(-cr+f(r))\sqrt{g}(r,\theta)dr\nonumber \\
   &\le c_2(n,p)\int^r_0\{(n-1)K
      -\mathrm{Ric}(\partial_r,\partial_r)\}_{+}^p
      \exp(-cr+f(r))\sqrt{g}(r,\theta)dr,
\end{align*}
where $c_2(n,p)$ is a constant depending only on $n$ and $p$, 
and $h_K$ is the solution to 
\begin{equation}
    h_K'(t)+\frac{h_K^2(t)}{n-1}=-(n-1)K,
\end{equation}
with initial condition $h_K(0)=a_0$, 
where $a_0\ge 0$ is a constant satisfying 
$\displaystyle a_0\ge \max_{\partial U} \Delta r$. 
\end{lem}
\begin{proof}
We set $\Delta r=h(r,\theta )$, 
then $h$ satisfies the differential inequality
$$
     \partial _r h+\frac{h^2}{n-1}\le -\mathrm{Ric}(\partial_r,\partial_r).
$$
For simplicity, we define
\begin{align*}
&\rho(r,\theta):=\max\{0,(n-1)K-\mathrm{Ric}(\partial_r,\partial_r)\},\\
&\psi (r,\theta):=\max\{0,h(r,\theta)-h_K(r)\}
\end{align*}
and set $\widetilde{\sqrt{g}}(r,\theta):=\exp(-cr+f(r))\sqrt{g}(r,\theta)$.
In the following, we fix $\theta \in \partial U$, therefore we simply 
write $\rho(r)$ and $\psi (r)$ for $\rho(r,\theta)$ and $\psi (r,\theta)$ 
respectively. 
Then we see that $\psi$ is absolutely continuous and satisfies 
\begin{equation}
     \psi '+\frac{\psi^2}{n-1}+2\frac{\psi h_K}{n-1}\le \rho.
\end{equation}
Indeed, if $(h-h_K)(r_0)>0$, then $\psi(r)$ coincides with 
$(h-h_K)(r)$ around $r_0$. 
Hence
\begin{align*}
   &\psi'+\frac{\psi^2}{n-1}+2\frac{\psi h_K}{n-1}
   =h'-h'_K+\frac{h^2}{n-1}-\frac{h_K^{2}}{n-1}\\
   \le &(n-1)K-\mathrm{Ric}(\partial r,\partial r)\le \rho
\end{align*}
and $(6)$ holds around $r_0$. 
When $(h-h_K)(r_0)=0$, $\psi(r_0)=0$ by definition. 
Moreover, in this case, we may take $\psi'(r_0)=0$ and 
$(6)$ also holds at these points because $\rho$ is nonnegative. 
When $(h-h_K)(r_0)<0$, $\psi(r)=0$ around $r_0$ and hence 
the left hand side of $(6)$ is zero. 
Therefore $(6)$ also holds at this point. 
Thus we conclude that $(6)$ holds for all $r$. 

Multiplying $(6)$ by $\psi^{2p-2}\exp\{-cr+f(r)\}\sqrt{g}(r,\theta)$ 
and integrating over $[0,r]$, we get 
\begin{align*}
   \int_0^r\rho\psi^{2p-2}\widetilde{\sqrt{g}}(t,\theta)dt
   \ge &\int_0^r\psi'\psi^{2p-2}\widetilde{\sqrt{g}}(t,\theta)dt
    +\frac{1}{n-1}\int_0^r\psi^{2p}\widetilde{\sqrt{g}}(t,\theta)dt\\
     &+\frac{2}{n-1}\int_0^rh_K\psi^{2p-1}\widetilde{\sqrt{g}}(t,\theta)dt.\\
\end{align*}
Integration by parts yields
\begin{align*}
   &\int_0^r\psi'\psi^{2p-2}\widetilde{\sqrt{g}}(t,\theta)dt\\
   =&\left[\frac{\psi^{2p-1}\widetilde{\sqrt{g}}(t,\theta)}
   {2p-1}\right]^r_0 -\frac{1}{2p-1}\int_0^r\psi^{2p-1}\{-c+f'(r)+h\}
   \widetilde{\sqrt{g}}(t,\theta)dt\\
   \ge &-\frac{1}{2p-1}\int^r_0\psi^{2p-1}(t)(f'(t)-c+h_K(t))
   \widetilde{\sqrt{g}}(t,\theta)dt\\
   &-\frac{1}{2p-1}\int_0^r\psi^{2p}\widetilde{\sqrt{g}}(t,\theta)dt,
\end{align*}
where we have used $\psi(0)=0$, which follows from our assumption 
$h_K(0)=a_0\ge \max \{\Delta r(0,\theta)~|~\theta\in \partial U\}$. 
Inserting this inequality in the previous one, we have
\begin{align*}
     \int_0^r\rho(t)\psi^{2p-2}(t)\widetilde{\sqrt{g}}(t,\theta)dt
\ge & \int_0^r\psi^{2p-1}(t)\frac{w(t)}{(n-1)(2p-1)}
       \widetilde{\sqrt{g}}(t,\theta)dt\\
    &  +\left(\frac{1}{n-1}-\frac{1}{2p-1}\right)\int_0^r\psi^{2p}(t)
       \widetilde{\sqrt{g}}(t,\theta)dt.
\end{align*}
Here, we set 
$w(t)=(n-1)(c-f'(t))+(4p-n-1)h_K(t)$. 
Since $h_K(t)$ satisfies $(5)$ and the initial condition $h_K(0)=a_0\ge 0$, 
we see that $h_K(t)\ge 0$ for $t\ge 0$ and, 
hence, $w(t)\ge 0$ for all $t\ge 0$. 
Thus we obtain 
\begin{align*}
&\left(\frac{1}{n-1}-\frac{1}{2p-1}\right)
\int_0^r\psi^{2p}(t)\widetilde{\sqrt{g}}(t,\theta)dt\\
&\le \int_0^r\rho(t)\psi^{2p-2}(t)\widetilde{\sqrt{g}}(t,\theta)dt\\
&\le \left(\int^r_0\rho^p(t)\widetilde{\sqrt{g}}(t,\theta)dt\right)^{1/p}
\cdot \left(\int^r_0\psi^{2p}(t)\widetilde{\sqrt{g}}(t,\theta)dt\right)^{1-\frac{1}{p}}.
\end{align*}
Hence we have 
$$
\left(\frac{1}{n-1}-\frac{1}{2p-1}\right)
\left(\int^r_0\psi^{2p}(t)\widetilde{\sqrt{g}}(t,\theta)dt\right)^{1/p}
\le \left(\int^r_0\rho^p(t)\widetilde{\sqrt{g}}(t,\theta)dt\right)^{1/p},
$$
that is, 
$$
\int^r_0\psi^{2p}(t)\widetilde{\sqrt{g}}(t,\theta)dt
\le \left(\frac{1}{n-1}-\frac{1}{2p-1}\right)^{-p}
\int^r_0\rho^p(t)\widetilde{\sqrt{g}}(t,\theta)dt.
$$
Thus, we obtain the desired inequality with 
$c_2(n,p)=\left(\frac{1}{n-1}-\frac{1}{2p-1}\right)^{-p}$.
\end {proof}
%
%%%%%%    REMARK 2.1    %%%%%%
%
\begin{rem}
We need not worry about 
the technical condition $c\ge f'(t)$ in Lemma $2.3$,
because our condition 
$\displaystyle \lim_{t\to \infty}\mu_f(A(0,t))=+\infty$ 
will enables us to disperse it. 
\end{rem}
%
%%%%%%    LEMMA 2.4    %%%%%%
%
\begin{lem}
Let $M,r,K,c=(n-1)\sqrt{-K}$ be as above and assume the 
following:\vspace{1mm}\\
\hspace{2mm}$(i)$~
$\displaystyle\lim_{t\to \infty}\mu_f(A(0,t))=+\infty ;$\vspace{1mm}\\
\hspace{2mm}$(ii)$~There exists a constant $p>n/2$ such that \vspace{1mm}\\
\hspace{20mm}
$
     \displaystyle \lim_{t\to \infty}
     \bint_{A(0,t)}\{(n-1)K-
     \mathrm{Ric}(\partial _r,\partial _r)\}_{+}^{p}d\mu_f=0.
$\\
Then we have
\begin{equation}
     \lim_{t\to \infty}\frac{d}{dt}\log (\mu_f(A(0,t))=0.
\end{equation}
In particular, for any monotone increasing sequence $\{b_i\}$ with 
$\displaystyle \lim_{i\to \infty}b_i=\infty$ and $\delta\in (0,1)$, 
define a sequence 
$\{a_i\}$ by $\mu_f(A(a_i,b_i))=\delta \cdot \mu_f(A(0,b_i))$. 
Then we have $\displaystyle \lim_{i\to \infty}(b_i-a_i)=\infty$. 
\end{lem}
\begin{proof}
We again set 
$\widetilde{\sqrt{g}}(r,\theta):=\exp(-cr+f(r))\sqrt{g}(r,\theta)$ 
for the sake of simplicity. 
First we note that $\displaystyle \lim_{r\to \infty}h_K(r)=(n-1)\sqrt{-K}$ 
because $h_K(0)=a_0\ge 0$.  
Let $\varepsilon >0$ be given and take sufficiently large $s_1>0$. 
For $t>s_1$, integrating the equation 
$\partial _r\widetilde{\sqrt{g}}(r,\theta)
=\{(\Delta r)(r,\theta)-c+f'(r)\}\widetilde{\sqrt{g}}(r,\theta)$ 
over $[s_1,t]\times \partial U$, we have 
\begin{align*}
     &  \int_{\partial U}\widetilde{\sqrt{g}}(t,\theta)d\theta
        -\int_{\partial U}\widetilde{\sqrt{g}}(s_1,\theta)d\theta\\
    =&  \int^t_{s_1}dr\int_{\partial U}
        \{\Delta r-c+f'(r)\}
        \widetilde{\sqrt{g}}(r,\theta)d\theta\\
    =&  \int^t_{s_1}dr\int_{\partial U}
        \{\Delta r-h_K(r)\}
        \widetilde{\sqrt{g}}(r,\theta)d\theta\\
     &  +\int^t_{s_1}dr\int_{\partial U}
        \left(h_K(r)-c+f'(r)\right)
        \widetilde{\sqrt{g}}(r,\theta)d\theta\\
  \le&  \int_{A(0,t)}
        \left\{\Delta r-h_K(r)\right\}_{+}d\mu_f
        +\varepsilon \mu_f(A(s_1,t))\\
  \le&  \mu_f(A(0,t))\left(\bint_{A(0,t)}
        \left\{\Delta r-h_K(r)\right\}_{+}^{2p}
        d\mu_f\right)^{\frac{1}{2p}}+\varepsilon \mu_f(A(s_1,t))\\
  \le&  \mu_f(A(0,t))c_3(n,p)\left(\bint_{A(0,t)}
        \left\{(n-1)K-\mathrm{Ric}(\partial_r,\partial_r)\right\}_{+}^{p}
        d\mu_f\right)^{\frac{1}{2p}}\\
     &  +\varepsilon \mu_f(A(s_1,t)),
\end{align*}
In the last inequality above, we have used Lemma $2.3$. 
Hence, 
\begin{align*}
     (0\le &)\frac{d}{dt}\log \mu_f\left(A(0,t)\right)
             =\mu_f(A(0,t))^{-1}
             \int_{S^{n-1}(1)}\widetilde{\sqrt{g}}(t,\theta)d\theta\\
     \le &\mu_f(A(0,t))^{-1}
     \int_{S^{n-1}(1)}\widetilde{\sqrt{g}}(s_1,\theta)d\theta\\
     &+c_3(n,p)\left(\bint_{A(0,t)}
       \{(n-1)K-\mathrm{Ric}(\partial_r,\partial_r)\}_{+}^{p}
       d\mu_f\right)^{\frac{1}{2p}}+\varepsilon
\end{align*}
and our assumptions $(i)$ and $(ii)$ imply that the equation $(7)$ holds. 

Next we shall show that $(7)$ implies 
$\displaystyle\lim_{i\to \infty}(b_i-a_i)=\infty$. 
For any $\varepsilon >0$, there exists $t_0>0$ such that 
\begin{equation}
     \frac{d}{dt}\log (\mu_f(A(0,t)))\le \varepsilon \qquad 
     \mathrm{for~all}~t\ge t_0.
\end{equation}
Since $\displaystyle\lim_{i\to \infty}a_i=\infty$ by Lemma $2.1$, 
there exits $i_0$ such that $a_i>t_0$ for all $i\ge i_0$. 
Thus, for any $i\ge i_0$, integrating this inequality $(7)$ over $[a_i,b_i]$, 
we obtain
$$
     0<-\log(1-\delta)
     =\log\frac{\mu_f(A(0,b_i))}{\mu_f(A(0,a_i))}
     \le \varepsilon(b_i-a_i).
$$
Hence, $\displaystyle\lim_{i\to \infty}(b_i-a_i)=\infty$. 
\end{proof}
%
%%%%%%    LEMMA 5    %%%%%%
%
\begin{lem}
We assume that the following three conditions hold:\vspace{1mm}\\
\hspace{2mm}$(i)$~$\displaystyle\lim_{t\to \infty}\mu_f(A(0,t))=+\infty ;$
\vspace{1mm}\\
\hspace{2mm}$(ii)$~there exists a constant $p>n/2$ such that \vspace{1mm}\\
\hspace{20mm}
$
     \displaystyle \lim_{t\to \infty}
     \bint_{A(0,t)}\{(n-1)K-
     \mathrm{Ric}(\partial _r,\partial _r)\}_{+}^{p}d\mu_f=0;
$\vspace{1mm}\\
\hspace{2mm}$(iii)$~there exists a constant 
$c_1\ge 0$ such that $\Delta r\ge -c_1$. \vspace{1mm}\\
If we define the sequence $\{a_i\}$ by 
$\mu_f(A(0,a_i))=\delta\cdot \mu_f(A(0,b_i))$ for any monotone 
increasing sequence $\{b_i\}$ 
with $\displaystyle\lim_{i\to \infty}b_i=\infty$ and $\delta \in (0,1)$, 
then the condition $(iii)$ in Lemma $2.2$ holds. 
\end{lem}
\begin{proof}
By Lemma $2.3$, we have 
\begin{align*}
       & \mu_f(A(a_i,b_i))^{-1}\int_{\partial U}d\theta 
         \int^{b_i}_{a_i}\{(\Delta r)(r,\theta)-h_K(r)\}^{2p}_{+}
         \widetilde{\sqrt{g}}(r,\theta)dr\\
   \le & \frac{c_2(n,p)}{(1-\delta)\mu_f(A(0,b_i))}
          \int_{\partial U}d\theta \int^{b_i}_{0}
          \{(n-1)K-\mathrm{Ric}(\partial _r,\partial _r)\}^p_{+}
          \widetilde{\sqrt{g}}(r,\theta)dr\\
     &   \to 0\qquad \mathrm{as}\quad i\to \infty.
\end{align*}
Hence, in view of the fact that 
$\displaystyle\lim_{i\to \infty}a_i=\infty$ and 
$\displaystyle\lim_{r\to \infty}h_K(r)=(n-1)\sqrt{-K}=c$, we see that
\begin{align}
     &  \bint_{A(a_i,b_i)}
       \{\Delta r-c\}_{+}^2d\mu_f\nonumber \\ 
  \le& \left(\bint_{A(a_i,b_i)}\{\Delta r-c\}_{+}^{2p}d\mu_f\right)
       ^{\frac{1}{p}}
       \to 0\qquad \mathrm{as}\quad i\to \infty.
\end{align}
On the other hand, integrating the equation 
$\displaystyle
     \widetilde{\sqrt{g}}(b_i,\theta)-\widetilde{\sqrt{g}}(a_i,\theta)
     =\int_{a_i}^{b_i}\partial _r\widetilde{\sqrt{g}}(r,\theta)dr
$ over $\partial U$, we have 
\begin{align*}
     &\mu_f(A(a_i,b_i))^{-1}\int_{\partial U}\widetilde{\sqrt{g}}
     (b_i,\theta)d\theta
     -\mu_f(A(a_i,b_i))^{-1}\int_{\partial U}
     \widetilde{\sqrt{g}}(a_i,\theta)d\theta\\
     =&\mu_f(A(a_i,b_i))^{-1}\int_{\partial U}d\theta
     \int_{a_i}^{b_i}\partial _r\widetilde{\sqrt{g}}(r,\theta)dr.
\end{align*}
Here, since
\begin{align*}
     &\mu_f(A(a_i,b_i))^{-1}\int_{\partial U}\widetilde{\sqrt{g}}
     (b_i,\theta)d\theta\\
     =&\frac{\mu_f(A(0,b_i))}{\mu_f(A(a_i,b_i))}
     \cdot \mu_f(A(0,b_i))^{-1}\int_{\partial U}\widetilde{\sqrt{g}}
     (b_i,\theta)d\theta\\
     =&(1-\delta)^{-1}\frac{d}{dt}\Big|_{t=b_i}\log\mu_f(A(0,t))
     \to 0\qquad \mathrm{as}\quad i\to \infty
\end{align*}
and similarly 
\begin{align*}
     &  \mu_f(A(a_i,b_i))^{-1}\int_{\partial U}
       \widetilde{\sqrt{g}}(a_i,\theta)d\theta\\
    =&\frac{\delta}{1-\delta}
     \cdot \frac{d}{dt}\Big|_{t=a_i}\log\mu_f(A(0,t))
     \to 0\qquad \mathrm{as}\quad i\to \infty,
\end{align*}
we have 
$$
     \lim_{i\to \infty}
       \bint_{A(a_i,b_i)}\{\Delta r-c\}d\mu_f
    =\lim_{i\to \infty}
       \bint_{A(a_i,b_i)}\{\Delta r-c+f'(r)\}d\mu_f=0.
$$
Therefore, in view of H\"older's inequality, $(9)$ implies that 
$$
     \lim_{i\to \infty}
     \bint_{A(a_i,b_i)}\{\Delta r-c\}_{-}d\mu_f=0.
$$
Here, since
$
     \{\Delta r-c\}_{-}=\max\{0,-(\Delta r-c)\}\le c_1+c
$
on $M-U$, 
\begin{equation}
     \lim_{i\to \infty}
     \bint_{A(a_i,b_i)}\{\Delta r-c\}_{-}^2d\mu_f=0.
\end{equation}
The equations $(9)$ and $(10)$ give the desired equality. 
\end{proof}
From five lemmas above and Remark $2.1$, we see that Theorem $1.2$ holds. 

\vspace{5mm}
\noindent{\it Proof of Proposition $1.1$}\vspace{1mm}\\
Since $(i)$ immediately follows from the comparison theorem in Riemannian 
geometry, we shall omit its proof.
Hence we shall only prove $(ii)$. 
Assuming that there exists a point $(r_0,\theta_0)\in M-U$ such that 
$(\Delta r)(r_0,\theta_0)<-(n-1)\sqrt{A}$, we shall obtain a contradiction. 
Let us set $h(r)=(\Delta r)(r,\theta_0)$ for $r\ge r_0$. 
Moreover let $f(r):[r_0,l)\to \mathrm{R}$ be the 
solution to the differential equation
$$
   f'+\frac{f^2}{n-1}-(n-1)A=0;~~f(r_0)=(\Delta r)(r_0,\theta_0),
$$
where $[r_0,l)$ is the maximal interval of existence of the solution. 
Then the comparison theorem in Riemannian geometry implies 
$h(r)\le f(r)$ on $[r_0,l)$. 
But the solution $f$ is expressed as follows:
$$
   f(r)=-(n-1)\sqrt{A}\frac{1+a_0e^{2\sqrt{A}r}}{1-a_0e^{2\sqrt{A}r}},
$$
where $a_0>0$ is the constant determined by 
$a_0e^{2\sqrt{A}r_0}=(B-1)/(B+1)$ and 
$B=-(\Delta r)(r_0,\theta_0)/(n-1)\sqrt{A}>1$. 
Hence 
$\displaystyle 
l=r_0+\frac{1}{2\sqrt{A}}\log \left(\frac{B+1}{B-1}\right)<\infty$ 
and $\displaystyle\lim_{r\to l-0}f(r)=-\infty$. 
This contradicts the fact that 
$h(r)$ is defined for all $r\ge r_0$. 
Therefore, $\Delta r\ge -(n-1)\sqrt{A}$ on $M-U$ and 
this completes the proof of Proposition $1.1$. \vspace{5mm}

Finally, we note that 
the condition $(i)$ can be replaced by a more familiar one for 
geometers:
\begin{lem}
Let us set $V(t)=\mathrm{Vol}(A(0,t))$. 
Then we see the following: 
\begin{enumerate}
	\item If $c>0$, 
$\displaystyle\lim_{t\to \infty}\exp(-ct+f(t))V(t)=\infty$ implies  
$\displaystyle\lim_{t\to \infty}\mu_f(A(0,t))=+\infty$. 
	\item If $c=0$, we add an assumption $f'\ge0$. 
Then $\displaystyle\lim_{t\to \infty}\exp(f(t))V(t)=\infty$ implies  
$\displaystyle\lim_{t\to \infty}\mu_f(A(0,t))=+\infty$. 
\end{enumerate}
\end{lem}
\begin{proof}
We write $S(r)=\mathrm{Vol}_{n-1}(\partial_{+} A(0,r))$, where 
$\partial_{+} A(0,r)=\{y\in M-U | $\\
$\mathrm{dist}(\partial U, y)=r\}$. 
Then integration by parts yields
\begin{align*}
     \mu_f(A(0,t))&=\int_0^t\exp(-cr+f(r))S(r)dr\\
     &=\left[\exp(-cr+f(r))V(r)\right]^t_0
     -\int_0^t(-c+f'(r))\exp(-cr+f(r))V(r)dr\\
     &=\exp(-ct+f(t))V(t)+\int_0^t(c-f'(r))\exp(-cr+f(r))V(r)dr.
\end{align*}
In view of our assumption $\displaystyle \lim_{t\to \infty}f'(t)=0$, 
we immediately see that Lemma $2.6$ holds. 
\end{proof}







%%%%%%    �Q�l����    %%%%%%
\begin{thebibliography}{99}
\bibitem{b1}~R.~Brooks, 
\textit{A relation between growth and the spectrum of the Laplacian},
 Math.~Z. \textbf{178}~(1981), 501-508.
%
\bibitem{b2}~R.~Brooks, 
\textit{On the spectrum of non-compact manifolds with finite volume},
 Math.~Z., \textbf{187}~(1984), 425-432.
%
\bibitem{c}~I.~Chavel, 
{\it Eigenvalues in Riemannian Geometry}, Academic Press, New York, 1984.

%
\bibitem{d}~H.~Donnelly,~
{\it On the essential spectrum of a complete Riemannian manifold},
 Topology~\textbf{20}~(1981), 1-14.
%
\bibitem{d-g}~H.~Donnelly~and~N.~Garofalo, 
{\it Riemannian manifolds whose Laplacians have purely continuous spectrum},
 Math.~Ann., \textbf{293}~(1992), 143-161.
%
\bibitem{d-l}~H.~Donnelly~and~P.~Li, 
{\it Pure point spectrum and negative curvature for noncompact manifolds},
 Duke~Math.~J.,~\textbf{46}~(1979),~497-503.
%
\bibitem{e-w}~K.~D.~Elworthy and F.-Y.~Wang, 
{\it On the essential spectrum of the Laplacian on Riemannian manifolds},
 preprint.
%
\bibitem{e}~J.~F.~Escobar, 
{\it On the spectrum of the Laplacian on complete Riemannian manifolds},
 Comm.~Partial~Differential Equations, \textbf{11}~(1986), 63-85.
%
\bibitem{e-f}~J.~F.~Escobar~and A.~Freire,
{\it The spectrum of the Laplacian of manifolds of positive curvature},
~Duke~Math.~J., \textbf{65}~(1992), 1-21.
%
\bibitem{g}~S.~Gallot, 
{\it Isoperimetric inequalities based on integral norms of Ricci curvature},
 Ast\'{e}risque 157-158~(1988), 191-216.
%
\bibitem{k-l}~L.~Karp, 
{\it Noncompact manifolds with purely continuous spectrum},
 Michigan~Math.~J., \textbf{31}~(1984), 339-347.
%
\bibitem{k}~H.~Kumura, 
{\it On the essential spectrum of the Laplacian on complete manifolds},
 J.~Math.~Soc.~Japan, \textbf{49}~(1997), 1-14.
%
\bibitem{l}~J.~Li,~
{\it Spectrum of the Laplacian on a complete Riemannian 
manifold 
with nonnegative Ricci curvature which possess(es) a pole},
 J.~Math.~Soc.~Japan, \textbf{46}~(1994), 213-216.
%
\bibitem{p-w}~P.~Petersen and G.~Wei, 
\textit{Relative volume comparison with integral curvature bounds},
 GAFA, Geom.~funct.~anal., \textbf{7}~(1997), 1031-1045.
%
\end{thebibliography}




\end{document}



math0505523.extracted_bib
\begin{thebibliography}{99}
\bibitem{b1}~R.~Brooks, 
\textit{A relation between growth and the spectrum of the Laplacian},
 Math.~Z. \textbf{178}~(1981), 501-508.
%
\bibitem{b2}~R.~Brooks, 
\textit{On the spectrum of non-compact manifolds with finite volume},
 Math.~Z., \textbf{187}~(1984), 425-432.
%
\bibitem{c}~I.~Chavel, 
{\it Eigenvalues in Riemannian Geometry}, Academic Press, New York, 1984.

%
\bibitem{d}~H.~Donnelly,~
{\it On the essential spectrum of a complete Riemannian manifold},
 Topology~\textbf{20}~(1981), 1-14.
%
\bibitem{d-g}~H.~Donnelly~and~N.~Garofalo, 
{\it Riemannian manifolds whose Laplacians have purely continuous spectrum},
 Math.~Ann., \textbf{293}~(1992), 143-161.
%
\bibitem{d-l}~H.~Donnelly~and~P.~Li, 
{\it Pure point spectrum and negative curvature for noncompact manifolds},
 Duke~Math.~J.,~\textbf{46}~(1979),~497-503.
%
\bibitem{e-w}~K.~D.~Elworthy and F.-Y.~Wang, 
{\it On the essential spectrum of the Laplacian on Riemannian manifolds},
 preprint.
%
\bibitem{e}~J.~F.~Escobar, 
{\it On the spectrum of the Laplacian on complete Riemannian manifolds},
 Comm.~Partial~Differential Equations, \textbf{11}~(1986), 63-85.
%
\bibitem{e-f}~J.~F.~Escobar~and A.~Freire,
{\it The spectrum of the Laplacian of manifolds of positive curvature},
~Duke~Math.~J., \textbf{65}~(1992), 1-21.
%
\bibitem{g}~S.~Gallot, 
{\it Isoperimetric inequalities based on integral norms of Ricci curvature},
 Ast\'{e}risque 157-158~(1988), 191-216.
%
\bibitem{k-l}~L.~Karp, 
{\it Noncompact manifolds with purely continuous spectrum},
 Michigan~Math.~J., \textbf{31}~(1984), 339-347.
%
\bibitem{k}~H.~Kumura, 
{\it On the essential spectrum of the Laplacian on complete manifolds},
 J.~Math.~Soc.~Japan, \textbf{49}~(1997), 1-14.
%
\bibitem{l}~J.~Li,~
{\it Spectrum of the Laplacian on a complete Riemannian 
manifold 
with nonnegative Ricci curvature which possess(es) a pole},
 J.~Math.~Soc.~Japan, \textbf{46}~(1994), 213-216.
%
\bibitem{p-w}~P.~Petersen and G.~Wei, 
\textit{Relative volume comparison with integral curvature bounds},
 GAFA, Geom.~funct.~anal., \textbf{7}~(1997), 1031-1045.
%
\end{thebibliography}