设$(M, F)$为一芬斯勒流形. Rademacher在文献[5]中定义了可反系数如下

$ \lambda_F=\sup\limits_{(x, y)\in TM\backslash0}\frac{F(x, -y)}{F(x, y)}. $

显然, $\lambda_F\in[1, +\infty]$. $\lambda_F=1$时当且仅当$~F$是可反的.当$\lambda_F < \infty$时,流形的向前完备性等价于向后完备性.特别地,紧致的芬斯勒流形必是完备的,此时$\lambda_F$是有限的.对于任意一点$x\in M$,定义$\lambda_F(x)=\sup\limits_{y\in T_xM\backslash0}\frac{F(x, -y)}{F(x, y)}.$则$\lambda_F=\sup\limits_{x\in M} \lambda_F(x).$

例1.1  在标准欧氏单位球${\Bbb B}^n(1)$中定义$\alpha=|y|, ~\beta=|x|y^1$,其中$|\cdot|$为标准欧氏范数.则$F(x, y)=\alpha+\beta$为一Randers度量.于是在点$x\in{\Bbb B}^n(1)$处, $\lambda_F(x)=\frac{1+|x|}{1-|x|}.$易见,当$|x|\to1$时, $\lambda_F(x)\to+\infty$.

设$\overleftarrow{F}(x, y)=F(x, -y)$及$F^*$分别为$F$的反向芬斯勒度量与对偶芬斯勒度量.定义

$ \begin{eqnarray} \overleftarrow{\lambda}_F=\sup\limits_{(x, y)\in TM\backslash0}\frac{\overleftarrow{F}(x, -y)}{\overleftarrow{F}(x, y)}, \quad \lambda^*_F=\sup\limits_{(x, \xi)\in T^*M\backslash0}\frac{F^*(x, -\xi)}{F^*(x, \xi)}.\end{eqnarray}$

则$\overleftarrow{\lambda}_F=\lambda_F$,且由文献[2]可知$\lambda^*_F=\lambda_F$.

可反系数在整体芬斯勒几何中具有重要的作用.文献[2, 7-9]中在可反系数有限的前提下给出了若干特征值估计,并得到Calabi-Yau型线性体积增长定理.本文考虑的问题是能否给出一些方法将"可反系数有限"这个条件去掉或减弱?

如上所述,可反系数揭示了度量$F$及其反向度量$\overleftarrow{F}$之间的关系.除此之外, $F$与$\overleftarrow{F}$对应的几何量与几何对象之间也具有重要的关系.如:

●   $F$的向前(向后)测地球就是$\overleftarrow{F}$的向后(向前)测地球;

●   $F$的向前(向后)距离函数就是$\overleftarrow{F}$的向后(向前)距离函数;

●  加权Ricci曲率满足$\overleftarrow{{\rm{Ric}}}_N(v)={\rm{Ric}}_N(-v), N\in[n, \infty)$;

●  若$f$为芬斯勒Laplacian $\Delta$的特征函数,则$-f$为向后Laplacian $\overleftarrow{\Delta}$的特征函数且$\Delta$与$\overleftarrow{\Delta}$具有相同的特征值.

利用这些关系及相应的讨论,往往可以避免引入可反系数.由此,本文给出了更加精细的特征值估计.特别地,我们去掉了"可反系数有限"这个条件.此外,在可反系数至多为多项式增长时,我们得到了Calabi-Yau型体积增长定理.

设$(M, F, {\rm d}\mu)$为$n$维芬斯勒流形. $B^+_{x_0}(r)$及$B^-_{x_0}(r)$分别表示以$x_0$为中心$r$为半径的向前(向后)测地球.对于正数$N$,定义$\overline{N}$为

(2.1) $ \begin{eqnarray}\label{1}\overline{N}:=\min\{m\in{\Bbb N}|m\geq N\}.\end{eqnarray} $

定理2.1  设$(M, F, {\rm d}\mu)$为$n$维完备芬斯勒流形.若加权Ricci曲率Ric$_{N}\geq (N-1)k, $$N\in(n, \infty)$,则芬斯勒Laplacian的第一Dirichlet特征值满足

$ \max\{\lambda_{1}(B^+_{x_0}(r)), \lambda_{1}(B^-_{x_0}(r))\}\leq \lambda_{1}(V_{\overline{N}}(k, r)), $

其中$V_{\overline{N}}(k, r)$表示具有旗曲率$k$与零$S$曲率的$\overline{N}$维单连通芬斯勒空间中半径为$r$的向前(或向后)测地球.

注2.1  在文献[6]中,相应的估计较为粗略,而文献[2]仅考虑Ricci曲率具有负下界情形.在文献[9,定理3.1]中,作者仅研究向前测地球情形,将之与黎曼空间形式中测地球加以比较.定理$2.1$改进或推广了以上结果,这为下面改进第一Neumann特征值估计作了准备.

证  若$V_{\overline{N}}(k, r)$为黎曼空间形式中测地球,则令$\varphi$为Laplacian在$\overline{V_{\overline{N}}(k, r)}$中第一Dirichlet特征函数.由于单连通空间形式是两点齐性的,故$\varphi$是径向函数,即$\varphi(\bar{x})=\varphi(d_R(\bar{x}_{0}, \bar{x}))$,其中$\bar{x}_{0}$为$\overline{V_{\overline{N}}(k, r)}$的中心, $d_R$为黎曼空间形式中距离函数.进而,我们有^[1]

(2.2) $\begin{eqnarray}\label{1.1} \left\{\begin{array}{l} \varphi^{\prime\prime}(t)+(\overline{N}-1){\rm{ct}}_{k}(t)\varphi^{\prime}(t) + \lambda_{1}(V_{\overline{N}}(k, r))\varphi(t)=0, \\\varphi(r)=0, \qquad\varphi^{\prime}(t)>0, \quad t\in(0, r). \end{array}\right. \end{eqnarray}$

此处

$ {\rm{ct}}_{k}(t)=\left\{\begin{array}{ll} \sqrt{k}\cdot\cot(\sqrt{k}t), & k>0, \\ \frac{1}{t}, & k=0, \\[2mm] \sqrt{-k}\cdot\coth(\sqrt{-k}t), & k <0.\end{array}\right. $

我们断言:若$V^+_{\overline{N}}(k, r)$为如上所述的芬斯勒空间形式中向前测地球, $\varphi(d_F)$为芬斯勒Laplacian第一特征函数,其中$d_F$为芬斯勒流形上距离函数,则方程(2.2)仍然成立.事实上,由Hessian比较定理^[7],可得$\Delta d_F=(\overline{N}-1){\rm{ct}}_{_{k}}(d_F)$.从而

$\begin{eqnarray*} \Delta\varphi(d_F)&=&{\rm{div}}(\nabla\varphi(d_F)) =\varphi^{\prime\prime}(d_F)+\varphi^{\prime}(d_F)\Delta d_F\\ &=&\varphi^{\prime\prime}(d_F)+(\overline{N}-1){\rm{ct}}_{k}(d_F)\varphi^{\prime}(d_F)\\ &=&-\lambda_{1}(V_{\overline{N}}(k, r))\varphi(d_F). \end{eqnarray*}$

下面我们证明$\lambda_{1}(V^+_{\overline{N}}(k, r))=\lambda_{1}(V^-_{\overline{N}}(k, r))$.注意到若$\varphi(d_F)$为$\Delta$的一个特征函数且特征值为$\lambda$,则$\varphi(d_{ \overleftarrow{F}})$必为$\overleftarrow{\Delta}$的一个特征函数,且特征值为$\lambda$.反之亦然.此处$\overleftarrow{\Delta}$及$d_{ \overleftarrow{F}}$为关于反向芬斯勒度量$\overleftarrow{F}$的Laplacian及距离函数.事实上,若$\lambda_{1}(V^+_{\overline{N}}(k, r)) < \lambda_{1}(V^-_{\overline{N}}(k, r))$,则对于$\lambda_{1}(V^+_{\overline{N}}(k, r))$,可以通过解方程(2.2)找到特征函数$\varphi(d_F)$,进一步得到$\varphi(d_{ \overleftarrow{F}})$.因此, $\varphi(d_{ \overleftarrow{F}})$是$V^-_{\overline{N}}(k, r)$中的第一特征函数,且第一特征值等于$\lambda_{1}(V^+_{\overline{N}}(k, r))$.这与前面的假定矛盾.类似地,相反的不等式也不成立.故两者必相等.

设$\rho(x)=d_{F}(x_0, x)$为$(M, F)$的向前距离函数, $u(x)=\varphi(\rho(x))$.因为d$u=\varphi^{\prime}{\rm d}\rho$且$\varphi^{\prime}>0$,所以, $\nabla u=\varphi^{\prime}\nabla\rho$.利用Laplacian比较定理^[4],我们有

$ \begin{eqnarray*} \Delta u&=&{\rm{div}}(\nabla u)={\rm{div}}(\varphi^{\prime}\nabla\rho) =\varphi^{\prime}(\rho)\Delta\rho+\nabla\rho(\varphi^{\prime})\\ &=&\varphi^{\prime\prime}(\rho)+\varphi^{\prime}(\rho)\Delta\rho \leq\varphi^{\prime\prime}(\rho)+(\overline{N}-1){\rm{ct}}_{_{k}}(\rho)\varphi^{\prime}(\rho)\\ &=&-\lambda_{1}(V_{\overline{N}}(k, r)) u. \end{eqnarray*} $

注意到$u|_{B^+_{x_0}(r)} < 0$且$u|_{\partial B^+_{x_0}(r)}=0$.从而有

(2.3) $ \begin{eqnarray}\label{2} \int_{B^+_{x_0}(r)}(F^{\ast}({\rm d}u))^{2}{\rm d}\mu&=&\int_{B^+_{x_0}(r)}du(\nabla u){\rm d}\mu =-\int_{B^+_{x_0}(r)}u\Delta u{\rm d}\mu \\ &\leq &\lambda_{1}(V_{\overline{N}}(k, r))\int_{B^+_{x_0}(r)}u^{2}{\rm d}\mu. \end{eqnarray} $

于是, $\lambda_{1}(B^+_{x_0}(r))\leq\frac{\int_{B^+_{x_0}(r)}(F^{\ast}({\rm d}u))^{2}{\rm d}\mu}{\int_{B^+_{x_0}(r)}u^{2}{\rm d}\mu} \leq \lambda_{1}(V_{\overline{N}}(k, r))$.

由于

$\inf\limits_{f}\frac{\int_{\Omega}|F^*(df)|^2{\rm d}\mu}{\int_{\Omega}f^2{\rm d}\mu}=\inf\limits_{f}\frac{\int_{\Omega}|F^*(-df)|^2{\rm d}\mu}{\int_{\Omega}f^2{\rm d}\mu} =\inf\limits_{f}\frac{\int_{\Omega}|\overleftarrow{F}^*(df)|^2{\rm d}\mu}{\int_{\Omega}f^2{\rm d}\mu}, $

所以, $\overleftarrow{\Delta}$的第一特征值与$\Delta$的第一特征值相等.故我们可以将$\lambda_{1}(B^-_{x_0}(r))$视为$\overleftarrow{\Delta}$的第一特征值.设$\overleftarrow{\rho}(x)=d_{F}(x, x_0)=d_{ \overleftarrow{F}}(x_0, x)$为$(M, F)$的反向距离函数且$v(x)=\varphi(\overleftarrow{\rho}(x))$.注意到$\overleftarrow{{\rm{Ric}}}_{N}\geq (N-1)k, N\in(n, \infty)$.于是利用关于$(M, \overleftarrow{F}, {\rm d}\mu)$的反向Laplacian比较定理以及相似的讨论,即可完成定理2.1证明.

推论2.1  设$(M, F, {\rm d}\mu)$为$n$维完备芬斯勒流形.若Ricci曲率Ric $\geq (n-1)k$且$S$曲率消失,则芬斯勒Laplacian第一Dirichlet特征值满足

$ \max\{\lambda_{1}(B^+_{x_0}(r)), \lambda_{1}(B^-_{x_0}(r))\}\leq \lambda_{1}(V_{n}(k, r)), $

其中$V_{n}(k, r)$表示具有旗曲率$k$与零$S$曲率的$n$维单连通芬斯勒空间中半径为$r$的向前(或向后)测地球.

定理2.2  设$(M, F, {\rm d}\mu)$为$n$维紧致芬斯勒流形.若加权Ricci曲率Ric$_{N}\geq (N-1)k, $$N\in(n, \infty)$,则芬斯勒Laplacian第一Neumann特征值满足

$\mu_{1}(M)\leq \lambda_{1}\left(V_{\overline{N}}\left(k, d/2\right)\right), $

其中$d$表示$M$的直径, $\lambda_{1}$表示第一Dirichlet特征值, $V_{\overline{N}}(k, d/2)$如上所述.

注2.2  与文献[9]中对应的结果相比,定理$2.2$在比较式中去掉了关于可反系数的限制.

证  设$x_{1}, x_{2}\in M$满足$d_{F}(x_{1}, x_{2})=d$.选取$r=\frac{d}{2}$,则$B^+_{x_{1}}(r)\cap B^-_{x_{2}}(r)=\emptyset$.事实上,若存在点$x\in B^+_{x_{1}}(r)\cap B^-_{x_{2}}(r)$,则导致以下矛盾:

$d=d_{F}(x_{1}, x_{2})\leq d_{F}(x_{1}, x)+d_{F}(x, x_{2})= d_{F}(x_{1}, x)+ d_{ \overleftarrow{F}}(x_{2}, x) < 2r=d. $

由(2.3)式及相似的讨论可得

$\int_{B^+_{x_{1}}(\frac{d}{2})}(F^{\ast}({\rm d}u_{1}))^{2}{\rm d}\mu\leq \lambda_{1}\left(V_{\overline{N}}\left(k, d/2\right)\right) \int_{B^+_{x_{1}}(\frac{d}{2})}u_{1}^{2}{\rm d}\mu, $

$\int_{B^-_{x_{2}}(\frac{d}{2})}(\overleftarrow{F}^{\ast}({\rm d}u_{2}))^{2}{\rm d}\mu\leq \lambda_{1}\left(V_{\overline{N}}\left(k, d/2\right)\right) \int_{B^-_{x_{2}}(\frac{d}{2})}u_{2}^{2}{\rm d}\mu. $

将$u_{1}$及$u_{2}$延拓至流形$M$上,使得在$B^+_{x_{1}}(\frac{d}{2})$及$B^-_{x_{2}}(\frac{d}{2})$外部分别为零.于是存在两个常数$a_{1}>0$及$a_{2} < 0$使得

$ u:=a_{1}u_{1}+a_{2}u_{2}\not\equiv 0, \quad\int_{M}u {\rm d}\mu=0. $

因此

$\begin{eqnarray*}\mu_{1}(M)\int_{M}u^{2} {\rm d}\mu&\leq&\int_{M}(F^{\ast}({\rm d}u))^{2} {\rm d}\mu=\int_{M}(F^{\ast}(a_1du_1+a_2du_2))^{2} {\rm d}\mu\\&=&a_{1}^{2}\int_{B^+_{x_{1}}(\frac{d}{2})}(F^{\ast}({\rm d}u_{1}))^{2}{\rm d}\mu+a_{2}^{2}\int_{B^-_{x_{2}}(\frac{d}{2})}(\overleftarrow{F}^{\ast}({\rm d}u_{2}))^{2}{\rm d}\mu\\&\leq&\lambda_{1}\left(V_{\overline{N}}\left(k, d/2\right)\right)\left(a_{1}^{2}\int_{B^+_{x_{1}}(\frac{d}{2})}u_{1}^{2}{\rm d}\mu+a_{2}^{2}\int_{B^-_{x_{2}}(\frac{d}{2})}u_{2}^{2}{\rm d}\mu\right)\\&= &\lambda_{1}\left(V_{\overline{N}}\left(k, d/2\right)\right)\int_{M}u^{2} {\rm d}\mu.\end{eqnarray*}$

定理2.2证毕.

引理2.1  设$(M, F, {\rm d}\mu)$为$n$维芬斯勒流形, $\Omega\subset M$为$M$上具有非空边界及紧致闭包的区域.假定$f$为芬斯勒Laplacian在$\Omega$中的第一Dirichlet特征函数, $X$为$\Omega$上一个向量场,且满足$\inf\limits_{\Omega}{\rm div}(X)>0$.则我们有下面的不等式:

(1)若存在一点$x_0\in\Omega$使得$f(x_0) < 0$,则

$\lambda_{1}(\Omega)\geq \left[\frac{\inf\limits_{\Omega}{\rm div}(X)}{2\sup\limits_{\Omega}F(X)}\right]^{2};$

(2)若存在一点$x_0\in\Omega$使得$f(x_0)>0$,则

$\lambda_{1}(\Omega)\geq \left[\frac{\inf\limits_{\Omega}{\rm div}(X)}{2\sup\limits_{\Omega}\overleftarrow{F}(X)}\right]^{2}.$

证  对任意的$\xi\in T^*M, X\in TM$,由基本不等式可得$\xi(X)\leq F^*(\xi)F(X)$.从而,我们有$-\xi(X)\leq F^*(-\xi)F(X)$及$\xi(-X)\leq F^*(\xi)F(-X)$.于是

$-F^*(-\xi)F(X)\leq\xi(X)\leq F^*(\xi)F(X), $

$-F^*(\xi)F(-X)\leq\xi(X)\leq F^*(\xi)F(X).$

如所知,若$f$为芬斯勒Laplacian在$\Omega$上的一个第一Dirichlet特征函数,令$f=f^++f^-$,则它的正部$f^+$与负部$f^-$均为第一Dirichlet特征函数.因此,我们不妨假定$f\leq0$或$f\geq0$.

情形Ⅰ  $f\leq0$.

因为$f\in C^{1, \alpha}_{0}(\Omega)$,所以向量场$f^{2}X$在$\Omega$内具有紧致支撑集.直接计算可得

(2.4) $\begin{eqnarray}\label{4.1}{\rm{div}}(f^{2}X)&=& X(f^2)+f^{2}{\rm{div}}(X)\\&\geq&- F^*(-df^2)F(X)+\inf\limits_{\Omega}{\rm{div}}(X)\cdot f^{2}\\&=&2f F^*(df)F(X)+\inf\limits_{\Omega}{\rm{div}}(X)\cdot f^{2}.\end{eqnarray}$

下面通过与文献[7]相似的讨论,我们有

$\int_{\Omega}F^*(df)^{2}{\rm d}\mu\geq\left[\frac{\inf\limits_{\Omega}{\rm{div}}(X)}{2\sup\limits_{\Omega}F(X)}\right]^{2}\int_{\Omega}f^{2}{\rm d}\mu.$

注意到这里的$f$为第一特征函数且满足$\frac{\int_{\Omega}F^*(df)^{2}{\rm d}\mu}{\int_{\Omega}f^{2}{\rm d}\mu}=\lambda_{1}(\Omega)$.从而结论(Ⅰ)成立.

情形Ⅱ   $f\geq0$.

对于这种情形, (2.4)式变成

$\begin{eqnarray}{\rm{div}}(f^{2}X)&=&X(f^2)+f^{2}{\rm{div}}(X)\\&\geq&-F^*(df^2)F(-X)+\inf\limits_{\Omega}{\rm{div}}(X)\cdot f^{2}\\&=&-2fF^*(df)\overleftarrow{F}(X)+\inf\limits_{\Omega}{\rm{div}}(X)\cdot f^{2}.\end{eqnarray}$

然后利用类似的方法可得

$\int_{\Omega}F^*(df)^{2}{\rm d}\mu\geq\left[\frac{\inf\limits_{\Omega}{\rm{div}}(X)}{2\sup\limits_{\Omega}\overleftarrow{F}(X)}\right]^{2}\int_{\Omega}f^{2}{\rm d}\mu.$

证毕.

对于情形Ⅰ,在引理2.1中令$X=\nabla\rho$,再利用文献[7]中定理5.1-5.2;对于情形Ⅱ,我们考虑反向度量$\overleftarrow{F}$并在文献[7]的定理5.1-5.2中采用反向Laplacian比较定理,此时,在引理2.1中令$X=\overleftarrow{\nabla}\overleftarrow{\rho}$.于是我们可以得到下面的McKean型不等式.与文献[7-8]中对应结果相比,我们在此去掉了关于可反系数的限制.

定理2.3  设$(M, F, {\rm d}\mu)$为$n$维完备非紧单连通的芬斯勒流形.若旗曲率$K\leq-a^{2}$$(a>0)$,且$\sup\limits_M\|S\| < (n-1)a$,则有

$\lambda_{1}(M)\geq\frac{\Big((n-1) a-\sup\limits_M\|S\|\Big)^{2} }{4}.$

定理2.4  设$(M, F, {\rm d}\mu)$为$n$维完备非紧单连通的芬斯勒流形.若旗曲率非正, Ricci曲率${\rm Ric}\leq-a^{2}(a>0)$,且$\sup\limits_M\|S\| < a$,则有

$\lambda_{1}(M)\geq\frac{\Big(a-\sup\limits_M\|S\|\Big)^{2}}{4}.$

设$(M, g)$为$n$维完备黎曼流形.若其截面曲率$K_M\geq-a^2$,则由文献[1]或[3]可知, Laplacian第一特征值满足

(2.5) $\begin{eqnarray}\label{2.5} \lambda_{1}(M)\leq\frac{(n-1)^2 a^2}{4}. \end{eqnarray}$

根据推论2.1,若芬斯勒流形的旗曲率$K\geq-a^2$且$S=0$,则(2.5)式在芬斯勒情形下仍然成立.因此,结合定理2.3易得:

推论2.2  设$(M, F, {\rm d}\mu)$为$n$维完备非紧单连通的芬斯勒流形.若$S=0$且旗曲率$K=-a^{2}(a>0), $则有

$\lambda_{1}(M)=\frac{(n-1)^2a^2}{4}.$

注2.3  对于$F$是可反的情形,该结果已被文献[2]及[9]先后得到.推论$2.2$表明,对于不可反情形,结论仍然正确.在这个意义下,我们推广了以前的结果.

在文献[9]中,作者将Calabi-Yau线性体积增长定理推广到芬斯勒流形上,但是可反系数必须是有界的.事实上,若将该条件减弱为"可反系数满足一定的增长条件",则体积就会以相应的"速度"增长.具体地说,我们得到如下的Calabi-Yau型体积增长定理.

定理2.5  设$(M, F, {\rm d}\mu)$为$n$维完备非紧的芬斯勒流形且加权Ricci曲率${\rm Ric}_{N}\geq0, $$ N\in(n, \infty)$.设$\rho^+(x)$及$\rho^-(x)$为从固定点$p\in M$出发的向前(向后)距离函数,且令$\rho^{\pm}(x)=\min\{\rho^+(x), \rho^-(x)\}$.若可反系数满足$\lambda_F(x)\leq C_0\rho^{\pm}(x)^k, $则有

$\min\{{\rm vol}_{F}^{{\rm d}\mu} (B^+_{p}(R)), {\rm vol}_{F}^{{\rm d}\mu} (B^-_{p}(R))\}\geq C(N, C_0, {\rm vol}_{F}^{{\rm d}\mu}(B^{\pm}_{p}(1)))R^{1/(k+1)}, $

其中$k\geq0, C_0>0$为常数, $B^+_{p}(R)$及$B^-_{p}(R)$分别表示以$p$为中心以$R$为半径的向前(向后)测地球, $B^{\pm}_{p}(1)=B^{+}_{p}(1)\cap B^{-}_{p}(1)$, $C$是依赖于$N, C_0, {\rm vol}_{F}^{{\rm d}\mu}(B^{\pm}_{p}(1))$的常数.

注2.4  定理$2.5$表明:若可反系数至多为距离函数的$k$次幂增长,则体积以$\frac{1}{k+1}$次幂增长.当$k=0$时,此即为文献[9]中结果.进一步,可以证明:若可反系数至多为指数函数的速度增长,则体积以至少为对数函数的速度增长.

证  设$x_{0}\in\partial B_{p}^{-}(R)$为一给定点, $\rho(x)=d(x_{0}, x)$为从$x_{0}$出发的向前距离函数.则$F(\nabla\rho)=1$.由文献[4]中Laplacian比较定理可知

$\Delta\rho\leq\frac{N-1}{\rho}.$

从而

$\Delta^{\nabla\rho}\rho^{2}=2\rho\Delta\rho+2F(\nabla\rho)^{2}\leq2(N-1)+2=2N.$

因此,对于任意非负函数$\varphi \in C_{0}^{\infty}(M)$,我们有

(2.6) $\begin{eqnarray}\label{3.3}\int_{M}\varphi\Delta^{\nabla\rho}\rho^{2}{\rm d}\mu\leq2N\int_{M}\varphi {\rm d}\mu.\end{eqnarray}$

对于任意$R>1$,令

$\psi(t):=\left\{\begin{array}{ll} 1, & 0\leq t\leq R-1 ;\\[2mm] \frac{R+1-t}{2}, \quad & R-1\leq t\leq R+1 ;\\[2mm] 0, & t\geq R+1.\end{array}\right. $

若$\varphi(x)=\psi(\rho(x))$,则$\varphi(x)$为Lipschitz连续函数,且${\rm{supp}}\varphi\subset B^+_{x_{0}}(R+1)$.由于Stokes公式对于Lipschitz连续函数仍然成立,我们有

$ \begin{eqnarray*} \int_{M}\varphi\Delta^{\nabla\rho}\rho^{2}{\rm d}\mu &=&-\int_{B^+_{x_{0}}(R+1)}g_{\nabla\rho}(\nabla^{\nabla\rho}\varphi, \nabla^{\nabla\rho}\rho^{2}){\rm d}\mu \\ &=&-2\int_{B^+_{x_{0}}(R+1)}\psi^{'}(\rho(x))\rho F(\nabla\rho)^{2}{\rm d}\mu\\ &=&\int_{B^+_{x_{0}}(R+1)\backslash B^+_{x_{0}}(R-1)}\rho {\rm d}\mu\\ &\geq&(R-1){\rm{vol}}_{F}^{{\rm d}\mu}(B^+_{x_{0}}(R+1)\backslash B^+_{x_{0}}(R-1)). \end{eqnarray*}$

结合(2.6)式得到

(2.7) $\begin{eqnarray}\label{3.5} (R-1){\rm{vol}}_{F}^{{\rm d}\mu}(B^+_{x_{0}}(R+1)\backslash B^+_{x_{0}}(R-1))&\leq&2N\int_{M}\varphi {\rm d}\mu \\&=&2N\int_{B^+_{x_{0}}(R+1)}\varphi {\rm d}\mu\\ &\leq& 2N\int_{B^+_{x_{0}}(R+1)} {\rm d}\mu \\&=&2N{\rm{vol}}_{F}^{{\rm d}\mu}(B^+_{x_{0}}(R+1)). \end{eqnarray} $

注意到$R>1$.故由三角不等式可得

(2.8) $ \begin{eqnarray}\label{3.6} B^{\pm}_{p}(1)\subset B^+_{x_{0}}(R+1)\backslash B^+_{x_{0}}(R-1). \end{eqnarray} $

因此,由(2.7)和(2.8)式得到

(2.9) $\begin{eqnarray}\label{3.7} 2N{\rm{vol}}_{F}^{{\rm d}\mu}(B^+_{x_{0}}(R+1))&\geq &(R-1) {\rm{vol}}_{F}^{{\rm d}\mu}(B^+_{x_{0}}(R+1)\backslash B^+_{x_{0}}(R-1))\\ &\geq& (R-1) {\rm{vol}}_{F}^{{\rm d}\mu}B^{\pm}_{p}(1). \end{eqnarray}$

另一方面,若存在点$y\in B^{+}_{x_{0}}(R+1)$,则有$d(x_{0}, y) < R+1$.注意到$x_{0}\in\partial B^{-}_p(R)$.于是

$d(p, y)\leq d(p, x_{0})+d(x_{0}, y)\leq \lambda_F d(x_{0}, p)+d(x_{0}, y) \leq C_0R^{k+1}+R+1.$

根据(2.9)式,我们有

${\rm{vol}}_{F}^{{\rm d}\mu}(B^+_{p}(C_0R^{k+1}+R+1))\geq\frac{(R-1)}{2N}{\rm{vol}}_{F}^{{\rm d}\mu}B^{\pm}_{p}(1).$

将$C_0R^{k+1}+R+1$替换为$R$,可得

$ \begin{eqnarray} {\rm{vol}}_{F}^{{\rm d}\mu}(B^+_{p}(R))\geq C(N, C_0, {\rm{vol}}_{F}^{{\rm d}\mu}(B^{\pm}_{p}(1)))R^{1/(k+1)}. \end{eqnarray}$

对于结论(2),我们考虑反向度量$\overleftarrow{F}(x, y)$.则对于任意的光滑函数$u$及切向量$v$, $F$与$\overleftarrow{F}$对应的量具有以下关系:

$\begin{eqnarray} d_{ \overleftarrow{F}}(x, y)=d_F(y, x), \quad \overleftarrow{\nabla} u=-\nabla (-u), \quad \overleftarrow{\Delta} u=-\Delta (-u), \quad \overleftarrow{{\rm{Ric}}}_N(v)={\rm{Ric}}_N(-v). \end{eqnarray} $

在定理的条件下, $\overleftarrow{{\rm{Ric}}}_N\geq0$.由于$F$的向后测地球即为$\overleftarrow{F}$的向前测地球,故利用与前面类似的讨论可得

$ \begin{eqnarray}{\rm{vol}}_{F}^{{\rm d}\mu}(B^-_{p}(R))\geq C(N, C_0, {\rm{vol}}_{F}^{{\rm d}\mu}(B^{\pm}_{p}(1)))R^{1/(k+1)}.\end{eqnarray}$

定理2.5证毕.