Heisenberg型群是一类Carnot群, 简称为H型群, 其在满足Hörmander条件的向量场理论研究中起着重要作用^[1-3].而漂移Laplace算子是一类重要的椭圆算子, 也被称为Witten-Laplace算子, 在几何分析、概率论等研究中发挥着重要作用(参见文献[4-6]).

本文研究具有加权测度的H型群$G$上漂移Laplace算子的特征值估计问题.具体来说, 具有加权测度$d\mu = {e^{ - \varphi }}dv$的$2n+m$维H型群$G$上漂移Laplace算子的形式如下

$ - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , $

其中${\Delta _G}$和${\nabla _G}$分别是H型群$G$上的次Laplace算子和梯度算子, $\varphi$为光滑函数.设$\Omega$是H型群$G$上的一个有界区域.考虑如下Dirichlet特征值问题

$ \begin{equation}\label{eq:1} \left\{ \begin{aligned} &- {\Delta _G}u + \left\langle {{\nabla _G}\varphi , {\nabla _G}u} \right\rangle = \lambda u, \quad \quad \text{在}\Omega \text{上} , \\ &{\left. u \right|_{\partial \Omega }} = 0.\\ \end{aligned}\right. \end{equation} $

(1.1)

由文献[7-8]可知, 该算子有离散谱$ 0 < {\lambda _1} \le {\lambda _2} \le \cdot \cdot \cdot \le {\lambda _i} \le \cdot \cdot \cdot \nearrow, $其中每个特征值按照其重数排列.

问题(1.1)包含了几种有趣的特征值问题.由H型群的定义可知:一方面, 当$m = 1$时, H型群即为Heisenberg群.此时, 问题(1.1)变为如下Heisenberg群$H^{n}$上漂移Laplace算子的Dirichlet特征值问题

$ \begin{equation}\label{eq:2} \left\{ \begin{aligned} & - {\Delta _{{H^n}}}u + \left\langle {{\nabla _{{H^n}}}\varphi , {\nabla _{{H^n}}}u} \right\rangle = \lambda u\quad \text{在}\Omega \text{上}, \\ & {\left. u \right|_{\partial \Omega }} = 0. \end{aligned}\right. \end{equation} $

(1.2)

另一方面, 当$\varphi$为常数时, 问题(1.1)变为如下H型群$G$上的次Laplace算子的Dirichlet特征值问题

$ \begin{equation}\label{eq:3} \left\{ \begin{aligned} & - {\Delta _G}u = \lambda u \quad \text{在}\Omega \text{上}, \\ & {\left. u \right|_{\partial \Omega }} = 0. \end{aligned}\right. \end{equation} $

(1.3)

因此当$m=1$时, 问题(1.3)进一步变为如下Heisenberg群$H^{n}$上次Laplace算子的Dirichlet特征值问题

$ \begin{equation}\label{eq:4} \left\{ \begin{aligned} & - {\Delta _{{H^n}}}u = \lambda u \quad \text{在}\Omega \text{上}, \\ & {\left. u \right|_{\partial \Omega }} = 0. \end{aligned}\right. \end{equation} $

(1.4)

随着黎曼流形上微分算子研究的深入, Heisenberg群、H型群上微分算子的特征值估计问题开始被学者们所关注(参见文献[9-10]). 2006年, 韩军强和钮鹏程^[11]获得了H型群上次Laplace算子相邻特征值之差的估计; 2015年, 谭沈阳和黄体仁^[12]建立了H型群上漂移Laplace算子问题(1.1)的Yang型特征值不等式.

本文的目标是对H型群上漂移Laplace算子的问题(1.1)建立Levitin-Parnovski型特征值不等式.对任意的正整数$j$, Ilias和Makhoul^[13]在2012年对Heisenberg群$H^{n}$上次Laplace算子的Dirichlet特征值问题(1.4)建立了如下Levitin-Parnovski型特征值不等式

$ \begin{equation}\label{eq:5} \sum\limits_{l = 1}^n {{\lambda _{j + l}}} \le \left( {n + 2} \right){\lambda _j}. \end{equation} $

(1.5)

当$j=1$时, 不等式(1.5)变为$ \sum\limits_{l = 1}^n {{\lambda _{l + 1}}} \le \left({n + 2} \right){\lambda _1}. $这就变为文献[14-15]等所获得类型的低阶特征值估计结果.即用第一特征值给出了从第2个到第$n+1$个特征值之和的上界估计.

在本文中, 首先建立了具有加权测度的H型群上漂移Laplace算子问题(1.1)的一个特征值一般不等式.

定理1  设$\Omega$是具有加权测度$d\mu = {e^{ - \varphi }}dv$的$2n + m$维H型群$G$上的有界区域, $\varphi$是区域$\Omega$闭包$\overset{-}{\Omega}$上的光滑函数, $\lambda_l$是$\Omega$上漂移Laplace算子$- {\Delta _G} + \left\langle {{\nabla _G}\varphi, {\nabla _G}\left(\cdot \right)} \right\rangle $特征值问题(1.1)的第$l$个特征值, $u_l$为对应于$\lambda_l$的单位正交特征函数, 且对应于不同特征值的特征函数相互正交.那么对任意正整数$j$, 有

$ \begin{equation}\label{th1:1} \sum\limits_{l = 1}^n {\left( {{\lambda _{j + l}} - {\lambda _j}} \right)} \le \frac{1}{2}\int_\Omega {\left( {u_j^2{{\left| {{\nabla _G}\varphi } \right|}^2} + 4{{\left| {{\nabla _G}{u_j}} \right|}^2} - 4{u_j}\left\langle {{\nabla _G}{u_j}, {\nabla _G}\varphi } \right\rangle } \right)d\mu }. \end{equation} $

(1.6)

进而获得了问题(1.1)的如下Levitin-Parnovski型特征值不等式.

定理2  设$\Omega$是具有加权测度$d\mu = {e^{ - \varphi }}dv$的$2n+m$维H型群$G$上的有界区域, $\varphi$是区域$\Omega$闭包$\overset{-}{\Omega}$上的光滑函数, $\lambda_l$为$\Omega$上漂移Laplace算子$- {\Delta _G} + \left\langle {{\nabla _G}\varphi, {\nabla _G}\left(\cdot \right)} \right\rangle $特征值问题(1.1)的第$l$个特征值.如果$\left| {{\nabla _G}\varphi } \right| \le c$, 则对任意正整数$j$, 有

$ \begin{equation}\label{th2:1} \sum\limits_{l = 1}^n {{\lambda _{j + l}}} \le (n + 2){\lambda _j} + 2c\lambda _j^{\frac{1}{2}} + \frac{{{c^2}}}{2}. \end{equation} $

(1.7)

不难看出, (1.7)式对问题(1.2)也成立.即有如下结论.

推论1  设$\Omega$是具有加权测度$d\mu = {e^{ - \varphi }}dv$的$n$维Heisenberg群$H^{n}$上的有界区域, $\varphi$是区域$\Omega$闭包$\overset{-}{\Omega}$上的光滑函数, $\lambda_l$为$\Omega$上漂移Laplace算子$- {\Delta _{{H^n}}} + \left\langle {{\nabla _{{H^n}}}\varphi, {\nabla _{{H^n}}}\left(\cdot \right)} \right\rangle $特征值问题(1.2)的第$l$个特征值.如果$\left| {{\nabla _{{H^n}}}\varphi } \right| \le c$, 则对任意正整数$j$, 有

$ \begin{equation*} \sum\limits_{l = 1}^n {{\lambda _{j + l}}} \le (n + 2){\lambda _j} + 2c\lambda _j^{\frac{1}{2}} + \frac{{{c^2}}}{2}. \end{equation*} $

另外, 当$\varphi$为常数时, 问题(1.1)变为问题(1.3).因此可由定理2得到如下推论.

推论2  设$\Omega$是$2n+m$维H型群$G$上的有界区域, $\lambda_l$为$\Omega$上次Laplace算子${\Delta _G}$特征值问题(1.3)的第$l$个特征值, 则对任意正整数$j$, 有$ \sum\limits_{l = 1}^n {{\lambda _{j + l}}} \le \left({n + 2} \right){\lambda _j}. $

当$m=1$时, 推论2即变为Ilias和Makhoul^[13]对Heisenberg群$H^{n}$上次Laplace算子的Dirichlet特征值问题(1.4)所获得结果.因此本文的结果推广并包含了Ilias和Makhoul^[13]所获得的结果.

本节给出H型群的一些基本概念与性质.设${U^1}, {U^2}, \cdot \cdot \cdot, {U^{2n}}$是满足下列条件的矩阵

(1) ${U^j}$是$m\times m$阶反对称正交矩阵, $\forall j = 1, 2, \cdot \cdot \cdot, 2n$;

(2) ${U^i}{U^j} + {U^j}{U^i} = 0$, $\forall i, j \in \left\{ {1, 2, \cdot \cdot \cdot, 2n} \right\}$, $i \ne j$.

在$2n+m$维欧氏空间$\mathbb{R}^{n}\times\mathbb{R}^{n}\times\mathbb{R}^{m}$定义如下群运算:

$ \left( {z, t} \right) \circ \left( {{z'}, {t'}} \right) = \left( {x, y, t} \right) \circ \left( {{x'}, {y'}, {t'}} \right) = \left( {{z_i} + z_i', {t_j} + t_j' + \frac{1}{2}\left\langle {z, {U^j}z'} \right\rangle } \right), $

其中$i = 1, 2, \cdot \cdot \cdot, 2n$; $j = 1, 2, \cdot \cdot \cdot, m$, $z = \left({x, y} \right) \in \mathbb{R}^{2n}$, $t\in \mathbb{R}^{m}$, $\left\langle { \cdot, \cdot } \right\rangle $表示欧氏内积.满足这种群运算的$2n+m$维欧氏空间称为H型群, 李代数$g$的基底为

$ {X_j} = \frac{\partial }{{\partial {x_j}}} + \frac{1}{2}\sum\limits_{k = 1}^m {\left( {\sum\limits_{l = 1}^{2n} {{z_l}U_{l, j}^{\left( k \right)}} } \right)\frac{\partial }{{\partial {t_k}}}} , \quad {Y_j} = \frac{\partial }{{\partial {y_j}}} + \frac{1}{2}\sum\limits_{k = 1}^m {\left( {\sum\limits_{l = 1}^{2n} {{z_l}U_{l, j + n}^{\left( k \right)}} } \right)\frac{\partial }{{\partial {t_k}}}} , \quad {T_k} = \frac{\partial }{{\partial {t_k}}}, $

其中${\left({U_{l, j}^{\left(k \right)}} \right)_{2n \times 2n}} = {U^{\left(k \right)}}$.当$m=1$时, H型群即为Heisenberg群. H型群$G$上的次Laplace算子和梯度算子定义为

$ {\Delta _G} = \sum\limits_{j = 1}^n {\left( {X_j^2 + Y_j^2} \right)} , \quad {\nabla _G} = \left( {{X_1}, {X_2}, \cdot \cdot \cdot , {X_n}, {Y_1}, {Y_2}, \cdot \cdot \cdot , {Y_n}} \right). $

在定理1的证明过程中, 需要使用Levitin和Parnovski^[16]获得的代数恒等式.

引理1  设$M$是一个给定内积$\langle\cdot, \cdot\rangle$的复Hilbert空间, $A:D \subset M \to M$是定义在有界稠密区域$D$上的一个自伴算子, 并且$A$有一组离散谱${\lambda _1} \le {\lambda _2} \le {\lambda _3} \le \cdot \cdot \cdot $.设$\left\{ {{B_l}:A\left(D \right) \to M} \right\}_{l = 1}^N$是由一组对称算子构成的集合, 且满足${B_l}\left(D \right) \subset D$.令$\left\{ {{u_i}} \right\}_{i = 1}^\infty $是算子$A$的正交特征向量构成的集合, $u_i$是第$i$个特征值$\lambda_i$对应的特征向量, 并且这组特征向量可构成$M$的一组正交基.那么, 对任意正整数$j$, 如下代数恒等式成立

$ \begin{equation}\label{lem} \sum\limits_{k = 1}^\infty {\frac{{{{\left| {\left\langle {\left[ {A, {B_l}} \right]{u_j}, {u_k}} \right\rangle } \right|}^2}}}{{{\lambda _k} - {\lambda _j}}} = - \frac{1}{2}\left\langle {\left[ {\left[ {A, {B_l}} \right], {B_l}} \right]{u_j}, {u_j}} \right\rangle }, \end{equation} $

(2.1)

其中$\left[{A, {B_l}} \right]: = A{B_l} - B{_l}A$是$A$和$B_l$的括号积.

本节给出定理1和定理2的证明.

定理1的证明  因为$u_i$为问题(1.1)的对应于第$i$个特征值$\lambda_i$的单位正交特征函数, 即$u_i$满足

$ \begin{equation}\label{th1:2} \left\{ \begin{aligned} & - {\Delta _G}{u_i} + \left\langle {{\nabla _G}\varphi , {\nabla _G}{u_i}} \right\rangle = {\lambda _i}{u_i} \quad \text{在} \Omega \text{上}, \\ & {\left. u \right|_{\partial \Omega }} = 0, \\ & \int_\Omega {{u_i}{u_j}d\mu = {\delta _{ij}}}. \end{aligned}\right. \end{equation} $

(3.1)

设${y_1}, \cdots, {y_n}$是$\mathbb{R}^n$上的一组标准坐标函数, 定义如下$n\times n$阶矩阵$T$

$ {\begin{equation*} T: =\\ \left( {\begin{array}{*{20}{c}} {\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {y_1}} \right]{u_j}, {u_{j + 1}}} \right\rangle }& \cdots &{\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {y_1}} \right]{u_j}, {u_{j + n}}} \right\rangle }\\ {\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {y_1}} \right]{u_j}, {u_{j + 1}}} \right\rangle }& \cdots &{\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {y_2}} \right]{u_j}, {u_{j + n}}} \right\rangle }\\ \cdots&\cdots&\cdots \\ {\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {y_n}} \right]{u_j}, {u_{j + 1}}} \right\rangle }& \cdots &{\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {y_n}} \right]{u_j}, {u_{j + n}}} \right\rangle } \end{array}} \right). \end{equation*}} $

根据QR -因式分解定理, 存在一个$n\times n$阶正交矩阵$Q = {\left({{q_{lr}}} \right)_{n \times n}}$使得$S = QT$, 其中$S$是一个上三角矩阵.因此有

$ \begin{equation*} \sum\limits_{r = 1}^n {{q_{lr}}\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {y_l}} \right]{u_j}, {u_{j + k}}} \right\rangle = 0} , \quad 1 \le k < l \le n. \end{equation*} $

令${x_l} = \sum\limits_{r = 1}^n {{q_{lr}}{y_r}} $, 可知$\mathbb{R}^n$上的一组标准坐标函数${x_1}, \cdots, {x_n}$满足如下等式

$ \begin{equation}\label{th1:3} \left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}, {u_{j + k}}} \right\rangle = 0, \quad 1 \le k < l \le n. \end{equation} $

(3.2)

从而, 根据(3.2)式, 可以得到

$ \begin{equation}\label{th1:4} \sum\limits_{k = 1}^{l - 1} {\frac{{{{\left| {\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}, {u_{j + k}}} \right\rangle } \right|}^2}}}{{{\lambda _{j + k}} - {\lambda _j}}}} = 0. \end{equation} $

(3.3)

在(2.1)式中取$A = - {\Delta _G} + \left\langle {{\nabla _G}\varphi, {\nabla _G}\left(\cdot \right)} \right\rangle $, ${B_l} = {x_l}$, $l = 1, 2, \cdot \cdot \cdot, n$, 有

$ \begin{equation}\label{th1:5} \begin{aligned} & \sum\limits_{k = 1}^\infty {\frac{{{{\left| {\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}, {u_k}} \right\rangle } \right|}^2}}}{{{\lambda _k} - {\lambda _j}}}} \\ =& - \frac{1}{2}\left\langle {\left[ {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right], {x_l}} \right]{u_j}, {u_j}} \right\rangle . \end{aligned} \end{equation} $

(3.4)

通过直接计算, 可得

$ \begin{equation}\label{th1:6} \begin{aligned} & \sum\limits_{k = 1}^\infty {\frac{{{{\left| {\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}, {u_k}} \right\rangle } \right|}^2}}}{{{\lambda _k} - {\lambda _j}}}} \\ =& \sum\limits_{k = 1}^{j - 1} {\frac{{{{\left| {\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}, {u_k}} \right\rangle } \right|}^2}}}{{{\lambda _k} - {\lambda _j}}}} + \sum\limits_{k = j + 1}^{j + l - 1} {\frac{{{{\left| {\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}, {u_k}} \right\rangle } \right|}^2}}}{{{\lambda _k} - {\lambda _j}}}} \\ & + \sum\limits_{k = j + l}^\infty {\frac{{{{\left| {\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}, {u_k}} \right\rangle } \right|}^2}}}{{{\lambda _k} - {\lambda _j}}}} . \end{aligned} \end{equation} $

(3.5)

根据特征值的单调性, 知道

$ \begin{equation}\label{th1:7} \sum\limits_{k = 1}^{j - 1} {\frac{{{{\left| {\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}, {u_k}} \right\rangle } \right|}^2}}}{{{\lambda _k} - {\lambda _j}}}} \le 0. \end{equation} $

(3.6)

并且根据(3.2)式, 有

$ \begin{equation}\label{th1:8} \begin{aligned} & \sum\limits_{k = j + 1}^{j + l - 1} {\frac{{{{\left| {\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}, {u_k}} \right\rangle } \right|}^2}}}{{{\lambda _k} - {\lambda _j}}}} \\ = &\sum\limits_{k = 1}^{l - 1} {\frac{{{{\left| {\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}, {u_{j + k}}} \right\rangle } \right|}^2}}}{{{\lambda _{j + k}} - {\lambda _j}}}} = 0. \end{aligned} \end{equation} $

(3.7)

结合(3.5), (3.6)和(3.7)式, 有

$ \begin{equation}\label{th1:9} \begin{aligned} & \sum\limits_{k = 1}^\infty {\frac{{{{\left| {\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}, {u_k}} \right\rangle } \right|}^2}}}{{{\lambda _k} - {\lambda _j}}}} \\ \le& \sum\limits_{k = j + l}^\infty {\frac{{{{\left| {\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}, {u_k}} \right\rangle } \right|}^2}}}{{{\lambda _k} - {\lambda _j}}}} \\ \le& \frac{1}{{{\lambda _{j + l}} - {\lambda _j}}}\sum\limits_{k = 1}^\infty {{{\left| {\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}, {u_k}} \right\rangle } \right|}^2}} . \end{aligned} \end{equation} $

(3.8)

由Parseval等式可知

$ \begin{equation}\label{th1:10} \sum\limits_{k = 1}^\infty {{{\left| {\left\langle {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}, {u_k}} \right\rangle } \right|}^2}} = {\left\| {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}} \right\|^2}. \end{equation} $

(3.9)

将(3.4)和(3.9)式代入(3.8)式中, 整理并对$l$从$1$到$n$求和, 可得

$ \begin{equation}\label{th1:11} \begin{aligned} & - \frac{1}{2}\sum\limits_{l = 1}^n {\left( {{\lambda _{j + l}} - {\lambda _j}} \right)\left\langle {\left[ {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right], {x_l}} \right]{u_j}, {u_j}} \right\rangle } \\ \le &\sum\limits_{l = 1}^n {{{\left\| {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}} \right\|}^2}} . \end{aligned} \end{equation} $

(3.10)

同理, 在(2.1)式中取$A = - {\Delta _G} + \left\langle {{\nabla _G}\varphi, {\nabla _G}\left(\cdot \right)} \right\rangle $, ${B_l} = {y_l}$, $l = 1, 2, \cdot \cdot \cdot, n$, 可得与(3.10)式类似的关于$y_l$的不等式.进而有

$ \begin{equation}\label{th1:12} \begin{aligned} & - \frac{1}{2}\sum\limits_{l = 1}^n {\left( {{\lambda _{j + l}} - {\lambda _j}} \right)} \left\langle {\left[ {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right], {x_l}} \right]{u_j}, {u_j}} \right\rangle \\ & - \frac{1}{2}\sum\limits_{l = 1}^n {\left( {{\lambda _{j + l}} - {\lambda _j}} \right)} \left\langle {\left[ {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {y_l}} \right], {y_l}} \right]{u_j}, {u_j}} \right\rangle \\ \le& \sum\limits_{l = 1}^n {{{\left\| {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}} \right\|}^2}} + \sum\limits_{l = 1}^n {{{\left\| {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {y_l}} \right]{u_j}} \right\|}^2}} . \end{aligned} \end{equation} $

(3.11)

直接计算可知

$ \begin{equation*} \begin{aligned} & \left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}\\ =& - {\Delta _G}\left( {{x_l}{u_j}} \right) + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( {{x_l}{u_j}} \right)} \right\rangle - {x_l}\left( { - {\Delta _G}{u_j} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( {{u_j}} \right)} \right\rangle } \right)\\ =& - \left( {X_l^2 + Y_l^2} \right)\left( {{x_l}{u_j}} \right) + {u_j}{X_l}\varphi + \left\langle {{\nabla _G}\varphi , {x_l}{\Delta _G}{u_j}} \right\rangle - {x_l}\left( { - {\Delta _G}{u_j} + \left\langle {{\nabla _G}\varphi , {\Delta _G}{u_j}} \right\rangle } \right)\\ =& - {X_l}\left( {{u_j} + {x_l}{X_l}{u_j}} \right) - {x_l}Y_l^2{u_j} + {u_j}{X_l}\varphi + {x_l}{\Delta _G}{u_j}\\ =& - 2{X_l}{u_j} + {u_j}{X_l}\varphi . \end{aligned} \end{equation*} $

同理, 可得

$ \begin{equation*} \left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {y_l}} \right]{u_j} = - 2{Y_l}{u_j} + {u_j}{Y_l}\varphi . \end{equation*} $

因此有

$ \begin{equation}\label{th1:13} \begin{aligned} & \sum\limits_{l = 1}^n {{{\left\| {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right]{u_j}} \right\|}^2}} + \sum\limits_{l = 1}^n {{{\left\| {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {y_l}} \right]{u_j}} \right\|}^2}} \\ & = \int_\Omega ( {u_j^2{{\left| {{\nabla _G}\varphi } \right|}^2} + 4{{\left| {{\nabla _G}{u_j}} \right|}^2} - 4{u_j} \left \langle {{\nabla _G}{u_j}, {\nabla _G}\varphi } \right\rangle ) d\mu } . \end{aligned} \end{equation} $

(3.12)

又因为

$ \begin{equation}\label{th1:14} \begin{aligned} \left[ {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right], {x_l}} \right]{u_j}& = \left[ { - 2{X_l} + {X_l}\varphi , {x_l}} \right]{u_j}\\ & = \left[ {\left( { - 2{X_l} + {X_l}\varphi } \right){x_l} - {x_l}\left( { - 2{X_l} + {X_l}\varphi } \right)} \right]{u_j}\\ & = - 2{X_l}\left( {{x_l}{u_j}} \right) + {x_l}{u_j}{X_l}\varphi + 2{x_l}{X_l}{u_j} - {x_l}{u_j}{X_l}\varphi \\ & = - 2{u_j}. \end{aligned} \end{equation} $

(3.13)

同理可得

$ \begin{equation}\label{th1:15} \left[ {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {y_l}} \right], {y_l}} \right]{u_j} = \left[ { - 2{Y_l} + {Y_l}\varphi , {y_l}} \right]{u_j} = - 2{u_j}. \end{equation} $

(3.14)

所以

$ \begin{eqnarray}\label{th1:16} &&\left\langle {\left[ {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {x_l}} \right], {x_l}} \right]{u_j}, {u_j}} \right\rangle = - 2\int_\Omega {u_j^2} d\mu = - 2, \end{eqnarray} $

(3.15)

$ \begin{eqnarray} \label{th1:17} &&\left\langle {\left[ {\left[ { - {\Delta _G} + \left\langle {{\nabla _G}\varphi , {\nabla _G}\left( \cdot \right)} \right\rangle , {y_l}} \right], {y_l}} \right]{u_j}, {u_j}} \right\rangle = - 2\int_\Omega {u_j^2} d\mu = - 2. \end{eqnarray} $

(3.16)

最后将(3.12), (3.15)和(3.16)式代入(3.11)式中, 就可以得到(1.6)式.从而完成定理1的证明.

定理2的证明  因为

$ \begin{equation}\label{th2:2} \begin{aligned} - \int_\Omega {{u_j}\left\langle {{\nabla _G}\varphi , {\nabla _G}{u_j}} \right\rangle d\mu }&\le \int_\Omega {\left| {{u_j}} \right|\left| {{\nabla _G}\varphi } \right|\left| {{\nabla _G}{u_j}} \right|d\mu } \\ & \le c{\left( {\int_\Omega {u_j^2d\mu } } \right)^{\frac{1}{2}}}{\left( {\int_\Omega {{{\left| {{\nabla _G}{u_j}} \right|}^2}d\mu } } \right)^{\frac{1}{2}}}\\ & = c{\left( {\int_\Omega {{{\left| {{\nabla _G}{u_j}} \right|}^2}d\mu } } \right)^{\frac{1}{2}}}. \end{aligned} \end{equation} $

(3.17)

并且注意到

$ \begin{equation}\label{th2:3} \begin{aligned} \int_\Omega {{{\left| {{\nabla _G}{u_j}} \right|}^2}d\mu } &= \int_\Omega {\left\langle {{\nabla _G}{u_j}, {\nabla _G}{u_j}} \right\rangle } d\mu \\ & = \int_\Omega {{u_j}( - {\Delta _G}{u_j} + \left\langle {{\nabla _G}\varphi , {\nabla _G}{u_j}} \right\rangle } )d\mu = {\lambda _j}. \end{aligned} \end{equation} $

(3.18)

由(3.17)和(3.18)式, 可得

$ \begin{equation}\label{th2:4} - \int_\Omega {{u_j}\left\langle {{\nabla _G}\varphi , {\nabla _G}{u_j}} \right\rangle d\mu } \le c\lambda _j^{\frac{1}{2}}. \end{equation} $

(3.19)

将(3.17)-(3.19)式代入(1.6)式中, 可获得(1.7)式.这就完成了定理2的证明.