令$ L = -\Delta_{{\Bbb H}^n}+V $是Heisenberg群$ {{\Bbb H}^n} $上Schrödinger算子, 其中非负位势$ V $属于逆Hölder类$ B_q $ ($ q\geq Q/2 $, $ Q>3 $), $ \Delta_{{\Bbb H}^n} $是$ {{\Bbb H}^{n}} $上次Laplace算子.

一些与$ L $有关的奇异积分算子已被许多学者研究过.在欧氏空间中, Fefferman^[2]和Zhong^[11]建立了一些满足逆Hölder不等式的Schrödinger算子的基本结果. Dong, Huang和Liu^[1]给出了$ H^p_L({{\Bbb R}} ^{n}) $的分子刻画, 并证明了算子$ {(-\Delta+V)^{{i}\gamma}} $和$ {\nabla}(-\Delta+V)^{-1/2} $在$ H_L^p({{\Bbb R}} ^{n}) $上有界. Wang和Li^[10]研究了与广义Schrödinger算子有关的Riesz变换在$ H_L^1({{\Bbb R}} ^{n}) $上的有界性.在Heisenberg群上, Liu和Tang^[7]证明了算子$ V(-\Delta_{{\Bbb H}^{n}}+V)^{-1} $和$ V^{1/2}{(-\Delta_{{\Bbb H}^{n}}+V)^{-1/2}} $在$ H^p_L({\Bbb H}^{n}) $到$ L^{p}({\Bbb H}^n) $上有界. Lin, Liu和Liu^[6]利用$ H^1_L({\Bbb H}^{n}) $上的原子分解证明了$ H^1_L({\Bbb H}^{n}) $可以用与$ L $相关的Riesz变换来刻画.

本文的目的是将文献[1]中的相关结果推广到Heisenberg群上, 通过$ H^p_L({\Bbb H}^{n}) $的分子刻画, 研究与$ L $相关的Riesz变换的$ H_{L}^{p} $ -有界性.与文献[1]不同的是, 文献[1]中所用到的一些欧氏空间上与$ L $相关的热半群的核函数估计已被Shen在文献[9]中证明过, 为了证明本文的主要结果, 我们需要将文献[9]中关于Riesz变换的核函数的相关估计推广到Heisenberg群的情形, 并利用得到的核函数估计, 建立Hardy型空间$ H_{L}^{p}({\Bbb H}^{n}) $的分子刻画以及算子在该函数空间上的有界性.

下面, 我们首先叙述Heisenberg群的一些基本知识(参见文献[6]). $ (2n+1) $ -维Heisenberg群$ {\Bbb H}^{n} $是具有底流形$ {{\Bbb R}} ^{2n}\times{{\Bbb R}} $的Lie群, 其乘法定义为

$ \begin{eqnarray*} (x, t)(y, s) = \Big(x+y, t+s+2\sum\limits_{j = 1}^{n}(x_{n+j}y_{j}-x_{j}y_{n+j})\Big). \end{eqnarray*} $

向量场

$ \begin{eqnarray*} X_{2n+1} = \frac{\partial}{\partial t}, \ X_{j} = \frac{\partial}{\partial x_{j}}+2x_{n+j}\frac{\partial}{\partial t}, \ X_{n+j} = \frac{\partial}{\partial x_{n+j}}-2x_{j}\frac{\partial}{\partial t}, \ j = 1, \cdots, n \end{eqnarray*} $

构成其Lie代数的一组基.所有非平凡交换子表示为$ [X_{j}, X_{n+j}] = -4X_{2n+1}, \ j = 1, \cdots, n $.次Laplace算子$ \Delta_{{\Bbb H}^{n}} $定义为$ \Delta_{{\Bbb H}^{n}} = \sum\limits_{j = 1}^{2n}{X_{j}}^{2} $, 梯度$ \nabla_{{\Bbb H}^{n}} $定义为$ \nabla_{{\Bbb H}^{n}} = (X_{1}, \cdots, X_{2n}) $. $ {\Bbb H}^{n} $上的伸缩变换为$ \delta_{r}(x, t) = (rx, r^{2}t) $, $ r>0 $.

Heisenberg群$ {\Bbb H}^{n} $上的Haar测度等同于$ {{\Bbb R}} ^{2n}\times{{\Bbb R}} $上的Lebesgue测度.任意可测集$ E $的测度系为$ |E| $. $ {\Bbb H}^{n} $上齐次范数定义为$ |g| = (|x|^{4}+|t|^{2})^{1/4} $, $ g = (x, t)\in{\Bbb H}^{n} $.该范数满足三角不等式且导出左不变距离$ d(g, h) = |g^{-1}h| $. $ {\Bbb H}^{n} $中以$ g $为球心, $ r $为半径的球定义为$ B(g, r) = \{h\in{\Bbb H}^{n}:|g^{-1}h|<r\} $.存在一个正常数$ \alpha_{1} $, 使得$ |B(g, r)| = \alpha_{1}r^{Q} $, 其中

$ \begin{eqnarray*} \alpha_{1} = |B(0, 1)| = \frac{2\pi^{n+1/2}\Gamma(n/2)}{(n+1)\Gamma(n)\Gamma((n+1)/2)}. \end{eqnarray*} $

令$ L = -\Delta_{{\Bbb H}^{n}}+V $为$ {\Bbb H}^{n} $上Schrödinger算子, 其中非负位势$ V\not\equiv0 $属于逆Hölder类$ B_{q} $, $ q\geq{Q}/{2} $.即存在$ C>0 $, 对$ {\Bbb H}^{n} $上任意球$ B $, 下述逆Hölder不等式成立

$ \begin{eqnarray*} \Big(\frac{1}{|B|}\int_{B}V(g)^{q}{\rm d}g\Big)^{1/q}\leq \frac{C}{|B|}\int_{B}V(g){\rm d}g. \end{eqnarray*} $

显然, 若$ V\in B_{q} $, 存在$ \varepsilon>0 $, $ V\in B_{q+\varepsilon} $.若$ {q_{1}}>{q_{2}} $, 则$ {B_{{q}_{1}}}\subset {B_{{q}_{2}}} $.

假设$ V\in B_{{q}_{0}} $, $ q_{0}>Q/2 $.当$ q_{0}>Q $时, 令$ \delta = $min$ \{1, 2-Q/q_{0}\} $.令$ H_{s}(\cdot) $为热半群$ \{T_{s}\}_{s>0} = \{e^{-s(-\Delta_{{\Bbb H}^{n}})}\}_{s>0} $的卷积核, $ \{T_s^L\}_{s>0} = \{e^{-sL}\}_{s>0} $是与算子$ L $相关的热半群, 其积分核记为$ {K_s^L}(\cdot, \cdot) $.由Trotter积公式(参见文献[3])可知, $ 0\leq{K_s^L}(\cdot, \cdot)\leq H_{s}(h^{-1}g) $.

如下辅助函数在本文中起到重要的作用

(1.1) $ \begin{equation} \rho(g, V) = \rho(g) = \frac{1}{m(g, V)} = \sup\limits_{r>0} \Big\{r:\ \frac{1}{r^{Q-2}}\int_{B(g.r)}V(h){\rm d}h\leq1 \Big\}\nonumber. \end{equation} $

与$ \{T_s^L\}_{s>0} $相关的极大函数定义为$ M^{L}f(g) = $$ \sup\limits_{s>0}|{T_s^L}f(g)| $.设$ f $是$ {\Bbb H}^{n} $上的局部可积函数, $ B = B(g, r) $是球心为$ g $半径为$ r $的球.令$ {f_B} = \frac{1}{|B|}\int_{B}f(h){\rm d}h $, 有

$ f(B, V) = \left\{\begin{array}{ll} {f_B}, & \mbox{$r< \rho(g), $}\\ 0, & \mbox{$r\geq \rho(g).$} \end{array}\right. $

令$ Q/(Q+\delta)<p<1 $, $ 1\leq q'\leq\infty $.一个局部可积函数$ f $属于Campanato型空间$ \Lambda^L_{1/p-1, q'}({\Bbb H}^{n}) $, 若

$ \begin{eqnarray*} \|f\|_{\Lambda^L_{1/p-1, q'}({\Bbb H}^{n})} = \sup\limits_{B\subseteq{{\Bbb H}}^{n}}\Big\{|B|^{1-1/p}{\Big(\int_{B}{|f-f(B, V)|}^{q'}\frac{{\rm d}h}{|B|}\Big)}^{1/q'}\Big\}<\infty. \end{eqnarray*} $

类似于$ {{\Bbb R}} ^{n} $中的情况, 在等价范数意义下, 所有$ \Lambda^L_{1/p-1, q'}({\Bbb H}^{n}) $空间等价范数是一致的, 而且是$ {H_L^p}({\Bbb H}^{n}) $的对偶空间(参见文献[7, p34]), 简记为$ \Lambda^L_{1/p-1}({\Bbb H}^{n}) $.根据文献[7], 对任意$ t>0 $, $ \sup\limits_{h\in{\Bbb H}^{n}}\|{H^L_t}(\cdot, h)\|_{\Lambda^L_{1/p-1}({\Bbb H}^{n})}<Ct^{-Q/2p} $, 对于$ f\in(\Lambda^L_{1/p-1}({\Bbb H}^{n}))^{\ast} $, 极大函数$ M^{L}f $是良定的. $ {\Bbb H}^{n} $上与$ L $相关的Hardy型空间定义为

(ⅰ) $ {H_L^p}({\Bbb H}^{n}) = \Big\{f\in(\Lambda^L_{1/p-1}({\Bbb H}^{n}))^{\ast}:\ M^{L}f\in L^{p}({\Bbb H}^{n})\Big\} $, $ Q/(Q+\delta)<p<1 $.其范数定义为$ \|f\|^p_{{H_L^p}({\Bbb H}^{n})} = \|M^{L}f\|^p_{L^{p}({\Bbb H}^{n})} $;

(ⅱ) $ {H_L^1}({\Bbb H}^{n}) $ = $ \Big\{f\in{L^{1}({\Bbb H}^{n})}:\ M^{L}f\in{L^{1}({\Bbb H}^{n})}\Big\} $.其范数定义为$ \|f\|_{H^1_L({\Bbb H}^{n})} = \|M^{L}f\|_{L^{1}({\Bbb H}^{n})} $.

与经典的Hardy空间类似, 与$ L $相关的Hardy空间具有原子刻画, 原子的定义如下.

定义1.1  令$ Q/(Q+\delta)<p\leq1\leq q\leq\infty $和$ p\neq q $.一个函数$ a $被称为一个与球$ B(g_{0}, r) $有关的$ H_L^{p, q}({\Bbb H}^{n}) $ -原子, 如果$ a $满足以下几个条件

(ⅰ) supp$ a \subset B(g_{0}, r) $;

(ⅱ) $ \|a\|_{{L^{q}}({\Bbb H}^{n})}\leq|B(g_{0}, r)|^{1/q-1/p} $;

(ⅲ)若$ r<\rho(g_{0}) $, 则$ \int a(g)dg = 0 $.

通过文献[7, 命题1], 我们可以得到下列$ H^{p}_{L}({\mathbb H}^{n}) $的原子刻画.此处省略证明.

命题1.1  令$ Q/(Q+\delta)<p\leq1\leq q\leq\infty $, $ p\neq q $, $ \delta = \min\Big\{1, 2-Q/q_{0}\Big\} $, 则$ f\in{H_L^p}({\Bbb H}^{n}) $当且仅当$ f $可以表示为$ f = \sum_{j}\lambda_{j}a_{j} $, 其中$ a_{j} $是$ H_L^{p, q}({\Bbb H}^{n}) $ -原子和$ \sum_{j}|\lambda_{j}|^{p}<\infty $, 和式$ \sum_{j} $按$ {H_L^p}({\Bbb H}^{n}) $范数收敛.此外, $ \|f\|_{H_L^p}\sim\|f\|_{H_L^{p, q}} = \inf\Big\{\Big({\sum\limits_j}|\lambda_{j}|^{p}\Big)^{1/p}\Big\} $, 其中下确界取遍所有原子分解$ f = \sum_{j}\lambda_{j}a_{j} $, 其中$ a_{j} $是$ H_L^{p, q}({\Bbb H}^{n}) $ -原子, $ \lambda_{j} $是标量.

在本节中, 我们叙述一些关于辅助函数的已知结果.假设非负位势$ V\in B_{q_{0}} $, $ q_{0}>Q/2 $.对于下面的引理2.1–2.5, 参见文献[5, 8].

引理2.1  测度$ V(h){\rm d}h $满足双倍条件, 即存在$ C>0 $, 使得

$ \begin{eqnarray*} \int_{B(g, 2r)}{V(h)}{\rm d}h\leq{C}\int_{B(g, r)}{V(h)}{\rm d}h \end{eqnarray*} $

对所有球$ B(g, r) $在$ {\Bbb H}^{n} $上成立.

引理2.2  存在$ C>0 $, 使得

$ \begin{eqnarray*} \frac{1}{r^{Q-2}}\int_{B(g, r)}{V(h)}{\rm d}h\leq{C}{\Big(\frac{R}{r}\Big)^{Q/q_{0}-2}}{\frac{1}{R^{Q-2}}}\int_{B(g, R)}V(h){\rm d}h, \ 0<r<R<\infty. \end{eqnarray*} $

引理2.3  若$ r = \rho(g) $, 则

$ \begin{eqnarray*} \frac{1}{r^{Q-2}}\int_{B(g, r)}{V(h)}{\rm d}h = 1. \end{eqnarray*} $

此外, $ \frac{1}{r^{Q-2}}\int_{B(g, r)}{V(h)}{\rm d}h\sim1 $当且仅当$ r\sim\rho(g) $.

引理2.4  存在$ l_{0}>0 $, 使得

$ \begin{eqnarray*} \frac{1}{C}\Big(1+ {|h^{-1}g|}/{\rho(g)}\Big)^{-l_{0}}\leq {\rho(h)}/{\rho(g)}\leq C\Big(1+ {|h^{-1}g|}/{\rho(g)}\Big)^{l_{0}/(l_{0}+1)}. \end{eqnarray*} $

特别地, 若$ |h^{-1}g|<C\rho(g) $, 则$ \rho(h)\sim\rho(g) $.

引理2.5  存在$ l_{1}>0 $, 使得

$ \begin{eqnarray*} \int_{B(g, R)}\frac{V(h)}{|h^{-1}g|^{Q-2}}{\rm d}h\leq\frac{C}{R^{Q-2}}\int_{B(g, R)}V(h){\rm d}h\leq C\Big(1+\frac{R}{\rho(g)}\Big)^{l_{1}}. \end{eqnarray*} $

引理2.6  对于$ 0\leq\sigma\leq\delta $, $ \delta = \min \{1, 2-Q/q_{0} \} $, 有

$ \begin{eqnarray*} \int_{B(g, R)}\frac{V(h)}{|h^{-1}g|^{Q-2+\sigma}}{\rm d}h\leq\frac{C}{R^{Q-2+\sigma}}\int_{B(g, R)}V(h){\rm d}h. \end{eqnarray*} $

证  由Hölder不等式和$ B_{q_{0}+\varepsilon} $条件, 我们得到

$ \begin{eqnarray*} &&\int_{B(g, R)}\frac{V(h)}{|h^{-1}g|^{Q-2+\sigma}}{\rm d}h\\ &\leq&\Big(\int_{B(g, R)}V(h)^{q_{0}+\varepsilon}{\rm d}h\Big)^{1/(q_{0}+\varepsilon)}\Big(\int_{B(g, R)}\frac{1}{|h^{-1}g|^{(Q-2+\sigma)\cdot\frac{q_{0}+\varepsilon}{q_{0}+\varepsilon-1}}}{\rm d}h\Big)^{\frac{q_{0}+\varepsilon-1}{q_{0}+\varepsilon}}\\ &\leq&\frac{C}{R^{Q-2+\sigma}}\int_{B(g, R)}V(h){\rm d}h. \end{eqnarray*} $

特别地, 当$ q_{0}>Q $时, $ \delta = 1 $, 可以取$ \sigma = 1 $, 引理仍然成立, 即

$ \begin{eqnarray*} \int_{B(g, R)}\frac{V(h)}{|h^{-1}g|^{Q-1}}{\rm d}h\leq\frac{C}{R^{Q-1}}\int_{B(g, R)}V(h){\rm d}h. \end{eqnarray*} $

引理2.6证毕.

本节中, 我们分别估计算子$ -\Delta_{{\Bbb H}^{n}}+{\rm i}\lambda $和$ L+{\rm i}\lambda $的基本解, 其中$ \lambda\in{{\Bbb R}} $.此外, 我们还分别得到了$ L^{i\gamma} $和$ \nabla_{{\Bbb H}^{n}, g}L^{-1/2} $的核的估计.

引理2.7(参见文献[6, (3) & (4)式])  令$ \Gamma(g, h, \lambda) $表示$ -\Delta_{{\Bbb H}^{n}}+{\rm i}\lambda $的基本解, 其中$ \lambda\in{{\Bbb R}} $.对任意的$ N>0 $, 存在$ C_{N}>0 $, 使得

$ \begin{eqnarray*} |\Gamma(g, h, \lambda)|\leq\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}}\cdot\frac{1}{|h^{-1}g|^{Q-2}} \end{eqnarray*} $

和

(2.1) $ \begin{equation} |\nabla_{{\Bbb H}^{n}, g}\Gamma(g, h, \lambda)|\leq\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}}\cdot\frac{1}{|h^{-1}g|^{Q-1}}. \end{equation} $

$ \nabla_{{\Bbb H}^{n}, g} $表示变量$ g $的梯度.将$ \nabla_{{\Bbb H}^{n}, g} $换成$ \nabla_{{\Bbb H}^{n}, h} $, 估计(2.1)仍然成立.

引理2.8(参见文献[8, 定理4.8])  令$ \Gamma^{L}(g, h, \lambda) $表示算子$ L+i\lambda $, $ \lambda\in{{\Bbb R}} $, 的基本解.对任意$ N>0 $, 存在$ C_{N}>0 $, 使得

$ \begin{eqnarray*} |\Gamma^{L}(g, h, \lambda)|\leq\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}(1+|h^{-1}g|\rho(h)^{-1})^{N}}\cdot\frac{1}{|h^{-1}g|^{Q-2}}. \end{eqnarray*} $

引理2.9 (参见文献[6, 引理6])  对任意的$ N>0 $, 存在$ C_{N}>0 $, 若$ |h^{-1}g|\leq\rho(h) $, 则有

$ \begin{eqnarray*} |\Gamma^{L}(g, h, \lambda)-\Gamma(g, h, \lambda)|\leq\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}|h^{-1}g|^{Q-2}}\Big(\frac{|h^{-1}g|}{\rho(h)}\Big)^{\delta}. \end{eqnarray*} $

引理2.10  对任意的$ N>0 $, 存在$ C_{N}>0 $, 如果$ |h^{-1}g|\leq\rho(h) $, 那么

$ \begin{eqnarray*} |\nabla_{{\Bbb H}^{n}, g}\Gamma^{L}(g, h, \lambda)-\nabla_{{\Bbb H}^{n}, g}\Gamma(g, h, \lambda)|\leq\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}|h^{-1}g|^{Q-1}}\Big(\frac{|h^{-1}g|}{\rho(h)}\Big)^{\delta}. \end{eqnarray*} $

证  令$ \nabla_{{\Bbb H}^{n}, g}\widetilde{\Gamma}(g, h.\lambda) = \nabla_{{\Bbb H}^{n}, g}\Gamma^{L}(g, h, \lambda)-\nabla_{{\Bbb H}^{n}, g}\Gamma(g, h, \lambda) $.由泛函演算可知

$ \begin{eqnarray*} \nabla_{{\Bbb H}^{n}, g}\widetilde{\Gamma}(g, h, \lambda) = -\int_{{\Bbb H}^{n}}\nabla_{{\Bbb H}^{n}, g}\Gamma(g, \omega, \lambda)V(\omega)\Gamma^{L}(\omega, h, \lambda){\rm d}\omega. \end{eqnarray*} $

若$ |h^{-1}g|\leq\rho(h) $, 通过引理2.7和引理2.8, 我们得到

$ \begin{eqnarray*} |\nabla_{{\Bbb H}^{n}, g}\widetilde{\Gamma}(g, h.\lambda)|\leq\int_{{\Bbb H}^{n}}|\nabla_{{\Bbb H}^{n}, g}\Gamma(g, \omega, \lambda)|V(\omega)|\Gamma^{L}(\omega, h, \lambda)|{\rm d}\omega\leq I_{1}+I_{2}+I_{3}, \end{eqnarray*} $

其中

$ \begin{eqnarray*} \left\{ \begin{array}{rl} I_{1} = &{ } \int_{|\omega^{-1}g|<|h^{-1}g|/2}\Big(C_{N}V(\omega)\Big((1+|\lambda|^{1/2}|\omega^{-1}g|)^{N}{|\omega^{-1}g|^{Q-1}}\Big)^{-1}{\rm d}\omega\Big)\\ &{ } \times\Big((1+|\lambda|^{1/2}|h^{-1}\omega|)^{N}(1+|h^{-1}\omega|\rho(h)^{-1})^{N}|h^{-1}\omega|^{Q-2}\Big)^{-1};\\ I_{2} = &{ } \int_{|h^{-1}\omega|<|h^{-1}g|/2}\Big(C_{N}V(\omega)\Big((1+|\lambda|^{1/2}|\omega^{-1}g|)^{N}{|\omega^{-1}g|^{Q-1}}\Big)^{-1}{\rm d}\omega\Big)\\ &\quad\times\Big((1+|\lambda|^{1/2}|h^{-1}\omega|)^{N}(1+|h^{-1}\omega|\rho(h)^{-1})^{N}|h^{-1}\omega|^{Q-2}\Big)^{-1};\\ I_{3} = &{ } \int_{|\omega^{-1}g|\geq|h^{-1}g|/2, |h^{-1}\omega|\geq|h^{-1}g|/2}\Big(C_{N}V(\omega)\Big((1+|\lambda|^{1/2}|\omega^{-1}g|)^{N}{|\omega^{-1}g|^{Q-1}}\Big)^{-1}{\rm d}\omega\Big)\\ &{ } \times\Big((1+|\lambda|^{1/2}|h^{-1}\omega|)^{N}(1+|h^{-1}\omega|\rho(h)^{-1})^{N}|h^{-1}\omega|^{Q-2}\Big)^{-1}. \end{array} \right. \end{eqnarray*} $

注意到$ \rho(g)\sim\rho(h) $.若$ |\omega^{-1}g|<|h^{-1}g|/2 $, 则$ |h^{-1}\omega|\sim|h^{-1}g| $.由引理2.2、引理2.5和引理2.6, 可得

$ \begin{eqnarray*} I_{1} &\leq&\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}|h^{-1}g|^{Q-2}}\int_{|\omega^{-1}g|<|h^{-1}g|/2}\frac{V(\omega)}{|\omega^{-1}g|^{Q-1}}{\rm d}\omega\\ &\leq&\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}|h^{-1}g|^{Q-1}}\cdot\frac{1}{|h^{-1}g|^{Q-2}}\int_{|\omega^{-1}g|<|h^{-1}g|/2}V(\omega){\rm d}\omega\\ &\leq&\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}|h^{-1}g|^{Q-1}}\Big(\frac{|h^{-1}g|}{\rho(h)}\Big)^{\delta}. \end{eqnarray*} $

类似地, 有

$ I_{2}\leq\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}|h^{-1}g|^{Q-1}}\Big(\frac{|h^{-1}g|}{\rho(h)}\Big)^{\delta}. $

对于$ I_{3} $, 有

$ \begin{eqnarray*} I_{3} &\leq&\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}}\int_{|h^{-1}\omega|\geq|h^{-1}g|/2}\frac{V(\omega)}{(1+|h^{-1}\omega|\rho(h)^{-1})^{N}|h^{-1}\omega|^{2Q-3}}{\rm d}\omega\\ &\leq&\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}}\bigg(\int_{|h^{-1}g|/2\leq|h^{-1}\omega|<\rho(h)}\frac{V(\omega){\rm d}\omega}{|h^{-1}\omega|^{2Q-3}}\\ &&+\rho(h)^{N}\int_{|h^{-1}\omega|\geq\rho(h)}\frac{V(\omega){\rm d}\omega}{|h^{-1}\omega|^{2Q-3+N}}\bigg). \end{eqnarray*} $

利用Hölder不等式和$ B_{q_{0}} $条件, 我们得到

$ \begin{eqnarray*} \int_{|h^{-1}g|/2\leq|h^{-1}\omega|<\rho(h)}\frac{V(\omega){\rm d}\omega}{|h^{-1}\omega|^{2Q-3}}&\leq & C\rho(h)^{Q/q_{0}-2}\cdot|h^{-1}g|^{3-2Q+Q/q_{0}'}\\ & = &\frac{C}{|h^{-1}g|^{Q-1}}\Big(\frac{|h^{-1}g|}{\rho(h)}\Big)^{\delta}. \end{eqnarray*} $

取$ N $足够大, 应用引理2.5, 可以推出

$ \begin{eqnarray*} \rho(h)^{N}\int_{|h^{-1}\omega|\geq\rho(h)}\frac{V(\omega){\rm d}\omega}{|h^{-1}\omega|^{2Q-3+N}} &\leq & C_{N}\rho(h)^{N}\sum\limits_{j = 1}^{\infty}\Big(2^{j}\rho(h)\Big)^{3-2Q-N}\int_{B(h, 2^{j}\rho(h))}V(\omega){\rm d}\omega\\ &\leq&\frac{C_{N}}{\rho(h)^{Q-1}}\sum\limits_{j = 1}^{\infty}2^{-(N-l_{1}+Q-1)j}\\ &\leq&\frac{C_{N}}{|h^{-1}g|^{Q-1}}\Big(\frac{|h^{-1}g|}{\rho(h)}\Big)^{\delta}. \end{eqnarray*} $

因此, 可得

$ \begin{eqnarray*} I_{3}\leq\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}|h^{-1}g|^{Q-1}}\Big(\frac{|h^{-1}g|}{\rho(h)}\Big)^{\delta}. \end{eqnarray*} $

引理2.10得证.

令$ K_{\gamma}^{L}(g, h) $, $ K_{\gamma}(g, h) $, $ R^{L}(g, h) $和$ R(g, h) $分别表示$ L^{i\gamma} $, $ (-\Delta_{{\Bbb H}^{n}})^{i\gamma} $, $ \nabla_{{\Bbb H}^{n}, g}L^{-1/2} $和$ \nabla_{{\Bbb H}^{n}, g}(-\Delta_{{\Bbb H}^{n}})^{-1/2} $的核.设$ \widetilde{K}_{\gamma}(g, h) = K_{\gamma}^{L}(g, h)-K_{\gamma}(g, h) $和$ \widetilde{R}(g, h) = R^{L}(g, h)-R(g, h) $.

引理2.11 (参见文献[4, 引理5], [6, (13)式])  对于$ Q/2<q_{0}<Q $, 有

$ |R^{L}(g, h)|\leq\frac{C_{N}}{(1+|h^{-1}g|\rho(h)^{-1})^{N}}\bigg(\frac{1}{|h^{-1}g|^{Q-1}}\int_{B(g, |h^{-1}g|)}\frac{V(\omega){\rm d}\omega}{|g^{-1}\omega|^{Q-1}}+\frac{1}{|h^{-1}g|^{Q}}\bigg) $

和

$ |\widetilde{R}(g, h)|\leq\frac{C}{|h^{-1}g|^{Q-1}}\int_{B(g, |h^{-1}g|/4)}\frac{V(\omega){\rm d}\omega}{|g^{-1}\omega|^{Q-1}}+\frac{1}{|h^{-1}g|^{Q-\delta}\rho(h)^{\delta}}. $

引理2.12   (ⅰ)当$ q_{0}>Q/2 $时, 对于$ |\xi|\leq|h^{-1}g|/2 $, 有

$ |K_{\gamma}^{L}(g, h\xi)-K_{\gamma}^{L}(g, h)|\leq\frac{C|\xi|^{\delta}}{|h^{-1}g|^{Q+\delta}}. $

(ⅱ)当$ q_{0}>Q $时, 对于$ |\xi|\leq|h^{-1}g|/2 $, 有

$ |R^{L}(g, h\xi)-R^{L}(g, h)|\leq\frac{C|\xi|^{\delta}}{|h^{-1}g|^{Q+\delta}}. $

证  通过泛函演算可知

(2.2) $ \begin{equation} \left\{\begin{array}{ll} { } (-\Delta_{{\Bbb H}^{n}}+V)^{i\gamma}f(g) = \int_{{\Bbb H}^{n}}K_{\gamma}^{L}(g, h)f(h){\rm d}h;\nonumber\\ { } \nabla_{{\Bbb H}^{n}, g}(-\Delta_{{\Bbb H}^{n}}+V)^{-1/2}f(g) = \int_{{\Bbb H}^{n}}R^{L}(g, h)f(h){\rm d}h;\nonumber\\ { } (-\Delta_{{\Bbb H}^{n}})^{i\gamma}f(g) = \int_{{\Bbb H}^{n}}K_{\gamma}(g, h)f(h){\rm d}h;\nonumber\\ { }\nabla_{{\Bbb H}^{n}, g}(-\Delta_{{\Bbb H}^{n}})^{-1/2}f(g) = \int_{{\Bbb H}^{n}}R(g, h)f(h){\rm d}h, \nonumber \end{array}\right. \end{equation} $

其中

$ \begin{eqnarray*} \left\{\begin{array}{ll} { } K_{\gamma}^{L}(g, h) = -\frac{1}{2\pi}\int_{{{\Bbb R}} }(-{\rm i}\lambda)^{i\gamma}\Gamma^{L}(g, h, \lambda){\rm d}\lambda;\\ { } R^{L}(g, h) = -\frac{1}{2\pi}\int_{{{\Bbb R}} }(-{\rm i}\lambda)^{-1/2}\nabla_{{\Bbb H}^{n}, g}\Gamma^{L}(g, h, \lambda){\rm d}\lambda;\\ { } K_{\gamma}(g, h) = -\frac{1}{2\pi}\int_{{{\Bbb R}} }(-{\rm i}\lambda)^{i\gamma}\Gamma(g, h, \lambda){\rm d}\lambda;\\ { } R(g, h) = -\frac{1}{2\pi}\int_{{{\Bbb R}} }(-{\rm i}\lambda)^{-1/2}\nabla_{{\Bbb H}^{n}, g}\Gamma(g, h, \lambda){\rm d}\lambda. \end{array}\right. \end{eqnarray*} $

我们可以得到

$ \begin{eqnarray*} |K_{\gamma}^{L}(g, h)|\leq\frac{C_{N}}{(1+|h^{-1}g|\rho(h)^{-1})^{N}}\cdot\frac{1}{|h^{-1}g|^{Q}} \end{eqnarray*} $

和

$ \begin{eqnarray*} |\widetilde{K}_{\gamma}(g, h)|\leq\frac{C_{N}}{|h^{-1}g|^{Q-\delta}\rho(h)^{\delta}}. \end{eqnarray*} $

若$ q_{0}>Q $, 则

(2.3) $ \begin{equation} |R^{L}(g, h)|\leq\frac{C_{N}}{(1+|h^{-1}g|\rho(h)^{-1})^{N}}\cdot\frac{1}{|h^{-1}g|^{Q}} \end{equation} $

和

(2.4) $ \begin{equation} |\widetilde{R}(g, h)|\leq\frac{C_{N}}{|h^{-1}g|^{Q-\delta}\rho(h)^{\delta}}. \end{equation} $

从而有

$ \begin{eqnarray*} |K_{\gamma}^{L}(g, h\xi)-K_{\gamma}^{L}(g, h)| &\leq&\frac{1}{2\pi}\int_{{{\Bbb R}} }|(-{\rm i}\lambda)^{i\gamma}||\Gamma^{L}(g, h\xi, \lambda)-\Gamma^{L}(g, h, \lambda)|{\rm d}\lambda\\ &\leq&C\int_{{{\Bbb R}} }|\Gamma^{L}(g, h\xi, \lambda)-\Gamma^{L}(g, h, \lambda)|{\rm d}\lambda. \end{eqnarray*} $

通过Morrey的嵌入定理, 对于$ 1/t = 1/q_{0}-1/Q $, 可以推出

$ \begin{eqnarray*} |\Gamma^{L}(g, h\xi, \lambda)-\Gamma^{L}(g, h, \lambda)| & = &|\Gamma^{L}(h\xi, g, -\lambda)-\Gamma^{L}(h, g, -\lambda)|\\ &\leq&C|\xi|^{1-Q/t}\Big(\int_{B(h, R)}|\nabla_{{\Bbb H}^{n}, g}\Gamma^{L}(u, g, -\lambda)|^{t}{\rm d}t\Big)^{1/t}\\ &\leq&C|\xi|^{1-Q/t}R^{Q/q_{0}-2}\Big(1+\frac{R}{\rho(g)}\Big)^{N_{0}}\sup\limits_{u\in B(h, 2R)}|\Gamma^{L}(u, g, -\lambda)|{\rm d}\lambda. \end{eqnarray*} $

令$ R = |h^{-1}g|/4 $.对于$ |u^{-1}h|\leq2R $, 易知$ |g^{-1}u|\geq c|h^{-1}g| $, 有

$ \begin{eqnarray*} \sup\limits_{u\in B(h, 2R)}|\Gamma^{L}(u, g, -\lambda)| &\leq&\sup\limits_{u\in B(h, 2R)}\frac{C_{N}}{(1+|\lambda|^{1/2}|g^{-1}u|)^{N}(1+|g^{-1}u|\rho(h)^{-1})^{N}}\cdot\frac{1}{|g^{-1}u|^{Q-2}}\\ &\leq&\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}(1+|h^{-1}g|\rho(h)^{-1})^{N}}\cdot\frac{1}{|h^{-1}g|^{Q-2}}. \end{eqnarray*} $

可得

(2.5) $ \begin{eqnarray} &&|\Gamma^{L}(g, h\xi, \lambda)-\Gamma^{L}(g, h, \lambda)|{}\\ &\leq &C\Big(\frac{\xi}{R}\Big)^{\delta}\Big(1+\frac{R}{\rho(h)}\Big)^{N_{0}}\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}(1+|h^{-1}g|\rho(h)^{-1})^{N}}\cdot\frac{1}{|h^{-1}g|^{Q-2}}{}\\ &\leq &\frac{C}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}(1+|h^{-1}g|\rho(h)^{-1})^{N}}\cdot\frac{|\xi|^{\delta}}{|h^{-1}g|^{Q-2+\delta}}. \end{eqnarray} $

最终, 我们得到

(2.6) $ \begin{eqnarray} &&|K_{\gamma}^{L}(g, h\xi)-K_{\gamma}^{L}(g, h)|{}\\ &\leq& C\int_{{{\Bbb R}} }\frac{|\xi|^{\delta}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}(1+|h^{-1}g|\rho(h)^{-1})^{N}|h^{-1}g|^{Q-2+\delta}}{\rm d}\lambda{}\\ &\leq& \frac{C|\xi|^{\delta}}{|h^{-1}g|^{Q-2+\delta}}\int_{{{\Bbb R}} }\frac{1}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}}{\rm d}\lambda \leq \frac{C|\xi|^{\delta}}{|h^{-1}g|^{Q+\delta}}. \end{eqnarray} $

若$ q_{0}>Q $, 利用类似于(2.6)式的证明中的方法可以得到

$ \begin{eqnarray*} |R^{L}(g, h\xi)-R^{L}(g, h)|\leq \frac{C|\xi|^{\delta}}{|h^{-1}g|^{Q+\delta}}. \end{eqnarray*} $

引理2.12得证.

引理2.13  对于$ q_{0}>Q $, 有

$ \begin{eqnarray*} |\widetilde{R}(g, h\xi)-\widetilde{R}(g, h)|\leq\frac{C}{|h^{-1}g|^{Q}}\Big(\frac{|\xi|}{\rho(h)}\Big)^{\delta}, \ |\xi|\leq|h^{-1}g|/2. \end{eqnarray*} $

对于$ Q/2<q_{0}<Q $, 有

$ \begin{eqnarray*} |\widetilde{K}_{\gamma}(g, h\xi)-\widetilde{K}_{\gamma}(g, h)|\leq\frac{C}{|h^{-1}g|^{Q}}\Big(\frac{|\xi|}{\rho(h)}\Big)^{\delta}, \ |\xi|\leq|h^{-1}g|/2. \end{eqnarray*} $

证  易知

$ \begin{eqnarray*} \left\{ \begin{array}{ll} { } |R(g, h\xi)-R(g, h)|\leq \frac{C|\xi|^{\delta}}{|h^{-1}g|^{Q+\delta}}, \\ { } |K_{\gamma}(g, h\xi)-K_{\gamma}(g, h)|\leq \frac{C|\xi|^{\delta}}{|h^{-1}g|^{Q+\delta}}. \end{array} \right. \end{eqnarray*} $

其中$ |\xi|\leq|h^{-1}g|/2 $.令$ |h^{-1}g|>\rho(h) $, 我们得到

(2.7) $ \begin{equation} \left\{ \begin{array}{ll} { } |\widetilde{R}(g, h\xi)-\widetilde{R}(g, h)|\leq \frac{C|\xi|^{\delta}}{|h^{-1}g|^{Q+\delta}}, \\ { } |\widetilde{K}_{\gamma}(g, h\xi)-\widetilde{K}_{\gamma}(g, h)|\leq \frac{C|\xi|^{\delta}}{|h^{-1}g|^{Q+\delta}}. \end{array} \right. \end{equation} $

显然, 引理2.13成立.下面我们假设$ |h^{-1}g|<\rho(h) $.当$ q_{0}>Q $时, 由引理2.7和(2.5)式可得$ |\nabla_{{\Bbb H}^{n}, g}\widetilde{\Gamma}(g, h\xi, \lambda)-\nabla_{{\Bbb H}^{n}, g}\widetilde{\Gamma}(g, h, \lambda)|\leq I_{1}+I_{2}+I_{3} $, 其中

$ \begin{eqnarray*} \left\{ \begin{array}{ll} I_{1} = &{ } \int_{|\omega^{-1}g|<|h^{-1}g|/2}\Big(C_{N}|\xi|^{\delta}V(\omega)\Big((1+|\lambda|^{1/2}|\omega^{-1}g|)^{N} {|\omega^{-1}g|^{Q-1}}\Big)^{-1}{\rm d}\omega\Big)\\ &{ } \times\Big((1+|\lambda|^{1/2}|h^{-1}\omega|)^{N}(1+|h^{-1}\omega|\rho(h)^{-1})^{N}|h^{-1}\omega|^{Q-2+\delta}\Big)^{-1};\\ I_{2} = &{ } \int_{|h^{-1}\omega|<|h^{-1}g|/2}\Big(C_{N}|\xi|^{\delta}V(\omega)\Big((1+|\lambda|^{1/2}|\omega^{-1}g|)^{N}{|\omega^{-1}g|^{Q-1}}\Big)^{-1}{\rm d}\omega\Big)\\ &\quad\times\Big((1+|\lambda|^{1/2}|h^{-1}\omega|)^{N}(1+|h^{-1}\omega|\rho(h)^{-1})^{N}|h^{-1}\omega|^{Q-2+\delta}\Big)^{-1};\\ I_{3} = &{ } \int_{|\omega^{-1}g|\geq|h^{-1}g|/2, |h^{-1}\omega|\geq|h^{-1}g|/2}\Big(C_{N}|\xi|^{\delta}V(\omega)\Big((1+|\lambda|^{1/2}|\omega^{-1}g|)^{N}{|\omega^{-1}g|^{Q-1}}\Big)^{-1}{\rm d}\omega\Big)\\ &{ }\times\Big((1+|\lambda|^{1/2}|h^{-1}\omega|)^{N}(1+|h^{-1}\omega|\rho(h)^{-1})^{N}|h^{-1}\omega|^{Q-2+\delta}\Big)^{-1}. \end{array} \right. \end{eqnarray*} $

对于$ |h^{-1}g|<\rho(h) $有$ \rho(g)\sim\rho(h) $.若$ |\omega^{-1}g|<|h^{-1}g|/2 $, 则$ |h^{-1}\omega|\sim|h^{-1}g| $.应用引理2.2、引理2.5和引理2.6可得

$ \begin{eqnarray*} I_{1} &\leq&\frac{C_{N}|\xi|^{\delta}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}|h^{-1}g|^{Q-2+\delta}}\int_{|\omega^{-1}g|<|h^{-1}g|/2}\frac{V(\omega)}{|\omega^{-1}g|^{Q-1}}{\rm d}\omega\\ &\leq&\frac{C_{N}|\xi|^{\delta}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}|h^{-1}g|^{Q-1+\delta}}\cdot\frac{1}{|h^{-1}g|^{Q-2}}\int_{|\omega^{-1}g|<|h^{-1}g|/2}V(\omega){\rm d}\omega\\ &\leq&\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}|h^{-1}g|^{Q-1}}\Big(\frac{|\xi|}{\rho(h)}\Big)^{\delta}. \end{eqnarray*} $

类似地, 若$ |h^{-1}\omega|<|h^{-1}g|/2 $, 则$ |\omega^{-1}g|\sim|h^{-1}g| $, 有

$ \begin{eqnarray*} I_{2}\leq\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}|h^{-1}g|^{Q-1}}\Big(\frac{|\xi|}{\rho(h)}\Big)^{\delta}. \end{eqnarray*} $

对于$ I_{3} $, 有

$ \begin{eqnarray*} I_{3} &\leq&\int_{|\omega^{-1}g|\geq|h^{-1}g|/2, |h^{-1}\omega|\geq|h^{-1}g|/2}\frac{C_{N}|\xi|^{\delta}V(\omega)}{(1+|\lambda|^{1/2}|h^{-1}\omega|)^{N}|\omega^{-1}g|^{Q-1}|h^{-1}\omega|^{Q-2+\delta}}{\rm d}\omega\\ &\leq&\frac{C_{N}|\xi|^{\delta}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}|h^{-1}g|^{Q-1}}\int_{|h^{-1}\omega|\geq|h^{-1}g|/2}\frac{V(\omega)}{|h^{-1}\omega|^{Q-2+\delta}}{\rm d}\omega. \end{eqnarray*} $

通过引理2.2和引理2.5, 我们有

$ \begin{eqnarray*} \int_{|h^{-1}\omega|\geq|h^{-1}g|/2}\frac{V(\omega)}{|h^{-1}\omega|^{Q-2+\delta}}{\rm d}\omega &\leq&\sum\limits_{k = -1}^{\infty}\frac{1}{(2^{k}|h^{-1}g|)^{Q-2+\delta}}\int_{|h^{-1}\omega|\leq2^{k+1}|h^{-1}g|}V(\omega){\rm d}\omega\\ &\leq & C\sum\limits_{k = -1}^{\infty}(2^{k+1}\rho(h))^{Q/q_{0}-Q}\int_{|h^{-1}\omega|\leq2^{k+1}\rho(h)}V(\omega){\rm d}\omega\\ &\leq & C\sum\limits_{k = -1}^{\infty}(2^{k+1}\rho(h))^{-\delta}\Big(1+\frac{2^{k+1}\rho(h)}{\rho(g)}\Big)^{l_{1}}, \end{eqnarray*} $

从而

$ \begin{eqnarray*} I_{3} &\leq&\frac{C_{N}|\xi|^{\delta}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}|h^{-1}g|^{Q-1}\rho(h)^{\delta}}\sum\limits_{k = -1}^{\infty}\frac{(1+2^{k+1})^{l_{1}}}{2^{k\delta}}\\ &\leq&\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}|h^{-1}g|^{Q-1}}\Big(\frac{|\xi|}{\rho(h)}\Big)^{\delta}, \end{eqnarray*} $

这里, 我们取$ 0<l_{1}<1 $.所以

(2.8) $ \begin{equation} |\nabla_{{\Bbb H}^{n}, g}\widetilde{\Gamma}(g, h\xi, \lambda)-\nabla_{{\Bbb H}^{n}, g}\widetilde{\Gamma}(g, h, \lambda)| \leq\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}|h^{-1}g|^{Q-1}}\Big(\frac{|\xi|}{\rho(h)}\Big)^{\delta}. \end{equation} $

对于$ \widetilde{\Gamma}(g, h, \lambda) = -\int_{{\Bbb H}^{n}}\Gamma(g, \omega, \lambda)V(\omega)\Gamma^{L}(\omega, h, \lambda){\rm d}\omega $.当$ Q/2<q_{0}<Q $时, 使用与(2.8)式相同的方法, 可以得到

$ \begin{eqnarray*} |\widetilde{\Gamma}(g, h\xi, \lambda)-\widetilde{\Gamma}(g, h, \lambda)| \leq\frac{C_{N}}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}|h^{-1}g|^{Q-2}}\Big(\frac{|\xi|}{\rho(h)}\Big)^{\delta}. \end{eqnarray*} $

最终得到

$ \begin{eqnarray*} |\widetilde{R}(g, h\xi)-\widetilde{R}(g, h)| &\leq&\frac{1}{2\pi}\int_{{{\Bbb R}} }|(-{\rm i}\lambda)^{-1/2}||\nabla_{{\Bbb H}^{n}, g}\widetilde{\Gamma}(g, h\xi, \lambda)-\nabla_{{\Bbb H}^{n}, g}\widetilde{\Gamma}(g, h, \lambda)|{\rm d}\lambda\\ &\leq&\frac{C|\xi|^{\delta}}{|h^{-1}g|^{Q-1}\rho(h)^{\delta}}\int_{{{\Bbb R}} }\frac{|(-{\rm i}\lambda)^{-1/2}|}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}}{\rm d}\lambda\\ &\leq&\frac{C}{|h^{-1}g|^{Q}}\Big(\frac{|\xi|}{\rho(h)}\Big)^{\delta} \end{eqnarray*} $

和

$ \begin{eqnarray*} |\widetilde{K}_{\gamma}(g, h\xi)-\widetilde{K}_{\gamma}(g, h)| &\leq&\frac{1}{2\pi}\int_{{{\Bbb R}} }|(-{\rm i}\lambda)^{i\gamma}||\widetilde{\Gamma}(g, h\xi, \lambda)-\widetilde{\Gamma}(g, h, \lambda)|{\rm d}\lambda\\ &\leq&\frac{C|\xi|^{\delta}}{|h^{-1}g|^{Q-2}\rho(h)^{\delta}}\int_{{{\Bbb R}} }\frac{|(-{\rm i}\lambda)^{i\gamma}|}{(1+|\lambda|^{1/2}|h^{-1}g|)^{N}}{\rm d}\lambda\\ &\leq&\frac{C}{|h^{-1}g|^{Q}}\Big(\frac{|\xi|}{\rho(h)}\Big)^{\delta}. \end{eqnarray*} $

引理2.13得证.

定义3.1  令$ Q/(Q+\delta)<p\leq1\leq q\leq\infty $, $ p\neq q $和$ \varepsilon>1/p-1 $.设$ a = 1-1/p+\varepsilon $, $ b = 1-1/q+\varepsilon $.函数$ M\in L^{q}({\Bbb H}^{n}) $被称为一个以$ g_{0} $为球心的$ H_L^{p, q, \varepsilon}({\Bbb H}^{n}) $ -分子, 若满足以下条件

(ⅰ) $ |g_{0}^{-1}\cdot|^{Qb}M \in L^{q}({\Bbb H}^{n}) $;

(ⅱ) $ {\mathfrak N}(M) = \alpha_1^{b-a}\|M\|_{L^{q}({\Bbb H}^{n})}^{a/b}\||g_{0}^{-1}\cdot|^{Qb}M\|_{L^{q}({\Bbb H}^{n})}^{1-a/b} $$ \leq1 $;

(ⅲ)若$ \|M\|_{L^{q}({\Bbb H}^{n})}^{1/(a-b)} $$ <\alpha_1\rho(g_{0})^{Q} $, 则$ \int M(g)dg = 0 $, 其中$ \alpha_1 $为$ {\Bbb H}^{n} $上单位球的体积.

引理3.1  任意一个$ H_{L}^{p, q}({\Bbb H}^{n}) $ -原子$ a $是一个常数因子小于或等于$ 1 $的$ H_{L}^{p, q, \varepsilon}({\Bbb H}^{n}) $ -分子.

证  若$ a $是一个$ H_{L}^{p, q}({\Bbb H}^{n}) $ -原子, 则有$ \|a\|_{L^{q}({\Bbb H}^{n})}\leq|B(g_{0}, r)|^{1/q-1/p} $.我们可得

$ \begin{eqnarray*} \||g_{0}^{-1}\cdot|^{Qb}a\|_{L^{q}({\Bbb H}^{n})}\leq r^{Qb}\|a\|_{L^{q}({\Bbb H}^{n})}\leq\alpha_{1}^{-b}|B(g_{0}, r)|^{a}, \end{eqnarray*} $

这意味着$ |g_{0}^{-1}\cdot|^{Qb}a\in L^{q}({\Bbb H}^{n}) $.从而$ \||g_{0}^{-1}\cdot|^{Qb}a\|_{L^{q}({\Bbb H}^{n})}^{1-a/b}\leq\alpha_{1}^{a-b}|B(g_{0}, r)|^{a-a^{2}/b} $.

易知$ \|a\|_{L^{q}({\Bbb H}^{n})}^{a/b}\leq|B(g_{0}, r)|^{a^{2}/b-a} $.则有$ {\mathfrak N}(a) = \alpha_{1}^{b-a}\|a\|_{L^{q}({\Bbb H}^{n})}^{a/b}\||g_{0}^{-1}\cdot|^{Qb}a\|_{L^{q}({\Bbb H}^{n})}^{1-a/b}\leq1 $.然后, 我们可以得到$ |B(g_{0}, r)|\leq\|a\|_{L^{q}({\Bbb H}^{n})}^{1/(a-b)}<\alpha_{1}\rho(g_{0})^{Q} $, 即$ \alpha_{1}r^{Q}<\alpha_{1}\rho(g_{0})^{Q} $.因此, $ r<\rho(g_{0}) $时, 有$ \int a(g)dg = 0 $.

命题3.1  令$ Q/(Q+\delta)<p\leq1\leq q\leq\infty $, $ p\neq q $, $ \varepsilon>1/p-1 $. $ f\in H_L^{p, q}({\Bbb H}^{n}) $当且仅当$ f $可以表示为$ f = \sum_{j}\lambda_{j}M_{j} $, 其中$ M_{j} $是$ H_L^{p, q, \varepsilon}({\Bbb H}^{n}) $ -分子, $ \sum_{j}|\lambda_{j}|^{p}<\infty $.和式$ \sum_{j} $按$ H_L^{p}({\Bbb H}^{n}) $范数收敛.此外, $ \|f\|_{H_L^p}\sim\|f\|_{H_L^{p, q, \varepsilon, M}} = \inf\Big\{\Big({\sum\limits_j}|\lambda_{j}|^{p}\Big)^{1/p}\Big\} $, 其中下确界取遍所有上述的分子分解.

证  命题1.1蕴含着任意$ H_{L}^{p, q, \varepsilon}({\Bbb H}^{n}) $ -分子$ M(g) $可以分解为原子的线性组合：$ M = \sum_{j}\lambda_{j}a_{j} $, 其中$ a_{j} $是$ H_{L}^{p, q}({\Bbb H}^{n}) $ -原子, $ \sum_{j}|\lambda_{j}|^{p}<C $.

假设$ M(g) $是一个以$ g_{0} $为球心的$ H_{L}^{p, q, \varepsilon}({\Bbb H}^{n}) $ -分子.令$ \mu = \|M\|_{L^{q}({\Bbb H}^{n})}^{1/(a-b)} $, $ a = 1-1/p+\varepsilon $和$ b = 1-1/q+\varepsilon $, 其中$ \varepsilon>1/p-1 $.若$ \mu<\alpha_{1}\rho(g_{0})^{Q} $, 则就回到了经典的分子理论.若$ \mu\geq\alpha_{1}\rho(g_{0})^{Q} $, 令$ B_{k} = \{g:|g_{0}^{-1}g|\leq2^{k}\alpha_{1}^{-1/Q}\mu^{1/Q}\}, \ k = 0, 1, 2, \cdots $; $ E_{0} = B_{0} $; $ E_{k} = B_{k}\setminus B_{k-1}, \ k = 1, 2, \cdots $.则有$ M(g) = \sum\limits_{k = 0}^{\infty}M(g)\chi_{E_{k}}(g) = \sum\limits_{k = 0}^{\infty}M_{k}(g) $.注意到supp$ M_{k}\subset B_{k} $和$ 2^{k}\alpha_{1}^{-1/Q}\mu^{1/Q}\geq\rho(g_{0}) $, $ k = 0, 1, 2, \cdots $.我们得到$ \|M_{0}\|_{L^{q}({\Bbb H}^{n})}\leq\|M\|_{L^{q}({\Bbb H}^{n})} = \mu^{a-b} = |B_{0}|^{1/q-1/p} $和

$ \begin{eqnarray*} \|M_{k}\|_{L^{q}({\Bbb H}^{n})} &\leq&\Big(\int_{2^{k-1}\alpha_{1}^{-1/Q}\mu^{1/Q}\leq|g_{0}^{-1}g|\leq2^{k}\alpha_{1}^{-1/Q}\mu^{1/Q}}|M(g)|^{q}{\rm d}g\Big)^{1/q}\\ &\leq&2^{-(k-1)Qb}\alpha_{1}^{b}\mu^{-b}\||g_{0}^{-1}g|^{Qb}M(g)\|_{L^{q}({\Bbb H}^{n})}\\ &\leq&2^{Qb-kQa}|B_{k}|^{1/q-1/p}. \end{eqnarray*} $

因此, $ M_{k}(g) = \lambda_{k}a_{k}(g) $, $ k = 0, 1, 2, \cdots $, 其中$ a_{k} $是$ H_{L}^{p, q}({\Bbb H}^{n}) $ -原子, $ \sum\limits_{k = 0}^{\infty}|\lambda_{k}|^{p}<C $, $ \sum\limits_{k = 0}^{\infty}M_{k}(g) $按$ H_{L}^{p}({\Bbb H}^{n}) $范数收敛.命题3.1证毕.

定理3.1  令$ \delta = \min\Big\{1, 2-Q/q_{0}\Big\} $, $ Q/(Q+\delta)<p\leq1 $.为简便起见, 我们记$ R^{L}: = \nabla_{{\Bbb H}^{n}, g}L^{-1/2} $和$ R: = \nabla_{{\Bbb H}^{n}, g}({-\Delta_{{\Bbb H}^{n}}})^{-1/2} $.

(ⅰ)当$ q_{0}>Q $时, $ R^{L} $在$ H_L^{p}({\Bbb H}^{n}) $上有界;

(ⅱ)当$ Q/2<q_{0}<Q $时, $ R^{L} $在$ H_L^{1}({\Bbb H}^{n}) $上有界.

证   (ⅰ)设$ a $是一个与球$ B(g_{0}, r) $有关的$ H_L^{p, q}({\Bbb H}^{n}) $ -原子.若$ r\geq\rho(g_{0}) $, 我们将证明$ R^{L}a(g) $小于或等于一个常数因子倍的$ H_L^{p, q, \varepsilon}({\Bbb H}^{n}) $ -分子.若$ r<\rho(g_{0}) $, $ R^{L}a(g) $可能不小于或等于一个常数因子倍的$ H_L^{p, q, \varepsilon}({\Bbb H}^{n}) $ -分子, 但$ Ra(g) $小于或等于一个常数因子倍的$ H_L^{p, q, \varepsilon}({\Bbb H}^{n}) $ -分子.所以, 我们将证明$ (R^{L}-R)a(g) $小于或等于一个常数因子倍的$ H_L^{p, q, \varepsilon}({\Bbb H}^{n}) $ -分子.

情形I   $ r\geq\rho(g_{0}) $.令$ \varepsilon>1/p-1 $, $ a = 1-1/p+\varepsilon $和$ b = 1-1/q+\varepsilon $.有$ \|R^{L}a(g)\|_{L^{q}({\Bbb H}^{n})}\leq C\|a(g)\|_{L^{q}({\Bbb H}^{n})}\leq C|B(g_{0}, r)|^{a-b} $.因此, $ \|R^{L}a(g)\|_{L^{q}({\Bbb H}^{n})}^{1/(a-b)}\geq\frac{1}{C}\rho(g_{0})^{Q} $.我们只需要估计$ {\mathfrak N}(R^{L}a(g)) $.

我们可以得到$ \||g_{0}^{-1}g|^{Qb}R^{L}a(g)\|_{L^{q}({\Bbb H}^{n})}\leq I_{1}+I_{2} $, 其中

$ \left\{\begin{array}{ll} { } I_{1} = \Big(\int_{B(g_{0}, 2r)}|g_{0}^{-1}g|^{Qbq}|R^{L}a(g)|^{q}{\rm d}g\Big)^{1/q}, \\ { } I_{2} = \Big(\int_{|g_{0}^{-1}g|\geq2r}|g_{0}^{-1}g|^{Qbq}|R^{L}a(g)|^{q}{\rm d}g\Big)^{1/q}. \end{array}\right. $

对于$ I_{1} $, 有$ I_{1}\leq(2r)^{Qb}\|R^{L}a(g)\|_{L^{q}({\Bbb H}^{n})}\leq C|B(g_{0}, r)|^{a} $.

当$ h\in B(g_{0}, r) $, $ \rho(h)>r $时, 由引理2.4可知$ \rho(h)\leq C\rho(g_{0})\leq Cr $.若$ g\not\in B(g_{0}, 2r) $, $ h\in B(g_{0}, r) $, 则有$ |h^{-1}g|\sim|g_{0}^{-1}g| $.令(2.2)中的$ N>Q\varepsilon $, 可得

$ \begin{eqnarray*} I_{2} &\leq&\Big(\int_{|g_{0}^{-1}g|\geq2r}|g_{0}^{-1}g|^{Qbq}\Big(\int_{B(g_{0}, r)}|R^{L}(g, h)||a(h)|{\rm d}h\Big)^{q}{\rm d}g\Big)^{1/q}\\ &\leq&\int_{B(g_{0}, r)}|a(h)|{\rm d}h\Big(\int_{|g_{0}^{-1}g|\geq2r}|g_{0}^{-1}g|^{Qbq}\frac{C_{N}}{(1+|h^{-1}g|\rho(h)^{-1})^{Nq}|h^{-1}g|^{Qq}}{\rm d}g\Big)^{1/q}\\ &\leq& C\int_{B(g_{0}, r)}|a(h)|{\rm d}h\Big(\int_{|g_{0}^{-1}g|\geq2r}|g_{0}^{-1}g|^{Qbq-Qq-Nq}\rho(h)^{Nq}{\rm d}g\Big)^{1/q}\\ &\leq& C r^{Qb-Q+Q/q}\int_{B(g_{0}, r)}|a(h)|{\rm d}h\leq C|B(g_{0}, r)|^{a}. \end{eqnarray*} $

所以, 我们得到$ \||g_{0}^{-1}g|^{Qb}R^{L}a(g)\|_{L^{q}({\Bbb H}^{n})}\leq C|B(g_{0}, r)|^{a} $和$ {\mathfrak N}(R^{L}a(g))\leq C $.

情形II  $ r<\rho(g_{0}) $.不难看出, $ \|(R^{L}-R)a(g)\|_{L^{q}({\Bbb H}^{n})}\leq J_{1}+J_{2}+J_{3} $, 其中

$ \left\{\begin{array}{ll} { } J_{1} = \Big(\int_{B(g_{0}, 2r)}|(R^{L}-R)a(g)|^{q}{\rm d}g\Big)^{1/q};\\ { } J_{2} = \Big(\int_{2r\leq|g_{0}^{-1}g|<2\rho(g_{0})}|(R^{L}-R)a(g)|^{q}{\rm d}g\Big)^{1/q};\\ { } J_{3} = \Big(\int_{|g_{0}^{-1}g|\geq2\rho(g_{0})}|(R^{L}-R)a(g)|^{q}{\rm d}g\Big)^{1/q}. \end{array}\right. $

如果$ h\in B(g_{0}, r) $, $ g\in B(g_{0}, 2r) $, 我们可以得到$ \rho(h)\sim\rho(g_{0}) $和$ |h^{-1}g|\sim|g_{0}^{-1}g| $.从而应用(2.4)式可以推出

$ \begin{eqnarray*} J_{1} &\leq&\int_{B(g_{0}, r)}|a(h)|{\rm d}h\Big(\int_{B(g_{0}, 2r)}|\widetilde{R}(g, h)|^{q}{\rm d}g\Big)^{1/q}\\ &\leq & r^{Q-Q/p}\Big(\int_{B(g_{0}, 2r)}\frac{C_{N}}{|h^{-1}g|^{(Q-\delta)q}}\cdot\rho(h)^{-\delta q}{\rm d}g\Big)^{1/q}\\ &\leq & Cr^{Q-Q/p}\rho(g_{0})^{-\delta}\Big(\int_{B(g_{0}, 2r)}\frac{1}{|g_{0}^{-1}g|^{(Q-\delta)q}}{\rm d}g\Big)^{1/q}\leq C\rho(g_{0})^{Q(a-b)}, \end{eqnarray*} $

这里我们取$ q $使得$ 1<q<Q/(Q-\delta) $和$ Q/q-Q/p+\delta>0 $成立.

若$ 2r\leq|g_{0}^{-1}g|<2\rho(g_{0}) $, 利用$ a $的消失性条件和引理2.13得到

$ \begin{eqnarray*} |(R^{L}-R)a(g)| &\leq&\int_{B(g_{0}, r)}|\widetilde{R}(g, h)-\widetilde{R}(g, g_{0})|a(h)|{\rm d}h\\ &\leq&\frac{C\rho(g_{0})^{-\delta}}{|g_{0}^{-1}g|^{Q}}\int_{B(g_{0}, r)}|g_{0}^{-1}h|^{\delta}|a(h)|{\rm d}h\\ &\leq&\frac{C\rho(g_{0})^{-\delta}}{|g_{0}^{-1}g|^{Q}}\cdot r^{Q-Q/p+\delta}. \end{eqnarray*} $

其中, 最后一个不等式用到了Hölder不等式.因此

$ \begin{eqnarray*} J_{2}\leq C\rho(g_{0})^{-\delta}r^{Q-Q/p+\delta}\Big(\int_{2r\leq|g_{0}^{-1}g|<2\rho(g_{0})}\frac{1}{|g_{0}^{-1}g|^{Qq}}{\rm d}g\Big)^{1/q}\leq C\rho(g_{0})^{Q(a-b)}. \end{eqnarray*} $

若$ |g_{0}^{-1}g|\geq2\rho(g_{0}) $, 由(2.7)式可得

(3.1) $ \begin{eqnarray} |(R^{L}-R)a(g)| &\leq&\int_{B(g_{0}, r)}|\widetilde{R}(g, h)-\widetilde{R}(g, g_{0})|a(h)|{\rm d}h{}\\ &\leq&\frac{C}{|g_{0}^{-1}g|^{Q+\delta}}\int_{B(g_{0}, r)}|g_{0}^{-1}h|^{\delta}|a(h)|{\rm d}h{}\\ &\leq&\frac{Cr^{Q-Q/p+\delta}}{|g_{0}^{-1}g|^{Q+\delta}}. \end{eqnarray} $

因此

$ \begin{eqnarray*} J_{3}\leq\Big(\int_{|g_{0}^{-1}g|\geq2\rho(g_{0})}\frac{Cr^{(Q-Q/p+\delta)q}}{|g_{0}^{-1}g|^{(Q+\delta)q}}{\rm d}g\Big)^{1/q}\leq Cr^{Q-Q/p+\delta}\rho(g_{0})^{Q/q-(Q+\delta)}\leq C\rho(g_{0})^{Q(a-b)}. \end{eqnarray*} $

所以, $ \|(R^{L}-R)a(g)\|_{L^{q}({\Bbb H}^{n})}\leq J_{1}+J_{2}+J_{3}\leq C\rho(g_{0})^{Q(a-b)} $.

接下来, 只需要证明$ {\mathfrak N}((R^{L}-R)a(g))\leq C $.我们有

$ \begin{eqnarray*} &\||g_{0}^{-1}g|^{Qb}|(R^{L}-R)a(g)|{\rm d}g\|_{L^{q}({\Bbb H}^{n})}\leq G_{1}+G_{2}, \end{eqnarray*} $

其中

$ \begin{eqnarray*} \left\{ \begin{array}{ll} { } G_{1} = \Big(\int_{B(g_{0}, 2\rho(g_{0}))}|g_{0}^{-1}g|^{Qbq}|(R^{L}-R)a(g)|^{q}{\rm d}g\Big)^{1/q}, \\ { } G_{2} = \Big(\int_{|g_{0}^{-1}g|\geq2\rho(g_{0})}|g_{0}^{-1}g|^{Qbq}|(R^{L}-R)a(g)|^{q}{\rm d}g\Big)^{1/q}. \end{array} \right. \end{eqnarray*} $

显而易见, $ G_{1}\leq C\rho(g_{0})^{Qb}\|(R^{L}-R)a(g)\|_{L^{q}({\Bbb H}^{n})}\leq C\rho(g_{0})^{Qa} $.对于$ G_{2} $, 由(3.1)式得

$ \begin{eqnarray*} G_{2} &\leq& Cr^{Q-Q/p+\delta}\Big(\int_{|g_{0}^{-1}g|\geq2\rho(g_{0})}|g_{0}^{-1}g|^{Qbq-(Q+\delta)q}{\rm d}g\Big)^{1/q}\\ &\leq& Cr^{Q-Q/p+\delta}\rho(g_{0})^{Qb+Q/q-Q-\delta}\leq C\rho(g_{0})^{Qa}. \end{eqnarray*} $

这里我们取$ \varepsilon $使得$ 1/p-1<\varepsilon<\delta/Q $, 这意味着$ (Qb-Q-\delta)q+Q<0 $.所以, $ \||g_{0}^{-1}g|^{Qb}|(R^{L}-R)a(g)|\|_{L^{q}({\Bbb H}^{n})}\leq G_{1}+G_{2}\leq C\rho(g_{0})^{Qa} $.因此, $ {\mathfrak N}((R^{L}-R)a(g))\leq C $.这就证明了, 当$ q_{0}>Q $时, $ R^{L} $在$ H_{L}^{p}({\Bbb H}^{n}) $上有界.

(ⅱ)现在, 我们证明当$ Q/2<q_{0}<Q $时, $ R^{L} $在$ H_L^{1}({\Bbb H}^{n}) $上有界.

情形I   $ r\geq\rho(g_{0}) $.考虑$ \|R^{L}a(g)\|_{L^{q}({\Bbb H}^{n})}\leq C\|a(g)\|_{L^{q}({\Bbb H}^{n})}\leq C|B(g_{0}, r)|^{a-b} $, 取$ 1<q<p_{0} $, 其中$ 1/p_{0} = 1/q_{0}-1/Q $, $ a = \varepsilon>0 $, $ b = 1-1/q+\varepsilon $.有$ \||g_{0}^{-1}g|^{Qb}R^{L}a(g)\|_{L^{q}({\Bbb H}^{n})}\leq I_{1}+I_{2} $, 其中

$ \begin{eqnarray*} \left\{ \begin{array}{ll} { } I_{1} = \Big(\int_{B(g_{0}, 2r)}|g_{0}^{-1}g|^{Qbq}|R^{L}a(g)|^{q}{\rm d}g\Big)^{1/q}, \\ { } I_{2} = \Big(\int_{|g_{0}^{-1}g|\geq2r}|g_{0}^{-1}g|^{Qbq}|R^{L}a(g)|^{q}{\rm d}g\Big)^{1/q}. \end{array} \right. \end{eqnarray*} $

显然, $ I_{1}\leq Cr^{Qb}\|R^{L}a(g)\|_{L^{q}({\Bbb H}^{n})}\leq C|B(g_{0}, r)|^{a} $.对于$ I_{2} $, 我们得到

$ \begin{eqnarray*} I_{2} &\leq&\Big(\int_{|g_{0}^{-1}g|\geq2r}|g_{0}^{-1}g|^{Qbq}\Big(\int_{B(g_{0}, r)}|R^{L}(g, h)||a(h)|{\rm d}h\Big)^{q}{\rm d}g\Big)^{1/q}\\ &\leq&\int_{B(g_{0}, r)}|a(h)|\Big(\int_{|g_{0}^{-1}g|\geq2r}|g_{0}^{-1}g|^{Qbq}|R^{L}(g, h)|^{q}{\rm d}g\Big)^{1/q}{\rm d}h\\ & = &\int_{B(g_{0}, r)}|a(h)|G(h){\rm d}h. \end{eqnarray*} $

若$ |g_{0}^{-1}g|\geq2r $和$ h\in B(g_{0}, r) $, 则$ |h^{-1}g|\sim|g_{0}^{-1}g| $.由引理2.4可知$ \rho(h)\leq C\rho(g_{0})\leq Cr $, 通过引理2.11可得$ G(h)\leq G_{1}+G_{2} $, 其中

$ \begin{eqnarray*} \left\{ \begin{array}{ll} { } G_{1} = C\rho(h)^{N}\bigg(\int_{|g_{0}^{-1}g|\geq2r}\frac{1}{|g_{0}^{-1}g|^{(Q-1+N-Qb)q}}\Big(\int_{B(g, |h^{-1}g|)}{\frac{V(\omega){\rm d}\omega}{|g^{-1}\omega|^{Q-1}}}\Big)^{q}{\rm d}g\bigg)^{1/q}, \\ { } G_{2} = C\rho(h)^{N}\bigg(\int_{|g_{0}^{-1}g|\geq2r}\frac{1}{|g_{0}^{-1}g|^{(Q+N-Qb)q}}{\rm d}g\bigg)^{1/q}. \end{array} \right. \end{eqnarray*} $

对于$ h\in B(g_{0}, r) $, $ \rho(h)\leq Cr $, 有$ G_{2}\leq Cr^{Q/q+Qb-Q}\leq C|B(g_{0}, r)|^{a} $, 其中$ N>Qa $, $ 1<q<p_{0} $, $ 1/p_{0} = 1/q_{0}-1/Q $.令$ 1/q = 1/s-1/Q $和$ s\leq q_{0} $.利用分数次积分的有界性, $ B_{s} $条件和引理2.5, 我们得到

$ \begin{eqnarray*} G_{1} &\leq & Cr^{N}\sum\limits_{j = 1}^{\infty}\Big(\int_{2^{j}r\leq|g_{0}^{-1}g|\leq2^{j+1}r}\frac{1}{|g_{0}^{-1}g|^{(Q-1+N-Qb)q}}\Big(\int_{B(g, |h^{-1}g|)}{\frac{V(\omega){\rm d}\omega}{|g^{-1}\omega|^{Q-1}}}\Big)^{q}{\rm d}g\Big)^{1/q}\\ &\leq & C\sum\limits_{j = 1}^{\infty}2^{-jN}(2^{j}r)^{-Q+Qb+1}\Big(\int_{B(g_{0}, 2^{j+1}r)}\Big(\int_{B(g, |h^{-1}g|)}{\frac{V(\omega){\rm d}\omega}{|g^{-1}\omega|^{Q-1}}}\Big)^{q}{\rm d}g\Big)^{1/q}\\ &\leq & C\sum\limits_{j = 1}^{\infty}2^{-jN}(2^{j}r)^{-Q+Qb+1}\Big(\int_{B(g_{0}, 2^{j+1}r)}{V(g)}^{s}{\rm d}g\Big)^{1/s}\\ &\leq & C\sum\limits_{j = 1}^{\infty}2^{-jN}(2^{j}r)^{-Q+Qb+1}(2^{j+1}r)^{Q/s-Q}\int_{B(g_{0}, 2^{j+1}r)}{V(g)}{\rm d}g \leq C|B(g_{0}, r)|^{a}. \end{eqnarray*} $

这里取$ N $足够的大.所以得到$ G(h)\leq G_{1}+G_{2}\leq C|B(g_{0}, r)|^{a} $和

(3.2) $ \begin{equation} I_{2}\leq\int_{B(g_{0}, r)}|a(h)|G(h){\rm d}h\leq C|B(g_{0}, r)|^{a}\|a(h)\|_{L^{1}({\Bbb H}^{n})}\leq C|B(g_{0}, r)|^{a}. \end{equation} $

因此, $ \||g_{0}^{-1}g|^{Qb}R^{L}a(g)\|_{L^{q}({\Bbb H}^{n})}\leq I_{1}+I_{2}\leq C|B(g_{0}, r)|^{a} $.进而得到$ {\mathfrak N}(R^{L}a(g))\leq C $.

情形II   $ r<\rho(g_{0}) $.我们有$ \|(R^{L}-R)a(g)\|_{L^{q}({\Bbb H}^{n})}\leq H_{1}+H_{2} $, 其中

$ \begin{eqnarray*} \left\{ \begin{array}{ll} { } H_{1} = \Big(\int_{B(g_{0}, 2\rho(g_{0}))}|(R^{L}-R)a(g)|^{q}{\rm d}g\Big)^{1/q}, \\ { } H_{2} = \Big(\int_{|g_{0}^{-1}g|\geq2\rho(g_{0})}|(R^{L}-R)a(g)|^{q}{\rm d}g\Big)^{1/q}. \end{array} \right. \end{eqnarray*} $

对于$ H_{1} $, 由Minkowski不等式, 有

$ \begin{eqnarray*} H_{1} &\leq&\Big(\int_{(g_{0}, 2\rho(g_{0}))}\Big(\int_{B(g_{0}, r)}|\widetilde{R}(g, h)||a(h)|{\rm d}h\Big)^{q}{\rm d}g\Big)^{1/q}\\ &\leq&\int_{B(g_{0}, r)}|a(h)|\Big(\int_{(g_{0}, 2\rho(g_{0}))}|\widetilde{R}(g, h)|^{q}{\rm d}g\Big)^{1/q}{\rm d}h\\ & = &\int_{B(g_{0}, r)}|a(h)|\widetilde{F}(h){\rm d}h. \end{eqnarray*} $

通过引理2.11, 得到

$ \begin{eqnarray*} \widetilde{F}(h) &\leq&\Big(\int_{(g_{0}, 2\rho(g_{0}))}\Big(\frac{C}{|h^{-1}g|^{Q-1}}\int_{B(g, |h^{-1}g|/4)}\frac{V(\omega){\rm d}\omega}{|g^{-1}\omega|^{Q-1}}+\frac{1}{|h^{-1}g|^{Q-\delta}\rho(h)^{\delta}}\Big)^{q}{\rm d}g\Big)^{1/q}\\ &\leq&\widetilde{F}_{1}(h)+\widetilde{F}_{2}(h), \end{eqnarray*} $

其中

$ \begin{eqnarray*} \left\{ \begin{array}{ll} { } \widetilde{F}_{1}(h) = \Big(\int_{(g_{0}, 2\rho(g_{0}))}\Big(\frac{C}{|h^{-1}g|^{Q-1}}\int_{B(g, |h^{-1}g|/4)}\frac{V(\omega){\rm d}\omega}{|g^{-1}\omega|^{Q-1}}\Big)^{q}{\rm d}g\Big)^{1/q}, \\ { } \widetilde{F}_{2}(h) = C\rho(h)^{-\delta}\Big(\int_{(g_{0}, 2\rho(g_{0}))}\frac{1}{|h^{-1}g|^{(Q-\delta)q}}{\rm d}g\Big)^{1/q}. \end{array} \right. \end{eqnarray*} $

如果$ h\in B(g_{0}, r) $, 那么$ \rho(h)\sim\rho(g_{0}) $, $ |h^{-1}g|\geq c\rho(g_{0}) $.所以有$ \widetilde{F}_{2}(h)\leq C\rho(g_{0})^{Q/q-Q} = C\rho(g_{0})^{Q(a-b)} $, 其中$ 1<q<Q/(Q-\delta) $.注意到$ 1/q = 1/s-1/Q $.由分数次积分有界性和$ B_{s} $条件($ s\leq q_{0} $), 然后利用引理2.5可推出

$ \begin{eqnarray*} \widetilde{F}_{1}(h) &\leq & C\sum\limits_{j = 0}^{\infty}\Big(2^{-j}\rho(g_{0})\Big)^{1-Q}\Big(\int_{|h^{-1}g|<2^{-j+3}\rho(g_{0})}\Big(\int_{B(g, |h^{-1}g|/4)}\frac{V(\omega){\rm d}\omega}{|g^{-1}\omega|^{Q-1}}\Big)^{q}{\rm d}g\Big)^{1/q}\\ &\leq & C\sum\limits_{j = 0}^{\infty}\Big(2^{-j}\rho(g_{0})\Big)^{1-Q}\Big(\int_{|h^{-1}g|<2^{-j+3}\rho(g_{0})}{V(g)}^{s}{\rm d}g\Big)^{1/s}\\ &\leq & C\sum\limits_{j = 0}^{\infty}\Big(2^{-j}\rho(g_{0})\Big)^{1-2Q+Q/s}\int_{|h^{-1}g|<2^{-j+3}\rho(g_{0})}{V(g)}{\rm d}g\\ &\leq &C\rho(g_{0})^{Q(a-b)}. \end{eqnarray*} $

从而, $ \widetilde{F}(h) = \widetilde{F}_{1}(h)+\widetilde{F}_{2}(h)\leq C\rho(g_{0})^{Q(a-b)} $.因此, $ H_{1}\leq\int_{B(g_{0}, r)}|a(h)|\widetilde{F}(h){\rm d}h\leq C\rho(g_{0})^{Q(a-b)} $.当$ |g_{0}^{-1}g|\geq2\rho(g_{0}) $和$ h\in B(g_{0}, r) $时, 有$ |h^{-1}g|\sim|g_{0}^{-1}g| $和$ |\widetilde{R}(g, h)|\leq{C}/{|h^{-1}g|^{Q}} $.则

$ \begin{eqnarray*} H_{2} &\leq&\Big(\int_{|g_{0}^{-1}g|\geq2\rho(g_{0})}\Big(\int_{B(g_{0}, r)}|\widetilde{R}(g, h)||a(h)|{\rm d}h\Big)^{q}{\rm d}g\Big)^{1/q}\\ &\leq & C\int_{B(g_{0}, r)}|a(h)|\Big(\int_{|g_{0}^{-1}g|\geq2\rho(g_{0})}\frac{1}{|g_{0}^{-1}g|^{Qq}}{\rm d}g\Big)^{1/q}{\rm d}h\\ &\leq & C\int_{B(g_{0}, r)}|a(h)|\rho(g_{0})^{Q(a-b)}{\rm d}h\leq C\rho(g_{0})^{Q(a-b)}. \end{eqnarray*} $

因此可得$ \|(R^{L}-R)a(g)\|_{L^{q}({\Bbb H}^{n})}\leq H_{1}+H_{2}\leq C\rho(g_{0})^{Q(a-b)} $.我们得到$ \||g_{0}^{-1}g|^{Qb}(R^{L}-R)a(g)\|_{L^{q}({\Bbb H}^{n})}\leq J_{1}+J_{2}+J_{3} $, 其中

$ \begin{eqnarray*} \left\{ \begin{array}{ll} { } J_{1} = \Big(\int_{B(g_{0}, 2\rho(g_{0}))}|g_{0}^{-1}g|^{Qbq}|(R^{L}-R)a(g)|^{q}{\rm d}g\Big)^{1/q};\\ { } J_{2} = \Big(\int_{|g_{0}^{-1}g|\geq2\rho(g_{0})}|g_{0}^{-1}g|^{Qbq}|R^{L}a(g)|^{q}{\rm d}g\Big)^{1/q};\\ { } J_{3} = \Big(\int_{|g_{0}^{-1}g|\geq2\rho(g_{0})}|g_{0}^{-1}g|^{Qbq}|Ra(g)|^{q}{\rm d}g\Big)^{1/q}. \end{array} \right. \end{eqnarray*} $

显然, $ J_{1}\leq C\rho(g_{0})^{Qb}\|(R^{L}-R)a(g)\|_{L^{q}({\Bbb H}^{n})}\leq C\rho(g_{0})^{Qa} $.对于$ J_{2} $, 有

$ \begin{eqnarray*} J_{2} &\leq&\Big(\int_{|g_{0}^{-1}g|\geq2\rho(g_{0})}|g_{0}^{-1}g|^{Qbq}\Big(\int_{B(g_{0}, r)}|R^{L}(g, h)||a(h)|{\rm d}h\Big)^{q}{\rm d}g\Big)^{1/q}\\ &\leq&\int_{B(g_{0}, r)}|a(h)|\Big(\int_{|g_{0}^{-1}g|\geq2\rho(g_{0})}|g_{0}^{-1}g|^{Qbq}|R^{L}(g, h)|^{q}{\rm d}g\Big)^{1/q}{\rm d}h\\ & = &\int_{B(g_{0}, r)}|a(h)|J_{0}(h){\rm d}h. \end{eqnarray*} $

若$ |g_{0}^{-1}g|\geq2\rho(g_{0}) $, $ h\in B(g_{0}, r) $, 则有$ |h^{-1}g|\sim|g_{0}^{-1}g| $和$ \rho(h)\sim\rho(g_{0}) $.由引理2.11得

$ \begin{eqnarray*} J_{0}(h) &\leq&\Bigg(\int_{|g_{0}^{-1}g|\geq2\rho(g_{0})}|g_{0}^{-1}g|^{Qbq}\Big(\frac{C_{N}}{(1+|h^{-1}g|\rho(h)^{-1})^{N}|h^{-1}g|^{Q-1}}\int_{B(g, |h^{-1}g|)}\frac{V(\omega){\rm d}\omega}{|g^{-1}\omega|^{Q-1}}\\ &&+\frac{C_{N}}{(1+|h^{-1}g|\rho(h)^{-1})^{N}|h^{-1}g|^{Q}}\Big)^{q}{\rm d}g\Bigg)^{1/q}\\ &\leq & C\rho(g_{0})^{N}\Bigg\{\Big(\int_{|g_{0}^{-1}g|\geq2\rho(g_{0})}\Big(\int_{B(g, |h^{-1}g|)}\frac{V(\omega){\rm d}\omega}{|g^{-1}\omega|^{Q-1}}\Big)^{q}\frac{{\rm d}g}{|g_{0}^{-1}g|^{(N-Qb+Q-1)q}}\Big)^{1/q}\\ &&+\Big(\int_{|g_{0}^{-1}g|\geq2\rho(g_{0})}\frac{1}{|g_{0}^{-1}g|^{(N-Qb+Q)q}}{\rm d}g\Big)^{1/q}\Bigg\}. \end{eqnarray*} $

类似于(3.2)式的证明中的$ G(h) $, 通过相同的方法, 我们可以得到$ J_{0}(h)\leq C\rho(g_{0})^{Qa} $.类似地, 我们还可以得到$ J_{2}\leq\int_{B(g_{0}, r)}|a(h)|J_{0}(h){\rm d}h\leq C\rho(g_{0})^{Qa} $.利用$ a $的消失性条件, 可得

$ \begin{eqnarray*} J_{3} &\leq&\int_{B(g_{0}, r)}|a(h)|\Big(\int_{|g_{0}^{-1}g|\geq2\rho(g_{0})}|g_{0}^{-1}g|^{Qbq}|R(g, h)-R(g, g_{0})|^{q}{\rm d}g\Big)^{1/q}{\rm d}h\\ &\leq & C\int_{B(g_{0}, r)}|a(h)|\Big(\int_{|g_{0}^{-1}g|\geq2\rho(g_{0})}\frac{|g_{0}^{-1}h|^{q}}{|g_{0}^{-1}g|^{(Q+1-Qb)q}}\Big)^{1/q}{\rm d}h\\ &\leq& C\int_{B(g_{0}, r)}|a(h)|\rho(g_{0})^{Qa}{\rm d}h\leq C\rho(g_{0})^{Qa}. \end{eqnarray*} $

其中$ 0<\varepsilon<1/n $.所以$ \||g_{0}^{-1}g|^{Qb}(R^{L}-R)a(g)\|_{L^{q}({\Bbb H}^{n})}\leq C\rho(g_{0})^{Qa} $.从而有$ {\mathfrak N}((R^{L}-R)a(g))\leq C $. (ⅱ)得证.

定理3.2  令$ \delta = \min\Big\{1, 2-Q/q_{0}\Big\} $, $ Q/(Q+\delta)<p\leq1 $.对于任意的$ \gamma\in{{\Bbb R}} $, $ L^{i\gamma} $在$ H_L^{p}({\Bbb H}^{n}) $上有界.

证   $ L^{i\gamma} $的核估计与$ \nabla_{{\Bbb H}^{n}, g}L^{-1/2} $在$ q_{0}>Q $情况下的核估计类似.因此, $ L^{i\gamma} $在$ H_{L}^{p}({\Bbb H}^{n}) $上的有界性可以用定理3.1中证明(ⅰ)的方法来证, 其中$ Q/(Q+\delta)<p\leq1 $.