球上双正则函数的增长性

Growth of Biregular Functions in Balls

摘要: Dirac算子零化的Clifford值函数称为正则函数，正则函数是全纯函数在高维空间中非交换领域的推广。双正则函数是双变量的正则函数。正则函数的增长性问题是Clifford分析中的重要问题之一。本文研究单位球上双正则函数的增长性问题。借鉴Wiman-Valiron理论，利用双正则函数的Taylor级数，研究双正则函数的增长阶，得到广义Lindelöf-Pringsheim定理，建立增长阶与Taylor级数的联系。

Abstract: The Clifford-valued functions of null-solutions of Dirac operator are called regular functions. A regular function is an extension of holomorphic functions in non-commutative domains in high-dimensional spaces. Biregular functions are regular functions of two variables. The growth problem of regular functions is one of the important problems in Clifford analysis. In this paper, we investigate the growth problem of biregular functions in unit balls. Drawing on Wiman-Valiron theory, the growth order of biregular functions is studied by using the Taylor series of biregular functions, and the generalization of Lindelöf-Pringsheim theorem is obtained. This theorem shows the relation between the growth order of biregular functions and the Taylor series.

1. 引言

Dirac算子零化的Clifford代数值函数，称为Clifford全纯函数。Clifford代数是高维空间上可结合非交换的几何代数。Clifford全纯函数又称为正则函数。正则函数是全纯函数(Cauchy-Riemann算子零化的复值函数)在非交换领域的推广[1]。Clifford分析是以正则函数为主要研究对象，广泛应用于偏微分方程、量子力学等[2] [3]。双正则函数是全纯函数在多复变量下的非交换领域的推广。Brackx等定义了Clifford分析中的双正则函数，研究了双正则函数的Cauchy积分公式和Taylor级数[4] [5]。运用不动点定理，Huang等研究了双正则函数的非线性边值问题[6] [7]。近年来，国内外许多学者进行了Clifford分析中双正则函数的相关研究：推广双正则函数的扩张定理到R²ⁿ中分形区域[8]；研究基于Pauli矩阵的双正则函数的性质[9]；构造带有Bergman核的slice双正则函数的变换公式[10]；研究双正则函数的隐函数定理[11]；研究Cayley-Dickson-Clifford分析中的双正则函数的Cauchy公式[12]。在以上工作的基础上，本文将研究Clifford分析中单位球上双正则函数的增长性问题。

全纯函数、亚纯函数的增长性问题是单复分析的重要问题之一。早期研究源于Lindelöf [13]和Pringsheim [14]。20世纪初，Wiman [15]和Valiron [16]研究了全纯整函数的增长级、增长型、最大项、中心指标等。随后，Nevanlinna研究了亚纯函数的增长性问题[17]。进一步，MacLane等进行了复偏微分方程解的增长性研究[18]-[20]。Clifford分析是复分析在高维空间中的推广。2015年以来，De Almeida等在Clifford分析框架下，引入增长级、增长型、最大项、中心指标，研究了正则整函数的增长性[21]。他们借鉴Wiman-Valiron理论，利用Taylor级数研究正则整函数、多项式正则整函数的增长级和型，给出了Lindelöf-Pringsheim定理[22] [23]，研究了正则整函数的近似阶的性质[24]。受De Almeida等研究工作的启发，我们定义球上双正则函数的增长阶，利用双正则函数的Cauchy积分公式和Taylor级数，研究单位球上双正则函数的增长性，给出广义Lindelöf-Pringsheim定理，建立增长阶与其Taylor级数的联系。

2. 预备知识

设$C{l}_n$ 是以$\left(e_1,\cdots ,e_n\right)$ 为正交基的实Clifford代数，其中，$e_i{e}_j+e_j{e}_i=0,i,j=1,2,\cdots ,n。$ 对于$a\in C{l}_n,$

定义$a={\displaystyle \sum _A{a}_A}{e}_A,\Vert a\Vert ={\left({\displaystyle \sum _A{\left|a_A\right|}^2}\right)}^{\frac{1}{2}},A=\left(h_1,\cdots ,h_r\right)\in P\left\{1,\cdots ,n\right\},1\le h_1<\cdots <h_r\le n。$

定义Clifford值二元函数$f\left(x,y\right):f\left(x,y\right)={\displaystyle \sum _A{f}_A\left(x,y\right)}{e}_A,\left(x,y\right)=\left(x_0,x_1,\cdots ,x_m;y_0,y_1,\cdots ,y_k\right)。$

设开集$\Omega \subset R^{m+1}\times R^{k+1},1<m,k\le n。$ 为了定义双正则函数，引入如下符号：

$U_y=\left\{x\in R^{m+1}:\left(x,y\right)\in \Omega \right\},y\in R^{k+1},$ $V_x=\left\{y\in R^{k+1}:\left(x,y\right)\in \Omega \right\},x\in R^{m+1}。$

若$\{\begin{array}{c}{D}_{x}f\left(x,y\right)={\displaystyle \sum _{i=0}^m{e}_i{\partial }_{x_i}}f\left(x,y\right)=0,f\left(x,y\right)\in C^1\left(U_y,C{l}_n\right),\\ f\left(x,y\right)D_y={\displaystyle \sum _{j=0}^k{e}_j{\partial }_{y_j}}f\left(x,y\right)=0,f\left(x,y\right)\in C^1\left(V_x,C{l}_n\right),\end{array}$ 则称$f\left(x,y\right)$ 是双正则函数。

引理1 设$S_m\times S_k\subset \Omega$ 。设$S_m^0=S_m/\partial S_m,S_k^0=S_k/\partial S_k$ ，其中$S_m\left(S_k\right)$ 是$m+1\left(k+1\right)$ 维紧致、可微、定向的带边界流形。若函数$f\left(x,y\right)$ 是在$\Omega$ 上的双正则函数，则对任意$\left(x,y\right)\in S_m^0\times S_k^0$ ，有

$f\left(x,y\right)={\displaystyle {\int }_{\partial S_m\times \partial S_k}{E}_m}\left(u-x\right)d{\sigma }_{u}f\left(u,v\right)d{\sigma }_{\nu }{E}_k\left(v-y\right),$

其中，Cauchy核函数

$E_m\left(x\right)=\frac{1}{{\omega }_{m+1}}\frac{\overline{x}}{{\left|x\right|}^{m+1}},E_k\left(y\right)=\frac{1}{{\omega }_{k+1}}\frac{\overline{y}}{{\left|y\right|}^{k+1}},$

${\omega }_{m+1}=\frac{2{\pi }^{\frac{m+1}{2}}}{\Gamma \left(\frac{m+1}{2}\right)},{\omega }_{k+1}=\frac{2{\pi }^{\frac{k+1}{2}}}{\Gamma \left(\frac{k+1}{2}\right)}.$

这里，$x={\displaystyle \sum _{i=0}^m{x}_i{e}_i},y={\displaystyle \sum _{j=0}^k{x}_j{e}_j},\overline{x}={\displaystyle \sum _{i=0}^m{x}_i}\overline{e_i},\overline{y}={\displaystyle \sum _{j=0}^k{y}_j\overline{e_j}},e_0=1,\overline{e_i}=-e_i,\overline{e_j}=-e_j,i,j=1,\cdots ,n$。

这些核函数满足$D_x{E}_m=E_m{D}_x=\delta ,D_y{E}_k=E_k{D}_y=\delta$ 。其中，$\delta$ 是Dirac测度。

为了叙述简便，引入符号：

${\left(l\right)}_m=\left(l_1,\cdots ,l_m\right),{\left(l\right)}_m!=l_1!\cdots l_m!,\left|{\left(l\right)}_m\right|=\left|l_1+\cdots +l_m\right|。$

${\left(h\right)}_k=\left(h_1,\cdots ,h_k\right),{\left(h\right)}_k!=h_1!\cdots h_k!,\left|{\left(h\right)}_k\right|=\left|h_1+\cdots +h_k\right|。$

引理2 若函数$f$ 是定义在球$B_m^{°}\left(0,R\right)\times B_k^{°}\left(0,R\right)$ 上的双正则函数，则函数$f$ 在球上可以唯一展成一致收敛的Taylor级数

$f\left(x,y\right)={\displaystyle \sum _{\left|{\left(l\right)}_m\right|=0}^{+\infty }{\displaystyle \sum _{\left|{\left(h\right)}_k\right|=0}^{+\infty }{V}_{{\left(l\right)}_m}\left(x\right)}}{\lambda }_{{\left(l\right)}_m{\left(h\right)}_k}{W}_{{\left(h\right)}_k}\left(y\right),$

其中，$V_{{\left(l\right)}_m}\left(x\right)=\frac{{\left(l\right)}_m!}{\left|{\left(l\right)}_m\right|!}{\displaystyle \sum _{\pi \in perm\left({\left(l\right)}_m\right)}{\xi }_{\pi \left(l_1\right)}}\cdots {\xi }_{\pi \left(l_m\right)},W_{{\left(h\right)}_k}\left(y\right)=\frac{{\left(h\right)}_k!}{\left|{\left(h\right)}_k\right|!}{\displaystyle \sum _{\pi \in perm\left({\left(h\right)}_k\right)}{\eta }_{\pi \left(h_1\right)}}\cdots {\eta }_{\pi \left(h_k\right)}$ ，$perm\left({\left(l\right)}_m\right)\left(perm\left({\left(h\right)}_k\right)\right)$ 是$\left(l_1,\cdots ,l_m\right)\left(\left(h_1,\cdots ,h_k\right)\right)$ 全排列集合。

超复变量${\xi }_s=x_s{e}_0-x_0{e}_s,s=1,\cdots ,m;{\eta }_t=y_t{e}_0-y_0{e}_t,t=1,\cdots ,k。$

这里，${\lambda }_{{\left(l\right)}_m{\left(h\right)}_k}={\displaystyle {\int }_{\partial B_m\left(R_m\right)\times \partial B_k\left(R_k\right)}{X}_{{\left(l\right)}_m}\left(u,y\right)d{\sigma }_{u}f\left(u,v\right)d{\sigma }_v{Y}_{{\left(h\right)}_k}\left(x,v\right)}$ 。

其中，$X_{{\left(l\right)}_m}\left(x\right)={\left(-1\right)}^m{\partial }_{x_{l_1}}\cdots {\partial }_{x_{l_m}}{E}_m\left(x\right),Y_{{\left(h\right)}_k}\left(y\right)={\left(-1\right)}^k{\partial }_{y_{h_1}}\cdots {\partial }_{y_{h_k}}{E}_k\left(y\right)。$

3. 球上双正则函数的增长阶

本节研究球上双正则函数的增长阶，得到广义Lindelöf-Pringsheim定理。在文献[19]中，作者定义了单复分析中球上全纯函数的增长阶。受其启发，我们定义Clifford分析中球上双正则函数的增长阶。

定义1 设$f\left(x,y\right):B_m^{°}\left(0,1\right)\times B_k^{°}\left(0,1\right)\to C{l}_n$ 为双正则函数，则函数$f$ 的增长阶为

$\rho =\underset{r\to 1^{-}}{\lim \sup }\frac{\log \log M\left(r,f\right)}{-\log (1-r)}，$

其中，$M\left(r,f\right)=\underset{\left(x,y\right)\in B_m^{°}\left(0,1\right)\times B_k^{°}\left(0,1\right)}{\sup }\left|f\left(x,y\right)\right|。$ $$

在本文中，我们假设$0<\rho <+\infty 。$

在文献[19]中，作者研究了关于单复分析中全纯函数的Lindelöf-Pringsheim定理，定理内容如下：

$若函数f\left(z\right)可展成泰勒级数f\left(z\right)={\displaystyle \sum _{n=0}^{\infty }{a}_n}{z}_n，则\rho =\underset{n\to +\infty }{\lim \sup }\frac{\log \log \left|a_n\right|}{\log n-\log \log \left|a_n\right|}。$

我们将这个定理推广到多元Clifford值函数——双正则函数，建立球上双正则函数的增长阶与Taylor级数的联系。

定理1 若$f\left(x,y\right):B_m^{°}\left(0,1\right)\times B_k^{°}\left(0,1\right)\to C{l}_n$ 为双正则函数，则函数$f$ 的增长阶为

$\rho =\underset{\left|{\left(l\right)}_m\right|,\left|{\left(h\right)}_k\right|\to +\infty }{\lim \sup }\frac{\log \log \left(\frac{\Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right)}{\log (\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|)-\log \log \left(\frac{\Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right)}，$ (1)

其中，$c\left(m,{\left(l\right)}_m\right)=\frac{m\left(m+1\right)\cdots \left(m+\left|{\left(l\right)}_m\right|-1\right)}{{\left(l\right)}_m!},c\left(k,{\left(h\right)}_k\right)=\frac{k\left(k+1\right)\cdots \left(k+\left|{\left(h\right)}_k\right|-1\right)}{{\left(h\right)}_k!}。$

证明：(1)式变形为$\rho =\underset{\left|{\left(l\right)}_m\right|,\left|{\left(h\right)}_k\right|\to +\infty }{\lim \sup }\frac{1}{\frac{\log \left(\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right)}{\log \log \left(\frac{\Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right)}-1}。$

下面只要证明：$\frac{\rho }{\rho +1}=\underset{\left|{\left(l\right)}_m\right|,\left|{\left(h\right)}_k\right|\to +\infty }{\lim \sup }\frac{\log \log \left(\frac{\Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right)}{\log \left(\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right)}。$

第一步，我们证明不等式：$\underset{\left|{\left(l\right)}_m\right|,\left|{\left(h\right)}_k\right|\to +\infty }{\lim \sup }\frac{\log \log \left(\frac{\Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right)}{\log \left(\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right)}\le \frac{\rho }{\rho +1}。$ 由引理1和引理2，得

$\begin{array}{l}\Vert {\lambda }_{{\left(l\right)}_m{\left(h\right)}_k}\Vert =\left|{\displaystyle {\int }_{\partial B_m\left(R_m\right)\times \partial B_k\left(R_k\right)}{X}_{{\left(l\right)}_m}\left(u,y\right)d{\sigma }_{u}f\left(u,v\right)d{\sigma }_v{Y}_{{\left(h\right)}_k}\left(x,v\right)}\right|\\ \le M\left(r,f\right){\displaystyle {\int }_{\partial B_m\left(R_m\right)\times \partial B_k\left(R_k\right)}\left|{\partial }_{x_{l_1}}\cdots {\partial }_{x_{l_m}}{E}_m\left(u\right)\right|\left|{\partial }_{y_{h_1}}\cdots {\partial }_{y_{h_k}}{E}_k\left(v\right)\right|d{\sigma }_{u}d{\sigma }_v}\\ =\frac{m\left(m+1\right)\cdots \left(m+\left|{\left(l\right)}_m\right|-1\right)k\left(k+1\right)\cdots \left(k+\left|{\left(h\right)}_k\right|-1\right)}{{\left(l\right)}_m!{\left(h\right)}_k!r^{\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|}}M\left(r,f\right)\\ =\frac{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}{r^{\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|}}M\left(r,f\right)。\end{array}$

上式两边取对数，得

$\log \left[\frac{\Vert {\lambda }_{{\left(l\right)}_m{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right]\le \log M\left(r,f\right)+\log r^{-\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}。$ (2)

由定义1，得$\rho =\underset{r\to 1^{-}}{\lim \sup }\frac{\log \log M\left(r,f\right)}{-\log (1-r)}。$

根据上确界定义$\exists \epsilon >0，\frac{\log \log M\left(r,f\right)}{-\log (1-r)}\le \rho +\epsilon ，$ 即

$\log M\left(r,f\right)\le {\left(1-r\right)}^{-\left(\rho +\epsilon \right)}。$

(2)式变为$\log \left[\frac{\Vert {\lambda }_{{\left(l\right)}_m{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right]\le {\left(1-r\right)}^{-\left(\rho +\epsilon \right)}+\log r^{-\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}。$

由于$\frac{1}{\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|}\to 0,0<1-r<1$ 。进一步，(2)式变形为

$\begin{array}{l}\log \left[\frac{\Vert {\lambda }_{{\left(l\right)}_m{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right]\le {\left[\frac{1}{\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|}\right]}^{-\frac{\rho +\epsilon }{1+\rho +\epsilon }}-\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]\log \left[1-{\left(\frac{1}{\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|}\right)}^{\frac{1}{1+\rho +\epsilon }}\right]\\ ={\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}^{\frac{\rho +\epsilon }{1+\rho +\epsilon }}-\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]\log \left[1-{\left(\frac{1}{\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|}\right)}^{\frac{1}{1+\rho +\epsilon }}\right]\\ ={\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}^{\frac{\rho +\epsilon }{1+\rho +\epsilon }}\left\{1-{\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}^{1-\frac{\rho +\epsilon }{1+\rho +\epsilon }}\log \left[1-{\left(\frac{1}{\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|}\right)}^{\frac{1}{1+\rho +\epsilon }}\right]\right\}。\end{array}$

上式两边取对数，取极限，得

$\begin{array}{c}\underset{\left|{\left(l\right)}_m\right|,\left|{\left(h\right)}_k\right|\to +\infty }{\lim \sup }\frac{\log \log \left(\frac{\Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right)}{\log \left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}\\ =\underset{\left|{\left(l\right)}_m\right|,\left|{\left(h\right)}_k\right|\to +\infty }{\lim \sup }\frac{\log {\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}^{\frac{\rho +\epsilon }{1+\rho +\epsilon }}}{\log \left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}\\ +\underset{\left|{\left(l\right)}_m\right|,\left|{\left(h\right)}_k\right|\to +\infty }{\lim \sup }\frac{\log \left\{1-{\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}^{1-\frac{\rho +\epsilon }{1+\rho +\epsilon }}\log \left[1-{\left(\frac{1}{\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|}\right)}^{\frac{1}{1+\rho +\epsilon }}\right]\right\}}{\log \left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}\\ =\frac{\rho +\epsilon }{1+\rho +\epsilon }。\end{array}$

令$\epsilon \to 0，$ 得$\underset{\left|{\left(l\right)}_m\right|,\left|{\left(h\right)}_k\right|\to +\infty }{\lim \sup }\frac{\log \log \left(\frac{\Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right)}{\log \left(\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right)}\le \frac{\rho }{\rho +1}。$

第二步，我们证明不等式：$\underset{\left|{\left(l\right)}_m\right|,\left|{\left(h\right)}_k\right|\to +\infty }{\lim \sup }\frac{\log \log \left(\frac{\Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right)}{\log \left(\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right)}\ge \frac{\rho }{\rho +1}。$

根据第一步的证明结果，$\exists \beta >0，\frac{\log \log \left(\frac{\Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right)}{\log \left(\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right)}\le \frac{\beta }{\beta +1}。$

设$\alpha =\frac{\beta }{\beta +1}$ ，则$\log \Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert \le {\left(\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right)}^{\alpha }$ ，即$\Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert \le e^{{\left(\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right)}^{\alpha }}$ 。由引理2可知

$f\left(x,y\right)={\displaystyle \sum _{\left|{\left(l\right)}_m\right|=0}^{+\infty }{\displaystyle \sum _{\left|{\left(h\right)}_k\right|=0}^{+\infty }{V}_{{\left(l\right)}_m}\left(x\right)}}{\lambda }_{{\left(l\right)}_m{\left(h\right)}_k}{W}_{{\left(h\right)}_k}\left(y\right)。$

在单位球上，$\left|V_{{\left(l\right)}_m}\left(x\right)\right|\le r^{\left|{(l)}_m\right|},\left|W_{{\left(h\right)}_k}\left(y\right)\right|\le r^{\left|{\left(h\right)}_k\right|}。$ 于是，有

$\begin{array}{l}M\left(r,f\right)\le {\displaystyle \sum _{\left|{\left(l\right)}_m\right|=0}^{+\infty }{\displaystyle \sum _{\left|{\left(h\right)}_k\right|=0}^{+\infty }\Vert {\lambda }_{{\left(l\right)}_m{\left(h\right)}_k}\Vert }}{r}^{\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|}\\ ={\displaystyle \sum _{\left|{\left(l\right)}_m\right|=0}^{N_0}{\displaystyle \sum _{\left|{\left(h\right)}_k\right|=0}^{N_0}\Vert {\lambda }_{{\left(l\right)}_m{\left(h\right)}_k}\Vert }}{r}^{\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|}+{\displaystyle \sum _{\left|{\left(l\right)}_m\right|=N_0}^{N_1}{\displaystyle \sum _{\left|{\left(h\right)}_k\right|=N_0}^{N_1}\Vert {\lambda }_{{\left(l\right)}_m{\left(h\right)}_k}\Vert }}{r}^{\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|}\\ +{\displaystyle \sum _{\left|{\left(l\right)}_m\right|=N_1}^{+\infty }{\displaystyle \sum _{\left|{\left(h\right)}_k\right|=N_1}^{+\infty }\Vert {\lambda }_{{\left(l\right)}_m{\left(h\right)}_k}\Vert }}{r}^{\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|}\\ \le A+{\displaystyle \sum _{\left|{\left(l\right)}_m\right|=N_0}^{N_1}{\displaystyle \sum _{\left|{\left(h\right)}_k\right|=N_0}^{N_1}{e}^{{\left(\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right)}^{\alpha }}}}{r}^{\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|}+{\displaystyle \sum _{\left|{\left(l\right)}_m\right|=N_1}^{+\infty }{\displaystyle \sum _{\left|{\left(h\right)}_k\right|=N_1}^{+\infty }{e}^{{\left(\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right)}^{\alpha }}}}{r}^{\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|}，\end{array}$

其中，A > 0为前N₁项的和。利用一元函数求导得到最大值的方法，容易得到以下等式：

$\underset{0\le x<\infty }{\max }\left(x^{\alpha }-x\log \frac{1}{r}\right)=\left({\alpha }^{\frac{\alpha }{1-\alpha }}-{\alpha }^{\frac{1}{1-\alpha }}\right){\left[\log \left(\frac{1}{r}\right)\right]}^{-\beta }=\varphi \left(\alpha \right){\left[\log \left(\frac{1}{r}\right)\right]}^{-\beta }，0<r<1。$

当$N_0\le \left|{\left(l\right)}_m\right|\le N_1,N_0\le \left|{\left(h\right)}_k\right|\le N_1$ 时，我们有

$\log \left[e^{{\left(\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right)}^{\alpha }}{r}^{\left(\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right)}\right]={\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}^{\alpha }+\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]\log r\le \varphi \left(\alpha \right){\left[\log \left(\frac{1}{r}\right)\right]}^{-\beta }。$

因此，${\displaystyle \sum _{\left|{\left(l\right)}_m\right|=N_0}^{N_1}{\displaystyle \sum _{\left|{\left(h\right)}_k\right|=N_0}^{N_1}{e}^{{\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}^{\alpha }}{r}^{\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}}}\le N_1{e}^{\varphi \left(\alpha \right){\left[\log \left(\frac{1}{r}\right)\right]}^{-\beta }}。$

现在考虑${\displaystyle \sum _{\left|{\left(l\right)}_m\right|=N_1+1}^{+\infty }{\displaystyle \sum _{\left|{\left(h\right)}_k\right|=N_1+1}^{+\infty }{e}^{{\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}^{\alpha }}{r}^{\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}}}$ 。我们不妨取$N_1=\left[{\left(\frac{1}{2}\log \left(\frac{1}{r}\right)\right)}^{\frac{-1}{1-\alpha }}\right]。$

对于$\left|{\left(l\right)}_m\right|\ge N_1+1,\left|{\left(h\right)}_k\right|\ge N_2+1,有{\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}^{1-\alpha }\ge \frac{2}{-\log r}。$

因此，${\displaystyle \sum _{\left|{\left(l\right)}_m\right|=N_1+1}^{+\infty }{\displaystyle \sum _{\left|{\left(h\right)}_k\right|=N_1+1}^{+\infty }{e}^{{\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}^{\alpha }}{r}^{\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}}\le {\displaystyle \sum _{\left|{\left(l\right)}_m\right|=N_1+1}^{+\infty }{\displaystyle \sum _{\left|{\left(h\right)}_k\right|=N_1+1}^{+\infty }{e}^{-\frac{\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}{2}\log r}{r}^{\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}}}}$

$\begin{array}{l}\le {\displaystyle \sum _{\left|{\left(l\right)}_m\right|=N_1+1}^{+\infty }{\displaystyle \sum _{\left|{\left(h\right)}_k\right|=N_1+1}^{+\infty }{r}^{\frac{\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}{2}}}}\\ ={\displaystyle \sum _{\left|{\left(l\right)}_m\right|=N_1+1}^{+\infty }{r}^{\frac{\left|{\left(l\right)}_m\right|}{2}}}{\displaystyle \sum _{\left|{\left(h\right)}_k\right|=N_1+1}^{+\infty }{r}^{\frac{\left|{\left(h\right)}_k\right|}{2}}}\le {\left(\frac{r^{\frac{N_1}{2}+1}}{1-r^{\frac{1}{2}}}\right)}^2。\end{array}$

设$\alpha =\frac{\beta }{\beta +1},则\beta =\frac{\alpha }{1-\alpha }。$ 于是，得$N_1+1\ge {\left(\frac{2}{-\log r}\right)}^{\frac{1}{1-\alpha }}$ 。

利用上式，我们可以计算$\log r^{\frac{N_1+1}{2}}\le -\frac{1}{2}{\left(\frac{2}{-\log r}\right)}^{\frac{1}{1-\alpha }}\log \frac{1}{r}\le -2^{\frac{1}{1-\alpha }-1}{\left(\log \left(\frac{1}{r}\right)\right)}^{1-\frac{1}{1-\alpha }}=-2^{\beta }{\left(\log \left(\frac{1}{r}\right)\right)}^{-\beta }$ 。当$r\to 1^{-}$ 时，有$\underset{r\to 1^{-}}{\lim }\frac{1-\sqrt{r}}{\frac{1}{2}\log \left(\frac{1}{r}\right)}=1$ ，即$1-\sqrt{r}=\frac{1}{2}\log \left(\frac{1}{r}\right)\left[1+o\left(1\right)\right]$ ，其中$o$ 表示高阶无穷小。

于是，存在$A_1>0,\frac{r^{\frac{N_1+1}{2}}}{1-\sqrt{r}}\le e^{-2^{\beta }{\left(\log \left(\frac{1}{r}\right)\right)}^{-\beta }+\log 2-\log \log \frac{1}{r}+o\left(\mathrm{l}\right)}\le A_1{e}^{-2^{\beta }{\left(\log \left(\frac{1}{r}\right)\right)}^{-\beta }-\log \log \frac{1}{r}}。$

当$r\to 1^{-}$ 时，存在$A_2>0$ ，有

${\displaystyle \sum _{\left|{\left(l\right)}_m\right|=N_1+1}^{+\infty }{\displaystyle \sum _{\left|{\left(h\right)}_k\right|=N_1+1}^{+\infty }{e}^{{\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}^{\alpha }}{r}^{\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}}}\le {\left[-A_1{e}^{-2^{\beta }{\left(\log \left(\frac{1}{r}\right)\right)}^{-\beta }-\log \log \frac{1}{r}}\right]}^2\le A_2。$

因此，$M\left(r,f\right)\le A+A_2+{\left(\frac{2}{\log \frac{1}{r}}\right)}^{\frac{1}{1-\alpha }}{e}^{\varphi \left(\alpha \right){\left[\log \left(\frac{1}{r}\right)\right]}^{-\beta }}\le A_3+{\left(\frac{2}{\log \frac{1}{r}}\right)}^{\frac{1}{1-\alpha }}{e}^{\varphi \left(\alpha \right){\left[\log \left(\frac{1}{r}\right)\right]}^{-\beta }}，A_3\ge A+A_2$ 。

由于$1-r=\log {\left(\frac{1}{r}\right)}^{-\beta }\left[1+o\left(1\right)\right]$ ，我们得到

$\begin{array}{l}\log M\left(r,f\right)\le \log A_3+\frac{1}{1-\beta }\log \left(\frac{2}{\log \frac{1}{r}}\right)+\varphi \left(\alpha \right){\left(\log \left(\frac{1}{r}\right)\right)}^{-\beta }\\ \le \varphi \left(\alpha \right){\left(\log \left(\frac{1}{r}\right)\right)}^{-\beta }\left[1+o\left(1\right)\right]\le \varphi \left(\alpha \right){\left(1-r\right)}^{-\beta }\left[1+o\left(1\right)\right]。\\ \end{array}$

进而得$\log \log M\left(r,f\right)\le \log \varphi \left(\alpha \right)+\log \left[{\left(1-r\right)}^{-\beta }\right]+\log \left[1+o\left(1\right)\right]。$

于是，$\rho =\underset{r\to 1^{-}}{\lim }\frac{{\log }^{+}\left({\log }^{+}M\left(r,f\right)\right)}{-\log (1-r)}\le \underset{r\to 1^{-}}{\lim }\frac{\varphi \left(\alpha \right)}{\log \frac{1}{1-r}}+\beta \underset{r\to 1^{-}}{\lim }\frac{\log \frac{1}{1-r}}{\log \frac{1}{1-r}}+\underset{r\to 1^{-}}{\lim }\frac{\log \left[1+o(1)\right]}{\log \frac{1}{1-r}}=\beta 。$

由上式容易得到$\frac{\rho }{1+\rho }\le \frac{\beta }{1+\beta }。$

由于$c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)\ge 1$ ，于是我们得到

$\begin{array}{l}\log \left[\frac{\Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right]\le \log \Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert -\log \left[c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)\right]\\ \le \log \Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert \le {\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}^{\alpha }={\left[\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right]}^{\frac{\beta }{1+\beta }}。\end{array}$

因此，$\frac{\log \log \left[\frac{\Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right]}{\log \left(\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right)}\le \frac{\beta }{1+\beta }$ ，使得$\frac{\log \log \left[\frac{\Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right]}{\log \left(\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right)}=\frac{\beta }{1+\beta }-\epsilon 。$

由于$\frac{\rho }{1+\rho }\le \frac{\beta }{1+\beta }$ ，可以得到以下不等式

$\frac{\log \log \left[\frac{\Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right]}{\log \left(\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right)}\ge \frac{\rho }{\rho +1}-\epsilon 。$

令$\epsilon \to 0$ ，取极限，得$\underset{\left|{\left(l\right)}_m\right|,\left|{\left(h\right)}_k\right|\to +\infty }{\lim \sup }\frac{\log \log \left[\frac{\Vert {\lambda }_{{\left(l\right)}_r{\left(h\right)}_k}\Vert }{c\left(m,{\left(l\right)}_m\right)c\left(k,{\left(h\right)}_k\right)}\right]}{\log \left(\left|{\left(l\right)}_m\right|+\left|{\left(h\right)}_k\right|\right)}\ge \frac{\rho }{\rho +1}。$

综合第一步证明和第二步证明，我们得出结论。

4. 结论

复分析向高维空间推广分为两个方面：一方面推广为多复变函数理论，实现单复变量向多复变量的推广；另一方面推广为Clifford分析，实现其向非交换领域的推广。在Clifford分析的背景下，本文以双正则函数为主要研究对象，双正则函数是含两个变量的正则函数。正则函数是取值于Clifford代数的广义的全纯函数。全纯函数的增长性问题是复分析的核心问题之一。近年来，Clifford值函数的增长性相关问题也逐渐成为Clifford分析的热点问题之一。本文研究球上双正则函数的增长性问题，建立球上双正则函数的增长阶与其Taylor级数的联系，推广Lindelöf-Pringsheim定理。因此，本文关于Clifford值多元函数的增长性问题研究，既实现了单变量向多变量的推广，又实现了其向非交换领域的推广。

基金项目

河北省自然科学基金资助(项目编号：A2022402007)；国家自然科学基金资助(项目编号：11426082)。

NOTES

^*通讯作者。