数学杂志  2015, Vol. 35 Issue (4): 987-994   PDF    
扩展功能
加入收藏夹
复制引文信息
加入引用管理器
Email Alert
RSS
本文作者相关文章
杨祺
曹月波
田宏根
一类解析的零级Laplace-Stieltjes变换
杨祺1, 曹月波2, 田宏根1    
1. 新疆师范大学数学科学学院, 新疆 乌鲁木齐 830054;
2. 石河子大学理学院, 新疆 石河子 832003
摘要:本文研究了Laplace-Stieltjes变换所定义的零级整函数的增长性.利用牛顿多边形, 得到了这类整函数关于增长性及正规增长性的充要条件, 推广了文献[3]中的结果.
关键词Laplace-Stieltjes变换        Newton多边形    
THE GROWTH OF A CLASS OF ANALYTIC ZERO ORDER LAPLACE-STIELTJES TRANSFORM
YANG Qi1, CAO Yue-bo2, TIAN Hong-gen1    
1. School of Mathematical Sciences, Xinjiang Normal University, Urumqi 830054, China;
2. College of Science, Shihezi University, Shihezi 832003, China
Abstract: The growth of entire function of zero order defined by Laplace-Stieltjes transform is studied in this paper.By using Newton polygon, sufficient and necessary conditions about the growth and the regular growth of this entire function are obtained, which improve the results in [3].
Key words: Laplace-Stieltjes transform     order     Newton polygon    
1 引言

关于Dirichlet级数

$f(s)=\sum\limits_{n = 0}^\infty {{b_n}} {e^{{\lambda _n}s}} ( s=\sigma+it,\sigma,t\in R,0=\lambda_{0}<\lambda_{1}<\lambda_{2}<\cdots<\lambda_{n}\uparrow+\infty)$

的增长性和值分布的研究, 文[1-3]中已经取得了一系列的结果.余家荣先生在文献[4]中首先对Laplace-Stieltjes变换的值分布的研究作了一些奠定性的工作, 得到了三种不同的收敛横坐标和Valiron-Knopp-Bohr公式, 并定义了全平面上收敛的Laplace-Stieltjes变换的最大模, 最大项和增长级等, 得到了类似Dirichlet级数的性质.最近, 关于Laplace-Stieltjes变换的研究, 已经有一些完美的结果[4-9].但关于全平面上零级Laplace-Stieltjes变换的研究还不是很多.本文定义了它的级, 并对全平面上的一类零级Laplace-Stieltjes变换的增长性进行了研究, 得到了关于它们的增长性和正规增长性的充要条件.对于文中采用的记号除特别说明外均与文献[4]中的保持一致.

2 相关定义及主要引理

设Laplace-Stieltjes变换

$ \begin{array}{rl} F(s)=\displaystyle\int_0^{\infty}e^{sy}d \alpha (y) (s=\sigma+it,t\in R) \end{array} $ (2.1)

(为了表述方便, 后面简称为$L-S$变换), 其中$\alpha(y)$是对于$y\geq0$有定义的实值或复值函数, 而且它在任何闭区间$[0,X](0<X<+\infty)$上是有界变差的.取序列$\{\lambda_n\}$, 满足

$0=\lambda_0<\lambda_1<\cdots<\lambda_n\uparrow+\infty,$ (2.2)
$\overline{\lim\limits_{n\to\infty}}{(\lambda_{n+1}-\lambda_n)}<+\infty,\overline{\lim\limits_{n\to\infty}}\frac {\log n}{\lambda_n}=d<+\infty,$ (2.3)

还假设$L-S$变换 (2.1) 满足

$ \begin{array}{rl} \overline{\lim\limits_{n\to\infty}}\frac{\log A_n^*}{\lambda_n}=-\infty, \end{array} $ (2.4)

其中

$A_n^*=\sup\limits_{\lambda_n<x\leq\lambda_{n+1},-\infty<t<\infty}|\int_{\lambda_{n}}^xe^{ity}d\alpha(y)|.$

由文[4]中的一致收敛横坐标公式可知$L-S$变换 (2.1) 的一致收敛横坐标是$-\infty$.因此$L-S$变换 (2.1) 在全平面上收敛.

定义2.1$L-S$变换 (2.1) 的最大模、最大项、最大项指标和增长级可以分别定义为

$M_u(\sigma,F)=\sup\limits_{0<x<\infty,-\infty<t<\infty}|\int_0^xe^{(\sigma+it)y}d\alpha(y)|,\\ \mu(\sigma,F)=\max\limits_{n\in N}\{A_n^*e^{\lambda_n\sigma}\},\\ N(\sigma,F)=\max\limits_{k}\{\lambda_k|\mu(\sigma,F)={A_k^*e^{\lambda_k\sigma}}\},\\ \tau_u=\overline{\lim\limits_{\sigma\to\infty}}\frac{\log^{+}\log^{+}M_u(\sigma,F)}{\sigma}.$

这里$\log^{+}x=\max\{0,\log x\}$, 当$\tau_u=0$时称$L-S$变换 (2.1) 为零级的.

$L-S$变换 (2.1) 满足 (2.2), (2.3), (2.4) 式时, 在$oxy$直角坐标平面上作点列$\{P_{n}=(\lambda_{n},-\log A_n^*)\}_{n=1}^{\infty}$, 任取$\sigma>0$, 过点$P_{n}=(\lambda_{n},-\log A_n^*)$, 作斜率是$\sigma$的直线

$L(\sigma):y-(-\log A_n^*)=\sigma(x-\lambda_{n}),$

该直线的纵截距为$y=-\log A_n^*-\lambda_{n}\sigma$, 即

$-y=\log A_n^*e^{\lambda_{n}\sigma},$

故对任意固定的$\sigma>0$, $L(\sigma)$$x$轴的交点越低, 对应项的正对数值越大.因此过最大项指标$\lambda_{n(\sigma)}$决定的点$P_{\lambda_{n(\sigma)}}=(\lambda_{\lambda_{n(\sigma)}},-\log A_{\lambda_{n(\sigma)}}^*)$, 斜率为$\sigma$的直线下方不会有集合$\{P_{n}=(\lambda_{n},-\log A_n^*)\}_{n=1}^{\infty}$中的点.记所有最大项指标的集合为$W(F)=\{\lambda_{n(\sigma)}|\sigma\in(-\infty,+\infty)\}.$记最大项指标所决定的点集

$H(F)=\{P_{n}=(\lambda_{n},-\log A_n^*)|\lambda_{n}\in W(F)\},$

依次连接$H(F)$中的点, 则可得到一个Newton多边形$\pi(F)$.

注意最大项指标$\lambda_{n(\sigma)}$是单调上升左连续的阶梯函数.记$\lambda_{n(\sigma)}$的所有间断点为$\{\sigma_{k}\}_{k=1}^{\infty}$, 它满足

$\sigma_{1}<\sigma_{2}<\cdots<\uparrow\ +\infty;\lambda_{n(\sigma)}=\lambda_{N_{k}};\sigma\in[\sigma_{k},\sigma_{k+1}),\nonumber\\ \sigma_{k}=\frac{-\log A_{{N_k}}^* +\log A_{N_{k-1}}^*} {\lambda_{N_{k}}-\lambda_{ N_{k-1}}}>0,k=1,2,\cdots.$ (2.5)

我们称$(2.5)$式中的$\{\lambda_{N_{k}}\}_{k=1}^{\infty}$为最大项指标序列.对应的$(\lambda_{N_{k}},-\log A_{{N_k}}^* )$是Newton多边形$\pi(F)$的顶点. $\sigma_{k}$$ {k}$是严格单调上升的.将不在Newton多边形$\pi(F)$边上的点$P_{n}$, 垂直下移至多边形的边上, 记为$\ P^{c}_{n}=(\lambda_n,-\log {A}^{c^*}_{n})$.若$P_n$$\pi(F)$的顶点或在其边上, 则$P_n$$P^{c}_n$重合.

定义

${f}^{c}(s)=\sum\limits_{n = 0}^\infty {A_n^{{c^*}}} {e^{{\lambda _n}s}},$

$L-S$变换$F(s)$和级数${F}^{c}(s)$有相等的最大项及最大项指标$N(\sigma,F)$.

引理2.1[5] 在以上规定下, 存在正整数$M$, 使得当$k\geq M$时有

1) $T_k=\frac{-\log A_{{N_k}}^*}{\lambda_{{N}_k}}>0$, 对$k$是严格单调上升的;

2) $\frac{-\log A_{{N}_k}^*+\log A_{{N}_{k-1}}^*}{\lambda_{{N}_{k}}-\lambda_{{N}_{k-1}}}>\frac{-\log A_{{N}_k}^*}{\lambda_{{N}_k}}.$

引理2.2[6] 设$L-S$变换 (2.1) 满足 (2.2), (2.3) 和 (2.4) 式, 则有

$\log \mu(\sigma,F)=\log \mu(\sigma_{1},F)+\int_{\sigma_{1}}^{\sigma}{N(\sigma,f)}d\sigma (\sigma_{1}>0).$

引理2.3[4, 6] 设$L-S$变换 (2.1) 满足 (2.2), (2.3) 和 (2.4) 式, 则对于任意的$\varepsilon\in(0,1)$和充分大的$\sigma$

$\frac{1}{2}\mu(\sigma,F)\leq M_u(\sigma,F)\leq C\mu((1+2\varepsilon)\sigma,F), $

其中$C$是常数.

引理2.4 设$L-S$变换 (2.1) 满足 (2.2), (2.3) 和 (2.4) 式, 则有

1) $\overline{\lim\limits_{\sigma\to\infty}}\frac{\log \log M_u(\sigma,F)}{\log \sigma}=\rho\Leftrightarrow\overline{\lim\limits_{\sigma\to\infty}}\frac{\log \log \mu(\sigma,F)}{\log \sigma}=\rho\Leftrightarrow\overline{\lim\limits_{\sigma\to\infty}}\frac{\log N(\sigma,F)}{\log \sigma}=\rho-1 (\rho>1).$

2) $\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log \log M_u(\sigma,F)}{\log \sigma}=\tau\Leftrightarrow\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log \log \mu(\sigma,F)}{\log \sigma}=\tau\Leftrightarrow\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log N(\sigma,F)}{\log \sigma}=\tau-1 (\tau>1).$

  1) 由引理2.3易得

$\overline{\lim\limits_{\sigma\to\infty}}\frac{\log \log M_u(\sigma,F)}{\log \sigma}=\rho\Leftrightarrow\overline{\lim\limits_{\sigma\to\infty}}\frac{\log \log \mu(\sigma,F)}{\log \sigma}=\rho.$

下证

$\overline{\lim\limits_{\sigma\to\infty}}\frac{\log \log \mu(\sigma,F)}{\log \sigma}=\rho\Leftrightarrow\overline{\lim\limits_{\sigma\to\infty}}\frac{\log N(\sigma,F)}{\log \sigma}=\rho-1.$

$\overline{\lim\limits_{\sigma\to\infty}}\frac{\log N(\sigma,F)}{\log \sigma}=\rho-1$, 则$\forall\varepsilon\in(0,\rho)$, 当$\sigma$充分大时, 有$N(\sigma,F)<{\sigma}^{\rho-1+\varepsilon},$从而

$\log\mu(\sigma,F)-\log\mu({\sigma_{1}},F)=\int_{\sigma_{1}}^{\sigma}{N(\sigma,F)}d\sigma \leq\int_{\sigma_{1}}^{\sigma}{{\sigma}^{\rho-1+\varepsilon}d\sigma\leq {\sigma}^{\rho+\varepsilon}}.$

不妨设$\sigma_{1}>0,$于是

$\overline{\lim\limits_{\sigma\to\infty}}\frac{\log \log \mu(\sigma,F)}{\log \sigma}\leq\rho+\varepsilon,$

$\varepsilon$的任意性知

$\overline{\lim\limits_{\sigma\to\infty}}\frac{\log \log \mu(\sigma,F)}{\log \sigma}\leq\rho.$

另一方面由$N(\sigma,F)$的单调性可得

$\sigma N(\sigma,F)< \int_{\sigma}^{2\sigma}{N(\sigma,F)}d\sigma=\log\mu(2\sigma,F)-\log\mu(\sigma,F)\leq\log\mu(2\sigma,F).$

取对数, 再同除以$\log \sigma$, 整理可得

$\overline{\lim\limits_{\sigma\to\infty}}\frac{\log N(\sigma,F)}{\log \sigma}\leq\overline{\lim\limits_{\sigma\to\infty}}\frac{\log \log \mu(\sigma,F)}{\log \sigma }-1.$

1) 得证.

2) 由引理$2.3$易得

$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log \log M_u(\sigma,F)}{\log \sigma}=\tau\Leftrightarrow\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log \log \mu(\sigma,F)}{\log \sigma}=\tau.$

下证

$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log \log \mu(\sigma,F)}{\log \sigma}=\tau\Leftrightarrow\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log N(\sigma,F)}{\log \sigma}=\tau-1.$

$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log N(\sigma,F)}{\log \sigma}=\tau-1,$$\forall\varepsilon\in(0,\tau),$$\sigma$充分大时, 有$N(\sigma,F)>{\sigma}^{\tau-1-\varepsilon},$从而

$\log\mu(\sigma,F)-\log\mu({\sigma_{1}},F)=\int_{\sigma_{1}}^{\sigma}{N(\sigma,f)}d\sigma >\int_{\sigma_{1}}^{\sigma}{{\sigma}^{\tau-1-\varepsilon}d\sigma \geq\frac{\sigma}{2}{\sigma}^{\tau-1-\varepsilon}},$

从而$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log \log \mu(\sigma,F)}{\log \sigma}\geq\tau-\varepsilon.$$\varepsilon$的任意性知$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log \log \mu(\sigma,F)}{\log \sigma}\geq\tau.$另一方面由$N(\sigma,F)$的单调性可得

$\log\mu(\sigma,F)-\log\mu({\sigma_{1}},F)=\int_{\sigma_{1}}^{\sigma}{N(\sigma,F)}d\sigma\leq\sigma N(\sigma,F).$

不妨设$\sigma_{1}>0$, 由上式两边取对数, 同除以$\log \sigma$, 再同取下极限可得

$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log \log \mu(\sigma,F)}{\log \sigma}-1\leq\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log N(\sigma,F)}{\log \sigma}.$

至此, 引理$2.4$得证.

3 主要结果

定理3.1 设$L-S$变换 (2.1) 满足条件 (2.2), (2.3), (2.4) 式, 则

$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log \log M_u(\sigma,F)}{\log \sigma}=\tau \Leftrightarrow \max\limits_{n_k}\mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{{n}_{k}}} {\log \log (A_{{n_{k+1}}}^*)^{-1}-\log \lambda_{{n}_{k+1}}}=\tau-1\\ \Leftrightarrow \mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{{N}_{k}}} {\log \log (A_{{N_{k+1}}}^*)^{-1}-\log \lambda_{{N}_{k+1}}}=\tau-1.$

其中最大值是对所有上升的正整数${n_k}$取的, 且最大值可以在最大项指标序列$\{\lambda_{N_{k}}\}$上达到.

 先证明

$\mathop {\underline {\lim } }\limits_{k \to \infty } \frac{{\log {\lambda _{{N_k}}}}}{{\log \log {{(A_{{N_{k + 1}}}^*)}^{ - 1}} - \log {\lambda _{{N_{k + 1}}}}}} = \tau - 1 \Rightarrow \mathop {\underline {\lim } }\limits_{\sigma \to \infty } \frac{{\log \log \mu (\sigma ,F)}}{{\log \sigma }} \ge \tau .$

由条件$\forall\varepsilon\epsilon(0,\tau-1)$, 当$k$充分大时, 有$\frac{\log \lambda_{{N}_{k}}} {\log \log (A_{{N_{k+1}}}^*)^{-1}-\log \lambda_{{N}_{k+1}}}\geq\tau-1-\varepsilon,$

$\log (A_{N_{k+1}}^{*})\geq-\lambda_{N_{k+1}}\lambda_{N_{k}}^{\frac{1}{\tau-1-\varepsilon}}.$

于是$\forall\sigma$

$\log \mu(\sigma,F)\geq\log (A_{N_{k+1}}^{*})e^{\lambda_{N_{k+1}}\sigma}\geq \lambda_{N_{k+1}}(\sigma-\lambda_{N_{k}}^{\frac{1}{\tau-1-\varepsilon}}).$

$\sigma_{k}$满足$\sigma_{k}=2\lambda_{N_{k}}^{\frac{1}{\tau-1-\varepsilon}},$$\sigma$充分大时, 存在$k$使得$\sigma\epsilon[\sigma_{k},\sigma_{k+1}),$

$\frac{\log \log \mu(\sigma,F)}{\log \sigma}\geq\frac{\log \lambda_{N_{k+1}}+\log \sigma+\log(1-\frac{\lambda_{N_{k}}^{\frac{1}{\tau-1-\varepsilon}}}{\sigma})}{\log \sigma}\\ \geq\frac{\log \lambda_{N_{k+1}} +\log(1-\frac{\lambda_{N_{k}}^{\frac{1}{\tau-1-\varepsilon}}}{\sigma_{k}})}{\log \sigma_{k+1}}+1=\frac{\log \lambda_{N_{k+1}}+\log \frac{1}{2}}{\log 2+\frac{1}{\tau-1-\varepsilon}\log \lambda_{N_{k+1}}}+1.$

故有$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log \log \mu (\sigma,F)}{\log \sigma}\geq\tau-\varepsilon.$$\varepsilon$的任意性知$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log \log \mu (\sigma,F)}{\log \sigma}\geq\tau.$

再证明

$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log \log \mu (\sigma,F)}{\log \sigma}=\tau\Rightarrow\mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{{N}_{k}}} {\log \log (A_{{N_{k+1}}}^*)^{-1}-\log \lambda_{{N}_{k+1}}}\geq\tau-1.$

由引理2.4知, 若$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log \log \mu (\sigma,F)}{\log \sigma}=\tau$, 则有

$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log N(\sigma,F)}{\log \sigma}=\tau-1.$

$\{\lambda_{N_k}\}=\{N(\sigma,F),\sigma>0\}$为最大项指标集合, 它随$k$单调上升, 且有

$\sigma_{k}=\frac{-\log A_{{N_k}}^* +\log A_{N_{k-1}}^*} {\lambda_{N_{k}}-\lambda_{ N_{k-1}}}>0,k=1,2,\cdots.$

对任意充分大的$\sigma>0,\exists k$使得$\sigma\in[\sigma_{k-1},\sigma_{k}),$此时$N(\sigma,F)=\lambda_{N_{k-1}},$故有

$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log^{+}N(\sigma,F)}{\log \sigma}=\mathop{\underline {\lim}}\limits_{k\to \infty}\frac{\log\lambda_{N_{k-1}}}{\log \sigma_k}=\tau-1.$

因此, $\forall\varepsilon\in(0,\tau-1),$存在正整数$p$, 使当$\sigma_k\geq\sigma_p$时, 有

$\lambda_{N_{k-1}}\geq {\sigma_{k}}^{\tau-1-\varepsilon}=(\frac{-\log A_{{N_k}}^* +\log A_{N_{k-1}}^*} {\lambda_{N_{k}}-\lambda_{ N_{k-1}}})^{\tau-1-\varepsilon}.$

由引理2.1得$\frac{-\log A_{{N}_k}^*}{\lambda_{{N}_k}}\leq {\lambda_{N_{k-1}}}^{\frac{1}{\tau-1-\varepsilon}}.$于是有

$\mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{{N}_{k}}} {\log \log (A_{{N_{k+1}}}^*)^{-1}-\log \lambda_{{N}_{k+1}}}\geq\tau-1-\varepsilon.$

$\varepsilon$的任意性知

$\mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{{N}_{k}}} {\log \log (A_{{N_{k+1}}}^*)^{-1}-\log \lambda_{{N}_{k+1}}}\geq\tau-1.$

结合引理2.4知定理3.1成立.

定理3.2 设$L-S$变换 (2.1) 满足条件 (2.2), (2.3), (2.4) 式, 则有

$\overline{\lim\limits_{\sigma\to\infty}}\frac{\log \log M_u(\sigma,F)}{\log \sigma}=\rho\Leftrightarrow\overline{\lim\limits_{n\to\infty}}\frac{\log \lambda_{n}} {\log \log (A_{n}^*)^{-1}-\log \lambda_{n}}=\rho-1\\ \Leftrightarrow\overline{\lim\limits_{k\to\infty}}\frac{\log \lambda_{{N}_{k}}} {\log \log (A_{{N_{k}}}^*)^{-1}-\log \lambda_{{N}_{k}}}=\rho-1.$

 先证

$\overline{\lim\limits_{\sigma\to\infty}}\frac{\log \log \mu(\sigma,F)}{\log \sigma}=\rho\Rightarrow\overline{\lim\limits_{n\to\infty}}\frac{\log \lambda_{n}} {\log \log (A_{n}^*)^{-1}-\log \lambda_{n}}\leq\rho-1.$

由条件$\forall\varepsilon>0$, 当$\sigma$充分大时, 有$\frac{\log \log \mu (\sigma,F)}{\log \sigma}<\rho+\varepsilon.$故有$\log A_{n}^* e^{\lambda_{n}\sigma}\leq\log \mu(\sigma,F)<\sigma^{\rho+\varepsilon},$从而$\log A_{n}^*\leq\sigma^{\rho+\varepsilon}-\lambda_{n}\sigma.$$\sigma=(\frac{\lambda_{n}}{\rho+\varepsilon})^{\frac{1}{\rho+\varepsilon-1}},$则有

$\log A_{n}^* \leq-\lambda_{n}^{\frac{\rho+\varepsilon}{\rho-1+\varepsilon}} [(\frac{1}{\rho+\varepsilon})^{\frac{1}{\rho-1+\varepsilon}}- (\frac{1}{\rho+\varepsilon})^{\frac{\rho+\varepsilon}{\rho-1+\varepsilon}}] =-\lambda_{n}^{\frac{\rho+\varepsilon}{\rho-1+\varepsilon}}c (c>0).$

所以

$\log \log (A_{n}^*)^{-1}\geq{\frac{\rho+\varepsilon}{\rho-1+\varepsilon}}\log \lambda_{n}+\log c,$

从而有

$\overline{\lim\limits_{n\to\infty}}\frac{\log \lambda_{n}} {\log \log (A_{n}^*)^{-1}-\log \lambda_{n}}\leq\rho-1+\varepsilon.$

$\varepsilon$的任意性知

$\overline{\lim\limits_{n\to\infty}}\frac{\log \lambda_{n}} {\log \log (A_{n}^*)^{-1}-\log \lambda_{n}}\leq\rho-1.$

再证明

$\overline{\lim\limits_{k\to\infty}}\frac{\log \lambda_{{N}_{k}}} {\log \log (A_{{N_{k}}}^*)^{-1}-\log \lambda_{{N}_{k}}}=\rho-1\Rightarrow\overline{\lim\limits_{\sigma\to\infty}}\frac{\log \log \mu (\sigma,F)}{\log \sigma}\leq\rho.$

由条件对$\forall\varepsilon>0$, 对任意充分大的$k$及任意充分大的$\sigma$

$\frac{\log \lambda_{{N}_{k}}} {\log \log (A_{{N_{k}}}^*)^{-1}-\log \lambda_{{N}_{k}}}<\rho-1+\varepsilon,$

从而有$\log A_{{N}_{k}}^*e^{\lambda_{{N}_{k}}\sigma}\leq -\lambda_{{N}_{k}}^{\frac{\rho+\varepsilon}{\rho-1+\varepsilon}}+\lambda_{{N}_{k}}\sigma.$取充分大的$k$$\sigma$使得$\sigma=\frac{\rho+\varepsilon}{\rho+\varepsilon-1}\lambda_{N_{k}}^{\frac{1}{\rho+\varepsilon-1}}$则有

$\log A_{{N}_{k}}^* e^{\lambda_{{N}_{k}}\sigma}\leq \sigma^{\rho+\varepsilon}(\frac{\rho-1+\varepsilon}{\rho+\varepsilon})^{\rho-1+\varepsilon} [1-\frac{\rho-1+\varepsilon}{\rho+\varepsilon}]=\sigma^{\rho+\varepsilon}c (c>0).$

于是$\log \mu(\sigma,F)\leq c\sigma^{\rho+\varepsilon},$从而有

$\overline{\lim\limits_{\sigma\to\infty}}\frac{\log \log \mu (\sigma,F)}{\log \sigma}\leq\rho+\varepsilon,$

$\varepsilon$的任意性知

$\overline{\lim\limits_{\sigma\to\infty}}\frac{\log \log \mu (\sigma,F)}{\log \sigma}\leq\rho.$

结合引理2.4知定理3.2成立.

定理3.3 设$L-S$变换 (2.1) 满足条件 (2.2), (2.3), (2.4) 式, 则下列条件等价:

1) $L-S$变换 (2.1) 的正规增长级为$\rho$, 即

$\lim\limits_{\sigma\rightarrow\infty}\frac{\log \log M_u(\sigma,F)}{\log \sigma }=\rho.$

2) $L-S$变换 (2.1) 满足

i) $\overline{\lim\limits_{\sigma\rightarrow\infty}}\frac{\log \log M_u(\sigma,F)}{\log \sigma }=\rho;$

ii) 存在上升的正整数列$\{n_k\}$, 使

$\lim\limits_{k\rightarrow\infty}\frac{\log \lambda_{n_{k-1}}}{\log \lambda_{n_k}}=1, \lim\limits_{k\rightarrow\infty}\frac{\log \lambda_{n_{k}}}{\log \log (A_{n_k}^{*})^{-1}-\log \lambda_{n_{k}}}=\rho-1;$

3) $\overline{\lim\limits_{k\rightarrow\infty}}\frac{\log \lambda_{{N}_{k}}}{\log \log (A_{N_k}^{*})^{-1}-\log \lambda_{N_{k}}} =\mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{{N}_{k}}}{\log \log (A_{N_{k+1}}^{*})^{-1}-\log \lambda_{N_{k+1}}}=\rho-1,$其中$\{\lambda_{N_k}\}$是最大项指标序列.

 先证1) 与2) 等价.

设2) 成立, 则由ii),

$\lim\limits_{k\rightarrow\infty}\frac{\log \lambda_{n_{k}}}{\log \log (A_{n_{k+1}}^{*})^{-1}-\log \lambda_{n_{k+1}}} =\lim\limits_{k\rightarrow\infty}\frac{\log \lambda_{n_{k+1}}}{\log \log (A_{n_{k+1}}^{*})^{-1}-\log \lambda_{n_{k+1}}}\lim\limits_{k\rightarrow\infty}{\frac{\log \lambda_{n_{k}}}{\log \lambda_{n_{k+1}}}}=\rho-1.$

从而$\mathop{\underline {\lim}}\limits_{k\rightarrow\infty}\frac{\log \lambda_{n_{k}}}{\log \log (A_{n_{k+1}}^{*})^{-1}-\log \lambda_{n_{k+1}}}=\rho-1$, 故

$ \max\limits_{n_k}\mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{n_{k}}}{\log \log (A_{n_{k+1}}^{*})^{-1}-\log \lambda_{n_{k+1}}}\geq\rho-1,$

由定理3.1知

$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log \log M_u(\sigma,F)}{\log \sigma}\geq\rho,$

结合i) 就得到1).

反之, 若1) 成立, 则说明

$\tau=\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log \log M_u(\sigma,F)}{\log \sigma} =\overline{\lim\limits_{\sigma\to\infty}}\frac{\log \log M_u(\sigma,F)}{\log \sigma}=\rho,$

从而可知i) 成立, 由定理3.1与定理3.2对最大项指标序列$\{\lambda_{N_k}\}$, 有

$1=\frac{\tau-1}{\rho-1}=\frac{\mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{{N}_{k-1}}} {\log \log (A_{{N_{k}}}^*)^{-1}-\log \lambda_{{N}_{k}}}}{\overline{\lim\limits_{\sigma\rightarrow\infty}}\frac{\log \lambda_{{N}_{k}}} {\log \log (A_{{N_{k}}}^*)^{-1}-\log \lambda_{{N}_{k}}}}\leq\mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{{N}_{k-1}}}{\log \lambda_{{N}_{k}}}\leq1.$

这说明对最大项指标序列ii) 的前一等式成立.注意到上式在第二个等号右边, 上面的下极限和下面的上极限相等, 这可以看出ii) 的后一等式成立.最后由定理3.1及定理3.2可看出1) 与3) 等价.综上所述知定理3.3成立.

参考文献
[1] 高宗升. Dirichlet级数表示整函数的增长性[J]. 数学学报, 1999, 42(4): 741–748.
[2] 杨祺, 田宏根. Dirichlet级数和随机Dirichlet级数的下级增长性[J]. 数学杂志, 2011, 31(6): 1079–1086.
[3] 杨祺, 曹月波, 田宏根. 一类零级Dirichlet级数的增长性[J]. 数学杂志, 2013, 33(5): 916–922.
[4] 余家荣. Laplace-Stieltjes变换所定义的整函数的Borel线[J]. 数学学报, 1963, 13(3): 471–484.
[5] 杨祺, 曹月波, 田宏根. 全平面上一类解析的零级和有限级Laplace-Stieltjes变换[J]. 数学物理学报, 2014, 34A(2): 454–462.
[6] 罗茜, 孔荫莹. 全平面上慢增长的Laplace-Stieltjes变换的级与型[J]. 数学物理学报, 2012, 32A(3): 601–607.
[7] 尚丽娜, 高宗升. Laplace-Stieltjes变换所表示的解析函数的值分布[J]. 数学学报, 2008, 51(5): 993–1000.
[8] 孔荫莹. 半平面解析的无穷级Laplace-Stieltjes变换[J]. 数学学报, 2012, 55(1): 141–148.
[9] Kong Yingying, Sun Daochun. On the growth of zero order Laplace-Stieltjes transform convergent in right half-plane[J]. Acta Math., 2008, 28B(2): 431–440.