关于Dirichlet级数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. Laplace-Stieltjes变换可以看成是Dirichlet级数的推广.余家荣先生于1963年在文献^[4]中首先对Laplace-Stieltjes变换的增长性和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, 推广了Dirichlet级数的相关结论.最近, 关于Laplace-Stieltjes变换的研究, 已经有一些完美的结果^[4-9].但关于全面上零级Laplace-Stieltjes变换的研究还不是很多.本文应用高宗升先生的零级型函数, 首先定义了它关于型函数的级, 下级和正规增长级, 然后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.

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

$\begin{eqnarray} 0=\lambda_0<\lambda_1<\cdots<\lambda_n\uparrow+\infty, \end{eqnarray}$

(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

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

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

引理2.1 ^{[1, 2]} 设$M_u(\sigma,F)$在$[a,\infty)$上正值连续且趋于$\infty$,

$\begin{array}{rl} \overline{\lim\limits_{\sigma\to\infty}}\frac{\log M_u(\sigma,F)}{\sigma}=\infty,\overline{\lim\limits_{\sigma\to\infty}}\frac{\log \log M_u(\sigma,F)-\log\sigma}{\log\sigma}=\beta (0<\beta<\infty),\end{array}$

(2.5)

则存在连续可微函数$U(\sigma)=\sigma^{\beta(\sigma)},\beta(\sigma)\rightarrow\beta (\sigma\rightarrow\infty).$满足如下条件:

(1) $\log M(\sigma)\leq U(\sigma)\sigma$, 存在$\sigma_{n}\uparrow\infty$, 使$\log M(\sigma_{n})= U(\sigma_{n})\sigma_{n};$

(2) $U(\sigma)$严格单调趋于$\infty$;

(3) 对于任何常数$\alpha >0,U(\alpha\sigma)=\alpha^{\beta+o(1)}(1+o(1))U(\sigma),$称$U(\sigma)$为型函数.

定义2.2 设$U(\sigma),M_u(\sigma,F)$为引理2.1中的函数, 若$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log^{+}\log^{+} M_u(\sigma,F)-\log\sigma}{\log U(\sigma)}=\tau$, 则称$F(s)$关于$U(\sigma)$的下级为$\tau$; 若$\overline{\lim\limits_{\sigma\to\infty}}\frac{\log^{+}\log^{+}M_u(\sigma,F)-\log\sigma}{\log U(\sigma)}=\rho$, 则称$F(s)$关于$U(\sigma)$的级为$\rho$.

当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$\sigma>0$, $L(\sigma)$与$x$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

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

(2.6)

称$(2.6)$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. $\sigma_{k}$对$ {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$是$\pi(F)$的顶点或在其边上, 则$P_n$与$P^{c}_n$重合.

定义

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

则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2.2 ^[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2.3 ^[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2.4 ^{[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$\sigma$有

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

其中$C$是常数.

引理2.5 设L-S变换$(2.1)$满足$(2.2),(2.3)$和$(2.4)$式, 则有

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

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

证(1) 由引理$2.4$并注意到型函数$U(\sigma)$的性质易得

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

下证

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

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

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

不妨设$\sigma_{1}>0,$于是

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

由$\varepsilon$的任意性知

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

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$\log U(\sigma)$, 整理可得

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

(1) 得证.

(2) 由引理$2.3$并注意到型函数$U(\sigma)$的性质易得

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

下证

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

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

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

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

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, 再同除以$\log U(\sigma)$, 整理可得

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

至此, 引理$2.5$得证.

定理3.1 设L-S变换$(2.1)$满足条件(2.2), (2.3), (2.4) 和(2.5) 式, 则

$\begin{eqnarray*}\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log^{+}\log^{+}M_u(\sigma, F)-\log\sigma}{\log U(\sigma)}=\tau & \Leftrightarrow & \max\limits_{n_k}\mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{{n}_{k-1}}} {{\log U(\frac{-\log A_{{n_k}}^*}{\lambda_{{n}_k}})}}=\tau\\ & \Leftrightarrow & \mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{{N}_{k-1}}}{{\log U(\frac{-\log A_{{N_k}}^*}{\lambda_{{N}_k}})}}=\tau, \end{eqnarray*}$

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.

证 由于$M_u(\sigma,F)$满足引理$2.1$中的$(2.5)$式, 则存在连续可微函数$U(\sigma)$满足引理$2.1$中的结论.任取上升的正整数序列${n_k}$, 设$\mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{{n}_{k-1}}}{{\log U(\frac{-\log A_{{n_k}}^*}{\lambda_{{n}_k}})}}=\tau_{1}>0.$由条件, $\forall\varepsilon\in(0,\tau_{1})$, 当$k$充分大时, $\frac{\log \lambda_{{n}_{k-1}}}{{\log U(\frac{-\log A_{{n_k}}^*}{\lambda_{{n}_k}})}}>\tau_{1}-\varepsilon$, 即$\lambda_{n_{k-1}}^{\frac{1}{\tau_{1}-\varepsilon}}>U(\frac{-\log A_{{n_k}}^*}{\lambda_{{n}_k}}).$

设$\nu=U(u)$与$u=\phi(\nu)$是两个互为反函数的函数, 故当$n>N$时, $\phi(\lambda_{n_{k-1}}^{\frac{1}{\tau_{1}-\varepsilon}})>\frac{-\log A_{{n_k}}^*}{\lambda_{{n}_k}}.$取$\sigma_{k}$满足:$\sigma_{k}=2\phi(\lambda_{n_{k}}^{\frac{1}{\tau_{1}-\varepsilon}}),$即$\lambda_{n_k}=U^{\tau_{1}-\varepsilon}(\frac{\sigma_k}{2})$.设$0<\sigma_{k-1}<\sigma<\sigma_k$时, 有

$$\log^{+}\mu(\sigma,F)\geq\log^{+}\mu(\sigma_{k-1},F)\geq-\lambda_{n_k}\phi(\lambda_{n_{k-1}}^{\frac{1}{\tau_{1}-\varepsilon}})+\lambda_{n_k}\sigma_{k-1}=\lambda_{n_k}\phi(\lambda_{n_{k-1}}^{\frac{1}{\tau_{1}-\varepsilon}})= U^{\tau_{1}-\varepsilon}(\frac{\sigma_k}{2}){\frac{\sigma_{k-1}}{2}}.$$

由$(2.3)$式中的条件$\overline{\lim\limits_{n\to\infty}}{(\lambda_{n+1}-\lambda_n)}<+\infty$知$\lim\limits_{n\rightarrow\infty}\frac {\log \lambda_{n+1}}{\log\lambda_n}=1,$从而$\lim\limits_{k\rightarrow\infty}\frac {\log \lambda_{n_{k+1}}}{\log\lambda_{n_k}}=1.$又因为

$$\lim\limits_{k\rightarrow\infty}\frac {\log \lambda_{n_{k+1}}}{\log\lambda_{n_k}}=\lim\limits_{k\rightarrow\infty}\frac{\log U^{\tau_{1}-\varepsilon}(\frac{\sigma_{k+1}}{2})}{\log U^{\tau_{1}-\varepsilon}(\frac{\sigma_k}{2})}=\lim\limits_{k\rightarrow\infty}\frac{\log U(\frac{\sigma_{k+1}}{2})}{\log U(\frac{\sigma_k}{2})},$$

故有$\lim\limits_{k\rightarrow\infty}\frac{\log U(\frac{\sigma_{k}}{2})}{\log U(\frac{\sigma_{k+1}}{2})}=1$.注意到型函数$U(\sigma)=\sigma^{\beta(\sigma)},\beta(\sigma)\rightarrow\beta (\sigma\rightarrow\infty)$以及引理$2.1$中的$(3)$有

$$\begin{aligned} \mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log^{+}\log^{+}\mu(\sigma,F)-\log\sigma}{\log U(\sigma)} &\geq\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log^{+}U^{\tau_{1}-\varepsilon}(\frac{\sigma_k}{2}){\frac{\sigma_{k-1}}{2}}-\log\sigma}{\log U(\sigma)}\\ &\geq\mathop{\underline {\lim}}\limits_{k\to \infty}\frac{\log^{+}U^{\tau_{1}-\varepsilon}(\frac{\sigma_k}{2}){\frac{\sigma_{k-1}}{2}}}{\log U(\sigma_{k})}-\frac{1}{\beta}\\ &\geq(\tau_{1}-\varepsilon)\mathop{\underline {\lim}}\limits_{k\to \infty}\frac{\log^{+}U(\frac{\sigma_k}{2})}{\log U(\sigma_{k})}+\mathop{\underline {\lim}}\limits_{k\to \infty}\frac{\log\frac{\sigma_{k-1}}{2}}{\log U(\sigma_{k})}-\frac{1}{\beta}\\ &=\tau_{1}-\varepsilon+\mathop{\underline {\lim}}\limits_{k\to \infty}\frac{\log\frac{\sigma_{k-1}}{2}}{\log U(\frac{\sigma_{k-1}}{2})} \cdot\frac{\log^{+}U(\frac{\sigma_{k-1}}{2})}{\log U(\frac{\sigma_{k}}{2})}\cdot\frac{\log^{+}U(\frac{\sigma_k}{2})}{\log U(\sigma_{k})}-\frac{1}{\beta}\\ &=\tau_{1}-\varepsilon. \end{aligned} $$

由$\varepsilon$的任意性知$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log^{+}\log^{+}\mu(\sigma,F)-\log\sigma}{\log U(\sigma)}\geq\tau_1.$结合引理$2.5$知

$\begin{array}{rl} \mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log^{+}\log^{+}M_u(\sigma,F)-\log\sigma}{\log U(\sigma)}\geq\max\limits_{n_k}\mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{{n}_{k-1}}}{{\log U(\frac{-\log A_{{n_k}}^*}{\lambda_{{n}_k}})}}\geq\mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{{N}_{k-1}}}{{\log U(\frac{-\log A_{{N_k}}^*}{\lambda_{{N}_k}})}}. \end{array}$

(3.1)

另一方面, 设$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log^{+}\log^{+}M_u(\sigma,F)-\log\sigma}{\log U(\sigma)}=\tau.$

由引理$2.5$知$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log^{+}N(\sigma,F)}{\log U(\sigma)}=\tau.$设$\{\lambda_{N_k}\}=\{N(\sigma,F),\sigma>0\}$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},\sigma_{k+1})$, 此时$N(\sigma,F)=\lambda_{N_{k-1}}$, 故有$\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log^{+}N(\sigma,F)}{\log U(\sigma)}=\mathop{\underline {\lim}}\limits_{k\to \infty}\frac{\log\lambda_{N_{k-1}}}{\log U(\sigma_k)}=\tau.$因此, $\forall\varepsilon\in(0,\tau),$存在正整数$p$, 使当$\sigma_k\geq\sigma_p$时, 有

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

由引理$2.2$得$\lambda_{N_{k-1}}\geq U^{\tau-\varepsilon}(\sigma_k)>U^{\tau-\varepsilon}(\frac{-\log A_{{N_k}}^*}{\lambda_{{N}_k}})$.所以$\mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{{N}_{k-1}}}{{\log U(\frac{-\log A_{{N_k}}^*}{\lambda_{{N}_k}})}}\geq\tau$, 即

$\begin{array}{rl} \mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{{N}_{k-1}}}{{\log U(\frac{-\log A_{{N_k}}^*}{\lambda_{{N}_k}})}}\geq\mathop{\underline {\lim}}\limits_{\sigma\to \infty}\frac{\log^{+}\log^{+}M_u(\sigma,F)-\log\sigma}{\log U(\sigma)}. \end{array}$

(3.2)

结合$(3.1),(3.2)$两式知定理$3.1$证毕.

定理3.2 设L-S变换$(2.1)$满足条件(2.2), (2.3), (2.4) 和(2.5) 式, 则$A_1=A_2=A_3$, 其中

$$A_1=\overline{\lim\limits_{\sigma\rightarrow\infty}}\frac{\log^{+}\log^{+}M_u(\sigma,F)-\log\sigma}{\log U(\sigma)},A_2=\overline{\lim\limits_{n\rightarrow\infty}}\frac{\log \lambda_n}{{\log U(\frac{-\log A_{{n}}^*}{\lambda_{n}})}},A_3=\overline{\lim\limits_{k\rightarrow\infty}}\frac{\log \lambda_{{N}_{k}}}{{\log U(\frac{-\log A_{{N_k}}^*}{\lambda_{{N}_k}})}}.$$

证 1) 首先$A_3\leq A_2$是显然的.

2) $A_2\leq A_1$, 用反证法, 假设$A_1<A_2$, 因为

$$\overline{\lim\limits_{\sigma\rightarrow\infty}}\frac{\log^{+}\log^{+}M_u(\sigma,F)-\log\sigma}{\log U(\sigma)}=A_1,$$

于是可选择$\varepsilon>0$, 使$A_1+2\varepsilon<A_2 . \exists\sigma_0$, 当$\sigma>\sigma_0$时有$\log^{+}M_u(\sigma,F)<U^{A_{1}+\varepsilon}(\sigma)\sigma$, 从而对于充分大的$\sigma$, $\forall n$有

$\begin{array}{rl} \log A^*_n+\lambda_n\sigma< U^{A_{1}+\varepsilon}(\sigma)\sigma,\end{array}$

(3.3)

对于固定的充分大的$n$, 取$\sigma$满足$ U^{A_{1}+\varepsilon}(\sigma)=\frac{1}{2}\lambda_n$, 由$(3.3)$式知$\log A^*_n+\lambda_n\sigma<\frac{1}{2}\lambda_n\sigma$, 故$U(\frac{1}{2}\sigma)<U(-\frac{\log A^*_n}{\lambda_n})$.由引理$2.1$中的$(3)$知$U(\sigma)<CU(-\frac{\log A^*_n}{\lambda_n})$从而

$$\lambda_n=2U^{A_{1}+\varepsilon}(\sigma)<CU^{A_1+\varepsilon}(-\frac{\log A^*_n}{\lambda_n}),$$

其中$C$是常数, 每次出现不一定相同.所以

$$\overline{\lim\limits_{n\rightarrow\infty}}\frac{\log \lambda_n}{{\log U(\frac{-\log A_{{n}}^*}{\lambda_{n}})}} \leq A_1+\varepsilon<A_2-\varepsilon,$$

这与已知$A_2=\overline{\lim\limits_{n\rightarrow\infty}}\frac{\log \lambda_n}{{\log U(\frac{-\log A_{{n}}^*}{\lambda_{n}})}}$矛盾!因此$A_2\leq A_1$.

3) 最后证$A_1\leq A_3$.由于$\overline{\lim\limits_{\sigma\rightarrow\infty}}\frac{\log \lambda_{{N}_{k}}}{{\log U(\frac{-\log A_{{N_k}}^*}{\lambda_{{N}_k}})}}=A_3$, 则$\forall\varepsilon\in(0,A_3)$, 对充分大的$k$有$\lambda_{N_k}<U^{A_{3}+\varepsilon}(\frac{-\log A_{{N_k}}^*}{\lambda_{N_k}})$.设$\nu=U(u)$与$u=\phi(\nu)$是两个互为反函数的函数, 则$\phi(\lambda_{N_{k}}^{\frac{1}{A_{3}+\varepsilon}})<\frac{-\log A_{{N_k}}^*}{\lambda_{N_k}}$.故有

$\begin{array}{rl} \log A_{{N_k}}^*e^{\lambda_{N_k}\sigma}\leq\lambda_{N_k}(\sigma-\phi(\lambda_{N_{k}}^{\frac{1}{A_{3}+\varepsilon}})). \end{array}$

(3.4)

对任意固定的充分大的$\sigma>0$, 取$K=K(\sigma)$使

$\begin{array}{rl} \label{11}\frac{K(\sigma)}{\sigma}=U^{A_{3}+\varepsilon}(\sigma+\frac{\sigma}{\log(\frac{-\log A_{{N_k}}^*}{\lambda_{N_k}})}),\phi[({\frac{K(\sigma)}{\sigma}})^{\frac{1}{A_{3}+\varepsilon}}]=\sigma+\frac{\sigma}{\log(\frac{-\log A_{{N_k}}^*}{\lambda_{N_k}})}. \end{array}$

(3.5)

(i)当$\lambda_{N_k}\sigma\leq K$时, 由$(3.4),(3.5)$两式及型函数的定义得

$$\log A_{{N_k}}^*e^{\lambda_{N_k}\sigma}\leq\lambda_{N_k}\sigma\leq K=\sigma U^{A_{3}+\varepsilon}(\sigma+\frac{\sigma}{\log(\frac{-\log A_{{N_k}}^*}{\lambda_{N_k}})})\leq\sigma U^{A_{3}+\varepsilon}(\sigma+\sigma)\leq \sigma U^{A_{3}+2\varepsilon}(\sigma).$$

(ii)当$\lambda_{N_k}\sigma>K$时, 结合$(3.4),(3.5)$两式得

$$\log A_{{N_k}}^*e^{\lambda_{N_k}\sigma}\leq\lambda_{N_k}(\sigma-\phi(\lambda_{N_{k}}^{\frac{1}{A_{3}+\varepsilon}}))\leq\lambda_{N_k}(\sigma-\phi(({\frac{K(\sigma)}{\sigma}})^{\frac{1}{A_{3}+\varepsilon}}))<0.$$

因此由(i), (ii)知, 对充分大的$k$有$\log\mu(\sigma,F)\leq \sigma U^{A_{3}+2\varepsilon}(\sigma)$, 由$\varepsilon$的任意性, 结合引理$2.5$有$A_1\leq A_3$.综上所述知定理$3.2$成立.

定理3.3 设L-S变换$(2.1)$满足条件$(2.2),(2.3),(2.4)$和$(2.5)$式, 则下列条件等价:

(1) L-S变换$(2.1)$关于$U(\sigma)$的正规增长级为$\rho$, 即$\lim\limits_{\sigma\rightarrow+\infty}\frac{\log^{+}\log^{+} M_u(\sigma,F)-\log \sigma}{\log U(\sigma)}=\rho .$

(2) L-S变换$(2.1)$满足

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

(ii)存在上升的正整数列$\{n_k\}$, 使

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

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

证 先证$(1)$与$(2)$等价

设$(2)$成立, 则由(ii)

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

从而$\mathop{\underline {\lim}}\limits_{k\rightarrow\infty}\frac{\log \lambda_{n_{k-1}}} {\log U(\frac{-\log A_{{n_{k}}}^*}{\lambda_{{n}_{k}}})}=\rho$, 故$ \max\limits_{n_k}\mathop {\underline {\lim}}\limits_{k\to\infty}\frac{\log \lambda_{{n}_{k-1}}} {{\log U(\frac{-\log A_{{n_k}}^*}{\lambda_{{n}_k}})}}\geq\rho$, 由定理$3.1$知从而$F(s)$关于$U(r)$的下级$\geq\rho$, 由(i)及定理$3.2$知$F(s)$的关于$U(r)$的级$=\rho$, 结合正规增长级的定义就得到$(1)$.反之, 若$(1)$成立, 则说明$F(s)$关于$U(r)$的下级与级相等.由定理$3.2$知(i)成立, 对最大项指标序列$\{\lambda_{N_k}\}$, 有

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

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, 即

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

由最大项指标序列的单调递增性和下极限的保不等式性知

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

故(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.