本文中,亚纯函数均指在复平面上的亚纯函数,并采用Nevanlinna理论中的标准记号和基本结果,见参考文献[1-2].对于非常数的亚纯函数$f(z)$, $N(r, f)$与$\overline{N} (r, f)$分别表示其极点的计数函数与精简计数函数, $\sigma(f)$和$\mu(f)$分别为其级和下级, $T(r, f)$为其特征函数.此外, $S(r, f)$表示满足$S(r, f)=o(T(r, f))(r \rightarrow \infty, r \notin E)$的量,其中例外集$E\subset(0, \infty)$的对数测度有穷,即$\int_E\frac{1}{t} \, {\rm{d}}t ＜ \infty$.

我们分别用$\lambda(\frac{1}{f})$与$\lambda(f)$表示$f$的极点收敛指数和零点收敛指数,定义如下

$\lambda(\frac{1}{f})=\overline{\mathop{\lim}\limits_{r \rightarrow \infty}}\frac{\log^{+}N(r, f)}{\log r}; \quad \lambda(f)=\overline{\mathop{\lim}\limits_{r \rightarrow \infty}}\frac{\log^{+}N(r, \frac{1}{f})}{\log r}.$

为了细致地研究计数函数与特征函数的关系,人们还定义了$f$在$a$处的亏量$\delta(a, f)$和精简亏量$\Theta(a, f)$:

$\delta(a, f)=1-\overline{\mathop{\lim}\limits_{r \rightarrow \infty}}\frac{N(r, \frac{1}{f-a})}{T(r, f)}; \quad \Theta(a, f)=1-\overline{\mathop{\lim}\limits_{r \rightarrow \infty}}\frac{\overline{N}(r, \frac{1}{f-a})}{T(r, f)}$

特别地, $f$在$\infty$处的精简亏量为$\Theta(\infty , f)=1-\overline{\mathop{\lim}\limits_{r \rightarrow \infty}}\frac{\overline{N}(r, f)}{T(r, f)}$.

本文中,我们将利用Nevanlinna理论讨论以下非线性方程

(1.1) $a_n(z)f(z+n)^{j_n}+\cdots+a_1(z)f(z+1)^{j_1}+a_0(z)f(z)^{j_0}=b(z), $

(1.2) $a_n(z)f(q^nz)^{j_n}+\cdots+a_1(z)f(qz)^{j_1}+a_0(z)f(z)^{j_0}=b(z), $

其中${a_i(z)(i=0, 1, \cdots, n)}$与${b(z)}$均为非零有理函数, ${j_i(i=0, 1, \cdots, n)}$为正整数, $q$为非零复数.当$j_i=1(i=0, 1, \cdots, n)$时,方程(1.1)和(1.2)分别退化为线性差分方程和线性$q$ -差分方程.线性差分方程超越亚纯解的存在性及其解析性质受到很大的关注,特别超越亚纯解的增长性已被学者们深入研究,其中蒋翼曼和冯绍继^[3]证明了

定理A  假设$ P_i(z)(i=0, 1, \cdots, n) $为多项式,并存在整数$l \in [0, n] $,使得

(1.3) ${\rm deg}(P_l)>\max\limits_{0\le j \le n , j \ne l}{{\rm deg}(P_j)}.$

如果$f$为方程

(1.4) $P_n(z)f(z+n)+\cdots+P_1(z)f(z+1)+P_0(z)f(z)=0$

的非零亚纯解,则我们有$\sigma(f)\ge 1$.

随后,陈宗煊在文献[4]中弱化了定理A中的条件(1.3),对(1.4)式及其非齐次线性方程的亚纯解进行了讨论,得到

定理B  假设$P_i(z)\ (i=0, 1, \cdots, n)$和$F(z)$均为多项式,满足$FP_nP_0 t\equiv 0$和

(1.5) ${\rm deg}(P_n+\cdots+P_0)=\max\{{\rm deg}(P_j):j=0, 1, \cdots, n\} \ge 1.$

如果$f$为以下方程的有穷级超越亚纯解

(1.6) $P_n(z)f(z+n)+\cdots+P_1(z)f(z+1)+P_0(z)f(z)=F(z).$

则我们有$\sigma(f)\ge 1$且$\sigma(f)=\lambda(f)$.

定理C  假设多项式$P_i(z)(i=0, 1, \cdots, n)$满足$P_nP_0 \ne 0$和(1.5)式.对于方程(1.4)的非零有穷级亚纯解$f(z)$,必然有$\sigma(f)\ge 1$.对于任意的非零复数$a$, $f$取$a$无穷多次且$\lambda(f-a)=\sigma(f)$.

如果方程(1.6)中$FP_nP_0 t\equiv 0$,当亚纯解$f(z)$具有无穷多个极点时,陈宗煊在文章^[4]中也得到$\lambda(\frac{1}{f})\ge 1$的结论.

针对形式更一般的方程(1.1)和(1.2),我们讨论其亚纯解的增长性,零点和极点分布,并得到以下两个定理.

定理1.1  假设$a_i(z)(i=0, 1, \cdots, n)$与$b(z)$均为非零有理函数, $j_i(i=0, 1, \cdots, n)$为正整数,如果$f(z)$为方程(1.1)的超越亚纯解,满足$\sigma_2(f)=\varsigma < 1$和$\Theta(\infty, f)=1$,则我们有$\delta(0, f) \le \frac{n}{n+1}$.这里$\sigma_2(f)$为$f$的超级,它的定义如下

$\sigma_2(f)=\overline{\mathop{\lim}\limits_{r \rightarrow \infty }}\frac{\log^{+}\log^{+}T(r, f)}{\log r}.$

定理1.2  假设$q$为非零复常数, $a_i(z), j_i(i=0, 1, \cdots, n)$与$b(z)$均满足定理$1.1$的条件.如果$f$为方程(1.2)的超越亚纯解,满足$\sigma_2(f)=\varsigma < 1$和$\Theta(\infty, f)=1$,则我们有$\delta(0, f) \le \frac{n}{n+1}$.

特别地,由定理1.1和定理1.2的结论,我们可继续得到方程(1.1)和(1.2)的这些亚纯解满足$\mu(f) \le \lambda(f) \le \sigma(f).$此外,我们利用陈宗煊在文献[4]中所用的思路,还得到以下结论:

定理1.3  假设$a_i(z), j_i(i=0, 1, \cdots, n)$与$b(z)$均满足定理$1.1$中的条件,如果$f(z)$为方程(1.1)的超越亚纯解,若$f$有无穷多个极点,则$\lambda(\frac{1}{f} )\ge 1$.

为了完成定理的证明,我们需要下面的引理.

引理2.1^[5]  假设$f(z)$为亚纯函数, $c$为非零复常数,当$r \rightarrow \infty$时,成立

$(1+o(1))T(r-|c|, f(z)) \le T(r, f(z+c)) \le (1+o(1))T(r+|c|, f(z)).$

引理2.2^[6]  假设$f(z)$为非常数亚纯函数,且$\varsigma:=\sigma_2(f) <1$,对于任意给定的正数$\varepsilon$和复数$c$,存在对数测度有限的集合$F\subset(0, \infty)$,使得

$m\left(r, \frac{f(z+c)}{f(z)}\right)=o\left(\frac{T(r, f)}{r^{1-\varsigma-\varepsilon}}\right), r\rightarrow \infty, r \notin F.$

引理2.3^[6]  假设$T(r):{\Bbb R}^{+}\rightarrow {\Bbb R}^{+}$为单调递增的连续函数, $s$为正实数.若

$\varsigma=\overline{\mathop{\lim}\limits_{r \rightarrow \infty }}\frac{\log^{+}\log^{+}T(r)}{\log r}＜1, $

则对于满足$0<\delta<1-\varsigma$的任意数$\delta$,均有

$T(r+s)=T(r)+o\left(\frac{T(r)}{r^{\delta}}\right), \quad r \rightarrow \infty, r \notin E, $

其中, $E \subset (0, \infty)$为某个对数测度有限的集合.

引理2.4^{[2,定理1.55]}  假设$g_j(z)\ (j=1, 2, \cdots, p)$为满足$\Theta(\infty, g_j)=1 $$ (j=1, 2, \cdots, p)$的非常数亚纯函数, $a_j\ (j=0, 1, 2, \cdots, p)$均为非零复数.若$\Sigma^p_{j=1} a_jg_j(z)=a_0$,则

$\Sigma^p_{j=1} \delta(0, g_j) \le p-1.$

引理2.5  假设$f(z)$为非常数亚纯函数, $c$为非零复常数.如果$\sigma_2(f)=\varsigma <1$,则存在对数测度有穷的集合$E \subset(0, \infty)$,使得当$r \rightarrow \infty $且$r \notin E$时,我们有

(i)   $T(r, f(z+c))=T(r, f(z))+S(r, f);$

(ii)   $N(r, f(z+c))=N(r, f(z))+S(r, f);$

(iii)   $\overline{N}(r, f(z+c))=\overline{N}(r, f(z))+S(r, f).$

证  不难看出, $T(r, f(z))$为关于$r$单调递增的连续函数.由引理2.3,我们可知:对于小于$1-\varsigma$的正数$\delta$,存在对数测度有穷的集合$E_1 \subset (0, \infty)$,使得

(2.1) $\begin{eqnarray}T(r\pm |c|, f(z))&=&T(r, f(z))+o\left(\frac{T(r, f(z))}{r^{\delta}}\right)\\&=&T(r, f(z))+S(r, f(z)), \quad r \rightarrow \infty , r \notin E_1. \end{eqnarray}$

另一方面,根据引理2.1, $r \rightarrow \infty $时,我们有

(2.2) $(1+o(1))T(r-|c|, f(z)) \le T(r, f(z+c)) \le (1+o(1))T(r+|c|, f(z)). $

结合(2.1)和(2.2)式易知

(2.3) $T(r, f(z+c))=T(r, f(z))+S(r, f), \quad r\rightarrow \infty, r \notin E_1.$

由于$\varsigma<1$,根据引理2.2,存在对数测度有穷的集合$E_2 \subset (0, \infty)$,使得

$m\left(r, \frac{f(z+c)}{f(z)}\right)=o\left(\frac{T(r, f)}{r^{1-\varsigma-\varepsilon}}\right), \quad r\rightarrow \infty, r \notin E_2;$

$m\left(r, \frac{f(z)}{f(z+c)}\right)=o\left(\frac{T(r, f(z+c))}{r^{1-\varsigma-\varepsilon}}\right), \quad r\rightarrow \infty, r \notin E_2.$

适当选取$\varepsilon >0$,满足$1-\varsigma-\varepsilon>0$时,利用(2.3)式可知,当$r\rightarrow \infty$且$ r \notin E_1 \cup E_2$时

(2.4) $m\left(r, \frac{f(z+c)}{f(z)}\right)=S(r, f), \quad m\left(r, \frac{f(z)}{f(z+c)}\right)=S(r, f). $

我们注意到

$\begin{eqnarray*} |m(r, f(z+c))-m(r, f(z))|&=&\frac{1}{2\pi}\left| \int^{2\pi}_0(\log^{+} \mid f(r{\rm e}^{{\rm i}\theta}+c)\mid -\log^{+} \mid f(r{\rm e}^{{\rm i}\theta})\mid)\, {\rm{d}}\theta \right| \\ &\le& \frac{1}{2\pi} \int^{2\pi}_0\left| \log^{+} \mid f(r{\rm e}^{{\rm i}\theta}+c)\mid -\log^{+} \mid f(r{\rm e}^{{\rm i}\theta})\mid \right| \, {\rm{d}}\theta\\ &\le &\frac{1}{2\pi}\int^{2\pi}_0 \log^{+} \left| \frac{f(r{\rm e}^{{\rm i}\theta}+c)}{f(r{\rm e}^{{\rm i}\theta})}\right| \, {\rm{d}}\theta + \frac{1}{2\pi}\int^{2\pi}_0 \log^{+} \left| \frac{f(r{\rm e}^{{\rm i}\theta})}{f(r{\rm e}^{{\rm i}\theta}+c)}\right|\, {\rm{d}}\theta\\ &=&m\left(r, \frac{f(z+c)}{f(z)}\right)+m\left(r, \frac{f(z)}{f(z+c)}\right).\end{eqnarray*}$

将(2.4)式代入上面的不等式中,我们推出

(2.5) $m(r, f(z+c))=m(r, f(z))+S(r, f), \quad r\rightarrow \infty, r \notin E_1 \cup E_2.$

记$E_0=E_1 \cup E_2$,显然$E_0$也为对数测度有穷的集合.由于$T(r, f)=m(r, f)+N(r, f)$,则利用(2.3)式与(2.5)式,我们可知

(2.6) $N(r, f(z+c))=N(r, f(z))+S(r, f), \quad r \rightarrow \infty , r \notin E_0.$

对于亚纯函数$f(z)$,我们知道

(2.7) $T(r, f')＜2T(r, f)+S(r, f).$

从而$r$充分大时,自然存在$T(r, f')<3T(r, f).$这意味着$\sigma_2(f')=\varsigma<1.$根据(2.6)式和(2.7)式,存在对数测度有限的集合$E_3 \subset (0, \infty)$,使得

(2.8) $N(r, f'(z+c))=N(r, f'(z))+S(r, f')=N(r, f'(z))+S(r, f) , \quad r \rightarrow \infty , r \notin E_3.$

注意到

(2.9) $N(r, f')=N(r, f)+\overline{N}(r, f).$

结合(2.6), (2.8)和(2.9)式可得

$\overline{N}(r, f(z+c))=\overline{N}(r, f)+S(r, f), \quad r \rightarrow \infty , r \notin E_0 \cup E_3.$

综上所述,引理2.5得证.

假设$f(z)$为方程(1.1)的超越亚纯解,满足$\sigma_2(f)=\varsigma<1$和$\Theta(\infty, f)=1.$令$c_i(z)=\frac{a_i(z)}{b(z)}$$(i=0, 1, \cdots, n)$,则我们对方程两边同时除以$b(z)$得到

(3.1) $c_n(z)f(z+n)^{j_n}+c_{n-1}(z)f(z+n-1)^{j_{n-1}}+\cdots+c_1(z)f(z+1)^{j_1}+c_0(z)f(z)^{j_0}=1.$

由于$a_i(z)(i=0, 1, 2, \cdots, n)$与$b(z)$为有理函数,从而

$N\left(r, c_i\right)=O(\log r), \quad N\left(r, \frac{1}{c_i}\right)=O(\log r) , \quad (i=0, 1, 2, \cdots, n).$

因为$f(z)$超越,显然

(3.2) $ N\left(r, c_i\right)=S(r, f(z)), \quad N\left(r, \frac{1}{c_i}\right)=S(r, f(z)), \quad (i=0, 1, 2, \cdots, n). $

利用计数函数的性质,我们不难看出

$N\left(r, \frac{1}{c_i(z)f(z+i)^{j_i}}\right) \le N\left(r, \frac{1}{c_i(z)}\right)+N\left(r, \frac{1}{f(z+i)^{j_i}}\right), \quad (i=0, 1, 2, \cdots, n);$

$N\left(r, \frac{1}{f(z+i)^{j_i}}\right) \le N\left(r, \frac{1}{c_i(z)f(z+i)^{j_i}}\right)+N\left(r, c_i(z)\right), \quad (i=0, 1, 2, \cdots, n).$

由这两个不等式,并利用(3.2)式和$N\left(r, \frac{1}{f(z+i)^{j_i}}\right)=j_iN\left(r, \frac{1}{f(z+i)}\right)$,可知

(3.3) $N\left(r, \frac{1}{c_i(z)f(z+i)^{j_i}}\right)=S(r, f(z))+j_iN\left(r, \frac{1}{f(z+i)}\right), \quad (i=0, 1, 2, \cdots, n).$

类似地,我们也可得到

$T\left(r, c_i(z)f(z+i)^{j_i}\right)=S(r, f(z))+j_iT(r, f(z+i)), \quad (i=0, 1, 2, \cdots, n);$

$\overline{N}\left(r, c_i(z)f(z+i)^{j_i}\right)=S(r, f(z))+\overline{N}(r, f(z+i)), \quad (i=0, 1, 2, \cdots, n).$

利用第一基本定理可知$\frac{1}{f}$的超级也为$\varsigma$,故对$f$和$\frac{1}{f}$都使用引理2.5.则存在$E\subset(0, \infty)$为对数测度有限集,当$r\rightarrow \infty, r \notin E$时,有

(3.4) $T(r, f(z+i))=T(r, f(z))+S(r, f(z)), \quad (i=0, 1, 2, \cdots, n);$

(3.5) $N\left(r, \frac{1}{f(z+i)}\right)=N\left(r, \frac{1}{f(z)}\right)+S(r, f(z)), \quad (i=0, 1, 2, \cdots, n). $

利用(3.3)-(3.5)式,我们得到

$\begin{eqnarray*} \overline{\mathop{\lim}\limits_{r \rightarrow \infty }}\frac{N\left(r, \frac{1}{c_i(z)f(z+i)^{j_i}}\right)}{T\left(r, c_i(z)f(z+i)^{j_i}\right)}&=&\overline{\mathop{\lim}\limits_{r \rightarrow \infty }}\frac{j_iN\left(r, \frac{1}{f(z)}\right)+S(r, f(z))}{j_iT(r, f(z))+S(r, f(z))}\\ &=&\overline{\mathop{\lim}\limits_{r \rightarrow \infty }}\frac{N\left(r, \frac{1}{f(z)}\right)}{T(r, f(z))}, \quad (i=0, 1, 2, \cdots, n). \end{eqnarray*}$

这意味着

(3.6) $\delta \left(0, c_i(z)f(z+i)^{j_i}\right)=\delta(0, f(z)), \quad (i=0, 1, 2, \cdots, n).$

类似地利用引理2.5,我们也可得到

$\overline{\mathop{\lim}\limits_{r \rightarrow \infty }}\frac{\overline{N}\left(r, c_i(z)f(z+i)^{j_i}\right)}{T\left(r, c_i(z)f(z+i)^{j_i}\right)}=\frac{1}{j_i} \, \overline{\mathop{\lim}\limits_{r \rightarrow \infty }}\frac{\overline{N}(r, f(z))}{T(r, f(z))}, \quad (i=0, 1, 2, \cdots, n).$

注意到条件$\Theta(\infty, f)=1$等价于$\overline{\mathop{\lim}\limits_{r \rightarrow \infty }}\frac{\overline{N}(r, f(z))}{T(r, f(z))}=0$,从而

(3.7) $\Theta\left(\infty, c_i(z)f(z+i)^{j_i}\right)=1-\frac{1}{j_i} \, \overline{\mathop{\lim}\limits_{r \rightarrow \infty }}\frac{\overline{N}(r, f(z))}{T(r, f(z))}=1$

对于$i=0, 1, 2\cdots, n$均成立.注意到(3.6)和(3.7)式,对方程(3.1)应用引理2.4,这必然推出

$\sum\limits_{i = 0}^n \delta\left(0, c_i(z)f(z+i)^{j_i}\right) \le n.$

将(3.6)式代入其中可知$(n+1)\delta(0, f) \le n$,即$\delta(0, f) \le \frac{n}{n+1}$,从而定理1.1得证.

假设$f(z)$为方程(1.2)的超越亚纯解,满足$\sigma_2(f)=\varsigma ＜1 $和$\Theta(\infty, f)=1$.类似于定理1.1的证明,我们将方程(1.2)改写为

(4.1) $c_n(z)f(q^nz)^{j_n}+c_{n-1}(z)f(q^{n-1}z)^{j_{n-1}}+\cdots+c_1(z)f(qz)^{j_1}+c_0(z)f(z)^{j_0}=1, $

并且对于$i=0, 1, 2, \cdots, n$,我们可以推出

(4.2) $T\left(r, c_i(z)f(q^iz)^{j_i}\right)=S(r, f(q^iz))+j_iT(r, f(q^iz)), $

(4.3) $N\left(r, \frac{1}{c_i(z)f(q^iz)^{j_i}}\right)=S(r, f(q^iz))+j_iN\left(r, \frac{1}{f(q^iz)}\right), $

(4.4) $\overline{N}\left(r, c_i(z)f(q^iz)^{j_i}\right)=S(r, f(q^iz))+\overline{N}(r, f(q^iz)), $

其中$S(r, f(q^iz))=o(T(r, f(q^iz)))$.通过几何观察,我们不难发现

(4.5) $T(r, f(qz))=T(r|q|, f(z)) , \, \overline{N}(r, f(qz))=\overline{N}(r|q|, f(z)), $

(4.6) $N\left(r, \frac{1}{f(qz)}\right)=N\left(r|q|, \frac{1}{f(z)}\right).$

结合(4.2), (4.3), (4.5)和(4.6)式,我们有

$\begin{eqnarray*} \overline{\mathop{\lim}\limits_{r \rightarrow \infty }}\frac{N\left(r, \frac{1}{c_i(z)f(q^iz)^{j_i}}\right)}{T\left(r, c_i(z)f(q^iz)^{j_i}\right)}&=&\overline{\mathop{\lim}\limits_{r \rightarrow \infty }}\frac{j_iN\left(|q|^ir, \frac{1}{f(z)}\right)+o(T(|q|^ir, f))}{(j_i+o(1))T(|q|^ir, f)}\\ &=&\overline{\mathop{\lim}\limits_{r \rightarrow \infty }}\frac{N\left(|q|^ir, \frac{1}{f(z)}\right)}{T(|q|^ir, f(z))}\\ &=&\overline{\mathop{\lim}\limits_{r \rightarrow \infty }}\frac{N\left(r, \frac{1}{f(z)}\right)}{T(r, f(z))}, \, (i=0, 1, 2, \cdots, n). \end{eqnarray*}$

这意味着

(4.7) $ \delta(0, c_i(z)f(q^iz)^{j_i})=\delta(0, f), \, (i=0, 1, 2, \cdots, n). $

同理可由条件$\Theta(\infty, f)=1$,我们得到

$\overline{\mathop{\lim}\limits_{r \rightarrow \infty }}\frac{\overline{N}\left(r, c_i(z)f(q^iz)^{j_i}\right)}{T\left(r, c_i(z)f(q^iz)^{j_i}\right)} =\overline{\mathop{\lim}\limits_{r \rightarrow \infty }}\frac{1}{j_i}\frac{\overline{N}(|q|^ir, f(z))}{T(|q|^ir, f(z))}=0, \, (i=0, 1, 2, \cdots, n).$

可推出

(4.8) $\Theta(\infty, c_i(z)f(q^iz)^{j_i})=1, \, (i=0, 1, 2, \cdots, n).$

注意到(4.7)和(4.8)式,对(4.1)式应用引理2.4,不难推出

$\sum\limits_{i = 0}^n \delta\left(0, c_i(z)f(q^iz)^{j_i}\right) \le n.$

再次利用(4.7)式,可知$(n+1)\delta(0, f) \le n$,故$\delta(0, f) \le \frac{n}{n+1}$,从而定理1.2得证.

假设$f(z)$为方程(1.1)的超越亚纯解,并且$f$有无穷多个的极点.因为$a_i (z)\, (i=0, 1, \cdots, n)$与$b(z)$均为非零有理函数,从而存在正数$M$,使得$ D=\{z\in {\Bbb C} : |{\rm Re}(z)| \le M, $$|{\rm Im}(z)| \le M\}$包含$a_i(z)$$(i=0, 1, \cdots, n)$和$b(z)$的所有零点和极点. $f$必存在不属于$D$的极点$z_0$.如若不然,则$f$的所有极点必都落在有界区域$D$中,从而必存在一列极点在$D$中有聚点,这是矛盾的.记$z_0$为$f$落在区域$D$外部的任一极点.我们由方程(1.1),可知

(5.1) $a_n(z_0)f(z_0+n)^{j_n}+\cdots+a_1(z_0)f(z_0+1)^{j_1}+a_0(z_0)f(z_0)^{j_0}=b(z_0). $

由此可知,必定存在$i_0 \in \{1, 2, \cdots, n\}$,使得$z_0+i_0$也是$f$的极点.

将$D$外的区域分成四个如下的子区域

$D_1=\{z \in C :{\rm Re}(z) >M \}, \quad D_2=\{z \in C :{\rm Re}(z) ＜-M \}, $

$D_3=\{z \in C :{\rm Im}(z) >M \}, \quad D_4=\{z \in C :{\rm Im}(z) ＜-M \}.$

由于$z_0 \notin D$,从而$z_0$必属于$D_i(i=1, 2, 3, 4)$之一.接下来我们分成三种情况讨论.

情况1  假设$z_0 \in D_1$.由$D_1$的定义,则$f$的极点$z_1:=z_0+i_0$也在区域$D_1$中.将(5.1)式中的$z_0$替换为$z_1$,同理可知:存在$i_1 \in \{1, 2, \cdots, n\}$,使得$z_2:=z_1+i_1$也是$f$的极点,并且$z_2$仍在区域$D_1$中.以此类推,我们可以得到$f$在区域$D_1$中的极点列$\{z_k\}(z_k=z_{k-1}+i_{k-1}).$记$r_k=|z_k|, k=0, 1, 2, \cdots$,我们不难看出$r_k$单调递增,且趋于无穷.

对于充分大的$r$,我们可知

(5.2) $N(2r, f)-N(r_0, f) \ge \int^{2r}_r \frac{n\left(t, f\right)}{t}\, {\rm{d}}t \ge n(r, f) \log 2 \ge \left[ \frac{r-r_0}{n}\right] \log 2. $

从而

$\lambda(\frac{1}{f}) \ge \overline{\mathop{\lim}\limits_{r \rightarrow \infty}}\frac{\log N(2r, f)}{\log 2r} \ge \overline{\mathop{\lim}\limits_{r \rightarrow \infty}} \frac{\log \left[ \frac{r-r_0}{n}\right]+\log \log 2}{\log r+\log 2}=1.$

情况2  假设$z_0 \in D_3 \cup D_4$.此时$z_1=z_0+i_0$仍然落在$D_3 \cup D_4$中.从而推出$D_3 \cup D_4$中存在一列极点$\{z_0, z_1, \cdots, z_k, \cdots\}.$类似于情况1中的推导,我们也可知$\lambda(\frac{1}{f}) \ge 1$.

情况3  假设$z_0 \in D_2$.显然Re$(z_0)<-M$.将方程(1.1)中的$z$替换为$z_0-n$得到

(5.3) $a_n(z_0-n)f(z_0)^{j_n}+\cdots+a_1(z_0-n)f(z_0-n+1)^{j_1}+a_0(z_0-n)f(z_0-n)^{j_0}=b(z_0-n).$

由于$z_0-n$仍在区域$D_2$中,根据(\ref{eq:e3})式可知:存在$i_0' \in \{1, 2, \cdots, n\}$,使得$z_1':=z_0-i_0'$也是$f$的极点,并且$z_1'$落在区域$D_2$中.重复上述步骤,我们得到$f$在区域$D_2$中的极点列$\{z_k'\}(z_k'=z_{k-1}'-i_{k-1}').$类似情况1中的推导,我们又可以推出$\lambda(\frac{1}{f} )\ge 1$.

综上所述,我们知道$\lambda(\frac{1}{f} )\ge 1$始终成立,从而定理1.3得证.