本文假定读者熟悉亚纯函数Nevanlinna值分布论中的常用记号和基本结果(参见文献[7, 17]).记${\Bbb C}$为复平面, $\widehat{{\Bbb C}}(={\Bbb C}\bigcup\{\infty\})$为扩充复平面, $\Omega(\subset{\Bbb C})$为一个角域.亚纯函数$f$的级定义为

$\rho = \rho (f) = \mathop {\lim \sup }\limits_{r \to \infty } \frac{{\log T(r,f)}}{{\log r}}.$

亚纯函数的幅角分布是复分析领域内一个有趣的研究课题, Borel方向在幅角分布论中扮演了一个重要的角色(参见文献[4, 12-14, 16, 18-19]). Valiron ^[17]证明了每个具有有限正级$\rho>0$的亚纯函数至少存在一条$\rho$级Borel方向.庄圻泰^[2-3]则研究了具有无穷级的亚纯函数的Borel方向的存在性问题.为叙述庄的结果, 首先给出如下定义.

定义1.1^[2]  假设$f$是一个无穷级的亚纯函数. $\rho(r)$是一个满足如下条件的实函数:

(i)对$r\geq r_0$, $\rho(r)$连续, 非降并且当$r\rightarrow\infty$时, $\rho(r)\rightarrow\infty$;

(ii)

$ \mathop {\lim }\limits_{r \to \infty } \frac{{\log U(R)}}{{\log U(r)}} = 1,R = r + \frac{r}{{\log U(r)}}, $

其中, $U(r)=r^{\rho(r)}$ $(r\geq r_0)$;

(iii)

$\mathop {\lim \sup }\limits_{r \to \infty } \frac{{\log T(r,f)}}{{\log U(r)}} = 1.$

则称$\rho(r)$为亚纯函数$f$的无穷级.这个定义由熊庆来在文献[2]中给出.

本文将研究亚纯函数的Borel方向与分担值的关系.为此, 先介绍一些基本的记号和结果.

对$a\in \widehat{{\Bbb C}}(={\Bbb C}\bigcup\{\infty\})$, 若$f(z)-a$和$g(z)-a$在$\Omega\subseteq {\Bbb C}$内具有计重数(不计重数)的相同的零点, 则称$f$和$g$在$\Omega\subseteq {\Bbb C}$内$CM$$(IM)$分担$a$.另外, 在$\Omega\subseteq {\Bbb C}$内用$f=a\rightleftharpoons g=a$表示$f$和$g$在$\Omega\subseteq {\Bbb C}$内$CM$分担$a$, 在$\Omega\subseteq {\Bbb C}$用$f=a\Longleftrightarrow g=a$表示表示$f$和$g$在$\Omega\subseteq {\Bbb C}$内$IM$分担$a$, 在$\Omega\subseteq {\Bbb C}$内用$f=a\Longrightarrow g=a$表示在$\Omega\subseteq {\Bbb C}$内由$f=a$可推出$g=a$.

Nevanlinna证明了如下著名的结果^[11].

定理1.1^[11]  若$f$和$g$是两个在$\Omega={\Bbb C}$内$IM$分担五个不同值$a_1, a_2, a_3, a_4, a_5$的非常值亚纯函数, 则$f(z)\equiv g(z)$.

定理1.2^[11]  若$f$和$g$是两个在${\Bbb C}$内$CM$分担四个不同值$a_1, a_2, a_3, a_4$的亚纯函数, 则$f$是$g$的一个Möbius变换, 其中有两个值$a_1$和$a_2$是例外值且交比$(a_1, a_2, a_3, a_4)=-1$.

之后, 有大量的涉及在全平面上分担值的亚纯函数的唯一性方面的研究工作^[20]. 2003年, 郑建华[22]首次研究了在${\Bbb C}$的某个角域内分担五个值的唯一性问题.之后, 郑^[23]又研究了在角域内分担四个值的唯一性问题.研究亚纯函数在角域内的分担值问题是一个有意义的课题^{[1, 9, 15, 21]}.曹廷彬和仪洪勋^[1], 张庆彩^[21], 徐洪焱和曹廷彬^[15]继续研究了分担五值和四值方面的工作; 林伟川、Mori和Tohge^[8]、林伟川、Mori和仪洪勋^[9]研究了亚纯函数和整函数在角域内分担集合的唯一性问题.

假设$f$为角域$\Omega(\alpha, \beta)=\{z: \alpha\leq\arg z\leq \beta\}$内的亚纯函数且$0<\beta-\alpha\leq 2\pi$.定义

$ A_{\alpha, \beta}(r, f)=\frac{\omega}{\pi}\int^r_1 \bigg(\frac{1}{t^\omega} -\frac{t^\omega}{r^{2\omega}}\bigg)\{\log^+|f(t{\rm e}^{{\rm i}\alpha})|+\log^+|f(t{\rm e}^{{\rm i}\beta})|\}\frac{{\rm d}t}{t}, $

$ {B_{\alpha ,\beta }}(r,f) = \frac{{2\omega }}{{\pi {r^\omega }}}\int_\alpha ^\beta {{{\log }^ + }} |f(r{{\rm{e}}^{{\rm{i}}\theta }})|\sin \omega (\theta - \alpha ){\rm{d}}\theta , $

$ C_{\alpha, \beta}(r, f) =2\sum_{1<|b_\mu|<r}\bigg(\frac{1}{|b_\mu|^\omega}-\frac{|b_\mu|^\omega}{r^{2\omega}}\bigg) \sin\omega(\theta_\mu-\alpha), $

$ S_{\alpha, \beta}(r, f)=A_{\alpha, \beta}(r, f)+B_{\alpha, \beta}(r, f)+C_{\alpha, \beta}(r, f), $

其中$\omega=\frac{\pi}{\beta-\alpha}$, $b_\mu=|b_\mu|{\rm e}^{{\rm i}\theta_\mu}\ (\mu=1, 2, \cdots)$是$f$在$\Omega(\alpha, \beta)$内计算重数的极点.称$S_{\alpha, \beta}(r, f)$为Nevanlinna角域特征函数, $C_{\alpha, \beta}(r, f)$为$f$在$\Omega(\alpha, \beta)$内极点的计数函数, $\overline{C}_{\alpha, \beta}(r, f)$是$C_{\alpha, \beta}(r, f)$的精简形式.

2012年, 龙见仁和伍鹏程^[10]研究了亚纯函数的Borel方向和唯一性的关系问题.为介绍他们的结果, 先介绍一些定义及记号.

设$\rho(r)$为亚纯函数$f$的一个无穷级, 用$M(\rho(r))$表示所有满足$\limsup\limits_{r\rightarrow\infty}\frac{\log T(r, g)}{\rho(r)\log r}\leq1$的亚纯函数$g$的集合, 即

$ M(\rho (r)): = \left\{ {g:\mathop {\lim \sup }\limits_{r \to \infty } \frac{{\log T(r,g)}}{{\rho (r)\log r}} \le 1} \right\}.$

令$\alpha<\beta, \beta-\alpha<2\pi, r>0$, $\Omega(\alpha, \beta, r):=\{z: \alpha\leq \arg z\leq \beta, 0<|z|\leq r\}$.无穷级为$\rho(r)$的亚纯函数$f$的Borel方向的定义为:

定义1.2^[2]  假设$f$为一个具有无穷级$\rho(r)$的亚纯函数.若对每个$\varepsilon(0<\varepsilon<\pi)$, 等式

$ \mathop {\lim \sup }\limits_{r \to \infty } \frac{{\log n(\Omega (\theta - \varepsilon ,\theta + \varepsilon ,r),f = a)}}{{\rho (r)\log r}} = 1 $

对任意的$a\in \widehat{{\Bbb C}}$都成立, 至多有两个例外, 其中$n(\Omega(\theta-\varepsilon, \theta+\varepsilon, r), f=a)$表示在角域内$\Omega(\theta-\varepsilon, \theta+\varepsilon)$内$f-a$的零点, 计重数.射线$\arg z=\theta$称为$f$的$\rho(r)$级Borel方向.

注1.1  庄圻泰^[2]证明了每个具有无穷级$\rho(r)$的亚纯函数$f$至少存在一条$\rho(r)$级的Borel方向.

定理1.3^{[10, 定理1.1]}  假设$f$是具有无穷级$\rho(r)$的亚纯函数, $g\in M(\rho(r))$, $\arg z=\theta$ $ (0\leq \theta<2\pi)$是$f$的一条$\rho(r)$级的Borel方向, $a_i\in \widehat{{\Bbb C}}\ (i=1, 2, 3, 4, 5)$是五个不同的复数.若$f$和$g$在角域$\Omega(\theta-\varepsilon, \theta+\varepsilon)\ (0<\varepsilon<\pi)$内$IM$分担$a_i\ (i=1, 2, 3, 4, 5)$, 则$f\equiv g$.

定理1.4^{[10, 定理1.2]}  假设$f$是具有无穷级$\rho(r)$的亚纯函数, $g\in M(\rho(r))$, $\arg z=\theta\ (0\leq \theta<2\pi)$是$f$的一条$\rho(r)$级的Borel方向, $a_i\in \widehat{{\Bbb C}}\ (i=1, 2, 3, 4)$是四个不同的复数.若$f$和$g$在角域$\Omega(\theta-\varepsilon, \theta+\varepsilon)\ (0<\varepsilon<\pi)$内$CM$分担$a_i\ (i=1, 2, 3, 4)$, 则$f$是$g$的一个Möbius变换.

本文继续研究亚纯函数的Borel方向与分担值的相关问题, 得到如下结果:

定理1.5  假设$f$是具有无穷级$\rho(r)$的超越亚纯函数, $g\in M(\rho(r))$, $\arg z=\theta\ (0\leq \theta<2\pi)$是$f$的一条$\rho(r)$级的Borel方向, 对每个$\varepsilon(0<\varepsilon<\pi)$, $\Omega:=\Omega(\theta-\varepsilon, \theta+\varepsilon)$.假设$f$和$g$在$\Omega$内$CM$分担两个不同的值$a_1, a_2$并且在$\Omega$内$f=a_3\Longrightarrow g=a_3$, $f=a_4\Longrightarrow g=a_4$, 则$f\equiv g$.

定理1.6假设$f$是具有无穷级$\rho(r)$的超越整函数, $g\in M(\rho(r))$, $\arg z=\theta\ (0\leq \theta<2\pi)$是$f$的一条$\rho(r)$级的Borel方向, 对每个$\varepsilon(0<\varepsilon<\pi)$, $\Omega:=\Omega(\theta-\varepsilon, \theta+\varepsilon)$.假定$f$和$g$在$\Omega$内$IM$分担两个不同的值$a_1, a_2$, 在$\Omega$内$f=a_3\Longrightarrow g=a_3$以及$g=a_4\Longrightarrow f=a_4$.则或者$f\equiv g$或者

$ f\equiv \frac{a_3g-a_1a_2}{g-a_4} $

且$a_1+a_2=a_3+a_4$, $a_3, a_4$分别是$f$和$g$在$\Omega$内的例外值.

为证明我们的结果, 需要以下引理.

引理2.1^{[6, 16]}  假设$f$是$\Omega(\alpha, \beta)$上的一个非常数的亚纯函数.则对于任意的复数$a$, 有

$ S_{\alpha, \beta}\left(r, \frac{1}{f-a}\right)=S_{\alpha, \beta}(r, f)+\varepsilon(r, a), $

其中当$r\rightarrow\infty$时, $\varepsilon(r, a)=O(1)$.

引理2.2^{[5, 23]}  假设$f$是角域$\Omega(\alpha, \beta)\ (0<\beta-\alpha\leq 2\pi)$上的一个非常数的亚纯函数, 则对任意$q$个不同的复数$a_j\in \widehat{{\Bbb C}}\ (1\leq j\leq q)$, 有

$ (q-2)S_{\alpha, \beta}(r, f)\leq \sum_{j=1}^q\overline{C}_{\alpha, \beta}\left(r, \frac{1}{f-a_j}\right)+R_{\alpha, \beta}(r, f), $

其中有某个$a_j=\infty$时, 用$\overline{C}_{\alpha, \beta}(r, f)$替换$\overline{C}_{\alpha, \beta}(r, \frac{1}{f-a_j})$, 且

$ \begin{eqnarray} R_{\alpha, \beta}(r, f)&=& A_{\alpha, \beta}\left(r, \frac{f'}{f}\right)+B_{\alpha, \beta}\left(r, \frac{f'}{f}\right)\\ &&+\sum_{j=1}^q\left\{A_{\alpha, \beta}\left(r, \frac{f'}{f-a_j}\right) +B_{\alpha, \beta}\left(r, \frac{f'}{f-a_j}\right)\right\}+O(1). \end{eqnarray} $

(2.1)

引理2.3^{[6, p138]}  假设$f$是复平面${\Bbb C}$上的一个非常数的亚纯函数.给定一个角域$\Omega(\alpha, \beta)$.则对任意的$1\leq r<R, $有

$ A_{\alpha, \beta}\left(r, \frac{f'}{f}\right) \leq K\left\{\left(\frac{R}{r}\right)^\omega\int_1^R\frac{\log^+T(r, f)}{t^{1+\omega}}{\rm d}t +\log^+\frac{r}{R-r}+\log \frac{R}{r}+1\right\}, $

$ B_{\alpha, \beta}\left(r, \frac{f'}{f}\right)\leq\frac{4\omega}{r^\omega}m\left(r, \frac{f'}{f}\right), $

其中$\omega=\frac{\pi}{\beta-\alpha}$, $K$是一个不依赖于$r$和$R$的一个正数.

注2.1  Nevanlinna猜想当$r$(除去一个有限线测度的例外集外)趋于$+\infty$时, 有

$ \begin{equation} A_{\alpha, \beta}\left(r, \frac{f'}{f}\right) +B_{\alpha, \beta}\left(r, \frac{f'}{f}\right)=o\left(S_{\alpha, \beta}\left(r, f\right)\right). \end{equation} $

(2.2)

他证明了当$f$为${\Bbb C}$上的亚纯函数且具有有限级时, $A_{\alpha, \beta}\big(r, \frac{f'}{f}\big) +B_{\alpha, \beta}\big(r, \frac{f'}{f}\big)=O(1)$. 1974年, Gol'dberg^[5]构造了(2.2)式的一个反例.

注2.2  由引理2.2-2.3, 我们可以得到以下结论

$ Q_{\alpha, \beta}\left(r, f\right)=\left\{\begin{array} {ll} O(1), & \rho(f)<+\infty, \\ O(\log U(r)), ~r\not\in E, & \rho(f)=+\infty, \end{array}\right. $

其中$Q_{\alpha, \beta}\left(r, f\right)$如(2.1)式所叙, $U(r)=r^{\rho(r)}, \rho(r)$是亚纯函数$f$的无穷级, $E$具有有限的线性测度.

引理2.4  假设$f$是一个具有无穷级$\rho(r)$的非常数亚纯函数, 射线$\arg z=\theta$是$f$的一条$\rho(r)$级Borel方向.令$P(f)=a_0f^p+a_1f^{p-1}+\cdots+a_p(a_0\neq0)$是$f$的次数为$p$的一个多项式, 其中系数$a_j(j=0, 1, \cdots, p)$为常数.令$b_j(j=1, 2, \cdots, q)$是$q(q\geq p+1)$个不同的有限复数.则对任意的$\varepsilon(0<\varepsilon<\pi/2)$,

$ (A+B)_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{P(f)\cdot f'}{(f-b_1)(f-b_2)\cdots(f-b_q)}\right)=R_{\theta-\varepsilon, \theta+\varepsilon}(r, f). $

证  用文献[20, 引理4.3]类似的方法可以证明.

引理2.5^[3]  假设$f$是一个无穷级$\rho(r)$的亚纯函数.则射线$\arg z=\theta$是$f$的$\rho(r)$级Borel方向的充要条件是对于任意的$\epsilon(0<\epsilon<\frac{\pi}{2})$, $f$满足等式

$ \mathop {\lim \sup }\limits_{r \to \infty } \frac{{\log {S_{\theta - \varepsilon ,\theta + \varepsilon }}(r,f)}}{{\rho (r)\log r}} = 1. $

(2.3)

证  因为$f$是一个无穷级$\rho(r)$的亚纯函数且$\arg z=\theta\ (0\leq \theta<2\pi)$是其一条$\rho(r)$级的Borel方向, 由引理2.4, 对于任意的$\varepsilon(0<\varepsilon<\pi)$, 有

$ \mathop {\lim \sup }\limits_{r \to \infty } \frac{{\log {S_{\theta - \varepsilon ,\theta + \varepsilon }}(r,f)}}{{\rho (r)\log r}} = 1. $

(3.1)

由$g\in M(\rho(r))$得

$ \mathop {\lim \sup }\limits_{r \to \infty } \frac{{\log {S_{\theta - \varepsilon ,\theta + \varepsilon }}(r,g)}}{{\rho (r)\log r}} \le 1. $

(3.2)

当$r\rightarrow\infty, r\not\in E$, $E$是一个有限线测度的集合, 记$R(r)=O(\rho(r)\log r)$, 则当$r\rightarrow\infty, r\not\in E$时, 有$R(r)=o(S_{\theta-\varepsilon, \theta+\varepsilon}(r, f))$.假设$f\not\equiv g$.令

$ \psi_1:=\frac{f'(f-a_3)}{(f-a_1)(f-a_2)}-\frac{g'(g-a_3)}{(g-a_1)(g-a_2)}, $

$ \psi_2:=\frac{f'(f-a_4)}{(f-a_1)(f-a_2)}-\frac{g'(g-a_4)}{(g-a_1)(g-a_2)}. $

根据引理2.4, 得到

$ \begin{equation} (A+B)_{\theta-\varepsilon, \theta+\varepsilon}(r, \psi_i)=R_{\theta-\varepsilon, \theta+\varepsilon}(r, f) +R_{\theta-\varepsilon, \theta+\varepsilon}(r, g), \quad i=1, 2. \end{equation} $

(3.3)

若$\psi_i(z)\not\equiv0 \ (i=1, 2)$, 我们将证明

$ \begin{equation} C_{\theta-\varepsilon, \theta+\varepsilon}(r, \psi_i)=O(1), ~~ (i=1, 2). \end{equation} $

(3.4)

事实上, 在$\Omega(\theta-\varepsilon, \theta+\varepsilon)$内$\psi_i$的极点只可能出现在$f-a_j$和$g-a_j$ $(i, j=1, 2)$在$\Omega(\theta-\varepsilon, \theta+\varepsilon)$的零点处.因为$f, g$在$\Omega(\theta-\varepsilon, \theta+\varepsilon)$内$CM$分担$a_1, a_2$, 可以得到如果$z_0\in \Omega(\theta-\varepsilon, \theta+\varepsilon)$是$f-a_j$的一个重数为$m(\geq1)$的零点, 则$z_0\in \Omega(\theta-\varepsilon, \theta+\varepsilon)$是$g-a_j$ $(j=1, 2)$的具有重数$m(\geq1)$的一个零点.假设

$ f-a_j=(z-z_0)^m\alpha_j(z), \qquad g-a_j=(z-z_0)^m\beta_j(z), $

其中$\alpha_j(z), \beta_j(z)$是$\Omega(\theta-\varepsilon, \theta+\varepsilon)$内的解析函数且$\alpha_j(z_0)\neq0, \beta_j(z_0)\neq0$.故

$\psi_1(z_0)=\frac{a_1-a_3}{a_1-a_2}\left(\frac{\alpha'_1(z_0)}{\alpha_1(z_0)} -\frac{\beta'_1(z_0)}{\beta_1(z_0)}\right), $

$\psi_2(z_0)=\frac{a_2-a_3}{a_2-a_1}\left(\frac{\alpha'_2(z_0)}{\alpha_2(z_0)} -\frac{\beta'_2(z_0)}{\beta_2(z_0)}\right). $

故能得到$\psi_i$$(i=1, 2)$在$\Omega(\theta-\varepsilon, \theta+\varepsilon)$内解析, 从而(3.4)式成立.由(3.3)和(3.4)式得

$ \begin{equation} S_{\theta-\varepsilon, \theta+\varepsilon}(r, \psi_i)=R_{\theta-\varepsilon, \theta+\varepsilon}(r, f) +R_{\theta-\varepsilon, \theta+\varepsilon}(r, g), ~~(i=1, 2). \end{equation} $

(3.5)

若$\psi_i\not\equiv 0, i=1, 2$, 则

$ \begin{equation} \overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_3}\right) \leq C_{\theta-\varepsilon, \theta+\varepsilon}\bigg(r, \frac{1}{\psi_1}\bigg)\leq S_{\theta-\varepsilon, \theta+\varepsilon}(r, \psi_1)+O(1), \end{equation} $

(3.6)

$ \begin{equation} \overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_4}\right) \leq C_{\theta-\varepsilon, \theta+\varepsilon}\bigg(r, \frac{1}{\psi_2}\bigg)\leq S_{\theta-\varepsilon, \theta+\varepsilon}(r, \psi_2)+O(1). \end{equation} $

(3.7)

因$f, g$是$\Omega(\theta-\varepsilon, \theta+\varepsilon)$内的两个超越解析函数且在$\Omega$内$IM$分担$a_1, a_2$, 在$\Omega$内$f=a_3\Longrightarrow g=a_3$, $f=a_4\Longrightarrow g=a_4$.再用引理2.2得

$ \begin{eqnarray} 3S_{\theta-\varepsilon, \theta+\varepsilon}(r, f) &\leq&\sum_{i=1}^4\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_i}\right)+R(r, f) \\ &\leq&\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-g}\right) +R_{\theta-\varepsilon, \theta+\varepsilon}(r, f) \\ &=&S_{\theta-\varepsilon, \theta+\varepsilon}(r, f-g)+R_{\theta-\varepsilon, \theta+\varepsilon}(r, f)\\ &\leq& S_{\theta-\varepsilon, \theta+\varepsilon}(r, f)+S_{\theta-\varepsilon, \theta+\varepsilon}(r, g) +R_{\theta-\varepsilon, \theta+\varepsilon}(r, f) \end{eqnarray} $

(3.8)

$ \begin{eqnarray} S_{\theta-\varepsilon, \theta+\varepsilon}(r, g) &\leq& \overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{g-a_1}\right) +\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{g-a_2}\right) +R_{\theta-\varepsilon, \theta+\varepsilon}(r, g) \\ &=&\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_1}\right) +\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_2}\right) +R_{\theta-\varepsilon, \theta+\varepsilon}(r, g)\\ &\leq&2S_{\theta-\varepsilon, \theta+\varepsilon}(r, f)+R_{\theta-\varepsilon, \theta+\varepsilon}(r, g). \end{eqnarray} $

(3.9)

由(3.8)和(3.9)式得

$ S_{\theta-\varepsilon, \theta+\varepsilon}(r, g)=2S_{\theta-\varepsilon, \theta+\varepsilon}(r, f) +R_{\theta-\varepsilon, \theta+\varepsilon}(r, f)+R_{\theta-\varepsilon, \theta+\varepsilon}(r, g), $

从而$R_{\theta-\varepsilon, \theta+\varepsilon}(r, g)=R_{\theta-\varepsilon, \theta+\varepsilon}(r, f)$.故$R(r)=R_{\theta-\varepsilon, \theta+\varepsilon}(r, g)=R_{\theta-\varepsilon, \theta+\varepsilon}(r, f)$.由(3.8)及(3.9)式也可得到

$ \begin{equation} S_{\theta-\varepsilon, \theta+\varepsilon}(r, f) =\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\bigg(r, \frac{1}{f-a_i}\bigg)+R(r), ~~i=1, 2. \end{equation} $

(3.10)

结合引理2.2, (3.6), (3.7)和(3.10)式, 得$S_{\theta-\varepsilon, \theta+\varepsilon}(r, f)\leq R(r)$.由$R(r)$的定义得到矛盾.假设$\psi_1$和$\psi_2$当中有一个恒为零, 不仿取$\psi_1\equiv0$, 则

$ \begin{equation} \overline{C}^{(2}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{g-a_3}\right) =\overline{C}^{(2}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_3}\right), \end{equation} $

(3.11)

其中$\overline{C}^{(2}_{\theta-\varepsilon, \theta+\varepsilon}\big(r, \frac{1}{f-a_3}\big)$是$f-a_3$在$\Omega$内重数满足$q\geq 2$的不同零点的计数函数, $\overline{C}^{(2}_{\theta-\varepsilon, \theta+\varepsilon}$ $\big(r, \frac{1}{g-a_3}\big)$类似.令

$ \begin{equation} \eta:=\frac{f'g'(f-g)}{(f-a_3)(f-a_4)(g-a_1)(g-a_2)}. \end{equation} $

(3.12)

根据(3.12)式及引理2.4, 得$S_{\theta-\varepsilon, \theta+\varepsilon}(r, \eta)=R(r)$.观察到$g(z_1)=a_3$可以推出对满足$\eta(z_1)\neq 0$的$z_1\in \Omega$有$f(z_1)=a_3$.得到

$ \begin{equation} \overline{C}^{1)}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{g-a_3}\right)= \overline{C}^{1)}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_3}\right)+R(r), \end{equation} $

(3.13)

其中$ \overline{C}^{1)}_{\theta-\varepsilon, \theta+\varepsilon}\big(r, \frac{1}{f-a_3}\big)$是$f-a_3$在$\Omega$内的不同的简单零点的计数函数, $ \overline{C}^{1)}_{\theta-\varepsilon, \theta+\varepsilon}\big(r, \frac{1}{g-a_3}\big)$类似.

由(3.11)及(3.13)式可得

$ \begin{equation} \overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{g-a_3}\right)= \overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_3}\right)+R(r). \end{equation} $

(3.14)

类似的, 当$\psi_2\equiv0$时, 有

$ \begin{equation} \overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{g-a_4}\right)= \overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_4}\right)+R(r). \end{equation} $

(3.15)

根据(3.8)-(3.10)式, (3.14)及(3.15)式, 有

$ 2S_{\theta-\varepsilon, \theta+\varepsilon}(r, f) =\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_3}\right)+R(r)$

或者

$ 2S_{\theta-\varepsilon, \theta+\varepsilon}(r, f) =\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_4}\right)+R(r).$

从而, 根据$f$是一个无穷级的解析函数及$R(r)$的定义, 可以得到矛盾.定理1.5得证.

证  假设$f\not\equiv g$.由引理2.2, 得

$ \begin{eqnarray*} & &2S_{\theta-\varepsilon, \theta+\varepsilon}(r, f) +\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{g-a_4}\right)\\ &\leq& \overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_1}\right) +\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_2}\right) + \overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_3}\right)\\ & & +\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{g-a_4}\right) +R_{\theta-\varepsilon, \theta+\varepsilon}(r, f)\\ &\leq&\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-g}\right) +R_{\theta-\varepsilon, \theta+\varepsilon}(r, f)\\ &\leq&S_{\theta-\varepsilon, \theta+\varepsilon}(r, f)+S_{\theta-\varepsilon, \theta+\varepsilon}(r, g) +R_{\theta-\varepsilon, \theta+\varepsilon}(r, f). \end{eqnarray*} $

从而, 有

$ \begin{equation} S_{\theta-\varepsilon, \theta+\varepsilon}(r, f) +\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{g-a_4}\right) \leq S_{\theta-\varepsilon, \theta+\varepsilon}(r, g)+R_{\theta-\varepsilon, \theta+\varepsilon}(r, f). \end{equation} $

(4.1)

类似的

$ \begin{equation} S_{\theta-\varepsilon, \theta+\varepsilon}(r, g) +\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_3}\right) \leq S_{\theta-\varepsilon, \theta+\varepsilon}(r, f)+R_{\theta-\varepsilon, \theta+\varepsilon}(r, g). \end{equation} $

(4.2)

由(4.1)及(4.2)式, 得$S_{\theta-\varepsilon, \theta+\varepsilon}(r, f)=S_{\theta-\varepsilon, \theta+\varepsilon}(r, g) +R_{\theta-\varepsilon, \theta+\varepsilon}(r, f)+R_{\theta-\varepsilon, \theta+\varepsilon}(r, g)$.根据定理1.6的假设, (3.1)及(3.2)式, 有$R(r)=R_{\theta-\varepsilon, \theta+\varepsilon}(r, f)=R_{\theta-\varepsilon, \theta+\varepsilon}(r, g)$.从而根据(4.1)及(4.2)式, 有

$ \begin{equation} \overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_3}\right)=R(r), \qquad \overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{g-a_4}\right)=R(r). \end{equation} $

(4.3)

故

$ \begin{equation} 2S_{\theta-\varepsilon, \theta+\varepsilon}(r, f)= \overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_1}\right) +\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_2}\right)+R(r). \end{equation} $

(4.4)

由(4.4)式, 能够得到$f-a_i$ $(i=1, 2)$在$\Omega$内``几乎所有的"零点都是简单零点.类似的, $g-a_i$ $(i=1, 2)$在$\Omega$内``几乎所有的"零点都是简单零点.置

$ \varphi_1:=\frac{(a_1-a_3)f'(f-a_2)}{(f-a_1)(f-a_3)}-\frac{(a_1-a_4)g'(g-a_2)}{(g-a_1)(g-a_4)} $

及

$ \varphi_2:=\frac{(a_2-a_3)f'(f-a_1)}{(f-a_2)(f-a_3)}-\frac{(a_2-a_4)g'(g-a_1)}{(g-a_2)(g-a_4)}.$

由引理2.4得$(A+B)_{\theta-\varepsilon, \theta+\varepsilon}(r, \varphi_i)=R(r), (i=1, 2)$.因为$f, g$在$\Omega$内$IM$分担$a_1, a_2$且由(4.3)式, 得到$C_{\theta-\varepsilon, \theta+\varepsilon}(r, \varphi_i)=R(r), (i=1, 2)$.故$S_{\theta-\varepsilon, \theta+\varepsilon}(r, \varphi_i)=R(r), (i=1, 2)$.

如果$\varphi_1\equiv0$, 则有

$\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\bigg(r, \frac{1}{f-a_2}\bigg)\leq \overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\bigg(r, \frac{1}{\varphi_1}\bigg)=R(r). $

从而由(4.4)式可推出矛盾.类似的, 当$\varphi_2\equiv0$时, 亦可得到矛盾.从而$\varphi_1, \varphi_2$恒为零.故$\frac{\varphi_1-\varphi_2}{a_1-a_2}\equiv0$, 即

$ \frac{f'}{f-a_3}-\frac{g'}{g-a_4}-\frac{f'}{f-a_1}+\frac{g'}{g-a_1} -\frac{f'}{f-a_2}+\frac{g'}{g-a_2}\equiv0, $

因此, 可得

$ \begin{equation} \frac{f-a_3}{g-a_4}\cdot\frac{(g-a_1)(g-a_2)}{(f-a_1)(f-a_2)}\equiv c, \end{equation} $

(4.5)

其中$c$是一个非零常数.将(4.5)式重新写为

$ \begin{equation} g^2-\left(a_1+a_2-\frac{c\gamma(f)}{f-a_3}\right)g +a_1a_2+\frac{ca_4\gamma(f)}{f-a_3}\equiv0, \end{equation} $

(4.6)

其中$\gamma(f):=(f-a_1)(f-a_2)$. (4.6)式的判别式为

$ \Delta(f)=\left(a_1+a_2-\frac{c\gamma(f)}{f-a_3}\right)^2 -4\left(a_1a_2+\frac{ca_4\gamma(f)}{f-a_3}\right):=\frac{Q(f)}{(f-a_3)^2}, $

其中$ Q(z):=((a_1+a_2)(z-a_3)-c\gamma(z))^2-4a_1a_2(z-a_3)^2-4ca_4\gamma(z)(z-a_3), $是$z$的一个4次多项式.若$a$是$Q(z)$在$\Omega$内的一个零点, 显然$a\neq a_3$.则由(4.6)式, $f(z)=a$可推出

$ \begin{equation} g(z)=\frac{1}{2}\left(a_1+a_2-\frac{c\gamma(a)}{a-a_3}\right):=b. \end{equation} $

(4.7)

置

$ \phi_1:=\frac{f'g'(f-g)}{(f-a_1)(g-a_2)(f-a_3)(g-a_4)}, $

$ \phi_2:=\frac{f'g'(f-g)}{(f-a_2)(g-a_1)(f-a_3)(g-a_4)}$

和

$ \phi=\frac{\phi_2}{\phi_1}=\frac{(f-a_1)(g-a_2)}{(f-a_2)(g-a_1)}. $

由引理2.4, 得$(A+B)_{\theta-\varepsilon, \theta+\varepsilon}(r, \phi_i)=R(r), (i=1, 2)$.另外, $C_{\theta-\varepsilon, \theta+\varepsilon}(r, \phi_i)=R(r), (i=1, 2)$.由$S_{\theta-\varepsilon, \theta+\varepsilon}(r, \phi_i)=R(r), (i=1, 2)$, 可得到$S_{\theta-\varepsilon, \theta+\varepsilon}(r, \phi)=R(r)$.

假设$f$不是$g$的一个Möbius变换, 则$\phi$是一个非常值函数.因为

$ Q(a_1)=(a_1-a_3)^2(a_1-a_2)^2\neq0, $

$ Q(a_2)=(a_2-a_3)^2(a_1-a_2)^2\neq0. $

由$a\neq a_i$ $(i=1, 2)$及(4.5)式, 得到

$ \begin{equation} \overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a}\right)\leq \overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{\phi-\xi}\right) \leq S_{\theta-\varepsilon, \theta+\varepsilon}(r, \vartheta)=R(r), \end{equation} $

(4.8)

其中$\vartheta=\frac{(a-a_1)(b-a_2)}{(a-a_2)(b-a_1)}$.根据$f$在$\Omega$内解析, 引理2.4及(4.4)式, 得

$ S_{\theta-\varepsilon, \theta+\varepsilon}(r, f)\leq \overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a_3}\right) +\overline{C}_{\theta-\varepsilon, \theta+\varepsilon}\left(r, \frac{1}{f-a}\right)+R(r)=R(r). $

由$f$是一个无穷级的解析函数可推出矛盾.从而$f$是$g$的一个Möbius变换.既然$f, g$是$\Omega$上的解析函数, 经过一个简单计算, 可得$a_1+a_2=a_3+a_4$及

$ f\equiv \frac{a_3g-a_1a_2}{g-a_4}, $

且$a_3, a_4$分别是$f$和$g$在$\Omega$的例外值.

定理1.6获证.