设$f(z)$与$g(z)$为开平面内的非常数亚纯函数, 本文采用亚纯函数Nevanlinna理论的常用符号(见文献[1, 2]), 如$T(r,f), m(r,f), N(r,f), \overline{N}(r,f), N(r,a,f), S(r,f)$等等，并假设读者已熟悉Nevanlinna基本理论(见文献[2]).设$a$, $b$为两个复数, 若$f-a$与$g-b$的零点相同, 则称$(a,b)$为$f$与$ g$的IM公共值对; 若$f-a$与$g-b$的零点相同, 且零点的重级也相同, 则称$(a,b)$为$f$与$g$的CM公共值对.若$(a,0)$为$f$和$1/g$的CM(IM)公共值对, 则$(a,\infty)$为$f$和$g$的CM(IM)公共值对.当$(a_{k},b_{k})(1\leq k\leq n,n\geq2)$为$f$与$g$的$n$个公共值对时, 我们要求$a_{i}\neq a_{j},b_{i}\neq b_{j}(i\neq j)$.当$a=b$时, 则称$a$为$f$和$g$的CM(IM)公共值.

设$k$为一正整数, 我们用$E_{k)}(a,f)$表示$f-a$的重级不超过$k$的零点的集合, 且重级零点按重数计算; $\overline{E}_{k)}(a,f)$表示重级零点仅计一次的情况.若$E_{+\infty)}(a,f)=E_{+\infty)}(a,g)$, 则$a$为$f$与$g$的CM公共值; 若$\overline{E}_{+\infty)}(a,f)=\overline{E}_{+\infty)}(a,g)$, 则$a$为$f$与$g$的IM公共值.我们用$N_{k)}(r,a,f)$表示$f-a$的重级不超过$k$的零点的计数函数; $N_{(k+1}(r,a,f)$表示$f-a$的重级超过$k$的零点的计数函数. $\overline{N}_{k)}(r,a,f)$和$\overline{N}_{(k+1}(r,a,f)$分别表示相应的精简计数函数.

$S(r,f)$表示$o(T(r,f))$($r\rightarrow\infty$, $r\not\in E$)型的量, 其中$E$为$R^{+}$上的一个线性测度有穷的集合, 若$S(r,f)=S(r,g)$, 则记$S(r)=S(r,f)=S(r,g)$.

我们用$\overline{N}_{E}(r,f-a=0=g-b)$表示$f-a$与$g-b$具有相同重级的公共零点的计数函数, 每个零点仅计一次; 用$\overline{N}(r,f-a=0=g-b)$表示$f-a$与$g-b$公共零点的计数函数, 每个零点仅计一次.如果

$\overline{N}(r,a,f)-\overline{N}_{E}(r,f-a=0=g-b)=S(r,f)$

及

$\overline{N}(r,a,g)-\overline{N}_{E}(r,f-a=0=g-b)=S(r,g),$

则称$(a,b)$为$f$与$g$的CM$^{\ast}$公共值对.如果

$\overline{N}(r,a,f)-\overline{N}(r,f-a=0=g-b)=S(r,f)$

及

$\overline{N}(r,a,g)-\overline{N}(r,f-a=0=g-b)=S(r,g),$

则称$(a,b)$为$f$与$g$的IM$^{\ast}$公共值对.

设$a(z)$为开平面内的亚纯函数, 若$T(r,a)=S(r,f)$, 则称$a$为$f$的小函数.当$a$, $b$为小函数时, 也有与上述类似的定义, 只需把“值”替换为“小函数”.

设$f(z)$与$g(z)$为两个非常数亚纯函数, 若存在$f$的四个小函数$\alpha_{i}(z)$, $i$=1, 2, 3, 4, 使得$g=(\alpha_{1}f+\alpha_{2})/(\alpha_{3}f+\alpha_{4})$ $(\alpha_{1}\alpha_{4}-\alpha_{2}\alpha_{3}\not\equiv0)$, 则称$g$为$f$的拟分式线性变换.

1929年, Nevanlinna (见文献[2])证明了下述四值定理:

定理A 设$f(z)$与$g(z)$为两个非常数亚纯函数, 以$a_{j}(j=1, 2, 3, 4)$为四个判别的CM公共值, 则$f$为$g$的分式线性变换.

1995年, Li和Yang (见文献[5])将“公共值”推广到“公共小函数”, 证明了下述定理:

定理B 设$f(z)$与$g(z)$为两个非常数亚纯函数, $a_{j}(z)(j=1, 2, 3, 4)$为$f$与$g$的四个判别的CM$^{\ast}$公共小函数, 则$f$为$g$的拟分式线性变换.

1999年, 张庆彩和杨连中(见文献[6])在考虑重值的情况下, 对定理A作了改进, 得到

定理C 设$f(z)$与$g(z)$为两个非常数亚纯函数, $a_{j}(j=1, 2, 3, 4)$为四个判别的复数, $k$为一正整数.若$E_{k)}(a_{j},f)=E_{k)}(a_{j},g)$(j=1, 2, 3, 4), $k\geq12$, 则$f$为$g$的分式线性变换.

2002年, Yao和Yu (见文献[7])在考虑重值的条件下, 改进了定理B, 得到

定理D 设$f(z)$与$g(z)$为两个非常数亚纯函数, $a_{j}(z)(j=1, 2, 3, 4)$为$f$与$g$的四个判别的小函数, $k$为一正整数.若$E_{k)}(a_{j},f)=E_{k)}(a_{j},g)$(j=1, 2, 3, 4), $k\geq15$, 则$f$为$g$的拟分式线性变换.

本文在定理C和定理D的基础上, 进一步得到如下结果:

定理1 设$f(z)$与$g(z)$为两个非常数亚纯函数, $a_{j}(z)(j=1, 2, 3, 4)$为$f$与$g$的四个判别的小函数, $k$为一正整数.若$E_{k)}(a_{j},f)=E_{k)}(a_{j},g)$(j=1, 2, 3, 4), $k\geq11$, 则$f$为$g$的拟分式线性变换.

对于上述四个值集的情况, 还可以予以精确化, 本文又证明了:

定理2 设$f(z)$与$g(z)$为两个非常数亚纯函数, $a_{j}(z)(j=1, 2, 3, 4)$为$f$与$g$的四个判别的小函数, $k$为一正整数.若$E_{11)}(a_{j},f)=E_{11)}(a_{j},g)(j=1,2,3)$, $E_{k)}(a_{4},f)=E_{k)}(a_{4},g)$, $k\geq10$, 则$f$为$g$的拟分式线性变换.

1997年, Czubiak和Gundersen (见文献[8])证明了下述定理:

定理E 设$f(z)$与$g(z)$为两个非常数亚纯函数, $(a_{k},b_{k})(1\leq k\leq6)$为$f$与$g$的六个IM公共值对, 且$a_{i}\neq a_{j},b_{i}\neq b_{j}(i\neq j)$, 则$f$为$g$的分式线性变换.

2003年, Hu, Li和Yang (见文献[9])对定理E作了改进, 证明了

定理F 设$f(z)$与$g(z)$为两个非常数亚纯函数, $(a_{k},b_{k})(1\leq k\leq5)$为$f$和$g$的五个IM$^{\ast}$公共值对, 且$a_{i}\neq a_{j},b_{i}\neq b_{j}(i\neq j)$, 若

$\overline{N}(r,a_{6},f)-\overline{N}(r,f-a_{6}=0=g-b_{6})=S(r),$

其中$a_{6}\neq a_{k},b_{6}\neq b_{k}, (1\leq k\leq5)$, 则$f$是$g$的分式线性变换.

本文得到了定理F的更一般结果:

定理3 设$f(z)$与$g(z)$为两个非常数亚纯函数, $(a_{k},b_{k})(1\leq k\leq5)$为$f$和$g$的五个IM$^{\ast}$公共值对, 且$a_{i}\neq a_{j},b_{i}\neq b_{j}(i\neq j)$, 若

$\begin{align} \overline{N}(r,a_{6},f)-\overline{N}(r,f-a_{6}=0=g-b_{6})\leq\lambda T(r,f)+S(r),\label{eqa1} \end{align}$

(1.1)

其中$\lambda\in[0,\frac{2}{5}), a_{6}\neq a_{k},b_{6}\neq b_{k}, (1\leq k\leq5)$, 则$f$是$g$的分式线性变换.

2011年, Gundersen (见文献[8])证明了下述定理和推论:

定理G 设$f(z)$与$g(z)$为两个非常数亚纯函数, $(a_{k},b_{k})(1\leq k\leq5)$为$f$和$g$的五个IM公共值对, 且$a_{i}\neq a_{j}, b_{i}\neq b_{j}(i\neq j)$, 则或者$f$为$g$的分式线性变换或者对任意不同的$i,j,k\in\{1, 2, 3, 4, 5\}$, 有以下五个不等式成立:

$\begin{eqnarray*} && \overline{N}(r,a_{i},f)+\overline{N}(r,a_{j},f)\leq\frac{3}{2}T(r,f)+S(r,f);\\ && T(r,f)\leq\overline{N}(r,a_{i},f)+\overline{N}(r,a_{j},f)+S(r,f);\\ && \overline{N}(r,a_{i},f)+\overline{N}(r,a_{j},f)+\overline{N}(r,a_{k},f)\leq2T(r,f)+S(r,f);\\ && \frac{3}{2}T(r,f)\leq\overline{N}(r,a_{i},f)+\overline{N}(r,a_{j},f)+\overline{N}(r,a_{k},f)+S(r,f);\\ && \overline{N}(r,a_{i},f)\leq2\overline{N}(r,a_{j},f)+S(r,f). \end{eqnarray*}$

推论A (见文献[8])设$f(z)$与$g(z)$为两个非常数亚纯函数, $(a_{k},b_{k})(1\leq k\leq5)$为$f$和$g$的五个IM公共值对, 且$a_{i}\neq a_{j}, b_{i}\neq b_{j}(i\neq j)$, 则或者$f$为$g$的分式线性变换或者$\overline{N}(r,a_{k},f)+S(r,f)\geq\frac{1}{3}T(r,f),1\leq k\leq5.$

推论B (见文献[8])设$f(z)$与$g(z)$为两个非常数整函数, $(a_{k},b_{k})(1\leq k\leq4)$为$f$和$g$的四个IM公共值对, 其中$a_{k}$和$b_{k}$均为有穷值, 且$a_{i}\neq a_{j}, b_{i}\neq b_{j}(i\neq j)$, 则$f$为$g$的分式线性变换.

本文对推论A作了改进, 得到:

定理4 设$f(z)$与$g(z)$为两个非常数亚纯函数, $(a_{k},b_{k})(1\leq k\leq4)$为$f$与$g$的四个IM$^{\ast}$公共小函数对, 且$a_{i}\neq a_{j}, b_{i}\neq b_{j}(i\neq j)$, 若

$\begin{align} \overline{N}(r,a_{5},f)\leq\lambda T(r,f)+S(r,f), \overline{N}(r,a_{5},g)\leq\lambda T(r,g)+S(r,g), \label{eqa2} \end{align}$

(1.2)

其中$\lambda\in[0,\frac{1}{3})$, $T(r,a_{5})=S(r,f)$, $T(r,b_{5})=S(r,g)$, $a_{5}\neq a_{k},b_{5}\neq b_{k}, (1\leq k\leq4)$, 则$f$是$g$的拟分式线性变换.

注1 由定理4, 可把推论B中的“公共值对”推广到“公共小函数对”.

引理1 (见文献[9])设$f(z)$为一个非常数亚纯函数, $a_{j}(z)(j=1,2,\cdots,q)$为$f$的$q$个判别的小函数, 则对于任意的$\varepsilon>0$, 有

$(q-2)T(r,f)\leq\sum\limits_{j=1}^{q} \overline{N}(r,a_{j},f)+\varepsilon T(r,f)+S(r,f),$

其中$r\not\in E, E\subset R$且$\int_{E} d\log\log r<\infty$.

引理2 (见文献[10])设$f(z)$, $a(z)$和$b(z)$为亚纯函数$(f(z),a(z),b(z)\not\equiv\infty)$, $a(z)$与$b(z)$为$f$的小函数, 且$a(z)\not\equiv b(z)$, 设

$\begin{equation} L(f,a,b)=\begin{matrix} \begin{vmatrix}f&f'&1\\a&a'&1\\b&b'&1\end{vmatrix} \end{matrix}, \end{equation}$

则$L(f,a,b)\not\equiv0$, 及

$m(r,\frac{L(f,a,b)f^{k}}{(f-a)(f-b)})=S(r,f)(k=0,1).$

引理3 设$f(z)$为一个非常数亚纯函数, $a_{j}(z)(j=1,2,\cdots,q)$为$f$的$q$个判别的小函数, $k$为正整数, 则

$\sum\limits_{j=1}^{q}\overline{N}_{(k+1}(r,a_{j},f)\leq\frac{2+\varepsilon}{k}T(r,f)+S(r,f).$

证 由引理1, 得

$\begin{align} (q-2)T(r,f)\leq\sum\limits_{j=1}^{q}\overline{N}(r,a_{j},f)+\varepsilon T(r,f)+S(r,f).\label{eqb1} \end{align}$

(2.1)

而

$\overline{N}(r,a_{j},f)+k\overline{N}_{(k+1}(r,a_{j},f)\leq N(r,a_{j},f)\leq T(r,f)+S(r,f),$

故有

$\begin{align} k\sum\limits_{j=1}^{q}\overline{N}_{(k+1}(r,a_{j},f)+\sum\limits_{j=1}^{q}\overline{N}(r,a_{j},f)\leq qT(r,f)+S(r,f).\label{eqb2} \end{align}$

(2.2)

由(2.1) 和(2.2) 式, 得

$(q-2)T(r,f)+k\sum\limits_{j=1}^{q}\overline{N}_{(k+1}(r,a_{j},f)\leq qT(r,f)+\varepsilon T(r,f)+S(r,f),$

即

$\sum\limits_{j=1}^{q}\overline{N}_{(k+1}(r,a_{j},f)\leq\frac{2+\varepsilon}{k}T(r,f)+S(r,f).$

引理3证毕.

引理4 设$f(z)$与$g(z)$为两个非常数亚纯函数, $a_{j}(z)(j=1, 2, 3, 4)$为$f$与$g$的四个判别的小函数, 若$f\not\equiv g$, 且$\overline{E}_{k)}(a_{j},f)=\overline {E}_{k)}(a_{j},g),(j=1, 2, 3, 4)$, $k\geq3$为正整数, 则

$\begin{eqnarray*} ({\hbox{a}}) &&T(r,f)\leq\frac{k}{k-2-\varepsilon-k\varepsilon}T(r,g)+S(r), T(r,g)\leq\frac{k}{k-2-\varepsilon-k\varepsilon}T(r,f)+S(r),\\ && S(r)=S(r,f)=S(r,g);\\ ({\hbox{b}}) &&(2-\frac{2+\varepsilon+k\varepsilon}{k})T(r,f)+S(r)\leq\sum\limits_{j=1}^{4}\overline{N}_{k)}(r,a_{j},f)\leq(1+\frac{k}{k-2-\varepsilon-k\varepsilon})T(r,f)+S(r),\\ &&(2-\frac{2+\varepsilon+k\varepsilon}{k})T(r,g)+S(r)\leq\sum\limits_{j=1}^{4}\overline{N}_{k)}(r,a_{j},g)\leq(1+\frac{k}{k-2-\varepsilon-k\varepsilon})T(r,g)+S(r). \end{eqnarray*}$

证 由引理1和引理3, 有

$\begin{align} 2T(r,f)&\leq\sum\limits_{j=1}^{4}\overline{N}(r,a_{j},f)+\varepsilon T(r,f)+S(r,f) \notag \\ &=\sum\limits_{j=1}^{4}\overline{N}_{k)}(r,a_{j},f) +\sum\limits_{j=1}^{4}\overline{N}_{(k+1}(r,a_{j},f)+\varepsilon T(r,f)+S(r,f) \label{eqb3}\\ &\leq T(r,f)+T(r,g)+\frac{2+\varepsilon+k\varepsilon}{k}T(r,f)+S(r,f), \notag \end{align}$

(2.3)

从而

$(k-2-\varepsilon-k\varepsilon)T(r,f)\leq kT(r,g)+S(r,f).$

因为$k\geq3$, 取$\varepsilon$充分小, 使$k-2-\varepsilon-k\varepsilon>0$, 所以有

$\begin{align} T(r,f)\leq\frac{k}{k-2-\varepsilon-k\varepsilon}T(r,g)+S(r,f).\label{eqb4} \end{align}$

(2.4)

同理

$\begin{align} T(r,g)\leq\frac{k}{k-2-\varepsilon-k\varepsilon}T(r,f)+S(r,g).\label{eqb5} \end{align}$

(2.5)

由(2.4) 和(2.5) 式, 得$S(r)=S(r,f)=S(r,g)$.此即得(a).

由(2.3) 式, 引理3和(a)得

$\begin{eqnarray*} 2T(r,f)&\leq&\sum\limits_{j=1}^{4}\overline{N}_{k)}(r,a_{j},f)+\frac{2+\varepsilon+k\varepsilon}{k}T(r,f)+S(r)\\ &\leq& T(r,f)+T(r,g)+\frac{2+\varepsilon+k\varepsilon}{k}T(r,f)+S(r)\\ &\leq&(1+\frac{k}{k-2-\varepsilon-k\varepsilon})T(r,f)+\frac{2+\varepsilon+k\varepsilon}{k}T(r,f)+S(r), \end{eqnarray*}$

即得

$(2-\frac{2+\varepsilon+k\varepsilon}{k})T(r,f)+S(r) \leq\sum\limits_{j=1}^{4}\overline{N}_{k)}(r,a_{j},f)\leq(1+\frac{k}{k-2-\varepsilon-k\varepsilon})T(r,f)+S(r).$

同理, 有

$(2-\frac{2+\varepsilon+k\varepsilon}{k})T(r,g)+S(r) \leq\sum\limits_{j=1}^{4}\overline{N}_{k)}(r,a_{j},g)\leq(1+\frac{k}{k-2-\varepsilon-k\varepsilon})T(r,g)+S(r).$

此即得(b).引理4证毕.

引理5 (见文献[2])设$f(z)$为一个非常数亚纯函数, $R(f)=\frac{P(f)}{Q(f)}$, 其中$P(f)=\sum\limits_{k=0}^{p}a_{k}f^{k}$和$Q(f)=\sum\limits_{j=0}^{q}b_{j}f^{j}$是两个互质的$f$的多项式, 系数$\{a_{k}(z)\}$和$\{b_{k}(z)\}$均为$f$的小函数, 且$a_{p}(z)\not\equiv0, b_{q}(z)\not\equiv0$, 则

$T(r,R(f))=\max\{p,q\}\cdot T(r,f)+S(r,f).$

引理6 (见文献[7])设$f(z)$和$g(z)$为两个非常数亚纯函数, $(a_{k},b_{k})(1\leq k\leq5)$为$f$和$g$的五个IM$^{\ast}$公共值对, 且$a_{i}\neq a_{j}, b_{i}\neq b_{j}(i\neq j)$, 若$f$不是$g$的分式线性变换, 则下列等式成立:

(a) $ T(r,f)=T(r,g)+S(r),\quad S(r)=S(r,f)=S(r,g);$

(b) $3T(r,f)=\sum\limits_{k=1}^{5}\overline{N}(r,a_{k},f)+S(r);$

(c) $T(r,f)=\overline{N}(r,a,f)+S(r),\quad a\neq a_{k},1\leq k\leq5.$

定理1的证明 不失一般性, 假设$a_{1}(z)\equiv \infty$, $a_{2}(z)\equiv 0$, $a_{3}(z)\equiv 1$, $a_{4}(z)\equiv a(z)(\not\equiv\infty,0,1)$, 且$f\not\equiv g$.令

$\begin{align} H_{1}=\begin{matrix}\begin{vmatrix}f(f-1)&f'\\a(a-1)&a' \end{vmatrix}\end{matrix}\Bigg/[f(f-1)(f-a)] -\begin{matrix}\begin{vmatrix}g(g-1)&g'\\a(a-1)&a' \end{vmatrix}\end{matrix}\Bigg/[g(g-1)(g-a)], \label{eqh1} \end{align}$

(3.1)

而

$\begin{matrix}\begin{vmatrix} f(f-1)&f'\\a(a-1)&a'\end{vmatrix}\end{matrix}=-f(f-1)(f'-a')+(1-a)(f-a)(f-1)f'+af(f-a)f',$

由对数导数引理, 有$m(r,H_{1})=S(r).$若$H_{1}\not\equiv0$, 则有

$\begin{align*} N(r,H_{1})\leq&\sum\limits_{j=2}^{4}\overline{N}_{(k+1}(r,a_{j},f)+\sum\limits_{j=2}^{4}\overline{N}_{(k+1}(r,a_{j},g)+S(r) \\ \leq&\frac{1}{k+1}\sum\limits_{j=2}^{4}N_{(k+1}(r,a_{j},f)+\frac{1}{k+1}\sum\limits_{j=2}^{4}N_{(k+1}(r,a_{j},g)+S(r) \\ =&\frac{1}{k+1}\bigg\{\sum\limits_{j=2}^{4}N(r,a_{j},f)+\sum\limits_{j=2}^{4}N(r,a_{j},g)\bigg\} \\ &-\frac{1}{k+1}\bigg\{\sum\limits_{j=2}^{4}N_{k)}(r,a_{j},f)+\sum\limits_{j=2}^{4}N_{k)}(r,a_{j},g)\bigg\}+S(r) \\ \leq&\frac{3}{k+1}\{T(r,f)+T(r,g)\} -\frac{2}{k+1}\sum\limits_{j=1}^{4}\overline{N}_{k)}(r,a_{j},f) +\frac{2}{k+1}N_{k)}(r,a_{1},f)+S(r). \end{align*}$

而

$N_{k)}(r,a_{1},f)\leq N(r,0,H_{1})\leq T(r,H_{1})+S(r)=N(r,H_{1})+S(r), $

所以

$ N(r,H_{1})\leq \frac{3}{k-1}\{T(r,f)+T(r,g)\}-\frac{2}{k-1}\sum\limits_{j=1}^{4}\overline{N}_{k)}(r,a_{j},f)+S(r). $

从而则根据上式和引理4(b)得

$\begin{align*} &\overline{N}_{k)}(r,a_{1},f)\leq N_{k)}(r,a_{1},f)\leq N(r,H_{1})+S(r) \\ \leq&\frac{3}{k-1}\{T(r,f)+T(r,g)\} -\frac{2}{k-1}\sum\limits_{j=1}^{4}\overline{N}_{k)}(r,a_{j},f)+S(r) \\ \leq&\frac{3}{k-1}\{T(r,f)+T(r,g)\}-\frac{2}{k-1}(2-\frac{2+\varepsilon+k\varepsilon}{k})T(r,f)+S(r). \end{align*}$

同理, 有

$\overline{N}_{k)}(r,a_{1},g) \leq\frac{3}{k-1}\{T(r,f)+T(r,g)\}-\frac{2}{k-1}(2-\frac{2+\varepsilon+k\varepsilon}{k})T(r,g)+S(r).$

故我们有

$\begin{align} \overline{N}_{k)}(r,a_{1},f)+\overline{N}_{k)}(r,a_{1},g) \leq \frac{2(k+2+\varepsilon+k\varepsilon)}{k(k-1)}\{T(r,f)+T(r,g)\}+S(r). \label{eqc7} \end{align}$

(3.2)

同理, 设

$\begin{align} H_{2}=L(f,1,a)f/[(f-1)(f-a)]-L(g,1,a)g/[(g-1)(g-a)], \label{eqh2} \end{align}$

(3.3)

$\begin{align} H_{3}=L(f,0,a)(f-1)/[f(f-a)]-L(g,0,a)(g-1)/[g(g-a)], \label{eqh3} \end{align}$

(3.4)

$\begin{align} H_{4}=L(f,0,1)(f-a)/[f(f-1)]-L(g,0,1)(g-a)/[g(g-1)]. \label{eqh4} \end{align}$

(3.5)

若$H_{j}\not\equiv0(j=2,3,4)$, 则由引理2, 当$j=2,3,4$时, 同样也有

$\begin{align} \overline{N}_{k)}(r,a_{j},f)+\overline{N}_{k)}(r,a_{j},g)\leq \frac{2(k+2+\varepsilon+k\varepsilon)}{k(k-1)}\{T(r,f)+T(r,g)\}+S(r).\label{eqc8} \end{align}$

(3.6)

由引理4(b), 可得

$(1-\frac{2+\varepsilon+k\varepsilon}{k})\{T(r,f)+T(r,g)\}\leq\sum\limits_{j=1,j\neq j_{0}}^{4}\{\overline{N}_{k)}(r,a_{j},f)+\overline{N}_{k)}(r,a_{j},g)\}+S(r).$

$j_{0}=1, 2, 3, 4$.可知$\overline{N}_{k)}(r,a_{j},f)+\overline{N}_{k)}(r,a_{j},g)(j=1, 2, 3, 4)$中至少有两个, 不失一般性, 可设$j=3,4$, 使得

$\begin{align} \overline{N}_{k)}(r,a_{3},f)+\overline{N}_{k)}(r,a_{3},g) &\geq\frac{1}{3}\{(1-\frac{2+\varepsilon+k\varepsilon}{k})+o(1)\}\{T(r,f)+T(r,g)\}(r\in I),\label{eqc9} \end{align}$

(3.7)

$\begin{align} \overline{N}_{k)}(r,a_{4},f)+\overline{N}_{k)}(r,a_{4},g) &\geq\frac{1}{3}\{(1-\frac{2+\varepsilon+k\varepsilon}{k})+o(1)\}\{T(r,f)+T(r,g)\}(r\in I),\label{eqc10} \end{align}$

(3.8)

其中$I$为无穷测度集.若$H_{3}\not\equiv0$, 则由(3.6) 和(3.7) 式, 得$\frac{1}{3}(1-\frac{2+\varepsilon+k\varepsilon}{k})\leq\frac{2(k+2+\varepsilon+k\varepsilon)}{k(k-1)}.$令$\varepsilon\rightarrow 0$, 有$ \frac{1}{3}(1-\frac{2}{k})\leq\frac{2(k+2)}{k(k-1)}. $解得$k\leq10$, 与定理2条件$k\geq11$矛盾, 故$H_{3}\equiv0$.同理由(3.6) 和(3.8) 式可得$H_{4}\equiv0$.即

$\begin{align} L(f,0,a)(f-1)/[f(f-a)]\equiv L(g,0,a)(g-1)/[g(g-a)],\label{eqc11} \end{align}$

(3.9)

$\begin{align} f'(f-a)/[f(f-1)]\equiv g'(g-a)/[g(g-1)].\label{eqc12} \end{align}$

(3.10)

设$z_{a}$为$f-a$的零点, 但不是$g-a$的零点.则$z_{a}$是(3.9) 式等号左边的一个极点, 由此对于等号右边, 我们有$g(z_{a})=0$或$g(z_{a})=\infty$.由(3.9) 和(3.10), 得

$\begin{align} L(f,0,a)f'/f^{2}\equiv L(g,0,a)g'/g^{2}.\label{eqc13} \end{align}$

(3.11)

$z_{a}$不是(3.11) 式等号左边的极点, 但却是等号右边的二重极点, 矛盾.故$z_{a}$为$g-a$的零点.由对称性, 并且比较(3.9) 式等号两边在点$z_{a}$的留数, 可得$a$为$f$与$g$的CM$^{\ast}$公共值.同理可得1为$f$与$g$的CM$^{\ast}$公共值.

根据式(3.9) 和(3.10), 我们可得

$\begin{align} &a(a-1)\frac{f'-a'}{f-a}+a\frac{f'}{f}\equiv a(a-1)\frac{g'-a'}{g-a}+a\frac{g'}{g},\label{eqc14} \end{align}$

(3.12)

$\begin{align} &(1-a)\frac{f'}{f-1}+a\frac{f'}{f}\equiv (1-a)\frac{g'}{g-1}+a\frac{g'}{g}.\label{eqc15} \end{align}$

(3.13)

消去$f'/f$和$g'/g$, 有

$\begin{align} a\frac{f'-a'}{f-a}+\frac{f'}{f-1}\equiv a\frac{g'-a'}{g-a}+\frac{g'}{g-1}.\label{eqc16} \end{align}$

(3.14)

由(3.14) 式, 显然有$\infty$为$f$与$g$的CM$^{\ast}$公共值.

由(3.12) 或(3.13) 式, 可得$0$也为$f$与$g$的CM$^{\ast}$公共值.故$0, 1, \infty, a$均为$f$与$g$的CM$^{\ast}$公共值, 从而由定理B, 得$f$是$g$的拟分式线性变换.定理1证毕.

定理2的证明 由定理1可知, 仅需考虑$k<11$的情况.考虑到拟分式线性变换, 不失一般性, 可设$a_{j}(z)\not\equiv \infty (j=1, 2, 3, 4)$, 且假设$f\not\equiv g$.由引理1, 有

$\begin{eqnarray*} &&2T(r,f)\leq\sum\limits_{j=1}^{4}\overline{N}(r,a_{j},f)+\varepsilon T(r,f)+S(r)\\ &=&\sum\limits_{j=1}^{3}\overline{N}_{11)}(r,a_{j},f)+\sum\limits_{j=1}^{3}\overline{N}_{(12}(r,a_{j},f)+ \overline{N}_{k)}(r,a_{4},f)+\overline{N}_{(k+1}(r,a_{4},f)+\varepsilon T(r,f)+S(r)\\ &\leq&\frac{11}{12}\sum\limits_{j=1}^{3}\overline{N}_{11)}(r,a_{j},f)+\frac{1}{12}\sum\limits_{j=1}^{3}N(r,a_{j},f)+ \frac{k}{k+1}\overline{N}_{k)}(r,a_{4},f)\\ &&+\frac{1}{k+1}N(r,a_{4},f)+\varepsilon T(r,f)+S(r)\\ &\leq&\frac{11}{12}\left[\sum\limits_{j=1}^{3}\overline{N}_{11)}(r,a_{j},f)+\overline{N}_{k)}(r,a_{4},f)\right] +\frac{1}{4}T(r,f)+\frac{1}{k+1}T(r,f)+\varepsilon T(r,f)+S(r)\\ &\leq&(\frac{7}{6}+\frac{1}{k+1}+\varepsilon)T(r,f)+\frac{11}{12}T(r,g)+S(r). \end{eqnarray*}$

从而根据上式$T(r,f)\leq\frac{11(k+1)}{2(5k-1-6k\varepsilon-6\varepsilon)}T(r,g)+S(r),$及

$\begin{align} \frac{3(7k+3-4k\varepsilon-4\varepsilon)}{11(k+1)}T(r,f)+S(r) \leq\sum\limits_{j=1}^{3}\overline{N}_{11)}(r,a_{j},f)+\overline{N}_{k)}(r,a_{4},f).\label{eqc17} \end{align}$

(3.15)

在接下来的证明中, 考虑到拟分式线性变换, 可设$a_{1}(z)\equiv \infty$, $a_{2}(z)\equiv 0$, $a_{3}(z)\equiv 1$, $a_{4}(z)\equiv a(z)(\not\equiv \infty, 0, 1)$.设$H_{1}$同(3.1) 式, 则有$m(r,H_{1})=S(r)$.若$H_{1}\not\equiv0$, 由(3.15) 式有

$\begin{align} &N(r,H_{1}) \notag\\ \leq&\sum\limits_{j=2}^{3}\overline{N}_{(12}(r,a_{j},f)+\sum\limits_{j=2}^{3}\overline{N}_{(12}(r,a_{j},g) +\overline{N}_{(k+1}(r,a_{4},f)+\overline{N}_{(k+1}(r,a_{4},g)+S(r) \notag\\ \leq&\frac{1}{12}\bigg\{\sum\limits_{j=2}^{3}N(r,a_{j},f)+\sum\limits_{j=2}^{3}N(r,a_{j},g)\bigg\} -\frac{1}{12}\bigg\{\sum\limits_{j=2}^{3}N_{11)}(r,a_{j},f)+\sum\limits_{j=2}^{3}N_{11)}(r,a_{j},g)\bigg\} \notag\\ &+\frac{1}{k+1}\{N(r,a_{4},f)+N(r,a_{4},g)\}-\frac{1}{k+1}\{N_{k)}(r,a_{4},f)+N_{k)}(r,a_{4},g)\}+S(r) \notag\\ \leq&(\frac{1}{6}+\frac{1}{k+1})\{T(r,f)+T(r,g)\}+\frac{1}{6}N_{11)}(r,a_{1},f) \notag\\ &-2\bigg\{\frac{1}{12}\sum\limits_{j=1}^{3}N_{11)}(r,a_{j},f)+\frac{1}{k+1}N_{k)}(r,a_{4},f)\bigg\}+S(r) \notag\\ \leq&(\frac{1}{6}+\frac{1}{k+1})\{T(r,f)+T(r,g)\}+\frac{1}{6}N_{11)}(r,a_{1},f)-\frac{7k+3-4k\varepsilon-4\varepsilon}{22(k+1)}T(r,f) +S(r). \label{eqc18} \end{align}$

(3.16)

而

$\begin{align} N_{11)}(r,a_{1},f)\leq N(r,0,H_{1})\leq T(r,H_{1})+S(r)=N(r,H_{1})+S(r), \label{eqc19} \end{align}$

(3.17)

则根据(3.16) 和(3.17) 式得

$\begin{eqnarray*} &&\overline{N}_{11)}(r,a_{1},f)\leq N_{11)}(r,a_{1},f)\\ &\leq&(\frac{k+7}{5(k+1)})\{T(r,f)+T(r,g)\}-\frac{3(7k+3-4k\varepsilon-4\varepsilon)}{55(k+1)}T(r,f)+S(r). \end{eqnarray*}$

同理, 有

$\overline{N}_{11)}(r,a_{1},g) \leq(\frac{k+7}{5(k+1)})\{T(r,f)+T(r,g)\}-\frac{3(7k+3-4k\varepsilon-4\varepsilon)}{55(k+1)}T(r,g)+S(r), $

所以有

$\begin{align} &\overline{N}_{11)}(r,a_{1},f)+\overline{N}_{11)}(r,a_{1},g)\notag\\ \leq&\{\frac{2(k+7)}{5(k+1)}-\frac{3(7k+3-4k\varepsilon-4\varepsilon)}{55(k+1)}\}\{T(r,f)+T(r,g)\}+S(r)\notag\\ =&\frac{k+145+12k\varepsilon+12\varepsilon}{55(k+1)}\{T(r,f)+T(r,g)\}+S(r). \label{eqc22} \end{align}$

(3.18)

同理, 设$H_{2}, H_{3}, H_{4}$分别同(3.3), (3.4), (3.5) 式, 若$H_{j}\not\equiv0(j=2,3,4)$, 则由引理2, 当$j=2,3,4$时, 同样也有

$\begin{align} \overline{N}_{11)}(r,a_{j},f)+\overline{N}_{11)}(r,a_{j},g) &\leq\frac{k+145+12k\varepsilon+12\varepsilon}{55(k+1)}\{T(r,f)+T(r,g)\}+S(r), \label{eqc23a} \end{align}$

(3.19)

$\begin{align} \overline{N}_{k)}(r,a_{4},f)+\overline{N}_{k)}(r,a_{4},g) &\leq\frac{2k+4+4k\varepsilon+4\varepsilon}{11(k+1)}\{T(r,f)+T(r,g)\}+S(r) \notag\\ &\leq\frac{k+145+12k\varepsilon+12\varepsilon}{55(k+1)}\{T(r,f)+T(r,g)\}+S(r), \label{eqc23b} \end{align}$

(3.20)

上式(3.20) 对$2\leq k <11$成立.为方便计我们设$k_{1}=k_{2}=k_{3}=11$, $k_{4}=k$, 由(3.15) 式可得

$\frac{2(5k-1-6k\varepsilon-6\varepsilon)}{11(k+1)}\{T(r,f)+T(r,g)\}\leq\sum\limits_{j=1,j\neq j_{0}}^{4}\{\overline{N}_{k_{j})}(r,a_{j},f)+\overline{N}_{k_{j})}(r,a_{j},g)\}+S(r), $

$j_{0}=1, 2, 3, 4$, 可知$\overline{N}_{k_{j})}(r,a_{j},f)+\overline{N}_{k_{j})}(r,a_{j},g)(j=1, 2, 3, 4)$中至少有两个, 不失一般性, 可设$j=3,4$, 使得

$\begin{align} \overline{N}_{11)}(r,a_{3},f)+\overline{N}_{11)}(r,a_{3},g) \geq\frac{1}{3}\{\frac{2(5k-1-6k\varepsilon-6\varepsilon)}{11(k+1)}+o(1)\}\{T(r,f)+T(r,g)\}(r\in I),\label{eqc24} \end{align}$

(3.21)

$\begin{align} \overline{N}_{k)}(r,a_{4},f)+\overline{N}_{k)}(r,a_{4},g) \geq\frac{1}{3}\{\frac{2(5k-1-6k\varepsilon-6\varepsilon)}{11(k+1)}+o(1)\}\{T(r,f)+T(r,g)\}(r\in I),\label{eqc25} \end{align}$

(3.22)

其中$I$为无穷测度集.若$H_{3}\not\equiv0$, 则由(3.19) 和(3.21) 式得

$\frac{2(5k-1-6k\varepsilon-6\varepsilon)}{33(k+1)}\leq\frac{k+145+12k\varepsilon+12\varepsilon}{55(k+1)}.$

令$\varepsilon\rightarrow 0$得$\frac{2(5k-1)}{33(k+1)}\leq\frac{k+145}{55(k+1)}.$解得$k\leq9$.当$k=10$时产生矛盾，故$H_{3}\equiv0$.同理由(3.20) 和(3.22) 式可得当$k=10$时$H_{4}\equiv0$.与定理2证明的讨论相同, 可得$0, 1, \infty, a$为$f$与$g$的CM$^{\ast}$公共值, 从而由定理B得$f$是$g$的拟分式线性变换.定理2证毕.

定理3的证明 假设$f$不是$g$的分式线性变换, 不失一般性, 设$a_{i}\neq\infty, b_{i}\neq\infty, 1\leq i\leq6$.存在不全为零的常数$A_{i}, 1\leq i\leq6$, 使得函数

$F(f,g)=A_{1}f^{2}g+A_{2}fg+A_{3}f^{2}+A_{4}f+A_{5}g+A_{6}$

满足$F(a_{i},b_{i})=0, i=1,\cdots,5.$若$F(f,g)\equiv 0$, 则$(A_{1}f^{2}+A_{2}f+A_{5})g=-A_{3}f^{2}-A_{4}f-A_{6}.$因为$f$不是常数, 所以$A_{1}f^{2}+A_{2}f+A_{5}\not\equiv 0$.故

$\begin{align} g=\frac{-A_{3}f^{2}-A_{4}f-A_{6}}{A_{1}f^{2}+A_{2}f+A_{5}}.\label{eqc28} \end{align}$

(3.23)

若存在一个$a_{i}$使得$A_{1}a_{i}^{2}+A_{2}a_{i}+A_{5}=0$, 那么$A_{3}a_{i}^{2}+A_{4}a_{i}+A_{6}=0$, 因此$f-a_{i}$为$-A_{3}f^{2}-A_{4}f-A_{6}$和$A_{1}f^{2}+A_{2}f+A_{5}$的公因式，所以$g$是$f$的分式线性变换, 这与假设矛盾.因此对任意的$a_{i}$, 有$A_{1}a_{i}^{2}+A_{2}a_{i}+A_{5}\neq0$.因为$F(a_{i},b_{i})=0$, 有

$\begin{align} b_{i}=\frac{-A_{3}a_{i}^{2}-A_{4}a_{i}-A_{6}}{A_{1}a_{i}^{2}+A_{2}a_{i}+A_{5}}.\label{eqc29} \end{align}$

(3.24)

由(3.23) 和(3.24) 式得

$\begin{align} g-b_{i}=\frac{(f-a_{i})(c_{i}f-d_{i})}{A_{1}f^{2}+A_{2}f+A_{5}}, \label{eqc30} \end{align}$

(3.25)

其中$c_{i},d_{i}$为常数.若$c_{i}= 0$, 由假设$g$不是$f$的分式线性变换，则此时$A_{1}\ne 0$.由引理5和引理6(a), 得$T(r,g)=2T(r,f)+S(r).$由第二基本定理, 有

$ 6T(r,f)=3T(r,g)+S(r)\leq\sum\limits_{i=1}^{5}\overline N(r,b_{i},g)+S(r) =\sum\limits_{i=1}^{5}\overline N(r,a_{i},f)+S(r)\leq5T(r,f)+S(r), $

得$T(r,f)=S(r)$, 矛盾.若$c_{i}\ne 0$, 由假设$g$不是$f$的分式线性变换，则此时$A_{1}\left(\frac{d_{i}}{c_{i}}\right)^{2}+A_{2}\left(\frac{d_{i}}{c_{i}}\right)+A_{5}\neq0$, 因此仍有$T(r,g)=2T(r,f)+S(r)$，同样可得矛盾$T(r,f)=S(r)$.故$F(f,g)\not\equiv0$.

由引理6(c), 可以得到$T(r,f)=\overline{N}(r,f)+S(r)$, 所以有$m(r,f)=S(r), m(r,g)=S(r)$.从而$m(r,F)=S(r)$.由引理6(a), 得

$\begin{align} \sum\limits_{i=1}^{5}\overline N(r,a_{i},f)&\leq\overline N(r,0,F)+S(r) \leq T(r,F)+S(r)=N(r,F)+S(r) \notag\\ &\leq2N(r,f)+N(r,g)+S(r)\leq2T(r,f)+T(r,g)+S(r) \notag\\ &=3T(r,f)+S(r). \label{eqc31a} \end{align}$

(3.26)

同理, 可得

$\begin{align} \sum\limits_{i=1}^{4}\overline N(r,a_{i},f)+\overline N(r,f-a_{6}=0=g-b_{6})\leq3T(r,f)+S(r). \label{eqc31b} \end{align}$

(3.27)

则由(1.1) 和(3.27) 式得

$\begin{align} &\sum\limits_{i=1}^{6}\overline N(r,a_{i},f)=\sum\limits_{i=1}^{4}\overline N(r,a_{i},f)+\overline N(r,a_{5},f)+ \overline N(r,a_{6},f) \notag\\ \leq&3T(r,f)+\overline N(r,a_{5},f)+ \overline N(r,a_{6},f)-\overline N(r,f-a_{6}=0=g-b_{6})+S(r) \notag\\ \leq&(3+\lambda)T(r,f)+\overline N(r,a_{5},f)+S(r). \label{eqc32} \end{align}$

(3.28)

同理, 有

$\begin{align} \sum\limits_{i=1}^{6}\overline N(r,a_{i},f)\leq(3+\lambda)T(r,f)+\overline N(r,a_{i},f)+S(r), \quad i=1, 2, 3, 4. \label{eqc33} \end{align}$

(3.29)

由第二基本定理, 得

$\begin{align} 4T(r,f)\leq\sum\limits_{i=1}^{6}\overline N(r,a_{i},f)+S(r).\label{eqc34} \end{align}$

(3.30)

由(3.28), (3.29), (3.30) 式和引理6(b), 得

$\begin{equation*} \begin{split} 20T(r,f)&\leq5\sum\limits_{i=1}^{6}\overline N(r,a_{i},f)+S(r)\leq(15+5\lambda)T(r,f)+\sum\limits_{i=1}^{5}\overline N(r,a_{i},f)+S(r)\\&=(18+5\lambda)T(r,f)+S(r), \end{split} \end{equation*}$

即

$\begin{align} (2-5\lambda)T(r,f)\leq S(r). \label{eqc35} \end{align}$

(3.31)

已知$\lambda\in[0,\frac{2}{5})$, 则(3.31) 式不成立, 所以假设不成立.故$f$为$g$的分式线性变换.定理3证毕.

定理4的证明 假设$f$不是$g$的拟分式线性变换, 不失一般性, 设$a_{k}\not\equiv\infty, b_{k}\not\equiv\infty, 1\leq k\leq5$.考虑$\{a_{1},a_{2},a_{3},a_{4}\}$中的任意三个小函数, 不失一般性, 假设$a_{1},a_{2},a_{3}$, 令$L(z)$为满足$L(a_{k})=b_{k}(k=1,2,3)$的拟分式线性变换, 则有

$\begin{align} \sum\limits_{k=1}^{3}\overline N(r,a_{k},f)\leq N(r,0,L(f)-g)\leq T(r,f)+T(r,g)+S(r).\label{eqc36} \end{align}$

(3.32)

显然从$\{a_{1},a_{2},a_{3},a_{4}\}$中任取三个小函数, 都可以得到相应的$L(f)$和(3.32) 式, 这样有四个不等式, 相加得

$\begin{align} 3\sum\limits_{k=1}^{4}\overline N(r,a_{4},f)\leq4T(r,f)+4T(r,g)+S(r).\label{eqc37} \end{align}$

(3.33)

由引理1, (1.2) 和(3.33) 式得

$\begin{equation*} \begin{split} 3T(r,f)&\leq\sum\limits_{k=1}^{5}\overline N(r,a_{k},f)+\varepsilon T(r,f)+S(r,f)\\&=\sum\limits_{k=1}^{4}\overline N(r,a_{k},f)+\overline N(r,a_{5},f)+\varepsilon T(r,f)+S(r,f) \\&\leq\frac{4}{3}T(r,f)+\frac{4}{3}T(r,g)+(\lambda+\varepsilon) T(r,f)+S(r,f), \end{split} \end{equation*}$

即

$\begin{align} (5-3\lambda-3\varepsilon)T(r,f)\leq4T(r,g)+S(r,f).\label{eqc38} \end{align}$

(3.34)

同理, 也有

$\begin{align} (5-3\lambda-3\varepsilon)T(r,g)\leq4T(r,f)+S(r,g).\label{eqc39} \end{align}$

(3.35)

(3.34) 与(3.35) 式相加得

$\begin{align} (1-3\lambda-3\varepsilon)\{T(r,f)+T(r,g)\}\leq S(r,f)+S(r,g).\label{eqc40} \end{align}$

(3.36)

已知$\lambda\in[0,\frac{1}{3})$, 取$\varepsilon$足够小, 则(3.36) 式不成立, 所以假设不成立.故$f$为$g$的拟分式线性变换.定理4证毕.