本文中的亚纯函数均指复平面上的亚纯函数. 设$ f $是非常数亚纯函数, 采用亚纯函数唯一性理论中的一些基本记号和结论^[1-2], 如$ T(r, f), \; N(r, f), \; \overline{N}(r, f), \; m(r, f) $等. 令$ S(r, f) $表示任意满足$ S(r, f)=o\{T(r, f)\} $ ($ r\rightarrow +\infty $, $ r\notin E $) 的量, 其中$ E $是一个有穷线性测度的集合, $ S(r, f) $每次出现时$ E $可能不相同. 若对亚纯函数$ a $, 有$ T(r, a) $ = $ S(r, f) $, 则称$ a $为$ f $的一个小函数. $ N_{k}\left(r, \frac{1}{f}\right) $表示$ f $的零点的计数函数, 其中当$ f $的零点重数$ m\leq k $时, 计$ m $次; 当$ m>k $时, 计$ k $次. $ N_{k)}\left(r, \frac{1}{f}\right) $表示$ f-a $的零点重数$ m\leq k $的计数函数, $ N_{(k}\left(r, \frac{1}{f}\right) $表示$ f-a $的零点重数$ m\geq k $的计数函数.

下面我们介绍由Lahiri I^[3-4]引进的权分担记号.

定义1.1  设$ f, \; g $是两个非常数亚纯函数, $ a\in \mathbb{C}\cup \{\infty\} $, $ k $为一正整数或$ \infty $. $ E_{k}(a, f) $表示$ f-a $的所有零点, 若零点重数$ m\leq k $时, 计$ m $次; 若$ m>k $时, 计$ k+1 $次. 若$ E_{k}(a, f)=E_{k}(a, g) $, 则称$ f $和$ g $以权$ k $分担$ a $.

这里记$ f $和$ g $分担$ (a, k) $表示$ f $和$ g $以权$ k $分担$ a $. 显然若$ f $和$ g $分担$ (a, k) $, 那么对任意的$ p\; (0\leq p < k) $, $ p $为整数, 都有$ f $和$ g $分担$ (a, p) $, 同时, 当且仅当$ f $和$ g $分担$ (a, 0) $ (或$ (a, \infty) $)时, $ f $和$ g $分担$ a $ IM (或$ a $ CM).

设$ S $是一个复数集合, $ f $和$ g $是两个非常数亚纯函数, 定义

$ \begin{aligned} E_{f}(S, k)=\mathop{\bigcup}\limits_{a\in S}E_{f}(a, k), \end{aligned} $

若$ E_{f}(S, k)=E_{g}(S, k) $, 则称$ f $和$ g $以权$ k $分担集合$ S $, 若$ E_{f}(S, \infty)=E_{g}(S, \infty) $, 则称$ S $为$ f $和$ g $的CM公共值集; 若$ E_{f}(S, 0)=E_{g}(S, 0) $, 则称$ S $为$ f $和$ g $的IM公共值集. 显然$ E_{f}(S, \infty)=E_{f}(S) $, $ E_{f}(S, 0)=\overline{E}_{f}(S) $.

在亚纯函数值分布理论中, 一个著名的问题是1959年由Hayman W K^[5]提出的, 即设$ f $是复平面上超越亚纯函数, $ n $为正整数, 则$ f^{n}f^{'} $取可能为零以外的任意复数无穷多次. 上述问题直到1995年才被陈怀惠和方明亮^[6], Zalcman L^[7]分别证得. 针对上述著名的Hayman问题, 杨重骏与华歆厚^[8]研究了微分单项式的唯一性并获得了下述定理.

定理1.1^[8]  设$ f $, $ g $是两个非常数整函数(亚纯函数), $ n>6\; (n>11) $是正整数, 若$ f^{n}f^{'} $与$ g^{n}g^{'} $分担$ 1 $ CM, 则或者$ f=c_{1}e^{cz}, \; g=c_{2}e^{-cz} $, 其中$ c_{1}, \; c_{2}, \; c $是非零常数, 且满足$ (c_{1}c_{2})^{n+1}c^{2}=-1 $, 或者$ f(z)\equiv tg(z) $且满足$ t^{n+1}=1 $.

近20年来, 许多复分析学者对微分多项式的唯一性问题开始了广泛的研究并获得了丰富的成果, 详见文献[9–13]. 注意到, $ f(z)^{n}f(z)^{'}=\frac{1}{n+1}(f(z)^{n+1})^{'} $, 进而方明亮^[9]考虑了定理1.1中$ k $阶导数的情形并获得了下述结果.

定理1.2^[9]  设$ f $, $ g $是两个非常数整函数, $ n, \; k $均为正整数且满足$ n>2k+4 $, 若$ (f^{n})^{(k)} $与$ (g^{n})^{(k)} $分担$ 1 $ CM, 则或者$ f=c_{1}e^{cz}, \; g=c_{2}e^{-cz} $, 其中$ c_{1}, \; c_{2}, \; c $是非零常数, 且满足$ (-1)^{k}(c_{1}c_{2})^{n}(nc)^{2k}=1 $, 或者$ f(z)\equiv tg(z) $且满足$ t^{n}=1 $.

定理1.3^[9]  设$ f $, $ g $是两个非常数整函数, $ n, \; k $均为正整数且满足$ n>2k+8 $, 若$ (f^{n}(f-1))^{(k)} $与$ (g^{n}(g-1))^{(k)} $分担$ 1 $ CM, 则$ f(z)\equiv g(z) $.

2008年, 张晓宇等人^[10]进一步将定理1.3中的$ (f^{n}(f-1))^{(k)} $推广到$ (f^{n}P(f))^{(k)} $, 其中$ P(z)=a_{m}z^{m}+a_{m-1}z^{m-1}+\cdots+a_{1}z+a_{0} $为$ m $次非零多项式, 得到了如下结果.

定理1.4^[10]  设$ f $, $ g $是两个非常数整函数, $ n, \; k, \; m $均为正整数且满足$ n>3m+2k+5 $, $ P(z)=a_{m}z^{m}+a_{m-1}z^{m-1}+\cdots+a_{1}z+a_{0} $或$ P(z)\equiv c_{0} $, 其中$ a_{0}\neq0, \; a_{1}, \; \cdots, \; a_{m-1}, \; a_{m}\neq0 $, $ c_{0}\neq0 $为常数, 若$ (f^{n}(P(f)))^{(k)} $与$ (g^{n}(P(g)))^{(k)} $分担$ 1 $ CM, 则

(Ⅰ) 若$ P(z)=a_{m}z^{m}+a_{m-1}z^{m-1}+\cdots+a_{1}z+a_{0} $, 则下述情况之一成立:

(Ⅰ.ⅰ) $ f(z)\equiv tg(z) $, 其中$ t $为非零常数且满足$ t^{d}=1 $, $ d=GCD(n+m, \cdots, n+m-i, \cdots, n) $, 且存在一个$ i\in\{0, \; 1, \; \cdots, \; m\} $, 使得$ a_{m-i}\neq0 $;

(Ⅰ.ⅱ) $ R(f, g)\equiv0 $, 其中$ R(\omega_{1}, \omega_{2})=\omega_{1}^{n}(a_{m}\omega_{1}^{m}+\cdots+a_{1}\omega_{1}+a_{0})-\omega_{2}^{n}(a_{m}\omega_{2}^{m}+\cdots+a_{1}\omega_{2}+a_{0}) $.

(Ⅱ) 若$ P(z)\equiv c_{0} $, 则下述情况之一成立:

(Ⅱ.ⅰ) $ f(z)\equiv tg(z) $, 其中$ t $为非零常数且满足$ t^{n}=1 $;

(Ⅱ.ⅱ) $ f=c_{1}/\sqrt[n]{c_{0}}e^{cz}, \; g=c_{2}/\sqrt[n]{c_{0}}e^{-cz} $, 其中$ c_{1}, \; c_{2}, \; c $是非零常数, 且满足$ (-1)^{k}(c_{1}c_{2})^{n}(nc)^{2k} $ $ =1 $.

近年来, 一些学者考虑了上述微分多项式中当分担值$ a $替换为集合$ S $时, 其中$ S=\{a\in\mathbb{C}:a^{s}=1\} $, 是否还能获得上述定理中$ f $与$ g $类似的关系. $ 2018 $年, An V H等人^[11]考虑了$ (f^{n})^{(k)} $与$ (g^{n})^{(k)} $ CM分担$ S $的情况, 得到了下述结果.

定理1.5^[11]  设$ f $, $ g $是非常数亚纯函数, $ n, \; k, \; s $均为正整数且满足$ n>2k+\frac{2k+8}{s} $, $ s\geq2 $, $ S=\{a\in\mathbb{C}:a^{s}=1\} $, 若$ E_{(f^{n})^{(k)}}(S, \infty)=E_{(g^{n})^{(k)}}(S, \infty) $, 则下述情况之一成立:

(ⅰ) $ f(z)\equiv tg(z) $, 其中$ t $为非零常数且满足$ t^{ns}=1 $, $ t\in\mathbb{C} $;

(ⅱ) $ f=c_{1}e^{cz}, \; g=c_{2}e^{-cz} $, 其中$ c_{1}, \; c_{2}, \; c $是非零常数, 且满足$ (-1)^{ks}(c_{1}c_{2})^{ns}(nc)^{2ks}=1 $.

2020年, Chao M等人^[12]考虑了$ (f^{n})^{(k)} $与$ (g^{n})^{(k)} $ IM分担集合$ S $与权1分担集合$ S $的情况, 得到了下述定理.

定理1.6^[12]  设$ f $, $ g $是非常数亚纯函数, $ n, \; k, \; s $均为正整数且满足$ n>2k+\frac{3k+9}{s} $, $ s\geq2 $, $ S=\{a\in\mathbb{C}:a^{s}=1\} $, 若$ E_{(f^{n})^{(k)}}(S, 1)=E_{(g^{n})^{(k)}}(S, 1) $, 则定理$ 1.5 $的结论成立.

定理1.7^[12]  设$ f $, $ g $是非常数亚纯函数, $ n, \; k, \; s $均为正整数且满足$ n>2k+\frac{8k+14}{s} $, $ s\geq2 $, $ S=\{a\in\mathbb{C}:a^{s}=1\} $, 若$ E_{(f^{n})^{(k)}}(S, 0)=E_{(g^{n})^{(k)}}(S, 0) $, 则定理$ 1.5 $的结论成立.

为了寻求这个方向上更多的结果, 本文结合多项式加权和的概念进一步探讨将定理1.5, 定理1.6以及定理1.7中$ (f^{n})^{(k)} $替换为$ (f^{n}(P(f))^{(k)} $时的情形, 为了方便本文的叙述, 我们引入下述定义.

定义1.2  设$ P(z)=a_{m}z^{m}+\cdots+a_{1}z+a_{0} $为$ m $次非零多项式, 其中$ v\; (1\leq\; v\; \leq\; m) $个不同的零点分别记为$ d_{1}, \; d_{2}, \; \cdots, \; d_{v} $, 相对应的零点重数分别记为$ p_{1}, \; p_{2}, \; \cdots, \; p_{v} $且满足$ p_{1}\; \leq\; p_{2}\; \leq\; \cdots\; \leq\; p_{v} $. 令

$ \begin{aligned} \gamma=\sum ^{v}_{i=1}\lambda\cdot[p_{i}-(k+1)], \end{aligned} $

其中$ \lambda=\begin{cases}0, \; p_{i}\leq k\\1, \; p_{i} >k\end{cases} $, 称$ \gamma $为$ P(z) $的零点重数关于$ k $的加权和.

当$ t $为$ P(z) $重数不超过$ k $的零点个数(不计重数) 时, 则

$ \begin{aligned} \gamma=\sum ^{v}_{i=t+1}[p_{i}-(k+1)], \end{aligned} $

且由定义可知$ 0\leq\gamma\leq m $. 特别地, 当$ \sum\limits_{i=1}^{t}p_{i}=m $时有$ \gamma=0 $.

结合加权和的定义, 本文得到了下述定理, 定理中的$ P(z) $相关符号与定义$ 1.2 $中的符号含义一致, 下文中不再一一叙述.

定理1.8  设$ f $, $ g $是非常数亚纯函数, $ n, \; k, \; s $均为正整数, $ s\geq2 $, $ P(z)=a_{m}z^{m}+a_{m-1}z^{m-1}+\cdots+a_{1}z+a_{0} $或$ P(z)\equiv c_{0} $, 其中$ a_{0}\neq0, \; a_{1}, \; \cdots, \; a_{m-1}, \; a_{m}\neq0 $, $ c_{0}\neq0 $为常数, $ P(z) $中相关符号如定义$ 1.2 $所设, $ S=\{a\in\mathbb{C}:a^{s}=1\} $, 若$ E_{(f^{n}P(f))^{(k)}}(S, k)=E_{(g^{n}P(g))^{(k)}}(S, k) $, 则

(Ⅰ) 当$ k=0 $时,

(Ⅰ.ⅰ) 若$ P(z)=a_{m}z^{m}+a_{m-1}z^{m-1}+\cdots+a_{1}z+a_{0} $且满足$ n> max\{2k+m+1, 2k+\frac{8k+14+7m-5\gamma}{s}-\gamma\} $, 则下述情况之一成立:

(Ⅰ.ⅰ.ⅰ) $ f(z)\equiv tg(z) $, 其中$ t $为常数且满足$ t^{sd}=1 $, $ d=GCD(n+m, \cdots, n+m-i, \cdots, n) $;

(Ⅰ.ⅰ.ⅱ) $ R(f, g)\equiv0 $, 其中$ R(\omega_{1}, \omega_{2})=\omega_{1}^{n}(a_{m}\omega_{1}^{m}+\cdots+a_{1}\omega_{1}+a_{0})-h\omega_{2}^{n}(a_{m}\omega_{2}^{m}+\cdots+a_{1}\omega_{2}+a_{0}) $且$ h^{s}=1 $;

(Ⅰ.ⅰ.ⅲ) $ (f^{n}P(f))^{(k)}(g^{n}P(g))^{(k)}\equiv h $, 其中$ h^{s}=1 $.

(Ⅰ.ⅱ) 若$ P(z)\equiv c_{0} $且满足$ n>2k+\frac{8k+14}{s} $, 则下述情况之一成立:

(Ⅰ.ⅱ.ⅰ) $ f(z)\equiv tg(z) $, 其中$ t $为非零常数且满足$ t^{ns}=1 $;

(Ⅰ.ⅱ.ⅱ) $ f=c_{1}e^{cz}, \; g=c_{2}e^{-cz} $, 其中$ c_{1}, \; c_{2}, \; c $是非零常数且满足$ c_{0}^{2s}(-1)^{ks}(c_{1}c_{2})^{ns}(nc)^{2ks}=1 $.

(Ⅱ) 当$ k=1 $时,

(Ⅱ.ⅰ) 若$ P(z)=a_{m}z^{m}+a_{m-1}z^{m-1}+\cdots+a_{1}z+a_{0} $且满足$ n>max\{2k+m+1, 2k+\frac{3k+9+\frac{9}{2}m-\frac{5}{2}\gamma}{s}-\gamma\} $, 则(Ⅰ.ⅰ) 的结论成立;

(Ⅱ.ⅱ) 若$ P(z)\equiv c_{0} $且满足$ n>2k+\frac{3k+9}{s} $, 则(Ⅰ.ⅱ) 的结论成立.

(Ⅲ) 当$ k\geq2 $时,

(Ⅲ.ⅰ) 若$ P(z)=a_{m}z^{m}+a_{m-1}z^{m-1}+\cdots+a_{1}z+a_{0} $且满足$ n> max\{2k+m+1, 2k+\frac{2k+8+4m-2\gamma}{s}-\gamma\} $, 则(Ⅰ.ⅰ) 的结论成立;

(Ⅲ.ⅱ) 若$ P(z)\equiv c_{0} $且满足$ n>2k+\frac{2k+8}{s} $, 则(Ⅰ.ⅱ) 的结论成立.

备注1.1  特别地, 当$ P(z)\equiv c_{0} $时, 由定理1.8可得到定理1.5 – 1.7, 因此定理1.8推广了定理1.5 – 1.7.

引理2.1^[2]  设$ f $, $ g $是两个非常数亚纯函数且$ a_{n}(z)(\not \equiv0), a_{n-1}(z), \; \cdots, \; a_{0}(z) $是满足$ T(r, a_{i})=S(r, f), \; i=0, 1, 2, \cdots, n $的亚纯函数, 则

$ \begin{aligned} T(r, a_{n}f^{n}+a_{n-1}f^{n-1}+\cdots+a_{1}f+a_{0})=nT(r, f)+S(r, f). \end{aligned} $

设$ f $, $ g $是两个非常数亚纯函数, 构造如下函数.

$ \begin{aligned} H=\left(\frac{f^{''}}{f'}-\frac{2f^{'}}{f-1}\right)-\left(\frac{g^{''}}{g^{'}}-\frac{2g^{'}}{g-1}\right). \end{aligned} $

引理2.2^[13]  设$ f $, $ g $是两个非常数亚纯函数, 若$ f $, $ g $分担$ (1, k), k\geq2 $且$ H\not\equiv0 $, 则

$ \begin{aligned} T(r, f)\leq N_{2}(r, f)+N_{2}\left(r, \frac{1}{f}\right)+N_{2}(r, g)+N_{2}\left(r, \frac{1}{g}\right)+S(r, f)+S(r, g). \end{aligned} $

$ T(r, g) $类同.

引理2.3^[14]  设$ f $, $ g $是两个非常数亚纯函数, 若$ f $, $ g $分担$ (1, 1) $且$ H\not\equiv0 $, 则

$ \begin{aligned} T(r, f)\leq& N_{2}(r, f)+N_{2}\left(r, \frac{1}{f}\right)+N_{2}(r, g)+N_{2}\left(r, \frac{1}{f}\right)\\ &+\frac{1}{2}\overline{N}\left(r, \frac{1}{f}\right)+\frac{1}{2}\overline{N}(r, f)+S(r, f)+S(r, g). \end{aligned} $

$ T(r, g) $类同.

引理2.4^[14]  设$ f $, $ g $是两个非常数亚纯函数, 若$ f $, $ g $分担$ (1, 0) $且$ H\not\equiv0 $, 则

$ \begin{aligned} T(r, f)\leq& N_{2}(r, f)+N_{2}\left(r, \frac{1}{f}\right)+N_{2}(r, g)+N_{2}\left(r, \frac{1}{g}\right)+2\overline{N}(r, f)+2\overline{N}\left(r, \frac{1}{f}\right)\\ &+\overline{N}\left(r, \frac{1}{g}\right)+\overline{N}(r, g)+S(r, f)+S(r, g). \end{aligned} $

$ T(r, g) $类同.

引理2.5^[15]  设$ f $是非常数亚纯函数, $ p, \; k $为正整数, 则

$ \begin{aligned} N_{p}\left(r, \frac{1}{f^{(k)}}\right)\leq N_{p+k}\left(r, \frac{1}{f}\right)+k\overline{N}(r, f)+S(r, f). \end{aligned} $

引理2.6  设$ f $是非常数亚纯函数, $ P(z) $中相关符号如定义$ 1.2 $所设, $ n, \; k $为正整数且$ n>k $, 则

$ \begin{aligned} (n-2k+\gamma)T(r, f)+kN(r, f)+N\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)\leq T(r, (f^{n}P(f))^{(k)})+S(r, f). \end{aligned} $

证明: 由对数导数引理可得

$ \begin{align} \begin{aligned} &(n+m-k)m(r, f)\leq m(r, f^{n-k}P(f))\leq m(r, (f^{n}P(f))^{(k)})+m\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)+O(1)\\ \leq& m(r, (f^{n}P(f))^{(k)})+T\left(r, \frac{(f^{n}P(f))^{(k)}}{f^{n-k}P(f)}\right)-N\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)+S(r, f)\\ =&m(r, (f^{n}P(f))^{(k)})+m\left(r, \frac{(f^{n}P(f))^{(k)}}{f^{n-k}P(f)}\right)+N\left(r, \frac{(f^{n}P(f))^{(k)}}{f^{n-k}P(f)}\right)\\ &-N\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)+S(r, f)\leq m(r, (f^{n}P(f))^{(k)})+kN(r, f)+k\overline{N}(r, f)\\ &+N_{k)}\left(r, \frac{1}{P(f)}\right)+kN_{(k+1}\left(r, \frac{1}{P(f)}\right)+km(r, f)-N\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)\\ &+S(r, f)\leq m(r, (f^{n}P(f))^{(k)})+kT(r, f)+k\overline{N}(r, f)+N_{k+1}\left(r, \frac{1}{P(f)}\right)\\ &-N\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)+S(r, f).\\ \end{aligned} \end{align} $

(2.1)

且

$ \begin{align} \begin{aligned} N(r, (f^{n}P(f))^{(k)})=(m+n)N(r, f)+k\overline{N}(r, f). \end{aligned} \end{align} $

(2.2)

将$ P(f) $改写为下述形式:

$ \begin{align} \begin{aligned} P(f)=a_{m}\prod ^{t}_{i=1}(f-d_{i})^{p_{i}}\prod ^{v}_{j=t+1}(f-d_{j})^{p_{j}}, \end{aligned} \end{align} $

(2.3)

其中$ p_{1}\; \leq\; p_{2}\; \leq\; \cdots\; p_{t}\; \leq\; k\; <\; k+1\; \leq\; p_{t+1}\; \leq\; \cdots\; \leq\; p_{v} $.

由定义1.2与(2.3) 式, 得

$ \begin{align} \begin{aligned} N_{k+1}\left(r, \frac{1}{P(f)}\right)&\leq \sum^{t}_{i=1}p_{i}N\left(r, \frac{1}{f-d_{j}}\right)+(k+1)\sum^{v}_{j=t+1}\overline{N}\left(r, \frac{1}{f-d_{j}}\right)+S(r, f)\\ & \leq \left[\sum^{t}_{i=1}p_{i}+(k+1)(v-t)\right]T(r, f)+S(r, f)\\ & \leq \{{m-\sum ^{v}_{i=t+1}[p_{i}-(k+1)]}\}T(r, f)+S(r, f)\\ & \leq (m-\tau)T(r, f)+S(r, f). \end{aligned} \end{align} $

(2.4)

结合(2.1), (2.2) 和(2.4) 式, 得

$ \begin{aligned} &(n+m-k)T(r, f)+kN(r, f)=(n+m-k)m(r, f)+(m+n)N(r, f)\\ \leq& N(r, (f^{n}P(f))^{(k)})+m(r, (f^{n}P(f))^{(k)})-k\overline{N}(r, f)+kT(r, f)\\ &+k\overline{N}(r, f)+N_{k+1}\left(r, \frac{1}{P(f)}\right)-N\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)+S(r, f)\\ \leq& T(r, (f^{n}P(f))^{(k)})+kT(r, f)+(m-\gamma)T(r, f)-N\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)+S(r, f). \end{aligned} $

因此

$ \begin{aligned} (n+\gamma-2k)&T(r, f)+kN(r, f)+N\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)\leq T(r, (f^{n}P(f))^{(k)})+S(r, f). \end{aligned} $

引理2.7^[16]  设$ f_{1}, \; f_{2} $是两个非常数亚纯函数, 若$ c_{1}f_{1}+c_{2}f_{2}=c_{3} $且$ c_{1}, \; c_{2}, \; c_{3} $为非零常数, 则

$ \begin{aligned} T(r, f_{1})\leq \overline{N}\left(r, f_{1}\right)+\overline{N}\left(r, \frac{1}{f_{1}}\right)+\overline{N}\left(r, \frac{1}{f_{2}}\right)+S(r, f_{1}). \end{aligned} $

引理2.8^[17]  设$ f $是非常数亚纯函数, $ a_{1}, \; a_{2}, \cdots\; a_{q} $为$ q $个判别的复数, 则

$ \begin{aligned} (q-1)T(r, f)<\overline{N}(r, f)+\sum ^{q}_{i=1}\overline{N}\left(r, \frac{1}{f-a_{i}}\right)-\text{log}r+O(1). \end{aligned} $

引理2.9  设$ f, \; g $是两个非常数亚纯函数, $ P(z)=a_{m}z^{m}+\cdots+a_{1}z+a_{0} $为$ m $次非零多项式, 若$ ((f^{n}P(f))^{(k)})^{s}=((g^{n}P(g))^{(k)})^{s} $, 则

(Ⅰ) 若$ P(z)=a_{m}z^{m}+a_{m-1}z^{m-1}+\cdots+a_{1}z+a_{0} $, 且$ n>2k+m+1 $, 则下述结论之一成立:

(Ⅰ.ⅰ) $ f(z)\equiv tg(z) $, 其中$ t $为常数且满足$ t^{sd}=1 $, $ d=GCD(n+m, \cdots, n+m-i, \cdots, n) $;

(Ⅰ.ⅱ) $ R(f, g)=0 $, 其中$ R(\omega_{1}, \omega_{2})=\omega_{1}^{n}(a_{m}\omega_{1}^{m}+\cdots+a_{1}\omega_{1}+a_{0})-h\omega_{2}^{n}(a_{m}\omega_{2}^{m}+\cdots+a_{1}\omega_{2}+a_{0}) $, $ h^{s}=1 $.

(Ⅱ) 若$ P(z)\equiv c_{0} $, 且$ n\geq2k+1 $, 则$ f(z)\equiv tg(z) $, 其中$ t $为非零常数且满足$ t^{ns}=1 $.

证  我们分两种情形讨论.

情形1  当$ P(z)=a_{m}z^{m}+a_{m-1}z^{m-1}+\cdots+a_{1}z+a_{0} $时, 因为$ ((f^{n}P(f))^{(k)})^{s}=((g^{n}P(g))^{(k)})^{s} $, 从而$ (f^{n}P(f))^{(k)}=h(g^{n}P(g))^{(k)}=(hg^{n}P(g))^{(k)} $, 其中$ h^{s}=1 $, 因此$ f^{n}P(f)=hg^{n}P(g)+Q(z) $, $ Q(z) $为$ deg(Q(z))\leq k-1 $多项式. 下证$ deg(Q(z))\equiv0 $, 假设$ Q(z)\not\equiv0 $, 则

$ \begin{align} \begin{aligned} \frac{f^{n}P(f)}{Q(z)}=h\frac{g^{n}P(g)}{Q(z)}+1. \end{aligned} \end{align} $

(2.5)

结合引理2.7与(2.5) 式, 得

$ \begin{align} \begin{aligned} T(r, \frac{f^{n}P(f)}{Q(z)})\leq\overline{N}\left(r, \frac{f^{n}P(f)}{Q(z)}\right)+\overline{N}\left(r, \frac{Q(z)}{f^{n}P(f)}\right)+\overline{N}\left(r, \frac{Q(z)}{g^{n}P(g)}\right)+S(r, f). \end{aligned} \end{align} $

(2.6)

由(2.6) 式, 得

$ \begin{align} \begin{aligned} T(r, f^{n}P(f))\leq& T\left(r, \frac{f^{n}P(f)}{Q(z)}\right)+T(r, Q(z))+O(1)\leq\overline{N}\left(r, \frac{f^{n}P(f)}{Q(z)}\right)\\ &+\overline{N}\left(r, \frac{Q(z)}{f^{n}P(f)}\right)+\overline{N}\left(r, \frac{Q(z)}{g^{n}P(g)}\right)+T(r, Q(z))+S(r, f)\leq\overline{N}\left(r, \frac{1}{f}\right)\\ &+\overline{N}\left(r, \frac{1}{P(f)}\right)+\overline{N}(r, f)+\overline{N}(r, \frac{1}{Q(z)})+\overline{N}\left(r, \frac{1}{g}\right)+\overline{N}\left(r, \frac{1}{P(g)}\right)\\ &+T(r, Q(z))+S(r, f)\leq\overline{N}\left(r, \frac{1}{f}\right)+\overline{N}(r, f)+\overline{N}\left(r, \frac{1}{g}\right)+\overline{N}\left(r, \frac{1}{P(f)}\right)\\ &+\overline{N}\left(r, \frac{1}{P(g)}\right)+2(k-1)\text{log}r+S(r, f).\\ \end{aligned} \end{align} $

(2.7)

同理

$ \begin{align} \begin{aligned} T(r, g^{n}P(g))\leq&\overline{N}\left(r, \frac{1}{g}\right)+\overline{N}(r, g)+\overline{N}\left(r, \frac{1}{f}\right)+\overline{N}\left(r, \frac{1}{P(f)}\right)\\ &+\overline{N}\left(r, \frac{1}{P(g)}\right)+2(k-1)\text{log}r+S(r, g).\\ \end{aligned} \end{align} $

(2.8)

因为$ f, g $为非常数亚纯函数, 故$ T(r, f)\geq \text{log}r+O(1) $, $ T(r, g)\geq \text{log}r+O(1) $, 结合(2.7) 与(2.8) 式, 得

$ \begin{align} \begin{aligned} (m+n)[T(r, f)+T(r, g)]\leq(2k+2m+1)[T(r, f)+T(r, g)]+S(r, f)+S(r, g). \end{aligned} \end{align} $

(2.9)

由(2.9) 式可知, 这与$ n>2k+m+1 $矛盾, 因此$ Q(z)\equiv0 $. 从而$ f^{n}P(f)=hg^{n}P(g) $, 即

$ \begin{align} \begin{aligned} f^{n}(a_{m}f^{m}+a_{m-1}f^{m-1}\cdots+a_{0})=hg^{n}(a_{m}g^{m}+a_{m-1}g^{m-1}\cdots+a_{0}). \end{aligned} \end{align} $

(2.10)

接下来, 我们再分两种子情形讨论.

情形1.1  若$ \frac{f}{g}=t $, 其中$ t $为非零常数, 将$ f=tg $带入(2.10) 式可得$ a_{m}g^{n+m}(t^{n+m}-h)+a_{m-1}g^{n+m-1}(t^{n+m-1}-h)+\cdots+a_{1}g^{n+1}(t^{n+1}-h)+a_{0}g^{n}(t^{n}-h)\equiv0 $. 因为$ g $为非常数亚纯函数, 从而$ t^{d}=h $, 其中$ d=GCD(n+m, \cdots, n+m-i, \cdots, n) $, 进一步有$ (t^{d})^{s}=h^{s}=1 $, 因此$ f(z)\equiv tg(z) $, 其中$ t $为常数且满足$ t^{sd}=1 $, $ d=GCD(n+m, \cdots, n+m-i, \cdots, n) $.

情形1.2  若$ \frac{f}{g}=t $, 其中$ t $不为常数, 将$ f=tg $带入(2.10) 式可得$ R(f, g)=0 $, 其中$ R(\omega_{1}, \omega_{2})=\omega_{1}^{n}(a_{m}\omega_{1}^{m}+\cdots+a_{1}\omega_{1}+a_{0})-h\omega_{2}^{n}(a_{m}\omega_{2}^{m}+\cdots+a_{1}\omega_{2}+a_{0}) $且$ h^{s}=1 $.

情形2  当$ P(z)\equiv c_{0} $时, 因为$ ((f^{n})^{(k)})^{s}=((g^{n})^{(k)})^{s} $, 从而$ (f^{n})^{(k)}=h(g^{n})^{(k)}=(hg^{n})^{(k)} $, 其中$ h^{s}=1 $, 因此$ f^{n}=hg^{n}+Q(z) $, $ Q(z) $为$ deg(Q(z))\leq k-1 $多项式. 下证$ deg(Q(z))\equiv0 $, 假设$ Q(z)\not\equiv0 $.

置

$ \begin{align} \begin{aligned} F=\frac{f^{n}}{Q(z)}, \; \; G=\frac{g^{n}}{Q(z)}. \end{aligned} \end{align} $

(2.11)

因此$ F=hG+1 $, 进一步, 有

$ \begin{align} \begin{aligned} \overline{N}\left(r, {\frac{1}{F-1}}\right)=\overline{N}\left(r, \frac{1}{G}\right). \end{aligned} \end{align} $

(2.12)

结合引理2.1与(2.11) 式, 得

$ \begin{align} \begin{aligned} nT(r, f)=T(r, f^{n})\leq T(r, F)+T(r, Q(z))\leq T(r, F)+(k-1)\text{log}r+O(1). \end{aligned} \end{align} $

(2.13)

$ \begin{align} \begin{aligned} \overline{N}\left(r, \frac{1}{F}\right)=\overline{N}\left(r, \frac{1}{f}\right)\leq T(r, f), \; \; \; \overline{N}\left(r, \frac{1}{G}\right)=\overline{N}\left(r, \frac{1}{g}\right)\leq T(r, g).\\ \end{aligned} \end{align} $

(2.14)

$ \begin{align} \begin{aligned} \overline{N}(r, F)&=\overline{N}(r, f^{n})+\overline{N}\left(r, \frac{1}{Q(z)}\right)\leq\overline{N}(r, f)+\overline{N}\left(r, \frac{1}{Q(z)}\right)\\ & \leq T(r, f)+(k-1)\text{log}r+O(1). \end{aligned} \end{align} $

(2.15)

由$ g^{n}=\frac{1}{h}f^{n}-\frac{1}{h}Q(z) $可知

$ \begin{aligned} nT(r, g)&=T(r, hg^{n})\leq nT(r, f)+T(r, Q(z))+O(1)\\ &\leq nT(r, f)+(k-1)\text{log}r+O(1). \end{aligned} $

进一步, 有

$ \begin{align} \begin{aligned} T(r, g)\leq T(r, f)+\frac{k-1}{n}\text{log}r+O(1)\leq T(r, f)+\frac{1}{2}\text{log}r+O(1). \end{aligned} \end{align} $

(2.16)

结合引理2.8与(2.12), (2.14) – (2.16) 式, 得

$ \begin{align} \begin{aligned} T(r, F)&\leq \overline{N}\left(r, \frac{1}{F}\right)+\overline{N}(r, F)+\overline{N}\left(r, \frac{1}{F-1}\right)-\text{log}r+O(1)\\ &\leq T(r, f)+T(r, f)+(k-1)\text{log}r+T(r, g)-\text{log}r+O(1)\\ &\leq T(r, f)+T(r, f)+(k-1)\text{log}r+T(r, f)+\frac{1}{2}\text{log}r-\text{log}r+O(1)\\ &\leq 3T(r, f)+(k-\frac{3}{2})\text{log}r+O(1). \end{aligned} \end{align} $

(2.17)

由(2.13) 与(2.17) 式, 得

$ \begin{aligned} nT(r, f)-(k-1)\text{log}r\leq 3T(r, f)+(k-\frac{3}{2})\text{log}r+O(1), \end{aligned} $

即

$ \begin{align} \begin{aligned} (n-3)T(r, f)-2(k-1)\text{log}r+\frac{1}{2}\text{log}r\leq O(1). \end{aligned} \end{align} $

(2.18)

因为$ f $为非常数亚纯函数, 故$ T(r, f)\geq \text{log}r+O(1) $, 结合(2.18) 式得

$ \begin{aligned} (n-2k-1)\text{log}r+\frac{1}{2}\text{log}r\leq O(1). \end{aligned} $

这与已知条件$ n\geq2k+1 $矛盾, 因此$ Q(z)\equiv0 $. 进而$ f^{n}=hg^{n} $, 从而$ f(z)\equiv tg(z) $, 其中$ t $为非零常数且满足$ t^{ns}=1 $. 引理2.9证毕.

引理2.10^[10]  设$ f, \; g $是两个非常数整函数, $ n, \; k $为正整数, 且$ n>k $, 如果$ (c_{0}f^{n})^{(k)}(c_{0}g^{n})^{(k)}\equiv h $, $ h $为非零常数, 则$ f=c_{1}e^{cz}, \; g=c_{2}e^{-cz} $, 其中$ c_{0}, \; c_{1}, \; c_{2}, \; c $是非零常数, 且满足$ c_{0}^{2}(-1)^{k}(c_{1}c_{2})^{n}(nc)^{2k}=h $.

设

$ \begin{align} \begin{aligned} F=((f^{n}P(f))^{(k)})^{s}\; , \; \; \; \; \; \; \; \; G=((g^{n}P(g))^{(k)})^{s}. \end{aligned} \end{align} $

(3.1)

置

$ \begin{aligned} H=\left(\frac{F^{''}}{F'}-\frac{2F^{'}}{F-1}\right)-\left(\frac{G^{''}}{G^{'}}-\frac{2G^{'}}{G-1}\right). \end{aligned} $

下面, 我们分三种情形讨论.

情形3  当$ k=0 $时, 即$ E_{(f^{n}P(f))^{(k)}}(S, 0)=E_{(g^{n}P(g))^{(k)}}(S, 0) $, 从而$ F, G $分担(1, 0), 假设$ H\not\equiv0 $, 由引理2.4, 得

$ \begin{align} \begin{aligned} T(r, F)\leq& N_{2}(r, F)+N_{2}\left(r, \frac{1}{F}\right)+N_{2}(r, G)+N_{2}\left(r, \frac{1}{G}\right)+2\overline{N}(r, F)\\ &+2\overline{N}\left(r, \frac{1}{F}\right)+\overline{N}\left(r, \frac{1}{G}\right)+\overline{N}(r, G)+S(r, F)+S(r, G). \end{aligned} \end{align} $

(3.2)

同理

$ \begin{align} \begin{aligned} T(r, G)\leq& N_{2}(r, F)+N_{2}\left(r, \frac{1}{F}\right)+N_{2}(r, G)+N_{2}\left(r, \frac{1}{G}\right)+2\overline{N}(r, G)\\ &+2\overline{N}\left(r, \frac{1}{G}\right)+\overline{N}\left(r, \frac{1}{F}\right)+\overline{N}(r, F)+S(r, F)+S(r, G). \end{aligned} \end{align} $

(3.3)

由引理$ 2.5 $, 得

$ \begin{align} \begin{aligned} N_{2}\left(r, \frac{1}{F}\right)&=N_{2}\left(r, \frac{1}{((f^{n}P(f))^{(k)})^{s}}\right)=2\overline{N}\left(r, \frac{1}{(f^{n}P(f))^{(k)}}\right)\\ &\leq2N_{k+1}\left(r, \frac{1}{f^{n}P(f)}\right)+2k\overline{N}(r, f^{n}P(f))+S(r, (f^{n}P(f))^{(k)})\\ &\leq2\left(N_{k+1}\left(r, \frac{1}{f^{n}}\right)+N_{k+1}\left(r, \frac{1}{P(f)}\right)\right)+2k\overline{N}(r, f^{n}P(f))+S(r, f)\\ &\leq2(k+1)\overline{N}\left(r, \frac{1}{f}\right)+2N_{k+1}\left(r, \frac{1}{P(f)}\right)+2k\overline{N}(r, f)+S(r, f). \end{aligned} \end{align} $

(3.4)

同理

$ \begin{align} \begin{aligned} N_{2}\left(r, \frac{1}{G}\right)\leq2(k+1)\overline{N}\left(r, \frac{1}{g}\right)+2N_{k+1}\left(r, \frac{1}{P(g)}\right)+2k\overline{N}(r, g)+S(r, g). \end{aligned} \end{align} $

(3.5)

由引理$ 2.5 $, 得

$ \begin{align} \begin{aligned} \overline{N}\left(r, \frac{1}{F}\right)&=\overline{N}\left(r, \frac{1}{((f^{n}P(f))^{(k)})^{s}}\right)=\overline{N}\left(r, \frac{1}{(f^{n}P(f))^{(k)}}\right)\\ &\leq N_{k+1}\left(r, \frac{1}{f^{n}P(f)}\right)+k\overline{N}(r, f^{n}P(f))+S(r, (f^{n}P(f))^{(k)})\\ &\leq N_{k+1}\left(r, \frac{1}{f^{n}}\right)+N_{k+1}\left(r, \frac{1}{P(f)}\right)+k\overline{N}(r, f^{n}P(f))+S(r, f)\\ &\leq(k+1)\overline{N}\left(r, \frac{1}{f}\right)+N_{k+1}\left(r, \frac{1}{P(f)}\right)+k\overline{N}(r, f)+S(r, f). \end{aligned} \end{align} $

(3.6)

同理

$ \begin{align} \begin{aligned} \overline{N}\left(r, \frac{1}{G}\right)\leq(k+1)\overline{N}\left(r, \frac{1}{g}\right)+N_{k+1}\left(r, \frac{1}{P(g)}\right)+k\overline{N}(r, g)+S(r, g). \end{aligned} \end{align} $

(3.7)

且

$ \begin{align} \begin{aligned} N_{2}\left(r, \frac{1}{F}\right)&=N_{2}\left(r, \frac{1}{((f^{n}P(f))^{(k)})^{s}}\right)=2\overline{N}\left(r, (f^{n}P(f))^{(k)})\right)\\ &\leq2\left(\overline{N}\left(r, \frac{1}{f^{n-k}P(f)}\right)+\overline{N}\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)\right)\\ &\leq2\overline{N}\left(r, \frac{1}{f}\right)+2\overline{N}\left(r, \frac{1}{P(f)}\right)+2\overline{N}\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right). \end{aligned} \end{align} $

(3.8)

同理

$ \begin{align} \begin{aligned} N_{2}\left(r, \frac{1}{G}\right)\leq2\overline{N}\left(r, \frac{1}{g}\right)+2\overline{N}\left(r, \frac{1}{P(g)}\right)+2\overline{N}\left(r, \frac{g^{n-k}P(g)}{(g^{n}P(g))^{(k)}}\right). \end{aligned} \end{align} $

(3.9)

根据定义1.2, 类同引理2.6的证明可得

$ \begin{align} \begin{aligned} N_{k+1}\left(r, \frac{1}{P(f)}\right)& \leq (m-\tau)T(r, f)+S(r, f). \end{aligned} \end{align} $

(3.10)

同理

$ \begin{align} \begin{aligned} N_{k+1}\left(r, \frac{1}{P(g)}\right)& \leq (m-\tau)T(r, g)+S(r, g). \end{aligned} \end{align} $

(3.11)

结合(3.2), (3.4), (3.6), (3.7), (3.9)–(3.11) 式, 得

$ \begin{aligned} T(r, F)\leq&(4k+4)\overline{N}(r, f)+(4k+4)\overline{N}\left(r, \frac{1}{f}\right)+(k+3)\overline{N}(r, g)\\ &+(k+3)\overline{N}\left(r, \frac{1}{g}\right)+4(m-\gamma)T(r, f)+(m-\gamma)T(r, g)\\ &+2\overline{N}\left(r, \frac{1}{p(g)}\right)+2\overline{N}\left(r, \frac{g^{n-k}P(g)}{(g^{n}P(g))^{(k)}}\right)+S(r, f)+S(r, g). \end{aligned} $

即

$ \begin{align} \begin{aligned} T(r, ((f^{n}P(f))^{(k)})^{s})\leq&(6k+8+4m-4\gamma)T(r, f)+(2k+6+3m-\gamma)T(r, g)\\ &+2k\overline{N}(r, f)+2\overline{N}\left(r, \frac{g^{n-k}P(g)}{(g^{n}P(g))^{(k)}}\right)+S(r, f)+S(r, g). \end{aligned} \end{align} $

(3.12)

同理

$ \begin{align} \begin{aligned} T(r, ((g^{n}P(g))^{(k)})^{s})\leq&(6k+8+4m-4\gamma)T(r, g)+(2k+6+3m-\gamma)T(r, f)\\ &+2k\overline{N}(r, g)+2\overline{N}\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)+S(r, f)+S(r, g). \end{aligned} \end{align} $

(3.13)

由引理2.1, 得

$ \begin{align} \begin{aligned} T(r, ((f^{n}P(f))^{(k)})^{s})&=sT(r, (f^{n}P(f))^{(k)})+S(r, (f^{n}P(f))^{(k)})\\ &=sT(r, (f^{n}P(f))^{(k)})+S(r, f). \end{aligned} \end{align} $

(3.14)

$ \begin{align} \begin{aligned} T(r, ((g^{n}P(g))^{(k)})^{s})&=sT(r, (g^{n}P(g))^{(k)})+S(r, (g^{n}P(g))^{(k)})\\ &=sT(r, (g^{n}P(g))^{(k)})+S(r, g). \end{aligned} \end{align} $

(3.15)

结合引理2.6与(3.14) 式, 得

$ \begin{align} \begin{aligned} &(n-2k+\gamma)sT(r, f)+ksN(r, f)+sN\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)\\ \leq& sT(r, (f^{n}P(f))^{(k)})+S(r, f). \end{aligned} \end{align} $

(3.16)

同理

$ \begin{align} \begin{aligned} &(n-2k+\gamma)sT(r, g)+ksN(r, g)+sN\left(r, \frac{g^{n-k}P(g)}{(g^{n}P(g))^{(k)}}\right)\\ \leq& sT(r, (g^{n}P(g))^{(k)})+S(r, g). \end{aligned} \end{align} $

(3.17)

结合(3.12), (3.13), (3.16) 与(3.17) 式, 得

$ \begin{align} \begin{aligned} &(n-2k+\gamma)sT(r, f)+(n-2k+\gamma)sT(r, g)+ksN(r, f)+sN\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)\\ &+ksN(r, g)+sN\left(r, \frac{g^{n-k}P(g)}{(g^{n}P(g))^{(k)}}\right)\leq(8k+14+7m-5\gamma)T(r, f)\\ &+(8k+14+7m-5\gamma)T(r, g)+2k\overline{N}(r, f)+2\overline{N}\left(r, \frac{g^{n-k}P(g)}{(g^{n}P(g))^{(k)}}\right)\\ &+2k\overline{N}(r, g)+2\overline{N}\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)+S(r, f)+S(r, g). \end{aligned} \end{align} $

(3.18)

由于$ s\geq2 $, 因此

$ \begin{align} \begin{aligned} sN\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)\geq2\overline{N}\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right), \end{aligned} \end{align} $

(3.19)

$ \begin{align} \begin{aligned} sN\left(r, \frac{g^{n-k}P(g)}{(g^{n}P(g))^{(k)}}\right)\geq2\overline{N}\left(r, \frac{g^{n-k}P(g)}{(g^{n}P(g))^{(k)}}\right), \end{aligned} \end{align} $

(3.20)

$ \begin{align} \begin{aligned} ksN(r, f)\geq2k\overline{N}(r, f), \end{aligned} \end{align} $

(3.21)

$ \begin{align} \begin{aligned} ksN(r, g)\geq2k\overline{N}(r, g). \end{aligned} \end{align} $

(3.22)

结合(3.18)–(3.22) 式, 得

$ \begin{aligned} &(ns-2ks+\gamma s-8k-14-7m+5\gamma)T(r, f)\\ &+(ns-2ks+\gamma s-8k-14-7m+5\gamma)T(r, g)\leq S(r, f)+S(r, g). \end{aligned} $

这与$ n>2k+\frac{8k+14+7m-5\gamma}{s}-\gamma $矛盾. 因此$ H\equiv 0 $, 即

$ \begin{align} \begin{aligned} \frac{F^{''}}{F'}-\frac{2F^{'}}{F-1}=\frac{G^{''}}{G^{'}}-\frac{2G^{'}}{G-1}. \end{aligned} \end{align} $

(3.23)

对(3.23) 式连续积分两次可得

$ \begin{align} \begin{aligned} \frac{1}{G-1}=\frac{A}{F-1}+B. \end{aligned} \end{align} $

(3.24)

其中$ A, B $为常数且$ A\neq0 $. 由(3.24) 式, 得

$ \begin{align} \begin{aligned} G=\frac{(B+1)F+(A-B-1)}{BF+(A-B)}, \end{aligned} \end{align} $

(3.25)

$ \begin{align} \begin{aligned} F=\frac{(B-A)G+(A-B-1)}{BG-(B+1)}. \end{aligned} \end{align} $

(3.26)

接下来我们分三种子情形讨论.

情形3.1  若$ B\neq0, \; -1 $, 由(3.26) 式, 得

$ \begin{align} \begin{aligned} \overline{N}(r, F)=\overline{N}\left(r, \frac{1}{G-\frac{B+1}{B}}\right). \end{aligned} \end{align} $

(3.27)

结合第二基本定理及$ (3.9) $与$ (3.27) $式, 得

$ \begin{align} \begin{aligned} T(r, G)\leq& \overline{N}(r, G)+\overline{N}\left(r, \frac{1}{G}\right)+\overline{N}\left(r, \frac{1}{G-\frac{B+1}{B}}\right)+S(r, G)\\ \leq& \overline{N}(r, G)+N_{2}\left(r, \frac{1}{G}\right)+\overline{N}(r, F)+S(r, G)\\ \leq& \overline{N}(r, g)+2\overline{N}\left(r, \frac{1}{g}\right)+2\overline{N}\left(r, \frac{1}{P(g)}\right)+2\overline{N}\left(r, \frac{g^{n-k}P(g)}{(g^{n}P(g))^{(k)}}\right)\\ &+\overline{N}(r, f)+S(r, g).\\ \end{aligned} \end{align} $

(3.28)

若$ A-B-1\neq0 $, 由(3.25) 式, 得

$ \begin{align} \begin{aligned} \overline{N}(r, \frac{1}{G})=\overline{N}\left(r, \frac{1}{F-\frac{B+1-A}{B+1}}\right). \end{aligned} \end{align} $

(3.29)

结合第二基本定理及(3.7), (3.8) 与$ (3.29) $式, 得

$ \begin{align} \begin{aligned} T(r, F)\leq& \overline{N}(r, F)+\overline{N}\left(r, \frac{1}{F}\right)+\overline{N}\left(r, \frac{1}{F-\frac{B+1-A}{B+1}}\right)+S(r, F)\\ \leq& \overline{N}(r, F)+N_{2}\left(r, \frac{1}{F}\right)+\overline{N}\left(r, \frac{1}{G}\right)+S(r, F)\\ \leq&\overline{N}(r, f)+2\overline{N}\left(r, \frac{1}{f}\right)+2\overline{N}\left(r, \frac{1}{P(f)}\right)+2\overline{N}\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)\\ &+(k+1)\overline{N}\left(r, \frac{1}{g}\right)+N_{k+1}\left(r, \frac{1}{P(g)}\right)+k\overline{N}(r, g)+S(r, f)+S(r, g). \end{aligned} \end{align} $

(3.30)

结合(3.16), (3.17), (3.28) 与(3.30) 式, 得

$ \begin{align} \begin{aligned} &(n-2k+\gamma)sT(r, f)+(n-2k+\gamma)sT(r, g)+ksN(r, f)+sN\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)\\ &+ksN(r, g)+sN\left(r, \frac{g^{n-k}P(g)}{(g^{n}P(g))^{(k)}}\right)\leq(2m+2)T(r, f)+(3m+3+k-\gamma)T(r, g)\\ &+2\overline{N}(r, f)+2\overline{N}\left(r, \frac{g^{n-k}P(g)}{(g^{n}P(g))^{(k)}}\right)+(k+1)\overline{N}(r, g)+2\overline{N}\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)\\ &+S(r, f)+S(r, g). \end{aligned} \end{align} $

(3.31)

由于$ s\geq2 $, 因此

$ \begin{align} \begin{aligned} ksN(r, g)\geq (k+1)\overline{N}(r, g). \end{aligned} \end{align} $

(3.32)

结合$ (3.18)-(3.20) $, (3.31) 与(3.32) 式, 得

$ \begin{aligned} &(ns-2ks+\gamma s-2m-2)T(r, f)+(ns-2ks+\gamma s-3m-3-k+\gamma)T(r, g)\\ \leq& S(r, f)+S(r, g). \end{aligned} $

因为$ 2k+\frac{8k+14+7m-5\gamma}{s}-\gamma=2k+\frac{8k+14+3m-\gamma+4(m-\gamma)}{s}-\gamma>2k+\frac{k+3+3m-\gamma}{s}-\gamma $, 且$ m\geq\gamma $, 这与$ n>2k+\frac{8k+14+7m-5\gamma}{s}-\gamma $矛盾, 因此$ A-B-1=0 $, 带入(3.25) 式可得

$ \begin{align} \begin{aligned} \overline{N}(r, G)=\overline{N}\left(r, \frac{1}{F+\frac{1}{B}}\right).\\ \end{aligned} \end{align} $

(3.33)

结合第二基本定理及(3.8) 与(3.33) 式, 得

$ \begin{align} \begin{aligned} T(r, F)\leq& \overline{N}(r, F)+\overline{N}\left(r, \frac{1}{F}\right)+\overline{N}\left(r, \frac{1}{F+\frac{1}{B}}\right)+S(r, F)\\ \leq& \overline{N}(r, F)+N_{2}\left(r, \frac{1}{F}\right)+\overline{N}(r, G))+S(r, f)\\ \leq&\overline{N}(r, f)+2\overline{N}\left(r, \frac{1}{f}\right)+2\overline{N}\left(r, \frac{1}{P(f)}\right)+2\overline{N}\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)\\ &+\overline{N}(r, g)+S(r, f). \end{aligned} \end{align} $

(3.34)

结合(3.16), (3.17), (3.28) 与(3.34) 式, 得

$ \begin{align} \begin{aligned} &(n-2k+\gamma)sT(r, f)+(n-2k+\gamma)sT(r, g)+ksN(r, f)+sN\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)\\ &+ksN(r, g)+sN\left(r, \frac{g^{n-k}P(g)}{(g^{n}P(g))^{(k)}}\right)\leq(2m+2)T(r, f)+(2m+2)T(r, g)\\ &+2\overline{N}(r, f)+2\overline{N}\left(r, \frac{g^{n-k}P(g)}{(g^{n}P(g))^{(k)}}\right)+2\overline{N}(r, g)+2\overline{N}\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)\\ &+S(r, f)+S(r, g). \end{aligned} \end{align} $

(3.35)

结合(3.19)–(3.22) 与(3.35) 式, 得

$ \begin{aligned} (ns-2ks+\gamma s-2m-2)T(r, f)+(ns-2ks+\gamma s-2m-2)T(r, g) \leq S(r, f)+S(r, g). \end{aligned} $

这与$ n>2k+\frac{8k+14+7m-5\gamma}{s}-\gamma $矛盾.

情形3.2  若$ B=-1 $, 由(3.25) 与(3.26) 式, 得

$ \begin{align} \begin{aligned} G=\frac{A}{A+1-F}, \end{aligned} \end{align} $

(3.36)

$ \begin{align} \begin{aligned} F=\frac{(1+A)G-A}{G}. \end{aligned} \end{align} $

(3.37)

若$ A+1\neq0 $, 由(3.36) 与(3.37) 式, 得

$ \begin{align} \begin{aligned} \overline{N}(r, G)=\overline{N}\left(r, \frac{1}{F-A-1}\right), \end{aligned} \end{align} $

(3.38)

$ \begin{align} \begin{aligned} \overline{N}\left(r, \frac{1}{F}\right)=\overline{N}\left(r, \frac{1}{G-\frac{A}{1+A}}\right). \end{aligned} \end{align} $

(3.39)

结合第二基本定理及(3.6), (3.9), (3.38) 与(3.39) 式, 得

$ \begin{align} \begin{aligned} T(r, G)\leq& \overline{N}(r, G)+\overline{N}\left(r, \frac{1}{G}\right)+\overline{N}\left(r, \frac{1}{G-\frac{A}{1+A}}\right)+S(r, G)\\ \leq& \overline{N}(r, G)+N_{2}\left(r, \frac{1}{G}\right)+\overline{N}\left(r, \frac{1}{F}\right)+S(r, G)\\ \leq& \overline{N}(r, g)+2\overline{N}\left(r, \frac{1}{g}\right)+2\overline{N}\left(r, \frac{1}{P(g)}\right)+2\overline{N}\left(r, \frac{g^{n-k}P(g)}{(g^{n}P(g))^{(k)}}\right)\\ &+(k+1)\overline{N}\left(r, \frac{1}{f}\right)+N_{k+1}\left(r, \frac{1}{P(f)}\right)+k\overline{N}(r, f)+S(r, f)+S(r, g).\\ \end{aligned} \end{align} $

(3.40)

$ \begin{align} \begin{aligned} T(r, F)\leq& \overline{N}(r, F)+\overline{N}\left(r, \frac{1}{F}\right)+\overline{N}\left(r, \frac{1}{F-A-1}\right)+S(r, F)\\ \leq& \overline{N}(r, F)+N_{2}\left(r, \frac{1}{F}\right)+\overline{N}(r, G)+S(r, F)\\ \leq&\overline{N}(r, f)+2\overline{N}\left(r, \frac{1}{f}\right)+2\overline{N}\left(r, \frac{1}{P(f)}\right)+2\overline{N}\left(r, \frac{f^{n-k}P(f)}{(g^{n}P(g))^{(k)}}\right)\\ &+\overline{N}(r, g)+S(r, f). \end{aligned} \end{align} $

(3.41)

结合(3.16), (3.17), (3.40) 与(3.41) 式, 得

$ \begin{align} \begin{aligned} &(n-2k+\gamma)sT(r, f)+(n-2k+\gamma)sT(r, g)+ksN(r, f)+sN\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)\\ &+ksN(r, g)+sN\left(r, \frac{g^{n-k}P(g)}{(g^{n}P(g))^{(k)}}\right)\leq(k+3+3m-\gamma)T(r, f)\\ &+(2m+2)T(r, g)+(k+1)\overline{N}(r, f)+2\overline{N}\left(r, \frac{g^{n-k}P(g)}{(g^{n}P(g))^{(k)}}\right)+2\overline{N}(r, g)\\ &+2\overline{N}\left(r, \frac{f^{n-k}P(f)}{(f^{n}P(f))^{(k)}}\right)+S(r, f)+S(r, g). \end{aligned} \end{align} $

(3.42)

由于$ s\geq2 $, 因此

$ \begin{align} \begin{aligned} ksN(r, f)\geq (k+1)\overline{N}(r, f). \end{aligned} \end{align} $

(3.43)

结合(3.19), (3.20), (3.22), (3.42) 与(3.43) 式, 得

$ \begin{aligned} (ns-2ks+\gamma-k-3-3m+\gamma)T(r, f)+(ns-2ks+\gamma s-2m-2)T(r, g) \leq S(r, f)+S(r, g). \end{aligned} $

这与$ n>2k+\frac{8k+14+7m-5\gamma}{s}-\gamma $矛盾, 因此$ A+1=0 $, 将其带入(3.36) 式可得$ FG\equiv1 $. 若$ P(z)=a_{m}z^{m}+a_{m-1}z^{m-1}+\cdots+a_{1}z+a_{0} $, 则$ (f^{n}P(f))^{(k)}(g^{n}P(g))^{(k)}\equiv h $, 其中$ h^{s}=1 $. 若$ P(z)\equiv c_{0} $, 则$ (c_{0}f^{n})^{(k)}(c_{0}g^{n})^{(k)}\equiv h $, 易见$ f\neq0, \; f\neq\infty $, $ g\neq0, \; g\neq\infty $, 故$ f, \; g $为非常数整函数. 从而由引理2.10得$ f=c_{1}e^{cz}, \; g=c_{2}e^{-cz} $, 其中$ c_{1}, \; c_{2}, \; c $是非零常数, 且满足$ c_{0}^{2s}(-1)^{ks}(c_{1}c_{2})^{ns}(nc)^{2ks}=1 $.

情形3.3  若$ B=0 $, 将其带入(3.25) 与(3.26) 式, 得

$ \begin{align} \begin{aligned} G=\frac{F+A-1}{A}, \end{aligned} \end{align} $

(3.44)

$ \begin{align} \begin{aligned} F=AG+1-A. \end{aligned} \end{align} $

(3.45)

与情况1.2的证明类同, 可得到$ A-1=0 $, 将其带入(3.44) 式可得$ F\equiv G $, 即$ ((f^{n}P(f))^{(k)})^{s}\equiv ((g^{n}P(g))^{(k)})^{s} $, 再应用引理2.9的结论, 这就完成了$ k=0 $时情况的证明. $ k=1 $与$ k\geq2 $情况的证明与$ k=0 $情况的证明完全类似, 不再赘述. 至此, 定理1.8证毕.