本文采用常用的Nevanlinna理论的标准记号和基本结果^[1-2].特别地, $ \lambda $表示为$ f(z) $的级, $ \sigma_{2}(f) $表示为$ f(z) $的超级.本世纪, 随着亚纯函数差分模拟的值分布理论的建立^[3-4], 国内外学者做了大量的研究, 得到了很多的研究成果^[5-8].特别, 值分布论中的一些经典结果也被相应的差分模拟.通常, 平移差分$ \Delta_{c} f(z) = f(z+c)-f(z) $被看作是$ f'(z) $的差分对应, $ f(z)^{n}\Delta_{c} f(z) $和$ f(z)^{n}+a\Delta_{c} f(z) $被看作是微分多项式$ f(z)^{n}f'(z) $和$ f(z)^{n}+af'(z) $的差分对应. 1959年, Hayman ^[9]证明了如下的两个定理.

定理A  设$ f(z) $为超越亚纯函数, $ n $为正整数, $ b $为非零有穷复数, 则$ n\geq3 $时, $ f(z)^{n}f'(z)-b $有无穷多个零点.

定理B  设$ f(z) $为超越亚纯函数, $ n $为正整数, $ a(\neq0) $和$ b $为两个有穷复数, 则$ n\geq5 $时, $ f'(z)+af(z)^{n}-b $有无穷多个零点.

后来, Ye和Fang等人^[10-11]将定理B中的$ f $和$ f' $交换位置, 得到了下面的定理C.

定理C  设$ f(z) $为超越亚纯函数, $ n $为正整数, $ a(\neq0) $和$ b $为两个有穷复数, 则$ n\geq2 $时, $ f(z)+af'(z)^{n}-b $有无穷多个零点.

2011年, Liu ^[12]等人对定理A中的$ f(z)^{n}f'(z) $进行差分模拟, 得到了下面的定理D.

定理D  设$ f(z) $为有限级超越亚纯函数, $ c $为非零复常数, $ n $为正整数, $ \alpha(z) $为$ f(z) $的小函数, 则$ n\geq6 $时, $ f(z)^{n}f(z+c)-\alpha(z) $有无穷多个零点.

2014年, Li ^[13]等人得到了更为细致的定理E.

定理E  设$ f(z) $为有限级超越亚纯函数, $ c $为非零复常数, $ n $为正整数, 多项式$ p(z)\not\equiv 0 $, 则

$ \begin{align*} \overline{N}(r, \frac{1}{f(z)^{n}f(z+c)-p(z)}) \geq& nT(r, f(z))+m(r, f(z))-2\overline{N}(r, f(z))\\ &-2\overline{N}(r, \frac{1}{f(z)})-N(r, \frac{1}{f(z)})+o(\frac{T(r, f(z))}{r^{1-\varepsilon}})+O(1). \end{align*} $

其中$ r\rightarrow\infty $, $ r\notin E $, $ E $为一个有限测度集.

本文主要研究了下面的问题.

(ⅰ)若将定理E中的$ f(z+c) $改为$ f(z) $的差分多项式, 可以得到怎样的结论？

注1.1  亚纯函数$ f(z) $的差分多项式$ H(z, f) $定义如下

$ \begin{equation} H(z, f) = \sum\limits_{i\in\lambda}a_{i}(z)\prod\limits_{j = 1}^{\tau_{i}}f(z+c_{i, j})^{\mu_{i, j}}, \end{equation} $

(1.1)

其中$ \lambda $为指标集, 包含$ m\; (\geq1) $个不同的$ i $, $ c_{i, j} $为复常数, $ \mu_{i, j} $为非负整数, 系数$ a_{i}(z)(\not\equiv 0) $为$ f(z) $的小函数, 对$ H(z, f) $的每一个单项式$ a_{i}(z)\prod\limits_{j = 1}^{\tau_{i}}f(z+c_{i, j})^{\mu_{i, j}} $, 定义其次数为$ d_{i} = \sum\limits_{j = 1}^{\tau_{i}}\mu_{i, j} $, 再将$ H(z, f) $所有单项式的最高次数定义为$ H(z, f) $的次数, 即

$ \begin{equation} d_{H} = \deg_{f}H(z, f) = \max\limits_{i\in\lambda}d_{i}. \end{equation} $

(1.2)

(ⅱ)定理C中的微分多项式$ f(z)+af'(z)^{n}-b $, 对应的差分模拟的零点情况如何？

对于问题(ⅰ), 首先在一般差分多项式的情况下, 证明了定理1.1.

定理1.1  设$ f(z) $为超越亚纯函数, $ \sigma_{2}(f)<1 $, $ c $为非零复常数, $ n $为正整数, $ \alpha(z)(\not\equiv 0) $为$ f(z) $的小函数, $ H(z, f) $是形如(1.1)式的差分多项式, 则

$ \begin{eqnarray*} &&\overline{N}\left(r, \frac{1}{f(z)^{n}H(z, f)-\alpha(z)}\right)\\ &\geq& (n-md_{H})T(r, f(z))-\overline{N}(r, f(z))-\overline{N}\left(r, \frac{1}{f(z)}\right)+S(r, f), \end{eqnarray*} $

其中$ r\rightarrow\infty $, $ r\notin E $, $ E $为一个有限测度集.

其次, 对比定理E, 下述推论1.1显著改进了其结果.

推论1.1  设$ f(z) $为超越亚纯函数, $ \sigma_{2}(f)<1 $, $ c $为非零复常数, $ n $为正整数, $ \alpha(z)(\not\equiv 0) $为$ f(z) $的小函数, 则

$ \begin{eqnarray*} &&\overline{N}\left(r, \frac{1}{f(z)^{n}f(z+c)-\alpha(z)}\right)\\ &\geq& (n-1)T(r, f(z))-\overline{N}(r, f(z))-\overline{N}\left(r, \frac{1}{f(z)}\right)+S(r, f), \end{eqnarray*} $

其中$ r\rightarrow\infty $, $ r\notin E $, $ E $为一个有限测度集.

推论1.2  设$ f(z) $为超越亚纯函数, $ \sigma_{2}(f)<1 $, $ c $为非零复常数, $ n $为正整数, $ \alpha(z)(\not\equiv 0) $为$ f(z) $的小函数, 则$ n\geq4 $时, $ f(z)^{n}f(z+c)-\alpha(z) $有无穷多个零点.

上述推论1.2将定理D的条件$ n\geq6 $改进为$ n\geq4 $.例1.1说明该结论最佳, 不能再改进.

例1.1  $ f(z) = \frac{e^{z}-1}{e^{z}+1} $, $ c = \pi i $, 则当$ n = $ 2、3时, $ f(z)^{n}f(z+c) = (\frac{e^{z}-1}{e^{z}+1})^{n-1}\neq1 $.

下面讨论问题(ⅱ)中微分多项式$ f(z)+af'(z)^{n}-b $的差分模拟的零点情况.

定理1.2  设$ f(z) $为超越亚纯函数, $ \sigma_{2}(f)<1 $, 且满足$ N(r, f) = S(r, f) $, $ c $为非零复常数, $ n $为正整数, $ \alpha(z)(\not\equiv 0) $为$ f(z) $的小函数, $ H(z, f) $是形如(1.1)式的差分多项式, 且$ H(z, f) $中仅有一个单项式具有最高次数, 则

$ \overline{N} \left(r, \frac{1}{f(z)+aH(z, f)^{n}-\alpha(z)}\right)\geq \left((n-1)d_{H}-1\right)T(r, f(z))+S(r, f), $

其中$ r\rightarrow\infty $, $ r\notin E $, $ E $为一个有限测度集.

例1.2  $ f(z) = \frac{e^{z}-1}{e^{z}+1} $, $ c = \pi i $, $ H(z, f) = f(z)f(z+\pi i) $, 则

$ f(z)+H(z, f)^{n}-2 = \frac{-2}{e^{z}+1}\neq0. $

上例说明定理1.2的条件$ N(r, f) = S(r, f) $不可缺.

推论1.3  设$ f(z) $为超越整函数, 则$ n\geq3 $时, $ f(z)+af(z+c)^{n}-\alpha(z) $有无穷多个零点.

例1.3  $ f(z) = e^{z}+1 $, $ c = \pi i $, $ n = 2 $, 则$ f(z)+\frac{1}{2}f(z+\pi i)^{2}-\frac{3}{2}\neq0. $

上例说明推论1.3结论最佳, 不能再改进.

定理1.3  设$ f(z) $为超越亚纯函数, $ \sigma_{2}(f)<1 $, $ c $为非零复常数, $ n $为正整数, $ \alpha(z)(\not\equiv 0) $为$ f(z) $的小函数, 若$ f(z) $满足

$ \lim\limits_{r\rightarrow\infty}\frac{T(r, f(z+c)-f(z))}{T(r, f(z))} = 1, $

则$ n\geq5 $时, $ f(z)+a(f(z+c)-f(z))^{n}-\alpha(z) $有无穷多个零点.

例1.4  $ f(z) = e^{z}+p(z) $, $ p(z) $为多项式, $ c = 2\pi i $, $ \alpha(z) = p(z)+(p(z+c)-p(z))^{n} $, 则

$ f(z)+(f(z+c)-f(z))^{n}-\alpha(z) = e^{z}\neq0. $

上例1.4说明定理1.3若要成立, 需对$ f(z+c)-f(z) $附加一些条件.

定理1.4  设$ f(z) $为超越亚纯函数, $ \sigma_{2}(f)<1 $, $ c $为非零复常数, $ n $为正整数, $ \alpha(z)(\not\equiv 0) $为$ f(z) $的小函数, 则

$ \begin{eqnarray*} &&\overline{N}\left(r, \frac{1}{f(z)+af(z+c)^{n}-\alpha(z)}\right)\\ &\geq& (n-1)T(r, f(z))-2\overline{N}(r, f(z))-\overline{N}\left(r, \frac{1}{f(z)}\right)+S(r, f), \end{eqnarray*} $

其中$ r\rightarrow\infty $, $ r\notin E $, $ E $为一个有限测度集.

推论1.4  设$ f(z) $为超越亚纯函数, $ \sigma_{2}(f)<1 $, $ c $为非零复常数, $ n $为正整数, $ \alpha(z)(\not\equiv 0) $为$ f(z) $的小函数, 则$ n\geq5 $时, $ f(z)+af(z+c)^{n}-\alpha(z) $有无穷多个零点.

引理2.1^[7]  设$ f(z) $为非常数亚纯函数, $ \sigma_{2}(f)<1 $, $ c $为非零复常数, 对任意的$ \varepsilon>0 $, 有

$ m\left(r, \frac{f(z+c)}{f(z)}\right) = o\left(\frac{T(r, f(z))}{r^{1-\sigma_{2}(f)-\varepsilon}}\right), $

其中$ r\rightarrow\infty $, $ r\notin E $, $ E $为一个有限测度集.

由文献[7, 引理8.3], [14, p. 66]及[15, 引理1], 得到下面的引理2.2.

引理2.2  设$ f(z) $为非常数亚纯函数, $ \sigma_{2}(f)<1 $, $ c $为非零复常数, 则

$ \begin{eqnarray*} && T(r, f(z+c)) = T(r, f(z))+S(r, f), \\ &&N(r, f(z+c)) = N(r, f(z))+S(r, f), \\ &&\overline{N}(r, f(z+c)) = \overline{N}(r, f(z))+S(r, f). \end{eqnarray*} $

引理2.3^[16]  设$ f(z) $为有限级亚纯函数, 满足$ N(r, f) = S(r, f) $, $ H(z, f) $是形如(1.1)式的差分多项式, 且$ H(z, f) $中仅有一个单项式具有最高次数$ d_{H} $, 则

$ T(r, H(z, f)) = d_{H}T(r, f(z))+S(r, f). $

由文献[16, 定理1]和[17, 引理2], 得到下面的引理.

引理2.4  设$ f(z) $为超越亚纯函数, $ \sigma_{2}(f)<1 $, $ H(z, f) $是形如(1.1)式的差分多项式, 则

$ T(r, H(z, f))\leq md_{H}T(r, f(z))+S(r, f). $

定理1.1的证明  记

$ \begin{equation} F(z) = f(z)^{n}H(z+c)-\alpha(z), \end{equation} $

(3.1)

则

$ \begin{align} nm(r, f) = &m(r, f^{n}) = m\left(r, \frac{F+\alpha(z)}{H(z, f)}\right)\\ \leq& m\left(r, F+\alpha(z)\right)+m\left(r, \frac{1}{H(z, f)}\right), \end{align} $

(3.2)

$ \begin{align} nN(r, f) = &N(r, f^{n}) = N\left(r, \frac{F+\alpha(z)}{H(z, f)}\right) \\ \leq& N(r, F+\alpha(z))+N\left(r, \frac{1}{H(z, f)}\right)-\overline{N}_{0}(r)-\overline{N}_{1}(r), \end{align} $

(3.3)

其中$ \overline{N}_{0}(r) $, $ \overline{N}_{1}(r) $分别为$ H(z, f) $和$ F+\alpha(z) $的公共零点及公共极点的精简密指量.由(3.2)–(3.3)式可得

$ \begin{align} nT(r, f) = &T(r, f^{n}) = T\left(r, \frac{F+\alpha(z)}{H(z, f)}\right) \\ \leq& T(r, F+\alpha(z))+T\left(r, \frac{1}{H(z, f)}\right)-\overline{N}_{0}(r)-\overline{N}_{1}(r). \end{align} $

(3.4)

由(3.1)式可知$ F+\alpha(z) = f^{n}H(z, f) $, 则有

$ \begin{eqnarray} &&\overline{N}\left(r, \frac{1}{F+\alpha(z)}\right)\leq\overline{N}\left(r, \frac{1}{f}\right)+\overline{N}_{0}(r), \end{eqnarray} $

(3.5)

$ \begin{eqnarray} &&\overline{N}(r, F+\alpha(z))\leq\overline{N}(r, f)+\overline{N}_{1}(r). \end{eqnarray} $

(3.6)

再由关于三个小函数的第二基本定理及(3.5)–(3.6)式, 可知

$ \begin{align} T(r, F+\alpha(z))\leq&\overline{N}\left(r, \frac{1}{F+\alpha(z)}\right)+\overline{N}(r, F+\alpha(z))+\overline{N}\left(r, \frac{1}{F}\right) +S(r, F) \\ \leq&\overline{N}\left(r, \frac{1}{f}\right)+\overline{N}(r, f)+\overline{N}\left(r, \frac{1}{F}\right)+\overline{N}_{0}(r)+\overline{N}_{1}(r) +S(r, f) . \end{align} $

(3.7)

又由引理2.4可知

$ \begin{equation} T\left(r, \frac{1}{H(z, f)}\right) = T(r, H(z, f))+O(1)\leq md_{H}T(r, f(z))+S(r, f), \end{equation} $

(3.8)

从而将(3.7)–(3.8)式代入(3.4)式可得

$ \begin{eqnarray*} &&\overline{N}\left(r, \frac{1}{f(z)^{n}H(z, f)-\alpha(z)}\right)\\ &\geq& (n-md_{H})T(r, f(z))-\overline{N}(r, f(z))-\overline{N}\left(r, \frac{1}{f(z)}\right)+S(r, f), \end{eqnarray*} $

其中$ r\rightarrow\infty $, $ r\notin E $, $ E $为一个有限测度集.

定理1.2的证明  记$ \varphi(z) = \frac{\alpha(z)-f(z)}{aH^{n}}, $则

$ \begin{align} nm(r, H) = &m(r, H^{n}) = m\left(r, \frac{\alpha(z)-f}{\varphi(z)}\right)\\ \leq& m(r, \alpha(z)-f)+m\left(r, \frac{1}{\varphi(z)}\right), \end{align} $

(3.9)

$ \begin{align} nN(r, H) = &N(r, H^{n}) = N\left(r, \frac{\alpha(z)-f}{\varphi(z)}\right) \\ \leq& N(r, \alpha(z)-f)+N\left(r, \frac{1}{\varphi(z)}\right)-\overline{N}_{0}(r)-\overline{N}_{1}(r), \end{align} $

(3.10)

其中$ \overline{N}_{0}(r) $、$ \overline{N}_{1}(r) $分别为$ \varphi(z) $和$ \alpha(z)-f $的公共零点及公共极点的精简密指量.由(3.9)–(3.10)式可得

$ \begin{equation} nT(r, H)\leq T(r, \alpha(z)-f)+T(r, \frac{1}{\varphi(z)})-\overline{N}_{0}(r)-\overline{N}_{1}(r). \end{equation} $

(3.11)

由$ N(r, f) = S(r, f) $, 可知$ N(r, H) = S(r, f) $.所以

$ \begin{align} \overline{N}(r, \varphi(z))\leq&\overline{N}(r, \frac{1}{H})+\overline{N}_{1}(r)+S(r, f), \end{align} $

(3.12)

$ \begin{align} \overline{N}\left(r, \frac{1}{\varphi(z)}\right)\leq&\overline{N}(r, H)+\overline{N}_{0}(r)+S(r, f)\leq\overline{N}_{0}(r) +S(r, f), \end{align} $

(3.13)

$ \begin{align} \overline{N}\left(r, \frac{1}{\varphi(z)-1}\right) = &\overline{N}\left(r, \frac{aH^{n}}{\alpha(z)-f-aH^{n}}\right) \leq\overline{N}\left(r, \frac{1}{f+ aH^{n} -\alpha(z)}\right)+S(r, f). \end{align} $

(3.14)

再由关于三个小函数的第二基本定理及(3.12)–(3.14)式可知

$ \begin{align} T\left(r, \frac{1}{\varphi(z)}\right)\leq&\overline{N}\left(r, \frac{1}{\varphi(z)}\right)+\overline{N}(r, \varphi(z))+\overline{N}\left(r, \frac{1}{\varphi(z)-1}\right) +S(r, \varphi) \\ \leq&T\left(r, \frac{1}{H}\right)+\overline{N}\left(r, \frac{1}{f+ aH^{n}-\alpha(z)}\right)+\overline{N}_{0}(r)+\overline{N}_{1}(r) +S(r, f) . \end{align} $

(3.15)

又由引理2.3可知

$ \begin{equation} T(r, H(z, f)) = d_{H}T(r, f(z))+S(r, f), \end{equation} $

(3.16)

从而将(3.15)–(3.16)式代入(3.11)式可得

$ \overline{N}\left(r, \frac{1}{f(z)+aH(z, f)^{n}-\alpha(z)}\right)\geq ((n-1)d_{H}-1)T(r, f(z))+S(r, f), $

其中$ r\rightarrow\infty $, $ r\notin E $, $ E $为一个有限测度集.

定理1.3的证明  记$ \Delta_{c} f(z) = f(z+c)-f(z) $, $ G(z) = \frac{\alpha(z)-f(z)}{a(\Delta_{c} f)^{n}} $, 则

$ \begin{align} nm(r, \Delta_{c} f) = &m\left(r, (\Delta_{c} f)^{n}\right) = m\left(r, \frac{\alpha(z)-f}{G(z)}\right)\\ \leq& m(r, \alpha(z)-f)+m\left(r, \frac{1}{G(z)}\right), \end{align} $

(3.17)

$ \begin{align} nN(r, \Delta_{c} f) = &N(r, (\Delta_{c} f)^{n}) = N\left(r, \frac{\alpha(z)-f}{G(z)}\right) \\ \leq& N(r, \alpha(z)-f)+N\left(r, \frac{1}{G(z)}\right)-\overline{N}_{0}(r)-\overline{N}_{1}(r), \end{align} $

(3.18)

其中$ \overline{N}_{0}(r) $、$ \overline{N}_{1}(r) $分别为$ G(z) $和$ \alpha(z)-f $的公共零点及公共极点的精简密指量.由(3.17)–(3.18)式可得

$ \begin{equation} nT(r, \Delta_{c} f)\leq T(r, \alpha(z)-f)+T\left(r, \frac{1}{G(z)}\right)-\overline{N}_{0}(r)-\overline{N}_{1}(r). \end{equation} $

(3.19)

由$ G(z) = \frac{\alpha(z)-f(z)}{a(\Delta_{c} f)^{n}} $, 可知

$ \begin{eqnarray} &&\overline{N}\left(r, \frac{1}{G(z)}\right)\leq\overline{N}(r, \Delta_{c} f)+\overline{N}_{0}(r)+S(r, f), \end{eqnarray} $

(3.20)

$ \begin{eqnarray} &&\overline{N}(r, G(z))\leq\overline{N}\left(r, \frac{1}{\Delta_{c} f}\right)+\overline{N}_{1}(r)+S(r, f). \end{eqnarray} $

(3.21)

又由$ \lim\limits_{r\rightarrow\infty}\frac{T(r, f(z+c)-f(z))}{T(r, f(z))} = 1 $, 可知

$ \begin{equation} T(r, \Delta_{c}f) = T(r, f)+S(r, f), \end{equation} $

(3.22)

再由关于三个小函数的第二基本定理及(3.20)–(3.22)式可知

$ \begin{align} T\left(r, \frac{1}{G(z)}\right)\leq&\overline{N}\left(r, \frac{1}{G(z)}\right)+\overline{N}(r, G(z))+\overline{N}\left(r, \frac{1}{G(z)-1}\right) +S(r, G) \\ \leq&\overline{N}(r, \Delta_{c}f)+\overline{N}\left(r, \frac{1}{\Delta_{c} f}\right)+\overline{N}\left(r, \frac{a(\Delta_{c}f)^{n}}{\alpha(z)-f-a(\Delta_{c}f)^{n}}\right)\\ &+\overline{N}_{0}(r)+\overline{N}_{1}(r)+S(r, f)\\ \leq&3T(r, \Delta_{c}f)+\overline{N}\left(r, \frac{1}{f(z)+a(\Delta_{c}f)^{n}-\alpha(z)}\right)\\ &+\overline{N}_{0}(r)+\overline{N}_{1}(r) +S(r, f) . \end{align} $

(3.23)

从而将(3.22)–(3.23)式代入(3.19)式, 可得

$ \overline{N}\left(r, \frac{1}{f(z)+a(\Delta_{c}f)^{n}-\alpha(z)}\right)\geq (n-4)T(r, f(z))+S(r, f), $

其中$ r\rightarrow\infty $, $ r\notin E $, $ E $为一个有限测度集.

所以$ n\geq5 $时, $ f(z)+a(f(z+c)-f(z))^{n}-\alpha(z) $有无穷多个零点.

定理1.4的证明  记$ \psi(z) = \frac{\alpha(z)-f(z)}{af(z+c)^{n}} $, 则

$ \begin{align} nm(r, f(z+c)) = &m(r, f(z+c)^{n}) = m\left(r, \frac{\alpha(z)-f}{\psi(z)}\right) \\ \leq& m(r, \alpha(z)-f)+m\left(r, \frac{1}{\psi(z)}\right), \end{align} $

(3.24)

$ \begin{align} nN(r, f(z+c)) = &N(r, f(z+c)^{n}) = N\left(r, \frac{\alpha(z)-f}{\psi(z)}\right) \\ \leq& N(r, \alpha(z)-f)+N\left(r, \frac{1}{\psi(z)}\right)-\overline{N}_{0}(r)-\overline{N}_{1}(r), \end{align} $

(3.25)

其中$ \overline{N}_{0}(r) $、$ \overline{N}_{1}(r) $分别为$ \psi(z) $和$ \alpha(z)-f $的公共零点及公共极点的精简密指量.

由(3.24)–(3.25)式, 可得

$ \begin{equation} nT(r, f(z+c))\leq T(r, \alpha(z)-f)+T\left(r, \frac{1}{\psi(z)}\right)-\overline{N}_{0}(r)-\overline{N}_{1}(r). \end{equation} $

(3.26)

由引理2.2, 可知

$ \begin{eqnarray} &&\overline{N}(r, f(z+c)) = \overline{N}(r, f)+S(r, f), \end{eqnarray} $

(3.27)

$ \begin{eqnarray} &&\overline{N}\left(r, \frac{1}{f(z+c)}\right) = \overline{N}\left(r, \frac{1}{f}\right)+S(r, f). \end{eqnarray} $

(3.28)

所以

$ \begin{eqnarray} &&\overline{N}\left(r, \frac{1}{\psi(z)}\right) = \overline{N}(r, f)+\overline{N}_{0}(r)+S(r, f), \end{eqnarray} $

(3.29)

$ \begin{eqnarray} &&\overline{N}(r, \psi(z)) = \overline{N}\left(r, \frac{1}{f}\right)+\overline{N}_{1}(r)+S(r, f). \end{eqnarray} $

(3.30)

再由关于三个小函数的第二基本定理及(3.29)–(3.30)式, 可知

$ \begin{align} T\left(r, \frac{1}{\psi(z)}\right)\leq&\overline{N}\left(r, \frac{1}{\psi(z)}\right)+\overline{N}(r, \psi(z))+\overline{N}\left(r, \frac{1}{\psi(z)-1}\right) +S(r, \psi) \\ \leq&\overline{N}(r, f)+\overline{N}\left(r, \frac{1}{ f}\right)+\overline{N}\left(r, \frac{1}{\psi(z)-1}\right)+\overline{N}_{0}(r)+\overline{N}_{1}(r) +S(r, f). \end{align} $

(3.31)

又由引理2.2可知

$ \begin{equation} T(r, f(z+c)) = T(r, f)+S(r, f), \end{equation} $

(3.32)

从而将(3.31)–(3.32)式代入(3.26)式可得

$ \overline{N}\left(r, \frac{1}{\psi(z)-1}\right)\geq (n-1)T(r, f)-\overline{N}(r, f)-\overline{N}\left(r, \frac{1}{ f}\right)+S(r, f), $

即

$ \begin{eqnarray*} &&\overline{N}\left(r, \frac{1}{f(z)+af(z+c)^{n}-\alpha(z)}\right)\\ &\geq& (n-1)T(r, f(z))-2\overline{N}(r, f(z))-\overline{N}\left(r, \frac{1}{f(z)}\right)+S(r, f), \end{eqnarray*} $

其中$ r\rightarrow\infty $, $ r\notin E $, $ E $为一个有限测度集.