本文将利用Nevanlinna理论研究亚纯函数差分的不动点.设$ f(z) $是定义在复平面$ \Bbb{C} $上的亚纯函数,我们使用Nevanlinna理论的标准记号: $ m(r, f), $$ N(r, f), m(r, \frac{1}{f-a}), $$ N(r, \frac{1}{f-a}), $$ T(r, f ), S(r, f ), \cdots, $并假设读者熟知这些基本记号(可参考Hayman^[1],杨乐^[2],仪洪勋、杨重骏^[3],郑建华^[4]等).此外,用$ \sigma(f) $表示函数$ f(z) $的级, $ \lambda(f, a) $表示$ f-a (a\in\Bbb{C}) $的零点收敛指数.特别地,记$ \lambda(f, 0) = \lambda(f) $.我们还用$ \lambda(\frac{1}{f}) $表示$ f(z) $的极点的收敛指数.如果$ \lambda(f, a)<\sigma(f) $ (或$ \lambda(\frac{1}{f})<\sigma(f)) $,则称$ a $ (或$ \infty) $为$ f(z) $的Borel例外值.函数$ f(z) $的$ a\in\Bbb{C}\cup\{\infty\} $值点的Nevanlinna亏量$ \delta(a, f) $定义为

$ \delta(a, f) = 1-\limsup\limits_{r\rightarrow\infty}\frac{N\left(r, \frac{1}{f-a}\right)}{T(r, f)} = \liminf\limits_{r\rightarrow\infty}\frac{m\left(r, \frac{1}{f-a}\right)}{T(r, f)}, $

$ \delta(\infty, f) = 1-\limsup\limits_{r\rightarrow\infty}\frac{N(r, f)}{T(r, f)} = \liminf\limits_{r\rightarrow\infty}\frac{m(r, f)}{T(r, f)}. $

对任意的$ z_{0}\in \Bbb{C} $,如果$ f(z_{0}) = z_{0} $,称$ z_{0} $为函数$ f(z) $的不动点^[5].陈宗煊^[6]给出了不动点的收敛指数的如下定义.记$ \tau(f) $为函数$ f $的不动点的收敛指数,则

$ \tau(f) = \limsup\limits_{r\rightarrow\infty}\frac{\log N\left(r, \frac{1}{f-z}\right)}{\log r}. $

关于亚纯函数的不动点的研究,可参考庄圻泰、杨重骏^[5].在1960年前后, Baker^[7]证明:对全平面上的超越整函数$ f $,如果存在$ a\in\Bbb{C} $使得$ \delta(a, f)>0 $,则函数$ f $有一阶不动点. 1993年, Lahiri^[8]部分推广了Baker的结果,得到定理A.

定理A  设$ f $是全平面上的超越亚纯函数,如果存在$ a\in\Bbb{C} $,使得$ \delta(a, f)>0 $且$ \delta(\infty, f) = 1 $.则$ f $有无穷多个不动点.

关于定理A,易得如下更一般的结果.

定理1.1  设$ f $是全平面上的超越亚纯函数,如果存在$ a\in\Bbb{C} $,使得$ \delta(a, f)>0 $且$ \delta(\infty, f) = 1 $,则对任意的正整数$ k $, $ f $和$ f^{(k)} $有无穷多个不动点,且$ \tau(f) = \sigma(f) = \tau(f^{(k)}). $

近二十年来,关于亚纯函数及其导函数的Nevanlinna理论的结果被推广到亚纯函数及其差分^[9-25].事实上,定理1.1的结果是平凡的,但作者至今尚未发现其完整证明.本文介绍它的主要目的是由此引入我们将要研究的亚纯函数差分的不动点的相关结果.设$ f(z) $是定义在复平面$ \Bbb{C} $上的亚纯函数,记

$ \Delta f = f(z+1)-f(z), \Delta^{k+1} f = \Delta^{k} f(z+1)-\Delta^{k} f, k = 1, 2, \cdots. $

关于亚纯函数差分的不动点,陈宗煊和Shon证明了如下结果.

定理B^[12]  设$ P(z), Q(z) $和$ R(z) $是不恒为零的多项式且满足

$ \deg P(z)\geq \max\{\deg R(z), \deg Q(z)\}, \deg P(z)\geq1. $

如果存在多项式$ h(z) $,使得

$ (-R(z)+Q(z)+zP(z))^{2}-4zP(z)Q(z) = h^{2}(z). $

则Pielou方程

$ f(z+1) = \frac{R(z)f(z)}{Q(z)+P(z)f(z)} $

的每一个有限级超越亚纯解$ f(z) $满足$ \tau(\Delta f) = \sigma(f). $

定理C^[10]  设$ P(z), Q(z) $和$ R(z) $是不恒为零的多项式且$ P(z)Q(z)R(z)\not\equiv0 $,则Pielou方程的每一个有限级超越亚纯解$ f(z) $满足$ \sigma(f)\geq1. $

因此,在定理B和定理C的条件下, Pielou方程的每一个有限级超越亚纯函数解$ f(z) $满足$ \tau(\Delta f) = \sigma(f)\geq1. $对于$ \sigma(f)\leq1 $的超越亚纯函数,陈宗煊、Shon^[11]证明了如下结果.

定理D^[11]  设$ f $是复平面上的超越亚纯函数, $ \sigma(f)\leq1 $.如果$ \lambda(\frac{1}{f})<\lambda(f)<1 $,或者$ \lambda(f) = 0 $,但$ f $有无限多个零点而仅有有限多个极点,则$ \Delta f $有无限多个不动点且$ \tau(\Delta f) = \sigma(f) $.

在定理D中,如果$ \lambda(\frac{1}{f})<\lambda(f)<\sigma(f)\leq 1 $,则$ 0, \infty $为$ f $的Borel例外值.由仪洪勋、杨重骏^[3]中定理2.11知$ \sigma(f) = 1. $对于级$ \sigma(f)<1 $的亚纯函数$ f $,吴昭君、徐洪焱^[21]证明了如下结果.

定理E^[21]  设$ f $是复平面上的超越亚纯函数, $ \sigma(f)<1 $.如果存在$ a\in\Bbb{C} $,使得$ \delta(a, f)>0 $且$ \delta(\infty, f) = 1 $.则$ \Delta f $有无限多个不动点.

本文将定理E的结论推广到亚纯函数的高阶差分,证明如下定理.

定理1.2  设$ f $是全平面上的超越亚纯函数, $ \sigma(f)<1 $.如果存在$ a\in\Bbb{C} $,使得$ \delta(a, f)>0 $且$ \delta(\infty, f) = 1 $.则对任意的正整数$ k $, $ \Delta^{k} f $有无穷多个不动点且$ \tau(\Delta^{k} f) = \sigma(f). $

定理1.2是定理1.1的差分模拟.

定理1.1的证明需要用到如下引理.

引理2.1 (对数导数引理)  设$ f(z) $是全平面上的超越亚纯函数,则当$ f(z) $为有穷级时

$ m\left(r, \frac{f^{(k)}}{f}\right) = O(\log r) = S(r, f) = o(T(r, f))(r\rightarrow\infty), $

当$ f(z) $为无穷级时

$ m\left(r, \frac{f^{(k)}}{f}\right) = O\left(\log (rT(r, f))\right) = S(r, f) = o(T(r, f)) (r\rightarrow\infty, r\notin E), $

其中$ k $为任意正整数, $ E $是一个有穷线测度的集合.

引理2.2^[26]  设$ f(z) $是全平面上的非常数亚纯函数, $ k $为任意正整数, $ \varphi(z)(\not\equiv0, \infty) $是$ f(z) $的小函数,则

$ T(r, f)\leq \overline{N}(r, f)+N\left(r, \frac{1}{f}\right)+N\left(r, \frac{1}{f^{(k)}-\varphi}\right) -N\left(r, \frac{1}{\left(\frac{f^{(k)}}{\varphi}\right)'}\right)+S(r, f). $

引理2.3^[3]  设$ f(z) $是全平面上的非常数亚纯函数, $ \varphi_{j}(z) (j = 1, 2, 3) $是$ f(z) $的相互判别的小函数,则

$ T(r, f)\leq \sum\limits_{j = 1}^{3}\overline{N}\left(r, \frac{1}{f-\varphi_{j}(z)}\right)+S(r, f). $

定理1.1的证明  由定理A, $ f $有无穷多个不动点.在引理2.3中,令$ \varphi_{1}(z) = \infty, $$ \varphi_{2}(z) = a, $$ \varphi_{3}(z) = z $,则

(2.1) $ \begin{equation} T(r, f)\leq N(r, f)+N\left(r, \frac{1}{f-a}\right)+N\left(r, \frac{1}{f-z}\right)+S(r, f). \end{equation} $

又因为$ \delta(\infty, f) = 1 $, $ \delta(a, f)>0 $,所以存在某个正数$ \theta<1 $,使得对任意充分大的$ r $,有

(2.2) $ \begin{equation} N\left(r, \frac{1}{f-a}\right)<\theta T(r, f), \end{equation} $

且

(2.3) $ \begin{equation} N(r, f) = o(T(r, f)). \end{equation} $

结合(2.1)–(2.3)式可得

$ (1-\theta-o(1))T(r, f)\leq N\left(r, \frac{1}{f-z}\right). $

因此$ \tau(f)\geq\sigma(f) $,故有$ \tau(f) = \sigma(f). $

此外,在引理2.2中,令$ \varphi(z) = z $, $ g = f-a, $则

$ \begin{eqnarray*} T(r, f)& = &T(r, g)+O(1)\\ &\leq& \overline{N}(r, g)+N\left(r, \frac{1}{g}\right)+N\left(r, \frac{1}{g^{(k)}-z}\right)+S(r, g)\\ &\leq &N(r, f)+N\left(r, \frac{1}{f-a}\right)+N\left(r, \frac{1}{f^{(k)}-z}\right)+S(r, f). \end{eqnarray*} $

由上式并结合(2.2), (2.3)式可得

$ (1-\theta-o(1))T(r, f)\leq N\left(r, \frac{1}{f^{(k)}-z}\right). $

所以$ \tau(f^{(k)})\geq\sigma(f) $,故有$ \tau(f^{(k)}) = \sigma(f). $证毕.

定理1.2的证明需要用到如下引理.

引理3.1^[14]  设$ f(z) $是复平面上的有限级超越亚纯函数,则$ \sigma(\Delta f)\leq\sigma(f). $

引理3.2^[21]  设$ f(z) $是复平面上的超越亚纯函数, $ \sigma(f)<1 $,则$ \Delta f $是超越亚纯函数.

引理3.3^[20]  设$ f(z) $是复平面上的超越亚纯函数, $ \sigma(f)<1 $, $ k $为任意的正整数.则

$ m\left(r, \frac{\Delta^{k}f(z)}{f(z)}\right) = O(\log r). $

且

$ N(r, f(z+c)) = N(r, f)+O(\log r), $

其中$ c $为任意非零复数.

引理3.4^[3]  设$ f(z) $为非常数周期亚纯函数,则$ f(z) $的级$ \sigma(f)\geq 1. $

引理3.5  设$ f(z) $是复平面上的超越亚纯函数, $ \sigma(f)<1 $, $ k $为正整数.则对任意的$ a\in\Bbb{C} $,都有

$ T(r, f)\leq (2k+3)N(r, f)+N\left(r, \frac{1}{f-a}\right)+N\left(r, \frac{1}{\Delta^{k}f-z}\right)+O(\log r). $

证  记$ F = \Delta^{k}f $,由引理3.1和引理3.2, $ F $是复平面上的超越亚纯函数, $ \sigma(F)<1 $.因此

$ \sigma(\frac{F}{z})<1. $

下证$ z\Delta F-F\not\equiv0. $如果$ z\Delta F-F\equiv0, $则

$ \begin{eqnarray*} (z+1)F(z)\equiv zF(z+1). \end{eqnarray*} $

因此$ \frac{F(z)}{z} $是一个以1为周期的亚纯函数.由引理3.4知$ \sigma(\frac{F}{z})\geq1 $,这与$ \sigma(\frac{F}{z})<1 $矛盾.因此$ z\Delta^{k+1}f-\Delta^{k}f\not\equiv0. $

因为

$ \frac{1}{f} = \frac{\Delta^{k}f}{zf}-\frac{z\Delta^{k+1}f-\Delta^{k}f}{zf}\frac{\Delta^{k}f-z}{z\Delta^{k+1}f-\Delta^{k}f}. $

所以由引理3.3可得

$ \begin{eqnarray*} m\left(r, \frac{1}{f}\right)&\leq& m\left(r, \frac{\Delta^{k}f}{zf}\right)+m\left(r, \frac{z\Delta^{k+1}f-\Delta^{k}f}{zf}\right)+m\left(r, \frac{\Delta^{k}f-z}{z\Delta^{k+1}f-\Delta^{k}f}\right)+\log2\\ &\leq& 2m\left(r, \frac{\Delta^{k}f}{f}\right)+m\left(r, \frac{\Delta^{k+1}f}{f}\right)+m\left(r, \frac{\Delta^{k}f-z}{z\Delta^{k+1}f-\Delta^{k}f}\right)+O(\log r)\\ & = & m\left(r, \frac{\Delta^{k}f-z}{z\Delta^{k+1}f-\Delta^{k}f}\right)+O(\log r). \end{eqnarray*} $

应用Nevanlinna第一基本定理,有

(3.1) $ \begin{equation} T(r, f)\leq N\left(r, \frac{1}{f}\right)+m\left(r, \frac{\Delta^{k}f-z}{z\Delta^{k+1}f-\Delta^{k}f}\right)+O(\log r), \end{equation} $

(3.2) $ \begin{eqnarray} m\left(r, \frac{\Delta^{k}f-z}{z\Delta^{k+1}f-\Delta^{k}f}\right)& = & m\left(r, \frac{z\Delta^{k+1}f-\Delta^{k}f}{\Delta^{k}f-z}\right)+N\left(r, \frac{z\Delta^{k+1}f-\Delta^{k}f}{\Delta^{k}f-z}\right)\\ &&-N\left(r, \frac{\Delta^{k}f-z}{z\Delta^{k+1}f-\Delta^{k}f}\right)+O(1) \\ &\leq& m\left(r, \frac{z\Delta^{k+1}f-\Delta^{k}f}{\Delta^{k}f-z}\right)+N\left(r, \frac{z\Delta^{k+1}f-\Delta^{k}f}{\Delta^{k}f-z}\right)+O(1). \qquad \end{eqnarray} $

注意到

$ \frac{z\Delta^{k+1}f-\Delta^{k}f}{\Delta^{k}f-z} = \frac{z\Delta^{k+1}f-z-\Delta^{k}f+z}{\Delta^{k}f-z} = z\frac{\Delta^{k+1}f-1}{\Delta^{k}f-z}-1 = z\frac{\Delta(\Delta^{k}f-z)}{\Delta^{k}f-z}-1. $

因此

(3.3) $ \begin{equation} m\left(r, \frac{z\Delta^{k+1}f-\Delta^{k}f}{\Delta^{k}f-z}\right)\leq m\left(r, \frac{\Delta(\Delta^{k}f-z)}{\Delta^{k}f-z}\right)+O(\log r). \end{equation} $

因为$ f(z) $是超越亚纯函数且$ \sigma(f)<1 $.由引理3.1和引理3.2知, $ \Delta(f) $也是超越亚纯函数且$ \sigma(\Delta(f))<1 $.从而$ \Delta^{k}f-z $也是超越亚纯函数且$ \sigma(\Delta^{k}f-z)<1 $.根据引理3.3,知

(3.4) $ \begin{equation} m\left(r, \frac{\Delta(\Delta^{k}f-z)}{\Delta^{k}f-z}\right) = O(\log r). \end{equation} $

又

(3.5) $ \begin{equation} N(r, \Delta^{k+1}f)\leq \sum\limits_{j = 0}^{k+1}N(r, f(z+j)). \end{equation} $

由引理3.3和(3.5)式可得

(3.6) $ \begin{eqnarray} N\left(r, \frac{z\Delta^{k+1}f-\Delta^{k}f}{\Delta^{k}f-z}\right) &\leq& N\left(r, z\Delta^{k+1}f-\Delta^{k}f\right)+N\left(r, \frac{1}{\Delta^{k}f-z}\right) \\ &\leq& N\left(r, \frac{1}{\Delta^{k}f-z}\right)+N(r, \Delta^{k+1}f)+N(r, \Delta^{k}f)\\ &\leq &N\left(r, \frac{1}{\Delta^{k}f-z}\right)+(2k+3)N(r, f)+O(\log r). \end{eqnarray} $

结合(3.1)–(3.6)式得

$ \begin{eqnarray*} T(r, f)\leq (2k+3)N(r, f)+N\left(r, \frac{1}{f}\right)+N\left(r, \frac{1}{\Delta^{k}f-z}\right)+O(\log r). \end{eqnarray*} $

令$ g = f-a, $则

$ \begin{eqnarray*} T(r, f)& = &T(r, g)+O(1)\\ &\leq& (2k+3)N(r, g)+N\left(r, \frac{1}{g}\right)+N\left(r, \frac{1}{\Delta^{k}g-z}\right)+O(\log r)\\ & = &(2k+3)N(r, f)+N\left(r, \frac{1}{f-a}\right)+N\left(r, \frac{1}{\Delta^{k}f-z}\right)+O(\log r). \end{eqnarray*} $

证毕.

定理1.2的证明  由引理3.5知

$ T(r, f)\leq (2k+3)N(r, f)+N\left(r, \frac{1}{f-a}\right)+N\left(r, \frac{1}{\Delta^{k}f-z}\right)+O(\log r). $

由上式并结合(2.2), (2.3)可得

$ (1-\theta-o(1))T(r, f)\leq N\left(r, \frac{1}{\Delta^{k}f-z}\right). $

因此$ \tau(\Delta^{k}f)\geq\sigma(f) $,故有$ \tau(\Delta^{k}f) = \sigma(f). $证毕.