﻿ 亚纯函数导数与其平移函数的唯一性定理 Uniqueness Theoremon Derivatives and Shifts of Meromorphic Functions

Pure Mathematics
Vol. 12  No. 06 ( 2022 ), Article ID: 53057 , 8 pages
10.12677/PM.2022.126116

Uniqueness Theoremon Derivatives and Shifts of Meromorphic Functions

Dengfeng Liu

School of Mathematics and Statistics, Fujian Normal University, Fuzhou Fujian

Received: May 16th, 2022; accepted: Jun. 21st, 2022; published: Jun. 28th, 2022

ABSTRACT

In this paper, we study the problem of uniqueness on derivatives and shifts of meromorphic functions. And we prove a uniqueness theorem for derivatives of meromorphic functions satisfies the condition of sharing values with its shift, which improves and extends the results given by Qi X.G. et al.

Keywords:Meromorphic Functions, Derivatives, Shift Functions, Uniqueness

Copyright © 2022 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言及主要结论

2018年，祁晓光等人 [8] 证明了：

2. 引理

$\text{ }H=\left(\frac{{F}^{″}}{{F}^{\prime }}-\frac{2{F}^{\prime }}{F-1}\right)-\left(\frac{{G}^{″}}{{G}^{\prime }}-\frac{2{G}^{\prime }}{G-1}\right)$.

$\text{ }V=\left(\frac{{F}^{\prime }}{F-1}-\frac{{F}^{\prime }}{F}\right)-\left(\frac{{G}^{\prime }}{G-1}-\frac{{G}^{\prime }}{G}\right)=\frac{{F}^{\prime }}{F\left(F-1\right)}-\frac{{G}^{\prime }}{G\left(G-1\right)}$.

$\text{ }N\left(r,1;F|=1\right)=N\left(r,1;G|=1\right)\le N\left(r,H\right)+S\left(r,F\right)+S\left(r,G\right)$.

$\begin{array}{l}N\left(r,H\right)\le N\left(r,0;F|\ge 2\right)+N\left(r,0;G|\ge 2\right)+{\stackrel{¯}{N}}_{*}\left(r,1;F,G\right)+{\stackrel{¯}{N}}_{*}\left(r,\infty ;F,G\right)\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }+{\stackrel{¯}{N}}_{0}\left(r,0,{F}^{\prime }\right)+{\stackrel{¯}{N}}_{0}\left(r,0,{G}^{\prime }\right)+S\left(r,F\right)+S\left(r,G\right).\end{array}$

${N}_{E}^{1\right)}\left(r,0,F-1\right)\le N\left(r,H\right)+S\left(r,F\right)+S\left(r,G\right)$.

$\begin{array}{l}\left(n-1-\frac{1}{k}\right)\stackrel{¯}{N}\left(r,\infty ,f\right)\le \frac{k+1}{k}\stackrel{¯}{N}\left(r,0,f\right)+\stackrel{¯}{N}\left(r,0,g\right)-\frac{1}{k}N\left(r,0;{f}^{\prime }|f\ne 0,1,\omega ,\cdots ,{\omega }^{n-1}\right)\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }+S\left(r,f\right)+S\left(r,g\right),\end{array}$

$\begin{array}{l}\left(n-1-\frac{1}{k}\right)\stackrel{¯}{N}\left(r,\infty ,g\right)\le \frac{k+1}{k}\stackrel{¯}{N}\left(r,0,g\right)+\stackrel{¯}{N}\left(r,0,f\right)-\frac{1}{k}N\left(r,0;{g}^{\prime }|g\ne 0,1,\omega ,\cdots ,{\omega }^{n-1}\right)\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }+S\left(r,f\right)+S\left(r,g\right).\end{array}$

$\text{ }T\left(r,P\left(f\right)\right)=nT\left(r,f\right)+S\left(r,f\right)$.

3. 定理的证明

$\varphi \left(z\right)=\left(\frac{{F}^{\prime }}{F-1}-\frac{{F}^{\prime }}{F}\right)-\left(\frac{{G}^{\prime }}{G-1}-\frac{{G}^{\prime }}{G}\right)=\frac{{F}^{\prime }}{F\left(F-1\right)}-\frac{{G}^{\prime }}{G\left(G-1\right)}$. (1)

$\frac{F-1}{F}=C\frac{G-1}{G}$. (2)

$C=1$，则由(2)式得 $F\equiv G$，即 ${f}^{\prime }\left(z\right)\equiv tf\left(z+c\right)$，其中 ${t}^{n}=1$

$C\ne 1$，则由(2)式得

$F=\frac{G}{\left(1-C\right)G+C}$. (3)

$N\left(r,\infty ,F\right)=N\left(r,0,G-\frac{C}{C-1}\right)$. (4)

$T\left(r,{f}^{\prime }\left(z\right)\right)=T\left(r,f\left(z+c\right)\right)+S\left(r,{f}^{\prime }\left(z\right)\right)$,$S\left(r,{f}^{\prime }\left(z\right)\right)=S\left(r,f\left(z+c\right)\right)$. (5)

$\begin{array}{l}nT\left(r,f\left(z+c\right)\right)+O\left(1\right)=T\left(r,G\right)\\ \text{ }\le \stackrel{¯}{N}\left(r,\infty ,G\right)+\stackrel{¯}{N}\left(r,0,G\right)+N\left(r,0,G-\frac{C}{C-1}\right)+S\left(r,G\right)\\ \le \stackrel{¯}{N}\left(r,\infty ,f\left(z+c\right)\right)+\stackrel{¯}{N}\left(r,0,f\left(z+c\right)\right)+\stackrel{¯}{N}\left(r,\infty ,{f}^{\prime }\left(z\right)\right)+S\left(r,f\left(z+c\right)\right)\\ \le 3T\left(r,f\left(z+c\right)\right)+S\left(r,f\left(z+c\right)\right).\end{array}$

$\left(n-3\right)T\left(r,f\left(z+c\right)\right)\le S\left(r,f\left(z+c\right)\right)$，这与已知条件 $n\ge 6$ 矛盾。

$\left(n-\frac{3}{2}\right)\stackrel{¯}{N}\left(r,\infty ,{f}^{\prime }\left(z\right)\right)\le \frac{3}{2}\stackrel{¯}{N}\left(r,0,{f}^{\prime }\left(z\right)\right)+\stackrel{¯}{N}\left(r,0,f\left(z+c\right)\right)+S\left(r,{f}^{\prime }\left(z\right)\right)+S\left(r,f\left(z+c\right)\right)$,

$\left(n-\frac{3}{2}\right)\stackrel{¯}{N}\left(r,\infty ,f\left(z+c\right)\right)\le \frac{3}{2}\stackrel{¯}{N}\left(r,0,f\left(z+c\right)\right)+\stackrel{¯}{N}\left(r,0,{f}^{\prime }\left(z\right)\right)+S\left(r,{f}^{\prime }\left(z\right)\right)+S\left(r,f\left(z+c\right)\right)$.

$\stackrel{¯}{N}\left(r,\infty ,{f}^{\prime }\left(z\right)\right)\le \frac{3}{2n-3}\stackrel{¯}{N}\left(r,0,{f}^{\prime }\left(z\right)\right)+\frac{2}{2n-3}\stackrel{¯}{N}\left(r,0,f\left(z+c\right)\right)+S\left(r,f\right)+S\left(r,f\left(z+c\right)\right)$, (6)

$\stackrel{¯}{N}\left(r,\infty ,f\left(z+c\right)\right)\le \frac{3}{2n-3}\stackrel{¯}{N}\left(r,0,f\left(z+c\right)\right)+\frac{2}{2n-3}\stackrel{¯}{N}\left(r,0,{f}^{\prime }\left(z\right)\right)+S\left(r,f\right)+S\left(r,f\left(z+c\right)\right)$. (7)

$\begin{array}{l}n\left(T\left(r,{f}^{\prime }\left(z\right)\right)+T\left(r,f\left(z+c\right)\right)\right)+O\left(1\right)\le T\left(r,F\right)+T\left(r,G\right)\\ \le \stackrel{¯}{N}\left(r,\infty ,F\right)+\stackrel{¯}{N}\left(r,0,F\right)+\stackrel{¯}{N}\left(r,0,F-1\right)+\stackrel{¯}{N}\left(r,\infty ,G\right)+\stackrel{¯}{N}\left(r,0,G\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\stackrel{¯}{N}\left(r,0,G-1\right)-\stackrel{¯}{{N}_{0}}\left(r,0,{F}^{\prime }\right)-\stackrel{¯}{{N}_{0}}\left(r,0,{G}^{\prime }\right)+S\left(r,F\right)+S\left(r,G\right).\end{array}$ (8)

$\phi \left(z\right)=\left(\frac{{F}^{″}}{{F}^{\prime }}-\frac{2{F}^{\prime }}{F-1}\right)-\left(\frac{{G}^{″}}{{G}^{\prime }}-\frac{2{G}^{\prime }}{G-1}\right)$.

$\begin{array}{l}\frac{n}{2}\left(T\left(r,{f}^{\prime }\left(z\right)\right)+T\left(r,f\left(z+c\right)\right)\right)\\ \le {N}_{2}\left(r,0,F\right)+{N}_{2}\left(r,0,G\right)+\stackrel{¯}{N}\left(r,\infty ,F\right)+\stackrel{¯}{N}\left(r,\infty ,G\right)+{\stackrel{¯}{N}}_{*}\left(r,\infty ;F,G\right)\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }-\left(l-\frac{3}{2}\right){\stackrel{¯}{N}}_{*}\left(r,1;F,G\right)+S\left(r,F\right)+S\left(r,G\right)\\ \le 2\stackrel{¯}{N}\left(r,0,{f}^{\prime }\left(z\right)\right)+2\stackrel{¯}{N}\left(r,0,f\left(z+c\right)\right)+\stackrel{¯}{N}\left(r,\infty ,{f}^{\prime }\left(z\right)\right)+\stackrel{¯}{N}\left(r,\infty ,f\left(z+c\right)\right)\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }+\frac{1}{2}\left(\stackrel{¯}{N}\left(r,\infty ,{f}^{\prime }\left(z\right)\right)+\stackrel{¯}{N}\left(r,\infty ,f\left(z+c\right)\right)\right)+S\left(r,{f}^{\prime }\left(z\right)\right)+S\left(r,f\left(z+c\right)\right).\end{array}$ (9)

$\left(2{n}^{2}-11n-3\right)\left[T\left(r,{f}^{\prime }\left(z\right)\right)+T\left(r,f\left(z+c\right)\right)\right]\le S\left(r,{f}^{\prime }\left(z\right)\right)+S\left(r,f\left(z+c\right)\right)$.

$\frac{{F}^{″}}{{F}^{\prime }}-\frac{2{F}^{\prime }}{F-1}=\frac{{G}^{″}}{{G}^{\prime }}-\frac{2{G}^{\prime }}{G-1}$. (10)

$\frac{1}{G-1}=\frac{{C}_{1}}{F-1}+{C}_{2}$. (11)

$G=\frac{\left({C}_{2}+1\right)F+\left({C}_{1}-{C}_{2}-1\right)}{{C}_{2}F+\left({C}_{1}-{C}_{2}\right)}$. (12)

$F=\frac{\left({C}_{2}-{C}_{1}\right)G+\left({C}_{1}-{C}_{2}-1\right)}{{C}_{2}G-\left({C}_{2}+1\right)}$. (13)

$T\left(r,{f}^{\prime }\left(z\right)\right)=T\left(r,f\left(z+c\right)\right)+S\left(r,{f}^{\prime }\left(z\right)\right)$,$S\left(r,{f}^{\prime }\left(z\right)\right)=S\left(r,f\left(z+c\right)\right)$. (14)

$\stackrel{¯}{N}\left(r,\infty ,F\right)=\stackrel{¯}{N}\left(r,0,G-\frac{{C}_{2}+1}{{C}_{2}}\right)$. (15)

$\begin{array}{l}nT\left(r,f\left(z+c\right)\right)+S\left(r,f\left(z+c\right)\right)=T\left(r,G\right)\\ \le \stackrel{¯}{N}\left(r,\infty ,G\right)+\stackrel{¯}{N}\left(r,0,G\right)+\stackrel{¯}{N}\left(r,0,G-\frac{{C}_{2}+1}{{C}_{2}}\right)+S\left(r,G\right)\\ \le \stackrel{¯}{N}\left(r,\infty ,G\right)+\stackrel{¯}{N}\left(r,0,G\right)+\stackrel{¯}{N}\left(r,\infty ,F\right)+S\left(r,G\right)\\ \le \stackrel{¯}{N}\left(r,\infty ,f\left(z+c\right)\right)+\stackrel{¯}{N}\left(r,0,f\left(z+c\right)\right)+\stackrel{¯}{N}\left(r,{f}^{\prime }\left(z\right)\right)+S\left(r,f\left(z+c\right)\right)\\ \le \text{3}T\left(r,f\left(z+c\right)\right)+S\left(r,f\left(z+c\right)\right).\end{array}$

$\left(n-3\right)T\left(r,f\left(z+c\right)\right)\le S\left(r,f\left(z+c\right)\right)$，这与已知条件 $n\ge 6$ 矛盾。

$G=\frac{{C}_{1}}{{C}_{1}+1-F}$. (16)

${C}_{1}\ne -1$，由(16)式得

$\stackrel{¯}{N}\left(r,\infty ,G\right)=\stackrel{¯}{N}\left(r,0,F-{C}_{1}-1\right)$. (17)

$\begin{array}{l}nT\left(r,{f}^{\prime }\left(z\right)\right)+S\left(r,{f}^{\prime }\left(z\right)\right)=T\left(r,F\right)\\ \le \stackrel{¯}{N}\left(r,\infty ,F\right)+\stackrel{¯}{N}\left(r,0,F\right)+\stackrel{¯}{N}\left(r,0,F-{C}_{1}-1\right)+S\left(r,F\right)\\ \le \stackrel{¯}{N}\left(r,\infty ,F\right)+\stackrel{¯}{N}\left(r,0,F\right)+\stackrel{¯}{N}\left(r,\infty ,G\right)+S\left(r,F\right)\\ \le \stackrel{¯}{N}\left(r,\infty ,{f}^{\prime }\left(z\right)\right)+\stackrel{¯}{N}\left(r,0,{f}^{\prime }\left(z\right)\right)+\stackrel{¯}{N}\left(r,\infty ,f\left(z+c\right)\right)+S\left(r,{f}^{\prime }\left(z\right)\right)\\ \le \text{3}T\left(r,{f}^{\prime }\left(z\right)\right)+S\left(r,{f}^{\prime }\left(z\right)\right).\end{array}$

$\left(n-3\right)T\left(r,{f}^{\prime }\left(z\right)\right)\le S\left(r,{f}^{\prime }\left(z\right)\right)$，这与已知条件 $n\ge 6$ 矛盾。因此 ${C}_{1}=-1$，将其带入(16)式得 $FG\equiv 1$，从而

${f}^{\prime }\left(z\right)f\left(z+c\right)\equiv t$.

$G=\frac{F+{C}_{1}-1}{{C}_{1}}$. (18)

${C}_{1}\ne 1$，由(18)式得

$\stackrel{¯}{N}\left(r,0,G\right)=\stackrel{¯}{N}\left(r,0,F+{C}_{1}-1\right)$. (19)

$\begin{array}{l}nT\left(r,{f}^{\prime }\left(z\right)\right)+S\left(r,{f}^{\prime }\left(z\right)\right)=T\left(r,F\right)\\ \le \stackrel{¯}{N}\left(r,\infty ,F\right)+\stackrel{¯}{N}\left(r,0,F\right)+\stackrel{¯}{N}\left(r,0,F+{C}_{1}-1\right)+S\left(r,F\right)\\ \le \stackrel{¯}{N}\left(r,\infty ,F\right)+\stackrel{¯}{N}\left(r,0,F\right)+\stackrel{¯}{N}\left(r,0,G\right)+S\left(r,F\right)\\ \le \stackrel{¯}{N}\left(r,\infty ,{f}^{\prime }\left(z\right)\right)+\stackrel{¯}{N}\left(r,0,{f}^{\prime }\left(z\right)\right)+\stackrel{¯}{N}\left(r,0,f\left(z+c\right)\right)+S\left(r,{f}^{\prime }\left(z\right)\right)\\ \le \text{3}T\left(r,{f}^{\prime }\left(z\right)\right)+S\left(r,{f}^{\prime }\left(z\right)\right).\end{array}$

$\left(n-3\right)T\left(r,{f}^{\prime }\left(z\right)\right)\le S\left(r,{f}^{\prime }\left(z\right)\right)$，这与已知条件 $n\ge 6$ 矛盾.因此 ${C}_{1}=1$，将其带入(18)式得 $F\equiv G$，从而

${f}^{\prime }\equiv tf\left(z+c\right)$.

$\begin{array}{l}\frac{n}{2}\left(T\left(r,{f}^{\prime }\left(z\right)\right)+T\left(r,f\left(z+c\right)\right)\right)\\ \le 2\stackrel{¯}{N}\left(r,0,{f}^{\prime }\left(z\right)\right)+2\stackrel{¯}{N}\left(r,0,f\left(z+c\right)\right)+\stackrel{¯}{N}\left(r,\infty ,{f}^{\prime }\left(z\right)\right)+\stackrel{¯}{N}\left(r,\infty ,f\left(z+c\right)\right)\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }+\frac{1}{2}\left(\stackrel{¯}{N}\left(r,\infty ,{f}^{\prime }\left(z\right)\right)+\stackrel{¯}{N}\left(r,\infty ,f\left(z+c\right)\right)\right)+S\left(r,{f}^{\prime }\left(z\right)\right)+S\left(r,f\left(z+c\right)\right).\end{array}$

$\begin{array}{l}n\left(T\left(r,{f}^{\prime }\left(z\right)\right)+T\left(r,f\left(z+c\right)\right)\right)\\ \le 4\stackrel{¯}{N}\left(r,0,{f}^{\prime }\left(z\right)\right)+4\stackrel{¯}{N}\left(r,0,f\left(z+c\right)\right)+S\left(r,{f}^{\prime }\left(z\right)\right)+S\left(r,f\left(z+c\right)\right).\end{array}$

Uniqueness Theoremon Derivatives and Shifts of Meromorphic Functions[J]. 理论数学, 2022, 12(06): 1059-1066. https://doi.org/10.12677/PM.2022.126116

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