﻿ 整函数与其差分多项式的唯一性

# 整函数与其差分多项式的唯一性Uniqueness of Entire Functions That Share Small Function with Their Difference Polynomials

Abstract: In this paper, we study the uniqueness of difference operators about transcendental entire function f(z) with a Borel entire exceptional function, which shares a small function a(z) with its difference polynomial. Furthermore, under the above assumption, we replace the condition “Borel entire exceptional function” by “δ(0,f)＞0”，and get the same result when f(z) shares a(z) CM with its difference polynomial.

1. 引言

$\begin{array}{c}\stackrel{¯}{N}\left(r,\frac{1}{F-1}\right)={\stackrel{¯}{N}}_{E}^{1\right)}\left(r,\frac{1}{F-1}\right)+{\stackrel{¯}{N}}_{L}\left(r,\frac{1}{F-1}\right)+{\stackrel{¯}{N}}_{L}\left(r,\frac{1}{G-1}\right)+{\stackrel{¯}{N}}_{E}^{\left(2}\left(r,\frac{1}{G-1}\right)\\ =\stackrel{¯}{N}\left(r,\frac{1}{G-1}\right)\end{array}$

$g\left(z\right)={m}_{1}\left(z\right)f\left(z+{c}_{1}\right)+{m}_{2}\left(z\right)f\left(z+{c}_{2}\right)+\cdots +{m}_{k}\left(z\right)f\left(z+{c}_{k}\right)$

$g\left(z\right)={m}_{1}\left(z\right)f\left(z+{c}_{1}\right)+{m}_{2}\left(z\right)f\left(z+{c}_{2}\right)+\cdots +{m}_{k}\left(z\right)f\left(z+{c}_{k}\right)$

$g\left(z\right)={m}_{1}\left(z\right)f\left(z+{c}_{1}\right)+{m}_{2}\left(z\right)f\left(z+{c}_{2}\right)+\cdots +{m}_{k}\left(z\right)f\left(z+{c}_{k}\right)$

2. 几个引理

$m\left(r,\frac{f\left(z+c\right)}{f\left(z\right)}\right)=S\left(r,f\right)$,

$N\left(r,\frac{1}{{f}^{\left(k\right)}}\right)\le N\left(r,\frac{1}{f}\right)+kN\left(r,f\right)+S\left(r,f\right)$.

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

$\lambda \left(f\cdot g\right)\le \mathrm{max}\left\{\lambda \left(f\right),\lambda \left(g\right)\right\}$.

$T\left(r,f\right)=o\left(T\left(r,g\right)\right)$.

$\lambda \left(H\right)<\lambda \left(f-\alpha \right)=\lambda \left(f\right)\le \mathrm{max}\left\{\lambda \left(H\right),\lambda \left({e}^{Q}\right)\right\}=\lambda \left({e}^{Q}\right)$,

$\lambda \left({e}^{Q}\right)\le \mathrm{max}\left\{\lambda \left(H\right),\lambda \left(f\right)\right\}=\lambda \left(f\right)$,

$\lambda \left(f\right)=\lambda \left({e}^{Q}\right)=\mu \left({e}^{Q}\right)$。又由引理5可得 。因此

$\begin{array}{c}T\left(r,f\right)=T\left(r,f-\alpha \right)+S\left(r,f\right)\\ =T\left(r,H{e}^{Q}\right)+S\left(r,f\right)\\ \le T\left(r,H\right)+T\left(r,{e}^{Q}\right)+S\left(r,f\right)\\ \le T\left(r,{e}^{Q}\right)+S\left(r,f\right)+S\left(r,{e}^{Q}\right)\end{array}$,

$\delta \left(\alpha ,f\right)=1-\stackrel{¯}{\underset{r\to \infty }{\mathrm{lim}}}\frac{N\left(r,\frac{1}{f-\alpha }\right)}{T\left(r,f\right)}\le 1-\stackrel{¯}{\underset{r\to \infty }{\mathrm{lim}}}\frac{N\left(r,\frac{1}{H}\right)}{T\left(r,{e}^{Q}\right)}=1$.

3. 定理1的证明

, (1)

$T\left(r,g\right)\le T\left(r,f\right)+S\left(r,f\right)=m\left(r,\frac{1}{f-\alpha }\right)+S\left(r,f\right)\le m\left(r,\frac{g-\beta }{f-\alpha }\right)+m\left(r,\frac{1}{g-\beta }\right)+S\left(r,f\right)$,

$T\left(r,g\right)=m\left(r,g\right)+S\left(r,f\right)$。故 $N\left(r,\frac{1}{g-\beta }\right)=S\left(r,f\right)$。从而有 $N\left(r,\frac{1}{F}\right)=S\left(r,f\right)$$N\left(r,\frac{1}{G}\right)=S\left(r,f\right)$

$\begin{array}{c}N\left(r,H\right)\le {\stackrel{¯}{N}}_{\left(2}\left(r,\frac{1}{F}\right)+{\stackrel{¯}{N}}_{\left(2}\left(r,\frac{1}{G}\right)+{N}_{L}\left(r,\frac{1}{F-1}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{N}_{L}\left(r,\frac{1}{G-1}\right)+{N}_{0}\left(r,\frac{1}{{F}^{\prime }}\right)+{N}_{0}\left(r,\frac{1}{{G}^{\prime }}\right)\\ \le {N}_{L}\left(r,\frac{1}{F-1}\right)+{N}_{L}\left(r,\frac{1}{G-1}\right)+{N}_{0}\left(r,\frac{1}{{F}^{\prime }}\right)+{N}_{0}\left(r,\frac{1}{{G}^{\prime }}\right)\end{array}$, (2)

. (3)

. (4)

$\begin{array}{l}\stackrel{¯}{N}\left(r,\frac{1}{F-1}\right)+\stackrel{¯}{N}\left(r,\frac{1}{G-1}\right)\\ ={\stackrel{¯}{N}}_{E}^{1\right)}\left(r,\frac{1}{F-1}\right)+3{N}_{L}\left(r,\frac{1}{F-1}\right)+3{N}_{L}\left(r,\frac{1}{G-1}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+2{N}_{E}^{\left(2}\left(r,\frac{1}{G-1}\right)+{N}_{0}\left(r,\frac{1}{{F}^{\prime }}\right)+{N}_{0}\left(r,\frac{1}{{G}^{\prime }}\right)+S\left(r,f\right)\end{array}$. (5)

${N}_{L}\left(r,\frac{1}{F-1}\right)+2{N}_{L}\left(r,\frac{1}{G-1}\right)+{N}_{E}^{1\right)}\left(r,\frac{1}{F-1}\right)+2{N}_{E}^{\left(2}\left(r,\frac{1}{G-1}\right)\le N\left(r,\frac{1}{G-1}\right)\le T\left(r,G\right)$. (6)

$\begin{array}{c}\stackrel{¯}{N}\left(r,\frac{1}{F-1}\right)+\stackrel{¯}{N}\left(r,\frac{1}{G-1}\right)\le 2{N}_{L}\left(r,\frac{1}{F-1}\right)+{N}_{L}\left(r,\frac{1}{G-1}\right)+T\left(r,G\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{N}_{0}\left(r,\frac{1}{{F}^{\prime }}\right)+{N}_{0}\left(r,\frac{1}{{G}^{\prime }}\right)+S\left(r,f\right)\end{array}$. (7)

$T\left(r,F\right)\le 2{N}_{L}\left(r,\frac{1}{F-1}\right)+{N}_{L}\left(r,\frac{1}{G-1}\right)+S\left(r,f\right)$. (8)

$\begin{array}{c}2{N}_{L}\left(r,\frac{1}{F-1}\right)+{N}_{L}\left(r,\frac{1}{G-1}\right)\le 2N\left(r,\frac{1}{{F}^{\prime }}\right)+N\left(r,\frac{1}{{G}^{\prime }}\right)\\ \le 2N\left(r,\frac{1}{F}\right)+N\left(r,\frac{1}{G}\right)\\ =S\left(r,f\right)\end{array}$. (9)

$\frac{1}{F-1}=\frac{A}{G-1}+B$, (10)

$F=\frac{\left(B+1\right)G+A-B-1}{BG+A-B},G=\frac{\left(B-A\right)F+A-B-1}{BF-B-1}$. (11)

$\begin{array}{c}T\left(r,f\right)=T\left(r,F\right)+S\left(r,f\right)\\ \le \stackrel{¯}{N}\left(r,\frac{1}{F}\right)+\stackrel{¯}{N}\left(r,\frac{1}{F-\frac{B+1}{B}}\right)+S\left(r,f\right)\\ =\stackrel{¯}{N}\left(r,G\right)+S\left(r,f\right)=S\left(r,f\right)\end{array}$, (12)

$T\left(r,f\right)=S\left(r,f\right)$。矛盾。

. (13)

$A\ne 1$，从(13)可得 $N\left(r,\frac{1}{F-\frac{A-1}{A}}\right)=N\left(r,G\right)$。类似于情形1的证明过程我们同样可得矛盾。因此 A=1 。由(10)可知 $F\equiv G$，即 $f\equiv g$

. (14)

$A\ne -1$，从(14)可得 $F\cdot G\equiv 1$，即

$\left(f-\alpha \right)\left(g-\beta \right)\equiv \left(a-\alpha \right)\left(a-\beta \right)$. (15)

$\begin{array}{c}2T\left(r,f\right)=2T\left(f-\alpha \right)+S\left(r,f\right)=T\left(r,\frac{1}{{\left(f-\alpha \right)}^{2}}\right)+S\left(r,f\right)\\ =m\left(r,\frac{1}{{\left(f-\alpha \right)}^{2}}\right)+N\left(r,\frac{1}{{\left(f-\alpha \right)}^{2}}\right)+S\left(r,f\right)\\ \le m\left(r,\frac{g-\beta }{f-\alpha }\frac{1}{\left(f-\alpha \right)\left(g-\beta \right)}\right)+S\left(r,f\right)\\ \le m\left(r,\frac{g-\beta }{f-\alpha }\right)+m\left(r,\frac{1}{\left(a-\alpha \right)\left(a-\beta \right)}\right)+S\left(r,f\right)\\ \le T\left(r,a-\alpha \right)+T\left(r,a-\beta \right)+S\left(r,f\right)=S\left(r,f\right)\end{array}$. (16)

$T\left(r,f\right)=S\left(r,f\right)$。矛盾。因此定理1得证。

4. 定理2的证明

$\frac{g-a}{f-a}={e}^{H}$, (17)

$g-a={e}^{H}\left(f-a\right)-\left(f-a\right)+f-a=\left({e}^{H}-1\right)\left(f-a\right)+f-a$,

$g=\left({e}^{H}-1\right)\left(f-a\right)+f$. (18)

$\frac{g}{\left({e}^{H}-1\right)fa}=\frac{f-a}{fa}+\frac{1}{\left({e}^{H}-1\right)a}=\frac{1}{a}-\frac{1}{f}+\frac{1}{\left({e}^{H}-1\right)a}$,

$\begin{array}{c}m\left(r,\frac{1}{f}\right)=m\left(r,\frac{1}{\left({e}^{H}-1\right)a}+\frac{1}{a}-\frac{g}{\left({e}^{H}-1\right)fa}\right)\\ \le 3m\left(r,\frac{1}{a}\right)+2m\left(r,\frac{1}{{e}^{H}-1}\right)+m\left(r,\frac{g}{f}\right)+O\left(1\right)\\ \le 2m\left(r,\frac{1}{{e}^{H}-1}\right)+S\left(r,f\right)\end{array}$. (19)

$T\left(r,{e}^{H}\right)\le N\left(r,\frac{1}{{e}^{H}-1}\right)+S\left(r,{e}^{H}\right)\le T\left(r,{e}^{H}\right)+S\left(r,{e}^{H}\right)$,

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