# Uniqueness Theorems of Difference Operator on Entire Functions

Authors: Jie Ding
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

We investigate the uniqueness questions of the difference operator on entire functions and obtain three uniqueness theorems using the idea of weight sharing.


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## Body

## 1. Introduction

A functionf(z) is called meromorphic, if it is analytic in the complex plane except at poles. It is assumed that the reader is familiar with the standard symbols and fundamental results of Nevanlinna theory such as the characteristic function T(r,f), and proximity function m(r,f), counting function N(r,f) (see [1, 2]). In addition we use S(r,f) denotes any quantity that satisfies the condition: S(r,f)=o(T(r,f)) as r→∞ possibly outside an exceptional set of finite logarithmic measure.Letf and g be two nonconstant meromorphic functions, a∈ℂ∪{∞}, we say that f and g share the value a IM (ignoring multiplicities) if f-a and g-a have the same zeros, they share the value a CM (counting multiplicities) if f-a and g-a have the same zeros with the same multiplicities. When a=∞ the zeros of f-a mean the poles of f (see [2]).Letp be a positive integer and a∈ℂ∪{∞}. We use Np)(r,1/(f-a)) to denote the counting function of the zeros of f-a (counted with proper multiplicities) whose multiplicities are not bigger than p, N(p+1(r,1/(f-a)) to denote the counting function of the zeros of f-a whose multiplicities are not less than p+1. N¯p)(r,1/(f-a)) and N¯(p+1(r,1/(f-a)) denote their corresponding reduced counting functions (ignoring multiplicities), respectively. We denote by E(a,f) the set of zeros of f-a with multiplicity, Ep)(a,f) the set of zeros of f-a (counted with proper multiplicities) whose multiplicities are not greater than p.In 1997, Yang and Hua (see [3]) studied the uniqueness of the differential monomials and obtained the following result.Theorem A.
Letf and g be nonconstant entire functions, and let n≥3 be an integer. If fnf′ and gng′ share 1 CM, then either f(z)=c1expcz, g(z)=c2exp-cz, where c1, c2, and c are constants satisfying (c1c2)n+1c2=-1 or f=tg for a constant such that tn+1=1.Recently, a number of papers (including [1, 3–17]) have focused on complex difference equations and differences analogues of Nevanlinna theory.In particular, Qi et al. (see [16]) proved Theorem B, which can be considered as a difference counterpart of Theorem A.Theorem B.
Letf and g be transcendental entire functions with finite order, c be a nonzero complex constant and n≥6 be an integer. If fnf(z+c) and gng(z+c) share 1 CM, then fg=t1 or f=t2g for some constant t1 and t2 which satisfy t1n+1=1 and t2n+1=1.In 2011, Zhang et al. (see [17]) investigated the distribution of zeros and shared values of the difference operator on meromorphic functions and uniqueness of difference polynomials with the same 1 points or fixed points. They obtained the following results.Theorem C.
Letf and g be nonconstant entire functions of finite order, and let n≥5 be an integer. Suppose that c is a nonzero complex constant such that f(z+c)-f(z)≢0 and g(z+c)-g(z)≢0. If fn[f(z+c)-f(z)] and gn[g(z+c)-g(z)] share 1 CM, and g(z+c) and g(z) share 0 CM, then f(z)=c1expaz and g(z)=c2exp-az, where c1, c2, and a are constants satisfying (c1c2)n+1(expaz+exp-az-2)=-1 or f=tg for a constant such that tn+1=1.Theorem D.
Letf and g be nonconstant entire functions of finite order, and let n≥5 be an integer. Suppose that c is a nonzero complex constant such that f(z+c)-f(z)≢0 and g(z+c)-g(z)≢0. If fn[f(z+c)-f(z)] and gn[g(z+c)-g(z)] share z CM, and g(z+c) and g(z) share 0 CM, then f=tg, where t is a constant satisfying tn+1=1.We investigate the uniqueness theorem of another differences polynomial and prove Theorem1.Theorem 1.
Letf and g be nonconstant transcendental entire functions of finite order, and let n≥9 be an integer. Suppose that c is a nonzero real constant such that f(z+2c)+f(z+c)+f(z)≠g(z+2c)+g(z+c)+g(z) and f(z+2c)+f(z+c)+f(z)≢0,g(z+2c)+g(z+c)+g(z)≢0. If E3)(z,fn[f(z+2c)+f(z+c)+f(z)])=E3)(z,gn[g(z+2c)+g(z+c)+g(z)]), then f(z)=tg(z), where t is a constant satisfying tn+1=1 except that t=1.In paper [15], Wang et al. improved the Theorem B and proved the following result.Theorem E.
Letf and g be transcendental entire functions with finite order, c be a nonzero complex constant and n≥6 be an integer. If E3)(1,fnf(z+c))=E3)(1,gng(z+c)), then fg=t1 or f=t2g for some constant t1 and t2 which satisfy t1n+1=1 and t2n+1=1.The purpose of this paper is to induce the idea of weight sharing to Theorems C and D, the results as follow.Theorem 2.
Letf and g be nonconstant entire functions of finite order, and let n≥6 be an integer. Suppose that c is a nonzero complex constant such that f(z+c)-f(z)≢0 and g(z+c)-g(z)≢0. If E3)(1,fn[f(z+c)-f(z)])=E3)(1,gn[g(z+c)-g(z)]) share 1 CM, g(z+c) and g(z) share 0 CM, then f(z)=c1expaz and g(z)=c2exp-az, where c1, c2, and a are constants satisfying (c1c2)n+1(expaz+exp-az-2)=-1 or f=tg for a constant such that tn+1=1.Theorem 3.
Letf and g be nonconstant entire functions of finite order, and let n≥6 be an integer. Suppose that c is a non-zero complex constant such that f(z+c)-f(z)≢0 and g(z+c)-g(z)≢0. If E3)(1,fn[f(z+c)-f(z)])=E3)(1,gn[g(z+c)-g(z)]) share z CM, g(z+c) and g(z) share 0 CM, then f=tg, where t is a constant satisfying tn+1=1.Remark 4.
Some ideas of this paper are based on [15, 17].


## 2. Some Lemmas

In order to prove our theorems, we need the following Lemmas.Lemma5 is a difference analogue of the logarithmic derivative lemma, given by Halburd and Korhonen [9] and Chiang and Feng [7] independently.Lemma 5 (see [9]).
Letf(z) be a meromorphic function of finite order, and let c∈ℂ and δ∈(0,1). Then
(1)m(r,f(z+c)f(z))+m(r,f(z)f(z+c))=O(T(r,f)rδ)=S(r,f),
for all r outside of a possibly exceptional set with finite logarithmic measure.Lemma 6 (see [2]).
Letf(z) be a nonconstant meromorphic function, and let P(f)=a0fn+a1fn-1+⋯an, where a0(≠0), a1,…,an are small functions of f. Then
(2)T(r,P(f))=nT(r,f)+S(r,f).Lemma 7 (see [18]).
Letf and g be two nonconstant meromorphic function satisfying Ek)(1,f)=Ek)(1,g) for some positive integer k∈ℕ. Define H as follow
(3)H=(f′′f′-2f′f-1)-(g′′g′-2g′g-1).
IfH≢0, then
(4)N(r,H)≤N¯(2(r,f)+N¯(2(r,1f)+N¯(2(r,g)+N¯(2(r,1g)+N¯0(r,1f′)+N¯0(r,1g′)+N¯(k+1(r,1f-1)+N¯(k+1(r,1g-1)+S(r,f)+S(r,g),
where N¯0(r,1/f′) denotes the counting function of zeros of f′ but not zeros of f(f-1), and N¯0(r,1/g′) is similarly defined.Lemma 8 (see [19]).
Under the conditions of Lemma7, we have
(5)N1)(r,1f-1)=N1)(r,1g-1)≤N(r,H)+S(r,f)+S(r,g).Lemma 9 (see [19]).
IfH=(f′′/f′-2f′/(f-1))-(g′′/g′-2g′/(g-1))≡0, then either f≡g or fg≡1 provided that
(6)limsupr→∞,r∈IN¯(r,f)+N¯(r,g)+N¯(r,1/f)+N¯(r,1/g)T(r)≤1,
where T(r):=max{T(r,f),T(r,g)} and I is a set with infinite linear measure.Lemma 10.
Letf(z) be a meromorphic function of finite order, c∈ℂ. Then
(7)N(r,f(z+c))=N(r,f(z))+S(r,f(z)).Proof.
Using Lemma5 and the formula (12) in [12]
(8)N(r,f(z+c))≤N(r,f(z))+S(r,f(z)).
Replacingf(z) with f(z-c), we have
(9)N(r,f(z))≤N(r,f(z-c))+S(r,f(z-c))=N(r,f(z-c))+S(r,f(z)),
for every c∈ℂ, so we deduce that
(10)N(r,f(z))≤N(r,f(z+c))+S(r,f(z)).
From (8) and (10), we obtain that
(11)N(r,f(z+c))=N(r,f(z))+S(r,f(z)).
Thus we completed the proof.Lemma 11 (see [9]).
LetT:(0,+∞)→(0,+∞) be a nondecreasing continuous function, s>0, 0<α<1, and let F⊂R+ be the set of all r satisfy
(12)T(r)≤αT(r+s).
If the logarithmic measure ofF is infinite, then
(13)limsupr→∞logT(r,f)logr=∞.


## 3. Proof of Theorems

Proof of Theorem1.
We define(14)F(z)=fn(z)[f(z+2c)+f(z+c)+f(z)]z,G(z)=gn(z)[g(z+2c)+g(z+c)+g(z)]z.
In Lemma7, we replace f and g, by F and G respectively, we claim that H≡0. If it is not true, then H≢0. From Lemma 8 we have that
(15)N1)(r,1F-1)=N1)(r,1G-1)≤N(r,H)+S(r,f)+S(r,g)≤N¯(2(r,1F)+N¯(2(r,1G)+N¯0(r,1F′)+N¯0(r,1G′)+N¯(4(r,1F-1)+N¯(4(r,1G-1)+S(r,f)+S(r,g).
From the Nevanlinna second foundational theorem, we can get that(16)T(r,F)+T(r,G)≤N¯(r,1F)+N¯(r,1G)+N¯0(r,1F′)+N¯0(r,1G′)+N¯(r,1F-1)+N¯(r,1G-1)+S(r,f)+S(r,g).
From the definitions ofNk) and N(k, the following inequalities are obvious:(17)N¯(r,1F-1)-12N1)(r,1F-1)+N¯(4(r,1F-1)≤12N(r,1F-1),N¯(r,1G-1)-12N1)(r,1G-1)+N¯(4(r,1G-1)≤12N(r,1G-1).
Combining (15), (16), and (17), we deduce that
(18)T(r,F)+T(r,G)≤2N¯(r,1F)+2N¯(2(r,1F)+2N¯(r,1G)+2N¯(2(r,1G)+S(r,f)+S(r,g)≤4N¯(r,1f)+4N¯(r,1g)+2N(r,1f(z+c))+2N(r,1g(z+c))+2N(r,1f(z+2c))+2N(r,1g(z+2c))+S(r,f)+S(r,g)≤8N(r,1f)+8N(r,1g)+S(r,f)+S(r,g)≤8T(r,1f)+8T(r,1g)+S(r,f)+S(r,g).
We can apply Lemma5, Lemma 6, and Lemma 10 to show that
(19)(n+1)T(r,f)=T(r,fn+1)≤T(r,F(z))+T(r,f(z+2)+f(z+1)+f(z)f(z))+S(r,f(z))=T(r,F(z))+N(r,f(z+2)+f(z+1)+f(z)f(z))+S(r,f(z))≤T(r,F(z))+T(r,f(z))+S(r,f(z)),
which implies
(20)T(r,F)≥nT(r,f)+S(r,f).
The same augment as above, we have that(21)T(r,G)≥nT(r,g)+S(r,g).
From (18), (20), and (21), we can deduce that
(22)(n-8)[T(r,f)+T(r,g)]≥S(r,f)+S(r,g).
which is a contraction. Therefore, H≡0.
Noting that(23)N¯(r,1F)+N¯(r,1G)≤4[T(r,f)+T(r,g)]+S(r,f)+S(r,g)≤T(r),
where T(r)=maxT(r,F),T(r,G).
Because of Lemma9, we have that F≡G or FG≡1. We will consider the following two cases.
Case1. Suppose that F(z)=G(z). Then
(24)fn(z)[f(z+2c)+f(z+c)+f(z)]=gn(z)[g(z+2c)+g(z+c)+g(z)].
Leth(z)=f(z)/g(z), we deduce that
(25)hn(z)[h(z+2c)g(z+2c)ahn(z)+h(z+c)g(z+c)+h(z)g(z)]=g(z+2c)+g(z+c)+g(z).
Ifh(z+c)≢h(z), by the hypothesis f(z+2c)+f(z+c)+f(z)≠g(z+2c)+g(z+c)+g(z), we get that h(z)≠1. So
(26)T(r,hn)≤N¯(r,1hn)+N¯(r,hn)+N¯(r,1hn-1)≤2T(r,h)+S(r,h),
which means h is a constant, because of n≥10.
Thenh(z)=t and t is a constant satisfying tn+1=1 except that t=1.
Case2. Suppose that F(z)·G(z)≡1. Then
(27)fn(z)[f(z+2c)+f(z+c)+f(z)]×gn(z)[g(z+2c)+g(z+c)+g(z)]=z2.
Note that zero is a Picard exceptional value off and g, then f(z)=eP(z) and g(z)=eQ(z), where P(z) and Q(z) are polynomials. In (27), we let z=0, then
(28)enP(0)[eP(2c)+eP(c)+eP(0)]enQ(0)[eQ(2c)+eQ(c)+eQ(0)]=0.
It is impossible, because ofc is a real number.Proof of Theorem2.
Denoting(29)F(z)=fn(z)[f(z+c)-f(z)],G(z)=gn(z)[g(z+c)-g(z)].
In Lemma7, we replace f and g, by F and G respectively. If H≢0, by Lemma 8 we deduce that
(30)N1)(r,1F-1)=N1)(r,1G-1)≤N(r,H)+S(r,f)+S(r,g)≤N¯(2(r,1F)+N¯(2(r,1G)+N¯0(r,1F′)+N¯0(r,1G′)+N¯(4(r,1F-1)+N¯(4(r,1G-1)+S(r,f)+S(r,g).
The same reasons as in the proof of Theorem1, we have that
(31)T(r,F)+T(r,G)≤N¯(r,1F)+N¯(r,1G)+N0¯(r,1F′)+N0¯(r,1G′)+N¯(r,1F-1)+N¯(r,1G-1)+S(r,f)+S(r,g),N¯(r,1F-1)-12N1)(r,1F-1)+N¯(4(r,1F-1)≤12N(r,1F-1),N¯(r,1G-1)-12N1)(r,1G-1)+N¯(4(r,1G-1)≤12N(r,1G-1).
Combining (30) and (31), we deduce that
(32)T(r,F)+T(r,G)≤2N¯(r,1F)+2N¯(2(r,1F)+2N¯(r,1G)+2N¯(2(r,1G)+S(r,f)+S(r,g)≤4N¯(r,1f)+4N¯(r,1g)+2N(r,1f(z+c)-f(z))+2N(r,1g(z+c)-g(z))+S(r,f)+S(r,g)≤4N¯(r,1f)+2T(r,f(z+c)-f(z))+6N(r,1g)+S(r,f)+S(r,g)=4N¯(r,1f)+2m(r,f(z+c)-f(z))+6N(r,1g)+S(r,f)+S(r,g)≤6T(r,1f)+6T(r,1g)+S(r,f)+S(r,g).
We can apply Lemmas5, 6, and 10 to show that
(33)(n+1)T(r,f)=T(r,fn+1)≤T(r,F(z))+T(r,f(z+c)-f(z)f(z))+S(r,f(z))=T(r,F(z))+N(r,f(z+c)-f(z)f(z))+S(r,f(z))≤T(r,F(z))+T(r,f(z))+S(r,f(z)),
which implies
(34)T(r,F)≥nT(r,f)+S(r,f).
We have thatN(r,(g(z+c)-g(z))/g(z)), Since g(z+c) and g(z) share 0 CM, then
(35)T(r,G)≥(n+1)T(r,g)+S(r,g).
From (32), (34), and (35), we can deduce that
(36)(n-6)T(r,f)+(n-5)T(r,g)≥S(r,f)+S(r,g).
which is a contraction. Therefore, H≡0.
Noting that(37)N¯(r,1F)+N¯(r,1G)≤3[T(r,f)+T(r,g)]+S(r,f)+S(r,g)≤T(r),
where T(r)=maxT(r,F),T(r,G).
Because of Lemma9, we have that F≡G or FG≡1.
By using the same methods as in the proof of Theorem  1.10 in [17], we can complete the proof of Theorem 2.Proof of Theorem3.
The proof is almost literally the same as the proof of Theorem2, with the methods as in the proof of Theorem  1.9 in [17] replacing the methods as in the proof of Theorem  1.10 in [17].



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