# Uniqueness Theorems on Difference Monomials of Entire Functions

Authors: Gang Wang; Deng-li Han; Zhi-Tao Wen
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

The aim of this paper is to discuss the uniqueness of the difference monomialsfnf(z+c). It assumed that f and g are transcendental entire functions with finite order and Ek)(1,fnf(z+c))=Ek)(1,gng(z+c)), where c is a nonzero complex constant and n, k are integers. It is proved that if one of the following holds (i) n≥6 and k=3, (ii) n≥7 and k=2, and (iii) n≥10 and k=1, then fg=t1 or f=t2g for some constants t2 and t3 which satisfy t2n+1=1 and t3n+1=1. It is an improvement of the result of Qi, Yang and Liu.


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## Body

## 1. Introduction and Main Results

In this paper, a meromorphic (respectively entire) function always means meromorphic (respectively, analytic) in the complex planeℂ. It is also assumed that the reader is familiar with the basic concepts of the Nevanlinna theory. We adopt the standard notations in the Nevanlinna value distribution theory of meromorphic functions as explained in [1, 2].Letf and g be two nonconstant meromorphic functions, and let a be a value in the extended plane. We say that f and g share the value a CM, provided that f and g have the same a-pints with the same multiplicities. We say that f and g share the value a IM, provided that f and g have the same a-points ignoring multiplicities. The order of f is defined by
(1.1)σ(f)=lim⁡sup⁡r→∞log⁡⁡T(r,f)log⁡⁡r.Letf be a nonconstant meromorphic function on ℂ, let a∈ℂ be a finite value, and let k be a positive integer or infinity. We denote by E(a,f) the set of zeros of f-a and count multiplicities, while by E¯(a,f) the set of zeros of f-a but ignore multiplicities. Also, we denote by Ek)(a,f) the set of zeros of f-a with multiplicities less than or equal to k and count multiplicities. For a∈ℂ⋃‍{∞}, we denote by Nk)(r,1/(f-a)) the counting function corresponding to the set Ek)(a,f) while by N(k+1(r,1/(f-a)) the counting function corresponding to the set E(k+1(a,f):=E(a,f)∖Ek)(a,f). If Ek(a,f)=Ek(a,g), we say that f, g share the value a with weight k.The definition implies that iff and g share a value a with weight k, then z0 is a zero of f-a with multiplicity m(≤k) if and only if it is a zero of g-a with multiplicity m(≤k) and z0 is a zero of f-a with multiplicity m(>k) if and only if it is a zero of g-a with multiplicity n(>k) where m is not necessarily equal to n.Also, we denote byN¯k)(r,1/(f-a)) and N¯(k+1(r,1/(f-a)) the reduced forms of Nk)(r,1/(f-a)) and N(k+1(r,1/(f-a)), respectively. At last, we set
(1.2)Nk(r,f)=N¯(r,f)+N¯(2(r,f)+⋯+N¯(k(r,f).Hayman proposed the well-known conjecture in [3].Hayman  Conjecture
If an entire functionf satisfies fnf′≠1 for all n∈ℕ, then f is a constant.
In fact, Hayman has proved that the conjecture holds in the casesn≥2 in [4] while Clunie proved the cases n=1 in [5], respectively. In 1997, Yang and Hua [6] studied the uniqueness theorem of the differential monomials and obtained the following result.Theorem A.
Letf and g be nonconstant entire function, and let n≥6 be an integer. If fnf′ and gng′ share 1 CM, then either f(z)=c1ecz, g(z)=c2e-cz where c1, c2 and c are constants satisfying (c1c2)n+1c2=-1, or f=tg for a constant t such that tn+1=1.In 2010, Qi et al. [7] studied the uniqueness of the difference monomials and obtained the following result.Theorem B.
Letf and g be transcendental entire functions with finite order, c a non-zero complex constant, and n≥6 an integer. If E(1,fnf(z+c))=E(1,gng(z+c)), then fg=t1 or f=t2g for some constants t1 and t2 which satisfy t1n+1=1 and t2n+1=1.In this paper, we will obtain the following results.Theorem 1.1.
Letf and g be transcendental entire functions with finite order, c a non-zero complex constant, and n≥6 an integer. If E3)(1,fnf(z+c))=E3)(1,gng(z+c)), then the assertion of Theorem B holds.Theorem 1.2.
Letf and g be transcendental entire functions with finite order, c a non-zero complex constant, and n≥7 an integer. If E2)(1,fnf(z+c))=E2)(1,gng(z+c)), then the assertion of Theorem B holds.Theorem 1.3.
Letf and g be transcendental entire functions with finite order, c a non-zero complex constant, and n≥10 an integer. If E1)(1,fnf(z+c))=E1)(1,gng(z+c)), then the assertion of Theorem B holds.


## 2. Auxiliary Results

Lemma 2.1 (see [8, Corollary 2.5]).
Letf(z) be a meromorphic function in the complex plane with finite order σ=σ(f), and let η be a fixed non-zero complex number. Then for each ε>0, one has
(2.1)m(r,f(z+η)f(z))+m(r,f(z)f(z+η))=O(rσ-1+ε).Lemma 2.2 (see [8, Theorem 2.1]).
Letf(z) be a meromorphic function in the complex plane with finite order σ=σ(f), and let η be a fixed non-zero complex number. Then for each ε>0, one has
(2.2)T(r,f(z+η))=T(r,f(z))+O(rσ-1+ε)+O(log⁡⁡r).Lemma 2.3.
Letf(z) be an entire function with finite order σ=σ(f), c a fixed non-zero complex number, and
(2.3)P(z)=anf(z)n+an-1f(z)n-1+⋯+a1f(z)+a0,
where aj(j=0,1,…,n) are constants. If F(z)=P(z)f(z+c), then
(2.4)T(r,F)=(n+1)T(r,f)+O(rσ(f)-1+ɛ)+O(log⁡r).Proof.
Sincef(z) is an entire transcendental function with finite order, we can deduce from Lemma 2.1 and the standard Valiron-Mohon'ko theorem that
(2.5)(n+1)T(r,f(z))=T(r,f(z)P(z))+O(1)=m(r,f(z)P(z))+O(1)≤m(r,f(z)P(z)F(z))+m(r,F(z))+O(1)=m(r,f(z)f(z+c))+m(r,F(z))+O(1)≤T(r,F(z))+O(rσ-1+ε)+O(1).
Therefore
(2.6)T(r,F)≥(n+1)T(r,f)+O(rσ-1+ε)+O(1).
On the other hand, Lemma 2.2 implies that
(2.7)T(r,F(z))≤T(r,P(z))+T(r,f(z+c))=nT(r,f(z))+T(r,f(z))+O(rσ-1+ε)+O(log⁡r)=(n+1)T(r,f)+O(rσ-1+ε)+O(log⁡r).
We will obtain the conclusion of Lemma 2.3.Remark 2.4.
The condition “entire” cannot be replaced by “meromorphic” in Lemma2.3, as is shown by the following example.Example 2.5.
Letf(z)=(ez-1)/(ez+1), c=πi, and F(z)=f(z)f(z+c), we can see
(2.8)T(r,F)≠2T(r,f)+O(rσ(f)-1+ε)+O(log⁡r)
for every set of {rn} with infinite measure.Lemma 2.6 (see [9, Lemma 2.1]).
Letf and g be two nonconstant meromorphic functions satisfying Ek)(1,f)=Ek)(1,g) for some positive integer k∈ℕ. Define H as follows:
(2.9)H=(f′′f′-2f′f-1)-(g′′g′-2g′g-1).
If H≢0, then
(2.10)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 2.7 (see [10]).
Under the condition of Lemma2.6, one has
(2.11)N1)(r,1f-1)=N1)(r,1g-1)≤N(r,H)+S(r,f)+S(r,g).Lemma 2.8 (see [10]).
LetH be defined as Lemma 2.6. If H≡0, then either f≡g or fg≡1 provided that
(2.12)lim⁡sup⁡r→∞,r∈IN¯(r,f)+N¯(r,g)+N¯(r,1/f)+N¯(r,1/f)T(r)<1,
where T(r):=max{T(r,f),T(r,g)} and I is a set with infinite linear measure.Lemma 2.9 (see [11, Lemma 2.2]).
LetT:(0,+∞)→(0,+∞) be a nondecreasing continuous function, s>0, 0<α<1, and let F⊂R+ be the set of all r such that
(2.13)T(r)≤αT(r+s).
If the logarithmic measure of F is infinite, then
(2.14)lim⁡sup⁡r→∞log⁡T(r,f)log⁡r=∞.


## 3. Proof of Theorem1.1

We define(3.1)F:=fnf(z+c),G:=gng(z+c).
First of all, suppose that H≢0. We replace f and g by F and G, respectively, in Lemma 2.7 and Lemma 2.8. Thus,
(3.2)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).
Applying the second main theorem to F and G jointly implies that
(3.3)T(r,F)+T(r,G)≤N¯(r,1F)+N¯(r,1F-1)+N¯(r,1G)+N¯(r,1G-1)-N¯0(r,1F′)-N¯0(r,1G′)+S(r,f)+S(r,g).
Noting that
(3.4)N¯(r,1F-1)-12N1)(r,1F-1)+N¯(4(r,1F-1)≤12N(r,1F-1)≤12T(r,F),N¯(r,1G-1)-12N1)(r,1G-1)+N¯(4(r,1G-1)≤12N(r,1G-1)≤12T(r,G).
According to Lemma 2.9 and (3.2)–(3.4), we can obtain that
(3.5)T(r,F)+T(r,G)≤2N2(r,1F)+2N2(r,1G)+S(r,f)+S(r,g)≤4N¯(r,1f)+4N¯(r,1g)+2N(r,1f(z+c))+2N(r,1g(z+c))+S(r,f)+S(r,g)≤6N(r,1f)+6N(r,1g)+S(r,f)+S(r,g)≤6T(r,1f)+6T(r,1g)+S(r,f)+S(r,g).
Lemma 2.3 shows that
(3.6)T(r,F)=(n+1)T(r,f)+O(rσ(f)-1+ε)+O(logr),T(r,G)=(n+1)T(r,g)+O(rσ(g)-1+ε)+O(log⁡r).
We can deduce that
(3.7)(n-5)(T(r,f)+T(r,g))≤O(rσ(f)-1+ε)+O(rσ(g)-1+ε)+S(r,f)+S(r,g),
which is impossible since n≥6. Therefore, we have H≡0. Noting that
(3.8)N(r,1F)+N(r,1G)≤3T(r,f)+3T(r,g)+S(r,f)+S(r,g)≤T(r),
where T(r)=max{T(r,F),T(r,G)}. Together with Lemma 2.8, it shows that either F≡G or FG≡1. We will consider the following two cases.Case 1.
Suppose thatF(z)=G(z). Therefore
(3.9)f(z)nf(z+c)=g(z)ng(z+c).
Let h1(z)=f(z)/g(z); we have
(3.10)h1(z)nh1(z+c)≡1.
If h1(z) is not a constant, then Lemma 2.2 and (3.10) imply that
(3.11)nT(r,h1)=T(r,h1(z+c))+O(1)=T(r,h1)+O(rσ(h1)-1+ε)+O(log⁡r),
which is a contraction with n≥6. Thus, h1(z)≡t1, where t1 is a constant. From (3.10), we have f(z)=t1g(z) and t1n+1=1.Case 2.
Suppose thatF(z)G(z)≡1. Therefore
(3.12)f(z)nf(z+c)g(z)ng(z+c)≡1.
Let h2(z)=f(z)g(z); we have
(3.13)h2(z)nh2(z+c)≡1.
By the same way as Case 1, we can obtain that h2 is a constant. Therefore, f(z)g(z)=t2 and t2n+1=1.


## 4. Proof of Theorem1.2

Noting that(4.1)N¯(r,1F-1)-25N1)(r,1F-1)+45N¯(3(r,1F-1)≤35N(r,1F-1)≤35T(r,F),N¯(r,1G-1)-25N1)(r,1G-1)+45N¯(3(r,1G-1)≤35N(r,1G-1)≤35T(r,G).
According to (3.1) and (4.1), we can obtain the conclusion of Theorem 1.2 by the same way as Section 3.


## 5. Proof of Theorem1.3

Noting that(5.1)N¯(r,1F-1)-14N1)(r,1F-1)+12N¯(2(r,1F-1)≤34N(r,1F-1)≤34T(r,F),N¯(r,1G-1)-14N1)(r,1G-1)+12N¯(2(r,1G-1)≤34N(r,1G-1)≤34T(r,G).
According to (3.1) and (5.1), we can obtain the conclusion of Theorem 1.2 by the same way as Section 3.



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