# Uniqueness Problems about Entire Functions with Their Difference Operator Sharing Sets

Authors: Fan Niu; Jianming Qi; Zhiyong Zhou
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this paper, we study the uniqueness questions of finite order transcendental entire functions and their difference operators sharing a set consisting of two distinct entire functions of finite smaller order. Our results in this paper improve the corresponding results from Liu (2009) and Li (2012).


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## Body

## 1. Introduction and Main Results

Before proceeding, we spare the reader for a moment and assume some familiarity with the basics of Nevanlinna theory of meromorphic functions inℂ such as the first and second main theorems and the usual notations such as the characteristic function Tr,f, the proximity function mr,f, and the counting function Nr,f. Sr,f denotes any quantity satisfying Sr,f=oTr,f as r⟶∞, except possibly on a set of finite logarithmic measure not necessarily the same at each occurrence, see e.g., [1–3].Letf be a meromorphic functions on ℂ. Here, the order ρf is defined by(1)ρf=limsupr⟶∞logTr,flogr,and the exponent of convergence of zeros λf is defined by(2)λf=limsupr⟶∞logNr,1/flogr.For a givena∈ℂ^=ℂ∪∞, we say that two meromorphic functions f and g share a CM (counting multiplicities) when f and g have the same a-points. Let S be a finite set of some entire functions and f an entire function. Then, a set EfS is defined as(3)EfS=∪a∈Sz|fz−a=0,counting multiplicities.Assume thatg is another entire function. We say that f and g share a set S, counting multiplicities (CM), provided that EfS=EgS.The uniqueness theory of meromorphic functions sharing sets generalizes that on sharing values and generally is more difficult. If meromorphic functions share a general set, it is not easy to determine these functions. In 1999, Li and Yang [4] deduced that if EfS=Ef′S with S contain two distinct constants, then f must have special forms. Fang and Zalcman [5] used the theory of normal family to solve the above problem by proving that there exists a finite set S containing three distinct elements such that if EfS=Ef′S, then f=f′.Recently years, Nevanlinna characteristic offz+c, the value distribution theory for difference analogue, Nevanlinna theory of the difference operator, and the difference analogue of the lemma on the logarithmic derivative had been built, see e.g., [1–4, 6–14]. For meromorphic functions fz, we define its shift by fcz=fz+c and its difference operators by(4)Δcf=fz+c−fz,Δcnf=Δcn−1Δcfz,n∈ℕ,n≥2.By Nevanlinna theory of the difference operator, a natural question to ask whether the derivativef′ can be replaced by the difference operator Δcfz=fz+c−fz in the above question ?In 2009, Liu [8] investigated the above question and proved the following result.Theorem 1.
Leta be a nonzero complex number and f be a transcendental entire function with finite order. If f and Δcf share a,−a CM, then Δcfz=fz for all z∈ℂ.In 2012, from Theorem1, considering the constant in set is replaced by the function, Li [9] proved the following.Theorem 2.
Ifa and b are two distinct entire functions, then f is a nonconstant entire function whose ρf≠1 and λf<ρf<∞ such that ρa<ρf and ρb<ρf. If f and Δcf share a,b CM, then fz=Δcfz for all z∈ℂ.After studying Theorem2, we propose some questions as follows.Question 1: from Theorem2, the condition ρf≠1 seems more stronger. So, one may ask whether it can be weakened or moved?Question 2: what will happen if the shiftΔcfz be replaced by Δcnfzn≥2 in Theorem 2?Fortunately, we have recently given a positive answer for Question 1 (see [14]). In this work, we also discuss the above problems and especially for Question 2. Finally, we derive the following results.Theorem 3.
Suppose thataandb are two distinct entire functions and f is a nonconstant entire function of finite order with λf<ρf<∞ such that ρa<ρf and ρb<ρf. If f and Δc2f share a,b CM, then fz must take one of the following conclusions:(1)
fz=Aeμz, where Aandμ are two nonzero constants satisfying eμc=2. Furthermore, fz=Δcfz.(2)
fz=HzeAz. Here, Hz is an entire function and λf=ρH<1.
Using the same method, we improve the above result from the shiftΔcfz to Δcnfzn≥3 in above theorem and obtain the following result.Theorem 4.
Suppose thataandb are two distinct entire functions and f is a nonconstant entire function of finite order with λf<ρf<∞ such that ρa<ρf and ρb<ρf. If f and Δcnfn≥3 share a,b CM, then fz=HzeAz, where Hz is an entire function such that λf=ρH<1.


## 2. Some Lemmas

We will introduce some lemmas for the proofs of our theorems in this section.Lemma 1.
(see [15]). Let f be a meromorphic function of finite order and let ω1andω2 be two arbitrary complex numbers such that ω1≠ω2. Assume that σ is the order of f, then for each ϵ>0, we have(5)mr,fz+ω1fz+ω2=Orσ−1+ϵ.Lemma 2 (see [16]).
Letg be a function transcendental and meromorphic in the plane with order less than 1. Set h>0. Then, there exists an ϵ-set E such that(6)gz+ωgz⟶1,whenz⟶∞inℂ∖E,uniformly in ω for ω≤h.Lemma 3.
(see [3]). Suppose that f1z,f2z,…,fnzn≥2 are meromorphic functions and g1z,g2z,…,gnz are entire functions satisfying the following conditions:(1)
∑j=1nfjzegjz=0.(2)
gjz−gkzare not constants for1≤j<k≤n.(3)
For1≤j≤n, 1≤h<k≤n, Tr,fj=oTr,egh−gk, r⟶∞,r∉E. Then, fjz=0,j=1,2,…,n.


## 3. Proof of Theorems

Proof of Theorem 1.
Due tof and Δc2f share a,b CM, so we set(7)Δc2f−aΔc2f−bf−af−b=eQ,where Q is an entire function. And then it follows from (7) and maxρa,ρb<ρf<∞ that Q is a polynomial.
Using Hadamard Factorization Theorem, we assume thatfz=hzePz, where h≡0 is an entire function and P is a polynomial which satisfied(8)λf=ρh<ρf=degP.
So,(9)Δc2f=fz+2c−2fz+c+fz=hz+2c−2hz+c+hzePz.
Put the forms off and Δc2f into (7) to yield(10)hz+2cePz+2c−Pz−2hz+cePz+c−Pz+hzePz−az,hz+2cePz+2c−Pz−2hz+cePz+c−Pz+hzePz−bz=hzePz−azhzePz−bzeQz.
Takew1=hz+2cePz+2c−Pz−2hz+cePz+c−Pz+hz. We assume that w1=0. Then,(11)hz+2cePz+2c−2hz+cePz+c+hzepz=0.
By Lemma3, if pz+2c−pz, pz+c−pz, and pz+2c−pz+c are not constants, then hz=0, a contradiction.
So,pz+2c−pz=a1, pz+c−pz=b1, and pz+2c−pz+c=c1, (where a1,b1,andc1 are three constants).
We can getp′z+2c−p′z=0, p′z+c−p′z=0, and p′z+2c−p′z+c=0.
Hence,pz is a periodic function. We also know pz is a polynomial. So, we get pz=Az+B (where A and B are two constants, and A≠0). So, we obtain fz=HzeAz. Here, Hz is an entire function and λf=ρH<1.
Ifw1≠0 below, obviously, w1 is a small function of eP. Rewrite (10) as(12)eQ=w12eP−a/w1eP−b/w1h2eP−a/heP−b/h.
Note thata≠b. Without loss of generality, we set a≠0. Suppose that z0 is a zero of eP−a/h, but not a zero of w1. From (12), we may easily obtain that z0 is a zero of eP−a/w1 or eP−b/w1. We denote by N1r,eP the reduced counting function of those common zeros of eP−a/h and ePz−a/w1. Similarly, we also denote N2r,eP the reduced counting function of those common zeros of eP−a/h and eP−b/w1. Then,(13)Tr,eP=N¯r,1ePz−a/h+Sr,eP=N1r,eP+N2r,eP+Sr,eP,which implies that either N1r,eP≠Sr,eP or N2r,eP≠Sr,eP. We distinguish the two cases as follows:Case 1:N1r,eP≠Sr,eP.We may assume thata0 is a common zero of eP−a/h and eP−a/w1. It is obvious that a0 is a zero of a/h−a/w1. If a/h−a/w1=0, then(14)Sr,eP≠N1r,eP≤Nr,1a/h−a/w1≤Tr,ah−aw1=Sr,eP,a contradiction. Hence,(15)h=w1.It deduces(16)2hz+chz+2c=ePz+2c−Pz+c.By Lemma1, for any ϵ>0,(17)mr,ePz+2c−Pz+c=mr,hz+chz+2c+O1=Orρh−1+ϵ+O1.We also getmr,ePz+2c−Pz+c=A+o1rρf−1, where A is a fixed positive constant.Ifρf>1, using ρf>ρh and the above estimates of mr,ePz+2c−Pz+c, It easily gets a contradiction. So, ρf≤1, this means that ePz+2c−Pz+c is a nonzero constant c0. Then, (16) changes to(18)2hz+chz+2c=c0.Also, noting that1≥ρf>ρh. Then, by Lemma 2, we get that there exists an ϵ-set E, as z∉E and z⟶∞ such that(19)hz+chz+2c⟶1.So,c0=2 and hz+c=hz+2c, and this also means that h is a periodic function. If h is a nonconstant function, then ρh≥1, a contradiction. Therefore, h is a constant. Noting that degP=ρf≤1 and f is a nonconstant entire function. Then, degP=1. Thus, we may set f=Aeμz, where Aandμ are two nonzero constants.Using the assumption of Case 1, one hasf−a and Δcf−a as common zeros, which are not zeros of a. Suppose that α0 is a common zero of f−a and Δc2f−a and not a zero of a. Then, z0 is a zero of Δc2f−f. Moreover,(20)fz0+2c−2fz0+c=0,this implies thateμc=2. Finally, we deduce Δcf=f, which is the desired result.Case 2:N2r,eP≠Sr,eP.Supposeb0 which is a common zero of eP−a/h and eP−b/w1. Then, it is obvious that b0 is a zero of a/h−b/w1. If a/h−b/w1≠0, then(21)Sr,eP≠N2r,eP≤Nr,1a/h−b/w1≤Tr,ah−bw1=Sr,eP,a contradiction. Hence,(22)ah−bw1=0.Ifb=0, then a/h=0, a contradiction. Thus, b=0.We may set thatc0 is a zero of eP−b/h, but not a zero of w1. It follows from (12) that c0 is a zero of eP−a/w1 or eP−b/w1. We take by N3r,eP the reduced counting function of those common zeros of eP−b/h and eP−a/w1. Similarly, we denote by N4r,eP the reduced counting function of those common zeros of eP−b/h and eP−b/w1. We obtain(23)Tr,eP=N¯r,1eP−b/h+Sr,eP=N3r,eP+N4r,eP+Sr,eP.
It implies that eitherN3r,eP≠Sr,eP or N4r,eP≠Sr,eP. If N4r,eP≠Sr,eP, likewise with Case 1, we deduce the desired result. Hence, we set that N3r,eP≠Sr,eP below. Similarly with Case 2, we also get that(24)bh−aw1=0.
Combining (22) with (24), we deduce that(25)a2=b2.
Note thata≠b. Thus, a=−b. Again using (24), we have w1=−h. We rewrite it as(26)hz+2cePz+2c−Pz−2hz+cePz+c−Pz+2hz=0.
Then,(27)hz+2cePz+2c−2hz+cePz+c+2hzePz=0.
By Lemma3, if pz+2c−pz, pz+c−pz, and pz+2c−pz+c are not constants, then hz=0; this is a contradiction.
So,pz+2c−pz=a1, pz+c−pz=b1, and pz+2c−pz+c=c1 (where a1,b1,andc1 are three constants).
We can getp′z+2c−p′z=0, p′z+c−p′z=0, and p′z+2c−p′z+c=0.
Hence,pz is a periodic function. We also know pz is a polynomial. So, we get pz=Az+B (where A and B are two constants, and A≠0). So, we obtain fz=HzeAz. Here, Hz is an entire function and λf=ρH<1.
Therefore, the proof of the main Theorem3 is finished.Proof of Theorem 2.
Note thatf and Δcnf share a,b CM. So, we also set(28)Δcnf−aΔcnf−bf−af−b=eQ,where Q is an entire function. Furthermore, it deduces from (28) and maxρa,ρb<ρf<∞ that Q is a polynomial.
Using Hadamard Factorization Theorem, we assume thatfz=hzePz, where h≠0 is an entire function and P is a polynomial satisfying(29)λf=ρh<ρf=degP.
Then,(30)Δcnf=fz+nc+bn−1fz+n−1c+⋯+b1fz+c+b0fz=hz+ncepz+nc+bn−1hz+n−1cepz+n−1c+⋯+b0hzepz,where b0,b1,…,bn−1 are constants. Substituting the forms of f and Δcnf into (28) yields(31)hz+ncepz+nc−pz+bn−1hz+n−1cepz+n−1c−pz+⋯+b0hzePz−az,hz+ncepz+nc−pz+bn−1hz+n−1cepz+n−1c−pz+⋯+b0hzePz−bz=hzePz−azhzePz−bzeQz.
Setw1=hz+ncepz+nc−pz+bn−1hz+n−1cepz+n−1c−pz+⋯+b0hz. Suppose that w1=0. Then,(32)hz+ncepz+nc+bn−1hz+n−1cepz+n−1c+⋯+b0hzepz0.
By Lemma3, if pz+nc−pz, pz+n−1c−pz,…,pz+c−pz are not constants, then hz=0; a contradiction.
So,pz+nc−pz=an, pz+n−1c−pz=an−1, and pz+c−pz=a1 (where an,an−1,…,a1 are three constants).
We can getp′z+nc−p′z=0, p′z+n−1c−p′z=0, and p′z+c−p′z=0.
Hence,pz is a periodic function. We also know pz is a polynomial. So, we get pz=Az+B (where A and B are two constants, and A≠0). So, we obtain fz=HzeAz. Here, Hz is an entire function and λf=ρH<1.
Ifw1≠0 below, obviously, w1 is a small function of eP. Rewrite (31) as(33)eQ=w12eP−a/w1eP−b/w1h2eP−a/heP−b/h.
Due toa≠b, without loss of generality, we set a≠0. Assume that z0 is a zero of eP−a/h, but not a zero of w1. It deduces from (33) that z0 is a zero of eP−a/w1 or eP−b/w1. We also take N1r,eP the reduced counting function of those common zeros of eP−a/h and eP−a/w1. Likewise, we denote by N2r,eP the reduced counting function of those common zeros of eP−a/h and eP−b/w1. Then,(34)Tr,eP=N¯r,1ePz−a/h+Sr,eP=N1r,eP+N2r,eP+Sr,eP,this implies that either N1r,eP≠Sr,eP or N2r,eP≠Sr,eP. We may distinguish the following two cases.Case 1:N1r,eP≠Sr,eP.We seta0 a common zero of eP−a/h and eP−a/w1. Then, it is obvious that a0 is a zero of a/h−a/w1. If a/h−a/w1≠0, then(35)Sr,eP≠N1r,eP≤Nr,1a/h−a/w1≤Tr,ah−aw1=Sr,eP,a contradiction. Hence,(36)h=w1.It leads to(37)hz+ncepz+nc+bn−1hz+n−1cepz+n−1c+⋯+b0−1hzepz=0.By Lemma3, if pz+nc−pz, pz+n−1c−pz,…pz+c−pz, are not constants, then hz=0, a contradiction.So,pz+nc−pz=an, pz+n−1c−pz=an−1,…,pz+c−pz=a1, (where an,an−1,a1 are three constants).We can getp′z+nc−p′z=0, p′z+n−1c−p′z=0,…, p′z+c−p′z=0.Here,pz is a periodic function. We also know pz is a polynomial. So, we get pz=Az+B (where A and B are two constants, and A≠0). Finally, we get fz=HzeAz, where Hz is an entire function and λf=ρH<1.Case 2:N2r,eP≠Sr,eP.Supposeb0 is a common zero of eP−a/h and eP−b/w1. Then, it is easy to see that b0 is a zero of a/h−b/w1. If a/h−b/w1≠0, then(38)Sr,eP≠N2r,eP≤Nr,1a/h−b/w1≤Tr,ah−bw1=Sr,eP,a contradiction. Thus,(39)ah−bw1=0.Ifb=0, then a/h=0, a contradiction. Thus, b≠0.We assume thatc0 is a zero of eP−b/h, but not a zero of w1. It deduces from (33) that c0 is a zero of eP−a/w1 or eP−b/w1. We take by N3r,eP the reduced counting function of those common zeros of eP−b/h and eP−a/w1. Similarly, we denote by N4r,eP the reduced counting function of those common zeros of eP−b/h and eP−b/w1. Then,(40)Tr,eP=N¯r,1eP−b/h+Sr,eP=N3r,eP+N4r,eP+Sr,eP,and this implies that either N3r,eP≠Sr,eP or N4r,eP≠Sr,eP. If N4r,eP≠Sr,eP. Similarly, as the same way in Case 1, we get the desired result. So, we assume that N3r,eP≠Sr,eP as follows. Similarly, as the way in Case 2, we can get that(41)bh−aw1=0.
It follows from (39) and (41) that(42)a2=b2.
Note thata≠b. Thus, a=−b. Again by (41), one has w1=−h. We also rewrite it as(43)hz+ncepz+nc+bn−1hz+n−1cepz+n−1c+⋯+b0+1hzepz=0.
By Lemma3, if pz+nc−pz, pz+n−1c−pz,…,pz+c−pz are not constants, then hz=0, a contradiction.
So,pz+nc−pz=an, pz+n−1c−pz=an−1,…,pz+c−pz=a1 (where an,an−1,…,a1 are three constants).
We also getp′z+nc−p′z=0, p′z+n−1c−p′z=0,…, p′z+c−p′z=0.
Hence,pz is a periodic function. We also know pz is a polynomial. So, we get pz=Az+B (where A and B are two constants, and A≠0). Finally, we get fz=HzeAz. Here, Hz is an entire function and λf=ρH<1.
Proof of Theorem 4 is completed.



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