# A Positive Answer for 3IM+1CM Problem with a General Difference Polynomial

Authors: Huicai Xu; Shugui Kang; Qingcai Zhang
Journal: Discrete Dynamics in Nature and Society
 (2021)

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## Abstract

In this paper, the 3IM+1CM theorem with a general difference polynomialLz,f will be established by using new methods and technologies. Note that the obtained result is valid when the sum of the coefficient of Lz,f is equal to zero or not. Thus, the theorem with the condition that the sum of the coefficient of Lz,f is equal to zero is also a good extension for recent results. However, it is new for the case that the sum of the coefficient of Lz,f is not equal to zero. In fact, the main difficulty of proof is also from this case, which causes the traditional theorem invalid. On the other hand, it is more interesting that the nonconstant finite-order meromorphic function f can be exactly expressed for the case f≡−Lz,f. Furthermore, the sharpness of our conditions and the existence of the main result are illustrated by examples. In particular, the main result is also valid for the discrete analytic functions.


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## Body

## 1. Introduction and Results

It is well known that any polynomial is uniquely determined by its zero points (the set on which the polynomial takes zeros) except for a nonconstant factor, but it is not true for transcendental entire or meromorphic functions. For example, functionsez and e−z have the same ±1,0 and ∞ points. Thus, it is interesting and complex to uniquely determine a meromorphic function. It is well known that the classical uniqueness results of the value distribution theory of meromorphic functions are the 5IM theorem (or five-point theorem) and the 4CM theorem (or four-point theorem) which had been obtained in the study by Nevanlinna [1], where IM means ignoring multiplicities and CM is counting multiplicities.Letf and g be two nonconstant meromorphic functions on the complex plane ℂ and a be a finite or infinite complex number in the whole complex plane ℂ^=ℂ∪∞. If f−a and g−a have the same zeros with the same multiplicities, we say that f and g share a CM (counting multiplicities). If not considering the multiplicities, we say that f and g share a IM (ignoring multiplicities). Clearly, they also share a IM if f and g share a CM. In this case, the famous five-point theorem and four-points theorem mean that if two nonconstant meromorphic functions f and g share five distinct values IM, then f=g, and if two nonconstant meromorphic functions f and g share four distinct values CM, then f is a factional linear transformation of g, respectively.The 4CM theorem in the study by Nevanlinna [1] can also be understood to be the 4CM+0 IM theorem which motivates us to consider the other cases as the 3CM+1IM theorem, the 2CM+2IM theorem, the 3IM+1CM theorem, or the 0CM+4IM theorem. Obviously, it is very good if the 0CM+4IM theorem is a fact. Unfortunately, Gundersen [2] has constructed counterexamples which show that the conclusion of the four-point theorem is invalid if CM is replaced by IM; that is, the 0CM+4IM theorem is false. However, the 2CM+2IM theorem is valid; please see the study by Gundersen [3]. It is known that the 2CM+2IM theorem implies that the 3CM+1IM theorem holds (see Gundersen [2, 3]). Thus, in [4], Gundersen proposed the following well-known question.Problem 1.
(3IM+1CM open problem [4]). If two nonconstant meromorphic functions share three values IM and share a fourth value CM, then do the functions necessarily share all four values CM?It is very difficult to completely solve Problem1 (see Gundersen [4]). In the past years, some researchers established 3IM+1CM theorems with some conditions, for example, the restriction of zero points for f−g (see Mues [5–7], Reinders [8, 9], Wang [10], Ueda [11, 12], Yi and Zhou [13], Czubiak and Gundersen [14], Qiu [15], Wang [16, 17], Huang [18, 19], Huang and Du [20], Ishizaki [21], Li [22], Yao [23], and Yang and Yi [24]), the periodicity hypothesis (see Lin and Ishizaki [25] and Banerjee and Ahamed [26]), or the differential limitation (see Lahiri and Sinha [27], Gundersen [28, 29], Mues and Steinmetz [30, 31], Mues and Reinders [32], and Frank and Hua [33]).In recent years, the difference variant of the Nevanlinna theory has been established in [34–37]. Using these theories, some mathematicians began to consider the uniqueness of meromorphic functions sharing values with their shifts or difference operators and produced many fine works; for example, see Banerjee and Bhattacharyya [38], Ahamed [39, 40], Ma et al. [41], Jiang et al. [42], Charak et al. [43], Lin et al. [44, 45], and Li et al. [46, 47].In this paper, we will establish the 3IM+1CM theorem with the difference condition. Actually, Li et al. [46] obtained the following uniqueness result when f and Δηnf fulfill the condition 3IM+1CM, where(1)Δηfz=fz+η−fz,Δηnfz=ΔηΔηn−1fz,for n=2,3,….Theorem 1 (see [46]).
Letf be a nonconstant meromorphic function with finite order, and let η be a nonzero complex number. Suppose that f and Δηnf share a1,a2,a3 IM and share ∞ CM, where a1,a2,anda3 are three distinct finite values. Then, fz=Δηnf for all z∈ℂ.Noting that(2)Δηnfz=∑i=0nnj−1n−jfz+jη,naturally, a more general case(3)Lz,f≜∑j=0nαjfz+dj,should be considered, where n≥1 is a positive integer, djj=0,…,n are distinct finite values, and αjj=0,…,n are nonzero constants. In the case, Problem 1 turns to be the following problem.Problem 2.
When a nonconstant meromorphic functionf and its general linear difference polynomial Lz,f satisfy the condition 3IM+1CM, does the uniqueness result still hold?In the following, we will give our main result, some corollaries, and remarks.Theorem 2.
Letf be a nonconstant meromorphic function with finite order; let Lz,f be defined as (3). If f and Lz,f share a1,a2,a3 IM and share ∞ CM, where a1,a2,anda3 are three distinct finite values, then for all z∈ℂ, one of the following results holds:(i)
f=Lz,f;(ii)
f=−Lz,f; furthermore,(4)fz=c1+eaz+b1−eaz+b,and ∑j=0nαj=−1, where c∈a1,a2,a3, and a≠0,b are constants.To explain a nonconstant meromorphic functionf with finite order, we need the following definition.Definition 1 (see [24, 48]).
For a nonconstant meromorphic functionf, the order of f, denoted by ρf, is defined as(5)ρf=limsupr⟶∞logTr,flogr,where(6)Tr,f=mr,f+Nr,f,mr,f=∫02πlog+freiθdθ,Nr,f=∫0rnt,f−n0,ftdt+n0,flogr,and nt,f denotes the number of poles of f (counting multiplicities) in z<t.A nonconstant meromorphic functionf with finite order means that ρf<∞.Remark 1.
The following example shows that Theorem2 (ii) can happen. Let(7)fz=a11+eaz+b1−eaz+b,Lz,fz=∑j=0nαja11+eaz+jη+b1−eaz−jη+b=a1∑j=0nαj1+eaz+b1−eaz+b=−a11+eaz+b1−eaz+b,where a1≠0, eaη=1, and αjj=0,1…,n are nonzero constants satisfying ∑j=0nαj=−1, and then f and Lz,f share a1,−a1,0,∞ CM, and f=−Lz,f.Remark 2.
The conditions of Theorem2 are sharp. From the example in the following, we know that the assumption 3IM+1CM cannot be relaxed to 4IM in Theorem 2. For example, see Steinmetz [49]. Let ℘ denote the Weierstrass ℘-function with a pair of primitive periods 2ω−ω′ and ω+ω′, and set(8)fz=℘z−℘ω/2℘z−℘3ω/22℘z−℘ω,Lz,f=fz+ω.
Then, the functionsf and Lz,f share the values 0,∞ and some values A and B IM, where A/B=1/21±3i. We note that it is showed that three meromorphic functions fz,Lz,f, and fz+ω′ share the four values 0,∞,A, and B IM in [49]. However, f≠Lz,f or f≠−Lz,f.Corollary 1.
Letf be a nonconstant meromorphic function with finite order; let Lz,f be defined as (3). Assume that f and Lz,f share a1,a2,a3 IM and share ∞ CM, where a1,a2,a3 are three distinct finite values. If ∑j=0nαj≠−1, then f=Lz,f for all z∈ℂ.Corollary 2.
Letf be a nonconstant meromorphic function with finite order; let Lz,f be defined as (3). Assume that f and Lz,f share a1,a2,a3 IM and share ∞ CM, where a1,a2,a3 are three distinct finite values. If ∑j=0nαj=0, then f=Lz,f for all z∈ℂ.Remark 3.
Note that(9)Δηnfz=∑i=0nnj−1n−jfz+jη,and the sum of its coefficients ∑i=0nnj−1n−j is zero. So, even if ∑j=0nαj=0 in Lz,f, Δηnfz is still a special case of Lz,f. Thus, Theorem 2 is a large extension of Theorem 1.2 in Li et al. [46]; naturally, the corresponding results in the study by Heittokangas et al. [50], Li et al. [46, 47], and so on are also the special cases of Theorem 2. However, our result is new when ∑j=0nαj≠0.Remark 4.
Choosing some specialαj and dj for j=0,1,…,n, we can obtain some difference equations of the form fz=∑j=0nαjfz+dj. Thus, the main result is also valid for the discrete analytic functions (see [51–55]). For example, let(10)fz+1+i−fzi+1=fz+i−fz+1i−1,(11)orfz=ifz+i−ifz−1+fz+1+i,which is the special case of (3). Thus, Theorem 2 is also valid for (10) or (11). When z=m+in and m,n∈ℤ, equation (10) is the definition of discrete analytic and harmonic functions (see Hundhausen [56]). At this time, we can obtain the uniqueness of (10) or (11). Another definition of discrete analytic and harmonic functions:(12)fz+1−fz=fz+i−fzi,(13)orfz=1−i2fz+1+1+i2fz+i,can be seen in Harman [57]. Clearly, our theorem is also valid for (12) or (13).We note that the main tool of proof for Theorem 1.2 in Li et al. [46] is the traditional Theorem 1.62 in [24]; however, it is invalid for Theorem 2 because ∑j=0nαj can be nonzero. We specially thank Corollary 1.105 in [58] which will be our key tool and listed in Section 2 as Lemma 8. On the other hand, our methods and technologies are also different with Li et al. [46]; for example, the proof is divided into the six cases in Li et al. [46], but it is only two cases in the proof of Theorem 2. For the other detail cases, please see the proof in Section 3. However, it is worth mentioning that the meromorphic function f can be exactly expressed by(14)fz=c1+eaz+b1−eaz+b,in some special cases. Certainly, some preliminaries of the difference value distribution theory [34–37] are also used in the proof of Theorem 2.In the final, the present paper will be organized as follows. In Section2, we will give some preliminaries which can be seen in the listed references. And the main result will be proved in the final section.


## 2. Some Lemmas

In this paper, the value distribution theory established by R. Nevanlinna is the main tool for the studies. For convenience of the reader who might not be familiar with Nevanlinna theory, we list here some results from Nevanlinna theory (see, e.g., [24, 48]).The following known results are important in the value distribution theory (see, e.g., [24, 48]).(i)
The arithmetic properties ofTr,f and mr,f are as follows:(15)Tr,fg≤Tr,f+Tr,g,Tr,f+g≤Tr,f+Tr,g+O1.The same inequalities hold formr,f.(ii)
The Nevanlinna first fundamental theorem is as follows:Tr,f=Tr,1/f+O1.(iii)
The logarithmic derivative lemma is as follows:mr,f′/f=Ologr, if the order ρf is finite.In the following, we present some lemmas, which will be needed in the sequel.Lemma 1 (see Theorem 3 in [59]).
Letf and g be two nonconstant rational functions. If f and g share a1,a2,a3,a4 IM, where a1,a2,a3,a4 are four distinct values in the extended complex plane, then f=g.Lemma 2 (see Lemma 1 in [4]).
Letf and g be two distinct nonconstant meromorphic functions, and let a1,a2,a3,a4 be four distinct values in the extended complex. If f and g share a1,a2,a3,a4 IM, then for a set with finite linear measure E,(i)
Tr,f=Tr,g+OlogrTr,f, as r∉E and r⟶∞(ii)
2Tr,f=∑j=14Nr,1/f−aj+OlogrTr,f, as as r∉E and r⟶∞, where Nr,1/f−∞=Nr,fRemark 5.
Under the assumptions of Lemma2, if f and g in Lemma 2 are of finite order, by the context of paper [26] and the proof of Lemma 1 in [26], we can find that the conclusion of Lemma 2 can be changed to(i)
Tr,f=Tr,g+Ologr,asr⟶∞(ii)
2Tr,f=∑j=14Nr,1/f−ajOlogr,asr⟶∞Lemma 3 (see Lemma 2 and Corollary 1 in [3]).
Letf and g be distinct nonconstant meromorphic functions that share four values a1,a2,a3, and a4 IM, where a4=∞. Then, for a set with finite linear measure E, the following statements hold:(i)
N0r,1/f′=OlogrTr,f and N0r,1/g′=OlogrTr,f as r∉E, r⟶∞, and E is a set with finite linear measure, where N0r,1/f′ and N0r,1/g′count, respectively, only those points in Nr,1/f′ and Nr,1/f′ which do not occur when fz=gz=aj for some j=1,2,3,4.(ii)
Forj=1,2,3,4, we next let N2r,aj refer only to those aj-points that are multiple for both f and g and count each such point the number of times of the smaller of the two multiplicities. Then, ∑j=14N2r,aj=OlogrTr,f as r∉E and r⟶∞.Remark 6.
Under the assumptions of Lemma3, if f and g in Lemma 3 are of finite order, by the context of paper [38] and the proof of Lemma 2 and Corollary 1 in [38], we can find that the conclusion of Lemma 3 can be changed into(i)
N0r,1/f′=Ologr and N0r,1/g′=Ologr,asr⟶∞(ii)
∑j=14N2r,aj=Ologr,asr⟶∞Lemma 4 (see Corollary 2.5 in [34]).
Letf be a nonconstant meromorphic function of finite order and η be a fixed nonzero complex number. Then, for each ε>0,(16)mr,fz+ηfz+mr,fzfz+η=Orρ−1+ε,asr⟶∞.Lemma 5 (see Theorem 1 in [4]).
Letf and g be two nonconstant meromorphic functions that share a1,a2,a3 IM and a4 CM, where a1,a2,a3,anda4 are four distinct values in the extended complex plane. Suppose that there exists some real constant μ>4/5 and some set I⊂0,+∞ that has infinite linear measure such that(17)Nr,1/f−a4Tr,f≥μ,for all r∈I, where Nr,1/f−a4=Nr,f, if a4=∞. Then, f and g share all four values CM.Lemma 6 (see Theorem 4.3 in [24]).
Letf and g be two distinct nonconstant meromorphic functions and ajj=1,2,3,4 be four distinct values in the extend complex plane. If f and g share ajj=1,2,3,4 CM, then fz=Tgz, where T is a Mobius transformation such that two of the four values are fixed points and another two (are Picard exceptional values of f and g) exchange each other under T.Lemma 7 (see Theorem 1.62 in [24]).
Letf1,f2,…,fn be nonconstant meromorphic functions, and let fn+1≢0 be a meromorphic function such that ∑j=1n+1fj=1. Suppose that there exists a subset I⊆R+ whose linear measure is ∞ such that(18)∑i=1n+1Nr,1fi+n∑i=1i≠jn+1N¯r,fi<λ+o1Tr,fj,j=1,2,…,n,
Asr∈I and r⟶∞, where 0≤λ<1, then fn+1≡1.Lemma 8 (see Corollary 1.105(iii) in [58]).
Assume that entire functionsf0,f1,…,fn vanish nowhere on ℂm such that(19)f0+f1+⋯+fn=0.
Partition the index setI=0,1,…,n into subsets Iα, I=∪α=0kIα, putting two indices i and j in the same subset Iα if and only if fi/fj is a constant. Then, we have(20)∑i∈Iαfi=0,α=0,1,…,k.


## 3. Proof of Theorem 2

IfLz,f is a constant, by the assumption that f and Lz,f share a1,a2,a3 IM and share ∞ CM, we know that f has more than two Picard values. And it is impossible, so Lz,f is nonconstant. Assume that f is a nonconstant rational function; according to the definition of Lz,f, we can find that Lz,f is also a nonconstant rational function, and then using Lemma 1, we deduce that Lz,f=f. Next, we suppose that f is a nonconstant transcendental meromorphic function. Since ρf<∞ and f and Lz,f share a1,a2,a3 IM and share ∞ CM, applying Remark 5, we obtain(21)Tr,f=Tr,Lz,f+Sr,f.Then, we can see thatLz,f is also a transcendental meromorphic function. Next, we assume Lz,f≢f.Under the assumption thatf and Lz,f share ∞ CM, Remark 5 (ii) leads to(22)N2r,f+N2r,Lz,f=2N2r,∞=Sr,f,where N2r,f denotes the counting function of the multiple poles of f in z<r, and each point in N2r,f is counted according to its multiplicity as a pole of f. N2r,Lz,f is similar with N2r,f. Hence,(23)Nr,Lz,f−f≤N¯1,1r,∞+Sr,f,where N¯1,1r,∞ is the counting function of the common simple poles of f and g. By the definition of Lz,f and Lemma 4, we deduce(24)mr,Lz,f−f=mr,Lz,f−fff≤mr,f+mr,Lz,ff+O1≤mr,f+Sr,f.From (23), (24), Remark 5 (ii), and the assumption that f and Lz,f share a1,a2,a3 IM and ∞ CM, we deduce(25)2Tr,f=N¯r,1f−a1+N¯r,1f−a2+N¯r,1f−a3+N¯r,f+Sr,f≤N¯r,1Lz,f−f+N¯1,1r,∞+Sr,f≤Tr,1Lz,f−f+N¯1,1r,∞+Sr,f=mr,Lz,f−f+Nr,Lz,f−f+N¯1,1r,∞+Sr,f≤mr,Lz,f−f+2N¯1,1r,∞+Sr,f≤Tr,f+N¯1,1r,∞+Sr,f.Hence, we get(26)limz⟶∞z∈INr,fTr,f=1,where I⊂0,+∞ with logarithmic measure logmesI=+∞. In view of Lemma 5, we deduce that f and Lz,f share a1,a2,a3 and ∞ CM. Then, by Lemma 6, we obtain that f is a Möbius transformation of Lz,f. We set(27)f=ALz,f+BCLz,f+D,where A,B,C, and D are constants and AD−BC≠0. We consider the following two cases.Case 1.
Suppose thatC=0. Then, from (27), we get(28)f=A1Lz,f+B1,where A1≠0 and B1 are two constants, and A1=A/D, and B1=B/D. If two of a1,a2,a3 are not the Picard exceptional value of f, we get f≡Lz,f, which is a contradiction, so two of a1,a2,a3 are Picard exceptional values of f. Without loss of generality, we set a1 and a2 as Picard exceptional values of f, and by the assumption of Theorem 2, we get a1 and a2 as Picard exceptional values of Lz,f. By rewriting (28), we obtain(29)f−a1=A1Lz,f−a1−B1A1,and noting f≠a1, we deduce a1−B1/A1=a1 or a2. Similarly, we get(30)f−a2=A1Lz,f−a2−B1A1,and a2−B1/A1=a1 or a2.
a1−B1/A1=a1 and a2−B1/A1=a1 imply that a1=a2, which is impossible. By a1−B1/A1=a1 and a2−B1/A1=a2, we obtain A1=1 and B1=0, that is, f≡Lz,f, which is a contradiction to the assumption. From a1−B1/A1=a2 and a2−B1/A1=a2, we have a1=a2 which is also impossible. Combining a1−B1/A1=a2 and a2−B1/A1=a1, we get A1=−1 and B1=a1+a2, and then(31)f+Lz,f=a1+a2.
Sincefz≠a1,a2, we set(32)f−a1f−a2=eγz,where γz=amzm+am−1zm−1+⋯+a1z+a0, a0,a1,…,am≠0, are constants, and 0<m≤ρf<∞. By a calculation, for j=0,1,2,…,n, we deduce that(33)γz+dj=γz+amCm1djzm−1+amCm1dj2+am−1Cm−11djzm−2+⋯.
Setam=ameiθm; then, when z=re−iθm/m⟶∞, we obtain(34)Reγz∼amrm,eγz⟶∞,and while z=re−iθm/m+π/m⟶∞, we know(35)Reγz∼−amrm,eγz⟶0.
Equation (32) implies(36)f=a1−a2eγz1−eγz.
So,(37)Lz,f=∑j=0nαja1−a2eγz+dj1−eγz+dj.
Substituting (36) and (37) to (31), we get(38)a1−a2eγz1−eγz+∑j=0nαja1−a2eγz+dj1−eγz+dj=a1+a2.
Whenz=re−iθm/m⟶∞, from (33), (34), and (38), we know(39)a2+a2∑j=0nαj=a1+a2,and when z=re−iθm/m+π/m⟶∞, we get(40)a1+a1∑j=0nαj=a1+a2.
Combining (39) and (40), we deduce that(41)∑j=0nαj+1=0,a2=−a1,B1=0,a3=0.
Then, by (31), we obtain(42)f+Lz,f=0.
Substituting (41) to (36) and (37), we have(43)f=a11+eγz1−eγz,(44)Lz,f=a1∑j=0nαj1+eγz+dj1−eγz+dj,(45)∑j=0nαj=−1.
Substituting (43) and (44) to (42), we deduce(46)1+eγz1−eγz+∑j=0nαj1+eγz+dj1−eγz+dj=0.
Denotingβn+1=1, dn+1=0, and βj=αj, where j=0,1,…,n, then(47)∑j=0n+1βj1+eγz+dj1−eγz+dj=0,where ∑j=0n+1βj=0. If there exist some j, 0≤j≤n, such that dj=0, we denote βn+1=1+dj and βj=0; then, we get a similar equation with the above one.
So,(48)∑j=0n+1βj1+eγz+dj∏i≠j1−eγz+dj=0.
Then, by a calculation, we get(49)∑j=0n+1βjeγz+dj+∑k=2n−1k−1∑i1,i2,…,ikβi1+⋯+βikeγz+di1+⋯+γz+dik=0,where Σi1,i2,…,ik denotes the sum for any k numbers in 0,1,…,n+1. Applying Lemma 8, we get that there exits nonnegative integer i,j and i≠j such that γz+di−γz+dj is a constant. From (33), we know that γz is a polynomial of degγ=1. We set(50)γz=az+b,where a≠0,b are constants. We substitute into (43) and get(51)fz=a11+eaz+b1−eaz+b,where a≠0,b are constants.Case 2.
Suppose thatC≠0. By (27), we get(52)f=A1Lz,f+B1Lz,f+D1,where A1=A/C, B1=B/C, D1=D/C, and A1D1−B1≠0.
Sincef and Lz,f share ∞ CM, from (52), we know ∞ is a Picard exceptional value of f and Lz,f. If f can take every one of a1,a2,a3, we get at least two of a1,a2,a3 are equal, which is a contradiction, so one of a1,a2,a3 is a Picard exceptional value of f and Lz,f. Without loss of generality, we set a3 is a Picard exceptional value of f; by assumption, a3 is also a Picard exceptional value of Lz,f. Noting that ∞ is Picard exceptional values of Lz,f and they share ∞ CM, by (52), we have A1 is a Picard exceptional value of f, so A1=a3. Since ∞ is Picard exceptional values of f and they share ∞ CM; by (52), we see that −D1 is a Picard exceptional value of Lz,f. Then, we have −D1=a3. Thus, (52) turns to be(53)f=a3Lz,f+B1Lz,f−a3,where a3 and B1 are not all zero.
Noting thatf and Lz,f share a1 and a2 CM and neither of them is a Picard exceptional value of f and Lz,f, from (53), we get(54)a1=a3a1+B1a1−a3,a2=a3a2+B1a2−a3.
So,a3=a1+a2/2 and B1=−a1a2. Then, by (53), we have(55)f=a1+a2/2Lz,f−a1a2Lz,f−a1+a2/2.
By rewriting it, we obtain(56)fLz,f−a1+a22f−a1+a22Lz,f=−a1a2.
Noting that both∞ and a3 are Picard exceptional values of f and Lz,f and a3=a1+a2/2, we set(57)f=a1+a22+eβz,where βz=bpzp+bp−1zp−1+⋯+b1z+b0, b0,b1,…,bp≠0, are constants, and p=ρf>0. Then, by the definition of Lz,f, we get(58)Lz,f=∑j=0nαjfz+dj=∑j=0nαja1+a22+eβz+dj=α0a1+a22+eβz+d0+α1a1+a22+eβz+d1+⋯+αna1+a22+eβz+dn.
Substituting (57) and (58) to the above equation, we have(59)α0+α1+⋯+αn−1a1+a22eβz+α0eβz+d0+βz+α1eβz+d1+βz+⋯+αneβz+dn+βz=a1−a224.
Then, using Lemma7, we get a contradiction. Thus, we complete the proof of Theorem 2.



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