Uniqueness Theorems on Difference Monomials of Entire Functions

Abstract

The aim of this paper is to discuss the uniqueness of the difference monomials 𝑓^𝑛𝑓(𝑧+𝑐). It assumed that 𝑓 and 𝑔 are transcendental entire functions with finite order and 𝐸_𝑘)(1,𝑓^𝑛𝑓(𝑧+𝑐))=𝐸_𝑘)(1,𝑔^𝑛𝑔(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 𝑡₂^𝑛+1=1 and 𝑡₃^𝑛+1=1. It is an improvement of the result of Qi, Yang and Liu.

1. Introduction and Main Results

In this paper, a meromorphic (respectively entire) function always means meromorphic (respectively, analytic) in the complex plane ℂ. 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].

Let 𝑓 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 𝜎𝑓=limsup_𝑟→∞log𝑇𝑟,𝑓.log𝑟 (1.1)

Let 𝑓 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 𝐸_𝑘)(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 𝑁_𝑘)(𝑟,1/(𝑓−𝑎)) the counting function corresponding to the set 𝐸_𝑘)(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)⧵𝐸_𝑘)(𝑎,𝑓). If 𝐸_𝑘(𝑎,𝑓)=𝐸_𝑘(𝑎,𝑔), we say that 𝑓, 𝑔 share the value a with weight 𝑘.

The definition implies that if 𝑓 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 by 𝑁_𝑘)(𝑟,1/(𝑓−𝑎)) and 𝑁_(𝑘+1(𝑟,1/(𝑓−𝑎)) the reduced forms of 𝑁_𝑘)(𝑟,1/(𝑓−𝑎)) and 𝑁_(𝑘+1(𝑟,1/(𝑓−𝑎)), respectively. At last, we set 𝑁_𝑘𝑟,𝑓=𝑁𝑟,𝑓+𝑁₍₂𝑟,𝑓+⋯+𝑁_(𝑘𝑟,𝑓. (1.2)

Hayman proposed the well-known conjecture in [3].

Hayman  Conjecture
If an entire function 𝑓 satisfies 𝑓^𝑛𝑓′≠1 for all 𝑛∈ℕ, then 𝑓 is a constant.
In fact, Hayman has proved that the conjecture holds in the cases 𝑛≥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. Let 𝑓 and 𝑔 be nonconstant entire function, and let 𝑛≥6 be an integer. If 𝑓^𝑛𝑓′ and 𝑔^𝑛𝑔′ share 1 CM, then either 𝑓(𝑧)=𝑐₁𝑒^𝑐𝑧, 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 𝑡^𝑛+1=1.

In 2010, Qi et al. [7] studied the uniqueness of the difference monomials and obtained the following result.

Theorem B. Let 𝑓 and 𝑔 be transcendental entire functions with finite order, 𝑐 a non-zero complex constant, and 𝑛≥6 an integer. If 𝐸(1,𝑓^𝑛𝑓(𝑧+𝑐))=𝐸(1,𝑔^𝑛𝑔(𝑧+𝑐)), then 𝑓𝑔=𝑡₁ or 𝑓=𝑡₂𝑔 for some constants 𝑡₁ and 𝑡₂ which satisfy 𝑡₁^𝑛+1=1 and 𝑡₂^𝑛+1=1.

In this paper, we will obtain the following results.

Theorem 1.1. Let 𝑓 and 𝑔 be transcendental entire functions with finite order, 𝑐 a non-zero complex constant, and 𝑛≥6 an integer. If 𝐸₃₎(1,𝑓^𝑛𝑓(𝑧+𝑐))=𝐸₃₎(1,𝑔^𝑛𝑔(𝑧+𝑐)), then the assertion of Theorem B holds.

Theorem 1.2. Let 𝑓 and 𝑔 be transcendental entire functions with finite order, 𝑐 a non-zero complex constant, and 𝑛≥7 an integer. If 𝐸₂₎(1,𝑓^𝑛𝑓(𝑧+𝑐))=𝐸₂₎(1,𝑔^𝑛𝑔(𝑧+𝑐)), then the assertion of Theorem B holds.

Theorem 1.3. Let 𝑓 and 𝑔 be transcendental entire functions with finite order, 𝑐 a non-zero complex constant, and 𝑛≥10 an integer. If 𝐸₁₎(1,𝑓^𝑛𝑓(𝑧+𝑐))=𝐸₁₎(1,𝑔^𝑛𝑔(𝑧+𝑐)), then the assertion of Theorem B holds.

2. Auxiliary Results

Lemma 2.1 (see [8, Corollary 2.5]). Let 𝑓(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 𝑚𝑓𝑟,𝑧+𝜂𝑓(𝑧)𝑓(𝑧)+𝑚𝑟,𝑓𝑟𝑧+𝜂=𝑂^{𝜎−1+𝜀}. (2.1)

Lemma 2.2 (see [8, Theorem 2.1]). Let 𝑓(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 𝑇𝑟,f𝑧+𝜂=𝑇𝑟,𝑓(𝑧)+𝑂𝑟^{𝜎−1+𝜀}+𝑂log𝑟. (2.2)

Lemma 2.3. Let 𝑓(𝑧) be an entire function with finite order 𝜎=𝜎(𝑓), 𝑐 a fixed non-zero complex number, and 𝑃(𝑧)=𝑎_𝑛𝑓(𝑧)^𝑛+𝑎_𝑛−1𝑓(𝑧)^𝑛−1+⋯+𝑎₁𝑓(𝑧)+𝑎₀, (2.3) where 𝑎_𝑗(𝑗=0,1,…,𝑛) are constants. If 𝐹(𝑧)=𝑃(𝑧)𝑓(𝑧+𝑐), then 𝑇(𝑟,𝐹)=(𝑛+1)𝑇𝑟,𝑓+𝑂𝑟^{𝜎(𝑓)−1+𝜀}+𝑂log𝑟. (2.4)

Proof. Since 𝑓(𝑧) is an entire transcendental function with finite order, we can deduce from Lemma 2.1 and the standard Valiron-Mohon'ko theorem that ()𝑇(𝑧)(𝑧)𝑃(𝑧)(1)(𝑧)𝑃(𝑧)(1)𝑓(𝑧)𝑃(𝑧)𝑛+1𝑟,𝑓=𝑇𝑟,𝑓+𝑂=𝑚𝑟,𝑓+𝑂≤𝑚𝑟,𝐹(𝑧)((𝑧(1)𝑓(𝑧)+𝑚𝑟,𝐹))+𝑂=𝑚𝑟,𝑓()((𝑧(1)((𝑧𝑟𝑧+𝑐+𝑚𝑟,𝐹))+𝑂≤𝑇𝑟,𝐹))+𝑂^{𝜎−1+𝜀}+𝑂(1). (2.5) Therefore 𝑇(𝑟,𝐹)≥(𝑛+1)𝑇𝑟,𝑓+𝑂𝑟^{𝜎−1+𝜀}+𝑂(1). (2.6) On the other hand, Lemma 2.2 implies that 𝑇((𝑧((𝑧()(𝑧)(𝑧)𝑟𝑟,𝐹))≤𝑇𝑟,𝑃))+𝑇𝑟,𝑓𝑧+𝑐=𝑛𝑇𝑟,𝑓+𝑇𝑟,𝑓+𝑂^{𝜎−1+𝜀}=()𝑇𝑟+𝑂log𝑟𝑛+1𝑟,𝑓+𝑂^{𝜎−1+𝜀}+𝑂log𝑟. (2.7) We will obtain the conclusion of Lemma 2.3.

Remark 2.4. The condition “entire” cannot be replaced by “meromorphic” in Lemma 2.3, as is shown by the following example.

Example 2.5. Let 𝑓(𝑧)=(𝑒^𝑧−1)/(𝑒^𝑧+1), 𝑐=𝜋𝑖, and 𝐹(𝑧)=𝑓(𝑧)𝑓(𝑧+𝑐), we can see 𝑇(𝑟,𝐹)≠2𝑇𝑟,𝑓+𝑂𝑟^{𝜎(𝑓)−1+𝜀}+𝑂log𝑟 (2.8) for every set of {𝑟_𝑛} with infinite measure.

Lemma 2.6 (see [9, Lemma 2.1]). Let 𝑓 and 𝑔 be two nonconstant meromorphic functions satisfying 𝐸_𝑘)(1,𝑓)=𝐸_𝑘)(1,𝑔) for some positive integer 𝑘∈ℕ. Define 𝐻 as follows: 𝑓𝐻=^′′𝑓^′−2𝑓^′−𝑔𝑓−1𝑔^′′′−2𝑔^′.𝑔−1 (2.9) If 𝐻≢0, then 𝑁(𝑟,𝐻)≤𝑁₍₂𝑟,𝑓+𝑁₍₂1𝑟,𝑓+𝑁₍₂𝑟,𝑔+N₍₂1𝑟,𝑔+𝑁₀1𝑟,+𝑓′𝑁₀1𝑟,𝑔^′+𝑁_(𝑘+11𝑟,+𝑓−1𝑁_(𝑘+11𝑟,,𝑔−1+𝑆𝑟,𝑓+𝑆𝑟,𝑔 (2.10) where 𝑁₀(𝑟,1/𝑓′) denotes the counting function of zeros of 𝑓′ but not zeros of 𝑓(𝑓−1) and 𝑁₀(𝑟,1/𝑔′) is similarly defined.

Lemma 2.7 (see [10]). Under the condition of Lemma 2.6, one has 𝑁₁₎1𝑟,𝑓−1=𝑁₁₎1𝑟,().𝑔−1≤𝑁𝑟,𝐻+𝑆𝑟,𝑓+𝑆𝑟,𝑔 (2.11)

Lemma 2.8 (see [10]). Let 𝐻 be defined as Lemma 2.6. If 𝐻≡0, then either 𝑓≡𝑔 or 𝑓𝑔≡1 provided that limsup_{𝑟→∞,𝑟∈𝐼}𝑁𝑟,𝑓+𝑁𝑟,𝑔+𝑁𝑟,1/𝑓+𝑁𝑟,1/𝑓𝑇(𝑟)<1, (2.12) 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]). Let 𝑇∶(0,+∞)→(0,+∞) be a nondecreasing continuous function, 𝑠>0, 0<𝛼<1, and let 𝐹⊂𝑅₊ be the set of all 𝑟 such that 𝑇(𝑟)≤𝛼𝑇(𝑟+𝑠). (2.13) If the logarithmic measure of 𝐹 is infinite, then limsup_𝑟→∞log𝑇𝑟,𝑓log𝑟=∞. (2.14)

3. Proof of Theorem 1.1

We define 𝐹∶=𝑓^𝑛𝑓(𝑧+𝑐),𝐺∶=𝑔^𝑛𝑔(𝑧+𝑐). (3.1) 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, 𝑁₁₎1𝑟,𝐹−1=𝑁₁₎1𝑟,()≤𝐺−1≤𝑁𝑟,𝐻+𝑆𝑟,𝑓+𝑆𝑟,𝑔𝑁₍₂1𝑟,𝐹+𝑁₍₂1𝑟,𝐺+𝑁₀1𝑟,+𝐹′𝑁₀1𝑟,+𝐺′𝑁₍₄1𝑟,+𝐹−1𝑁₍₄1𝑟,.𝐺−1+𝑆𝑟,𝑓+𝑆𝑟,𝑔 (3.2) Applying the second main theorem to 𝐹 and 𝐺 jointly implies that 𝑇(𝑟,𝐹)+𝑇(𝑟,𝐺)≤𝑁1𝑟,𝐹+𝑁1𝑟,+𝐹−1𝑁1𝑟,𝐺+𝑁1𝑟,−𝐺−1𝑁₀1𝑟,𝐹^′−𝑁₀1𝑟,𝐺^′+𝑆𝑟,𝑓+𝑆𝑟,𝑔. (3.3) Noting that 𝑁1𝑟,−1𝐹−12𝑁₁₎1𝑟,+𝐹−1𝑁₍₄1𝑟,≤1𝐹−12𝑁1𝑟,≤1𝐹−12𝑇(𝑟,𝐹),𝑁1𝑟,−1𝐺−12𝑁₁₎1𝑟,+𝐺−1𝑁₍₄1𝑟,≤1𝐺−12𝑁1𝑟,≤1𝐺−12𝑇(𝑟,𝐺). (3.4) According to Lemma 2.9 and (3.2)–(3.4), we can obtain that 𝑇(𝑟,𝐹)+𝑇(𝑟,𝐺)≤2𝑁₂1𝑟,𝐹+2𝑁₂1𝑟,𝐺+𝑆𝑟,𝑓+𝑆𝑟,𝑔≤4𝑁1𝑟,𝑓+4𝑁1𝑟,𝑔1+2𝑁𝑟,𝑓()1𝑧+𝑐+2𝑁𝑟,𝑔()1𝑧+𝑐+𝑆𝑟,𝑓+𝑆𝑟,𝑔≤6𝑁𝑟,𝑓1+6𝑁𝑟,𝑔1+𝑆𝑟,𝑓+𝑆𝑟,𝑔≤6𝑇𝑟,𝑓1+6𝑇𝑟,𝑔+𝑆𝑟,𝑓+𝑆𝑟,𝑔. (3.5) Lemma 2.3 shows that 𝑇(𝑟,𝐹)=(𝑛+1)𝑇𝑟,𝑓+𝑂𝑟^{𝜎(𝑓)−1+𝜀},𝑇()=()𝑇𝑟+𝑂log𝑟𝑟,𝐺𝑛+1𝑟,𝑔+𝑂^{𝜎(𝑔)−1+𝜀}+𝑂log𝑟. (3.6) We can deduce that (𝑛−5)𝑇𝑟,𝑓+𝑇𝑟,𝑔≤𝑂𝑟^{𝜎(𝑓)−1+𝜀}+𝑂𝑟^{𝜎(𝑔)−1+𝜀}+𝑆𝑟,𝑓+𝑆𝑟,𝑔, (3.7) which is impossible since 𝑛≥6. Therefore, we have 𝐻≡0. Noting that 𝑁1𝑟,𝐹1+𝑁𝑟,𝐺≤3𝑇𝑟,𝑓+3𝑇𝑟,𝑔+𝑆𝑟,𝑓+𝑆𝑟,𝑔≤𝑇(𝑟), (3.8) 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 that 𝐹(𝑧)=𝐺(𝑧). Therefore 𝑓(𝑧)^𝑛𝑓(𝑧+𝑐)=𝑔(𝑧)^𝑛𝑔(𝑧+𝑐). (3.9) Let ℎ₁(𝑧)=𝑓(𝑧)/𝑔(𝑧); we have ℎ₁(𝑧)^𝑛ℎ₁(𝑧+𝑐)≡1. (3.10) If ℎ₁(𝑧) is not a constant, then Lemma 2.2 and (3.10) imply that 𝑛𝑇𝑟,ℎ₁=𝑇𝑟,ℎ₁(𝑧+𝑐)+𝑂(1)=𝑇𝑟,ℎ₁+𝑂𝑟^{𝜎(ℎ1)−1+𝜀}+𝑂log𝑟, (3.11) 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 𝑡₁^𝑛+1=1.

Case 2. Suppose that 𝐹(𝑧)𝐺(𝑧)≡1. Therefore 𝑓(𝑧)^𝑛𝑓(𝑧+𝑐)𝑔(𝑧)^𝑛𝑔(𝑧+𝑐)≡1. (3.12) Let ℎ₂(𝑧)=𝑓(𝑧)𝑔(𝑧); we have ℎ₂(𝑧)^𝑛ℎ₂(𝑧+𝑐)≡1. (3.13) By the same way as Case 1, we can obtain that ℎ₂ is a constant. Therefore, 𝑓(𝑧)𝑔(𝑧)=𝑡₂ and 𝑡₂^𝑛+1=1.

4. Proof of Theorem 1.2

Noting that 𝑁1𝑟,−2𝐹−15𝑁₁₎1𝑟,+4𝐹−15𝑁₍₃1𝑟,≤3𝐹−15𝑁1𝑟,≤3𝐹−15𝑇(𝑟,𝐹),𝑁1𝑟,−2𝐺−15𝑁₁₎1𝑟,+4𝐺−15𝑁₍₃1𝑟,≤3𝐺−15𝑁1𝑟,≤3𝐺−15𝑇(𝑟,𝐺). (4.1) 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 Theorem 1.3

Noting that 𝑁1𝑟,−1𝐹−14𝑁₁₎1𝑟,+1𝐹−12𝑁₍₂1𝑟,≤3𝐹−14𝑁1𝑟,≤3𝐹−14𝑇(𝑟,𝐹),𝑁1𝑟,−1𝐺−14𝑁₁₎1𝑟,+1𝐺−12𝑁₍₂1𝑟,≤3𝐺−14𝑁1𝑟,≤3𝐺−14𝑇(𝑟,𝐺). (5.1) According to (3.1) and (5.1), we can obtain the conclusion of Theorem 1.2 by the same way as Section 3.

Acknowledgment

The third author was supported in part by China Scholarship Council (CSC).