Normal Families of Zero-Free Meromorphic Functions

Abstract

Let 𝑎(≠0),𝑏∈ℂ, and 𝑛 and 𝑘 be two positive integers such that 𝑛≥2. Let ℱ be a family of zero-free meromorphic functions defined in a domain 𝒟 such that for each 𝑓∈ℱ, 𝑓+𝑎(𝑓^(𝑘))^𝑛−𝑏 has at most 𝑛𝑘 zeros, ignoring multiplicity. Then ℱ is normal in 𝒟.

1. Introduction and Main Results

Let 𝒟 be a domain in ℂ, and let ℱ be a family of meromorphic functions defined in the domain 𝒟. ℱ is said to be normal in 𝒟, in the sense of Montel, if for every sequence {𝑓_𝑛}⊆ℱ contains a subsequence {𝑓_𝑛𝑗} such that 𝑓_𝑛𝑗 converges spherically uniformly on compact subsets of 𝒟 (see [1, Definition  3.1.1]).

ℱ is said to be normal at a point 𝑧₀∈𝒟 if there exists a neighborhood of 𝑧₀ in which ℱ is normal. It is well known that ℱ is normal in a domain 𝒟 if and only if it is normal at each of its points (see [1, Theorem  3.3.2]).

Let 𝑓 be a meromorphic function in the complex plane. We use the standard notations and results of value distribution theory as presented in [2–4]. In particular, 𝑇(𝑟,𝑓) is Nevanlinna’s characteristic function and 𝑆(𝑟,𝑓) denotes a function with the property 𝑆(𝑟,𝑓)=𝑜(𝑇(𝑟,𝑓)) as 𝑟→∞ (outside an exceptional set of finite linear measure).

In 1959, Hayman [5] proved the following well-known result.

Theorem A. Let 𝑓 be a transcendental meromorphic function on the complex plane C, let 𝑎 be a nonzero finite complex number, and let 𝑛 be a positive integer. If 𝑛≥5, then 𝑓′+𝑎𝑓^𝑛 assumes each value 𝑏∈𝐶 infinitely often.

There are some examples constructed by Mues [6] which show that Theorem A is not true when 𝑛=3,4. Corresponding to Theorem A, Ye [7, Theorem  2.1] proved the following interesting result.

Theorem B. Let 𝑓 be a transcendental meromorphic function. If 𝑎≠0 is a finite complex number and 𝑛≥3 is an positive integer, then 𝑓+𝑎𝑓′^𝑛 assumes all finite complex number infinitely often.

In [7, Theorem  2.2], Ye also obtained the following result, which may be considered as a normal family analogue of Theorem B.

Theorem C. Let ℱ be a family of meromorphic functions defined in a domain 𝒟, 𝑓≠𝑏 and 𝑓+𝑎𝑓′^𝑛≠𝑏 for every 𝑓∈ℱ, where 𝑛≥2 is an integer and 𝑎≠0,𝑏 are two finite complex numbers. Then, ℱ is normal.

Ye [7] asked whether Theorem B remains valid for 𝑛=2. Recently, Fang and Zalcman showed that Theorem B holds for 𝑛=2. In [8], the condition in Theorem C that 𝑓≠𝑏 can be relaxed to that all zeros of each function in ℱ are of multiplicity at least 2. Actually. they obtained the following results.

Theorem D. Let 𝑓 be a transcendental meromorphic function. If 𝑎≠0 is a finite complex number and 𝑛≥2 is an positive integer, then 𝑓+𝑎𝑓′^𝑛 assumes all finite complex number infinitely often.

Theorem E. Let ℱ be a family of meromorphic functions on the plane domain 𝒟, let 𝑛≥2 be a positive integer, and let 𝑎≠0,𝑏 be complex numbers. If, for each 𝑓∈ℱ, all zeros of 𝑓 are multiple and 𝑓+𝑎𝑓′^𝑛≠𝑏 on D, then ℱ is normal on D.

A natural problem arises: what can we say if 𝑓′ in Theorems E is replaced by the 𝑘th derivative 𝑓^(𝑘)? In [9], Xu et al. proved the following result.

Theorem F. Let 𝑎(≠0),𝑏∈ℂ and 𝑛 and 𝑘 be two positive integers such that 𝑛≥𝑘+1. Let ℱ be a family of meromorphic functions defined on a domain 𝒟. If, for every function 𝑓∈ℱ, 𝑓 has only zeros of multiplicity at least 𝑘+1, and 𝑓+𝑎(𝑓^(𝑘))^𝑛≠𝑏 in D, then ℱ is normal.

Xu et al. [9] asked whether Theorem F remains valid for 𝑛=2. We partially answer this question. If 𝑓≠0, we generalize Theorem F by allowing 𝑓+𝑎(𝑓^(𝑘))^𝑛−𝑏 to have zeros but restricting their numbers.

Theorem 1.1. Let 𝑎(≠0),𝑏∈ℂ, and 𝑛 and 𝑘 be two positive integers such that 𝑛≥2. Let ℱ be a family of zero-free meromorphic functions defined in a domain 𝒟 such that for each 𝑓∈ℱ, 𝑓+𝑎(𝑓^(𝑘))^𝑛−𝑏 has at most 𝑛𝑘 zeros, ignoring multiplicity. Then, ℱ is normal in 𝒟.

Remark 1.2. Here, 𝑓≠0 can be replaced by 𝑓≠𝑐, where 𝑐 is any finite complex numbers.

Example 1.3. Let 𝒟={𝑧∶|𝑧|<1}. Let ℱ={𝑓_𝑚}, where 𝑓_𝑚∶=𝑒^𝑚𝑧. Then, 𝑓_𝑚+𝑎𝑓^𝑚=(1+𝑎𝑚)𝑒^𝑚𝑧≠0 in 𝒟 for every function 𝑓∈ℱ. However, it is easily obtained that ℱ is not normal at the point 𝑧=0.

Example 1.4. Let 𝒟={𝑧∶|𝑧|<1}. Let ℱ={𝑓_𝑚}, where 𝑓_𝑚∶=1/𝑚𝑧. Then, 𝑓_𝑚+𝑎(𝑓^𝑚)²=(𝑚𝑧³+1)/𝑚²𝑧⁴ has 3 zeros in 𝒟 for every function 𝑓∈ℱ. However, it is easily obtained that ℱ is not normal at the point 𝑧=0.

Example 1.5. Let 𝒟={𝑧∶|𝑧|<1}. Let ℱ={𝑓_𝑚}, where 𝑓_𝑚∶=𝑚𝑧. It follows that 𝑓_𝑚+𝑎(𝑓^𝑚)²=𝑚𝑧+𝑚² has no zero in 𝒟 for every function 𝑓∈ℱ. However, it is easily obtained that ℱ is not normal at the point 𝑧=0.

Examples 1.3 and 1.4 show that the conditions that 𝑛≥2 and 𝑓+𝑎(𝑓^(𝑘))^𝑛−𝑏 have at most 𝑛𝑘 distinct zeros in Theorem 1.1 are shape. Example 1.5 shows the condition that 𝑓≠0 cannot be omitted.

2. Some Lemmas

To prove our results, we need some preliminary results.

Lemma 2.1 ([9], Lemma 2.2). Let 𝑛≥2,𝑘 be positive integers, let 𝑎 be a nonzero constant and let 𝑃(𝑧) be a polynomial. Then, the solution of the differential equation 𝑎(𝑊^(𝑘)(𝑧))^𝑛+𝑊(𝑧)=𝑃(𝑧) must be polynomial.

Lemma 2.2. Let 𝑓 be a nonzero transcendental meromorphic function. If 𝑎 be a nonzero finite complex number and let 𝑛≥2 and 𝑘 be two positive integers. Then, 𝑓+𝑎(𝑓^(𝑘))^𝑛 assumes each value 𝑏∈ℂ infinitely often.

Proof. Set 𝜙=𝐹𝐹′=𝑓+𝑎𝑓^(𝑘)^𝑛−𝑏,(2.1)𝐹=𝑓′+𝑎𝑛𝑓^(𝑘)^𝑛−1𝑓^(𝑘+1)𝜓=𝑛𝑓_(𝑘+1)𝑓+𝑎𝑓^(𝑘)^𝑛−𝑏,(2.2)𝑓^(𝑘)−𝐹^𝐹=𝑛𝑓^(𝑘+1)𝑓−𝑏𝑛𝑓^(𝑘)−𝑓^𝑓^(𝑘)𝑓^(𝑘)𝑓+𝑎𝑓^(𝑘)^𝑛−𝑏.(2.3)
We claim that 𝜙𝜓≢0. If 𝜙≡0, then 𝐹≡0. We can deduce that 𝐹≡𝑐, where 𝑐 is a finite complex number. We conclude from (2.1) and Lemma 2.1 that, 𝑓 must be a polynomial, which is a contradiction.
If 𝜓≡0, from (2.3), we can obtain 𝑐𝑓^(𝑘)^𝑛=𝑓+𝑎𝑓^(𝑘)^𝑛−𝑏,(2.4) where 𝑐 is a finite complex number, that is, (𝑎−𝑐)𝑓^(𝑘)^𝑛+𝑓=𝑏.(2.5) If 𝑎−𝑐=0, we can get that 𝑓≡𝑏, which is a contradiction. If 𝑎−𝑐≠0, we conclude from (2.5) and Lemma 2.1 that 𝑓 must be a polynomial, which is a contradiction.
By elementary Nevanlinna theory and (2.1), we have 𝑇(𝑟,𝐹)=𝑂(𝑇(𝑟,𝑓)). Thus, from (2.2) and (2.3), we have 𝑚(𝑟,𝜙)=𝑆(𝑟,𝑓),𝑚(𝑟,𝜓)=𝑆(𝑟,𝑓).(2.6)
It follows from (2.2), (2.3) and Nevanlinna’s First Fundamental Theorem that 𝑁𝑟,1𝜙⩽⩽𝑁𝑚((𝑟𝑟,,𝜙𝜙))++𝑁𝑆((𝑟𝑟,,𝜙𝑓))−⩽𝑚₋𝑟,𝜙1+O(1)𝑁(𝑟,𝑓)+𝑁⁻𝑟,1𝑁𝑟,1𝐹+𝑆(𝑟,𝑓),(2.7)𝜓⩽⩽𝑚𝑁((𝑟𝑟,,𝜓𝜓))++𝑁𝑆((𝑟𝑟,,𝜓𝑓))⩽−𝑚₋𝑟,𝜓1+O(1)𝑁𝑟,1𝑓^(𝑘)+𝑁𝑟,𝐹1+𝑆(𝑟,𝑓).(2.8) By (2.2) and (2.3), we get 𝜙(𝑓−𝑏)−𝑓^=𝑎𝑓^(𝑘)^𝑛𝜓.(2.9) We have by (2.6)-(2.7) 𝑇𝑟,𝜙(𝑓−𝑏)−𝑓^=𝑇𝑟,(𝑓−𝑏)𝜙−𝑓^⩽𝑇(𝑟,𝑓−𝑏)+𝑇𝑟,𝜙−𝑓_𝑓−𝑏⩽𝑚(𝑟,𝑓−𝑏)+𝑁(𝑟,𝑓−𝑏𝑓)+−𝑚𝑏𝑟+,𝜙𝑆−(𝑟,𝑓𝑓)_⩽𝑚(𝑟,𝑓)+𝑁(𝑟,𝑓)+𝑚(𝑟,𝜙)+𝑚𝑟,𝑓𝑓_−𝑏+𝑁𝑟,𝜙−𝑓𝑓−^𝑏+𝑆(𝑟,𝑓)⩽𝑇(𝑟,𝑓)+₋𝑓−𝑏+𝑁𝑟,𝜙−𝑓𝑓−^𝑏+𝑆(𝑟,𝑓)𝑁(𝑟,𝑓)+𝑁⁻𝑟,1𝐹+𝑆(𝑟,𝑓).(2.10)
It follows from (2.6)–(2.10) that 𝑛𝑇𝑟,𝑓^(𝑘)⩽⩽𝑇𝑚((𝑟𝑟,,𝜓𝜓))++𝑁𝑇(𝑟𝑟,,𝜙𝜓()𝑓+−𝑇𝑏()𝑟−,𝑓𝑓)^++₋𝑆(𝑟,𝑓)𝑁(𝑟,𝑓)+𝑁𝑟,1⩽𝐹+𝑆(𝑟,𝑓)₋𝑁𝑟,1𝑓^(𝑘)+𝑁𝑟,𝐹1+𝑚𝑟,𝑓1+𝑁𝑟,𝑓1+⁻+𝑁(𝑟,𝑓)₋𝑁𝑟,1⩽𝐹+𝑆(𝑟,𝑓)₋𝑁𝑟,1+₋𝑓^(𝑘)+2𝑁𝑟,𝐹1+𝑚𝑟,𝑓𝑓^(𝑘)+𝑚𝑟,𝑓1^(𝑘)+𝑁𝑟,𝑓1⩽𝑇𝑁𝑟,(𝑟,1𝑓)+𝑆(𝑟,𝑓)𝑓^(𝑘)+2𝑁𝑟,𝐹1+𝑁𝑟,𝑓1+⁻⩽𝑇𝑟,𝑓_(𝑘)𝑁(𝑟,𝑓)+𝑆(𝑟,𝑓)+2𝑁𝑟,𝐹1+𝑁𝑟,𝑓1+𝑁⁻(𝑟,𝑓)+𝑆(𝑟,𝑓).(2.11)
So, we have (𝑛−1)𝑇𝑟,𝑓^(𝑘)⩽2𝑁𝑟,1𝐹+𝑁𝑟,𝑓1+⁻𝑁(𝑟,𝑓)+𝑆(𝑟,𝑓).(2.12)
We have (𝑛−1)𝑇𝑟,𝑓^(𝑘)≥(𝑛−1)𝑁𝑟,𝑓^(𝑘)≥(𝑛−1)𝑁(𝑟,𝑓)+(𝑛−1)𝑁⁻(𝑟,𝑓).(2.13)
Since 𝑓≠0, if 𝑓+𝑎(𝑓^(𝑘))^𝑛 assumes the value 𝑏 only finitely often, we by (2.12) can get 𝑁(𝑟,𝑓)=𝑆(𝑟,𝑓).(2.14) Hence, (𝑛−1)𝑇𝑟,𝑓^(𝑘)⩽2𝑁𝑟,1𝐹+𝑆(𝑟,𝑓).(2.15) So 𝑓+𝑎(𝑓^(𝑘))^𝑛 assumes each value 𝑏∈ℂ infinitely often.
We complete the proof of Lemma 2.2.

Using the method of Chang [10, Lemma  4], we obtain the following lemma.

Lemma 2.3. Let 𝑓 be a nonconstant zero-free rational function, 𝑛≥2,  let 𝑘 be two positive integers, and 𝑎≠0,𝑏 be two complex constants. Then, the function 𝑓+𝑎(𝑓^(𝑘))^𝑛−𝑏 has at least 𝑛𝑘+1 distinct zeros in ℂ.

Proof. Since 𝑓(𝑧) is a nonconstant zero-free rational function, 𝑓(𝑧) is not a polynomial, and hence it has at least one finite pole. Thus, we can write 𝑓(𝑧)=𝐶¹∏^𝑚𝑖=1𝑧+𝑧_𝑖^𝑝𝑖,(2.16) where 𝐶₁ is a nonzero constant, 𝑚 and 𝑝_𝑖 are positive integers, the 𝑧_𝑖 (when 1≤𝑖≤𝑚) are distinct complex numbers, and denote 𝑝=∑^𝑚𝑖=1𝑝_𝑖.
By induction, we deduce from (2.16) that 𝑓^(𝑘)(𝑧)=𝑃^{(𝑚−1)𝑘}∏^𝑚𝑖=1𝑧+𝑧_𝑖^{𝑝𝑖+𝑘},(2.17) where 𝑃_{(𝑚−1)𝑘} is polynomial of degree (𝑚−1)𝑘.
So the degree of numerator of the function 𝑓+𝑎(𝑓^(𝑘))^𝑛 is equal to ∑^𝑚𝑖=1(𝑛−1)𝑝_𝑖+𝑛𝑘. By calculation, 𝑓+𝑎(𝑓^(𝑘))^𝑛−𝑏 has at least one zero in ℂ. Thus, we can write 𝑓+𝑎𝑓^(𝑘)^𝑛−𝑏=𝐶²∏^𝑠𝑖=1𝑧+𝛼_𝑖^𝑙𝑖∏^𝑚𝑖=1𝑧+𝑧_𝑖^{𝑛(𝑝𝑖+𝑘)},(2.18) where 𝐶₂ is a nonzero constant, 𝑙_𝑖 are positive integers, 𝛼_𝑖 (when 1≤𝑖≤𝑠), and 𝑧_𝑖 (when 1≤𝑖≤𝑚) are distinct complex numbers. Thus, by (2.16), (2.17), and (2.18), we get 𝐶_1𝑚_𝑖=1𝑧+𝑧_𝑖^{(𝑛−1)𝑝𝑖+𝑛𝑘}+𝑎𝑃_{(𝑚−1)𝑘}^𝑛=𝑏^𝑚_𝑖=1𝑧+𝑧_𝑖^{𝑛(𝑝𝑖+𝑘)}+𝐶_2𝑠_𝑖=1𝑧+𝛼_𝑖^𝑙𝑖.(2.19)Case 1. If 𝑏=0, it follows that ∑^𝑚𝑖=1[(𝑛−1)𝑝_𝑖+𝑛𝑘]=∑^𝑠𝑖=1𝑙_𝑖 and 𝐶₁=𝐶₂. Thus, it follows from (2.19) that ^𝑚_𝑖=11+𝑧_𝑖𝑡^{(𝑛−1)𝑝𝑖+𝑛𝑘}−^𝑠_𝑖=11+𝛼_𝑖𝑡^𝑙𝑖=𝑡^{(𝑛−1)𝑝+𝑛𝑘}𝑄(𝑡),(2.20) where 𝑄(𝑡)=(−𝑎/𝐶₁)𝑡^{(𝑚−1)𝑛𝑘}(𝑃_{(𝑚−1)𝑘}(1/𝑡))^𝑛 is a polynomial. Then, 𝑄(𝑡) is a polynomial of degree less than (𝑚−1)𝑛𝑘, and it follows that ∏^𝑚𝑖=11+𝑧_𝑖𝑡^{(𝑛−1)𝑝𝑖+𝑛𝑘}∏^𝑠𝑖=11+𝛼_𝑖𝑡^𝑙𝑖=1+𝑡^{(𝑛−1)𝑝+𝑛𝑘}𝑄(𝑡)∏^𝑠𝑖=11+𝛼_𝑖𝑡^𝑙𝑖=1+𝑂𝑡^{(𝑛−1)𝑝+𝑛𝑘}(2.21) as 𝑡→0.
Logarithmic differentiation of both sides of (2.21) shows that ^𝑚_𝑖=1(𝑛−1)𝑝_𝑖+𝑛𝑘𝑧_𝑖1+𝑧_𝑖𝑡−^𝑠_𝑖=1𝑙_𝑖𝛼_𝑖1+𝛼_𝑖𝑡=𝑂𝑡^{(𝑛−1)𝑝+𝑛𝑘−1}(2.22) as 𝑡→0.
Comparing the coefficient of (2.22) for 𝑡^𝑗, 𝑗=0,1,…,(𝑛−1)𝑝+𝑛𝑘−2, we have ^𝑚_𝑖=1(𝑛−1)𝑝_𝑖+𝑛𝑘𝑧^𝑗𝑖−^𝑠_𝑖=1𝑙_𝑖𝛼^𝑗𝑖=0(2.23) for 𝑗=1,…,(𝑛−1)𝑝+𝑛𝑘−1.
Set 𝑧_𝑚+𝑖=−𝛼_𝑖 when 1≤𝑖≤𝑠. Noting that ∑^𝑚𝑖=1[(𝑛−1)𝑝_𝑖+𝑛𝑘]=∑^𝑠𝑖=1𝑙_𝑖, then it follows from (2.23) that the system of linear equations, ^𝑚+𝑠_𝑖=1𝑧^𝑗𝑖𝑥_𝑖=0,(2.24) where 0≤𝑗≤(𝑛−1)𝑝+𝑛𝑘−1, has a nonzero solution 𝑥₁,…,𝑥_𝑚,𝑥_𝑚+1,…,𝑥_𝑚+𝑠=(𝑛−1)𝑝₁+𝑛𝑘,…,(𝑛−1)𝑝_𝑚+𝑛𝑘,𝑙₁,…,𝑙_𝑠.(2.25)
If (𝑛−1)𝑝+𝑛𝑘≥𝑚+𝑠, then the determinant det(𝑧^𝑗𝑖)_{(𝑚+𝑠)×(𝑚+𝑠)} of the coefficients of the system of (2.24), where 0≤𝑗≤(𝑛−1)𝑝+𝑛𝑘−1, is equal to zero, by Cramer’s rule (see, e.g., [11]). However, the 𝑧_𝑖 are distinct complex numbers when 1≤𝑖≤𝑚+𝑠, and the determinant is a Vandermonde determinant, so it cannot be 0 (see [11]), which is a contradiction.
Hence, we conclude that (𝑛−1)𝑝+𝑛𝑘<𝑚+𝑠. Noting that 𝑛≥2, it follows from this and 𝑝=∑^𝑚𝑖=1𝑝_𝑖≥𝑚 that 𝑠≥𝑛𝑘+1.

Case 2. If 𝑏≠0, set 𝑏^𝑚_𝑖=1𝑧+𝑧_𝑖^{𝑛(𝑝𝑖+𝑘)}−𝐶_1𝑚_𝑖=1𝑧+𝑧_𝑖^{(𝑛−1)𝑝𝑖+𝑛𝑘}=𝑏^𝑚_𝑖=1𝑧+𝑧_𝑖^{(𝑛−1)𝑝𝑖𝑞+𝑛𝑘}_𝑖=1𝑧+𝛽_𝑖^𝑡𝑖,(2.26) where 𝑡_𝑖 are positive integers. It follows that 𝛽_𝑖 (when 1≤𝑖≤𝑞) and 𝑧_𝑖 (when 1≤𝑖≤𝑚) are distinct complex numbers, and ∑^𝑞𝑖=1𝑡_𝑖=𝑝.

By (2.19), we have 𝑏^𝑚_𝑖=1𝑧+𝑧_𝑖^{(𝑛−1)𝑝𝑖𝑞+𝑛𝑘}_𝑖=1𝑧+𝛽_𝑖^𝑡𝑖+𝐶_2𝑠_𝑖=1𝑧+𝛼_𝑖^𝑙𝑖=𝑎𝑃_{(𝑚−1)𝑘}^𝑛.(2.27)

It follows that ^𝑚_𝑖=1(𝑛−1)𝑝_𝑖+𝑛𝑘+^𝑞_𝑖=1𝑡_𝑖=𝑛𝑝+𝑛𝑚𝑘=^𝑠_𝑖=1𝑙_𝑖,(2.28) and 𝐶₂=−𝑏. Thus, by (2.27), ^𝑚_𝑖=11+𝑧_𝑖𝑡^{(𝑛−1)𝑝𝑖𝑞+𝑛𝑘}_𝑖=11+𝛽_𝑖𝑡^𝑡𝑖−^𝑠_𝑖=11+𝛼_𝑖𝑡^𝑙𝑖=𝑡^{𝑛(𝑝+𝑘)}𝑄(𝑡),(2.29) where 𝑄(𝑡)=(𝑎/𝑏)𝑡^{(𝑚−1)𝑛𝑘}(𝑃_{(𝑚−1)𝑘}(1/𝑡))^𝑛 is a polynomial. Then, 𝑄(𝑡) is a polynomial of degree less than (𝑚−1)𝑛𝑘, and it follows that ∏^𝑚𝑖=11+𝑧_𝑖𝑡^{(𝑛−1)𝑝𝑖+𝑛𝑘}∏^𝑞𝑖=11+𝛽_𝑖𝑡^𝑡𝑖∏^𝑠𝑖=11+𝛼_𝑖𝑡^𝑙𝑖=1+𝑡^{𝑛(𝑝+𝑘)}𝑄(𝑡)∏^𝑠𝑖=11+𝛼_𝑖𝑡^𝑙𝑖=𝑂𝑡^{𝑛(𝑝+𝑘)}(2.30) as 𝑡→0.

Thus, by taking logarithmic derivatives of both sides of (2.12), we get ^𝑚_𝑖=1(𝑛−1)𝑝_𝑖+𝑛𝑘𝑧_𝑖1+𝑧_𝑖𝑡+^𝑞_𝑖=1𝑡_𝑖𝛽_𝑖1+𝛽_𝑖𝑡−^𝑠_𝑖=1𝑙_𝑖𝛼_𝑖1+𝛼_𝑖𝑡=𝑂𝑡^{𝑛(𝑝+𝑘)−1}.(2.31)

We consider two cases.

Subcase 2.1 ({𝛼₁,…,𝛼_𝑠}∩{𝛽₁,…,𝛽_𝑞}=∅). Applying the reasoning of Case 1 and noting that 𝑝≥𝑞, we deduce that 𝑠≥𝑛𝑘.

Subcase 2.2 ({𝛼₁,…,𝛼_𝑠}∩{𝛽₁,…,𝛽_𝑞}≠∅). Without loss of generality, we may assume that 𝛼_𝑞−𝑖=𝛽_𝑖,for(1≤𝑖≤𝑀). Denote 𝑧_𝑖=⎧⎪⎨⎪⎩𝑧^𝑖𝛽for1≤𝑖≤𝑚,_𝑖−𝑚𝛼for𝑚+1≤𝑖≤𝑚+𝑞,_{𝑀+𝑖−𝑚−𝑞}𝑁for𝑚+𝑞+1≤𝑖≤𝑚+𝑞+𝑠−𝑀,_𝑖=⎧⎪⎨⎪⎩(𝑛−1)𝑝^𝑖𝑡+𝑛𝑘for1≤𝑖≤𝑚,_𝑖−𝑚𝑡for𝑚+1≤𝑖≤𝑚+𝑠−𝑀,_𝑖−𝑚𝑙−𝑙_{𝑖−𝑚−𝑠+𝑀}for𝑚+𝑠−𝑀+1≤𝑖≤𝑚+𝑞,_{𝑖−𝑚−𝑞+𝑀}for𝑚+𝑞+1≤𝑖≤𝑚+𝑞+𝑠−𝑀.(2.32)

The formula (2.31) can be rewritten: ^𝑚+𝑞^+𝑠−𝑀_𝑖=1𝑁_𝑖𝑧_𝑖1+𝑧_𝑖𝑡=𝑂𝑡^{𝑛(𝑝+𝑘)−1}.(2.33)

Applying the reasoning of Case 1, and noting that 𝑝≥𝑞, we deduce that 𝑠≥𝑛𝑘+1.

This completes the proof of Lemma 2.3.

Lemma 2.4 ([10], Lemma 4). Let 𝑓 be a nonconstant zero-free rational function, let 𝑎≠0 be a complex constant, and let 𝑘 be a positive integer. Then 𝑓^(𝑘)−𝑎 has at least 𝑘+1 distinct zeros in ℂ.

Lemma 2.5 (see [12], Lemma  2, Zalcman’s lemma). Let ℱ be a family of functions meromorphic on a domain 𝒟, all of whose zeros have multiplicity at least 𝑘. Suppose that there exists 𝐴⩾1 such that |𝑓^(𝑘)(𝑧)|⩽𝐴 whenever 𝑓(𝑧)=0. Then, if ℱ is not normal at 𝑧₀∈𝒟, there exist, for each 0⩽𝛼⩽𝑘,(a)points 𝑧_𝑛,𝑧_𝑛→𝑧₀;(b)functions 𝑓_𝑛∈ℱ;(c)positive numbers 𝜌_𝑛→0⁺; such that 𝜌^𝑛−𝛼𝑓_𝑛(𝑧_𝑛+𝜌_𝑛𝜉)=𝑔_𝑛(𝜉)→𝑔(𝜉) locally uniformly with respect to the spherical metric, where 𝑔(𝜉) is a nonconstant meromorphic function on ℂ, all of whose zeros of 𝑔(𝜉) are of multiplicity at least 𝑘, such that 𝑔^#(𝜉)≤𝑔^#(0)=𝑘𝐴+1.

Here, as usual, 𝑔^#(𝜉)=|𝑔′(𝜉)|/(1+|𝑔(𝜉)|²) is the spherical derivative.

3. Proof of Theorem

Suppose that ℱ is not normal in 𝒟. Then, there exists at least one point 𝑧₀ such that ℱ is not normal at the point 𝑧₀∈𝒟. Without loss of generality, we assume that 𝑧₀=0. We consider two cases.

Case 1 (𝑏=0). By Zalcman’s lemma, there exist:(a)points 𝑧_𝑛,𝑧_𝑛→𝑧₀;(b)functions 𝑓_𝑛∈ℱ; (c)positive numbers 𝜌_𝑛→0⁺; such that 𝑔_𝑗(𝜉)=𝜌^{𝑗−𝑛𝑘/(𝑛−1)}𝑓_𝑗𝑧_𝑗+𝜌_𝑗𝜉⟶𝑔(𝜉),(3.1) spherically uniformly on compact subsets of ℂ, where 𝑔(𝜉) is a nonconstant meromorphic function in ℂ. Since 𝑓_𝑗≠0, by Hurwitz’s theorem, it implies that 𝑔(𝜉)≠0.
On every compact subset of ℂ which contains no poles of 𝑔, from (3.1), we get 𝑔_𝑗(𝜉)+𝑎𝑔^𝑘𝑗(𝜉)^𝑛=𝜌^{𝑗−𝑛𝑘/(𝑛−1)}𝑓_𝑗𝑧_𝑗+𝜌_𝑗𝜉+𝑎𝑓^𝑘𝑗𝑧_𝑗+𝜌_𝑗𝜉^𝑛⟶𝑔(𝜉)+𝑎𝑔^𝑘(𝜉)^𝑛,(3.2) also locally uniformly with respect to the spherical metric.
We claim that 𝑔(𝜉)+𝑎(𝑔^𝑘(𝜉))^𝑛 has at most 𝑛𝑘 distinct zeros.
Suppose that 𝑔(𝜉)+𝑎(𝑔^𝑘(𝜉))^𝑛 has 𝑛𝑘+1 distinct zeros 𝜉_𝑖, 1≤𝑖≤𝑛𝑘+1, and choose 𝛿(>0) small enough such that ⋂^{𝑛𝑘+1𝑖=1}𝐷(𝜉_𝑖,𝛿)=∅, where 𝐷(𝜉₀,𝛿)={𝜉∣|𝜉−𝜉_𝑖|<𝛿}.
From (3.2), by Hurwitz’s theorem, there exist points 𝜉^𝑗𝑖∈𝐷(𝜉_𝑖,𝛿) (1≤𝑖≤𝑛𝑘+1) such that for sufficiently large 𝑗, 𝑓_𝑗𝑧_𝑗+𝜌_𝑗𝜉^𝑗𝑖+𝑎𝑓^𝑘𝑗𝑧_𝑗+𝜌_𝑗𝜉^𝑗𝑖^𝑛=0,(3.3) for 1≤𝑖≤𝑛𝑘+1.
Since 𝑧_𝑗→0 and 𝜌_𝑗→0⁺, we have 𝑧_𝑗+𝜌_𝑗𝜉^𝑗𝑖∈𝐷(0,𝜎)  (𝜎 is a positive constant) for sufficiently large 𝑗, so 𝑓_𝑗(𝑧)+𝑎(𝑓^𝑘𝑗(𝑧))^𝑛 has 𝑛𝑘+1 distinct zeros, which contradicts the fact that 𝑓_𝑗(𝑧)+𝑎(𝑓^𝑘𝑗(𝑧))^𝑛 has at most 𝑛𝑘 zero.
However, by Lemmas 2.2 and 2.3, there do not exist nonconstant meromorphic functions that have the above properties. This contradiction shows that ℱ is normal in 𝒟.

Case 2 (𝑏≠0). By Zalcman’s lemma, there exist:(a)points 𝑧_𝑛,𝑧_𝑛→𝑧₀;(b)functions 𝑓_𝑛∈ℱ; (c)positive numbers 𝜌_𝑛→0⁺; such that 𝑔_𝑗(𝜉)=𝜌^𝑗−𝑘𝑓_𝑗𝑧_𝑗+𝜌_𝑗𝜉⟶𝑔(𝜉)(3.4) spherically uniformly on compact subsets of ℂ, where 𝑔(𝜉) is a nonconstant meromorphic function in ℂ. Since 𝑓_𝑗≠0, by Hurwitz’s theorem, it implies that 𝑔(𝜉)≠0.
On every compact subset of ℂ which contains no poles of 𝑔, from (3.4), we get 𝜌^𝑘𝑗𝑔_𝑗(𝜉)+𝑎𝑔^𝑘𝑗(𝜉)^𝑛−𝑏⟶𝑎𝑔^𝑘(𝜉)^𝑛−𝑏(3.5) also locally uniformly with respect to the spherical metric.
Noting that 𝜌^𝑘𝑗𝑔_𝑗(𝜉)+𝑎𝑔^𝑘𝑗(𝜉)^𝑛−𝑏=𝑓_𝑗𝑧_𝑗+𝜌_𝑗𝜉+𝑎𝑓^𝑘𝑗𝑧_𝑗+𝜌_𝑗𝜉^𝑛−𝑏,(3.6) we claim that 𝑎(𝑔^𝑘(𝜉))^𝑛−𝑏 has at most 𝑛𝑘 distinct zeros.
Suppose that 𝑔(𝜉)+𝑎(𝑔^𝑘(𝜉))^𝑛−𝑏 has 𝑛𝑘+1 distinct zeros 𝜉_𝑖, 1≤𝑖≤𝑛𝑘+1, and choose 𝛿(>0) small enough such that ⋂^{𝑛𝑘+1𝑖=1}𝐷(𝜉_𝑖,𝛿)=∅, where 𝐷(𝜉₀,𝛿)={𝜉∣|𝜉−𝜉_𝑖|<𝛿}.
From (3.2), by Hurwitz’s theorem, there exist points 𝜉^𝑗𝑖∈𝐷(𝜉_𝑖,𝛿) (1≤𝑖≤𝑛𝑘+1) such that for sufficiently large 𝑗𝑓_𝑗𝑧_𝑗+𝜌_𝑗𝜉^𝑗𝑖+𝑎𝑓^𝑘𝑗𝑧_𝑗+𝜌_𝑗𝜉^𝑗𝑖^𝑛−𝑏=0,(3.7) for 1≤𝑖≤𝑛𝑘+1.
Since 𝑧_𝑗→0 and 𝜌_𝑗→0⁺, we have 𝑧_𝑗+𝜌_𝑗𝜉^𝑗𝑖∈𝐷(0,𝜎)  (𝜎 is a positive constant) for sufficiently large 𝑗, so 𝑓_𝑗(𝑧)+𝑎(𝑓^𝑘𝑗(𝑧))^𝑛−𝑏 has 𝑛𝑘+1 distinct zeros, which contradicts the fact that 𝑓_𝑗(𝑧)+𝑎(𝑓^𝑘𝑗(𝑧))^𝑛−𝑏 has at most 𝑛𝑘 zero.
Denote 𝑐₁,𝑐₂,…,𝑐_𝑛 by the different roots of 𝜔^𝑛=𝑏/𝑎, then 𝑎𝑔^𝑘(𝜉)^𝑛−𝑏=𝑎^𝑛_𝑖=1𝑔^𝑘(𝜉)−𝑐_𝑖.(3.8)Subcase 2.1 (If 𝑔(𝜉) is a rational function). By Lemma 2.4 and (3.8), we can deduce that 𝑎(𝑔^𝑘(𝜉))^𝑛−𝑏 has at least 𝑛𝑘+𝑛 distinct zeros. This contradicts the claim that 𝑎(𝑔^𝑘(𝜉))^𝑛−𝑏 has at most 𝑛𝑘 distinct zeros.

Subcase 2.2 (If 𝑔(𝜉) is a transcendental meromorphic function). By Nevanlinnas second main theorem, we have 𝑇𝑟,𝑔^(𝑘)≤𝑁⁻𝑟,𝑔^(𝑘)+^𝑛=₋_−𝑖=1𝑁𝑟,𝑔^(𝑘)1−𝑐_𝑖+𝑆𝑟,𝑔^(𝑘)𝑁𝑟,𝑔^(𝑘)+𝑁⁻𝑟,1𝑎𝑔^(𝑘)^𝑛−𝑏+𝑆𝑟,𝑔^(𝑘)≤1≤𝑘+11𝑁𝑟,𝑔^(𝑘)+𝑆𝑟,𝑔^(𝑘)𝑘+1𝑇𝑟,𝑔^(𝑘)+𝑆𝑟,𝑔^(𝑘).(3.9)

It follows that 𝑇(𝑟,𝑔^(𝑘))≤𝑆(𝑟,𝑔^(𝑘)), which is a contradiction. This contradiction shows that ℱ is normal in 𝒟.

Hence, Theorem 1.1 is proved.

Acknowledgment

The authors thank the referee(s) for reading the manuscript very carefully and making a number of valuable and kind comments which improved the presentation.