Abstract and Applied Analysis

Abstract and Applied Analysis / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 908123 | https://doi.org/10.1155/2012/908123

Yuntong Li, "Normal Families of Zero-Free Meromorphic Functions", Abstract and Applied Analysis, vol. 2012, Article ID 908123, 12 pages, 2012. https://doi.org/10.1155/2012/908123

Normal Families of Zero-Free Meromorphic Functions

Academic Editor: Sergey V. Zelik
Received26 Mar 2012
Accepted07 Aug 2012
Published30 Aug 2012

Abstract

Let ğ‘Ž(≠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+ğ‘Ž(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 {𝑓𝑛}⊆ℱ contains a subsequence {𝑓𝑛𝑗} such that 𝑓𝑛𝑗 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 ğ‘Ÿâ†’âˆž (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 ğ‘Ž be a nonzero finite complex number, and let 𝑛 be a positive integer. If 𝑛≥5, then 𝑓′+ğ‘Žğ‘“ğ‘› 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 ğ‘Žâ‰ 0 is a finite complex number and 𝑛≥3 is an positive integer, then 𝑓+ğ‘Žğ‘“â€²ğ‘› 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+ğ‘Žğ‘“â€²ğ‘›â‰ ğ‘ for every 𝑓∈ℱ, where 𝑛≥2 is an integer and ğ‘Žâ‰ 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 ğ‘Žâ‰ 0 is a finite complex number and 𝑛≥2 is an positive integer, then 𝑓+ğ‘Žğ‘“â€²ğ‘› 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 ğ‘Žâ‰ 0,𝑏 be complex numbers. If, for each 𝑓∈ℱ, all zeros of 𝑓 are multiple and 𝑓+ğ‘Žğ‘“â€²ğ‘›â‰ ğ‘ 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 ğ‘Ž(≠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+ğ‘Ž(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+ğ‘Ž(𝑓(𝑘))𝑛−𝑏 to have zeros but restricting their numbers.

Theorem 1.1. Let ğ‘Ž(≠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+ğ‘Ž(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={𝑓𝑚}, where 𝑓𝑚∶=𝑒𝑚𝑧. Then, 𝑓𝑚+ğ‘Žğ‘“î…žğ‘š=(1+ğ‘Žğ‘š)𝑒𝑚𝑧≠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={𝑓𝑚}, where 𝑓𝑚∶=1/𝑚𝑧. Then, 𝑓𝑚+ğ‘Ž(ğ‘“î…žğ‘š)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={𝑓𝑚}, where 𝑓𝑚∶=𝑚𝑧. It follows that 𝑓𝑚+ğ‘Ž(ğ‘“î…žğ‘š)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+ğ‘Ž(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 ğ‘Ž be a nonzero constant and let 𝑃(𝑧) be a polynomial. Then, the solution of the differential equation ğ‘Ž(𝑊(𝑘)(𝑧))𝑛+𝑊(𝑧)=𝑃(𝑧) must be polynomial.

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

Proof. Set 𝑓𝐹=𝑓+ğ‘Ž(𝑘)𝑛−𝑏,(2.1)𝜙=𝐹′𝐹=𝑓𝑓′+ğ‘Žğ‘›(𝑘)𝑛−1𝑓(𝑘+1)𝑓𝑓+ğ‘Ž(𝑘)𝑛,𝑓−𝑏(2.2)𝜓=𝑛(𝑘+1)𝑓(𝑘)âˆ’ğ¹î…žğ¹=𝑛𝑓(𝑘+1)𝑓−𝑏𝑛𝑓(𝑘)âˆ’ğ‘“î…žğ‘“(𝑘)𝑓(𝑘)𝑓𝑓+ğ‘Ž(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=𝑓+ğ‘Ž(𝑘)𝑛−𝑏,(2.4) where 𝑐 is a finite complex number, that is, 𝑓(ğ‘Žâˆ’ğ‘)(𝑘)𝑛+𝑓=𝑏.(2.5) If ğ‘Žâˆ’ğ‘=0, we can get that 𝑓≡𝑏, which is a contradiction. If ğ‘Žâˆ’ğ‘â‰ 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1𝑟,𝜙1⩽𝑚(𝑟,𝜙)+𝑁(𝑟,𝜙)−𝑚𝑟,𝜙+O(1)⩽𝑁(𝑟,𝜙)+𝑆(𝑟,𝑓)⩽−𝑁(𝑟,𝑓)+−𝑁1𝑟,𝐹𝑁1+𝑆(𝑟,𝑓),(2.7)𝑟,𝜓1⩽𝑚(𝑟,𝜓)+𝑁(𝑟,𝜓)−𝑚𝑟,𝜓+O(1)⩽𝑁(𝑟,𝜓)+𝑆(𝑟,𝑓)⩽−𝑁1𝑟,𝑓(𝑘)1+𝑁𝑟,𝐹+𝑆(𝑟,𝑓).(2.8) By (2.2) and (2.3), we get 𝜙(𝑓−𝑏)âˆ’ğ‘“î…žî€·ğ‘“=ğ‘Ž(𝑘)𝑛𝜓.(2.9) We have by (2.6)-(2.7) 𝑇𝑟,𝜙(𝑓−𝑏)âˆ’ğ‘“î…žî€¸î‚µî‚µğ‘“=𝑇𝑟,(𝑓−𝑏)ğœ™âˆ’î…žî‚µğ‘“ğ‘“âˆ’ğ‘î‚¶î‚¶â©½ğ‘‡(𝑟,𝑓−𝑏)+𝑇𝑟,ğœ™âˆ’î…žî‚¶î‚µğ‘“ğ‘“âˆ’ğ‘+𝑆(𝑟,𝑓)⩽𝑚(𝑟,𝑓−𝑏)+𝑁(𝑟,𝑓−𝑏)+𝑚𝑟,ğœ™âˆ’î…žî‚¶î‚µğ‘“ğ‘“âˆ’ğ‘+𝑁𝑟,ğœ™âˆ’î…žî‚¶î‚µğ‘“ğ‘“âˆ’ğ‘+𝑆(𝑟,𝑓)⩽𝑚(𝑟,𝑓)+𝑁(𝑟,𝑓)+𝑚(𝑟,𝜙)+𝑚𝑟,î…žî‚¶î‚µğ‘“ğ‘“âˆ’ğ‘+𝑁𝑟,ğœ™âˆ’î…žî‚¶ğ‘“âˆ’ğ‘+𝑆(𝑟,𝑓)⩽𝑇(𝑟,𝑓)+−𝑁(𝑟,𝑓)+−𝑁1𝑟,𝐹+𝑆(𝑟,𝑓).(2.10)
It follows from (2.6)–(2.10) that 𝑛𝑇𝑟,𝑓(𝑘)⩽𝑇(𝑟,𝜓)+𝑇𝑟,𝜙(𝑓−𝑏)âˆ’ğ‘“î…žî€¸+𝑆(𝑟,𝑓)⩽𝑚(𝑟,𝜓)+𝑁(𝑟,𝜓)+𝑇(𝑟,𝑓)+−𝑁1(𝑟,𝑓)+𝑁𝑟,𝐹⩽+𝑆(𝑟,𝑓)−𝑁1𝑟,𝑓(𝑘)1+𝑁𝑟,𝐹1+𝑚𝑟,𝑓1+𝑁𝑟,𝑓+−𝑁+(𝑟,𝑓)−𝑁1𝑟,𝐹⩽+𝑆(𝑟,𝑓)−𝑁1𝑟,𝑓(𝑘)1+2𝑁𝑟,𝐹𝑓+𝑚𝑟,(𝑘)𝑓1+𝑚𝑟,𝑓(𝑘)1+𝑁𝑟,𝑓+−𝑁1(𝑟,𝑓)+𝑆(𝑟,𝑓)⩽𝑇𝑟,𝑓(𝑘)1+2𝑁𝑟,𝐹1+𝑁𝑟,𝑓+−𝑁(𝑟,𝑓)+𝑆(𝑟,𝑓)⩽𝑇𝑟,𝑓(𝑘)1+2𝑁𝑟,𝐹1+𝑁𝑟,𝑓+−𝑁(𝑟,𝑓)+𝑆(𝑟,𝑓).(2.11)
So, we have (𝑛−1)𝑇𝑟,𝑓(𝑘)1⩽2𝑁𝑟,𝐹1+𝑁𝑟,𝑓+−𝑁(𝑟,𝑓)+𝑆(𝑟,𝑓).(2.12)
We have (𝑛−1)𝑇𝑟,𝑓(𝑘)≥(𝑛−1)𝑁𝑟,𝑓(𝑘)≥(𝑛−1)𝑁(𝑟,𝑓)+(𝑛−1)−𝑁(𝑟,𝑓).(2.13)
Since 𝑓≠0, if 𝑓+ğ‘Ž(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)𝑇𝑟,𝑓(𝑘)1⩽2𝑁𝑟,𝐹+𝑆(𝑟,𝑓).(2.15) So 𝑓+ğ‘Ž(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 ğ‘Žâ‰ 0,𝑏 be two complex constants. Then, the function 𝑓+ğ‘Ž(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 𝐶𝑓(𝑧)=1∏𝑚𝑖=1𝑧+𝑧𝑖𝑝𝑖,(2.16) where 𝐶1 is a nonzero constant, 𝑚 and 𝑝𝑖 are positive integers, the 𝑧𝑖 (when 1≤𝑖≤𝑚) are distinct complex numbers, and denote ∑𝑝=𝑚𝑖=1𝑝𝑖.
By induction, we deduce from (2.16) that 𝑓(𝑘)𝑃(𝑧)=(𝑚−1)𝑘∏𝑚𝑖=1𝑧+𝑧𝑖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+ğ‘Ž(𝑓(𝑘))𝑛 is equal to ∑𝑚𝑖=1(𝑛−1)𝑝𝑖+𝑛𝑘. By calculation, 𝑓+ğ‘Ž(𝑓(𝑘))𝑛−𝑏 has at least one zero in ℂ. Thus, we can write 𝑓𝑓+ğ‘Ž(𝑘)𝑛𝐶−𝑏=2∏𝑠𝑖=1𝑧+𝛼𝑖𝑙𝑖∏𝑚𝑖=1𝑧+𝑧𝑖𝑛(𝑝𝑖+𝑘),(2.18) where 𝐶2 is a nonzero constant, 𝑙𝑖 are positive integers, 𝛼𝑖 (when 1≤𝑖≤𝑠), and 𝑧𝑖 (when 1≤𝑖≤𝑚) are distinct complex numbers. Thus, by (2.16), (2.17), and (2.18), we get 𝐶1𝑚𝑖=1𝑧+𝑧𝑖(𝑛−1)𝑝𝑖+𝑛𝑘𝑃+ğ‘Ž(𝑚−1)𝑘𝑛=𝑏𝑚𝑖=1𝑧+𝑧𝑖𝑛(𝑝𝑖+𝑘)+𝐶2𝑠𝑖=1𝑧+𝛼𝑖𝑙𝑖.(2.19)Case 1. If 𝑏=0, it follows that ∑𝑚𝑖=1[(𝑛−1)𝑝𝑖∑+𝑛𝑘]=𝑠𝑖=1𝑙𝑖 and 𝐶1=𝐶2. Thus, it follows from (2.19) that 𝑚𝑖=11+𝑧𝑖𝑡(𝑛−1)𝑝𝑖+𝑛𝑘−𝑠𝑖=11+𝛼𝑖𝑡𝑙𝑖=𝑡(𝑛−1)𝑝+𝑛𝑘𝑄(𝑡),(2.20) where 𝑄(𝑡)=(âˆ’ğ‘Ž/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+𝑧𝑖𝑡(𝑛−1)𝑝𝑖+𝑛𝑘∏𝑠𝑖=11+𝛼𝑖𝑡𝑙𝑖𝑡=1+(𝑛−1)𝑝+𝑛𝑘𝑄(𝑡)∏𝑠𝑖=11+𝛼𝑖𝑡𝑙𝑖𝑡=1+𝑂(𝑛−1)𝑝+𝑛𝑘(2.21) as 𝑡→0.
Logarithmic differentiation of both sides of (2.21) shows that 𝑚𝑖=1(𝑛−1)𝑝𝑖𝑧+𝑛𝑘𝑖1+𝑧𝑖𝑡−𝑠𝑖=1𝑙𝑖𝛼𝑖1+𝛼𝑖𝑡𝑡=𝑂(𝑛−1)𝑝+𝑛𝑘−1(2.22) as 𝑡→0.
Comparing the coefficient of (2.22) for 𝑡𝑗, 𝑗=0,1,…,(𝑛−1)𝑝+𝑛𝑘−2, we have 𝑚𝑖=1(𝑛−1)𝑝𝑖𝑧+𝑛𝑘𝑗𝑖−𝑠𝑖=1𝑙𝑖𝛼𝑗𝑖=0(2.23) for 𝑗=1,…,(𝑛−1)𝑝+𝑛𝑘−1.
Set 𝑧𝑚+𝑖=−𝛼𝑖 when 1≤𝑖≤𝑠. Noting that ∑𝑚𝑖=1[(𝑛−1)𝑝𝑖∑+𝑛𝑘]=𝑠𝑖=1𝑙𝑖, then it follows from (2.23) that the system of linear equations, 𝑚+𝑠𝑖=1𝑧𝑗𝑖𝑥𝑖=0,(2.24) where 0≤𝑗≤(𝑛−1)𝑝+𝑛𝑘−1, has a nonzero solution 𝑥1,…,𝑥𝑚,𝑥𝑚+1,…,𝑥𝑚+𝑠=(𝑛−1)𝑝1+𝑛𝑘,…,(𝑛−1)𝑝𝑚+𝑛𝑘,𝑙1,…,𝑙𝑠.(2.25)
If (𝑛−1)𝑝+𝑛𝑘≥𝑚+𝑠, then the determinant det(𝑧𝑗𝑖)(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 𝑧𝑖 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=1𝑝𝑖≥𝑚 that 𝑠≥𝑛𝑘+1.

Case 2. If 𝑏≠0, set 𝑏𝑚𝑖=1𝑧+𝑧𝑖𝑛(𝑝𝑖+𝑘)−𝐶1𝑚𝑖=1𝑧+𝑧𝑖(𝑛−1)𝑝𝑖+𝑛𝑘=𝑏𝑚𝑖=1𝑧+𝑧𝑖(𝑛−1)ğ‘ğ‘–ğ‘ž+𝑛𝑘𝑖=1𝑧+𝛽𝑖𝑡𝑖,(2.26) where 𝑡𝑖 are positive integers. It follows that 𝛽𝑖 (when 1â‰¤ğ‘–â‰¤ğ‘ž) and 𝑧𝑖 (when 1≤𝑖≤𝑚) are distinct complex numbers, and âˆ‘ğ‘žğ‘–=1𝑡𝑖=𝑝.

By (2.19), we have 𝑏𝑚𝑖=1𝑧+𝑧𝑖(𝑛−1)ğ‘ğ‘–ğ‘ž+𝑛𝑘𝑖=1𝑧+𝛽𝑖𝑡𝑖+𝐶2𝑠𝑖=1𝑧+𝛼𝑖𝑙𝑖𝑃=ğ‘Ž(𝑚−1)𝑘𝑛.(2.27)

It follows that 𝑚𝑖=1(𝑛−1)𝑝𝑖++ğ‘›ğ‘˜ğ‘žî“ğ‘–=1𝑡𝑖=𝑛𝑝+𝑛𝑚𝑘=𝑠𝑖=1𝑙𝑖,(2.28) and 𝐶2=−𝑏. Thus, by (2.27), 𝑚𝑖=11+𝑧𝑖𝑡(𝑛−1)ğ‘ğ‘–ğ‘ž+𝑛𝑘𝑖=11+𝛽𝑖𝑡𝑡𝑖−𝑠𝑖=11+𝛼𝑖𝑡𝑙𝑖=𝑡𝑛(𝑝+𝑘)𝑄(𝑡),(2.29) where 𝑄(𝑡)=(ğ‘Ž/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+𝑧𝑖𝑡(𝑛−1)𝑝𝑖+ğ‘›ğ‘˜âˆğ‘žğ‘–=11+𝛽𝑖𝑡𝑡𝑖∏𝑠𝑖=11+𝛼𝑖𝑡𝑙𝑖𝑡=1+𝑛(𝑝+𝑘)𝑄(𝑡)∏𝑠𝑖=11+𝛼𝑖𝑡𝑙𝑖𝑡=𝑂𝑛(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)𝑝𝑖𝑧+𝑛𝑘𝑖1+𝑧𝑖𝑡+ğ‘žî“ğ‘–=1𝑡𝑖𝛽𝑖1+𝛽𝑖𝑡−𝑠𝑖=1𝑙𝑖𝛼𝑖1+𝛼𝑖𝑡𝑡=𝑂𝑛(𝑝+𝑘)−1.(2.31)

We consider two cases.

Subcase 2.1 ({𝛼1,…,𝛼𝑠}∩{𝛽1,…,ğ›½ğ‘ž}=∅). Applying the reasoning of Case 1 and noting that ğ‘â‰¥ğ‘ž, we deduce that 𝑠≥𝑛𝑘.

Subcase 2.2 ({𝛼1,…,𝛼𝑠}∩{𝛽1,…,ğ›½ğ‘ž}≠∅). Without loss of generality, we may assume that ğ›¼ğ‘žâˆ’ğ‘–=𝛽𝑖,for(1≤𝑖≤𝑀). Denote 𝑧𝑖=âŽ§âŽªâŽ¨âŽªâŽ©ğ‘§ğ‘–ğ›½for1≤𝑖≤𝑚,𝑖−𝑚𝛼for𝑚+1≤𝑖≤𝑚+ğ‘ž,𝑀+ğ‘–âˆ’ğ‘šâˆ’ğ‘žğ‘for𝑚+ğ‘ž+1≤𝑖≤𝑚+ğ‘ž+𝑠−𝑀,𝑖=⎧⎪⎨⎪⎩(𝑛−1)𝑝𝑖𝑡+𝑛𝑘for1≤𝑖≤𝑚,𝑖−𝑚𝑡for𝑚+1≤𝑖≤𝑚+𝑠−𝑀,𝑖−𝑚−𝑙𝑖−𝑚−𝑠+𝑀𝑙for𝑚+𝑠−𝑀+1≤𝑖≤𝑚+ğ‘ž,ğ‘–âˆ’ğ‘šâˆ’ğ‘ž+𝑀for𝑚+ğ‘ž+1≤𝑖≤𝑚+ğ‘ž+𝑠−𝑀.(2.32)

The formula (2.31) can be rewritten: 𝑚+ğ‘ž+𝑠−𝑀𝑖=1𝑁𝑖𝑧𝑖1+𝑧𝑖𝑡𝑡=𝑂𝑛(𝑝+𝑘)−1.(2.33)

Applying the reasoning of Case 1, and noting that ğ‘â‰¥ğ‘ž, 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 ğ‘Žâ‰ 0 be a complex constant, and let 𝑘 be a positive integer. Then 𝑓(𝑘)âˆ’ğ‘Ž 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 𝑧𝑛,𝑧𝑛→𝑧0;(b)functions 𝑓𝑛∈ℱ;(c)positive numbers 𝜌𝑛→0+; such that 𝜌𝑛−𝛼𝑓𝑛(𝑧𝑛+𝜌𝑛𝜉)=𝑔𝑛(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 𝑧𝑛,𝑧𝑛→𝑧0;(b)functions 𝑓𝑛∈ℱ; (c)positive numbers 𝜌𝑛→0+; such that 𝑔𝑗(𝜉)=𝜌𝑗−𝑛𝑘/(𝑛−1)𝑓𝑗𝑧𝑗+𝜌𝑗𝜉⟶𝑔(𝜉),(3.1) spherically uniformly on compact subsets of ℂ, where 𝑔(𝜉) is a nonconstant meromorphic function in ℂ. Since 𝑓𝑗≠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 𝑔𝑗𝑔(𝜉)+ğ‘Žğ‘˜ğ‘—î‚(𝜉)𝑛=𝜌𝑗−𝑛𝑘/(𝑛−1)𝑓𝑗𝑧𝑗+𝜌𝑗𝜉𝑓+ğ‘Žğ‘˜ğ‘—î€·ğ‘§ğ‘—+𝜌𝑗𝜉𝑛𝑔⟶𝑔(𝜉)+ğ‘Žğ‘˜î€¸(𝜉)𝑛,(3.2) also locally uniformly with respect to the spherical metric.
We claim that 𝑔(𝜉)+ğ‘Ž(𝑔𝑘(𝜉))𝑛 has at most 𝑛𝑘 distinct zeros.
Suppose that 𝑔(𝜉)+ğ‘Ž(𝑔𝑘(𝜉))𝑛 has 𝑛𝑘+1 distinct zeros 𝜉𝑖, 1≤𝑖≤𝑛𝑘+1, and choose 𝛿(>0) small enough such that ⋂𝑛𝑘+1𝑖=1𝐷(𝜉𝑖,𝛿)=∅, where 𝐷(𝜉0,𝛿)={𝜉∣|𝜉−𝜉𝑖|<𝛿}.
From (3.2), by Hurwitz’s theorem, there exist points 𝜉𝑗𝑖∈𝐷(𝜉𝑖,𝛿) (1≤𝑖≤𝑛𝑘+1) such that for sufficiently large 𝑗, 𝑓𝑗𝑧𝑗+𝜌𝑗𝜉𝑗𝑖𝑓+ğ‘Žğ‘˜ğ‘—î€·ğ‘§ğ‘—+𝜌𝑗𝜉𝑗𝑖𝑛=0,(3.3) for 1≤𝑖≤𝑛𝑘+1.
Since 𝑧𝑗→0 and 𝜌𝑗→0+, we have 𝑧𝑗+𝜌𝑗𝜉𝑗𝑖∈𝐷(0,ğœŽ)  (ğœŽ is a positive constant) for sufficiently large 𝑗, so 𝑓𝑗(𝑧)+ğ‘Ž(𝑓𝑘𝑗(𝑧))𝑛 has 𝑛𝑘+1 distinct zeros, which contradicts the fact that 𝑓𝑗(𝑧)+ğ‘Ž(𝑓𝑘𝑗(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 𝑧𝑛,𝑧𝑛→𝑧0;(b)functions 𝑓𝑛∈ℱ; (c)positive numbers 𝜌𝑛→0+; such that 𝑔𝑗(𝜉)=𝜌𝑗−𝑘𝑓𝑗𝑧𝑗+𝜌𝑗𝜉⟶𝑔(𝜉)(3.4) spherically uniformly on compact subsets of ℂ, where 𝑔(𝜉) is a nonconstant meromorphic function in ℂ. Since 𝑓𝑗≠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 𝜌𝑘𝑗𝑔𝑗𝑔(𝜉)+ğ‘Žğ‘˜ğ‘—î‚(𝜉)ğ‘›î€·ğ‘”âˆ’ğ‘âŸ¶ğ‘Žğ‘˜î€¸(𝜉)𝑛−𝑏(3.5) also locally uniformly with respect to the spherical metric.
Noting that 𝜌𝑘𝑗𝑔𝑗𝑔(𝜉)+ğ‘Žğ‘˜ğ‘—î‚(𝜉)𝑛−𝑏=𝑓𝑗𝑧𝑗+𝜌𝑗𝜉𝑓+ğ‘Žğ‘˜ğ‘—î€·ğ‘§ğ‘—+𝜌𝑗𝜉𝑛−𝑏,(3.6) we claim that ğ‘Ž(𝑔𝑘(𝜉))𝑛−𝑏 has at most 𝑛𝑘 distinct zeros.
Suppose that 𝑔(𝜉)+ğ‘Ž(𝑔𝑘(𝜉))𝑛−𝑏 has 𝑛𝑘+1 distinct zeros 𝜉𝑖, 1≤𝑖≤𝑛𝑘+1, and choose 𝛿(>0) small enough such that ⋂𝑛𝑘+1𝑖=1𝐷(𝜉𝑖,𝛿)=∅, where 𝐷(𝜉0,𝛿)={𝜉∣|𝜉−𝜉𝑖|<𝛿}.
From (3.2), by Hurwitz’s theorem, there exist points 𝜉𝑗𝑖∈𝐷(𝜉𝑖,𝛿) (1≤𝑖≤𝑛𝑘+1) such that for sufficiently large 𝑗𝑓𝑗𝑧𝑗+𝜌𝑗𝜉𝑗𝑖𝑓+ğ‘Žğ‘˜ğ‘—î€·ğ‘§ğ‘—+𝜌𝑗𝜉𝑗𝑖𝑛−𝑏=0,(3.7) for 1≤𝑖≤𝑛𝑘+1.
Since 𝑧𝑗→0 and 𝜌𝑗→0+, we have 𝑧𝑗+𝜌𝑗𝜉𝑗𝑖∈𝐷(0,ğœŽ)  (ğœŽ is a positive constant) for sufficiently large 𝑗, so 𝑓𝑗(𝑧)+ğ‘Ž(𝑓𝑘𝑗(𝑧))𝑛−𝑏 has 𝑛𝑘+1 distinct zeros, which contradicts the fact that 𝑓𝑗(𝑧)+ğ‘Ž(𝑓𝑘𝑗(𝑧))𝑛−𝑏 has at most 𝑛𝑘 zero.
Denote 𝑐1,𝑐2,…,𝑐𝑛 by the different roots of 𝜔𝑛=𝑏/ğ‘Ž, then ğ‘Žî€·ğ‘”ğ‘˜î€¸(𝜉)𝑛−𝑏=ğ‘Žğ‘›î‘ğ‘–=1𝑔𝑘(𝜉)−𝑐𝑖.(3.8)Subcase 2.1 (If 𝑔(𝜉) is a rational function). By Lemma 2.4 and (3.8), we can deduce that ğ‘Ž(𝑔𝑘(𝜉))𝑛−𝑏 has at least 𝑛𝑘+𝑛 distinct zeros. This contradicts the claim that ğ‘Ž(𝑔𝑘(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1𝑟,𝑔(𝑘)−𝑐𝑖+𝑆𝑟,𝑔(𝑘)=−𝑁𝑟,𝑔(𝑘)+−𝑁1𝑟,ğ‘Žî€·ğ‘”(𝑘)𝑛−𝑏+𝑆𝑟,𝑔(𝑘)≤1𝑁𝑘+1𝑟,𝑔(𝑘)+𝑆𝑟,𝑔(𝑘)≤1𝑇𝑘+1𝑟,𝑔(𝑘)+𝑆𝑟,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.

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Copyright © 2012 Yuntong Li. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


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