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Abstract and Applied Analysis
VolumeΒ 2012, Article IDΒ 908123, 12 pages
http://dx.doi.org/10.1155/2012/908123
Research Article

Normal Families of Zero-Free Meromorphic Functions

Department of Basic Courses, Shaanxi Railway Institute, Weinan 714000, Shaanxi, China

Received 26 March 2012; Accepted 7 August 2012

Academic Editor: Sergey V.Β Zelik

Copyright Β© 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.

Abstract

Let π‘Ž(β‰ 0),π‘βˆˆβ„‚, and 𝑛 and π‘˜ be two positive integers such that 𝑛β‰₯2. Let β„± be a family of zero-free meromorphic functions defined in a domain π’Ÿ such that for each π‘“βˆˆβ„±, 𝑓+π‘Ž(𝑓(π‘˜))π‘›βˆ’π‘ has at most π‘›π‘˜ zeros, ignoring multiplicity. Then β„± is normal in π’Ÿ.

1. Introduction and Main Results

Let π’Ÿ be a domain in β„‚, and let β„± be a family of meromorphic functions defined in the domain π’Ÿ. β„± is said to be normal in π’Ÿ, in the sense of Montel, if for every sequence {𝑓𝑛}βŠ†β„± contains a subsequence {𝑓𝑛𝑗} such that 𝑓𝑛𝑗 converges spherically uniformly on compact subsets of π’Ÿ (see [1, Definition  3.1.1]).

β„± is said to be normal at a point 𝑧0βˆˆπ’Ÿ if there exists a neighborhood of 𝑧0 in which β„± is normal. It is well known that β„± is normal in a domain π’Ÿ 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(π‘Ÿ,𝑓) is Nevanlinna’s characteristic function and 𝑆(π‘Ÿ,𝑓) denotes a function with the property 𝑆(π‘Ÿ,𝑓)=π‘œ(𝑇(π‘Ÿ,𝑓)) as π‘Ÿβ†’βˆž (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 π‘Ž be a nonzero finite complex number, and let 𝑛 be a positive integer. If 𝑛β‰₯5, then 𝑓′+π‘Žπ‘“π‘› assumes each value π‘βˆˆπΆ 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 π‘Žβ‰ 0 is a finite complex number and 𝑛β‰₯3 is an positive integer, then 𝑓+π‘Žπ‘“β€²π‘› 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 β„± be a family of meromorphic functions defined in a domain π’Ÿ, 𝑓≠𝑏 and 𝑓+π‘Žπ‘“β€²π‘›β‰ π‘ for every π‘“βˆˆβ„±, where 𝑛β‰₯2 is an integer and π‘Žβ‰ 0,𝑏 are two finite complex numbers. Then, β„± 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 β„± are of multiplicity at least 2. Actually. they obtained the following results.

Theorem D. Let 𝑓 be a transcendental meromorphic function. If π‘Žβ‰ 0 is a finite complex number and 𝑛β‰₯2 is an positive integer, then 𝑓+π‘Žπ‘“β€²π‘› assumes all finite complex number infinitely often.

Theorem E. Let β„± be a family of meromorphic functions on the plane domain π’Ÿ, let 𝑛β‰₯2 be a positive integer, and let π‘Žβ‰ 0,𝑏 be complex numbers. If, for each π‘“βˆˆβ„±, all zeros of 𝑓 are multiple and 𝑓+π‘Žπ‘“β€²π‘›β‰ π‘ on D, then β„± is normal on D.

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

Theorem F. Let π‘Ž(β‰ 0),π‘βˆˆβ„‚ and 𝑛 and π‘˜ be two positive integers such that 𝑛β‰₯π‘˜+1. Let β„± be a family of meromorphic functions defined on a domain π’Ÿ. If, for every function π‘“βˆˆβ„±, 𝑓 has only zeros of multiplicity at least π‘˜+1, and 𝑓+π‘Ž(𝑓(π‘˜))𝑛≠𝑏 in D, then β„± 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+π‘Ž(𝑓(π‘˜))π‘›βˆ’π‘ to have zeros but restricting their numbers.

Theorem 1.1. Let π‘Ž(β‰ 0),π‘βˆˆβ„‚, and 𝑛 and π‘˜ be two positive integers such that 𝑛β‰₯2. Let β„± be a family of zero-free meromorphic functions defined in a domain π’Ÿ such that for each π‘“βˆˆβ„±, 𝑓+π‘Ž(𝑓(π‘˜))π‘›βˆ’π‘ has at most π‘›π‘˜ zeros, ignoring multiplicity. Then, β„± is normal in π’Ÿ.

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

Example 1.3. Let π’Ÿ={π‘§βˆΆ|𝑧|<1}. Let β„±={π‘“π‘š}, where π‘“π‘šβˆΆ=π‘’π‘šπ‘§. Then, π‘“π‘š+π‘Žπ‘“ξ…žπ‘š=(1+π‘Žπ‘š)π‘’π‘šπ‘§β‰ 0 in π’Ÿ for every function π‘“βˆˆβ„±. However, it is easily obtained that β„± is not normal at the point 𝑧=0.

Example 1.4. Let π’Ÿ={π‘§βˆΆ|𝑧|<1}. Let β„±={π‘“π‘š}, where π‘“π‘šβˆΆ=1/π‘šπ‘§. Then, π‘“π‘š+π‘Ž(π‘“ξ…žπ‘š)2=(π‘šπ‘§3+1)/π‘š2𝑧4 has 3 zeros in π’Ÿ for every function π‘“βˆˆβ„±. However, it is easily obtained that β„± is not normal at the point 𝑧=0.

Example 1.5. Let π’Ÿ={π‘§βˆΆ|𝑧|<1}. Let β„±={π‘“π‘š}, where π‘“π‘šβˆΆ=π‘šπ‘§. It follows that π‘“π‘š+π‘Ž(π‘“ξ…žπ‘š)2=π‘šπ‘§+π‘š2 has no zero in π’Ÿ for every function π‘“βˆˆβ„±. However, it is easily obtained that β„± is not normal at the point 𝑧=0.

Examples 1.3 and 1.4 show that the conditions that 𝑛β‰₯2 and 𝑓+π‘Ž(𝑓(π‘˜))π‘›βˆ’π‘ have at most π‘›π‘˜ 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 𝑛β‰₯2,π‘˜ be positive integers, let π‘Ž be a nonzero constant and let 𝑃(𝑧) be a polynomial. Then, the solution of the differential equation π‘Ž(π‘Š(π‘˜)(𝑧))𝑛+π‘Š(𝑧)=𝑃(𝑧) must be polynomial.

Lemma 2.2. Let 𝑓 be a nonzero transcendental meromorphic function. If π‘Ž be a nonzero finite complex number and let 𝑛β‰₯2 and π‘˜ be two positive integers. Then, 𝑓+π‘Ž(𝑓(π‘˜))𝑛 assumes each value π‘βˆˆβ„‚ infinitely often.

Proof. Set 𝑓𝐹=𝑓+π‘Ž(π‘˜)ξ€Έπ‘›βˆ’π‘,(2.1)πœ™=𝐹′𝐹=𝑓𝑓′+π‘Žπ‘›(π‘˜)ξ€Έπ‘›βˆ’1𝑓(π‘˜+1)𝑓𝑓+π‘Ž(π‘˜)𝑛,π‘“βˆ’π‘(2.2)πœ“=𝑛(π‘˜+1)𝑓(π‘˜)βˆ’πΉξ…žπΉ=𝑛𝑓(π‘˜+1)π‘“βˆ’π‘π‘›π‘“(π‘˜)βˆ’π‘“ξ…žπ‘“(π‘˜)𝑓(π‘˜)𝑓𝑓+π‘Ž(π‘˜)𝑛.βˆ’π‘(2.3)
We claim that πœ™πœ“β‰’0. If πœ™β‰‘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 πœ“β‰‘0, from (2.3), we can obtain 𝑐𝑓(π‘˜)𝑛𝑓=𝑓+π‘Ž(π‘˜)ξ€Έπ‘›βˆ’π‘,(2.4) where 𝑐 is a finite complex number, that is, 𝑓(π‘Žβˆ’π‘)(π‘˜)𝑛+𝑓=𝑏.(2.5) If π‘Žβˆ’π‘=0, we can get that 𝑓≑𝑏, which is a contradiction. If π‘Žβˆ’π‘β‰ 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(π‘Ÿ,𝐹)=𝑂(𝑇(π‘Ÿ,𝑓)). Thus, from (2.2) and (2.3), we have π‘š(π‘Ÿ,πœ™)=𝑆(π‘Ÿ,𝑓),π‘š(π‘Ÿ,πœ“)=𝑆(π‘Ÿ,𝑓).(2.6)
It follows from (2.2), (2.3) and Nevanlinna’s First Fundamental Theorem that 𝑁1π‘Ÿ,πœ™ξ‚Άξ‚΅1β©½π‘š(π‘Ÿ,πœ™)+𝑁(π‘Ÿ,πœ™)βˆ’π‘šπ‘Ÿ,πœ™ξ‚Ά+O(1)⩽𝑁(π‘Ÿ,πœ™)+𝑆(π‘Ÿ,𝑓)β©½βˆ’π‘(π‘Ÿ,𝑓)+βˆ’π‘ξ‚€1π‘Ÿ,𝐹𝑁1+𝑆(π‘Ÿ,𝑓),(2.7)π‘Ÿ,πœ“ξ‚Άξ‚΅1β©½π‘š(π‘Ÿ,πœ“)+𝑁(π‘Ÿ,πœ“)βˆ’π‘šπ‘Ÿ,πœ“ξ‚Ά+O(1)⩽𝑁(π‘Ÿ,πœ“)+𝑆(π‘Ÿ,𝑓)β©½βˆ’π‘ξ‚΅1π‘Ÿ,𝑓(π‘˜)ξ‚Άξ‚€1+π‘π‘Ÿ,𝐹+𝑆(π‘Ÿ,𝑓).(2.8) By (2.2) and (2.3), we get πœ™(π‘“βˆ’π‘)βˆ’π‘“ξ…žξ€·π‘“=π‘Ž(π‘˜)ξ€Έπ‘›πœ“.(2.9) We have by (2.6)-(2.7) π‘‡ξ€·π‘Ÿ,πœ™(π‘“βˆ’π‘)βˆ’π‘“ξ…žξ€Έξ‚΅ξ‚΅π‘“=π‘‡π‘Ÿ,(π‘“βˆ’π‘)πœ™βˆ’ξ…žξ‚΅π‘“π‘“βˆ’π‘ξ‚Άξ‚Άβ©½π‘‡(π‘Ÿ,π‘“βˆ’π‘)+π‘‡π‘Ÿ,πœ™βˆ’ξ…žξ‚Άξ‚΅π‘“π‘“βˆ’π‘+𝑆(π‘Ÿ,𝑓)β©½π‘š(π‘Ÿ,π‘“βˆ’π‘)+𝑁(π‘Ÿ,π‘“βˆ’π‘)+π‘šπ‘Ÿ,πœ™βˆ’ξ…žξ‚Άξ‚΅π‘“π‘“βˆ’π‘+π‘π‘Ÿ,πœ™βˆ’ξ…žξ‚Άξ‚΅π‘“π‘“βˆ’π‘+𝑆(π‘Ÿ,𝑓)β©½π‘š(π‘Ÿ,𝑓)+𝑁(π‘Ÿ,𝑓)+π‘š(π‘Ÿ,πœ™)+π‘šπ‘Ÿ,ξ…žξ‚Άξ‚΅π‘“π‘“βˆ’π‘+π‘π‘Ÿ,πœ™βˆ’ξ…žξ‚Άπ‘“βˆ’π‘+𝑆(π‘Ÿ,𝑓)⩽𝑇(π‘Ÿ,𝑓)+βˆ’π‘(π‘Ÿ,𝑓)+βˆ’π‘ξ‚€1π‘Ÿ,𝐹+𝑆(π‘Ÿ,𝑓).(2.10)
It follows from (2.6)–(2.10) that ξ€·π‘›π‘‡π‘Ÿ,𝑓(π‘˜)⩽𝑇(π‘Ÿ,πœ“)+π‘‡π‘Ÿ,πœ™(π‘“βˆ’π‘)βˆ’π‘“ξ…žξ€Έ+𝑆(π‘Ÿ,𝑓)β©½π‘š(π‘Ÿ,πœ“)+𝑁(π‘Ÿ,πœ“)+𝑇(π‘Ÿ,𝑓)+βˆ’π‘ξ‚€1(π‘Ÿ,𝑓)+π‘π‘Ÿ,𝐹⩽+𝑆(π‘Ÿ,𝑓)βˆ’π‘ξ‚΅1π‘Ÿ,𝑓(π‘˜)ξ‚Άξ‚€1+π‘π‘Ÿ,𝐹1+π‘šπ‘Ÿ,𝑓1+π‘π‘Ÿ,𝑓+βˆ’π‘+(π‘Ÿ,𝑓)βˆ’π‘ξ‚€1π‘Ÿ,𝐹⩽+𝑆(π‘Ÿ,𝑓)βˆ’π‘ξ‚΅1π‘Ÿ,𝑓(π‘˜)ξ‚Άξ‚€1+2π‘π‘Ÿ,𝐹𝑓+π‘šπ‘Ÿ,(π‘˜)𝑓1+π‘šπ‘Ÿ,𝑓(π‘˜)ξ‚Άξ‚΅1+π‘π‘Ÿ,𝑓+βˆ’π‘ξ‚΅1(π‘Ÿ,𝑓)+𝑆(π‘Ÿ,𝑓)β©½π‘‡π‘Ÿ,𝑓(π‘˜)ξ‚Άξ‚€1+2π‘π‘Ÿ,𝐹1+π‘π‘Ÿ,𝑓+βˆ’π‘ξ€·(π‘Ÿ,𝑓)+𝑆(π‘Ÿ,𝑓)β©½π‘‡π‘Ÿ,𝑓(π‘˜)ξ€Έξ‚€1+2π‘π‘Ÿ,𝐹1+π‘π‘Ÿ,𝑓+βˆ’π‘(π‘Ÿ,𝑓)+𝑆(π‘Ÿ,𝑓).(2.11)
So, we have ξ€·(π‘›βˆ’1)π‘‡π‘Ÿ,𝑓(π‘˜)ξ€Έξ‚€1β©½2π‘π‘Ÿ,𝐹1+π‘π‘Ÿ,𝑓+βˆ’π‘(π‘Ÿ,𝑓)+𝑆(π‘Ÿ,𝑓).(2.12)
We have ξ€·(π‘›βˆ’1)π‘‡π‘Ÿ,𝑓(π‘˜)ξ€Έξ€·β‰₯(π‘›βˆ’1)π‘π‘Ÿ,𝑓(π‘˜)ξ€Έβ‰₯(π‘›βˆ’1)𝑁(π‘Ÿ,𝑓)+(π‘›βˆ’1)βˆ’π‘(π‘Ÿ,𝑓).(2.13)
Since 𝑓≠0, if 𝑓+π‘Ž(𝑓(π‘˜))𝑛 assumes the value 𝑏 only finitely often, we by (2.12) can get 𝑁(π‘Ÿ,𝑓)=𝑆(π‘Ÿ,𝑓).(2.14) Hence, ξ€·(π‘›βˆ’1)π‘‡π‘Ÿ,𝑓(π‘˜)ξ€Έξ‚€1β©½2π‘π‘Ÿ,𝐹+𝑆(π‘Ÿ,𝑓).(2.15) So 𝑓+π‘Ž(𝑓(π‘˜))𝑛 assumes each value π‘βˆˆβ„‚ 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, 𝑛β‰₯2,  let π‘˜ be two positive integers, and π‘Žβ‰ 0,𝑏 be two complex constants. Then, the function 𝑓+π‘Ž(𝑓(π‘˜))π‘›βˆ’π‘ has at least π‘›π‘˜+1 distinct zeros in β„‚.

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 𝐢𝑓(𝑧)=1βˆπ‘šπ‘–=1𝑧+𝑧𝑖𝑝𝑖,(2.16) where 𝐢1 is a nonzero constant, π‘š and 𝑝𝑖 are positive integers, the 𝑧𝑖 (when 1β‰€π‘–β‰€π‘š) are distinct complex numbers, and denote βˆ‘π‘=π‘šπ‘–=1𝑝𝑖.
By induction, we deduce from (2.16) that 𝑓(π‘˜)𝑃(𝑧)=(π‘šβˆ’1)π‘˜βˆπ‘šπ‘–=1𝑧+𝑧𝑖𝑝𝑖+π‘˜,(2.17) where 𝑃(π‘šβˆ’1)π‘˜ is polynomial of degree (π‘šβˆ’1)π‘˜.
So the degree of numerator of the function 𝑓+π‘Ž(𝑓(π‘˜))𝑛 is equal to βˆ‘π‘šπ‘–=1(π‘›βˆ’1)𝑝𝑖+π‘›π‘˜. By calculation, 𝑓+π‘Ž(𝑓(π‘˜))π‘›βˆ’π‘ has at least one zero in β„‚. Thus, we can write 𝑓𝑓+π‘Ž(π‘˜)ξ€Έπ‘›πΆβˆ’π‘=2βˆπ‘ π‘–=1𝑧+π›Όπ‘–ξ€Έπ‘™π‘–βˆπ‘šπ‘–=1𝑧+𝑧𝑖𝑛(𝑝𝑖+π‘˜),(2.18) where 𝐢2 is a nonzero constant, 𝑙𝑖 are positive integers, 𝛼𝑖 (when 1≀𝑖≀𝑠), and 𝑧𝑖 (when 1β‰€π‘–β‰€π‘š) are distinct complex numbers. Thus, by (2.16), (2.17), and (2.18), we get 𝐢1π‘šξ‘π‘–=1𝑧+𝑧𝑖(π‘›βˆ’1)𝑝𝑖+π‘›π‘˜ξ€·π‘ƒ+π‘Ž(π‘šβˆ’1)π‘˜ξ€Έπ‘›=π‘π‘šξ‘π‘–=1𝑧+𝑧𝑖𝑛(𝑝𝑖+π‘˜)+𝐢2𝑠𝑖=1𝑧+𝛼𝑖𝑙𝑖.(2.19)Case 1. If 𝑏=0, it follows that βˆ‘π‘šπ‘–=1[(π‘›βˆ’1)π‘π‘–βˆ‘+π‘›π‘˜]=𝑠𝑖=1𝑙𝑖 and 𝐢1=𝐢2. Thus, it follows from (2.19) that π‘šξ‘π‘–=1ξ€·1+𝑧𝑖𝑑(π‘›βˆ’1)𝑝𝑖+π‘›π‘˜βˆ’π‘ ξ‘π‘–=1ξ€·1+𝛼𝑖𝑑𝑙𝑖=𝑑(π‘›βˆ’1)𝑝+π‘›π‘˜π‘„(𝑑),(2.20) where 𝑄(𝑑)=(βˆ’π‘Ž/𝐢1)𝑑(π‘šβˆ’1)π‘›π‘˜(𝑃(π‘šβˆ’1)π‘˜(1/𝑑))𝑛 is a polynomial. Then, 𝑄(𝑑) is a polynomial of degree less than (π‘šβˆ’1)π‘›π‘˜, and it follows that βˆπ‘šπ‘–=1ξ€·1+𝑧𝑖𝑑(π‘›βˆ’1)𝑝𝑖+π‘›π‘˜βˆπ‘ π‘–=1ξ€·1+𝛼𝑖𝑑𝑙𝑖𝑑=1+(π‘›βˆ’1)𝑝+π‘›π‘˜π‘„(𝑑)βˆπ‘ π‘–=1ξ€·1+𝛼𝑖𝑑𝑙𝑖𝑑=1+𝑂(π‘›βˆ’1)𝑝+π‘›π‘˜ξ€Έ(2.21) as 𝑑→0.
Logarithmic differentiation of both sides of (2.21) shows that π‘šξ“π‘–=1ξ€·(π‘›βˆ’1)𝑝𝑖𝑧+π‘›π‘˜π‘–1+π‘§π‘–π‘‘βˆ’π‘ ξ“π‘–=1𝑙𝑖𝛼𝑖1+𝛼𝑖𝑑𝑑=𝑂(π‘›βˆ’1)𝑝+π‘›π‘˜βˆ’1ξ€Έ(2.22) as 𝑑→0.
Comparing the coefficient of (2.22) for 𝑑𝑗, 𝑗=0,1,…,(π‘›βˆ’1)𝑝+π‘›π‘˜βˆ’2, we have π‘šξ“π‘–=1ξ€·(π‘›βˆ’1)𝑝𝑖𝑧+π‘›π‘˜π‘—π‘–βˆ’π‘ ξ“π‘–=1𝑙𝑖𝛼𝑗𝑖=0(2.23) for 𝑗=1,…,(π‘›βˆ’1)𝑝+π‘›π‘˜βˆ’1.
Set π‘§π‘š+𝑖=βˆ’π›Όπ‘– when 1≀𝑖≀𝑠. Noting that βˆ‘π‘šπ‘–=1[(π‘›βˆ’1)π‘π‘–βˆ‘+π‘›π‘˜]=𝑠𝑖=1𝑙𝑖, then it follows from (2.23) that the system of linear equations, π‘š+𝑠𝑖=1𝑧𝑗𝑖π‘₯𝑖=0,(2.24) where 0≀𝑗≀(π‘›βˆ’1)𝑝+π‘›π‘˜βˆ’1, has a nonzero solution ξ€·π‘₯1,…,π‘₯π‘š,π‘₯π‘š+1,…,π‘₯π‘š+𝑠=ξ€·(π‘›βˆ’1)𝑝1+π‘›π‘˜,…,(π‘›βˆ’1)π‘π‘š+π‘›π‘˜,𝑙1,…,𝑙𝑠.(2.25)
If (π‘›βˆ’1)𝑝+π‘›π‘˜β‰₯π‘š+𝑠, then the determinant det(𝑧𝑗𝑖)(π‘š+𝑠)Γ—(π‘š+𝑠) of the coefficients of the system of (2.24), where 0≀𝑗≀(π‘›βˆ’1)𝑝+π‘›π‘˜βˆ’1, is equal to zero, by Cramer’s rule (see, e.g., [11]). However, the 𝑧𝑖 are distinct complex numbers when 1β‰€π‘–β‰€π‘š+𝑠, and the determinant is a Vandermonde determinant, so it cannot be 0 (see [11]), which is a contradiction.
Hence, we conclude that (π‘›βˆ’1)𝑝+π‘›π‘˜<π‘š+𝑠. Noting that 𝑛β‰₯2, it follows from this and βˆ‘π‘=π‘šπ‘–=1𝑝𝑖β‰₯π‘š that 𝑠β‰₯π‘›π‘˜+1.

Case 2. If 𝑏≠0, set π‘π‘šξ‘π‘–=1𝑧+𝑧𝑖𝑛(𝑝𝑖+π‘˜)βˆ’πΆ1π‘šξ‘π‘–=1𝑧+𝑧𝑖(π‘›βˆ’1)𝑝𝑖+π‘›π‘˜=π‘π‘šξ‘π‘–=1𝑧+𝑧𝑖(π‘›βˆ’1)π‘π‘–π‘ž+π‘›π‘˜ξ‘π‘–=1𝑧+𝛽𝑖𝑑𝑖,(2.26) where 𝑑𝑖 are positive integers. It follows that 𝛽𝑖 (when 1β‰€π‘–β‰€π‘ž) and 𝑧𝑖 (when 1β‰€π‘–β‰€π‘š) are distinct complex numbers, and βˆ‘π‘žπ‘–=1𝑑𝑖=𝑝.

By (2.19), we have π‘π‘šξ‘π‘–=1𝑧+𝑧𝑖(π‘›βˆ’1)π‘π‘–π‘ž+π‘›π‘˜ξ‘π‘–=1𝑧+𝛽𝑖𝑑𝑖+𝐢2𝑠𝑖=1𝑧+𝛼𝑖𝑙𝑖𝑃=π‘Ž(π‘šβˆ’1)π‘˜ξ€Έπ‘›.(2.27)

It follows that π‘šξ“π‘–=1ξ€Ί(π‘›βˆ’1)𝑝𝑖++π‘›π‘˜π‘žξ“π‘–=1𝑑𝑖=𝑛𝑝+π‘›π‘šπ‘˜=𝑠𝑖=1𝑙𝑖,(2.28) and 𝐢2=βˆ’π‘. Thus, by (2.27), π‘šξ‘π‘–=1ξ€·1+𝑧𝑖𝑑(π‘›βˆ’1)π‘π‘–π‘ž+π‘›π‘˜ξ‘π‘–=1ξ€·1+π›½π‘–π‘‘ξ€Έπ‘‘π‘–βˆ’π‘ ξ‘π‘–=1ξ€·1+𝛼𝑖𝑑𝑙𝑖=𝑑𝑛(𝑝+π‘˜)𝑄(𝑑),(2.29) where 𝑄(𝑑)=(π‘Ž/𝑏)𝑑(π‘šβˆ’1)π‘›π‘˜(𝑃(π‘šβˆ’1)π‘˜(1/𝑑))𝑛 is a polynomial. Then, 𝑄(𝑑) is a polynomial of degree less than (π‘šβˆ’1)π‘›π‘˜, and it follows that βˆπ‘šπ‘–=1ξ€·1+𝑧𝑖𝑑(π‘›βˆ’1)𝑝𝑖+π‘›π‘˜βˆπ‘žπ‘–=1ξ€·1+π›½π‘–π‘‘ξ€Έπ‘‘π‘–βˆπ‘ π‘–=1ξ€·1+𝛼𝑖𝑑𝑙𝑖𝑑=1+𝑛(𝑝+π‘˜)𝑄(𝑑)βˆπ‘ π‘–=1ξ€·1+𝛼𝑖𝑑𝑙𝑖𝑑=𝑂𝑛(𝑝+π‘˜)ξ€Έ(2.30) as 𝑑→0.

Thus, by taking logarithmic derivatives of both sides of (2.12), we get π‘šξ“π‘–=1ξ€·(π‘›βˆ’1)𝑝𝑖𝑧+π‘›π‘˜π‘–1+𝑧𝑖𝑑+π‘žξ“π‘–=1𝑑𝑖𝛽𝑖1+π›½π‘–π‘‘βˆ’π‘ ξ“π‘–=1𝑙𝑖𝛼𝑖1+𝛼𝑖𝑑𝑑=𝑂𝑛(𝑝+π‘˜)βˆ’1ξ€Έ.(2.31)

We consider two cases.

Subcase 2.1 ({𝛼1,…,𝛼𝑠}∩{𝛽1,…,π›½π‘ž}=βˆ…). Applying the reasoning of Case 1 and noting that 𝑝β‰₯π‘ž, we deduce that 𝑠β‰₯π‘›π‘˜.

Subcase 2.2 ({𝛼1,…,𝛼𝑠}∩{𝛽1,…,π›½π‘ž}β‰ βˆ…). Without loss of generality, we may assume that π›Όπ‘žβˆ’π‘–=𝛽𝑖,for(1≀𝑖≀𝑀). Denote 𝑧𝑖=⎧βŽͺ⎨βŽͺβŽ©π‘§π‘–π›½for1β‰€π‘–β‰€π‘š,π‘–βˆ’π‘šπ›Όforπ‘š+1β‰€π‘–β‰€π‘š+π‘ž,𝑀+π‘–βˆ’π‘šβˆ’π‘žπ‘forπ‘š+π‘ž+1β‰€π‘–β‰€π‘š+π‘ž+π‘ βˆ’π‘€,𝑖=⎧βŽͺ⎨βŽͺ⎩(π‘›βˆ’1)𝑝𝑖𝑑+π‘›π‘˜for1β‰€π‘–β‰€π‘š,π‘–βˆ’π‘šπ‘‘forπ‘š+1β‰€π‘–β‰€π‘š+π‘ βˆ’π‘€,π‘–βˆ’π‘šβˆ’π‘™π‘–βˆ’π‘šβˆ’π‘ +𝑀𝑙forπ‘š+π‘ βˆ’π‘€+1β‰€π‘–β‰€π‘š+π‘ž,π‘–βˆ’π‘šβˆ’π‘ž+𝑀forπ‘š+π‘ž+1β‰€π‘–β‰€π‘š+π‘ž+π‘ βˆ’π‘€.(2.32)

The formula (2.31) can be rewritten: π‘š+π‘ž+π‘ βˆ’π‘€ξ“π‘–=1𝑁𝑖𝑧𝑖1+𝑧𝑖𝑑𝑑=𝑂𝑛(𝑝+π‘˜)βˆ’1ξ€Έ.(2.33)

Applying the reasoning of Case 1, and noting that 𝑝β‰₯π‘ž, we deduce that 𝑠β‰₯π‘›π‘˜+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 π‘Žβ‰ 0 be a complex constant, and let π‘˜ be a positive integer. Then 𝑓(π‘˜)βˆ’π‘Ž has at least π‘˜+1 distinct zeros in β„‚.

Lemma 2.5 (see [12], Lemma  2, Zalcman’s lemma). Let β„± be a family of functions meromorphic on a domain π’Ÿ, all of whose zeros have multiplicity at least π‘˜. Suppose that there exists 𝐴⩾1 such that |𝑓(π‘˜)(𝑧)|⩽𝐴 whenever 𝑓(𝑧)=0. Then, if β„± is not normal at 𝑧0βˆˆπ’Ÿ, there exist, for each 0β©½π›Όβ©½π‘˜,(a)points 𝑧𝑛,𝑧𝑛→𝑧0;(b)functions π‘“π‘›βˆˆβ„±;(c)positive numbers πœŒπ‘›β†’0+; such that πœŒπ‘›βˆ’π›Όπ‘“π‘›(𝑧𝑛+πœŒπ‘›πœ‰)=𝑔𝑛(πœ‰)→𝑔(πœ‰) locally uniformly with respect to the spherical metric, where 𝑔(πœ‰) is a nonconstant meromorphic function on β„‚, all of whose zeros of 𝑔(πœ‰) are of multiplicity at least π‘˜, such that 𝑔#(πœ‰)≀𝑔#(0)=π‘˜π΄+1.

Here, as usual, 𝑔#(πœ‰)=|𝑔′(πœ‰)|/(1+|𝑔(πœ‰)|2) is the spherical derivative.

3. Proof of Theorem

Suppose that β„± is not normal in π’Ÿ. Then, there exists at least one point 𝑧0 such that β„± is not normal at the point 𝑧0βˆˆπ’Ÿ. 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 π‘“π‘›βˆˆβ„±; (c)positive numbers πœŒπ‘›β†’0+; such that 𝑔𝑗(πœ‰)=πœŒπ‘—βˆ’π‘›π‘˜/(π‘›βˆ’1)𝑓𝑗𝑧𝑗+πœŒπ‘—πœ‰ξ€ΈβŸΆπ‘”(πœ‰),(3.1) spherically uniformly on compact subsets of β„‚, where 𝑔(πœ‰) is a nonconstant meromorphic function in β„‚. Since 𝑓𝑗≠0, by Hurwitz’s theorem, it implies that 𝑔(πœ‰)β‰ 0.
On every compact subset of β„‚ which contains no poles of 𝑔, from (3.1), we get 𝑔𝑗𝑔(πœ‰)+π‘Žπ‘˜π‘—ξ‚(πœ‰)𝑛=πœŒπ‘—βˆ’π‘›π‘˜/(π‘›βˆ’1)𝑓𝑗𝑧𝑗+πœŒπ‘—πœ‰ξ€Έξ€·π‘“+π‘Žπ‘˜π‘—ξ€·π‘§π‘—+πœŒπ‘—πœ‰ξ€Έξ€Έπ‘›ξ€Έξ€·π‘”βŸΆπ‘”(πœ‰)+π‘Žπ‘˜ξ€Έ(πœ‰)𝑛,(3.2) also locally uniformly with respect to the spherical metric.
We claim that 𝑔(πœ‰)+π‘Ž(π‘”π‘˜(πœ‰))𝑛 has at most π‘›π‘˜ distinct zeros.
Suppose that 𝑔(πœ‰)+π‘Ž(π‘”π‘˜(πœ‰))𝑛 has π‘›π‘˜+1 distinct zeros πœ‰π‘–, 1β‰€π‘–β‰€π‘›π‘˜+1, and choose 𝛿(>0) small enough such that β‹‚π‘›π‘˜+1𝑖=1𝐷(πœ‰π‘–,𝛿)=βˆ…, where 𝐷(πœ‰0,𝛿)={πœ‰βˆ£|πœ‰βˆ’πœ‰π‘–|<𝛿}.
From (3.2), by Hurwitz’s theorem, there exist points πœ‰π‘—π‘–βˆˆπ·(πœ‰π‘–,𝛿) (1β‰€π‘–β‰€π‘›π‘˜+1) such that for sufficiently large 𝑗, 𝑓𝑗𝑧𝑗+πœŒπ‘—πœ‰π‘—π‘–ξ€Έξ€·π‘“+π‘Žπ‘˜π‘—ξ€·π‘§π‘—+πœŒπ‘—πœ‰π‘—π‘–ξ€Έξ€Έπ‘›=0,(3.3) for 1β‰€π‘–β‰€π‘›π‘˜+1.
Since 𝑧𝑗→0 and πœŒπ‘—β†’0+, we have 𝑧𝑗+πœŒπ‘—πœ‰π‘—π‘–βˆˆπ·(0,𝜎)  (𝜎 is a positive constant) for sufficiently large 𝑗, so 𝑓𝑗(𝑧)+π‘Ž(π‘“π‘˜π‘—(𝑧))𝑛 has π‘›π‘˜+1 distinct zeros, which contradicts the fact that 𝑓𝑗(𝑧)+π‘Ž(π‘“π‘˜π‘—(𝑧))𝑛 has at most π‘›π‘˜ 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 β„± is normal in π’Ÿ.

Case 2 (𝑏≠0). By Zalcman’s lemma, there exist:(a)points 𝑧𝑛,𝑧𝑛→𝑧0;(b)functions π‘“π‘›βˆˆβ„±; (c)positive numbers πœŒπ‘›β†’0+; such that 𝑔𝑗(πœ‰)=πœŒπ‘—βˆ’π‘˜π‘“π‘—ξ€·π‘§π‘—+πœŒπ‘—πœ‰ξ€ΈβŸΆπ‘”(πœ‰)(3.4) spherically uniformly on compact subsets of β„‚, where 𝑔(πœ‰) is a nonconstant meromorphic function in β„‚. Since 𝑓𝑗≠0, by Hurwitz’s theorem, it implies that 𝑔(πœ‰)β‰ 0.
On every compact subset of β„‚ which contains no poles of 𝑔, from (3.4), we get πœŒπ‘˜π‘—π‘”π‘—ξ‚€π‘”(πœ‰)+π‘Žπ‘˜π‘—ξ‚(πœ‰)π‘›ξ€·π‘”βˆ’π‘βŸΆπ‘Žπ‘˜ξ€Έ(πœ‰)π‘›βˆ’π‘(3.5) also locally uniformly with respect to the spherical metric.
Noting that πœŒπ‘˜π‘—π‘”π‘—ξ‚€π‘”(πœ‰)+π‘Žπ‘˜π‘—ξ‚(πœ‰)π‘›βˆ’π‘=𝑓𝑗𝑧𝑗+πœŒπ‘—πœ‰ξ€Έξ€·π‘“+π‘Žπ‘˜π‘—ξ€·π‘§π‘—+πœŒπ‘—πœ‰ξ€Έξ€Έπ‘›βˆ’π‘,(3.6) we claim that π‘Ž(π‘”π‘˜(πœ‰))π‘›βˆ’π‘ has at most π‘›π‘˜ distinct zeros.
Suppose that 𝑔(πœ‰)+π‘Ž(π‘”π‘˜(πœ‰))π‘›βˆ’π‘ has π‘›π‘˜+1 distinct zeros πœ‰π‘–, 1β‰€π‘–β‰€π‘›π‘˜+1, and choose 𝛿(>0) small enough such that β‹‚π‘›π‘˜+1𝑖=1𝐷(πœ‰π‘–,𝛿)=βˆ…, where 𝐷(πœ‰0,𝛿)={πœ‰βˆ£|πœ‰βˆ’πœ‰π‘–|<𝛿}.
From (3.2), by Hurwitz’s theorem, there exist points πœ‰π‘—π‘–βˆˆπ·(πœ‰π‘–,𝛿) (1β‰€π‘–β‰€π‘›π‘˜+1) such that for sufficiently large 𝑗𝑓𝑗𝑧𝑗+πœŒπ‘—πœ‰π‘—π‘–ξ€Έξ€·π‘“+π‘Žπ‘˜π‘—ξ€·π‘§π‘—+πœŒπ‘—πœ‰π‘—π‘–ξ€Έξ€Έπ‘›βˆ’π‘=0,(3.7) for 1β‰€π‘–β‰€π‘›π‘˜+1.
Since 𝑧𝑗→0 and πœŒπ‘—β†’0+, we have 𝑧𝑗+πœŒπ‘—πœ‰π‘—π‘–βˆˆπ·(0,𝜎)  (𝜎 is a positive constant) for sufficiently large 𝑗, so 𝑓𝑗(𝑧)+π‘Ž(π‘“π‘˜π‘—(𝑧))π‘›βˆ’π‘ has π‘›π‘˜+1 distinct zeros, which contradicts the fact that 𝑓𝑗(𝑧)+π‘Ž(π‘“π‘˜π‘—(𝑧))π‘›βˆ’π‘ has at most π‘›π‘˜ zero.
Denote 𝑐1,𝑐2,…,𝑐𝑛 by the different roots of πœ”π‘›=𝑏/π‘Ž, then π‘Žξ€·π‘”π‘˜ξ€Έ(πœ‰)π‘›βˆ’π‘=π‘Žπ‘›ξ‘π‘–=1ξ€·π‘”π‘˜(πœ‰)βˆ’π‘π‘–ξ€Έ.(3.8)Subcase 2.1 (If 𝑔(πœ‰) is a rational function). By Lemma 2.4 and (3.8), we can deduce that π‘Ž(π‘”π‘˜(πœ‰))π‘›βˆ’π‘ has at least π‘›π‘˜+𝑛 distinct zeros. This contradicts the claim that π‘Ž(π‘”π‘˜(πœ‰))π‘›βˆ’π‘ has at most π‘›π‘˜ distinct zeros.

Subcase 2.2 (If 𝑔(πœ‰) is a transcendental meromorphic function). By Nevanlinnas second main theorem, we have π‘‡ξ€·π‘Ÿ,𝑔(π‘˜)ξ€Έβ‰€βˆ’π‘ξ€·π‘Ÿ,𝑔(π‘˜)ξ€Έ+π‘›ξ“βˆ’π‘–=1𝑁1π‘Ÿ,𝑔(π‘˜)βˆ’π‘π‘–ξ‚Άξ€·+π‘†π‘Ÿ,𝑔(π‘˜)ξ€Έ=βˆ’π‘ξ€·π‘Ÿ,𝑔(π‘˜)ξ€Έ+βˆ’π‘ξƒ©1π‘Ÿ,π‘Žξ€·π‘”(π‘˜)𝑛ξƒͺξ€·βˆ’π‘+π‘†π‘Ÿ,𝑔(π‘˜)≀1π‘ξ€·π‘˜+1π‘Ÿ,𝑔(π‘˜)ξ€Έξ€·+π‘†π‘Ÿ,𝑔(π‘˜)≀1π‘‡ξ€·π‘˜+1π‘Ÿ,𝑔(π‘˜)ξ€Έξ€·+π‘†π‘Ÿ,𝑔(π‘˜)ξ€Έ.(3.9)

It follows that 𝑇(π‘Ÿ,𝑔(π‘˜))≀𝑆(π‘Ÿ,𝑔(π‘˜)), which is a contradiction. This contradiction shows that β„± is normal in π’Ÿ.

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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