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Journal of Applied Mathematics
Volume 2012, Article ID 312324, 7 pages
http://dx.doi.org/10.1155/2012/312324
Research Article

Normal Criterion Concerning Shared Values

1School of Mathematical Sciences, Xinjiang Normal University, Xinjiang, Urumqi 830054, China
2School of Mathematical Sciences, Xiamen University, Xiamen, Fujian 361105, China

Received 12 June 2012; Accepted 21 August 2012

Academic Editor: Yansheng Liu

Copyright © 2012 Wei Chen et al. 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

We study normal criterion of meromorphic functions shared values, we obtain the following. Let 𝐹 be a family of meromorphic functions in a domain 𝐷, such that function 𝑓𝐹 has zeros of multiplicity at least 2, there exists nonzero complex numbers 𝑏𝑓,𝑐𝑓 depending on 𝑓 satisfying (i)𝑏𝑓/𝑐𝑓 is a constant;  (ii)min{𝜎(0,𝑏𝑓),𝜎(0,𝑐𝑓),𝜎(𝑏𝑓,𝑐𝑓)𝑚} for some 𝑚>0;  (iii)(1/𝑐𝑓𝑘1)(𝑓)𝑘(𝑧)+𝑓(𝑧)𝑏𝑘𝑓/𝑐𝑓𝑘1 or (1/𝑐𝑓𝑘1)(𝑓)𝑘(𝑧)+𝑓(𝑧)=𝑏𝑘𝑓/𝑐𝑓𝑘1𝑓(𝑧)=𝑏𝑓, then 𝐹 is normal. These results improve some earlier previous results.

1. Introduction and Main Results

We use 𝐶 to denote the open complex plane, 𝐶(=𝐶{}) to denote the extended complex plane and 𝐷 to denote a domain in 𝐶. A family 𝐹 of meromorphic functions defined in 𝐷𝐶 is said to be normal, if for any sequence {𝑓𝑛}𝐹 contains a subsequence which converges spherically, and locally, uniformly in 𝐷 to a meromorphic function or . Clearly 𝐹 is said to be normal in 𝐷 if and only if it is normal at every point of 𝐷 see [1].

Let 𝐷 be a domain in 𝐶. For 𝑓 meromorphic on 𝐶 and 𝑎𝐶, set 𝐸𝑓(𝑎)=𝑓1({𝑎})𝐷={𝑧𝐷𝑓(𝑧)=𝑎}.(1.1)

Two meromorphic functions 𝑓 and 𝑔 on 𝐷 are said to share the value 𝑎 if 𝐸𝑓(𝑎)=𝐸𝑔(𝑎). Let 𝑎 and 𝑏 be complex numbers. If 𝑔(𝑧)=𝑏 whenever 𝑓(𝑧)=𝑎, we write 𝑓(𝑧)=𝑎𝑔(𝑧)=𝑏.(1.2) If 𝑓(𝑧)=𝑎𝑔(𝑧)=𝑏 and 𝑔(𝑧)=𝑏𝑓(𝑧)=𝑎, we write 𝑓(𝑧)=𝑎𝑔(𝑧)=𝑏.(1.3)

According to Bloch’s principle [2], every condition which reduces a meromorphic function in the plane 𝐶 to 𝑎 constant forces a family of meromorphic functions in 𝑎 domain 𝐷 normal. Although the principle is false in general (see [3]), many authors proved normality criterion for families of meromorphic functions by starting from Liouville-Picard type theorem (see [4]). It is also more interesting to find normality criteria from the point of view of shared values. In this area, Schwick [5] first proved an interesting result that a family of meromorphic functions in a domain is normal if in which every function shares three distinct finite complex numbers with its first derivative. And later, more results about normality criteria concerning shared values have emerged [69]. In recent years, this subject has attracted the attention of many researchers worldwide.

In this paper, we use 𝜎(𝑥,𝑦) to denote the spherical distance between 𝑥 and 𝑦 and the definition of the spherical distance can be found in [10].

In 2008, Fang and Zalcman [11] proved the following results.

Theorem 1.1 (see [11]). Let 𝑓 be a transcendental function. Let 𝑎(0) and 𝑏 be complex numbers, and let 𝑛(2),𝑘 be positive integers, then 𝑓+𝑎(𝑓)𝑛 assumes every value 𝑏𝐶 infinitely often.

Theorem 1.2 (see [11]). Let 𝐹 be a transcendental function. Let 𝑎(0) and 𝑏 be complex numbers, and let 𝑛(2),𝑘 be positive integers. If for every 𝑓𝐹 has multiple zeros, and 𝑓+𝑎(𝑓)𝑛𝑏, then 𝐹 is normal in 𝐷.

In 2009, Xu et al. [12] proved the following results.

Theorem 1.3 (see [12]). Let 𝑓 be a transcendental function. Let 𝑎(0) and let 𝑏 be complex numbers, and 𝑛,𝑘 be positive integers, which satisfy 𝑛𝑘+1, then 𝑓+𝑎(𝑓(𝑘))𝑛 assumes each value 𝑏𝐶 infinitely often.

Theorem 1.4 (see [12]). Let 𝑓 be a transcendental function. Let 𝑎(0) and 𝑏 be complex numbers, and let 𝑛,𝑘 be positive integers, which satisfy 𝑛𝑘+1. If for every 𝑓𝐹 has only zeros of multiplicity at least 𝑘+1, and satisfies 𝑓+𝑎(𝑓(𝑘))𝑛𝑏, then 𝐹 is normal in 𝐷.

In Theorems 1.2 and 1.4, the constants are the same for each 𝑓𝐹. Now we will prove the condition for the constants be the same can be relaxed to some extent.

Theorem A. Let 𝐹 be a family of meromorphic functions in the unit disc Δ, and 𝑘 be a positive integer and 𝑘3. For every 𝑓𝐹, such that all zeros of 𝑓 have multiplicity at least 2, there exist finite nonzero complex numbers 𝑏𝑓,𝑐𝑓 depending on 𝑓 satisfying that(i)𝑏𝑓/𝑐𝑓 is a constant; (ii)min{𝜎(0,𝑏𝑓),𝜎(0,𝑐𝑓),𝜎(𝑏𝑓,𝑐𝑓)𝑚} for some 𝑚>0; (iii)(1/𝑐𝑓𝑘1)(𝑓)𝑘(𝑧)+𝑓(𝑧)𝑏𝑘𝑓/𝑐𝑓𝑘1.Then 𝐹 is normal in Δ.

Theorem B. Let 𝐹 be a family of meromorphic functions in the unit disc Δ, and 𝑘(3) be a positive integer. For every 𝑓𝐹, such that all zeros of 𝑓 have multiplicity at least 2, there exist finite nonzero complex numbers 𝑏𝑓,𝑐𝑓 depending on 𝑓 satisfying that(i)𝑏𝑓/𝑐𝑓is a constant; (ii)min{𝜎(0,𝑏𝑓),𝜎(0,𝑐𝑓),𝜎(𝑏𝑓,𝑐𝑓)𝑚} for some 𝑚>0; (iii)(1/𝑐𝑓𝑘1)(𝑓)𝑘(𝑧)+𝑓(𝑧)=𝑏𝑘𝑓/𝑐𝑓𝑘1𝑓(𝑧)=𝑏𝑓.Then 𝐹 is normal in Δ.

2. Some Lemmas

In order to prove our theorems, we require the following results.

Lemma 2.1 (see [7]). Let 𝐹 be a family of meromorphic functions in a domain 𝐷, and 𝑘 be a positive integer, such that each function 𝑓𝐹 has only zeros of multiplicity at least 𝑘, and suppose that there exists 𝐴1 such that |𝑓(𝑘)(𝑧)|𝐴 whenever 𝑓(𝑧)=0,𝑓𝐹. If 𝐹 is not normal at 𝑧0𝐷, then for each 0𝛼𝑘, there exist a sequence of points 𝑧𝑛𝐷,𝑧𝑛𝑧0, a sequence of positive numbers 𝜌𝑛0+, and a subsequence of functions 𝑓𝑛𝐹 such that 𝑔𝑛𝑓(𝜁)=𝑛𝑧𝑛+𝜌𝑛𝜍𝜌𝛼𝑛𝑔(𝜁)(2.1) locally uniformly with respect to the spherical metric in 𝐶, where 𝑔 is a nonconstant meromorphic function, all of whose zeros have multiplicity at least 𝑘, such that 𝑔#(𝜁)𝑔#(0)=𝑘𝐴+1. Morever,  𝑔 has order at most 2.
Here as usual, 𝑔#(𝜁)=|𝑔(𝜁)|/(1+|𝑔(𝜁)|2) is the spherical derivative.

Lemma 2.2 (see [10]). Let 𝑚 be any positive number. Then, Möbius transformation 𝑔 satisfies 𝜎(𝑔(𝑎),𝑔(𝑏))𝑚,𝜎(𝑔(𝑏),𝑔(𝑐))𝑚,𝜎(𝑔(𝑐),𝑔(𝑎))𝑚, for some constants 𝑎,𝑏, and 𝑐 also satisfy the uniform Lipschitz condition 𝜎(𝑔(𝑧),𝑔(𝑤))𝑘𝑚𝜎(𝑧,𝑤),(2.2) where 𝑘𝑚 is a constant depending on 𝑚.

3. Proof of Theorems

Proof of Theorem A. Let 𝑀=𝑏𝑓/𝑐𝑓. We can find nonzero constants 𝑏 and 𝑐 satisfying 𝑀=𝑏/𝑐. For each 𝑓𝐹, define a Möbius map 𝑔𝑓 by 𝑔𝑓=𝑐𝑓𝑧/𝑐, thus 𝑔𝑓1=𝑐𝑧/𝑐𝑓.
Next we will show 𝐺={(𝑔𝑓1𝑓)𝑓𝐹} is normal in Δ. Suppose to the contrary, 𝐺 is not normal in Δ. Then by Lemma 2.1. We can find 𝑔𝑛𝐺,𝑧𝑛Δ, and 𝜌𝑛0+, such that 𝑇𝑛(𝜁)=𝑔𝑛(𝑧𝑛+𝜌𝑛𝜁)/𝜌𝑛1/(𝑘+1) converges locally uniformly with respect to the spherical metric to a nonconstant meromorphic function 𝑇(𝜁) whose zeros of multiplicity at least 2 and spherical derivative is limited and 𝑇 has order at most 2.
We now consider three cases.
Case 1. If (1/𝑐𝑘1)(𝑇)𝑘(𝜁)𝑏𝑘/𝑐𝑘1, then 𝑇(𝜁) is a polynomial with degree at most 1, a contradiction.
Case 2. If there exists 𝜁0 such that (1/𝑐𝑘1)(𝑇)𝑘(𝜁0)=𝑏𝑘/𝑐𝑘1. Noting that 𝜌𝑛𝑇𝑛(𝜁)+(1/𝑐𝑘1)(𝑇𝑛)𝑘(𝜁)(𝑏𝑘/𝑐𝑘1)(1/𝑐𝑘1)(𝑇)𝑘(𝜁)(𝑏𝑘/𝑐𝑘1). By Hurwitz’s theorem, there exist a sequence of points 𝜁𝑛𝜁0 such that (for large enough 𝑛) 0=𝜌𝑛𝑇𝑛𝜁𝑛+1𝑐𝑘1𝑇𝑛𝑘𝜁𝑛𝑏𝑘𝑐𝑘1=𝑔𝑛𝑧𝑛+𝜌𝑛𝜁𝑛+1𝑐𝑘1𝑔𝑛𝑘𝑧𝑛+𝜁𝑛𝑏𝑘𝑐𝑘1=𝑐𝑐𝑓𝑓𝑛𝑧𝑛+𝜌𝑛𝜁𝑛+1𝑐𝑘1𝑐𝑘𝑐𝑘𝑓𝑓𝑛𝑘𝑧𝑛+𝜁𝑛𝑏𝑘𝑐𝑘1.(3.1) Hence 𝑓𝑛(𝑧𝑛+𝜌𝑛𝜁𝑛)+(1/𝑐𝑓𝑘1)(𝑓𝑛)𝑘(𝑧𝑛+𝜁𝑛)=𝑏𝑘𝑓/𝑐𝑓𝑘1. This contradicts with the suppose of Theorem A.
Case 3. If (1/𝑐𝑘1)(𝑇)𝑘(𝜁)𝑏𝑘/𝑐𝑘1. Let 𝑐1,𝑐2,,𝑐𝑘 be the solution of the equation 𝑤𝑘=𝑐𝑘, then 𝑇(𝜁)𝑐𝑖(𝑖=1,2,,𝑘). When 𝑇(𝜁) is a rational function, then 𝑇(𝜁) is also a rational function. By Picard Theorem we can deduce that 𝑇(𝜁) is a constant (𝑘3). Hence 𝑇(𝜁) is a polynomial with degree at most 1. This contradicts with 𝑇(𝜁) has zeros of multiplicity at least 2. When 𝑇(𝜁) is a transcendental function, combining with the second main theorem, we have 𝑇𝑟,𝑇𝑁𝑟,𝑇+𝑘𝑖=1𝑁1𝑟,𝑇𝑐𝑖+𝑠𝑟,𝑇𝑁𝑟,𝑇+𝑠𝑟,𝑇12𝑁𝑟,𝑇+𝑠𝑟,𝑇12𝑇𝑟,𝑇+𝑠𝑟,𝑇.(3.2) Hence, 𝑇(𝑟,𝑇)𝑠(𝑟,𝑇), a contradiction.
Hence 𝐺={(𝑔𝑓1𝑓)𝑓𝐹} is normal and equicontinuous in Δ. There given (𝜀/𝑘𝑚>0), where 𝑘𝑚 is the constant of Lemma 2.2, there exists 𝛿>0 such that for the spherical distance 𝜎(𝑥,𝑦)<𝛿, 𝜎𝑔𝑓1𝑔𝑓(𝑥),𝑓1<𝜀(𝑦)𝑘𝑚(3.3) for each 𝑓𝐹. Hence by Lemma 2.2. 𝑔𝜎(𝑓(𝑥),𝑓(𝑦))=𝜎𝑓𝑔𝑓1𝑔𝑓(𝑥),𝑓𝑔𝑓1𝑓(𝑦)=𝑘𝑚𝜎𝑔𝑓1𝑔𝑓(𝑥),𝑓1𝑓(𝑦)<𝜀.(3.4) Therefore, the family is equicontinuous in Δ. This completes the proof of Theorem A.

Proof of Theorem B. Let 𝑀=𝑏𝑓/𝑐𝑓. We can find nonzero constants 𝑏 and 𝑐 satisfying 𝑀=𝑏/𝑐. For each 𝑓𝐹, define a Möbius map 𝑔𝑓 by 𝑔𝑓=𝑐𝑓𝑧/𝑐, thus 𝑔𝑓1=𝑐𝑧/𝑐𝑓.
Next we will show 𝐺={(𝑔𝑓1𝑓)𝑓𝐹} is normal in Δ. Suppose to the contrary, 𝐺 is not normal in Δ. Then by Lemma 2.1. We can find 𝑔𝑛𝐺,𝑧𝑛Δ, and 𝜌𝑛0+, such that 𝑇𝑛(𝜁)=𝑔𝑛(𝑧𝑛+𝜌𝑛𝜁)/𝜌𝑛1/(𝑘+1) converges locally uniformly with respect to the spherical metric to a nonconstant meromorphic function 𝑇(𝜁) whose spherical derivate is limited and 𝑇 has order at most 2.
We will also consider three cases.
Case 1. If (1/𝑐𝑘1)(𝑇)𝑘(𝜁)𝑏𝑘/𝑐𝑘1, then 𝑇(𝜁) is a polynomial with degree at most 1, a contradiction.
Case 2. If there exists 𝜁0 such that (1/𝑐𝑘1)(𝑇)𝑘(𝜁0)=𝑏𝑘/𝑐𝑘1. Noting that 𝜌𝑛𝑇𝑛(𝜁)+(1/𝑐𝑘1)(𝑇𝑛)𝑘(𝜁)(𝑏𝑘/𝑐𝑘1)(1/𝑐𝑘1)(𝑇)𝑘(𝜁)(𝑏𝑘/𝑐𝑘1). By Hurwitz’s theorem, there exist a sequence of points 𝜁𝑛𝜁0 such that (for large enough 𝑛) 0=𝜌𝑛𝑇𝑛𝜁𝑛+1𝑐𝑘1𝑇𝑛𝑘𝜁𝑛𝑏𝑘𝑐𝑘1=𝑔𝑛𝑧𝑛+𝜌𝑛𝜁𝑛+1𝑐𝑘1𝑔𝑛𝑘𝑧𝑛+𝜁𝑛𝑏𝑘𝑐𝑘1=𝑐𝑐𝑓𝑓𝑛𝑧𝑛+𝜌𝑛𝜁𝑛+1𝑐𝑘1𝑐𝑘𝑐𝑘𝑓𝑓𝑛𝑘𝑧𝑛+𝜁𝑛𝑏𝑘𝑐𝑘1.(3.5) Hence 𝑓𝑛(𝑧𝑛+𝜌𝑛𝜁𝑛)+(1/𝑐𝑓𝑘1)(𝑓𝑛)𝑘(𝑧𝑛+𝜁𝑛)=𝑏𝑘𝑓/𝑐𝑓𝑘1, then we have 𝑓𝑛(𝑧𝑛+𝜌𝑛𝜁𝑛)=𝑏𝑓 by the condition (iii)(1/𝑐𝑓𝑘1)(𝑓)𝑘(𝑧)+𝑓(𝑧)=𝑏𝑘𝑓/𝑐𝑓𝑘1𝑓(𝑧)=𝑏𝑓.
Thus 𝑇𝜁0=lim𝑛𝑔𝑛𝑧𝑛+𝜌𝑛𝜁𝑛𝜌𝑛=lim𝑛𝑧𝑐𝑓𝑛+𝜌𝑛𝜁𝑛𝑐𝑓𝜌𝑛=lim𝑛𝑏𝜌𝑛=.(3.6) This is a contradiction.
Case 3. If (1/𝑐𝑘1)(𝑇)𝑘(𝜁)𝑏𝑘/𝑐𝑘1. Let 𝑐1,𝑐2,,𝑐𝑘 be the solution of the equation 𝑤𝑘=𝑐𝑘, then 𝑇(𝜁)𝑐𝑖(𝑖=1,2,,𝑘). When 𝑇(𝜁) is a rational function, then 𝑇(𝜁) is also a rational function. By Picard theorem we can deduce that 𝑇(𝜁) is a constant (𝑘3). Hence 𝑇(𝜁) is a polynomial with degree at most 1. This contradicts with 𝑇(𝜁) has zeros of multiplicity at least 2. When 𝑇(𝜁) is a transcendental function, combining with the second main theorem, we have 𝑇𝑟,𝑇𝑁𝑟,𝑇+𝑘𝑖=1𝑁1𝑟,𝑇𝑐𝑖+𝑠𝑟,𝑇𝑁𝑟,𝑇+𝑠𝑟,𝑇12𝑁𝑟,𝑇+𝑠𝑟,𝑇12𝑇𝑟,𝑇+𝑠𝑟,𝑇.(3.7) Hence, 𝑇(𝑟,𝑇)𝑠(𝑟,𝑇), a contradiction.
Hence 𝐺={(𝑔𝑓1𝑓)𝑓𝐹} is normal and equicontinuous in Δ. There given (𝜀/𝑘𝑚>0), where 𝑘𝑚 is the constant of Lemma 2.2, there exists 𝛿>0 such that for the spherical distance 𝜎(𝑥,𝑦)<𝛿, 𝜎𝑔𝑓1𝑔𝑓(𝑥),𝑓1<𝜀(𝑦)𝑘𝑚(3.8) for each 𝑓𝐹. Hence by Lemma 2.2. 𝑔𝜎(𝑓(𝑥),𝑓(𝑦))=𝜎𝑓𝑔𝑓1𝑔𝑓(𝑥),𝑓𝑔𝑓1𝑓(𝑦)=𝑘𝑚𝜎𝑔𝑓1𝑔𝑓(𝑥),𝑓1𝑓(𝑦)<𝜀.(3.9) Therefore, the family is equicontinuous in Δ. This completes the proof of Theorem B.

Remark 3.1. Using the similar argument, if the condition (iii)𝑓(𝑧)=𝑏𝑓 when (1/𝑐𝑓𝑘1)(𝑓)𝑘(𝑧)+𝑓(𝑧)=𝑏𝑘𝑓/𝑐𝑓𝑘1 is replaced by (iii)|𝑓(𝑧)||𝑏𝑓| when (1/𝑐𝑓𝑘1)(𝑓)𝑘(𝑧)+𝑓(𝑧)=𝑏𝑘𝑓/𝑐𝑓𝑘1, then 𝐹 is normal too.

Authors’ Contribution

W. Chen performed the proof and drafted the paper. All authors read and approved the final paper.

Conflict of Interests

The authors declare that they have no conflict of interests.

Acknowledgment

This paper is supported by Nature Science Foundation of Fujian Province (2012J01022). The authors wish to thank the referee for some valuable corrections.

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