Abstract

We obtain a normal criterion of meromorphic functions concerning, shared values. Let be a family of meromorphic functions in a domain and let be positive integers. Let be two finite complex constants. If, for each , all zeros of have multiplicity at least and and share in for every pair of functions , then is normal in . This result generalizes the related theorem according to Xu et al. and Qi et al., respectively. There is a gap in the proofs of Lemma 3 by Wang (2012) and Theorem 1 by Zhang (2008), respectively. They did not consider the case of being zerofree. We will fill the gap in this paper.

1. Introduction and Main Results

We use to denote the open complex plane, to denote the extended complex plane, and to denote a domain in . A family of meromorphic functions defined in is said to be normal, if for any sequence contains a subsequence which converges spherically, and locally, uniformly in to a meromorphic function or . Clearly, is said to be normal in if and only if it is normal at every point of (see [1, 2]).

Let be a meromorphic function in a domain . We say that is a normal function if there exists a positive number such that for all , where denotes the spherical derivative of .

Let be a domain in and let and be two nonconstant meromorphic functions in . Let and be two complex numbers. If whenever , we write that If and , we write that

If , we say that and share (ignoring multiplicities) on . When means (see [2]).

Influenced from Bloch's principle [3], every condition which reduces a meromorphic function in the plane to a constant makes a family of meromorphic functions in a domain normal. Although the principle is false in general (see [4]), many authors proved normality criterion for families of meromorphic functions corresponding to Liouville-Picard type theorem (see [2, 5, 6]).

It is also more interesting to find normality criteria from the point of view of shared values. In this area, Schwick [7] first proved an interesting result that a family of meromorphic functions in a domina is normal if in which every function shares three distinct finite complex numbers with its first derivative. And later, Sun [8] proved that a family of meromorphic functions in a domina is normal if in which each pair of functions share three fixed distinct values, which is an improvement of the famous Montel's Normal Criterion [9] by the ideas of shared values. More results about normality criteria concerning shared values can be found, for instance, see [1013] and so on.

In 1959, Hayman [14] proved that let be a meromorphic function in , if , where is a positive integer and , are two finite complex numbers such that and , then is a constant. On the other hand, Mues [15] showed that for , the conclusion is not valid.

The following theorem confirmed a Hayman's well-known conjecture about normal families in [16].

Theorem A. Let be a family of meromorphic functions in , a positive integer, and , two finite complex numbers such that . If and for each function , , then is normal in .

In 2008, by the ideas of shared values, Zhang [13] proved the following result.

Theorem B. Let be a family of meromorphic functions in , a positive integer, and , two finite complex numbers such that . If and for every pair of functions and in , if and share the value , then is normal in .

In 1994, Ye [17] considered a similar problem and obtained that if is a transcendental meromorphic function and is a nonzero finite complex number, then assumes every finite complex value infinitely often for . Ye [17] also asked whether the conclusion remains valid for .

In 2008, Fang and Zalcman [18] solved this problem and obtained the following result.

Theorem C. Let be a transcendental meromorphic function and let be a nonzero complex number. Then, assumes every complex value infinitely often for each positive integer .

Remark 1. By a special example, Fang and Zalcman [18] showed Theorem C is not valid for .

On the basis of the previous results, Fang and Zalcman [18] obtained the normality criterion corresponding to Theorem C.

Theorem D. Let be a family of meromorphic functions in . Let be an integer and let be two finite complex numbers. If, for each , all zeros of are multiple and on , then is normal in .

With the ideas of shared values, Qi and Zhu [19] and Wang [20] extended Theorem D and obtained the following result.

Theorem E. Let be a family of meromorphic functions in , all of whose zeros are multiple. Let be an integer and , two nonzero finite complex numbers. If and share on for every pair of functions , then is normal in .

It is natural to ask whether the condition can be replaced by in the previous theorems.

In 2009, Xu et al. [21] considered the case that is replaced by the th derivative and proved the following results.

Theorem F. Let be a transcendental meromorphic function on , be a nonzero finite complex number, and and two positive integers. If , then assumes each value infinitely often.

Theorem G. Let , and and be two positive integers such that . Let be a family of meromorphic functions defined on a domain . If has only zeros of multiplicity at least and in for every function , then is normal.

In this paper, we study the previous problem and get the following results.

Theorem 2. Let be a family of meromorphic functions in the plane domain and positive integers. Let be two finite complex constants. If all zeros of have multiplicity at least for each and and share in for every pair of functions , then is normal in .

Theorem 3. Let be a family of meromorphic functions in the plane domain and positive integers. Let be finite complex constants. If all zeros of have multiplicity at least and for each and share in for every pair of functions , then is normal in .

Example 4. Let and , where then Clearly, for each pair of functions, , and share the value 0 in ; however, is not normal in .

Example 5. Let , , and , where then Clearly, for each pair of functions, , and share the value 0 in ; however, is not normal in .

Remark 6. Example 4 shows that the condition in Theorem 2 is sharp. And Example 5 shows that the condition in which all zeros of have multiplicity at least in Theorem 2 is sharp.

Remark 7. There is a gap in the proofs of Theorems 1.2 and 1.5 by Zhang [13] and Wang [20], respectively. They did not consider the case of being zero free in the proof of both Lemma 3 in [20] and Theorem 1 in [13], respectively. In Section 2, both Lemmas 9 and 11 fill the gap in their proofs of both Lemma 3 in [20] and Theorem 1 in [13], respectively.

2. Preliminary Lemmas

In order to prove our theorems, we need the following lemmas.

First, we need the following well-known Pang-Zalcman lemma, which is the local version of [11, 22].

Lemma 8. Let be a family of meromorphic functions in a domain and a positive integer, such that each function has only zeros of multiplicity at least , and suppose that there exists such that whenever . If is not normal at , then for each , there exist a sequence of points , a sequence of positive numbers , and a subsequence of functions such that locally uniformly with respect to the spherical metric in . Here, is a nonconstant meromorphic function such that and all of whose zeros have multiplicity at least . Moreover, has an order at most 2. If is a holomorphic function, then is of exponential type and has an order at most 1.

Lemma 9 (see [23]). Let be a nonconstant zero-free rational function, , two positive integers, and two complex constants. Then, the function has at least distinct zeros in .

Lemma 10. Let be two positive integers and let be a constant. Let be a nonconstant rational meromorphic function, all zeros of have multiplicity at least , then has at least two distinct zeros.

Proof. When is a nonconstant polynomial, noting that and all zeros of have multiplicity at least , we know that must have zero. Hence, has exactly one zero . Set that where is a nonzero constant and is a positive integer such that .
It follows from (8) that where . Since all zeros of have multiplicity at least and has only one zero , we can deduce from (9) that has only zero . Writing that , where is an integer such that , then where . Equation (9) combing inequality that gives that has at least two distinct zeros, a contradiction.
When is rational but not a polynomial, we consider two cases.
Case  1. If has exactly one zero, then suppose that has only zero with multiplicity at least . If , by Lemma 9, we get a contradiction. So has zeros, then we can deduce that is the only zero of . Otherwise, has at least two zeros. Set that where is a nonzero constant and . For simplicity, we denote that From (11), we have where is polynomial and is a constant.
By (11) and (13), we have
Since , we deduce that , and then
By the assumption that has exactly one zero with multiply and (15), we get where is a nonzero constant. Both (15) and (16) imply that
Case  1.1. If , from (17), we can deduce that is a zero of , a contradiction.
Case  1.2. If , from (17), it follows that
For simplicity, we denote that
Case  1.2.1. If , then
Thus, . Therefore, is a non-constant polynomial, a contradiction.
Case  1.2.2. If , then Hence,
Therefore, is a non-constant polynomial, a contradiction.
Case  2. If has no zeros, then for (16). It is the same argument as the proof of Case  1 that we have a contradiction.
The proof is complete.

Lemma 11 (see [24]). Let be a nonconstant zero-free rational function and a positive integer. Then, has at least distinct zeros.

3. The Proofs of Theorems

Proof of Theorem 2. Without loss of generality, we may assume that . Suppose, on the contrary, that is not normal at .
Case  1. When , by Lemma 8, there exist a sequence of complex numbers with and a sequence of positive numbers with such that locally uniformly on compact subsets of , where is a non-constant meromorphic function in , all of whose zeros have multiplicity at least . Moreover, has an order at most .
If , then has no poles, so is an entire function. Since all zeros of have multiplicity at least , if is a zero of with multiply , then is a zero of with multiplicity , by . It follows that , so which implies that , then , which contradicts with . So has no zeros, then we can deduce that , where are constants. Then, Hence, which is impossible because .
Theorems F, G, and Lemma 10 imply that has at least two distinct zeros; we may assume that there exist two zeros , . We choose a positive number small enough such that and has no other zeros in except for and , where Since by Hurwitz's theorem, we know that for sufficiently large there exist points such that Note that and share , it follows that Fix , let , and note that , then we obtain Since the zeros of have no accumulation point, for sufficiently large , we have Hence,
This contradicts with the facts .
Case  2. When , by Lemma 8, there exist a sequence of complex numbers with and a sequence of positive numbers with such that locally uniformly on compact subsets of , where is a non-constant meromorphic function in , all of whose zeros have multiplicity at least . Moreover, has an order at most 2.
Case  2.1. If , then is a polynomial with a degree at most , this contradicts with the fact that has only zeros with multiplicity at least .
Case  2.2. If , let be the (distinct) solutions of , then , and then is a constant by Picard theorem. It follows that is a polynomial with a degree at most , this contradicts with the fact that has only zeros with multiplicity at least . Therefore, has solutions.
We claim that has just one zero. Otherwise, we may assume that there exist two zeros , . We choose a positive number small enough such that and such that has no other zeros in except for and , where Combing and Hurwitz's theorem, we have that for sufficiently large there exist points such that
From the hypothesis that and share , it follows that By letting and noting that , we obtain Since the zeros of have no accumulation point, for sufficiently large , we have Hence, which contradicts with the facts , , .
Therefore, has just one zero. Thus, has at most one zero for the only one of . We may assume that has at most one zero, then . From the second fundamental theorem of Nevanlinna, It follows that , which is impossible.
The proof of Theorem 2 is proved.

Proof of Theorem 3. Suppose, to the contrary, that is not normal in . Then, there exists at least one point such that is not normal at the point . Without loss of generality, we assume that . Then, by Lemma 8, we can find that , and , such that converges locally uniformly with respect to the spherical metric to a nonconstant meromorphic function , and all zeros of are of multiplicity at least , and the spherical derivative of is bounded.
Then, we have also locally uniformly with respect to the spherical metric.
We claim that all zeros of are of multiplicity at least . Suppose that is a zero of , then obviously, . From Hurwitz's theorem, there exist such that then we have by the condition that .
If , then all zeros of each are of multiplicity at least near 0 by the hypothesis. Hence, Theorem 2 tells us that is normal at 0, which contradicts our supposition.
If , then which is also a contradiction.
The proof of Theorem 3 is completed.

Acknowledgment

This work was supported by the Visiting Scholar Program of Chern Institute of Mathematics at Nankai University. Zo Yingying would like to express his hearty thanks to Chern Institute of Mathematics which provided him with very comfortable research environments while working as a visiting scholar. This word was also supported by Nature Science Foundation of China (11271090); Nature Science Foundation of Guangdong Province (S2012010010121); Academic technology innovation Foundation of Xinjiang Normal University.