Research Article | Open Access

Volume 2014 |Article ID 654294 | https://doi.org/10.1155/2014/654294

Wei Chen, Qi Yang, Wen-jun Yuan, Hong-gen Tian, "The Normality Criteria of Meromorphic Functions Concerning Shared Fixed-Points", Discrete Dynamics in Nature and Society, vol. 2014, Article ID 654294, 8 pages, 2014. https://doi.org/10.1155/2014/654294

# The Normality Criteria of Meromorphic Functions Concerning Shared Fixed-Points

Accepted17 Dec 2013
Published06 Apr 2014

#### Abstract

We study the normality criteria of meromorphic functions concerning shared fixed-points; we obtain the following: Let be a family of meromorphic functions defined in a domain and a positive integer. For every , all zeros of are of multiplicity at least and all poles of are multiple. If and share in for each pair of functions and , then is normal.

#### 1. Introduction and Main Results

Let be a family of meromorphic functions defined in the domain . If any sequence contains a subsequence that converges spherically locally uniformly in , to a meromorphic function or , we say that is normal in (see [1, 2]).

Let be a meromorphic function in a domain , and . We say is a fixed-point of when . Let and be two meromorphic functions in ; if and have the same zeros (ignoring multiplicity), then we say and share fixed-points.

In , Lu and Gu  proved the following results.

Theorem 1. Let be a positive integer and a transcendental. If all zeros of have multiplicity at least , then assumes every finite nonzero value infinitely often.

Theorem 2. Let be a family of meromorphic functions defined in a domain . Suppose that is a positive integer and is a finite complex number. If, for each , all zeros of are of multiplicity at least , and , then is normal in .

In 2011, Hu and Meng  extended Theorem 2 as follows.

Theorem 3. Let be a family of meromorphic functions defined in a domain . Let be a positive integer and be a finite complex number. If, for every , and share in , and, for every pair of functions , , all zeros of are of multiplicity at least , then is normal in .

A natural question is the following: what can be said if the finite complex number in Theorem 3 is replaced by the fixed-point ? In this paper, we answer this question by proving the following theorems.

Theorem 4 (main theorem). Let be a family of meromorphic functions defined in a domain . Let be a positive integer. If, for every such that , and share in for every pair of function , , then is normal in .

Theorem 5 (main theorem). Let be a family of meromorphic functions defined in a domain . Let be a positive integer. For every , all zeros of have multiplicity at least and all poles of are multiple. If and share in for every pair of function , , then is normal in .

Theorem 6 (main theorem). Let be a family of meromorphic functions defined in a domain . Let be a positive integer. For every , has only zeros with multiplicity at least and has poles at most . If and share in for every pair of function , , then is normal in .

#### 2. Some Lemmas

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

Lemma 7 (see [5, 6]). Let be a family of meromorphic functions in a domain D, and let be 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 , where is a nonconstant meromorphic function, whose zeros all have multiplicity at least , such that . Moreover, has order at most .
Here as usual, is the spherical derivative.

Lemma 8. Let be a integer and a nonconstant rational meromorphic function such that ; then has at least two distinct zeros.

Proof. Since , set where is a nonzero constant and are positive integers.
For simplicity, we denote . Obviously, .
From (2), we have where is a polynomial and .
Differentiating (3), we get where is a polynomial and .
Next, we assume, to the contrary, that has at most one zero. We distinguish two cases.
Case 1. If has exactly one zero . From (3), we obtain where is a nonconstant. Differentiating (5), we have where is a polynomial and .
From (4) and (6), we obtain then ; this is a contradiction.
Case 2. If has no zero. Then we have from (5), which is a contradiction.
The proof is complete.

Lemma 9. Let be an integer and a nonconstant rational meromorphic function. If has only zeros with multiplicity at least and all poles of are multiple, then has at least two distinct zeros.

Proof. Assume, to the contrary, that has at most one zero.
Case 1. If is a polynomial, since all zeros of are of multiplicity at least , then we can know that has at least one zero with multiplicity . So has at least one zero and has zeros with multiplicity at least . According to the assumption, we obtain that has only a zero ; then there exists a nonzero constant and an integer such that So we have which, however, has only simple zeros. This is a contraction.
Case 2. If is a rational but not a polynomial. We see where is a nonzero constant and   .
For simplicity, we denote From (10), we have where is a polynomial and .
Differentiating (13), we have where ,   are polynomials and , .
Now we distinguish two subcases.
Case 2.1. Supposing that has exactly one zero , from (13), we obtain Differentiating (16), we have where ,   are polynomials, , and . From (14) and (17), we have .
Further, we distinguish two subcases.
Case 2.1.1. . From (16), it is easily obtained that . Thus, (13) implies So ; that is, . From (13) and (16), noting that , we have It follows that ; combining this inequality with (11), we obtain which is impossible.
Case 2.1.2. . Next we distinguish two subcases: and .
When , similar to the Case , it follows that ; from (13) and (16), we get a contradiction.
If , by combining (15) and (18), we may give the following inequality: and hence This is a contradiction.
Case 2.2. Suppose that has no zero; then for (16). Similarly with the proof of Case 2.1, we also obtain a contradiction.

Lemma 10. Let be an integer and a nonconstant rational meromorphic function. If has only zeros with multiplicity at least and has poles at most , then has at least two distinct zeros.

Proof. Assume, to the contrary, that has at most one zero.
Case 1. If is a polynomial, since all zeros of are of multiplicity at least , then we can know that has at least one zero with multiplicity . So we see that has at least one zero and has zeros with multiplicity at least . According to the assumption, we obtain that has only a zero ; then there exists a nonzero constant and an integer such that So we have which, however, has only simple zeros. This is a contraction.
Case 2. If is a rational but not a polynomial, we set where is a nonzero constant and .
For simplicity, we denote From (26), we have where is a polynomial and .
Differentiating (29), we have where , are polynomials and and .
Now we distinguish two subcases.
Case 2.1. Supposing that has exactly one zero , from (13), we obtain Differentiating (32), we have where , are polynomials, , and . From (30) and (33), we have . From (31) and (34), we see So . Since , we have , so .
Therefore, has no zeros. According to Lemma 8, we have a contradiction.
Case 2.2. Suppose that has no zero; then for (32). Similarly with the proof of Case 2.1, we also obtain a contradiction.

Lemma 11 (see ). Let be a transcendental meromorphic function, and let be a small function such that ; then

Lemma 12. Let be a transcendental meromorphic function; let be a positive integer; let be a polynomial. If all zeros of have multiplicity at least , then has infinitely many zeros.

Proof. We denote then Since is a polynomial, we see . Now we distinguish two cases as follows.(i)If , we have By Lemma 11, we have Thus, (ii)If , then By Lemma 11, we have
So
From (41) and (44), we can deduce that has infinitely many zeros; thus, has infinitely many zeros.

Lemma 13 (see ). Take a positive integer and a nonzero complex number . If is a nonconstant meromorphic function such that has only zeros of multiplicity at least , then has at least two distinct zeros.

#### 3. Proof of Theorems

Proof of Theorem 5. Consider the following.
Case 1. For , let . If is not normal at , by Lemma 7, 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 nonconstant meromorphic function in .
Here we distinguish two cases.
Case 1.1. Supposing that , is a finite complex number. Then locally uniformly on compact subsets of disjoint from the poles of , where is a nonconstant meromorphic function in , all of whose zeros have multiplicity at least and all poles of which have multiplicity at least . Hence spherically locally uniformly in disjoint from the poles of .
If , since has zeros with multiplicity at least , obviously there is a contradiction. Hence, . Since the multiplicity of all zeros of is at least , so by Lemmas 9 and 12, has at least two distinct zeros.
Suppose that , are two distinct zeros of . We choose a positive number small enough such that and such that has no other zeros in except for and , where
By Hurwitz’s Theorem, for sufficiently large , there exist points , such that
By the assumption in Theorem 5 that and share , it follows that Fix , take , and note , ; we obtain Since the zeros of has no accumulation point, for sufficiently large , we have Therefore, when is large enough, . This contradicts with the facts , , . Thus is normal at 0.
Case 1.2. We may suppose that . We have where . Thus we have Since , , we have On the other hand, for , we have
Thus, we have spherically locally uniformly in disjoint from the poles of .
If , then has no zeros. Of course, also has no poles. Since is a nonconstant meromorphic function of order at most , then there exists constant , , and ; obviously, this is contrary to the case . Hence, .
Since the multiplicity of all zeros of is at least and all poles of which have multiplicity at least , thus, by Lemma 13, has at least two distinct zeros.
Suppose that , are two distinct zeros of . We choose a positive number small enough such that and such that has no other zeros in except for and , where
By Hurwitz’s Theorem, for sufficiently large there exist points , such that
Similar to the proof of Case 1.1, we get a contradiction. Then, is normal at 0.
From Cases 1.1 and 1.2, we know that is normal at 0; there exists and a subsequence of , and we may still denote it as , such that converges spherically locally uniformly to a meromorphic function or in .
Here we distinguish two cases.
Case i. When is large enough, . Then . Thus, for each , there exists such that if , then for all . Thus, for sufficiently large , , is holomorphic in . Therefore, for all , when , we have By maximum Principle and Montel’s Theorem, is normal at .
Case ii. There exists a subsequence of , and we may still denote it as such that . Since , the multiplicity of all zeros of is at least ; then . Thus, there exists such that is holomorphic in and has a unique zero in . Then converges spherically locally uniformly to a holomorphic function in ; converges spherically locally uniformly to a holomorphic function in . Hence is normal at .
By Cases and , is normal at .
Case 2. For , suppose, to the contrary, that is not normal in . Then there exists at least one point such that is not normal at the point . Then by Lemma 7, 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 nonconstant meromorphic function in , whose zeros all have multiplicity at least . Moreover, has order at most .
From (61) we get also locally uniformly with respect to the spherical metric besides the poles of .
If , then has no zeros. Of course, also has no poles. Since is a nonconstant meromorphic function of order at most , then there exists constant , and . Obviously, this is contrary to the case . Hence .
By Lemmas 12 and 13, we deduce that has at least two distinct zeros. Next we show that it is impossible. Let and be two distinct zeros of . We choose a positive number small enough such that and such that has no other zeros in expect for and , where
Similar to the proof of Case 1, we get a contradiction. This finally completes the proof of Theorem 5.

Similar to the proof of Theorem 5, combining with Lemmas 8 and 10 and Lemma 12, we can prove Theorems 4 and 6 easily; here we omit the proof.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This work was supported by the Visiting Scholar Program of Chern Institute of Mathematics at Nankai University. The third author would like to express his hearty thanks to Chern Institute of Mathematics that provided very comfortable research environments to him while working as visiting scholar. The authors thank the referees for reading the paper very carefully and making a number of valuable suggestions to improve the readability of the paper. Foundation item: Nature Science Foundation of China (11271090); Nature Science Foundation of Guangdong Province (S2012010010121); Graduate Research and Innovation Projects (XJGRI2013131) of Xinjiang Province.

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