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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

Accepted07 Aug 2012
Published30 Aug 2012

#### Abstract

Let , and and be two positive integers such that . Let be a family of zero-free meromorphic functions defined in a domain such that for each , has at most zeros, ignoring multiplicity. Then is normal in .

#### 1. Introduction and Main Results

Let be a domain in , and let be a family of meromorphic functions defined in the domain . is said to be normal in , in the sense of Montel, if for every sequence contains a subsequence such that converges spherically uniformly on compact subsets of (see [1, Definitionββ3.1.1]).

is said to be normal at a point if there exists a neighborhood of in which is normal. It is well known that is normal in a domain if and only if it is normal at each of its points (see [1, Theoremββ3.3.2]).

Let be a meromorphic function in the complex plane. We use the standard notations and results of value distribution theory as presented in [2β4]. In particular, is Nevanlinnaβs characteristic function and denotes a function with the property as (outside an exceptional set of finite linear measure).

In 1959, Hayman [5] proved the following well-known result.

Theorem A. Let be a transcendental meromorphic function on the complex plane C, let be a nonzero finite complex number, and let be a positive integer. If , then assumes each value infinitely often.

There are some examples constructed by Mues [6] which show that Theorem A is not true when . Corresponding to Theorem A, Ye [7, Theoremββ2.1] proved the following interesting result.

Theorem B. Let be a transcendental meromorphic function. If is a finite complex number and is an positive integer, then assumes all finite complex number infinitely often.

In [7, Theoremββ2.2], Ye also obtained the following result, which may be considered as a normal family analogue of Theorem B.

Theorem C. Let be a family of meromorphic functions defined in a domain , and for every , where is an integer and are two finite complex numbers. Then, is normal.

Ye [7] asked whether Theorem B remains valid for . Recently, Fang and Zalcman showed that Theorem B holds for . In [8], the condition in Theorem C that can be relaxed to that all zeros of each function in are of multiplicity at least 2. Actually. they obtained the following results.

Theorem D. Let be a transcendental meromorphic function. If is a finite complex number and is an positive integer, then assumes all finite complex number infinitely often.

Theorem E. Let be a family of meromorphic functions on the plane domain , let be a positive integer, and let be complex numbers. If, for each , all zeros of are multiple and on D, then is normal on D.

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

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

Xu et al. [9] asked whether Theorem F remains valid for . We partially answer this question. If , we generalize Theorem F by allowing to have zeros but restricting their numbers.

Theorem 1.1. Let , and and be two positive integers such that . Let be a family of zero-free meromorphic functions defined in a domain such that for each , has at most zeros, ignoring multiplicity. Then, is normal in .

Remark 1.2. Here, can be replaced by , where is any finite complex numbers.

Example 1.3. Let . Let , where . Then, in for every function . However, it is easily obtained that is not normal at the point .

Example 1.4. Let . Let , where . Then, has zeros in for every function . However, it is easily obtained that is not normal at the point .

Example 1.5. Let . Let , where . It follows that has no zero in for every function . However, it is easily obtained that is not normal at the point .

Examples 1.3 and 1.4 show that the conditions that and have at most distinct zeros in Theorem 1.1 are shape. Example 1.5 shows the condition that cannot be omitted.

#### 2. Some Lemmas

To prove our results, we need some preliminary results.

Lemma 2.1 ([9], Lemma 2.2). Let be positive integers, let be a nonzero constant and let be a polynomial. Then, the solution of the differential equation must be polynomial.

Lemma 2.2. Let be a nonzero transcendental meromorphic function. If be a nonzero finite complex number and let and be two positive integers. Then, assumes each value infinitely often.

Proof. Set
We claim that . If , then . We can deduce that , where is a finite complex number. We conclude from (2.1) and Lemma 2.1 that, must be a polynomial, which is a contradiction.
If , from (2.3), we can obtain where is a finite complex number, that is, βIf , we can get that , which is a contradiction.βIf , we conclude from (2.5) and Lemma 2.1 that must be a polynomial, which is a contradiction.
By elementary Nevanlinna theory and (2.1), we have . Thus, from (2.2) and (2.3), we have
It follows from (2.2), (2.3) and Nevanlinnaβs First Fundamental Theorem that By (2.2) and (2.3), we get We have by (2.6)-(2.7)
It follows from (2.6)β(2.10) that
So, we have
We have
Since , if assumes the value only finitely often, we by (2.12) can get Hence, So assumes each value infinitely often.
We complete the proof of Lemma 2.2.

Using the method of Chang [10, Lemmaββ4], we obtain the following lemma.

Lemma 2.3. Let be a nonconstant zero-free rational function, ,β let be two positive integers, and be two complex constants. Then, the function has at least distinct zeros in .

Proof. Since is a nonconstant zero-free rational function, is not a polynomial, and hence it has at least one finite pole. Thus, we can write where is a nonzero constant, and are positive integers, the (when ) are distinct complex numbers, and denote .
By induction, we deduce from (2.16) that where is polynomial of degree .
So the degree of numerator of the function is equal to . By calculation, has at least one zero in . Thus, we can write where is a nonzero constant, are positive integers, (when ), and (when ) are distinct complex numbers. Thus, by (2.16), (2.17), and (2.18), we get Case 1. If , it follows that and . Thus, it follows from (2.19) that where is a polynomial. Then, is a polynomial of degree less than , and it follows that as .
Logarithmic differentiation of both sides of (2.21) shows that as .
Comparing the coefficient of (2.22) for , , we have for .
Set when . Noting that , then it follows from (2.23) that the system of linear equations, where , has a nonzero solution
If , then the determinant of the coefficients of the system of (2.24), where , is equal to zero, by Cramerβs rule (see, e.g., [11]). However, the are distinct complex numbers when , and the determinant is a Vandermonde determinant, so it cannot be 0 (see [11]), which is a contradiction.
Hence, we conclude that . Noting that , it follows from this and that .

Case 2. If , set where are positive integers. It follows that (when ) and (when ) are distinct complex numbers, and .

By (2.19), we have

It follows that and . Thus, by (2.27), where is a polynomial. Then, is a polynomial of degree less than , and it follows that as .

Thus, by taking logarithmic derivatives of both sides of (2.12), we get

We consider two cases.

Subcase 2.1 (). Applying the reasoning of Case 1 and noting that , we deduce that .

Subcase 2.2 (). Without loss of generality, we may assume that . Denote

The formula (2.31) can be rewritten:

Applying the reasoning of Case 1, and noting that , we deduce that .

This completes the proof of Lemma 2.3.

Lemma 2.4 ([10], Lemma 4). Let be a nonconstant zero-free rational function, let be a complex constant, and let be a positive integer. Then has at least distinct zeros in .

Lemma 2.5 (see [12], Lemmaββ2, Zalcmanβs lemma). Let be a family of functions meromorphic on a domain , all of whose zeros have multiplicity at least . Suppose that there exists such that whenever . Then, if is not normal at , there exist, for each ,(a)points ;(b)functions ;(c)positive numbers ; such that locally uniformly with respect to the spherical metric, where is a nonconstant meromorphic function on , all of whose zeros of are of multiplicity at least , such that .

Here, as usual, is the spherical derivative.

#### 3. Proof of Theorem

Suppose 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 . We consider two cases.

Case 1 (). By Zalcmanβs lemma, there exist:(a)points ;(b)functions ; (c)positive numbers ; such that spherically uniformly on compact subsets of , where is a nonconstant meromorphic function in . Since , by Hurwitzβs theorem, it implies that .
On every compact subset of which contains no poles of , from (3.1), we get also locally uniformly with respect to the spherical metric.
We claim that has at most distinct zeros.
Suppose that has distinct zeros , , and choose small enough such that , where .
From (3.2), by Hurwitzβs theorem, there exist points () such that for sufficiently large , for .
Since and , we have ββ( is a positive constant) for sufficiently large , so has distinct zeros, which contradicts the fact that has at most 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 is normal in .

Case 2 (). By Zalcmanβs lemma, there exist:(a)points ;(b)functions ; (c)positive numbers ; such that spherically uniformly on compact subsets of , where is a nonconstant meromorphic function in . Since , by Hurwitzβs theorem, it implies that .
On every compact subset of which contains no poles of , from (3.4), we get also locally uniformly with respect to the spherical metric.
Noting that we claim that has at most distinct zeros.
Suppose that has distinct zeros , , and choose small enough such that , where .
From (3.2), by Hurwitzβs theorem, there exist points () such that for sufficiently large for .
Since and , we have ββ( is a positive constant) for sufficiently large , so has distinct zeros, which contradicts the fact that has at most zero.
Denote by the different roots of , then Subcase 2.1 (If is a rational function). By Lemma 2.4 and (3.8), we can deduce that has at least distinct zeros. This contradicts the claim that has at most distinct zeros.

Subcase 2.2 (If is a transcendental meromorphic function). By Nevanlinnas second main theorem, we have

It follows that , which is a contradiction. This contradiction shows that is normal in .

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.

#### References

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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.