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`Abstract and Applied AnalysisVolume 2014 (2014), Article ID 310251, 10 pageshttp://dx.doi.org/10.1155/2014/310251`
Research Article

## Derivatives of Meromorphic Functions with Multiple Zeros and Small Functions

1College of Applied Mathematics, Chengdu University of Information Technology, Chengdu 610225, China
2Department of Mathematics, Anhui Science and Technology University, Chuzhou 233100, China

Received 21 September 2013; Accepted 16 December 2013; Published 29 January 2014

Copyright © 2014 Pai Yang and Peiyan Niu. 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 be a meromorphic function in , and let , where is a nonconstant elliptic function and is a rational function. Suppose that all zeros of are multiple except finitely many and as . Then has infinitely many solutions.

#### 1. Introduction

The value distribution theory of meromorphic functions occupies one of the central places in complex analysis which now have been applied to complex dynamics, complex differential and functional equations, Diophantine equations, and others.

In his excellent paper [1], Hayman studied the value distribution of certain meromorphic functions and their derivatives under various conditions. Among other important results, he proved that if is a transcendental meromorphic function in the plane, then either assumes every finite value infinitely often or every derivative of assumes every finite nonzero value infinitely often. This result is known as Hayman’s alternative. Thereafter, the value distribution of derivatives of transcendental functions continued to be studied.

In 1998, Wang and Fang proved the following results.

Theorem A (see [2, Theorem 3]). Let be a transcendental meromorphic function in , all of whose zeros have multiplicity at least . Then assumes each nonzero complex value infinitely often.

In 2006, Pang et al. proved the following result, which is a significant improvement of Theorem A.

Theorem B (see [3, Theorem 1]). Let be a transcendental meromorphic function in , all but finitely many of whose zeros are multiple, and let be a rational function. Then has infinitely many zeros.

Relative to , is a small function in Theorem B. Specifically, as in Theorem B. A natural problem arises: what can we say if the rational function in Theorem B is replaced by a more general small function ? In this direction, we obtain the following result.

Theorem 1. Let be a meromorphic function in , and let , where is a nonconstant elliptic function and is a rational function. Suppose that all zeros of are multiple except finitely many and as . Then has infinitely many solutions (including the possibility of infinitely many common poles of and ).

#### 2. Notation and Preliminary Lemmas

We use the following notation. Let be complex plane and let be a domain in . For and , , , , and . We write in to indicate that the sequence converges to in the spherical metric uniformly on compact subsets of and in if the convergence is in the Euclidean metric. Let denote the number of poles of ) in (counting multiplicities), and let .

For meromorphic in , we denote The Ahlfors-Shimizu characteristic is defined by

Remark 2. Let denote the usual Nevanlinna characteristic function. Since is bounded as a function of , we can replace with in the paper.

Recall that an elliptic function [4] is a meromorphic function defined in for which there exist two nonzero complex numbers and with not real such that for all in .

Lemma 3 (see [5, Lemma 2]). Let be a family of functions meromorphic in , all of whose zeros have multiplicity at least , and 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 in , where is a nonconstant meromorphic function in , all of whose zeros have multiplicity at least .

Lemma 4 (see [6, Lemma 3.2]). Let be a positive integer, and let be a rational function satisfying in . If all zeros of are multiple, then where is a nonnegative integer, , and   .

Lemma 5 (see [7, Lemma 6]). Let be a positive integer, and let be a rational function and all of whose zeros are multiple. If in , then is a constant function.

Lemma 6 (see [6, Lemma 3.6]). Let be a sequence of functions meromorphic in . Suppose that in , where is a nonconstant meromorphic function or in . If there exists such that for each , , then there exists such that .

Lemma 7 (see [8, Corollary 2]). If is a nonconstant elliptic function with primitive periods , , where is not real, then as , where is a constant.

Lemma 8 (see [6, Lemma 3.4]). Let and be two sequences of meromorphic functions in , and let and be two meromorphic functions in . Suppose that(a) and in ;(b) in .Then, either or in .

Lemma 9 (see [9, Lemma 3.1]). Let be a sequence of meromorphic functions in , and let be a sequence of holomorphic functions in such that , where in . If for each , and for all , then is normal in .

Using the same proving method of Theorem 1.1 in [10], we can prove the following result without any difficulties. In fact, there is no essential distinction between Theorem 1.1 in [10] and the following result.

Lemma 10. Let be a family of meromorphic functions in , all of whose zeros and poles are multiple, and let be a sequence of meromorphic functions in such that in , where is meromorphic and zero-free in . Suppose that and have the same poles with the same multiplicity and for all . Then is normal in .

Lemma 11 (see [6, Lemma 3.8]). Let be a sequence of meromorphic functions in , all of whose zeros are multiple, and let be a sequence of meromorphic functions in such that in , where is a nonvanishing holomorphic function in . Let be a (countable) discrete set in which has no accumulation points in . Suppose that (a) in ;(b)for some , no subsequence of is normal at ;(c)for all , in .Then, (d) there exists such that for sufficiently large , has a single zero of order and a single pole of order in , where as , ;(e) .

Lemma 12 (see [6, Lemma 3.9]). Let be a family of meromorphic functions in , all of whose zeros are multiple, and let be a sequence of meromorphic functions in such that in , where in . If for each , for all , then is quasinormal in .

#### 3. Auxiliary Lemmas

Lemma 13. Let be a family of meromorphic functions in , all of whose zeros are multiple. Let be a sequence of meromorphic functions in such that in , where is a meromorphic function and . Suppose that (a) and have the same zeros and poles with the same multiplicity;(b)for all and all , ;(c)there exist points in such that and as ;(d) in , where is a meromorphic function in .Then in .

Proof. Set . By , and hence . Since all zeros of are multiple and , we have . Hence, and for sufficiently large . Since and for sufficiently large , is not equicontinuous at and hence is not normal at .
By , we have in , where . By Hurwitz's theorem, either or in . Suppose first that in . Obviously, is a meromorphic function in , So in . Suppose that in . If or , then . Hence in . Suppose that in . By the assumptions, there exists such that has no poles on and has no zeros on . Thus, we have By the maximum principle, (4) holds in and then is normal at . A contradiction. Thus, in .

Lemma 14. Let be a meromorphic function in satisfying . Then there exist and such that

Proof. We claim that there exist and such that Otherwise, there would exist and such that for all . From this follows and hence Now, there exists such that , and hence which contradicts the hypothesis   .
By (6), there exists a sequence such that and as . Set . Obviously, and , and hence as .

Lemma 15. Let be an integer, and let be a transcendental meromorphic function, all of whose zeros are multiple. Set with if . If , then there exist sequences and such that as ,

Proof. By standard results in Nevanlinna theory, and as . Thus, . By Lemma 14, there exist and such that Set . Then and hence is not normal at . Since all zeros of are multiple in , all zeros of are multiple in for sufficiently large . Using Lemma 3 for , there exist points and positive numbers and a subsequence of (still denoted by ) such that in , where is a nonconstant meromorphic function in , all of whose zeros are multiple.
is not a constant function (otherwise, either is a constant function, or the zero of is not multiple). Thus, we can assume is not a zero or pole of . Set . Now, we have where . Since and is not a zero or pole of , we have , , and as .
Now, we have and Set . Obviously, and , and hence as .

Lemma 16. Let be a sequence of meromorphic functions in , and let be a sequence of meromorphic functions in such that in , where . If and for all in , then is normal in .

Proof. By Lemma 9, it suffices to prove that is normal at points where has poles or zeros. Without loss of generality, we assume that ,  , where in , and is an integer. Then is normal in .
Suppose is not normal at . Since in , we have that there exists such that and in . By Argument Principle, for sufficiently large , we have
Since , . Obviously, has poles (otherwise in ) which are different from the poles of , so . A contradiction.

Lemma 17. Let be a family of meromorphic functions in , all of whose zeros are multiple. Let be a sequence of meromorphic functions in such that in , where . Let be a set which has no accumulation points in . Suppose that (a) and have the same zeros and poles with the same multiplicity;(b) for all and all , ;(c) for each , no subsequence of is normal at ;(d) in .Then(e) for each , ;(f) for each , there exist and such that for sufficiently large , , where and only depend on ;(g) for each , in .

#### 4. Proof of Lemma 17

Proof. It suffices to prove that each subsequence of has a subsequence which satisfies that , and prove that and hold. So suppose we have a subsequence of , which (to avoid complication in notation) we again call .
Without loss of generality, for each , we may assume that , , , and where and in .
We consider the following three cases.
Case 1 (). We will derive a contradiction in the case, and hence holds. For convenience, we set . Thus, , where is a positive integer. Clearly, we have in , in , and .
Subcase 1.1 (For sufficiently large , ). We claim that for each , there exists at least one zero of in for sufficiently large . Otherwise, there exists a subsequence of (still denoted by ) such that in . Since , in for sufficiently large . By Lemma 16, is normal at . A contradiction.
Taking a subsequence and renumbering if necessary, we may assume that is the zero of of the smallest modulus. Obviously, as . Set . We have(A1) in ;(A2)all zeros of are multiple and ;(A3) and in .By Lemmas 16 and 12, is normal in and quasinormal in . Thus, there exists a subsequence of (still denoted by ) and such that (B1) has no accumulation point in ;(B2)for each , no subsequence of is normal at ;(B3) in .Obviously, and all zeros of are multiple.
Subcase 1.1.1 . By , . Let . By Lemma 11, which contradicts . Hence, is an empty set. Since , is a meromorphic function in . By Lemma 8 and , either or in . If , we have which contradicts . If , then by Theorem B and Lemma 5, is a constant function. Since , in . Now,
We claim that for each , there exists at least one pole of in for sufficiently large . Otherwise, there exist and a subsequence of (still denoted by ) such that has no poles in . Since and , we have . Thus, is a sequence of holomorphic functions in . By Lemma 10, is normal at . A contradiction.
Taking a subsequence and renumbering if necessary, we may assume that is the pole of of the smallest modulus. Obviously, as . By Hurwitz's theorem and (16), we have as . Set , and we have (C1) is holomorphic in ;(C2);(C3)all zeros of are multiple;(C4) and in .By Lemma 10 and Lemma 12, is normal in and quasinormal in . Thus, there exists a subsequence of (still denoted by ) and such that (D1) has no accumulation point in ;(D2)for each , no subsequence of is normal at ;(D3) in .Obviously, and all zeros of are multiple in .
Clearly, , so is meromorphic in . By Lemma 8 and , either or in .
(1) ( is an empty set.) By , we have . If in , then by Theorem B and Lemma 5, we have which contradicts that . If in , we have which contradicts that .
(2) ( is not an empty set.) Let . Since , by Lemma 11, we have in . Clearly, and are meromorphic functions in , so we have in which contradicts .
Subcase 1.1.2 (). By Lemma 11, we have . If , is a multivalued function. A contradiction. Thus, and we have Let be the th root of the equation , where .
We claim that . Suppose that and . Obviously, we have and which contradicts that all of zeros of are multiple. Suppose that and . Since , by Lemma 11, Comparing the coefficients of (17) and (18), we obtain that . A contradiction.
Now, we have By Hurwitz's theorem, there exist such that and , where . Observing that , we have .
Set , where is one of of the largest modulus. Then, there exists a subsequence of (still denoted by ) such that (E1)for each , in for sufficiently large ;(E2);(E3) and in ;(E4) has only poles on , where .In fact, holds by (19). By Lemma 16, , and , we obtain that is normal in . We assume that in . Obviously, by .
(1) ( is a meromorphic function in .) By Lemma 8 and , either or in . If , we have which contradicts that . If , then by Theorem B and Lemma 5, is a constant function which contradicts .
(2) ( in .) Set By the maximum principle applied to , we get that Set By (19), for sufficiently large , has no pole in , and by the maximum principle, in . Thus, we have By (21) and (22), in . Equation (24) implies that as which contradicts (23).
Subcase 1.2. There exists a subsequence of (still denoted by ) such that for each .
Doing as in Subcase 1.1.1, we may assume that is the pole of of the smallest modulus and as . Set . We have (F1) is holomorphic function in ;(F2);(F3)all zeros of are multiple;(F4) and in .By Lemma 10 and Lemma 12, is normal in and quasinormal in . Thus, there exists a subsequence of (still denoted by ) and such that (G1) has no accumulation point in ;(G2)for each , no subsequence of is normal at ;(G3) in .Obviously, and all zeros of are multiple in .
Clearly, , so is a meromorphic function in . By Lemma 8 and , either or in .
Subcase 1.2.1 ( is an empty set). By , . If , then by Theorem B and Lemma 5, is a constant function which contradicts that . If , we have which contradicts that .
Subcase 1.2.2 ( is not an empty set). Let . Since , by Lemma 11, we have which contradicts .
Case  2 (). In this case, we will show that and hold. Clearly, we have in , in , and .
We claim that for each , there exists at least one zero of in for sufficiently large . Otherwise, there exist and a subsequence of (still denoted by ) such that in . Since and all the zeros of are multiple, we have , and hence in . By Lemma 16, is normal at which contradicts the condition .
Taking a subsequence and renumbering if necessary, we may assume that is the zero of of the smallest modulus. Obviously, as . Set . We have that (a1) in ;(a2)all zeros of are multiple and ;(a3) and in .By Lemma 16 and Lemma 12, is normal in and quasinormal in . Thus, there exists a subsequence of (still denoted by ) and such that (b1) has no accumulation point in ;(b2)for each , no subsequence of is normal at ;(b3) in .Obviously, and all zeros of are multiple in .
Subcase 2.1 (). By , , and hence is a meromorphic function in .
We claim that . Otherwise, let . Since , by Lemma 11, which contradicts that .
By Lemma 8 and , either or in . Since , we have in . By Theorem B, must be rational, and then by Lemma 4, where is a nonnegative integer, , and . Now, we have By Hurwitz's theorem, there exist sequences and as (counting multiplicities of zeros and poles, resp.) such that for sufficiently large , and , where and . Write . Thus, and as . Set .
Subcase 2.1.1. For each , has at least zeros (counting multiplicities) in for sufficiently large .
Taking a subsequence and renumbering if necessary, we may assume that is the zero of of the smallest modulus in . Obviously, as . Observing that and , where , by Hurwitz's theorem and (26), we have as . Let . We have that for sufficiently large ,(c1) has only zeros in . Obviously, as ;(c2)all zeros of are multiple and ;(c3) and in .By Lemma 16 and Lemma 12, is normal in and quasinormal in . Thus, there exists a subsequence of (still denoted by ) and such that (d1) has no accumulation point in ;(d2)for each , no subsequence of is normal at ;(d3) in .Obviously, and all zeros of are multiple in .
Let By (26), in . Hence
(1) ( in .) Obviously, has no zeros in for sufficiently large . Applying the maximum principle to the sequence of analytic functions, we see that in which contradict (29).
(2) is a meromorphic function in .
We claim that in . By Lemma 13, , where is a constant. Since has no zeros in for sufficiently large , applying the maximum principle to the sequence of analytic functions, we have in . Hence, , and then we get that by (29).
Suppose that . By , which contradicts . Suppose that . By Lemma 11, which contradicts .
Subcase 2.1.2. There exists such that has exactly zeros (counting multiplicities) in for sufficiently large .
Now, holds with and . Next, we will show that also holds.
Set By (26), in , and hence we have
(1) ( in .) Since has exactly zeros in for sufficiently large , has no zeros in for sufficiently large . By the maximum principle applied to , we have which contradicts (31).
(2) ( is a meromorphic function in .) By Lemma 13, in , and hence in , where is a constant. Since has no zeros in for sufficiently large , by the maximum principle applied to , in , and hence . By (31), . Now, in .
Subcase 2.2 (). By Lemma 11, Let be the th root of the equation , where .
We claim that . Suppose that , where . Obviously, we have and which contradict that all of zeros of are multiple. Suppose that , where . By Lemma 11, in . By (32), . A contradiction.
By Lemma 11, there exists such that for sufficiently large , has a single zero of order 2 and a single pole of order in . Set . Thus, and as , where . Set .
Subcase 2.2.1. For each , has at least zeros (not counting multiplicities) in for sufficiently large .
Taking a subsequence and renumbering if necessary, we may assume that is the zero of of the smallest modulus in . Obviously, . Since , we have , where . Since has a single zero of order 2 in , by Hurwitz's theorem and (32), we have as . Let . We have that for sufficiently large (e1) has only zeros of order 2 and at least poles of order 1 in , and obviously, and as ;(e2)all zeros of are multiple and ;(e3) and in .By Lemma 16 and Lemma 12, is normal in and quasinormal in . Thus, there exist a subsequence of (still denoted by ) and such that(f1) has no accumulation point in ;(f2)for each , no subsequence of is normal at ;(f3) in .Obviously, and all zeros of are multiple in .
Let By (32), Hence
(1) ( in .) Obviously, has no zeros in for sufficiently large . Applying the maximum principle to the sequence of analytic functions, we have that in which contradict (35).
(2) ( is a meromorphic function in .) We claim that in . By Lemma 13, , where is a constant. Since has no zeros in for sufficiently large , applying the maximum principle to the sequence of analytic functions, we have in . Hence, , and then by (35).
Suppose that . By , which contradicts that . Suppose that . By Lemma 11, which contradicts that .
Subcase 2.2.2. There exists such that has exactly zeros (not counting multiplicities) for sufficiently large .
Similar to the previous treatment in Subcase , we finally can show that and hold.
Case 3 (). Obviously, and hold by Lemma 11.

#### 5. Proof of Theorem 1

Proof. We assume that has at most finitely many solutions and derive a contradiction. Let as , where and .
Clearly, and as . By Lemma 7, as , where is a constant. By standard results in Nevanlinna theory, and as . Thus, as . Since as , we obtain that .
Set . By Lemma 15, there exist sequences and such that
Let be the two fundamental periods of and let be a fundamental parallelogram of . There exist integers and such that , where . There exists a subsequence of (still denoted by ) such that as . Set Clearly, we have , , and . By (36), we have