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Journal of Function Spaces

Volume 2016 (2016), Article ID 9315248, 5 pages

http://dx.doi.org/10.1155/2016/9315248

## Filling Disks of Hayman Type of Meromorphic Functions

^{1}Department of Mathematics, School of Science, China University of Mining and Technology, Beijing 100083, China^{2}Beijing Key Laboratory of Information Service Engineering, Department of General Education, Beijing Union University, No. 97 Bei Si Huan Dong Road, Chaoyang District, Beijing 100101, China

Received 4 January 2016; Accepted 10 April 2016

Academic Editor: Jaeyoung Chung

Copyright © 2016 Nan Wu and Zuxing Xuan. 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 obtain the existence of the filling disks with respect to Hayman directions. We prove that, under the condition , there exists a sequence of filling disks of Hayman type, and these filling disks can determine a Hayman direction. Every meromorphic function of positive and finite order has a sequence of filling disks of Hayman type, which can also determine a Hayman direction of order .

#### 1. Introduction and Results

For a meromorphic function , disks of the formwhere and , are called filling disks of if takes every extended complex value with at most two exceptions infinitely often in any infinite subcollection of them, which are also called the filling disks of Julia type. We can prove that if a meromorphic function satisfies then must possess a sequence of filling disks of Julia type, and these filling disks can determine a Julia direction. A direction is said to be a Julia direction for a meromorphic function , if, given , takes all complex values infinitely often in the region except possibly two exceptions.

For a meromorphic function of order , disks of the form (1), where and , are called filling disks of Borel type of if takes every extended complex value at least times, except some complex values which can be contained in a disk whose spherical radius is , where . A meromorphic function with positive and finite order must possess a sequence of Borel type filling disks, which determine a -order Borel direction, and a -order Borel direction also determines a sequence of Borel type filling disks. A direction is said to be a Borel direction for a meromorphic function of order , if, given , for all complex values , at most with two possible exceptions. Here and throughout the paper, denotes the numbers of the roots of in the region .

For the case of Hayman direction, we pose a question whether there exist filling disks of Hayman type. In this paper, we mainly obtain the following two theorems.

Theorem 1. *Let be a meromorphic function in the plane satisfying Then, there exists a sequence of filling disks with the form (1), where and ; in each disk, takes all complex values at least times or else there exists an integer number such that takes all complex values, except possibly zero at least times, where . Moreover, these filling disks can determine a Hayman direction , such that for arbitrary small , positive integer , and complex numbers and we have*

Theorem 2. *Let be a meromorphic function in the plane of order . Then, there exists a sequence of filling disks with the form (1), where and ; in each disk, takes all complex values at least times or else there exists an integer number such that takes all complex values, except possibly zero at least times, where . Moreover, these filling disks can determine a Hayman direction , such that, for arbitrary small , positive integer , and complex numbers and , we have*

*Remark*. The filling disks in Theorem 1 are called the filling disks of Hayman-Julia type. Moreover, if we add the growth condition to , we can obtain the filling disks of Hayman-Borel type in Theorem 2. Hayman-Borel type filling disks are more precise than Hayman-Julia type filling disks.

#### 2. Proof of Theorem 1

First of all, let us recall the definition of Ahlfors-Shimizu characteristic in an angle (see [1]). Let be a meromorphic function defined in an angle . Set . Define To prove our theorems, we need the following lemma which was first established by Chen and Guo [2] and essentially comes from the Hayman inequality and the estimation of primary values appeared in the inequality and was used to confirm the existence of Hayman directions of meromorphic functions by Zheng and the first author [3].

Lemma 3. *Let be meromorphic in and let for two complex numbers and with . Then, we have where is a positive constant depending only on .*

Lemma 3 can be obtained by using Lemma 8 of Chen and Guo [2] to the function in the unit disk and noticing the following: where is the area element on the Riemann sphere and is the chordal distance between and .

The following lemma comes from [4], which is very useful for our study.

Lemma 4. *Let be meromorphic in the plane, and let , , and be arbitrary positive numbers with and . If satisfies then there exists a positive number such thatwhere *

In view of Lemmas 3 and 4, we can prove Theorem 1.

*Proof of Theorem 1. *For each , let , , and . Then, there exists , such that Divide into domains , where Then, there exists at least one , such that where . In view of for large enough , we have Here we choose the proper , making . The proof is complete.

Here we point out that, under the current technology, we cannot replace the growth condition (4) with (2). Rossi [4] also obtained that if the growth condition (4) was replaced by (2), then the inequality in Lemma 4 should be replaced by . Unfortunately, the inequality of Lemma 3 has the term ; owing to this term, we cannot replace the growth condition, because we notice that if the growth condition is (2) then the term should be bounded. If we choose , then the term can be bounded but we cannot assure that as .

#### 3. Proof of Theorem 2

The following lemma can be proved by the same method of Lemma 4, and we prove it for the completeness.

Lemma 5. *Let be meromorphic in the plane with order , and let , , and be arbitrary positive numbers with and . Then, for any , there exists a positive number such that *

*Proof. *Suppose that the lemma is not true. Then, for every , Choose larger than and set . Then, there exists such that Let and . Combining (19) with (20), we have It follows from (20) that Substituting (22) into (21) leads to the order of being by the fact that and are fixed.

*Proof of Theorem 2. *For each , let , , , and . Then, there exists , such that Divide into domains , where Then, there exists at least one , such that where . In view of when is large enough, we have Choosing , we can obtain the result. Hence, we complete the proof of Theorem 2. It is not difficult to see that these filling disks can determine a Borel direction of order .

Zhang and Yang [5] also obtained a sequence of filling disks, whose result is as follows: let be a meromorphic function of positive and finite order . Then, there exists a sequence such that at least one of the following holds:(1) in takes all complex values at least times, with some exceptional values which can be contained in a spherical disk with center and radius .(2)For any fixed positive integer , in takes all complex values at least times, with some exceptional values which can be contained in two spherical disks with center and and radius , where .

We can see that our result is different from theirs.

#### 4. Filling Disks on an Angular Region

In Yang’s book [6], he said that a Borel direction of can determine a sequence of Borel type filling disks. In this paper, we can see that a order Borel direction can determine not only a sequence of Borel type filling disks but also a sequence of Hayman type filling disks. Actually, if a meromorphic function on any angular region satisfies some growth conditions, it possesses a sequence of filling disks. Zhang [7] obtained the following lemma, which established the growth condition on any angular domain containing the Borel direction .

Lemma 6. *Let be a meromorphic function in the plane of order . Then, a half line is a -order Borel direction if and only if it satisfies *

In view of Lemma 6, we have the following result.

Theorem 7. *Let be a meromorphic function in the plane and let be a Borel direction of order . Then, there exists a sequence of disks of the form where , . In , takes all but possibly two extended complex values with exponent and takes and takes with exponent , where and .*

*Proof. *For each , let , , , and . It follows from (29) and Lemma 5 that there exists , such that Divide into domains , where Then, there exists at least one , such that where . In view of when is large enough, we have Choosing , we can obtain the result.

At last, we pose a question: *Does a Hayman direction of order ** have a sequence of filling disks of Hayman type*?

#### Competing Interests

The authors declare that they have no competing interests.

#### Acknowledgments

Nan Wu is supported in part by the grants of NSF of China (nos. 11231009, 11326086, and 11371363). Zuxing Xuan is supported in part by NNSFC (no. 91420202), Beijing Natural Science Foundation (no. 1132013), and the Project of Construction of Innovative Teams and Teacher Career Development for Universities and Colleges Under Beijing Municipality (CIT and TCD201504041, IDHT20140508).

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