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Abstract and Applied Analysis
Volume 2012 (2012), Article ID 407351, 8 pages
doi:10.1155/2012/407351
Uniqueness Theorems on Difference Monomials of Entire Functions
1Shandong Transport Vocational College, Weifang, Shandong 261206, China
2Laiwu Vocational and Technical College, Laiwu, Shandong 271100, China
3Department of Physics and Mathematics, University of Eastern Finland, P.O. Box 111, 80101 Joensuu, Finland
Received 5 April 2012; Accepted 11 June 2012
Academic Editor: Matti Vuorinen
Copyright © 2012 Gang Wang et al. 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
The aim of this paper is to discuss the uniqueness of the difference monomials . It assumed that and are transcendental entire functions with finite order and , where is a nonzero complex constant and , are integers. It is proved that if one of the following holds (i) and , (ii) and , and (iii) and , then or for some constants and which satisfy and . It is an improvement of the result of Qi, Yang and Liu.
1. Introduction and Main Results
In this paper, a meromorphic (respectively entire) function always means meromorphic (respectively, analytic) in the complex plane . It is also assumed that the reader is familiar with the basic concepts of the Nevanlinna theory. We adopt the standard notations in the Nevanlinna value distribution theory of meromorphic functions as explained in [1, 2].
Let and be two nonconstant meromorphic functions, and let be a value in the extended plane. We say that and share the value CM, provided that and have the same -pints with the same multiplicities. We say that and share the value IM, provided that and have the same -points ignoring multiplicities. The order of is defined by
Let be a nonconstant meromorphic function on , let be a finite value, and let be a positive integer or infinity. We denote by the set of zeros of and count multiplicities, while by the set of zeros of but ignore multiplicities. Also, we denote by the set of zeros of with multiplicities less than or equal to and count multiplicities. For , we denote by the counting function corresponding to the set while by the counting function corresponding to the set . If , we say that , share the value a with weight .
The definition implies that if and share a value with weight , then is a zero of with multiplicity if and only if it is a zero of with multiplicity and is a zero of with multiplicity if and only if it is a zero of with multiplicity where is not necessarily equal to .
Also, we denote by and the reduced forms of and , respectively. At last, we set
Hayman proposed the well-known conjecture in [3].
Hayman Conjecture
If an entire function satisfies for all , then is a constant.
In fact, Hayman has proved that the conjecture holds in the cases in [4] while Clunie proved the cases in [5], respectively. In 1997, Yang and Hua [6] studied the uniqueness theorem of the differential monomials and obtained the following result.
Theorem A. Let and be nonconstant entire function, and let be an integer. If and share 1 CM, then either , where , and are constants satisfying , or for a constant such that .
In 2010, Qi et al. [7] studied the uniqueness of the difference monomials and obtained the following result.
Theorem B. Let and be transcendental entire functions with finite order, a non-zero complex constant, and an integer. If , then or for some constants and which satisfy and .
In this paper, we will obtain the following results.
Theorem 1.1. Let and be transcendental entire functions with finite order, a non-zero complex constant, and an integer. If , then the assertion of Theorem B holds.
Theorem 1.2. Let and be transcendental entire functions with finite order, a non-zero complex constant, and an integer. If , then the assertion of Theorem B holds.
Theorem 1.3. Let and be transcendental entire functions with finite order, a non-zero complex constant, and an integer. If , then the assertion of Theorem B holds.
2. Auxiliary Results
Lemma 2.1 (see [8, Corollary 2.5]). Let be a meromorphic function in the complex plane with finite order , and let be a fixed non-zero complex number. Then for each , one has
Lemma 2.2 (see [8, Theorem 2.1]). Let be a meromorphic function in the complex plane with finite order , and let be a fixed non-zero complex number. Then for each , one has
Lemma 2.3. Let be an entire function with finite order , a fixed non-zero complex number, and where are constants. If , then
Proof. Since is an entire transcendental function with finite order, we can deduce from Lemma 2.1 and the standard Valiron-Mohon'ko theorem that Therefore On the other hand, Lemma 2.2 implies that We will obtain the conclusion of Lemma 2.3.
Remark 2.4. The condition “entire” cannot be replaced by “meromorphic” in Lemma 2.3, as is shown by the following example.
Example 2.5. Let , , and , we can see for every set of with infinite measure.
Lemma 2.6 (see [9, Lemma 2.1]). Let and be two nonconstant meromorphic functions satisfying for some positive integer . Define as follows: If , then where denotes the counting function of zeros of but not zeros of and is similarly defined.
Lemma 2.7 (see [10]). Under the condition of Lemma 2.6, one has
Lemma 2.8 (see [10]). Let be defined as Lemma 2.6. If , then either or provided that where and is a set with infinite linear measure.
Lemma 2.9 (see [11, Lemma 2.2]). Let be a nondecreasing continuous function, , , and let be the set of all such that If the logarithmic measure of is infinite, then
3. Proof of Theorem 1.1
We define First of all, suppose that . We replace and by and , respectively, in Lemma 2.7 and Lemma 2.8. Thus, Applying the second main theorem to and jointly implies that Noting that According to Lemma 2.9 and (3.2)–(3.4), we can obtain that Lemma 2.3 shows that We can deduce that which is impossible since . Therefore, we have . Noting that where . Together with Lemma 2.8, it shows that either or . We will consider the following two cases.
Case 1. Suppose that . Therefore Let ; we have If is not a constant, then Lemma 2.2 and (3.10) imply that which is a contraction with . Thus, , where is a constant. From (3.10), we have and .
Case 2. Suppose that . Therefore Let ; we have By the same way as Case 1, we can obtain that is a constant. Therefore, and .
4. Proof of Theorem 1.2
Noting that According to (3.1) and (4.1), we can obtain the conclusion of Theorem 1.2 by the same way as Section 3.
5. Proof of Theorem 1.3
Noting that According to (3.1) and (5.1), we can obtain the conclusion of Theorem 1.2 by the same way as Section 3.
Acknowledgment
The third author was supported in part by China Scholarship Council (CSC).
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