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Discrete Dynamics in Nature and Society
Volume 2018, Article ID 4673165, 12 pages
https://doi.org/10.1155/2018/4673165
Research Article

Uniqueness of L-Functions Concerning Certain Differential Polynomials

1Department of Mathematics, Fujian Normal University, Fuzhou 350117, China
2Key Laboratory of Applied Mathematics (Putian University), Fujian Province University, Fujian Putian 351100, China

Correspondence should be addressed to Jun-Fan Chen; moc.361@nehcnafnuj

Received 21 June 2018; Accepted 18 September 2018; Published 18 October 2018

Academic Editor: Cengiz Çinar

Copyright © 2018 Wen-Jie Hao and Jun-Fan Chen. 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

Relying on Nevanlinna theory and the properties of L-functions in the extended Selberg class, we mainly study the uniqueness problems on L-functions concerning certain differential polynomials. This generalizes some results of Steuding, Li, Fang, and Liu-Li-Yi.

1. Introduction

The Riemann hypothesis as one of the millennium problems has been given a lot of attention by mathematical workers for a long time. Selberg guessed that the Riemann hypothesis is also true for L-functions in the Selberg class. Such an L-function based on Riemann zeta function as the prototype is defined to be a Dirichlet seriesof a complex variable satisfying the following axioms:

(i) Ramanujan hypothesis: for every

(ii) Analytic continuation: there is a nonnegative integer such that is an entire function of finite order

(iii) Functional equation: satisfies a functional equation of type where with positive real numbers Q, , and complex numbers , with and

(iv) Euler product: , where unless is a positive power of a prime and for some

It is mentioned that there are many Dirichlet series only satisfying axioms (i)-(iii) [1] and are regarded as the extended Selberg class. All the L-functions are studied in this article from the extended Selberg class. Therefore, the conclusions obtained in this article are also true for L-functions in the Selberg class. The uniqueness of two L-functions was firstly studied by Steuding [2], as seen from Theorem 1.

Theorem 1 (see [2]). Suppose that is a finite complex number. If two L-functions and with share CM, then .

In 2016, Hu and Li [3] gave an example and . From this we can know the above theorem is false when .

Due to the complication to study the distribution of public zero of two L-functions, researchers take up study of the relationship of an L-function and a meromorphic function. Since L-function itself can be analytically continued as a meromorphic function in the whole complex plane, therefore, L-functions will be taken as special meromorphic functions, with the help of Nevanlinna’s value distribution theory, in order to study the uniqueness of L-functions. Suppose that and are two nonconstant meromorphics in the whole complex plane; denotes a values in the extended complex plane. If and have the same zeros counting multiplicities, we say that and share CM. If and have the same zeros ignoring multiplicities, then we say that and share IM. One nonconstant meromorphic function in the whole complex plane can be determined by five such preimages or four such preimages [4, 5]. In 2010, Li [6] considered a meromorphic function and a nonconstant L-function and he obtained the following.

Theorem 2 (see [6]). Let and be two distinct finite values and be a meromorphic function in the complex plane with finitely many poles. If and a nonconstant L-function share CM and IM, then .

In 1997, the following question was raised by Lahiri [7]: what is the relationship between function and function , when two differential polynomials have the same nonzero finite value? The two differential polynomials are generated by and , respectively. In this direction, Fang [8] proved the following theorem.

Theorem 3 (see [8]). Let and be two nonconstant entire functions, and let , be two positive integers. Suppose that and share 1 CM. If , then .

Recently, Liu-Li-Yi [9, 10] considered an L-function and a meromorphic function whose certain differential polynomials share one finite nonzero value. The following conclusions were obtained.

Theorem 4 (see [9]). Let be a nonconstant meromorphic function, let be an L-function, and let and be two positive integers. Suppose that and share 1 CM. If , then for a constant satisfying .

Theorem 5 (see [10]). Let be a nonconstant meromorphic function, let be an L-function, and let and be two positive integers. Suppose that and share 1 CM. If and , then .

Naturally, is it still set up if it can be generalized to the general differential polynomials, for instance, or ? For simplicity, we use the notations and , where

In this paper, we have the results as follows.

Theorem 6. Let be a nonconstant meromorphic function, let be an L-function, and let , , and be three positive integers and , be two constants satisfying . Suppose that and share 1 CM. If , then , where(i)when , is a constant such that (ii)when , , is a constant such that

Remark 7. In Theorem 6, if , , we can get Theorem 4. If , , we can get Theorem 5. Moreover, if the condition substitutes in Theorem 5, then we can obtain , which implies the conclusion remains valid in Theorem 5. Therefore, Theorem 6 is the generalization of Theorems 4 and 5.

Remark 8. In Theorem 6, the condition cannot be dropped. Let , , , , , . Then , . This shows that and share 1 CM. But .

Theorem 9. Let be a nonconstant meromorphic function, let be an L-function, and let , , and be three positive integers. Suppose that and share 1 CM. If and , then or .

Remark 10. In Theorem 9, if , we can get Theorem 5. Actually, if and , we can get a contradiction by subcase 3.2 in the proof of Theorem 6. We can also see that Theorem 5 is included in Theorem 9.

Moreover, we consider what will happen if the CM becomes the IM in Theorems 6 and 9. The theorems are then established.

Theorem 11. Let be a nonconstant meromorphic function, let be an L-function, and let , , and be three positive integers and , be two constants satisfying . Suppose that and share 1 IM. If , then , where(i)when , is a constant such that (ii)when , , is a constant such that

Theorem 12. Let be a nonconstant meromorphic function, let be an L-function, and let , , and be three positive integers. Suppose that and share 1 IM. If , , then or .

Corollary 13. Let be a nonconstant meromorphic function, let be an L-function, and let , be two positive integers. Suppose that and share 1 IM. If , then for a constant satisfying .

Corollary 14. Let be a nonconstant meromorphic function, let be an L-function, and let , be two positive integers. Suppose that and share 1 IM. If and , then .

To prove the main theorems, the order of a meromorphic function will be needed. It is defined to be a superior limit Next, we introduce some definitions.

Definition 15. Suppose that is a positive integer and is a value in the extended complex plane. Then is defined as the counting function of those zeros of of order . is defined as the counting function of those zeros of of order . , are defined as the corresponding reduced counting functions.

Definition 16. Suppose that is a common c-point of and with multiplicity and , respectively. We denote by the reduced counting function of those c-points of and where and denote by the reduced counting function of those c-points of and where . Similarly, we can define and .

Definition 17. Suppose that is a value in the extended complex plane and is a positive integer. We define

2. Some Lemmas

In order to facilitate the proofs of the theorems, we list some important lemmas which will be employed in this paper.

Lemma 18 (see [11]). Let be a nonconstant meromorphic function and let be an irreducible rational function in with constant coefficients and , where and . Then , where .

Lemma 19 (see [4], Theorem 3.2). Let be a nonconstant meromorphic function, let be a positive integer, and let be a nonzero finite complex number. Then where is the counting function which only counts those points such that but .

Lemma 20 (see [5]). Let be a nonconstant meromorphic function and be a positive integer. Then

Lemma 21. Let be a nonconstant meromorphic function, let be an L-function, and let , , and be three positive integers and , be two constants satisfying . SetIf and share 1 CM and , then , where , are two constants.

Proof. LetIf is a constant, we have from (12) that is a constant, . Since and share 1 CM, we obtain . Using Lemma 18, we deduceMoreover, from Lemma 19, we know Combining (13) with , we get that is a nonconstant L-function. Similarly, using Lemma 18, we deduceSince , we getSuppose , is a zero of of order , and is a zero of of order , in view of and sharing 1 CM. By checking the Laurent expansion of , we have when and when . ThusSimilarly, assume that is a pole of of order 1; we get . If is a pole of of order 1, we get . Therefore, we get, by a calculation and (11), thatwhere is the reduced counting function of those zeros of not that of . From the definition, we obtainBy Lemma 20, we obtainCombining (19) with (20), we haveWe can getin view of the assumption that and share 1 CM. Combining (17)-(22), the second fundamental theorem yieldsFrom the first fundamental theorem, we know whereLet be a zero of of order , . Since , , we can deduce the zeros of of order . ThusAlso, for , we can deduce thatCombining (24)-(27), we getBy Lemma 20, we obtainUsing the first fundamental theorem, we haveFrom(13), (16), (28)-(30), we getSimilarly, we haveSince at most has one pole, we know . At the same time, we know , , . By (31) and (32), we can obtain the following results, respectively:Assume that there exists some subset with its linear measure satisfying , as and . Then it follows from (33) that , which contradicts . Assume that there exists some subset with its linear measure such that , as and . Then it follows from (34) that , which contradicts . Therefore, . That is, Integrating this gives , where , are two constants.
This completes the proof of Lemma 21.

Remark 22. Suppose that and are replaced by and , respectively, in Lemma 21. That is still true. In fact, is replaced by in (31) and (32); we still get in (33) and (34).

Lemma 23 (see [12], Theorem 1.2). Suppose that is a meromorphic of finite order in the complex plane and that has finitely many zeros for some . Then has finitely many poles in the complex plane.

Lemma 24 (see [4]). Let be a transcendental meromorphic function, and let , be two distinct meromorphic functions such that , . Then

Lemma 25 (see [13]). Let and be relatively prime integers, and let be a finite complex number such that . Then there exists one and only one common zero of and .

Lemma 26 (see [14, 15]). Let be a nonconstant meromorphic function, and let , be two positive integers. Then

Lemma 27. Let be a nonconstant meromorphic function, let be an L-function, and let , , and be three positive integers and , be two constants satisfying . SetIf and share 1 IM, , then , where , are two constants.

Proof. LetIn the same manner as Lemma 21, we know is a nonconstant L-function. By Lemma 18, we haveSince , we getSuppose ; is a zero of of order , and is a zero of of order , which is possibly different from , in view of and sharing 1 IM. By checking the Laurent expansion of , we have when . ThusSimilarly, if is a pole of of order 1, we get . If is a pole of of order 1, we get . Therefore, we get, by a calculation and (39), thatwhere is the reduced counting function of those zeros of not that of . Note that and share 1 IM. We haveCombining (44)-(46), we getNote thatSubstituting (48) into (47), we obtainBy the second fundamental theorem and (49), we get Namely,Note that Let be a zero of of order , . Since , , we can deduce the zeros of of order . Therefore,Also, for , we haveCombining (51)-(54) and the first fundamental theorem shows thatFrom (43), we see thatBy (55), (56), and the first fundamental theorem, we haveBy Lemma 20, we haveBy Lemma 26, we haveIn addition, we know thatThe same inequality (58)-(60) holds for . Substituting (58)-(60) into (57), and with (41) we deduce thatLikewise, we haveSince at most has one pole, we get . At the same time, we have , , . By (61) and (62), we can obtain the following results, respectively:Assume that there exists some subset with its linear measure satisfying , as and . Then it follows from (63) that , which contradicts . Assume that there exists some subset with its linear measure satisfying , as and . Then it follows from (64) that , which contradicts . Therefore, . That is, Integrating this gives , where , are two constants.
This completes the proof of Lemma 27.

Remark 28. In Theorem 11, let . By Lemma 27, we can get , where , are two constants. We can get that and share 1 CM. Then we can get Theorem 11 by Theorem 6. Similarly, we get Theorem 12 by Theorem 9.

3. Proof of Theorem 6

LetBy Lemma 21, we havewhere , are two constants.

We discuss three cases.

Case 1. and . Then (67) can be written asSubcase 1.1. . From (68) we have ,Let be the degree of . Then , where and are the numbers of the axiom (iii) of the definition of L-function. Thus, by Steuding [2], p.150, we getNext we distinguish two cases.
Subcase 1.1.1.. By (69), (70), Lemma 18, and a result from Whittaker [16], p.82, we haveSince at most has one pole , we deduce by (69) that at most has one zero . By (71), the assumption , and Lemma 23, we have that and so has at most finitely many poles. This together with (69) implies that has at most finitely many zeros. Moreover, by the assumption , we deduce that has at most finitely many zeros. Thus,where is a rational function such that has neither a pole nor a zero, , and are constants. In view of (72) and Hayman [4], p.7, we have which contradicts (70).
Subcase 1.1.2. . By , we can consider two subcases.
Subcase 1.1.2.1., . Then (69) becomes . Assume that is a zero of of order . Then, we can get that is a pole of of order , satisfying , that is, , and we have , which contradicts the assumption . Hence, we prove that has no zeros,where is a rational function satisfying that has no poles and , are two constants. In view of (74) and Hayman [4], p.7, we have which contradicts (70).
Subcase 1.1.2.2. , . By using the argument as in Subcase 1.1.2.1, we obtain that , and so , which contradicts the assumption . Hence, we know that has no zeros. Similarly, we get a contradiction.
Subcase 1.2.. Then it follows from (68) thatNoting that has at most one pole, then has at most one pole; from (76) we have that has at most one zero. By Lemma 19, we obtain