Uniqueness theorems for L-functions in the extended Selberg class

Wen-Jie Hao: College of Mathematics and Informatics/Fujian Key Laboratory of Mathematical Analysis and Applications, Fujian Normal University, Fuzhou 350117, China, E-mail: wenjiehao1993@163.com *Corresponding Author: Jun-Fan Chen: College of Mathematics and Informatics/Fujian Key Laboratory of Mathematical Analysis and Applications, Fujian Normal University, Fuzhou 350117, China and Key Laboratory of Applied Mathematics (Putian University), Fujian Province University, Fujian Putian, 351100, China, E-mail: junfanchen@163.com


Introduction
The Riemann hypothesis as one of the millennium problems has been given a lot of attention by many scholars for a long time. Selberg guessed that the Riemann hypothesis also holds for the L-function in the Selberg class. Such an L-function based on the Riemann zeta function as a prototype is de ned to be a Dirichlet series of a complex variable s = σ + it satisfying the following axioms [1]: (i) Ramanujan hypothesis: a(n) ≪ n ε for every ε > . with positive real numbers Q, λ j , and complex numbers ν j , ω with Reν j ≥ and ω = .
(iv) Euler product: log L(s) = ∑ Selberg class. Therefore, the conclusions proved in this article are also true for L-functions in the Selberg class. Theorems in this paper will be proved by means of Nevanlinna's Value distribution theory. Suppose that F and G are two nonconstant meromorphic functions in the complex plane C, c denotes a value in the extended complex plane C ∪ {∞}. If F − c and G − c have the same zeros counting multiplicities, we say that F and G share c CM. If F − c and G − c have the same zeros ignoring multiplicities, then we say that F and G share c IM. It is well known that two nonconstant meromorphic functions in C are identically equal when they share ve distinct values IM [3,4]. The following uniqueness theorem of two L-functions was proved by Steuding [1]. Theorem 1.1 (see [1]). If two L-functions with a( ) = share a complex value c ≠ ∞ CM, then they are identically equal.

Remark 1.2.
In [5], the authors gave an example that L = + s and L = + s , which showed that Theorem 1.1 is actually false when c = .
In 2011, Li [6] considered values which are shared IM and got Theorem 1.3 (see [6]). Let L and L be two L-functions satisfying the same functional equation with a( ) = and let a , a ∈ C be two distinct values.
In 2001, Lahiri [7] put forward the concept of weighted sharing as follows.
Let k be a nonnegative integer or ∞, , we say that f and g share the value c with weight k (see [7]).
In 2015, Wu and Hu [8] removed the assumption that both L-functions satisfy the same functional equation in Theorem 1.3. By including weights, they had shown the following result. Theorem 1.4 (see [8]). Let L and L be two L-functions, and let a , a ∈ C be two distinct values. Take two positive integers k , k with k k > . If E k j (a j , L ) = E k j (a j , L ), j = , , then L ≡ L .
In 2003, the following question was posed by C.C. Yang [9]. Question 1.5 (see [9]). Let f be a meromorphic function in the complex plane and a, b, c are three distinct values, where c ≠ , ∞. If f and the Riemann zeta function ζ share a, b CM and c IM, will then f ≡ ζ?
The L-function is based on the Riemann zeta function as the model. It is then valuable that we study the relationship between an L-function and an arbitrary meromorphic function [10][11][12][13][14]. This paper concerns the problem of how meromorphic functions and L-functions are uniquely determined by their c-values. Firstly, we introduced the following theorem. Then, using the idea of weighted sharing, we will prove the following theorem. Theorem 1.7. Let f be a meromorphic function in the complex plane with nitely many poles, let L be a nonconstant L-function, and let a , a ∈ C be two distinct values. Take two positive integers k , k with k k > . If E k j (a j , f ) = E k j (a j , L), j = , , then L ≡ f . Remark 1.8. Note that an L-function itself can be analytically continued as a meromorphic function in the complex plane. Therefore, an L-function will be taken as a special meromorphic function. We can also see that Theorem 1.4 is included in Theorem 1.7.
In 1976, the following question was mentioned by Gross in [15].

Question 1.9 (see [15]). Must two nonconstant entire functions f and f be identically equal if f and f share a nite set S?
Recently, Yuan, Li and Yi [16] considered this question leading to the theorem below. Theorem 1.10 (see [16]).
Here l is a positive integer satisfying ≤ l ≤ n, n and m are relatively prime positive integers with n ≥ and n > m, and a, b, c are nonzero nite constants, where c ≠ ω j for ≤ j ≤ l. Let f be a nonconstant meromorphic function such that f has nitely many poles in C, and let L be a nonconstant L-function. If f and L share S CM and c IM, then f ≡ L.
Concerning shared set, we prove the following theorem. Furthermore, we obtain a result which is similar to Theorem 1.10 by di erent means.

Some lemmas
In this section, we present some important lemmas which will be needed in the sequel. Firstly, let f be a meromorphic function in C. The order ρ (f ) is de ned as follows: Lemma 2.1 (see [4], Lemma 1.22). Let f be a nonconstant meromorphic function and let k ≥ be an integer.
which imply that the order and the lower order of T (r) are not greater than the order and the lower order of T (r) respectively. Lemma 2.4 (see [4], Theorem 1.14). Let f and g be two nonconstant meromorphic functions. If the order of f and g is ρ (f ) and ρ (g) respectively, then Because f and L share a , a weighted k , k respectively, by (3), from the rst and second fundamental theorems we have Then from (4) and Lemma 2.3 we obtain ρ(f ) ≤ ρ(L).
We introduce two auxiliary functions below.
Next, we assume that F ≡ and F ≡ . By (8) By the assumption L and f share (a , k ), (a , k ), from (3), (9) and (11) we have Similarly, from (3), (10) and (11) we have Combining (12) with (13) yields Since k k > , from (14) we obtain Substituting (15) into (12) implies Noting L and f share (a , k ), (a , k ), combining (15) with (16) yields Clearly, where Q is a rational function satisfying that G is a zero-free entire function. From (17) and (18), it is easy to see that such a Q does exist. By Lemma 2.2 and Lemma 2.4 we get By the Hadamard factorization theorem [19], p.384, we know where ϕ is a polynomial of degree at most deg(ϕ) ≤ . We may write ϕ = a s + b for some complex numbers a , b . In view of (20) and Hayman [3], p.7, we have By (19), the assumption that L and f share a , we get that every a -point of L has to be 1-point of G Q − . Now (20), (21) and the rst fundamental theorem yield Similarly, set We also get This contradicts (2). Thus, F ≡ or F ≡ . By integration, we have from (9) that where A(≠ ) is a constant. This implies that L and f share a CM. Hence by Theorem 1.6 we deduce Theorem 1.7 holds. If F ≡ , using the same manner, we also have the conclusion. This completes the proof of Theorem 1.7.

. Proof of Theorem 1.11
First we consider the following function where is a rational function satisfying that G has no zeros and no poles in C; A is a nonzero nite value; m is the nonnegative integer in the axiom (ii) of the de nition of L-functions.
We claim that such a Q does exist. By the condition that f and L share S CM, set We can see that there can be only a pole of f or L such that F = or F = ∞. Since f has no pole and L has only one possible pole at s = , it follows that F has no zero and only one possible pole at s = . Hence such a Q does exist. Next, assume that a , a , ⋯, a n are all distinct roots of R(a). Using the rst fundamental theorem we get Noting n ≥ , by the second fundamental theorem we have which gives This together with Lemma 2.3 yields Similarly, ρ(L) ≤ ρ(f ). From the Hadamard factorization theorem [19], p.384 we see where h(s) is a polynomial of degree deg(h(s)) ≤ . One can write a polynomial in σ with α(t), β(t) being polynomials in t. Now the claim is α(t) ≡ . From (25), (27) and (32) we get Since lim σ→+∞ L(s) = , lim σ→+∞ f (s) = k(k ≠ ∞), R(k) ≠ and R( ) ≠ , it follows that where C ≠ is a nite value. If α(t) ≡ , we obtain α(t ) ≠ for some value t . If α(t ) > , from (34) we know that Thus from (26), (33), (35) and (36) we can deduce that, C = ∞ when σ → +∞ with t = t , which is a contradiction. Similarly, if α(t ) < , we have that, C = when σ → +∞ with t = t , which is also a contradiction. Therefore α(t) ≡ . Now by (33) and (36) we get Combining (35) with (37) yields for a xed t. Considering that the limit of Q as σ → +∞ is a nonzero nite constant for some value t and n ≥ , in view of (26) we see that m = , and then Q(s) ≡ A. From (38) we have e β(t) = A C . Thus it follows by (37) that Since C ≠ is a nite complex number, from (39) we deduce that R(L) R(f ) is a constant. Then by (35) we know that From the assumption in the theorem we have f (s ) = L(s ) = b for some s ∈ C. It now follows from (40) that This completes the proof of Theorem 1.11.

. Proof of Theorem 1.12
First, we have that the algebraic equation ω n+m + αω n + β = has at least n + m − > m + ≥ distinct roots in view of Lemma 2.5. By Theorem 1.11, we get Set H = f L . Then by (42) we deduce We discuss two cases: Case 1. H is a constant. If H n+m ≠ , by (43), we get that L is a constant, which contradicts the assumption that L is a nonconstant L-function. Therefore, H n+m = , and so it follows by (43)   where ζ , ζ , ⋯, ζ n− are n − distinct nite complex numbers satisfying ζ n i = , ζ i ≠ , ≤ i ≤ n − ; τ , τ , ⋯, τ n+m− are n + m − distinct nite complex numbers satisfying τ n+m j = , τ j ≠ , ≤ j ≤ n + m − . Let m = . By Lemma 2.6 we see H n − and H n+ − have only one common zero, so H cannot be equal to any n + m − values of {τ , τ , ⋯, τ n+m− }. From n > m + it follows that H is a constant, contradicting the assumption.
Let m ≥ . If any 1-point of H n is a 1-point of H n+m , then any 1-point of H n is a 1-point of H m . Note that n > m + . This contradicts the assumption that H is nonconstant. If there is at least one ζ i ≠ τ j , ≤ i ≤ n − , ≤ j ≤ n + m − , then H cannot be equal to any m + values of {τ , τ , ⋯, τ n+m− }. From m ≥ , we know H is a constant, contradicting the assumption.
This completes the proof of Theorem 1.12.