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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1

# 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
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• Other articles by this author:
/ Jun-Fan Chen
• Corresponding author
• Key Laboratory of Applied Mathematics (Putian University), Fujian Province University, Fujian Putian, 351100, China
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• Other articles by this author:
Published Online: 2018-11-10 | DOI: https://doi.org/10.1515/math-2018-0107

## Abstract

In this paper, we obtain uniqueness theorems of L-functions from the extended Selberg class, which generalize and complement some recent results due to Li, Wu-Hu, and Yuan-Li-Yi.

MSC 2010: 11M36; 30D35; 30D30

## 1 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 defined to be a Dirichlet series

$L(s)=∑n=1∞a(n)ns$(1)

of a complex variable s = σ + it satisfying the following axioms [1]:

• (i)

Ramanujan hypothesis: a(n) ≪ nϵ for every ϵ > 0.

• (ii)

Analytic continuation: There exists a nonnegative integer m such that (s − 1)mL(s) is an entire function of finite order.

• (iii)

Functional equation: L satisfies a functional equation of type

$ΛL(s)=ωΛL(1−s¯)¯,$

where

$ΛL(s)=L(s)Qs∏j=1KΓ(λjs+νj)$

with positive real numbers Q, λj, and complex numbers νj, ω with Reνj ≥ 0 and |ω| = 1.

• (iv)

Euler product: $\mathrm{log}L\left(s\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{b\left(n\right)}{{n}^{s}}$where b(n) = 0 unless n is a positive power of a prime and b(n) ≪ nØ for some $\theta <\frac{1}{2}$

It is mentioned that there are many Dirichlet series but only those satisfying the axioms (i)-(iii) are regarded as the extended Selberg class [1, 2]. All the L-functions which are studied in this article are from the extended

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 Fc and Gc have the same zeros counting multiplicities, we say that F and G share c CM. If Fc and Gc 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 five distinct values IM [3, 41].

Theorem 1.1 (see [1]). If two L-functions with a(1) = 1 share a complex value c ≠ ∞ CM, then they are identically equal.

Remark 1.2. In [5], the authors gave an example that ${L}_{1}=1+\frac{2}{{4}^{s}}and{L}_{2}=1+\frac{3}{{9}^{s}}$ which showed that Theorem 1.1 is actually false when c = 1.

In 2011, Li [6] considered values which are shared IM and got

Theorem 1.3 (see [6]). Let L1 and L2 be two L-functions satisfying the same functional equation with a(1) = 1 and let a1, a2 ϵ ℂ be two distinct values. If ${L}_{1}^{-1}\left({a}_{j}\right)={L}_{2}^{-1}\left({a}_{j}\right),j=1,2$then L1L2.

In 2001, Lahiri [7] put forward the concept of weighted sharing as follows.

Let k be a nonnegative integer or∞, c ϵ ℂ∪{∞}. We denote by Ek(c, f) the set of all zeros of fc, where a zero of multiplicity m is counted m times if mk and k +1 times if m > k. If Ek(c, f) = Ek(c, g), 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 L1 and L2 be two L-functions, and let a1, a2 ϵ ℂ be two distinct values. Take two positive integers k1, k2 with k1k2 > 1. If Ekj (aj, L1) = Ekj (aj, L2), j = 1, 2, then L1L2.

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 ≠ 0,∞. If f and the Riemann zeta function ζ share a, b CM and c IM, will then fζ?

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, 1314]. 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.

Theorem 1.6 (see [10]). Let a and b be two distinct finite values and f be a meromorphic function in the complex plane with finitely many poles. If f and a nonconstant L-function L share a CM and b IM, then Lf.

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 finitely many poles, let L be a nonconstant L-function, and let a1, a2 ϵ ℂ be two distinct values. Take two positive integers k1, k2 with k1k2 > 1. If Ekj (aj, f) = Ekj (aj, L), j = 1, 2, then Lf.

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 f1 and f2 be identically equal if f1 and f2 share a finite set S?

Recently, Yuan, Li and Yi [16] considered this question leading to the theorem below.

Theorem 1.10 (see [16]). Let S = {ω1, ω2,⋯, ωl}, where ω1, ω2,⋯, ωl are all distinct roots of the algebraic equation ωn + m + b = 0. Here l is a positive integer satisfying 1 ≤ ln, n and m are relatively prime positive integers with n ≥ 5 and n > m, and a, b, c are nonzero finite constants, where cωj for 1 ≤ jl. Let f be a nonconstant meromorphic function such that f has finitely many poles in, and let L be a nonconstant L-function. If f and L share S CM and c IM, then fL.

Concerning shared set, we prove the following theorem.

Theorem 1.11. Let f be an entire function with limR(s)→+∞ f (s) = k (k ≠ ∞) and let R(a) = 0 be a algebraic equation with n ≥ 2 distinct roots, and R(k), R(b), R(1) ≠ 0. Suppose that f (s0) = L(s0) = b for some s0 ϵ ℂ. If f and a nonconstant L-function L share S CM, where S = {a : R(a) = 0}, then R(L) ≡ R(f).

Furthermore, we obtain a result which is similar to Theorem 1.10 by different means.

Theorem 1.12. Let f be an entire function with limR(s)→+∞ f (s) = k (k ≠ ∞). Let S = {ω1, ω2,⋯, ωi} ⊂ ℂ\{1, k, b}, where ω1, ω2,⋯, ωi are all distinct roots of the algebraic equation ωn+m + αωn + β = 0, 1 ≤ in+m, n, m are two positive integers with n > m +2, α, β are finite nonzero constants. If f and a nonconstant L-function L share S CM and f (s0) = L(s0) = b for some s0 ϵ ℂ, then ftL, where t is a constant such that td = 1, d = GCD(n, m).

## 2 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 defined as follows:

$ρf=lim supr→∞log⁡Tr,flog⁡r.$

Lemma 2.1 (see [4], Lemma 1.22). Let f be a nonconstant meromorphic function and let k ≥ 1 be an integer. Then m $\left(r,\frac{{f}^{\left(k\right)}}{f}\right)=S\left(r,f\right)$Further if ρ(f) < +∞, then

$mr,f(k)f=O(log⁡r).$

Lemma 2.2 (see [4], Corollary of Theorem 1.5). Let f be a nonconstant meromorphic function. Then f is a rational function if and only if $\underset{r\to \mathrm{\infty }}{lim inf}\frac{T\left(r,f\right)}{\mathrm{log}r}<\mathrm{\infty }.$

Lemma 2.3 (see [4], Theorem 1.19). Let T1(r) and T2(r) be two nonnegative, nondecreasing real functions defined in r > r0 > 0. If T1(r) = O (T2(r)) (r → ∞, rE), where E is a set with finite linear measure, then

$lim supr→∞log+⁡T1(r)log⁡r≤lim supr→∞log+⁡T2(r)log⁡r$

and

$lim infr→∞log+⁡T1(r)log⁡r≤lim infr→∞log+⁡T2(r)log⁡r,$

which imply that the order and the lower order of T1(r) are not greater than the order and the lower order of T2(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

$ρf⋅g≤maxρf,ρg,$$ρf+g≤maxρf,ρg.$

Lemma 2.5 (see [17], Lemma 2.7). Let R(ω) = ωn + m + b, where n, m are positive integers satisfying n > m, a, b are finite nonzero complex numbers. Then the algebraic equation R(ω) = 0 has at least n −1 distinct roots.

Lemma 2.6 (see [18], Lemma 8). Let s > 0 and t be relatively prime integers, and let c be a finite complex number such that cs = 1. Then there exists one and only one common zero of ωs − 1 and ωtc.

## 3.1 Proof of Theorem 1.7

First of all, we denote by d the degree of L. Then $d=2\sum _{j=1}^{k}{\lambda }_{j}>0$, where k and λj are respectively the positive integer and the positive real number in the functional equation of the axiom (iii) of the definition of L-functions. According to a result due to Steuding [1], p.150, we have

$T(r,L)=dπrlog⁡r+O(r).$(2)

Therefore (L) = λ and S(r, L) = O(log r).

Noting that f has finitely many poles and L at most has one pole at s = 1 in the complex plane, it follows that

$N(r,f)=O(log⁡r),N(r,L)=O(log⁡r).$(3)

Because f and L share a1, a2 weighted k1, k2 respectively, by (3), from the first and second fundamental theorems we have

$T(r,f)≤N¯r,1f−a1+N¯r,1f−a2+N¯r,f+S(r,f)=N¯r,1L−a1+N¯r,1L−a2+O(log⁡r)+S(r,f)≤Tr,1L−a1+Tr,1L−a2+O(log⁡r)+S(r,f)=2Tr,L+O(log⁡r)+S(r,f).$(4)

Then from (4) and Lemma 2.3 we obtain

$ρ(f)≤ρ(L).$(5)

Similarly,

$ρ(L)≤ρ(f).$(6)

Combining (6) yields

$ρ(f)=ρ(L).$(7)

Thus

$S(r,f)=O(log⁡r).$(8)

We introduce two auxiliary functions below.

$F1=L′L−a1−f′f−a1,$(9)$F2=L′L−a2−f′f−a2.$(10)

Next, we assume that F1 ≠ 0 and F 2 ≠ 0. By (8) and Lemma 2.1 we get

$m(r,F1)=O(log⁡r).$(11)

By the assumption L and f share (a1, k1), (a2, k2), from (3), (11) we have

$k2N¯(k2+1r,1L−a2≤Nr,1F1≤T(r,F1)+O(1)≤N(r,F1)+m(r,F1)+O(1)≤N¯(k1+1r,1L−a1+N¯(r,L)+N¯(r,f)+O(log⁡r)≤N¯(k1+1r,1L−a1+O(log⁡r).$(12)

Similarly, from (10) and (11) we have

$k1N¯(k1+1r,1L−a1≤Nr,1F2≤T(r,F2)+O(1)≤N(r,F2)+m(r,F2)+O(1)≤N¯(k2+1r,1L−a2+N¯(r,L)+N¯(r,f)+O(log⁡r)≤N¯(k2+1r,1L−a2+O(log⁡r).$(13)

Combining (12) with (13) yields

$N¯(k1+1r,1L−a1≤1k1N¯(k2+1r,1L−a2+O(log⁡r)≤1k1k2N¯(k1+1r,1L−a1+O(log⁡r).$(14)

Since k1k2 > 1, from (14) we obtain

$N¯(k1+1r,1L−a1=O(log⁡r).$(15)

Substituting (15) into (12) implies

$N¯(k2+1r,1L−a2=O(log⁡r).$(16)

Set

$G=L−a1f−a1.$

Noting L and f share (a1, k1), (a2, k2), combining (15) with (16) yields

$N¯(k1+1r,1L−a1=N¯(k1+1r,1f−a1=O(log⁡r),$$N¯(k2+1r,1L−a2=N¯(k2+1r,1f−a2=O(log⁡r).$

Clearly,

$N¯(r,G)≤N(r,L)+N¯(k1+1r,1f−a1=O(log⁡r),$(17)$N¯r,1G≤N(r,f)+N¯(k1+1r,1L−a1=O(log⁡r).$(18)

Set

$G1=Q(L−a1)f−a1,$(19)

where Q is a rational function satisfying that G1 is a zero-free entire function. From (17) and (10), it is easy to see that such a Q does exist. By Lemma 2.2 and Lemma 2.4 we get

$ρ(G1)≤max{ρ(Q),ρ(L),ρ(f)}=1.$

By the Hadamard factorization theorem [19], p.384, we know

$G1=Q(L−a1)f−a1=eφ,$(20)

where φ is a polynomial of degree at most deg() ≤ 1. We may write φ = a0s + b0 for some complex numbers a0, b0. In view of (20) and Hayman [3], p.7, we have

$T(r,G1)=T(r,ea0s+b0)=O(r).$(21)

By (19), the assumption that L and f share a2, we get that every a2-point of L has to be 1-point of $\frac{{G}_{1}}{Q}-1$Now (20), (21) and the first fundamental theorem yield

$N¯r,1L−a2≤Nr,1G1Q−1≤Tr,1G1Q−1=Tr,G1Q−1+O(1)≤T(r,G1)+T(r,Q)+O(1)=O(r).$(22)

Similarly, set

$G2=L−a2f−a2.$

We also get

$N¯r,1L−a1=O(r).$(23)

By (22), (23) and the second fundamental theorem it follows that

$T(r,L)≤N¯r,1L−a1+N¯r,1L−a2+N¯(r,L)+O(log⁡r)=O(r).$(24)

This contradicts (2). Thus, F1 ≡ 0 or F 2 ≡ 0. By integration, we have from (9) that

$L−a1≡A(f−a1),$

where A(≠ 0) is a constant. This implies that L and f share a1 CM. Hence by Theorem 1.6 we deduce Theorem 1.7 holds. If F 2 ≡ 0, using the same manner, we also have the conclusion.

This completes the proof of Theorem 1.7.

## 3.2 Proof of Theorem 1.11

First we consider the following function

$G=QR(L)R(f),$(25)

where

$Q(s)=A(s−1)nm$(26)

is a rational function satisfying that G has no zeros and no poles in ℂ; A is a nonzero finite value; m is the nonnegative integer in the axiom (ii) of the definition of L-functions.

We claim that such a Q does exist. By the condition that f and L share S CM, set

$F=R(L)R(f).$(27)

We can see that there can be only a pole of f or L such that F = 0 or F = ∞. Since f has no pole and L has only one possible pole at s = 1, it follows that F has no zero and only one possible pole at s = 1. Hence such a Q does exist.

Next, assume that a1, a2,∞, an are all distinct roots of R(a). Using the first fundamental theorem we get

$Tr,L−ai=T(r,L)+O(1),i=1,2,⋯,n.$

Noting n ≥ 2, by the second fundamental theorem we have

$(n−1)T(r,f)≤∑i=1nN¯r,1f−ai+N¯(r,f)+S(r,f)=∑i=1nN¯r,1L−ai+N¯(r,f)+S(r,f)≤∑i=1nTr,1L−ai+S(r,f)=nTr,L+S(r,f),$(28)

which gives

$T(r,f)≤nn−1Tr,L+S(r,f).$

This together with Lemma 2.3 yields

$ρ(f)≤ρ(L).$(29)

Similarly,

$ρ(L)≤ρ(f).$(30)

By (29), (30) and (2) we obtain

$ρ(f)=ρ(L)=1.$(31)

Also, from the first fundamental theorem we get

$ρ1f−ai=ρ(f)=1,$

and then by Lemma 2.2 and Lemma 2.4 we deduce

$ρ(G)≤max{ρ(Q),ρ(L),ρ(f)}=1.$

From the Hadamard factorization theorem [19], p.384 we see

$G=eh(s),$(32)

where h(s) is a polynomial of degree deg(h(s)) ≤ 1. One can write

$ℜh(σ+it)=α(t)σ+β(t),$(33)

a polynomial in σ with α(t), β(t) being polynomials in t. Now the claim is α(t) ≡ 0. From (25), (27) and (32) we get

$F=R(L)R(f)=eh(s)Q−1.$(34)

Since limσ→+∞ L(s) = 1, limσ→+∞ f (s) = k(k ≠ ∞), R(k) ≠ 0 and R(1) ≠ 0, it follows that

$limσ→+∞R(L)R(f)=C,$(35)

where ≠ 0 is a finite value. If α(t) ≠ 0, we obtain α(t0) ≠ 0 for some value t0. If α(t0) > 0, from (34) we know that

$R(L)R(f)=Q−1eℜh(σ+it).$(36)

Thus from (26), (35) and (36) we can deduce that, |C| = ∞ when σ → +∞ with t = t0, which is a contradiction. Similarly, if α(t0) < 0, we have that, |C| = 0 when σ → +∞ with t = t0, which is also a contradiction. Therefore α(t) ≡ 0. Now by (33) and (36) we get

$R(L)R(f)=Q−1eβ(t).$(37)

Combining (37) yields

$limσ→+∞|Q|=eβ(t)|C|$(38)

for a fixed t. Considering that the limit of |Q| as σ → +∞ is a nonzero finite constant for some value t and n ≥ 2, in view of (26) we see that m = 0, and then Q(s) ≡ A. From (38) we have eβ(t) = |A||C|. Thus it follows by (37) that

$R(L)R(f)=|C|.$(39)

Since c ≠ 0 is a finite complex number, from (35) we know that

$R(L)R(f)≡C.$(40)

From the assumption in the theorem we have f (s0) = L(s0) = b for some s0 ϵ ℂ. It now follows from (40) that = 1. Thus

$R(L)R(f)≡1.$(41)

That is R(L) ≡ R(f).

This completes the proof of Theorem 1.11.

## 3.3 Proof of Theorem 1.12

First, we have that the algebraic equation ωn+m + αωn + β = 0 has at least n +m −1 > 3m+1 ≥ 4 distinct roots in view of Lemma 2.5. By Theorem 1.11, we get

$Ln+m+αLn≡fn+m+αfn.$(42)

Set $H=\frac{f}{L}$Then by (42) we deduce

$1αLm=Hn−1Hn+m−1.$(43)

We discuss two cases:

Case 1. H is a constant. If H n+m ≠ 1, by (43), we get that L is a constant, which contradicts the assumption that L is a nonconstant L-function. Therefore, H n+m = 1, and so it follows by (43) that H m = H n = 1, that is fn = Ln and fm = Lm. We get fd = Ld.

Case 2. H is a nonconstant meromorphic function. Note that L has at most one pole. Now we discuss the following two subcases again.

Subcase 2.1. L has no poles. Then, from (43) we get that every 1-point of H n+m has to be 1-point of Hn. Since H n+m = H nHm, we have any 1-point of H n+m to be a 1-point of Hm. Because n > m + 2, it follows that H is a constant, contradicting the assumption.

Subcase 2.2. L has one and only one pole. Then by (43) we know every zero of H n+m − 1 has to be zero of Hn − 1 with one exception. Put

$Hn−1=(H−1)(H−ζ1)⋯(H−ζn−1),$$Hn+m−1=(H−1)(H−τ1)⋯(H−τn+m−1),$

where ζ1, ζ2,⋯, ζn−1 are n − 1 distinct finite complex numbers satisfying ${\zeta }_{i}^{n}=1,{\zeta }_{i}\ne 1,1\le i\le n-1;{\tau }_{1},{\tau }_{2},\cdots ,{\tau }_{n+m-1}$ are n + m − 1 distinct finite complex numbers satisfying ${\tau }_{j}^{n+m}=1,{\tau }_{j}\ne 1,1\le j\le n+m-1$

Let m = 1. By Lemma 2.6 we see H n − 1 and H n+1 − 1 have only one common zero, so H cannot be equal to any n +m −2 values of {τ1, τ2,⋯, τn+m−1}. From n > m +2 it follows that H is a constant, contradicting the assumption.

Let m ≥ 2. If any 1-point of H n is a 1-point of Hn+m, then any 1-point of Hn is a 1-point of Hm. Note that n > m + 2. This contradicts the assumption that H is nonconstant. If there is at least one ζiτj, 1 ≤ in − 1, 1 ≤ jn + m − 1, then H cannot be equal to any m + 1 values of {τ1, τ2,⋯, τn+m−1}. From m ≥ 2, we know H is a constant, contradicting the assumption.

This completes the proof of Theorem 1.12.

## Acknowledgement

The authors would like to thank the referees for their thorough comments and helpful suggestions.

Project supported by the National Natural Science Foundation of China (Grant No. 11301076), the Natural Science Foundation of Fujian Province, China (Grant No. 2018J01658) and Key Laboratory of Applied Mathematics of Fujian Province University (Putian University) (Grant No. SX201801).

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Accepted: 2018-10-01

Published Online: 2018-11-10

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 1291–1299, ISSN (Online) 2391-5455,

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