# Uniqueness of L-Functions Concerning Certain Differential Polynomials

Authors: Wen-Jie Hao; Jun-Fan Chen
Journal: Discrete Dynamics in Nature and Society
 (2018)

---

## 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.


---

## Body

## 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 series(1)Ls=∑n=1∞annsof a complex variable s=σ+it satisfying the following axioms:(i) Ramanujan hypothesis:a(n)≪nε for every ε>0(ii) Analytic continuation: there is a nonnegative integerm such that (s-1)mL(s) is an entire function of finite order(iii) Functional equation:L satisfies a functional equation of type (2)ΛLs=ωΛL1-s¯¯,where (3)ΛLs=LsQs∏j=1KΓλjs+νjwith positive real numbers Q, λj, and complex numbers νj, ω with Reνj≥0 and ω=1(iv) Euler product:logL(s)=∑n=1∞b(n)/ns, where b(n)=0 unless n is a positive power of a prime and b(n)≪nθ for some θ<1/2It 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 thatc is a finite complex number. If two L-functions L1 and L2 with a(1)=1 share c CM, then L1≡L2.In 2016, Hu and Li [3] gave an example L1=1+2/4s and L2=1+3/9s. From this we can know the above theorem is false when c=1.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 thatF and G are two nonconstant meromorphics in the whole complex plane; c denotes a values in the extended complex plane. 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. 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]).
Leta 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 L≡f.In 1997, the following question was raised by Lahiri [7]: what is the relationship between function f and function g, when two differential polynomials have the same nonzero finite value? The two differential polynomials are generated by f and g, respectively. In this direction, Fang [8] proved the following theorem.Theorem 3 (see [8]).
Letf and g be two nonconstant entire functions, and let n, k be two positive integers. Suppose that fnf-1(k) and gng-1(k) share 1 CM. If n≥2k+8, then f≡g.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]).
Letf be a nonconstant meromorphic function, let L be an L-function, and let n and k be two positive integers. Suppose that fn(k) and Ln(k) share 1 CM. If n>3k+6, then f≡tL for a constant t satisfying tn=1.Theorem 5 (see [10]).
Letf be a nonconstant meromorphic function, let L be an L-function, and let n and k be two positive integers. Suppose that fnf-1(k) and LnL-1(k) share 1 CM. If n>3k+9 and k≥2, then f≡L.Naturally, is it still set up if it can be generalized to the general differential polynomials, for instance,[fn(fm-1)](k) or [fn(f-1)m](k)? For simplicity, we use the notations d=GCD(n,m) and m~:=m~(α), where (4)m~α=0,α=0,m,α≠0.In this paper, we have the results as follows.Theorem 6.
Letf be a nonconstant meromorphic function, let L be an L-function, and let n, m, and k be three positive integers and α, β be two constants satisfying α+β≠0. Suppose that [fn(αfm+β)](k) and [Ln(αLm+β)](k) share 1 CM. If n>3k+m~+6, then f≡tL, where(i)
whenαβ=0, t is a constant such that tn+m~=1(ii)
whenαβ≠0, k≥2, t is a constant such that td=1Remark 7.
In Theorem6, if α=0, β=1, we can get Theorem 4. If α=m=1, β=-1, we can get Theorem 5. Moreover, if the condition n>3k+7 substitutes n>3k+9 in Theorem 5, then we can obtain f≡L, which implies the conclusion remains valid in Theorem 5. Therefore, Theorem 6 is the generalization of Theorems 4 and 5.Remark 8.
In Theorem6, the condition n>3k+m~+6 cannot be dropped. Let α=0, β=1, f=2ζ-z, L=ζ, n=1, k=1. Then f′-1=2ζ′-2, L′-1=ζ′-1. This shows that f′ and L′ share 1 CM. But f≢L.Theorem 9.
Letf be a nonconstant meromorphic function, let L be an L-function, and let n, m, and k be three positive integers. Suppose that [fn(f-1)m](k) and [Ln(L-1)m](k) share 1 CM. If n>3k+m+6 and k≥2, then f≡L or fn(f-1)m≡Ln(L-1)m.Remark 10.
In Theorem9, if m=1, we can get Theorem 5. Actually, if f≢L and fn(f-1)≡Ln(L-1), 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 Theorems6 and 9. The theorems are then established.Theorem 11.
Letf be a nonconstant meromorphic function, let L be an L-function, and let n, m, and k be three positive integers and α, β be two constants satisfying α+β≠0. Suppose that [fn(αfm+β)](k) and [Ln(αLm+β)](k) share 1 IM. If n>7k+4m~+11, then f≡tL, where(i)
whenαβ=0, t is a constant such that tn+m~=1(ii)
whenαβ≠0, k≥2, t is a constant such that td=1Theorem 12.
Letf be a nonconstant meromorphic function, let L be an L-function, and let n, m, and k be three positive integers. Suppose that [fn(f-1)m](k) and [Ln(L-1)m](k) share 1 IM. If n>7k+4m+11, k≥2, then f≡L or fn(f-1)m≡Ln(L-1)m.Corollary 13.
Letf be a nonconstant meromorphic function, let L be an L-function, and let n, k be two positive integers. Suppose that (fn)(k) and (Ln)(k) share 1 IM. If n>7k+11, then f≡tL for a constant t satisfying tn=1.Corollary 14.
Letf be a nonconstant meromorphic function, let L be an L-function, and let n, k be two positive integers. Suppose that [fn(f-1)](k) and [Ln(L-1)](k) share 1 IM. If n>7k+15 and k≥2, then f≡L.To prove the main theorems, the orderρf of a meromorphic function f will be needed. It is defined to be a superior limit (5)ρf=limsupr→∞logTr,flogr.Next, we introduce some definitions.Definition 15.
Suppose thatp is a positive integer and c is a value in the extended complex plane. Then Np)(r,1/(F-c)) is defined as the counting function of those zeros of F-c of order ≤p. N(p(r,1/(F-c)) is defined as the counting function of those zeros of F-c of order ≥p. N¯p)(r,1/(F-c)), N¯(p(r,1/(F-c)) are defined as the corresponding reduced counting functions.Definition 16.
Suppose thatz0 is a commonc-point of F and G with multiplicity k and l, respectively. We denote by N¯L(r,1/(F-c)) the reduced counting function of thosec-points of F and G where k>l and denote by N¯E(2(r,1/(F-c)) the reduced counting function of thosec-points of F and G where k=l≥2. Similarly, we can define N¯L(r,1/(G-c)) and N¯E(2(r,1/(G-c)).Definition 17.
Suppose thatc is a value in the extended complex plane and k is a positive integer. We define (6)Nkr,1F-c=N¯r,1F-c+N¯(2r,1F-c+⋯+N¯(kr,1F-c.


## 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]).
Letf be a nonconstant meromorphic function and let (7)F=∑k=0pakfk∑j=0qbjfjbe an irreducible rational function in f with constant coefficients {ak} and {bj}, where ap≠0 and bq≠0. Then T(r,F)=d0T(r,f)+O(1), where d0=max{p,q}.Lemma 19 (see [4], Theorem 3.2).
Letf be a nonconstant meromorphic function, let k be a positive integer, and let c be a nonzero finite complex number. Then (8)Tr,f≤N¯r,f+Nr,1f+Nr,1fk-c-Nr,1fk+1+Sr,f≤N¯r,f+Nk+1r,1f+N¯r,1fk-c-N0r,1fk+1+Sr,f, where N0r,1/f(k+1) is the counting function which only counts those points such that f(k+1)=0 but f(f(k)-c)≠0.Lemma 20 (see [5]).
Letf be a nonconstant meromorphic function and k be a positive integer. Then (9)Nr,1fk≤Nr,1f+kN¯r,f+Sr,f.Lemma 21.
Letf be a nonconstant meromorphic function, let L be an L-function, and let n, m, and k be three positive integers and α, β be two constants satisfying α+β≠0. Set(10)F=fnαfm+βk,G=LnαLm+βk.If F and G share 1 CM and n>3k+m~+6, then 1/F-1=VG+U-V/G-1, where U≠0, V are two constants.Proof.
Let(11)W=F′′F′-2F′F-1-G′′G′-2G′G-1,(12)F1=fnαfm+β,G1=LnαLm+β.
IfL is a constant, we have from (12) that G1 is a constant, G≡0. Since F and G share 1 CM, we obtain F≠1. Using Lemma 18, we deduce(13)Tr,F1=n+m~Tr,f+Sr,f.Moreover, from Lemma 19, we know (14)Tr,F1≤N¯r,F1+Nk+1r,1F1+N¯r,1F-1+Sr,f≤N¯r,f+Nk+1r,1fn+Nk+1r,1αfm+β+Sr,f≤k+m~+2Tr,f+Sr,f. Combining (13) with n>3k+m~+6, we get that L is a nonconstant L-function. Similarly, using Lemma 18, we deduce(15)Tr,G1=n+m~Tr,L+Sr,L.Since (F1)(k)=F, we get(16)mr,1F1≤mr,1F+Sr,f.Suppose W≢0, z0 is a zero of F-1 of order j, and z0 is a zero of G-1 of order j, in view of F and G sharing 1 CM. By checking the Laurent expansion of W, we have W(z0)=0 when j=1 and W(z0)≠∞ when j≥2. Thus(17)N1)r,1F-1=N1)r,1G-1≤Nr,1W≤Tr,W+O1≤Nr,W+Sr,f+Sr,L.Similarly, assume that z1 is a pole of F of order 1; we get W(z1)≠∞. If z2 is a pole of G of order 1, we get W(z2)≠∞. Therefore, we get, by a calculation and (11), that(18)Nr,W≤N¯(2r,1F+N¯(2r,1G+N¯(2r,F+N¯(2r,G+N¯0r,1F′+N¯0r,1G′.where N¯0r,1/F′ is the reduced counting function of those zeros of F′ not that of F(F-1). From the definition, we obtain(19)N¯0r,1G′+N¯(2r,1G-1+N(2r,1G-N¯(2r,1G≤Nr,1G′.By Lemma 20, we obtain(20)Nr,1G′≤Nr,1G+N¯r,G+Sr,L.Combining (19) with (20), we have(21)N¯0r,1G′+N¯(2r,1G-1≤N¯r,1G+N¯r,G+Sr,L.We can get(22)N¯r,1F-1=N1)r,1F-1+N¯(2r,1G-1,in view of the assumption that F and G share 1 CM. Combining (17)-(22), the second fundamental theorem yields(23)Tr,F≤N¯r,F+N¯r,1F+N¯r,1F-1-N0r,1F′+Sr,f≤N¯r,F+N¯r,1F+N¯(2r,1F+N¯(2r,1G+N¯(2r,F+N¯(2r,G+N¯r,1G+N¯r,G+Sr,f+Sr,L.From the first fundamental theorem, we know(24)Tr,1F≤N2r,F+N2r,G+N2r,1F+N2r,1G+Sr,f+Sr,L, where(25)N2r,1F=Nr,1F-N(2r,1F-2N¯(2r,1F,N2r,1G=Nr,1G-N(2r,1G-2N¯(2r,1G.Let z3 be a zero of f of order i, i≥1. Since n>3k+m~+6>k+2, ni-k>(k+2)i-k=(i-1)k+2i≥2, we can deduce the zeros of F of order ≥3. Thus(26)N(2r,1F-2N¯(2r,1F≥n-k-2Nr,1f.Also, for G, we can deduce that(27)N(2r,1G-2N¯(2r,1G≥n-k-2Nr,1L.Combining (24)-(27), we get(28)Tr,1F-Nr,1F≤Nr,1G+N2r,F+N2r,G-n-k-2Nr,1f-n-k-2Nr,1L+Sr,f+Sr,L.By Lemma 20, we obtain(29)Nr,1G≤Nr,1G1+kN¯r,G1+Sr,L.Using the first fundamental theorem, we have(30)Tr,F1=Tr,1F1+O1=mr,1F1+Nr,1F1+O1.From(13), (16), (28)-(30), we get(31)n+m~Tr,f≤Nr,1F1+Nr,1G1+kN¯r,G1+N2r,F+N2r,G-n-k-2Nr,1f-n-k-2Nr,1L+Sr,f+Sr,L≤nNr,1f+Nr,1αfm+β+nNr,1L+Nr,1αLm+β+kN¯r,G1+N2r,F+N2r,G-n-k-2Nr,1f+Nr,1L+Sr,f+Sr,L≤Nr,1αfm+β+Nr,1αLm+β+kN¯r,G1+N2r,F+N2r,G+k+2Nr,1f+k+2Nr,1L+Sr,f+Sr,L.Similarly, we have(32)n+m~Tr,L≤Nr,1αfm+β+Nr,1αLm+β+kN¯r,F1+N2r,F+N2r,G+k+2Nr,1f+k+2Nr,1L+Sr,f+Sr,L.Since L at most has one pole, we know N(r,L)=O(logr). At the same time, we know N¯(r,F)=N¯(r,F1)=N¯(r,f), N¯(r,G)=N¯(r,G1)=N¯(r,L), N2(r,F)≤2N¯(r,F). By (31) and (32), we can obtain the following results, respectively:(33)n+m~Tr,f≤k+m~+4Tr,f+k+m~+2Tr,L+Sr,f+Sr,L,(34)n+m~Tr,L≤2k+m~+4Tr,f+k+m~+2Tr,L+Sr,f+Sr,L.Assume that there exists some subset E⊆R+ with its linear measure mesE=∞ satisfying T(r,L)≤T(r,f), as r∈E and r→∞. Then it follows from (33) that n≤2k+m~+6, which contradicts n>3k+m~+6. Assume that there exists some subset E⊆R+ with its linear measure mesE=∞ such that T(r,f)≤T(r,L), as r∈E and r→∞. Then it follows from (34) that n≤3k+m~+6, which contradicts n>3k+m~+6. Therefore, W≡0. That is, (35)F′′F′-2F′F-1=G′′G′-2G′G-1. Integrating this gives 1/F-1=VG+U-V/G-1, where U≠0, V are two constants.
This completes the proof of Lemma21.Remark 22.
Suppose thatF=[fn(αfm+β)](k) and G=[Ln(αLm+β)](k) are replaced by F=[fn(f-1)m](k) and G=[Ln(L-1)m](k), respectively, in Lemma 21. That is still true. In fact, Nr,1/αfm+β is replaced by Nr,1/f-1m in (31) and (32); we still get mT(r,f) in (33) and (34).Lemma 23 (see [12], Theorem 1.2).
Suppose thatf is a meromorphic of finite order in the complex plane and that f(k) has finitely many zeros for some k≥2. Then f has finitely many poles in the complex plane.Lemma 24 (see [4]).
Letf be a transcendental meromorphic function, and let a1, a2 be two distinct meromorphic functions such that T(r,ai)=S(r,f), i=1,2. Then (36)Tr,f≤N¯r,f+N¯r,1f-a1+N¯r,1f-a2+Sr,f.Lemma 25 (see [13]).
Lets>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 ωt-c.Lemma 26 (see [14, 15]).
Letf be a nonconstant meromorphic function, and let p, k be two positive integers. Then (37)Npr,1fk≤Np+kr,1f+kN¯r,f+Sr,f.Lemma 27.
Letf be a nonconstant meromorphic function, let L be an L-function, and let n, m, and k be three positive integers and α, β be two constants satisfying α+β≠0. Set(38)F=fnαfm+βk,G=LnαLm+βk.If F and G share 1 IM, n>7k+4m~+11, then 1/F-1=VG+U-V/G-1, where U≠0, V are two constants.Proof.
Let(39)W=F′′F′-2F′F-1-G′′G′-2G′G-1,(40)F1=fnαfm+β,G1=LnαLm+β.
In the same manner as Lemma21, we know L is a nonconstant L-function. By Lemma 18, we have(41)Tr,F1=n+m~Tr,f+Sr,f,(42)Tr,G1=n+m~Tr,L+Sr,L.Since (F1)(k)=F, we get(43)mr,1F1≤mr,1F+Sr,f.Suppose W≢0; z0 is a zero of F-1 of order q, and z0 is a zero of G-1 of order j, which is possibly different from q, in view of F and G sharing 1 IM. By checking the Laurent expansion of W, we have W(z0)=0 when q=j=1. Thus(44)N1)r,1F-1=N1)r,1G-1≤Nr,1W≤Tr,W+O1≤Nr,W+Sr,f+Sr,L.Similarly, if z1 is a pole of F of order 1, we get W(z1)≠∞. If z2 is a pole of G of order 1, we get W(z2)≠∞. Therefore, we get, by a calculation and (39), that(45)Nr,W≤N¯(2r,F+N¯(2r,G+N¯(2r,1F+N¯(2r,1G+N¯Lr,1F-1+N¯Lr,1G-1+N¯0r,1F′+N¯0r,1G′,where N¯0r,1/F′ is the reduced counting function of those zeros of F′ not that of F(F-1). Note that F and G share 1 IM. We have(46)N¯r,1F-1≤N1)r,1F-1+N¯E(2r,1F-1+N¯Lr,1F-1+N¯Lr,1G-1.Combining (44)-(46), we get(47)N¯r,1F-1+N¯r,1G-1≤N¯(2r,F+N¯(2r,G+N¯(2r,1F+N¯(2r,1G+3N¯Lr,1F-1+3N¯Lr,1G-1+N¯0r,1F′+N¯0r,1G′+N1)r,1G-1+2N¯E(2r,1G-1.Note that(48)N1)r,1G-1+2N¯E(2r,1G-1+N¯Lr,1F-1+2N¯Lr,1G-1≤Nr,1G-1≤Tr,G+O1.Substituting (48) into (47), we obtain(49)N¯r,1F-1+N¯r,1G-1≤N¯(2r,F+N¯(2r,G+N¯(2r,1F+N¯(2r,1G+2N¯Lr,1F-1+N¯Lr,1G-1+N¯0r,1F′+N¯0r,1G′+Tr,G+O1.By the second fundamental theorem and (49), we get (50)Tr,F+Tr,G≤N¯r,F+N¯r,G+N¯r,1F+N¯r,1G+N¯(2r,F+N¯(2r,G+N¯(2r,1F+N¯(2r,1G+2N¯Lr,1F-1+N¯Lr,1G-1+Tr,G+Sr,f+Sr,L. Namely,(51)Tr,F≤N2r,F+N2r,G+N2r,1F+N2r,1G+2N¯Lr,1F-1+N¯Lr,1G-1+Sr,f+Sr,L.Note that (52)N2r,1F=Nr,1F-N(2r,1F-2N¯(2r,1F,N2r,1G=Nr,1G-N(2r,1G-2N¯(2r,1G.Let z3 be a zero of f of order i, i≥1. Since n>3k+m~+6>k+2, ni-k>(k+2)i-k=(i-1)k+2i≥2, we can deduce the zeros of F of order ≥3. Therefore,(53)N(2r,1F-2N¯(2r,1F≥n-k-2Nr,1f.Also, for G, we have(54)N(2r,1G-2N¯(2r,1G≥n-k-2Nr,1L.Combining (51)-(54) and the first fundamental theorem shows that(55)Tr,1F≤N2r,F+N2r,G+Nr,1F+Nr,1G+2N¯Lr,1F-1+N¯Lr,1G-1-n-k-2Nr,1f-n-k-2Nr,1L+Sr,f+Sr,L.From (43), we see that(56)Tr,1F1=mr,1F1+Nr,1F1≤mr,1F+Nr,1F1=Tr,1F-Nr,1F+Nr,1F1.By (55), (56), and the first fundamental theorem, we have(57)Tr,F1≤Nr,1F1+N2r,F+N2r,G+Nr,1G+2N¯Lr,1F-1+N¯Lr,1G-1-n-k-2Nr,1f-n-k-2Nr,1L+Sr,f+Sr,L.By Lemma 20, we have(58)Nr,1G≤Nr,1G1+kN¯r,G1+Sr,L≤nNr,1L+Nr,1αLm+β+kN¯r,G1+Sr,L.By Lemma 26, we have(59)N¯r,1G≤Nk+1r,1G1+kN¯r,G1+Sr,L≤Nk+1r,1Ln+Nk+1r,1αLm+β+kN¯r,G1+Sr,L≤k+1N¯r,1L+Nk+1r,1αLm+β+kN¯r,G1+Sr,L.In addition, we know that(60)N¯Lr,1G-1≤Nr,1G-1-N¯r,1G-1≤Nr,GG′≤Tr,GG′=Tr,G′G+O1≤Nr,G′G+mr,G′G+O1≤N¯r,G+N¯r,1G+Sr,L.The same inequality (58)-(60) holds for F. Substituting (58)-(60) into (57), and with (41) we deduce that(61)n+m~Tr,f≤N2r,F+N2r,G+2kN¯r,G1+2kN¯r,F1+2N¯r,F+N¯r,G+Nr,1αfm+β+Nr,1αLm+β+2Nk+1r,1αfm+β+Nk+1r,1αLm+β+3k+4Nr,1f+2k+3Nr,1L+Sr,f+Sr,L.Likewise, we have(62)n+m~Tr,L≤N2r,F+N2r,G+2kN¯r,F1+2kN¯r,G1+2N¯r,G+N¯r,F+Nr,1αfm+β+Nr,1αLm+β+2Nk+1r,1αLm+β+Nk+1r,1αfm+β+3k+4Nr,1L+2k+3Nr,1f+Sr,f+Sr,L.Since L at most has one pole, we get N(r,L)=O(logr). At the same time, we have N¯(r,F)=N¯(r,F1)=N¯(r,f), N¯(r,G)=N¯(r,G1)=N¯(r,L), N2(r,F)≤2N¯(r,F). By (61) and (62), we can obtain the following results, respectively:(63)n+m~Tr,f≤5k+3m~+8Tr,f+2k+2m~+3Tr,L+Sr,f+Sr,L,(64)n+m~Tr,L≤4k+2m~+6Tr,f+3k+3m~+4Tr,L+Sr,f+Sr,L.Assume that there exists some subset E⊆R+ with its linear measure mesE=∞ satisfying T(r,L)≤T(r,f), as r∈E and r→∞. Then it follows from (63) that n≤7k+4m~+11, which contradicts n>7k+4m~+11. Assume that there exists some subset E⊆R+ with its linear measure mesE=∞ satisfying T(r,L)≤T(r,f), as r∈E and r→∞. Then it follows from (64) that n≤7k+4m~+10, which contradicts n>7k+4m~+11. Therefore, W≡0. That is, (65)F′′F′-2F′F-1=G′′G′-2G′G-1. Integrating this gives 1/F-1=VG+U-V/G-1, where U≠0, V are two constants.
This completes the proof of Lemma27.Remark 28.
In Theorem11, let F=[fn(αfm+β)](k),G=[Ln(αLm+β)](k). By Lemma 27, we can get 1/F-1=VG+U-V/G-1, where U≠0, V are two constants. We can get that F and G share 1 CM. Then we can get Theorem 11 by Theorem 6. Similarly, we get Theorem 12 by Theorem 9.


## 3. Proof of Theorem6

Let(66)F=fnαfm+βk,G=LnαLm+βk.By Lemma 21, we have(67)1F-1=VG+U-VG-1,where U≠0, V are two constants.We discuss three cases.Case 1.
V ≠ 0 and U=V. Then (67) can be written as(68)1F-1=VGG-1.
Subcase 1.1. V=-1. From (68) we have FG=1,(69)fnαfm+βkLnαLm+βk≡1.Let d1 be the degree of L. Then d1=2∑j=1Kλj>0, where K and λj are the numbers of the axiom (iii) of the definition of L-function. Thus, by Steuding [2], p.150, we get(70)Tr,L=d1πrlogr+Or.
Next we distinguish two cases.
Subcase 1.1.1. α β ≠ 0. By (69), (70), Lemma 18, and a result from Whittaker [16], p.82, we have(71)ρf=ρfnαfm+β=ρfnαfm+βk=ρLnαLm+βk=ρLnαLm+β=ρL=1.Since L at most has one pole z=1, we deduce by (69) that [fn(αfm+β)](k) at most has one zero z=1. By (71), the assumption k≥2, and Lemma 23, we have that fn(αfm+β) and so f has at most finitely many poles. This together with (69) implies that [Ln(αLm+β)](k) has at most finitely many zeros. Moreover, by the assumption n>3k+m~+6, we deduce that L has at most finitely many zeros. Thus,(72)L=Q1eA1z+B1,where Q1 is a rational function such that L has neither a pole nor a zero, A1≠0, and B1 are constants. In view of (72) and Hayman [4], p.7, we have (73)Tr,L=Tr,Q1eA1z+B1≤A1rπ1+o1+Ologr, which contradicts (70).
Subcase 1.1.2. αβ=0. By α+β≠0, we can consider two subcases.
Subcase 1.1.2.1. α = 0, β≠0. Then (69) becomes (βfn)(k)(βLn)(k)≡1. Assume that z0 is a zero of L of order λ. Then, we can get that z0 is a pole of f of order μ, satisfying nλ-k=nμ+k, that is, (λ-μ)n=2k, and we have n≤2k, which contradicts the assumption n>3k+m~+6. Hence, we prove that L has no zeros,(74)L=Q2eA2z+B2,where Q2 is a rational function satisfying that L has no poles and A2≠0, B2 are two constants. In view of (74) and Hayman [4], p.7, we have (75)Tr,L=Tr,Q2eA2z+B2≤A2rπ1+o1+Ologr, which contradicts (70).
Subcase 1.1.2.2. α ≠ 0, β=0. By using the argument as in Subcase 1.1.2.1, we obtain that (n+m~)(λ-μ)=2k, and so n≤2k-m~, which contradicts the assumption n>3k+m~+6. Hence, we know that L has no zeros. Similarly, we get a contradiction.
Subcase 1.2. V ≠ - 1. Then it follows from (68) that(76)F-1+1V=-1VG.Noting that L has at most one pole, then G has at most one pole; from (76) we have that F-1+1/V has at most one zero. By Lemma 19, we obtain (77)n+m~Tr,f=Tr,fnαfm+β+Sr,f≤N¯r,f+Nk+1r,1fnαfm+β+N¯r,1F-1+1/V+Sr,f≤N¯r,f+k+1N¯r,1f+Nk+1r,1αfm+β+Ologr+Sr,f≤k+m~+2Tr,f+Sr,f, which contradicts n>3k+m~+6.Case 2.
V ≠ 0 and U≠V. Then from (67), we have(78)G+U-VV=-UV2·1F-1+1/V.Similarly, noting that L has at most one pole, from (78) we have that F-1+1/V has at most one zero. By using the same method as in Subcase 1.2, we know it is a paradox.Case 3.
V = 0 and U≠0. From (67) we get(79)G=UF-U+1,so that(80)LnαLm+β=Ufnαfm+β+1-Uγz,where γ(z) is a polynomial of degree at most k. By (70), we get that L is a transcendental meromorphic function, and so Ln(αLm+β) is a transcendental meromorphic function. Then, we obtain T(r,γ)=o{T(r,L)}. If U≠1, then (1-U)γ(z)≢0. Considering L has at most one pole, we get from (80) and Lemma 24 that(81)n+m~Tr,L=Tr,LnαLm+β+Sr,L≤N¯r,L+N¯r,1LnαLm+β+N¯r,1fnαfm+β+Sr,L≤N¯r,1L+N¯r,1f+N¯r,1αLm+β+N¯r,1αfm+β+Ologr+Sr,L≤m~+1Tr,f+Tr,L+Sr,L.In addition, from (80) and Lemma 18 we have T(r,L)=T(r,f)+S(r,L). Using this in (81) we have (82)n+m~Tr,L≤2m~+1Tr,L+Sr,L, we get that n≤m~+2, a contradiction. Hence, we have U=1. By (80), we obtain that(83)fnαfm+β=LnαLm+β.
Consider two subcases as follows.
Subcase 3.1. α β = 0, by α+β≠0; we deduce f=tL, where t is a constant satisfying tn+m~=1.
Subcase 3.2. α β ≠ 0; set H=f/L; by (83) we deduce(84)-αβLmHn+m-1=Hn-1.
We discuss two subcases.
Subcase 3.2.1. H is a constant. If Hn+m≠1, by (84), we get that L is a constant, which contradicts the assumption that L is a nonconstant L-function. Therefore, Hn+m=1, and so it follows by (84) that Hm=Hn=1; that is, fn=Ln and fm=Lm. We get fd=Ld.
Subcase 3.2.2. H is a nonconstant meromorphic function. From (84), we have(85)-αβLm=Hn-1Hn+m-1.
SinceL has at most one pole, we divide this case into two subcases again.
Subcase 3.2.2.1. L has no poles. Then, from (85) we get every 1-point of Hn+m has to be 1-point of Hn. Since Hn+m=HnHm, we have any 1-point of Hn+m be a 1-point of Hm. In view of n>3k+m~+6, we deduce that H is a constant, which contradicts the assumption.
Subcase 3.2.2.2. L has one and only one pole. Then, from (85) we get every zero of Hn+m-1 has to be zero of Hn-1 with one exception. Set (86)Hn-1=H-1H-δ1⋯H-δn-1,Hn+m-1=H-1H-η1⋯H-ηn+m-1, where δ1,δ2,…,δn-1 are n-1 distinct finite complex numbers satisfying δin=1, δi≠1, 1≤i≤n-1; η1,η2,…,ηn+m-1 are n+m-1 distinct finite complex numbers satisfying ηjn+m=1, ηj≠1, 1≤j≤n+m-1.
Letm=1. From Lemma 25 we know Hn-1 and Hn+1-1 have only one common zero; then H cannot be equal to any n+m-2 values of {η1,η2,…,ηn+m-1}. By n>3k+m~+6, we deduce H is a constant, which contradicts the assumption.
Letm≥2. If any 1-point of Hn is a 1-point of Hn+m, then any 1-point of Hn is a 1-point of Hm, since n>3k+m~+6, a contradiction to the assumption that H is nonconstant. If there is at least one δi≠ηj, 1≤i≤n-1, 1≤j≤n+m-1, then H cannot be equal to any m+1 values of {η1,η2,…,ηn+m-1}. By m≥2, we deduce H is a constant, which contradicts the assumption.
This completes the proof of Theorem6.


## 4. Proof of Theorem9

Set(87)F=fnf-1mk,G=LnL-1mk,(88)F1=fnf-1m,G1=LnL-1m.By Lemma 18, we get(89)Tr,F1=n+mTr,f+Sr,f,(90)Tr,G1=n+mTr,L+Sr,L.By Lemma 21, we have(91)1F-1=VG+U-VG-1,where U≠0 and V are two constants.We discuss three cases.Case 1.
V ≠ 0 and U=V. By (91), we have(92)1F-1=VGG-1.
Subcase 1.1. V=-1; then from (92) FG=1,(93)fnf-1mkLnL-1mk≡1.Let d1 be the degree of L. Then d1=2∑j=1Kλj>0, where K and λj are the numbers of the axiom (iii) of the definition of L-function. Thus, by Steuding [2], p150, we get(94)Tr,L=d1πrlogr+Or.By (93), (94), Lemma 18, and a result from Whittaker [16], p.82, we have(95)ρf=ρfnf-1m=ρfnf-1mk=ρLnL-1mk=ρLnL-1m=ρL=1.Since L at most has one pole z=1, we deduce by (93) that [fn(αfm+β)](k) at most has one zero z=1. By (95), the assumption k≥2, Lemma 23, we get fn(f-1)m and so f has at most finitely many poles. This together with (93) implies that [Ln(L-1)m](k) has at most finitely many zeros. Moreover, by the assumption n>3k+m+6, we deduce that L has at most finitely many zeros. Thus,(96)L=QeAz+B,where Q is a rational function satisfying L has no zeros and poles, A≠0, and B are constants. By (96) and Hayman [4], we get (97)Tr,L=Tr,QeAz+B≤Arπ1+o1+Ologr. which contradicts (94).
Subcase 1.2.  V≠-1; then from (92) we have(98)F-1+1V=-1VG.In view of L having at most one pole z=1, we know that G has at most one pole z=1, from (98) we have that F-1+1/V has at most one zero. By (89) and Lemma 19 we obtain(99)n+mTr,f=Tr,F1+Sr,f≤N¯r,f+Nk+1r,1F1+N¯r,1F-1+1/V+Sr,f≤N¯r,f+k+1Nr,1f+Nk+1r,1f-1m+Ologr+Sr,f≤k+m+2Tr,f+Sr,f,which contradicts n>3k+m+6.Case 2.
V ≠ 0 and U≠V. From (91), we have(100)G+U-VV=-UV2·1F-1+1/V.Similarly, noting that L has at most one pole z=1, by (100) we have that F-1+1/V has at most one zero. By using the same method as in Subcase 1.2, we know it is a paradox.Case 3.
V = 0 and U≠0. Then, from (91), we have(101)G=UF-U+1,so that(102)G1=UF1+1-Uγz,where γ(z) is a polynomial of degree at most k. By (94), we get L is a transcendental meromorphic function, and so G1 is a transcendental meromorphic function. Then, we get T(r,γ)=o{T(r,L)}. If γ≠1, then (1-U)γ(z)≢0. From (102) and Lemma 24 we have(103)n+mTr,L=Tr,G1+Sr,L≤N¯r,G1+N¯r,1G1+N¯r,1G1-1-UPz+Sr,L=N¯r,L+N¯r,1G1+N¯r,1F1+Sr,L≤N¯r,1L+N¯r,1f+N¯r,1L-1+N¯r,1f-1+Ologr+Sr,L≤2Tr,f+Tr,L+Sr,L.In addition, by (102) and Lemma 18 we have T(r,L)=T(r,f)+S(r,L). Using this in (103) we have (104)n+mTr,L≤4Tr,L+Sr,L, which contradicts n>3k+m+6. Hence, we get U=1. By (102), we obtain F1=G1; that is,(105)fnf-1m=LnL-1m.That is,(106)fnfm+⋯+-1iCmm-ifm-i+⋯+-1m=LnLm+⋯+-1iCmm-iLm-i+⋯+-1m.Set H=f/L. If H is a nonconstant meromorphic function, from (105) we get fn(f-1)m≡Ln(L-1)m. If H is a constant, from (106) we get (107)Ln+mHn+m-1+⋯+-1iCmm-iLn+m-iHn+m-i-1+⋯-LnHn-1=0. which implies H=1. Therefore, f≡L.
This completes the proof of Theorem9.



---

*Source: 4673165-2018-10-18.xml*