Uniqueness theorem on meromorphic functions and their difference operators

R E S E A R C H

Open Access

Uniqueness theorem on meromorphic

functions and their difference operators

Dan Liu

_{, Bingmao Deng}

_{and Mingliang Fang}

*_{Correspondence:} [email protected]

1_{Institute of Applied Mathematics,}

South China Agricultural University, Guangzhou, China

Abstract

In this paper, we study the uniqueness problems of meromorphic functions and their difference operators. Our main result is a difference analogue of a result of

Jank–Mues–Volkmann, which is concerned with the uniqueness of an entire function sharing one finite value with its derivatives. Some recent papers studied the case of entire functions of finite order sharing a periodic small function tof. We consider the case of meromorphic functions of finite order sharing a polynomial, which is a more popular case. Examples are provided for our results.

MSC: Primary 30D35; secondary 39B32

Keywords: Uniqueness; Meromorphic functions; Difference operators

1 Introduction and main results

LetCdenote the complex plane andf a meromorphic function in the whole complex planeC. We use the following standard notations of the Nevanlinna value distribution theory (see [11,15,16]):

m(r,f),N(r,f),N(r,f),T(r,f),ρ(f),S(r,f), . . . ,

where

m(r,f) = 1 2π

2π

0

log+freiθdθ,

N(r,f) = r

0

n(t,f) –n(0,f)

t dt+n(0,f)logr, N(r,f) =

r

0

n(t,f) –n(0,f)

t dt+n(0,f)logr, T(r,f) =m(r,f) +N(r,f).

Among them,m(r,f) is the average of the positive logarithm of|f(z)|on the circle|z|= r,N(r,f) is called the counting function of poles off, andN(r,f) is called the reduced counting function of poles off.T(r,f) is called the characteristic function off, and it plays a cardinal role in the whole theory of meromorphic functions.

The order of growth off is denoted byρ(f) as follows:

ρ(f) = lim

r→∞

log+T(r,f)

logr .

Ifρ(f) <∞, then we say thatf is a meromorphic function of finite order.

For a meromorphic functiona, ifT(r,a) =S(r,f), whereS(r,f) =o(T(r,f)), asr→ ∞, possibly outside of an exceptional set of finite logarithmic measure, then we say thatais a small function off. We useS(f) to denote the set of the small functions off.

For a meromorphic functionf(z), we define its shift byfc(z) =f(z+c) and its difference

operators by

cf(z) =f(z+c) –f(z), ncf(z) =nc–1

cf(z)

, n∈N,n≥2.

Letf andgbe two meromorphic functions, and letP(z) be a polynomial. We say thatf andgshareP(z) CM, provided thatf(z) –P(z) andg(z) –P(z) have the same zeros counting multiplicities.

In 1986, Jank et al. proved the following.

Theorem 1([12]) Let f be a nonconstant meromorphic function,and let a be a nonzero finite complex number.If f,f,and fshare a CM,then f ≡f.

Later on, the uniqueness theorem of entire functions sharing a constant with itskth and (k+ 1)th derivatives was proved by Li and Yang in 2001.

Theorem 2([14]) Let f be a nonconstant entire function;let a be a finite nonzero constant, and let k be a positive integer.If f,f(k)_,f(k+1)_{share the value a CM,then f ≡f.}

In 2004, Chang and Fang extended Theorem2to the case of sharing small functions.

Theorem 3([2]) Let f be a nonconstant entire function; let a≡0 be a small function

related to f;and let k≥2be a positive integer.If f,f(k),and f(k+1)share a CM,then f ≡f.

In recent years, the value distribution of meromorphic functions with respect to differ-ence has become a subject of some interests (see [1,3–5,7–10,13]). Chen et al. [3] proved a similar result analogue of Theorem1concerning difference. In 2015, Latreuch et al. stud-ied the case of entire function with its difference analogue of Theorem2and Theorem3 and proved the following.

Theorem 4([13]) Let f be a nonconstant entire function of finite order,and let a(z) (≡0) ∈S(f)be a periodic entire function with period c.If f,n

cf,andnc+1f (n≥1)share a(z)

CM,thenn+1

c f ≡ncf.

Theorem 5([13]) Let f be a nonconstant entire function of finite order.If f,n_cf,andn_c+1f (n≥1)share0CM,thenn_c+1f ≡Cn

cf,where C is a nonzero constant.

In 2016, El Farissi et al. successfully proved the relation betweenf andcf under the

condition thata(z) is a periodic function.

Theorem 6([8]) Let f(z)be a nonconstant entire function of finite order such thatn_cf(z)≡ 0,and let a(z) (≡0)∈S(f)be a periodic entire function with period c.If f,n_cf,andn_c+1f (n≥1)share a(z)CM,thencf(z)≡f(z).

For other related results, see Latreuch, El Farissi, Belaïdi [13], El Farissi, Latreuch, Asiri [7], El Farissi et al. [8].

By Theorems4and6, we naturally have the following problems:

(i) Can we get rid of the condition thata(z)is a periodic entire function with periodc

in Theorem6?

(ii) Since the casen= 1has been proved by Deng, Liu, and Fang [6] under the condition thata(z)is a polynomial, can we still deduce thatf ≡cf forn≥2?

Theorem 7([6]) Let f be a nonconstant meromorphic function of finite order,and let p(z) (≡0)be a polynomial.If f,cf,and2cf share∞and p(z)CM,then either f ≡cf or

f(z) =eAz+B+b,where p(z)≡b(= 0),and A= 0satisfying eAc= 1.

In this paper, we study these problems and give an affirmative answer to them.

Theorem 8 Let f be a nonconstant meromorphic function of finite order,let n∈N+be a positive integer,and let P(z) (≡0)be a polynomial.If f,n_cf,andn_c+1f share∞and P(z) CM,then one of the following three cases must occur:

(i) f≡cf;

(ii) f= (1 –eα(z)_)P(z),whereeα(z)_{is a periodic function of periodcwithdegα= 1,and} P(z)is a polynomial of degree less thann;

(iii) f(z) =AP(z),whereA(= 0, 1)is a constant,andP(z)is a polynomial of degree less thann.

Corollary 1 Let f be a transcendental entire function of finite order,and let P(z) (≡0)be a polynomial.If f,n

cf,andnc+1f share P(z)CM,then either f ≡cf,or f = (1 –eα(z))P(z),

where eα(z)_{is a periodic function of period c withdegα= 1,and P(z)is a polynomial with}

degP≤n.

Problem 1 In this paper, we study the uniqueness of meromorphic functions of finite order sharing a polynomial with its differences. We have the following question: What can we say if sharing a meromorphic function? What is more, does this result still hold for meromorphic functions of arbitrary order?

In the following, we give three cases to show that all these cases may occur.

Example1 LetA,cbe two nonzero finite complex numbers satisfyingeAc_{= 2, and let}

f(z) =eAz_{. Thenf(z) =}n

cf=nc+1f=eAz(n≥1), and for any polynomialP(z), we havef,

n_cf, andn_c+1f share∞andP(z) CM. This example satisfies Case (i) of Theorem8.

n+1

c P≡0. Therefore,f–P=eπizP,cnf–P= –P, andnc+1f–P= –Phave the same zeros

with the same multiplicities, andf = (1 +eπiz_{)P(z), where e}πiz _{is a periodic function of}

periodc= 2. This example shows that Case (ii) in Theorem8exists.

Example3 Letf(z) = 2zn–1_{,P(z) =z}n–1_{. It is obvious thatf(z) –P(z) =z}n–1_,n

cf(z) –P(z) =

n_c+1f(z) –P(z) = –zn–1 _{have the same zeros. Thusf(z),}n

cf(z), andnc+1f(z) shareP(z),

∞CM. This example shows that Case (iii) in Theorem8exists.

2 Some lemmas

Lemma 1([5,9]) Let f(z)be a meromorphic function of finite order,and let c be a nonzero complex constant.Then

Tr,f(z+c)=T(r,f) +S(r,f).

Lemma 2([9,10]) Let c∈C,k be a positive integer,and let f(z)be a meromorphic function of finite order.Then

m

r,

k cf(z)

f(z)

=S(r,f).

Lemma 3([16]) Suppose that f(z)is a meromorphic function in the complex plane and P(z) =a0fn+a1fn–1+· · ·+an,where a0(≡0),a1, . . . ,anare small functions of f(z).Then

Tr,P(f)=nT(r,f) +S(r,f).

Lemma 4([15,16]) Suppose that fi(z) (i= 1, 2, . . . ,n)are meromorphic functions and gi(z)

(i= 1, 2, . . . ,n),n≥2are entire functions satisfying

(i) n_i=1fi(z)egi(z)≡0,

(ii) the orders offiare less than those ofegk–gl for1≤i≤n,1≤k<l≤n.

Then fi(z)≡0 (i= 1, 2, . . . ,n).

Lemma 5([15,16]) Suppose that f1(z),f2(z), . . . ,fn(z) (n≥3)are meromorphic functions

which are not constants except for fn(z).Furthermore,let

n

j=1fj(z) = 1.If fn(z)≡0and

n

j=1 N

r,1

fj

+ (n– 1)

n

j=1

N(r,fj) <

λ+o(1)T(r,fk),

whereλ< 1and k= 1, 2, . . . ,n– 1,then fn≡1.

Lemma 6([16]) Suppose that f(z)and g(z)are two nonconstant meromorphic functions in the complex plane withρ(f)andρ(g)as their orders,respectively.Then

ρ(f·g)≤maxρ(f),ρ(g), and

Using the ideas of Chang and Fang [2] and El Farissi et al. [7,8], we prove the following lemma.

Lemma 7 Let f be a nonconstant meromorphic function of finite order,and let P(z) (≡0) be a polynomial,n be a positive integer.Suppose that

n_cf(z) –P(z) f(z) –P(z) =e

α(z)_, nc+1f(z) –P(z)

f(z) –P(z) =e

β(z)_{, (2.1)}

whereα(z)andβ(z)are two polynomials.If

Tr,eα+Tr,eβ=S(r,f),

then either f(z) =AP(z),whereα,β,A=eα_eα–1 are constants and P(z)is a polynomial with degree less than n,or f ≡cf.

Proof Two cases will be discussed in the following. Case 1. f(z) is a rational function.

Firstly, we claim thatf(z) cannot be a non-polynomial rational function. Otherwise, sup-pose thatf(z) = a_b(₍z_z)₎, wherea(z) andb(z) are two co-prime polynomials. By equation (2.1), we can get thatf,n_cf, andn_c+1f share∞CM. Thenn_cf andn_c+1f are rational func-tions, too. Suppose thatn_cf =g_b((_zz)₎, whereg(z) andb(z) are two co-prime polynomials. We claim thatb(z) is a constant. Otherwise, suppose that there existsz0such thatb(z0+c) = 0. Sincen

cf andnc+1f share∞CM, and

n_c+1f=n_cf(z+c) –n_cf(z) = g(z+c) b(z+c)–

g(z) b(z)=

g(z+c)b(z) –g(z)b(z+c) b(z)b(z+c) ,

then all zeros ofb(z+c) must be the zeros ofb(z). Otherwise, if there existsz1such that b(z1+c) = 0 butb(z1)= 0, thenz1is one pole ofnc+1f but not the pole ofncf, a

contra-diction. So we get

b(z0+c) = 0 ⇒ b(z0) = 0 ⇒ b(z0–c) = 0 ⇒ · · · ⇒ b(z0–kc) = 0,

k∈N+,

which implies thatb(z) has infinitely many zeros, a contradiction. Therefore,f(z) is a nonconstant polynomial.

Without loss of generality, we assume that

f(z) =alzl+al–1zl–1+· · ·+a1z+a0 (al= 0),

P(z) =bmzm+bm–1zm–1+· · ·+b1z+b0 (bm= 0).

By equation (2.1), it is easy to prove thatα(z),β(z) are two constants, then ⎧

⎨ ⎩

n_cf–P(z) =eα_f–eα_P(z),

Ifeα_{= 1 (ore}β_{= 1), thenf ≡}n

cf (orf ≡nc+1f), which contradictsdegf >degncf (or

degf >degn+1

c f). Thuseα,eβ= 1.

Comparing the degree of two equations, it follows thatdegf =l≤m=degP. Ifdegf=l<m=degP, theneα_=eβ_{= 1, thenf ≡}n

cf≡nc+1f, which is a contradiction.

Ifdegf =l=m=degP, comparing the coefficients ofzm_{of both sides of equation (2.1),}

we immediately have

aleα=

eα– 1bm, aleβ=

eβ– 1bm,

it follows thateα_=eβ_{, then we get}n

cf ≡nc+1f.

Sincef(z) is a polynomial, thenn_cf ≡n_c+1f ≡0, which meansf(z) is a polynomial with degree less thann.

By equation (2.1), we have

–P(z) =eαf(z) –eαP(z),

thenf(z) =AP(z), whereA=eα_e–1α (= 0, 1) is a constant,P(z) is a polynomial with degree less thann.

Case 2. f(z) is a transcendental meromorphic function, thenT(r,P) =S(r,f). By equation (2.1), we have

n_cf =eαf(z) –eα– 1P(z); (2.2)

n_c+1f=eβ_{f(z) –e}β_{– 1P(z), (2.3)}

then

n_c+1f=n_cf(z+c) –n_cf(z)

=eα(z+c)f(z+c) –eα(z+c)– 1P(z+c) –eα(z)f+eα– 1P

=eβ(z)f–eβ– 1P. (2.4)

Therefore,

fc(z) =f(z+c)

=eα–αc_+eβ–αc_{f +1 –e}–αc_P

c–

eα–αc_+eβ–αc_{– 2e}–αc_P

=A(z)f(z) +B(z),

whereA(z) = (eα–αc_+eβ–αc_{),B(z) = (1 –e}–αc_)P

c– (eα–αc+eβ–αc– 2e–αc)P.

Then

f2c(z) =fc(z+c)

=A(z+c)f(z+c) +B(z+c)

=A(z+c)A(z)f(z) +B(z)+B(z+c)

=A(z+c)A(z)f(z) +A(z+c)B(z) +B(z+c)

whereA2(z) =A(z+c)A(z) = (eαc–α2c+eβc–α2c)(eα–αc+eβ–αc) =e–(αc+α2c)(eαc+eβc)(eα+eβ), andB2(z) =A(z+c)B(z) +B(z+c).

By mathematical induction, we can deduce that

fic(z) =Ai(z)f(z) +Bi(z) (i≥1), (2.5)

We rewrite (2.7) as follows:

S(z)f(z) +T(z) = 0, (2.8) can deduce from equation (2.8) that

n

i=1

Ci_n(–1)n–ie–ik=1αkc_eik=1α(k–1)c

i

k=1

1 +eβ(k–1)c–α(k–1)c

=eα– (–1)n,

and

n

i=1

Ci_n(–1)n–ieα–αic

i–1

k=0

1 +eβkc–αkc_=eα_{– (–1)}n_{. (2.9)}

Letγkc(z) =βkc(z) –αkc(z),k∈ {1, 2, . . . ,n}, andγ(z) =β(z) –α(z) whenk= 0.

So equation (2.9) can be written as follows:

n

i=1

Ci_n(–1)n–ieα–αic

i–1

k=0

1 +eγkc_=eα_{– (–1)}n_{. (2.10)}

For

i–1

k=0

1 +eγkc

=1 +eγ_{1 +e}γc_{1 +e}γ2c· · ·_{1 +e}γ(i–1)c

= 1 +eγ +eγc₊· · ·_+eγ(i–1)c_+eγ+γc_+eγ+γ2c₊· · ·_+eγ(i–2)c+γ(i–1)c

+· · ·+eγ+γc+···+γ(i–1)c_.

Next, we discuss two cases.

Case 2.1.γ(z) is a nonconstant polynomial.

Setγ(z) =amzm+am–1zm–1+· · ·+a1z+a0, andam= 0.

Sinceγ(z) =β(z) –α(z), we claim thatdegγ(z)≥degα(z). Whendegα=degβ, thendegγ =max{degα,degβ}; Whendegα=degβ, thendegγ ≤degα.

Ifdegγ <degα, by equation (2.10), the order of the left-hand side

ρ

C_n1(–1)n–1eα–αic_{1 +e}γ_+C2

n(–1)n–2e

α–α2c_{1 +e}γ_{1 +e}γc

+· · ·+C_nneα–αnc

n–1

k=0

1 +eγkc

≤degα– 1,

but the order of the right-hand sideρ(eα_{– (–1)}n_{) =degα, a contradiction.}

Thusm=degγ ≥degα. By equation (2.10), we have

α0+α1eamz

m

+α2e2amz

m

+· · ·+αnenamz

m

where α0 = (–1)n + n

i=1Cin(–1)n–ieα–αic = eα[(–1)n +

n

i=1Cni(–1)n–ie–αic] = eα ×

n

i=0Cni(–1)n–ie–αic=eαnce–α, andαn=eα–αnc. We consider two subcases as follows.

Case 2.1.1.degα<m.

It is obvious thatαn≡0, by Lemma1and (2.11), we can get

degα=ρeα=ρα0+α1eamz

m

+α2e2amz

m

+· · ·+αnenamz

m

=m,

which is a contradiction. Case 2.1.2.degα=m.

Setα(z) =bmzm+α∗(z), wherebm= 0, anddegα∗(z)≤m– 1. We claim that there exists

at leastj∈ {1, 2, . . . ,n}such thatbm=jam.

Otherwise,d=iam,i∈ {1, 2, . . . ,n}, then by (2.11),

α1eamz

m

+α2e2amz

m

+· · ·+αnenamz

m

–ebmzm_eα∗(z)_{= –α} 0.

Then by Lemma4, we deduce that

αi=eα

∗_(z)

= 0 (i= 1, 2, . . . ,n),

a contradiction.

Thus there existsj∈ {1, 2, . . . ,n}such thatbm=jam. Without loss of generality, we

as-sume thatbm=nam, then

α1eamz

m

+α2e2amz

m

+· · ·+αn–eα

∗_(z)

enamzm_+α

0= 0.

By Lemma4, we haveα0≡0, that is,eαnce–α≡0, then

n_ce–α₌ n

i=0

C_ni(–1)n–ie–αic≡_{0. (2.12)}

We claim thatdegα=degγ=m= 1. Otherwise, suppose thatdegα=degγ =m> 1. Let α(z) =bmzm+bm–1zm–1+· · ·+b1z+b0, andbm= 0.

Then

α(z+jc) =bm(z+jc)m+bm–1(z+jc)m–1+· · ·+b1(z+jc) +b0

=bmzm+ (bm–1+mbmjc)zm–1+μj(z),

whereμj(z) is a polynomial withdegμj<m– 1.

Hence equation (2.12) can be written as follows:

e–μn(z)_e–bmzm–(bm–1+mbmnc)zm–1_–C1

ne–

μn–1(z)_e–bmzm–(bm–1+mbm(n–1)c)zm–1

+· · ·+ (–1)ne–μ0(z)_e–bmzm–bm–1zm–1_{= 0.}

It is obvious that

ρe–bmzm–(bm–1+mbmkc)zm–1+bmzm+(bm–1+mbmlc)zm–1_=ρembm(l–k)zm–1_{=m– 1}

Sinceρ(e–μj(z)_{) <m– 1 forj= 1, 2, . . . ,n, then by Lemma4}

e–μ0(z)₌· · ·_=e–μn(z)≡_0,

which is a contradiction. We thus proved thatdegα=degγ = 1. Thendegβ≤1 forβ= α+γ.

Without loss of generality, we may assume that

α(z) =λ1z+η1, β(z) =λ2z+η2,

whereλ1(= 0),λ2,η1,η2are constants. Thusα(z+jc) =α(z) +jλ1c,

and

n_ce–α₌ n

i=1

C_ni(–1)n–ie–αic

=

n

i=1

C_ni(–1)n–ie–αe–iλ1c

=e–α n

i=1

C_ni(–1)n–ie–iλ1c

=e–αe–λ1c– 1n.

Forn

ce–α≡0, that is,e–α(e–λ1c– 1)n≡0, theneλ1c≡1.

Therefore,

eα(z+ic)_=eα(z)+iλ1c_=eα(z)_,

which implies thateα(z)_{is a periodic function with periodc.}

We claim thateβ(z)_{is also a periodic function with periodc, ande}λ2c_{= 1.}

Otherwise, suppose thateλ2c_{= 1.}

Sinceγ(z) =β(z) –α(z) = (λ2–λ1)z+ (η2–η1), andγjc(z) =γ(z) +jc(λ2–λ1). Then by equation (2.11), we have

α1e(λ2–λ1)z+α2e2(λ2–λ1)z+· · ·+αnen(λ2–λ1)z=eλ1z+η1, (2.13)

whereα1,α2, . . . ,αn(=Cnn= 1) are constants.

Noticing thateλ1c_{= 1, thus}

α1=

C1_n(–1)n–1

+C_n2(–1)n–2+C_n2(–1)n–2eλ2c

+C_n3(–1)n–3+C_n3(–1)n–3eλ2c+C_n3(–1)n–3e2λ2c₊· · ·

+C_nn(–1)n–n+C_nn(–1)n–neλ2c+· · ·+C_nn(–1)n–ne(n–1)λ2ceη2–η1

=

C_n1(–1)n–1e

λ2c_{– 1}

eλ2c_{– 1}+C

2

n(–1)n–2

e2λ2c_{– 1}

+C_nn(–1)n–ne

nλ2c–1

eλ2c_{– 1}

eη2–η1

= _n

i=0

C_ni(–1)n–ieiλ2c– (–1)n–

n

i=1

C_ni(–1)n–i

eη2–η1

eλ2c_{– 1}

= _n

i=0

C_ni(–1)n–ieiλ2c–

n

i=0

C_ni(–1)n–i

eη2–η1

eλ2c_{– 1}

=eλ2c– 1n e

η2–η1

eλ2c_{– 1}

=eλ2c– 1n–1eη2–η1.

Ifλ1=λ2, theneβ(z+jc)=eβ(z), a contradiction. Ifλ1=λ2, we can rewrite equation (2.13) as follows:

α1e(λ2–2λ1)z+α2e(2λ2–3λ1)z+· · ·+αne(nλ2–(n+1)λ1)z=eη1. (2.14)

Ifiλ2– (i+ 1)λ1= 0 for eachi= 1, 2, . . . ,n, it is obvious that, for 1≤l<m≤n,

ρe[mλ2–(m+1)λ1–lλ2+(l+1)λ1]z_=ρe(m–l)(λ2–λ1)z_{= 1,}

then by Lemma4, we get

αi=eη1= 0, i= 1, 2, . . . ,n,

which is a contradiction.

So there exists one integerj∈ {1, 2, . . . ,n}such thatjλ2– (j+ 1)λ1= 0. Without loss of generality, we assume thatnλ2– (n+ 1)λ1= 0.

Thus (2.14) can be rewritten as follows:

α1e(λ2–2λ1)z+α2e(2λ2–3λ1)z+· · ·+αn–1e((n–1)λ2–nλ1)z=eη1–αn.

Sinceαn= 1, thuseη1–αn= 0. We also can deduce by Lemma4thatα1= 0, which means (eλ2c_{– 1)}n–1_eη2–η1_{= 0, thuse}λ2c_{= 1, a contradiction.}

Thus

eβ(z+jc)_=eβ(z)_ejλ2c_=eβ(z)_,

soeβ(z)_{is also a periodic function with periodc.} Therefore, we can deduce by equation (2.1) that

n_c+1f=n_cf(z+c) –n_cf(z)

=eαcf–

eα– 1cP

=eβ_{f –e}β_{– 1P, (2.15)}

thus

cf =eβ–αf +

1 –e–αcP–

By mathematical induction, we have

2_cf =e2(β–α)f +1 –e–α2_cP+eβ–α1 –e–α–eβ–α–e–α ×cP–eβ–α

eβ–α–e–αP,

3_cf =e3(β–α)f +1 –e–α3_cP+eβ–α1 –e–α–eβ–α–e–α ×2_cP+eβ–α_eβ–α_–e–α_–e2(β–α)_{1 –e}–α

cP

–e2(β–α)eβ–α–e–αP, · · ·,

n_cf =en(β–α)f+1 –e–αn_cP+eβ–α1 –e–α–eβ–α–e–α ×n_c–1P+· · ·+e(n–1)(β–α)_{1 –e}–α_–e(n–2)(β–α)_eβ–α_–e–α

×cP–e(n–1)(β–α)

eβ–α–e–αP

=en(β–α)f+1 –e–α

n–1

i=0

ei(β–α)n_c–iP

–eβ–α–e–α

n–1

i=0

ei(β–α)n_c–i–1P.

And by (2.1)

n_cf =eαf–eα– 1P.

Therefore

eαf –eα– 1P=en(β–α)f+1 –e–α

n–1

i=0

ei(β–α)n_c–iP

–eβ–α–e–α

n–1

i=0

ei(β–α)n_c–i–1P. (2.16)

Hence

eα_–en(β–α)_{f=1 –e}–α n–1

i=0

ei(β–α)n–i c P

–eβ–α–e–α

n–1

i=0

ei(β–α)_cn–i–1P+eα– 1P. (2.17)

Ifeα_–en(β–α)_{≡0, by the above equation,T(r,f) =S(r,f), a contradiction.} Thuseα_≡en(β–α)_{, that is,e}nβ_≡e(n+1)α_.

At the same time,

1 –e–α

n–1

i=0

ei(β–α)n–i c P–

1 –e–β

n–1

i=0

e(i+1)(β–α)n–i–1

c P

1 –e–αn_cP+1 –e–α

n–1

i=1

ei(β–α)_cn–iP–1 –e–β (2.18)

×

n–1

i=1

ei(β–α)_cn–iP–en(β–α)1 –e–βP+eα– 1P= 0,

1 –e–αn cP+

e–β_–e–α n–1

i=1

ei(β–α)n–i c P

+eα– 1–en(β–α)1 –e–βP= 0.

IfP(z) is a constant, theni_cP= 0 for anyi∈N+_{, then by (2.18),}

eα– 1–en(β–α)1 –e–βP= 0,

thus

eα– 1 =en(β–α)1 –e–β.

Noticing thateα_≡en(β–α)_{, we havee}α–β_{= 1, which implies thate}α_=en(β–α)_{= 1, a} con-tradiction.

SoP(z) is a nonconstant polynomial, we deduce by Nevanlinna’s first fundamental the-orem that

T(r,P) =Sr,eα_;

T(r,P) =Sr,eβ;

degi_cP<degP

for eachi∈N+.

Sinceα(z) andβ(z) are both polynomials satisfyinge(n+1)α_=enβ_{, set}

α1(z) = α(z)

n , β1(z) = β(z) n+ 1,

henceα1(z),β1(z) are polynomials of degree 1 satisfying

eα1_=eβ1_{, e}α_=enα1_{, e}β_=e(n+1)β1_,

and

T(r,P) =Sr,eα1_{, T(r,P) =Sr,e}β1_.

Rewrite (2.18) as follows:

enα–e(n–1)αn_cP+enα–β–e(n–1)α

n–1

i=1

thus,

enαn_cP–P

=e(n–1)αn_cP–n_c–1P+eβ+(n–2)αn_c–1P–n_c–2P+· · ·

+eα+(n–2)β2

cP–cP

+e(n–1)β_(P– cP),

en2α1n_cP–P

=e(n2–n)α1n

cP–nc–1P

+e(n2–n+1)α1n–1

c P–nc–2P

+· · ·+e(n2–2)α12 cP–cP

+e(n2–1)α1_(P– cP).

(2.19)

SinceP(z)≡i

cPfor eachi∈N+, by Nevanlinna’s first fundamental theorem and (2.19),

n2Tr,eα1_+Sr,eα1

=Tr,en2α1n cP–P

=Tr,e(n2–n)α1n cP–n

–1

c P

+· · ·+e(n2–2)α12

cP–cP

+e(n2–1)α1_(P– cP)

≤n2– 1Tr,eα1_+Sr,eα_,

a contradiction.

Case 2.2.γ(z) is a constant.

We claim thatα(z) is also a constant.

Suppose thatdegα(z)≥1, thusdeg(α–αic)≤degα– 1. Notice thatβ=α+γ andγkc=

βkc–αkc=γ. Then, by equation (2.10),

degα≤degα– 1,

which is impossible.

Henceαis a constant, so isβforβ=α+γ.

Using the same method as the previous proof, we can deduce by equation (2.1) that

cf =eβ–αf +

1 –e–αcP–

eβ–α–e–αP. (2.20)

Thus

n_cf =en(β–α)_{f+1 –e}–α n–1

i=0

ei(β–α)n–i c P

–eβ–α–e–α

n–1

i=0

ei(β–α)n_c–i–1P,

=eα_f–eα_{– 1P(z), (2.21)}

n_c+1f=e(n+1)(β–α)f+1 –e–α

n

i=0

=1 –e–α

n–1

i=0

e(i+1)(β–α)_cn–iP–eβ–α–e–α

n–1

i=0

e(i+1)(β–α)n_c–i–1P

+1 –eβ–αP.

Hence,

1 –e–α_cn+1P–eβ–α–e–αn_cP=1 –eβ–αP. (2.24)

We claim that 1 –eβ–α_{= 0.}

Suppose 1 –eβ–α_{= 0.}

IfP(z) is a constant, by (2.24), we haveP≡0, a contradiction.

IfP(z) is a nonconstant polynomial, the degree of the left-hand side of (2.24) is less than that of the right-hand side, a contradiction.

Thuseβ–α_{= 1, thene}α_=en(β–α)_{= 1,e}β_=e(n+1)(β–α)_{= 1.}

Then, by (2.23), we have provedn_cf ≡n_c+1f, which is a contradiction to our assump-tion.

Thusn_cf ≡_cn+1f, andeα_=en(β–α)_{= 1,e}β_=e(n+1)(β–α)_{= 1.} By (2.20), we have

f ≡cf,

hence Lemma7is proved.

3 Proof of Theorem8

Proof Sincef,n_cf, and_cn+1f share∞andP(z) CM, and the order off is finite, then there exist two polynomialsα(z) andβ(z) such that

⎧ ⎨ ⎩

ncf(z)–P(z)

f(z)–P(z) =e

α(z)_,

nc+1f(z)–P(z)

f(z)–P(z) =e

β(z)_.

(3.1)

By Lemma7, two cases will be considered in the following. Case 1. f is a rational function.

Using the same method as Lemma7, we have

f(z) =AP(z),

whereA= 0, 1 is a constant, andP(z) is a polynomial withdegP(z)≤n– 1. Case 2. f is a transcendental meromorphic function, thenT(r,P) =S(r,f). Set

F(z) =f(z) –P(z),

It follows that

_cnf =n_cF+n_cP, _cn+1f =n_c+1F+n_c+1P.

Then (3.1) can be written as ⎧

Two cases will be discussed in the following. Case 2.1.φ(z)≡0.

then we can rewriteφ(z) as follows:

φ(z) =P(z) –n_c+1P(z)eα(z)–P(z) –n_cP(z)eβ(z), (3.4)

thus by Nevanlinna’s second fundamental theorem,

Noticing that T(r,φ) =S(r,F) and T(r,P) =S(r,F), we have T(r,eα(z)_{) =S(r,F), so} T(r,eα(z)_{) =S(r,f).}

And by (3.5), we also getT(r,eβ(z)_{) =S(r,f).}

Therefore, by Lemma7, we havef ≡cf sincef is a transcendental meromorphic

func-tion.

Case 2.2.φ(z)≡0.

By the definition ofφ(z), we have

P(z) –n_c+1P(z)n_cF(z) =P(z) –n_cP(z)n_c+1F(z). (3.6)

ForF(z) =f(z) –P(z), thus

P(z) –n_c+1P(z)n_cf(z) –n_cP(z)

=P(z) –n_cP(z)_cn+1f(z) –n_c+1P(z), ×P(z) –n_c+1P(z)n_cf–Pn_cP+P2

=P(z) –n_cP(z)_cn+1f(z) –Pn_c+1P+P2,

×P(z) –n_c+1P(z)n_cf–P=P(z) –n_cP(z)n_c+1f–P, ×P(z) –n_c+1P(z)eα=P(z) –n_cP(z)eβ,

which implies

eα–β₌ P(z) –ncP(z)

P(z) –n+1

c P(z)

. (3.7)

SinceP(z) is a polynomial, theneα–β_{is a constant. Assume thate}α–β_{=A, then}

P(z) –n_cP(z) =AP(z) –n_c+1P(z). (3.8)

Comparing the coefficients of both sides of equation (3.8),A= 1, thenn_cP≡n_c+1P. SinceP(z) is a polynomial, thusn_cP≡n_c+1P≡0, andP(z) is a polynomial of degree less thann.

By (3.6), we haven_cF(z) =n_c+1F(z). Thus by (3.1),eα(z)_=eβ(z)_{, then (3.2) can be} rewrit-ten as follows:

⎧ ⎨ ⎩

ncF–P

F =e α(z)_,

nc+1F–P

F =e

α(z)_.

(3.9)

Hence ⎧ ⎨ ⎩

n

cF=eα(z)F+P,

n_c+1F=eα(z)_F+P. (3.10)

By the definition ofnth difference,

=eαc_F

c+Pc–eαF–P

=eα(z)F+P.

Thus

Fc= 2eα–αcF+ (2P–Pc)e–αc. (3.11)

Then we can deduce that

F2c= 22eα–α2cF+

Using the same method, we have

Hence,

2n+1n_cν+

n

i=0

n_cνCi_n+1+n_c+1νC_ni(–1)n+1–i2ieα(n+1)c–αic

=eα(n+1)cn

cν–nc+1ν

. (3.15)

Set

γi(z) = (–1)n+1–i2ieα(n+1)c–αic, i= 1, 2, . . . ,n,

and

γn+1(z)

=n_cν–_cn+1νeα(n+1)c

=

n

i=0

C_ni(–1)n–iPiceα(n+1)c–αic– n+1

i=0

C_ni₊₁(–1)n+1–iPiceα(n+1)c–αic.

It is obvious thatρ(γi(z))≤degα– 1 for eachi∈ {1, 2, . . . ,n,n+ 1}. Let

M(z) =

n

i=0

C_ni₊₁γi(z) + 2n+1, N(z) = n

i=0

C_niγi(z),

thenρ(M(z))≤degα– 1, andρ(N(z))≤degα– 1. Thus by (3.15)

M(z)

n

i=0

Ci_n(–1)n–iPice–αic+N(z) n+1

i=0

C_ni₊₁(–1)n+1–iPice–αic=γn+1(z). (3.16)

Then we rewrite (3.16) as follows:

dn+1(z)e–α(z+(n+1)c)+dn(z)e–α(z+nc)+· · ·+d0(z)e–α(z)=γn+1(z), (3.17)

where dn+1(z) =N(z)P(n+1)c,dn(z) = (M(z) – (n + 1)N(z))Pnc, . . . ,d0(z) = ((–1)n+1N(z) + (–1)n_M(z))P.

Next, we consider three cases. Case 2.2.1.degα≥2.

By the definition ofM(z),N(z), it can be proved thatρ(di(z))≤degα– 1,i= 0, 1, . . . ,n,

n+ 1. Since

γn+1(z)

=n_cν–_cn+1νeα(n+1)c

=

n

i=0

C_ni(–1)n–iPiceα(n+1)c–αic– n+1

i=0

= 2(–1)nPeα(n+1)c–α_+C1

n+1+C1n

(–1)n–1Pceα(n+1)c–αc+

C_n2₊₁+C_n2(–1)n–2 ×P2ceα(n+1)c–α2c+· · ·+

Cn_n+1+C_nnPnceα(n+1)c–αnc–P(n+1)c,

we claim thatγn+1(z)≡0. Otherwise,

P(n+1)c= 2(–1)nPeα(n+1)c–α+

C_n1₊₁+C1_n(–1)n–1Pceα(n+1)c–αc

+C2_n+1+C_n2(–1)n–2P2ceα(n+1)c–α2c+· · ·

+Cn_n+1+C_nnPnceα(n+1)c–αnc.

Thus

2(–1)n P P(n+1)c

eα(n+1)c–α_{+ (–1)}n–1_C1

n+1+C1n

Pc

P(n+1)c

eα(n+1)c–αc

+· · ·+Cn_n+1+C_nn Pnc P(n+1)c

eα(n+1)c–αnc_{= 1. (3.18)}

It is obvious that (–1)n–i(C_ni₊₁+C_ni) Pic

P(n+1)ce

α_(n+1)c–αic _{cannot be constantsC}

i(= 0), and note

thatρ(eα(n+1)c–αic₎≥_{1, hence by Lemma3,}

Cn

n+1+Cnn

Pnc

P(n+1)c

eα(n+1)c–αnc≡_1,

which implies thateα(n+1)c–αnc ₌ P(n+1)c

(C_nn₊₁+Cnn)Pnc, but it contradicts with the fact thate

α(n+1)c–αnc

is a transcendental entire function. Thusγn+1(z)≡0. Therefore, we rewrite (3.17) as follows:

dn+1(z) γn+1(z)

e–α(z+(n+1)c)₊ dn(z)

γn+1(z)

e–α(z+nc)_{+· · ·+} d1(z) γn+1(z)

e–α(z+c)₊ d0(z) γn+1(z)

e–α(z)_{= 1.}

Obviously,d_γj(z)e–αic

n+1(z) cannot be constantCj(= 0), thus by Lemma3, we have

d0(z)e–α(z) γn+1(z) ≡

0,

which meanse–α(z)_≡γn+1(z)

d0(z) , a contradiction. Case 2.2.2.degα= 1.

Ifγn+1(z)≡0, by (3.17), by Lemma3, and by the same method as Case 2.2.1, we also deduce a contradiction.

Ifγn+1(z)≡0, without loss of generality, we assume thatα(z) =λz+μ,λ= 0, then

α(n+1)c–αic= (n+ 1 –i)λc.

Thus

γn+1(z)

= 2(–1)nPe(n+1)λc+C1_n+1+C_n1(–1)n–1Pcenλc+

×(–1)n–2P2ce(n–1)λc+· · ·+

C_nn₊₁+C_nnPnceλc–P(n+1)c

=

n

i=0

(–1)n–iC_ni +Ci_n+1Pice(n+1–i)λc–P(n+1)c.

SincedegP≤n– 1, without loss of generality, we assume that

P(z) =szn–t+P∗(z),

wheretis a positive integer satisfying 1≤t<n, ands(= 0)∈C. We assume that the coefficient ofzn–t_inγ

n+1(z) isA, then

A=s _n

i=0

(–1)n–iC_ni₊₁+Ci_ne(n+1–i)λc– 1

,

hence

A/s=

n

i=0

(–1)n–i_Ci n+1+Cin

e(n+1–i)λc_{– 1}

=

n

i=0

(–1)n–iC_ni₊₁e(n+1–i)λc– 1 +

n

i=0

(–1)n–iC_nie(n+1–i)λc

= –1 –eλcn+1+eλc

n

i=0

(–1)n–iC_nie(n–i)λc

=1 –eλcn+1+eλc1 –eλcn

=1 –eλcn2eλc– 1.

Sinceγn+1(z)≡0, thenA= 0, two subcases will be considered in the following. Case 2.2.2.1.1 –eλc= 0, theneλc= 1.

Hence

eα(z+ic)=eα(z)+iλc=eα(z),

which meanseα(z)_{is a periodic function of periodc, ande}αic_=eα_.

Now we have

An= n

i=0

Ci_n(–1)n–i2i–eα= 1 –eα,

Bn= n

i=0

Ci_n(–1)n–i2iP–Pic

e–α–P

=e–α

P

n

i=0

Ci_n(–1)n–i2i–

n

i=0

Ci_n(–1)n–iPic

–P

By (3.13),

0 =1 –eαF+e–α– 1P,

F=1 –e –α

1 –eα P

= –e–αP,

then

f =F+P=1 –e–αP.

Thus,

n_cF=e–αn cP≡0,

n_c+1F=e–αn_c+1P≡0.

Therefore,

n_cf ≡0, n_c+1f ≡0.

It is obvious thatf,n

cf,nc+1f share∞andP(z) CM, thus

f =1 –e–α(z)P(z),

whereeα(z)_{is a periodic function of periodcwithdegα= 1, andP(z) is a polynomial with}

degP≤1.

Case 2.2.2.2.1 – 2eλc= 0, theneλc=1₂, and

eαic_=eα_eiλc_{= 2}–i_eα_.

Hence

An= n

i=0

Ci_n(–1)n–i2ieα_e–αic_–eα

=

n

i=0

Ci_n(–1)n–i4i–eα

= 3n_–eα_,

Bn= n

i=0

Ci_n(–1)n–i2iP–Pic

e–αic_–P

=e–α

P

n

i=0 Ci

n(–1)n

–i₄i_– n

i=0 Ci

n(–1)n

–i_P ic2i

–P

= 3nPe–α–e–α

n

i=0

and

n

i=0

Ci_n(–1)n–iPic2i= n+1

i=0

C_ni₊₁(–1)n+1–iPic2i≡0. (3.20)

By our assumption,P=szn–t_+P∗_{, wheres= 0 anddegP}∗_{<n–t. Then the coefficient of}

zn–tof (3.20) is

s

n+1

i=0 Ci

n+1(–1)n+1–i2i≡0.

It follows that

s= 0,

a contradiction.

Case 2.2.3.degα= 0, thenαis a constant. Ifeα_{= 1, thene}β_{= 1. By (3.1), we have}

f ≡n_cf≡n_c+1f,

thencf ≡cn+1f, hencef ≡cf.

Ifeα_{= 1, by (3.13), we have}

F= –e–αP,

thus

f =AP(z),

whereP(z) is a polynomial withdegP(z)≤n– 1, andA= 1 –e–α_{(= 0, 1). But it contradicts}

the assumption thatf(z) is a transcendental function.

Thus, Theorem8is proved.

Acknowledgements

The authors thank the referee for his/her valuable suggestions to improve the present article.

Funding

Research was supported by the NNSF of China (Grant No. 11701188; 11371149).

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All the authors drafted the manuscript, and read and approved the final manuscript.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

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