Meromorphic functions that share four or three small functions with their difference operators

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R E S E A R C H

Open Access

Meromorphic functions that share four or

three small functions with their difference

operators

Huifang Liu

1*

_{and Zhiqiang Mao}

2*

*_{Correspondence:}

[email protected];

[email protected]

1_{College of Mathematics and}

Information Science, Jiangxi Normal University, Nanchang, China 2_{School of Mathematics and}

Computer, Jiangxi Science and Technology Normal University, Nanchang, China

Abstract

In this paper, we prove that non-constant meromorphic functions of finite order and their difference operators are identical, if they share four small functions “IM”, or share two small functions and∞CM. Our results show that a conjecture posed by Chen–Yi in 2013 is still valid for shared small functions, and improve some earlier results obtained by Li–Yi, Lü et al. We also study the uniqueness of a meromorphic function partially sharing three small functions with their difference operators.

MSC: Primary 30D35; secondary 39B32

Keywords: Meromorphic function; Small function; Diﬀerence operator; Uniqueness

1 Introduction and main results

In this paper, a meromorphic function always means meromorphic in the complex plane. We adopt the standard notations in Nevanlinna theory; see, e.g. [11,21]. In addition, we use the notationsσ(f),σ2(f) to denote the order and the hyper-order off(z), respectively, where

σ(f) =lim sup r→∞

log+T(r,f)

logr , σ2(f) =lim supr→∞

log+log+T(r,f)

logr .

A meromorphic functionα(≡ ∞) is called a small function off provided thatT(r,α) =

o(T(r,f)) asr→ ∞, possibly outside a set ofrof ﬁnite logarithmic measure. We useS(f) to denote the family of all meromorphic functions which are small functions off, and denoteSˆ(f) =S(f)∪ {∞}.

Letf andgbe two non-constant meromorphic functions, and letαbe a meromorphic function. We say thatf andg shareαCM (IM), provided thatf –α andg–αhave the same zeros counting multiplicities (ignoring multiplicities). If1

f and

1

g share 0 CM (IM),

then we say thatf andgshare∞CM (IM).

Nevanlinna’s four-value theorem shows that if two non-constant meromorphic func-tionsf andgshare four distinct values CM, thenf is a Möbius transformation ofg. In [4], Gundersen constructed a counterexample to show that four-value theorem is not valid if 4 CM is replaced by 4 IM. But whengis the derivative off, Gundersen and Mues–Steinmetz, respectively, obtained the following result.

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Theorem A([5,18]) If a non-constant meromorphic function f and its derivative fshare three distinct ﬁnite values a1,a2,a3IM,then f ≡f.

Remark1.1 Observe that a meromorphic functionf andftrivially share∞IM. So, in this sense, the four-value theorem is valid forf andfsharing four values IM.

Furthermore, Gundersen and Mues–Steinmetz improved TheoremAas follows.

Theorem B([6,19]) If a non-constant meromorphic function f and its derivative fshare two distinct ﬁnite values a1,a2CM,then f ≡f.

Recently, the diﬀerence analog of Nevanlinna theory has been established; see, e.g. [2,

7–10,14]. Many researchers ([1,12,13,15–17], etc.) started to consider the uniqueness of meromorphic functions sharing values with their shifts or their difference operators. For a nonzero finite valueη,f(z+η) is called a shift off(z), its difference operators are defined as

ηf(z) =f(z+η) –f(z) and nηf(z) =

n–1

η

ηf(z)

, n∈N,n≥2.

It is well known thatηf can be regarded as the diﬀerence counterpart off. So,

consid-ering the diﬀerence analog of TheoremsAandB, the following results are obtained.

Theorem C([15]) Let f be a non-constant meromorphic function ofσ(f) <∞.If f and

ηf share four distinct values a1,a2,a3,a4IM,then f ≡ηf.

Theorem D([1]) Let f be a transcendental meromorphic function such thatσ(f)is ﬁnite but not an integer.If f andηf(≡0)share three distinct values a1,a2,∞CM,then f ≡ηf.

In [1], the authors conjecture that the condition “order of growthσ(f) is not an integer or inﬁnite” can be removed. Lü [17] considered this conjecture and obtained the following result.

Theorem E([17]) Let f be a transcendental meromorphic function ofσ(f) <∞.If f and

ηf share three distinct values a1,a2,∞CM,then f ≡ηf.

It is natural to pose the question: what can be said on replacingshared values in Theo-remsC–Eby shared small functions. Concerning this question, we obtain the following results which extend TheoremsC–E. For the convenience of statement, we need the fol-lowing deﬁnition; see [21].

Letf,gandαbe three distinct meromorphic functions,N0(r,α,f,g) denote the counting function of commonzeros of f(z) –α(z)and g(z) –α(z), each counted only once. If

N

r, 1

f –α

–N0(r,α,f,g) =S(r,f)

and

N

r, 1

g–α

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whereS(r,f) =o(T(r,f)) asr→ ∞, possibly outsidea set of r of ﬁnitelogarithmic measure, then we say thatf andgshareα“IM”. Obviously, iff andgshareαIM, thenf andgshare α“IM”. But the reverse is not true.

Theorem 1.1 Let f be a transcendental meromorphic function ofσ2(f) < 1,αj∈S(f) (j=

1, 2, 3, 4),and letηbe a nonzero ﬁnite value.If f andηf shareα1,α2,α3,α4 “IM”,then

f ≡ηf.

Remark1.2 Obviously, Theorem1.1is an improvement of TheoremC.

Theorem 1.2 Let f be a non-constant meromorphic function ofσ(f) <∞,α1,α2∈S(f),

and letηbe a nonzero ﬁnite value.If f andηf shareα1,α2,∞CM,and if f andα1,α2

have no common poles with the same multiplicity,then f ≡ηf.

Remark1.3 Obviously, TheoremsDandEare direct results of Theorem1.2. By Theorem1.2, we get the following corollary.

Corollary 1.1 Let f be a non-constant entire function ofσ(f) <∞,α1,α2∈S(f),and letη

be a nonzero ﬁnite value.If f andηf shareα1,α2CM,then f ≡ηf.

We do not know whether Theorem1.2is valid, iffandηfshare three distinct functions

α1,α2,α3∈S(f). But under some additional restriction onαj, we get the following result.

Theorem 1.3 Let f be a non-constant meromorphic function ofσ2(f) < 1,αj∈S(f) (j=

1, 2, 3),and letηbe a nonzero ﬁnite value.If,for j= 1, 2, 3,

E(αj,f)⊂E(αj,ηf), ηαj≡αj,

where E(αj,f)is the set of zeros of f –αj,counting multiplicity,then f ≡ηf.

Remark1.4 The conditionηαj≡αj(j= 1, 2, 3) in Theorem1.3is necessary. For

exam-ple, letf(z) = 1

eπiz₊₁,η= 1,α1= 0,α2= 1,α3=3₄, it is obvious thatE(αj,f)⊂E(αj,ηf) (j=

1, 2, 3). Butηαj≡αj(j= 2, 3), andηf(z) = 2e πiz

1–e2πiz ≡f(z). 2 Lemmas

Lemma 2.1([10]) Let f be a non-constant meromorphic function,ε> 0,andηbe a ﬁnite value.If f is of ﬁnite order,then there exists a set E=E(f,ε)satisfying

lim sup r→∞

E∩[1,r)dt/t

logr ≤ε,

i.e.of logarithmic density at mostε,such that

m

r,f(z+η)

f(z)

=O

logr r T(r,f)

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for all r outside the set E.Ifσ2(f) < 1andε> 0,then

m

r,f(z+η)

f(z)

=o

T(r,f)

r1–σ2(f)–ε

for all r outside of a set of ﬁnite logarithmic measure.

Lemma 2.2([10]) Let T: [0, +∞)→[0, +∞)be a non-decreasing continuous function and let s> 0.If

lim sup r→∞

log logT(r)

logr =ς< 1 andδ∈(0, 1 –ς),then

T(r+s) =T(r) +o

T(r)

rδ

,

where r runs to inﬁnity outside of a set of ﬁnite logarithmic measure.

Letf be a meromorphic function, it is shown in [3], p. 66, that, for an arbitrary complex numberc= 0, the inequalities

1 +o(1)Tr–|c|,f(z)≤Tr,f(z+c)≤1 +o(1)Tr+|c|,f(z) hold asr→ ∞. Similarly, we have

1 +o(1)Nr–|c|,f(z)≤Nr,f(z+c)≤1 +o(1)Nr+|c|,f(z), (r→ ∞). So combining the above inequalities and Lemma2.2,we get the following result.

Lemma 2.3 Let f be a non-constant meromorphic function ofσ2(f) < 1.Then,for an

arbi-trary complex number c= 0,

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

Lemma 2.4([20]) Let f be a transcendental meromorphic function andαj(j= 1, . . . ,q)be

q distinct small functions of f.Then,forε> 0,

(q– 2 –ε)T(r,f)≤

q

j=1

N

r, 1

f –αj

+oT(r,f)

as r∈/E→ ∞for a set E of ﬁnite linear measure.

Remark2.1 In [23], Zheng pointed out that theεin the above inequality can be removed. Using a similar argument to thatof[21], Theorem 4.4, we obtain the following result.

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(i) T(r,f) =T(r,g) +S(r,f),T(r,g) =T(r,f) +S(r,g). (ii) 4_j₌₁N(r,_f_–1_α

j) = 2T(r,f) +S(r,f).

Lemma 2.6 ([22]) Let f and g be non-constant meromorphic functions and let αj (j=

1, . . . , 5)be ﬁve distinct elements inSˆ(f)∩ ˆS(g).If f ≡g,then

N0(r,α5,f,g)≤ 4

j=1

N12(r,αj,f,g) +S(r,f) +S(r,g),

where N12(r,α,f,g) =N(r,_f_–1α) +N(r,

1

g–α) – 2N0(r,α,f,g).

Lemma 2.7([21]) Let f1, . . . ,fn(n≥2)be meromorphic functions,and g1, . . . ,gnbe entire

functions satisfying the following conditions. (i) n_j₌₁fj(z)egj(z)≡0.

(ii) gj(z) –gk(z)are not constants for1≤j<k≤n.

(iii) For1≤j≤n, 1≤t<k≤n,T(r,fj) =o(T(r,egt–gk)) (r→ ∞,r∈/E),whereE⊂(1,∞) has ﬁnite linear measure or ﬁnite logarithmic measure.

Then fj(z)≡0 (j= 1, . . . ,n).

Lemma 2.8 Letα(≡0)be a meromorphic function,and let c,ηbe nonzero ﬁnite values.If

α(z+η) =cα(z),then T(r,α)≥dr–O(1)holds for suﬃciently large r,where d is a positive number.

Proof It follows fromα(z+η) =cα(z) thatα(z) is transcendental. If 0,∞are the Picard exceptional values ofα(z), then there exists a non-constant entire functionh(z), such that α(z) =eh(z)_{. This implies that}_T₍_r_,_α)_≥_dr_–_O_{(1) holds for suﬃciently large}_r_{and some} positive numberd. Ifα(z) has at least one zero or one polez0, thenz0+jη,j∈Zare zeros or poles ofα(z). This implies thatN(r,_α1)≥drorN(r,α)≥drholds for suﬃciently larger

and some positive numberd. So we getT(r,α)≥dr–O(1) holds for suﬃciently larger.

Lemma 2.9([9]) LetMbe the set of all meromorphic functions in the complex plane,N

be a subﬁeld ofM,and let f∈N\ker(L),where L:M→Mis a linear operator such that m(r,L(_ff)) =S(r,f).If a1, . . . ,aqare q≥1diﬀerent elements ofker(L)∩S(f),then

(q– 1)T(r,f) +NL(f)(r,f)≤N(r,f) +

q

j=1

N

r, 1

f–aj

+S(r,f),

where NL(f)(r,f) = 2N(r,f) –N(r,L(f)) +N(r,_L1₍_f₎).

3 Proofs of the results

Proof of Theorem1.1 Suppose thatf≡ηf, from the fact thatfandηf shareα1,α2,α3,α4 “IM” and Lemma2.5, we get

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from which we deduce thatηf is transcendental and

S(r,ηf) =S(r,f). (1)

By Lemmas2.4and2.5, we get

3T(r,f)≤N(r,f) + 4

j=1

N

r, 1

f–αj

+S(r,f)

≤N(r,f) + 2T(r,f) +S(r,f) ≤N(r,f) + 2T(r,f) +S(r,f),

from which we deduce that

T(r,f) =N(r,f) +S(r,f) =N(r,f) +S(r,f). (2) Sincen(2)(r,f)≤2(n(r,f) –n(r,f)), it follows from (2) that

N(2)(r,f) =S(r,f), (3) wheren(2)(r,f) denotesthe number of multiple polesoff in|z| ≤r, counting multiplicity,

N(2)(r,f) denotes its corresponding counting function. Similarly, we get

N(2)(r,ηf) =S(r,ηf) =S(r,f). (4)

On the other hand, from the fact thatf andηfshareα1,α2,α3,α4“IM” and Lemma2.6, (1), we get

N0(r,∞;f,ηf)≤

4

j=1

N12(r,αj;f,ηf) +S(r,f) +S(r,ηf) =S(r,f). (5)

LetN(r,f(z) =a,g(z)=b) denote the reduced counting function of those points in|z| ≤r, which area-points off, notb-points ofg(z), (5) and Lemma2.3imply that

N(r,ηf –f)≤N

r,ηf=∞,f(z)=∞

+Nr,f(z) =∞,ηf =∞

+N0(r,∞;f,ηf)

=Nr,f(z+η) =∞,f(z)=∞+Nr,f(z) =∞,ηf =∞

+S(r,f) ≤Nr,f(z) =∞,ηf =∞

+S(r,f)

≤N(r,f) +S(r,f). (6)

Hence by (3), (4) and (6), we get

N(r,ηf –f)≤N(r,ηf–f) +N(2)(r,ηf –f)

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≤N(r,f) +S(r,f). (7)

Then, by Lemma2.1, Lemma2.4and (7), we get

3T(r,f)≤N(r,f) + 4

j=1

N

r, 1

f–αj

+S(r,f)

≤N(r,f) +N

r, 1

f –ηf

+S(r,f)

≤N(r,f) +N(r,f –ηf) +m(r,f –ηf) +S(r,f)

≤2N(r,f) +m(r,f) +S(r,f) ≤2T(r,f) +S(r,f),

which impliesT(r,f) =S(r,f). This is absurd. So we getf ≡ηf.

Proof of Theorem1.2 It follows from Lemma2.3thatηf is of ﬁnite order. Sincef and

ηf shareα1,α2,∞CM, we get ηf –α1

f –α1

=eP, ηf–α2

f –α2

=eQ, (8)

whereP,Qare polynomials.

Suppose thatηf ≡f, theneP≡1,eQ≡1 andeP≡eQ. By (8), we get

f(z) =α1(z) +

α2(z) –α1(z)

eQ(z)_{– 1}

eQ(z)_–_eP(z) (9) and

ηf(z) =α1(z) +

α2(z) –α1(z)

eP(z)+Q(z)_–_ep(z)

eQ(z)_–_eP(z) . (10) On the other hand, (9) also implies

ηf(z) =ηα1(z) +

α2(z+η) –α1(z+η)

eQ(z+η)_{– 1}

eQ(z+η)_–_eP(z+η) –α2(z) –α1(z)

eQ(z)_{– 1}

eQ(z)_–_eP(z). (11) Now we discuss the following three cases.

Case1. Suppose that botheP_and_eQ_{are constants, then, by (}₉_{), we get}_T₍_r_,_f_{) =}_S₍_r_,_f_).

This is absurd.

Case2. Suppose that only one betweeneP_and_eQ_{is a constant, without loss of generality,}

we assume thateP≡c, by (9), we get

T(r,f) =Tr,eQ+S(r,f), S(r,f) =Sr,eQ. (12)

Subcase2.1. IfeQ(z+η)_≡_eQ(z)_{, then}_deg_Q_{= 1. (}₁₀_{) and (}₁₁_{) imply that} (1 –c)α1(z) +cα2(z) –ηα2(z)

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By (12) and (13), we get

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Subcase3.1.degP>degQ. By (9) we get

S(r,f) =Sr,eP. (19) Equation (17) implies that

ψ1e2P+ψ2eP=ψ3, (20)

where

ψ1=H2p+H2p+qeQ, ψ2=Hp+2qe2Q+Hp+qeQ+Hp, ψ3= –H2qe2Q–HqeQ,

such thatT(r,ψj) =S(r,eP) (j= 1, 2, 3). Then, by (20) and Lemma2.7, we getψj≡0 (j=

1, 2, 3). From this and (18), we get ⎧

⎨ ⎩

(α2–ηα1) – (α2–α1)eQ= 0,

(α2–α1)eηQ+2Q+ (ηα1–α2)eηP+Q+ (α2–α1)eηP= (α2–α1)eηQ.

(21)

Solving (21) deduce

(α1–ηα1)(2α2–α1–ηα1)

eηQ_–_eηP≡_0.

Sincedeg(ηP) =degP– 1 >degQ– 1 =deg(ηQ), we getα1≡ηα1orηα1≡2α2–α1. From this and (21), we geteQ_≡_{1 or}_eQ_≡_{–1, which contradicts that}_eQ_{is not a constant.}

Subcase3.2.degP<degQ. By (9) we getS(r,f) =S(r,eQ_{). Using a similar argument to}

subcase 3.1, we geteP_≡_{1 or}_eP_≡_{–1, which contradicts that}_eP_{is not a constant.}

Subcase3.3.degP=degQ=m≥1. By (9), we get

S(r,f) =Sr,ezm. (22) Set

P(z) =azm+am–1zm–1+· · ·+a0, Q(z) =bzm+bm–1zm–1+· · ·+b0, (23) wherea,am–1, . . . ,a0,b,bm–1, . . . ,b0are constants such thatab= 0. By (17) and (23), we get

j∈Λ

ϕj(z)ejz

m

= 0, (24)

where

Λ={2a, 2a+b,a+ 2b, 2b,a+b,a,b},

ϕ2a=H2pγ2, ϕ2a+b=H2p+qγ2η, ϕa+2b=Hp+2qγ η2, ϕ2b=H2qη2,

ϕa+b=Hp+qγ η, ϕa=Hpγ, ϕb=Hqη,

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(11)

≤N(r,f) +N

r, 1

f –ηf

–N

r, 1

L(f)

+S(r,f) ≤N(r,f) +S(r,f),

which impliesT(r,f) =S(r,f). This is absurd. Theorem1.3is thus proved.

Funding

This work is supported by the National Natural Science Foundation of China (No. 11661044, 11201195, 11171119).

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

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

Publisher’s Note

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

Received: 1 May 2018 Accepted: 23 April 2019

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