Meromorphic Function Sharing Sets with Its Diff erence Operator or Shifts?

Bingmao DENG Chunlin LEI Mingliang FANG

Abstract Let f be a nonconstant meromorphic function,c∈C,and let a(z)(?0)∈S(f)be a meromorphic function.If f(z)and P(z,f(z))share the sets{a(z),?a(z)},{0}CM almost and share{∞}IM almost,where P(z,f(z))is defined as(1.1),then f(z)≡±P(z,f(z))or f(z)P(z,f(z))≡±a2(z).This extends the results due to Chen and Chen(2013),Liu(2009)and Yi(1987).

Keywords Meromorphic function,Diff erence operator,Shared sets

1 Introduction

In this paper,a meromorphic function always means meromorphic in the whole complex plane,and we assume that the reader is familiar with Nevanlinna theory of meromorphic functions.For a meromorphic function f(z),we denote by S(f)the set of all meromorphic functions a(z)such that T(r,a)=o(T(r,f))for all r outside of a set with finite logarithmic measure(see[6,8]).

For a meromorphic function f and a set S?C,we define

If Ef(S)=Eg(S),then we say that f and g share S CM.

Let a(z)be a common small function of both f(z)and g(z),and set N(r,a)be a counting function of both zeros of f(z)?a(z)and g(z)?a(z)with same multiplicity.If

then we call that f(z)and g(z)share a(z)CM almost(see[3]).

Set N(r,a)be a counting function of both zeros of f(z)?a(z)and g(z)?a(z)ignoring multiplicity.If

then we call that f(z)and g(z)share a(z)IM almost(see[3]).

Specially,N(r,1)(N(r,1))denote the counting function of both zeros of f(z)?1 and g(z)?1 with same multiplicity(ignoring multiplicity).

For a meromorphic function f(z),c∈C,we denote its shift and difference operator by f(z+c)and?cf:=f(z+c)?f(z),respectively.

The classical results in the uniqueness theory of meromorphic functions are the five values and four values theorems due to Nevanlinna(see[6,8]).Corresponding to sharing sets,Gross and Osgood[4]obtained the following result.

Theorem 1.1 Let f and g be two nonconstant entire functions of finite order.If f and g share the sets{1,?1}and{0}CM,then f≡±g or fg≡±1.

In 1987,Yi[9]improved Theorem 1.1 as follows.

Theorem 1.2 Let f and g be two nonconstant meromorphic functions.If f and g share the sets{1,?1},{0}and{∞}CM,then f≡±g or fg≡±1.

Recently,a number of papers(including[1,2,5,7,10])have focused on value distribution of difference analogues of meromorphic functions.Liu[7]investigated the cases that f(z)shares sets with its shift f(z+c)or difference operator?cf:=f(z+c)?f(z),and proved the following result.

Theorem 1.3 Let f be a nonconstant entire function of finite order,c∈C,and let a(z)∈S(f)be a non-vanishing periodic entire function with period c.If f(z)and f(z+c)share the sets{a(z),?a(z)}and{0}CM,then f(z)≡ ±f(z+c).

In 2013,Chen and Chen[1]extended Theorem 1.3 as follows.

Theorem 1.4 Let f be a nonconstant entire function of finite order,c∈C,let a(z)∈S(f)be a non-vanishing periodic entire function with period c,and let

where bk(z)? 0,b0(z),···,bk(z)∈ S(f)and k is a nonnegative integer.If f(z)and P(z,f(z))share the sets{a(z),?a(z)}and{0}CM,then f(z)≡±P(z,f(z)).

Now one may ask the following questions which are the motivation of the paper:

(I)In Theorem 1.2,can 3CM be replaced by 2CM+1IM?

(II)In Theorems 1.3–1.4,is the condition “f(z)has finite order” necessary?

(III)What will happen in Theorems 1.3–1.4 if f(z)is a meromorphic function?

(IV)In Theorems 1.3–1.4,can the condition “a(z)∈ S(f)be a non-vanishing periodic entire function with period c” be replaced by “a(z)∈ S(f)”?

In this paper we investigate the above problems,and prove the following results.

Theorem 1.5 Let f and g be two nonconstant meromorphic functions,c∈C,and let a(z)(?0)be a common small function related to f and g.If f(z)and g(z)share the sets{a(z),?a(z)},{0}CM almost and share{∞}IM almost,then f(z)≡ ±g(z)or f(z)g(z)≡±a2(z).

With Theorem 1.5,it is easy to get Theorem 1.6.

Theorem 1.6 Let f be a nonconstant meromorphic function,c∈C,and let a(z)(?0)∈S(f)be a meromorphic function.If f(z)and P(z,f(z))share the sets{a(z),?a(z)},{0}CM almost and share{∞}IM almost,where P(z,f(z))is defined as(1.1),then f(z)≡ ±P(z,f(z))or f(z)P(z,f(z))≡ ±a2(z).

From Theorem 1.6,we have the corollary as follows.

Corollary 1.1 Let f be a nonconstant entire function,c∈ C,and let a(z)(? 0)∈ S(f)be a meromorphic function.If f(z)and P(z,f(z))share the sets{a(z),?a(z)},{0}CM almost,where P(z,f(z))is defined as(1.1),then f(z)≡±P(z,f(z))or f(z)P(z,f(z))≡±a2(z).

For the meromorphic function share three sets with its shift,we obtain the following result.

Theorem 1.7 Let f be a nonconstant meromorphic function,c∈C.If f(z)and?cf share the sets{1,?1},{0}CM and share{∞}IM almost,then f(z+c)≡ 2f(z).

For the meromorphic function with finite order,we prove the following result.

Theorem 1.8 Let f be a nonconstant meromorphic function of finite order,c∈C,and let a(z)(? 0)∈ S(f)be a meromorphic function. If f(z)and P(z,f(z))share the sets{a(z),?a(z)},{0}CM almost and share{∞}IM almost,where P(z,f(z))is defined as(1.1),then f(z)≡±P(z,f(z)).

From Theorem 1.8,we can deduce Theorems 1.3–1.4 immediately.

Example 1.1 Let f(z)=eez,and P(z,f(z))=f(z+πi),a(z)≡ 1,then P(z,f(z))=e?ez.Obviously f(z)and P(z,f(z))share the sets{a(z),?a(z)},{0}CM almost and share{∞}IM almost,and f(z)P(z,f(z))≡ 1.Thus,the case“f(z)P(z,f(z))≡ ±a2(z)”in Theorem 1.6 can not be deleted.

2 Some Lemmas

For the proof of our results,we need the following results.

Lemma 2.1 Let F(z)and G(z)be two nonconstant meromorphic functions with(r,F)=S(r,F).Supposed that F(z),G(z)share 0,1 CM almost,and share∞IM almost.If

then F(z)≡G(z).

Proof Let

If φ(z)? 0,then m(r,φ)=S(r,F)+S(r,G).Since F(z),G(z)share 0,1 CM almost,share ∞IM almost,and(r,F)=S(r,F),we get N(r,φ)≤(r,F)+S(r,F)+S(r,G)≤ S(r,F)+S(r,G).So from(2.1)we have

a contradiction.Thus φ(z)≡ 0.From(2.1),it is easy to obtain F(z)≡ cG(z),where c is a constant.Since N(r,1)?(r,1)S(r,F)+S(r,G),there exists z0such that F(z0)=G(z0)=1.So c=1,that is F(z)≡G(z).

This completed the proof of Lemma 2.1.

Lemma 2.2 Let F(z)and G(z)be two nonconstant meromorphic functions with(r,F)=S(r,F).Supposed that F(z),G(z)share 0,1 CM almost,and share∞IM almost.If F(z)is not a Mobius transformation of G(z),then

Proof By Lemma 2.1,we have

Set

If ?(z) ≡ 0,then from(2.4),we know that F(z)is a Mobius transformation of G(z),a contradiction.Thus ?(z)0.From(2.3)–(2.4)and the fact that F(z),G(z)share 0,1 CM almost,share∞IM almost,we obtain

where N1)(r,1)denotes the counting function of both simple zeros of F(z)? 1 and G(z)? 1.

This completed the proof of Lemma 2.2.

Lemma 2.3(see[5]) Let f(z)be a non-constant meromorphic function of finite order,c∈C,then

for all r outside of a possible exceptional set E with finite logarithmicmeasure.

3 Proof of Theorem 1.5

Obviously f2(z)and g2(z)share 0,a2(z)CM almost,and share∞IM almost.Set F(z)=.Then F(z)and G(z)share 0,1 CM almost,and share∞IM almost.Thus

So S(r,F)=S(r,G).Define S(r)=S(r,F)+S(r,G).

Obviously F(z)and G(z)have no simple zeros and poles.

Now we assume that both F(z)?G(z)and F(z)G(z)?1.We claim that

Firstly,we prove N(r,F)=N(r,G)=S(r).Set

Since F(z)and G(z)share 0,1 CM almost,and share∞IM,we get

From(3.1)we get

From(3.3)and the fact that F and G share∞IM almost,we obtain that all the multiple poles of F and G must be the multiple zeros of ??1.Noting that F and G have no simple pole,if ?′? 0,we get

Combining(3.2)and(3.4),we get N(r,F)=S(r).

If ?′(z)≡ 0,then ?(z)≡ c,where c is a constant.If c=1,form(3.1)we get F(z)≡ G(z),a contradiction.If c ≠1,then it follows from(3.3)that N(r,F)=S(r).

Since F and G share∞IM almost,

If φ(z)≡ 0,then we get F(z)≡ G(z),a contradiction.

If φ(z)? 0,then m(r,φ)=S(r,F).Since F(z)and G(z)share 0,1 CM almost,and share ∞IM almost,from(3.5),we know that the pole of ?(z)must be the pole of F(z)and all the poles of φ(z)are simple.Thus N(r,φ) ≤(r,F)=S(r).So T(r,φ)=m(r,φ)+N(r,φ)=S(r).From(3.5)we also get that the multiple pole of F(z)must be the zeros of φ(z).Since F(z)have no simple zero,

Since F and G share 0 CM almost,Thus the claim is proved.

If F(z)is not a Mobius transformation of G(z),then from Lemma 2.2,we get

and

On the other hand,by Nevanlinna’s second fundamental theorem and

we have

and

By(3.6)–(3.9),we get

Then T(r,F)+T(r,G)≤S(r),a contradiction.Thus F(z)is a Mobius transformation of G(z),that is

where A,B,C,D are constants,and AD?BC ≠0.

Next we discuss following two cases.

Case 1 C=0.Thus AD ≠0.From(3.10),we have.If B ≠0,it follows fromthat,then we get a contradiction by Nevanlinna’s second fundamental theorem.Hence.If F(z)≠1,it is easy to get a contradiction by Nevanlinna’s second fundamental theorem.So there exists z0such that F(z0)=G(z0)=1.Thus we get=1,that is F(z)≡G(z).

Case 2 C ≠0.We consider two subcases.

Case 2.1 D ≠0.Then from(3.10),we obtain F(z)≠∞,G(z)≠∞,G(z)≠?.By Nevanlinna’s second fundamental theorem,we get a contradiction.

Case 2.2 D=0.Then B ≠0.From(3.10),we have CF(z)G(z)=AG(z)+B.It is easy to get F(z)≠ ∞,and G(z)≠ ∞.If A ≠0,we get G ≠ ?,which contradicts Nevanlinna’s second fundamental theorem.So A=0.Then F(z)G(z)=.In the same way as in Case 1,we can get=1.

Thus we get F(z)≡G(z)or F(z)G(z)≡1.

If F(z)≡ G(z),then f(z)≡ ±g(z).If F(z)G(z)≡ 1,then f(z)g(z)≡ ±a2(z).

This completed the proof of Theorem 1.5.

4 Proof of Theorem 1.7

By Theorem 1.5,we get f(z)≡±?cf or f(z)?cf≡±1.

If f(z)?cf≡ ±1,that is

From(4.1)and the fact that f(z),?cf share 0 CM almost and share∞ IM almost,we obtain f(z)≠0 and f(z)≠∞.Thus f(z)=eh(z),where h(z)be a nonconstant entire function.

By(4.1),we get

where t2=1.

From(4.2)and f(z)=eh(z),we obtain

That is

Since eh(z)h(z+c)≠0,we easily get e2h(z)≠ ?t,and obviously e2h(z)≠0,∞.Then by Picard theorem,we get e2h(z)≡C1,then h≡C2,where C1,C2are constants.A contradiction.

So we get f(z)≡ ±[f(z+c)?f(z)],that is f(z)≡ f(z+c)?f(z)or f(z)≡ f(z)?f(z+c).If f(z)≡ f(z)?f(z+c),then f(z+c)≡ 0.So f(z)≡ 0,a contradiction.So f(z)≡ f(z+c)?f(z).Thus f(z+c)≡2f(z).

This completed the proof of Theorem 1.7.

5 Proof of Theorem 1.8

By Theorem 1.5,we get f(z)≡ ±P(z,f(z))or f(z)P(z,f(z))≡ ±a2(z).

If f(z)P(z,f(z))≡±a2(z),we get

Since f(z),P(z,f(z))share 0 CM almost,share∞IM almost and T(r,a(z))=S(r),N?r,?=S(r).

By(5.1)and Lemma 2.3,we get

This completed the proof of Theorem 1.8.

Acknowledgement The authors thank the referees and editors for several helpful suggestions.