Lian Gui,Chen Jun-fanand Cai Xiao-hua

(Department of Mathematics,Fujian Normal University,Fuzhou,350117)

Communicated by Ji You-qing

Abstract:We mainly study the periodicity theorems of meromorphic functions having truncated or partial sharing values with their shifts,where meromorphic functions are of hyper order less than 1 and N(r,f)= αT(r,f)for some positive number α.

Key words:meromorphic function,periodicity,truncated sharing value,partial sharing value

1 Introduction and Results

Throughout the paper,meromorphic functions always mean non-constant meromorphic functions in the complex plane.The whole paper uses the standard notation of Nevanlinna theory(see[1]–[3]),such as T(r,f),N(r,f),m(r,f),(r,f)and so on.For such a meromorphic function f,we denote by S(f)the all quantities satisfying S(r,f)=o(T(r,f)),as r tends to infinity outside of a possible exceptional set of finite logarithmic measure and possibly different each time.Moreover,S(f)contains constant functions and?S(f)means S(f)∪{∞}.In addition,the order ρ(f)and hyper order ρ2(f)of a meromorphic function are defined in turn as follows:

We say that two meromorphic functions f and g share a IM(ignoring multiplicities)if(a,f)=(a,g).Moreover,if(a,f)=(a,g)and the multiplicities of the zeros are pairwise the same,then we say f and g share a CM(counting multiplicities).

Furthermore,let a∈?S(f)and k be a positive integer.We useto denote the set of zeros of f(z)?a(z)with multiplicities no greater than k and each zero is counted only once.Letbe the counting function of zeros of f(z)?a(z)whose multiplicities are no more than k andbe the reduced counting function correspondingly.Similarly,denotes the set of zeros of f(z)?a(z)with multiplicities no less than k ignoring multiplicities,denotes the corresponding counting function anddenotes the corresponding reduced counting function.

For two meromorphic functions f,g and a positive integer k,we say that a is a truncated sharing value of them if

Obviously,f and g share a IM also means

which shows that truncated sharing values are more general to some extent.We define that a meromorphic function f shares a partially with a meromorphic function g,ifˉE(a,f(z))?ˉE(a,g(z))and here a is called partial sharing value.It is also easy to see that f and g share a IM which means

It follows that partial sharing values are more general.

This paper generalize the definition,called the pseudo-deficiency put forward by Yang[2],by replacing the constant with the small function.

Definition 1.1Let f(z)be a transcendental meromorphic function,a∈?S(f)and k be a positive number.Then we define

In 2011,Heittokangas et al.[4]obtained the following sufficient condition for periodicity of a meromorphic function of finite order from the view of sharing values.

Theorem 1.1[4]Let f(z)be a nonconstant meromorphic function of finite order,c∈C{0}and a1,a2,a3∈(f)be three distinct periodic functions with period c.If f(z)and f(z+c)share a1,a2,a3CM,then f(z)≡f(z+c).

Recently,Heittokangas et al.[5]improved Theorem 1.1 to the next theorem by replacing“3CM” sharing values with “2CM+1IM”.

Theorem 1.2[5]Let f(z)be a nonconstant meromorphic function of finite order,c∈C{0}and a1,a2,a3∈(f)be three distinct periodic functions with period c.If f(z)and f(z+c)share a1,a2CM and a3IM,then f(z)≡f(z+c).

In 2017,on the basis of an exact difference analogue proved by Chiang and Feng[6],Chen and Lin[7]obtained the following theorem,which is an improvement of Theorem 1.2.

Theorem 1.3[7]Let f(z)be a nonconstant meromorphic function satisfying ρ2(f)<1,c∈C{0}and a1,a2,a3∈?S(f)be three distinct periodic functions with period c.If f(z)and f(z+c)share a1,a2CM and a3IM,then f(z)≡f(z+c).

It leads us to consider the following question:

Question 1.1How can we change the sharing values to more general ones to study the periodicity problem of meromorphic functions?

This paper is based on the results of complex differences and uses standard definitions and notations of difference operators.

Definition 1.2Let f(z)be a meromorphic function in the complex plane and c∈C{0}.We say that△cf=f(z+c)?f(z)is the first order difference operator of f(z),f=is the n-th order difference operator of f(z).All of them are called difference operators of f(z).

Using the difference analogue of the lemma on the logarithmic derivative established by Halburd et al.[8],we change the sharing values in Theorem 1.3 to partial or truncated sharing values,and firstly obtain the following result:

Theorem 1.4Let f(z)be a nonconstant meromorphic function satisfying ρ2(f)<1 and N(r,f)=αT(r,f),0≤α<1,c∈C{0},k1,k2be positive integers satisfying k1+k2>2,a1,a2∈S(f){0}be distinct.Suppose that f(z)and f(z+c)share 0 CM.If

Remark 1.1The condition(1.2)is necessary.We take f(z)=ezas a counterexample,correspondingly,c=iπ,k1=k2=2,a1=1,a2= ?1.After some simple calculation,we deduce that(1,f(z))=(1,f(z+c)),(?1,f(z))=(?1,f(z+c)),and δ1)(1,f)+ δ1)(?1,f)=0,but f(z)f(z+c).

As the consequence of Theorem 1.4,we have the following corollary:

Corollary 1.1Let f(z)be a nonconstant entire function satisfying ρ2(f)<1,c ∈ C{0},k1,k2be positive integers satisfying k1+k2>2,a1,a2∈S(f){0}be distinct.Suppose that f(z)and f(z+c)share 0 CM.If(a1,f(z))?a1,f(z+c)),(a2,f(z))?(a2,f(z+c)),and δk1?1)(a1,f)+δk2?1)(a2,f)>1,then f(z)≡f(z+c).

As a special case,given k1=1,k2≥2,we can draw the next conclusion.

Corollary 1.2Let f(z)be a nonconstant meromorphic function satisfying ρ2(f)<1 and N(r,f)=αT(r,f),0≤α<1,c∈C{0},k be a positive integer no less than 2,a,b∈S(f){0}be distinct.Suppose that f(z)and f(z+c)share 0 CM.If(a,f(z))?

It is natural to pose the following question:

Question 1.2Given k1=k2=1,what conclusion can we draw from Theorem 1.4?

In this paper,Theorem 1.5 can be viewed as an answer to Question 1.2.Furthermore,the condition of Theorem 1.5 is better.

Theorem 1.5Let f(z)be a nonconstant meromorphic function satisfying ρ2(f)<1 andbe three distinct periodic functions with period c.If

then f(z)≡f(z+c).

Given α=0,we have

Corollary 1.3Let f(z)be a nonconstant entire function satisfying ρ2(f)<1,c ∈ C{0},a1,a2,a3∈S(f)be three distinct periodic functions with period c.If

then f(z)≡f(z+c).

In fact,we draw a more general conclusion as follows.

Theorem 1.6Let f(z)be a nonconstant meromorphic function satisfying ρ2(f)<1 andbe three distinct periodic functions with period c.If

then f(z)≡f(z+c).

2Some Lemmas

To prove Theorems 1.4 and 1.6,we need to recall the following results for the use in the proof.

Lemma 2.1[2]Let f(z)be a nonconstant rational function in the complex plane.Then f(z)has only one deficient value.

Lemma 2.2[8]If f(z)is a nonconstant meromorphic function satisfying ρ2(f)<1,c ∈C{0},then

where S(r,f)=o(T(r,f))as r→∞outside of a possible exceptional set of finite logarithmic measure.

Lemma 2.3[8]Let f(z)be a nonconstant meromorphic function in the complex plane of ρ2(f)<1 and c∈ C.Then

where S(r,f)=o(T(r,f))as r→∞outside of a possible exceptional set of finite logarithmic measure.

Lemma 2.4[8]Let f(z)be an nonconstant meromorphic function satisfying ρ2(f)<1 and a1,a2,···,apbe p(≥ 2)distinct periodic small functions of f(z)with period c.If c is a nonzero complex number such that △cf≡ 0,then

and the exceptional set with S(r,f)is of finite logarithmic measure.

Lemma 2.5[9]Suppose that f(z)is a meromorphic function.If a1,a2,···,apare p(≥ 3)distinct small functions of f(z)and one of them can be in finite,then for each ε>0,we have

where E is of finite logarithmic measure.

Lemma 2.6[1]Let f(z)be a meromorphic function.where a,b,c,d∈S(f)and ad?bc≡0,then T(r,g)=T(r,f)+S(r,f).

3 Proof of Two Theorems

3.1 Proof of Theorem 1.4

We distinguish two cases to discuss.

Case 1.Suppose that f(z)is a nonconstant rational function.We set

where P(z)and Q(z)are relatively prime polynomials.It follows that

From above two equations,we can get

Since f(z)and f(z+c)share 0 CM,we have

We claim that P(z)is a non-zero constant.Suppose that P(z)is a polynomial of degree n.Then P(z)has n zeros according to fundamental theorem of algebra.If z0is one of them,we deduce from(3.4)that z0+mc are also zeros of P(z),where m∈Z+.It implies that P(z)has in finite zeros,which is a contradiction.Therefore,P(z)is constant.Consequently,

where C is a non-zero constant,which leads to the fact that 0 is a Picard exceptional value of f.Noting that N(r,f)=αT(r,f),0≤α<1,it is obvious that∞is also a deficient value.Hence,f(z)has two deficient values,which contradicts Lemma 2.1.So f(z)cannot be rational.

Case 2.Assume that f(z)is transcendental and suppose on the contrary that f(z)≡f(z+c).We set

According to assumptions of Theorem 1.4,it is obvious that g≡0,1.Since

applying Lemma 2.2 to f(z),we have

Noting that f(z)and f(z+c)share 0 CM,combining(3.6)with(3.7),we deduce that

Besides,by Lemma 2.3,it is easy to see that

It follows from(3.8)and(3.9)that

By the assumption(1.1)in Theorem 1.4 and(3.10),we have

By Lemma 2.5,we have that

holds for all sufficiently small ε>0.According to De fi nition 1.1 and arbitrariness of ε,we get

that is,

a contradiction.Hence,the conclusion of Theorem 1.4 is true.This completes the proof of Theorem 1.4.

3.2 Proof of Theorem 1.6

Suppose on the contrary that f(z)≡ f(z+c).From the assumption(1.3),we have

Hence,by Lemma 2.5,

holds for each ε>0.Noting thatthis implies that

holds for the above ε and all sufficiently large r.Therefore,

Combining(3.14),(3.15)and(3.16),applying Lemma 2.4,we deduce that

This is impossible since ε is arbitrary andThis completes the proof of Theorem 1.6.