Uniqueness problems on entire functions that share a small function with their difference operators

R E S E A R C H

Open Access

Uniqueness problems on entire functions

that share a small function with their

difference operators

BaoQin Chen and Sheng Li

*_{Correspondence: [email protected]}

College of Science, Guangdong Ocean University, Zhangjiang, 524088, P.R. China

Abstract

In this paper, we consider uniqueness problems on entire functions that share a small periodic entire function with their two difference operators and obtain some results. Our first theorem provides a difference analogue of a result of Li and Yang (J. Math. Anal. Appl. 253(1):50-57, 2001).

MSC: Primary 30D35; secondary 39B32

Keywords: uniqueness; entire functions; difference operators

1 Introduction and main results

Throughout this paper, a meromorphic function always means meromorphic in the whole complex plane, andcalways means a nonzero constant. We use the basic notations of the Nevanlinna theory of meromorphic functions such asT(r,f),m(r,f),N(r,f) andN(r,f) as explained in [–]. In addition, we say that a meromorphic functiona(z) is a small function off(z) ifT(r,a) =S(r,f), whereS(r,f) =o(T(r,f)), asr→ ∞outside of a possible exceptional set of finite logarithmic measure.

For a meromorphic functionf(z), we define its shift byf(z+c), and define its difference operators by

cf(z) =f(z+c) –f(z) and ncf(z) =nc–

cf(z)

, n∈N,n≥.

In particular,n

cf(z) =nf(z) for the casec= .

Letf(z) andg(z) be two meromorphic functions, and leta(z) be a small function off(z) andg(z). We say thatf(z) andg(z) sharea(z) IM, provided thatf(z) –a(z) andg(z) –a(z) have the same zeros ignoring multiplicities. Similarly, we say thatf(z) andg(z) sharea(z) CM, provided thatf(z) –a(z) andg(z) –a(z) have the same zeros counting multiplicities. The problem on meromorphic functions sharing small functions with their derivatives is an important topic of uniqueness of meromorphic functions.

In , Jank, Mues and Volkmann [] proved the following result.

Theorem A([]) Let f be a nonconstant meromorphic function,and let a≡be a finite constant.If f,f,and fshare the value a CM,then f ≡f.

Many authors have been considering about some related cases, and got some interesting results (see,e.g., [, ]). In , Li and Yang [] obtained the following result for a special case thatf(z) is an entire function, andf,f, andf(n)share one value.

Theorem B([]) Let f(z)be an entire function,let a be a finite nonzero constant,and let

n(≥)be a positive integer.If f,f,and f(n)share the value a CM,then f assumes the form

f(z) =becz–a( –c)

c ,

where b,c are nonzero constants and cn–_{= .}

Recently, a number of papers (including [–]) have focused on difference analogues of Nevanlinna theory. In addition, many papers have been devoted to the investigation of the uniqueness problems related to meromorphic functions and their shifts or their difference operators and got a lot of results (see,e.g., [–]).

Our aim in this paper is to investigate uniqueness problems on entire functions that share a small periodic entire function with their two difference operators and provide a difference analogue of Theorem B. We now state the following theorem, which is the main result of this paper.

Theorem . Let f(z)be a nonconstant entire function of finite order,and let a(z)(≡)∈ S(f)be a periodic entire function with period c.If f(z),cf,andncf (n≥)share a(z)CM,

thenn

cf≡cf.

Examples

() Letf(z) =e(πi+ln

√

)z_{+  +i, thenf≡}_{f =ie}(π_i+ln√)z_{, and hencef(z),f, and}

_{f share  CM, butf(z)≡f. This example shows that the conclusion}n cf ≡cf

in Theorem . cannot be extended tof(z)≡cf in general.

() Letf(z) =ezln_{, thenf ≡f(z),}n_{f ≡}n_{f(z), and hencef(z),f and}n_{f share }

CM, butnf≡n–_{f ≡f (n≥). This example shows that the restriction}

a(z)≡in Theorem . is necessary.

Remark In the above example (), f(z) =e(π_i+ln√)z _{+  +i can be changed to f(z) =}

g(z)e(πi+ln

√

)z_{+  +i, whereg(z) is a periodic entire function with period , and the result}

still holds. This shows that the order of the functionf(z) in Theorem . is not always one.

As a continuation of Theorem . and example () above, we prove the following result.

Theorem . Let f(z)be a nonconstant entire function of finite order.If f(z),cf,andncf

(n≥)shareCM,thenn_cf ≡Ccf,where C is a nonzero constant.

2 Proof of Theorem 1.1

Firstly, we present some lemmas which will be needed in the proof of Theorem ..

m

where the exceptional set associated with S(r,f)is of at most finite logarithmic measure.

Now we divide this proof into the following two steps.

Step . Suppose thatβ(z) is not a constant. Now we rewrite the second equation in (.) as

cf =a(z)f(z) +b(z), (.)

and

f(z+c) =a(z)f(z) +b(z), (.)

wherea(z) =eβ(z)+ ,a(z) =eβ(z),b(z) =b(z) =a(z)( –eβ(z)).

We deduce that

_cf =cf(z+c) –cf(z)

=a(z+c)f(z+c) +b(z+c) –a(z)f(z) –b(z)

=eβ(z+c)_{f(z+c) +a(z+c) –e}β(z+c)_–eβ(z)_{f(z) –a(z) –e}β(z)

=eβ(z+c)eβ(z)+ f(z) +a(z) –eβ(z)+a(z) –eβ(z+c) –eβ(z)f(z) –a(z) –eβ(z)

=eβ(z+c)+β(z)+eβ(z+c)–eβ(z)f(z) +a(z)eβ(z)–eβ(z+c)+β(z), _cf =_cf(z+c) –_cf(z)

=eβ(z+c)+β(z+c)+eβ(z+c)–eβ(z+c)f(z+c) +a(z+c)eβ(z+c)–eβ(z+c)+β(z+c) –eβ(z+c)+β(z)+eβ(z+c)–eβ(z)f(z) –a(z)eβ(z)–eβ(z+c)+β(z)

=eβ(z+c)+β(z+c)+eβ(z+c)–eβ(z+c)eβ(z)+ f(z) +a(z) –eβ(z) +a(z)eβ(z+c)_–eβ(z+c)+β(z+c)_–eβ(z+c)+β(z)_+eβ(z+c)_–eβ(z)_f(z)

–a(z)eβ(z)–eβ(z+c)+β(z)

=eβ(z+c)+β(z+c)+β(z)+eβ(z+c)+β(z+c)+eβ(z+c)+β(z)– eβ(z+c)+β(z) +eβ(z+c)– eβ(z+c)+eβ(z)f(z) +a(z)–eβ(z+c)+β(z+c)+β(z) –eβ(z+c)+β(z)+ eβ(z+c)+β(z)+eβ(z+c)–eβ(z).

That is,

_cf =a(z)f(z) +b(z), cf =a(z)f(z) +b(z), (.)

where

a(z) =eβ(z+c)+β(z)+eβ(z+c)–eβ(z),

a(z) =eβ(z+c)+β(z+c)+β(z)+eβ(z+c)+β(z+c)+eβ(z+c)+β(z)

– eβ(z+c)+β(z)_+eβ(z+c)_{– e}β(z+c)_+eβ(z)_,

b(z) =a(z)

eβ(z)–eβ(z+c)+β(z), b(z) =a(z)

Let={, , . . . ,n–}be a finite set ofnelements, and denoteP() ={∅,{},{}, . . . ,{n–

are nonzero constants. In particular,λn,= , and

n–

By the above equation and (.), we obtain

an(z) =enlmz

Rewrite the first equation in (.) as

Notice thata(z)∈S(f),T(r,eα_{) =S(r,f), andT(r,e}β_{) =S(r,f). Ifa}

n(z) –eα(z)≡, (.)

yields

T(r,f) +S(r,f) =Tr,an(z) –eα(z)

f(z)

=Tr,a(z) –eα(z)–bn(z)

=S(r,f).

That is impossible.

Hencean(z) –eα(z)≡. This together with (.) gives

ePn,(z)_enlmzm_+λ

n–,ePn–,(z)+· · ·+λn–,n–ePn–,n–(z)

·e(n–)lmzm₊· · ·_+λ

,eP,(z)+· · ·+λ,n–eP,n–(z)

elmzm

–eα(z)≡. (.)

Now we distinguish three cases as follows:

Case (i). Suppose thatdegα(z) >m. Then, for any ≤j≤n, we see that

ρeα(z)–jlmzm_=ρeα(z)_{=degα(z) >m,}

and for ≤h<k≤n, we have

ρehlmzm–klmzm_{=m. (.)}

SincePs,t(z), fors= ,  . . . ,n,t= , , . . . ,Cns– , are polynomials with degree less thanm,

it is easy to see that, fors= , , . . . ,n– ,

ρ

Cs n–

t=

λs,tePs,t(z)

≤m– , ρePn,(z)≤_{m– . (.)}

By Lemma ., we haveePn,(z)≡_{, which is impossible.}

Case (ii). Suppose thatdegα(z) <m. Then, by a similar argument to above, we can also get a contradiction.

Case (iii). Now suppose that degα(z) =m. Set α(z) =dzm _+P∗_{(z), with d =  and}

degP∗(z) <m. Rewrite (.) as

ePn,(z)_enlmzm_+λ

n–,ePn–,(z)+· · ·+λn–,n–ePn–,n–(z)

·e(n–)lmzm₊· · ·_+λ

,eP,(z)+· · ·+λ,n–eP,n–(z)

elmzm

–eP∗(z)_edzm_{≡. (.)}

Subcase (i). Ifd=jlm, for anyj= , , . . . ,n, then we have

ρedzm–jlmzm_=ρe(d–jlm)zm_{=m, ρe}P∗(z)≤_{m– .}

Subcase (ii). Ifd=jlm, for somej= , , . . . ,n– . Without loss of generality, we assume

thatj=n– . Then we rewrite (.) as

ePn,(z)_enlmzm_+λ

n–,ePn–,(z)+· · ·+λn–,n–ePn–,n–(z)–eP

∗_(z) ·e(n–)lmzm₊· · ·_+λ

,eP,(z)+· · ·+λ,n–eP,n–(z)

elmzm≡_{. (.)}

By a similar method as the above, we can also getePn,(z)_{≡. That is impossible.}

Subcase (iii). Ifd=nlm, then we rewrite (.) as

ePn,(z)_–eP∗(z)_enlmzm_+λ

n–,ePn–,(z)+· · ·+λn–,n–ePn–,n–(z)

·e(n–)lmzm₊· · ·_+λ

,eP,(z)+· · ·+λ,n–eP,n–(z)

elmzm≡_{. (.)}

By a similar argument to the above and Lemma ., we can get

λ,eP,(z)+· · ·+λ,n–eP,n–(z)≡,

which implies

λ,eP,(z)+· · ·+λ,n–eP,n–(z)

elmzm≡_.

By (.) and (.), we get

n_c–eβ(z)₌

n–

j=

n–  j

(–)n––jeβ(z+jc)_{≡. (.)}

Suppose thatm> . Note that forj= , , . . . ,n– , we have

β(z+jc) =lmzm+ (lm–+mlmjc)zm–+Qj(z),

whereQj(z) are polynomials with degree less thanm– .

Rewrite (.) as

eQn–(z)_elmzm+(lm–+mlm(n–)c)zm–_{– (n– )e}Qn–(z)

·elmzm+(lm–+mlm(n–)c)zm–_{+· · ·+ (–)}n–_eQ(z)_elmzm+lm–zm–≡_{. (.)}

For any ≤h<k≤n– , we have

ρelmzm+(lm–+mlmhc)zm––(lmzm+(lm–+mlmkc)zm–)_=ρemlm(h–k)czm–_{=m– ,}

and forj= , , . . . ,n– , we see that

ρeQj(z)≤_{m– .}

Suppose thatm= , thenβ(z) =lz+l, withl= . It is easy to see that

ceβ(z)=eβ(z+c)–eβ(z)=elz+lc+l–elz+l=

elc_{– e}β(z)_,

_ceβ(z)=elc_{– }

ceβ(z)=

elc_{– }_eβ(z)_.

By induction,

n_c–eβ(z)=elc_{– }n–_eβ(z)_.

This together with (.) gives

elc_{– }n–≡_,

which yieldselc≡_{. Therefore, for anyj}∈Z_,

eβ(z+jc)=elz+jlc+l_=elz+l_elcj_=eβ(z)_{. (.)}

By the second equation in (.) and (.), we have

cf =eβ(z)f(z) +

 –eβ(z)a(z),

_cf =eβ(z+c)_{f(z+c) + –e}β(z+c)_{a(z+c) –e}β(z)_{f(z) + –e}β(z)_a(z)

=eβ(z)_{f(z+c) + –e}β(z)_{a(z) –e}β(z)_{f(z) – –e}β(z)_a(z)

=eβ(z)cf,

_cf =eβ(z+c)

cf(z+c) –eβ(z)cf=eβ(z)cf =eβ(z)cf.

By induction,

n_cf =e(n–)β(z)

cf. (.)

Rewriting (.), and combining it with the second equation in (.), we obtain

n_cf –a(z) =e(n–)β(z)cf –a(z)

=e(n–)β(z)eβ(z)f(z) –a(z)+a(z)–a(z)

=enβ(z)_{f(z) –a(z)+a(z)e}(n–)β(z)_{– . (.)}

Substituting the first equation in (.) into (.), we get

eα(z)_–enβ(z)_{f(z) –a(z)=a(z)e}(n–)β(z)_{– . (.)}

Ifeα(z)_–enβ(z)_{≡, (.) yields}

T(r,f) +S(r,f) =Tr,eα(z)_–enβ(z)_{f(z) –a(z)=Tr,a(z)e}(n–)β(z)_{– =S(r,f).}

We get a contradiction again.

Hence,eα(z)_–enβ(z) _{≡. By (.), we see that a(z)(e}(n–)β(z)_{– )≡, which implies}

Step . Suppose thatβ(z) is a constant. Now we rewrite the second equation in (.) as By Lemma . and the first equation in (.), we deduce that

a(z)

According to our assumption thatn

3 Proof of Theorem 1.2

Note thatf(z) is a nonconstant entire function of finite order. Sincef(z),cf, andncf

(n≥) share  CM, we have

n_cf f(z) =e

α(z)_, cf

f(z)=e

β(z)_{, (.)}

whereα(z) andβ(z) are polynomials.

Ifβ(z) is a constant, then we can easily get from (.)

n_cf =e(n–)βcf:=Ccf.

This completes our proof.

Ifβ(z) is not a constant, by assuming that ncf

cf is not a constant, with a similar arguing

as in the proof of Theorem ., we can deduce that the casedegβ(z) >  is impossible. For the casedegβ(z) = , from (.), we can obtain that

f(z+c) =eβ(z)_{+ f(z). (.)}

Letzbe a zero ofeβ(z)+ . Now we can find that (.) still holds here. Then from (.),

we haveeβ(z+kc)_{+  = , for allk∈Z. Therefore, from (.), we see thatz}

k=z+kcis a zero

off(z+c), thenzk+=z+ (k+ )cis a zero off(z), for allk∈Z. Suppose thatz=z+cis

a zero off(z) of orderk, then we get from (.) and (.)

f(z) =eβ(z–c)_{+ f(z–c) =e}β(z)_{+ }

f(z– c) =· · ·

=eβ(z)+ k+fz– (k+ )c

.

This indicates thatzis a zero off(z) of order at leastk+ , which is impossible.

Theo-rem . is thus proved.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

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

Acknowledgements

This work was supported by the NNSFC (No. 11301091) and the Guangdong Natural Science Foundation (No. S2013040014347) and the Foundation for Distinguished Young Talents in Higher Education of Guangdong (No. 2013LYM_0037).

Received: 26 July 2014 Accepted: 25 November 2014 Published:08 Dec 2014

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