R E S E A R C H
Open Access
Entire functions sharing a small function
with their two difference operators
Feng Lü
1, Yanfeng Wang
1and Junfeng Xu
2**Correspondence:
2Department of Mathematics, Wuyi
University, Jiangmen, Guangdong 529020, P.R. China
Full list of author information is available at the end of the article
Abstract
In this article, we deduce a uniqueness result of entire functions that share a small entire function with their two difference operators, generalizing some previous theorems of (Farissi et al. in Complex Anal. Oper. Theory 10:1317-1327, 2015, Theorem 1.1) and (Chen and Li in Adv. Differ. Equ. 2014:311, 2014, Theorem 1.1) by omitting the assumption that the shared small entire function is periodic.
MSC: 30D35; 30D30; 39A10
Keywords: uniqueness; entire functions; difference operators; Nevanlinna theory
1 Introduction and main result
Nevanlinna theory of value distributions is concerned with the density of points where a meromorphic function takes a certain value in the complex plane. Nowadays, there has been recent interest in connections between the Nevanlinna theory and the difference op-erator. 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 one got a lot of results (see,e.g., [–]).
In order to state the main result, we give the following definition. For a meromorphic functionf(z), we define its shift byfc=f(z+c) and its difference operators by
cf(z) =f(z+c) –f(z),
cnf(z) =cn–cf(z)
, n∈N,n≥.
A meromorphic functiona(z) is said to be a small function with respect tof(z) if and only ifT(r,a) =S(r,f), whereS(r,f) =o(T(r,f)), asr→ ∞outside of a possible exceptional set of finite logarithmic measure. Denote the set of all the small functions off(z) byS(f). Let
f(z) andg(z) be two meromorphic functions and leta(z) be a small entire 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. Recently, Chenet al.[, ] investigated two uniqueness problems on entire functions that share a small periodic entire function with their two difference operators as follows.
Theorem A(see [], Theorem .) Let f(z)be a nonconstant entire function of finite order,
let a(z)(≡)∈S(f)be a periodic entire function with period c.If f(z),cf(z),cf(z)share
a(z)CM,thencf(z)≡cf(z).
Theorem B(see [], Theorem .) Let f(z)be a nonconstant entire function of finite order.
If f(z),cf(z),cf(z)shareCM,thencf(z)≡Ccf(z),where C is a nonzero constant.
In , El Farissi, Latreuch and Asiri further studied the above problem and obtained
Theorem C(see [], Theorem .) Let f(z)be a nonconstant entire function of finite order,
let a(z)(≡)∈S(f)be a periodic entire function with period c.If f(z),cf(z),cf(z)share
a(z)CM,then f(z)≡cf(z).
Remark It is necessary to point out that Theorems A and B have been generalized from
cf(z) to ncf(z) by Chen, Chen and Li in []. There are also some interesting results related the above theorems (see,e.g., [, ]).
In the previous results, we find that the shared small functiona(z) is a periodic function with periodc. So, it is natural to ask what will happen if the periodic condition ofa(z) is omitted. In this paper, we focus on this problem and we obtain the following result.
Theorem Let f(z)be a nonconstant entire function of finite order,and let a(z)∈S(f)be an entire function.If f(z),cf(z),cf(z)share a(z)CM,then one of the following assertions
holds:
(i) Ifca(z)≡a(z),thencf(z) –a(z) =C(cf(z) –a(z)),whereCis a nonzero
constant.
(ii) Ifca(z)≡a(z),thencf(z) =f(z)orcf(z) –a(z) =eγ(cf(z) –a(z)),whereγ is a
polynomial withdegγ <ρ(a).
Remark We point out that Theorem is a generalization of the previous theorems.
If a(z)≡, then ca(z) =a(z). Then it follows from (i) of Theorem thatcf(z) =
Ccf(z), whereCis a nonzero constant.
Ifa(z)≡ is a periodic function with periodc, thenca(z)≡a(z). It follows from (ii)
of Theorem thatcf(z) =f(z) orcf(z) =cf(z). Furthermore, by Theorem C we can
deduce thatcf(z) =f(z).
As an application of Theorem , we can obtain an interesting result, wherea(z) is a slow growth small function.
Theorem Let f(z)be a nonconstant entire function of finite order,and let a(z)(≡)∈S(f)
be an entire function withρ(a) < .If f(z),cf(z),cf(z)share a(z)CM,thencf(z) =f(z).
For convenience of the reader, we list here some notations. For a meromophic function
f, we use the basic notations of the Nevanlinna theory of meromorphic functions such as
2 Some lemmas
In this section, we state some results that we employ in our proofs.
Lemma .([], Theorem .) Let c∈C,n∈N,and let f be a meromorphic function with a finite order.Then for all small periodic functions a(z)∈S(f)
m
r,
n cf(z)
f(z) –a(z)
=S(r,f),
where S(r,f) =o(T(r,f))for all r outside of a possible exceptional set E with finite logarith-mic measure.
Lemma .([], Lemma .) Let g be a nonconstant meromorphic function in the plane of order less than,and let h> .Then there exists a-set E such that
g(z+η)
g(z) →, as z→ ∞inC\E,
uniformly inηfor|η|<h.
Lemma . plays an important role in the proof of Theorem .
3 Proof of Theorem 1
Note thatf(z) is a nonconstant entire function of finite order. Thencf(z) andcf(z) are
also two entire functions of finite order. Setg(z) =f(z) –a(z). Then
cg(z) =cf(z) –ca(z), cg(z) =cf(z) –ca(z).
Sincef(z),cf(z),cf(z) sharea(z) CM, we have
cf(z) –a(z)
f(z) –a(z) =
cg(z) +ca(z) –a(z)
g(z) =e
P(z), () cf(z) –a(z)
f(z) –a(z) =
cg(z) +ca(z) –a(z)
g(z) =e
Q(z), ()
whereP(z) andQ(z) are two polynomials.
Suppose thatca(z)≡a(z). Obviously, we can getca(z)≡a(z). It is clear thatg(z),
cg(z), andcg(z) share CM. By Theorem B, we can obtaincg(z)≡Ccg(z), where
C is a nonconstant. So
cf(z) –ca(z)≡C(cf(z) –ca(z)). That is,cf(z) –a(z)≡
C(cf(z) –a(z)).
In the following, we assume thatca(z)≡a(z). We consider into two cases.
Case .ca(z)≡a(z). Set
ϕ(z) = (ca(z) –a(z))(
cf(z) –ca(z)) – (ca(z) –a(z))(cf(z) –ca(z))
f(z) –a(z) ()
=(ca(z) –a(z))(
cf(z) –a(z)) – (ca(z) –a(z))(cf(z) –a(z))
From () and (), we can rewrite the above function as
ϕ(z) =ca(z) –a(z)
eQ(z)–ca(z) –a(z)eP(z), ()
which implies thatϕis an entire function.
By () and Lemma ., we deduce thatϕ(z)∈S(f).
By the second main theorem, we deduce
Combining equations () and () yields
f(z) +eP(z+c) +eP(z)– +eP(z)+ =f(z)eQ(z), ()
a(z+c) –eP(z+c)+a(z) +eP(z+c) –eP(z)– –eP(z)=a(z) –eQ(z). () From (), we have
eP(z)eP(z+c)+eP(z+c)–P(z)– =eQ(z). ()
Suppose thateP(z)is a nonconstant function. Denoteη=eP(z+c)–P(z)– . Obviously,ηis a
small function ofeP(z). It follows from () thateP(z+c)+ηhave no zeros. Note thateP(z+c)
has only one Picard value, say . Thenη≡, which implieseP(z+c)–P(z)= . Again by (),
we have
eP(z)=eQ(z).
IfeP(z)is a constant, theneP(z+c)=eP(z). By () we also get (eP(z))=eQ(z).
Furthermore, it follows from (),eP(z+c)=eP(z)and (eP(z))=eQ(z)that
a(z+c) – a(z) –eP(z)= ,
ca(z) –a(z)
–eP(z)= .
Note thata(z)=ca(z), theneP(z)≡, soeQ(z)≡. Hence, we obtaincf(z) =f(z).
Subcase ..We assume thatϕ(z)≡. Then it follows from () and () that
cf(z) –ca(z)
cf(z) –ca(z)
=
ca(z) –a(z)
ca(z) –a(z)
=eQ–P=eγ,
whereγ =Q–Pis a polynomial anddegγ<ρ(a). From this, we also havecf(z) –a(z) =
eγ(
cf(z) –a(z)).
Case .ca(z)≡a(z).
By () and Lemma ., we geteQ∈S(f). Rewrite () and () as
cg(z) =g(z)eP(z)+a(z) –ca(z), ()
cg(z) =g(z)eQ(z). () Then
eQ(z)g(z) =eP(z+c)g(z+c) –eP(z)g(z) +ca(z) –a(z). ()
Rewrite () asg(z+c) =g(z)[ +eP(z)] +a(z) –
ca(z).
Now, substitute the form ofg(z+c) into () yields
Suppose thatn=degP≤degQ. SinceeQ∈S(f), we haveeP∈S(f).
Similar to the discussion of Subcase ., we obtaincf(z) =f(z).
Now, we assume thatdegP>degQ. Suppose that
P(z) =anzn+an–zn–+· · ·+a, an= .
Seth=eanzn. TheneP(z)=βh,eP(z+c)=βh, whereβ
,βare two small functions ofh.
Note thatdegP>degQ, soeQis also a small function ofh.
From (), we have
ca(z) –a(z)
g(z) =
eP(z)eP(z+c)+eP(z+c)–eP(z)–eQ(z)
eP(z+c)– . ()
We claim that
eP(z)eP(z+c)+eP(z+c)–eP(z)–eQ(z) eP(z+c)–
is irreducible except the factor which is a small function ofh.
Suppose thatzis a common zero ofeP(z+c)– andeP(z)eP(z+c)+eP(z+c)–eP(z)–eQ(z). It is
easy to deduce thateQ(z)= . Note thateQ(z)– ∈S(h). Thus, the claim holds. Rewrite () as
ca(z) –a(z)
g(z) =
ββh+ (β–β)h–eQ(z)
βh– . ()
DenoteH=ββh+ (β–β)h–eQ(z)=αh+αh+α, whereαj∈S(h),j= , , . By the
above equation, we see thatT(r,g) =O(T(r,h)), which implies thata∈S(h). Next we can prove that
N
r,
H
≤N
r,
ca(z) –a(z)
+S(r,h).
In fact, from () we have
H=ca(z) –a(z)
g(z)
eP(z+c)– .
Note thatg(z) is an entire function. All the zeros ofHcome from the zeros ofca(z) –a(z)
andeP(z+c)– . ByνF(z) we denote the multiplicity of the zero of meromorphic functionF
at the pointz.
Suppose thatzis a zero ofH. We claimνH(z)≤νca–a(z) +degP·νeQ–(z). Now we
split into two cases.
Case A. zis not a zero ofeP(z+c)– . Thenzmust be a zero ofca–a. HenceνH(z)≤ νca–a(z).
Case B. zis a zero ofeP(z+c)– . SetP(z) =P(z+c). Obviously,ν
eP–(z)≤degP. Then H(z) = andeP(z+c)– = , which leads toeQ(z)= .
Assume thateQ≡. Then we can rewrite () as
ca(z) –a(z)
g(z) =
eP(z)eP(z+c)+eP(z+c)–eP(z)– eP(z+c)– =e
Further,
a contradiction. Thus, the case cannot occur. Hence, we finish the proof of Theorem .
4 Proof of Theorem 2
Ifa(z) is a constant, then it follows from Theorem C thatf(z)≡cf(z). In the following,
we assume thata(z) is a nonconstant entire function. Suppose thatca(z)≡a(z). Then we have
a(z+c)
a(z) = .
Then there exists a-setEof finite logarithmic measure, so that
a(z+c)
a(z) →,
for allz→ ∞inC\E. It is absurd. Thus,ca(z)≡a(z). It follows from (ii) of Theorem
thatcf(z) =f(z) or
cf(z) –ca(z)
cf(z) –ca(z)
=
ca(z) –a(z)
ca(z) –a(z)
=eγ,
whereγ is a polynomial. Suppose that
cf(z) –ca(z)
cf(z) –ca(z)
=
ca(z) –a(z)
ca(z) –a(z)
=eγ.
Note thatρ(a) < . We get
ρeγ≤ρ
ca(z) –a(z)
ca(z) –a(z)
≤ρ(a) < ,
which implies thateγ is a nonzero constant, sayC. Furthermore, we get
ca(z) –a(z) =Cca(z) –a(z)
.
Rewrite it as
a(z+ c) – ( +C)a(z+c) + Ca(z) = .
Applying Lemma . again, we deduce that – ( +C) + C= , which implies thatC= . Then the above difference equation reduces to
a(z+ c) – a(z+c) + a(z) = .
Setb(z) =a(z+c) – a(z). Then the above equation can be rewritten as
b(z+ ) =b(z),
which implies that b(z) is a periodic function. If b(z) is a nonconstant function, then
ρ(b)≥. It contradictsρ(b)≤ρ(a) < . Thus,bis a constant. Hence
a(z+c) – a(z) =b. ()
Then applying Lemma . again, there exists a-setEof finite logarithmic measure, so that
a(z+c)
for allz→ ∞inC\E. Rewrite () as
a(z+c)
a(z) – =
b a(z).
Then choose a sequence{zk}such that |zk|=rk,zk∈/ E and|a(zk)|=M(rk,a),rk→ ∞
ask→ ∞. Substitutingzkinto the above function yields a contradiction by Lemma ..
Thus, this case cannot occur. Then we deduce the desired result. Hence, we finish the proof of Theorem .
Acknowledgements
The research was supported by NNSF of China Project (No. 11601521), the Fundamental Research Fund for Central Universities in China (Nos. 15CX05061A, 15CX05063A and 15CX08011), NSF of Guangdong Province (Nos. 2016A030313002, 2015A030313644) and Funds of Education Department of Guangdong (2016KTSCX145).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors drafted the manuscript, read and approved the final manuscript.
Author details
1College of Science, China University of Petroleum, Qingdao, Shandong 266580, P.R. China.2Department of Mathematics,
Wuyi University, Jiangmen, Guangdong 529020, P.R. China.
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Received: 12 December 2016 Accepted: 17 July 2017 References
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