# Uniqueness Theorems on Entire Functions and Their Difference Operators or Shifts

Authors: Baoqin Chen; Zongxuan Chen; Sheng Li
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We study the uniqueness problems on entire functions and their difference operators or shifts. Our main result is a difference analogue of a result of Jank-Mues-Volkmann, which is concerned with the uniqueness of the entire function sharing one finite value with its derivatives. Two relative results are proved, and examples are provided for our results.


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## Body

## 1. Introduction and Main Results

Throughout this paper, we assume the reader is familiar with the standard notations and fundamental results of Nevanlinna theory of meromorphic functions (see, e.g., [1–3]). In what follows, a meromorphic function always means meromorphic in the whole complex plane, and c always means a nonzero complex constant. For a meromorphic function f(z), we define its shift by f(z+c) and its difference operators by(1.1)Δcf(z)=f(z+c)-f(z),Δcnf(z)=Δcn-1(Δcf(z)),n∈N,n≥2.For a meromorphic functionf(z), we use S(f) to denote the family of all meromorphic functions a(z) that satisfy T(r,a)=S(r,f), where S(r,f)=o(T(r,f)), as r→∞ outside of a possible exceptional set of finite logarithmic measure. Functions in the set S(f) are called small functions with respect to f(z).Letf(z) and g(z) be two meromorphic functions, and let a(z) be a small function with respect to f(z) and g(z). We say that f(z) and g(z) share a(z) IM, provided that f(z)-a(z) and g(z)-a(z) have the same zeros (ignoring multiplicities), and we say that f(z) and g(z) share a(z) CM, provided that f(z)-a(z) and g(z)-a(z) have the same zeros with the same multiplicities.Uniqueness theory of meromorphic functions is an important part of the Nevanlinna theory. In the past 40 years, a very active subject is the investigation on the uniqueness of the entire function sharing values with its derivatives, which was initiated by Rubel and Yang [4]. We first recall the following result by Jank et al. [5].Theorem A (see [5]).
Letf be a nonconstant meromorphic function, and let  a≢0 be a finite constant. If  f,  f′, and  f′′ share the value  a CM, then  f≡f′.Recently, value distribution in difference analogues of meromorphic functions has become a subject of some interest (see, e.g., [6–11]). In particular, a few authors started to consider the uniqueness of meromorphic functions sharing small functions with their shifts or difference operators (see, e.g., [12, 13]).In this paper, we consider difference analogues of Theorem A.Theorem 1.1.
Letf(z)  be a nonconstant entire function of finite order, and let a(z)(≢0)∈S(f)  be a periodic entire function with period c. If  f(z), Δcf, and Δc2f  share a(z)  CM, then Δc2f=Δcf.Example 1.2.
Letf(z)=ezln2 and c=1.Then, for any a∈ℂ, we notice that f(z), Δcf, and Δc2f share a CM and can easily see that Δc2f=Δcf. This example satisfies Theorem 1.1.Remark 1.3.
In Example1.2, we have Δc2f=Δcf=f. However, it remains open whether the claim Δc2f=Δcf in Theorem 1.1 can be replaced by Δcf=f in general. In fact, the next example resulted from our efforts to find an entire function f(z) satisfying Theorem 1.1, while Δcf≠f.Example 1.4.
Letf(z)=ezln2-2,a=-1,b=1,and c=1.Then we observe that f(z)-a=ezln2-1,Δcf-b=ezln2-1,and Δc2f-b=ezln2-1 share 0 CM. Here, we also get Δc2f=Δcf.
From this example, it is natural to ask what happens iff(z)-a(z), Δcf-b(z), and Δc2f-b(z) share 0 CM, where a(z) and b(z) are two (not necessarily distinct) small periodic entire functions. Considering this question, we prove the following Theorem 1.5, whose proof is omitted as it is similar to the proof of Theorem 1.1.Theorem 1.5.
Letf(z) be a nonconstant entire function of finite order, and let a(z), b(z)(≢0)∈S(f) be periodic entire functions with period c. If f(z)-a(z), Δcf-b(z), and Δc2f-b(z) share 0 CM, then Δc2f=Δcf.Now it would be interesting to know what happens if the difference operators off(z) are replaced by shifts of f(z) in Theorem 1.5. We prove the following result concerning this question.Theorem 1.6.
Letf(z)  be a nonconstant entire function of finite order, let a(z), b(z)∈S(f)  be two distinct periodic entire functions with period  c, and let  n and  m be positive integers  satisfying n>m. If f(z)-a(z), f(z+mc)-b(z), and f(z+nc)-b(z)  share 0 CM, then f(z+mc)=f(z+nc) for all z∈ℂ.Example 1.7.
Letf(z)=sinz+1,a=0,b=2,and c=π.Then we notice that f(z)-a=sinz+1,f(z+c)-b=-sinz-1,and f(z+3c)-b=-sinz-1 share 0 CM and can easily see that f(z+c)=f(z+3c) for all z∈ℂ.This example satisfies Theorem 1.6.Example 1.8.
Letf(z)=ez2+a(z),where a(z)∈S(f) is a periodic entire function with period 1.Then f(z)-a(z)=ez2,f(z+1)-a(z)=e(z+1)2,and f(z+3)-a(z)=e(z+3)2 share 0 CM, while f(z+c)-f(z+3c)≢0.This example shows that the condition that a(z) and b(z) are distinct in Theorem 1.6 cannot be deleted.


## 2. Proof of Theorem1.1

Lemma 2.1 (see [8, Theorem  2.1]).
Letf(z)  be a meromorphic function of finite order  ρ and let  c be a nonzero complex constant. Then, for each  ε>0,
(2.1)T(r,f(z+c))=T(r,f(z))+O(rρ-1+ε)+O(log⁡⁡r).Lemma 2.2 (see [10, Lemma  2.3]).
Letc∈ℂ,  n∈ℕ, and let  f(z) be a meromorphic function of finite order. Then for any small periodic function  a(z) with period  c, with respect to f(z),
(2.2)m(r,Δcnff-a)=S(r,f),
where the exceptional set associated with  S(r,f) is of at most finite logarithmic measure.Proof of Theorem1.1.
Suppose, on the contrary, the assertion thatΔc2f≠Δcf. Note that f(z) is a nonconstant entire function of finite order. By Lemma 2.1, Δcf and Δc2f are entire functions of finite order.
Sincef(z), Δcf, and Δc2f share a(z) CM, then we have
(2.3)Δc2f-a(z)f(z)-a(z)=eα(z),Δcf-a(z)f(z)-a(z)=eβ(z),
where α(z) and β(z) are polynomials.
Set(2.4)φ(z)=Δc2f-Δcff(z)-a(z).
From (2.3), we get φ(z)=eα(z)-eβ(z). Then by supposition and (2.4), we see that φ(z)≢0. By Lemma 2.2, we deduce that
(2.5)T(r,φ)=m(r,φ)≤m(r,Δc2ff(z)-a(z))+m(r,Δcff(z)-a(z))+log⁡⁡2=S(r,f).
Note thateα/φ-eβ/φ=1. By using the second main theorem and (2.5), we have
(2.6)T(r,eαφ)≤N̅(r,eαφ)+N̅(r,φeα)+N̅(r,1eα/φ-1)+S(r,eαφ)=N̅(r,eαφ)+N̅(r,φeα)+N̅(r,φeβ)+S(r,eαφ)=S(r,f)+S(r,eαφ).
Thus, by (2.5) and (2.6), we have T(r,eα)=S(r,f). Similarly, T(r,eβ)=S(r,f).
By Lemma2.2 and the first equation in (2.3), we deduce that a(z)/(f(z)-a(z))=Δc2f/(f(z)-a(z))-eα(z) and
(2.7)m(r,1f(z)-a(z))=m(r,1a(z)(Δc2ff(z)-a(z)-eα(z)))≤m(r,Δc2ff(z)-a(z))+m(r,eα(z))+S(r,f)=S(r,f).
From (2.7), we see that
(2.8)N(r,1f(z)-a(z))=T(r,f(z))-m(r,1f(z)-a(z))+S(r,f)=T(r,f(z))+S(r,f).
Now we rewrite the second equation in (2.3) as Δcf=eβ(z)(f(z)-a(z))+a(z) and deduce that
(2.9)Δc2f=Δc(eβ(z)(f(z)-a(z))+a(z))=eβ(z+c)(f(z+c)-a(z+c))+a(z+c)-eβ(z)(f(z)-a(z))-a(z)=eβ(z+c)(f(z+c)-a(z))-eβ(z)(f(z)-a(z)).
This together with the first equation in (2.3) gives
(2.10)f(z+c)=(eα(z)-β(z+c)+eβ(z)-β(z+c))f(z)-a(z)(eα(z)-β(z+c)+eβ(z)-β(z+c)-1-e-β(z+c)),
that is,
(2.11)Δcf=(eα(z)-β(z+c)+eβ(z)-β(z+c)-1)f(z)-a(z)(eα(z)-β(z+c)+eβ(z)-β(z+c)-1-e-β(z+c)).
Thus, (2.11) can be rewritten as
(2.12)Δcf=γ(z)f(z)+δ(z),
where
(2.13)γ(z)=eα(z)-β(z+c)+eβ(z)-β(z+c)-1,δ(z)=-a(z)(eα(z)-β(z+c)+eβ(z)-β(z+c)-1-e-β(z+c))=-a(z)γ(z)+a(z)e-β(z+c),
which satisfy T(r,γ)=S(r,f) and T(r,δ)=S(r,f).
Now we rewriteΔcf=γ(z)f(z)+δ(z) as
(2.14)Δcf-a(z)-γ(z)(f(z)-a(z))=γ(z)a(z)+δ(z)-a(z).
Suppose thatγ(z)a(z)+δ(z)-a(z)≢0. Let z0 be a zero of f(z)-a(z) with multiplicity k. Since f(z), Δcf share a(z) CM, then z0 is a zero of Δcf-a(z) with multiplicity k. Thus, z0 is a zero of Δcf-a(z)-γ(z)(f(z)-a(z)) with multiplicity at least k. Then, by (2.8) and (2.14), we see that
(2.15)N(r,1γ(z)a(z)+δ(z)-a(z))=N(r,1Δcf-a(z)-γ(z)(f(z)-a(z)))≥N(r,1f(z)-a(z))=T(r,f(z))+S(r,f).
On the other hand, we have(2.16)N(r,1γ(z)a(z)+δ(z)-a(z))≤T(r,1γ(z)a(z)+δ(z)-a(z))=S(r,f).
Then, by (2.15) and (2.16), we get T(r,f)≤S(r,f), which is a contradiction.
Thus,γ(z)a(z)+δ(z)-a(z)≡0. Noting that δ(z)=-a(z)γ(z)+a(z)e-β(z+c), we deduce that e-β(z+c)≡1. So, eβ(z)≡eβ(z+c)≡1, since β(z) is a polynomial.
By the second equation in (2.3), we obtain Δcf=f, which leads to Δc2f=Δcf. This is a contradiction. The proof is thus completed.


## 3. Proof of Theorem1.6

Lemma 3.1 (see [9, Corollary  2.2]).
Letf(z) be a nonconstant meromorphic function of finite order,  c∈ℂ and  δ<1. Then
(3.1)m(r,f(z+c)f(z))=o(T(r+|c|,f)rδ),
for all  r outside of a possible exceptional set with finite logarithmic measure.Proof of Theorem1.6.
Suppose, on the contrary, the assertion thatf(z+mc)-f(z+nc)≢0. Since f(z)-a(z), f(z+mc)-b(z), and f(z+nc)-b(z) share 0 CM, then we have
(3.2)f(z+nc)-b(z)f(z)-a(z)=eα(z),f(z+mc)-b(z)f(z)-a(z)=eβ(z),
where α(z) and β(z) are polynomials.
By (3.2), we obtain
(3.3)f(z+nc)-f(z+mc)f(z)-a(z)=eα(z)-eβ(z).
Setψ(z)=eα(z)-eβ(z). Then by supposition, we see that ψ(z)≢0. By Lemma 3.1, we deduce that
(3.4)T(r,ψ)=m(r,f(z+nc)-f(z+mc)f(z)-a(z))≤m(r,f(z+nc)-a(z+nc)f(z)-a(z))+m(r,f(z+mc)-a(z+mc)f(z)-a(z))+log⁡⁡2=S(r,f).
Note thateα/ψ-eβ/ψ=1. Thus, using a similar method as in the proof of Theorem 1.1, we get T(r,eα)=S(r,f) and T(r,eβ)=S(r,f).
By Lemma3.1 and the first equation in (3.2), we deduce that
(3.5)m(r,1f(z)-a(z))=m(r,1b(z)-a(z)(f(z+nc)-a(z)f(z)-a(z)-eα(z)))≤m(r,f(z+nc)-a(z+nc)f(z)-a(z))+m(r,eα)+S(r,f)=S(r,f).
From (3.5), we see that
(3.6)N(r,1f(z)-a(z))=T(r,f(z))+S(r,f).
Now we rewrite the second equation in (3.2) as f(z+mc)=eβ(z)(f(z)-a(z))+b(z) and deduce that
(3.7)f(z+nc)=eβ(z+(n-m)c)(f(z+(n-m)c)-a(z+(n-m)c))+b(z+(n-m)c).
This together with the first equation in (3.2) gives
(3.8)f(z+(n-m)c)=eα(z)-β(z+(n-m)c)f(z)-a(z)eα(z)-β(z+(n-m)c)+a(z+(n-m)c),
that is,
(3.9)f(z+nc)=eα(z+mc)-β(z+nc)f(z+mc)-a(z+mc)eα(z+mc)-β(z+nc)+a(z+nc)=eα(z+mc)-β(z+nc)f(z+mc)-a(z)eα(z+mc)-β(z+nc)+a(z).
Now we rewrite (3.9) as
(3.10)f(z+nc)-b(z)-eα(z+mc)-β(z+nc)(f(z+mc)-b(z))=(b(z)-a(z))(eα(z+mc)-β(z+nc)-1).
Suppose thateα(z+mc)-β(z+nc)≡1; then, by (3.9), we get f(z+nc)=f(z+mc), which is a contradiction.
Now we haveeα(z+mc)-β(z+nc)-1≢0. Then using a similar method as in the proof of Theorem 1.1, we can also get a contradiction and obtain that f(z+mc)=f(z+nc) for all z∈ℂ. Thus, Theorem 1.6 is proved.



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