# Value Distribution and Uniqueness Results of Zero-Order Meromorphic Functions to Theirq-Shift

Authors: Haiwa Guan; Gang Wang; Qiuqin Luo
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

We investigate value distribution and uniqueness problems of meromorphic functions with theirq-shift. We obtain that if f is a transcendental meromorphic (or entire) function of zero order, and Q(z) is a polynomial, then afn(qz)+f(z)−Q(z) has infinitely many zeros, where q∈ℂ∖{0}, a is nonzero constant, and n≥5 (or n≥3). We also obtain that zero-order meromorphic function share is three distinct values IM with its q-difference polynomial P(f), and if limsupr→∞(N(r,f)/T(r,f))<1, then f≡P(f).


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## Body

## 1. Introduction and Main Results

A functionf(z) is called meromorphic function if it is analytic in the complex plane except at isolated poles. It is assumed that the reader is familiar with the standard symbols and fundamental results of Nevanlinna theory such as the characteristic function T(r,f), proximity function m(r,f), and the counting function N(r,f), see [1–3]. Let us recall the definition of the order and the zeros exponent convergence of function f. The order of meromorphic function f is said by
(1.1)ρ(f)=limsupr→∞logT(r,f)logr.
The zeros of exponent convergence of meromorphic function f is said by
(1.2)λ(f)=limsupr→∞logN(r,f)logr.In 1959, Hayman proved the following Theorems.Theorem A (see [4], Theorem 8).
Letf be a transcendental entire function, and let n≥3 be an integer and a be a nonzero constant. Then f'(z)-af(z) assumes all finite values infinitely often.Theorem B (see [4], Theorem 9).
Letf be a transcendental meromorphic function, and let n≥5 be an integer and a be a nonzero constant. Then f'(z)-af(z) assumes all finite values infinitely often.Recently the difference variant of the Nevanlinna theory has been established independently in [5, 6]. Using these theories, value distribution theory uniqueness theory of difference polynomials of finite order transcendental meromorphic functions has been studied as well. We recall the following result by Liu and Laine.Theorem C (see [7], Theorem 1.1).
Letf be a transcendental entire function of finite order not of period c, where c is a nonzero constant, and let s(z) be a nonzero small function of f. Then the difference polynomial fn(z)+f(z+c)-f(z)-s(z) has infinitely many zeros in the complex plane provided that n≥3.In 2010, Chen considered the difference counterpart of Hayman's theorem and porved an almost direct difference analogue of Hayman's theorem.Theorem D (see [8], Theorem 1.1).
Letf be a transcendental entire function of finite order not of period c, and let a(≠0),b,c(≠0) be three complex numbers. Then Ψn(z)=f(z+c)-f(z)-afn(z) assumes all finite values infinitely often, provided that n≥3 and for every b one has λ(Ψn(z)-b)=ρ(f).In this paper, we consider the value distribution of zero-order meromorphic functions with theirq-shirt and prove the following results.Theorem 1.1.
Letf be a transcendental meromorphic function of zero order, and let Q(z) be a polynomial. If n is an integer and n≥4, then afn(qz)+f(z)-Q(z) has infinitely many zeros, where q∈ℂ∖{0} and a is nonzero constant.Theorem 1.2.
Letf be a transcendental entire function of zero order, and let Q(z) be a polynomial. If n is an integer and n≥3, then afn(qz)+f(z)-Q(z) has infinitely many zeros, where q∈ℂ∖{0}, and a is nonzero constant.It is well known that two meromorphic functions must be equal, if they share five distinct values. Recently, Heittokangas et al. research the uniqueness of meromorphic functions with their shifts in [6]. They got that if f(z) and f(z+c) share three distinct values, where f(z) is finite order, then f(z)=f(z+c). In this paper, we want to get some results on uniqueness of f(z) and f(qz), where f(z) is zero order and q∈ℂ∖{0,1}. Let us recall the notation of q-difference which is written by ∇q=f(qz)-f(z).Theorem 1.3.
Letf be a meromorphic function of zero order, let q∈ℂ∖{0,1}, and let a1,a2,a3∈ℂ be three distinct values.  (a)
Iff(z) and f(qz) share a1,a2,a3 CM, then f(z)=f(qz) for all z∈ℂ.(b)
Iff(z) and f(qz) share a1,a2 CM, and if
(1.3)Limsupr→∞N(r,f)T(r,f)<1,
then f(z)=f(qz) for all z∈ℂ.Corollary 1.4.
Letf be an entire function of zero order, let q∈ℂ∖{0,1}, and let a1,a2∈ℂ be two distinct values.  (a)
Iff(z) and f(qz) share a1,a2 CM, then f(z)=f(qz) for all z∈ℂ.(b)
Iff(z) and f(qz) share a1 CM, and if
(1.4)limsupr→∞N(r,f)T(r,f)<1,
then f(z)=f(qz) for all z∈ℂ.Corollary 1.5.
Letf be a meromorphic function of zero order, and let q∈ℂ∖{0,1}. If f(z) and f(qz) share ∞ CM and a constant a∈ℂ CM, and if there exists a constant b∈ℂ∖{a} such that
(1.5)limsupr→∞N(r,1/(f-b))T(r,f)<1,
then f(z)=f(qz) for all z∈ℂ.Theorem 1.6.
Letf be a meromorphic function of zero order, let q∈ℂ∖{0,1}, and let
(1.6)P(f)=bk(z)f(qkz)+⋯+b1(z)f(qz)+b0(z)f(z),
where bk,…,b0 are constants. Let n∈{1,…,k+1} be the number of nonzero coefficients of the q-difference polynomial P(f)-f. If f and P(f) share three distinct finite values a1,a2,a3 IM, and if
(1.7)limsupr→∞N(r,f)T(r,f)<1,
then f≡P(f).Corollary 1.7.
Letf be a meromorphic function of zero order, let q∈ℂ∖{0,1}, and let a1,a2,a3∈ℂ be three distinct finite values.  (a)
Iff(z) and f(qz) share a1,a2,a3 IM, and if
(1.8)limsupr→∞N(r,f)T(r,f)<12,
then f(z)=f(qz) for all z∈ℂ.(b)
Iff(z) and f(qz) share ∞ IM and two constants a1,a2∈ℂ IM, and if there exists a constant b∈ℂ∖{a1,a2} such that
(1.9)limsupr→∞N(r,1/(f-b))T(r,f)<12,
then f(z)=f(qz) for all z∈ℂ.Corollary 1.8.
Letf be an entire function of zero order, let q∈ℂ∖{0,1}, and let a1,a2,a3∈ℂ be three distinct finite values.  (a)
Iff(z) and f(qz) share a1,a2,a3 IM, then f(z)=f(qz) for all z∈ℂ.(b)
Iff(z) and f(qz) share a1,a2∈ℂ IM, and if there exists a constant b∈ℂ∖{a1,a2} such that
(1.10)limsupr→∞N(r,1/(f-b))T(r,f)<12,
then f(z)=f(qz) for all z∈ℂ.


## 2. Auxiliary Results

The following auxiliary results will be instrumental in proving the theorems.Lemma 2.1 (see [9], Theorem 1.2).
Letf(z) be a nonconstant zero-order meromorphic function, and q∈ℂ∖{0}. Then
(2.1)m(r,f(qz)f(z))=Sq(r,f).Lemma 2.2 (see [9], Theorem 3.1).
Letf be a non-constant meromorphic functions of zero order, let q∈ℂ∖{0,1}, and let a1,…,ap∈ℂ, where p≥2, be distinct points. Then
(2.2)m(r,f)+∑k=1p‍m(r,1f-ak)≤2T(r,f)-Npair(r,f)+Sq(r,f),
where
(2.3)Npair(r,f):=2N(r,f)-N(r,∇qf)+N(r,1∇qf).Lemma 2.3 (see [10], Theorem 1.1).
Letf(z) be a non-constant zero-order meromorphic function, and q∈ℂ∖{0}. Then
(2.4)T(r,f(qz))=T(r,f)+Sq(r,f).Lemma 2.4 (see [10], Theorem 1.3).
Letf(z) be a non-constant zero-order meromorphic function, and q∈ℂ∖{0}. Then
(2.5)N(r,f(qz))=N(r,f)+Sq(r,f).Lemma 2.5 (see [11], Lemma 4).
IfT:ℝ+→ℝ+ is a piecewise continuous increasing function such that
(2.6)limr→∞logT(r)logr=0,
then the set
(2.7)E:={r:T(C1r)≥C2T(r)}
has logarithmic density 0 for all C1>1 and C2>1.


## 3. Proof of Theorem1.1

Let us put(3.1)Ψ(z):=afn(qz)+f(z)-Q(z).Hence,Ψ(z) is not a constant identity. If not, let us suppose that Ψ(z)≡c, where c is a constant, and then
(3.2)afn(qz)≡Q(z)+c-f(z),
which give us
(3.3)nT(r,f(qz))=T(r,f)+O(logr).By using Lemma2.3, we have n≤1, which contradicts the assumption n≥5. Hence Ψ(z)≢c. By taking logarithmic derivative on two sides of (3.1), we have
(3.4)Ψ′(z)Ψ(z)=(afn(qz)+f(z)-Q(z))′afn(qz)+f(z)-Q(z).If(3.5)(fn(qz))′fn(qz)-Ψ′(z)Ψ(z)≡0,
then by integrating two sides of which, we have Ψ(z)=bfn(qz), where b is a nonzero constant, and hence
(3.6)(b-a)fn(qz)=f(z)-Q(z).
If b=a, then f(z)=Q(z), which is contradiction with f(z), is transcendental function. If b≠a, then Lemma 2.3 implies that n=1, which is impossible. Therefore, we can write (3.4) as
(3.7)afn(qz)=(Ψ′(z)/Ψ(z))(f(z)-Q(z))-(f(z)-Q(z))′(fn(qz))′/fn(qz)-Ψ′(z)/Ψ(z).Let us put(3.8)Ψ1(z):=Ψ′(z)Ψ(z)(f(z)-Q(z))-(f(z)-Q(z))′.Now we consider the poles ofΨ1(z). The poles of Ψ1(z) come from the zeros of Ψ(z) and the poles of f(qz), f(z), and Q(z). If z0 is a zero of Ψ(z) or a pole of f(qz), but not a pole of f(z), then z0 is a simple pole of Ψ(z). If z0 is a common pole of f(qz) and f(z) with multiplicities of k and l, respectively, then z0 is a pole of Ψ1(z) with multiplicity no more than l+1. If z0 is a pole of f(z) but not a pole of f(qz), we obtain that z0 is at most a simple pole of Ψ1(z) by using (3.7). Hence, Lemma 2.5 implies that
(3.9)N(r,Ψ1)≤N¯(r,1Ψ)+N¯(r,f(qz))+N(r,f(z))+O(logr)≤N¯(r,1Ψ)+N¯(r,f)+N(r,f(z))+Sq(r,f).Let us put(3.10)Ψ2(z):=(fn(qz))′fn(qz)-Ψ′(z)Ψ(z).Now, let us consider the pole ofΨ2(z). The poles of Ψ2(z) come from the poles of f(qz) and f(z) and the zeros of f(qz) and Ψ(z). If z0 is a zero of Ψ(z), zero of f(qz), or pole of f(z), then z0 is a simple pole of Ψ2(z). If z0 is a pole of f(qz) but not a pole of f(z), by using Laurent series, we obtain that Ψ2(z) is analytic at z0. Therefore, we have
(3.11)N(r,Ψ2)≤N¯(r,1Ψ)+N¯(r,1f(qz))+N¯(r,f(z))+O(logr)≤N¯(r,1Ψ)+N¯(r,1f(z))+N¯(r,f(z))+Sq(r,f),
according to Lemma 2.5. In the coming (3.7) and Lemma 2.1, it implies that
(3.12)nT(r,f)=T(r,afn(qz))+Sq(r,f)=T(r,Ψ1Ψ2)+Sq(r,f)≤m(r,Ψ1)+m(r,Ψ2)+N(r,Ψ1)+N(r,Ψ2)+Sq(r,f)≤m(r,f)+m(r,Ψ′Ψ-(f-Q)′f-Q)+m(r,Ψ2)+N(r,Ψ1)+N(r,Ψ2)+Sq(r,f)≤2N¯(r,1Ψ)+N¯(r,1f)+N¯(r,f(z))+2N(r,f)+m(r,f)+Sq(r,f).Therefore, we have(3.13)(n-3)T(r,f)≤2N¯(r,1Ψ)+Sq(r,f),
which shows that afn(qz)+f(z)-Q(z) has infinite zeros by n≥4.


## 4. Proof of Theorem1.2

In the same manner as in the proof of Theorem1.1, we have (3.1)–(3.7), noting that
(4.1)Ψ1(z):=Ψ′(z)Ψ(z)(f(z)-Q(z))-(f(z)-Q(z))′.Now consider the poles ofΨ1(z). The poles of Ψ1(z) come from the zeros of Ψ(z) and the poles of Q(z). If z0 is a zero of Ψ(z), then z0 is a simple pole of Ψ(z). Hence, Lemma 2.4 implies that
(4.2)N(r,Ψ1)≤N¯(r,1Ψ)+O(logr)≤N¯(r,1Ψ)+Sq(r,f).Let us put(4.3)Ψ2(z):=(fn(qz))′fn(qz)-Ψ′(z)Ψ(z).Now, let us consider the pole ofΨ2(z). The poles of Ψ2(z) come from the poles of Q(z) and the zeros of f(qz) and Ψ(z). If z0 is a zero of Ψ(z), zero of f(qz), then z0 is a simple pole of Ψ2(z). Therefore, we have
(4.4)N(r,Ψ2)≤N¯(r,1Ψ)+N¯(r,1f(qz))+O(logr)≤N¯(r,1Ψ)+N¯(r,1f(z))+Sq(r,f),
according to Lemma 2.5. In the coming (3.7) and Lemma 2.1, it implies that
(4.5)nT(r,f)=T(r,afn(qz))+Sq(r,f)=T(r,Ψ1Ψ2)+Sq(r,f)≤m(r,Ψ1)+m(r,Ψ2)+N(r,Ψ1)+N(r,Ψ2)+Sq(r,f)≤m(r,f)+m(r,Ψ′Ψ-(f-Q)′f-Q)+m(r,Ψ2)+N(r,Ψ1)+N(r,Ψ2)+Sq(r,f)≤2N¯(r,1Ψ)+N¯(r,1f)+m(r,f)+Sq(r,f).Therefore, we have(4.6)(n-2)T(r,f)≤2N¯(r,1Ψ)+Sq(r,f),
which shows that afn(qz)+f(z)-Q(z) has infinite zeros by n≥3.


## 5. Proof of Theorem1.3

(a) Suppose first thata1,a2,a3 are three distinct values, and assume conversely to the assertion that ∇q≢0. Then Lemma 2.2 yields
(5.1)∑k=13‍m(r,1f-ak)≤2T(r,f)+N(r,∇qf)-2N(r,f)-N(r,1∇qf)+Sq(r,f)
and so
(5.2)T(r,f)≤∑k=13‍N(r,1f-ak)+N(r,f(qz))-N(r,f)-N(r,1∇qf).Sincef(z) and f(qz) share a1,a2, and a3 CM, it follows that
(5.3)∑k=13‍N(r,1f-ak)≤N(r,1∇qf)In addition, sincef is zero-order meromorphic function, from Lemma 2.4, Lemma 2.5 and equations (5.2) and (5.3), we have
(5.4)T(r,f)=Sq(r,f)
which is impossible. This contradiction is only avoided when ∇qf≡0.Suppose thata3=∞ while a1 and a2 are distinct finite values. Similarly as above, f(z)=f(qz) can be obtained. Therefore, f(z)=f(qz) for all z∈ℂ.(b) Assume that∇qf≢0. Similarly as above, Lemma 2.2 yields
(5.5)m(r,f)+∑k=12‍m(r,1f-ak)≤2T(r,f)+N(r,∇qf)-2N(r,f)-N(r,1∇qf)+Sq(r,f),
and therefore m(r,f)=Sq(r,f). This together with the condition results in a contradiction. Hence ∇qf≢0.


## 6. Proof of Theorem1.6

Assume on the contrary to the assertion thatf≢P(f). In what follows, ε>0 is small enough and R>0 is large enough. From the condition, we have
(6.1)N(r,f)≤1-2εnT(r,f),r≥R,
and this together with Lemma 2.4 and Lemma 2.1 gives
(6.2)N(r,1P(f)-f)≤T(r,P(f)-f)+O(1)≤m(r,P(f)-f)+N(r,P(f)-f)+O(1)=m(r,f(P(f)f-1))+nN(r,f)+Sq(r,f)≤T(r,f)+(n-1)N(r,f)+Sq(r,f),r≥R.Therefore, by the sharing assumption,(6.3)∑j=13‍N¯(r,1f-aj)≤N(r,1P(f)-f)≤T(r,f)+(n-1)N(r,f)+Sq(r,f),r≥R
from above, it follows that
(6.4)(2-ε)T(r,f)≤N¯(r,f)+∑j=13‍N¯(r,1f-aj)≤(2-2ε)T(r,f)+Sq(r,f),
which is impossible. This contradiction yields f≡P(f).



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