# Value Sharing Results forq-Shifts Difference Polynomials

Authors: Yong Liu; Yinhong Cao; Xiaoguang Qi; Hongxun Yi
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

We investigate the zero distribution ofq-shift difference polynomials of meromorphic functions with zero order and obtain some results that extend previous results of K. Liu et al.


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## Body

## 1. Introduction and Main Results

In this paper, we assume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna theory (see, e.g., [1, 2]). Let f(z) and g(z) be two nonconstant meromorphic functions in the complex plane. By S(r,f), we denote any quantity satisfying S(r,f)=o(T(r,f)) as r→∞, possibly outside a set of r with finite linear measure. Then the meromorphic function α is called a small function of f(z), if T(r,α)=S(r,f). If f(z)-α and g(z)-α have same zeros, counting multiplicity (ignoring multiplicity), then we say that f(z) and g(z) share the small function α CM (IM). The logarithmic density of a set Fn is defined as follows:
(1)limsupr→∞1logr∫[1,r]∩Fn‍1tdt.Currently, many articles have focused on value distribution in difference analogues of meromorphic functions (see, e.g., [3–11]). In particular, there has been an increasing interest in studying the uniqueness problems related to meromorphic functions and their shifts or their difference operators (see, e.g., [8, 12–15]). Our aim in this article is to investigate the uniqueness problems of q-difference polynomials.Recently, Liu et al. [13] considered uniqueness of difference polynomials of meromorphic functions, corresponding to uniqueness theorems of meromorphic functions sharing values (see, e.g., [9, 16]). They got the following.Theorem A.
Letf(z) and g(z) be two transcendental meromorphic functions with finite order. Suppose that c is a nonzero complex constant and n is an integer. If n≥14 and fn(z)f(z+c) and gn(z)g(z+c) share 1CM, then f(z)≡tg(z) or f(z)g(z)=t, where tn+1=1.Theorem B.
Under the conditions of Theorem A, ifn≥26and fn(z)f(z+c) and gn(z)g(z+c) share 1IM, then f(z)≡tg(z) or f(z)g(z)=t, where tn+1=1.In this paper, we consider the case ofq-shift difference polynomials and extend Theorem A as follows:Theorem 1.
Letf(z) and g(z) be two transcendental meromorphic functions with ρ(f)=ρ(g)=0. Suppose that q and c are two nonzero complex constants and n is an integer. If n≥14 and fn(z)f(qz+c) and gn(z)g(qz+c) share 1CM, then f(z)≡tg(z) or f(z)g(z)=t, where tn+1=1.It is natural to ask whether Theorem1 holds if fn(z)f(qz+c) and gn(z)g(qz+c) share 1 IM. Corresponding to this question, we get the following result.Theorem 2.
Under the conditions of Theorem1, if n≥26 and fn(z)f(qz+c) and gn(z)g(qz+c) share 1IM, then f(z)≡tg(z) or f(z)g(z)=t, where tn+1=1.Next, we consider the uniqueness ofq-difference products of entire functions and obtain the following results.Theorem 3.
Letf(z) and g(z) be two transcendental entire functions with ρ(f)=ρ(g)=0, and let q and c be two nonzero complex constants, and let P(z)=anzn+an-1zn-1+⋯+a1z+a0 be a nonzero polynomial, where an(≠0), an-1,…,a0 are complex constants, and k denotes the number of the distinct zero of P(z). If n>2k+1 and P(f(z))f(qz+c) and P(g(z))g(qz+c) share 1CM, then one of the following results holds: (1)
f(z)≡tg(z) for a constant t such that td=1, where d=GCD{λ0,λ1,…,λn} and
(2)λj={n+1,aj=0,j+1,aj≠0,j=0,1,…,n;(2)
f(z) and g(z) satisfy the algebraic equation R(f(z),g(z))=0, where
(3)R(w1,w2)=P(w1)w1(qz+c)-P(w2)w2(qz+c).Remark 4.
A similar result can be found in [15], but the method of this paper is more concise, and the condition of this paper is better.


## 2. Preliminary Lemmas

The following lemma is aq-difference analogue of the logarithmic derivative lemma.Lemma 5 (see [14]).
Letf(z) be a meromorphic function of zero order, and let c and q be two nonzero complex numbers. Then one has
(4)m(r,f(qz+c)f(z))=S(r,f)
on a set of logarithmic density 1.Lemma 6 (see [7]).
IfT:ℝ+→ℝ+ is an increasing function such that
(5)limsupr→∞logT(r)logr=0,
then the set
(6)E:={r:T(C1r)≥C2T(r)}
has logarithmic density 0 for all C1>1 and C2>1.The following lemma is essential in our proof and is due to Heittokangas et al., see [12, Theorems 6 and 7].Lemma 7.
Letf(z) be a meromorphic function of finite order, and let c≠0 be fixed. Then
(7)N¯(r,f(z+c))≤N¯(r,f(z))+S(r,f),N(r,1f(z+c))≤N(r,1f(z))+S(r,f),N(r,f(z+c))≤N(r,f(z))+S(r,f),N¯(r,1f(z+c))≤N¯(r,1f(z))+S(r,f).Lemma 8.
Letf(z) be a meromorphic function with ρ(f)=0, and let c and q be two nonzero complex numbers. Then
(8)N¯(r,f(qz+c))≤N¯(r,f(z))+S(r,f),N(r,1f(qz+c))≤N(r,1f(z))+S(r,f),N(r,f(qz+c))≤N(r,f(z))+S(r,f),N¯(r,1f(qz+c))≤N¯(r,1f(z))+S(r,f).Proof of Lemma8.
We only prove the case|q|≥1. For the case |q|≤1, we can use the same method in the proof. By a simple geometric observation, we obtain
(9)N(r,1f(qz+c))≤N(|q|r,1f(z+(c/q))).
Combining ρ(f)=0 with Lemma 6, we obtain
(10)N(|q|r,1f(z+(c/q)))≤N(r,1f(z+(c/q)))+S(r,f)
on a set of logarithmic density 1. On the other hand, we have
(11)N(r,1f(z+(c/q)))≤N(|qr|,1f(z+(c/q))).
Therefore,
(12)N(|q|r,1f(z+(c/q)))=N(r,1f(z+(c/q)))+S(r,f)
on a set of logarithmic density 1. From (9) and (12), we have
(13)N(r,1f(qz+c))≤N(r,1f(z+(c/q)))+S(r,f).By Lemma 7, we have
(14)N(r,1f(qz+c))≤N(r,1f(z+(c/q)))+S(r,f)≤N(r,1f(z))+S(r,f).
Similarly, we have(15)N¯(r,f(qz+c))≤N¯(r,f(z))+S(r,f),N(r,f(qz+c))≤N(r,f(z))+S(r,f),N¯(r,1f(qz+c))≤N¯(r,1f(z))+S(r,f).Lemma 9.
Letf be a nonconstant meromorphic function of zero order, and let c and q be two nonzero complex numbers. Then
(16)T(r,f(qz+c))≤T(r,f(z))+S(r,f)
on a set of logarithmic density 1.Proof of Lemma9.
By Lemmas5 and 8, we have
(17)T(r,f(qz+c))=m(r,f(qz+c))+N(r,f(qz+c))≤m(r,f(qz+c)f(z))+m(r,f(z))+N(r,f(z))+S(r,f)=T(r,f(z))+S(r,f)
on a set of logarithmic density 1.Lemma 10.
Letf(z) be an entire function with ρ(f)=0, let q and c be two fixed nonzero complex constants, and let P(z)=anzn+an-1zn-1+⋯+a1z+a0 be a nonzero polynomial, where an(≠0),an-1,…,a0 are complex constants. Then
(18)T(r,P(f(z))f(qz+c))=T(r,P(f(z))f(z))+S(r,f).Proof of Lemma10.
Byρ(f)=0 and Lemma 5, we obtain
(19)T(r,P(f(z))f(qz+c))=m(r,P(f(z))f(qz+c))≤m(r,P(f(z))f(z))+m(r,f(qz+c)f(z))=T(r,P(f(z))f(z))+S(r,f)
on a set of logarithmic density 1. Using the similar method as above, we also get
(20)T(r,P(f(z))f(z))≤T(r,P(f(z))f(qz+c))+S(r,f)
on a set of logarithmic density 1.
Hence, we haveT(r,P(f(z))f(z))=T(r,P(f(z))f(qz+c))+S(r,f) on a set of logarithmic density 1.Lemma 11 (see [17]).
LetF and G be two nonconstant meromorphic functions. If F and G share 1CM, then one of the following three cases holds:(i)
(21)max{T(r,F),T(r,G)}≤N2(r,1F)+N2(r,F)+N2(r,1G)+N2(r,G)+S(r,F)+S(r,G),(ii)
F=G,(iii)
FG≡1,
whereN2(r,1/F) denotes the counting function of zero of F, such that simple zero are counted once and multiple zeros are counted twice.In order to prove Theorem2, we need the following lemma.Lemma 12 (see [16]).
LetF and G be two nonconstant meromorphic functions, and let F and G share 1IM. Let
(22)H=F′′F′-2F′F-1-G′′G′+2G′G-1.
IfH≢0, then
(23)T(r,F)+T(r,G)≤2(N2(r,1F)+N2(r,F)+N2(r,1G)+N2(r,G))+3(N¯(r,F)+N¯(r,G)+N¯(r,1F)+N¯(r,1G))+S(r,F)+S(r,G).


## 3. Proof of Theorem1

LetF(z)=fn(z)f(qz+c) and G(z)=gn(z)g(qz+c). Thus, F and G share 1 CM. Combining the first main theorem with Lemma 9, we obtain
(24)nT(r,f(z))≤T(r,fn(z)f(qz+c))+T(r,f(z))+O(1).Hence, we obtain(25)(n-1)T(r,f(z))≤T(r,F(z))+S(r,f).
Using the similar method as above, we have
(26)(n-1)T(r,g(z))≤T(r,G(z))+S(r,g).From Lemma9, we have
(27)T(r,F)≤(n+1)T(r,f)+S(r,f),(28)T(r,G)≤(n+1)T(r,g)+S(r,g).
By the second main theorem, Lemma 9, and (28), we obtain
(29)T(r,F)≤N¯(r,F)+N¯(r,1F)+N¯(1F-1)+S(r,F)≤N¯(r,f)+N¯(r,f(qz+c))+N¯(r,1f)+N¯(r,1f(qz+c))+N¯(r,1G-1)+S(r,f)≤4T(r,f)+T(r,G)+S(r,f)≤4T(r,f)+(n+1)T(r,g)+S(r,g)+S(r,f).
Hence, (25) and (29) imply that
(30)(n-5)T(r,f)≤(n+1)T(r,g)+S(r,f)+S(r,g).
Similarly, we have
(31)(n-5)T(r,g)≤(n+1)T(r,f)+S(r,f)+S(r,g).
Equations (30) and (31) imply that S(r,f)=S(r,g). Together the definition of F with Lemma 9, we have
(32)N2(r,1F)≤2N¯(r,1f)+N(r,1f(qz+c))+S(r,f)≤3T(r,f)+S(r,f).Similarly,(33)N2(r,1G)≤3T(r,g)+S(r,f),N2(r,F)≤3T(r,f)+S(r,f),N2(r,G)≤3T(r,g)+S(r,f).
Thus, together (21) with (32)-(33), we obtain
(34)T(r,F)+T(r,G)≤2N2(r,1F)+2N2(r,F)+2N2(r,1G)+2N2(r,G)+S(r,f)+S(r,g)≤12(T(r,f)+T(r,g))+S(r,f)+S(r,g).Then, by (25), (26), and (34), we obtain
(35)(n-13)[T(r,f)+T(r,g)]≤S(r,f)+S(r,g),
which is a contradiction since n≥14. By Lemma 11, we have F≡G or FG≡1. If F≡G, that is, fn(z)f(qz+c)=gn(z)g(qz+c). Set H(z)=f(z)/g(z). Suppose that H(z) is not a constant. Then we obtain
(36)Hn(z)H(qz+c)=1.Lemma9 and (36) imply that
(37)nT(r,H(z))=T(r,Hn(z))=T(r,1H(qz+c))≤T(r,H(z))+S(r,H).
Hence, H(z) must be a nonzero constant, since n≥14. Set H(z)=t. By (36), we know tn+1=1. Thus, f(z)=tg(z), where tn+1=1.IfFG=1, that is,
(38)fn(z)f(qz+c)gn(z)g(qz+c)=1.LetL(z)=f(z)g(z). Using the similar method as above, we also obtain that L(z) must be a nonzero constant. Thus, we have fg=t, where tn+1=1.


## 4. Proof of Theorem2

LetF(z)=fn(z)f(qz+c) and G(z)=gn(z)g(qz+c), and let H be defined in Lemma 12. Using the similar proof as the proof of Theorem 1, we prove that (25)–(33) hold. By Lemma 9, we obtain
(39)N¯(r,F(z))≤N¯(r,f(z))+N¯(r,f(qz+c))+S(r,f)≤2T(r,f)+S(r,f).Similarly, we obtain
(40)N¯(r,1F(z))≤2T(r,f)+S(r,f),N¯(r,G(z))≤2T(r,g)+S(r,g),N¯(r,1G(z))≤2T(r,g)+S(r,g).Together Lemma12 with (32), (33), (39), and (40), we have
(41)T(r,F(z))+T(r,G(z))≤24(T(r,f)+T(r,g))+S(r,f)+S(r,g).
By (25), (26) and (41) yield that
(42)(n-1)(T(r,f(z))+T(r,g(z)))≤24(T(r,f)+T(r,g))+S(r,f)+S(r,g),
which is impossible, since n≥26. Hence, we have H≡0.By integrating (22) twice, we have
(43)F=(b+1)G+(a-b-1)bG+(a-b),
which yields that T(r,F)=T(r,G)+O(1). From (25)–(28), we obtain
(44)(n-1)T(r,f)≤(n+1)T(r,g)+S(r,f)+S(r,g),(45)(n-1)T(r,g)≤(n+1)T(r,f)+S(r,f)+S(r,g).Next, we will prove thatF=G or FG=1.Case 1 (b≠0,-1).
Ifa-b-1≠0, by (43), we obtain
(46)N¯(r,1F)=N¯(r,1G-(a-b-1)/(b+1)).
Together the Nevanlinna second main theorem with Lemma9, (28), and (44), we obtain
(47)(n-1)T(r,g)≤T(r,G)+S(r,g)≤N¯(r,1G)+N¯(r,G)+N¯(r,1G-(a-b-1)/(b+1))+S(r,G)+S(r,g)≤N¯(r,1G)+N¯(r,G)+N¯(r,1F)+S(r,g)+S(r,f)≤N¯(r,1g)+N¯(r,1g(qz+c))+N¯(r,g)+N¯(r,g(qz+c))+N¯(r,1f)+N¯(r,1f(qz+c))+S(r,g)≤4T(r,g)+2T(r,f)+S(r,g)≤(4+2n+1n-1)T(r,g)+S(r,g),
which yields thatn2-8n+3≤0, which is impossible, since n≥26. Hence, we obtain a-b-1=0, so
(48)F(z)=(b+1)G(z)bG(z)+1.
Using the similar method as above, we obtain
(49)(n-1)T(r,g)≤T(r,G)+S(r,g)≤N¯(r,1G)+N¯(r,G)+N¯(r,1G+1/b)+S(r,G)≤N¯(r,1G)+N¯(r,G)+N¯(r,F)+S(r,G)≤(4+2n+1n-1)T(r,g)+S(r,g),
which is impossible.Case 2.
Ifb=-1 and a=-1, then FG=1 follows trivially. Therefore, we may consider the case b=-1 and a≠-1. By (43), we have
(50)F=aa+1-G.
Similarly, we get a contradiction.Case 3.
Ifb=0, a=1 and then F=G  follows trivially. Therefore, we may consider the case b=0 and a≠1. By (43), we obtain
(51)F=G+a-1a.
Similarly, we get a contradiction.


## 5. Proof of Theorem3

SinceP(f(z))f(qz+c) and P(g(z))g(qz+c) share 1 CM, we obtain
(52)P(f(z))f(qz+c)-1P(g(z))g(qz+c)-1=el(z),
where l(z) is an entire function. by ρ(f)=0 and ρ(g)=0, we have el(z)≡η as a constant. We can rewrite (52) as follows:
(53)ηP(g(z))g(qz+c)=P(f(z))f(qz+c)-1+η.Ifη≠1, by the first main theory, the second main theory, and Lemma 9, we have
(54)T(r,P(f(z)))f(qz+c)≤N¯(r,1P(f(z))f(qz+c))+N¯(r,1P(f(z))f(qz+c)-1+η)+S(r,f)=N¯(r,1P(f(z))f(qz+c))+N¯(r,1P(g(z))g(qz+c))+S(r,f)≤(k+1)T(r,f(z))+(k+1)T(r,g(z))+S(r,f)+S(r,g).
By Lemma 10 and (54), we have
(55)(n+1)T(r,f(z))=T(r,P(f(z))f(z))=T(r,P(f(z)))f(qz+c)+S(r,f)≤(k+1)T(r,f(z))+(k+1)T(r,g(z))+S(r,f)+S(r,g).
Hence, we have
(56)(n-k)T(r,f(z))≤(k+1)T(r,g(z))+S(r,f)+S(r,g).
Similarly, we have
(57)(n-k)T(r,g(z))≤(k+1)T(r,f(z))+S(r,f)+S(r,g).Equations (56) and (57) imply that
(58)(n-2k-1)[T(r,f(z))+T(r,g(z))]≤S(r,f)+S(r,g),
which is impossible, since n>2k+1. Hence, we have η=1. We can rewrite (52) as follows:
(59)P(f(z))f(qz+c)=P(g(z))g(qz+c).Seth(z)=f(z)/g(z). We break the rest of the proof into two cases.Case 1.
Suppose thath(z) is a constant. Then by substituting f=gh into (59), we obtain
(60)g(qz+c)[angn(hn+1-1)+an-1gn-1(hn-1)+⋯+a0(h-1)angn(hn+1-1)]≡0,
where an(≠0),an-1,…,a0 are complex constants. By the fact that g is transcendental entire function, we have g(qz+c)≢0. Hence, we obtain
(61)angn(hn+1-1)+an-1gn-1(hn-1)+⋯+a0(h-1)≡0.
Equation (61) implies that hn+1=1 and hi+1=1 when ai≠0 for i=0,1,…,n-1. Therefore, hd=1, where d is defined as the assumption of Theorem 3.Case 2.
Suppose thath is not a constant, then we know by (59) that f and g satisfy the algebraic equation R(f,g)=0, where R(w1,w2)=P(w1)w1(qz+c)-P(w2)w2(qz+c).



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