# On Value Distribution of Difference  Polynomials  of Meromorphic Functions

Authors: Zong-Xuan Chen
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We study the value distribution of the difference counterpartΔf(z)−af(z)n of f′(z)−af(z)n and obtain an almost direct difference analogue of results of Hayman.


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## Body

## 1. Introduction and Results

Hayman proved the following Theorem A.Theorem A (see [1]).
Iff(z) is a transcendental entire function, n≥3 is an integer, and a(≠0) is a constant, then f′(z)-af(z)n assumes all finite values infinitely often.Recently, many papers (see [2–7]) have focused on complex differences, giving many difference analogues in value distribution theory of meromorphic functions.It is well known thatΔf(z)=f(z+c)-f(z) (where c∈ℂ∖{0} is a constant satisfying f(z+c)-f(z)≢0) is regarded as the difference counterpart of f′(z), so that Δf(z)-af(z)n is regarded as the difference counterpart of f′(z)-af(z)n, where a∈ℂ∖{0} is a constant.Liu and Laine [7] obtain the followingTheorem B.
Letf be a transcendental entire function of finite order ρ, not of period c, where c is a nonzero complex constant, and let s(z) be a nonzero function, small compared to f. Then the difference polynomial f(z)n+f(z+c)-f(z)-s(z) has infinitely many zeros in the complex plane, provided that n≥3.We use the basic notions of Nevanlinna's theory (see [8, 9]) and in addition use σ(f) to denote the order of growth of the meromorphic f(z) and λ(f) to denote the exponent of convergence of the zeros of f(z).In this paper, we consider the difference counterpart of Theorem A. Whenn≥3 is an integer, we prove the following Theorem 1.1. Compared with Theorem B, Theorem 1.1 is an almost direct difference analogue of of Theorem A and gives an estimate of numbers of b-points, namely, λ(Ψn(z)-b)=σ(f) for every b∈ℂ. Our method of the proof is also different from the method of the proof in Theorem B.Theorem 1.1.
Letf(z) be a transcendental entire function of finite order, and let a,c∈ℂ∖{0} be constants, with c such that f(z+c)≢f(z). Set Ψn(z)=Δf(z)-af(z)n, where Δf(z)=f(z+c)-f(z) and n≥3 is an integer. Then Ψn(z) assumes all finite values infinitely often, and for every b∈ℂ one has λ(Ψn(z)-b)=σ(f).Example 1.2.
Forf(z)=exp{exp{z}}, c=log3 and a=1, we have
(1.1)Ψ3(z)=Δf(z)-af(z)3=-exp⁡⁡{exp⁡⁡{z}}.
Here Ψ3(z)≠0, which shows that Theorem 1.1 may fail for entire functions of infinite order.Example 1.3.
Forf(z)=exp{z}+1,c=log3,a=1, we have Ψ2(z)=Δf(z)-af(z)2=-exp{2z}-1. Here Ψ2(z)≠-1, which shows that Theorem 1.1 may fail for n=2 and that the condition n≥3 in Theorem 1.1 is sharp.Example 1.4.
Forf(z)=exp{z},c=log3,a=1, we have Ψ2(z)=Δf(z)-af(z)2=exp{z}(2-exp{z}), which assumes all finite values infinitely often.What can we say aboutΨ2(z) when n=2? We consider this question and obtain the following Theorems 1.5 and 1.6.Theorem 1.5.
Letf(z) be a transcendental entire function of finite order with a Borel exceptional value 0, and let a,c∈ℂ∖{0} be constants, with c such that f(z+c)≢f(z). Then Ψ2(z) assumes all finite values infinitely often, and for every b∈ℂ one has λ(Ψ2(z)-b)=σ(f).Theorem 1.6.
Letf(z) be a transcendental entire function of finite order with a finite nonzero Borel exceptional value d, and let a,c∈ℂ∖{0} be constants, with c such that f(z+c)≢f(z). Then for every b∈ℂ with b≠-ad2, Ψ2(z) assumes the value b infinitely often, and λ(Ψ2(z)-b)=σ(f).Remark 1.7.
From Theorems1.5 and 1.6, we see that if f(z) has the Borel exceptional value 0, then Ψ2(z) has not any finite Borel exceptional value, but if f(z) has a nonzero Borel exceptional value, then Ψ2(z) may have a finite Borel exceptional value. From Theorem 1.6, this possible Borel exceptional value is -ad2. Example 1.3 shows that this Borel exceptional value -ad2(=-1) may arise, and thus the conclusion of Theorem 1.6 is sharp.


## 2. Proof of Theorem1.1

We need the following lemmas.Lemma 2.1 (see [3, 4]).
Letf(z) be a meromorphic function of finite order, and let  c∈ℂ∖{0}. Then
(2.1)m(r,f(z+c)f(z))=S(r,f),
where S(r,f)=o{T(r,f)}.Lemma 2.2 (see [3]).
Letf(z) be a meromorphic function with order σ(f)=σ<∞, and let c be a nonzero constant. Then, for each ɛ>0, one has
(2.2)T(r,f(z+c))=T(r,f)+O(rσ-1+ɛ)+O(log⁡⁡r).Lemma 2.3.
Suppose thatn, a, c, f(z), Ψn(z) satisfy the conditions of Theorem 1.1. If b∈ℂ, then Ψn(z)-b is transcendental.Proof.
Suppose thatΨn(z)-b=p(z), where p(z) is a polynomial. Then
(2.3)-af(z)n=b-Δf(z)+p(z).
By Lemma 2.2, for each ɛ>0, we have
(2.4)T(r,Δf(z))≤2T(r,f)+S(r,f)+O(rσ-1+ɛ),
where σ=σ(f). By an identity due to Valiron-Mohon'ko (see [10, 11]), we have
(2.5)T(r,af(z)n)=nT(r,f)+S(r,f)=T(r,b-Δf(z)+p(z))≤2T(r,f)+S(r,f)+O(rσ-1+ɛ).
This contradicts the fact that n≥3. Hence Ψn(z)-b is transcendental.Lemma 2.4.
Suppose thatn, a, c, f(z), Ψn(z) satisfy the conditions of Theorem 1.1. Suppose also that b∈ℂ, q(z)≢0 is a polynomial, and p(z)≢0 is an entire function with σ(p)<σ(f). If
(2.6)Ψn(z)-b=p(z)exp⁡⁡{q(z)},
then
(2.7)P(z,f)=np(z)f′(z)-(p′(z)+q′(z)p(z))f(z)≢0.Proof.
Suppose that(2.8)np(z)f′(z)-(p′(z)+q′(z)p(z))f(z)≡0.
Integrating (2.8) results in
(2.9)f(z)n=dp(z)exp⁡⁡{q(z)},
where d(≠0) is a constant. Therefore, by (2.6), (2.9), and the definition of Ψn(z), we obtain
(2.10)Ψn(z)-b=f(z+c)-f(z)-af(z)n-b=1df(z)n,
and so
(2.11)d(f(z+c)-f(z))=(ad+1)f(z)n+bd.
We must have ad+1≠0. In fact, if ad+1=0, then by (2.11) and d≠0, we have that
(2.12)f(z+c)-f(z)=b,
so f′(z) is periodic. Then, write (2.8) as
(2.13)f′(z)=R(z)f(z),
where
(2.14)R(z)=p′(z)+q′(z)p(z)np(z).
Clearly, R(z)≢0 and σ(R)≤σ(p)<σ(f). We obtain from (2.12) and (2.13)
(2.15)(R(z+c)-R(z))f(z)=-bR(z+c).
If R(z+c)-R(z)≡0, then b=0 by (2.15) and R(z)≢0. Thus, by (2.12), we have f(z+c)≡f(z), which contradicts our condition. If R(z+c)-R(z)≢0, then by (2.15), we have
(2.16)σ(f)=σ(-bR(z+c)R(z+c)-R(z))≤σ(R)<σ(f).
This is also a contradiction. Hence ad+1≠0.
Differentiating (2.11), and then dividing by f′(z) result in
(2.17)d(f′(z+c)f′(z)-1)=n(ad+1)f(z)n-1.
Therefore, by Lemma 2.1, we get that
(2.18)(n-1)T(r,f)=(n-1)m(r,f)=S(r,f′)=S(r,f),
a contradiction for n≥2. Hence P(z,f)≢0.Halburd and Korhonen obtained the following difference analogue of the Clunie lemma [4, Corollary 3.3].Lemma 2.5.
Letf(z) be a nonconstant, finite order meromorphic solution of
(2.19)fnP1(z,f)=Q1(z,f),
where P1(z,f), Q1(z,f) are difference polynomials in f(z) with small meromorphic coefficients, and let δ<1. If the degree of Q1(r,f) as a polynomial in f(z) and its shifts is at most n, then
(2.20)m(r,P1(z,f))=o(T(r+|c|,f)rδ)+o(T(r,f))
for all r outside an exceptional set of finite logarithmic measure.Remark 2.6.
If coefficients ofP1, Q1 are aj(z)(j=1,…,s) satisfying σ(aj)<σ(f), then using the same method as in the proof of Lemma 2.5 (see [4]), we have
(2.21)m(r,P1(z,f))=o(T(r+|c|,f)rδ)+o(T(r,f))+O(∑j=1sm(r,aj))
for all r outside an exceptional set of finite logarithmic measure.We are now able to prove Theorem1.1. We only prove the case σ(f)>0. For the case σ(f)=0, we can use the same method in the proof. Suppose that b∈ℂ and λ(Ψn(z)-b)<σ(f). Then, by Lemma 2.3, we see that Ψn(z)-b is transcendental. Thus, Ψn(z)-b can be written as (2.22)Ψn(z)-b=p(z)exp⁡⁡{q(z)},
where q(z)≢0 is a polynomial, p(z)≢0 is an entire function with σ(p)<σ(f).Differentiating (2.22) and eliminating exp{q(z)}, we obtain (2.23)f(z)n-1P(z,f)=Q(z,f),
where (2.24)P(z,f)=anp(z)f′(z)-a(p′(z)+q′(z)p(z))f(z)Q(z,f)=p(z)f′(z)-p(z)f′(z+c)+Δf(z)(p′(z)+q′(z)p(z))-b(p′(z)+q′(z)p(z)).By Lemma2.4, we see that P(z,f)≢0. Since n≥3 and the total degree of Q(z,f) as a polynomial in f(z) and its shifts, degfQ(z,f)=1, by (2.23), Lemma 2.5, and Remark 2.6, we obtain that for δ<1(2.25)T(r,P(z,f))=m(r,P(z,f))=o(T(r+|c|,f)rδ)+o(T(r,f))+O(m(r,p)),T(r,fP(z,f))=m(r,fP(z,f))=o(T(r+|c|,f)rδ)+o(T(r,f))+O(m(r,p))
for all r outside of an exceptional set of finite logarithmic measure.Thus, (2.25) give that (2.26)T(r,f)=o(T(r+|c|,f)rδ)+o(T(r,f))+O(m(r,p))
for all r outside of an exceptional set of finite logarithmic measure. This is a contradiction. Hence Ψn(z)-b has infinitely many zeros and λ(Ψn(z)-b)=σ(f), which proves Theorem 1.1.


## 3. Proof of Theorem1.5

We need the following lemma.Lemma 3.1 (see [12, page 69–70], [13, page 79–80], or [14]).
Suppose thatn≥2, and let fj(z), j=1,…,n, be meromorphic functions and gj(z), j=1,…,n, entire functions such that (i)
∑j=1nfj(z)exp{gj(z)}≡0;(ii)
when1≤j<k≤n, gj(z)-gk(z) is not constant;(iii)
when1≤j≤n, 1≤h<k≤n,
(3.1)T(r,fj)=o{T(r,exp⁡⁡{gh-gk})}(r⟶∞,r∉E),
whereE⊂(1,∞) is of finite linear measure or finite logarithmic measure.  Then fj(z)≡0, j=1,…,n.To prove Theorem1.5, note first that f(z) has a Borel exceptional value 0, we can write f(z) as (3.2)f(z)=g(z)exp⁡⁡{αzk},f(z+c)=g(z+c)g1(z)exp⁡⁡{αzk},
where α(≠0) is a constant, k(≥1) is an integer satisfying σ(f)=k, and g(z),g1(z) are entire functions such that g(z)g1(z)≢0, σ(g)<k, σ(g1)=k-1.First, we proveΨ2(z)-b=Δf(z)-af(z)2-b is transcendental. If Ψ2(z)-b=p(z), where p(z) is a polynomial, then (3.3)af(z)2=Δf(z)-b+p(z).
Thus by Lemma 2.1 and an identity due to Valiron-Mohon'ko (see [10, 11]), we have (3.4)T(r,af2)=2T(r,f)+S(r,f)T(r,Δf(z)-b+p(z))=m(r,Δf(z)-b+p(z))≤m(r,f)+m(r,f(z+c)f(z)-1)+O(log⁡⁡r)≤T(r,f)+S(r,f).
By (3.4), we see that (3.3) is a contradiction.Secondly, we proveσ(Ψ2-b)=σ(f)=k≥1. By the expression of Ψ2(z), we have σ(Ψ2-b)≤k. Set G(z)=Ψ2(z)-b. If σ(G)=k1<k, then by (3.2), we have (3.5)(g(z+c)g1(z)-g(z))exp⁡⁡{αzk}-ag(z)2exp⁡⁡{2αzk}-(b+G(z))=0.
Since σ(G)=k1<k and σ(g)<k, we see that the left hand side of (3.5) is of order =k by applying the general form of the Valiron-Mohon'ko lemma in [10], a contradiction. So, σ(G)=k.Thirdly, we proveλ(G)=k(≥1). If λ(G)<k, then G(z) can be written as (3.6)G(z)=g*(z)exp⁡⁡{βzk},
where β(≠0) is a constant, g*(z)(≢0) is an entire function satisfying σ(g*)<k. Thus by (3.2), (3.6), and G(z)=Ψ2(z)-b, we have (3.7)(g(z+c)g1(z)-g(z))exp⁡⁡{αzk}-ag(z)2exp⁡⁡{2αzk}-b-g*exp⁡⁡{βzk}=0.
In (3.7), there are three cases for β: (i)
β≠α and β≠2α;(ii)
β=α;(iii)
β=2α.Applying Lemma3.1 to (3.7), we have in case (i) (3.8)g(z+c)g1(z)-g(z)≡ag(z)2≡g*(z)≡0;
in case (ii), we have (3.9)ag(z)2≡0;
in case (iii), we have (3.10)g(z+c)g1(z)-g(z)≡0.
Since f(z+c)-f(z)≢0 and f(z) is transcendental, we see that any one of (3.8)–(3.10) is a contradiction. Hence λ(Ψ2-b)=σ(f).


## 4. Proof of Theorem1.6

Sincef(z) has a nonzero Borel exceptional value d, we can write f(z) as (4.1)f(z)=d+φ(z)exp⁡⁡{αzk},f(z+c)=d+φ(z+c)φ1(z)exp⁡⁡{αzk},
where α(≠0) is a constant, k(≥1) is an integer satisfying σ(f)=k, and φ(z), φ1(z) are entire functions such that φ(z)φ1(z)≢0, σ(φ)<k, σ(φ1)=k-1.Using the same method as in the proof of Theorem1.5, we can show that (4.2)Ψ2(z)-b=Δf(z)-af(z)2-b
is transcendental and σ(Ψ2-b)=σ(f)=k≥1.Now we show thatλ(Ψ2(z)-b)=k(≥1). Set G(z)=Ψ2(z)-b. If λ(G)<k, then G can be written as (4.3)G(z)=φ*(z)exp⁡⁡{szk},
where s(≠0) is a constant and φ*(z)(≢0) is an entire function satisfying σ(φ*)<k. Thus by (4.1) and (4.3), we have (4.4)(φ(z+c)φ1(z)-φ(z)-2adφ(z))exp⁡⁡{αzk}-aφ(z)2exp⁡⁡{2αzk}-φ*(z)exp⁡⁡{szk}-(ad2+b)=0.
In (4.4), there are three cases for s: (i)
s≠α and s≠2α;(ii)
s=α;(iii)
s=2α.Applying the same method as in the proof of Theorem1.5 to these three cases, we obtain (4.5)ad2+b=0,
which contradicts our supposition that b≠-ad2. Hence λ(Ψ2-b)=σ(f).



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