Kyungpook Mathematical Journal 2018; 58(4): 689-695
On Zeros and Fixed Points of Differences of Meromorphic Functions
School of Mathematics and Statistics, Anyang Normal University, Anyang, Henan, 455000, China, e-mail : herrzgw@foxmail.com, Department of Mathematics and Physics, Shanghai Dianji University, Shanghai 200240, China, e-mail : qijianmingsdju@163.com, Class 2015, National Economy Major, Wenlan School of Business, Zhongnan University of Economics and Law, Wuhan 430073, China, e-mail : 931730916@qq.com
*Corresponding Author.
Received: December 13, 2017; Revised: October 8, 2018; Accepted: October 16, 2018; Published online: December 23, 2018.

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Abstract

In this paper, we give some results on the zeros and fixed points of the difference and the divided difference of transcendental meromorphic functions. This improves on results of Langley.

Keywords: zeros, fixed points, difference, entire function, meromorphic function.
1. Introduction

We assume that the reader is familiar with the basic notions of Nevanlinna’s value distribution theory (see [7, 9, 10]). Let f be a function that is transcendental and meromorphic in the plane. The forward differences Δnf are defined in the standard way [8] by

$Δf(z)=f(z+1)-f(z), Δn+1f(z)=Δnf(z+1)-Δnf(z)$

and the divided difference is defined by

$Gn(z)=Δnf(z)f(z).$

Recently, a number of papers focus on complex difference equations and differences analogues of Nevanlinna’s theory. Bergweiler and Langley [1] firstly investigated the existed of zeros of Δf(z) and $Δf(z)f(z)$, and obtained many profound results. The results may be viewed as discrete analogues of the following existing theorem on the zeros of f′.

### Theorem A.([BL])

Let f be transcendental and meromorphic in the plane with$lim infr→∞T(r,f)r=0$. Then f′ has infinitely many zeros.

In [1], Bergweiler and Langley proved the following theorems.

### Theorem B.([1])

Let n ∈ ℕ and f be transcendental entire function of order$σ(f)=σ<12$and Gn(z) is defined by (1.2). If Gn is transcendental, then Gn(z) has infinitely many zeros. In particular, if f has order less than$min[1n,12]$, then Gn(z) is transcendental and has infinitely many zeros.

For the first difference Δf(z) and divided difference $Δf(z)f(z)$, they also have the following theorem.

### Theorem C.([1])

Let f be a function transcendental and meromorphic function in the plane which satisfies$lim_r→∞T(r,f)r=0$, then Δf(z), $Δf(z)f(z)$are both transcendental.

But for some n ≥ 2, Gn(z) fails to be transcendental if f is an entire function of order less than $12$; see [5].

In [4], Langley extended Theorem A. In fact, he got the following theorem.

### Theorem D.([4])

Let n ∈ ℕ and f be a transcendental meromorphic function of order of growth σ(f) = σ < 1 in the plane and assume that Gn(z) as defined by (1.2) is transcendental.

If Gn(z) has lower order μ < α < 1/2, which holds in particular if σ < 1/2, then δ(0,Gn(z)) ≤ 1 − cos πα or$δ(∞,f)≤μα$.

If σ = 1/2, then either Gn(z) has infinitely many zeros or δ(∞, f) < 1.

If f is entire and$σ<12+δ0$, then Gn(z) has infinitely many zeros: here δ0is a small positive absolute constant.

In [2], Chen and Shon studied the fixed points of differences and divided differences of meromorphic functions and got some results. One of them is the following.

### Theorem E.([2])

Let n ∈ ℕ, c ∈ ℂ and f be a function transcendental meromorphic function of order of growth σ(f) = σ < 1. If$Gn(z)=Δnf(z)f(z)$is transcendental, then Gn(z) has infinitely many fixed points.

Comparing Theorem B with Theorem D (1), we know that the latter give a precise estimation of the zeros of Gn(z). For the fixed points of Gn(z), can we estimate it precisely like Theorem D (1)? In fact, Theorem E is extended as following.

### Theorem 1

Let n ∈ ℕ, let f be a transcendental and meromorphic function of order σ(f) < 1 in the plane and assume that Gn(z) as defined by (1.2) is transcendental. Then 2δ(0,Gn(z)–z)+δ(0,Gn(z)) ≤ 2. Furthermore, if Gn(z) has lower order$μ<α<12$, which holds in particular if$σ(f)<12$, and δ(0,Gn(z)) > 1 − cos πα, then$δ(0,Gn(z)-z)<12(1+cosπα)$and$δ(∞,f)≤μα$.

In Theorem D, there is a condition that Gn(z) is transcendental. Can we remove it for the case (2) and (3)? The question is studied and the following results are obtained.

### Theorem 2

Let n be odd, let f be a transcendental meromorphic function of order$σ(f)=12$and f has finite many poles in the plane, Gn(z) is defined by (1.2). Then Gn(z) has infinitely many zeros.

### Theorem 3

Let n ∈ ℕ, let f be a transcendental meromorphic function of order$σ(f)<12+δ0$and$σ(f)≠n-kn$, k ∈ ℕ, $n(12-δ0), where δ0is a small positive absolute constant, and Gn(z) is defined by (1.2). Then Gn(z) has infinitely many zeros.

The final theorem from [1] showed that for transcendental meromorphic function satisfying the very strong growth restriction T(r, f) = O(log r)2 as r → ∞, either the first differential or the first divided difference has infinitely many zeros. Langley improved it in [4] as following.

### Theorem F.([4])

Let f be a transcendental meromorphic function in the plane of order less than 1/6, and define G = Δf/f. Then at least one of G and Δf has infinitely many zeros.

Obviously, we have the following corollary by Theorem A, C and E.

### Corollary 1

Under the hypothesis of Theorem F, G has infinity many fixed points.

Thus, there exists a natural question that how about that fixed points of Δf under the hypothesis of Theorem F. It deserves for further study.

2. Lemmas for the Proofs of Theorems

### Remark 1

Following Hayman ([11], p75–p76), we define an ɛset to be a countable union of open discs not containing the origin and subtending angles at the origin whose sum is finite. If E is an ɛset, then the set of r ≥ 1 for which the circle S(0, r) meets E has finite logarithmic measure, and for almost all real θ the intersection of E with the ray arg z = θ is bounded.

### Lemma 1.([1])

Let n ∈ ℕ and f be transcendental and meromorphic of order less than 1 in the plane. Then there exists an ɛset En such that

$Δnf(z)~f(n)(z)$

as z → ∞ inEn.

### Lemma 2.([6])

Let f be a function transcendental and meromorphic function with order σ(f) = σ < ∞, H = {(k1, j1), (k2, j2), · · ·, (kq, jq)} be a finite set of distinct pairs of integers that satisfy ki > ji ≥ 0, for i = 1, · · ·, q and let ɛ > 0 be a given constant. Then there exists a set E ⊂ (1,∞) with finite logarithmic measure such that for all z satisfying |z| ∉ E ∪ [0, 1] and for all (k, j) ∈ H, we have

$|f(k)(z)f(j)(z)|≤ ∣z∣(k-j)(σ-1+ɛ).$

### Lemma 3.([9])

Suppose that f is a meromorphic function in the complex plane, and a1(z), a2(z) and a3(z) are three distinct small functions of f(z). Then

$T(r,f)<∑j=13N¯(r,1f-aj(z))+S(r,f).$

### Lemma 4.([3])

Suppose that$f(z)=g(z)d(z)$is a meromorphic function with σ(f) = σ, where g(z) is an entire function and d(z) is a polynomial. Then there exists a sequence {rj}, rj → ∞, such that for all z satisfying |z| = rj, |g(z)| = M(rj, g), when j sufficiently large, we have

$f(n)(z)f(z)=(vg(z)z)n (1+o(1)), n≥1,$

and

$σ(f)=limj→∞logvg(rj)logrj.$
3. Proof of Theorems

### Proof of Theorem 1

Assume n, c, σ, f,Gn(z) are as in the hypotheses. Set $Gn*(z)=Gn(z)-z$, then $σ(Gn*(z))=σ(Gn(z))≤σ(f)<1,Gn*$ is transcendental. By Lemma 1, there exists an ɛset En, such that, as z →∞ in CEn,

$Gn(z)=Δnf(z)f(z)~f(n)(z)f(z),$

where En contains all zeros and poles of Gn(z). So, there exists a subset F1 ⊂ (1,∞) of finite logarithmic measure such that for large |z| = r not in F1, zEn and

$Gn*(z)~f(n)(z)f(z)-z.$

By Lemma 2, for any given ɛ(0 < 2ɛ < 1 − σ), there exists a subset F2 ⊂ (1,∞) of finite logarithmic measure such that for large |z| = r not in F2,

$|f(n)(z)f(z)|≤ ∣z∣n(σ-1+ɛ).$

Set an $En*$ consists of all zeros and poles of $Gn*(z)$, then there exists a subset F3 ⊂ (1,∞) of finite logarithmic measure such that if $z∈En*$, then |z| = rF3. Thus, by (3.2) and (3.3), we see that for large |z| = r ∉ [0, 1] ∪ F1F2F3, $Gn*(z)$ has no zero and pole on |z| = r, and

$∣Gn*(z)+z∣ =|f(n)(z)f(z)(1+o(1))|≤ ∣z∣ɛ< ∣Gn*(z)∣+∣z∣$

holds on |z| = r. Applying the Rouché’s theorem to function z and $Gn*(z)$, we obtain that

$n (r,1Gn*)-n(r,Gn*)=n (r,1z)-n(r,z)=1.$

Applying Lemma 3 (Generation of second fundamental theorem) to function Gn(z), we have

$T(r,Gn(z))

Since $N¯(r,Gn(z))=N¯(r,Gn*(z)),T(r,Gn*(z))=T(r,Gn(z))+S(r,Gn)$, by (3.5) and (3.6), we have

$T(r,Gn*(z))

Thus, by the definition of deficiency, we have

$2δ(0,Gn*(z))+δ(0,Gn(z))≤2.$

Assume further that δ(0,Gn) > 1–cos πα and by the proof of Theorem D(1)(see[4]), we have

$δ(0,Gn*)<12(1+cos πα), δ(∞,f)>μα.$

### Proof of Theorem 2

Since f is a transcendental meromorphic function of order of growth $σ(f)=12$ and f has finite many poles, we set $f(z)=g(z)d(z)$, where g(z) is an entire function and d(z) is a polynomial. By Lemma 1, we know that there exists an ɛset En, such that

$Δnf(z)~f(n)(z)$

as z →∞in C En. By Lemma 4, there exists a sequence {rj}, rj →∞, such that for all z satisfying |z| = rj, |g(z)| = M(rj, g), when j sufficiently large, we have

$f(n)(z)f(z)=(vg(rj)z)n (1+o(1)), n≥1,$$σ(f)=limj→∞logvg(rj)logrj,$

where υg(r) is the central index of g(z). By (3.10) and (3.11), we have

$Gn(z)=(vg(rj)z)n (1+o(1))$

for the sequence {rj}, rj → ∞. Assume that Gn(z) is a rational function. Set H = {|z| = r : rEn}. Then by Remark 1, H is of finite logarithmic measure. Set the logarithmic measure of H by lm(H) = logκ < ∞, then for the above sequence {rj}, there is a point $rj′∈[rj,(1+κ)rj]H$. Since

$log vg(rj′)log rj′≥log vg(rj)log[(1+κ)rj]=log vg(rj)log rj[1+log(1+κ)log rj].$

We have

$σ(f)=limrj′→∞log vg(rj′)log rj′.$

By (3.14), for any given ɛ(0 < ɛ < 1 − σ), we get that for sufficiently large j,

$(rj′)(σ-1-ɛ)n≤(vg(rj′)rj′)n≤(rj′)(σ-1+ɛ)n.$

Since (σ −1+ɛ)n < 0 and Gn(z) is rational function, by (3.12) and (3.15), we can deduce that, as z →∞,

$Gn(z)~βz-k$

where β ≠ 0 is a constant and k is a positive integer. Since ɛ is arbitrary, by (3.13), (3.15) and (3.16), we have

$σ=1-kn=12.$

Thus, n = 2k. Since k is a positive integer, n is even, it is contradicts the hypothesis. Hence, Gn(z) is transcendental. Since the poles of f is finite, we have δ(∞, f) = 1. By Theorem D(2), Gn(z) has infinitely many zeros.

### Proof of Theorem 3

Using the Wiman-Valiron theory and by the same argument of the proof of Theorem 2, we can prove it.

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