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.
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 Δ^{n}f are defined in the standard way [8] by
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 ${\scriptstyle \frac{\mathrm{\Delta}f(z)}{f(z)}}$, 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$\text{lim\hspace{0.17em}}{\text{inf}}_{r\to \infty}{\scriptstyle \frac{T(r,f)}{r}}=0$. Then f′ has infinitely many zeros.
In [1], Bergweiler and Langley proved the following theorems.
Let n ∈ ℕ and f be transcendental entire function of order$\sigma (f)=\sigma <{\scriptstyle \frac{1}{2}}$and G_{n}(z) is defined by (1.2). If G_{n} is transcendental, then G_{n}(z) has infinitely many zeros. In particular, if f has order less than$\text{min}[{\scriptstyle \frac{1}{n}},{\scriptstyle \frac{1}{2}}]$, then G_{n}(z) is transcendental and has infinitely many zeros.
For the first difference Δf(z) and divided difference ${\scriptstyle \frac{\mathrm{\Delta}f(z)}{f(z)}}$, they also have the following theorem.
Let f be a function transcendental and meromorphic function in the plane which satisfies${\underset{\_}{\text{lim}}}_{r\to \infty}{\scriptstyle \frac{T(r,f)}{r}}=0$, then Δf(z), ${\scriptstyle \frac{\mathrm{\Delta}f(z)}{f(z)}}$are both transcendental.
But for some n ≥ 2, G_{n}(z) fails to be transcendental if f is an entire function of order less than ${\scriptstyle \frac{1}{2}}$; see [5].
In [4], Langley extended Theorem A. In fact, he got the following theorem.
Let n ∈ ℕ and f be a transcendental meromorphic function of order of growth σ(f) = σ < 1 in the plane and assume that G_{n}(z) as defined by (1.2) is transcendental.
If G_{n}(z) has lower order μ < α < 1/2, which holds in particular if σ < 1/2, then δ(0,G_{n}(z)) ≤ 1 − cos πα or$\delta (\infty ,f)\le {\scriptstyle \frac{\mu}{\alpha}}$.
If σ = 1/2, then either G_{n}(z) has infinitely many zeros or δ(∞, f) < 1.
If f is entire and$\sigma <{\scriptstyle \frac{1}{2}}+{\delta}_{0}$, then G_{n}(z) has infinitely many zeros: here δ_{0}is 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.
Let n ∈ ℕ, c ∈ ℂ and f be a function transcendental meromorphic function of order of growth σ(f) = σ < 1. If${G}_{n}(z)={\scriptstyle \frac{{\mathrm{\Delta}}^{n}f(z)}{f(z)}}$is transcendental, then G_{n}(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 G_{n}(z). For the fixed points of G_{n}(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 G_{n}(z) as defined by (1.2) is transcendental. Then 2δ(0,G_{n}(z)–z)+δ(0,G_{n}(z)) ≤ 2. Furthermore, if G_{n}(z) has lower order$\mu <\alpha <{\scriptstyle \frac{1}{2}}$, which holds in particular if$\sigma (f)<{\scriptstyle \frac{1}{2}}$, and δ(0,G_{n}(z)) > 1 − cos πα, then$\delta (0,{G}_{n}(z)-z)<{\scriptstyle \frac{1}{2}}(1+cos\pi \alpha )$and$\delta (\infty ,f)\le {\scriptstyle \frac{\mu}{\alpha}}$.
In Theorem D, there is a condition that G_{n}(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$\sigma (f)={\scriptstyle \frac{1}{2}}$and f has finite many poles in the plane, G_{n}(z) is defined by (1.2). Then G_{n}(z) has infinitely many zeros.
Theorem 3
Let n ∈ ℕ, let f be a transcendental meromorphic function of order$\sigma (f)<{\scriptstyle \frac{1}{2}}+{\delta}_{0}$and$\sigma (f)\ne {\scriptstyle \frac{n-k}{n}}$, k ∈ ℕ, $n({\scriptstyle \frac{1}{2}}-{\delta}_{0})<k<n$, where δ_{0}is a small positive absolute constant, and G_{n}(z) is defined by (1.2). Then G_{n}(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.
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 ɛ–set 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 ɛ–set, 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.
Let f be a function transcendental and meromorphic function with order σ(f) = σ < ∞, H = {(k_{1}, j_{1}), (k_{2}, j_{2}), · · ·, (k_{q}, j_{q})} be a finite set of distinct pairs of integers that satisfy k_{i} > j_{i} ≥ 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
Suppose that$f(z)={\scriptstyle \frac{g(z)}{d(z)}}$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 {r_{j}}, r_{j} → ∞, such that for all z satisfying |z| = r_{j}, |g(z)| = M(r_{j}, g), when j sufficiently large, we have
Assume n, c, σ, f,G_{n}(z) are as in the hypotheses. Set ${G}_{n}^{*}(z)={G}_{n}(z)-z$, then $\sigma ({G}_{n}^{*}(z))=\sigma ({G}_{n}(z))\le \sigma (f)<1,{G}_{n}^{*}$ is transcendental. By Lemma 1, there exists an ɛ – set E_{n}, such that, as z →∞ in CE_{n},
where E_{n} contains all zeros and poles of G_{n}(z). So, there exists a subset F_{1} ⊂ (1,∞) of finite logarithmic measure such that for large |z| = r not in F_{1}, z ∉ E_{n} and
$${G}_{n}^{*}(z)~\frac{{f}^{(n)}(z)}{f(z)}-z.$$
By Lemma 2, for any given ɛ(0 < 2ɛ < 1 − σ), there exists a subset F_{2} ⊂ (1,∞) of finite logarithmic measure such that for large |z| = r not in F_{2},
Set an ${E}_{n}^{*}$ consists of all zeros and poles of ${G}_{n}^{*}(z)$, then there exists a subset F_{3} ⊂ (1,∞) of finite logarithmic measure such that if $z\in {E}_{n}^{*}$, then |z| = r ∈ F_{3}. Thus, by (3.2) and (3.3), we see that for large |z| = r ∉ [0, 1] ∪ F_{1} ∪ F_{2} ∪ F_{3}, ${G}_{n}^{*}(z)$ has no zero and pole on |z| = r, and
Since f is a transcendental meromorphic function of order of growth $\sigma (f)={\scriptstyle \frac{1}{2}}$ and f has finite many poles, we set $f(z)={\scriptstyle \frac{g(z)}{d(z)}}$, where g(z) is an entire function and d(z) is a polynomial. By Lemma 1, we know that there exists an ɛ – set E_{n}, such that
$${\mathrm{\Delta}}^{n}f(z)~{f}^{(n)}(z)$$
as z →∞in CE_{n}. By Lemma 4, there exists a sequence {r_{j}}, r_{j} →∞, such that for all z satisfying |z| = r_{j}, |g(z)| = M(r_{j}, g), when j sufficiently large, we have
for the sequence {r_{j}}, r_{j} → ∞. Assume that G_{n}(z) is a rational function. Set H = {|z| = r : r ∈ E_{n}}. 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 {r_{j}}, there is a point ${r}_{j}^{\prime}\in [{r}_{j},(1+\kappa ){r}_{j}]H$. Since
Since (σ −1+ɛ)n < 0 and G_{n}(z) is rational function, by (3.12) and (3.15), we can deduce that, as z →∞,
$${G}_{n}(z)~\beta {z}^{-k}$$
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
$$\sigma =1-\frac{k}{n}=\frac{1}{2}.$$
Thus, n = 2k. Since k is a positive integer, n is even, it is contradicts the hypothesis. Hence, G_{n}(z) is transcendental. Since the poles of f is finite, we have δ(∞, f) = 1. By Theorem D(2), G_{n}(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.
References
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