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Fixed points of differences of meromorphic functions

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Abstract

Let f be a transcendental meromorphic function of finite order and c be a nonzero complex number. Define \(\Delta _{c}f=f(z+c)-f(z)\). The authors investigate the existence on the fixed points of \(\Delta _{c}f\). The results obtained in this paper may be viewed as discrete analogues on the existing theorem on the fixed points of \(f'\). The existing theorem on the fixed points of \(\Delta _{c}f\) generalizes the relevant results obtained by (Chen in Ann. Pol. Math. 109(2):153–163, 2013; Zhang and Chen in Acta Math. Sin. New Ser. 32(10):1189–1202, 2016; Cui and Yang in Acta Math. Sci. 33B(3):773–780, 2013) et al.

Introduction

Let \(f(z)\) be a function meromorphic in the complex plane C. We use the general notation of the Nevanlinna theory (see [12, 20, 23]) such as \(m(r, f) \), \(N(r, f)\), \(T(r,f )\), \(m (r, \frac{1}{f-a} )\), \(N (r, \frac{1}{f-a} )\), … , and assume that the reader is familiar with these notations. We also use \(S(r, f)\) to denote any quantity of \(S(r, f)=o(T(r, f))\) (\(r \rightarrow \infty \)), possibly outside a set with finite logarithmic measure. The order and the lower order of \(f(z)\) are denoted by \(\sigma (f)\) and \(\mu (f)\) respectively.

For any \(a\in {C}\), the exponent of convergence of zeros of \(f(z)-a\) (or poles of \(f(z)\)) is denoted by \(\lambda (f, a)\) (or \(\lambda (\frac{1}{f} )\)). Especially, we denote \(\lambda (f, 0)\) by \(\lambda (f)\). If \(\lambda (f, a)<\sigma (f)\) (or \(\lambda (\frac{1}{f} )<\sigma (f)\)), then a (or ∞) is said to be a Borel exceptional value of \(f(z)\). Nevanlinna’s deficiency of f with respect to complex number \(a\in C\cup \{\infty \}\) is defined by

$$ \delta (a, f)=\liminf_{r\rightarrow \infty }\frac{m (r, \frac{1}{f-a} )}{T(r, f)} =1-\limsup _{r\rightarrow \infty }\frac{N (r, \frac{1}{f-a} )}{T(r, f)}. $$

If \(a=\infty \), then one should replace \(N (r, \frac{1}{f-a} )\) in the above formula by \(N(r, f)\).

A point \(z_{0}\in {C}\cup \{\infty \}\) is said to be a fixed point of \(f(z)\) if \(f(z_{0})=z_{0}\). There is a considerable number of results on the fixed points of meromorphic functions, we refer the reader to Chuang and Yang [7]. It follows Chen and Shon [2, 4], we use the notation \(\tau (f)\) to denote the exponent of convergence of fixed points of f, i.e.,

$$ \tau (f)=\limsup_{r\rightarrow \infty }\frac{\log N (r, \frac{1}{f-z} )}{\log r}. $$

In 1993, Lahiri [13] proved the following theorem.

Theorem A

Let f be a transcendental meromorphic function in the plane. Suppose that there exists \(a\in {C}\) with \(\delta (a, f)>0\) and \(\delta (\infty , f)=1\). Then f has infinitely many fixed points.

In this paper, we shall study the fixed points of the differences of meromorphic functions. For each \(c\in {C}\backslash \{0 \}\), the forward difference \(\Delta _{c}^{k+1} f(z)\) is defined (see [1]) by

$$ \Delta _{c} f(z)=f(z+c)-f(z), \Delta _{c}^{2} f(z)=\Delta _{c} f(z+c)- \Delta _{c}f(z). $$

Especially, we denote \(\Delta _{1} f(z)\) by \(\Delta f(z)\).

Recently, some well-known facts of the Nevanlinna theory have been extended for the differences of meromorphic functions (see [5, 6, 9,10,11, 14,15,16,17,18]). For the existence on the fixed points of differences, Cui and Yang [8] have proved the following theorems.

Theorem B

([8])

Let f be a function transcendental and meromorphic in the plane with the order \(\sigma (f)=1\). If f has finitely many poles and infinitely many zeros with exponent of convergence of zeros \(\lambda (f)\neq 1\), then Δf has infinitely many zeros and fixed points.

Theorem C

([8])

Let f be a non-periodic function transcendental and meromorphic in the plane with the order \(\sigma (f)=1\), \(\max \{\lambda (f), \lambda (\frac{1}{f} ) \} \neq 1\). If f has infinitely many zeros, then Δf has infinitely many zeros and fixed points.

The conditions of Theorems B and C imply that 0, ∞ are Borel exceptional values. If ∞ and \(d\in {C}\) are Borel exceptional values of f, Chen [3] obtains the following theorem.

Theorem D

([3])

Let f be a finite order meromorphic function such that \(\lambda (\frac{1}{f} )< \sigma (f)\), and let \(c\in {C}\backslash \{0\}\) be a constant such that \(f(z+c)\not \equiv f(z)\). If \(f(z)\) has a Borel exceptional value \(d\in {C}\), then \(\tau (\Delta _{c} f)=\sigma (f)\).

In [22], Zhang and Chen showered that the condition \(\lambda (\frac{1}{f} )<\sigma (f)\) in Theorem D cannot be omitted. Moreover, they obtained the following theorem.

Theorem E

([22])

Let f be a finite order meromorphic function, and let \(c\in {C}\backslash \{0 \}\) be a constant such that \(f(z+c)\not \equiv f(z)\). If \(f(z)\) has two Borel exceptional values, then \(\tau (\Delta _{c} f)=\sigma (f)\).

In [19], Yi and Yang have proved the following theorem.

Theorem F

([19])

Let f be a transcendental meromorphic function in C with a positive order. If f has two distinct Borel exceptional values, say \(a_{1}\) and \(a_{2}\), then the order of f is a positive integer orand \(\sigma (f)=\mu (f)\), \(\delta (a_{1}, f)=\delta (a_{2}, f)=1\).

By Theorem F, we can derive that the order of f in Theorems D and E is a positive integer. Is it necessary to ask if the order of f is an integer?, i.e., Can we get similar results as those in Theorems B, C, D, and E if the order of f is not a positive integer? The main purpose of this paper is to study this question. In fact, we shall prove the following theorems.

Theorem 1.1

Let f be a transcendental meromorphic function of finite order in the plane. Suppose that \(c\in {C}\setminus \{0\}\) such that \(\Delta _{c} f\not \equiv 0\). If there is \(a\in {C}\) with \(\delta (a, f)>0\) and \(\delta (\infty , f)=1\), then \(\Delta _{c} f\) have infinitely many fixed points and \(\tau ( \Delta _{c} f)=\sigma (f)\).

Theorem 1.2

Let f be a transcendental meromorphic function of finite order in the plane. Suppose that \(c\in {C}\setminus \{0\}\) such that \(\Delta _{c} f\not \equiv 0\). If \(\delta (\infty , f)=1\), \(\delta (0, f)=1\), then

$$ T(r, \Delta _{c} f)\sim T(r, f)\sim N \biggl(r, \frac{1}{(\Delta _{c} f)-z} \biggr), $$

as \(r\rightarrow \infty \), \(r\notin E \), where E is a possible exception set of r with finite logarithmic measure.

Let \(f(z)=\frac{e^{z}}{z}\), then \(N(r, f)=\log r=S(r, f)\), \(N (r, \frac{1}{f} )=0\) and \(\Delta _{c} f=\frac{(e^{c}-1)z-1}{z(z+c)}e ^{z}\not \equiv 0\). By the second fundmental theorem, we have

$$ T(r, \Delta _{c} f)\sim T(r, f)\sim N \biggl(r, \frac{1}{(\Delta _{c} f)-z} \biggr)\quad (r\rightarrow \infty ), $$

and \(\tau (\Delta _{c} f)=\sigma (f)\).

Proof of Theorems 1.1 and 1.2

Lemma 2.1

([6])

Let \(f(z)\) be a finite order meromorphic function, then, for each \(k \in {N}\), \(\sigma (\Delta _{c}^{k} f)\leq \sigma (f)\).

Lemma 2.2

([9])

Let f be a transcendental meromorphic function of finite order. Then, for any positive integer n, we have

$$ m \biggl(r, \frac{\Delta _{c}^{n}f(z)}{f(z)} \biggr)=S(r, f). $$

Lemma 2.3

Let f be a transcendental meromorphic function of finite order. Suppose that \(c\in {C}\setminus \{0\}\) such that \(\Delta _{c} f\not \equiv 0\) and \(\delta (0, f)>0\). Then \(\Delta _{c} f\) is a transcendental and meromorphic function of finite order.

Proof

From Lemma 2.1, we know that \(\sigma (\Delta _{c} f)\leq \sigma (f)<+\infty \). If \(\Delta _{c} f\) is not a transcendental meromorphic function, then there is a rational function \(R(z)\) such that \(R(z)\Delta _{c} f\equiv 1\), i.e.,

$$ \frac{1}{f}\equiv R(z)\frac{\Delta _{c} f}{f}. $$

Applying Lemma 2.2 and noticing that \(f(z)\) is transcendental, we have

$$\begin{aligned} m \biggl(r, \frac{1}{f} \biggr)\leq m\bigl(r, R(z)\bigr)+m \biggl(r, \frac{\Delta _{c} f}{f} \biggr)=S(r, f). \end{aligned}$$

This contradicts \(\delta (0, f)>0\). Thus \(\Delta _{c} f\) is a transcendental and meromorphic function of finite order. □

Lemma 2.4

([11])

Let \(f(z)\) be a transcendental meromorphic function of finite order, then

$$ m \biggl(r,\frac{f(z+c)}{f} \biggr)=S(r, f). $$

Lemma 2.5

([14, 21])

Let f be a transcendental meromorphic function of finite order. Then

$$\begin{aligned}& N\bigl(r, f(z+c)\bigr)= N(r, f)+S(r, f), \\& T\bigl(r, f(z+c)\bigr)= T(r, f)+S(r, f). \end{aligned}$$

Lemma 2.6

Let f be a finite order transcendental meromorphic function. Suppose that \(c\in {C}\setminus \{0\}\) such that \(\Delta _{c} f\not \equiv 0\). If \(\delta (0, f)>0\), then

$$ T(r, f)\leq 4N(r, f)+N \biggl(r, \frac{1}{f} \biggr)+N \biggl(r, \frac{1}{( \Delta _{c}f)-z} \biggr)+S(r, f). $$

Proof

By Lemma 2.3, we know that \(\Delta _{c} f\) is a transcendental meromorphic function. Put \(F=\Delta _{c}f\), then there is \(\eta \in {C}\setminus \{0\}\) such that \(z\Delta _{\eta }F-\eta F(z)\not \equiv 0\). If not, then

$$ \frac{F(z)}{z}\equiv \frac{F(z+\eta )}{z+\eta } $$

holds for any \(\eta \in {C}\setminus \{0\}\). Hence \(\frac{F(z)}{z}\) is a constant, which contradicts \(F=\Delta _{c}f\) is a transcendental meromorphic function. Hence there is \(\eta \in \setminus \{0\}\) such that \(z\Delta _{\eta }F-\eta F(z)\not \equiv 0\), i.e.,

$$\begin{aligned}& z\Delta _{\eta }F-\eta F(z) \\& \quad =z\Delta _{\eta }(\Delta _{c}f)-\eta \Delta _{c}f \\& \quad = z\Delta _{\eta }\bigl((\Delta _{c}f)-z\bigr)-\eta \bigl((\Delta _{c}f)-z\bigr) \\& \quad =zf(z+c+\eta )-zf(z+\eta )-(z+\eta )f(z+c)+(z+\eta )f(z)\not \equiv 0. \end{aligned}$$
(1)

Noticing

$$\begin{aligned} \frac{1}{f}=\frac{\Delta _{c}f}{zf}-\frac{z\Delta _{\eta }(\Delta _{c}f)- \eta \Delta _{c}f}{zf} \frac{(\Delta _{c}f)-z}{z\Delta _{\eta }(\Delta _{c}f)-\eta \Delta _{c}f}. \end{aligned}$$
(2)

Combining (1), (2) and Lemmas 2.2, 2.4, we can get

$$\begin{aligned}& m \biggl(r, \frac{1}{f} \biggr) \\& \quad \leq m \biggl(r, \frac{\Delta _{c}f}{zf} \biggr)+m \biggl(r, \frac{z \Delta _{\eta }(\Delta _{c}f)-\eta \Delta _{c}f}{zf} \biggr) \\& \qquad {} +m \biggl(r, \frac{(\Delta _{c}f)-z}{z\Delta _{\eta }(\Delta _{c}f)-\eta \Delta _{c}f} \biggr)+\log 2 \\& \quad \leq m \biggl(r, \frac{\Delta _{c}f}{f} \biggr)+ m \biggl(r, \frac{f(z+c+ \eta )}{f} \biggr)+ m \biggl(r, \frac{f(z+c)}{f} \biggr) \\& \qquad {} +m \biggl(r, \frac{f(z+\eta )}{f} \biggr)+m \biggl(r, \frac{( \Delta _{c}f)-z}{z\Delta _{\eta }(\Delta _{c}f)-\eta \Delta _{c}f} \biggr)+O( \log r) \\& \quad = m \biggl(r, \frac{(\Delta _{c}f)-z}{z\Delta _{\eta }(\Delta _{c}f)- \eta \Delta _{c}f} \biggr)+S(r, f). \end{aligned}$$

Applying the first fundamental theorem of Nevanlinna theory, we have

$$\begin{aligned} T(r, f)\leq N \biggl(r, \frac{1}{f} \biggr)+m \biggl(r, \frac{(\Delta _{c}f)-z}{z\Delta _{\eta }(\Delta _{c}f)-\eta \Delta _{c}f} \biggr)+S(r, f), \end{aligned}$$
(3)

and we get

$$\begin{aligned}& m \biggl(r, \frac{(\Delta _{c}f)-z}{z\Delta _{\eta }(\Delta _{c}f)- \eta \Delta _{c}f} \biggr) \\& \quad \leq m \biggl(r, \frac{z\Delta _{\eta }(\Delta _{c}f)-\eta \Delta _{c}f}{(\Delta _{c}f)-z} \biggr)+N \biggl(r, \frac{z\Delta _{\eta }( \Delta _{c}f)-\eta \Delta _{c}f}{(\Delta _{c}f)-z} \biggr)+O(1). \end{aligned}$$
(4)

It follows from (1) that

$$\begin{aligned} m \biggl(r, \frac{z\Delta _{\eta }(\Delta _{c}f)-\eta \Delta _{c}f}{( \Delta _{c}f)-z} \biggr)\leq m \biggl(r, \frac{\Delta _{\eta }((\Delta _{c}f)-z)}{(\Delta _{c}f)-z} \biggr)+S(r, f). \end{aligned}$$
(5)

Applying Lemma 2.3 and Lemma 2.5, we know that \((\Delta _{c}f)-z\) is a transcendental meromorphic function of finite order and

$$ T\bigl(r, (\Delta _{c}f)-z\bigr)\leq 2T(r, f)+S(r, f). $$

Therefore,

$$\begin{aligned} S\bigl(r, (\Delta _{c}f)-z\bigr)=S(r, f). \end{aligned}$$
(6)

It follows from Lemma 2.2 and (6) that

$$\begin{aligned} m \biggl(r,\frac{z\Delta _{\eta }(\Delta _{c}f)-\eta \Delta _{c}f}{(\Delta _{c}f)-z} \biggr)=S(r, f). \end{aligned}$$
(7)

By Lemma 2.5 and (1), we derive

$$\begin{aligned}& N \biggl(r, \frac{z\Delta _{\eta }(\Delta _{c}f)-\eta \Delta _{c}f}{(\Delta _{c}f)-z} \biggr) \\& \quad \leq N\bigl(r, z\Delta _{\eta }(\Delta _{c}f)-\eta \Delta _{c}f\bigr)+N \biggl(r, \frac{1}{(\Delta _{c}f)-z} \biggr) \\& \quad \leq N \biggl(r, \frac{1}{(\Delta _{c}f)-z} \biggr)+4N(r, f)+S(r, f). \end{aligned}$$
(8)

Combining (3)–(5) and (7)–(8), we have

$$\begin{aligned} T(r, f)\leq 4N(r, f)+N \biggl(r, \frac{1}{f} \biggr)+N \biggl(r, \frac{1}{( \Delta _{c}f)-z} \biggr)+S(r, f). \end{aligned}$$

 □

Proof of Theorem 1.1

Denoting \(g=f-a\), by Lemma 2.6, we have

$$\begin{aligned} T(r, f) =&T(r, g)+O(1) \\ \leq & 4N(r, g)+N \biggl(r, \frac{1}{g} \biggr)+N \biggl(r, \frac{1}{( \Delta _{c}g)-z} \biggr)+S(r, g) \\ =& 4N(r, f)+N \biggl(r, \frac{1}{f-a} \biggr)+N \biggl(r, \frac{1}{( \Delta _{c} f)-z} \biggr)+S(r, f). \end{aligned}$$
(9)

Since \(\delta (a, f)>0\) and \(\delta (\infty , f)=1\), then there is a positive number \(\theta <1\) such that

$$\begin{aligned}& N\biggl(r, \frac{1}{f-a}\biggr)< \theta T(r, f), \end{aligned}$$
(10)
$$\begin{aligned}& N(r, f)\leq o(1)T(r, f). \end{aligned}$$
(11)

Combining (9)–(11), we can get

$$\begin{aligned} \bigl(1-o(1)-\theta \bigr)T(r, f)\leq N \biggl(r, \frac{1}{(\Delta _{c}f)-z} \biggr). \end{aligned}$$
(12)

Note that f is transcendental, we can get that \(\Delta _{c} f\) has infinitely many fixed points and \(\tau (\Delta _{c} f)=\sigma (f)\) from (12).

Proof of Theorem 1.2

Since

$$\begin{aligned} m \biggl(r, \frac{1}{f} \biggr) =& m \biggl(r, \frac{\Delta _{c}f}{f}\frac{1}{ \Delta _{c}f} \biggr)\leq m \biggl(r,\frac{\Delta _{c}f}{f} \biggr)+ m \biggl(r, \frac{1}{\Delta _{c}f} \biggr) \\ \leq & m \biggl(r, \frac{1}{\Delta _{c}f} \biggr)+S(r, f). \end{aligned}$$
(13)

By the first fundamental theorem of Nevanlinna theory and (13), we can get

$$\begin{aligned} T(r, f)\leq T(r, \Delta _{c}f)+N \biggl(r, \frac{1}{f} \biggr)+S(r, f). \end{aligned}$$
(14)

Hence

$$\begin{aligned} 1 \leq &\liminf_{r\rightarrow \infty }\frac{T(r, \Delta _{c}f)}{T(r, f)}+\limsup _{r\rightarrow \infty }\frac{N (r, \frac{1}{f} )}{T(r ,f)} \\ =&\liminf_{r\rightarrow \infty }\frac{T(r, \Delta _{c}f)}{T(r, f)}+\bigl(1-\delta (0, f)\bigr) \\ =&\liminf_{r\rightarrow \infty }\frac{T(r, \Delta _{c}f)}{T(r, f)}. \end{aligned}$$
(15)

On the other hand, we have

$$\begin{aligned} T(r, \Delta _{c}f) =& m(r, \Delta _{c}f)+N(r, \Delta _{c}f) \\ =& m \biggl(r, \frac{f\Delta _{c}f}{f} \biggr)+N(r, \Delta _{c}f) \\ \leq & m \biggl(r, \frac{\Delta _{c}f}{f} \biggr)+m(r, f)+N(r, f)+N\bigl(r, f(z+c) \bigr). \end{aligned}$$

It follows from Lemma 2.2 and Lemma 2.5 that

$$\begin{aligned} T(r, \Delta _{c}f)\leq T(r, f)+N(r, f)+S(r, f). \end{aligned}$$

As \(\delta (\infty , f)=1\), so

$$\begin{aligned} \limsup_{r\rightarrow \infty }\frac{T(r, \Delta _{c}f)}{T(r, f)}\leq 1+ \limsup _{r\rightarrow \infty }\frac{N(r, f)}{T(r, f)}=1. \end{aligned}$$

Therefore

$$\begin{aligned} \lim_{r\rightarrow +\infty }\frac{T(r, \Delta _{c}f)}{T(r, f)}=1. \end{aligned}$$
(16)

Since \(\delta (0, f)=1\) and \(\delta (\infty , f)=1\), then

$$\begin{aligned} N \biggl(r, \frac{1}{f} \biggr)=S(r,f), N(r, f)=S(r, f). \end{aligned}$$
(17)

By (17) and Lemma 2.6, we have

$$\begin{aligned} T(r, f) \leq & N \biggl(r, \frac{1}{(\Delta _{c}f)-z} \biggr)+S(r, f) \\ \leq & T \biggl(r, \frac{1}{(\Delta _{c}f)-z} \biggr)+S(r, f) \\ \leq & T (r, \Delta _{c}f )+S(r, f). \end{aligned}$$
(18)

Combining (16) and (18) implies

$$ T(r, \Delta _{c} f)\sim T(r, f)\sim N \biggl(r, \frac{1}{(\Delta _{c} f)-z} \biggr), $$

as \(r\rightarrow \infty \).

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This work was supported by the Natural Science Foundation of Hubei Provincial Department of Education(Grant No. D20182801) and by Hubei Key Laboratory of Applied Mathematics (Hubei University).

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All authors drafted the manuscript, read and approved the final manuscript. All authors contributed equally.

Correspondence to Zhaojun Wu.

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Wu, Z., Wu, J. Fixed points of differences of meromorphic functions. Adv Differ Equ 2019, 453 (2019) doi:10.1186/s13662-019-2386-8

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Keywords

  • Difference operator
  • Fixed point
  • Borel exceptional values
  • Deficiency