Theory and Modern Applications

# Fixed points of differences of meromorphic functions

## 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 . 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  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 ) 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  have proved the following theorems.

### Theorem B

()

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

()

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  obtains the following theorem.

### Theorem D

()

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 , 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

()

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 , Yi and Yang have proved the following theorem.

### Theorem F

()

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

()

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

()

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

()

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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Not applicable.

## Funding

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.

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Correspondence to Zhaojun Wu.

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