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.

1 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 or ∞ and $\sigma (f)=\mu (f)$, $\delta (a_{1}, f)=\delta (a_{2}, f)=1$.

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

2 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

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

 □

2.1 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).

2.2 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 $r\rightarrow \infty$.