Sharing values of q-difference-differential polynomials

Abstract

This paper is devoted to the uniqueness of q-difference-differential polynomials of different types. Using the idea of common zeros and common poles (Chin. Ann. Math., Ser. A 35:675–684, 2014), we improve the conditions of the former theorems and obtain some new results on the uniqueness of q-difference-differential polynomials of meromorphic functions.

1 Introduction and main results

In this paper, a meromorphic function is assumed meromorphic in the whole complex plane. We assume that the reader is familiar with the basic symbols and fundamental results of Nevanlinna theory; see, for example, [2, 3, 10]. We say that two meromorphic functions f and g share a point a CM (IM) if $f(z)-a$ and $g(z)-a$ have the same zeros counting multiplicities (ignoring multiplicities). The logarithmic density of the set E is defined by

$$\begin{aligned} \limsup_{r\rightarrow\infty}\frac{1}{\log r} \int_{[1,r]\cap E}\frac{1}{t}\,dt. \end{aligned}$$

Denote by $S(r,f)$ a quantity of $o(T(r,f))$ as $r\rightarrow\infty$ outside a possible exceptional set E of logarithmic density 0.

Yang and Hua [9] obtained an important result on the uniqueness when the differential polynomials $f^{n}f'$ and $g^{n}g'$ share one value CM. Recently, many studies are devoted to the uniqueness of difference and q-difference polynomials; see [4–6, 11–14]. Zhang [12] obtained the following result.

Theorem A

([12])

Let$f(z)$and$g(z)$be transcendental entire functions of zero order, and letn, m, dbe positive integers. If$n \geq m+5d$and$f(z)^{n}(f(z)^{m}-1)\prod_{i=1}^{d}f(q_{i}z)$and$g(z)^{n} (g(z)^{m}-1)\prod_{i=1}^{d}g(q_{i}z)$share 1 CM, then$f(z)\equiv tg(z)$, $t^{n+d}=t^{m}=1$.

Liu, Liu, and Cao [4] and Zhang and Korhonen [11] obtained the following two theorems.

Theorem B

([4, Theorem 1.5])

Let$f(z)$and$g(z)$be transcendental zero-order entire functions, and letmbe a positive integer. If$n \geq m+5$and$f(z)^{n}(f(z)^{m}-a)f(qz+c)$and$g(z)^{n}(g(z)^{m}-a)g(qz+c)$share a nonzero polynomial$p(z)$CM, then$f(z)\equiv g(z)$.

Theorem C

([11, Theorem 5.1])

Let$f(z)$and$g(z)$be transcendental zero-order meromorphic functions. If$n\geq8$and$f(z)^{n}f(qz)$and$g(z)^{n}g(qz)$share 1 and ∞ CM, then$f(z)\equiv tg(z)$, $t^{n+1}=1$.

Zhao and Zhang [13] proved the following theorem.

Theorem D

([13, Theorem 1.4])

Let$f(z)$and$g(z)$be transcendental zero-order entire functions, and letkbe a positive integer. If$n \geq2k+6$and$(f(z)^{n}f(qz+c))^{(k)}$and$(g(z)^{n}g(qz+c))^{(k)}$share 1 CM, then$f(z)\equiv tg(z)$, where$t^{n+1}=1$.

Wang and Ye [8] improved the conditions of Theorems B and C to $n \geq m+4$ and $n \geq6$, respectively, by using the idea of common zeros and common poles. Here we give the main idea of common zeros and common poles. Let f, g be two nonconstant meromorphic functions. Denote by $\overline{n}_{0}(r)$ or $\overline {n}_{1}(r)$ the numbers of common zeros or poles of fg and g, ignoring multiplicities. Let p, q be positive integers. We assume that the Laurent series of f and g at $z_{0}$ are as follows:

$$f(z)=\frac{1}{(z-z_{0})^{p}}f_{1}(z), \qquad g(z)=(z-z_{0})^{q}g_{1}(z), $$

where $f_{1}(z)$ and $g_{1}(z)$ are analytic functions at $z_{0}$, and $f_{1}(z_{0})\neq0$, $g_{1}(z_{0})\neq0$; the other cases can be discussed in a similar way. So $z_{0}$ is a zero of $g(z)$ with multiplicities q. If $q>p$, then $z_{0}$ is a zero of $f(z)g(z)$ with multiplicity $q-p$, and thus the contribution to $\overline{n}_{0}(r)$ is 1 at $z_{0}$. If $q\leq p$, then $z_{0}$ is a pole of $f(z)g(z)$ with multiplicity $p-q$ or an analytic point of $f(z)g(z)$, and thus the contribution to $\overline{n}_{0}(r)$ is 0 at $z_{0}$. A similar method can be discussed for $\overline{n}_{1}(r)$. As usual, denote by $\overline{N}_{0}(r)$ or $\overline{N}_{1}(r)$ the counting functions of the common zeros or poles of fg and g, ignoring multiplicities. Thus we have $\overline{N}(r,\frac{1}{fg})\leq\overline{N}(r,\frac {1}{f})+\overline{N}_{0}(r)$ and $\overline{N}(r,fg)\leq\overline{N}(r,f)+\overline{N}_{1}(r)$. In this paper, we continue to consider the uniqueness of q-difference-differential polynomials. Firstly, we improve the condition $n\geq m+5d$ in Theorem A to $n\geq m+d+3$ in Theorem 1.1. Set

$$\begin{aligned} L(z,f)=\prod_{i=1}^{d}f(q_{i}z+c_{i}), \end{aligned}$$

where $c_{i}$ and $q_{i}\neq0$ ($i=1,\ldots, d$) are constants, and d is a positive integer.

Theorem 1.1

Let$f(z)$and$g(z)$be transcendental zero-order entire functions, and letmbe a positive integer. If$n \geq m+d+3$and$f(z)^{n}(f(z)^{m}-1)L(z,f)$and$g(z)^{n}(g(z)^{m}-1)L(z,g)$share 1 CM, then$f(z)\equiv c_{1}g(z)$, $c_{1}^{n+d}=c_{1}^{m}=1$.

In the following theorem, we improve the condition $n\geq2k+6$ in Theorem D to $n\geq6$.

Theorem 1.2

Let$f(z)$and$g(z)$be transcendental zero-order meromorphic functions, and letkbe a positive integer. If$n \geq6$and$(f(z)^{n}f(qz+c))^{(k)}$and$(g(z)^{n}g(qz+c))^{(k)}$share 1 and ∞ CM, then$f(z)\equiv c_{2}g(z)$, $c_{2}^{n+1}=1$.

We also consider the following theorems for q-difference polynomials of different types. The following theorem is also an improvement of Theorem C.

Theorem 1.3

Let$f(z)$and$g(z)$be transcendental zero-order meromorphic functions, and letsbe a positive integer. If$n\geq(d+1)s+4$and$f(z)^{n}L(z,f)^{s}$and$g(z)^{n}L(z,g)^{s}$share 1 and ∞ CM, then$f(z)\equiv c_{3}g(z)$, $c_{3}^{n+sd}=1$.

Theorem 1.4

Let$f(z)$and$g(z)$be transcendental zero-order meromorphic functions, $q, c\in\mathbb{C}$, and$q\neq0$. If$n\geq7$, and$f(z)^{n}(f(qz+c)-f(z))$and$g(z)^{n}(g(qz+c)-g(z))$share 1 and ∞ CM, then

$$f(z)^{n}\bigl(f(qz+c)-f(z)\bigr)= g(z)^{n} \bigl(g(qz+c)-g(z)\bigr). $$

If$\frac{g(z)}{g(qz+c)}$is transcendental with only finitely many zeros, then$f(z)\equiv c_{4}g(z)$, where$c_{4}^{n+1}=1$.

2 Lemmas

Combining [11, Theorem 1.1] and [1, Theorem 2.1], we easily get the following lemma.

Lemma 2.1

Let$f(z)$be a transcendental zero-order meromorphic function, $q, c\in \mathbb{C}$, and$q\neq0$. Then

$$\begin{aligned} T\bigl(r,f(qz+c)\bigr)=T(r,f)+S(r,f) \end{aligned}$$

on a set of logarithmic density 1.

Lemma 2.2

([7])

Let$f(z)$be a zero-order meromorphic function$q, c\in\mathbb{C}$, and$q\neq0$. Then

$$\begin{aligned} m \biggl(r,\frac{f(qz+c)}{f(z)} \biggr)=S(r,f) \end{aligned}$$

on a set of logarithmic density 1.

Lemma 2.3

Iffis a transcendental zero-order entire function, then

$$\begin{aligned} T\bigl(r, f(z)^{n}\bigl(f(z)^{m}-1\bigr)L(z,f) \bigr)=(m+n+d)T(r,f)+S(r,f) \end{aligned}$$

on a set of logarithmic density 1.

Proof

Set $F(z)=f(z)^{n}(f(z)^{m}-1)L(z,f)$. By Lemma 2.2 and the standard Valiron–Mohon’ko theorem, if f is a transcendental zero-order entire function, then

$$\begin{aligned} (n+m+d)T(r,f) =& T\bigl(r,f^{n+d}\bigl(f^{m}-1\bigr) \bigr)+S(r,f) \\ =& m\bigl(r,f^{n+d}\bigl(f^{m}-1\bigr)\bigr)+S(r,f) \\ \leq& m \biggl(r,\frac{f^{n+d}(f^{m}-1)}{f^{n}(f^{m}-1)L(z,f)} \biggr)+m(r,F)+S(r,f) \\ \leq& m \biggl(r,\frac{f^{d}}{L(z,f)} \biggr)+m(r,F)+S(r,f) \\ \leq& T(r,F)+S(r,f) \end{aligned}$$

on a set of logarithmic density 1. On the other hand, combining Lemma 2.1 with the fact that f is a transcendental zero-order function, we have

$$\begin{aligned} T(r,F) \leq& T\bigl(r,f^{n}\bigl(f^{m}-1\bigr)\bigr)+T \bigl(r,L(z,f)\bigr) \\ \leq& (n+m+d)T(r,f)+S(r,f) \end{aligned}$$

on a set of logarithmic density 1. □

3 Proofs of theorems

Proof of Theorem 1.1

Let $F(z)=f^{n}(f^{m}-1)L(z,f)$ and $G(z)=g^{n}(g^{m}-1)L(z,g)$. Since $F(z)$ and $G(z)$ share 1 and ∞ CM, we have that $\frac{F-1}{G-1}=B$, that is,

$$\begin{aligned} F=BG+1-B, \end{aligned}$$

(1)

where B is a nonzero constant.

If $B\neq1$, then from the second main theorem of Nevanlinna theory, Lemma 2.1, and Lemma 2.3 we obtain

$$\begin{aligned} (n+m+d)T(r,f) =&T(r,F)+S(r,f) \\ \leq& \overline{N}(r,F)+\overline{N} \biggl(r,\frac{1}{F} \biggr) + \overline{N} \biggl(r,\frac{1}{F-1+B} \biggr)+S(r,f) \\ \leq& \overline{N} \biggl(r,\frac{1}{f^{n}} \biggr)+\overline{N} \biggl(r,\frac{1}{f^{m}-1} \biggr) +\overline{N} \biggl(r, \frac{1}{L(z,f)} \biggr) +\overline{N} \biggl(r,\frac{1}{G} \biggr)+S(r,f) \\ \leq& (m+d+1) \bigl(T(r,f)+T(r,g)\bigr)+S(r,f)+S(r,g). \end{aligned}$$

(2)

Using the same method, we have

$$ (n+m+d)T(r,g)\leq(m+d+1) \bigl(T(r,f)+T(r,g) \bigr)+S(r,f)+S(r,g). $$

(3)

Combining (2) with (3), we have

$$\begin{aligned} (n-m-d-2) \bigl(T(r,f)+T(r,g)\bigr)\leq S(r,f)+S(r,g), \end{aligned}$$

which contradicts to $n\geq m+d+3$. Thus $B=1$, and from (1) we have

$$ f^{n}\bigl(f^{m}-1\bigr)L(z,f)=g^{n} \bigl(g^{m}-1\bigr)L(z,g). $$

(4)

Let $h(z)=\frac{f(z)}{g(z)}$. So $\frac{L(z,f)}{L(z,g)}=\prod_{i=1}^{d}\frac{f(q_{i}z+c_{i})}{g(q_{i}z+c_{i})} =\prod_{i=1}^{d}h(q_{i}z+c_{i})=L(z,h)$, and then (4) can be written as

$$ g^{m}\bigl(h^{n+m}L(z,h)-1 \bigr)=h^{n}L(z,h)-1. $$

(5)

Next, we prove that $h(z)\equiv c_{1}$ and $c_{1}^{n+d}=c_{1}^{m}=1$, where $c_{1}$ is a constant. Assume on the contrary that $h(z)$ is not a constant. From Lemma 2.1 we have

$$T\bigl(r,h^{n+m}L(z,h)\bigr)\leq(n+m+d)T(r,h)+S(r,h). $$

We also have

$$\begin{aligned} (n+m)T(r,h) =&T\bigl(r,h^{n+m}\bigr) \\ \leq& T \biggl(r,\frac{1}{h^{n+m}L(z,h)} \biggr) +T\bigl(r,L(z,h)\bigr))+O(1) \\ \leq& T \bigl(r,h^{n+m}L(z,h) \bigr) +dT(r,h)+S(r,h). \end{aligned}$$

Since $n\geq m+d+3$, from the last two inequalities it follows that $S(r,h^{n+m}L(z,h))=S(r,h)$. Denote by $\overline{N}_{1}(r)$ the counting function of the common poles of $h^{n+m}L(z,h)$ and $L(z,h)$ ignoring multiplicities. Then

$$\overline{N}\bigl(r,h^{n+m}L(z,h)\bigr)\leq\overline{N}(r,h)+ \overline{N}_{1}(r). $$

Here we should remark that the poles of $L(z,h)$ may be the zeros of h and the zeros of $L(z,h)$ may be the poles of h. Similarly, denote by $\overline{N}_{0}(r)$ the counting function of the common zeros of $h^{n+m}L(z,h)$ and $L(z,h)$ ignoring multiplicities, and then

$$\overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)} \biggr)\leq\overline {N}\biggl(r, \frac{1}{h}\biggr)+\overline{N}_{0}(r). $$

From the second main theorem of Nevanlinna theory and the last two inequalities we have

$$\begin{aligned} T\bigl(r,h^{n+m}L(z,h)\bigr)&\leq\overline{N} \bigl(r,h^{n+m}L(z,h)\bigr) +\overline{N} \biggl(r, \frac{1}{h^{n+m}L(z,h)} \biggr) \\ &\quad+\overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)-1} \biggr)+{S \bigl(r,h^{n+m}L(z,h)\bigr)} \\ &\leq\overline{N}(r,h)+\overline{N}_{1}(r)+\overline{N} \biggl(r, \frac {1}{h} \biggr) +\overline{N}_{0}(r) \\ &\quad+\overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)-1} \biggr)+S(r,h). \end{aligned}$$

(6)

On the other hand,

$$\begin{aligned}& (n+m)m(r,h)=m\bigl(r,h^{n+m}\bigr)\leq m \bigl(r,h^{n+m}L(z,h)\bigr)+m \biggl(r,\frac {1}{L(z,h)} \biggr)+O(1), \end{aligned}$$

(7)

$$\begin{aligned}& \begin{aligned}[b](n+m)N(r,h)&=N\bigl(r,h^{n+m}\bigr)= N \biggl( \frac{h^{n+m}L(z,h)}{L(z,h)} \biggr) \\ &\leq N\bigl(r,h^{n+m}L(z,h)\bigr)+N \biggl(r,\frac{1}{L(z,h)} \biggr)-\overline {N}_{1}(r)-\overline{N}_{0}(r).\end{aligned} \end{aligned}$$

(8)

From (7) and (8) we get

$$\begin{aligned} (n+m)T(r,h)&\leq T\bigl(r,h^{n+m}L(z,h)\bigr)+T \biggl(r,\frac{1}{L(z,h)} \biggr) \\ &\quad-\overline{N}_{1}(r)-\overline{N}_{0}(r)+O(1). \end{aligned}$$

(9)

From (6) and (9) we get

$$\begin{aligned} (n+m)T(r,h) \leq& \overline{N}(r,h)+\overline{N} \biggl(r, \frac {1}{h} \biggr)+\overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)-1} \biggr) \\ &{}+T \biggl(r,\frac{1}{L(z,h)} \biggr)+S(r,h) \\ \leq& (d+2)T(r,h)+\overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)-1} \biggr)+S(r,h). \end{aligned}$$

(10)

Since $n\geq m+d+3$, the value 1 is not the Picard exceptional value of $h^{n+m}L(z,h)$ from (10). Furthermore, we prove $h^{n+m}L(z,h)\equiv1$ and $h(z)\equiv c_{1}$ is a nonzero constant. If $h^{n+m}L(z,h)\not\equiv1$,then since 1 is not the Picard exceptional value of $h^{n+m}L(z,h)$, there exists a point $z_{0}$ satisfying $h(z_{0})^{n+m}L(z,h(z_{0}))=1$. From the condition that $g(z)$ is an entire function and (5) we have $h(z_{0})^{m}=1$ and

$$ \overline{N} \biggl(r,\frac{1}{h^{n+m}L(z,h)-1} \biggr) \leq \overline{N} \biggl(r,\frac{1}{h^{m}-1} \biggr) \leq mT(r,h)+S(r,h). $$

(11)

Substituting (11) into (10), we get a contradiction to $n\geq m+d+3$, so $h^{n+m}L(z,h)\equiv1$, that is, $h^{n+m}=\frac {1}{L(z,h)}$. From Lemma 2.1 we have

$$\begin{aligned} (n+m)T(r,h)=T\bigl(r,L(z,h)\bigr)\leq dT(r,h)+S(r,h), \end{aligned}$$

which also contradicts to $n\geq m+d+3$, so $h(z)\equiv c_{1}$, where $c_{1}$ is a nonzero constant, that is, $f(z)\equiv c_{1}g(z)$, and $L(z,h)=c_{1}^{d}$. From (5) we can get $c_{1}^{m}=c_{1}^{n+d}=1$. Thus the theorem is proved. □

Proof of Theorem 1.2

Let $F(z)=f(z)^{n}f(qz+c)$ and $G(z)=g(z)^{n}g(qz+c)$. From the condition in Theorem 1.2 we know that $F^{(k)}$ and $G^{(k)}$ share 1 and ∞ CM, so

$$\begin{aligned} \frac{F^{(k)}-1}{G^{(k)}-1}=C, \end{aligned}$$

where C is a nonzero constant, that is,

$$ F^{(k)}=CG^{(k)}-C+1. $$

(12)

Integrating both sides of (12), we have

$$ F=CG+\frac{1-C}{k!}z^{k}+p_{1}(z), $$

(13)

where $p_{1}(z)$ is a polynomial of degree at most $k-1$. Denote $\frac {1-C}{k!}z^{k}+p_{1}(z)$ by $p(z)$. If $p(z)\not\equiv0$, then by the second main theorem of Nevanlinna theory, Lemma 2.1, and (13) we obtain

$$\begin{aligned} T(r,F) \leq& \overline{N}(r,F)+\overline{N} \biggl(r, \frac{1}{F} \biggr) +\overline{N} \biggl(r,\frac{1}{F-p} \biggr)+S(r,f) \\ \leq& \overline{N}(r,f)+\overline{N}_{1}(r)+\overline{N} \biggl(r, \frac {1}{f} \biggr) +\overline{N}_{0}(r)+\overline{N} \biggl(r,\frac{1}{G} \biggr)+S(r,f) \\ \leq& 2T(r,f)+2T(r,g)+\overline{N}_{1}(r)+\overline{N}_{0}(r)+S(r,f)+S(r,g), \end{aligned}$$

(14)

where $\overline{N}_{0}(r)$ denotes the counting function ignoring multiplicities of the common zeros of $F(z)$ and $f(qz+c)$, and $\overline{N}_{1}(r)$ denotes the counting function ignoring multiplicities of the common poles of $F(z)$ and $f(qz+c)$. On the other hand,

$$\begin{aligned}& nm(r,f)=m\bigl(r,f^{n}\bigr)\leq m(r,F)+m \biggl(r, \frac{1}{f(qz+c)} \biggr)+O(1). \end{aligned}$$

(15)

$$\begin{aligned}& \begin{aligned}[b]nN(r,f)&=N\bigl(r,f^{n}\bigr)=N \biggl( \frac{F(z)}{f(qz+c)} \biggr) \\ &\leq N(r,F)+N \biggl(r,\frac{1}{f(qz+c)} \biggr)-\overline{N}_{1}(r)- \overline {N}_{0}(r).\end{aligned} \end{aligned}$$

(16)

From (15), (16), and Lemma 2.1 we have

$$ (n-1)T(r,f)\leq T(r,F)-\overline{N}_{1}(r)- \overline{N}_{0}(r)+O(1). $$

(17)

Substituting (14) into (17), we obtain

$$ (n-3)T(r,f)\leq2T(r,g)+S(r,f)+S(r,g). $$

(18)

Using the same method, we also get

$$ (n-3)T(r,g)\leq2T(r,f)+S(r,f)+S(r,g). $$

(19)

Combining (18) with (19), we have

$$\begin{aligned} (n-5) \bigl(T(r,g)+T(r,f)\bigr)\leq S(r,f)+S(r,g), \end{aligned}$$

which contradicts to $n\geq6$, and thus $p(z)\equiv0$. Since the degree of $p_{1}(z)$ is at most $k-1$, we have $C=1$ and $p_{1}(z)\equiv 0$. From (13) we get

$$\begin{aligned} f^{n}f(qz+c)=g^{n}g(qz+c). \end{aligned}$$

Assume that $h(z)=\frac{f(z)}{g(z)}$. Then $h(qz+c)h(z)^{n}=1$, that is, $h(z)^{n}=\frac{1}{h(qz+c)}$, and from Lemma 2.1 we have

$$\begin{aligned} nT(r,h)=T\bigl(r,h(qz+c)\bigr)\leq T(r,h)+S(r,h), \end{aligned}$$

which also contradicts to $n\geq6$, so $h(z)$ is a nonzero constant, say $c_{2}$. So $f(z)\equiv c_{2}g(z)$, and $c_{2}^{n+1}=1$. Thus the theorem is proved. □

Proof of Theorem 1.3

Since $f(z)$ and $g(z)$ are transcendental zero-order meromorphic functions and $f(z)^{n}L(z,f)^{s}$ and $g(z)^{n}L(z,g)^{s}$ share 1 and ∞ CM, we have

$$ \frac{f(z)^{n}L(z,f)^{s}-1}{g(z)^{n}L(z,g)^{s}-1}=E, $$

(20)

where E is a nonzero constant. Then (20) can be rewritten as

$$ Eg(z)^{n}L(z,g)^{s}=f(z)^{n}L(z,f)^{s}-1+E. $$

(21)

Let $F(z)=f(z)^{n}L(z,f)^{s}$ and $G(z)=g(z)^{n}L(z,g)^{s}$. We affirm that $E=1$. On the contrary, assume that $E\neq1$. Using the second main theorem of Nevanlinna theory and Lemma 2.1 for (21), we get

$$\begin{aligned} T(r,F) \leq& \overline{N}(r,F)+\overline{N} \biggl(r, \frac{1}{F} \biggr)+\overline{N} \biggl(r,\frac{1}{F-1+E} \biggr) +S(r,f) \\ \leq& \overline{N}(r,f)+\overline{N}_{1}(r)+\overline{N} \biggl(r, \frac {1}{f} \biggr) +\overline{N}_{0}(r)+\overline{N} \biggl(r,\frac{1}{G} \biggr)+S(r,f) \\ \leq& 2T(r,f)+\overline{N} \biggl(r,\frac{1}{g} \biggr)+\overline{N} \biggl(r,\frac{1}{L(z,g)} \biggr) +\overline{N}_{1}(r) + \overline{N}_{0}(r)+S(r,f) \\ \leq& 2T(r,f)+(d+1)T(r,g)+\overline{N}_{1}(r)+\overline {N}_{0}(r)+S(r,f)+S(r,g), \end{aligned}$$

(22)

where $\overline{N}_{0}(r)$ denotes the counting function ignoring multiplicities of the common zeros of $F(z)$ and $L(z,f)$, and $\overline{N}_{1}(r)$ denotes the counting function ignoring multiplicities of the common poles of $F(z)$ and $L(z,f)$.

Since $F(z)=f(z)^{n}L(z,f)^{s}$, we have

$$\begin{aligned}& nm(r,f)=m\bigl(r,f^{n}\bigr)\leq m(r,F)+sm \biggl(r, \frac{1}{L(z,f)} \biggr)+O(1), \\& \begin{aligned}nN(r,f)&=N\bigl(r,f^{n}\bigr)=N \biggl(r,F(z) \biggl[ \frac{1}{L(z,f)} \biggr]^{s} \biggr) \\ &\leq N(r,F)+sN \biggl(r,\frac{1}{L(z,f)} \biggr)-\overline{N}_{1}(r)- \overline{N}_{0}(r).\end{aligned} \end{aligned}$$

So

$$ \begin{aligned} nT(r,f)&\leq T(r,F)+sT \biggl(r, \frac{1}{L(z,f)} \biggr)-\overline {N}_{1}(r)- \overline{N}_{0}(r)+O(1) \\ &\leq T(r,F)+dsT(r,f)-\overline{N}_{1}(r)-\overline{N}_{0}(r)+S(r,f). \end{aligned} $$

(23)

Substituting (22) into (23), we obtain

$$ (n-ds-2)T(r,f)\leq(d+1)T(r,g)+S(r,f)+S(r,g). $$

(24)

Using the same method, we can get

$$ (n-ds-2)T(r,g)\leq(d+1)T(r,f)+S(r,f)+S(r,g). $$

(25)

Combining (24) with (25), it follows

$$\begin{aligned} (n-d-ds-3) \bigl(T(r,g)+T(r,f)\bigr)\leq S(r,f)+S(r,g), \end{aligned}$$

which contradicts to $n\geq(d+1)s+4$, and thus $E=1$. From (21) we get

$$\begin{aligned} f(z)^{n}L(z,f)^{s}=g(z)^{n}L(z,g)^{s}. \end{aligned}$$

Let $h(z)=\frac{f(z)}{g(z)}$. So $\frac{L(z,f)}{L(z,g)}=L(z,h)$. Then

$$ h(z)^{n}\bigl[L(z,h)\bigr]^{s}=1. $$

(26)

So $nT(r,h)=sT(r,L(z,h))\leq sdT(r,h)+S(r,h)$. Since $n\geq(d+1)s+4$, $h(z)$ must be a constant, say $c_{3}$, that is, $f(z)\equiv c_{3}g(z)$. Then from (26) it follows that $c_{3}^{n+sd}=1$. □

Proof of Theorem 1.4

Letting $L(z,f)=f(qz+c)-f(z)$ and $L(z,g)=g(qz+c)-g(z)$, $s=1$ in Theorem 1.3, we obtain that if $n\geq7$, then

$$\begin{aligned} f(z)^{n}\bigl(f(qz+c)-f(z)\bigr)=g(z)^{n} \bigl(g(qz+c)-g(z)\bigr). \end{aligned}$$

Let $h(z)=\frac{f(z)}{g(z)}$ and $H(z)=h(qz+c)h(z)^{n}$. The last equation implies that

$$ g(qz+c) \bigl(H(z)-1 \bigr)=g(z) \bigl(h(z)^{n+1}-1 \bigr). $$

(27)

We know that $T(r,H)\leq(n+1)T(r,h)+S(r,h)$ from the expression of $H(z)$ and Lemma 2.1. Thus $S(r,H)=S(r,h)$. Next, we prove that $h(z)\equiv c_{4}$, where $c_{4}^{n+1}=1$, when $\frac {g(z)}{g(qz+c)}$ is transcendental with only finitely many zeros. Obviously, $h(z)$ is neither a constant except $c_{4}$ nor a rational function from (27) since $\frac{g(z)}{g(qz+c)}$ is transcendental. Thus we assume that $h(z)$ is a transcendental meromorphic function.

First, we affirm that $H(z)-1$ has infinitely many zeros. Otherwise, by the second main theorem of Nevanlinna theory and Lemma 2.1

$$\begin{aligned} T\bigl(r,H(z)\bigr) \leq& \overline{N}\bigl(r,H(z)\bigr)+ \overline{N} \biggl(r,\frac {1}{H(z)} \biggr) +\overline{N} \biggl(r, \frac{1}{H(z)-1} \biggr)+S(r,H) \\ \leq& 2T\bigl(r,h(z)\bigr)+2T\bigl(r,h(qz+c)\bigr)+S(r,h) \\ \leq& 4T\bigl(r,h(z)\bigr)+S(r,h). \end{aligned}$$

(28)

From the Valiron–Mohon’ko theorem, Lemma 2.1, and (28) we obtain

$$\begin{aligned} nT(r,h)&=T\bigl(r,h(z)^{n}\bigr)\leq T\bigl(r,H(z) \bigr)+T \biggl(r,\frac{1}{h(qz+c)} \biggr)+S(r,h) \\ &\leq5T\bigl(r,h(z)\bigr)+S(r,h), \end{aligned}$$

which contradicts to $n\geq7$. Thus $H(z)-1$ has infinitely many zeros.

Then we prove that $H(z)\equiv1$ and $h(z)\equiv c_{4}$ is a nonzero constant, that is, $c_{4}^{n+1}=1$. If $H(z)\not\equiv1$, then since $H(z)-1$ has infinitely many zeros, we can choose a point $z_{0}$ satisfying $H(z_{0})=1$, and $z_{0}$ is not the zero of $\frac{g(z)}{g(qz+c)}$. From (27) we have $h(qz_{0}+c)=h(z_{0})$. By Lemma 2.1

$$\begin{aligned} \overline{N} \biggl(r,\frac{1}{H(z)-1} \biggr) \leq \overline{N} \biggl(r,\frac{1}{h(qz+c)-h(z)} \biggr)\leq2T\bigl(r,h(z) \bigr)+S(r,h). \end{aligned}$$

(29)

Using the second main theorem of Nevanlinna theory, Lemma 2.1, and (29), we obtain

$$\begin{aligned} T\bigl(r,H(z)\bigr) \leq& \overline{N}\bigl(r,H(z)\bigr)+ \overline{N} \biggl(r,\frac{1}{H(z)} \biggr) +\overline{N} \biggl(r, \frac{1}{H(z)-1} \biggr)+S(r,h) \\ \leq& \overline{N}(r,h)+\overline{N}_{1}(r)+\overline{N} \biggl(r, \frac {1}{h} \biggr)+\overline{N}_{0}(r) +\overline{N} \biggl(r,\frac{1}{H(z)-1} \biggr)+S(r,h) \\ \leq& 4T(r,h)+\overline{N}_{1}(r)+\overline{N}_{0}(r)+S(r,h), \end{aligned}$$

(30)

where $\overline{N}_{0}(r)$ denotes the counting function ignoring multiplicities of the common zeros of $H(z)$ and $h(qz+c)$, and $\overline{N}_{1}(r)$ denotes the counting function ignoring multiplicities of the common poles of $H(z)$ and $h(qz+c)$.

On the other hand, from $H(z)=h(qz+c)h(z)^{n}$ we have

$$\begin{aligned}& nm(r,h)=m\bigl(r,h^{n}\bigr)\leq m\bigl(r,H(z) \bigr)+m \biggl(r,\frac{1}{h(qz+c)} \biggr)+O(1), \end{aligned}$$

(31)

$$\begin{aligned}& nN(r,h)=N\bigl(r,h^{n}\bigr)\leq N\bigl(r,H(z) \bigr)+N \biggl(r,\frac{1}{h(qz+c)} \biggr)-\overline{N}_{1}(r)- \overline{N}_{0}(r). \end{aligned}$$

(32)

From (31), (32), and Lemma 2.1 we have

$$ \begin{aligned}[b] nT(r,h)&\leq T\bigl(r,H(z)\bigr)+T \biggl(r,\frac{1}{h(qz+c)} \biggr)-\overline {N}_{1}(r)- \overline{N}_{0}(r)+O(1) \\ &\leq T\bigl(r,H(z)\bigr)+T(r,h)-\overline{N}_{1}(r)- \overline{N}_{0}(r)+S(r,h). \end{aligned} $$

(33)

Substituting (30) into (33), we obtain

$$\begin{aligned} nT(r,h)\leq5T(r,h)+S(r,h), \end{aligned}$$

which contradicts to $n\geq7$, so $H(z)=h(qz+c)h(z)^{n}\equiv1$, that is, $h(z)^{n}=\frac{1}{h(qz+c)}$. From Lemma 2.1 we have

$$\begin{aligned} nT(r,h)=T\bigl(r,h(qz+c)\bigr)\leq T(r,h)+S(r,h), \end{aligned}$$

which also contradicts to $n\geq7$, so $h(z)$ is a nonzero constant, say $c_{4}$, and from (27) we get $c_{4}^{n+1}=1$. Thus the theorem is proved. □