Theory and Modern Applications

# Meromorphic functions that share a polynomial with their difference operators

## Abstract

In this paper, we prove the following result: Let f be a nonconstant meromorphic function of finite order, p be a nonconstant polynomial, and c be a nonzero constant. If f, $$\Delta _{c}f$$, and $$\Delta_{c}^{n}f$$ ($$n\ge 2$$) share ∞ and p CM, then $$f\equiv \Delta_{c}f$$. Our result provides a difference analogue of the result of Chang and Fang in 2004 (Complex Var. Theory Appl. 49(12):871–895, 2004).

## Introduction and main results

In this paper, we use the base notations of the Nevanlinna theory of meromorphic functions which are defined as follows [9, 18, 19].

Let f be a meromorphic function. Throughout this paper, a meromorphic function always means meromorphic in the whole complex plane.

### Definition 1

$$m(r,f)=\frac{1}{2\pi } \int_{0}^{2\pi }\log^{+} \bigl\vert f \bigl(re^{i\theta } \bigr) \bigr\vert \,d \theta .$$

$$m(r,f)$$ is the average of the positive logarithm of $$\vert f(z) \vert$$ on the circle $$\vert z \vert =r$$.

### Definition 2

$$\begin{gathered} N(r,f) = \int_{0}^{r}\frac{n(t,f)-n(0,f)}{t}\,dt+n(0,f)\log r, \\ \overline{N}(r,f)= \int_{0}^{r} \frac{\overline{n}(t,f)-\overline{n}(0,f)}{t}\,dt+ \overline{n}(0,f) \log r, \end{gathered}$$

where $$n(t,f)$$ ($$\overline{n}(t,f)$$) denotes the number of poles of f in the disc $$\vert z \vert \le t$$, multiples poles are counted according to their multiplicities (ignore multiplicity). $$n(0,f)$$ ($$\overline{n}(0,f)$$) denotes the multiplicity of poles of f at the origin (ignore multiplicity).

$$N(r,f)$$ is called the counting function of poles of f, and $$\overline{N}(r,f)$$ is called the reduced counting function of poles of f.

### Definition 3

$$T(r,f)=m(r,f)+N(r,f).$$

$$T(r,f)$$ is called the characteristic function of f. It plays a cardinal role in the whole theory of meromorphic functions.

### Definition 4

Let f be a meromorphic function. The order of growth of f is defined as follows:

$$\rho (f)=\overline{\lim_{r\to \infty }}\frac{\log^{+} T(r,f)}{\log r}.$$

If $$\rho (f)< \infty$$, then we say that f is a meromorphic function of finite order.

### Definition 5

Let a, f be two meromorphic functions. If $$T(r,a)=S(r,f)$$, where $$S(r,f)=o(T(r,f))$$, as $$r\to \infty$$ outside of a possible exceptional set of finite logarithmic measure. Then we say that a is a small function of f. And we use $$S(f)$$ to denote the family of all small functions with respect to f.

### Definition 6

Let f and g be two meromorphic functions, and p be a polynomial. We say that f and g share p CM, provided that $$f(z)-p(z)$$ and $$g(z)-p(z)$$ have the same zeros counting multiplicity. And if f and g have the same poles counting multiplicity, then we say that f and g shareCM.

In this paper, we also use some known properties of the characteristic function $$T(r,f)$$ as follows [9, 18, 19].

### Property 1

Let $$f_{j}$$ ($$j=1,2,\ldots,q$$) be q meromorphic functions in $$\vert z \vert < R$$ and $$0< r< R$$. Then

$$T \Biggl(r,\prod_{j=1}^{q}T(r,f_{j}) \Biggr)\le \sum_{j=1}^{q}T(r,f_{j}), \quad\quad T \Biggl(r, \sum_{j=1}^{q}f_{j} \Biggr)\le \sum_{j=1}^{q}T(r,f_{j})+ \log q$$

hold for $$1\le r< R$$.

### Property 2

Suppose that f is meromorphic in $$\vert z \vert < R$$ ($$R\le \infty$$) and a is any complex number. Then, for $$0< r< R$$, we have

$$T \biggl(r,\frac{1}{f-a} \biggr)=T(r,f)+O(1).$$

Property 2 is the first fundamental theorem.

### Property 3

Suppose that f is a nonconstant meromorphic function and $$a_{1}, a _{2},\ldots, a_{n}$$ are $$n\ge 3$$ distinct values in the extended complex plane. Then

$$(n-2)T(r,f)< \sum_{j=1}^{n}\overline{N} \biggl(r,\frac{1}{f-a_{j}} \biggr)+S(r,f).$$

Property 3 is the second fundamental theorem. For more properties about $$T(r,f)$$, please see [9, 18, 19].

For a meromorphic function $$f(z)$$, we define its shift by $$f_{c}(z) = f(z + c)$$ and its difference operators by

$$\Delta_{c}f(z)=f(z+c)-f(z),\quad\quad \Delta_{c}^{n}f(z)= \Delta _{c}^{n-1} \bigl(\Delta_{c}f(z) \bigr). \bigl(\Delta_{c}f(z) \bigr).$$

In  the following result was proved.

### Theorem 1

Let f be a nonconstant meromorphic function, and a be a nonzero finite complex number. If f, $$f'$$, and $$f''$$ share a CM, then $$f\equiv f'$$.

In 2001, Li and Yang  considered the case when f, $$f'$$, and $$f^{(n)}$$ share one value.

### Theorem 2

Let f be an entire function, a be a finite nonzero constant, and $${n\ge 2}$$ be a positive integer. If f, $$f'$$, and $$f^{(n)}$$ share a CM, then f assumes the form

$$f(z)=be^{wz}-\frac{a(1-w)}{w},$$
(1.1)

where b, w are two nonzero constants satisfying $$w^{n-1}=1$$.

### Remark 1

It is easy to see that the functions in ( 1.1 ) really share value a, since when $$b\neq 0$$ and $$w^{n-1}=1$$, from $$f^{(j)}(z)=a$$, $$j=0,1,n$$, it follows that $$bwe^{wz}=a$$ for each $$j=0,1,n$$. So, the functions $$f^{(j)}-a$$, $$j=0, 1, n$$, have the same zeros counting multiplicity.

In 2004, Chang and Fang  considered the case when f, $$f'$$, and $$f^{(n)}$$ share a small function.

### Theorem 3

Let f be an entire function, a be a nonzero small function of f, and $$n\ge 2$$ be a positive integer. If f, $$f'$$, and $$f^{(n)}$$ share a CM, then $$f\equiv f'$$.

Recently, value distribution in difference analogue of meromorphic functions has become a subject of some interest, see, e.g., [28, 11].

In 2012 and 2014, Chen et al. [2, 3] considered difference analogue of Theorem 1 and Theorem 2, and established the following result.

### Theorem 4

Let f be a nonconstant entire function of finite order, and a (0) $$\in S(f)$$ be a periodic entire function with period c. If f, $$\Delta_{c}f$$, and $$\Delta_{c}^{n}f$$ ($$n\ge 2$$) share a CM, then $$\Delta_{c}f\equiv \Delta_{c}^{n}f$$.

For other related results, the reader is referred to the references due to Latreuch, El Farissi, Belaïdi , El Farissi, Latreuch, Asiri , El Farissi, Latreuch, Belaïdi and Asiri .

### Remark 2

There are examples in  which show that the conclusion $$\Delta_{c}f\equiv \Delta_{c}^{n}f$$ in Theorem 4 cannot be replaced by $$f\equiv \Delta_{c}f$$, and the condition $$a(z)\not \equiv 0$$ is necessary.

By Theorems 3 and 4, it is natural to ask: Can we provide a difference analogue of Theorem 3? Or, can we delete the condition that ‘$$a(z)$$ is a periodic entire function with period c’ in Theorem 4?

In this paper, we study the problem and prove the following result.

### Theorem 5

Let f be a nonconstant meromorphic function of finite order, and p be a nonconstant polynomial. If f, $$\Delta_{c}f$$, and $$\Delta_{c} ^{n}f$$ ($$n\ge 2$$) share p andCM, then $$f\equiv \Delta_{c}f$$.

If f is an entire function, then f, $$\Delta_{c}f$$, and $$\Delta_{c}^{n}f$$ have no poles, obviously f, $$\Delta_{c}f$$ and $$\Delta_{c}^{n}f$$ share ∞ CM. By Theorem 5, we consequently get the following result.

### Corollary 1

Let f be a nonconstant entire function of finite order, and $$n \ge 2$$ be a positive integer. If f, $$\Delta_{c}f$$, and $$\Delta_{c} ^{n}f$$ share z CM, then $$f\equiv \Delta_{c}f$$.

### Example 1

Let A, a, b, c be four finite nonzero complex numbers satisfying $$a\neq b$$, n (≥2) $$\in \mathbb{N}$$ satisfying $$[e^{Ac}-1]^{n-1}=1$$, $$e^{Ac}-1=\frac{a}{a-b}$$, and $$g(z)$$ be a periodic entire function with period c, and let $$f(z)=g(z)e^{Az}+b$$. By simple calculation, we obtain

$$\Delta_{c}^{n}f(z)=\Delta_{c}f(z)= \bigl[e^{Ac}-1 \bigr]f(z)+a \bigl[1-e^{Ac}+1 \bigr].$$

It is easy to see that f, $$\Delta_{c}f$$, and $$\Delta_{c} ^{n}f$$ ($$n\ge 2$$) share a CM, and $$f\neq \Delta_{c}f$$ when $$e^{Ac} \neq 2$$. This example shows that ‘$$p(z)$$ cannot be a constant’ in Theorem 5.

### Example 2

Let A, b, c be three nonzero finite complex numbers satisfying $$e^{Ac}=1$$, and $$f(z)=e^{Az}+b$$, $$p(z)=b$$. It is easy to see that f, $$\Delta_{c}f$$, and $$\Delta_{c}^{n}f$$ share $$p(z)$$ CM. But $$\Delta_{c}f\equiv 0 \not \equiv f$$. This example also shows that ‘$$p(z)$$ cannot be a constant’ in Theorem 5.

### Example 3

Let A, c be two nonzero finite complex numbers satisfying $$e^{Ac}=2$$ and $$f(z)=e^{Az}\cot (\frac{\pi z}{c})$$. By simple calculation, we obtain

$$f(z)=\Delta_{c}f=\Delta_{c}^{n}f=e^{Az} \cot \biggl(\frac{\pi z}{c} \biggr).$$

Obviously, for any polynomial p, f, $$\Delta_{c}f$$, and $$\Delta_{c}^{n}f$$ share p and ∞ CM. This example satisfies Theorem 5.

In Examples 1 and  2 , we have $$\Delta_{c}f\equiv tf+a(1-t)$$ and $$f(z)=e^{Az+B}+a$$, respectively, when f, $$\Delta_{c}f$$, and $$\Delta_{c}^{n}f$$ ($$n\ge 2$$) share a nonzero constant a CM. Hence we posed the following problem.

### Problem 1

Assume that f is a nonconstant entire function of finite order, a is a nonzero constant, and that f, $$\Delta_{c}f$$, and $$\Delta_{c}^{n}f$$ ($$n\ge 2$$) share a CM. Whether or not, one of the following two cases occurs:

1. (1)

$$\Delta_{c}f\equiv tf+a(1-t)$$, where t is a constant satisfying $$t^{n-1}=1$$,

2. (2)

$$f(z)=e^{Az+B}+a$$, where A (≠0), B are two constants satisfying $$e^{Ac}=1$$.

## Some lemmas

### Lemma 1

([4, 7])

Let f be a meromorphic function of finite order, and c be a nonzero complex constant. Then

$$T \bigl(r,f(z+c) \bigr)=T(r,f)+S(r,f).$$

### Lemma 2

([7, 8])

Let $$c\in \mathbb{C}$$, k be a positive integer, and f be a meromorphic function of finite order. Then

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

### Lemma 3

([18, 19])

Let $$n\ge 2$$ be a positive integer. Suppose that $$f_{i}(z)$$ ($$i = 1,2,\ldots,n$$) are meromorphic functions and $$g_{i}(z)$$ ($$i = 1,2,\ldots,n$$) are entire functions satisfying

1. (i)

$$\sum_{i=1}^{n}f_{i}(z)e^{g_{i}(z)}\equiv 0$$,

2. (ii)

the orders of $$f_{i}$$ are less than those of $$e^{g_{k}-g_{l}}$$ for $$1\le i \le n$$, $$1\le k < l \le n$$.

Then $$f_{i}(z)\equiv 0$$ ($$i=1,2,\ldots,n$$).

The following lemma is well known.

### Lemma 4

Let the function f satisfy the following difference equation:

$$f(w+1)=\alpha (w)f(w)+\beta (w)$$

in the complex plane.

Then the following formula holds:

$$f(w+k)=f(w)\prod_{j=0}^{k-1} \alpha (w+j)+\sum_{l=0}^{k-1}\beta (w+l) \prod_{j=l+1}^{k-1}\alpha (w+j)$$
(2.1)

for every $$w\in \mathbb{C}$$ and $$k\in \mathbb{N}^{+}$$.

Formula (2.1) has many applications. For example, many solvable difference equations are essentially solved by using it (see ), and by using such obtained formulas the behavior of their solution can be studied (see, for example, recent papers [13, 14]; see also many related references therein). As another simple application, by using a linear change of variables, the following corollary is obtained:

### Corollary 2

Let the function f satisfy the following difference equation:

$$f(w+c)=\alpha (w)f(w)+\beta (w)$$

in the complex plane.

Then the following formula holds:

$$f(w+kc)=f(w)\prod_{j=0}^{k-1} \alpha (w+jc)+\sum_{l=0}^{k-1}\beta (w+lc) \prod_{j=l+1}^{k-1}\alpha (w+jc)$$

for every $$w\in \mathbb{C}$$ and $$k\in \mathbb{N}^{+}$$.

From the ideas of Chang and Fang  and Chen and Li , we prove the following lemma.

### Lemma 5

Let f be a nonconstant meromorphic function of finite order, p (0) be a polynomial, and $$n\ge 2$$ be an integer. Suppose that

$$\frac{\Delta_{c}^{n}f(z)-p(z)}{f(z)-p(z)}=e^{\alpha (z)},\quad\quad \frac{ \Delta_{c}f(z)-p(z)}{f(z)-p(z)}=e^{\beta (z)},$$
(2.2)

where α and β are two polynomials, and that

$$T \bigl(r,e^{\alpha } \bigr)+T \bigl(r,e^{\beta } \bigr)=S(r,f).$$
(2.3)

Then $$\Delta_{c}f\equiv tf+b(1-t)$$, where t, b are constants satisfying $$t^{n-1}=1$$ and $$b\neq 0$$. Moreover, if $$t\neq 1$$, then $$p(z)\equiv b$$.

### Proof

Firstly, we prove that f cannot be a rational function. Otherwise, suppose that $$f(z)=P(z)/Q(z)$$, where $$P(z)$$ and $$Q(z)$$ are two co-prime polynomials. It follows from (2.2) that $$f(z)$$ and $$\Delta_{c}f(z)$$ share ∞ CM. We claim that $$Q(z)$$ is a constant. Otherwise, suppose that there exists $$z_{0}$$ such that $$Q(z_{0}+c)=0$$. Since $$f(z)$$ and $$\Delta_{c}f(z)$$ share ∞ CM, and

$$\Delta_{c}f(z)=\frac{P(z+c)}{Q(z+c)}-\frac{P(z)}{Q(z)}= \frac{P(z+c)Q(z)-P(z)Q(z+c)}{Q(z)Q(z+c)}.$$
(2.4)

We deduce that all zeros of $$Q(z+c)$$ must be the zeros of $$Q(z)$$. Otherwise, suppose that there exists $$z_{1}$$ such that $$Q(z_{1}+c)=0$$ but $$Q(z_{1})\neq 0$$, then it follows from (2.4) that $$z_{1}+c$$ is a pole of $$\Delta_{c}f$$ but not the pole of f, which contradicts with $$f(z)$$ and $$\Delta_{c}f(z)$$ share ∞ CM. Then we get

$$Q(z_{0}+c)=0\quad \Rightarrow \quad Q(z_{0})=0 \quad \Rightarrow\quad Q(z_{0}-c)=0\quad \Rightarrow \quad \cdots \quad \Rightarrow \quad Q(z_{0}-lc)=0.$$

This implies that $$Q(z)$$ has infinitely many zeros, which is a contradiction. Thus, the claim is proved.

Then f is a nonconstant polynomial, suppose that

$$f(z)=a_{k}z^{k}+a_{k-1}z^{k-1}+\cdots +a_{1}z+a_{0}, \quad\quad p(z)=b_{m}z ^{m}+\cdots +b_{1}z+b_{0}.$$

Then we get $$\Delta_{c}f(z)=f(z+c)-f(z)$$, and obviously, $$\deg \Delta_{c}^{n}f(z)\le \deg \Delta_{c}f(z) < \deg f(z)$$. Then it follows from (2.2) that $$\alpha (z)$$, $$\beta (z)$$ are constants, and we let $$e^{\alpha (z)}=a$$, $$e^{\beta (z)}=b$$. So we have

$$\Delta_{c}^{n}f(z)-p(z)=a \bigl(f(z)-p(z) \bigr), \quad \quad \Delta_{c}f(z)-p(z)=b \bigl(f(z)-p(z) \bigr).$$

Then we get $$\deg f=k\le \deg p=m$$ since $$\deg \Delta_{c}^{n}f(z) \le \deg \Delta_{c}f(z) < \deg f(z)$$. If $$k< m$$, then we get $$a=b=1$$. This implies $$f\equiv \Delta_{c}f(z) \equiv \Delta_{c}^{n}f$$, which contradicts with $$\deg \Delta_{c}f(z) < \deg f(z)$$. If $$k= m$$ then we get $$a=b=b_{k}/(a_{k}-b_{k})$$, $$\Delta_{c}f(z) \equiv \Delta_{c}^{n}f$$, and hence $$f(z)$$ is a constant, which is a contradiction.

Hence, f is a transcendental meromorphic function. Thus $$T(r,p)=S(r,f)$$.

Next, we consider two cases.

Case 1. $$\beta (z)$$ is a nonconstant polynomial. It follows from the second equation in (2.2) that $$\Delta_{c}f(z)=e^{\beta (z)}(f(z)-p(z))+p(z)$$, and that

$$f(z+c)=a_{1}(z)f(z)+b_{1}(z),$$

where $$a_{1}(z)=e^{\beta (z)}+1$$, $$b_{1}(z)=p(z)[1-e^{\beta (z)}]$$.

By Corollary  2 , it is easy to get, for any $$k\in \mathbb{N^{+}}$$,

$$f(z+kc)=a_{k}(z)f(z)+b_{k}(z),$$
(2.5)

where

$$a_{k}(z)=\prod_{j=0}^{k-1} \bigl(e^{\beta (z+jc)}+1 \bigr), \quad\quad b_{k}(z)=\sum _{l=0}^{k-1}b_{1}(z+lc)\prod _{j=l+1}^{k-1}a_{1}(z+jc).$$
(2.6)

It follows from (2.5) and (2.6) that

\begin{aligned}[b] \Delta_{c}^{n}f(z) &=\sum _{i=0}^{n}(-1)^{n-i}C_{n}^{i}f(z+ic) \\ &= \sum_{i=0}^{n}(-1)^{n-i}C_{n}^{i} \bigl[a_{i}(z)f(z)+b_{i}(z) \bigr] \\ &= \Biggl[ (-1)^{n}+ \sum_{i=1}^{n}(-1)^{n-i}C_{n}^{i} \prod_{j=0}^{i-1} \bigl(e^{\beta (z+jc)}+1 \bigr) \Biggr] f(z)+ \sum_{i=0}^{n}(-1)^{n-i}C_{n}^{i}b_{i}(z), \\ &=\mu_{n}(z)f(z)+\nu_{n}(z), \end{aligned}
(2.7)

where

$$\begin{gathered} \mu_{n}(z) =\prod_{j=0}^{n-1}e^{\beta (z+jc)}+ \sum_{t=0}^{n-1} \lambda_{n-1,t}\prod _{j=0,j\neq t}^{n-1}e^{\beta (z+jc)}+\cdots + \sum _{t=0}^{n-1}\lambda_{1,t}e^{\beta (z+tc)}, \\ \nu_{n}(z)=\sum_{i=0} ^{n}(-1)^{n-i}C_{n}^{i}b_{i}(z). \end{gathered}$$
(2.8)

In particular, $$\lambda_{1,t}=(-1)^{n-1-t}C_{n-1}^{t}$$, which implies

$$\sum_{t=0}^{n-1}\lambda_{1,t}e^{\beta (z+tc)}= \Delta_{c}^{n-1}e^{ \beta (z)}.$$
(2.9)

By ( 2.3 ), ( 2.8 ), and Lemma  1 , it is easy to get

$$T(r,\mu_{n})+T(r,\nu_{n})=S(r,f).$$
(2.10)

Since $$\beta (z)$$ is a nonconstant polynomial, we have that

$$\beta (z)=l_{m}z^{m}+l_{m-1}z^{m-1}+ \cdots +l_{0},$$

where $$l_{i}$$ ($$0\le i \le m$$) are constants satisfying $$l_{m}\neq 0$$ and $$m\ge 1$$. Obviously, for any $$j\in \{0, 1,\ldots, n-1\}$$, we have

$$\beta (z+jc)=l_{m}z^{m}+(l_{m-1}+ml_{m}jc)z^{m-1}+ \cdots +\sum_{t=0} ^{m}l_{t}(jc)^{t}.$$
(2.11)

From (2.8) and (2.11), we get

\begin{aligned}[b] \mu_{n}(z)&= e^{nl_{m}z^{m}+P_{n,0}(z)}+\lambda_{n-1,0}e^{(n-1)l_{m}z ^{m}+P_{n-1,0}(z)}+ \cdots +\lambda_{n-1,n-1}e^{(n-1)l_{m}z^{m}+P_{n-1,n-1}(z)} \\ &\quad {} +\cdots +\lambda_{1,0}e^{l_{m}z^{m}+P_{1,0}(z)}+\cdots + \lambda_{1,n-1}e^{l_{m}z^{m}+P_{1,n-1}(z)}, \end{aligned}
(2.12)

where $$P_{i,j}(z)$$ are polynomials with degree less than m for $$i\in \{1, 2,\ldots,n\}$$, $$j\in \{0,1,\ldots, {C_{n}^{i}-1}\}$$.

It follows from the first equation in (2.2) that

$$\Delta_{c}^{n}f(z)-e^{\alpha (z)}f(z)=p(z) \bigl(1-e^{\alpha (z)} \bigr).$$
(2.13)

From (2.7) and (2.13), we have

$$\bigl(\mu_{n}(z)-e^{\alpha (z)} \bigr)f(z)=p(z) \bigl(1-e^{\alpha (z)} \bigr)-\nu_{n}(z).$$
(2.14)

From ( 2.3 ), ( 2.10 ) and since $$T(r,p)=S(r,f)$$, we have that

$$T(r,p)+T \bigl(r,e^{\alpha } \bigr)+T \bigl(r,e^{\beta } \bigr)+T(r,\mu_{n})+T(r,\nu_{n})=S(r,f).$$
(2.15)

If $$\mu_{n}(z)-e^{\alpha (z)}\not \equiv 0$$, by ( 2.14 ), ( 2.15 ), Property  1 , and Property  2 , we obtain

$$T(r,f)=T \biggl(r,\frac{p(z)(1-e^{\alpha (z)})-\nu_{n}(z)}{\mu_{n}(z)-e^{ \alpha (z)}} \biggr)=S(r,f),$$

Hence $$\mu_{n}(z)-e^{\alpha (z)}\equiv 0$$. Combining this with (2.12), we get

\begin{aligned}[b] &e^{P_{n,0}(z)}e^{nl_{m}z^{m}}+ \bigl(\lambda_{n-1,0}e^{P_{n-1,0}(z)}+ \cdots +\lambda_{n-1,n-1}e^{P_{n-1,n-1}(z)} \bigr)e^{(n-1)l_{m}z^{m}} \\ &\quad{} +\cdots + \bigl( \lambda_{1,0}e^{P_{1,0}(z)}+\cdots + \lambda_{1,n-1}e^{P_{1,n-1}(z)} \bigr)e ^{l_{m}z^{m}}-e^{\alpha (z)} \equiv 0. \end{aligned}
(2.16)

Next, we consider three subcases.

Case 1.1. $$\deg \alpha (z)> m$$. Then, for any $$1\le i \le n$$, $$1\le k< j \le n$$, we have

$$\rho \bigl(e^{\alpha (z)-il_{m}z^{m}} \bigr)=\rho \bigl(e^{\alpha (z)} \bigr)=\deg \alpha (z)> m, \quad\quad \rho \bigl(e^{jl_{m}z^{m}-kl_{m}z^{m}} \bigr)=m.$$

Since $$P_{i,j}(z)$$ are polynomials with degree less than m for $$i\in \{1, 2,\ldots,n\}$$, $$j\in \{0,1,\ldots, C_{n}^{i}-1\}$$, then for $$i=1,2,\ldots,n-1$$,

$$\rho \Biggl( \sum_{j=0}^{C_{n}^{i}-1} \lambda_{i,j}e^{P_{i,j}(z)} \Biggr) \le m-1, \quad\quad \rho \bigl( e^{P_{n,0}(z)} \bigr) \le m-1.$$

By ( 2.16 ) and using Lemma 3, we obtain $$e^{P_{n,0}}\equiv 0$$, which is a contradiction.

Case 1.2. $$\deg \alpha (z)< m$$. Then, for any $$1\le i \le n$$, $$1\le k< j \le n$$, we have

$$\rho \bigl(e^{\alpha (z)-il_{m}z^{m}} \bigr)=\rho \bigl(e^{-il_{m}z^{m}} \bigr)= m, \quad \quad \rho \bigl(e^{jl_{m}z^{m}-kl_{m}z^{m}} \bigr)=m.$$

Since $$P_{i,j}(z)$$ are polynomials with degree less than m for $$i\in \{1, 2,\ldots,n\}$$, $$j\in \{0,1,\ldots, C_{n}^{i}-1\}$$, then for $$i=1,2,\ldots,n-1$$,

$$\rho \Biggl( \sum_{j=0}^{C_{n}^{i}-1} \lambda_{i,j}e^{P_{i,j}(z)} \Biggr) \le m-1, \quad\quad \rho \bigl( e^{P_{n,0}(z)} \bigr) \le m-1.$$

By ( 2.16 ) and using Lemma 3, we obtain $$e^{P_{n,0}}\equiv 0$$, which is a contradiction.

Case 1.3. $$\deg \alpha (z)= m$$. Set $$\alpha (z)=dz^{m}+\alpha^{*}(z)$$, where $$d\neq 0$$ and $$\deg \alpha^{*}(z)< m$$. Rewrite (2.16) as

\begin{aligned}[b] &e^{P_{n,0}(z)}e^{nl_{m}z^{m}}+ \bigl(\lambda_{n-1,0}e^{P_{n-1,0}(z)}+ \cdots +\lambda_{n-1,n-1}e^{P_{n-1,n-1}(z)} \bigr)e^{(n-1)l_{m}z^{m}} \\ &\quad{} +\cdots + \bigl( \lambda_{1,0}e^{P_{1,0}(z)}+\cdots + \lambda_{1,n-1}e^{P_{1,n-1}(z)} \bigr)e ^{l_{m}z^{m}}-e^{\alpha^{*}(z)}e^{dz^{m}} \equiv 0. \end{aligned}
(2.17)

If $$d\neq jl_{m}$$, for any $$j=1,2,\ldots,n$$, we have

$$\rho \bigl(e^{dz^{m}-jl_{m}z^{m}} \bigr)=\rho \bigl(e^{(d-jl_{m})z^{m}} \bigr)=m,\quad \quad \rho \bigl(e^{\alpha^{*}(z)} \bigr)< m.$$

Combining this with (2.17), by using Lemma 3, we get a contradiction.

If $$d= jl_{m}$$, for some $$j=1,2,\ldots,n-1$$, without loss of generality, we assume that $$j=1$$, then (2.17) can be rewritten as

\begin{aligned} &e^{P_{n,0}(z)}e^{nl_{m}z^{m}}+ \bigl(\lambda_{n-1,0}e^{P_{n-1,0}(z)}+ \cdots +\lambda_{n-1,n-1}e^{P_{n-1,n-1}(z)} \bigr)e^{(n-1)l_{m}z^{m}} \\ &\quad{} +\cdots + \bigl( \lambda_{1,0}e^{P_{1,0}(z)}+\cdots + \lambda_{1,n-1}e^{P_{1,n-1}(z)}-e ^{\alpha^{*}(z)} \bigr)e^{l_{m}z^{m}} \equiv 0. \end{aligned}

And then, by using the same argument as above, we get a contradiction.

Hence, $$d= nl_{m}$$. Rewrite (2.17) as

\begin{aligned} &\bigl(e^{P_{n,0}(z)}-e^{\alpha^{*}(z)} \bigr)e^{nl_{m}z^{m}}+ \bigl(\lambda_{n-1,0}e ^{P_{n-1,0}(z)}+\cdots +\lambda_{n-1,n-1}e^{P_{n-1,n-1}(z)} \bigr)e^{(n-1)l _{m}z^{m}} \\ &\quad{} +\cdots + \bigl(\lambda_{1,0}e^{P_{1,0}(z)}+\cdots +\lambda _{1,n-1}e^{P_{1,n-1}(z)} \bigr)e^{l_{m}z^{m}}\equiv 0. \end{aligned}

Using the same argument as in Case 1.1 and using Lemma 3, we obtain

$$\lambda_{1,0}e^{P_{1,0}(z)}+\cdots +\lambda_{1,n-1}e^{P_{1,n-1}(z)} \equiv 0.$$
(2.18)

Then, by (2.9) and (2.18), we get

$$\sum_{t=0}^{n-1}\lambda_{1,t}e^{\beta (z+tc)}= \Delta_{c}^{n-1}e^{ \beta (z)}=\sum _{t=0}^{n-1}(-1)^{n-1-t}C_{n-1}^{t}e^{\beta (z+tc)} \equiv 0.$$
(2.19)

If $$m\ge 2$$, then for any $$t=0,1,\ldots,n-1$$, we have

$$\beta (z+tc)=l_{m}z^{m}+(l_{m-1}+ml_{m}tc)z^{m-1}+q_{t}(z),$$
(2.20)

where $$q_{t}(z)$$ are polynomials with $$\deg q_{t}(z)< m-1$$.

From (2.19) with (2.20), we get

\begin{aligned} &e^{q_{n-1}(z)}e^{l_{m}z^{m}+(l_{m-1}+ml_{m}(n-1)c)z^{m-1}}-(n-1)e^{q _{n-2}(z)}e^{l_{m}z^{m}+(l_{m-1}+ml_{m}(n-2)c)z^{m-1}} \\ &\quad{}+\cdots +(-1)^{n-1}e ^{q_{0}(z)}e^{l_{m}z^{m}+(l_{m-1}+ml_{m}c)z^{m-1}}\equiv 0. \end{aligned}

Using the same argument as Case 1.1, we obtain a contradiction.

Hence $$m=1$$. Thus $$\beta (z)=l_{1}z+l_{0}$$, where $$l_{1}\neq 0$$. Then, for any $$n\ge 1$$, we deduce that

$$\Delta_{c}^{n-1}e^{\beta (z)}= \bigl(e^{l_{1}c}-1 \bigr)^{n-1}e^{\beta (z)}.$$

Hence, it follows from (2.19) that $$(e^{l_{1}c}-1)^{n-1}\equiv 0$$, which yields $$e^{l_{1}c}=1$$. Then, for any $$t\in \mathbb{N^{+}}$$, we have

$$e^{\beta (z+tc)}=e^{l_{1}z+tl_{1}c+l_{0}}=e^{l_{1}z+l_{0}} \bigl(e^{l_{1}c} \bigr)^{t}=e ^{\beta (z)}.$$
(2.21)

By the second equation in (2.2) and (2.21), we get

\begin{aligned}& \Delta_{c}f(z) =e^{\beta (z)}f(z)+p(z) \bigl(1-e^{\beta (z)} \bigr)=e^{\beta (z)}f(z)+ \bigl(1-e ^{\beta (z)} \bigr)b_{1}(z), \\& \begin{aligned} \Delta_{c}^{2}f(z)&=e^{\beta (z)}\Delta_{c}f(z)+ \Delta_{c}p(z) \bigl(1-e^{\beta (z)} \bigr) \\ &=e^{\beta (z)} \bigl[e^{\beta (z)}f(z)+p(z) \bigl(1-e ^{\beta (z)} \bigr) \bigr]+\Delta_{c}p(z) \bigl(1-e^{\beta (z)} \bigr) \\ &=e^{2\beta (z)}f(z)+ \bigl(1-e ^{\beta (z)} \bigr) \bigl[p(z)e^{\beta (z)}+ \Delta_{c}p(z) \bigr] \\ &=e^{2\beta (z)}f(z)+ \bigl(1-e ^{\beta (z)} \bigr)b_{2}(z). \end{aligned} \end{aligned}

By mathematical induction, it is easy to get, for any integer $$t\ge 2$$,

$$\Delta_{c}^{t}f(z)=e^{t\beta (z)}f(z)+ \bigl(1-e^{\beta (z)} \bigr)b_{t}(z),$$

where $$b_{1}(z)=p(z)$$, $$b_{t}(z)=p(z)e^{(t-1)\beta (z)}+\Delta_{c}b _{t-1}=\sum_{i=0}^{t-1}e^{(t-1-i)\beta (z)}\Delta_{c}^{i}p(z)$$.

Hence,

$$\Delta_{c}^{n}f(z)=e^{n\beta (z)}f(z)+ \bigl(1-e^{\beta (z)} \bigr)b_{n}(z),$$
(2.22)

where $$b_{n}(z)=\sum_{i=0}^{n-1}e^{(n-1-i)\beta (z)}\Delta_{c}^{i}p(z)$$.

From the first equation in (2.2) and (2.22), we have

$$\bigl(e^{\alpha (z)}-e^{n\beta (z)} \bigr)f(z)= \bigl(1-e^{\beta (z)} \bigr)b_{n}(z)-p(z) \bigl(1-e ^{\alpha (z)} \bigr).$$
(2.23)

If $$e^{\alpha (z)}-e^{n\beta (z)}\not \equiv 0$$, then by ( 2.15 ), ( 2.23 ), Property  1 , and Property  2 , we have

$$T(r,f)=T \biggl(r,\frac{(1-e^{\beta (z)})b_{n}(z)-p(z)(1-e^{\alpha (z)})}{e ^{\alpha (z)}-e^{n\beta (z)}} \biggr)=S(r,f),$$

Hence $$e^{\alpha (z)}-e^{n\beta (z)}\equiv 0$$. It follows from (2.22) and (2.23) that

\begin{aligned} \bigl(1-e^{\beta (z)} \bigr)b_{n}(z) &= \bigl(1-e^{\beta (z)} \bigr) \Biggl( \sum_{i=0}^{n-1}e ^{(n-1-i)\beta (z)}\Delta_{c}^{i}p(z) \Biggr) \\ &=\sum_{i=0}^{n-1}e ^{(n-1-i)\beta (z)} \Delta_{c}^{i}p(z)-\sum_{i=0}^{n-1}e^{(n-i)\beta (z)} \Delta_{c}^{i}p(z) \\ &=-p(z)e^{n\beta (z)}+\sum_{i=0}^{n-2}e^{(n-1-i) \beta (z)} \bigl( \Delta_{c}^{i}p(z)-\Delta_{c}^{i+1}p(z) \bigr) +\Delta _{c}^{n-1}p(z) \\ &\equiv p(z) \bigl(1-e^{\alpha (z)} \bigr). \end{aligned}

That is,

$$\sum_{i=0}^{n-2}e^{(n-1-i)\beta (z)} \bigl( \Delta_{c}^{i}p(z)-\Delta_{c} ^{i+1}p(z) \bigr)+\Delta_{c}^{n-1}p(z)-p(z)\equiv 0.$$
(2.24)

If $$p(z)$$ is a constant, as $$\Delta_{c}^{i}p(z)=0$$ for any $$i\in \mathbb{N^{+}}$$. It follows from (2.24) that

$$p(z)e^{(n-1)\beta (z)}-p(z)\equiv 0.$$

Hence, $$e^{(n-1)\beta (z)}\equiv 1$$, which is a contradiction.

If $$p(z)$$ is a nonconstant polynomial, then $$p(z)-\Delta_{c}^{i}p(z) \not \equiv 0$$ for any $$i\in \mathbb{N^{+}}$$. It follows from (2.24) that

$$e^{(n-1)\beta (z)} \bigl(p(z)-\Delta_{c}p(z) \bigr)\equiv -\sum _{i=1}^{n-2}e^{(n-1-i) \beta (z)} \bigl( \Delta_{c}^{i}p(z)- \Delta_{c}^{i+1}p(z) \bigr)-\Delta_{c}^{n-1}p(z)+p(z).$$

Thus we have

\begin{aligned} (n-1)T \bigl(r,e^{\beta } \bigr) &=T \bigl(r,e^{(n-1)\beta (z)} \bigl(p(z)-\Delta_{c}p(z) \bigr) \bigr)+S \bigl(r,e ^{\beta } \bigr) \\ &=T \Biggl(r,-\sum_{i=1}^{n-2}e^{(n-1-i)\beta (z)} \bigl(\Delta_{c} ^{i}p(z)-\Delta_{c}^{i+1}p(z) \bigr)-\Delta_{c}^{n-1}p(z)+p(z) \Biggr) \\ &\quad{} +S \bigl(r,e^{ \beta } \bigr) \\ &\le (n-2)T \bigl(r,e^{\beta } \bigr)+S \bigl(r,e^{\beta } \bigr), \end{aligned}

Case 2. $$\beta (z)=\beta \in \mathbb{C}$$ is a constant. By the second equation in (2.2), we get

\begin{aligned}& \Delta_{c}f(z) =e^{\beta }f(z)+p(z) \bigl(1-e^{\beta } \bigr), \\& \Delta_{c}^{2}f(z)=e ^{\beta }\Delta_{c}f(z)+ \Delta_{c}p(z) \bigl(1-e^{\beta } \bigr)=e^{\beta }\Delta _{c}f(z)+ \bigl(1-e^{\beta } \bigr)b_{2}(z). \end{aligned}

By mathematical induction, it is easy to get, for any integer $$t\ge 2$$,

$$\Delta_{c}^{t}f(z)=e^{(t-1)\beta }\Delta_{c}f(z)+ \bigl(1-e^{\beta } \bigr)b_{t}(z),$$

where $$b_{2}(z)=\Delta_{c}p(z)$$, $$b_{t}(z)=\Delta_{c}p(z)e^{(t-1) \beta }+\Delta_{c}b_{t-1}=\sum_{i=1}^{t-1}\Delta_{c}^{i}p(z)e^{(t-1-i) \beta }$$.

Hence,

$$\Delta_{c}^{n}f(z)=e^{(n-1)\beta }\Delta_{c}f(z)+ \bigl(1-e^{\beta } \bigr)b_{n}(z),$$
(2.25)

where $$b_{n}(z)=\sum_{i=1}^{n-1}\Delta_{c}^{i}p(z)e^{(n-1-i)\beta }$$.

Using the same argument as the above, it is easy to get $$e^{\alpha }=e ^{n\beta }$$. Then it follows from (2.2) and $$e^{\alpha }=e^{n\beta }$$ that

$$\Delta_{c}^{n}f(z)=e^{(n-1)\beta }\Delta_{c}f(z)+ \bigl(1-e^{(n-1)\beta } \bigr)p(z).$$
(2.26)

If $$\Delta_{c}f(z) \not \equiv \Delta_{c}^{n}f(z)$$, it follows from (2.26) that $$e^{(n-1)\beta }\neq 1$$. Combining (2.25) and (2.26), we have

$$\bigl(1-e^{\beta } \bigr)\sum_{i=1}^{n-1} \Delta_{c}^{i}p(z)e^{(n-1-i)\beta }= \bigl(1-e ^{\beta } \bigr)b_{n}(z)= \bigl(1-e^{(n-1)\beta } \bigr)p(z).$$
(2.27)

If $$p(z)$$ is a constant, then the left-hand side of equation (2.27) is equal to 0, and hence $$p(z)\equiv 0$$, which is a contradiction.

If $$p(z)$$ is a nonconstant polynomial, let $$d=\deg p(z)\ge 1$$, then the left-hand side of equation (2.27) is a polynomial with degree less than d, but the right-hand side of the equation is a polynomial with degree d, which is a contradiction.

Hence $$\Delta_{c}f(z)\equiv \Delta_{c}^{n}f(z)$$, and $$e^{(n-1)\beta }= 1$$.

If $$e^{\beta }\neq 1$$ and $$p(z)$$ is a nonconstant polynomial, then it follows from (2.25)–(2.26) that $$b_{n}(z)\equiv 0$$. Thus

$$\sum_{i=1}^{n-1}\Delta_{c}^{i}p(z)e^{(n-1-i)\beta } \equiv 0.$$
(2.28)

Let $$p(z)=a_{m}z^{m}+a_{m-1}z^{m-1}+\cdots +a_{0}$$. It follows that $$\deg \Delta_{c}^{i}p(z)=m-i$$ if $$m\ge i$$. If $$m\ge 2$$, then the left-hand side of (2.28) is a polynomial with degree $$m-1\ge 1$$, which is a contradiction.

Hence $$m=1$$, that is, $$p(z)=a_{1}z+a_{0}$$. Thus $$\Delta_{c}p(z)=a_{1}c \neq 0$$. It follows from (2.28) that $$a_{1}ce^{(n-2)\beta }=0$$, which is a contradiction.

From the above discussion, we obtain that if $$e^{\beta }\neq 1$$, then $$p(z)$$ (≡b) is a nonzero constant, hence

$$\Delta_{c}f(z)=e^{\beta }f(z)+p(z) \bigl(1-e^{\beta } \bigr)=e^{\beta }f(z)+b \bigl(1-e^{\beta } \bigr)=tf(z)+b(1-t),$$

where $$t=e^{\beta }$$ satisfying $$t^{n-1}=1$$.

Thus, Lemma 5 is proved. □

### Lemma 6

Let f be an entire function of finite order $$\rho (f)$$ with zeros $$\{z_{1}, z_{2},\ldots \}\subset \mathbb{C}\backslash \{0\}$$ and a k-fold zero at the origin. Then

$$f(z)=z^{k}\alpha (z)e^{\beta (z)},$$

where α is the canonical product of f formed with the non-null zeros of f, and β is a polynomial of degree $$\le \rho (f)$$.

## Proof of Theorem 5

### Proof

Since the order of f is finite, and f, $$\Delta_{c}f$$, $$\Delta_{c}^{n}f$$ share ∞ and $$p(z)$$ CM, obviously $$(\Delta _{c}^{n}f(z)-p(z))/(f(z)-p(z))$$ and $$(\Delta_{c}f(z)-p(z))/(f(z)-p(z))$$ have no zeros and poles. By Lemmas 1 and  6 , we have

$$\frac{\Delta_{c}^{n}f(z)-p(z)}{f(z)-p(z)}=e^{\alpha (z)}, \quad\quad \frac{ \Delta_{c}f(z)-p(z)}{f(z)-p(z)}=e^{\beta (z)},$$
(3.1)

where $$\alpha (z)$$ and $$\beta (z)$$ are two polynomials with degree $$\le \rho (f)$$.

Using the same discussion as in Lemma 5, we deduce that f cannot be a rational function. Hence, f is a transcendental meromorphic function, and $$T(r,p)=S(r,f)$$.

Set $$F(z):=f(z)-p(z)$$, then $$T(r,f)=T(r,F)+S(r,f)$$ and $$T(r,p)=S(r,F)$$.

Obviously, we have

$$f(z)=F(z)+p(z),\quad\quad \Delta_{c}f(z)=\Delta_{c}F(z)+ \Delta_{c}p(z), \quad\quad \Delta_{c}^{n}f(z)= \Delta_{c}^{n}F(z)+\Delta_{c}^{n}p(z).$$

Rewrite (3.1) as

$$\frac{\Delta_{c}^{n}F(z)+\Delta_{c}^{n}p(z)-p(z)}{F(z)}=e^{\alpha (z)}, \quad\quad \frac{\Delta_{c}F(z)+\Delta_{c}p(z)-p(z)}{F(z)}=e^{\beta (z)}.$$
(3.2)

Since $$p(z)$$ is a nonconstant polynomial, it follows that $$\Delta_{c}^{n}p(z)-p(z)\not \equiv 0$$ and $$\Delta_{c}p(z)-p(z) \not \equiv 0$$. Set

$$\phi (z):=\frac{(p(z)-\Delta_{c}^{n}p(z))\Delta_{c}F(z)-(p(z)-\Delta _{c}p(z))\Delta_{c}^{n}F(z)}{F(z)}.$$
(3.3)

Next, we consider two cases.

Case 1. $$\phi (z)\not \equiv 0$$. Then, by $$T(r,p)=S(r,F)$$, Lemma 1, and Lemma 2, we get

$$m(r,\phi )=S(r,F).$$
(3.4)

By (3.2)–(3.3), we can rewrite $$\phi (z)$$ as

$$\phi (z)= \bigl(p(z)-\Delta_{c}^{n}p(z) \bigr)e^{\beta (z)}- \bigl(p(z)-\Delta_{c}p(z) \bigr)e ^{\alpha (z)}.$$
(3.5)

Since $$p(z)$$ is a polynomial, we deduce that $$N(r,\phi )=S(r,F)$$. Hence, we get

$$T(r,\phi )=m(r,\phi )+N(r,\phi )=S(r,F).$$
(3.6)

Since $$\phi (z)\not \equiv 0$$, by (3.5) we have

$$\bigl(p(z)-\Delta_{c}^{n}p(z) \bigr)\frac{e^{\beta (z)}}{\phi (z)}=1+ \bigl(p(z)-\Delta _{c}p(z) \bigr)\frac{e^{\alpha (z)}}{\phi (z)}.$$
(3.7)

Then by (3.6), (3.7), $$T(r,p)=S(r,F)$$, Property 2, and Property 3, we have

\begin{aligned} T \biggl( r, \bigl(p-\Delta_{c}^{n}p \bigr) \frac{e^{\beta }}{\phi } \biggr) &\le \overline{N} \biggl( r, \bigl(p- \Delta_{c}^{n}p \bigr)\frac{e^{\beta }}{\phi } \biggr) + \overline{N} \biggl( r,\frac{\phi }{(p-\Delta_{c}^{n}p)e^{\beta }} \biggr) \\ & \quad {} + \overline{N} \biggl( r, \frac{1}{(p-\Delta_{c}^{n}p)(e^{\beta }/\phi )-1} \biggr) +S \biggl( r, \bigl(p- \Delta_{c}^{n}p \bigr)\frac{e^{\beta }}{\phi } \biggr) \\ &= \overline{N} \biggl( r, \bigl(p-\Delta_{c}^{n}p \bigr) \frac{e^{\beta }}{\phi } \biggr) + \overline{N} \biggl( r,\frac{\phi }{(p-\Delta_{c}^{n}p)e^{\beta }} \biggr) \\ & \quad {} + \overline{N} \biggl( r,\frac{\phi }{(p-\Delta_{c}p)e^{\alpha }} \biggr) +S \biggl( r, \bigl(p- \Delta_{c}^{n}p \bigr)\frac{e^{\beta }}{\phi } \biggr) \\ &\le S(r,F)+S \biggl( r, \bigl(p-\Delta_{c}^{n}p \bigr) \frac{e^{\beta }}{\phi } \biggr) . \end{aligned}

Hence by ( 3.6 ), Property  1 , and the previous inequality, we get

$$T \bigl(r,e^{\beta } \bigr)=S(r,F).$$
(3.8)

Thus by (3.5)–(3.8), we have

$$T \bigl(r,e^{\alpha } \bigr)=T \biggl( r,\frac{(p-\Delta_{c}^{n}p)e^{\beta }-\phi }{p- \Delta_{c}p} \biggr) =S(r,F).$$
(3.9)

Hence, by Lemma 5 and since $$p(z)$$ is a nonconstant polynomial, we obtain $$f\equiv \Delta_{c}{f}$$.

Case 2. $$\phi (z)\equiv 0$$. That is,

$$\bigl(p(z)-\Delta_{c}^{n}p(z) \bigr)\Delta_{c}F(z)= \bigl(p(z)-\Delta_{c}p(z) \bigr)\Delta _{c}^{n}F(z).$$
(3.10)

By simple calculation, we can rewrite (3.10) as follows:

$$\bigl(p(z)-\Delta_{c}^{n}p(z) \bigr) \bigl( \Delta_{c}f(z)-p(z) \bigr)= \bigl(p(z)-\Delta_{c}p(z) \bigr) \bigl( \Delta_{c}^{n}f(z)-p(z) \bigr).$$
(3.11)

From (3.1) and (3.11), we get

$$\frac{\Delta_{c}^{n}f(z)-p(z)}{\Delta_{c}f(z)-p(z)}=e^{\alpha (z)- \beta (z)}=\frac{p(z)-\Delta_{c}^{n}p(z)}{p(z)-\Delta_{c}p(z)}.$$
(3.12)

Since $$p(z)$$ is a polynomial, it follows from (3.12) that $$e^{\alpha (z)-\beta (z)}$$ is a constant. Suppose that $$e^{\alpha (z)- \beta (z)}=A$$, then we get $$p(z)-\Delta_{c}^{n}p(z)=A(p(z)-\Delta_{c}p(z))$$. It follows that $$A=1$$ and $$p(z)$$ is a constant, which is a contradiction.

This completes the proof of Theorem  5 . □

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## Acknowledgements

Research supported by the NNSF of China (Grant No. 11371149; 11701188) and the Graduate Student Overseas Study Program from South China Agricultural University (Grant No. 2017LHPY003).

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

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Correspondence to Mingliang Fang.

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