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).

1 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 share ∞ CM.

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 [10] 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 [12] 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 [1] 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., [2–8, 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 [11], El Farissi, Latreuch, Asiri [5], El Farissi, Latreuch, Belaïdi and Asiri [6].

Remark 2

There are examples in [3] 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 and ∞ CM, 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) $\Delta_{c}f\equiv tf+a(1-t)$, where t is a constant satisfying $t^{n-1}=1$,

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

2 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

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

• (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 [15–17]), 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 [1] and Chen and Li [3], 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

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

which is a contradiction.

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

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

which is a contradiction.

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

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

a contradiction.

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

(Hadamard’s factorization theorem [18])

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 $\le \rho (f)$.

3 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 . □