# A note on entire functions and their differences

- Chuang-Xin Chen
^{1}and - Zong-Xuan Chen
^{1}Email author

**2013**:587

https://doi.org/10.1186/1029-242X-2013-587

© Chen and Chen; licensee Springer. 2013

**Received: **17 April 2013

**Accepted: **19 November 2013

**Published: **17 December 2013

## Abstract

In this paper, we prove that for a transcendental entire function $f(z)$ of finite order such that $\lambda (f-a(z))<\sigma (f)$, where $a(z)$ is an entire function and satisfies $\sigma (a(z))<1$, *n* is a positive integer and if ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ and $f(z)$ share the function $a(z)$ CM, where *η* ($\in \mathbb{C}$) satisfies ${\mathrm{\Delta}}_{\eta}^{n}f(z)\not\equiv 0$, then

where *c*, ${c}_{1}$ are two nonzero constants.

**MSC:**39A10, 30D35.

## Keywords

## 1 Introduction and results

*f*, and $\sigma (f)$ to denote the order growth of

*f*. For a nonzero constant

*η*, the forward differences ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ are defined (see [4, 5]) by

Throughout this paper, we denote by $S(r,f)$ any function satisfying $S(r,f)=o(T(r,f))$ as $r\to \mathrm{\infty}$, possibly outside a set of *r* of finite logarithmic measure. A meromorphic function $\alpha (z)$ is said to be a small function of $f(z)$ if $T(r,\alpha (z))=S(r,f)$, and we denote by $S(f)$ the set of functions which are small compared to $f(z)$.

Let *f* and *g* be two nonconstant meromorphic functions, and let $a\in \mathbb{C}$. We say that *f* and *g* share the value *a* CM (IM) provided that $f-a$ and $g-a$ have the same zeros counting multiplicities (ignoring multiplicities), that *f* and *g* share the value ∞ CM (IM) provided that *f* and *g* have the same poles counting multiplicities (ignoring multiplicities). Using the same method, we can define that *f* and *g* share the function $a(z)$ CM (IM), where $a(z)\in S(f)\cap S(g)$. Nevanlinna’s four values theorem [6] says that if two nonconstant meromorphic functions *f* and *g* share four values CM, then $f\equiv g$ or *f* is a Möbius transformation of *g*. The condition ‘*f* and *g* share four values CM’ has been weakened to ‘*f* and *g* share two values CM and two values IM’ by Gundersen [7, 8], as well as by Mues [9]. But whether the condition can be weakened to ‘*f* and *g* share three values IM and another value CM’ is still an open question.

In the special case, we recall a well-known conjecture by Brück [10].

**Conjecture**

*Let*

*f*

*be a nonconstant entire function such that hyper order*${\sigma}_{2}(f)<\mathrm{\infty}$

*and*${\sigma}_{2}(f)$

*is not a positive integer*.

*If*

*f*

*and*${f}^{\prime}$

*share the finite value*

*a*

*CM*,

*then*

*where* *c* *is a nonzero constant*.

The conjecture has been verified in the special cases when $a=0$ [10], or when *f* is of finite order [12], or when ${\sigma}_{2}(f)<\frac{1}{2}$ [13].

Recently, many authors [14–17] started to consider sharing values of meromorphic functions with their shifts. Heittokangas *et al.* proved the following theorems.

**Theorem A** (See [15])

*Let*

*f*

*be a meromorphic function with*$\sigma (f)<2$,

*and let*$c\in \mathbb{C}$.

*If*$f(z)$

*and*$f(z+c)$

*share the values*

*a*($\in \mathbb{C}$)

*and*∞

*CM*,

*then*

*for some constant* *τ*.

In [15], Heittokangas *et al.* give the example $f(z)={e}^{{z}^{2}}+1$ which shows that $\sigma (f)<2$ cannot be relaxed to $\sigma (f)\le 2$.

**Theorem B** (See [16])

*Let* *f* *be a meromorphic function of finite order*, *let* $c\in \mathbb{C}$. *If* $f(z)$ *and* $f(z+c)$ *share three distinct periodic functions* ${a}_{1},{a}_{2},{a}_{3}\in \stackrel{\u02c6}{S}(f)$ *with period* *c* *CM* (*where* $\stackrel{\u02c6}{S}(f)=S(f)\cup \{\mathrm{\infty}\}$), *then* $f(z)=f(z+c)$ *for all* $z\in \mathbb{C}$.

Recently, many results of complex difference equations have been rapidly obtained (see [18–25]). In the present paper, we utilize a complex difference equation to consider uniqueness problems.

The main purpose of this paper is to utilize a complex difference equation to study problems concerning sharing values of meromorphic functions and their differences. It is well known that ${\mathrm{\Delta}}_{\eta}f(z)=f(z+\eta )-f(z)$ (where *η* ($\in \mathbb{C}$) is a constant satisfying $f(z+\eta )-f(z)\not\equiv 0$) is regarded as the difference counterpart of ${f}^{\prime}$. So, Chen and Yi [20] considered the problem that ${\mathrm{\Delta}}_{\eta}f(z)$ and $f(z)$ share one value *a* CM and proved the following theorem.

**Theorem C** (See [20])

*Let*

*f*

*be a finite order transcendental entire function which has a finite Borel exceptional value*

*a*,

*and let*

*η*($\in \mathbb{C}$)

*be a constant such that*$f(z+\eta )\not\equiv f(z)$.

*If*${\mathrm{\Delta}}_{\eta}f(z)=f(z+\eta )-f(z)$

*and*$f(z)$

*share the value*

*a*

*CM*,

*then*

*where* *A* *is a nonzero constant*.

**Question 1** What can be said if we consider the forward difference ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ and $f(z)$ share one value or one small function?

In this paper, we answer Question 1 and prove the following theorem.

**Theorem 1.1**

*Let*$f(z)$

*be a finite order transcendental entire function such that*$\lambda (f-a(z))<\sigma (f)$,

*where*$a(z)$

*is an entire function and satisfies*$\sigma (a)<1$.

*Let*

*n*

*be a positive integer*.

*If*${\mathrm{\Delta}}_{\eta}^{n}f(z)$

*and*$f(z)$

*share*$a(z)$

*CM*,

*where*

*η*($\in \mathbb{C}$)

*satisfies*${\mathrm{\Delta}}_{\eta}^{n}f(z)\not\equiv 0$,

*then*

*where* *c*, ${c}_{1}$ *are two nonzero constants*.

In the special case, if we take $a(z)\equiv a$ in Theorem 1.1, we can get the following corollary.

**Corollary 1.1**

*Let*$f(z)$

*be a finite order transcendental entire function which has a finite Borel exceptional value*

*a*.

*Let*

*n*

*be a positive integer*.

*If*${\mathrm{\Delta}}_{\eta}^{n}f(z)$

*and*$f(z)$

*share value*

*a*

*CM*,

*where*

*η*($\in \mathbb{C}$)

*satisfies*${\mathrm{\Delta}}_{\eta}^{n}f(z)\not\equiv 0$,

*then*

*where* *c*, ${c}_{1}$ *are two nonzero constants*.

**Remark 1.1** From Corollary 1.1, we know that $\frac{{\mathrm{\Delta}}_{\eta}^{n}f(z)}{f(z)}={({e}^{{c}_{1}\eta}-1)}^{n}$ and it shows that the quotient of ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ and $f(z)$ is related to *η*, *n* and ${c}_{1}$, but not related to *c*. On the other hand, Corollary 1.1 shows that if *f* has a nonzero finite Borel exceptional value ${b}^{\ast}$, then, for any constant *η* satisfying ${\mathrm{\Delta}}_{\eta}^{n}f(z)\not\equiv 0$, the value ${b}^{\ast}$ is not shared CM by ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ and $f(z)$. See the following example.

**Example 1.1** Suppose that $f(z)={e}^{z}+{b}^{\ast}$, where ${b}^{\ast}$ is a nonzero finite value. Then *f* has a nonzero finite Borel exceptional value ${b}^{\ast}$. For any $\eta \ne 2k\pi i$, $k\in \mathbb{Z}$, the value ${b}^{\ast}$ is not shared CM by ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ and $f(z)$. Observe that ${\mathrm{\Delta}}_{\eta}^{n}f(z)={\sum}_{j=0}^{n}{(-1)}^{j}{C}_{n}^{j}f(z+(n-j)\eta )$, where ${C}_{n}^{j}$ are the binomial coefficients. Thus, for any $\eta \ne 2k\pi i$, $k\in \mathbb{Z}$, we have ${\mathrm{\Delta}}_{\eta}^{n}f(z)={({e}^{\eta}-1)}^{n}\cdot {e}^{z}$. Thus, we can see that $f(z)-{b}^{\ast}={e}^{z}$ has no zero, but ${\mathrm{\Delta}}_{\eta}^{n}f(z)-{b}^{\ast}={({e}^{\eta}-1)}^{n}{e}^{z}-{b}^{\ast}$ has infinitely many zeros. Hence, the value ${b}^{\ast}$ is not shared CM by ${\mathrm{\Delta}}_{\eta}^{n}f(z)$ and $f(z)$.

In the special case, if we take $n=1$ in Theorem 1.1 and $n=1$ in Corollary 1.1, we can obtain the following corollaries.

**Corollary 1.2**

*Let*$f(z)$

*be a finite order transcendental entire function such that*$\lambda (f-a(z))<\sigma (f)$,

*where*$a(z)$

*is an entire function and satisfies*$\sigma (a)<1$.

*If*${\mathrm{\Delta}}_{\eta}f(z)=f(z+\eta )-f(z)$

*and*$f(z)$

*share*$a(z)$

*CM*,

*where*

*η*($\in \mathbb{C}$)

*satisfies*$f(z+\eta )\not\equiv f(z)$,

*then*

*where* *c*, ${c}_{1}$ *are two nonzero constants*.

**Corollary 1.3**

*Let*$f(z)$

*be a finite order transcendental entire function which has a finite Borel exceptional value*

*a*.

*If*${\mathrm{\Delta}}_{\eta}f(z)=f(z+\eta )-f(z)$

*and*$f(z)$

*share value*

*a*

*CM*,

*where*

*η*($\in \mathbb{C}$)

*satisfies*$f(z+\eta )\not\equiv f(z)$,

*then*

*where* *c*, ${c}_{1}$ *are two nonzero constants*.

**Remark 1.2** The Corollary 1.2 shows that if a nonzero polynomial $a(z)$ satisfies $\lambda (f-a)<\sigma (f)$, then $a(z)$ is not shared CM by $\mathrm{\Delta}f(z)$ and $f(z)$. For example, if we take $a(z)\equiv z$, and $\lambda (f-z)<\sigma (f)$ holds, then $\mathrm{\Delta}f(z)$ and $f(z)$ do not have any common fixed point (counting multiplicities). See the following example.

**Example 1.2** Suppose that $f(z)={e}^{z}+z$. Then $f(z)$ satisfies $\lambda (f(z)-z)=0<1=\sigma (f)$ and has no fixed point. But for any $\eta \ne 2k\pi i$, $k\in \mathbb{Z}$, the function ${\mathrm{\Delta}}_{\eta}f(z)=f(z+\eta )-f(z)=({e}^{\eta}-1){e}^{z}+\eta $ has infinitely many fixed points by Milloux’s theorem (see [1, 3]). Hence, the nonzero polynomial $a(z)\equiv z$ is not shared CM by ${\mathrm{\Delta}}_{\eta}f(z)$ and $f(z)$.

**Remark 1.3** From Corollary 1.3, we can see that under the hypothesis of Theorem C, we can get the expression of $f(z)$, that is, $f(z)=c{e}^{{c}_{1}z}$. Thus, we know that the constant *A* in Theorem C is related to *η* and ${c}_{1}$, but not related to *c*. Actually, from the proof of Lemma 2.9, we have $A={e}^{{c}_{1}\eta}-1$ (obviously, we can obtain $A\ne -1$). Hence, Corollary 1.3 contains and improves Theorem C. Obviously, Theorem 1.1 generalizes Theorem C.

## 2 Lemmas for the proof of theorems

**Lemma 2.1** (See [21])

*Let*

*f*

*be a meromorphic function with a finite order*

*σ*,

*η*

*be a nonzero constant*.

*Let*$\epsilon >0$

*be given*,

*then there exists a subset*$E\subset (1,\mathrm{\infty})$

*with finite logarithmic measure such that for all*

*z*

*satisfying*$|z|=r\notin E\cup [0,1]$,

*we have*

*Suppose that*$n\ge 2$

*and let*${f}_{1}(z),\dots ,{f}_{n}(z)$

*be meromorphic functions and*${g}_{1}(z),\dots ,{g}_{n}(z)$

*be entire functions such that*

- (i)
${\sum}_{j=1}^{n}{f}_{j}(z)exp\{{g}_{j}(z)\}=0$;

- (ii)
*when*$1\le j<k\le n$, ${g}_{j}(z)-{g}_{k}(z)$*is not constant*; - (iii)
*when*$1\le j\le n$, $1\le h<k\le n$,$T(r,{f}_{j})=o\left\{T(r,exp\{{g}_{h}-{g}_{k}\})\right\}\phantom{\rule{1em}{0ex}}(r\to \mathrm{\infty},r\notin E),$

*where* $E\subset (1,\mathrm{\infty})$ *has finite linear measure or logarithmic measure*.

*Then* ${f}_{j}(z)\equiv 0$, $j=1,\dots ,n$.

**ε-set** Following Hayman [1], we define an *ε*-set to be a countable union of open discs not containing the origin and subtending angles at the origin whose sum is finite. If *E* is an *ε*-set, then the set of $r\ge 1$, for which the circle $S(0,r)$ meets *E*, has finite logarithmic measure, and for almost all real *θ*, the intersection of *E* with the ray $argz=\theta $ is bounded.

**Lemma 2.3** (See [4])

*Let*

*f*

*be a function transcendental and meromorphic in the plane of order*<1.

*Let*$h>0$.

*Then there exists an*

*ε*-

*set*

*E*

*such that*

*uniformly in* *c* *for* $|c|\le h$.

**Lemma 2.4** (See [25])

*Let*

*f*

*be a transcendental meromorphic solution of finite order*

*ρ*

*of a difference equation of the form*

*where*$U(z,f)$, $P(z,f)$, $Q(z,f)$

*are difference polynomials such that the total degree*$degU(z,f)=n$

*in*$f(z)$

*and its shifts*,

*and*$degQ(z,f)\le n$.

*Moreover*,

*we assume that*$U(z,f)$

*contains just one term of maximal total degree in*$f(z)$

*and its shifts*.

*Then*,

*for each*$\epsilon >0$,

*possibly outside of an exceptional set of finite logarithmic measure*.

**Remark 2.1** From the proof of Lemma 2.4 in [25], we can see that if the coefficients of $U(z,f)$, $P(z,f)$, $Q(z,f)$, namely ${\alpha}_{\lambda}(z)$, satisfy $m(r,{\alpha}_{\lambda})=S(r,f)$, then the same conclusion still holds.

**Lemma 2.5** (See [27])

*Let*${P}_{n}(z),\dots ,{P}_{0}(z)$

*be polynomials such that*${P}_{n}{P}_{0}\not\equiv 0$

*and satisfy*

*Then every finite order transcendental meromorphic solution*$f(z)$ (≢0)

*of the equation*

*satisfies* $\sigma (f)\ge 1$, *and* $f(z)$ *assumes every nonzero value* $a\in \mathbb{C}$ *infinitely often and* $\lambda (f-a)=\sigma (f)$.

**Remark 2.2** If equation (2.2) satisfies condition (2.1) and all ${P}_{j}(z)$ are constants, we can easily see that equation (2.2) does not possess any nonzero polynomial solution.

**Lemma 2.6** (See [27])

*Let*$F(z),{P}_{n}(z),\dots ,{P}_{0}(z)$

*be polynomials such that*$F{P}_{n}{P}_{0}\not\equiv 0$.

*Then every finite order transcendental meromorphic solution*$f(z)$ (≢0)

*of the equation*

*satisfies* $\lambda (f)=\sigma (f)\ge 1$.

**Remark 2.3** From the proof of Lemma 2.5 in [27], we can see that if we replace $f(z+j)$ by $f(z+j\eta )$ ($j=1,\dots ,n$) in equation (2.2) or (2.3), then the corresponding conclusion still holds.

**Lemma 2.7** (See [4])

*Let* *f* *be a function transcendental and meromorphic in the plane which satisfies* ${\underline{lim}}_{r\to \mathrm{\infty}}\frac{T(r,f)}{r}=0$. *Then* $g(z)=f(z+1)-f(z)$ *and* $G(z)=\frac{f(z+1)-f(z)}{f(z)}$ *are both transcendental*.

**Remark 2.4** From the proof of Lemma 2.7 in [4], we can see that, under the same hypotheses of Lemma 2.7, we can obtain the following conclusion: ${\mathrm{\Delta}}_{\eta}f(z)=f(z+\eta )-f(z)$ and $G(z)=\frac{{\mathrm{\Delta}}_{\eta}f(z)}{f(z)}=\frac{f(z+\eta )-f(z)}{f(z)}$ are both transcendental.

**Lemma 2.8**

*Let*$f(z)=H(z){e}^{{c}_{1}z}$,

*where*$H(z)$ (≢0)

*is an entire function such that*$\sigma (H)<1$

*and*${c}_{1}$

*is a nonzero constant*.

*If*${\mathrm{\Delta}}_{\eta}^{n}f(z)\not\equiv 0$

*for some constant*

*η*,

*and*

*holds*, *where* *A* *is a constant*, *then* $H(z)$ *is a constant*.

*Proof*From ${\mathrm{\Delta}}_{\eta}^{n}f(z)\not\equiv 0$, we can see that $A\ne 0$. In order to prove that $H(z)$ is a constant, we only need to prove ${H}^{\prime}(z)\equiv 0$. Substituting $f(z)=H(z){e}^{{c}_{1}z}$ into (2.4), we can obtain

*ε*-set

*E*such that for $j=1,2,\dots ,n$,

*K*is a constant and satisfies

Thus, we can obtain from the equality above that ${e}^{{c}_{1}\eta}=1$ since $n-1\ge 1$. By (2.6) we have $A={({e}^{{c}_{1}\eta}-1)}^{n}=0$, which contradicts $A\ne 0$. Hence $K\ne 0$ and (2.9) implies ${H}^{\prime}(z)\not\equiv 0$. Thus, we can know that $H(z)$ is a nonzero constant.

Thus, Lemma 2.8 is proved. □

**Lemma 2.9**

*Suppose that*$f(z)$

*is a finite order transcendental entire function such that*$\lambda (f-a(z))<\sigma (f)$,

*where*$a(z)$

*is an entire function and satisfies*$\sigma (a)<1$.

*Let*

*n*

*be a positive integer*.

*If*${\mathrm{\Delta}}_{\eta}^{n}f(z)\not\equiv 0$

*for some constant*

*η*($\in \mathbb{C}$),

*and*

*holds*,

*where*

*A*

*is a constant*,

*then*

*where* *c*, ${c}_{1}$ *are two nonzero constants*.

*Proof*Since $f(z)$ is a transcendental entire function of finite order and satisfies $\lambda (f-a(z))<\sigma (f)$, we can write $f(z)$ in the form

*H*(≢0) is an entire function,

*h*is a polynomial with $degh=k$ ($k\ge 1$),

*H*and

*h*satisfy

*k*, and the order of growth of the right-hand side of (2.14) is equal to

*k*. This is a contradiction. Hence, $b(z)\equiv {\mathrm{\Delta}}_{\eta}^{n}a(z)-a(z)\equiv 0$. Namely,

Secondly, we prove that $A\ne 0$. In fact, if $A=0$, we obtain from (2.16) that ${\mathrm{\Delta}}_{\eta}^{n}f(z)\equiv 0$, which contradicts our hypothesis.

Thirdly, we prove that $\sigma (f)=k=1$. On the contrary, we suppose that $\sigma (f)=k\ge 2$. Thus, we will deduce a contradiction for cases $A={(-1)}^{n}$ and $A\ne {(-1)}^{n}$, respectively.

Case 1. Suppose that $A={(-1)}^{n}$. Thus, for a positive integer *n*, there are three subcases: (1) $n=1$; (2) $n=2$; (3) $n\ge 3$.

a contradiction.

*z*satisfying $|z|=r\notin [0,1]\cup {E}_{1}$, we have

so that $\sigma ({Q}_{1}(z))\le {\sigma}_{1}-1+{\epsilon}_{1}<k-1$. Thus, by $deg(h(z+\eta )-h(z))=k-1$ and $\sigma ({Q}_{1})<k-1$, we can see that the order of growth of the left-hand side of (2.17) is equal to $k-1$, and the order of growth of the right-hand side of (2.17) is less than $k-1$. This is a contradiction.

*z*satisfying $|z|=r\notin [0,1]\cup {E}_{2}$, we have

a contradiction.

Case 2. Suppose that $A\ne {(-1)}^{n}$. Thus, for a positive integer *n*, there are two subcases: (1) $n=1$; (2) $n\ge 2$.

Noting the $A+1\ne 0$, we can use the same method as in the proof of Subcase 1.2 and deduce a contradiction.

We can see that ${U}_{3}(z,{Q}_{3}(z))\not\equiv 0$ since $A-{(-1)}^{n}\ne 0$. Noting that ${deg}_{{Q}_{3}}{U}_{3}(z,{Q}_{3}(z))=n-1\ge 1$, we can use the same method as in the proof of Subcase 1.3 and deduce a contradiction.

where *c*, ${c}_{1}$ are two nonzero constants.

Thus, Lemma 2.9 is proved. □

**Remark 2.5** From the proof of Lemma 2.9 or Remark 1.3, we can see that $A\ne -1$ in Lemma 2.9 when $n=1$ and Theorem C. Unfortunately, we cannot obtain $A\ne {(-1)}^{n}$ when $n\ge 2$ in Lemma 2.9. This is because we can get a contradiction from the equality ${e}^{{c}_{1}\eta}-1=-1$, but we cannot obtain a contradiction from the equality ${({e}^{{c}_{1}\eta}-1)}^{n}={(-1)}^{n}$ when $n\ge 2$.

## 3 Proof of Theorem 1.1

where $P(z)$ is a polynomial and $b(z)={\mathrm{\Delta}}_{\eta}^{n}a(z)-a(z)$. Obviously, $\sigma (b(z))\le \sigma (a(z))<1$.

where *A* (≠0) is a constant. If $P(z)\equiv 0$, then, by (3.1), we see that (3.2) holds and $A=1$.

where $k=\sigma (f)\ge 1$, ${a}_{k}\phantom{\rule{0.25em}{0ex}}(\ne 0),{a}_{k-1},\dots ,{a}_{0}$, ${b}_{s}\phantom{\rule{0.25em}{0ex}}(\ne 0),{b}_{s-1},\dots ,{b}_{0}$ are constants. By (3.1), we can see that $0\le degP=s\le degh=k$.

In this case, we prove that $P(z)$ is a constant, that is, $s=0$. To this end, we will deduce a contradiction for the cases $s=k$ and $1\le s<k$, respectively.

Case 1. Suppose that $1\le s=k$. Thus, there are two subcases: (1) $a(z)\not\equiv 0$; (2) $a(z)\equiv 0$.

Thus, applying Lemma 2.2 to (3.4), by (3.5), we can obtain ${g}_{1m}(z)\equiv 0$ ($m=1,2,3$). Clearly, this is a contradiction.

First, we suppose that $H(z)$ is transcendental. Then, noting that $\sigma (H)<1$ implies ${\underline{lim}}_{r\to \mathrm{\infty}}\frac{T(r,H)}{r}=0$, by Lemma 2.7 and Remark 2.4, we know that ${\mathrm{\Delta}}_{\eta}H(z)=H(z+\eta )-H(z)$ is transcendental. Moreover, $\sigma ({\mathrm{\Delta}}_{\eta}H(z))\le \sigma (H(z))<1$ implies ${\underline{lim}}_{r\to \mathrm{\infty}}\frac{T(r,{\mathrm{\Delta}}_{\eta}H)}{r}=0$. Repeating the process above $n-1$ times, we can see that ${\mathrm{\Delta}}_{\eta}^{n}H(z)$ is transcendental. That is, the left-hand side of (3.9) is a transcendental function. Hence (3.9) is impossible.

Secondly, we suppose that $H(z)$ is a nonzero polynomial. Then, noting that ${b}_{k}=-{a}_{k}$, we can see that ${e}^{p(z)+h(z)}$ is a nonzero constant. Thus, from $b(z)=H(z){e}^{p(z)+h(z)}$, we can know that $b(z)$ is a nonzero polynomial. Thus, applying Lemma 2.6 to the equation ${\mathrm{\Delta}}_{\eta}^{n}a(z)-a(z)=b(z)$ and by Remark 2.3, we have $\sigma (a)\ge 1$, a contradiction. Hence, $H(z){e}^{P(z)+h(z)}-b(z)\not\equiv 0$. Thus, since $deg(P+h)\le k-1$, $deg(-h)=k$, $deg(h(z+j\eta )-h(z))=k-1$ ($j=1,2,\dots ,n$) and $\sigma (H)<k$, we see that the order of growth of the left-hand side of (3.6) is equal to *k*, and the order of growth of the right-hand side of (3.6) is less than *k*. This is a contradiction.

Since $H(z)\not\equiv 0$, $\sigma (H)<k$, $degP=s=k$ and $deg(h(z+j\eta )-h(z))=k-1$ ($j=1,2,\dots ,n$), we can see that the order of growth of the left-hand side of (3.10) is equal to *k*, and the order of growth of the right-hand side of (3.10) is less than *k*. This is a contradiction.

Case 2. Suppose that $1\le s<k$. Thus, there are two subcases: (1) $a(z)\not\equiv 0$; (2) $a(z)\equiv 0$.

We assert that $b(z)\not\equiv 0$. In fact, if $b(z)\equiv 0$, then (2.15) obviously holds. Hence, using the same method as in the proof of Lemma 2.9, by Lemma 2.5 and Remarks 2.2-2.3, we can get that $\sigma (a)\ge 1$, a contradiction. Hence, $b(z)\not\equiv 0$. Since $degh=k$, $deg(h(z+j\eta )-h(z))=k-1$ ($j=1,2,\dots ,n$), $degP=s<k$ and $\sigma (H)<k$, we see that the order of growth of the left-hand side of (3.11) is less than *k*, and the order of growth of the right-hand side of (3.11) is equal to *k*. This is a contradiction.

Thus, there are two subcases: (1) $n=1$; (2) $n\ge 2$.

*z*satisfying $|z|=r\notin [0,1]\cup {E}_{4}$, we have

Since $s<k$, we can see that $degP\le k-1$. If $degP<k-1$, then, by (3.15) and $deg(h(z+\eta )-h(z))=k-1$, we can see that the order of growth of the left-hand side of (3.13) is equal to $k-1$, and the order of growth of the right-hand side of (3.13) is equal to deg*P* which is less than $k-1$. This is a contradiction.

If $degP=k-1$, then since $\frac{H(z+\eta )}{H(z)}$ is an entire function and $deg(h(z+\eta )-h(z))=k-1\ge 1$, by (3.15), we can see that the entire function $\frac{H(z+\eta )}{H(z)}{e}^{h(z+\eta )-h(z)}$ has a Borel exceptional value 0, thus the value 1 must be not its Borel exceptional value. Hence, the left-hand side of (3.13), $\frac{H(z+\eta )}{H(z)}{e}^{h(z+\eta )-h(z)}-1$, has infinitely many zeros, but the right-hand side of (3.13), ${e}^{P(z)}$, has no zero. This is a contradiction.

Subcase 2.2.2. Now we suppose that $n\ge 2$. Thus, for *s* ($=degP$), there are two subcases: (1) $s<k-1$; (2) $s=k-1$.

Clearly, this is a contradiction.

where ${P}_{j,t}(z)$ is a polynomial with degree at most $k-2$.

Clearly, this is a contradiction.

In fact, if $nk|\eta {a}_{k}|=|{b}_{k-1}|$, then, by ${b}_{k-1}\ne nk\eta {a}_{k}$, we know that $cos((k-1){\theta}_{0}+{\theta}_{2})\ne 1$, that is, $cos((k-1){\theta}_{0}+{\theta}_{2})<1$, and hence $nk|\eta {a}_{k}|cos((k-1){\theta}_{0}+{\theta}_{2})<nk|\eta {a}_{k}|=|{b}_{k-1}|$. If $nk|\eta {a}_{k}|<|{b}_{k-1}|$, then we have $nk|\eta {a}_{k}|cos((k-1){\theta}_{0}+{\theta}_{2})\le nk|\eta {a}_{k}|<|{b}_{k-1}|$.

a contradiction. Thus, we have proved that *P* is only a constant and (3.2) holds.

Second step. Applying Lemma 2.9 to (3.2), we can obtain the conclusion.

Thus, Theorem 1.1 is proved.

## Declarations

### Acknowledgements

This research was supported by the National Natural Science Foundation of China (Nos. 11171119, 11226090) and supported by the Natural Science Foundation of Guangdong Province, China (No. S2013040014347).

## Authors’ Affiliations

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