A note on entire functions and their differences

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

$$a(z)\equiv 0\text{and}f(z)=c{e}^{c_{1}z},$$

where c, $c_1$ are two nonzero constants.

MSC:39A10, 30D35.

1 Introduction and results

In this paper, we assume that the reader is familiar with the fundamental results and the standard notations of Nevanlinna’s value distribution theory of meromorphic functions (see [1–3]). In addition, we use the notation $\lambda (f)$ for the exponent of convergence of the sequence of zeros of a meromorphic function 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

$$\begin{array}{c}{\mathrm{\Delta }}_{\eta }f(z)={\mathrm{\Delta }}_{\eta }^{1}f(z)=f(z+\eta )-f(z)\text{and}\hfill \\ {\mathrm{\Delta }}_{\eta }^{n+1}f(z)={\mathrm{\Delta }}_{\eta }^{n}f(z+\eta )-{\mathrm{\Delta }}_{\eta }^{n}f(z),n=1,2,\dots .\hfill \end{array}$$

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

$$f^{\prime }-a=c(f-a),$$

where c is a nonzero constant.

The notation ${\sigma }_2(f)$ denotes hyper-order (see [11]) of $f(z)$ which is defined by

$${\sigma }_2(f)=\underset{r\to \mathrm{\infty }}{\overline{lim}}\frac{loglogT(r,f)}{logr}.$$

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

$$f(z+c)-a=\tau (f(z)-a)$$

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 $a_1,a_2,a_3\in \hat{S}(f)$ with period c CM (where $\hat{S}(f)=S(f)\cup \{\mathrm{\infty }\}$), 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

$$a=0\mathit{\text{and}}\frac{f(z+\eta )-f(z)}{f(z)}=A,$$

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

$$a(z)\equiv 0\mathit{\text{and}}f(z)=c{e}^{c_{1}z},$$

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

$$a=0\mathit{\text{and}}f(z)=c{e}^{c_{1}z},$$

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

$$a(z)\equiv 0\mathit{\text{and}}f(z)=c{e}^{c_{1}z},$$

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

$$a=0\mathit{\text{and}}f(z)=c{e}^{c_{1}z},$$

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

$$exp\{-r^{\sigma -1+\epsilon }\}\le \left|\frac{f(z+\eta )}{f(z)}\right|\le exp\left\{r^{\sigma -1+\epsilon }\right\}.$$

Lemma 2.2 (See [11, 26])

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 $1\le j\le n$, $1\le h<k\le n$,

  $$T(r,f_j)=o\left\{T(r,exp\{g_h-g_k\})\right\}(r\to \mathrm{\infty },r\notin E),$$

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

$$f(z+c)-f(z)=c{f}^{\prime }(z)(1+o(1))\mathit{\text{as }}z\to \mathrm{\infty }\mathit{\text{ in }}\mathbb{C}\mathrm{\setminus }E,$$

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

$$U(z,f)P(z,f)=Q(z,f),$$

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 $\epsilon >0$,

$$m(r,P(z,f))=O\left(r^{\rho -1+\epsilon }\right)+S(r,f),$$

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

$$P_n(z)+\cdots +P_0(z)\not\equiv 0.$$

(2.1)

Then every finite order transcendental meromorphic solution $f(z)$ (≢0) of the equation

$$P_n(z)f(z+n)+P_{n-1}(z)f(z+n-1)+\cdots +P_0(z)f(z)=0$$

(2.2)

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

$$P_n(z)f(z+n)+P_{n-1}(z)f(z+n-1)+\cdots +P_0(z)f(z)=F$$

(2.3)

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

$$\frac{{\mathrm{\Delta }}_{\eta }^{n}f(z)}{f(z)}=A$$

(2.4)

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

$$\sum _{j=0}^{n-1}{(-1)}^j{C}_n^j{e}^{(n-j)c_1\eta }H(z+(n-j)\eta )+({(-1)}^n-A)H(z)=0.$$

(2.5)

First, we assert that the sum of all coefficients of equation (2.5) is equal to zero, that is,

$$e^{n{c}_1\eta }-C_n^1{e}^{(n-1)c_1\eta }+\cdots +{(-1)}^{n-1}{C}_n^{n-1}{e}^{c_1\eta }+({(-1)}^n-A)=0.$$

(2.6)

On the contrary, we suppose that

$$e^{n{c}_1\eta }-C_n^1{e}^{(n-1)c_1\eta }+\cdots +{(-1)}^{n-1}{C}_n^{n-1}{e}^{c_1\eta }+({(-1)}^n-A)\ne 0.$$

Thus, applying Lemma 2.5 and Remarks 2.2-2.3 to (2.5), we have $\sigma (H)\ge 1$, a contradiction. Hence, (2.6) holds. Thus, by (2.6) and (2.5), we have

$$\sum _{j=0}^{n-1}{(-1)}^j{C}_n^j{e}^{(n-j)c_1\eta }(H(z+(n-j)\eta )-H(z))=0.$$

(2.7)

By Lemma 2.3, we see that there exists an ε-set E such that for $j=1,2,\dots ,n$,

$$H(z+j\eta )-H(z)=j\eta H^{\prime }(z)(1+o(1))\text{as }z\to \mathrm{\infty }\text{ in }\mathbb{C}\mathrm{\setminus }E.$$

(2.8)

Substituting (2.8) into (2.7), we can get

$$\eta H^{\prime }(z)\cdot K+\eta H^{\prime }(z)\cdot K\cdot o(1)=0\text{as }z\to \mathrm{\infty }\text{ in }\mathbb{C}\mathrm{\setminus }E,$$

(2.9)

where K is a constant and satisfies

$$K=n{e}^{n{c}_1\eta }-C_n^1(n-1)e^{(n-1)c_1\eta }+\cdots +{(-1)}^{n-2}{C}_n^{n-2}2e^{2c_1\eta }+{(-1)}^{n-1}{C}_n^{n-1}{e}^{c_1\eta }.$$

Secondly, we assert that $K\ne 0$. If $n=1$, then $K=e^{c_1\eta }\ne 0$; if $n\ge 2$, on the contrary, we suppose that $K=0$. Then, for $j=0,1,\dots ,n-1$, noting that

$$C_n^j\cdot (n-j)=\frac{n!\cdot (n-j)}{(n-j)!j!}=\frac{(n-1)!\cdot n}{(n-1-j)!j!}=n{C}_{n-1}^j,$$

we have

$$\sum _{j=0}^{n-1}{(-1)}^j{C}_n^j(n-j)e^{(n-j)c_1\eta }=n{e}^{c_1\eta }{(e^{c_1\eta }-1)}^{n-1}=0.$$

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

$$\frac{{\mathrm{\Delta }}_{\eta }^{n}f(z)-a(z)}{f(z)-a(z)}=A$$

(2.10)

holds, where A is a constant, then

$$a(z)\equiv 0,A\ne 0\mathit{\text{and}}f(z)=c{e}^{c_{1}z},$$

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

$$f(z)=a(z)+H(z)e^{h(z)},$$

(2.11)

where H (≢0) is an entire function, h is a polynomial with $degh=k$ ($k\ge 1$), H and h satisfy

$$\lambda (H)=\sigma (H)=\lambda (f-a(z))<\sigma (f)=degh.$$

(2.12)

First, we assert that $a(z)\equiv 0$. Substituting (2.11) into (2.10), we can get that

$$\frac{{\mathrm{\Delta }}_{\eta }^{n}f(z)-a(z)}{f(z)-a(z)}=\frac{{\sum }_{j=0}^n{(-1)}^j{C}_n^{j}H(z+(n-j)\eta )e^{h(z+(n-j)\eta )}+b(z)}{H(z)e^{h(z)}}=A,$$

(2.13)

where $b(z)={\mathrm{\Delta }}_{\eta }^{n}a(z)-a(z)$. Rewrite (2.13) in the form

$$\sum _{j=0}^{n-1}{(-1)}^j{C}_n^{j}H(z+(n-j)\eta )e^{h(z+(n-j)\eta )-h(z)}+({(-1)}^n-A)H(z)=-b(z)e^{-h(z)}.$$

(2.14)

Suppose that $b(z)\not\equiv 0$. Then, from $\sigma (H(z+(n-j)\eta ))=\sigma (H(z))<degh(z)=k$ ($j=0,1,\dots ,n-1$), $deg(h(z+(n-j)\eta )-h(z))=k-1$ and $\sigma (b(z))\le \sigma (a(z))<1\le k$, we can see that the order of growth of the left-hand side of (2.14) is less than 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,

$$a(z+n\eta )-C_n^{1}a(z+(n-1)\eta )+\cdots +{(-1)}^{n-1}{C}_n^{n-1}a(z+\eta )+({(-1)}^n-1)a(z)=0.$$

(2.15)

Suppose that $a(z)\not\equiv 0$. Note that the sum of all coefficients of (2.15) does not vanish. Then we can apply Lemma 2.5 and Remarks 2.2-2.3 to (2.15) and obtain $\sigma (a(z))\ge 1$, which contradicts our hypothesis. Hence, $a(z)\equiv 0$. Thus, (2.13) can be rewritten as

$$\frac{{\mathrm{\Delta }}_{\eta }^{n}f(z)}{f(z)}=\frac{{\sum }_{j=0}^n{(-1)}^j{C}_n^{j}H(z+(n-j)\eta )e^{h(z+(n-j)\eta )-h(z)}}{H(z)}=A.$$

(2.16)

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) $n\ge 3$.

Subcase 1.1. Suppose that $n=1$. Then, by $A=-1$, we can obtain from (2.16) that

$$e^{h(z+\eta )-h(z)}=(1+A)\cdot \frac{H(z)}{H(z+\eta )}\equiv 0,$$

a contradiction.

Subcase 1.2. Suppose that $n=2$. Then, by $A={(-1)}^2=1$ and (2.16), we have

$$e^{h(z+2\eta )-h(z+\eta )}=\frac{2H(z+\eta )}{H(z+2\eta )}.$$

(2.17)

Set $Q_1(z)=\frac{2H(z+2\eta )}{H(z+\eta )}$. Then, from (2.17), we can know that $Q_1(z)$ is a nonconstant entire function. Set $\sigma (H)={\sigma }_1$. Then ${\sigma }_1<\sigma (f)=k$. By Lemma 2.1, we see that for any given ${\epsilon }_1$ ($0<3{\epsilon }_1<k-{\sigma }_1$), there exists a set $E_1\subset (1,\mathrm{\infty })$ of finite logarithmic measure such that for all z satisfying $|z|=r\notin [0,1]\cup E_1$, we have

$$exp\{-r^{{\sigma }_1-1+{\epsilon }_1}\}\le \left|\frac{2H(z+\eta )}{H(z+2\eta )}\right|\le exp\left\{r^{{\sigma }_1-1+{\epsilon }_1}\right\}.$$

(2.18)

Since $Q_1(z)$ is an entire function, by (2.18), we have

$$T(r,Q_1(z))=m(r,Q_1(z))\le m(r,\frac{2H(z+\eta )}{H(z+2\eta )})+O(1)\le r^{{\sigma }_1-1+{\epsilon }_1},$$

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.

Subcase 1.3. Suppose that $n\ge 3$. Then we can obtain from (2.16) that

$$\sum _{j=0}^{n-2}{(-1)}^j{C}_n^j\frac{H(z+(n-j)\eta )}{H(z+\eta )}{e}^{h(z+(n-j)\eta )-h(z+\eta )}+{(-1)}^{n-1}{C}_n^{n-1}=0.$$

(2.19)

Set $Q_2(z)=e^{h(z+2\eta )-h(z+\eta )}$. Then $Q_2(z)$ is a transcendental entire function since $\sigma (Q_2(z))=k-1\ge 1$. For $j=3,4,\dots ,n$, we have

$$e^{h(z+j\eta )-h(z+\eta )}=Q_2(z+(j-2)\eta )Q_2(z+(j-3)\eta )\cdots Q_2(z).$$

Thus, (2.19) can be rewritten as

$$U_2(z,Q_2(z))\cdot Q_2(z)={(-1)}^n{C}_n^{n-1},$$

(2.20)

where

$$\begin{array}{rcl}{U}_2(z,Q_2(z))& =& \frac{H(z+n\eta )}{H(z+\eta )}{Q}_2(z+(n-2)\eta )Q_2(z+(n-3)\eta )\cdots Q_2(z+\eta )\\ -C_n^1\frac{H(z+(n-1)\eta )}{H(z+\eta )}{Q}_2(z+(n-3)\eta )Q_2(z+(n-4)\eta )\cdots Q_2(z+\eta )\\ +\cdots +{(-1)}^{n-2}{C}_n^{n-2}\frac{H(z+2\eta )}{H(z+\eta )}.\end{array}$$

Noting that ${(-1)}^n{C}_n^{n-1}\ne 0$, we can see that $U_2(z,Q_2(z))\not\equiv 0$. Set $\sigma (H)={\sigma }_2$. Then ${\sigma }_2<k$. Since $Q_2(z)$ is of regular growth and $\sigma (Q_2(z))=k-1$, for any given ${\epsilon }_2$ ($0<3{\epsilon }_2<k-{\sigma }_2$) and all $r>r_0$ (>0), we have

$$T(r,Q_2(z))>r^{k-1-{\epsilon }_2}.$$

(2.21)

By Lemma 2.1, we see that for ${\epsilon }_2$, there exists a set $E_2\subset (1,\mathrm{\infty })$ of finite logarithmic measure such that for all z satisfying $|z|=r\notin [0,1]\cup E_2$, we have

$$exp\{-r^{{\sigma }_2-1+{\epsilon }_2}\}\le \left|\frac{H(z+j\eta )}{H(z+\eta )}\right|\le exp\left\{r^{{\sigma }_2-1+{\epsilon }_2}\right\}(j=2,3,\dots ,n).$$

(2.22)

Thus, from (2.21) and (2.22), we can get that for $j=2,3,\dots ,n$,

$$\frac{m(r,\frac{H(z+j\eta )}{H(z+\eta )})}{T(r,Q_2(z))}\le \frac{r^{{\sigma }_2-1+{\epsilon }_2}}{r^{k-1-{\epsilon }_2}}\to 0(r\to \mathrm{\infty }\text{ and }r\notin [0,1]\cup E_2),$$

that is,

$$m(r,\frac{H(z+j\eta )}{H(z+\eta )})=S(r,Q_2)(j=2,3,\dots ,n).$$

(2.23)

Noting that ${deg}_{Q_2}U(z,Q_2)=n-2\ge 1$ and by Lemma 2.4 and Remark 2.1, we have

$$T(r,Q_2)=m(r,Q_2)=S(r,Q_2),$$

a contradiction.

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

Subcase 2.1. Suppose that $n=1$. Thus, (2.16) can be rewritten as

$$\frac{H(z+\eta )}{H(z)}=(A-{(-1)}^n)e^{h(z)-h(z+\eta )}=(A+1)e^{h(z)-h(z+\eta )}.$$

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

Subcase 2.2. Suppose that $n\ge 2$. Then we can obtain from (2.16) that

$$\sum _{j=0}^{n-1}{(-1)}^j{C}_n^j\frac{H(z+(n-j)\eta )}{H(z)}{e}^{h(z+(n-j)\eta )-h(z)}+{(-1)}^n-A=0.$$

(2.24)

Set $Q_3(z)=e^{h(z+\eta )-h(z)}$. Then $Q_3(z)$ is a transcendental entire function since $\sigma (Q_3(z))=k-1\ge 1$. For $j=1,2,\dots ,n$, we have

$$e^{h(z+j\eta )-h(z)}=Q_3(z+(j-1)\eta )Q_3(z+(j-2)\eta )\cdots Q_3(z).$$

Thus, (2.24) can be rewritten as

$$U_3(z,Q_3(z))\cdot Q_3(z)=A-{(-1)}^n,$$

(2.25)

where

$$\begin{array}{rcl}{U}_3(z,Q_3(z))& =& \frac{H(z+n\eta )}{H(z)}{Q}_3(z+(n-1)\eta )Q_3(z+(n-2)\eta )\cdots Q_3(z+\eta )\\ -C_n^1\frac{H(z+(n-1)\eta )}{H(z)}{Q}_3(z+(n-2)\eta )Q_3(z+(n-3)\eta )\cdots Q_3(z+\eta )\\ +\cdots +{(-1)}^{n-1}{C}_n^{n-1}\frac{H(z+\eta )}{H(z)}.\end{array}$$

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.

Thus, we have proved that $\sigma (f)=k=1$. And $f(z)$ can be written as

$$f(z)=H(z)e^{c_{1}z+c_0}=H^{\ast }(z)e^{c_{1}z},$$

(2.26)

where $c_0$, $c_1$ (≠0) are two constants and $H^{\ast }(z)=e^{c_0}H(z)$ (≢0) is an entire function and satisfies

$$\sigma (H^{\ast }(z))=\lambda (H^{\ast }(z))=\lambda (f)<\sigma (f)=1.$$

(2.27)

Thus, by (2.26), (2.27), (2.16) and Lemma 2.8, we can get that $H^{\ast }(z)$ is a nonzero constant, and so, $f(z)$ can be written as

$$f(z)=c{e}^{c_{1}z},$$

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

By the hypotheses of Theorem 1.1, we can write $f(z)$ in the form (2.11), and (2.12) holds. Since ${\mathrm{\Delta }}_{\eta }^{n}f(z)$ and $f(z)$ share an entire function $a(z)$ CM, then

$$\frac{{\mathrm{\Delta }}_{\eta }^{n}f(z)-a(z)}{f(z)-a(z)}=\frac{{\sum }_{j=0}^n{(-1)}^{n-j}{C}_n^{j}H(z+j\eta )e^{h(z+j\eta )}+b(z)}{H(z)e^{h(z)}}=e^{P(z)},$$

(3.1)

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

First step. We prove

$$\frac{{\mathrm{\Delta }}_{\eta }^{n}f(z)-a(z)}{f(z)-a(z)}=A,$$

(3.2)

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

Now suppose that $P(z)\not\equiv 0$ and $degP(z)=s$. Set

$$h(z)=a_k{z}^k+a_{k-1}{z}^{k-1}+\cdots +a_0,P(z)=b_s{z}^s+b_{s-1}{z}^{s-1}+\cdots +b_0,$$

(3.3)

where $k=\sigma (f)\ge 1$, $a_k(\ne 0),a_{k-1},\dots ,a_0$, $b_s(\ne 0),b_{s-1},\dots ,b_0$ 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) $a(z)\equiv 0$.

Subcase 1.1. Suppose that $a(z)\not\equiv 0$. First we suppose that $b_k\ne -a_k$. Then (3.1) is rewritten as

$$g_{11}(z)e^{P(z)}+g_{12}{e}^{-h(z)}+g_{13}{e}^{h_0(z)}=0,$$

(3.4)

where $h_0(z)\equiv 0$ and

$$g_{11}(z)=-H(z);g_{12}(z)=b(z);g_{13}(z)=\sum _{j=0}^n{(-1)}^{n-j}{C}_n^{j}H(z+j\eta )e^{h(z+j\eta )-h(z)}.$$

Since $\sigma (H)<k$, $\sigma (b)<1\le k$ and $deg(h(z+j\eta )-h(z))=k-1$ ($j=1,2,\dots ,n$), we can see that $\sigma (g_{1m}(z))<k$ ($m=1,2,3$). On the other hand, by $b_k\ne -a_k$, we can see that $deg(P-(-h))=deg(P-h_0)=deg(-h-h_0)=k$. Since $e^{P-(-h)}$, $e^{P-h_0}$ and $e^{-h-h_0}$ are of regular growth, and $\sigma (g_{1m})<k$ ($m=1,2,3$), we can see that for $m=1,2,3$,

$$T(r,g_{1m})=o\left(T(r,e^{P-(-h)})\right)=o\left(T(r,e^{P-h_0})\right)=o\left(T(r,e^{-h-h_0})\right).$$

(3.5)

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.

Now we suppose that $b_k=-a_k$. Then (3.1) is rewritten as

$$[H(z)e^{P(z)+h(z)}-b(z)]e^{-h(z)}=\sum _{j=0}^n{(-1)}^{n-j}{C}_n^{j}H(z+j\eta )e^{h(z+j\eta )-h(z)}.$$

(3.6)

We affirm that $H(z)e^{P(z)+h(z)}-b(z)\not\equiv 0$. In fact, if $H(z)e^{P(z)+h(z)}-b(z)\equiv 0$, then, by (3.6), we can obtain

$$\sum _{j=1}^n{(-1)}^{n-j}{C}_n^{j}H(z+j\eta )e^{h(z+j\eta )-h(z)}+{(-1)}^{n}H(z)\equiv 0,$$

(3.7)

this is the special case of (2.14) when $b(z)\equiv 0$ and $A=0$. Hence, using the same method as in the proof of Case 2 in the proof of Lemma 2.9, we can get that $\sigma (f)=k=1$. Hence, substituting $h(z)=c_{1}z+c_0$ into (3.7), we have

$$\sum _{j=0}^n{(-1)}^j{C}_n^j{e}^{(n-j)c_1\eta }H(z+(n-j)\eta )=0.$$

(3.8)

On this occasion, we assert that ${(e^{c_1\eta }-1)}^n=0$. On the contrary, we suppose that ${(e^{c_1\eta }-1)}^n\ne 0$. Then the sum of all coefficients of (3.8) is ${(e^{\eta }-1)}^n$, which does not vanish. By Lemma 2.5 and Remarks 2.2-2.3, we have $\sigma (H)\ge 1$, a contradiction. Hence, ${(e^{c_1\eta }-1)}^n=0$. Thus, $e^{c_1\eta }=1$. Substituting it into (3.8), we have

$$\sum _{j=0}^n{(-1)}^j{C}_n^{j}H(z+(n-j)\eta )=0.$$

(3.9)

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.

Subcase 1.2. Suppose that $a(z)\equiv 0$. Then (3.1) is rewritten as

$$H(z)e^{P(z)}=\sum _{j=0}^n{(-1)}^{n-j}{C}_n^{j}H(z+j\eta )e^{h(z+j\eta )-h(z)}.$$

(3.10)

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) $a(z)\equiv 0$.

Subcase 2.1. Suppose that $a(z)\not\equiv 0$. Then, by (3.1), we can obtain

$$\sum _{j=0}^n{(-1)}^{n-j}{C}_n^{j}H(z+j\eta )e^{h(z+j\eta )-h(z)}-H(z)e^{P(z)}=b(z)e^{-h(z)}.$$

(3.11)

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.

Subcase 2.2. Suppose that $a(z)\equiv 0$. Then, by (3.1), we obtain

$$\sum _{j=1}^n{(-1)}^{n-j}{C}_n^j\frac{H(z+j\eta )}{H(z)}{e}^{h(z+j\eta )-h(z)}+{(-1)}^n=e^{P(z)}.$$

(3.12)

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

Subcase 2.2.1. Suppose that $n=1$. Then (3.12) can be rewritten as

$$\frac{H(z+\eta )}{H(z)}{e}^{h(z+\eta )-h(z)}-1=e^{P(z)}.$$

(3.13)

By (3.13), we see that $\frac{H(z+\eta )}{H(z)}$ is a nonzero entire function. Set $\sigma (H)={\sigma }_4$. Then ${\sigma }_4<\sigma (f)=k$. By Lemma 2.1, we see that for any given ${\epsilon }_4$ ($0<3{\epsilon }_4<k-{\sigma }_4$), there exists a set $E_4\subset (1,\mathrm{\infty })$ of finite logarithmic measure such that for all z satisfying $|z|=r\notin [0,1]\cup E_4$, we have

$$exp\{-r^{{\sigma }_4-1+{\epsilon }_4}\}\le \left|\frac{H(z+\eta )}{H(z)}\right|\le exp\left\{r^{{\sigma }_4-1+{\epsilon }_4}\right\}.$$

(3.14)

Since $\frac{H(z+\eta )}{H(z)}$ is an entire function, by (3.13), we have

$$T(r,\frac{H(z+\eta )}{H(z)})=m(r,\frac{H(z+\eta )}{H(z)})\le r^{{\sigma }_4-1+{\epsilon }_4},$$

so that

$$\sigma \left(\frac{H(z+\eta )}{H(z)}\right)\le {\sigma }_4-1+{\epsilon }_4<k-1.$$

(3.15)

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 degP 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) $s=k-1$.

Subcase 2.2.2.1. Now we suppose that $s<k-1$. Set $Q_5(z)=e^{h(z+\eta )-h(z)}$. Since $\sigma (Q_5)=k-1\ge 1$, $Q_5(z)$ is a transcendental entire function. Thus, (3.12) can be rewritten as

$$U_5(z,Q_5(z))\cdot Q_5(z)=e^{P(z)}-{(-1)}^n,$$

(3.16)

where

$$\begin{array}{rcl}{U}_5(z,Q_5(z))& =& \frac{H(z+n\eta )}{H(z)}{Q}_5(z+(n-1)\eta )Q_5(z+(n-2)\eta )\cdots Q_5(z+\eta )\\ -C_n^1\frac{H(z+(n-1)\eta )}{H(z)}{Q}_5(z+(n-2)\eta )Q_5(z+(n-3)\eta )\cdots Q_5(z+\eta )\\ +\cdots +{(-1)}^{n-1}{C}_n^{n-1}\frac{H(z+\eta )}{H(z)}.\end{array}$$

(3.17)

Thus, using the same method as in the proof of Subcase 1.3 in the proof of Lemma 2.9 and noting that $\sigma (e^{P(z)}-{(-1)}^n)=degP<k-1$, we have

$$m(r,e^{P(z)}-{(-1)}^n)=S(r,Q_5).$$

Noting that $n\ge 2$ and so $deg{U}_5(z,Q_5)=n-1\ge 1$. Using the same method as in the proof of Subcase 1.3 in the proof of Lemma 2.9, we can obtain

$$T(r,Q_5)=m(r,Q_5)=S(r,Q_5).$$

Clearly, this is a contradiction.

Subcase 2.2.2.2. Now we suppose that $s=k-1$. Thus, (3.12) is written as

$$\sum _{j=1}^n{(-1)}^{n-j}{C}_n^j\frac{H(z+j\eta )}{H(z)}{e}^{T_j(z)}+{(-1)}^n-e^{P(z)}=0,$$

(3.18)

where $T_j(z)=h(z+j\eta )-h(z)$ ($j=1,2,\dots ,n$). Thus, by (3.3), we have

$$T_j(z)=jk\eta a_k{z}^{k-1}+P_{k-2,j}(z),$$

(3.19)

where $P_{k-2,j}(z)$ is a polynomial with degree at most $k-2$. Thus, we have

$$T_j(z)-T_t(z)=(j-t)k\eta a_k{z}^{k-1}+P_{j,t}(z)(1\le j\ne t\le n),$$

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

First, we suppose that there is some $j_0$ ($1\le j_0\le n$) such that $j_{0}k\eta a_k=b_{k-1}$, that is, $deg(T_{j_0}(z)-P(z))\le k-2$. Thus, (3.18) can be written as

$$\sum _{1\le j\le n,j\ne j_0}{(-1)}^{n-j}{C}_n^j\frac{H(z+j\eta )}{H(z)}{e}^{h(z+j\eta )-h(z)}+B_{j_0}(z)e^{h(z+j_0\eta )-h(z)}={(-1)}^{n+1},$$

(3.20)

where

$$B_{j_0}(z)={(-1)}^{n-j_0}{C}_n^{n-j_0}\frac{H(z+j_0\eta )}{H(z)}-e^{P(z)+h(z)-h(z+j_0\eta )}.$$

Set $Q_6(z)=e^{h(z+\eta )-h(z)}$ and $\sigma (H)={\sigma }_6$. Then (3.20) can be rewritten as

$$U_6(z,Q_6(z))\cdot Q_6(z)={(-1)}^{n+1},$$

(3.21)

where

$$\begin{array}{r}{U}_6(z,Q_6(z))\\ =\sum _{1\le j\le n,j\ne j_0}{(-1)}^{n-j}{C}_n^{n-j}\frac{H(z+j\eta )}{H(z)}{Q}_6(z+(j-1)\eta )Q_6(z+(j-2)\eta )\cdots Q_6(z+\eta )\\ +B_{j_0}(z)Q_6(z+(j_0-1)\eta )Q_6(z+(j_0-2)\eta )\cdots Q_6(z+\eta )(j_0\ge 2),\end{array}$$

(3.22)

or

$$\begin{array}{r}{U}_6(z,Q_6(z))\\ =\sum _{2\le j\le n}{(-1)}^{n-j}{C}_n^{n-j}\frac{H(z+j\eta )}{H(z)}{Q}_6(z+(j-1)\eta )Q_6(z+(j-2)\eta )\cdots Q_6(z+\eta )\\ +B_{j_0}(z)(j_0=1).\end{array}$$

(3.23)

Noting that ${(-1)}^{n+1}\ne 0$, we can see that $U_6(z,Q_6(z))\not\equiv 0$. Since $\sigma (H)<k$ and $\sigma (e^{P(z)+h(z)-h(z+j_0\eta )})\le k-2<k-1$, using the same method as in the proof of Subcase 1.3 in the proof of Lemma 2.9, we have

$$m(r,B_{j_0}(z))=S(r,Q_6).$$

(3.24)

Noting that $n\ge 2$ and so $deg{U}_6(z,Q_6)=n-1\ge 1$. Combining (3.21)-(3.24), using the same method as in the proof of Subcase 1.3 in the proof of Lemma 2.9, we can obtain

$$T(r,Q_6)=m(r,Q_6)=S(r,Q_6).$$

Clearly, this is a contradiction.

Secondly, we suppose that $jk\eta a_k\ne b_{k-1}$ for any $1\le j\le n$. Thus, equation (3.18) can be rewritten as

$$e^{P(z)}=e^{b_{k-1}{z}^{k-1}}\cdot e^{P_{k-2}(z)}=\sum _{j=0}^n{(-1)}^{n-j}{C}_n^j\frac{H(z+j\eta )}{H(z)}{e}^{h(z+j\eta )-h(z)},$$

(3.25)

where $P_{k-2}(z)=P(z)-b_{k-1}{z}^{k-1}=b_{k-2}{z}^{k-2}+b_{k-3}{z}^{k-3}+\cdots +b_0$. For dealing with equation (3.25), we just compare $|b_{k-1}|$ with $nk|\eta a_k|$ since $nk|\eta a_k|>(n-1)k|\eta a_k|>\cdots >k|\eta a_k|$. Without loss of generality, we suppose that $nk|\eta a_k|\le |b_{k-1}|$. Let $arg{b}_{k-1}={\theta }_1$, $arg(\eta a_k)={\theta }_2$ and $\sigma (H)={\sigma }_7<k$. Take ${\theta }_0$ such that $cos((k-1){\theta }_0+{\theta }_1)=1$. By Lemma 2.1, we see that for any given ${\epsilon }_7$ ($0<3{\epsilon }_7<k-{\sigma }_7$), there exists a set $E_7\subset (1,\mathrm{\infty })$ of finite logarithmic measure such that for all $z=r{e}^{i{\theta }_0}$ satisfying $|z|=r\notin [0,1]\cup E_7$, we have

$$exp\{-r^{{\sigma }_7-1+{\epsilon }_7}\}\le \left|\frac{H(z+j\eta )}{H(z)}\right|\le exp\left\{r^{{\sigma }_7-1+{\epsilon }_7}\right\}(j=1,\dots ,n).$$

(3.26)

Thus, noting that $e^{P_{k-2}(z)}$ is of regular growth, we can deduce from (3.25) and (3.26) that

$$\begin{array}{rl}\left|e^{b_{k-1}{z}^{k-1}}\right|& =\left|\frac{e^{P(z)}}{e^{P_{k-2}(z)}}\right|\\ \le \frac{|{\sum }_{j=0}^n{(-1)}^j{C}_n^j\frac{H(z+(n-j)\eta )}{H(z)}{e}^{h(z+(n-j)\eta )-h(z)}|}{|e^{b_{k-2}{z}^{k-2}+b_{k-3}{z}^{k-3}+\cdots +b_0}|}\\ \le \frac{(n+1)n!exp\{r^{{\sigma }_7-1+{\epsilon }_7}\}exp\{nk|\eta a_k|cos((k-1){\theta }_0+{\theta }_2)r^{k-1}+O(r^{k-2})\}}{exp\{\frac{|b_{k-2}|}{2}{r}^{k-2}\}},\end{array}$$

that is,

$$\begin{array}{r}exp\{|b_{k-1}|r^{k-1}\}\\ \le exp\{nk|\eta a_k|cos((k-1){\theta }_0+{\theta }_2)r^{k-1}+r^{{\sigma }_7-1+{\epsilon }_7}+O\left(r^{k-2}\right)-\frac{|b_{k-2}|}{2}{r}^{k-2}\}\\ \le exp\{nk|\eta a_k|cos((k-1){\theta }_0+{\theta }_2)r^{k-1}+o\left(r^{k-1}\right)\}.\end{array}$$

(3.27)

We assert that

$$nk|\eta a_k|cos((k-1){\theta }_0+{\theta }_2)<|b_{k-1}|.$$

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 $nk|\eta a_k|cos((k-1){\theta }_0+{\theta }_2)\le nk|\eta a_k|<|b_{k-1}|$.

Thus, taking a positive constant ${\epsilon }_8$ ($0<{\epsilon }_8<\frac{|b_{k-1}|-nk|\eta a_k|cos((k-1){\theta }_0+{\theta }_2)}{3}$), we can deduce from (3.27) that

$$\begin{array}{rl}exp\{|b_{k-1}|r^{k-1}\}& \le exp\{nk|\eta a_k|cos((k-1){\theta }_0+{\theta }_2)r^{k-1}+o\left(r^{k-1}\right)\}\\ \le exp\left\{(|b_{k-1}|-{\epsilon }_8)r^{k-1}\right\},\end{array}$$

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.