Unicity on Entire Function Concerning Its Differential-Difference Operators

Abstract

In this paper, we study the uniqueness of entire function and its differential-difference operators. We prove the following result: let f be a transcendental entire function of finite order, let $\eta$ be a non-zero complex number, $n\ge 1, k\ge 0$ two integers and let a and b be two distinct finite complex numbers. If f and $(\Delta _{\eta }^{n}f)^{(k)}$ share a CM and share b IM, then $f\equiv (\Delta _{\eta }^{n}f)^{(k)}$.

1 Introduction and Main Results

In this paper, we use the standard denotations in the Nevanlinna value distribution theory, (see [6, 18, 19]). Throughout this paper, f(z) is a meromorphic function on the whole complex plane. S(r, f) means that $S(r, f) = o(T(r, f))$, as $r\rightarrow \infty$ outside of a possible exceptional set of finite logarithmic measure. Define

$$\begin{aligned} \rho _{1}(f)= & {} \varlimsup _{r\rightarrow \infty }\frac{log^{+}T(r,f)}{logr},\\ \rho _{2}(f)= & {} \varlimsup _{r\rightarrow \infty }\frac{log^{+}log^{+}T(r,f)}{logr} \end{aligned}$$

by the order and the hyper-order of f, respectively.

Let f(z) be a meromorphic function, and a finite complex number $\eta$, we define its difference operators by

$$\begin{aligned} \Delta _\eta f(z)=f(z+\eta )-f(z), \quad \Delta _\eta ^{n}f(z)=\Delta _{\eta }^{n-1}(\Delta _\eta f(z)). \end{aligned}$$

From above definition, we have

$$\begin{aligned} \Delta _{\eta }^{n}f(z)=\sum _{j=0}^{n}(-1)^{n-j}C_{n}^{j}f(z+j\eta ), \end{aligned}$$

(1.1)

where $C_{n}^{j}$ is a combinatorial number.

Let f(z) and g(z) be two meromorphic functions, and let a be a complex value. We say that f(z) and g(z) share a CM(IM), if $f(z)-a$ and $g(z)-a$ have the same zeros counting multiplicities(ignoring multiplicities).

In 1929, Nevanlinna [18] proved the following celebrated five-value theorem, which stated that two nonconstant meromorphic functions must be identity equal if they share five distinct values in the extended complex plane.

Next, Rubel and Yang [17] considered the uniqueness of an entire function and its derivative. They proved.

Theorem A

Let f(z) be a non-constant entire function, and let a, b be two finite distinct complex values. If f(z) and $f'(z)$ share a, b CM, then $f(z)\equiv f'(z)$.

Li and Yang [10] improved Theorem A and proved

Theorem B

Let f(z) be a non-constant entire function, and let a, b be two finite distinct complex values. If f(z) and $f^{(k)}(z)$ share a CM, and share b IM. Then $f(z)\equiv f^{(k)}(z)$.

In recent years, there has been tremendous interests in developing the value distribution of meromorphic functions with respect to difference analogue, see [1,2,3,4, 7,8,9, 11,12,16, 20]. Heittokangas et al. [7] proved a similar result analogue of Theorem A concerning shift.

Theorem C

Let f(z) be a nonconstant entire function of finite order, let $\eta$ be a nonzero finite complex value, and let a, b be two finite distinct complex values. If f(z) and $f(z+\eta )$ share a, b CM, then $f(z)\equiv f(z+\eta ).$

Chen and Yi [3] proved

Theorem D

Let f(z) be a transcendental entire function of finite non-integer order, let $\eta$ be a non-zero complex number and let a and b be two distinct complex values. If f(z) and $\Delta _{\eta }f(z)$ share a, b CM, then $f(z)\equiv \Delta _{\eta }f(z)$.

They conjectured that the condition “non-integer” of Theorem D can be removed. Zhang and Liao [20] and Liu et al. [12] confirmed the conjecture. They proved

Theorem E

Let f(z) be a transcendental entire function of finite order, let $\eta$ be a non-zero complex number, n be a positive integer, and let a, b be two finite distinct complex values. If f(z) and $\Delta _{\eta }^{n}f(z)$ share a, b CM, then $f(z)\equiv \Delta _{\eta }^{n}f(z)$.

Li et al. [11] proved

Theorem F

Let f be a transcendental entire function of finite order, let $\eta$ be a non-zero complex number, n a positive integer and let a be a nonzero complex number. If f(z) and $\Delta _{\eta }^{n}f(z)$ share 0 CM and share a IM, then $f(z)\equiv \Delta _{\eta }^{n}f(z)$.

The authors posed a question:

Question 1 Let f(z) be a transcendental entire function of finite order, let $\eta \ne 0$ be a finite complex number, n a positive integer and let a, b be two finite distinct complex values. If f(z) and $\Delta _{\eta }^{n}f(z)$ share a CM and share b IM, is $f(z)\equiv \Delta _{\eta }f(z)$?

Recently, Liu and Dong [13] first studied the complex differential-difference equation $f'(z)=f(z+\eta )$, where $\eta \ne 0$ is a finite constant. In [15], Qi et al. investigated the value sharing problem related to $f'(z)$ and $f(z+\eta )$, and proved

Theorem G

Let f be a nonconstant entire function of finite order, and let $a, \eta$ be two nonzero finite complex values. If $f'(z)$ and $f(z+\eta )$ share 0, a CM, then $f'(z)\equiv f(z+\eta )$.

Recently, Qi and Yang [16] improved Theorem G and proved

Theorem H

Let f(z) be a nonconstant entire function of finite order, and let $a, \eta$ be two nonzero finite complex values. If $f'(z)$ and $f(z+\eta )$ share 0 CM and a IM, then $f'(z)\equiv f(z+\eta ).$

A question is that

Question 2 Let f(z) be a transcendental entire function of finite order, let $\eta \ne 0$ be a finite complex number, $n\ge 1, k\ge 0$ two integers and let a, b be two distinct finite complex values. If f(z) and $(\Delta _{\eta }^{n}f(z))^{(k)}$ share a CM and share b IM, is $f(z)\equiv (\Delta _{\eta }^{n}f(z))^{(k)}$?

We give a positive answer to above question. We prove.

Theorem 1

Let f(z) be a transcendental entire function of finite order, let $\eta \ne 0$ be a finite complex number, $n\ge 1, k\ge 0$ two integers and let a, b be two distinct finite complex values. If f(z) and $(\Delta _{\eta }^{n}f(z))^{(k)}$ share a CM and share b IM, then $f(z)\equiv (\Delta _{\eta }^{n}f(z))^{(k)}$.

Immediately, we obtain following result.

Corollary 1

Let f(z) be a transcendental entire function of finite order, let $\eta \ne 0$ be a finite complex number, n a positive integer and let a, b be two distinct finite complex values. If f(z) and $\Delta _{\eta }^{n}f(z)$ share a CM and share b IM, then $f(z)\equiv \Delta _{\eta }^{n}f(z)$.

2 Some Lemmas

Lemma 2.1

[4]. Let f be a nonconstant meromorphic function of finite order, and let $\eta$ be a non-zero complex number. Then

$$\begin{aligned} m\left( r,\frac{f(z+\eta )}{f(z)}\right) =S(r, f), \end{aligned}$$

for all r outside of a possible exceptional set E with finite logarithmic measure.

Lemma 2.2

[18]. Suppose $f_{1},f_{2}$ are two nonconstant meromorphic functions in the complex plane, then

$$\begin{aligned} N(r,f_{1}f_{2})-N\left( r,\frac{1}{f_{1}f_{2}}\right) =N(r,f_{1})+N(r,f_{2})-N\left( r,\frac{1}{f_{1}}\right) -N\left( r,\frac{1}{f_{2}}\right) . \end{aligned}$$

Lemma 2.3

[4, 5]. Let f be a nonconstant meromorphic function of finite order, and let $\eta \ne 0$ be a finite complex number. Then

$$\begin{aligned} T(r,f(z+\eta ))=T(r, f(z))+S(r,f). \end{aligned}$$

Lemma 2.4

[18]. Let f be a nonconstant meromorphic function, and let $P(f)=a_{0}f^{p}+a_{1}f^{p-1}+\cdots +a_{p}(a_{0}\ne 0)$ be a polynomial of degree p with constant coefficients $a_{j}(j=0,1,\ldots ,p)$. Suppose that $b_{j}(j=0,1,\ldots ,q)(q>p)$. Then

$$\begin{aligned} m(r,\frac{P(f)f'}{(f-b_{1})(f-b_{2})\cdots (f-b_{q})})=S(r,f). \end{aligned}$$

Lemma 2.5

Let f and g be two nonconstant entire functions, and let a, b be two finite distinct complex values. If

$$\begin{aligned} H=\frac{f'}{(f-a)(f-b)}-\frac{g'}{(g-a)(g-b)}\equiv 0, \end{aligned}$$

and f and g share a CM, and share b IM, then either $2T(r,f)\le \overline{N}(r,\frac{1}{f-a})+\overline{N}(r,\frac{1}{f-b})+S(r,f)$, or $f\equiv g$.

Proof

Integrating H which leads to

$$\begin{aligned} \frac{g-b}{g-a}=C\frac{f-b}{f-a}, \end{aligned}$$

where C is a nonzero constant.

If $C=1$, then $f\equiv g$. If $C\ne 1$, then from above, we have

$$\begin{aligned} \frac{a-b}{g-a}\equiv \frac{(C-1)f-Cb+a}{f-a}, \end{aligned}$$

and

$$\begin{aligned} T(r,f)=T(r,g)+S(r,f)+S(r,g). \end{aligned}$$

Obviously, $\frac{Cb-a}{C-1}\ne a$ and $\frac{Cb-a}{C-1}\ne b$. It follows that $N(r,\frac{1}{f-\frac{Cb-a}{C-1}})=0$. Then by the Second Fundamental Theorem,

$$\begin{aligned} \begin{aligned} 2T(r,f)&\le \overline{N}(r,f)+\overline{N}(r,\frac{1}{f-a})+\overline{N}(r,\frac{1}{f-b})+\overline{N}(r,\frac{1}{f-\frac{Cb-a}{C-1}})+S(r,f)\\&\le \overline{N}(r,\frac{1}{f-a})+\overline{N}(r,\frac{1}{f-b})+S(r,f), \end{aligned} \end{aligned}$$

that is $2T(r,f)\le \overline{N}(r,\frac{1}{f-a})+\overline{N}(r,\frac{1}{f-b})+S(r,f)$. $\square$

Lemma 2.6

Let f be a transcendental entire function of finite order, let $\eta \ne 0$ be a finite complex number, $n\ge 1 ,k\ge 0$ two integers, and let a be a nonzero complex value. If f and $(\Delta _{\eta }^{n}f)^{(k)}$ share a CM, and $N(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}})=S(r,f)$, then there is a polynomial p such that either $T(r,e^{p})=S(r,f)$, or $(\Delta _{\eta }^{n}f)^{(k)}=He^{p}$, where $H\not \equiv 0$ is a small function of $e^{p}$.

Proof

Since f is a transcendental entire function of finite order, f and $(\Delta _{\eta }^{n}f)^{(k)}$ share a CM, then there is a polynomial p such that

$$\begin{aligned} f-a=e^{p}(\Delta _{\eta }^{n}f)^{(k)}-ae^{p}. \end{aligned}$$

(2.1)

Set $g=(\Delta _{\eta }^{n}f)^{(k)}$. It follows by (2.1) that

$$\begin{aligned} g=(\Delta _{\eta }^{n}ge^{p})^{(k)}-(\Delta _{\eta }^{n}ae^{p})^{(k)}. \end{aligned}$$

(2.2)

Then we rewrite (2.2) as

$$\begin{aligned} 1+\frac{(\Delta _{\eta }^{n}ae^{p})^{(k)}}{g}=De^{p}, \end{aligned}$$

(2.3)

where

$$\begin{aligned} D=\frac{(\Delta _{\eta }^{n}ge^{p})^{(k)}}{ge^{p}}. \end{aligned}$$

(2.4)

Note that $N(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}})=N(r,\frac{1}{g})=S(r,f)$, then by Lemma 2.1 we have

$$\begin{aligned} T(r,D)&= T(r,\frac{(\Delta _{\eta }^{n}ge^{p})^{(k)}}{ge^{p}})\le \sum _{i=0}^{n}T(r,\frac{[g(z+i\eta )e^{p(z+i\eta )}]^{(k)}}{ge^{p}})\nonumber \\&\le \sum _{i=0}^{n}m(r,\frac{[g(z+i\eta )e^{p(z+i\eta )}]^{(k)}}{ge^{p}})+\sum _{i=0}^{n}N(r,\frac{[g(z+i\eta )e^{p(z+i\eta )}]^{(k)}}{ge^{p}})\nonumber \\&\quad +\,S(r,f)\le \sum _{i=0}^{n}N(r,\frac{[g(z+i\eta )e^{p(z+i\eta )}]^{(k)}}{ge^{p}})+S(r,f)=S(r,f). \end{aligned}$$

(2.5)

Next we discuss two cases.

Case 1    $e^{-p}-D\not \equiv 0$. Rewrite (2.3) as

$$\begin{aligned} ge^{p}(e^{-p}-D)=-(\Delta _{\eta }^{n}ae^{p})^{(k)}. \end{aligned}$$

(2.6)

When $D\equiv 0$, (2.6) implies

$$\begin{aligned} g=He^{p}, \end{aligned}$$

(2.7)

where $H\not \equiv 0$ is a small function of $e^{p}$.

When $D\not \equiv 0$, it follows from (2.6) that $N(r,\frac{1}{e^{-p}-D})=S(r,f)$. Then using the Second Fundamental Theorem to $e^{p}$ we can obtain

$$\begin{aligned} T(r,e^{p})&=T(r,e^{-p})+O(1)\nonumber \\&\le \overline{N}(r,e^{-p})+\overline{N}(r,\frac{1}{e^{-p}})+\overline{N}(r,\frac{1}{e^{-p}-D})\nonumber \\&\quad +\, O(1)=S(r,f). \end{aligned}$$

(2.8)

Case 2    $e^{-p}-D\equiv 0$. It implies that $T(r,e^{p})=T(r,e^{-p})+O(1)=S(r,f)$. $\square$

Lemma 2.7

[18]. Let f be a nonconstant meromorphic function, and $R(f)=\frac{P(f)}{Q(f)}$, where

$$\begin{aligned} P(f)=\sum _{k=0}^{p}a_{k}f^{k} \quad and \quad Q(f)=\sum _{j=0}^{q}a_{j}f^{q} \end{aligned}$$

are two mutually prime polynomials in f. If the coefficients ${a_{k}}$ and ${b_{j}}$ are small functions of f and $a_{p}\not \equiv 0$, $b_{q}\not \equiv 0$, then

$$\begin{aligned} T(r,R(f))=max\{p,q\}T(r,f)+S(r,f). \end{aligned}$$

Lemma 2.8

[6, 18, 19]. Suppose that f(z) is a meromorphic function in the complex plane and $p(f)= a_{0}f^{n}(z)+a_{1}f^{n-1}(z)+\cdots +a_{n}$ , where $a_{0}(\not \equiv 0)$, $a_{1}$,$\ldots$,$a_{n}$ are small functions of f(z). Then

$$\begin{aligned} T(r,p(f))= nT(r,f(z))+S(r,f). \end{aligned}$$

Lemma 2.9

[18]. Suppose $f_{1}, f_{2},\ldots , f_{n}(n\ne 2)$ are meromorphic functions and $g_{1}, g_{2},\ldots , g_{n}$ are entire functions such that

• (i) $\sum _{j=1}^{n}f_{j}e^{g_{j}}=0$,

• (ii) $g_{j}-g_{k}$ are not constants for $1\le j<k\le n$,

• (iii) For $1\le j\le n$ and $1\le h<k\le n$,

  $$\begin{aligned} T(r,f_{j})=S(r,e^{g_{j}-g_{k}})(r\rightarrow \infty , r\not \in E). \end{aligned}$$

  Then $f_{j}\equiv 0$ for all $1\le j\le n$.

3 The Proof of Theorem 1

If $f\equiv (\Delta _{\eta }^{n}f)^{(k)}$, there is nothing to prove. Suppose $f\not \equiv (\Delta _{\eta }^{n}f)^{(k)}$. Since f is a transcendental entire function of finite order, f and $(\Delta _{\eta }^{n}f)^{(k)}$ share a CM, then we get

$$\begin{aligned} \frac{(\Delta _{\eta }^{n}f)^{(k)}-a}{f-a}=e^{h}, \end{aligned}$$

(3.1)

where h is a polynomial, and (2.1) implies $h=-p$.

Because f and $(\Delta _{\eta }^{n}f)^{(k)}$ share a CM and share b IM, then by the Second Fundamental Theorem and Lemma 2.1 we have

$$\begin{aligned} \begin{aligned} T(r,f)&\le \overline{N}(r,\frac{1}{f-a})+\overline{N}(r,\frac{1}{f-b})+S(r,f)= \overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-a})\\&\quad +\, \overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-b})\le N(r,\frac{1}{f-(\Delta _{\eta }^{n}f)^{(k)}})+S(r,f)\\&\le T(r,f-(\Delta _{\eta }^{n}f)^{(k)})+S(r,f)\le m(r,f-(\Delta _{\eta }^{n}f)^{(k)})+S(r,f)\\&\le m(r,f)+m(r,1-\frac{(\Delta _{\eta }^{n}f)^{(k)}}{f})+S(r,f)\le T(r,f)+S(r,f). \end{aligned} \end{aligned}$$

That is

$$\begin{aligned} T(r,f)=\overline{N}(r,\frac{1}{f-a})+\overline{N}(r,\frac{1}{f-b})+S(r,f). \end{aligned}$$

(3.2)

According to Lemma 2.1, (3.1) and (3.2) we have

$$\begin{aligned} T(r,f)=T(r,f-(\Delta _{\eta }^{n}f)^{(k)})+S(r,f)=N(r,\frac{1}{f-(\Delta _{\eta }^{n}f)^{(k)}})+S(r,f).\nonumber \\ \end{aligned}$$

(3.3)

and

$$\begin{aligned} T(r,e^{h})=m(r,e^{h})=m(r,\frac{(\Delta _{\eta }^{n}f)^{(k)}-a}{f-a})\le m(r,\frac{1}{f-a})+S(r,f). \end{aligned}$$

(3.4)

Then it follows from (3.1) and (3.3) that

$$\begin{aligned} m(r,\frac{1}{f-a})&=m(r,\frac{e^{h}-1}{f-(\Delta _{\eta }^{n}f)^{(k)}})\nonumber \\&\le m(r,\frac{1}{f-(\Delta _{\eta }^{n}f)^{(k)}})+m(r,e^{h}-1)\nonumber \\&\le T(r,e^{h})+S(r,f). \end{aligned}$$

(3.5)

Then by (3.4) and (3.5)

$$\begin{aligned} T(r,e^{h})= m(r,\frac{1}{f-a})+S(r,f). \end{aligned}$$

(3.6)

On the other hand, we rewrite (3.1) as

$$\begin{aligned} \frac{(\Delta _{\eta }^{n}f)^{(k)}-f}{f-a}=e^{h}-1, \end{aligned}$$

(3.7)

which implies

$$\begin{aligned} \overline{N}(r,\frac{1}{f-b})\le \overline{N}(r,\frac{1}{e^{h}-1})=T(r,e^{h})+S(r,f). \end{aligned}$$

(3.8)

Thus, by (3.2), (3.6) and (3.8)

$$\begin{aligned} \begin{aligned} m(r,\frac{1}{f-a})+N(r,\frac{1}{f-a})&= \overline{N}(r,\frac{1}{f-a})+\overline{N}(r,\frac{1}{f-b})+S(r,f)\\&\le \overline{N}(r,\frac{1}{f-a})+\overline{N}(r,\frac{1}{e^{h}-1})+S(r,f)\\&\le \overline{N}(r,\frac{1}{f-a})+m(r,\frac{1}{f-a})+S(r,f), \end{aligned} \end{aligned}$$

that is

$$\begin{aligned} N(r,\frac{1}{f-a})=\overline{N}(r,\frac{1}{f-a})+S(r,f). \end{aligned}$$

(3.9)

And then

$$\begin{aligned} \overline{N}(r,\frac{1}{f-b})=T(r,e^{h})+S(r,f). \end{aligned}$$

(3.10)

Set

$$\begin{aligned} \varphi =\frac{f'(f-(\Delta _{\eta }^{n}f)^{(k)})}{(f-a)(f-b)}, \end{aligned}$$

(3.11)

and

$$\begin{aligned} \psi =\frac{(\Delta _{\eta }^{n}f)^{(k+1)}(f-(\Delta _{\eta }^{n}f)^{(k)})}{((\Delta _{\eta }^{n}f)^{(k)}-a)((\Delta _{\eta }^{n}f)^{(k)}-b)}. \end{aligned}$$

(3.12)

Easy to know that $\varphi \not \equiv 0$ because of $f\not \equiv (\Delta _{\eta }^{n}f)^{(k)}$, and $\varphi$ is an entire function. By Lemmas 2.1 and 2.4 we have

$$\begin{aligned} \begin{aligned} T(r,\varphi )&=m(r,\varphi )=m(r,\frac{f'(f-(\Delta _{\eta }^{n}f)^{(k)})}{(f-a)(f-b)})+S(r,f)\nonumber \\&\le m(r,\frac{f'f}{(f-a)(f-b)})+m(r,1-\frac{(\Delta _{\eta }^{n}f)^{(k)}}{f})+S(r,f)=S(r,f), \end{aligned} \end{aligned}$$

that is

$$\begin{aligned} T(r,\varphi )=S(r,f). \end{aligned}$$

(3.13)

Let $d=a-j(a-b)(j\ne 0,1)$. Obviously, by Lemmas 2.1 and 2.4, we obtain

$$\begin{aligned} m(r,\frac{1}{f})&=m(r,\frac{1}{(b-a)\varphi }(\frac{f'}{f-a}-\frac{f'}{f-b})(1-\frac{(\Delta _{\eta }^{n}f)^{(k)}}{f}))\nonumber \\&\le m(r,\frac{1}{\varphi })+m(r,\frac{f'}{f-a}-\frac{f'}{f-b})\nonumber \\&\quad +\, m(r,1-\frac{(\Delta _{\eta }^{n}f)^{(k)}}{f})+S(r,f)=S(r,f), \end{aligned}$$

(3.14)

and

$$\begin{aligned} m(r,\frac{1}{f-d})&=m(r,\frac{f'(f-(\Delta _{\eta }^{n}f)^{(k)})}{\varphi (f-a)(f-b)(f-d)})\le m(r,1-\frac{(\Delta _{\eta }^{n}f)^{(k)}}{f})\nonumber \\&\quad +\, m(r,\frac{ff'}{(f-a)(f-b)(f-d)})+S(r,f)=S(r,f). \end{aligned}$$

(3.15)

Set

$$\begin{aligned} \phi =\frac{(\Delta _{\eta }^{n}f)^{(k+1)}}{((\Delta _{\eta }^{n}f)^{(k)}-a)((\Delta _{\eta }^{n}f)^{(k)}-b)}-\frac{f'}{(f-a)(f-b)}. \end{aligned}$$

(3.16)

We discuss two cases.

Case 1   $\phi \equiv 0$. By (3.16), we have

$$\begin{aligned} \frac{f-b}{f-a}=C\frac{(\Delta _{\eta }^{n}f)^{(k)}-b}{(\Delta _{\eta }^{n}f)^{(k)}-a}, \end{aligned}$$

(3.17)

where C is a nonzero constant. Since $f\not \equiv (\Delta _{\eta }^{n}f)^{(k)}$, then by Lemma 2.5 we get

$$\begin{aligned} 2T(r,f)\le \overline{N}(r,\frac{1}{f-a})+\overline{N}(r,\frac{1}{f-b})+S(r,f), \end{aligned}$$

(3.18)

which contradicts with (3.2).

Case 2    $\phi \not \equiv 0$. By (3.3), (3.13) and (3.16), we can obtain

$$\begin{aligned} m(r,f)&=m(r,f-(\Delta _{\eta }^{n}f)^{(k)})+S(r,f)\nonumber \\&=m(r,\frac{\phi (f-(\Delta _{\eta }^{n}f)^{(k)})}{\phi })+S(r,f)=m(r,\frac{\psi -\varphi }{\phi })+S(r,f)\nonumber \\&\le T(r,\frac{\phi }{\psi -\varphi })+S(r,f)\le T(r,\psi -\varphi )+T(r,\phi )+S(r,f)\nonumber \\&\le T(r,\psi )+T(r,\phi )+S(r,f)\nonumber \\&\le T(r,\psi )+\overline{N}(r,\frac{1}{f-b})+S(r,f), \end{aligned}$$

(3.19)

on the other hand,

$$\begin{aligned} T(r,\psi )&=T(r,\frac{(\Delta _{\eta }^{n}f)^{(k+1)}(f-(\Delta _{\eta }^{n}f)^{(k)})}{((\Delta _{\eta }^{n}f)^{(k)}-a)((\Delta _{\eta }^{n}f)^{(k)}-b)})\nonumber \\&=m(r,\frac{(\Delta _{\eta }^{n}f)^{(k+1)}(f-(\Delta _{\eta }^{n}f)^{(k)})}{((\Delta _{\eta }^{n}f)^{(k)}-a)((\Delta _{\eta }^{n}f)^{(k)}-b)})+S(r,f)\nonumber \\&\le m(r,\frac{(\Delta _{\eta }^{n}f)^{(k+1)}}{(\Delta _{\eta }^{n}f)^{(k)}-b})+m(r,\frac{f-(\Delta _{\eta }^{n}f)^{(k)}}{(\Delta _{\eta }^{n}f)^{(k)}-a})\nonumber \\&\le m(r,\frac{1}{f-a})+S(r,f)=\overline{N}(r,\frac{1}{f-b})+S(r,f). \end{aligned}$$

(3.20)

Hence combining (3.19) and (3.20), we obtain

$$\begin{aligned} T(r,f)\le 2\overline{N}(r,\frac{1}{f-b})+S(r,f). \end{aligned}$$

(3.21)

Next, Case 2 is divided into three subcases.

Subcase 2.1   $a=0$. Then by (3.1) and Lemma 2.1 we can get

$$\begin{aligned} m(r,e^{h})=m(r,\frac{(\Delta _{\eta }^{n}f)^{(k)}}{f})=S(r,f). \end{aligned}$$

(3.22)

Then by (3.10), (3.21) and (3.22) we can have $T(r,f)=S(r,f)$, a contradiction.

Subcase 2.2    $b=0$. Then by (3.6), (3.10), (3.21) and Lemma 2.1, we get

$$\begin{aligned} T(r,f)&\le m(r,\frac{1}{f-a})+\overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}})+S(r,f)\nonumber \\&\le m(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}})+\overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}})+S(r,f)\nonumber \\&\le T(r,(\Delta _{\eta }^{n}f)^{(k)})+S(r,f). \end{aligned}$$

(3.23)

From the fact that

$$\begin{aligned} T(r,(\Delta _{\eta }^{n}f)^{(k)})\le T(r,f)+S(r,f), \end{aligned}$$

(3.24)

which follows from (3.23) that

$$\begin{aligned} T(r,f)=T(r,(\Delta _{\eta }^{n}f)^{(k)})+S(r,f). \end{aligned}$$

(3.25)

By the Second Nevanlinna Fundamental Theorem, Lemma 2.1, (3.2) and (3.25), we have

$$\begin{aligned} \begin{aligned} 2T(r,f)&\le 2T(r,(\Delta _{\eta }^{n}f)^{(k)})+S(r,f)\\&\le \overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-a})+\overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}})+\overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-d})+S(r,f)\\&\le \overline{N}(r,\frac{1}{f-a})+\overline{N}(r,\frac{1}{f})+T(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-d})-m(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-d})+S(r,f)\\&\le T(r,f)+T(r,(\Delta _{\eta }^{n}f)^{(k)})-m(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-d})+S(r,f)\\&\le 2T(r,f)-m(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-d})+S(r,f). \end{aligned} \end{aligned}$$

Thus

$$\begin{aligned} m(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-d})=S(r,f). \end{aligned}$$

(3.26)

From the First Fundamental Theorem, Lemmas 2.1, 2.2, (3.14), (3.15), (3.25), (3.26) and that f is a transcendental entire function of finite order, we obtain

$$\begin{aligned} \begin{aligned} m(r,\frac{f-d}{(\Delta _{\eta }^{n}f)^{(k)}-d})&\le m(r,\frac{f}{(\Delta _{\eta }^{n}f)^{(k)}-d})+m(r,\frac{d}{(\Delta _{\eta }^{n}f)^{(k)}-d})+S(r,f)\\&\le T(r,\frac{f}{(\Delta _{\eta }^{n}f)^{(k)}-d})-N(r,\frac{f}{(\Delta _{\eta }^{n}f)^{(k)}-d})+S(r,f)\\&=m(r,\frac{(\Delta _{\eta }^{n}f)^{(k)}-d}{f})+N(r,\frac{(\Delta _{\eta }^{n}f)^{(k)}-d}{f})-N(r,\frac{f}{(\Delta _{\eta }^{n}f)^{(k)}-d})\\&\quad +\, S(r,f)\le N(r,\frac{1}{f})-N(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-d})+S(r,f)\\&=T(r,\frac{1}{f})-T(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-d})+S(r,f)\\&=T(r,f)-T(r,(\Delta _{\eta }^{n}f)^{(k)})+S(r,f)=S(r,f). \end{aligned} \end{aligned}$$

Thus we get

$$\begin{aligned} m(r,\frac{f-d}{(\Delta _{\eta }^{n}f)^{(k)}-d})=S(r,f). \end{aligned}$$

(3.27)

It’s easy to see that $N(r,\psi )=S(r,f)$ and (3.12) can be rewritten as

$$\begin{aligned} \psi =\left[ \frac{a-d}{a}\frac{(\Delta _{\eta }^{n}f)^{(k+1)}}{(\Delta _{\eta }^{n}f)^{(k)}-a}+\frac{d}{a}\frac{(\Delta _{\eta }^{n}f)^{(k+1)}}{(\Delta _{\eta }^{n}f)^{(k)}}\right] \left[ \frac{f-d}{(\Delta _{\eta }^{n}f)^{(k)}-d}-1\right] . \end{aligned}$$

(3.28)

Then by (3.27) and (3.28) we can get

$$\begin{aligned} T(r,\psi )=m(r,\psi )+N(r,\psi )=S(r,f). \end{aligned}$$

(3.29)

By (3.2), (3.19), and (3.29) we get

$$\begin{aligned} \overline{N}(r,\frac{1}{f-a})=S(r,f). \end{aligned}$$

(3.30)

Moreover, by (3.2), (3.25) and (3.30), we have

$$\begin{aligned} m(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}})=S(r,f), \end{aligned}$$

(3.31)

which implies

$$\begin{aligned} \overline{N}(r,\frac{1}{f})=m(r,\frac{1}{f-a})\le m(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}})=S(r,f). \end{aligned}$$

(3.32)

Then by (3.2), (3.30) and (3.32), we obtain $T(r,f)=S(r,f)$, a contradiction.

Subcase 2.3 $ab\ne 0$. So by (3.6), (3.10), (3.21) and the Second Fundamental Theorem of Nevanlinna, we can get

$$\begin{aligned} \begin{aligned} T(r,f)&\le 2m(r,\frac{1}{f-a})+S(r,f)\le 2m(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}})\\&\quad +\, S(r,f)=2T(r,(\Delta _{\eta }^{n}f)^{(k)})-2N(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}})+S(r,f)\\&\le \overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-a})+\overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-b})+\overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}})\\&\quad -\, 2N(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}})+S(r,f)\\&\le T(r,f)-N(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}})+S(r,f), \end{aligned} \end{aligned}$$

which deduces that

$$\begin{aligned} N(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}})=S(r,f). \end{aligned}$$

(3.33)

It follows from the Second Fundamental Theorem of Nevanlinna that

$$\begin{aligned} \begin{aligned} T(r,(\Delta _{\eta }^{n}f)^{(k)})&\le \overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}})+\overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-a})+S(r,f)\\&\le \overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-a})+S(r,f)\\&\le T(r,(\Delta _{\eta }^{n}f)^{(k)})+S(r,f), \end{aligned} \end{aligned}$$

which implies that

$$\begin{aligned} T(r,(\Delta _{\eta }^{n}f)^{(k)})=\overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-a})+S(r,f). \end{aligned}$$

(3.34)

Similarly

$$\begin{aligned} T(r,(\Delta _{\eta }^{n}f)^{(k)})=\overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-b})+S(r,f). \end{aligned}$$

(3.35)

Then by (3.2), (3.34), (3.35) and the fact that f and $(\Delta _{\eta }^{n}f)^{(k)}$ share a CM, and b IM, we get

$$\begin{aligned} T(r,f)=2T(r,(\Delta _{\eta }^{n}f)^{(k)})+S(r,f). \end{aligned}$$

(3.36)

Easy to see from (3.16) that

$$\begin{aligned} T(r,\phi )=N(r,\phi )+S(r,f)\le \overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-b})+S(r,f). \end{aligned}$$

(3.37)

We claim that

$$\begin{aligned} T(r,\phi )=\overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-b})+S(r,f). \end{aligned}$$

(3.38)

Otherwise,

$$\begin{aligned} T(r,\phi )<\overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-b})+S(r,f). \end{aligned}$$

(3.39)

We can deduce from (3.2), (3.12) and Lemma 2.2 that

$$\begin{aligned} \begin{aligned} T(r,\psi )&=T(r,\frac{(\Delta _{\eta }^{n}f)^{(k+1)}(f-(\Delta _{\eta }^{n}f)^{(k)})}{((\Delta _{\eta }^{n}f)^{(k)}-a)((\Delta _{\eta }^{n}f)^{(k)}-b)})\nonumber \\&=m(r,\frac{(\Delta _{\eta }^{n}f)^{(k+1)}(f-(\Delta _{\eta }^{n}f)^{(k)})}{((\Delta _{\eta }^{n}f)^{(k)}-a)((\Delta _{\eta }^{n}f)^{(k)}-b)})+S(r,f)\nonumber \\&\le m(r,\frac{(\Delta _{\eta }^{n}f)^{(k+1)}}{(\Delta _{\eta }^{n}f)^{(k)}-a})+m(r,\frac{f-b}{(\Delta _{\eta }^{n}f)^{(k)}-b}-1)\nonumber \\&\le m(r,\frac{(\Delta _{\eta }^{n}f)^{(k)}-b}{f-b})+N(r,\frac{(\Delta _{\eta }^{n}f)^{(k)}-b}{f-b})-N(r,\frac{f-b}{(\Delta _{\eta }^{n}f)^{(k)}-b})+S(r,f)\\&\le m(r,\frac{1}{f-b})+N(r,\frac{1}{f-b})-N(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-b})+S(r,f)\\&\le T(r,f)-\overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-b})+S(r,f)\le \overline{N}(r,\frac{1}{f-a})+S(r,f), \end{aligned} \end{aligned}$$

which is

$$\begin{aligned} T(r,\psi )\le \overline{N}(r,\frac{1}{f-a})+S(r,f). \end{aligned}$$

(3.40)

Then combining (3.2), (3.39)–(3.40) and the proof of (3.19), we obtain

$$\begin{aligned} \begin{aligned}&\overline{N}(r,\frac{1}{f-a})+\overline{N}(r,\frac{1}{f-b})=T(r,f)+S(r,f)\nonumber \\&\le \overline{N}(r,\frac{1}{f-a})+T(r,\phi )+S(r,f), \end{aligned} \end{aligned}$$

that is

$$\begin{aligned} \overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-b})\le T(r,\phi )+S(r,f), \end{aligned}$$

(3.41)

a contradiction. Similarly, we can also obtain

$$\begin{aligned} T(r,\psi )=\overline{N}(r,\frac{1}{(\Delta _{\eta }^{n}f)^{(k)}-a})+S(r,f). \end{aligned}$$

(3.42)

By Lemma 2.6, if $T(r,e^{p})=S(r,f)$, then we can obtain $T(r,f)=S(r,f)$ from (3.10) and (3.21), a contradiction. Hence,

$$\begin{aligned} (\Delta _{\eta }^{n}f)^{(k)}=He^{p}, \end{aligned}$$

(3.43)

where $H\not \equiv 0$ is a small function of $e^{p}$.

Rewrite (3.16) as

$$\begin{aligned} \phi \equiv \frac{(\Delta _{\eta }^{n}f)^{(k+1)}(f-a)(f-b)-f'((\Delta _{\eta }^{n}f)^{(k)}-a)((\Delta _{\eta }^{n}f)^{(k)}-b)}{(f-a)(f-b)((\Delta _{\eta }^{n}f)^{(k)}-a)((\Delta _{\eta }^{n}f)^{(k)}-b)}. \end{aligned}$$

(3.44)

Combing (2.1) with (3.38), we can set

$$\begin{aligned} P&=(\Delta _{\eta }^{n}f)^{(k+1)}(f-a)(f-b)-f'((\Delta _{\eta }^{n}f)^{(k)}-a)((\Delta _{\eta }^{n}f)^{(k)}-b)\nonumber \\&=\sum _{i=0}^{5}\alpha _{i}e^{ip}, \end{aligned}$$

(3.45)

and

$$\begin{aligned} Q&=(f-a)(f-b)((\Delta _{\eta }^{n}f)^{(k)}-a)((\Delta _{\eta }^{n}f)^{(k)}-b)\nonumber \\&=\sum _{j=0}^{6}\beta _{j}e^{jp}, \end{aligned}$$

(3.46)

where $\alpha _{i} (i=0,1,2,3,4,5)$ and $\beta _{j} (j=0,1,2,3,4,5,6)$ are small functions of $e^{p}$, and $\alpha _{5}\not \equiv 0$, $\beta _{6}\not \equiv 0$.

If P and Q are two mutually prime polynomials in $e^{p}$, then by Lemma 2.8 we can get $T(r,\phi )=6T(r,e^{p})+S(r,f)$. It follows from (3.10), (3.38) and (3.44)–(3.46) that $T(r,f)=S(r,f)$, a contradiction.

If P and Q are not two mutually prime polynomials in $e^{p}$, it is easy to see that the degree of Q is large than P.

According to (3.38), (3.44) and by simple calculation, we must have

$$\begin{aligned} \phi =\frac{c}{(\Delta _{\eta }^{n}f)^{(k)}-b}, \end{aligned}$$

(3.47)

where $c\not \equiv 0$ is a small function of $e^{p}$.

Put (3.47) into (3.16) we have

$$\begin{aligned} \frac{c(\Delta _{\eta }^{n}f)^{(k)}-(\Delta _{\eta }^{n}f)^{(k+1)}-ca}{((\Delta _{\eta }^{n}f)^{(k)}-a)((\Delta _{\eta }^{n}f)^{(k)}-b)}\equiv \frac{-f'}{(f-a)(f-b)}. \end{aligned}$$

(3.48)

We claim that $c(\Delta _{\eta }^{n}f)^{(k)}\equiv (\Delta _{\eta }^{n}f)^{(k+1)}$. Otherwise, (2.1), (3.43) and (3.48) deduce $\sum _{i=0}^{5}\gamma _{i}e^{ip}\equiv 0$, where $\gamma _{5}\not \equiv 0$ and $\gamma _{i} (i=1,2,3,4)$ are small functions of $e^{p}$. Then by Lemma 2.8, we can get $T(r,e^{p})=S(r,f)$. It follows from (3.10) and (3.21) that $T(r,f)=S(r,f)$, a contradiction.

Hence $c(\Delta _{\eta }^{n}f)^{(k)}\equiv (\Delta _{\eta }^{n}f)^{(k+1)}$, and by (2.1), (3.43) and (3.48), we can get

$$\begin{aligned} \begin{aligned} ace^{p}(e^{p}(He^{p})-a)+a-b)=(He^{p}-b)((2p'H+H')e^{2p}-ap'e^{p}), \end{aligned} \end{aligned}$$

which is

$$\begin{aligned}&(ac-2p'H-H')He^{3p}+(a^{2}c-aHp'\\&-b(2p'H+H'))e^{2p}+((a-b)ac-abp')e^{p}=0. \end{aligned}$$

It follows from above and Lemma 2.8 that

$$\begin{aligned} ac=2p'H+H', a^{2}c=aHp'+b(2p'H+H'), (a-b)c=bp', \end{aligned}$$

(3.49)

which implies

$$\begin{aligned} H=b, c=p' \end{aligned}$$

(3.50)

Hence,

$$\begin{aligned} f=be^{2p}-ae^{p}+a. \end{aligned}$$

(3.51)

Since f and $(\Delta _{\eta }^{n}f)^{(k)}$ share b IM, and by (3.35)–(3.36) and (3.51), we get

$$\begin{aligned} be^{2p}-ae^{p}+a-b=b(e^{p}-1)^{2}, \end{aligned}$$

(3.52)

that is

$$\begin{aligned} a=2b. \end{aligned}$$

(3.53)

We claim that deg$p=1$. Otherwise, by (2.6), (3.51) and (3.53), we have

$$\begin{aligned} \sum _{i=1}^{n}C_{i}e^{p(z+i\eta )-p(z)}-C_{n+1}e^{Q}\equiv 0, \end{aligned}$$

(3.54)

where $e^{Q}=1$, and $C_{i}(i=1,\ldots , n+1)$ are small functions of $e^{p(z+i\eta )-p(z)-Q}$ for all $i=1,\ldots , n+1$. Then by Lemma 2.9, we know that $C_{i}\equiv 0$ for all $i=1,\ldots , n+1$, a contradiction. Hence, according to $c(\Delta _{\eta }^{n}f)^{(k)}\equiv (\Delta _{\eta }^{n}f)^{(k+1)}$, we know that

$$\begin{aligned} (\Delta _{\eta }^{n}f)^{(k)}=be^{cz+d}, \end{aligned}$$

(3.55)

where d is a finite constant. It follows from above and (2.6) and (3.55) that

$$\begin{aligned} H=-a(e^{c\eta }-1)^{n}=b. \end{aligned}$$

(3.56)

It follows from (3.53) and (3.56) that

$$\begin{aligned} e^{c\eta }=(-2)^{-\frac{1}{n}}+1. \end{aligned}$$

(3.57)

But we can not get (2.2) from (3.57), a contradiction.

This completes the proof of Theorem 1.