\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 32, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2016/32\hfil Entire functions]
{Entire functions that share a small function with
their difference operators}

\author[A. El Farissi, Z. Latreuch, B. Bela\"idi, A. Asiri \hfil EJDE-2016/32\hfilneg]
{Abdallah El Farissi, Zinel\^aabidine Latreuch, \\
Benharrat Bela\"idi, Asim Asiri}

\address{Abdallah El Farissi \newline
Department of Mathematics and Informatics,
Faculty of Exact Sciences,
University of Bechar, Algeria}
\email{elfarissi.abdallah@yahoo.fr}

\address{Zinel\^aabidine Latreuch \newline
Department of Mathematics,
Laboratory of Pure and Applied Mathematics,
University of Mostaganem (UMAB),
B. P. 227 Mostaganem, Algeria}
\email{z.latreuch@gmail.com}

\address{Benharrat Bela\"idi \newline
Department of Mathematics,
Laboratory of Pure and Applied Mathematics,
University of Mostaganem (UMAB),
B. P. 227 Mostaganem, Algeria}
\email{benharrat.belaidi@univ-mosta.dz}

\address{Asim Asiri \newline
Department of Mathematics,
Faculty of Science, King Abdulaziz University,
P.O. Box 80203, Jeddah 21589, Saudi Arabia}
\email{amkasiri@kau.edu.sa}

\thanks{Submitted July 27, 2015. Published January 21, 2016.}
\subjclass[2010]{30D35, 39A32}
\keywords{Uniqueness; entire functions; difference operators}

\begin{abstract}
 In this article, we study the uniqueness of entire functions
 that share small functions of finite order with their difference 
 operators. In particular, we give a generalization of results in 
 \cite{c2,c3,l3}.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction and statement of results}

 In this article, by meromorphic functions we  mean
meromorphic functions in the complex plane. In what follows, we assume that
the reader is familiar with the fundamental results and the standard
notations of the Nevanlinna's value distribution theory of meromorphic
functions \cite{h3,l1,y1}. 
In addition, we will use $\rho (f) $ to denote the order of growth
of $f$ and $\lambda (f)$ to denote the exponent of convergence
of zeros of $f$, we say that a meromorphic function $\varphi (z)
$ is a small function of $f(z) $ if $T(r,\varphi)=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, we use $S(f) $ to denote the
family of all small functions with respect to $f(z)$. For a
meromorphic function $f(z)$, we define its shift by $f_c(z) =f(z+c) $
and its difference operators by
\begin{equation*}
\Delta _cf(z) =f(z+c) -f(z) ,\quad
\Delta _c^nf(z) =\Delta _c^{n-1}(\Delta_cf(z)) ,\quad
 n\in\mathbb{N},\; n\geq 2.
\end{equation*}
In particular, $\Delta _c^nf(z) =\Delta ^nf(z)$ for the case $c=1$. 


 Let $f$ and $g$ be two meromorphic functions and let $a$ be
a finite nonzero value. We say that $f$ and $g$ share the value $a$ CM
provided that $f-a$ and $g-a$ have the same zeros counting multiplicities.
Similarly, we say that $f$ and $g$ share $a$ IM provided that $f-a$ and $g-a$
have the same zeros ignoring multiplicities. It is well-known that if $f$
and $g$ share four distinct values CM, then $f$ is a M\"{o}bius
transformation of $g$.  Rubel and Yang \cite{r1} proved that if
an entire function $f$ shares two distinct complex numbers CM with its
derivative $f'$, then $f\equiv f'$ . In 1986, Jank et al \cite{j1} proved that for 
a nonconstant meromorphic function $f$, if $f$, $f'$ and $f^{\prime \prime }$
share a finite nonzero value CM, then $f'\equiv f$ . This result
suggests the following question:
\begin{quote}
Question $1$ in \cite{y1}. Let $f$
be a nonconstant meromorphic function, let $a$ be a finite nonzero
constant, and let $n$ and $m$ $(n<m)$
be positive integers. If $f$, $f^{(n) }$
and $f^{(m) }$ share $a$ CM, then
can we get the result $f^{(n) }\equiv f$?
\end{quote}

 The following example from \cite{y2}
shows that the answer to the above question is, in general, negative. 
Let $n$ and $m$ be positive integers satisfying $m>n+1$, and let $b$ be a constant
satisfying $b^n=b^{m}\neq 1$. Set $a=b^n$ and 
$f(z) =e^{bz}+a-1$. Then $f$, $f^{(n) }$ and $f^{(m) }$
share the value $a$ CM, and $f^{(n) }\not\equiv f$. However,
when $f$ is an entire function of finite order and $m=n+1$, the answer to
Question 1 is  positive. In fact, P. Li and C. C. Yang proved the
following:

\begin{theorem}[\cite{l4}] \label{thmA}  Let $f$
be a nonconstant entire function, let $a$ be a finite nonzero
constant, and let $n$ be a positive integer. If $f$, 
$f^{(n) }$ and $f^{(n+1) }$ share
the value $a$ CM, then $f\equiv f'$.
\end{theorem}


 Recently several papers have focussed on the Nevanlinna
theory with respect to difference operators see, e.g.
 \cite{b1,c4,h1,h2}. Many authors started to investigate the
uniqueness of meromorphic functions sharing values with their shifts or
difference operators.  Chen et al \cite{c2,c3} proved a
difference analogue of result of Jank et al and obtained the following
results.

\begin{theorem}[\cite{c2}] \label{thmB}
Let $f(z) $ be a nonconstant entire function of finite order, and let
$a(z)\in S(f)$ ($\not\equiv 0$)
be a periodic entire function with period $c$. 
If $f(z)$, $\Delta _cf(z) $ and $\Delta _c^{2}f(z) $ share 
$a(z) $ CM, then $\Delta _cf\equiv \Delta _c^{2}f$.
\end{theorem}

\begin{theorem}[\cite{c3}] \label{thmC}
Let $f(z)$ be a nonconstant entire function of finite order, and let
$a(z)\in S(f)$ ($\not\equiv 0$)
 be a periodic entire function with period $c$. 
If $f(z)$, $\Delta _cf(z) $  and $\Delta _c^nf(z) $ 
$(n\geq 2) $ share $a(z) $ CM, then $\Delta _cf\equiv \Delta _c^nf$.
\end{theorem}

 \begin{theorem}[\cite{c3}] \label{thmD} 
Let $f(z) $  be a nonconstant entire function of finite order. 
If $f(z)$, $\Delta _cf(z) $  and  $\Delta _c^nf(z) $
 share $0$  CM, then $\Delta _c^nf(z) =C\Delta _cf(z)$,
where $C$ is a nonzero constant.
\end{theorem}

 Recently   Latreuch et al \cite{l3} proved
the following results.


 \begin{theorem}[\cite{l3}] \label{thmE}
Let $f(z) $ be a nonconstant entire function of finite order,
and let $a(z) \in S(f) $ ($\not\equiv 0$)
 be a periodic entire function with period $c$. 
If $f(z) $, $\Delta _c^nf(z) $ and
$\Delta _c^{n+1}f(z) $ ($n\geq 1$) 
share $a(z) $ CM, then $\Delta_c^{n+1}f(z) \equiv \Delta _c^nf(z) $.
\end{theorem}

\begin{theorem}[\cite{l3}] \label{thmF} 
Let $f(z) $ be a nonconstant entire function of finite order. If 
$f(z)$, $\Delta _c^nf(z)$ and $\Delta _c^{n+1}f(z) $ share 
$0$ CM, then $\Delta _c^{n+1}f(z) =C\Delta _c^nf(z)$,
 where $C$ is a nonzero constant.
\end{theorem}

For the case $n=1$,  El Farissi and others gave the
following result.

\begin{theorem}[\cite{e1}] \label{thmG}
Let $f(z) $ be a non-periodic entire function of finite order, and
let $a(z)\in S(f) $ ($\not\equiv 0$) 
 be a periodic entire function with period $c$. If 
$f(z)$, $\Delta _cf(z) $ and $\Delta _c^{2}f(z) $ share $a(z) $
 CM, then $\Delta _cf(z) \equiv f(z) $.
\end{theorem}

We remark that Theorem \ref{thmG} is essentially known in \cite{e1}. For
the convenience of readers, we give his proof in the Lemma \ref{lem2.4}.
Now It is natural to ask the following question:
\begin{quote}
Under the hypotheses of Theorem \ref{thmE}, can we obtain   
$\Delta _cf(z) \equiv f(z) $? 
\end{quote}
The aim of this article is to answer this question and to give a difference 
analogue of result of  Li and Yang \cite{l4}. In fact we obtain the
following results:


\begin{theorem} \label{thm1.1} 
Let $f(z) $ be a nonconstant entire function of finite order such that 
$\Delta_c^nf(z) \not\equiv 0$, and let 
$a(z)\in S(f) $ ($\not\equiv 0$)  be a periodic
entire function with period $c$. 
If $f(z) $, $ \Delta _c^nf(z) $ and 
$\Delta _c^{n+1}f( z) $ $(n\geq 1) $ share 
$a(z) $ CM, then $\Delta _cf(z) \equiv f(z) $.
\end{theorem}

The condition $\Delta_c^nf(z) \not\equiv 0$ is necessary.
 Let us take for example the entire function $f(z) =1+e^{2\pi iz}$ and 
$c=a=1$, then $f-a$ and $\Delta ^nf-a=\Delta ^{n+1}f-a=-1$ have the 
same zeros but $\Delta f\neq f$. On the other hand, under the conditions 
of Theorem \ref{thm1.1}, $\Delta _c^nf(z) \not\equiv 0$ can not be a periodic entire 
function with periodic $c$ because 
$\Delta _c^{n+1}f(z) \equiv \Delta_c^nf(z) $  \cite[Theorem 1.5]{l3}.

 \begin{example} \label{examp1.1}\rm
Let $f(z) =e^{z\ln 2}$ and $c=1$.
Then, for any $a\in\mathbb{C}$, we notice that $f(z) $, $\Delta _c^nf(z) $
and $\Delta _c^{n+1}f(z) $ share $a$ CM for all $n\in\mathbb{N}$ and we can 
easily see that $\Delta _cf(z) \equiv f(z) $. This example satisfies 
Theorem \ref{thm1.1}.
\end{example}

\begin{theorem} \label{thm1.2}
Let $f(z) $ be a nonconstant entire function of finite order such that 
$\Delta_c^nf(z) \not\equiv 0$, and let $a(z) $,
$b(z) \in S(f) $ ($\not\equiv 0$)
such that $b(z) $ is a periodic entire function with
period $c$ and $\Delta _c^{m}a(z) \equiv 0$ ($1\leq m\leq n$).
 If $f(z) -a(z)$, $\Delta _c^nf(z) -b(z) $ and $\Delta _c^{n+1}f(z) -b(z) $
 share $0$ CM, then $\Delta _cf(z) \equiv f(z)
+b(z) +\Delta _ca(z) -a(z) $.
\end{theorem}

The condition $b(z) \not\equiv 0$
is necessary in the proof of Theorem \ref{thm1.2}, for the case $b(z)
\equiv 0$, please see Theorem \ref{thm1.4}.
The condition $\Delta _c^{m}a(z) \equiv 0$ in Theorem \ref{thm1.2} is
 more general than the condition ``periodic entire
function of period $c$''.
 For the case $m=1$, we deduce the following result.

\begin{corollary} \label{coro1.1}
Let $f(z) $ be a nonconstant entire function of finite order 
such that $\Delta_c^nf(z) \not\equiv 0$, and let $a(z) $,
$b(z) \in S(f) $ ($\not\equiv 0$) 
be periodic entire functions with period $c$. If 
$f(z)-a(z)$, $\Delta _c^nf(z) -b(z) $  and 
$\Delta _c^{n+1}f(z) -b(z) $ share $0$  CM, then 
$\Delta _cf(z) \equiv f(z) +b(z) -a(z) $.
\end{corollary}

 \begin{example} \label{examp1.2} \rm
Let $f(z) =e^{z\ln 2}-2$, $a=-1$ and $b=1$. It is clear that $f(z) -a$, 
$\Delta ^nf( z) -b$ and $\Delta ^{n+1}f(z) -b$ share $0$ CM. Here, we
also get $\Delta f(z) =f(z) +b-a$.
\end{example}


 \begin{example} \label{examp1.3} \rm
Let $f(z) =e^{z\ln 2}+z^{3}-1$, $a(z) =z^{3}$ and $b=1$. It is clear that 
$f(z)-z^{3}$, $\Delta ^{4}f(z) -1$ and 
$\Delta ^{5}f(z) -1 $ share $0$ CM. On the other hand, we can verify that 
$\Delta f(z) =f(z) +1+\Delta z^{3}-z^{3}$ which satisfies Theorem \ref{thm1.2}.
\end{example}

\begin{theorem} \label{thm1.3}
Let $f(z) $ be a nonconstant entire function of finite order such that 
$\Delta_c^nf(z) \not\equiv 0$. If $f(z)$,
$\Delta _c^nf(z) $ and $\Delta_c^{n+1}f(z) $ share 
$0$ CM, then $\Delta_cf(z) \equiv Cf(z)$,  where 
$C$ is a nonzero constant.
\end{theorem}



\begin{example} \label{examp1.4} \rm
Let $f(z) =e^{az}$ and $c=1$
where $a\neq 2k\pi i$ $(k\in\mathbb{Z}) $, it is clear that 
$\Delta _c^nf(z) =(e^{a}-1) ^ne^{az}$ for any integer $n\geq 1$. 
So, $f(z)$, $\Delta _c^nf(z) $ and $\Delta_c^{n+1}f(z) $ share $0$ CM 
for all $n\in\mathbb{N}$ and we can easily see that
 $\Delta _cf(z) \equiv Cf(z) $ where $C=e^{a}-1$. This example satisfies 
Theorem \ref{thm1.3}.
\end{example}

\begin{corollary} \label{coro1.2}
Let $f(z) $ be a nonconstant entire function of finite order such that 
$f(z)$, $\Delta _c^nf(z) $ ($\not\equiv0$) and $\Delta _c^{n+1}f(z) $
($n\geq 1$) share $0$  CM. If there exists a
point $z_0$  and an integer $m\geq 1$ such that
$\Delta _c^{m}f(z_0) =f(z_0) \neq 0$,
then $\Delta _c^{m}f(z) \equiv f(z) $.
\end{corollary}

 By combining Theorem \ref{thm1.2} and Theorem \ref{thm1.3} we can prove the
following result.

 \begin{theorem} \label{thm1.4} 
Let $f(z) $ be a nonconstant entire function of finite order such that 
$\Delta _c^nf(z) \not\equiv 0$, and let 
$a(z)\in S(f) $  such that $\Delta _c^{m}a(z)\equiv 0$ ($1\leq m\leq n$). 
If $f(z)-a(z)$, $\Delta _c^nf(z) $
and $\Delta _c^{n+1}f(z) $ share $0$ CM,
then $\Delta _cf(z) \equiv Cf(z) +\Delta
_ca(z) -a(z)$,  where $C$ is a
nonzero constant.
\end{theorem}

\section{Some lemmas}

 \begin{lemma}[\cite{c4}] \label{lem2.1}
 Let $\eta _1,\eta _2$ be two arbitrary complex numbers such that
$\eta _1\neq \eta _2$ and let $f(z) $ be a finite order
meromorphic function. Let $\sigma $ be the order of 
$f(z) $, then for each $\varepsilon >0$, we have
\begin{equation*}
m\Big(r,\frac{f(z+\eta _1) }{f(z+\eta _2) }
\Big) =O(r^{\sigma -1+\varepsilon }).
\end{equation*}
\end{lemma}

 By combining \cite[Theorem 1.4]{c1}  and \cite[Theorem 2.2]{l2},
 we can prove the following lemma.


 \begin{lemma} \label{lem2.2} 
Let $a_0(z), a_1(z) ,\dots ,a_n(z) (\not\equiv 0)$,
$F(z)$ ($\not\equiv 0$) be finite order
meromorphic functions, $c_k$ ($k=0,\dots ,n$) be
constants, unequal to each other. 
If $f$ is a finite order meromorphic solution of the equation
\begin{equation}
a_n(z) f(z+c_n) +\dots +a_1(z)f(z+c_1) +a_0(z) f(z+c_0) 
=F(z)  \label{e2.1}
\end{equation}
with 
\begin{equation*}
\max \{ \rho (a_{i}) ,(i=0,\dots ,n) ,\rho
(F) \} <\rho (f) ,
\end{equation*}
then $\lambda (f) =\rho (f) $.
\end{lemma}

 \begin{proof}
By \eqref{e2.1}  we have
\begin{equation}
\frac{1}{f(z+c_0) }=\frac{1}{F}\Big(a_n\frac{f(
z+c_n) }{f(z+c_0) }+\dots +a_1\frac{f(
z+c_1) }{f(z+c_0) }+a_0\Big) .  \label{e2.2}
\end{equation}
Set $\max \{ \rho (a_j) \text{ }(j=0,\dots
,n) ,\rho (F) \} =\beta <\rho (f) =\rho$.  Then, for any given 
$\varepsilon $ $(0<\varepsilon <\frac{\rho -\beta }{2}) $, we have
\begin{equation}
\sum_{j=0}^n T(r,a_j) +T(r,F)
\leq (n+2) \exp \{ r^{\beta +\varepsilon }\} 
=o( 1) \exp \{ r^{\rho -\varepsilon }\} .  \label{e2.3}
\end{equation}
By \eqref{e2.2}, \eqref{e2.3} and Lemma \ref{lem2.1}, we obtain
\begin{equation}
\begin{aligned}
T(r,f) &=T\big(r,\frac{1}{f}\big) +O(1) \\
&=m( r,\frac{1}{f}) +N\big(r,\frac{1}{f}\big) +O(1) \\
&\leq N\big(r,\frac{1}{f}\big) +m\big(r,\frac{1}{F}\big)
+\sum_{j=0}^n m(r,a_j) \\
&\quad +\sum_{j=1}^n m(r,\frac{f(z+c_j) }{f(z+c_0) }) +O(1) \\
&\leq N\big(r,\frac{1}{f}\big) +T(r,\frac{1}{F}) 
+\sum_{j=0}^n T(r,a_j)
+\sum_{j=1}^n m(r,\frac{f(z+c_j) }{f(z+c_0) }) +O(1) \\
&\leq N\big(r,\frac{1}{f}\big) +O(r^{\rho -1+\varepsilon })
+o(1) \exp \{ r^{\rho -\varepsilon }\} .  
\end{aligned} \label{e2.4}
\end{equation}
From this this inequality we obtain that $\rho (f) \leq \lambda (f) $ and since 
$\lambda (f) \leq \rho (f) $ for every meromorphic function, we deduce that 
$\lambda (f) =\rho (f) $.
\end{proof}

Recently,  Wu and Zheng \cite{w1}
obtained Lemma \ref{lem2.2} by using a different proof.

\begin{lemma}[\cite{y1}] \label{lem2.3} 
Suppose $f_j(z) $ $(j=1,2,\dots ,n+1)$  and 
$g_j(z) $ $(j=1,2,\dots ,n)$ $(n\geq 1)$
are entire functions satisfying the following conditions:
\begin{itemize}
\item[(i)] $\sum_{j=1}^n f_j(z) e^{g_j(z) }\equiv f_{n+1}(z)$;

\item[(ii)] The order of $f_j(z) $\ is less than the order of 
$e^{g_k(z) }$  for $1\leq j\leq n+1$,  $1\leq k\leq n$.
Furthermore, the order of $f_j(z) $ is less than the
order of $e^{g_{h}(z) -g_k(z) }$ for $n\geq 2$  and 
$1\leq j\leq n+1$, $1\leq h<k\leq n$.
\end{itemize}
Then $f_j(z) \equiv 0$, $(j=1,2,\dots n+1)$.
\end{lemma}


 \begin{lemma}[\cite{e1}] \label{lem2.4}
  Let $f(z) $ be a non-periodic entire function of finite order, and
let $a(z)\in S(f) $ ($\not\equiv 0$) 
 be a periodic entire function with period $c$.
 If $f(z)$, $\Delta _cf(z) $and $\Delta _c^{2}f(z) $ share 
$a(z) $  CM, then $\Delta _cf(z) \equiv f(z) $.
\end{lemma}

 \begin{proof} 
Suppose that $\Delta _cf(z)\not\equiv f(z) $. Since $f$, $\Delta _cf$ and 
$\Delta_c^{2}f$ share $a(z)$ CM, we have
\[
\frac{\Delta _cf(z) -a(z) }{f(z) -a(z) }=e^{P(z) },\quad
\frac{\Delta _c^{2}f(z) -a(z) }{f(z) -a(z) }=e^{Q(z) }
\]
where $P$ $(e^{P}\not\equiv 1) $ and $Q$ are polynomials. 
Using Theorem \ref{thmB}, we obtain that $\Delta _c^{2}f\equiv \Delta _cf$, which
means that
\begin{equation}
\alpha (z) =\Delta _cf(z) -f(z) \label{e2.5}
\end{equation}
is entire periodic function of period $c$. By \eqref{e2.5}  we have
\begin{equation*}
\Delta _cf(z) -a(z) =f(z) -a(z) +\alpha (z) ,
\end{equation*}
then
\begin{equation*}
\frac{\Delta _cf(z) -a(z) }{f(z)-a(z) }
=1+\frac{\alpha (z) }{f(z)-a(z) }=e^{P(z) },
\end{equation*}
which is equivalent to
\begin{equation}
f(z) -a(z) =\frac{\alpha (z) }{e^{P(z) }-1}.  \label{e2.6}
\end{equation}
Since $\alpha (z) $ and $a(z) $ are periodic
functions of period $c$,  we have
\begin{gather}
\Delta _cf(z) =\alpha (z) \Delta _c(\frac{1}{e^{P(z) }-1}),  \label{e2.7} \\
\Delta _c^{2}f(z) =\alpha (z) \Delta_c^{2}(\frac{1}{e^{P(z) }-1}) .  \label{e2.8}
\end{gather}
We have the following two subcases:

 (i)  If $P\equiv K$ ($K\neq 2k\pi i$, $K\in\mathbb{Z}$), then by 
\eqref{e2.7}  we have $\Delta _cf(z) =0$. On the other hand, by using 
\eqref{e2.5}, \eqref{e2.6} and $\Delta _cf(z) =0$, we deduce that
\[
f(z) -a(z) =\frac{-f(z) }{e^{K}-1}, \quad K\in\mathbb{C}-\{ 2k\pi i,k\in\mathbb{Z}\}.
\]
So,
\begin{equation*}
f(z) =\frac{e^{K}-1}{e^{K}}a(z) .
\end{equation*}
Hence
\begin{equation*}
T(r,f) =S(r,f) ,
\end{equation*}
which is a contradiction.
\smallskip

(ii)  If $P$ is nonconstant and since 
$\Delta_c^{2}f(z) =\Delta _cf(z) $, then
\begin{equation*}
e^{P_c(z) +P(z) }-3e^{P_{2c}(z)
+P(z) }+2e^{P_{2c}(z) +P_c(z)
}+e^{P_{2c}(z) }-3e^{P_c(z) }+2e^{P(z)}=0
\end{equation*}
which is equivalent to
\begin{equation}
e^{P_c(z) }+(2e^{\Delta _cP(z) }-3)
e^{P_{2c}(z) }=-e^{\Delta _cP_c(z) +\Delta
_cP(z) }+3e^{\Delta _cP(z) }-2.  \label{e2.9}
\end{equation}
Since $\deg \Delta _cP=\deg P-1$, we have
\begin{equation}
\rho (e^{P_c}+(2e^{\Delta _cP}-3) e^{P_{2c}})
=\rho (-e^{\Delta _cP_c+\Delta _cP}+3e^{\Delta _cP}-2)
\leq \deg P-1.  \label{e2.10}
\end{equation}
On the other hand,
\begin{equation}
\rho (e^{P_c}+(2e^{\Delta _cP}-3) e^{P_{2c}})
=\rho (e^{P_c}) =\deg P  \label{e2.11}
\end{equation}
because if we have the contrary
\begin{equation*}
\rho (e^{P_c}+(2e^{\Delta _cP}-3) e^{P_{2c}})
<\rho (e^{P_c}) ,
\end{equation*}
we obtain the following contradiction
\begin{equation*}
\deg P=\rho (\frac{e^{P_c}+(2e^{\Delta _cP}-3)
e^{P_{2c}}}{e^{P_c}}) =\rho (1+(2e^{\Delta
_cP}-3) e^{\Delta P_c}) \leq \deg P-1.
\end{equation*}
By using \eqref{e2.10}  and \eqref{e2.11}, we obtain
$\deg P\leq \deg P-1$
which is a contradiction. This leads to 
$\Delta _cf(z)=f(z) $. Thus, the proof is complete.
\end{proof}

\section{Proof of main results}

\begin{proof}[Proof of the Theorem \ref{thm1.1}]
 Obviously, suppose that $\Delta_cf(z) \not\equiv f(z) $. 
By using Theorem \ref{thmE},
 we have
\begin{gather}
\frac{\Delta _c^nf(z) -a(z) }{f(z)-a(z) }=e^{P(z) },  \label{e3.1} \\
\frac{\Delta _c^{n+1}f(z) -a(z) }{f(z) -a(z) }=e^{P(z) },  \label{e3.2}
\end{gather}
where $P$ $(e^{P}\not\equiv 1) $ is a polynomial. We divide into two cases:
\smallskip


\noindent\textbf{Case 1.} $P$ is a nonconstant polynomial. Setting
now $g(z) =f(z) -a(z) $, 
from \eqref{e3.1} and \eqref{e3.2} we have
\begin{gather}
\Delta _c^ng(z) =e^{P(z) }g(z)+a(z),  \label{e3.3} \\
\Delta _c^{n+1}g(z) =e^{P(z) }g(z)+a(z) .  \label{e3.4}
\end{gather}
By \eqref{e3.3}  and \eqref{e3.4}, we have
\begin{equation*}
g_c(z) =2e^{P-P_c}g(z) +a(z)e^{-P_c}.
\end{equation*}
Using the principle of mathematical induction, we obtain
\begin{equation}
g_{ic}(z) =2^{i}e^{P-P_{ic}}g(z) +a(z)
(2^{i}-1) e^{-P_{ic}},\quad i\geq 1.  \label{e3.5}
\end{equation}
Now, we can rewrite \eqref{e3.3} as
\begin{align*}
\Delta _c^ng(z) 
&=\overset{n}{\underset{i=1}{\sum }}
C_n^{i}(-1) ^{n-i}(2^{i}e^{P-P_{ic}}g(z)
+a(z) (2^{i}-1) e^{-P_{ic}})
+(-1) ^ng(z) \\
&=e^{P}g(z) +a(z) ,
\end{align*}
which implies
\begin{align*}
&\Big(\sum_{i=0}^n C_n^{i}(-1) ^{n-i}2^{i}e^{P-P_{ic}}-e^{P}\Big) g(z)\\
&+a(z) \Big(\sum_{i=0}^n C_n^{i}(
-1) ^{n-i}(2^{i}-1) e^{-P_{ic}}-1\Big) =0.
\end{align*}
Hence
\begin{equation}
A_n(z) g(z) +B_n(z) =0,  \label{e3.6}
\end{equation}
where
\begin{gather*}
A_n(z) =\sum_{i=0}^n C_n^{i}(-1) ^{n-i}2^{i}e^{P-P_{ic}}-e^{P},\\
B_n(z) =a(z) \Big(\sum_{i=0}^n C_n^{i}(-1) ^{n-i}(2^{i}-1)e^{-P_{ic}}-1\Big) .
\end{gather*}
By the same method, we can rewrite \eqref{e3.4} as
\begin{equation}
A_{n+1}(z) g(z) +B_{n+1}(z) =0, \label{e3.7}
\end{equation}
where
\begin{gather*}
A_{n+1}(z) =\sum_{i=0}^{n+1} C_{n+1}^{i}(-1) ^{n+1-i}2^{i}e^{P-P_{ic}}-e^{P},\\
B_{n+1}(z) =a(z) \Big(\sum_{i=0}^{n+1} C_{n+1}^{i}(-1) ^{n+1-i}(2^{i}-1)
e^{-P_{ic}}-1\Big) .
\end{gather*}
We can see easily from the equations \eqref{e3.6}  and 
\eqref{e3.7} that
\begin{equation}
h(z) =A_n(z) B_{n+1}(z) -A_{n+1}(z) B_n(z) \equiv 0.  \label{e3.8}
\end{equation}
On the other hand, we remark that
\begin{align*}
e^{P}B_n(z) 
&=a(z) e^{P}\Big(\sum_{i=0}^n C_n^{i}(-1) ^{n-i}2^{i}e^{-P_{ic}}
-\sum_{i=0}^n C_n^{i}(-1) ^{n-i}e^{-P_{ic}}-1\Big) \\
&=a(z) e^{P}\Big(\sum_{i=0}^n
C_n^{i}(-1) ^{n-i}2^{i}e^{-P_{ic}}-1-\Delta _c^n(e^{-P})\Big) \\
&=a(z) (A_n(z) -e^{P}\Delta _c^n(e^{-P})) .
\end{align*}
Then
\begin{equation}
B_n(z) =a(z) (e^{-P}A_n(z)-\Delta _c^n(e^{-P})) .  \label{e3.9}
\end{equation}
By the same method, we obtain
\begin{equation}
B_{n+1}(z) =a(z) (e^{-P}A_{n+1}(z)-\Delta _c^{n+1}(e^{-P})) .  \label{e3.10}
\end{equation}
Now we return equation \eqref{e3.8}, by using \eqref{e3.9} and \eqref{e3.10}, 
we obtain
\begin{align*}
h(z) &=A_n(z) B_{n+1}(z) -A_{n+1}(z) B_n(z) \\
&=A_n(z) [ a(z) (e^{-P}A_{n+1}(z) -\Delta _c^{n+1}(e^{-P})) ]\\
&-A_{n+1}(z) [ a(z) (e^{-P}A_n(z) -\Delta _c^n(e^{-P})) ] \\
&=a(z) [ A_{n+1}(z) \Delta _c^n(
e^{-P}) -A_n(z) \Delta _c^{n+1}(e^{-P})] \equiv 0.
\end{align*}
Hence
\begin{equation*}
A_{n+1}(z) \Delta _c^n(e^{-P}) -A_n(z) \Delta _c^{n+1}(e^{-P}) \equiv 0.
\end{equation*}
Therefore,
\begin{align*}
&\Delta _c^n(e^{-P}) 
\Big(\sum_{i=0}^{n+1} C_{n+1}^{i}(-1) ^{n+1-i}2^{i}e^{-P_{ic}}-1\Big) \\
&-\Delta _c^{n+1}(e^{-P}) 
\Big(\sum_{i=0} ^n C_n^{i}(-1) ^{n-i}2^{i}e^{-P_{ic}}-1\Big) =0.
\end{align*}
Thus
\begin{align*}
&\Delta _c^n(e^{-P}) \sum_{i=0}^{n+1}
C_{n+1}^{i}(-1) ^{n+1-i}2^{i}e^{-P_{i}c}-\Delta _c^{n+1}(
e^{-P}) \sum_{i=0}^n C_n^{i}(-1)^{n-i}2^{i}e^{-P_{ic}} \\
&=\Delta _c^n(e^{-P}) -\Delta _c^{n+1}(e^{-P})
=\Delta _c^n(2e^{-P}-e^{-P_c}) .
\end{align*}
Then
\begin{align*}
&\sum_{i=0}^n (\Delta _c^n(e^{-P})
C_{n+1}^{i}(-1) ^{n+1-i}-\Delta _c^{n+1}(e^{-P})
C_n^{i}(-1) ^{n-i}) 2^{i}e^{-P_{ic}} \\
&+\Delta _c^n(e^{-P}) 2^{n+1}e^{-P_{(n+1)c}}\\
&=\Delta _c^n(2e^{-P}-e^{-P_c}) ,
\end{align*}
which yields
\begin{equation}
\begin{aligned}
&\sum_{i=0}^n (\Delta _c^n(e^{-P})
C_{n+1}^{i}+\Delta _c^{n+1}(e^{-P}) C_n^{i}) (
-1) ^{n+1-i}2^{i}e^{P_{(n+1) c}-P_{ic}} \\
&+\Delta _c^n(e^{-P}) 2^{n+1}=e^{P_{(n+1)
c}}\Delta _c^n(2e^{-P}-e^{-P_c}) .
\end{aligned}  \label{e3.11}
\end{equation}
Let us denote
\begin{equation*}
\alpha _{i}(z) =(-1) ^{n+1-i}2^{i}e^{P_{(n+1) c}-P_{ic}},\text{ }i=0,\dots ,n
\end{equation*}
and
\begin{equation*}
\alpha _{n+1}(z) =e^{P_{(n+1) c}}\Delta_c^n(2e^{-P}-e^{-P_c}) .
\end{equation*}
It is clear that $\rho (\alpha _{i}) \leq \deg P-1$ for all
 $i=0,2,\dots ,n+1$. Then  \eqref{e3.11} becomes
\begin{equation}
\begin{aligned}
&\sum_{i=0}^n (\Delta _c^n(e^{-P})
C_{n+1}^{i}+\Delta _c^{n+1}(e^{-P}) C_n^{i}) \alpha
_{i}(z) +\Delta _c^n(e^{-P}) 2^{n+1} \\
&=\Big(\sum_{i=0}^n C_{n+1}^{i}\alpha _{i}(
z) +2^{n+1}\Big) \Delta _c^n(e^{-P}) \\
&\quad +\Big(\sum_{i=0}^n C_n^{i}\alpha _{i}(
z)\Big) \Delta _c^{n+1}(e^{-P}) =\alpha _{n+1}(z) . 
\end{aligned} \label{e3.12}
\end{equation}
For convenience, we denote 
\begin{equation*}
M(z) =\sum_{i=0}^n C_{n+1}^{i}\alpha_{i}(z) +2^{n+1},\quad
N(z) =\sum_{i=0}^n C_n^{i}\alpha _{i}(z) .
\end{equation*}
Then \eqref{e3.12} is equivalent to
\begin{equation}
\begin{aligned}
&M(z) \sum_{i=0}^n C_n^{i}(-1)^{n-i}e^{-P_{ic}}+N(z) 
\sum_{i=0}^{n+1} C_{n+1}^{i}(-1) ^{n+1-i}e^{-P_{ic}} \\
&=\sum_{i=0}^n (C_n^{i}M(z)
-C_{n+1}^{i}N(z)) (-1)
^{n-i}e^{-P_{ic}}+N(z) e^{-P_{(n+1) c}}\\
&=\alpha_{n+1}(z) . 
\end{aligned} \label{e3.13}
\end{equation}
As a conclusion,  \eqref{e3.13} can be written as
\begin{equation}
a_{n+1}(z) e^{-P(z+(n+1) c)
}+a_n(z) e^{-P(z+nc) }+\dots +a_0(
z) e^{-P(z) }=\alpha _{n+1}(z) ,  \label{e3.14}
\end{equation}
where $a_0(z) ,\dots ,a_{n+1}(z) $ and $\alpha
_{n+1}(z) $ are entire functions. We distingue the following two
subcases.
\smallskip


\noindent\textbf{(i)}  If $\deg P>1$, then 
\begin{equation}
\max \{ \rho (a_{i}) \text{ }(i=0,\dots ,n+1)
,\rho (\alpha _{n+1}) \} <\deg P.  \label{e3.15}
\end{equation}
To prove that $\alpha _{n+1}(z) \not\equiv 0$, it
suffices to show that $\Delta _c^n(2e^{-P}-e^{-P_c})
\not\equiv 0$. Suppose the contrary. Thus
\begin{equation}
\sum_{i=0}^n C_n^{i}(-1) ^{n-i}(
2e^{-P_{ic}}-e^{-P_{(i+1) c}}) \equiv 0.  \label{e3.16}
\end{equation}
The equation \eqref{e3.16}  can be written as
\begin{equation*}
\sum_{i=0}^{n+1} b_{i}e^{-P_{ic}}\equiv 0,
\end{equation*}
where
\begin{equation*}
b_{i}=\begin{cases}
2(-1) ^n, & \text{if }i=0 \\
(2C_n^{i}+C_n^{i-1}) (-1) ^{n-i}, &\text{if } 1\leq i\leq n \\
-1, &\text{if }i=n+1.
\end{cases}
\end{equation*}
Since $\deg P=m>1$, then for any two integers $j$ and $k$ such that 
$0\leq j<k\leq n+1$, we have
\begin{equation*}
\rho (e^{-P_{kc}+P_{jc}}) =\deg P-1.
\end{equation*}
It is clear now that all the conditions of Lemma \ref{lem2.3} are satisfied. 
So, by Lemma \ref{lem2.3} we obtain $b_{i}\equiv 0$ for all $i=0,\dots ,n+1$, which is
impossible. Then, $\alpha _{n+1}(z) \not\equiv 0$. By Lemma \ref{lem2.2},
\eqref{e3.14}  and \eqref{e3.15} , we deduce that $\lambda
(e^{P}) =\deg P>1$, which is a contradiction.
\smallskip

\noindent\textbf{(ii)}  $\deg P=1$. Suppose now that 
$P(z) =\mu z+\eta $ $(\mu \neq 0) $. Assume that 
$\alpha_{n+1}(z) \equiv 0$. It easy to see that
\begin{equation*}
\Delta _c^n(2e^{-P}-e^{-P_c}) =(2-e^{-\mu c})
\Delta _c^n(e^{-P}) .
\end{equation*}
In the following two subcases, we prove that both of 
$(2-e^{-\mu c}) $ and $\Delta _c^n(e^{-P}) $ are not vanishing.
\smallskip

\noindent\textbf{(A)} Suppose that $2=e^{-\mu c}$. Then for any integer $i$, 
we have $e^{-i\mu c}=2^{i}$ and $e^{-P_{ic}}=2^{i}e^{-P}$, applying that
on  \eqref{e3.6}, we obtain
\begin{gather*}
A_n(z) =\sum_{i=0}^n C_n^{i}(-1) ^{n-i}2^{i}e^{-i\mu c}-e^{P}=3^n-e^{P}, \\
\begin{aligned}
B_n(z) &=a(z) \Big(\sum_{i=0}^n C_n^{i}(-1) ^{n-i}(2^{i}-1)e^{-P_{ic}}-1\Big) \\
&=a(z) (\sum_{i=0}^n C_n^{i}( -1) ^{n-i}(4^{i}-2^{i}) e^{-P}-1) =a(z)
((3^n-1) e^{-P}-1) .
\end{aligned}
\end{gather*}
Then
\begin{equation*}
(3^n-e^{P}) g(z) +a(z) ((3^n-1) e^{-P}-1) =0,
\end{equation*}
which is equivalent to
\begin{equation}
g(z) =a(z) \frac{e^{P}-(3^n-1) }{e^{P}(3^n-e^{P}) }.  \label{e3.17}
\end{equation}
By the same argument as before and  \eqref{e3.7}, we obtain
\begin{equation*}
g(z) =a(z) \frac{e^{P}-(3^{n+1}-1) }{e^{P}(3^{n+1}-e^{P}) },
\end{equation*}
which contradicts \eqref{e3.17}.
\smallskip

\noindent\textbf{(B)} Suppose now that $\Delta _c^n(e^{-P}) \equiv 0$. Then
\begin{align*}
\Delta _c^n(e^{-P})
& =\sum_{i=0}^n C_n^{i}(-1) ^{n-i}e^{-\mu (z+ic) -\eta }\\
&=e^{-P} \sum_{i=0}^n C_n^{i}(-1) ^{n-i}e^{-\mu ic} \\
&=e^{-P}(e^{-\mu c}-1) ^n.
\end{align*}
This together with $\Delta _c^ne^{-P}\equiv 0$ gives 
$(e^{-\mu c}-1) ^n\equiv 0$, which yields $e^{\mu c}\equiv 1$. Therefore, for
any $j\in \mathbb{Z}$,
\begin{equation}
e^{P(z+jc) }=e^{\mu z+\mu jc+\eta }=(e^{\mu c})
^{j}e^{P(z) }=e^{P(z) }.  \label{e3.18}
\end{equation}
On the other hand,  from \eqref{e3.1} we have
\begin{equation}
\Delta _c^nf(z) =e^{P(z) }(f(z) -a(z)) +a(z) .  \label{e3.19}
\end{equation}
By \eqref{e3.18} and \eqref{e3.19}, we have
\begin{equation}
\Delta _c^{n+1}f(z) =e^{P(z) }\Delta _cf(z)  \label{e3.20}
\end{equation}
Combining \eqref{e3.2} and \eqref{e3.20}, we obtain
\begin{equation*}
\Delta _cf(z) =(f(z) -a(z)) +a(z) e^{-P(z) }
\end{equation*}
which means that 
$\Delta _c^{n+1}f(z) =\Delta _c^nf(z) $ for all $n\geq 1$. 
Therefore, $f(z)$, $\Delta _cf(z) $ and 
$\Delta _c^{2}f(z) $ share $a(z) $ CM and by Lemma \ref{lem2.4}
we obtain $\Delta _cf(z) =f(z) $, which
contradicts the hypothesis. Then 
$\Delta _c^n(e^{-P})\not\equiv 0$. From the subcases (A) and (B), 
we can deduce that $\alpha _{n+1}(z) \not\equiv 0$. It is clear that
\begin{equation*}
\max \{ \rho (a_{i}) ,\rho (\alpha _{n+1}),i=0,\dots ,n+1\} <\deg P=1.
\end{equation*}
By using Lemma \ref{lem2.2}, we obtain $\lambda (e^{P}) =\deg P=1$, which
is a contradiction, and $P$ must be a constant.
\smallskip

\noindent\textbf{Case 2.} $P(z) \equiv K$, 
$K\in\mathbb{C}-\{ 2k\pi i,k\in\mathbb{Z}\} $. From \eqref{e3.1} we have
\begin{equation*}
\Delta _c^nf(z) =e^{K}(f(z) -a(z)) +a(z) .
\end{equation*}
Hence
\begin{equation}
\Delta _c^{n+1}f(z) =e^{K}\Delta _cf(z) .
\label{e3.21}
\end{equation}
Combining \eqref{e3.2}  and \eqref{e3.21}, we obtain
\begin{equation*}
\Delta _cf(z) =(f(z) -a(z)) +a(z) e^{-K}
\end{equation*}
which means that $\Delta _c^{n+1}f(z) =\Delta _c^nf(z) $ for all $n\geq 1$. 
Therefore, $f(z)$, $\Delta _cf(z) $ and
 $\Delta _c^{2}f(z) $ share $a(z) $ CM and by Lemma \ref{lem2.4}
we obtain $\Delta _cf(z) =f(z) $, which
contradicts the hypothesis. Then $e^{P}\equiv 1$ and the proof 
is complete.
\end{proof}


\begin{proof}[Proof of the Theorem \ref{thm1.2}]
Setting $g(z) =f(z) +b(z) -a(z) $. Since 
$\Delta_c^{m}a(z) \equiv 0$ ($1\leq m\leq n$),  we
can remark that
\begin{gather*}
g(z) -b(z) =f(z) -a(z) , \\
\Delta _c^ng(z) -b(z) =\Delta _c^nf(z) -b(z), \\
\Delta _c^{n+1}g(z) -b(z) =\Delta _c^nf(z) -b(z) ,\quad n\geq 2.
\end{gather*}
Since $f(z) -a(z)$, $\Delta_c^nf(z) -b(z) $ and 
$\Delta_c^{n+1}f(z) -b(z) $ share $0$ CM, then 
$g(z)$,  $\Delta _c^ng(z) $ and $\Delta _c^{n+1}g(z) $ share $b(z) $ CM.
 By using Theorem \ref{thm1.1}, we deduce that 
$\Delta _cg(z) \equiv g(z) $, which leads to 
$\Delta _cf(z) \equiv f(z) +b(z) +\Delta _ca(z) -a(z) $
and the proof  is complete.
\end{proof}


\begin{proof}[Proof of the Theorem \ref{thm1.3}]
Note that $f(z) $ is a nonconstant entire function of finite order. 
Since $f(z)$,  $\Delta _c^nf(z) $ and 
$\Delta_c^{n+1}f(z) $ share $0$ CM, it follows from
Theorem \ref{thmF} that $\Delta _c^{n+1}f(z) =C\Delta _c^nf(z) $, where $C$ 
is a nonzero constant. Then we have
\begin{gather}
\frac{\Delta _c^nf(z) }{f(z) }=e^{P(z) }, \label{e3.22} \\
\frac{\Delta _c^{n+1}f(z) }{f(z) }=Ce^{P(z) },  \label{e3.23}
\end{gather}
where $P$ is a polynomial. By \eqref{e3.22}  and \eqref{e3.23} 
we obtain
\begin{equation}
f_{ic}(z) =(C+1) ^{i}e^{P-P_{ic}}f(z). \label{e3.24}
\end{equation}
Then
\begin{equation}
\Delta _c^nf(z) 
=\Big(\sum_{i=0}^nC_n^{i}(-1) ^{n-i}(C+1) ^{i}e^{P-P_{ic}}\Big)f(z) 
=e^{P(z) }f(z) .  \label{e3.25}
\end{equation}
This equality  leads to $\deg P=0$. Hence 
$P(z) -P_{ic}(z) \equiv 0$ and \eqref{e3.25}  will be
\begin{equation}
\sum_{i=0}^n C_n^{i}(-1) ^{n-i}(C+1) ^{i}=C^n=e^{P(z) }.  \label{e3.26}
\end{equation}
By \eqref{e3.22}, \eqref{e3.23}  and \eqref{e3.26}  we
deduce that
\begin{gather*}
\Delta _c^nf(z) =C^nf(z), \\
\Delta _c^{n+1}f(z) =C^{n+1}f(z) .
\end{gather*}
Then
\begin{equation*}
\Delta _c^{n+1}f(z) =\Delta _c(\Delta _c^nf(z)) =\Delta _c(C^nf(z))
=C^n\Delta _cf(z) =C^{n+1}f(z) ,
\end{equation*}
which implies $\Delta _cf(z) =Cf(z) $. Thus, the
proof  is complete.
\end{proof}


\begin{proof}[Proof of Corollary \ref{coro1.2}]
By Theorem \ref{thm1.3} we have $\Delta _cf(z) =Cf(z) $, where $C$ is a nonzero constant.
Then
\begin{equation}
\Delta _c^{m}f(z) =C\Delta _c^{m-1}f(z)
=C^{m}f(z) ,\text{ }m\geq 1.  \label{e3.27}
\end{equation}
On the other hand, for $z_0\in\mathbb{C}$ we have
\begin{equation}
\Delta _c^{m}f(z_0) =f(z_0) .  \label{e3.28}
\end{equation}
By \eqref{e3.27}  and \eqref{e3.28}  we deduce that $C^{m}=1$.
Hence $\Delta _c^{m}f(z) =f(z) $.
\end{proof}

\begin{proof}[Proof of the Theorem \ref{thm1.4}]
Setting $g(z)=f(z) -a(z) $, we have
\begin{gather*}
g(z) =f(z) -a(z) , \\
\Delta _c^ng(z) =\Delta _c^nf(z) -b(z), \\
\Delta _c^{n+1}g(z) =\Delta _c^nf(z) -b(z) ,\quad n\geq 2.
\end{gather*}
Since $f(z) -a(z)$, $\Delta _c^nf(z) -b(z) $
and $\Delta _c^{n+1}f(z) -b(z) $ share $0$ CM, it follows that
$g(z)$, $\Delta _c^ng(z) $ and $\Delta _c^{n+1}g(z) $ share $0$ CM. 
Using Theorem \ref{thm1.3}, we deduce that $\Delta _cg(z) \equiv Cg(z) $, where 
$C $ is a nonzero constant, which leads to 
$\Delta _cf(z) \equiv Cf(z) +\Delta _ca(z) -a(z) $ and the proof
is complete.
\end{proof}

\section{Open Problem}

 It has been proved in \cite{e1} that

\begin{theorem}[{\cite[Corollary 1.1]{e1}}] \label{thmH}
Let $f(z) $ be a non-periodic entire function of finite order, 
and let $a(z) \in S(f) $ ($\not\equiv 0$)  be a periodic entire function with
period $c$. If $f(z)$, $\Delta _cf(z) $ and 
$\Delta _c^{3}f(z) $ share $a(z) $ CM, then $\Delta _cf(z) \equiv f(z) $.
\end{theorem}

 It is an open question to see under what conditions
 Theorem \ref{thmH} holds for entire functions share a small function with 
$\Delta _c^nf(z) $ and $\Delta _c^{n+2}f(z) $ $(n\geq 1)$. We believe that:
\begin{quote} Let $f(z) $
be a nonconstant entire function of finite order such that 
$\Delta_c^nf(z) \not\equiv 0$, and let 
$a(z) \in S(f) $ ($\not\equiv 0$) be a periodic
entire function with period $c$. If $f(z) $, 
$\Delta _c^nf(z) $ and $\Delta _c^{n+2}f(z) $
$(n\geq 1) $ share $a(z) $ CM, then 
$\Delta _cf(z) \equiv f(z) $.
\end{quote}
 Unfortunately, we have not succeed in proving this.

 \subsection*{Acknowledgements}
The authors would like to thank to anonymous referees for their helpful comments.

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