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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 132, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/132\hfil Entire functions sharing small functions]
{Entire functions sharing small functions with their difference
operators}

\author[Z. Latreuch, A. EL Farissi, B. Bela\"idi \hfil EJDE-2015/132\hfilneg]
{Zinel\^aabidine Latreuch, Abdallah EL Farissi, Benharrat Bela\"idi}

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

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

\thanks{Submitted February 16, 2015. Published May 10, 2015.}
\subjclass[2010]{30D35, 39A32}
\keywords{Uniqueness; entire functions; difference operators}

\begin{abstract}
 We study the uniqueness for entire functions that share small
 functions of finite order with difference operators applied 
 to the entire functions.  In particular, we generalize of a result 
 in \cite{c1}.
\end{abstract}

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

\section{Introduction and Main Results}

In this article, we assume that the reader is familiar with
the fundamental results and the standard notation of the Nevanlinna's value
distribution theory \cite{h3,l1,y1}.
In addition, we will use $\rho (f) $ to denote the
order of growth of $f$ and $\tau (f) $ to denote the type of
growth of $f$, we say that a meromorphic function $a(z) $ is a
small function of $f(z) $ if $T(r,a) =S(r,f) $,
where $S(r,f) =o(T(r,f) )$, as $r\to \infty $ outside of a possible exceptional
set of finite logarithmic measure, 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) $
 (Resp. $f_0(z) =f(z)$) and its difference operators by
\begin{equation*}
\Delta _cf(z) =f(z+c) -f(z) ,\quad
\Delta _c^{n}f(z) =\Delta _c^{n-1}(\Delta_cf(z) ) ,\quad
n\in \mathbb{N},\; n\geq 2.
\end{equation*}
In particular, $\Delta _c^{n}f(z) =\Delta ^{n}f(z)
$ for the case $c=1$.

 Let $f(z) $ and $g(z) $ be two
meromorphic functions, and let $a(z) $ be a small function with
respect to $f(z) $ and $g(z) $. We say that $f(z) $ and $g(z) $ share 
$a(z)$ counting multiplicity (for short CM), provided that $f(z) -a(z) $ and 
$g(z) -a(z) $ have the 
same zeros including multiplicities.


 The problem of meromorphic functions sharing small
functions with their differences is an important topic of uniqueness theory
of meromorphic functions (see \cite{b1,c3,h1,h2}).
In 1986, Jank, Mues and Volkmann \cite{j1} proved the following result.

 \begin{theorem} \label{thmA} 
Let $f$ be a nonconstant meromorphic function, and let $a\neq 0$
 be a finite constant. If  $f$, $f'$ and $f''$
share the value $a$ CM, then $f\equiv f'$.
\end{theorem}


 Li and Yang \cite{l3} gave the following generalization of Theorem \ref{thmA}.

 \begin{theorem} \label{thmB} 
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}



Chen et al \cite{c1}  proved a difference
analogue of result of Theorem \ref{thmA} and obtained the following results.

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

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


 Recently  Chen and Li \cite{c2}
generalized Theorem \ref{thmC} and proved the following results.

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

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

 It is interesting to see what happen when 
$f(z) $, $\Delta _c^{n}f(z) $ and 
$\Delta_c^{n+1}f(z) $ $(n\geq 1) $ share $a(z) $ CM. 
The aim of this article is to give a difference analogue of
result of Theorem \ref{thmB}. In fact, we prove that the conclusion of 
Theorems \ref{thmE} and \ref{thmF} remain valid when we replace $\Delta _cf(z) $ by 
$\Delta _c^{n+1}f(z) $. We obtain the following results.

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

 \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^{n}f(z) $
and $\Delta _c^{n+1}f(z) $ share $a$ CM for all $n\in \mathbb{N}$ 
and we can easily see that $\Delta _c^{n+1}f(z) \equiv \Delta _c^{n}f(z) $. 
This example satisfies Theorem \ref{thm1.1}.
\end{example}

 \begin{remark} \label{rmk1.1}\rm
In Example \ref{examp1.1}, we have $\Delta_c^{m}f(z) \equiv \Delta _c^{n}f(z) $ for any
integer $m>n+1$. However, it remains open when $f(z) $, 
$\Delta _c^{n}f(z) $ and 
$\Delta _c^{m}f(z) $ $(m>n+1) $ share $a(z) $ CM, the claim 
$\Delta _c^{n+1}f(z) \equiv \Delta _c^{n}f(z) $ in Theorem \ref{thm1.1} can be 
replaced by $\Delta_c^{m}f(z) \equiv \Delta _c^{n}f(z) $ in general.
\end{remark}

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

\begin{theorem} \label{thm1.3}
Let $f(z) $ be a nonconstant entire function of finite order. 
If $f(z)$, $\Delta _c^{n}f(z) $  and 
$\Delta _c^{n+1}f(z) $ share $0$ CM, then 
$\Delta _c^{n+1}f(z) \equiv C\Delta _c^{n}f(z)$, 
 where $C$ is a nonzero constant.
\end{theorem}

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

\section{Some lemmas}

 \begin{lemma}[\cite{l2}] \label{lem2.1} 
Let $f$ and $g$ be meromorphic functions such that 
$0<\rho (f)$, $\rho (g) <\infty $ and $0<\tau (f) ,\tau (g) <\infty $.
 Then we have
\begin{itemize}
\item[(i)]   If $\rho (f) >\rho (g) $, then we obtain
\begin{equation*}
\tau (f+g) =\tau (fg) =\tau (f) .
\end{equation*}

\item[(ii)]  If $\rho (f) =\rho (g) $  and 
$\tau (f) \neq \tau (g) $, then
\begin{equation*}
\rho (f+g) =\rho (fg) =\rho (f) =\rho (g) .
\end{equation*}
\end{itemize}
\end{lemma}

\begin{lemma}[\cite{y1}] \label{lem2.2}
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 two 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{h1}] \label{lem2.3} 
Let $c\in \mathbb{C}$, $n\in\mathbb{N}$, and let 
$f(z) $ be a meromorphic function of finite order. 
Then for any small periodic function $a(z)$  with period $c$, with respect to 
$f(z)$, 
\begin{equation*}
m(r,\frac{\Delta _c^{n}f}{f-a}) =S(r,f) ,
\end{equation*}
where the exceptional set associated with $S(r,f) $
is of at most finite logarithmic measure.
\end{lemma}

\section{Proof of the Theorems}

\begin{proof}[Proof of the Theorem \ref{thm1.1}]
Suppose on the contrary to the assertion that 
$\Delta _c^{n}f(z) \not\equiv \Delta _c^{n+1}f(z) $. Note that $f(z) $ 
is a nonconstant entire function of finite order. 
By Lemma \ref{lem2.3}, for $n\geq 1$, we have
\begin{equation*}
T(r,\Delta _c^{n}f) =m(r,\Delta _c^{n}f) \leq
m\big(r,\frac{\Delta _c^{n}f}{f}\big) +m(r,f) \leq T(
r,f) +S(r,f) .
\end{equation*}
Since $f(z) $, $\Delta ^{n}f(z) $ and $\Delta^{n+1}f(z) $ $(n\geq 1) $ 
share $a(z) $ CM, then
\begin{gather}
\frac{\Delta _c^{n}f(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^{Q(z) },  \label{e3.2}
\end{gather}
where $P$ and $Q$ are polynomials. Set
\begin{equation}
\varphi (z) =\frac{\Delta _c^{n+1}f(z) -\Delta
_c^{n}f(z) }{f(z) -a(z) }.  \label{e3.3}
\end{equation}
From \eqref{e3.1}  and \eqref{e3.2}, we obtain
$\varphi (z) =e^{Q(z) }-e^{P(z) }$. Then, by supposition
and \eqref{e3.3}, we see that $\varphi (z) \not\equiv 0$. 
By Lemma \ref{lem2.3}, we deduce that
\begin{equation}
T(r,\varphi ) =m\big(r,\varphi \big) 
\leq m\big(r,\frac{\Delta _c^{n+1}f}{f-a}\big) 
+m\big(r,\frac{\Delta _c^{n}f}{f-a} \big) +O(1) =S(r,f) .  \label{e3.4}
\end{equation}
Note that $\frac{e^{Q(z) }}{\varphi (z) }-\frac{
e^{P(z) }}{\varphi (z) }=1$. By using the second
main theorem and \eqref{e3.4}, we have
\begin{equation}
\begin{aligned}
T(r,\frac{e^{Q}}{\varphi })
& \leq \overline{N}\big(r,\frac{e^{Q}}{\varphi }\big) 
+\overline{N}\big(r,\frac{\varphi }{e^{Q}}\big)
+ \overline{N}\big(r,\frac{1}{\frac{e^{Q}}{\varphi }-1}\big) 
+S\big(r,\frac{e^{Q}}{\varphi }\big)\\
&=\overline{N}\big(r,\frac{e^{Q}}{\varphi }\big) 
+\overline{N}\big(r,\frac{\varphi }{e^{Q}}\big) 
+\overline{N}\big(r,\frac{\varphi }{e^{P}}\big) 
 +S\big( r,\frac{e^{Q}}{\varphi }\big) \\
&=S(r,f) +S\big(r,\frac{e^{Q}}{\varphi }\big) .
\end{aligned}  \label{e3.5}
\end{equation}
Thus, by \eqref{e3.4} and \eqref{e3.5}, we have $T(r,e^{Q})$ $=S(r,f)$. 
Similarly, $T(r,e^{P})=S(r,f)$. Setting now $g(z) =f(z) -a(
z) $,  from \eqref{e3.1} and \eqref{e3.2} we have
\begin{gather}
\Delta _c^{n}g(z) =g(z) e^{P(z)}+a(z),   \label{e3.6} \\
\Delta _c^{n+1}g(z) =g(z) e^{Q(z)}+a(z) .  \label{e3.7}
\end{gather}
By \eqref{e3.6} and \eqref{e3.7}, we have
\[
g(z) e^{Q(z) }+a(z) =\Delta _c(\Delta _c^{n}g(z) )
 =\Delta _c(g(z) e^{P(z) }+a(z) ) .
\]
Thus
\begin{equation*}
g(z) e^{Q(z) }+a(z) =g_c(z) e^{P_c(z) }-g(z) e^{P(z) },
\end{equation*}
which implies
\begin{equation}
g_c(z) =M(z) g(z) +N(z) ,
\label{e3.8}
\end{equation}
where $M(z) =e^{-P_c(z) }(e^{P(z) }+e^{Q(z) }) $ and 
$N(z) =a(z) e^{-P_c(z) }$. From \eqref{e3.8}, we have
\begin{equation*}
g_{2c}(z) =M_c(z) g_c(z)+N_c(z) =M_c(z) (M(z) g(z) +N(z) ) +N_c(z) ,
\end{equation*}
hence
\begin{equation*}
g_{2c}(z) =M_c(z) M_0(z) g(z) +N^{1}(z) ,
\end{equation*}
where $N^{1}(z) =M_c(z) N_0(z)+N_c(z) $. By the same method, we can deduce that
\begin{equation}
g_{ic}(z) =(\prod_{k=0}^{i-1} M_{kc}(z) ) g(z) +N^{i-1}(z) \quad
(i\geq 1) ,  \label{e3.9}
\end{equation}
where $N^{i-1}(z) $ $(i\geq 1) $ is an entire
function depending on $a(z) ,e^{P(z) },e^{Q(z) }$ and their differences. 
Now, we can rewrite \eqref{e3.6} as
\begin{equation}
\sum_{i=1}^n C_{n}^{i}(-1) ^{n-i}g_{ic}(z)
 =(e^{P(z) }-(-1) ^{n}) g(z) +a(z) .  \label{e3.10}
\end{equation}
By \eqref{e3.9} and \eqref{e3.10}, we have
\begin{equation*}
\sum_{i=1}^n C_{n}^{i}(-1) ^{n-i}\Big(
\Big(\prod_{k=0}^{i-1}M_{kc}(z) \Big) g(z) +N^{i-1}(z) \Big) 
-(e^{P(z) }-(-1) ^{n}) g(z) =a(z)
\end{equation*}
which implies
\begin{equation}
A(z) g(z) +B(z) =0,  \label{e3.11}
\end{equation}
where
\begin{gather*}
A(z) =\sum_{i=1}^n C_{n}^{i}(-1) ^{n-i}\prod_{k=0}^{i-1}M_{kc}(z)
-e^{P(z) }+(-1) ^{n}, \\
B(z) =\sum_{i=1}^n C_{n}^{i}(-1) ^{n-i}N^{i-1}(z) -a(z) .
\end{gather*}
It is clear that $A(z) $ and $B(z) $ are small
functions with respect to $f(z) $. 
If $A(z) \not\equiv 0$, then \eqref{e3.11} yields the contradiction
\begin{equation*}
T(r,f) =T(r,g) =T(r,\frac{B}{A}) =S(r,f) .
\end{equation*}

Suppose now that $A(z) \equiv 0$, rewrite the equation $A(z) \equiv 0$ as
\begin{equation*}
\sum_{i=1}^n C_{n}^{i}(-1) ^{n-i}\underset{
k=0}{\overset{i-1}{\prod }}e^{-P_{(k+1) c}}(
e^{P_{kc}}+e^{Q_{kc}}) =e^{P}-(-1) ^{n}.
\end{equation*}
We can rewrite the left side of above equality as
\begin{align*}
&\sum_{i=1}^n C_{n}^{i}(-1) ^{n-i}e^{-
\overset{i}{\underset{k=1}{\sum }}P_{kc}}\underset{k=0}{\overset{i-1}{\prod }
}(e^{P_{kc}}+e^{Q_{kc}})\\
&=\sum_{i=1}^n C_{n}^{i}(-1) ^{n-i}e^{-
\overset{i}{\underset{k=1}{\sum }}P_{kc}}e^{\overset{i-1}{\underset{k=0}{
\sum }}P_{kc}}\prod_{k=0}^{i-1}(
1+e^{Q_{kc}-P_{kc}}) \\
&=\sum_{i=1}^n C_{n}^{i}(-1) ^{n-i}e^{P-P_{ic}}\prod_{k=0}^{i-1}(
1+e^{Q_{kc}-P_{kc}}) .
\end{align*}
So
\begin{equation}
\sum_{i=1}^n C_{n}^{i}(-1)^{n-i}e^{P-P_{ic}}\prod_{k=0}^{i-1}(
1+e^{h_{kc}}) =e^{P}-(-1) ^{n},  \label{e3.12}
\end{equation}
where $h_{kc}=Q_{kc}-P_{kc}$. On the other hand, let
 $\Omega _{i}=\{0,1,\dots ,i-1\} $ be a finite set of $i$ elements,
and
\begin{equation*}
P(\Omega _{i}) =\{\emptyset ,\{ 0\} ,\{1\} ,\dots ,\{ i-1\} ,\{ 0,1\} ,\{
0,2\} ,\dots ,\Omega _{i}\},
\end{equation*}
where $\emptyset $ is the empty set. It is easy to see that
\begin{equation}
\begin{aligned}
\prod_{k=0}^{i-1}(1+e^{h_{kc}}) 
&=1+\sum_{A\in P(\Omega _{i}) \backslash \{\emptyset \}}
\exp \Big(\sum_{j\in A} h_{jc}\Big) \\
&=1+[ e^{h}+e^{h_c}+\dots +e^{h_{(i-1) c}}] \\
&\quad +[e^{h+h_c}+e^{h+h_{2c}}+\dots ] +\dots 
+[ e^{h+h_c+\dots +h_{(i-1) c}}] .
\end{aligned} \label{e3.13}
\end{equation}
We divide the proof into two parts:
\smallskip

\noindent\textbf{Part (1).} $h(z) $ is non-constant polynomial.
Suppose that $h(z) =a_{m}z^{m}+\dots +a_0$ $(a_{m}\neq
0) $, since $P(\Omega _{i}) \subset P(\Omega
_{i+1}) $, then by \eqref{e3.12} and \eqref{e3.13} we
have
\begin{equation*}
\sum_{i=1}^n C_{n}^{i}(-1)
^{n-i}e^{P-P_{ic}}+\alpha _{1}e^{a_{m}z^{m}}+\alpha
_{2}e^{2a_{m}z^{m}}+\dots +\alpha _{n}e^{na_{m}z^{m}}=e^{P}-(
-1) ^{n}
\end{equation*}
which is equivalent to
\begin{equation}
\alpha _0+\alpha _{1}e^{a_{m}z^{m}}+\alpha _{2}e^{2a_{m}z^{m}}+\dots
+\alpha _{n}e^{na_{m}z^{m}}=e^{P},  \label{e3.14}
\end{equation}
where $\alpha _{i}$ $(i=0,\dots ,n) $ are entire functions of
order less than $m$. Moreover,
\begin{align*}
\alpha _0&=\sum_{i=1}^n C_{n}^{i}(-1)^{n-i}e^{P-P_{ic}}+(-1) ^{n}\\
&=e^{P}(\sum_{i=1}^n C_{n}^{i}(-1)
^{n-i}e^{-P_{ic}}+(-1) ^{n}e^{-P}) \\
&=e^{P}\Delta _c^{n}e^{-P}.
\end{align*}

(i)  If $\deg P>m$, then we obtain from \eqref{e3.14} that
$\deg P\leq m$ which is a contradiction.

(ii)  If $\deg P<m$, then by using Lemma \ref{lem2.1}
and \eqref{e3.14} we obtain
\begin{equation*}
\deg P=\rho (e^{P}) 
=\rho \Big(\alpha _0+\alpha
_{1}e^{a_{m}z^{m}}+\alpha _{2}e^{2a_{m}z^{m}}+\dots +\alpha
_{n}e^{na_{m}z^{m}}\Big) =m,
\end{equation*}
which is also a contradiction.

(iii)  If $\deg P=m$, then we suppose that $P(z) =dz^{m}+P^{\ast }(z) $
 where $\deg P^{\ast }<m$.
We have to study two subcases:

 $(\ast)$ If $d\neq ia_{m}$ $(i=1,\dots,n) $, then 
\begin{equation*}
\alpha _{1}e^{a_{m}z^{m}}+\alpha _{2}e^{2a_{m}z^{m}}+\dots +\alpha
_{n}e^{na_{m}z^{m}}-e^{P^{\ast }}e^{dz^{m}}=-\alpha _0.
\end{equation*}
By using Lemma \ref{lem2.2}, we obtain $e^{P^{\ast }}\equiv 0$, 
which is impossible.

 $(\ast \ast ) $ Suppose now that there exists at most $
j\in \{ 1,2,\dots ,n\} $ such that $d=ja_{m}$. Without loss of
generality, we assume that $j=n$. Then we rewrite \eqref{e3.14} as
\begin{equation*}
\alpha _{1}e^{a_{m}z^{m}}+\alpha _{2}e^{2a_{m}z^{m}}+\dots +(\alpha
_{n}-e^{P^{\ast }}) e^{na_{m}z^{m}}=-\alpha _0.
\end{equation*}
By using Lemma \ref{lem2.2}, we have $\alpha _0\equiv 0$, so 
$\Delta _c^{n}e^{-P}=0$. Thus
\begin{equation}
\sum_{i=0}^n C_{n}^{i}(-1)^{n-i}e^{-P_{ic}}\equiv 0.  \label{e3.15}
\end{equation}
Suppose that $\deg P=\deg h=m>1$ and
\begin{equation*}
P(z) =b_{m}z^{m}+b_{m-1}z^{m-1}+\dots +b_0,\quad (b_{m}\neq 0) .
\end{equation*}
Note that for $j=0,1,\dots ,n$, we have
\begin{equation*}
P(z+jc) =b_{m}z^{m}+(b_{m-1}+mb_{m}jc) z^{m-1}+\beta_{j}(z) ,
\end{equation*}
where $\beta _{j}(z) $ are polynomials with degree less than $m-1$. 
Rewrite \eqref{e3.15} as
\begin{equation}
\begin{aligned}
&e^{-\beta _{n}(z) }e^{-b_{m}z^{m}-(b_{m-1}+mb_{m}nc)z^{m-1}}\\
&-ne^{-\beta _{n-1}(z) }e^{-b_{m}z^{m}-(b_{m-1}+mb_{m}(n-1) c) z^{m-1}}
+\dots \\
&+(-1) ^{n}e^{-\beta _0(z)}e^{-b_{m}z^{m}-b_{m-1}z^{m-1}}\equiv 0.
\end{aligned}  \label{e3.16}
\end{equation}
For any $0\leq l<k\leq n$, we have
\begin{align*}
&\rho (e^{-b_{m}z^{m}-(b_{m-1}+mb_{m}lc) z^{m-1}-(
-b_{m}z^{m}-(b_{m-1}+mb_{m}kc) z^{m-1}) }) \\
&=\rho(e^{-mb_{m}(l-k) cz^{m-1}})
=m-1,
\end{align*}
and for $j=0,1,\dots ,n$, we see that
\begin{equation*}
\rho (e^{\beta _{j}}) \leq m-2.
\end{equation*}
By this, together with \eqref{e3.16} and Lemma \ref{lem2.2}, we obtain 
$e^{-\beta _{n}(z) }\equiv 0$, which is impossible. Suppose now
that $P(z) =\mu z+\eta $ $(\mu \neq 0) $ and 
$Q(z) =\alpha z+\beta $ because if $\deg Q>1$, then we go back to
 case (ii). It easy to see that
\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 $\Delta _c^{n}e^{-P}\equiv 0$ 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) }.
\end{equation*}
To prove that $e^{Q(z) }$ is also periodic entire
function with period $c$, we suppose the contrary, which means that 
$e^{\alpha c}\neq 1$. Since $e^{P(z) }$ is of period $c$, then by
\eqref{e3.14}, we obtain
\begin{equation}
\alpha _{1}e^{(\alpha -\mu ) z}+\alpha _{2}e^{2(\alpha
-\mu ) z}+\dots +\alpha _{n}e^{n(\alpha -\mu ) z}
=e^{\mu z+\eta },  \label{e3.17}
\end{equation}
where $\alpha _{i}$ $(i=1,\dots ,n) $ are constants. In
particular,
\begin{equation*}
\alpha _{n}=e^{n(\beta -\eta ) +\alpha c\frac{n(n-1)}{2}}
\end{equation*}
and
\begin{align*}
\alpha _{1}
&=\Big[ \sum_{i=1}^n C_{n}^{i}(-1) ^{n-i}
+\sum_{i=2}^n C_{n}^{i}(-1) ^{n-i}e^{\alpha c} \\
&\quad+ \sum_{i=3}^n C_{n}^{i}(-1)^{n-i}e^{2\alpha c}
 +\dots +e^{(n-1) \alpha c}] e^{(\beta -\eta ) }
 \\
&=\Big[C_{n}^{1}(-1) ^{n-1}+C_{n}^{2}(-1) ^{n-2}(1+e^{\alpha c})
 +C_{n}^{3}(-1) ^{n-3}(1+e^{\alpha c}+e^{2\alpha c}) \\
&\quad +\dots +C_{n}^{n}(-1) ^{n-n}(1+e^{\alpha c}+\dots
+e^{(n-1) \alpha c}) \Big]e^{(\beta -\eta ) } \\
&=\Big[C_{n}^{1}(-1) ^{n-1}\frac{e^{\alpha c}-1}{e^{\alpha c}-1}
+C_{n}^{2}(-1) ^{n-2}\frac{e^{2\alpha c}-1}{e^{\alpha c}-1}
+C_{n}^{3}(-1) ^{n-3}\frac{e^{3\alpha c}-1}{e^{\alpha c}-1} \\
&\quad +\dots +C_{n}^{n}(-1) ^{n-n}\frac{e^{n\alpha c}-1}{e^{\alpha
c}-1}]e^{(\beta -\eta ) } \\
&=\Big[C_{n}^{1}(-1) ^{n-1}(e^{\alpha c}-1)
+C_{n}^{2}(-1) ^{n-2}(e^{2\alpha c}-1)
+C_{n}^{3}(-1) ^{n-3}(e^{3\alpha c}-1) \\
&\quad +\dots +C_{n}^{n}(-1) ^{n-n}(e^{n\alpha c}-1) \Big]
\frac{e^{(\beta -\eta ) }}{e^{\alpha c}-1} \\
&=\Big[ \underset{i=0}{\overset{n}{\sum }}C_{n}^{i}(-1)
^{n-i}e^{i\alpha c}-(-1) ^{n}-\sum_{i=1}^n
C_{n}^{i}(-1) ^{n-i}\Big] \frac{e^{(\beta -\eta ) }
}{e^{\alpha c}-1} \\
&=(e^{\alpha c}-1) ^{n-1}e^{(\beta -\eta ) }.
\end{align*}
Rewrite \eqref{e3.17} as
\begin{equation}
\alpha _{1}e^{(\alpha -2\mu ) z}+\alpha _{2}e^{(2\alpha
-3\mu ) z}+\dots +\alpha _{n}e^{(n\alpha -(n+1) \mu
) z}=e^{\eta },  \label{e3.18}
\end{equation}
it is clear that for each $1\leq l<m\leq n$, we have
\begin{equation*}
\rho \big(e^{(m\alpha -(m+1) \mu -l\alpha +(l+1) \mu ) z}\big) 
=\rho \big(e^{(m-l) (\alpha -\mu ) z}\big) =1.
\end{equation*}
We have the following two cases:

 (i1)  If $j\alpha -(j+1) \mu \neq 0$ for all $j\in \{ 1,2,\dots ,n\} $, 
which means that
\begin{equation*}
\rho \big(e^{(j\alpha -(j+1) \mu ) z}\big) =1, \quad  1\leq j\leq n
\end{equation*}
then, by applying Lemma \ref{lem2.2} we obtain $e^{\eta }\equiv 0$, which is a
contradiction.

 (i2)  If there exists (at most one)  an integer 
$j\in \{ 1,2,\dots ,n\} $ such
that $j\alpha -(j+1) \mu =0$. Without loss of generality, assume
that $e^{(n\alpha -(n+1) \mu ) z}=1$, the equation \eqref{e3.18} will be
\begin{equation*}
\alpha _{1}e^{(\alpha -2\mu ) z}+\alpha _{2}e^{(2\alpha
-3\mu ) z}+\dots +\alpha _{n-1}e^{((n-1) \alpha
-n\mu ) z}=e^{\eta }-e^{n(\beta -\eta ) +\alpha c\frac{
n(n-1) }{2}}
\end{equation*}
and by applying Lemma \ref{lem2.2}, we obtain 
$\alpha _{1}=(e^{\alpha c}-1) ^{n-1}e^{(\beta -\eta ) }\equiv 0$, which is
impossible. So, by (i1)  and (i2), we deduce that $e^{\alpha c}\equiv 1$.
 Therefore, for any $
j\in \mathbb{Z}$ we have
\begin{equation*}
e^{Q(z+jc) }=e^{\alpha z+\beta }(e^{\alpha c})^{j}=e^{Q(z) },
\end{equation*}
which implies that $e^{Q}$ is periodic of period $c$. Since 
$e^{P(z) }$ is of period $c$, then by \eqref{e3.1}, we obtain
\begin{equation}
\Delta _c^{n+1}f(z) =e^{P}\Delta _cf(z) ,
\label{e3.19}
\end{equation}
then $\Delta _c^{n+1}f(z) $ and $\Delta _cf(z) $
share $0$ CM. Substituting \eqref{e3.19} into the second equation 
\eqref{e3.2}, we obtain
\begin{equation}
e^{P(z) }\Delta _cf(z) =e^{Q(z) }(f(z) -a(z) ) +a(z) .
\label{e3.20}
\end{equation}
Since $e^{P(z) }$ and $e^{Q(z) }$ are of period $c$,
then by \eqref{e3.20}, we obtain
\begin{equation}
\Delta _c^{n+1}f(z) =e^{Q-P}\Delta _c^{n}f(z) . \label{e3.21}
\end{equation}
So, $\Delta ^{n+1}f(z) $ and $\Delta ^{n}f(z) $
share $0,a(z) $ CM, combining \eqref{e3.1}, \eqref{e3.2}
 and \eqref{e3.21}, we deduce that
\begin{equation*}
\frac{\Delta ^{n+1}f(z) -a(z) }{\Delta ^{n}f(z) -a(z) }
=\frac{\Delta ^{n+1}f(z) }{\Delta ^{n}f(z) },
\end{equation*}
and we obtain
\begin{equation*}
\Delta ^{n+1}f(z) =\Delta ^{n}f(z)
\end{equation*}
which is a contradiction. Suppose now that $P=c_{1}$ and $Q=c_{2}$ are
constants $(e^{c_{1}}\neq e^{c_{2}}) $. By \eqref{e3.8}
we have
\begin{equation*}
g_c(z) =(e^{c_{2}-c_{1}}+1) g(z)+a(z) e^{-c_{1}}
\end{equation*}
by the same,
\begin{equation*}
g_{2c}(z) =(e^{c_{2}-c_{1}}+1) ^{2}g(z)
+a(z) e^{-c_{1}}((e^{c_{2}-c_{1}}+1) +1).
\end{equation*}
By induction, we obtain
\begin{align*}
g_{nc}(z) 
&=(e^{c_{2}-c_{1}}+1) ^{n}g(z)
+a(z) e^{-c_{1}}\underset{i=0}{\overset{n-1}{\sum }}(
e^{c_{2}-c_{1}}+1) ^{i} \\
&=(e^{c_{2}-c_{1}}+1) ^{n}g(z) +a(z)
e^{-c_{2}}((e^{c_{2}-c_{1}}+1) ^{n}-1) .
\end{align*}
Rewrite the equation \eqref{e3.6} as
\begin{align*}
\Delta _c^{n}g(z) 
&=\sum_{i=0}^n C_{n}^{i}(-1) ^{n-i}[ (e^{c_{2}-c_{1}}+1)
^{i}g(z) +a(z) e^{-c_{2}}\big((e^{c_{2}-c_{1}}+1) ^{i}-1\big) ]\\
&=e^{c_{1}}g(z) +a(z) .
\end{align*}
Since $A(z) \equiv 0$,  we have
\begin{gather*}
\sum_{i=0}^n C_{n}^{i}(-1) ^{n-i}(e^{c_{2}-c_{1}}+1) ^{i}=e^{c_{1}},\\
\sum_{i=0}^n C_{n}^{i}(-1) ^{n-i}((e^{c_{2}-c_{1}}+1) ^{i}-1) =e^{c_{2}}
\end{gather*}
which are equivalent to
\begin{gather*}
e^{n(c_{2}-c_{1}) }=e^{c_{1}},\\
e^{n(c_{2}-c_{1}) }=e^{c_{2}}
\end{gather*}
which is a contradiction.
\smallskip

\noindent\textbf{Part (2).} $h(z) $ is a constant. We show
first that $P(z) $ is a constant. If $\deg P>0$, from the
equation \eqref{e3.12}, we see
\begin{equation*}
\deg P\leq \deg P-1,
\end{equation*}
which is a contradiction. Then $P(z) $ must be a constant and
since $h(z) =Q(z) -P(z) $ is a constant,
we deduce that both of $P(z) $ and $Q(z) $ is
constant. This case is impossible too (the last case in Part (1)), and we
deduced that $h(z) $ can not be a constant. Thus, the proof  complete.
\end{proof}

\begin{proof}[Proof of the Theorem \ref{thm1.2}]
Setting $g(z)=f(z) +b(z) -a(z) $, we can remark that
\begin{gather*}
g(z) -b(z) =f(z) -a(z) , \\
\Delta _c^{n}g(z) -b(z) =\Delta _c^{n}f(z) -b(z), \\
\Delta _c^{n+1}g(z) -b(z) =\Delta _c^{n}f(z) -b(z) ,\; n\geq 2.
\end{gather*}
Since $f(z) -a(z)$, $\Delta _c^{n}f(z) -b(z) $ and 
$\Delta _c^{n+1}f(z) -b(z) $ share $0$ CM, it follows that
 $g(z)$, $\Delta _c^{n}g(z) $ and 
$\Delta _c^{n+1}g(z) $ share $b(z) $ CM. By using
Theorem \ref{thm1.1}, we deduce that $\Delta _c^{n+1}g(z) \equiv \Delta
_c^{n}g(z) $, which leads to $\Delta _c^{n+1}f(z) \equiv \Delta _c^{n}f(z) $ 
and the proof 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^{n}f(z) $ and $\Delta _c^{n+1}f(z) $ share $0$
CM, it follows that 
\begin{gather}
\frac{\Delta _c^{n}f(z) }{f(z) }=e^{P(z) },  \label{e3.22} \\
\frac{\Delta _c^{n+1}f(z) }{f(z) }=e^{Q(z) },  \label{e3.23}
\end{gather}
where $P$ and $Q$ are polynomials. If $Q-P$ is a constant, then we can get
easily from \eqref{e3.22} and \eqref{e3.23}
\begin{equation*}
\Delta _c^{n+1}f(z) =e^{Q(z) -P(z)}\Delta _c^{n}f(z) :\equiv C\Delta _c^{n}f(z) .
\end{equation*}
This completes the proof.
 If $Q-P$ is a not constant, with a similar arguing
as in the proof of Theorem \ref{thm1.1}, we can deduce that the case 
$\deg P=\deg (Q-P) >1$ is impossible.
 For the case $\deg P=\deg (Q-P) =1$, we can obtain that $e^{P(z) }$ 
is periodic entire function with period $c$. This together with \eqref{e3.22}
yields
\begin{equation}
\Delta _c^{n+1}f(z) =e^{P(z) }\Delta _cf(z)   \label{e3.24}
\end{equation}
which means that $f(z)$, $\Delta _cf(z) $ and 
$\Delta _c^{n+1}f(z) $ share $0$ CM. 
Thus, by Theorem \ref{thmF}, we
obtain
\begin{equation*}
\Delta _c^{n+1}f(z) \equiv C\Delta _cf(z)
\end{equation*}
which is a contradiction to \eqref{e3.22} and $\deg P=1$. 
Theorem \ref{thm1.3} is thus proved.
\end{proof}

 \subsection*{Acknowledgements.} 
The authors are grateful to the anonymous  referees
for their valuable comments which lead to the improvement of this paper.

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\end{document}