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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 09, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/09\hfil Existence of non-oscillatory solutions]
{Existence of non-oscillatory solutions to higher-order
mixed difference equations}

\author[Q. Li,  H. Liang, Z. Zhang \hfil EJDE-2007/09\hfilneg]
{Qiaoluan Li,  Haiyan Liang, Wenlei Dong, Zhenguo Zhang}  % in alphabetical order

\address{Qiaoluan Li, Haiyan Liang, Wenlei Dong,  Zhenguo Zhang \newline
 College of Mathematics and Information Science,
 Hebei Normal University,
 Shijiazhuang, 050016, China}
 \email[Qiaoluan Li]{qll71125@163.com} 
 \email[Haiyan Liang]{Liang730110@eyou.com}

 \address{ Zhenguo Zhang \newline
 Information College, Zhejiang Ocean University,
 Zhoushan, Zhejiang, 316000, China}
\email[Zhenguo Zhang]{Zhangzhg@mail.hebtu.edu.cn}

\thanks{Submitted April 30, 2006. Published January 2, 2007.}
\thanks{Supported by the Natural Science Foundation of Hebei Province
 and by the \hfill\break\indent
Main Foundation of Hebei Normal University}
\subjclass[2000]{39A05, 39A10}
\keywords{Nonoscillatory; existence; neutral equation}

\begin{abstract}
 In this paper, we consider the higher order neutral nonlinear
 difference equation
 \begin{gather*}
 \Delta^{m}(x(n)+p(n)x(\tau(n)))+f_1(n,x(\sigma_{1}(n)))
 -f_2(n,x(\sigma_{2}(n)))=0, \\
 \Delta^{m}(x(n)+p(n)x(\tau(n)))+f_1(n,x(\sigma_{1}(n)))
 -f_2(n,x(\sigma_{2}(n)))=g(n), \\
 \Delta^{m}(x(n)+p(n)x(\tau(n)))+\sum_{i=1}^{l}b_i(n)x(\sigma_i(n))=0.
 \end{gather*}
 We obtain  sufficient conditions for the existence of non-oscillatory
 solutions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

 Consider the  difference equations
\begin{gather}
\Delta^{m}(x(n)+p(n)x(\tau(n)))+f_1(n,x(\sigma_{1}(n)))
-f_2(n,x(\sigma_{2}(n)))=0,  \label{e1.1}\\
\Delta^{m}(x(n)+p(n)x(\tau(n)))+f_1(n,x(\sigma_{1}(n)))
-f_2(n,x(\sigma_{2}(n)))=g(n), \label{e1.2}\\
\Delta^{m}(x(n)+p(n)x(\tau(n)))+\sum_{i=1}^{l}b_i(n)x(\sigma_i(n))=0,
\label{e1.3}
\end{gather}
for $n\geq n_0$, where  $\tau(n),\sigma_{i}(n)$ are sequences of
positive integers with  $\tau(n)\leq n$,
$\lim_{n\to\infty}\tau(n)=\infty$,
$\lim_{n\to\infty}\sigma_{i}(n)=\infty$, $i=1,2,\dots, l$. Also
where $p(n), g(n), b_j(n)$, $j=1,2,\dots, l$ are sequences of real
numbers, $f_i(n,x)$, $i=1,2$ are continuous and nondecreasing for
$x$, $f_1(n,x)f_2(n,x)>0$.  There exists $b\neq 0$ such that
\begin{gather}
\sum_{s=n}^{\infty}(s-n)^{(m-1)}|f_i(s,b)|<\infty,\quad i=1,2,\label{e1.4}\\
\sum_{s=n}^{\infty}(s-n)^{(m-1)}|g(s)|<\infty,\label{e1.5} \\
\sum_{s=n}^{\infty}(s-n)^{(m-1)}|b_j(s)|<\infty.\label{e1.6}
\end{gather}

 Recently, there has been an increasing interest in the
study of existence and oscillation of  solutions to
differential and difference equations. The papers \cite{e1,l2,w1,y1}
discussed the existence of non-oscillatory solutions of
differential equations. The papers \cite{s1,t1} discussed the oscillation
of difference equations.  But there are relatively few which
guarantee the existence of non-oscillatory solutions of difference
equations, see \cite{j1,l1}.

  This paper is motivated by the recent paper \cite{z1}, where the authors
  gave sufficient conditions for the existence of
  non-oscillatory solutions of some first-order neutral delay
  differential equations. The purpose of this paper is to present
  some new criteria for the existence of non-oscillatory solution of
  \eqref{e1.1}-\eqref{e1.3}.

A solution of \eqref{e1.1} (\eqref{e1.2} \eqref{e1.3}) is said to
be oscillatory if it has arbitrarily large zeros;
 otherwise it is said to be non-oscillatory.

\section{Main Results}



 To obtain our main results, we need the following
lemma.

\begin{lemma}[\cite{c1}] \label{lem1}
 Let $K$ be a closed bounded and convex subset of $l^\infty$,
the Banach space consisting of all bounded real sequences.
 Suppose $\Gamma$ is a continuous map such that
$\Gamma(K) \subset K$, and suppose further that
$\Gamma(K)$ is uniformly Cauchy. Then $\Gamma$ has a fixed point in $K$.
\end{lemma}

In the sequel, without loss of generality, we assume that
$f_i(n,x)>0$, $i=1,2$ and \eqref{e1.4} holds for $b>0$.

\begin{theorem} \label{thm1}
Assume that $0\leq p(n)\leq p<1$, \eqref{e1.4} holds, then \eqref{e1.1}
has a bounded non-oscillatory solution which is bounded away from zero.
\end{theorem}

\begin{proof}
Choose $N>n_0$, such that
$$
N_{0}:=\min\{\inf_{n\geq N}\{\tau(n)\}, \inf_{n\geq N}
\{\sigma_{1}(n)\}, \inf_{n\geq N}\{\sigma_{2}(n)\}\}\geq n_{0}.
$$
Let $BC$ be the collection of bounded real sequence in Banach space
$l^{\infty}$ and $\|x(n)\|=\sup_{n\geq N}|x(n)|$.
 Define a set $ \Omega \subset BC$ as follows:
\[
 \Omega=\{x(n)\in BC,\,0<M_1\leq x(n)\leq M_2<b, n\geq n_0\},
\]
where $M_1<(1-p)M_2$.
Then $\Omega $ is a closed bounded and convex subset of $BC$. Set
$c=\min\{M_2-\alpha, \alpha-pM_2-M_1\}$, where
$pM_2+M_1<\alpha<M_2$. From \eqref{e1.4}, we get that there exists
$N_1>N$, such that for $n>N_1$,
$$
\sum_{s=n}^{\infty}\frac{(s-n+1)^{(m-1)}}{(m-1)!}
f_i(s,b)\leq c,\quad i=1,2.
$$
Define two maps  $\Gamma_1$ and
$\Gamma_2$ on $\Omega $ as follows:
 $$
(\Gamma_1x)(n)=\begin{cases}
\alpha-p(n)x(\tau(n)), & n\geq N_1,\\
(\Gamma_1x)(N_1), & N_{0}\leq n\leq N_1
\end{cases}
$$
$$
(\Gamma_2x)(n)=\begin{cases}
\frac{(-1)^{m-1}}{(m-1)!}
 \sum_{s=n}^{\infty}(s-n+1)^{(m-1)}\\
\times [f_1(s, x(\sigma_{1}(s)))- f_2(s, x(\sigma_{1}(s)))],& n\geq N_1\\
(\Gamma_2x)(N_1),& N_{0}\leq n\leq N_1
\end{cases}
$$
For any $x,y\in \Omega$, we have
\begin{gather*}
(\Gamma_1x)(n)+(\Gamma_2y)(n)  \leq \alpha+c\leq M_2,\\
(\Gamma_1 x)(n)+(\Gamma_2y)(n) \geq \alpha-pM_2-c\geq M_1.
\end{gather*}
That is $\Gamma_1x+\Gamma_2y \in \Omega$.
Since $0\leq p(n)\leq p<1$, it is easy to check that
$\Gamma_1$  is a contraction mapping.

Now we show that $\Gamma_2$ is continuous. For any
$\varepsilon>0$, we can choose $n_{2}>N_1$, such that
 $$
\sum_{s=n_{2}}^{\infty}\frac{(s-n_{0}+1)^{(m-1)}}{(m-1)!}f_i(s,b)
 <\varepsilon,\quad i=1,2.
$$
 Let $\{x_{k}(n)\}$ be a sequence in $\Omega$, such that
 $\lim_{k\to\infty}\|x_{k}-x\|=0$.
 Since  $\Omega $ is a closed set, we get that $x\in \Omega$ and
 \begin{align*}
&|(\Gamma_2x_{k})(n)-(\Gamma_2x)(n)|\\
&\leq \big|\sum_{s=n}^{n_{2}-1}\frac{(s-n+1)^{(m-1)}}{(m-1)!}
(f_1(s,x_k(\sigma_1(s)))-f_1(s,x(\sigma_1(s))))\big|
\\
&\quad +|\sum_{s=n}^{n_{2}-1}
\frac{(s-n+1)^{(m-1)}}{(m-1)!}(f_2(s,x_k(\sigma_2(s)))-f_2(s,x(\sigma_2(s))))|
+4\varepsilon.
\end{align*}
Since $f_i$ is continuous for $x$, we get that
$\lim_{k\to\infty}\|\Gamma_2x_k-\Gamma_2x\|=0$.
We also know that $\Gamma_2$ is uniformly bounded and for
 for all
$\varepsilon>0$, there exists $N_2$ such that for $m_1>m_2\geq N_2$ and
for  all $x(n)\in \Omega$,
 \begin{align*}
& |\Gamma_2 x(m_1)-\Gamma_2 x(m_2)|\\
&\leq  \sum_{s=m_2}^{m_1-1}\frac{(s-n_0+1)^{(m-1)}}{(m-1)!}
 |f_1(s,x(\sigma_1(s)))- f_2(s,x(\sigma_2(s)))|\leq \varepsilon.
 \end{align*}
 From the discrete Krasnoselskii's fixed point theorem, there exists
 $x\in \Omega$, such that $x=\Gamma x$, i.e.
\begin{align*}
x(n)&=\alpha-p(n)x(\tau(n))\\
&\quad +(-1)^{m-1}\sum_{s=n}^{\infty}
\frac{(s-n+1)^{(m-1)}}{(m-1)!}
\Big(f_1(s,x(\sigma_1(s)))-f_2(s,x(\sigma_2(s)))\Big).
\end{align*}
Note that $x(n)$ is a bounded non-oscillatory solution of \eqref{e1.1}
which is bounded away from zero.
\end{proof}

\begin{theorem} \label{thm2}
Assume that $1<p_1\leq p(n)\leq p_2$, \eqref{e1.4}
 holds, $\tau(n)$ is strictly increasing, then \eqref{e1.1}
  has a bounded non-oscillatory solution which is bounded away from zero.
\end{theorem}

\begin{proof}
 We choose $N_1>n_0$, such that
 $$
N_{0}=\min\{\tau(N_1), \inf_{n\geq N_1}\{\sigma_{1}(n)\},
\inf_{n\geq N_1}\{\sigma_2(n)\}\}\geq n_{0}.
$$
Let $BC$ be the collection of
bounded real sequences in the Banach space $l^{\infty}$ and
$\|x(n)\|=\sup_{n\geq N_1}|x(n)|$.
 Define a set $ X \subset BC$ as follows:
\begin{align*}
X=\big\{&x(n)\in BC: \Delta x(n)\leq 0,
0<M_1\leq x(n)\leq p_{1}M_1<b \text{ for } n\geq N_1\\
& x_{(n)}=x_{(N_1)} \text{ for } N_{0}\leq n\leq N_1\big\}
\end{align*}
Then $X$ is a closed bounded and convex subset of $BC$.

Let $c=\min\{\alpha-M_1, p_1M_1-\alpha\}$, where
$M_1<\alpha<p_1M_1 $.
We choose $N\geq N_1$, such that for $n\geq N$,
\[
\sum_{s=n}^{\infty}\frac{(s-n+1)^{(m-1)}}{(m-1)!}f_i(s,b)\leq  c.
\]
 For  $x\in X$, define
\[
 \psi(n)=\begin{cases}
 \sum_{i=1}^{\infty}\frac{(-1)^{i-1}x(\tau^{-i}(n))}{H_{i}
 (\tau^{-i}(n))},& n\geq N\\
 \psi(N), & N_{0}\leq n\leq N
\end{cases}
\]
where $\tau^{0}(n)=n$, $\tau^{i}(n)=\tau(\tau^{i-1}(n))$,
$\tau^{-i}(n)=\tau^{-1}(\tau^{-(i-1)}(n))$,
$H_{0}(n)=1$,
 $H_{i}(n)= \prod_{j=0}^{i-1}p(\tau^{j}(n))$, $i=1,2,\dots$.
 From $M_1\leq x(n)\leq p_{1}M_1$, we know
$0<\psi (n)\leq p_1M_1$, $n\geq N$.


 Define a mapping  $\Gamma$ on $X$ as follows
$$
\Gamma x(n)=\begin{cases}
 \alpha+(-1)^{m-1}\sum_{s=n}^{\infty}\frac{(s-n+1)^{(m-1)}}{(m-1)!}\\
 \times [f_1(s, \psi(\sigma_{1}(s)))-f_2(s, \psi(\sigma_{2}(s)))],
& n\geq N\\[4pt]
 \Gamma x(N),&  N_{0}\leq n\leq N
\end{cases}
$$
Note that $\Gamma$ satisfies the following three conditions:
\begin{itemize}
\item[(a)]  $\Gamma(X)\subseteq X$.
 In fact, for any $x\in X$, $\Gamma x(n)\geq \alpha-c\geq
 M_1$,
 $\Gamma x(n)\leq \alpha+c\leq
 p_1M_1$.

\item[(b)]  $\Gamma$  is continuous.
  Let $\{x_{k}(n)\}$ be a sequence in $X$, such that
 $\lim_{k\to\infty}\|x_{k}-x\|=0$.
 Since  $X$  is a closed set, we know $x\in X$.
 For any  $\varepsilon>0$, we can choose $n_{2}>N$, such that
 $$
\sum_{s=n_{2}}^{\infty}\frac{(s-n_{0}+1)^{(m-1)}}{(m-1)!}f_i(s,b)
 <\varepsilon, \quad i=1,2.
$$
 \begin{align*}
&|\Gamma x_{k}(n)-\Gamma x(n)|\\
&\leq  \sum_{s=n}^{n_{2}-1}
\frac{(s-n+1)^{(m-1)}}{(m-1)!}\sum_{i=1}^{2}|f_i(s,\psi_k(\sigma_i(s)))-
f_i(s,\psi(\sigma_i(s)))|
 +4\varepsilon.
 \end{align*}
So
$\lim_{k\to\infty}\|\Gamma x_{k}-\Gamma x\|=0$.

\item[(c)] $\Gamma X$ is uniformly Cauchy.
For all $\varepsilon>0$, there exists $ n_3$ such that for
$m_1>m_2\geq n_3$ and for  all $x(n)\in X$,
 \begin{align*}
&|\Gamma x(m_{1})-\Gamma x(m_{2})|\\
&\leq \sum_{s=m_2}^{m_1-1}\frac{(s-n_0+1)^{(m-1)}}{(m-1)!}
|f_1(s,\psi(\sigma_1(s)))-f_2(s,\psi(\sigma_2(s)))|
\leq  \varepsilon.
\end{align*}
 This shows that $\Gamma X$ is uniformly Cauchy.
\end{itemize}

 From Lemma \ref{lem1}, there exists
 $x\in X$, such that $x=\Gamma x$, i.e.
 $$
x(n)=\alpha+(-1)^{m-1}\sum_{s=n}^{\infty}
\frac{(s-n+1)^{(m-1)}}{(m-1)!}[f_1(s,
\psi(\sigma_{1}(s)))-f_2(s,\psi(\sigma_{2}(s)))],
$$
for $n\geq N$.
 Since $\psi(n)+p(n)\psi(\tau(n))=x(n)$, we obtain
\begin{align*}
&\psi(n)+p(n)\psi(\tau(n))\\
&=\alpha+(-1)^{m-1}\sum_{s=n}^{\infty}
\frac{(s-n+1)^{(m-1)}}{(m-1)!}[f_1(s,
\psi(\sigma_{1}(s)))-f_2(s,\psi(\sigma_{2}(s)))].
\end{align*}
 So $\psi(n)$ satisfies \eqref{e1.1} for $n\geq N$, and
 $\frac{p_{1}-1}{p_{1}p_{2}}x(\tau^{-1}(n))\leq \psi(n)\leq
 x(n)$.
 \end{proof}

\begin{theorem} \label{thm3}
 Assume that $-1<p\leq p(n)\leq 0$, and \eqref{e1.4}
 holds. Then \eqref{e1.1} has a bounded non-oscillatory solution
which is bounded away from zero.
\end{theorem}

\begin{proof}
Let $BC$ be the set of bounded real sequence in
the  Banach space $l^{\infty}$ and  $\|x(n)\|=\sup_{n\geq n_0}|x(n)|$.
We choose $M_1, M_2, \alpha$ such that $0<M_1<\alpha<(1+p)M_2$.
Define $\Omega=\{x\in BC, M_1\leq x(n)\leq M_2,\,\,n\geq n_0\}$.
Let $c=\min\{\alpha-M_1, M_2-\alpha\}$, from \eqref{e1.4} we get that
there exists $N$ such that for $n\geq N$,
$$
\frac{1}{(m-1)!}\sum_{s=n}^{\infty}(s-n_0+1)^{(m-1)}f_i(s,b)\leq c,\quad
i=1,2.
$$
For  $x\in \Omega$, define:
\[
\varphi(n)=\begin{cases}
 \sum_{i=0}^{k_n-1}(-1)^ip_n^{(i)}x(\tau_n^{(i)})+(-1)^{k_n}p_n^{(k_n)}\frac{
 x_N}{1+p_N},& n\geq N\\
 \frac{x_N}{1+p_N}, & n_{0}\leq n\leq N
\end{cases}
\]
 where we take $k_n$ such that $n_0\leq \tau_n^{(k_n)}\leq N$,
$\tau_n^{(0)}=n$,
$\tau_n^{(1)}=\tau_n$,
$\tau_n^{(2)}=\tau_{\tau_n},\dots,\tau_n^{(k)}=\tau_{\tau_n^{(k-1)}}$,
$p_n^{(0)}=1$, $p_n^{(1)}=p_n,\dots,p_n^{(s)}=p_np_{\tau_n}
\dots p_{\tau_n^{(s-1)}}$.
  It is easy to prove that
 $x(n)=\varphi(n)+p(n)\varphi(\tau(n))$, $n\geq N$ and
$M_1\leq x(n)\leq  \varphi(n)\leq \frac{M_2}{1+p}$.
Define a mapping  $\Gamma $ on $\Omega $ as follows:
 $$
\Gamma x(n)=\begin{cases}
 \alpha+\sum_{s=n}^{\infty}\frac{(-1)^{m-1}(s-n+1)^{(m-1)}}{(m-1)!}\\
 \times [f_1(s, x(\sigma_{1}(s)))-f_2(s, x(\sigma_{2}(s)))],&
n\geq N\\[4pt]
 \Gamma x(N),&  N_{0}\leq n\leq N
\end{cases}
$$
For any $x \in \Omega$,
$M_1\leq \alpha-c\leq \Gamma x(n)\leq \alpha+c\leq M_2$,
we get $\Gamma\Omega\subseteq \Omega$. Similar
to the proof of Theorem \ref{thm1}, we can obtain $\Gamma$ is continuous
and uniformly Cauchy. So there exists $x\in \Omega$ such that
$x=\Gamma x$. The proof is complete.
\end{proof}

\begin{theorem} \label{thm4}
 Assume that $p_1\leq p(n)\leq p_2<-1$, and \eqref{e1.4}
 holds. Then \eqref{e1.1} has a bounded non-oscillatory solution
 which is bounded away from zero.
\end{theorem}

\begin{proof}
We choose positive constants $M_1, M_2, \alpha$ such
that $-p_1M_1<\alpha<(-p_2-1)M_2$. Let
$\Omega=\{x\in BC, M_1\leq x(n)\leq M_2$, $n\geq n_0\}$,
$c=\min\{\frac{(\alpha+M_1p_1)p_2}{p_1}, (-p_2-1)M_2-\alpha\}$.
Choosing $N$ sufficiently large such that for $ n\geq N$,
$$
\frac{1}{(m-1)!}\sum_{s=n}^{\infty}(s-n+1)^{(m-1)}f_i(s,b)\leq
c,\quad i=1,2.
$$
Define two maps $\Gamma_1$, $\Gamma_2$
on $\Omega$ as follows:
\begin{gather*}
\Gamma_1 x(n)=\begin{cases}
 -\frac{\alpha}{p(\tau^{-1}(n))}-\frac{x(\tau^{-1}(n))}{p(\tau^{-1}(n))}
 ,& n\geq N\\
 \Gamma_1 x(N),  & n_{0}\leq n\leq N
\end{cases}
\\
\Gamma_2 x(n)=\begin{cases}
\sum_{s=\tau^{-1}(n)}^{\infty}
\frac{(-1)^{m-1}(s-\tau^{-1}(n)+1)^{(m-1)}}{(m-1)!p(\tau^{-1}(n))}\\
\times [f_1(s, x(\sigma_{1}(s)))-f_2(s, x(\sigma_{2}(s)))],& n\geq N\\[4pt]
 \Gamma_2 x(N),&  N_{0}\leq n\leq N
\end{cases}
\end{gather*}
For each $x, y \in \Omega$,
\begin{gather*}
\Gamma_1 x(n)+\Gamma_2 y(n)\geq
\frac{-\alpha}{p_1}+\frac{c}{p_2}\geq M_1,\quad
\Gamma_1 x(n)+\Gamma_2 y(n)\leq
\frac{-\alpha}{p_2}-\frac{M_2}{p_2}-\frac{c}{p_2}\leq M_2.
\end{gather*}
So that $\Gamma_1 x(n)+\Gamma_2 y(n) \in \Omega$.
Since $p_1\leq p(n)\leq p_2\leq -1$, we get $\Gamma_1$ is a
contraction mapping. We also
can prove that $\Gamma_2$ is uniformly bounded and continuous.
Further we know $\Gamma_2$ is uniformly Cauchy. So by discrete
Krasnoselskii's fixed point theorem, there exists $x\in \Omega$
such that $\Gamma_1 x+\Gamma_2 x=x$. i.e.
\begin{align*}
x(n)&= -\frac{\alpha}{p(\tau^{-1}(n))}-\frac{x(\tau^{-1}(n))}{p(\tau^{-1}(n))}
+\frac{(-1)^{m-1}}{(m-1)!p(\tau^{-1}(n))}\\
&\quad\times  \sum_{s=\tau^{-1}(n)}^{\infty}(s-\tau^{-1}(n)+1)^{(m-1)}
 [f_1(s, x(\sigma_{1}(s)))-f_2(s, x(\sigma_{2}(s)))].
\end{align*}
The proof is complete.
\end{proof}

\begin{theorem} \label{thm5}
 Assume that $p(n)$ satisfies the
conditions in one of Theorems \ref{thm1}--\ref{thm4},
and \eqref{e1.4}, \eqref{e1.5} hold. Then \eqref{e1.2}
 has a bounded non-oscillatory solution which is bounded away from zero.
\end{theorem}

\begin{proof}
 Set $g_+(n)=\max\{g(n),0\}$, $g_-(n)=\max\{-g(n),0\}$. Then
$g(n)=g_+(n)-g_-(n)$. Also \eqref{e1.2} can be written as
$$
\Delta^{m}(x(n)+p(n)x(\tau(n)))+[f_1(n,x(\sigma_{1}(n)))+g_-(n)]-
[f_2(n,x(\sigma_{2}(n)))+g_{+}(n)]=0.
$$
Let
$F_1(n,x(\sigma_1(n)))=f_1(n,x(\sigma_{1}(n)))+g_-(n)$,
$F_2(n,x(\sigma_2(n)))=f_2(n,x(\sigma_{2}(n)))+g_+(n)$. Similar to
the proof of Theorems  \ref{thm1}--\ref{thm4}, we obtain the conclusion.
\end{proof}

\begin{theorem} \label{thm6}
Assume that $p(n)$ satisfies the conditions in one of the
Theorems \ref{thm1}--\ref{thm4}, and \eqref{e1.6} holds.
Then \eqref{e1.3} has a bounded non-oscillatory solution which
is bounded away from zero.
\end{theorem}

\begin{proof}
 We  prove only the case  $0\leq p(n)\leq p<1$.
Let $BC$ be the set of bounded real sequence in the Banach space $l^{\infty}$
and  $\|x(n)\|=\sup_{n\geq n_0}|x(n)|$. We choose $M_1,
M_2, \alpha$ such that $pM_2+M_1<\alpha <M_2$. Define
$\Omega=\{x\in BC, M_1\leq x(n)\leq M_2\}$,
$c=\min\{\frac{\alpha-pM_2-M_1}{lM_2}, \frac{M_2-\alpha}{lM_2}\}$.
$N$ is sufficiently large such that for $n\geq N$
$$
\frac{1}{(m-1)!}\sum_{s=n}^{\infty}(s-n+1)^{(m-1)}|b_i(s)|\leq
c,\quad i=1,2,\dots,l.
$$
 Define two maps  $\Gamma_1$, $\Gamma_2 $ on $\Omega $ as follows
\begin{gather*}
\Gamma_1 x(n)=\begin{cases}
 \alpha-p(n)x(\tau(n)),& n\geq N\\
 \Gamma x_1(N), & n_{0}\leq n\leq N ,
\end{cases}
\\
\Gamma_2 x(n)=\begin{cases}
(-1)^{m-1}\sum_{s=n}^{\infty}\frac{(s-n+1)^{(m-1)}}{(m-1)!}
 \sum_{i=1}^{l}b_i(s)x(\sigma_i(s)),& n\geq N\\
 \Gamma_2 x(N), &  n_{0}\leq n\leq N
\end{cases}
\end{gather*}
For each $x,y \in \Omega$,
$\Gamma_1 x(n)+\Gamma_2 y(n)\geq \alpha-pM_2-lM_2c\geq M_1$,
$\Gamma_1 x(n)+\Gamma_2 y(n)\leq \alpha+lM_2c\leq M_2$,
that is $\Gamma_1 x(n)+\Gamma_2 y(n) \in \Omega$.
$\Gamma_1$ is a contraction mapping and $\Gamma_2$ is
continuous and uniformly Cauchy. So there exists $x\in \Omega$
such that $ \Gamma_1 x+\Gamma_2x=x$. The proof is complete.
\end{proof}

\begin{thebibliography}{00}

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\end{document}