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\AtBeginDocument{{\noindent\small
2005-Oujda International Conference on Nonlinear Analysis.
\newline {\em Electronic Journal of Differential Equations},
Conference 14, 2006, pp. 227--229.\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 2006 Texas State University - San Marcos.}
\vspace{9mm}}
\setcounter{page}{227}

\begin{document}

\title[\hfilneg EJDE/Conf/14 \hfil Uniformly ergodic theorem]
{Uniformly ergodic theorem for commuting  multioperators}

\author[S. Lahrech, A. Mbarki,  A. Ouahab, S. Rais \hfil EJDE/Conf/14 \hfilneg]
{Samir Lahrech, Abderrahim Mbarki,  Abdelmalek Ouahab, Said Rais} 
 % in alphabetical order

\address{Samir Lahrech \newline
D\'epartement de Math\'ematiques, Universit\'e Oujda, 60000 Oujda,  Morocco}
\email{lahrech@sciences.univ-oujda.ac.ma}

\address{Abderrahim Mbarki \newline
 Current address: National school of Applied Sciences, P.O. Box 669, Oujda University,  Morocco}
\email{ambarki@ensa.univ-oujda.ac.ma}

\address{Abdelmalek Ouahab \newline
D\'epartement de Math\'ematiques, Universit\'e Oujda, 60000 Oujda,  Morocco}
\email{ouahab@sciences.univ-oujda.ac.ma}

\address{Said Rais \newline
D\'epartement de Math\'ematiques, Universit\'e Oujda, 60000 Oujda,  Morocco}
 \email{said\_rais@yahoo.fr}


\date{}
\thanks{Published September 20, 2006.}
\subjclass[2000]{47A35, 47A13}
\keywords{Average; E-k condition; finite descent; uniform ergodicity}

\begin{abstract}
 In this paper, we established some uniformly Ergodic theorems by using
 multioperators satisfying the E-k condition introduce in \cite{l1}.
 One consequence, is that if $I-T$ is quasi-Fredholm and satisfies
 E-k condition then $T$ is uniformly ergodic.
 Also we give some conditions for uniform ergodicity of a commuting
 multioperators satisfies condition E-k.
 These results are of interest in view of analogous results for unvalued
 operators (see, for example \cite{k1}) also in view  of the recent developments
 in the ergodic theory and its applications.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction and main results}

Throughout this paper, $X$ is a complex Banach space, and $L(X)$ is
the algebra of linear continuous operators acting in $X$.
If there is an integer $n$ for which $T^{n+1}X=T^nX,$ then we say that
$T$ has finite descent and the smallest integer
$d(T)$ for which equality occurs is called the descent of $T$.
If there is exists an integer $m$ for which $kerT^{m+1}=kerT^m$,
then $T$ is said to have finite ascent
and the smallest integer $a(T)$ for this equality occurs is called
ascent of $T$. If both $a(T)$ and $d(T)$ are finite, then they are
equal \cite[38.3]{h1}. We say that $T$ is chain-finite and that its chain
length is this common minimal value. Moreover
\cite[38.4]{h1}, in this case there is a decomposition of the vector space
$$
X=T^{d(T)}X\oplus \ker T^{d(T)}.
$$
We now focus on the topological situation: For every $T\in L(X)$ we set
\begin{equation} \label{e1.1}
M_i(T)=i^{-1}(I+T+T^2+\dots +T^{i-1}), \quad i=1,2,3,\dots ,
 \end{equation}
i.e. the averages associated with $T$, where $I=id_X$ is the identity of $X$.
If $T=(T_1, T_2,\dots , T_n)\in L(X)^n$ is commuting multioperator
(briefly, c.m.), we also set
\begin{equation} \label{e1.2}
  M_v(T)= M_{v_1}(T_1)M_{v_2}(T_2)\dots   M_{v_n}(T_n), \quad
  v\in Z^{n}_{+}, v\geq e,
\end{equation}
where $Z^{n}_{+}$ is the family of multi-indices of length $n$
(i.e. n-tuples of nonnegative integers) and $e:=(1,1, \dots , 1)\in Z^{n}_{+}$.
 In other words, \eqref{e1.2}
defines the averages associated with $T$.

\begin{definition} \label{def1.1} \rm
A commuting multioperator $T\in L(X)^n$
is said to be uniformly ergodic if the limit
\begin{equation} \label{e1.3}
 \lim_v M_v(T)
\end{equation}
exists in the uniform topology of $L(X)$.
\end{definition}

\begin{remark} \label{rmk1.1}\rm
(a) If $n=1$, then \eqref{e1.3} is automatically fulfilled, and therefore the above
definition extends the usual concept of uniformly ergodic operator (see, for example
\cite{k1}).

(b) If $T= (I,\dots, T_j , I,\dots ,I)\in L(X)^n$, then $T$ is
uniformly ergodic if and only  the $\lim_{v_j} M_{v_j}(T_j)$ exists
in the uniform topology of $L(X)$.
\end{remark}

\begin{definition}[\cite{l1}] \label{def3}\rm
Let $k=(k_1, \dots k_n)\in Z^{n}_{+}$ and $T\in L(X)^n$ be a c.m.
 We say that $T$ satisfies condition E-k if
$\lim_v(I-T_j)^{k_j} M_v(T)= 0$ for each $j\in\{1,\dots , n\}$.
\end{definition}

 It is clear that condition E-k implies condition E-n for any $n\geq k$
Thus we see that the example $T=(T_1, I, \dots , I )\in Z^{n}_{+}$ with
$$
T_1=\begin{pmatrix}
 1 & 1 \\
 0 & 1 \end{pmatrix}.
$$
This shows that E-2e is strictly weaker than E-e.

\begin{theorem} \label{thm1.1}
 Let $k \in Z^{n}_{+}$. Suppose $T\in L(X)^n$ satisfies condition
E-k and $\sum_{j=1}^{n}(I-T_j)^{k_j}X$,
$\sum_{j=1}^{n}(I^*-T_{j}^*)^{k_j}X^* $
are closed in $X$ and $X^*$ respectively.
If $\big[\sum_{j=1}^{n}(I-T_j)^{k_j}X\big]\cap
\big[\cap_{j=1}^{n}\ker(I-T_j)^{k_j}\big]=\{0\}$. Then $T$ is uniformly
ergodic
\end{theorem}

\begin{proof} Arguing exactly as in \cite[Theorem 1]{m1}, with
$\delta_T$ and $\gamma_T$ given by
$$
\oplus^{n}_{j=1}x_j \to \delta_T(\oplus^{n}_{j=1}x_j)
= \sum_{j=1}^{n}(I-T_j)^{k_j}x_j \ \mbox{and} \ x\to
\gamma_T(x)= \oplus^{n}_{j=1}(I-T_j)^{k_j}x.
$$
\end{proof}

\begin{theorem} \label{thm2.1}
Let $T\in L(X)$ satisfy condition E-r, and one of the following
nine conditions:
\begin{itemize}
\item[(a)] $I-T$ has chain length at most $r$
\item[(b)] $1$ is a pole of the resolvent of order at most $r$
\item[(c)] $I-T$ is quasi-Fredholm operator
\item[(d)] $(I-T)^rX$ is closed and $\ker(I-T)^r$ has a closed
 $T$-invariant complement
\item[(e)] $(I-T)^rX \bigoplus \ker(I-T)^r=(I-T)^rX + \ker(I-T)^r$
\item[(f)] $(I-T)^mX$ is closed for all $m\geq r$
\item[(g)] $(I-T)^rX$ is closed
\item[(h)] $(I-T)^mX$ is closed for some $m\geq r$
\item[(i)] $I-T$ has finite descent.
\end{itemize}
Then $T$ is uniformly ergodic.
\end{theorem}

\begin{proof}
Firstly, from \cite[Theorem 6]{l1}, the above statements (a)--(i) are equivalent.
Then, take $G=(T, I, \dots I) \in L(X)^n$ and $k=(r, 1,\dots ,1)\in  Z^{n}_{+}$;
Therefore, we have $\sum_{j=1}^{n}(I-G_j)^{k_j}X=(I-T)^rX$ is closed, it follows
that $\sum_{j=1}^{n}(I^*-G_{j}^*)^{k_j}X^* =(I^*-T^*)^rX^* $ is closed, which
implies since $(I-T)^rX \cap \ker(I-T)^r=\{0\}$, that $G$ is uniformly ergodic in
$L(X)^n $. From Theorem 1.4 this means that $T$ is uniformly ergodic in $L(X)$
\end{proof}

\begin{theorem} \label{thm3.1}
Let $k \in Z^{n}_{+}$. If $T\in L(X)^n$ A c.m. satisfies
condition E-k, such that $\sum_{j=1}^{n}(I-T_j)$ has chain length at
most 1 and $\ker(\sum_{j=1}^{n}(I-T_j))= \cap_{j=1}^{n}\ker(I-T_j)$.
Then $T$ is uniformly.
\end{theorem}

\begin{proof} There are two cases

\noindent\textbf{Case 1: $d\big(\sum_{j=1}^{n}(I-T_j)\big)=0$.}
Then $\sum_{j=1}^{n}(I-T_j)$ is
bijective then  $X=\sum_{j=1}^{n}(I-T_j)X\oplus \ker(\sum_{j=1}^{n}(I-T_j))$,
which implies, since
$\cap_{j=1}^{n}\ker(I-T_j)\subset \ker(\sum_{j=1}^{n}(I-T_j))=\{0\}$ that
$X=\sum_{j=1}^{n}(I-T_j)X\bigoplus \cap_{j=1}^{n}\ker(I-T_j)$,
from the \cite[Theorem 10]{l1} we obtain  $T$ is uniformly ergodic.

\noindent\textbf{Case 2: $d(\big(\sum_{j=1}^{n}(I-T_j)\big))=1$.}
Then $\big(\sum_{j=1}^{n}(I-T_j)\big)X= \big(\sum_{j=1}^{n}(I-T_j)\big)^{2}X$,
so $\big(\sum_{j=1}^{n}(I-T_j)\big)X=\big(\sum_{j=1}^{n}(I-T_j)\big)^{nr}
\big(\sum_{j=1}^{n}(I-T_j)\big)X$,
with $r=\max_{1\leq j\leq n}k_j$. so $\big(\sum_{j=1}^{n}(I-T_j)\big)^{nr}$
is a bijection of $\big(\sum_{j=1}^{n}(I-T_j)\big)X$ onto itself.
Which implies, since $T$ satisfies condition E-k,
that $M_v(T)|\big(\sum_{j=1}^{n}(I-T_j)\big)X\to 0$
and since $M_v(T)|\cap_{j=1}^{n}\ker(I-T_j)=I|\cap_{j=1}^{n}\ker(I-T_j)$,
it follows that $T$ is uniformly ergodic.
\end{proof}

\begin{thebibliography}{0}
\bibitem{h1} Heuser H. G., \emph{Functional analysis}, Wiley, Chichester, 1982.

\bibitem{k1} Krengel U., \emph{Ergodic theorems}, De Gruyter,
Berlin and New York, 1985.

\bibitem{l1} Lahrech S., A. Azizi, A. Mbarki, A. Ouahab;
\emph{Uniformly Ergodic theorem and finite chain length for multioperators}
International Journal of Pure and Applied Mathematics,
\textbf{22, No. 2} (2005), 167-172.

\bibitem{l2} Laursen K. B. and Mbekhta M.;
\emph{Operators with finite chain length and the ergodic theorem},
Proc. Amer. Math. Soc. \textbf{123} (1995), 3443-3448.

\bibitem{m1} Mbekhta M. and Vasilescu F.-H.,
\emph{Uniformly ergodic multioperators}, Trans. Amer. Math. Soc. \textbf{347},
(1995), 1847- 1854.

\end{thebibliography}

\end{document}
