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\noindent\parbox{2.95cm}{\includegraphics*[keepaspectratio=true,scale=0.125]{AFA.jpg}}
\noindent\parbox{4.85in}{\hspace{0.1mm}\\[1.5cm]\noindent
\qquad Ann. Funct. Anal. 2 (2011), no. 2, 10--21\\
{\footnotesize
\qquad \textsc{\textbf{$\mathscr{A}$}nnals of \textbf{$\mathscr{F}$}unctional
\textbf{$\mathscr{A}$}nalysis}\\
\qquad ISSN: 2008-8752 (electronic)\\
\qquad URL: \textcolor[rgb]{0.00,0.00,0.99}{www.emis.de/journals/AFA/}
}\\[.5in]}

\title[Iterative algorithm for nonexpansive mappings]{A general iterative algorithm for nonexpansive mappings in Banach spaces}

\author[B. Ali, G.C. Ugwunnadi, Y. Shehu]{ Bashir Ali$^{1}$, Godwin C. Ugwunnadi$^{2}$ and Yekini Shehu$^*$$^{2}$ }

\address{$^{1}$ Department of Mathematical Sciences, Bayero University, Kano.}

\email{\textcolor[rgb]{0.00,0.00,0.84}{bashiralik@yahoo.com}}

\address{$^{2}$ Department of Mathematics, University of Nigeria, Nsukka.}

\email{\textcolor[rgb]{0.00,0.00,0.84}{ugwunnadi4u@yahoo.com}}
\email{\textcolor[rgb]{0.00,0.00,0.84}{deltanougt2006@yahoo.com}}

\dedicatory{{\rm Communicated by H.-K. Xu}}


\subjclass[2010]{Primary 47H09; Secondary 47H10, 47J20.}

\keywords{$\eta-$strongly accretive maps, $\kappa-$Lipschitzian
maps, nonexpansive maps, $q-$uniformly smooth Banach spaces. }

\date{Received: 22 September 2010; Revised: 1 April 2011; Accepted: 30 May 2011.
\newline \indent $^{*}$ Corresponding author}

\begin{abstract}
Let $E$ be a real $q$-uniformly smooth Banach space whose duality map is weakly sequentially continuous. Let $T:E\to E$ be a nonexpansive mapping with $F(T)\neq\emptyset.$ Let $A:E\to E$ be an
$\eta$-strongly accretive map which is also $\kappa$-Lipschitzian.
Let $f:E\to E$ be a contraction map with coefficient $0<\alpha<1.$
Let a sequence $\{y_{n}\}$ be defined
iteratively by $y_{0}\in E,~~ y_{n+1}=\alpha_n\gamma
f(y_n)+(I-\alpha_n\mu A)Ty_n,n\geq0,$ where $\{\alpha_n\},~~\gamma$ and $\mu$ satisfy some appropriate conditions.
Then, we prove that $\{y_{n}\}$ converges strongly to the unique solution $x^{*} \in F(T)$ of the variational
inequality  $\langle(\gamma f-\mu A)x^{*},j(y-x^{*})\rangle\leq0,~\forall~y\in F(T).$ Convergence of the correspondent implicit scheme is also proved without the assumption that $E$ has weakly sequentially continuous duality map. Our results are applicable in $l_{p}$ spaces, $1< p<\infty$.
\end{abstract}
\maketitle

\section{Introduction}

Let $E$ be a real Banach space and $E^{*}$ be the dual space of $E.$
A mapping $\varphi:[0,\infty)\to [0,\infty)$ is called a gauge
function if it is strictly increasing, continuous and
$\varphi(0)=0.$ Let $\varphi$ be a gauge function, a generalized
duality mapping with respect to $\varphi,$ $J_{\varphi}:E\to
2^{E^{*}}$ is defined by, $x\in E,$
$$J_{\varphi}x=\{x^{*}\in E^{*}:\langle x,x^{*}
\rangle=\Vert x \Vert \varphi(\Vert x\Vert),\Vert x^{*}\Vert=\varphi(\Vert x\Vert)\},$$ where $\langle . ,.\rangle$ denotes the duality
pairing between element of $E$ and that of $E^{*}$. If $\varphi(t)=t,$ then $J_{\varphi}$ is simply called the normalized duality mapping and is denoted by $J.$ For any $x\in E,$ an element of $J_{\varphi}x$ is denoted by $j_{\varphi}(x)$.\\
If however $\varphi(t)=t^{q-1},$ for some $q>1,$ then $J_{\varphi}$ is still called the generalized duality mapping and is denoted by $J_{q}$ (see, for example \cite{yongfusu, afa}). \\[2mm]
The space $E$ is said to have weakly (sequentially) continuous duality map if there exists a gauge function $\varphi$ such that $J_{\varphi}$ is singled valued and (sequentially) continous from $E$ with weak topology to $E^{*}$ with weak$^{*}$ topology. It is well known that all $l_{p}$ spaces, $(1<p<\infty)$ have weakly sequentially continuous duality mappings.
It is well known (see, for example, \cite{bxu}) that
$J_q(x)= ||x||^{q-2}J(x)$ if $x\neq 0$, and that if $E^*$ is
strictly convex then $J_q$ is single valued.\\[2mm]
\noindent A mapping $A:D(A)\subset E \to E$ is said to be {\it
accretive} if $\forall x,y \in D(A),$ there exists $j_{q}(x-y)\in
J_{q}(x-y)$ such that
\begin{equation}\label{31}
\langle Ax-Ay,j_{q}(x-y)\rangle \ge 0,
\end{equation}
where $D(A)$ denotes the domain of $A.$ \noindent $A$ is called {\it $\eta-$strongly accretive} if
$\forall x,y \in D(A),$ there exists $j_{q}(x-y)\in J_{q}(x-y)$ and $\eta \in (0,1)$ such
that
\begin{equation}\label{accretive}
\langle Ax-Ay,j_{q}(x-y)\rangle \ge \eta \Vert x-y \Vert^{q}.
\end{equation}
$A$ is $\kappa-${\it Lipschitzian} if for some $\kappa>0,$ $\Vert A(x)-A(y)\Vert \le \kappa \Vert x-y \Vert$
$\forall~~x,y \in D(A)$. A mapping $T:E\rightarrow E$ is called {\it nonexpansive} if
\begin{eqnarray*}
||T x-T y|| \leq ||x-y||,~~\forall x, y \in E.
\end{eqnarray*}
A point $x \in E$ is called {\it a fixed point} of $T$ if $Tx=x$.
The set of fixed points of $T$ is denoted by $F(T):=\{x \in E:Tx=x\}$.
In Hilbert spaces, accretive operators are called {\it monotone}
where inequalities \eqref{31} and \eqref{accretive} hold with $j_{q}$ replaced by the identity map on $H.$ \\

\noindent Moudafi \cite{moudafi} introduced the viscosity approximation method
for nonexpansive mappings. Let $f$ be a contraction on $H$, starting
with an arbitrary $x_{0}\in H$, define a sequence $\{x_{n}\}$
recursively by
\begin{eqnarray}\label{mouda}
x_{n+1}=\alpha_{n}f(x_{n})+(1-\alpha_{n})Tx_{n},~~n\geq0,
\end{eqnarray}
where $\{\alpha_{n}\}$ is a sequence in (0,1). Xu \cite{xuk} proved
that under certain appropriate conditions on $\{\alpha_{n}\}$, the
sequence $\{x_{n}\}$ generated by \eqref{mouda} strongly converges
to the unique solution $x^{*}$ in $F(T)$ of the variational inequality
\begin{eqnarray*}
\langle(I-f)x^{*},x-x^{*}\rangle\geq0,~~{\rm for}~x\in F(T).
\end{eqnarray*}
In \cite{xu4}, it is proved, under some conditions on the real sequence $\{\alpha_{n}\},$ that the sequence $\{x_{n}\}$ defined by
$x_{0}\in H$ chosen
arbitrary,
\begin{eqnarray}\label{05}
 x_{n+1}=\alpha_{n}b+(I-\alpha_{n}A)Tx_{n},~~n\geq0,
 \end{eqnarray}
 converges strongly to  $x^{*} \in F(T)$ which is the unique solution of the minimization
 problem
 \begin{eqnarray*}
 \underset{x\in F(T)}{\min}\frac{1}{2}\langle Ax,x\rangle-\langle
 x,b\rangle,
\end{eqnarray*}
where $A$ is a strongly positive bounded linear operator. That is, there is
a constant $\bar{\gamma}>0$ with the property
\begin{eqnarray*}
\langle Ax,x\rangle\geq\bar{\gamma}||x||^{2},~~\forall x\in H.
\end{eqnarray*}
Combining the iterative method \eqref{mouda} and \eqref{05}, Marino
and Xu \cite{xu6} consider the following general iterative method:
\begin{eqnarray}\label{06}
x_{n+1}=\alpha_{n}\gamma f(x_{n})+(I-\alpha_{n}A)Tx_{n},~~~n\geq0.
\end{eqnarray}
They proved that if the sequence $\{\alpha_{n}\}$ of parameters
satisfies appropriate conditions, then the sequence $\{x_{n}\}$
generated by \eqref{06} converges strongly to $x^{*} \in F(T)$ which solves
the variational inequality
\begin{eqnarray*}
\langle(\gamma f-A)x^{*},x-x^{*}\rangle\leq0,~~x\in F(T),
\end{eqnarray*}
which is the optimality condition for the minimization problem
\begin{eqnarray*}
 \underset{x\in F(T)}{\min}\frac{1}{2}\langle Ax,x\rangle-h(x),
\end{eqnarray*}
where $h$ is a potential function for $\gamma f$ (i.e. $h'(x)=\gamma
f(x)$ for $x\in H$).\\

\noindent Let $K$ be a nonempty, closed and convex subset of a real Hilbert space $H$. The variational inequality problem: Find a point $x^{*}\in K$ such that
\begin{eqnarray*}
\langle Ax^{*},y-x^{*}\rangle \ge 0, ~~\forall y\in K
\end{eqnarray*}
is equivalent to the following fixed point equation
\begin{equation}\label{fixpoint}
x^{*}=P_{K}(x^{*}-\delta Ax^{*}),
\end{equation}
where $\delta>0$ is an arbitrary fixed constant, $A$ is a nonlinear operator on $K$ and $P_{K}$ is the {\it nearest point projection map} from $H$ onto $K,$ i.e., $P_{K}x=y$ where $\Vert x-y \Vert = \underset{u\in K}{inf}\Vert x-u \Vert$ for $x\in H$. Consequently, under appropriate conditions on $A$ and $\delta,$ fixed point methods can be used to find or
 approximate a solution of the variational inequality. Considerable efforts have been
 devoted to this problem (see, for example, \cite{xukim, yamada} and the references contained therein). For instance, if $A$ is strongly monotone and Lipschitz then, a mapping $B:H\to H$ defined by $Bx= P_{K}(x-\delta Ax),$ $x\in H$ with $\delta>0$ sufficiently small is a strict contraction. Hence,
the {\it Picard iteration,} $x_{0}\in H,$ $x_{n+1}=Bx_{n},~~n\ge 0$ of the classical Banach contraction mapping
principle converges to the unique solution of the variational inequality. It has been observed that the projection operator $P_{K}$ in the
fixed point formulation \eqref{fixpoint} may make the computation of
the iterates difficult due to possible complexity of the convex set
$K.$ In order to reduce the possible difficulty with the use of
$P_{K},$ Yamada \cite{yamada} introduced the following  hybrid descent
method for solving the variational inequality:
\begin{equation}\label{recur-1}
x_{n+1}=Tx_{n}-\lambda_n\mu A(Tx_{n}), ~~~n\geq 0,
\end{equation}
where $T$ is a nonexpansive mapping, $A$ is an $\eta-$strongly
monotone and $\kappa-$Lipschitz operator with $\eta > 0,~~\kappa> 0,~~0<\mu <\frac{2\eta}{\kappa^2}$. He proved that if
$\{\lambda_n\}$ satisfies appropriate conditions then, $\{x_n\}$ converges strongly to the unique solution $x^{*}$
of the variational inequality
\begin{eqnarray*}
\langle Ax^{*},x-x^{*}\rangle\geq0,~~x\in F(T).
\end{eqnarray*}
\noindent Very recently, Tian \cite{tian} combined the Yamada's method
\eqref{recur-1} with the iterative method \eqref{06} and introduced
the following general iterative method in Hilbert spaces:
\begin{eqnarray}\label{09}
x_{n+1}=\alpha_{n}\gamma f(x_{n})+(I-\alpha_{n}\mu
A)Tx_{n},~~n\geq0.
\end{eqnarray}
Then, he proved that the sequence $\{x_{n}\}$ generated by
\eqref{09} converges strongly to the unique solution $x^{*}\in F(T)$
of the variational inequality
\begin{eqnarray*}
\langle(\gamma f-\mu A)x^{*},x-x^{*}\rangle\leq0,~~x\in F(T).
\end{eqnarray*}


\noindent We remark immediately here that the results of Tian
\cite{tian} improved the results of Yamada \cite{yamada}, Moudafi
\cite{moudafi}, Xu \cite{xuk} and Marino and Xu \cite{xu4} in Hilbert spaces.\\


\noindent In this paper, motivated and inspired by the above research results,
our purpose is to extend the result of Tian \cite{tian} to $q$-uniformly smooth
Banach space whose duality mapping is weakly sequentially continuous. Thus, our results are applicable in $l_{p}$ spaces, $1< p<\infty$. Furthermore, our results extend the results of Moudafi \cite{moudafi}, Xu \cite{xuk} and Marino and Xu \cite{xu4} to Banach spaces much more general than Hilbert.


 \section{Preliminaries}
\noindent Let $E$ be a real Banach space.
\noindent Let $K$ be a nonempty closed convex and bounded subset of
a Banach space $E$ and let the diameter of $K$ be defined by
$d(K):=sup\{\|x-y\|: x, y\in K\}$. For each $x\in K,$ let $r(x,
K):=sup\{\|x-y\|: y\in K\}$ and let $r(K):=inf\{r(x, K): x \in K\}$
denote the Chebyshev radius of $K$ relative to itself. The
\emph{normal structure coefficient} $N(E)$ of $E$ (see, for example,
\cite{bynum}) is defined by $N(E):=inf\Big \{\frac{d(K)}{r(K)}: K $
is a closed convex and bounded
 subset of E with $ d(K)>0\Big \}.$ A space $E$ such that $N(E)>1$ is said to
have uniform normal structure. It is known that all uniformly convex
and uniformly
smooth Banach spaces have uniform normal structure (see, for example, \cite{chiliudo, limxu}).\\

\noindent Let $\mu$ be a linear continuous functional on $\ell^{\infty}$ and let $a=(a_1,a_2,\ldots) \in \ell^{\infty}$. We will sometimes write
$\mu_n(a_n)$ in place of the value $\mu(a)$. A linear continuous functional $\mu$ such that $||\mu||=1=\mu(1)$ and $\mu_n(a_n)=\mu_n(a_{n+1})$ for every $a=(a_1,a_2,\ldots) \in \ell^{\infty}$ is called a {\it Banach limit}. It is known that if $\mu$ is a Banach limit, then
$$\underset{n\rightarrow \infty}\liminf a_n \leq \mu_n(a_n) \leq \underset{n\rightarrow \infty}\limsup a_n$$
\noindent for every $a=(a_1,a_2,\ldots) \in \ell^{\infty}$ (see, for example, \cite{chiliudo, chidmono}).\\[2mm]
Let $E$ be a normed space  with ${\rm dimE} ~ \geq 2$. The {\it
modulus of smoothness} of $E$  is the function $\rho_{E}: [0, \infty
) \rightarrow [0, \infty)$  defined by
$$\rho_{E}(\tau):= \sup \left\{ \frac {\Vert x+y\Vert \ + \Vert x-y\Vert}{2} -1: \Vert x\Vert
 = 1; \Vert y \Vert = \tau \right \}.$$  The space $E$ is called {\it uniformly
  smooth} if and only if $\underset{t \rightarrow 0^{+}}
{lim}\frac{\rho_{E}(t)}{t} = 0$. For some positive constant $q,$ $E$
is called {\it $q-$uniformly smooth} if there exists a constant
$c>0$ such that $\rho_{E}(t)\le ct^{q},~~~~t>0.$ It is known that
\begin{displaymath}
L_{p} (or~~ l_{p})~~~{\rm spaces~~~are}~~~ \left\{ \begin{array}{ll}
2-~~~{\rm uniformly~~ smooth}, ~~~{\rm if,}~~~ 2\le p<\infty \\[3mm]
 p-~~~{\rm uniformly~~ smooth}, ~~~{\rm if,}~~~ 1< p\le 2.
\end{array} \right.
\end{displaymath}
 It is well known that if $E$ is smooth
then the duality mapping is singled-valued, and if $E$ is uniformly
smooth then the duality mapping is norm-to-norm uniformly continuous
on bounded subset
of E. \\[2mm]
\noindent We shall make use of the following well known results.
\begin{lemma}\label{lm21}
Let $E$ be a real normed space. Then
$$||x+y||^2 \leq ||x||^2+ 2\langle y, j(x+y) \rangle,$$
\noindent for all $x, y \in E$ and for all $j(x+y) \in J(x+y).$
\end{lemma}
\begin{lemma} {\rm (Xu, \cite{xu3})}\label{xu}
Let $E$ be a real $q$-uniformly smooth Banach space for some $q>1,$
then there exists some positive constant $d_{q}$ such that
\begin{eqnarray*}
\Vert x+y \Vert^{q} \le \Vert x \Vert^{q}+ q\langle y,j_{q}(x)
\rangle + d_{q} \Vert y \Vert^{q}~~~~\forall~ x,y \in E ~{\rm and}~
j_{q}(x) \in J_{q}(x).
\end{eqnarray*}
\end{lemma}

 \begin{lemma}{\rm (Xu, \cite{xu2})}\label{xu2}
 Let $\{a_{n}\}$ be a sequence of
 nonnegative real numbers satisfying the following relation:
 $$a_{n+1} \leq (1-\alpha_{n})a_{n} + \alpha_{n} \sigma_{n} + \gamma_{n}, n\geq 0,$$
where,
 $(i) ~\{\alpha_{n}\} \subset [0,1], ~\sum \alpha_{n} = \infty;$
 $(ii)~ \limsup ~\sigma_{n} \leq 0;$ $(iii)~\gamma_{n} \geq
 0;~(n\geq 0),\\~\sum\gamma_{n} < \infty.$ Then,
$a_{n}\rightarrow 0$  as $ n\rightarrow \infty.$
 \end{lemma}


\begin{lemma} {\rm (Lim and Xu, \cite{limxu})}\label{limxu}
Suppose $E$ is a Banach space with uniform normal structure, $K$
is a nonempty bounded subset of $E,$ and $T:K\to K$ is uniformly
$k-$Lipschitzian mapping with $k<N(E)^{\frac{1}{2}}.$ Suppose also
there exists a nonempty bounded closed convex subset $C$ of $K$
with the following property $(P):$
$$(P)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x\in C~~~~~~{\rm implies}~~~~~~~~\omega_{w}(x) \subset C,$$  where $\omega_{w}(x)$ is the $\omega-$limit set of $T$ at $x,$ i.e., the set $$\{y\in E: y={\rm weak}-\underset{j}{lim}T^{n_{j}}x~~~~~{\rm for~~ some}~~~ n_{j}\to \infty\}.$$ Then, $T$ has a fixed point in $C.$
\end{lemma}



\begin{lemma} {\rm (Jung, \cite{jung})}\label{jung}
Let $C$ be a nonempty, closed and convex subset of a reflexive Banach space $E$
which satisfies Opial's condition and suppose $T:C\rightarrow E$ is nonexpansive. Then $I-T$ is demiclosed at zero, i.e., $x_n\rightharpoonup x,~~x_n-Tx_n\rightarrow 0$ implies that $x =Tx$.
\end{lemma}


\begin{lemma}\label{accre}
Let $E$ be a real Banach space, $f:E\to E$
a contraction with coefficient $0<\alpha<1$, and $A:E\to E$ a
$\kappa$-Lipschitzian and $\eta$-strongly
accretive operator with $\kappa>0,~~\eta \in (0,1)$. Then for $\gamma\in(0,\frac{\mu \eta}{\alpha})$,
$$\langle(\mu A-\gamma f)x-(\mu A-\gamma f)y,j(x-y)\rangle\geq(\mu \eta-\gamma\alpha)||x-y||^{2},~\forall x,y\in E.$$
\noindent That is, $\mu A-\gamma f$ is strongly accretive with coefficient $\mu \eta-\gamma\alpha$.
\end{lemma}



 \section{Main Results}
\noindent We begin with the following lemma.
\begin{lemma}\label{l1}
Let $E$ be a $q$-uniformly smooth real Banach space with constant
$d_{q},q >1.$ Let $f:E\to E$ be a contraction mapping with constant of contraction $\alpha \in (0,1)$. Let $T:E\to E$ be a nonexpansive
mapping such that $F(T) \neq \emptyset$ and $A:E\to E$ be an $\eta$-strongly accretive mapping which is also
$\kappa$-Lipschitzian. Let $\mu\in\Big(0,\min\Big\{1,(\frac{q\eta}{d_{q}\kappa^{q}})^\frac{1}{q-1}\Big\}\Big)$ and $\tau:=\mu \big(\eta-\frac{\mu^{q-1}d_{q}\kappa^{q}}{q}\big).$ For each $t\in (0,1)$ and $\gamma \in (0,\frac{\tau}{\alpha})$
define a map $T_t:E\to E$ by
\begin{eqnarray*}
T_tx=t\gamma f(x)+(I-t\mu A)Tx,~x\in E.
\end{eqnarray*}
 Then,
$T_t$ is a strict contraction. Furthermore
\begin{eqnarray*}
||T_tx-T_ty||\leq[1-t(\tau-\gamma\alpha)]||x-y||.
\end{eqnarray*}
\end{lemma}
\begin{proof}
Without loss of generality, assume $\eta <\frac{1}{q}.$ Then, as $\mu<(\frac{q\eta}{d_{q}\kappa^{q}})^\frac{1}{(q-1)},$ we have $0<q\eta-\mu^{q-1}d_{q}\kappa^{q}.$ Furthermore, from
$\eta<\frac{1}{q}$ we have $q\eta-\mu^{q-1}d_{q}\kappa^{q}<1$ so that $0<q\eta-\mu^{q-1}d_{q}\kappa^{q}<1.$
Also as $\mu<1$ and $t \in (0,1)$ we obtained that $0<t \mu(q\eta-\mu^{q-1}d_{q}\kappa^{q})<1.$\\

\noindent For each $t\in (0,1),$ define $S_{t}x=(I-t\mu A)Tx,~~x\in E$, then for $x,y \in K$
\begin{eqnarray}\label{ugw}
||S_{t}x-S_{t}y||^{q}&=&||(I-t\mu A)Tx-(I-\mu A)Ty||^{q}\nonumber\\
&=&||(Tx-Ty)-t\mu(A(Tx)-A(Ty))||^{q}\nonumber\\
&\leq&||Tx-Ty||^{q}-qt\mu\langle A(Tx)-A(Ty),j_{q}(Tx-Ty)\rangle\nonumber\\
& &+t^{q}\mu^{q}d_{q}||A(Tx)-A(Ty)||^{q}\nonumber\\
&\leq&||Tx-Ty||^{q}-qt\mu\eta||Tx-Ty||^{q}\nonumber\\
& &+ t^{q}\mu^{q}\kappa^{q}d_{q}||Tx-Ty||^{q}\nonumber\\
&\leq&[1-t\mu(q\eta-t^{q-1}\mu^{q-1}\kappa^{q}d_{q})]||x-y||^{q}\nonumber\\
&\leq&\Big[1-qt\mu \big(\eta-\frac{\mu^{q-1}\kappa^{q}d_{q}}{q}\big)\Big]||x-y||^{q}
\nonumber\\
&\leq&\Big[1-t\mu \big(\eta-\frac{\mu^{q-1}\kappa^{q}d_{q}}{q}\big)\Big]^{q}||x-y||^{q}
\nonumber\\ &=&(1-t\tau)^{q}||x-y||^{q}.
\end{eqnarray}
\noindent It then follows from \eqref{ugw} that,
\begin{eqnarray*}
||S_{t}x-S_{t}y||\leq(1-t\tau)||x-y||.
\end{eqnarray*}
Using the fact that $T_tx=t\gamma f(x)+S_{t}x,~~x \in E$, we obtain for all $x,y \in E$ that
\begin{eqnarray*}
||T_tx-T_ty||&=&||t\gamma(f(x)-f(y))+(S_{t}x-S_{t}y)||\\
&\leq&t\gamma||f(x)-f(y)||+||S_{t}x-S_{t}y||\\
&\leq& t\gamma\alpha||x-y||+(1-t\tau)||x-y||\\
&=&[1-t(\tau-\gamma\alpha)]||x-y||.
\end{eqnarray*}
Therefore
\begin{eqnarray*}
||T_tx-T_ty||\leq[1-t(\tau-\gamma\alpha)]||x-y||,
\end{eqnarray*}
which implies that $T_t$ is a strict contraction. Therefore, by Banach contraction mapping principle, there exists a unique
fixed point $x_{t}$ of $T_t$ in $E$. That is,
\begin{equation}\label{f}
x_{t}=t\gamma f(x_{t})+(I-t\mu A)Tx_{t}.
\end{equation}
\end{proof}

\begin{proposition}\label{prop}
Let $\{x_{t}\}$ be defined by \eqref{f}, then
\begin{itemize}
\item [(i)] $\{x_{t}\}$ is bounded for $t \in (0,\frac{1}{\tau})$.
\item [(ii)] $\underset{t\to 0}{\lim}||x_{t}-Tx_{t}||=0$.
\end{itemize}
\end{proposition}

\begin{proof}
(i) For any $p\in F(T)$, we have
\begin{eqnarray*}
||x_{t}-p||&=&||(I-t\mu A)Tx_{t}-(I-t\mu A)p+t(\gamma f(x_{t})-\mu A(p))||\\
&\leq&(1-t\tau)||x_{t}-p||+t\gamma\alpha||x_{t}-p||+t||\gamma f(p)-\mu A(p)||\\
&=&[1-t(\tau-\gamma\alpha)]||x_{t}-p||+t||\gamma f(p)-\mu A(p)||.
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
||x_{t}-p||\leq\frac{1}{\tau-\gamma\alpha}||\gamma f(p)-\mu A(p)||.
\end{eqnarray*}
Hence, $\{x_{t}\}$ is bounded. Furthermore $\{f(x_{t})\}$ and
$\{A(Tx_{t})\}$ are also bounded. \\
(ii) From \eqref{f}, we have
\begin{eqnarray}\label{semee}
||x_{t}-Tx_{t}||=t||\gamma f(x_{t})-\mu A(Tx_{t})||\to 0~~{\rm as}~~t\to 0.
\end{eqnarray}
\end{proof}

\noindent Next, we show that $\{x_t\}$ is relatively norm compact as $t\rightarrow 0$. Let $\{t_n\}$ be
 a sequence in (0,1) such that $t_n\rightarrow 0$ as $n\rightarrow \infty$. Put $x_n:=x_{t_n}$. From \eqref{semee}, we obtain that
\begin{eqnarray*}
||x_n-Tx_n||\to 0~~{\rm as}~~n\to \infty.
\end{eqnarray*}


\begin{theorem}\label{tm1}
Assume that $\{x_{t}\}$ is  defined by \eqref{f}, then $\{x_{t}\}$
converges strongly as $t\to 0$ to a fixed point  $\tilde{x}$ of $T$
which solves the variational inequality problem:
\begin{eqnarray}\label{2}
\langle (\mu A-\gamma f)\tilde{x},j(\tilde{x}-z)\rangle\leq0,~~z\in F(T).
\end{eqnarray}
\end{theorem}

\begin{proof}
By Lemma \ref{accre}, $(\mu A-\gamma f)$ is strongly accretive, so
the variational inequality \eqref{2} has a unique solution in $F(T).$ Below we
use $x^{*}\in F(T)$ to denote the unique solution of \eqref{2}. \\

\noindent We next prove that $x_t\rightarrow x^{*}~~(t\rightarrow 0)$.
Now, define a map $\phi:E\rightarrow \mathbb{R}$ by
$$\phi(x):=\mu_n||x_n-x||^2,~~\forall x \in E,$$
\noindent where $\mu_n$ is a Banach limit for each $n$. Then, $\phi(x)\rightarrow \infty$ as $||x||\rightarrow \infty$, $\phi$ is continuous and convex, so as $E$ is reflexive, it follows that there exits $y^* \in E$ such that $\phi(y^*)=\underset{u \in E}\min \phi(u)$. Hence, the set
$$K^*:=\{x \in E:\phi(x)=\underset{u \in E}\min \phi(u)\}\neq \emptyset.$$
\noindent We now show that $T$ has a fixed point in $K^*$. We shall make use of Lemma \ref{limxu}. If $x$ is in $K^*$ and $y:=\omega-\lim_jT^{m_j}x$, then from the  weak lower semi-continuity of $\phi$ (since $\phi$ is lower semi-continuous and convex)  and $\underset{n \rightarrow \infty}\lim||x_n-Tx_n||=0$,
we have (since $\underset{n \rightarrow \infty}\lim ||x_n-Tx_n||=0$ implies $\underset{n \rightarrow \infty}\lim ||x_n-T^{m}x_n||=0,~~m \geq 1$, this is easily proved by induction),
\begin{eqnarray*}
\phi(y)&\leq&\underset{j \rightarrow \infty}\liminf \phi \Big(T^{m_j}x\Big) \leq \underset{m \rightarrow \infty}\limsup \phi \Big(T^mx\Big)\\
&=&\underset{m \rightarrow \infty}\limsup  \Big(\mu_n||x_n-T^mx||^2\Big)\\
&=&\underset{m \rightarrow \infty}\limsup  \Big(\mu_n||x_n-T^{m}x_n+T^{m}x_n-T^mx||^2\Big)\\
&\leq&\underset{m \rightarrow \infty}\limsup  \Big(\mu_n||T^{m}x_n-T^mx||^2\Big)\leq\underset{m \rightarrow \infty}\limsup  \Big(\mu_n||x_n-x||^2\Big)
=\phi(x)\\
&=&\underset{u \in E}\min \phi(u).
\end{eqnarray*}
So, $y \in K^*$. By Lemma \ref{limxu}, $T$ has a fixed point in $K^*$ and so $K^* \cap F(T) \neq \emptyset$. Now let $y\in K^{*}\cap F(T).$ Then, it follows that $\phi(y)\leq \phi(y+t(\gamma f-\mu A)y)$ and using Lemma  \ref{lm21}, we obtain that
$$||x_n-y-t(\gamma f-\mu A)y||^2\leq ||x_n-y||^2-2t\langle (\gamma f-\mu A)y,j(x_n-y-t(\gamma f-\mu A)y)\rangle.$$
\noindent This implies that $\mu_n\langle (\gamma f-\mu A)y,j(x_n-y-t(\gamma f-\mu A)y)\rangle \leq 0.$
Moreover,\\
$
\mu_{n}\langle(\gamma f-\mu A)y,j(x_{n}-y)\rangle=\mu_{n}\langle(\gamma f-\mu A)y,j(x_{n}-y)-j(x_{n}-y+t(\mu A-\gamma
f)y)\rangle \\
+\mu_{n}\langle(\gamma f-\mu A)y, j(x_{n}-y+t(\mu A-\gamma f)y)\rangle
\leq \mu_{n}\langle(\gamma f-\mu A)y,j(x_{n}-y)-j(x_{n}-y+t(\mu A-\gamma f)y)\rangle.
$\\
Since $j$ is norm-to-norm uniformly continuous on bounded subsets of
$E$, we obtain as $t\rightarrow 0$ that
\begin{eqnarray*}
\mu_{n}\langle(\gamma f-\mu A)y,j(x_n-y)\rangle\leq0.
\end{eqnarray*}
\noindent Now, using \eqref{f}, we have
\begin{align*}
||x_n-y||^2=\,&t_n\langle \gamma f(x_n)-\mu Ay, j(x_{n}-y)\rangle+\langle (I-t_n\mu A)(Tx_{n}-y),j(x_{n}-y)\rangle \\
=\,&t_n\langle \gamma f(x_n)-\mu A y,j(x_{n}-y)\rangle +\langle(I-\mu A)Tx_n-(I-\mu A)y,j(x_{n}-y)\rangle\\
\leq\,&[1-t_{n}(\tau-\gamma\alpha)]||x_{n}-y||^{2}+t_n\langle(\gamma f-\mu A)y,j(x_{n}-y)\rangle.
\end{align*}
So,
\begin{eqnarray*}
||x_n-y||^2\leq\frac{1}{\tau-\gamma\alpha}\langle (\gamma f-\mu A)y,j(x_n-y)\rangle.
\end{eqnarray*}
Again, taking Banach limit, we obtain
$$\mu_n||x_n-y||^2\leq \frac{1}{\tau-\gamma\alpha}\mu_n \langle (\gamma f-\mu A)y,j(x_n-y)\rangle \leq 0,$$
\noindent which implies that $\mu_n||x_n-y||^2=0$. Hence, there exists a subsequence of $\{x_n\}_{n=1}^{\infty}$ which we still denoted by $\{x_n\}_{n=1}^{\infty}$ such that $\underset{n \rightarrow \infty}\lim x_n=y$. We now show that $y$ solves the variational inequality \eqref{2}.
Since
$$x_{t}=t\gamma f(x_{t})+(I-t\mu A)Tx_{t},$$
\noindent we can derive that
$$(\mu A-\gamma f)(x_t)=-\frac{1}{t}(I-T)x_t+\mu (Ax_t-ATx_t).$$
\noindent It follows that for $z \in F(T)$,
\begin{eqnarray}\label{ihun}
\langle (\mu A-\gamma f)(x_t),j(x_t-z)\rangle&=&-\frac{1}{t}\langle (I-T)x_t-(I-T)z,j(x_t-z)\rangle\nonumber\\
& &+\mu \langle(Ax_t-ATx_t),j(x_t-z)\rangle \nonumber\\
&\leq& \mu \langle(Ax_t-ATx_t),j(x_t-z)\rangle.
\end{eqnarray}
Since $T$ is nonexpansive, then, $I-T$ is accretive, which implies, $\langle (I-T)x_t-(I-T)z,j(x_t-z)\rangle\geq 0$. Now replacing $t$ in \eqref{ihun} with $t_n$ and letting $n\rightarrow \infty$, noticing that $(Ax_{t_n}-ATx_{t_n})\rightarrow (Ay-Ay)$ we obtain
$$\langle (\mu A-\gamma f)y,j(y-z)\rangle\leq0,$$ since $z\in F(T)$ is arbitrary, we get $y=x^{*}.$\\
Assume now that there exists another subsequence $\{x_m\}_{m=1}^{\infty}$ of $\{x_n\}_{n=1}^{\infty}$ such that $\underset{m \rightarrow \infty}\lim x_m=u^{*}$. Then, since $\underset{n \rightarrow \infty}\lim||x_n-Tx_n||=0,$ we have that $u^{*} \in F(T).$ Repeating the argument above with $y$ replaced by $u^{*}$ we will get that $u^{*}$ solves the variational inequality \eqref{2}, and so by uniqueness, we obtain $x^{*}=y=u^{*}.$ This complete the proof.
\end{proof}
\begin{theorem}\label{tm2}
Let $E$ be a real $q$-uniformly smooth Banach space with
whose duality map is weakly sequentially continuous. Let $T:E\to E$ be a nonexpansive mapping with
$F(T)\neq\emptyset.$ Let $A:E\to E$ be an
$\eta$-strongly accretive map which is also $\kappa$-Lipschitzian.
Let $f:E\to E$ be a contraction map with coefficient $0<\alpha<1.$
Let $\{\alpha_{n}\}_{n=1}^{\infty}$ be a real sequence in [0,1]
satisfying:\\
$(C1)~~\lim\alpha_{n}=0,$\\
$(C2)~~\sum\alpha_{n}=\infty$ and \\
$(C3)~~\sum |\alpha_{n+1}-\alpha_n|<\infty$.\\
Let $\mu,~~\gamma$ and $\tau$ be as in Lemma \ref{l1}. Define a sequence $\{y_n\}_{n=1}^{\infty}$ iteratively in $E$ by $y_0\in E$,
\begin{eqnarray}\label{namdi}
y_{n+1}=\alpha_{n}\gamma f(y_n)+(I-\alpha_{n}\mu A)Ty_n.
\end{eqnarray}
Then, $\{y_n\}_{n=1}^{\infty}$ converges strongly to $x^{*}\in F(T)$ which is also a solution to the following
variational inequality
\begin{eqnarray}\label{namdi2}
\langle(\gamma f-\mu A)x^{*},j(y-x^{*})\rangle\leq 0,~\forall~y\in F(T).
\end{eqnarray}
\end{theorem}

\begin{proof}
Since the mapping $T:E\to E$ is nonexpansive,
then from Theorem \ref{tm1}, the variational inequality \eqref{namdi2}
has a unique solution $x^{*}$ in $F(T)$. Furthermore, the sequence $\{y_n\}$
satisfies
\begin{eqnarray*}
||y_{n}-x^{*}||\leq\max\Big\{||y_{0}-x^{*}||,\frac{||\gamma
f(x^{*})-\mu Ax^{*}||}{\tau-\gamma\alpha}\Big\},~\forall n\geq0.
\end{eqnarray*}
It is obvious that this is true for $n=0$. Assume it is true for
$n=k$ for some $k\in \mathbb{N},$ from the recursion formula \eqref{namdi}, we have
\begin{eqnarray*}
||y_{k+1}-x^{*}||&=&||\alpha_{k}\gamma f(y_{k})+(I-\alpha_{k}\mu
A)Ty_{k}-x^{*}||\\&=&||\alpha_{k}(\gamma f(y_{k})-\mu Ax^{*})+(I-\alpha_{k}\mu A)Ty_{k}-(I-\alpha_{k}\mu A)x^{*}||\\
&\leq&[1-\alpha_{k}(\tau-\gamma\alpha)]||y_{k}-x^{*}||+\alpha_{k}(\tau-\gamma\alpha)\frac{||\gamma f(x^{*})-\mu Ax^{*}||}{\tau-\gamma\alpha}\\
&\leq&\max\Big\{||y_{k}-x^{*}||,\frac{||\gamma f(x^{*})-\mu Ax^{*}||}{\tau-\gamma\alpha}\Big\}
\end{eqnarray*}
and the claim follows by induction. Thus, the sequence $\{y_n\}_{n=1}^{\infty}$ is
bounded and so are $\{f(y_n)\}_{n=1}^{\infty}$ and
$\{Ty_n\}_{n=1}^{\infty}$.  Also from \eqref{namdi}, we have
\begin{eqnarray*}
||y_{n+1}-y_n||&=&||\alpha_n\gamma(f(y_n)-f(y_{n-1}))+\gamma (\alpha_n-\alpha_{n-1})f(y_{n-1})\nonumber \\
& &+(I-\mu \alpha_nA)Ty_n-(I-\mu \alpha_nA)Ty_{n-1}+\mu(\alpha_n-\alpha_{n-1})ATy_{n-1}||\nonumber \\
&\leq&(1-\alpha_n(\tau-\gamma\alpha))||y_n-y_{n-1}||+M|\alpha_n-\alpha_{n-1}|,
\end{eqnarray*}
for some $M> 0$. By Lemma \ref{xu2}, we have $\underset{n\rightarrow \infty}\lim ||y_{n+1}-y_n||=0$.
Furthermore, we obtain
\begin{eqnarray}\label{demiclose}
||y_n-Ty_n||&\leq&||y_n-y_{n+1}||+||y_{n+1}-Ty_n||\\
&=&||y_n-y_{n+1}||+\alpha_n||\gamma f(y_n)-\mu ATy_n||\rightarrow 0~{\rm as}~n\to \infty.\nonumber
\end{eqnarray}
\noindent Let $\{y_{n_{j}}\}$ be a subsequence of $\{y_{n}\}$ such that $$\underset{n\to\infty}{limsup}\langle(\gamma f-\mu A)x^{*},j(y_{n}-x^{*})\rangle=\underset{j\to\infty}{lim}\langle(\gamma f-\mu A)x^{*},j(y_{n_{j}}-x^{*})\rangle. $$
Assume also $y_{n_{j}}\rightharpoonup z$ as $ j\to\infty,$ for some $z\in E.$ Then, using this, \eqref{demiclose} and the demiclosedness of $(I-T)$ at zero, we have $z\in F(T).$ Since $j$ is weakly sequentially continuous, we have

\begin{eqnarray*}
\underset{n\to\infty}{limsup}\langle(\gamma f-\mu A)x^{*},j(y_{n}-x^{*})\rangle&=&\underset{j\to\infty}{lim}\langle(\gamma f-\mu A)x^{*},j(y_{n_{j}}-x^{*})\rangle\\ &=& \langle(\gamma f-\mu A)x^{*},j(z-x^{*})\rangle \le 0.
\end{eqnarray*}
\noindent Finally, we show that $y_n\to x^{*}$. From the recursion formula
\eqref{namdi}, let
$$T_ny_{n}:=\alpha_{n}\gamma f(y_{n})+(I-\alpha_{n}\mu
A)Ty_{n},$$
\noindent and from Lemma \ref{l1}, we have
\begin{eqnarray*}
||y_{n+1}-x^{*}||^{2}&=&||T_ny_{n}-T_nx^{*}+T_nx^{*}-x^{*}||^{2}\\
&=& ||T_ny_{n}-T_nx^{*}+\alpha_{n}(\gamma f-\mu A)x^{*}||^{2}\\
&\leq&||T_ny_{n}-T_nx^{*}||^2+2\alpha_n\langle(\gamma f-\mu A)x^{*},j(y_{n+1}-x^{*})\rangle\\
&\leq&[1-\alpha_{n}(\tau-\gamma\alpha)]||y_{n}-x^{*}||^2\\
& &+2\alpha_{n}(\tau-\gamma\alpha)\frac{\langle(\gamma f-\mu A)x^{*},j(y_{n+1}-x^{*})\rangle}{\tau-\gamma\alpha}
\end{eqnarray*}
and by Lemma \ref{xu2} we have that $ y_{n} \to x^{*}~~{\rm
as}~~n\to \infty.$ This completes the proof.
\end{proof}


\noindent We have the following corollaries.

\begin{corollary}
Let $E=l_{p}$ space, $(1<p<\infty)$ and $\{x_n\}_{n=1}^{\infty}$ be generated by $x_0\in E$,
\begin{eqnarray*}
x_{n+1}=\alpha_{n}\gamma f(x_n)+(I-\alpha_{n}\mu A)Tx_n.
\end{eqnarray*}
Assume that $\{\alpha_{n}\}_{n=1}^{\infty}$ is a sequence in [0,1]
satisfying $(C1)-(C3)$, then $\{x_n\}_{n=1}^{\infty}$ converges strongly to $x^{*}\in F(T)$ which solves the
variational inequality
\begin{eqnarray}\label{ttian}
\langle(\gamma f-\mu A)x^{*},y-x^{*}\rangle\leq 0,~\forall~y\in F(T).
\end{eqnarray}
 \end{corollary}
\begin{corollary}{\rm(Tian \cite{tian})}
Let $E=H$ be a real Hilbert space and $\{x_n\}_{n=1}^{\infty}$ be generated by $x_0\in H$,
\begin{eqnarray*}
x_{n+1}=\alpha_{n}\gamma f(x_n)+(I-\alpha_{n}\mu A)Tx_n.
\end{eqnarray*}
Assume that $\{\alpha_{n}\}_{n=1}^{\infty}$ is a sequence in [0,1]
satisfying $(C1)-(C3)$, then $\{x_n\}_{n=1}^{\infty}$ converges strongly to $x^{*}\in F(T)$ which solves the
variational inequality \eqref{ttian}
 \end{corollary}

\begin{corollary}{\rm (Marino and Xu \cite{xu6})}
Let $\{x_n\}_{n=1}^{\infty}$ be generated by $x_0\in H$,
\begin{eqnarray*}
x_{n+1}=\alpha_{n}\gamma f(x_n)+(I-\alpha_{n}A)Tx_n.
\end{eqnarray*}
Assume that $\{\alpha_{n}\}_{n=1}^{\infty}$ is a sequence in [0,1]
satisfying $(C1)-(C3)$, then $\{x_n\}_{n=1}^{\infty}$ converges strongly to $x^{*}\in F(T)$ which solves the
variational inequality \eqref{ttian}.
 \end{corollary}

\begin{corollary}{\rm (Xu \cite{xuk})}
Let $\{x_n\}_{n=1}^{\infty}$ be generated by $x_0\in H$,
\begin{eqnarray*}
x_{n+1}=\alpha_{n} f(x_n)+(1-\alpha_{n})Tx_n,~~n\geq 0.
\end{eqnarray*}
Assume that $\{\alpha_{n}\}_{n=1}^{\infty}$ is a sequence in [0,1]
satisfying $(C1)-(C3)$, then $\{x_n\}_{n=1}^{\infty}$ converges strongly to $x^{*}\in F(T)$.
 \end{corollary}

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