text:
\section{Introduction}
In this paper, we consider the following equilibrium problem ($EP$)
in the sense of Blum and Otteli \cite{Blum}, which consists in
finding a point $x^{*}\in C$ such that
\begin{equation*}\label{vthoe1}
f(x^{*},y)\geq 0, \: \:\ \forall \ y\in C,
\end{equation*}
where $C$ is a nonempty, closed and convex subset of a real Banach
space $E$ and \linebreak
$f:C\times C\rightarrow \mathbb{R}$ is an equilibrium
bifunction, i.e., $f(x,x)=0$ for all $x\in C$. The solution set of
($EP$) is denoted by $E(f)$. The equilibrium problem which also
known under the name of Ky Fan inequality \cite{Fan} covers, as special
cases, many well-known problems, such as the optimization problem,
the variational inequality problem and nonlinear complementarity
problem, the saddle point problem, the generalized Nash equilibrium
problem in game theory, the fixed point problem and others; (see
\cite{Mu,Stro}). Also numerous problems in physic and
economic reduce to find a solution of an equilibrium problem. Many
methods have been proposed to solve the equilibrium problems see for
instance \cite{Blum,Mou,Mu,Van,Vuo}. In 1980,\linebreak Cohen \cite{Coh}
introduced a useful tool for solving optimization problem which is
known as auxiliary problem principle and extended it to
variational inequality \cite{Cohe}. In auxiliary problem principle a
sequence $\{x_k\}$ is generated as follows: $x_{k+1} \in C$ is a
unique solution of the following strongly convex problem
\begin{equation}\label{equ.1}
\min\Big\{c_k f(x_k, y) + \frac{1}{2} \|x_k - y\|\Big\},
\end{equation}
where $c_k >0$. Recently, Mastroeni \cite{Mas} extended the auxiliary
problem principle to equilibrium problems under the assumptions that
the equilibrium function $f$ is strongly monotone on $C\times C$ and
that $f$ satisfies the following Lipschitz- type condition:
\begin{equation}\label{equ.2}
f(x,y)+f(y,z)\geq f(x,z)-c_{1}\|y-x\|^{2}-c_{2}\|z-y\|^{2},
\end{equation}
for all $x,y,z \in C$ where $c_1, c_2>0$. To avoid the monotonicity
of $f$, motivated by Antipin \cite{Ant}, Tran et al. \cite{Tr} have used
an extrapolation step in each iteration after solving (\ref{equ.1})
and suppose that $f$ is pseudomonotone on $C\times C$ which is
weaker than monotonicity assumption. They assumed $y_k$ was the
unique solution of (\ref{equ.1}) and the unique solution of the
following strongly convex problem
\begin{equation*}
\min\Big\{c_k f(y_k, y) + \frac{1}{2} \|y - x_k\|\Big\},
\end{equation*}
is denoted by $\{x_{k+1}\}$. In special case , when the problem
$(EP)$ is a variational inequality problem, this method reduces to
the classical extragradient method which has been introduced by
Korpelevich \cite{korp}.
The extragradient method is well known because of its efficiency
in numerical tests. In the recent years, many authors obtained
extragradient algorithms for solving ($EP$) in Hilbert spaces
where convergence of the proposed algorithms was required $f$ to
satisfy a certain Lipschitz-type condition \cite{Ng,Tr,Vuo}.
Lipschitz-type condition depends on two positive parameters
$c_{1}$ and $c_{2}$ which in some cases, they are unknown or
difficult to approximate. In other to avoid this requirement,
authors used the linesearch technique in a Hilbert space to obtain
convergent algorithms for solving equilibrium problem
\cite{Ng,Tr,Vuo}.
\par
In this paper, we consider the following auxiliary equilibrium
problem ($AUEP$) for finding $x^{*}\in C$ such that
\begin{equation}\label{vthoe2}
\rho f(x^{*},y)+L(x^{*},y)\geq 0,
\end{equation}
for all $y\in C$, where $\rho>0$ is a regularization parameter and
$L:C\times C\rightarrow \mathbb{R}$ be a nonnegative differentiable
convex bifunction on $C$ with respect to the second argument $y$,
for each fixed $x\in C$, such that
 \begin{enumerate}
\item[(i)]$L(x,x)=0$ for all $x\in C$,
\item[(ii)]$\nabla_{2}L(x,x)=0$ for all $x\in C$.
\end{enumerate}
Where $\nabla_{2}L(x,x)$ denotes the gradient of the function
$L(x,.)$ at $x$.
\par
In the recent years, many authors studied the problem of finding a
common element of the set of fixed points of a nonlinear mapping
and the set of solutions of an equilibrium problem in the
framework of Hilbert spaces and Banach spaces, see for instance
\cite{Ceng,Qin,tak,Ta,Vuo}. In all of these methods, authors have
used metric projection in Hilbert spaces and generalized metric
projection in Banach spaces.
\par
In this paper, motivated D. Q. Tran et al. \cite{Tr} and P. T.
vuong et al. \cite{Vuo}, we introduce new extragradient and
linesearch algorithms for finding a common element of the set of
solutions of an equilibrium problem and the set of fixed points of
a relatively nonexpansive mapping in Banach spaces, by using sunny
generalized nonexpansive retraction. Using this method, we prove
strong convergence theorems under suitable conditions.
\section{Preliminaries}
 We denote by $J$ the normalized duality mapping from $E$ to $2^{E^{*}}$ defined by
\[Jx=\lbrace x^{*}\in E^{*} :\langle x , x^{*} \rangle=\Vert x\Vert^{2}=\Vert x^{*}\Vert^{2}\rbrace, \  \forall  \ x\in E.\]
Also, denote the strong convergence and the weak convergence of a sequence $\{x^{k}\}$ to $x$ in $E$ by $x^{k}\rightarrow x$ and $x^{k}\rightharpoonup x$, respectively.\\

Let $S(E)$ be the unite sphere centered at the origin of $E$. A Banach space $E$ is strictly convex if $\|\frac{x+y}{2}\|<1$, whenever $x, y \in S(E)$ and $x\neq y$.
Modulus of convexity of $E$ is defined by $$\delta_{E}(\epsilon)=inf \{1-\frac{1}{2}\|(x+y)\|:\ \|x\|, \|y\|\leq 1, \ \|x-y\|\geq \epsilon\}$$ for all $\epsilon\in [0, 2]$. $E$ is said to be uniformly convex if $\delta_{E}(0)=0$ and $\delta_{E}(\epsilon)>0$ for all $0<\epsilon\leq2$. Let $p$ be a fixed real number with $p\geq2$. A Banach space $E$ is said to be $p$-uniformly convex \cite{kato} if there exists a constant $c>0$ such that $\delta_{E}\geq c\epsilon^{p}$ for all $\epsilon\in [0, 2]$.
The Banach space $E$ is called smooth if the limit
 \begin{equation}\label{eq1}
 \lim_{t\rightarrow0}\frac{\|x+ty\|-\|x\|}{t},
 \end{equation}
 exists for all $x,y\in S(E)$. It is also said to be uniformly smooth if the limit
  $(\ref{eq1})$ is attained uniformly for all $x,y\in S(E)$. Every uniformly smooth Banach space $E$ is smooth.
If a Banach space $E$ uniformly convex, then $E$ is reflexive and strictly convex \cite{Agarwal,Takahashi}.
  \par
  Many properties of the normalized duality mapping $J$ have been given in \cite{Agarwal,Takahashi}.\\ We give some of those in the following:
  \begin{enumerate}
\item For every $x\in E$, $Jx$ is nonempty closed convex and bounded subset of $E^*$.
\item If $E$ is smooth or $E^*$ is strictly convex, then $J$ is single-valued.
\item If $E$ is strictly convex, then $J$ is one-one.
\item If $E$ is reflexive, then $J$ is onto.
\item If $E$ is strictly convex, then $J$ is strictly monotone, that is,
$$\langle x-y, Jx-Jy\rangle>0,$$
for all  $x, y\in E$ such that $x\neq y$.
\item If $E$ is smooth, strictly convex  and reflexive  and $J^*:E^*\rightarrow 2^E$ is the normalized duality mapping on $E^*$, then $J^{-1}=J^*$, $JJ^*=I_{E^*}$ and $J^*J=I_{E}$, where $I_{E}$ and $I_{E^*}$ are the identity mapping on $E$ and $E^*$, respectively.
\item If $E$ is  uniformly convex and uniformly smooth, then $J$ is uniformly norm-to-norm continuous on bounded sets of $E$ and $J^{-1}=J^*$ is also uniformly norm-to-norm continuous on bounded sets of $E^*$, i.e., for $\varepsilon>0$ and $M>0$, there is a $\delta>0$ such that
\begin{equation}\label{eq30}
\|x\|\leq M, \ \|y\|\leq M \ \ \text{and} \ \ \|x-y\|<\delta \ \ \Rightarrow \ \ \|Jx-Jy\|<\varepsilon,
\end{equation}
\begin{equation}\label{eq18}
\|x^{*}\|\leq M, \ \|y^{*}\|\leq M \ \ \text{and} \ \ \|x^{*}-y^{*}\|<\delta \ \ \Rightarrow \ \ \|J^{-1}x^{*}-J^{-1}y^{*}\|<\varepsilon.
\end{equation}
\end{enumerate}
\par
Let $E$ be a smooth Banach space, we define the function $\phi: E\times E\rightarrow \mathbb{R}$ by \[\phi(x, y)=\Vert x\Vert^{2}-2\langle x,Jy \rangle+\Vert y\Vert^{2},\]
for all $x, y\in E$. Observe that, in a Hilbert space $H$, $\phi(x,y)=\|x-y\|^{2}$ for all $x,y\in H.$\\
It is clear from definition of $\phi$ that for all $x, y, z, w\in E$,
\begin{equation}\label{eq14}
(\Vert x\Vert -\Vert y\Vert)^{2} \leq \phi(x,y)\leq (\Vert x\Vert+\Vert y\Vert)^{2},
\end{equation}
\begin{equation}\label{eq36}
\phi(x,y)=\phi(x,z)+\phi(z,y)+2\langle x-z,Jz-Jy\rangle,
\end{equation}
\begin{equation}\label{equa6}
2\langle x-y,Jz-Jw\rangle=\phi(x,w)+\phi(y,z)-\phi(x,z)-\phi(y,w).
\end{equation}
If $E$ additionally assumed to be strictly convex, then
\begin{equation}\label{equa21}
\phi(x,y)=0\:\:\:\Longleftrightarrow\:\:\: x=y.
\end{equation}
Also, we define the function $V:E\times E^{*}\rightarrow \mathbb{R}$ by $V(x, x^{*})=\| x\|^{2}-2<x, x^{*}>+\| x^{*} \|^{2}$,\\ for all $x\in E$ and $x^{*}\in E$. That is, $V(x,x^{*})=\phi(x,J^{-1}x^{*})$ for all $x\in E$ and $x\in E^{*}$.\\
It is well known that, if $E$ is a reflexive strictly convex and smooth Banach space with $E^{*}$ as its dual, then
\begin{equation}\label{eq5}
 V(x, x^{*})+2\langle J^{-1}x^{*}-x,y^{*}\rangle \leq V(x, x^{*}+y^{*}),
 \end{equation}
 for all $x\in E$ and all $x^{*},y^{*}\in E^{*}$ \cite{Rockfellar}.
\par
Let $E$ be a smooth Banach space and $C$ be a nonempty subset of $E$. A mapping $T: C\rightarrow C$ is called generalized nonexpansive \cite{Ibaraki} if $F(T)\neq \emptyset$ and $$\phi(y,Tx)\leq \phi(y,x),$$
for all $x\in C$ and all $y\in F(T).$\\
Let $C$ be a closed convex subset of $E$ and $T:C\rightarrow C$ be a mapping. A point $p$ in $C$ is said to be an asymptotic fixed point of $T$ if $C$ contains a sequence $\{x^{k}\}$ which converges weakly to $p$ such that $\lim\limits_{k\rightarrow\infty}(Tx^{k}-x^{k})=0$. The set of asymptotic fixed points of $T$ will be denoted by $\hat{F}(T)$.
A mapping $T:C\rightarrow C$ is called relatively nonexpansive if $\hat{F}(T)=F(T)$ and $\phi(p,Tx)\leq\phi(p,x)$ for all $x\in C$ and $p\in F(T)$. The asymptotic behavior of relatively nonexpansive mappings was studied in \cite{But}. $T$ is said to be relatively quasi-nonexpansive if $F(T)\neq\emptyset$ and $\phi(p,Tx)\leq\phi(p,x)$ for all $x\in C$ and all $p\in F(T)$. The class of relatively quasi-nonexpansive mapping is broader than the class of relatively nonexpansive mappings which requires $\hat{F}(T)=F(T)$.\\
It is well known that, if $E$ is a strictly convex and smooth Banach space, $C$ is a nonempty closed convex subset of $E$ and $T:C\rightarrow C$ is a relatively quasi-nonexpansive mapping, then $F(T)$ is a closed convex subset of $C$ \cite{Q}.
\par
Let $D$ be a nonempty subset of a Banach space $E$. A mapping $R:E\rightarrow D$ is said to be sunny \cite{Ibaraki} if $$R(Rx+t(x-Rx))=Rx,$$
for all $x\in E$ and all $t\geq 0$. A mapping $R:E\rightarrow D$ is said to be a retraction if $Rx=x$ for all $x\in D$.
$R$ is a sunny nonexpansive retraction from $E$ onto $D$ if $R$ is a retraction which is also sunny and nonexpansive.
A nonempty subset $D$ of a smooth Banach space $E$ is said to be a generalized nonexpansive retract (resp. sunny generalized nonexpansive retract) of $E$ if there exists a generalized nonexpansive retraction (resp. sunny generalized nonexpansive retraction) $R$ from $E$ onto $D$.\\
If $E$ is a smooth, strictly convex and reflexive Banach space,
$C^{*}$ be a nonempty closed convex subset of $E^{*}$ and
$\Pi_{C^{*}}$ be the generalized metric projection of $E^{*}$ onto
$C^{*}$. Then the $R=J^{-1}\Pi_{C^{*}}J$ is a sunny generalized
nonexpansive retraction of $E$ onto $J^{-1}C^{*}$ \cite{Kohsa}.
\begin{rk}
If $E$ is a Hilbert space. Then $R_{C}=\Pi_{C}=P_{C}$.
\end{rk}
We need the following lemmas for the proof of our main results.\\
If $C$ is a convex subset of Banach space $E$, then we denote by $N_{C}(\nu)$ the normal cone for $C$ at a point $\nu\in C$, that is \[N_{C}(\nu):=\{x^{*}\in E^{*}: \langle \nu-y,x^{*}\rangle\geq0,\ \forall  \ y\in C\}.\]
Suppose that $E$ is a Banach space and let $f:E\rightarrow (-\infty,+\infty]$ be a paper function. For $x_{0}\in D(f)$, we define the subdifferential of $f$ at $x_{0}$ as the subset of $E^{*}$ given by
$$\partial f(x_{0})=\{x^{*}\in E^{*}: \: f(x)\geq f(x_{0})+\langle x^{*},x-x_{0}\rangle, \: \forall x\in E\}.$$
If $\partial f(x_{0})\neq \emptyset$, then we say $f$ is subdifferentiable at $x_{0}$.
\begin{lm}\label{lem7}
Let $C$ be a nonempty convex subset of a Banach space $E$ and $f:E\rightarrow\mathbb{R}$ be a convex and subdifferentiable function, then $f$ is minimized at $x\in E$ if and only if $$0\in \partial f(x)+N_{C}(x).$$
\end{lm}
\begin{lm}\label{lem8}\cite{Au}
Let $E$ be a reflexive Banach space. If $f:E\rightarrow \mathbb{R}\cup\{+\infty\}$ and $g:E\rightarrow \mathbb{R}\cup\{+\infty\}$ are nontrivial, convex and lower continuous functions and if $\:0\in Int (Dom f-Dom g)\:$, then
$$\partial(f+g)(x)=\partial f(x)+\partial g(x).$$
\end{lm}
\begin{lm}\label{lem20}\cite{Au}
Suppose that a convex function $f$ is continuouse on the interior of its domain. Then, for all $x\in Int\: (Dom f)$, $\partial f(x)$
is non-empty and bounded.
\end{lm}
\begin{lm}\label{lem13}
Let $C$ be a nonempty convex subset of a Banach space $E$ and let $f:C\times C\rightarrow\mathbb{R}$ be an equilibrium bifunction and convex respect to the second variable. then the following statements are equivalent:
\begin{enumerate}
\item[(i)] $x^{*}$ is a solution to $E(f)$,
\item[(ii)] $x^{*}$ is a solution to the problem
$$\min_{y\in C}f(x^{*},y).$$
\end{enumerate}
\end{lm}
\begin{proof}
Using Lemma \ref{lem7}, we get desired results.
\end{proof}
Equivalence between $E(f)$ and ($AUEP$) is stated in the following lemma.
\begin{lm}\label{lem2}
Let $C$ nonempty, convex and closed subset of a reflexive Banach space $E$ and $f:C\times C\rightarrow\mathbb{R}$ be an equilibrium bifunction and let $x^{*}\in C$. suppose that $f(x^{*},.):C\rightarrow\mathbb{R}$ is convex and subdifferentiable on $C$. Let $L:C\times C\rightarrow\mathbb{R}_{+}$ be a differentiable convex function on $C$ with respect to the second argument $y$ such that
\begin{enumerate}
\item[(i)] $L(x^{*},x^{*})=0$,
\item[(ii)] $\nabla_2L(x^{*},x^{*})=0$.
\end{enumerate}
Then $x^{*}\in C$ is a solution to $E(f)$ if and only if $x^{*}$ is a solution to ($AUEP$).
\end{lm}
\begin{proof}
It is clear from lemmas \ref{lem8} and \ref{lem13}.
\end{proof}
\begin{lm}\cite{Ibaraki}
Let $C$ be a nonempty closed sunny generalized nonexpansive retract of a smooth and strictly convex Banach space $E$. Then the sunny generalized nonexpansive retraction from $E$ onto $C$ is uniquely determined.
\end{lm}
\begin{lm}\cite{Ibaraki}\label{lem4}
Let $C$ be a nonempty closed subset of a smooth and strictly convex Banach space $E$ such that there exists a sunny generalized nonexpansive retraction $R$ from $E$ onto $C$ and let $(x,z)\in E\times C$. Then the following hold:
\begin{enumerate}
\item $z=Rx $ if and only if $\langle x-z,Jy-Jz\rangle\leq 0$ for all $y\in C$,
\item $\phi(z,Rx)+\phi(Rx,x)\leq\phi(z,x).$
\end{enumerate}
\end{lm}
\begin{lm}\cite{Kohsa}\label{eq38}
Let $E$ be a smooth, strictly convex and reflexive Banach space and $C$ be a nonempty closed sunny generalized nonexpansive retract of $E$. Let $R$ be the sunny generalized nonexpansive retraction from $E$ onto $C$ and $(x,z)\in E\times C$. Then the following are equivalent:
\begin{enumerate}
\item $z=Rx$,
\item $\phi(z,x)=\min_{y\in C}\phi(y,x)$
\end{enumerate}
\begin{lm}\cite{Ibaraki}\label{Ibaraki}
Let $E$ be a smooth, strictly convex and reflexive Banach space and let $C$ be a nonempty closed subset of $E$. Then the following are equivalent:
\begin{enumerate}
\item $C$ is a sunny generalized nonexpansive retract of $E$,
\item $C$ is a generalized retract of $E$,
\item $JC$ is closed and convex.
\end{enumerate}
\end{lm}
\end{lm}
\begin{lm}\cite{Yao}\label{eq6}
Let $E$ be a $2$-uniformly convex and smooth Banach space. Then, for all $x, y\in E$, we have $$\|x-y\|\leq\frac{2}{c^{2}}\|Jx-Jy\|,$$
where $\frac{1}{c}(0\leq c\leq 1)$ is the $2$-uniformly convex constant of $E$.
\end{lm}
\begin{lm}\label{lem16}\cite{ch}
Suppose $p>1$ is a real number, then the following are equivalent
\begin{enumerate}
\item[(i)] $E$ is a $p$-uniformly convex Banach space,
\item[(ii)] there exits $\tau>0$, such that for each $f_{x}\in J_{p}(x)$ and $f_{y}\in J_{p}(y)$, we have
$$\langle x-y,f_{x}-f_{y}\rangle\geq\tau\|x-y\|^{p}.$$
\end{enumerate}
Where $J_{p}(x)=\{x^{*}\in E^{*}: \langle x,x^{*}\rangle=\|x\|\|x^{*}\|, \: \|x^{*}\|=\|x\|^{p-1}\}$.
\end{lm}
\begin{lm}\label{kamimra}\cite{Kamimura}
Let $E$ be a uniformly convex Banach space and let $r>0$. Then there exists a strictly increasing, continuous and convex function $g:[0, 2r]\rightarrow [0, \infty)$, $g(0)=0$ such that $$g(\|x-y\|)\leq\phi(x,y),$$ for all $x,y\in B_{r}(0)=\{z\in E:\|z\|\leq r\}$.
\end{lm}
\begin{lm}\cite{Cho}\label{lem15}
Let $E$ be a uniformly convex Banach space. Then there exists a continuous strictly increasing convex function $g:[0, 2r]\rightarrow [0, \infty)$, $g(0)=0$ such that $$\|\lambda x+\mu y+\gamma z\|^{2}\leq \lambda\|x\|^{2}+\mu\|y\|^{2}+\gamma\|z\|^{2}-\lambda\mu g(\|x-y\|),$$
for all $x,y,z\in B_{r}(0)=\{z\in E:\|z\|\leq r\}$ and all $\lambda, \mu, \gamma\in [0, 1]$ with $\lambda+\mu+\gamma=1$.
\end{lm}
\begin{lm}\cite{Kamimura}\label{lem5}
Let $E$ be a uniformly convex and smooth Banach space and let $\{x^{k}\}$ and $\{y^{k}\}$ be two sequences of $E$. If $\phi(x^{k}, y^{k})\rightarrow 0$ and either $\{x^{k}\}$ or $\{y^{k}\}$ is bounded, then $x^{k}-y^{k}\rightarrow 0.$
\end{lm}
\begin{lm}\label{lem10}\cite{ka}
Let $\{\alpha_{n}\}$ and $\{\beta_{n}\}$ be two positive and bounded sequences in $\mathbb{R}$, then
\begin{equation*}
\begin{aligned}
(\liminf_{n\rightarrow \infty}\alpha_{n})\times(\liminf_{n\rightarrow\infty}\beta_{n})&\leq\liminf_{n\rightarrow\infty}(\alpha_{n}\beta_{n})\\
&\leq(\liminf_{n\rightarrow\infty}\alpha_{n})\times(\limsup_{n\rightarrow\infty}\beta_{n})\\
&\leq\limsup_{n\rightarrow\infty}(\alpha_{n}\beta_{n})\\
&\leq(\limsup_{n\rightarrow\infty}\alpha_{n})\times(\limsup_{n\rightarrow\infty}\beta_{n}).
\end{aligned}
\end{equation*}
\end{lm}
\section{An Extragradient Algorithm}
In this section, we present an algorithm for finding a solution of the ($EP$) which
is also the common element of the set of fixed points of a relatively nonexpansive mapping.\\
Here, we assume that bifunction $f:C\times C\rightarrow\mathbb{R}$ satisfies in following conditions which $C$ is nonempty, convex and closed subset of uniformly convex and uniformly
smooth Banach space $E$,
 \begin{enumerate}
\item[(A1)]$f(x,x)=0$ for all $x\in C,$
\item[(A2)]$f$ is pseudomonotone on $C$, i.e., $f(x,y)\geq0\Longrightarrow f(y,x)\leq0$ for all $x,y\in C,$
\item[(A3)]$f$ is jointly weakly continuous on $C\times C$, i.e., if $x,y\in C$ and $\{x_{n}\}$ and $\{y_{n}\}$ are two sequences in $C$ converging weakly to $x$ and $y$, respectively, then $f(x_{n},y_{n})\rightarrow f(x,y),$
\item[(A4)]$f(x,.)$ is convex, lower semicontinuous and subdifferentiable on $C$ for every $x\in C,$
\item[(A5)]$f$ satisfies $\phi$-Lipschitz-type condition: $\exists c_{1}>0, \exists c_{2}>0,$ such that for every $x,y,z\in C$
\begin{equation}\label{equa17}
 f(x,y)+f(y,z)\geq f(x,z)-c_{1}\phi(y,x)-c_{2}\phi(z,y).
 \end{equation} \end{enumerate}
It is easy to see that if $f$ satisfies the properties $(A_{1})-(A_{4})$, then the set $E(f)$ of solutions of an equilibrium problem is closed and convex. Indeed, when $E$ is a Hilbert space, $\phi$-Lipschitz-type condition reduces to Lipschitz-type condition (\ref{equ.2}).\\
Throughout the paper $S$ is a relatively nonexpansive self-mapping of $C$.
\par
{\bf Algorithm $1$}
\begin{description}
\item[Step 0.] Suppose that $\{\alpha_{n}\}\subseteq [a,e]$ for some $0<a<e<1$, $\{\beta_{n}\}\subseteq [d,b]$ for some $0<d<b<1$ and $\{\lambda_{n}\}\subseteq (0,1]$ such that $\{\lambda_{n}\}\subseteq [\lambda_{min},\lambda_{max}]$, where
 $$0<\lambda_{min}\leq\lambda_{max}<\min\{\frac{1}{2c_{1}},\frac{1}{2c_{2}}\}.$$
\item[Step 1.] Let $x_{0}\in C$. Set n=0.\\
\item[Step 2.] Compute $y_{n}$ and $x_{n}$, such that
$$y_{n}=\arg\min_{y\in C}\{\lambda_{n}f(x_{n},y)+\frac{1}{2}\phi(y,x_{n})\},$$
$$z_{n}=\arg\min_{y\in C}\{\lambda_{n}f(y_{n},y)+\frac{1}{2}\phi(y,x_{n})\}.$$
\item[Step 3.] Compute $t_{n}=J^{-1}(\alpha_{n}Jx_{n}+(1-\alpha_{n})(\beta_{n}Jz_{n}+(1-\beta_{n})JSz_{n}))$.\\
If $y_{n}=x_{n}$ and $t_{n}=x_{n}$, then $x_{n}\in E(f)\cap F(S)$ and go to step 4.\\
\item[Step 4.] Compute $x_{n+1}=R_{C_{n}\cap D_{n}}x_{0}$, where $R_{C_{n}\cap D_{n}}$ is the sunny generalized nonexpansive retraction from $E$
onto $C_{n}\cap D_{n}$ and
$$C_{n}=\{z\in C: \phi(z,t_{n})\leq\phi(z,x_{n})\},$$
$$D_{n}=\{z\in C: \langle Jx_{n}-Jz,x_{0}-x_{n}\rangle\geq 0\}.$$
\item[Step 5.] set $n:=n+1$ and go to Step 2.
\end{description}
Before proving the strong convergence of the iterates generated by Algorithm 1, we prove following lemmas.
\begin{lm}\label{lem1}
 For every $x^{*}\in E(f)$ and $n\in \mathbb{N}$, we obtain
\begin{enumerate}
\item[(i)] $\langle Jx_{n}-Jy_{n},y-y_{n}\rangle\leq\lambda_{n}f(x_{n},y)-\lambda_{n}f(x_{n},y_{n}),\:\:\: \ \forall y\in C,$
\item[(ii)] $\phi(x^{*},z_{n})\leq\phi(x^{*},x_{n})-(1-2\lambda_{n}c_{1})\phi(y_{n},x_{n})-(1-2\lambda_{n}c_{2})\phi(z_{n},y_{n})$.
\end{enumerate}
 \end{lm}
 \begin{proof}
By the condition ($A4$) for $f(x,.)$ and from lemmas \ref{lem7} and \ref{lem8}, we obtain
$$z_{n}=\arg\min_{y\in C}\{\lambda_{n}f(y_{n},y)+\frac{1}{2}\phi(y,x_{n})\}$$
if and only if
$$0\in \lambda_{n}\partial_{2}f(y_{n},z_{n})+\frac{1}{2}\nabla_{1}\phi(z_{n},x_{n})+N_{C}(z_{n}).$$
This implies that $w\in \partial_{2}f(y_{n},z_{n})$ and $\overline{w}\in N_{C}(z_{n})$ exist such that
\begin{equation}\label{equa18}
0=\lambda_{n}w+Jz_{n}-Jx_{n}+\overline{w},
\end{equation}
so, from definition of $\partial_{2}f(y_{n},z_{n})$, we obtain
$$\langle w, y-z_{n}\rangle\leq f(y_{n},y)-f(y_{n},z_{n}),$$
for all $y\in C$. Set $y=x^{*}$, we have
$$\langle w, x^{*}-z_{n}\rangle\leq f(y_{n},x^{*})-f(y_{n},z_{n}).$$
So, by definition of the $N_{C}(z_{n})$ and equality (\ref{equa18}), we get
\begin{equation}\label{eq40} \lambda_{n}\langle w, z_{n}-y\rangle\leq \langle Jz_{n}-Jx_{n},y-z_{n}\rangle,
\end{equation} for all $y\in C$. Put $y=x^{*}$ in inequality \ref{eq40}, we have
\begin{equation}\label{equa1}
\langle Jz_{n}-Jx_{n},x^{*}-z_{n}\rangle\geq\lambda_{n}\{f(y_{n},z_{n})-f(y_{n},x^{*})\}
\geq\lambda_{n}f(y_{n},z_{n}),
\end{equation}
since $f(x^{*},y_{n})\geq0$ and $f$ is pseudomonotone on $C$. Replacing $x, y$ and $z$ by $x_{n}$, $y_{n}$ and $z_{n}$ in inequality (\ref{equa17}), respectively, we get
\begin{equation}\label{equa2}
f(y_{n},z_{n})\geq f(x_{n},z_{n})-f(x_{n},y_{n})-c_{1}\phi(y_{n},x_{n})-c_{2}\phi(z_{n},y_{n}).
\end{equation}
In a similar way, since $y_{n}=\arg\min_{y\in C}\{\lambda_{n}f(x_{n},y)+\frac{1}{2}\phi(y,x_{n})\}$, we have
$$\langle Jx_{n}-Jy_{n},y-y_{n}\rangle\leq\lambda_{n}\{f(x_{n},y)-f(x_{n},y_{n})\},$$
for all $y\in C$, hence ($i$) is proved. Let $y=z_{n}$ in above inequality, we obtain
\begin{equation}\label{equa3}
\langle Jx_{n}-Jy_{n},z_{n}-y_{n}\rangle\leq\lambda_{n}\{f(x_{n},z_{n})-f(x_{n},y_{n})\}.
\end{equation}
Combining inequalities (\ref{equa1}), (\ref{equa2}) and (\ref{equa3}), we get
\begin{equation}\label{equa19}
2\langle Jz_{n}-Jx_{n},x^{*}-z_{n}\rangle\geq 2\langle Jy_{n}-Jx_{n},y_{n}-z_{n}\rangle-2\lambda_{n}c_{1}\phi(y_{n},x_{n})-2\lambda_{n}c_{2}\phi(z_{n},y_{n}).
\end{equation}
From inequality (\ref{equa19}) and (\ref{eq36}), we have
\begin{equation*}
\begin{aligned}
\phi(x^{*},x_{n})-\phi(x^{*},z_{n})\geq\phi(z_{n},y_{n})+\phi(y_{n},x_{n})
-2\lambda_{n}c_{1}\phi(y_{n},x_{n})-2\lambda_{n}c_{2}\phi(z_{n},y_{n}).
\end{aligned}
\end{equation*}
Hence, ($ii$) is proved.
\end{proof}
\begin{rk}
In a real Hilbert space $E$, Lemma \ref{lem1} is reduced to Lemma
$3.1$ in \cite{Anh}.
\end{rk}
\begin{lm}
In Algorithm $1$, we obtain the unique optimal solutions $y_{n}$ and $z_{n}$.
\end{lm}
\begin{proof}
Let $y_{n},\acute{y}_{n}\in\arg\min_{y\in C}\{\lambda_{n}f(x_{n},y)+\frac{1}{2}\phi(y,x_{n})\}$, then using Lemma \ref{lem1}($i$), we have
\begin{equation}\label{equa22}
\langle Jx_{n}-Jy_{n},y-y_{n}\rangle\leq\lambda_{n}f(x_{n},y)-\lambda_{n}f(x_{n},y_{n}),\:\:\: \ \forall y\in C,
\end{equation}
\begin{equation}\label{equa23}
\langle Jx_{n}-J\acute{y}_{n},y-\acute{y}_{n}\rangle\leq\lambda_{n}f(x_{n},y)-\lambda_{n}f(x_{n},\acute{y}_{n}),\:\:\: \ \forall y\in C.
\end{equation}
Set $y=\acute{y}_{n}$ in inequality (\ref{equa22}) and $y=y_{n}$ in inequality (\ref{equa23}). Hence, we get
$$\langle J\acute{y}_{n}-Jy_{n},\acute{y}_{n}-y_{n}\rangle\leq0.$$
Since $J$ is monotone and one-one, we obtain $y_{n}=\acute{y}_{n}.$ In a similar way, also $z_{n}$ is unique.
\end{proof}
\begin{rk}
If $E$ is a real Hilbert space, then Algorithm $1$ is the same
Extragradient Algorithm in \cite{Vuo} provided that the sequence
$\{\alpha_{n}\}$ satisfies the conditions of Step $0$ of Algorithm
$1$.
\end{rk}
\begin{lm}\label{lem3}
For every $x^{*}\in E(f)\cap F(S)$ and $n\in \mathbb{N}$, we get
$$\phi(x^{*},t_{n})\leq\phi(x^{*},x_{n}).$$
\end{lm}
\begin{proof}
From Lemma \ref{lem1}, by the convexity of $\|.\|^{2}$ and by the definition of functions $\phi$ and $S$, we have
\begin{equation}\label{equa37}
\begin{aligned}
\phi(x^{*},t_{n})&=\phi(x^{*},J^{-1}(\alpha_{n}Jx_{n}+(1-\alpha_{n})(\beta_{n}Jz_{n}+(1-\beta_{n})JSz_{n})))\\
&=\|x^{*}\|^{2}+\|\alpha_{n}Jx_{n}+(1-\alpha_{n})(\beta_{n}Jz_{n}+(1-\beta_{n})JSz_{n})\|\\
&-2\langle x^{*},\alpha_{n}Jx_{n}+(1-\alpha_{n})(\beta_{n}Jz_{n}+(1-\beta_{n})JSz_{n})\rangle\\
&\leq\|x^{*}\|^{2}+\alpha_{n}\|x_{n}\|^{2}+(1-\alpha_{n})\beta_{n}\|z_{n}\|^{2}+(1-\alpha_{n})(1-\beta_{n})\|Sz_{n}\|^{2}\\
&-2\alpha_{n}\langle x^{*},Jx_{n}\rangle-2(1-\alpha_{n})\beta_{n}\langle x^{*},Jz_{n}\rangle-2(1-\alpha_{n})(1-\beta_{n})\langle x^{*},JSz_{n}\rangle\\
&=\alpha_{n}\phi(x^{*},x_{n})+(1-\alpha_{n})\beta_{n}\phi(x^{*},z_{n})+(1-\alpha_{n})(1-\beta_{n})\phi(x^{*},Sz_{n})\\
&\leq\phi(x^{*},x_{n}).
\end{aligned}
\end{equation}
\end{proof}
We examine the stopping condition in the following lemma.
\begin{lm}\label{lem6}
Let $y_{n}=x_{n}$, then $x_{n}\in E(f)$. If $y_{n}=x_{n}$ and $t_{n}=x_{n}$, then $x_{n}\in E(f)\cap F(S)$.
\end{lm}
\begin{proof}
Suppose $y_{n}=x_{n}$, then by the definition of $y_{n}$, condition ($A1$), property of $\phi$ (\ref{equa21}) and since $0<\lambda_{min}\leq\lambda_{n}\leq\lambda_{max}\leq1$ , we have
$$0\leq \lambda_{n}f(x_{n},y_{n})+\frac{1}{2}\phi(y_{n},x_{n})\leq f(x_{n},y)+\frac{1}{2}\phi(y,x_{n}),$$
for all $y\in C$. Set $\phi(y,x_{n})=L(x_{n},y)$, hence Lemma \ref{lem2} implies that $x_{n}\in E(f)$.\\
Let $y_{n}=x_{n}$ and $t_{n}=x_{n}$, we have that $z_{n}=x_{n}$ and since $J^{-1}$ is one-one, we get
$$Jx_{n}=\alpha_{n}Jx_{n}+(1-\alpha_{n})(\beta_{n}Jx_{n}+(1-\beta_{n})JSx_{n}).$$
Since $1-\alpha_{n}>0$ and $1-\beta_{n}>0$, it follows that $Jx_{n}=JSx_{n}$ and since $J$ is one-one, we get $x_{n}=Sx_{n}$. So $x_{n}\in F(S)$
\end{proof}
\begin{rk}
In a real Hilbert space $E$, Lemma \ref{lem6} is the same
Proposition $3.5$ in \cite{Vuo} with different proof, provided
that the sequence $\{\alpha_{n}\}$ satisfies the conditions of
Step $0$ of Algorithm $1$.
\end{rk}
\begin{thm}\label{eq25}
Let $C$ be a nonempty closed convex subset of a uniformly
convex, uniformly smooth Banach space $E$.
 Assume that $f:C\times
C\rightarrow \mathbb{R}$ is a bifunction which satisfies
conditions $(A1)-(A5)$ and $S:C\rightarrow C$
is a relatively nonexpansive mapping such that $$\Omega:=E(f)\cap F(S)\neq\emptyset.$$
Then sequences
$\{x_{n}\}_{n=1}^{\infty}$, $\{y_{n}\}_{n=1}^{\infty}$, $\{z_{n}\}_{n=1}^{\infty}$ and $\{t_{n}\}_{n=1}^{\infty}$ generated
by Algorithm $1$ converge strongly to the some solution $u^{*}\in
\Omega$, where $u^{*}=R_{\Omega}{x_{0}}$, and $R_{\Omega}$ is sunny generalized nonexpansive retraction from $E$ onto $\Omega$.
\end{thm}
\begin{proof}
At First, using induction we show that $\Omega\subseteq C_{n}\cap D_{n}$ for all $n\geq 0$. Let $x^{*}\in\Omega$, from Lemma \ref{lem3}, we get $\Omega\subseteq C_{n}$ for all $n\geq 0$. Now, we show that $\Omega\subseteq D_{n}$ for all $n\geq0$. It is clear that $\Omega\subseteq D_{0}$. Suppose that $\Omega\subseteq D_{n}$, i.e $\langle Jx_{n}-Jx^{*},x_{0}-x_{n}\rangle\geq0,$ for all $x^{*}\in \Omega$. Since $x_{n+1}=R_{C_{n}\cap D_{n}}x_{0}$, using Lemma \ref{lem4}, we get $\langle Jx_{n+1}-Jz,x_{0}-x_{n+1}\rangle\geq0,$ for all $z\in C_{n}\cap D_{n}$. This implies that $x^{*}\in D_{n+1}$. Therefore $\Omega\subseteq D_{n+1}$.
\par
Let $x^{*}\in\Omega\subseteq D_{n+1}$. Since $x_{n+1}\in D_{n}$,
using successively equality (\ref{eq36}),  it is easy to see that
the $\{\phi(x_{0},x_{n})\}$ is increasing and bounded from above
by $\phi(x_{0},x^{*})$, so
 $\lim\limits_{n\rightarrow\infty}\phi(x_{0},x_{n})$ exists.
 This yields that $\{\phi(x_{0},x_{n})\}$ is bounded. From inequality (\ref{eq14}),
 we know that $\{x_{n}\}$ is bounded. It is clear that $\lim_{n\rightarrow\infty}\phi(x_{n},x_{n+1})=0$,
 so Lemma \ref{lem5} implies that $\lim\limits_{n\rightarrow\infty}\|x_{n}-x_{n+1}\|=0$ and
 therefore $\{x_{n}\}$ converges strongly to $\bar{x}\in C.$ Since $J$ is uniformly norm-to-norm
 continuous on bounded sets, from equality (\ref{equa6}),
 we obtain that $\lim_{n\rightarrow\infty}\phi(x_{n+1},x_{n})=0$,
 and $\lim\limits_{n\rightarrow\infty}\phi(\nu,x_{n})$ exists for all $\nu \in C$. Since $x_{n+1}\in C_{n}$,
 we have $\lim\limits_{n\rightarrow\infty}\phi(x_{n+1},t_{n})=0$ and from Lemma \ref{lem5},
 we deduce that $\lim\limits_{n\rightarrow\infty}\|x_{n+1}-t_{n}\|=0$,
 thus $\lim\limits_{n\rightarrow\infty}\|x_{n}-t_{n}\|=0$ which implies that $\{t_{n}\}$
  converges strongly to $\bar{x}$. Using norm-to-norm continuity of $J$
  on bounded sets, we conclude that $\lim\limits_{n\rightarrow\infty}\|Jx_{n}-Jt_{n}\|=0$ and therefore
\begin{equation}\label{equa9}
\lim\limits_{n\rightarrow\infty}\phi(x^{*},x_{n})=\lim\limits_{n\rightarrow\infty}\phi(x^{*},t_{n}).
\end{equation}
Using Lemma \ref{lem1} $(ii)$, we obtain $\phi(x^{*},z_{n})\leq\phi(x^{*},x_{n})$. From inequality (\ref{eq14}) and the definition of $S$, we derive that $\{z_{n}\}$ and $\{Sz_{n}\}$ are bounded.
Let $r_{1}=\sup_{n\geq0}\{\|x_{n}\|,\|z_{n}\|\}$ and\linebreak $r_{2}=\sup_{n\geq0}\{\|z_{n}\|,\|Sz_{n}\|\}$. So, by Lemma \ref{lem15}, there exists a continuous, strictly increasing and convex function $g_{1}:[0,2r_{1}]\rightarrow\mathbb{R}$ with $g_{1}(0)=0$ such that for all $x^{*}\in \Omega$, we get
\begin{equation}
\begin{aligned}
\phi(x^{*},t_{n})&=\phi(x^{*},J^{-1}(\alpha_{n}Jx_{n}+(1-\alpha_{n})(\beta_{n}Jz_{n}+(1-\beta_{n})JSz_{n})))\\
&\leq\|x^{*}\|^{2}+\alpha_{n}\|x_{n}\|^{2}+(1-\alpha_{n})\beta_{n}\|z_{n}\|^{2}+(1-\alpha_{n})(1-\beta_{n})\|Sz_{n}\|^{2}
\\
&\hspace{.5cm}-2\alpha_{n}\langle
x^{*},Jx_{n}\rangle-2(1-\alpha_{n})\beta_{n}\langle x^{*},Jz_{n}
\rangle-2(1-\alpha_{n})(1-\beta_{n})\langle x^{*},JSz_{n}\rangle\\
&\hspace{.5cm}-\alpha_{n}(1-\alpha_{n})\beta_{n}g_{1}(\|Jz_{n}-Jx_{n}\|)\\
&\leq\phi(x^{*},x_{n})-\alpha_{n}(1-\alpha_{n})\beta_{n}g_{1}(\|Jz_{n}-Jx_{n}\|),
\end{aligned}
\end{equation}
and using the same argument, there exists a continuous, strictly increasing and convex function $g_{2}:[0,2r_{2}]\rightarrow\mathbb{R}$ with $g_{2}(0)=0$ such that for all $x^{*}\in \Omega$, we have
$$\phi(x^{*},t_{n})\leq\phi(x^{*},x_{n})-(1-\alpha_{n})^{2}\beta_{n}(1-\beta_{n})g_{2}(\|Jz_{n}-JSz_{n}\|)=0,$$
which imply
\begin{equation}\label{equa8}
\alpha_{n}(1-\alpha_{n})\beta_{n}g_{1}(\|Jz_{n}-Jx_{n}\|)\leq\phi(x^{*},x_{n})-\phi(x^{*},t_{n}),
\end{equation}
\begin{equation}\label{equa11}
(1-\alpha_{n})^{2}\beta_{n}(1-\beta_{n})g_{2}(\|Jz_{n}-JSz_{n}\|)\leq\phi(x^{*},x_{n})-\phi(x^{*},t_{n}).
\end{equation}
By letting $n\rightarrow\infty$ in inequalities (\ref{equa8}) and
(\ref{equa11}),
 using Lemma \ref{lem10} and equality (\ref{equa9}), we obtain
\begin{equation*}
\lim\limits_{n\rightarrow\infty}g_{1}(\|Jz_{n}-Jx_{n}\|)=0\quad \& \quad
\lim\limits_{n\rightarrow\infty}g_{2}(\|Jz_{n}-JSz_{n}\|)=0.
\end{equation*}
From the properties of $g_{1}$ and $g_{2}$, we have
\begin{equation}\label{equa33}
\lim\limits_{n\rightarrow\infty}\|Jz_{n}-Jx_{n}\|=0\quad \& \quad
\lim\limits_{n\rightarrow\infty}\|Jz_{n}-JSz_{n}\|=0.
\end{equation}
So from (\ref{eq18}), we obtain
\begin{equation}\label{equa12}
\lim\limits_{n\rightarrow\infty}\|z_{n}-x_{n}\|=0\quad \& \quad
\lim\limits_{n\rightarrow\infty}\|z_{n}-Sz_{n}\|=0,
\end{equation}
since $J^{-1}$ is uniformly norm-to-norm continuous on bounded sets.
By the same reason as in the proof of (\ref{equa9}), we can conclude
from (\ref{equa33}) and (\ref{equa12}) that
\begin{equation}\label{equa34}
\lim_{n\rightarrow\infty}\phi(x^{*},x_{n})=\lim_{n\rightarrow\infty}\phi(x^{*},z_{n}),
\end{equation}
for all $x^{*} \in \Omega$. Using Lemma \ref{lem1} $(ii)$, we have
\begin{equation}\label{equa35}
(1-2\lambda_{n}c_{1})\phi(y_{n},x_{n})+(1-2\lambda_{n}c_{2})\phi(z_{n},y_{n})\leq\phi(x^{*},x_{n})-\phi(x^{*},z_{n}).
\end{equation}
for all $x^{*} \in \Omega$. Taking the limits as
$n\rightarrow\infty$ in inequality (\ref{equa35}), using equality
(\ref{equa34}), we get
\begin{equation}\label{equa10}
\lim\limits_{n\rightarrow\infty}\phi(y_{n},x_{n})=0\quad \& \quad
\lim\limits_{n\rightarrow\infty}\phi(z_{n},y_{n})=0.
\end{equation}
Since $\{x_{n}\}$ and $\{z_{n}\}$ are bounded, it follows from Lemma \ref{lem5} that
\begin{equation*}
\lim\limits_{n\rightarrow\infty}\|y_{n}-x_{n}\|=0\quad \& \quad
\lim\limits_{n\rightarrow\infty}\|z_{n}-y_{n}\|=0,
\end{equation*}
which imply $\{y_{n}\}$ and $\{z_{n}\}$ converges strongly to
$\bar{x}\in C.$
\par
Now, we prove that $\bar{x}\in E(f)$. It follows from the definition
of $y_{n}$ that
\begin{equation}\label{equa36}
\lambda_{n}f(x_{n},y_{n})+\frac{1}{2}\phi(y_{n},x_{n})\leq
\lambda_{n}f(x_{n},y)+\frac{1}{2}\phi(y,x_{n})
\end{equation}
for all $y\in C$. By letting $n\rightarrow\infty$ in inequality
(\ref{equa36}), it follows from equality (\ref{equa10}), conditions
$(A1)$ and $(A3)$ and uniformly norm-to-norm continuity of $J$  on
bounded sets that
$$0\leq f(\bar{x},y)+\phi(y,\bar{x}),$$
because of $0<\lambda_{min}\leq\lambda_{n}\leq\lambda_{max}\leq1$.
Hence, letting
 $\phi(y,\bar{x})=L(\bar{x},y)$, Lemma \ref{lem2} implies that $\bar{x}\in E(f)$.
 \par
Now, since $z_{n}\rightharpoonup\bar{x}$, from (\ref{equa12}), we get $\bar{x}\in \hat{F}(S)$. So,
 using the definition of $S$, we have $\bar{x}\in F(S).$ Setting $z=u^{*}$ in Lemma \ref{lem4}, since $x_{n+1}=R_{C_{n}\cap D_{n}}x_{0}$ and $\phi$ is continuous respect to the first argument, we obtain
$$\phi(u^{*},x_{0})\geq\lim\limits_{n\rightarrow\infty}\phi(x_{n+1},x_{0})=\lim\limits_{n\rightarrow\infty}\phi(x_{n},x_{0})=\phi(\bar{x},x_{0}),$$
also, using Lemma \ref{eq38}, we have
$$\phi(u^{*},x_{0})\leq\phi(y,x_{0}),$$
for all $y\in \Omega,$ because of $u^{*}=R_{\Omega}x_{0}$.
Therefore $\bar{x}=u^{*}$ and consequently the sequences
$\{x_{n}\}_{n=1}^{\infty}$, $\{y_{n}\}_{n=1}^{\infty}$,
$\{z_{n}\}_{n=1}^{\infty}$ and $\{t_{n}\}_{n=1}^{\infty}$ converge
strongly to $R_{\Omega}{x_{0}}$.
\end{proof}
\begin{rk}
If $E$ is a real Hilbert space, then Theorem \ref{eq25} is the
same Theorem $3.1$ in \cite{Vuo} for a nonexpansive mapping $S$
with different proof, provided that the sequence $\{\alpha_{n}\}$
satisfies the conditions of Step $0$ of Algorithm $1$.
\end{rk}
\section{A Linesearch Algorithm}
As we see in the previous section, $\phi$-Lipschitz-type condition
$(A 5)$ depends on two positive parameters $c_{1}$ and $c_{2}$. In
some cases, these parameters  are unknown or difficult to
approximate. To avoid this difficulty, using linesearch method, we
modify Extragradient Algorithm. We prove strong convergence of
this new algorithm without assuming the $\phi$-Lipschitz-type
condition. linesearch method has a good efficiency in numerical
tests.

\par
Here, we assume that bifunction $f:\Delta\times \Delta\rightarrow\mathbb{R}$ satisfies in conditions $(A1)$, $(A2)$ and $(A4)$ and also in following condition which $C$ is nonempty, convex and closed of $2$-uniformly convex, uniformly
smooth Banach space $E$ and $\Delta$ is an open convex set containing $C$,\\
($A3^{*}$) $f$ is jointly weakly continuous on $\Delta\times\Delta$, i.e., if $x,y\in C$ and $\{x_{n}\}$ and $\{y_{n}\}$ are two sequences in $\Delta$ converging weakly to $x$ and $y$, respectively, then $f(x_{n},y_{n})\rightarrow f(x,y).$\\
\par
{\bf Algorithm $2$}
\begin{description}
\item[Step 0.] Let $\alpha\in(0,1)$ , $\gamma \in (0,1)$ and suppose that
$\{\alpha_{n}\}\subseteq [a,e]$ for some $0<a < e<1$,
$\{\beta_{n}\}\subseteq [d,b]$ for some $0<d < b<1$,
$\{\lambda_{n}\}\subseteq [\lambda,1]$, where
 $0<\lambda\leq1$ and $0<\nu<\frac{c^{2}}{2}$, where $\frac{1}{c}$ ($0<c\leq 1$) is the $2-$uniformly convexity constant of $E$.
 \item[Step 1.] Let $x_{0}\in C$. set $n=0.$
 \item[Step 2.] Obtain the unique optimal solution $y_{n}$ by Solving the following convex problem
 \begin{equation}\label{equa38}
 \min_{y\in C}\{\lambda_{n}f(x_{n},y)+\dfrac{1}{2}\phi(x_{n},y)\}
 \end{equation}
 \item[Step 3.] If $y_{n}=x_{n}$, then set $z_{n}=x_{n}$. Otherwise
 \begin{description}
 \item[Step 3.1.] Find $m$ the smallest nonnegative integer such that
 \begin{equation}\label{equa41}
\begin{cases}
f(z_{n,m},x_{n})-f(z_{n,m},y_{n})\geq\frac{\alpha}{2\lambda_{n}}\phi(y_{n},x_{n})\: where\\
z_{n,m}=(1-\gamma^{m})x_{n}+\gamma^{m}y_{n}.
\end{cases}
\end{equation}
 \item[Step 3.2.] Set $\rho_{n}=\gamma^{m}$, $z_{n}=z_{n,m}$ and go to Step 4.
  \end{description}
  \item[Step 4.] Choose $g_{n}\in\partial _{2} f(z_{n},x_{n})$ and compute $w_{n}=R_{C}J^{-1}(Jx_{n}-\sigma_{n}g_{n})$. If $y_{n}\neq x_{n}$, then $\sigma_{n}=\dfrac{\nu f(z_{n},x_{n})}{\Vert g_{n}\Vert^{2}}$ and $\sigma_{n}=0$ otherwise.
 \item[Step 5.] Compute $t_{n}=J^{-1}(\alpha_{n}Jx_{n}+(1-\alpha_{n})(\beta_{n}Jw_{n}+(1-\beta_{n})JSw_{n})).$
 If $y_{n}=x_{n}$ and $t_{n}=x_{n},$ then STOP: $x_{n}\in E(f)\cap F(S)$. Otherwise, go to Step 6.
 \item[Step 6.] Compute $x_{n+1}=R_{C_{n}\cap D_{n}}x_{0}$, where
 $$C_{n}=\{z\in C: \phi(z,t_{n})\leq\phi(z,x_{n})\},$$
$$D_{n}=\{z\in C: \langle Jx_{n}-Jz,x_{0}-x_{n}\rangle\geq 0\}.$$
 \item[Step 7.] Set n:=n+1, and go to Step 2.
 \end{description}
 The following lemma shows that linesearch corresponding to $x_{n}$ and $y_{n}$ (Step 3.1) is well defined.
\begin{lm}\label{lem18}
  Assume that  $y_{n}=x_{n}$ for some $n\in\mathbb{N}$. Then
\begin{enumerate}
\item[(i)]
There exists a nonnegative integer $m$ such that the inequality in (\ref{equa41}) holds.
\item[(ii)]
$f(z_{n},x_{n})>0.$
\item[(iii)]
$0\notin \partial_{2}f(z_{n},x_{n}).$
 \end{enumerate}
 \end{lm}
 \begin{proof}

 Suppose that $n\in \mathbb{N}$. Assume towards a contradiction that for each nonnegative integer $m$,
 \begin{equation}\label{equa15}
f(z_{n,m},x_{n})-f(z_{n,m},y_{n})<\frac{\alpha}{2\lambda_{n}}\phi(y_{n},x_{n}),
 \end{equation}
 where $z_{n,m}=(1-\gamma^{m})x_{n}+\gamma^{m}y_{n}.$
It is easy to see that $z_{n,m}\rightarrow x_{n}$ as $m\rightarrow\infty$. Using condition $(A3^{*})$, we obtain
 \begin{equation}
  f(z_{n,m},x_{n})\rightarrow f(x_{n},x_{n})\quad \& \quad
   f(z_{n,m},y_{n})\rightarrow f(x_{n},y_{n}).
  \end{equation}
  Since $f(x_{n},x_{n})=0$, letting $m\rightarrow\infty$ in inequality (\ref{equa15}), we get
\begin{equation}\label{equa16}
0\leq f(x_{n},y_{n})+\frac{\alpha}{2\lambda_{n}}\phi(y_{n},x_{n}).
\end{equation}
Because of  $y_{n}$ is a solution of the convex optimization problem (\ref{equa38}), we deduce
 $$\lambda_{n}f(x_{n},y)+\frac{1}{2}\phi(y,x_{n})\geq\lambda_{n}f(x_{n},y_{n})+\frac{1}{2}\phi(y_{n},x_{n}),$$
 for all $y\in C$. If $y=x_{n}$, then
\begin{equation}\label{equa40}
\lambda_{n}f(x_{n},y_{n})+\frac{1}{2}\phi(y_{n},x_{n})\leq 0.
\end{equation}
 It follows from inequalities (\ref{equa16}) and (\ref{equa40}) that
 \begin{equation}\label{equa39}
f(x_{n},y_{n})+\frac{1}{2\lambda_{n}}\phi(y_{n},x_{n})\leq f(x_{n},y_{n})+\frac{\alpha}{2\lambda_{n}}\phi(y_{n},x_{n}).
\end{equation}
 Therefore from inequality (\ref{equa39}), we obtain
 $$\frac{1-\alpha}{2}\phi(y_{n},x_{n})\leq 0,$$
 since $\lambda_{n}\leq1$.
It follows from (\ref{equa21}) that $\phi(y_{n},x_{n})>0$, because of $y_{n}\neq x_{n}$.
Thus, $1-\alpha\leq0$ or $\alpha\geq 1$ where contradict the assumption $\alpha\in(0,1)$. So, $(i)$ is proved.
\par
 Now, we prove $(ii)$. Since $f$ is convex, we obtain
 \begin{equation}\label{equa42}
\rho_{n}f(z_{n},y_{n})+(1-\rho_{n})f(z_{n},x_{n})\geq f(z_{n},z_{n})=0.
\end{equation}
consequently from inequality (\ref{equa42}), we get
 $$f(z_{n},x_{n})\geq \rho_{n}[f(z_{n},x_{n})-f(z_{n},y_{n})]\geq\frac{\alpha\rho_{n}}{\lambda_{n}}\phi(y_{n},x_{n})>0,$$
because of $y_{n}\neq x_{n}$. Therefore $f(z_{n},x_{n})>0.$\\
The proof $(iii)$ can be found in \cite{Tr} (Lemma $4.5$).
\end{proof}
\begin{rk}
If $E$ is a real Hilbert space, then Lemma \ref{lem18} is reduced
to Proposition $4.1$ in \cite{Vuo} when $\alpha\in (0,1).$
\end{rk}
 We examine the stopping condition in the following lemma where its proof is similar to the proof Lemma \ref{lem6}.
\begin{lm}\label{lem17}
Let $y_{n}=x_{n}$, then $x_{n}\in E(f)$. If $y_{n}=x_{n}$ and $t_{n}=x_{n}$, then $w_{n}=x_{n}$ and $x_{n}\in E(f)\cap F(S)$.
\end{lm}
\begin{rk}
If $E$ is a real Hilbert space, then Lemma \ref{lem17} is the same
Proposition $4.2$ in \cite{Vuo} with different proof, provided
that the sequence $\{\alpha_{n}\}$ satisfies the conditions of
Step $0$ of Algorithm $2$.
\end{rk}

 \begin{lm}\label{lem14}
  Suppose that $f:\Delta\times \Delta\rightarrow\mathbb{R}$ is a bifunction satisfying conditions $(A3^{*})$ and $(A4)$. Let $\{x_{n}\}$ and $\{z_{n}\}$ be two sequences in $\Delta$ such that Suppose  $x_{n} \rightarrow \bar{x}$ and $z_{n}\rightharpoonup \bar{z}$, where $\bar{x}, \bar{z} \in \Delta$. Then we have

 $$\partial_{2}f(z_{n},x_{n})\subseteq\partial_{2}f(\bar{z},\bar{x}).$$
 \end{lm}
 \begin{proof}
 Let $x^{*}\in \partial_{2}f(z_{n},x_{n})$, It follows from condition $(A4)$ and the definition of $\partial_{2}f$ that
 \begin{equation*}\label{equa51}
 f(z_{n},x)\geq f(z_{n},x_{n})+\langle x^{*},x-x_{n}\rangle,
 \end{equation*}
 for all $x \in \Delta$. Taking the limits as $n\rightarrow\infty$, using the condition $(A3^{*})$, we give
 $$f(\bar{z},x)\geq f(\bar{z},\bar{x})+\langle x^{*},x-\bar{x}\rangle,$$
 for all $x \in \Delta$. Hence, $x^{*}\in \partial_{2}f(\bar{z},\bar{x})$.
 \end{proof}
 Now, we prove the following proposition for Algorithm $2$, which have important role in the proof of main result in this section.
 \begin{prop}\label{lem9}
 For all $x^{*}\in E(f)\cap F(S)$ and all $n\in \mathbb{N}$, we get
 \begin{enumerate}
 \item[(i)]
$\phi(x^{*},w_{n})\leq\phi(x^{*},x_{n})-(\frac{2}{\nu}-\frac{4}{c^{2}})\sigma_{n}^{2}\|g_{n}\|^{2},$
\item[(ii)]
$\phi(x^{*},t_{n})\leq\phi(x^{*},x_{n})-(1-\alpha_{n})(\frac{2}{\nu}-\frac{4}{c^{2}})\sigma_{n}^{2}\|g_{n}\|^{2}.$
 \end{enumerate}
 \end{prop}
 \begin{proof}
Using Lemma \ref{lem4}, the definition of $V$ and inequality (\ref{eq5}), we have
 \begin{equation}
 \begin{aligned}
 \phi(x^{*},w_{n})&=\phi(x^{*},R_{C}J^{-1}(Jx_{n}-\sigma_{n}g_{n}))\\
 &\leq\phi(x^{*},J^{-1}(Jx_{n}-\sigma_{n}g_{n}))\\
 &=V(x^{*},Jx_{n}-\sigma_{n}g_{n})\\
 &\leq V(x^{*},Jx_{n}-\sigma_{n}g_{n}+\sigma_{n}g_{n})-2\langle J^{-1}(Jx_{n}-\sigma_{n}g_{n})-x^{*},\sigma_{n}g_{n}\rangle\\
 &=\phi(x^{*},x_{n})-2\langle J^{-1}(Jx_{n}-\sigma_{n}g_{n})-x^{*},\sigma_{n}g_{n}\rangle\\
 &=\phi(x^{*},x_{n})-2\sigma_{n}\langle x_{n}-x^{*},g_{n}\rangle+2\langle J^{-1}(Jx_{n}-\sigma_{n}g_{n})-x_{n},-\sigma_{n}g_{n}\rangle.
 \end{aligned}
 \end{equation}
 Since $g_{n}\in \partial_{2}f(z_{n},x_{n})$, we get
 \begin{equation*}
\langle x_{n}-x^{*},g_{n}\rangle\geq f(z_{n},x_{n})-f(z_{n},x^{*})\geq f(z_{n},x_{n})=\frac{\sigma_{n}\|g_{n}\|^{2}}{\nu}.
 \end{equation*}
 Therefore, we obtain
 \begin{equation}\label{equa24}
 -\frac{2}{\nu}\sigma_{n}^{2}\|g_{n}\|^{2}\geq -2\sigma_{n}\langle x_{n}-x^{*},g_{n}\rangle.
 \end{equation}
 On the other hand, from Lemma \ref{eq6}, we get
 \begin{equation}\label{equa25}
 \begin{aligned}
 2\langle J^{-1}(Jx_{n}-&\sigma_{n}g_{n})-x_{n},-\sigma_{n}g_{n}\rangle\\
 &=2\langle J^{-1}(Jx_{n}-\sigma_{n}g_{n})-J^{-1}(Jx_{n}), -\sigma_{n}g_{n}\rangle\\
 &\leq 2\|J(J^{-1}(Jx_{n}-\sigma_{n}g_{n}))-J(J^{-1}Jx_{n})\|\|\sigma_{n}g_{n}\|\\
 &\leq\frac{4}{c^{2}}\sigma_{n}^{2}\|g_{n}\|^{2}.
 \end{aligned}
 \end{equation}
Where $\frac{1}{c}(0< c\leq 1)$ is the $2$-uniformly convex constant of $E$. Thus, combining inequalities (\ref{equa24}) and (\ref{equa25}), we can derive $(i)$. A similar argument as in the proof of Lemma \ref{lem3} shows that
 \begin{equation*}
 \phi(x^{*},t_{n})\leq\alpha_{n}\phi(x^{*},x_{n})+(1-\alpha_{n})\phi(x^{*},w_{n}).
 \end{equation*}
 Using (i), we see that
 \begin{equation*}
 \begin{aligned}
 \phi(x^{*},t_{n}) &\leq\alpha_{n}\phi(x^{*},x_{n})+(1-\alpha_{n})\phi(x^{*},x_{n})-(1-\alpha_{n})(\frac{2}{\nu}-\frac{4}{c^{2}})\sigma_{n}^{2}\|g_{n}\|^{2}\\
 &=\phi(x^{*},x_{n})-(1-\alpha_{n})(\frac{2}{\nu}-\frac{4}{c^{2}})\sigma_{n}^{2}\|g_{n}\|^{2}.
 \end{aligned}
 \end{equation*}
Therefore $(ii)$ is proved.
 \end{proof}
 \begin{thm}\label{thm1}

Assume that $S:C\rightarrow C$
is a relatively nonexpansive mapping such that $$\Omega:=E(f)\cap F(S)\neq\emptyset.$$
Then the sequences
$\{x_{n}\}_{n=1}^{\infty}$, $\{y_{n}\}_{n=1}^{\infty}$, $\{z_{n}\}_{n=1}^{\infty}$, $\{w_{n}\}_{n=1}^{\infty}$ and $\{t_{n}\}_{n=1}^{\infty}$ generated
by Algorithm $2$ converge strongly to the some solution $u^{*}\in
\Omega$, where $u^{*}=R_{\Omega}{x_{0}}$ and $R_{\Omega}$ is the sunny generalized nonexpansive retraction from $E$ onto $\Omega$.
\end{thm}
 \begin{proof}
Let $x^{*}\in \Omega$. Similar to the proof of Theorem \ref{eq25}, we have
\begin{equation}
 \lim\limits_{n\rightarrow\infty}\|x_{n+1}-x_{n}\|=0\quad \& \quad
 \lim\limits_{n\rightarrow\infty}\|x_{n}-t_{n}\|=0,
 \end{equation}
 which imply that $\{x_{n}\}$ and consequently $\{t_{n}\}$ converge strongly to $\bar{x}\in C$ and
 \begin{equation}\label{equa43}
 \lim\limits_{n\rightarrow\infty}(\phi(x^{*},x_{n})-\phi(x^{*},t_{n}))=0.
 \end{equation}
 Since $(1-\alpha_{n})(\frac{2}{\nu}-\frac{4}{c^{2}})>0$, it follows from Lemma \ref{lem9} $(ii)$ that
 $\:\lim\limits_{n\rightarrow\infty}\sigma_{n}\|g_{n}\|=0.$
 \par
 Now, we prove that $\{y_{n}\}$, $\{z_{n}\}$ and $\{g_{n}\}$ are bounded and $f(z_{n},x_{n})\rightarrow 0$, as $n\rightarrow\infty$. Suppose that
 $$A(y)=\lambda_{n}f(x_{n},y)+\frac{1}{2}\phi(y,x_{n}).$$
 Since $\phi$ is lower semicontinuous respect to the first argument $y$, from Lemma \ref{lem8}, we deduce
 $$\partial A(y)=\lambda_{n}\partial_{2}f(x_{n},y)+\frac{1}{2}\nabla_{1}\phi(y,x_{n}).$$
 Let $u_{1}, u_{2}\in C$, $w_{1}\in\partial_{2}f(x_{n},u_{1})$ and $w_{2}\in \partial_{2}f(x_{n},u_{2})$, we have
 \begin{equation}\label{equa5}
 \langle w_{1},u_{1}-y\rangle\geq f(x_{n},u_{1})-f(x_{n},y), \:\: \: \forall y\in C,
 \end{equation}
 \begin{equation}\label{equa7}
 \langle -w_{2},y-u_{2}\rangle\geq f(x_{n},u_{2})-f(x_{n},y), \:\: \: \forall y\in C.
 \end{equation}
 Set $y=u_{2}$ in inequality (\ref{equa5}) and $y=u_{1}$ in inequality (\ref{equa7}), we get
 \begin{equation}\label{equa4}
\langle w_{1}-w_{2},u_{1}-u_{2}\rangle\geq 0.
\end{equation}
On the other hand, we have
\begin{equation}
\frac{1}{2}\nabla_{1}\phi(u_{i},x_{n})=Ju_{i}-Jx_{n},\:\:\: i=1,2.
\end{equation}
Therefore, using Lemma \ref{lem16}, there exists $\tau>0 $ such that
\begin{equation}\label{equa13}
\langle Ju_{1}-Ju_{2},u_{1}-u_{2}\rangle\geq \tau \|u_{1}-u_{2}\|^{2}.
\end{equation}
From inequalities (\ref{equa4}) and (\ref{equa13}), we obtain
\begin{equation}\label{equa26}
\langle (\lambda_{n} w_{1}+Ju_{1}-Jx_{n})-(\lambda_{n}w_{2}+Ju_{2}-Jx_{n}),u_{1}-u_{2}\rangle\geq\tau\|u_{1}-u_{2}\|^{2}.
\end{equation}
For all $u \in C$, put $T_{n}(u) := \lambda_{n}w + Ju - Jx_{n}$, where $w \in \partial_{2}f(x_{n},u)$. So $T_{n}(u) \subseteq \partial A(u)$ for all $u\in C$. Therefore it follows from inequality (\ref{equa26}) that
\begin{equation*}
\langle t^{n}(u_{1}) - t^{n}(u_{2}), u_{1} - u_{2} \rangle \geq\tau\| u_{1} - u_{2}\|,
\end{equation*}
 for all $u_{1}, u_{2} \in C$, all $t^{n}(u_{1}) \in T_{n}(u_{1})$ and $t^{n}(u_{2}) \in T_{n}(u_{2})$, this means $T_{n}$ is multivalued monotone.
\par
Using Lemma \ref{lem7} and Lemma \ref{lem8}, we have
$$y_{n}=\min_{y\in C}{\lambda_{n}f(x_{n},y)+\frac{1}{2}\phi(y,x_{n})},$$
if only if
$$0\in \partial A(y_{n})+N_{C}(y_{n}).$$
Which implies that $-T_{n}(y_{n})\subseteq N_{C}(y_{n})$, thus
\begin{equation}\label{equa27}
\langle t^{n}(y_{n}),y-y_{n}\rangle\geq 0,
\end{equation}
for all $y\in C$ and all $t^{n}(y_{n}) \in T_{n}(y_{n})$. Replacing $u_{1}$ and $u_{2}$ by $x_{n}$ and $y_{n}$ in inequality (\ref{equa26}), respectively  and interchanging $y$ by $x_{n}$  in inequality (\ref{equa27}), we have
\begin{equation}\label{equa28}
\langle t^{n}(x_{n}),x_{n}-y_{n}\rangle\geq\langle t^{n}(y_{n}),x_{n}-y_{n}\rangle+\tau\|x_{n}-y_{n}\|^{2}\geq\tau\|x_{n}-y_{n}\|^{2},
\end{equation}
for all $t^{n}(x_{n}) \in T_{n}(x_{n})$ and  all $t^{n}(y_{n}) \in T_{n}(y_{n})$. So, $t^{n}(x_{n})\in \partial A(x_{n})$, for all $t^{n}(x_{n}) \in T_{n}(x_{n})$ and since $\frac{1}{2}\nabla_{1}\phi(x_{n},x_{n})=0$, we obtain $t^{n}(x_{n})\in \lambda_{n}\partial_{2} f(x_{n},x_{n}).$ Since $x_{n}\rightarrow\bar{x}$, it follows from Lemma \ref{lem14} that
$$\partial_{2}f(x_{n},x_{n})\subset\partial_{2}f(\bar{x},\bar{x}).$$
Thus, we can deduce from Lemma \ref{lem20} that $t^{n}(x_{n})$ is bounded, because of $\: 0<\lambda\leq\lambda_{n}\leq 1$. So, from inequality (\ref{equa28}), we obtain
$\tau\|x_{n}-y_{n}\|\leq\|t^{n}(x_{n})\|.$
Therefore,  we can conclude that $\{y_{n}\}$ is bounded. Since $\{z_{n}\}$ is a convex combination of $\{x_{n}\}$ and $\{y_{n}\}$, it is also bounded, hence, there exists a subsequence of $\{z_{n}\}$, again denoted by $\{z_{n}\}$, which converges weakly to $\bar{z}\in C$. In a similar way, it follows from lemmas \ref{lem20} and \ref{lem14} that the sequence $\{g_{n}\}$ is bounded.\\
If $x_{n}=y_{n}$ then we have $f(z_{n},x_{n})=0$ and $\sigma_{n}=0$ and if $x_{n}\neq y_{n}$, by the definition of $\sigma_{n}$, we obtain
$$\nu f(z_{n},x_{n})=\sigma_{n}\|g_{n}\|\|g_{n}\|\rightarrow0\:\:\:\Longrightarrow\:\:\: f(z_{n},x_{n})\rightarrow0,$$
since $\sigma_{n}\|g_{n}\|\rightarrow 0$ and $0<\nu<\frac{c^{2}}{2}$.
\par
Now, we show that $\bar{x}\in E(f)$ and $\|x_{n}-y_{n}\|\rightarrow 0$.
If $y_{n}=x_{n}$ then it follows from Lemma \ref{lem1} (i), that
\begin{equation}\label{equa44}
 \lambda_{n}f(x_{n},y)\geq0,
 \end{equation}
for all $y\in C.$ By letting $n\rightarrow\infty$ in inequality (\ref{equa44}) and using condition $(A3^{*})$, we get $\: f(\bar{x},y)\geq0$, because of $0<\lambda\leq\lambda_{n}\leq1$, i.e, $\bar{x}\in E(f).$
\par
Now, we let that $y_{n}\neq x_{n}$, since $f(z_{n},.)$ is convex, we obtain
\begin{equation*}
\begin{aligned}
\rho_{n}&f(z_{n},y_{n})+(1-\rho_{n})f(z_{n},x_{n})\\
&\geq f(z_{n},\rho_{n}y_{n}+(1-\rho_{n})x_{n})=f(z_{n},z_{n})=0.
\end{aligned}
\end{equation*}
Therefore,
\begin{equation}\label{equa45}
\rho_{n}[f(z_{n},x_{n})-f(z_{n},y_{n})]\leq f(z_{n},x_{n})\rightarrow0,
\end{equation}
as $n\rightarrow\infty$. By the Step $3.1$ of Algorithm $2$ and inequality (\ref{equa45}), we have
\begin{equation}\label{equa46}
\frac{\alpha\rho_{n}}{2\lambda_{n}}\phi(y_{n},x_{n})\leq\rho_{n}[f(z_{n},x_{n})-f(z_{n},y_{n})]\leq f(z_{n},x_{n})\rightarrow0.
\end{equation}
Now, we consider two cases:\\
{\bf Case $1$:} $\limsup_{n\rightarrow\infty}\rho_{n}>0$.\\
 In this case, there exists $\bar{\rho}>0$ and a subsequence of $\{\rho_{n}\}$, again denoted by $\{\rho_{n}\}$, such that $\rho_{n}\rightarrow\bar{\rho}$ and since $0<\lambda\leq\lambda_{n}\leq1$, from inequality (\ref{equa46}), we can conclude that
$$\phi(y_{n},x_{n})\rightarrow0.$$
Thus, from Lemma \ref{lem5}, we have $\|y_{n}-x_{n}\|\rightarrow0,$ which implies $y_{n}\rightarrow\bar{x}.$\\
{\bf Case $2$:} $\lim\limits_{n}\rho_{n}\rightarrow0$. \\
Let $m$ be the smallest nonnegative integer such that the Step 3.1 of Algorithm $2$ is satisfied, i.e.,
$$f(z_{n,m},x_{n})-f(z_{n,m},y_{n})\geq\frac{\alpha}{2\lambda}\phi(y_{n},x_{n}),$$
where $z_{n,m}=(1-\gamma^{m})x_{n}+\gamma^{m}y_{n}$. So,
\begin{equation}\label{equa47}
f(z_{n,m-1},x_{n})-f(z_{n,m-1},y_{n})<\frac{\alpha}{2\lambda_{n}}\phi(y_{n},x_{n}).
\end{equation}
On the other hand, setting $y=x_{n}$ in Lemma \ref{lem1} (i), condition $(A1)$ and equality (\ref{equa6}) imply that
$$-\lambda_{n}f(x_{n},y_{n})\geq\langle Jy_{n}-Jx_{n},y_{n}-x_{n}\rangle=\frac{1}{2}\phi(y_{n},x_{n})+\frac{1}{2}\phi(x_{n},y_{n}).$$
Therefore,
\begin{equation}\label{equa48}
\frac{1}{2}\phi(y_{n},x_{n})\leq-\lambda_{n}f(x_{n},y_{n}).
\end{equation}
From inequalities (\ref{equa47}) and (\ref{equa48}), we have
\begin{equation}\label{equa49}
f(z_{n,m-1},x_{n})-f(z_{n,m-1}, y_{n})<-\alpha f(x_{n},y_{n}).
\end{equation}
Taking the limits as $n\rightarrow\infty$ in above inequality,  we obtain $z_{n,m-1}\rightarrow\bar{x}$, since $\gamma^{m}=\rho_{n}\rightarrow0$. Because of $\{y_{n}\}$ is bounded,  there exists a subsequence of $\{y_{n}\}$, again denoted by $\{y_{n}\}$, which converges weakly to $\bar{y}\in C$. By letting $n \to \infty$ in inequality (\ref{equa49}) and using conditions $(A1)$ and $(A3^{*})$, we get
\begin{equation}
-f(\bar{x},\bar{y})\leq-\alpha f(\bar{x},\bar{y}).
\end{equation}
Which implies that $f(\bar{x},\bar{y})\geq0$, because of $\alpha\in (0,1)$. So, If we take the limits as $n\rightarrow\infty$ in inequality (\ref{equa48}), then we can conclude that
$\phi(y_{n},x_{n})\rightarrow0.$
Thus, from Lemma \ref{lem5}, we have
$\|y_{n}-x_{n}\|\rightarrow 0$,
which implies that $y_{n}\rightarrow\bar{x}$. By Lemma \ref{lem1}, we have
\begin{equation}\label{equa50}
\lambda_{n}[f(x_{n},y)-f(x_{n},y_{n})]\geq\langle Jy_{n}-Jx_{n},y_{n}-y\rangle,
\end{equation}
for all $y\in C$. By letting $n\rightarrow\infty$ in inequality (\ref{equa50}), it follows that $f(\bar{x},y)\geq0$, because of $0<\lambda\leq\lambda_{n}\leq1$, this means $\bar{x}\in E(f).$
\par
Now, we show that $\bar{x}\in F(S)$. Let $r_{1}=\sup_{n\geq0}\{\|x_{n}\|,\|w_{n}\|\}$ and $r_{2}=\sup_{n\geq0}\{\|w_{n}\|,\|Sw_{n}\|\}$. Using Lemma \ref{lem15}, there exists a continuous, strictly increasing and convex function\linebreak $g_{1}:[0,2r_{1}]\rightarrow\mathbb{R}$ with $g_{1}(0)=0$ such that for $x^{*}\in \Omega$, we get
\begin{equation}\label{eq.n1}
\begin{aligned}
\phi(x^{*},t_{n})&=\phi(x^{*},J^{-1}(\alpha_{n}Jx_{n}+(1-\alpha_{n})(\beta_{n}Jw_{n}+(1-\beta_{n})JSw_{n})))\\
&\leq\|x^{*}\|^{2}+\alpha_{n}\|x_{n}\|^{2}+(1-\alpha_{n})\beta_{n}\|w_{n}\|^{2}+(1-\alpha_{n})(1-\beta_{n})\|Sw_{n}\|^{2}\\
&\hspace{.5cm}-2\alpha_{n}\langle x^{*},Jx_{n}\rangle-2(1-\alpha_{n})\beta_{n}\langle x^{*},Jw_{n}
\rangle-2(1-\alpha_{n})(1-\beta_{n})\langle x^{*},JSw_{n}\rangle\\
&\hspace{.5cm}-\alpha_{n}(1-\alpha_{n})\beta_{n}g_{1}(\|Jw_{n}-Jx_{n}\|)\\
&\leq\phi(x^{*},x_{n})-\alpha_{n}(1-\alpha_{n})\beta_{n}g_{1}(\|Jw_{n}-Jx_{n}\|),
\end{aligned}
\end{equation}
and in a similar way, there exists a continuous, strictly increasing and convex function $g_{2}:[0,2r_{2}]\rightarrow\mathbb{R}$ with $g_{2}(0)=0$ such that for $x^{*}\in \Omega$, we obtain
\begin{equation}\label{eq.n2}
   \phi(x^{*},t_{n})\leq\phi(x^{*},x_{n})-(1-\alpha_{n})^{2}\beta_{n}(1-\beta_{n})g_{2}(\|Jw_{n}-JSw_{n}\|)=0.
\end{equation}
It follows from inequalities (\ref{eq.n1}) and (\ref{eq.n2}) that
\begin{equation}\label{equa30}
\alpha_{n}(1-\alpha_{n})\beta_{n}g_{1}(\|Jw_{n}-Jx_{n}\|)\leq\phi(x^{*},x_{n})-\phi(x^{*},t_{n}),
\end{equation}
\begin{equation}\label{equa31}
(1-\alpha_{n})^{2}\beta_{n}(1-\beta_{n})g_{2}(\|Jw_{n}-JSw_{n}\|)\leq\phi(x^{*},x_{n})-\phi(x^{*},t_{n}).
\end{equation}
Taking the limits as $n\rightarrow\infty$ in inequalities (\ref{equa30}) and (\ref{equa31}), using Lemma \ref{lem10} and equality (\ref{equa43}), we obtain
\begin{equation*}
\lim\limits_{n\rightarrow\infty}g_{1}(\|Jw_{n}-Jx_{n}\|)=0\quad \& \quad
\lim\limits_{n\rightarrow\infty}g_{2}(\|Jw_{n}-JSw_{n}\|)=0.
\end{equation*}
From the properties of $g_{1}$ and $g_{2}$, we have
\begin{equation*}
\lim\limits_{n\rightarrow\infty}\|Jw_{n}-Jx_{n}\|=0\quad \& \quad
\lim\limits_{n\rightarrow\infty}\|Jw_{n}-JSw_{n}\|=0.
\end{equation*}
Since $J^{-1}$ is uniformly norm-to-norm continuous on bounded sets, we obtain
\begin{equation}\label{equa32}
\lim\limits_{n\rightarrow\infty}\|w_{n}-x_{n}\|=0\quad \& \quad
\lim\limits_{n\rightarrow\infty}\|w_{n}-Sw_{n}\|=0.
\end{equation}
So, we  get $\bar{x}\in \hat{F}(S)$, because of $w_{n}\rightharpoonup\bar{x}$, therefore
using the definition of $S$, we have that $\bar{x}\in F(S).$
\par
  Setting $z=u^{*}$ in Lemma \ref{lem4}, since $x_{n+1}=R_{C_{n}\cap D_{n}}x_{0}$ and $\phi$ is continuous respect to the first argument, we obtain
$$\phi(u^{*},x_{0})\geq\lim\limits_{n\rightarrow\infty}\phi(x_{n+1},x_{0})=
\lim\limits_{n\rightarrow\infty}\phi(x_{n},x_{0})=\phi(\bar{x},x_{0}).$$
Also, since $u^{*}=R_{\Omega}x_{0}$, it follows from Lemma
\ref{eq38} that
$$\phi(u^{*},x_{0})\leq\phi(y,x_{0}),$$
for all $y\in \Omega$. Therefore $\bar{x}=u^{*}$ and consequently sequences
$\{x_{n}\}_{n=1}^{\infty}$, $\{y_{n}\}_{n=1}^{\infty}$, $\{z_{n}\}_{n=1}^{\infty}$, $\{w_{n}\}_{n=1}^{\infty}$ and $\{t_{n}\}_{n=1}^{\infty}$ converge strongly to $R_{\Omega}{x_{0}}$.
\end{proof}
\section{Numerical Example}
Now, we demonstrate theorems \ref{eq25} and \ref{thm1} with an example. Also, we compare the behavior of the sequence $\{x_{n}\}$ generated by algorithms $1$ and $2$.
\begin{exa}
Let $E=\mathbb{R}$ and $C=[-100,100]$. Define $f(x,y):=y^{2}-4x^{2}+3xy.$\\
We see that $f$ satisfies the conditions $(A1)-(A5)$ as follows:
 \begin{enumerate}
 \item[(A1)] $f(x,x):=x^{2}-4x^{2}+3x^{2}=0$ for all $x\in C$,
 \item[(A2)] If $f(x,y)=(y-x)(y+4x)\geq0$, then $$f(y,x)=(x-y)(x+4y)=(x-y)((y+4x)-3(x-y))=-f(x,y)-3(x-y)^{2}\leq0,$$
 for all $x, y\in C$, i.e., $f$ is pseudomonotone, while $f$ is not monotone.
 \item[(A3)]
 If $x_{n}\rightharpoonup\bar{x}$ and $y_{n}\rightharpoonup\bar{y}$, then $$f(x_{n},y_{n})=y_{n}^{2}-4x_{n}^{2}+3x_{n}y_{n}\rightarrow\bar{y}^{2}-4\bar{x}^{2}+3\bar{x}\bar{y}=f(\bar{x},\bar{y}),$$
 i.e., $f$ is jointly weakly continuous on $C\times C$.
\item[(A4)] Let $\theta\in (0,1)$. Since
\begin{equation*}
 \begin{aligned}
 f(x,\theta y_{1}+(1-\theta)y_{2})&=(\theta y_{1}+(1-\theta) y_{2})^{2}-4x^{2}+3x(\theta y_{1}+(1-\theta) y_{2})\\
 &\leq\theta(y_{1}^{2}-4x^{2}+3xy_{1})+(1-\theta)(y_{2}^{2}-4x^{2}+3xy_{2})\\
&=\theta f(x,y_{1})+(1-\theta)f(x,y_{2}),
 \end{aligned}
\end{equation*}
 so $f(x,.)$ is convex, also, $\liminf_{y\rightarrow\bar{y}}f(x,y)=f(x,\bar{y})$, hence $f(x,.)$ is lower semicontinuous. Since $\partial_{2}f(x,y)=2y+3x$, thus $f(x,.)$ is subdifferentiable on $C$ for each $x\in C.$
\item[(A5)] Since $\phi(y,x)=(y-x)^{2}$, we get
 \begin{equation}
\begin{aligned}
 f(x,y)+f(y,z)&=z^{2}-4x^{2}-3y^{2}+3xy+3zy\\
&= f(x,z)-\frac{3}{2}(y-x)^{2}-\frac{3}{2}(z-y)^{2}+\frac{3}{2}(x-z)^{2}\\
&\geq f(x,z)-\frac{3}{2}(y-x)^{2}-\frac{3}{2}(z-y)^{2},
 \end{aligned}
\end{equation}
i.e., $f$ satisfies the $\phi$-Lipschitz-type condition with $c_{1},c_{2}=\frac{3}{2}.$
 \end{enumerate}
Now, define $S:C\rightarrow C$ by $Sx=\frac{x}{5}$ for all $x\in C$, so $F(S)=\{0\}$ and
\begin{equation}
\begin{aligned}
\phi(0,Sx)&=\phi(0,\frac{x}{5})\\
&=0-2\langle 0,\frac{x}{5}\rangle+|\frac{x}{5}|^{2}\\
&\leq|x|^{2}\\
&=\phi(0,x),
\end{aligned}
\end{equation}
for all $x\in C$. Let $x_{n}\rightharpoonup p$ such that $\lim\limits_{n\rightarrow\infty}(Sx_{n}-x_{n})=0$,
this implies that $\hat{F}(S)=\{0\}$.\\
\begin{figure}[!h]
\includegraphics[totalheight=2.5in]{1}\vspace{-.75cm}\caption{Extragradient Algorithm}
\end{figure}
\begin{figure}[!h]
\includegraphics[totalheight=2.5in]{2}\vspace{-.75cm}\caption{Linesearch Algorithm}
\end{figure}
Thus, $\hat{F}(S)=F(S)$, i.e., $S$ is relatively nonexpansive
mapping. On the other hand, if for each $y\in C$, $f(x,y)\geq0$,
then $x=0$, i.e., $E(f)=\{0\}$ and consequently $\Omega=E(f)\cap
F(S)=\{0\}.$
Also, assume that $\alpha_{n}=\frac{1}{2}+\frac{1}{3+n}$,
$\beta_{n}=\frac{1}{3}+\frac{1}{4+n}$ for all $n\geq 0$ and
$x_{0}=100$.
\par
In Extragradient Algorithm, if $\lambda_{n}=\frac{1}{6}$, then we have
 $$\frac{1}{6}f(x_{n},y_{n})+\frac{1}{2}(y_{n}-x_{n})^{2}=\min_{y\in C}\{\frac{1}{6}f(x_{n},y)+\frac{1}{2}(y-x_{n})^{2}\},$$
 i.e., $y_{n}=\frac{3}{8}x_{n},$ also $$ \frac{1}{6}f(y_{n},z_{n})+\frac{1}{2}(z_{n}-x_{n})^{2}=\min_{y\in C}\{\frac{1}{6}f(y_{n},y)+\frac{1}{2}(y-x_{n})^{2}\},$$
i.e., $z_{n}=\frac{39}{24}y_{n}=\frac{39}{64}x_{n},$ therefore
\begin{equation*}
t_{n}=\alpha_{n}x_{n}+(1-\alpha_{n})[\beta_{n}z_{n}+\frac{1}{5}(1-\beta_{n})z_{n}],
\end{equation*}
and $x_{n+1}=R_{C_{n}\cap D_{n}}x_{0}$ or $|x_{n+1}-x_{0}|=\min_{z\in C_{n}\cap D_{n}}|z-x_{0}|$, where
\begin{equation*}
\begin{cases}
C_{n}=\{z\in C: |t_{n}-z|\leq |x_{n}-z|\},\\
D_{n}=\{z\in C: (x_{n}-z)(x_{0}-x_{n})\geq0\}.
\end{cases}
\end{equation*}
\begin{center}
\vspace{3mm}
\begin{tabular}{ c c c c c}
\hline Table1 & {\small Numerical Results for Algorithm $1$} &
\\
\cline{1-5}
n\hspace{0 cm}  & \hspace{-3 cm} {$x_{n}$} & \hspace{-4 cm} {$y_{n}$} &\hspace{0 cm} {$z_{n}$} &\hspace{1.5 cm} {$t_{n}$}\\
\hline\hline
0\hspace{0 cm} &\hspace{-3 cm}$100$&\hspace{-4 cm} $37.5$ &\hspace{0 cm}$60.94$ & \hspace{1.5 cm} $90.104$ \\
1\hspace{0 cm} &\hspace{-3 cm}$95.05$&\hspace{-4 cm}$35.65$ & \hspace{0 cm}$57.92$ & \hspace{1.5 cm}$80.36$\\
2\hspace{0 cm} &\hspace{-3 cm}$87.71$&\hspace{-4 cm}$32.89$ & \hspace{0 cm}$53.47$ & \hspace{1.5 cm}$71.016$\\
3\hspace{0 cm} &\hspace{-3 cm}$79.36$&\hspace{-4 cm}$29.61$ & \hspace{0 cm}$48.36$ &\hspace{1.5 cm}$62.277$\\
\hspace{0 cm} &\hspace{-3 cm}$\vdots$&\hspace{-4 cm}$\vdots$ & \hspace{0 cm}$\vdots$ & \hspace{1.5 cm}$\vdots$\\
37\hspace{0 cm} &\hspace{-3 cm}$0.2826$&\hspace{-4 cm} $0.1059$ &\hspace{0 cm}$0.1722$ & \hspace{1.5 cm} $0.1881$ \\
38\hspace{0 cm} &\hspace{-3 cm}$0.2353$&\hspace{-4 cm}$0.0882$ & \hspace{0 cm}$0.1434$ & \hspace{1.5 cm}$0.1565$\\
39\hspace{0 cm} &\hspace{-3 cm}$0.1959$&\hspace{-4 cm} $0.0734$ &\hspace{0 cm}$0.1194$ & \hspace{1.5 cm} $0.1302$ \\
40\hspace{0 cm} &\hspace{-3 cm}$0.163$&\hspace{-4 cm}$0.0611$ & \hspace{0 cm}$0.0993$ & \hspace{1.5 cm}$0.1082$\\
 \hspace{0 cm} &\hspace{-3 cm}$\vdots$&\hspace{-4 cm}$\vdots$ & \hspace{0 cm}$\vdots$ & \hspace{1.5 cm}$\vdots$\\
70\hspace{0 cm} &\hspace{-3 cm}$0.0003$ & \hspace{-4 cm}$0.000113$ & \hspace{0 cm}$0.000183$ &\hspace{1.5 cm}$0.000197$\\
71\hspace{0 cm} &\hspace{-3 cm}$0.0002 $&\hspace{-4 cm}$7.50e-05$ & \hspace{0 cm}$0.000122$ &\hspace{1.5 cm}$0.000131$\\
72\hspace{0 cm} &\hspace{-3 cm}$0.0001$&\hspace{-4 cm}$3.75e-05$ & \hspace{0 cm}$6.09e-05$ & \hspace{1.5 cm}$6.55e-05$\\
73\hspace{0 cm} &\hspace{-3 cm}$0$&\hspace{-4 cm}$0$ & \hspace{0 cm}$0$ & \hspace{1.5 cm}$0$\\
\hline
\end{tabular}
\end{center}
\begin{center}
\vspace{3mm}
\begin{tabular}{ c c c c c c}
\hline Table2 &  {\small  Numerical Results for Algorithm $2$}&
\\
\cline{1-6}
n\hspace{0 cm} & \hspace{-4cm} {$x_{n}$} & \hspace{-6 cm} {$y_{n}$} & \hspace{-1 cm} {$z_{n}$} & \hspace{1 cm} {$t_{n}$}  & \hspace{1.5 cm}{$w_{n}$}\\
\hline\hline
0\hspace{0 cm} &\hspace{-4 cm}$100$ & \hspace{-6 cm} $-25$ &\hspace{-1 cm} $75$ & \hspace{1 cm} $72.22$ & \hspace{1.5 cm} $-100$\\
1\hspace{0 cm} &\hspace{-4 cm}$86.11$&\hspace{-6 cm}$-21.53$ & \hspace{-1 cm}$64.53$ & \hspace{1 cm}$48.91$& \hspace{1.5 cm}$-100$\\
2\hspace{0 cm} &\hspace{-4 cm}$67.51$&\hspace{-6 cm}$-16.88$ & \hspace{-1 cm}$50.63$ & \hspace{1 cm}$29.26$& \hspace{1.5 cm}$-100$\\
3\hspace{0 cm} &\hspace{-4 cm}$48.38$&\hspace{-6 cm}$-12.097$ & \hspace{-1 cm}$36.29$ &\hspace{1 cm}$12.89$& \hspace{1.5 cm}$-100$\\
\hspace{0 cm} &\hspace{-4 cm}$\vdots$&\hspace{-6 cm}$\vdots$ & \hspace{-1 cm} $\vdots$ & \hspace{1 cm}$\vdots$& \hspace{1.5 cm}$\vdots$\\
37\hspace{0 cm} &\hspace{-4 cm}$0.0561$&\hspace{-6 cm} $-0.01403$ &\hspace{-1 cm} $0.0421$ & \hspace{1 cm} $0.0422$& \hspace{1.5 cm}$0.0553$ \\
38\hspace{0 cm} &\hspace{-4 cm}$0.0491$&\hspace{-6 cm}$-0.01228$ & \hspace{-1 cm} $0.0368$ & \hspace{1 cm}$0.0369$& \hspace{1.5 cm}$0.0485$\\
39\hspace{0 cm} &\hspace{-4 cm}$0.043$&\hspace{-6 cm} $-0.01075$ &\hspace{-1 cm} $0.0322$ & \hspace{1 cm} $0.0329$& \hspace{1.5 cm}$0.0426$ \\
40\hspace{0 cm} &\hspace{-4 cm}$0.0376$&\hspace{-6 cm}$-0.0094$ & \hspace{-1 cm} $0.0282$ & \hspace{1 cm}$0.0283$& \hspace{1.5 cm}$0.0373$\\
 \hspace{0 cm} &\hspace{-4 cm}$\vdots$&\hspace{-6 cm}$\vdots$ & \hspace{-1 cm}$\vdots$ & \hspace{1 cm}$\vdots$& \hspace{1.5 cm}$\vdots$\\
70\hspace{0 cm} &\hspace{-4 cm}$0.0003$&\hspace{-6 cm}$-7.50e-05$ &  \hspace{-1 cm}$0.00022$ &\hspace{1 cm}$0.00022$& \hspace{1.5 cm}$0.0003$\\
71\hspace{0 cm} &\hspace{-4 cm}$0.0002 $&\hspace{-6 cm}$-5.00e-05$ & \hspace{-1 cm}$0.00015$ &\hspace{1 cm}$0.00015$& \hspace{1.5 cm}$0.0002$\\
72\hspace{0 cm} &\hspace{-4 cm}$0.0001$&\hspace{-6 cm}$-2.50e-05$ & \hspace{-1 cm}$7.50e-5$ & \hspace{1 cm}$7.46e-5$ & \hspace{1.5 cm}$1.00e-04$\\
73\hspace{0 cm} & \hspace{-4 cm}$0$ &\hspace{-6 cm}$0$ & \hspace{-1 cm}$0$ & \hspace{1 cm}$0$ & \hspace{1.5 cm}$0$\\
\hline
\end{tabular}
\end{center}

 Since $\Omega=\{0\}$, we get $R_{\Omega}(x_{0})=0$.
  Moreover, numerical results for Algorithm $1$ show that the sequences $\{x_{n}\}$, $\{y_{n}\}$, $\{z_{n}\}$ and $\{t_{n}\}$ converge strongly to $0$.
\par
  In Linesearch Algorithm, $x_{n}$ is the same in Extragradient Algorithm.
  Assume that $\lambda_{n}=\frac{1}{2}$, $\alpha=\frac{1}{2}$, $\gamma=0.2$, $\nu=\frac{1}{4}$ and $c=1$. So,
   $$\frac{1}{2}f(x_{n},y_{n})+\frac{1}{2}(y_{n}-x_{n})^{2}=\min_{y\in C}\frac{1}{2}\{f(x_{n},y)+(y-x_{n})^{2}\},$$
 i.e., $y_{n}=-\frac{1}{4}x_{n},$ and $m$ is the smallest nonnegative integer such that
  $$(x_{n}-y_{n})(\frac{1}{2}x_{n}+\frac{3}{2}y_{n}+3z_{n})\geq0,$$
  where
 $$z_{n}=z_{n,m}=(1-(0.2)^{m})x_{n}+(0.2)^{m}y_{n}.$$
  Also, $g_{n}=2x_{n}+3z_{n}$ and $|w_{n}-(J^{-1}(Jx_{n}-\sigma_{n}g_{n}))|=\min_{z\in C}|z-(J^{-1}(Jx_{n}-\sigma_{n}g_{n})|$. \linebreak
  Since $y_{n}\neq x_{n}$, then $$\sigma_{n}=\frac{\frac{1}{4}(x_{n}^{2}-4z_{n}^{2}+3x_{n}z_{n})}{|g_{n}|},$$
  and
  \begin{equation*}
t_{n}=\alpha_{n}x_{n}+(1-\alpha_{n})[\beta_{n}w_{n}+\frac{1}{5}(1-\beta_{n})w_{n}].
\end{equation*}
Furthermore,  numerical results for Algorithm $2$ show that the
sequences $\{x_{n}\}$, $\{y_{n}\}$, $\{z_{n}\}$, $\{t_{n}\}$ and
$\{w_{n}\}$ converge strongly to $0$. By comparing Figure1 and
Figure2, we see that the speed of convergence of the sequence
$\{x_{n}\}$ generated by Linesearch Algorithm is equal to
Extragradient Algorithm. The computations associated with example
were performed using MATLAB (Step:$10^{-4}$) software.

\end{exa}

{\footnotesize}

redpajama_set_name: RedPajamaArXiv


text:
<?php

return [
    new Symfony\Bundle\FrameworkBundle\FrameworkBundle(),
    new Symfony\Bundle\SecurityBundle\SecurityBundle(),
    new \Symfony\Bundle\SwiftmailerBundle\SwiftmailerBundle(),
    new Symfony\Bundle\TwigBundle\TwigBundle(),
    new \Sensio\Bundle\FrameworkExtraBundle\SensioFrameworkExtraBundle(),
    new Kunstmaan\UtilitiesBundle\KunstmaanUtilitiesBundle(),
    new \Kunstmaan\AdminBundle\KunstmaanAdminBundle(),
    new \Kunstmaan\TranslatorBundle\KunstmaanTranslatorBundle(),
    new \Symfony\Bundle\AclBundle\AclBundle(),
    new \Doctrine\Bundle\DoctrineBundle\DoctrineBundle(),
];

redpajama_set_name: RedPajamaGithub


Is Decimal not good enough for the cases the OS integer is not enough?
INT (default) primary key used by outsystems.
DECIMAL To get the equivalent range of BIGINT in decimal, you would need a precision of 19 and a scale of 0. ie DECIMAL(19,0), this will now support the BIGINT range, but with a greater range.
However DECIMAL with a precision greater or equal to 10 uses up 9 bytes of storage, unlike bigint which uses 8 bytes of storage, therefore the difference here is that DECIMAL(19,0) uses 1 extra byte per value stored.
Outsystems confirmed support for BIGINT.
"It is in our backlog, but word from our Product Management is that we are indeed planning to support bigint as an identifier in the future, but unfortunately there isn't a date for it yet."
Currently i.e. Large / Bigint is unsupported.
It would be nice if these were to be added as a datatype.
Thank you for the feedback and context.


text:
She Said: Breaking the Sexual Harassment Story That Helped Ignite a Movement
Jodi Kantor (Author) Megan Twohey (Author)
FORMAT
Paperback$18.00$16.74(English)

Hardcover$28.00$26.04(English)

Now a major motion picture, starring Carey Mulligan and Zoe Kazan
"An instant classic of investigative journalism...'All the President's Men' for the Me Too era." -- Carlos Lozada, The Washington Post
From Pulitzer Prize-winning journalists Jodi Kantor and Megan Twohey, the untold story of their investigation of Harvey Weinstein and its consequences for the #MeToo movement
For years, reporters had tried to get to the truth about Harvey Weinstein's treatment of women. Rumors of wrongdoing had long circulated, and in 2017, when Jodi Kantor and Megan Twohey began their investigation for the New York Times, his name was still synonymous with power. But during months of confidential interviews with actresses, former Weinstein employees, and other sources, many disturbing and long-buried allegations were unearthed, and a web of onerous secret payouts and nondisclosure agreements was revealed. When Kantor and Twohey were finally able to convince sources to go on the record, a dramatic final showdown between Weinstein and the New York Times was set in motion.
In the tradition of great investigative journalism, She Said tells a thrilling story about the power of truth and reveals the inspiring and affecting journeys of the women who spoke up--for the sake of other women, for future generations, and for themselves.
Jodi Kantor and Megan Twohey are investigative reporters at the New York Times. Kantor has focused on the workplace in her reporting, and particularly the treatment of women, covered two presidential campaigns, and is the author of The Obamas. Twohey has focused much of her attention on the treatment of women and children, and, in 2014, as a reporter with Reuters News, was a finalist for the Pulitzer Prize for Investigative Reporting. Kantor and Twohey shared numerous honors for breaking the Harvey Weinstein story, including a George Polk Award, and, along with colleagues, the Pulitzer Prize for Public Service.
"A binge-read of a book, propelled, for the most part, by a clear, adrenaline-spiking ticktock of how their stories came together, and studded with all manner of new astonishing details...This is Kantor and Twohey's story and one everyone should read for a panoply of reasons. By simply recounting their reporting, the two offer a masterful explanation of how a man like Weinstein is allowed to abuse his power and many women for so long in something approaching plain sight...In many ways, "She Said" is more significant than "All the President's Men," and not just because journalism is currently under siege, financially and politically, in a way it was not in the 1970s. There was a finite number of people responsible for the crimes of the Nixon administration; the alleged crimes of Harvey Weinstein are also the crimes of our culture, and they continue to be committed every day by many men all around the world. Although now, one hopes, without as much silence, secrecy and cultural complacency." -- Los Angeles Times Review
"'She Said, ' a new book detailing the astonishing behind-the-scenes of the New York Times's bombshell Harvey Weinstein exposé, is an instant classic of investigative journalism. If your jaw dropped at the newspaper's original allegations against the predatory movie mogul, prepare for it to hit the floor as authors Jodi Kantor and Megan Twohey recount how they uncovered the story: secret meetings, harrowing phone calls, private text exchanges with A-list actresses agonizing over whether to go on the record. Ashley Judd plays the stoic warrior; Gwyneth Paltrow, the circumspect liaison who tries to help the reporters find other sources." - Monica Hesse, The Washington Post
"She Said, the journalists' clear-eyed record of that effort, reads at some moments as a thriller, and at others as an indictment of a system full of rot. But it is ultimately about the women, bonded in their pain, who refused to be silent any longer." - The Atlantic
"'She Said, ' a chronicle of the #MeToo era by Jodi Kantor and Megan Twohey, reveals the power of women who, together, refused to stay silent." - The New Republic
"Kantor and Twohey even-handedly assess the impact of the #MeToo movement thus far while also turning a perceptive, hopeful eye on the way forward." -- Esquire
"She Said is the story behind the story that changed the world....Come for the shocking reveals, stay for how the book expertly parses the current moment."--Refinery29
"'She Said' is riveting and, crafted by two of the country's most talented journalists, a vibrant, cinematic read."-- CNN
"Jodi Kantor and Megan Twohey are two names that future generations will read about in history books." -- Mashable
"Kantor and Twohey have crafted their news dispatches into a seamless and suspenseful account of their reportorial journey, a gripping blow-by-blow of how they managed, "working in the blank spaces between the words," to corroborate allegations that had been chased and abandoned by multiple journalists before them. . . Watching Kantor and Twohey pursue their goal while guarding each other's back is as exhilarating as watching Megan Rapinoe and Crystal Dunn on the pitch. . . It turns out we did need to hear more about Weinstein -- and the "more" that Kantor and Twohey give us draws an important distinction between the trendy ethic of hashtag justice and the disciplined professionalism and institutional heft that actually got the job done."--Susan Faludi, New York Times Book Review
"An instant classic of investigative journalism. The book is packed with reluctant sources, emotional interviews, clandestine meetings, impatient editors, secret documents, late-night door knocks, toady lawyers and showdowns with Weinstein himself. The cumulative effect is almost cinematic, a sort of "All the President's Men" for the Me Too era, except the men are women, and they don't protect the boss, they take him down."-- Carlos Lozada, The Washington Post
"She Said is first and foremost an account of incredible reporting, the kind that takes time, diligence and the kind of institutional support many newspapers can no longer afford. For journalist readers, it is a chance to watch experts at work. And this book is a rare view for nonjournalists into the exacting and rigorous process of quality reporting, and it acts as an implicit counterargument to rising, ambient skepticism of the press. Kantor and Twohey show the background research they ran on sources, to protect both them and the paper, the careful way they documented and substantiated information, and their extraordinary precision in acquiring proof. . . Deeply suspenseful, a kind of less swaggering All the President's Men."--NPR
"Kantor and Twohey, writing a professional memoir that often reads as a riveting work of true crime, offer damning evidence for what is by now a familiar theme: a legal system that promises blindness and balance--the mechanisms through which truth might be finally determined--and too often comes up short. She Said finds Kantor and Twohey (and the extensive team of editors, lawyers, and fact-checkers who bolster their work) exposing not merely Weinstein, but also the system that kept him, at the expense of so many others, safe... So while there is triumph in She Said--the book is a tale of investigative reporting's power to nudge the world out of its complacencies--there is also a sense of necessary open-endedness... The journalist offers the evidence; it is for the rest of us to decide what the justice looks like."--The Atlantic
"Kantor and Twohey are godsends...their reporting and their commitment were pivotal in the #MeToo movement, and this history is flush with gratitude for the women who trusted them with their stories." - The Oregonian
Read It Before You See It 2022
by Reads by Rachael Book Recs View all (10)

redpajama_set_name: RedPajamaCommonCrawl