\documentclass[11pt,twoside]{article} \usepackage{amsmath, amsthm, amscd, amsfonts, amssymb, graphicx, color} \usepackage[bookmarksnumbered, colorlinks]{hyperref} \usepackage{float} \usepackage{lipsum} \usepackage{cite} \usepackage{afterpage} \usepackage[labelfont=bf]{caption} \usepackage[nottoc,notlof,notlot]{tocbibind} %\renewcommand\bibname{References} \def\bibname{\Large \bf References} \usepackage{lipsum} \usepackage{fancyhdr} \pagestyle{fancy} \fancyhf{} \renewcommand{\headrulewidth}{0pt} \fancyhead[LE,RO]{\thepage} \thispagestyle{empty} %\afterpage{\lhead{new value}} \fancyhead[CE]{H.A. Rizvi, R. Ahmad and M. Rahaman} \fancyhead[CO]{\footnotesize{$H(\cdot,\cdot)$-mixed relaxed co-$\eta$-monotone mapping and Resolvent equation problem}} %\topmargin=-1.6cm \textheight 17.5cm% \textwidth 12cm % \topmargin 8mm % \oddsidemargin 20mm % \evensidemargin 20mm % \footskip=24pt % \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{defi}[theorem]{Definition} \newtheorem{algo}[theorem]{Algorithm} \newtheorem{example}[theorem]{Example} \newtheorem{xca}[theorem]{Exercise} %\theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \renewenvironment{proof}{{\bfseries \noindent Proof.}}{~~~~$\square$} \makeatletter \def\th@newremark{\th@remark\thm@headfont{\bfseries}} \makeatletter %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % If you want to insert other packages. Insert them here %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %\long\def\symbolfootnote[#1]#2{\begingroup% %\def\thefootnote{\fnsymbol{footnote}}\footnote[#1]{#2}\endgroup} \def \thesection{\arabic{section}} \begin{document} %\baselineskip 9mm {\noindent Journal of Mathematical Extension \\ ‎\vspace*{9mm} ‎ \begin{center} {\Large \bf $H(\cdot,\cdot)$-mixed relaxed co-$\eta$-monotone mapping with an application for solving a resolvent equation problem\\} \let\thefootnote\relax\footnote{\scriptsize Received: 10-Dec-2014} {\bf Haider Abbas Rizvi}\vspace*{-2mm}\\ \vspace{2mm} {\small Aligarh Muslim University, Aligarh, India} \vspace{2mm} {\bf Rais Ahmad$^*$\let\thefootnote\relax\footnote{$^*$Corresponding Author}}\vspace*{-2mm}\\ \vspace{2mm} {\small Aligarh Muslim University, Aligarh, India} \vspace{2mm} {\bf Mijanur Rahaman}\vspace*{-2mm}\\ \vspace{2mm} {\small Aligarh Muslim University, Aligarh, India} \vspace{2mm} \end{center} \vspace{4mm} {\footnotesize \begin{quotation} {\noindent \bf Abstract.} In this paper, we define an $H(\cdot,\cdot)$-mixed relaxed co-$\eta$-monotone mapping and we apply the same for solving a resolvent equation problem in Hilbert spaces. We also prove some of the properties of $H(\cdot,\cdot)$-mixed relaxed co-$\eta$-monotone mapping. A mann-type iterative algorithm is developed to approximate the solution of resolvent equation problem. Convergence of the iterative sequences is also demonstrated. \end{quotation} \begin{quotation} \noindent{\bf AMS Subject Classification:} 19C33; 49J40 \noindent{\bf Keywords and Phrases:} Relaxed, Space, Algorithm, Sequence, Solution \end{quotation}} \section{Introduction} Since the theory of variational inequalities (inclusions) is quite application oriented and thus developed a lot in recent past in many different directions. This theory provides us, a framework to understand and solve many problems occurring in economics, optimization, transportation, elasticity and applied sciences, etc., see \cite{1,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21} and references therein. Using the concept of resolvent operator technique Noor \cite{22} introduced and studied the resolvent equations. He developed equivalence between variational inequalities and resolvent equations.\vskip 2mm For simplicity and elegance, almost all the problems related to variational inequalities (inclusions) are solved by using different kinds of monotonicities and convexities. The purpose of this communication is to define a new generalized monotone mapping and we call it $H(\cdot,\cdot)$-mixed relaxed co-$\eta$-monotone mapping. We prove some of the properties of this new generalized monotone mapping and apply them to solve a resolvent equation problem. For related work, see also \cite{2,3,4,5,6}. \section{Preliminaries} We begin this section by recalling required definitions and concepts to prove the main result of this paper. Their details can be traced in \cite{3}. \begin{defi} Let $H:X\times X\longrightarrow X$, $\eta:X\times X\longrightarrow X$ and $A, B:X\longrightarrow X$ be the mappings. Then \begin{enumerate} \item[{$(i)$}]$H(A,\cdot)$ is said to be $\eta$-cocoercive with respect to $A$ if for a fixed $u\in X$, there exists a constant $\nu>0$ such that \begin{equation*} \langle H(Ax,u)-H(Ay,u),\eta(x,y)\rangle\geq\nu\|Ax-Ay\|^2,~~\forall x,y\in X; \end{equation*} \item[{$(ii)$}]$H(A,\cdot)$ is said to be relaxed $\eta$-cocoercive with respect to $A$ if for a fixed $u\in X$, there exists a constant $\mu>0$ such that \begin{equation*} \langle H(Ax,u)-H(Ay,u),\eta(x,y)\rangle\geq(-\mu)\|Ax-Ay\|^2,~~\forall x,y\in X; \end{equation*} \item[{$(iii)$}]$H(\cdot,B)$ is said to be $\alpha$-$\xi$-relaxed $\eta$-cocoercive with respect to $B$ if for a fixed $u\in X$, there exist constants $\alpha, \xi>0$ such that \begin{equation*} \langle H(u,Bx)-H(u,By),\eta(x,y)\rangle\geq(-\alpha)\|Bx-By\|^2+\xi\|x-y\|^2,~~\forall x,y\in X; \end{equation*} \item[{$(iv)$}]$H(A,\cdot)$ is said to be $\rho$-Lipschitz continuous with respect to $A$ if for a fixed $u\in X$, there exists a constant $\rho>0$ such that \begin{equation*} \|H(Ax,u)-H(Ay,u)\|\leq\rho\|x-y\|,~~\forall x,y\in X. \end{equation*} Similarly, we can define the Lipschitz continuity of $H$ with respect to $B$ in the second argument. \end{enumerate} \end{defi} \begin{defi} A multi-valued mapping $M:X\longrightarrow2^X$ is said to be $m$-relaxed $\eta$-monotone if, there exists a constant $m>0$ such that \begin{equation*} \langle u-v,\eta(x,y)\rangle\geq-m\|u-v\|^2,~~\forall u,v\in X, x\in M(u), y\in M(v). \end{equation*} \end{defi} \begin{defi} Let $A,B:X \longrightarrow X$, $H,\eta:X\times X\longrightarrow X$ be the single valued mappings such that $H$ is $\mu$-relaxed $\eta$-cocercive with respect to $A$ and $\alpha$-$\xi$-relaxed $\eta$-cocoercive with respect to $B$. Then a multi-valued mapping $M:X\longrightarrow2^X$ is said to be $H(\cdot,\cdot)$-mixed relaxed co-$\eta$-monotone mapping with respect to $A$ and $B$, if $M$ is $m$-relaxed $\eta$-monotone and $(H(A,B)+\lambda M)(X)=X,$ for every $\lambda>0.$ \end{defi} \begin{theorem} Let $H(\cdot,\cdot)$ be a $\mu$-relaxed $\eta$-cocoercive with respect to $A$ and $\alpha$-$\xi$-relaxed $\eta$-cocoercive with respect to $B$, $A$ is $\beta$-Lipschitz continuous and $B$ is $\gamma$-Lipschitz continuous. Let $M$ be $H(\cdot,\cdot)$-mixed relaxed co-$\eta$-monotone with respect to $A$ and $B$. Then the operator $(H(A,B)+\lambda M)^{-1}$ is single-valued for $0<\lambda<\displaystyle\frac{\xi-(\mu\beta^2+\alpha\gamma^2)}{m}$. \end{theorem} Proof: For any given $u\in X$, let $x,~y\in(H(A,B)+\lambda M)^{-1}(u)$. It follows that \begin{equation}\label{eq0} \begin{aligned} -H(Ax,Bx)+u&\in\lambda Mx;\\ -H(Ay,By)+u&\in\lambda My. \end{aligned} \end{equation} As $M$ is $m$-relaxed $\eta$-monotone, we have \begin{eqnarray}\label{eq1} -m\|x-y\|^2&\leq&\frac{1}{\lambda}\langle-H(Ax,Bx)+u-(-H(Ay,By)+u),\eta(x,y)\rangle\nonumber\\ &=&-\frac{1}{\lambda}\langle H(Ax,Bx)-H(Ay,By),\eta(x,y)\rangle\nonumber\\ &=&-\frac{1}{\lambda}\langle H(Ax,Bx)-H(Ay,Bx)+H(Ay,Bx)-H(Ay,By),\eta(x,y)\rangle\nonumber\\ &=&-\frac{1}{\lambda}\langle H(Ax,Bx)-H(Ay,Bx),\eta(x,y)\rangle\nonumber\\ &&-\frac{1}{\lambda}\langle H(Ay,Bx)-H(Ay,By),\eta(x,y)\rangle. \end{eqnarray} Since $H$ is $\mu$-relaxed $\eta$- cocoercive with respect to $A$ and $\alpha$-$\xi$-relaxed $\eta$-cocoercive with respect to $B$, $A$ is $\beta$-Lipschitz continuous and $B$ is $\gamma$-Lipschitz continuous,thus (\ref{eq1}) becomes \begin{eqnarray*} -m\lambda\|x-y\|^2&\leq&\mu\beta^2\|x-y\|^2+\alpha\gamma^2\|x-y\|^2-\xi\|x-y\|^2\\ &=&(\mu\beta^2+\alpha\gamma^2-\xi)\|x-y\|^2, \end{eqnarray*} which implies that \begin{equation} m\lambda\|x-y\|^2\geq-(\mu\beta^2+\alpha\gamma^2-\xi)\|x-y\|^2. \end{equation} If $x\neq y$, then $\lambda\geq\displaystyle\frac{\xi-\mu\beta^2-\alpha\gamma^2}{m}$, which contradicts that $0<\lambda<\displaystyle\frac{\xi-(\mu\beta^2+\alpha\gamma^2)}{m}$. Thus, we have $x=y$, i.e., $(H(A,B)+\lambda M)^{-1}$ is single-valued. \begin{defi}\label{defi1} Let $H(A,B)$ be a $\mu$-relaxed $\eta$- cocoercive with respect to $A$ and $\alpha$-$\xi$-relaxed $\eta$-cocoercive with respect to $B$, $A$ is $\beta$-Lipschitz continuous and $B$ is $\gamma$-Lipschitz continuous. Let $M$ be an $H(\cdot,\cdot)$-mixed relaxed co-$\eta$-monotone mapping with respect to $A$ and $B$. The resolvent operator $R^{H(\cdot,\cdot)-\eta}_{\lambda,M}:X\longrightarrow X$ associated with $H$ and $M$ is defined by \begin{equation}\label{eq2} R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(u)=\left[H(A,B)+\lambda M\right]^{-1}(u),~~\forall u\in X. \end{equation} \end{defi} \begin{theorem} Let $H(A,B)$ be a $\mu$-relaxed $\eta$-cocoercive with respect to $A$ and $\alpha$-$\xi$-relaxed $\eta$-cocoercive with respect to $B$, $\eta:X\times X\longrightarrow X$ be $\sigma$-Lipschitz continuous, $A$ is $\beta$-Lipschitz continuous and $B$ is $\gamma$-Lipschitz continuous. Let $M$ be an $H(\cdot,\cdot)$-mixed relaxed co-$\eta$-monotone mapping with respect to $A$ and $B$. Then the resolvent operator defined by $(\ref{eq2})$ is $\displaystyle\frac{\sigma}{\left[\xi-m\lambda-(\mu\beta^2+\alpha\gamma^2)\right]}$-Lipschitz continuous for $0<\lambda<\displaystyle\frac{\xi-(\mu\beta^2+\alpha\gamma^2)}{m}$, i.e., \begin{equation} \|R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(u)-R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(v)\|\leq\displaystyle\frac{\sigma}{\left[\xi-m\lambda-(\mu\beta^2+\alpha\gamma^2)\right]}\|u-v\|,~~\forall u,v\in X. \end{equation} \end{theorem} \noindent{\bf Proof: } Let $u$ and $v$ be any given points in $X$. It follows from (\ref{eq2}) that \begin{equation}\label{eq3} \begin{aligned} R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(u)&=[H(A,B)+\lambda M]^{-1}(u);\\ R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(v)&=[H(A,B)+\lambda M]^{-1}(v). \end{aligned} \end{equation} For the sake of clarity, we denote \begin{equation*} t_1=R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(u);~~~t_2=R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(v). \end{equation*} Then (\ref{eq0}) implies that \begin{equation}\label{eq4} \begin{aligned} \frac{1}{\lambda}\Big(u-H\big(A(t_1),B(t_1)\big)\Big)&\in M(t_1);\\ \frac{1}{\lambda}\Big(v-H\big(A(t_2),B(t_2)\big)\Big)&\in M(t_2). \end{aligned} \end{equation} Since $M$ is $m$-relaxed $\eta$-monotone, we have \begin{eqnarray*} -m\|t_1-t_2\|^2&\leq&\frac{1}{\lambda}\bigg\langle\Big(u-H\big(A(t_1),B(t_1)\big)\Big)-\Big(v-H\big(A(t_2),B(t_2)\big)\Big),\eta(t_1,t_2)\bigg\rangle\\ &=&\frac{1}{\lambda}\Big\langle u-v-H\big(A(t_1),B(t_1)\big)+H\big(A(t_2),B(t_2)\big),\eta(t_1,t_2)\Big\rangle, \end{eqnarray*} which implies that \begin{equation}\label{eq5} -m\lambda\|t_1-t_2\|^2\leq\langle u-v,\eta(t_1,t_2)\rangle+\Big\langle-H\big(A(t_1),B(t_1)\big)+H\big(A(t_2),B(t_2)\big),\eta(t_1,t_2)\Big\rangle. \end{equation} By Cauchy-Schwartz inequality and (\ref{eq5}), we have \begin{eqnarray} \|u-v\|\|\eta(t_1,t_2)\|&\geq&\langle u-v,\eta(t_1,t_2)\rangle\nonumber\\ &\geq&-\Big\langle-H\big(A(t_1),B(t_1)\big)+H\big(A(t_2),B(t_2)\big),\eta(t_1,t_2)\Big\rangle-\nonumber\\ &&m\lambda\|t_1-t_2\|^2\nonumber\\ &=&\Big\langle H\big(A(t_1),B(t_1)\big)-H\big(A(t_2),B(t_2)\big),\eta(t_1,t_2)\Big\rangle-\nonumber\\ &&m\lambda\|t_1-t_2\|^2\nonumber\\ &=&\Big\langle H\big(A(t_1),B(t_1)\big)-H\big(A(t_2),B(t_1)\big),\eta(t_1,t_2)\Big\rangle-\nonumber\\ &&m\lambda\|t_1-t_2\|^2+\Big\langle H\big(A(t_2),B(t_1)\big)-H\big(A(t_2),B(t_2)\big),\eta(t_1,t_2)\Big\rangle.\nonumber\\ \end{eqnarray} As $H$ is $\mu$-relaxed $\eta$-cocoercive with respect to $A$, $\alpha$-$\xi$-relaxed $\eta$-cocoercive with respect to $B$, $\eta:X\times X\longrightarrow X$ is $\sigma$-Lipschitz continuous, $A$ is $\beta$-Lipschitz continuous and $B$ is $\gamma$-Lipschitz continuous, we have \begin{equation*} \sigma\|u-v\|\|t_1-t_2\|\geq[-\mu\beta^2-\alpha\gamma^2+\xi-m\lambda]\|t_1-t_2\|^2. \end{equation*} Thus, we have \begin{equation*} \|t_1-t_2\|\leq\frac{\sigma}{[\xi-m\lambda-(\mu\beta^2+\alpha\gamma^2)]}\|u-v\|, \end{equation*} i.e., \begin{equation*} \|R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(u)-R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(v)\|\leq\theta_1\|u-v\|,~~\forall u,v\in X, \end{equation*} where $\theta_1=\displaystyle\frac{\sigma}{\left[\xi-m\lambda-(\mu\beta^2+\alpha\gamma^2)\right]}.$ This completes the proof. \section{Resolvent equation problem and Existence theory} In this section, we deal with a variational inclusion problem and its corresponding resolvent equation problem. We establish an equivalence relation between both the problems. An iterative algorithm is also defined to approximate the solution of the resolvent equation problem.\vskip 2mm Let $T,F:X\rightarrow CB(X)$, $M:X\rightarrow 2^X$ be the multi-valued mappings, $S:X\times X\rightarrow X$ and $g:X\rightarrow X$ are the single valued mappings. Then we consider the problem of finding $u\in X$, $x\in T(u)$, $y\in F(u)$ such that \begin{equation}\label{eq6} 0\in S(x,y)+M(g(u)). \end{equation} Problem (\ref{eq6}) is called variational inclusion problem, studied by many authors, see e.g. \cite{22,23}. \vskip 2mm In connection with the variational inclusion problem (\ref{eq6}), we consider the following resolvent equation problem:\vskip 2mm Find $z,u\in X,$ $x\in T(u)$, $y\in F(u)$ such that \begin{equation}\label{eq7} S(x,y) + \lambda^{-1}J^{H(\cdot,\cdot)-\eta}_{M,\lambda}(z)=0, \end{equation} where $\lambda > 0$ is a constant and $J^{H(\cdot,\cdot)-\eta}_{M,\lambda}(z)=[I-H(A(R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(z)),B(R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(z)))]$, where $I$ is the identity operator, $R^{H(\cdot,\cdot)-\eta}_{\lambda,M}$ is the resolvent operator and $$H[A(R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(z)),B(R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(z))]=[H(A(R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(\cdot)),B(R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(\cdot)))](z).$$ \begin{lemma} The triplet $(u,x,y)$, where $u\in X, x\in T(u), y\in F(u)$, is a solution of variational inclusion problem $(\ref{eq6})$ if and only if it satisfies the equation: \begin{equation} g(u)=R^{H(\cdot,\cdot)-\eta}_{\lambda,M}[H(A(g(u)),B(g(u)))-\lambda S(x,y)], \end{equation} where $\lambda > 0$ is a constant. \end{lemma} \noindent{\bf Proof:} By using the definition of resolvent operator $R^{H(\cdot,\cdot)-\eta}_{\lambda,M}$, the conclusion follows directly.\vskip 2mm Based on Lemma (3.1), we prove the following lemma which may be treated as an equivalence lemma for variational inclusion problem (\ref{eq6}) and resolvent equation problem (\ref{eq7}) \begin{lemma}\label{lemma2} The variational inclusion problem $(\ref{eq6})$ has a solution $(u,x,y)$, where $u\in X$, $x\in T(u)$, $y\in F(u),$ if and only if the resolvent equation problem $(\ref{eq7})$ has a solution $(z,u,x,y)$, where $z,u\in X$, $x\in T(u)$, $y\in F(u)$, where \begin{equation}\label{eq8} g(u)= R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(z), \end{equation} and $z=H(A(g(u)),B(g(u)))-\lambda S(x,y)$. \end{lemma} \noindent{\bf Proof:} Let $(u,x,y)$ where $u\in X, x\in T(u), y\in F(u)$ is a solution of variational inclusion problem (\ref{eq6}). Then by lemma (3.1), we have $$g(u)=R^{H(\cdot,\cdot)-\eta}_{\lambda,M}[H(A(g(u)),B(g(u)))-\lambda S(x,y)].$$ Using the fact that $J^{H(\cdot,\cdot)-\eta}_{M,\lambda}=I-H(A(R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(\cdot)),B(R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(\cdot)))$ and equation (\ref{eq8}), we obtain \begin{eqnarray*} &&J^{H(\cdot,\cdot)-\eta}_{M,\lambda}[H(A(g(u)),B(g(u)))-\lambda S(x,y)]\\ &=&H(A(g(u)),B(g(u)))-\lambda S(x,y)\\ &-&H[(A(R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(H(A(g(u)),B(g(u)))-\lambda S(x,y))),\nonumber\\ &&B(R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(H(A(g(u)),B(g(u)))-\lambda S(x,y)))]\nonumber\\ &=&-\lambda S(x,y), \end{eqnarray*} \noindent which implies that $S(x,y) + \lambda^{-1}J^{H(\cdot,\cdot)-\eta}_{\lambda,M}(z)=0$, with $z= H(A(g(u)),B(g(u)))-\lambda S(x,y)$, that is $(z,u,x,y)$ is a solution of resolvent equation problem (\ref{eq7}).\\ Conversely, let $(z,u,x,y)$ is a solution of the resolvent equation problem (\ref{eq7}), then $$S(x,y) + \lambda^{-1}J_{\lambda,M}^{H(\cdot,\cdot)-\eta}(z)=0 ,$$ i.e., \ \ \ \ $J_{\lambda,M}^{H(\cdot,\cdot)-\eta}(z)=-\lambda S(x,y).$\\ Using the definition of $J_{\lambda,M}^{H(\cdot,\cdot)-\eta}$, we have \begin{align} [I-H(A(R^{H(\cdot,\cdot)-\eta}_{\lambda,M}),B(R^{H(\cdot,\cdot)-\eta}_{\lambda,M}))](z)= -\lambda S(x,y),\nonumber\\ z-H(A(R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(z)),B(R^{H(\cdot,\cdot)-\eta}_{\lambda,M})(z))= -\lambda S(x,y),\nonumber \\ z-H(A(g(u)),B(g(u)))= -\lambda S(x,y),\nonumber \\ R_{\lambda,M}^{H(\cdot,\cdot)-\eta}(z)=R_{\lambda,M}^{H(\cdot,\cdot)-\eta}[H(A(g(u)),B(g(u)))-\lambda S(x,y)],\nonumber \\ g(u)= R_{\lambda,M}^{H(\cdot,\cdot)-\eta}[H(A(g(u)),B(g(u)))-\lambda S(x,y)],\nonumber \end{align} that is , $(u,x,y)$ is a solution of the variational inclusion problem (\ref{eq6}).\vskip 2mm Based on above discussion, we now define the following algorithm for solving resolvent equation problem (\ref{eq7}).\vskip 2mm \begin{algo}\label{algo1} For any initial points $(z_o,u_o,x_o,y_o)$, where $z_o, u_o\in X$, $x_o\in T(u_o)$, $y_o\in F(u_o)$, compute the sequences $\{z_n\}$, $\{u_n\}$, $\{x_n\}$ and $\{y_n\}$ by the iterative scheme: \begin{enumerate} \item [(i)] $g(u_n)=R_{\lambda,M}^{H(\cdot,\cdot)-\eta}(z_n)$; \item [(ii)]$\|x_n-x_{n+1}\|$ $\leq D(T(u_n),T(u_{n+1})),$ $x_n\in T(u_n)$, $x_{n+1}\in T(u_{n+1})$; \item [(iii)]$\|y_n-y_{n+1}\|$ $\leq D(F(u_n),F(u_{n+1})),$ $y_n\in F(u_n)$, $y_{n+1}\in F(u_{n+1});$ \item [(iv)] $z_{n+1}$ = $H(A(g(u_n)),B(g(u_n)))-\lambda S(x_n,y_n),$ where $n=0,1,2,.....,$ and $\lambda >0$ is a constant, $D(\cdot,\cdot)$ is the Hausdorff metric on $CB(X).$ \end{enumerate} \end{algo} \begin{theorem} Let $X$ be a real Hilbert space and $ A,B,g:X\rightarrow X$, $S,H:X\times X\rightarrow X$, $\eta: X\times X\rightarrow X$ be the single valued mappings. Let $T,F:X\rightarrow CB(X)$ be the multi-valued mappings and $M:X\rightarrow 2^X$ be $H(\cdot,\cdot)$-mixed relaxed co-$\eta$-monotone mapping. Assume that \begin{enumerate} \item[{$($i$)$}]$H(A,B)$ is $\mu$-relaxed $\eta$-cocoercive with respect to $A$ and $\alpha$-$\xi$-relaxed $\eta$-cocorecive with respect to $B$, $r_1$-Lipschitz continuous with respect to $A$ and $r_2$-Lipschitz continuous with respect to $B$; \item[{$($ii$)$}] $T$ and $F$ are $D$-Lipschitz continuous mappings with constant $\delta_T$ and $\delta_F,$ respectively; \item[{$($iii$)$}] $A$ is $\beta$-Lipschitz continuous and $B$ is $\gamma$-Lipschitz continuous; \item[{$($iv$)$}] $g$ is $\lambda_g$-Lipschitz continuous and $t$-strongly monotone; \item[{$($v$)$}] $S$ is $\lambda_{s_1}$-Lipschitz continuous in first argument and $\lambda_{s_2}$ Lipschitz continuous in second argument. \end{enumerate} If for some $\lambda>0$, the following condition is satisfied: \begin{gather}\label{eq9} \sqrt{1-2t+\lambda_g^2}<[1- \theta\theta_1],\\ where~~\theta=\displaystyle\frac{\sigma}{\left[\xi-m\lambda-(\mu\beta^2+\alpha\gamma^2)\right]}~~,\theta_1=[(r_1+r_2)\lambda_g+\lambda(\lambda_{s_1}\delta_T +\lambda_{s_2}\delta_F)].\nonumber \end{gather} Then, there exist $z,u\in X , x \in T(u), y \in F(u)$ satisfying resolvent equation problem (\ref{eq6}) and the iterative sequences $\{u_n\}$, $\{z_n\}$, $\{x_n\},$ $\{y_n\}$ generated by the Algorithm 3.1 strongly converge to $u,z,x$ and $y$, respectively. \end{theorem} \noindent{\bf Proof:} From (iv) of Algorithm \ref{algo1}, we have \begin{eqnarray}\label{eq10} \|z_{n+1}-z_n\|&=&\|[H(A(g(u_n)),B(g(u_n)))-\lambda S(x_n,y_n)]-[H(A(g(u_{n-1})),\nonumber\\ &&B(g(u_{n-1}))-\lambda S(x_{n-1},y_{n-1})]\|\nonumber\\ &\leq& \|H(A(g(u_n)),B(g(u_n)))-H(A(g(u_{n-1})),B(g(u_{n-1})))\|\nonumber\\ &&-\lambda\| S(x_n,y_n)-S(x_{n-1},y_{n-1})\|\nonumber\\ &\leq& \|H(A(g(u_n)),B(g(u_n)))-H(A(g(u_{n-1})),B(g(u_n)))\|\nonumber\\ &&+ \|H(Ag(u_{n-1}),Bg(u_n))-H(Ag(u_{n-1}),Bg(u_{n-1}))\|\nonumber\\ &&+\lambda\|S(x_n,y_n)-S(x_{n-1},y_n)\|\nonumber\\ &&+\lambda\|S(x_{n-1},y_n)-S(x_{n-1},y_{n-1})\|. \end{eqnarray} \noindent As $H(A,B)$ is $r_1$-Lipschitz continuous with respect to $A$ and $r_2$-Lipschitz with respect to $B$ and $g$ is $\lambda_g$-Lipschitz continuous, we have \begin{eqnarray}\label{eq11} &&\|H(A(g(u_n)),B(g(u_n)))-H(A(g(u_{n-1})),B(g(u_{n-1})))\| \nonumber\\ &&+\|H(A(g(u_{n-1})),B(g(u_n)))\nonumber\\ &&-H(A(g(u_{n-1})),B(g(u_{n-1})))\| \nonumber \\ &\leq &r_1\lambda_g\|u_n-u_{n-1}\|+ r_2 \lambda_g\|u_n-u_{n-1}\|. \end{eqnarray} \noindent Since $S$ is $\lambda_{s_1}$-Lipschitz continuous in the first argument and $\lambda_{s_2}$-Lipschitz continuous in the second argument, $T$ is $D$-Lipschitz continuous with constant $\delta_T$ and $F$ is $D$-Lipschitz continuous with constant $\delta_F$, we have \begin{eqnarray}\label{eq12} &&\|S(x_n,y_n)-S(x_{n-1},y_n)\| +\|S(x_{n-1},y_n)\nonumber\\ &&- S(x_{n-1},y_{n-1})\|\nonumber\\ &\leq& \lambda_{s_1}\|x_n-x_{n-1}\| + \lambda_{s_2}\|y_n-y_{n-1}\| \nonumber\\ &\leq& \lambda_{s_1}\delta_T\|u_n-u_{n-1}\|\nonumber\\ && + \lambda_{s_2}\delta_F\|u_n-u_{n-1}\|. \end{eqnarray} \noindent Using (\ref{eq11}) and (\ref{eq12}), (\ref{eq10}) becomes \begin{eqnarray}\label{eq13} \|z_{n+1}-z_n\| &\leq&[r_1\lambda_g + r_2\lambda_g +\lambda\lambda_{s_1}\delta_T + \lambda\lambda_{s_2}\delta_F]\|u_n-u_{n-1}\|\nonumber\\ &=&\theta_1\|u_n-u_{n-1}\|, \end{eqnarray} \noindent where \ \ $\theta_1$=$[(r_1+r_2)\lambda_g + \lambda(\lambda_{s_1}\delta_T +\lambda_{s_2}\delta_F)].$\\ By (i) of Algorithm \ref{algo1}, we have \begin{eqnarray}\label{eq14} \|u_n-u_{n-1}\|&=&\|u_n-u_{n-1}-g(u_n)+g(u_{n-1})+ R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(z_n)-\nonumber\\ &&R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(z_{n-1})\| \nonumber\\ &\leq& \|u_n-u_{n-1}-(g(u_n)-g(u_{n-1}))\|\nonumber\\ &&+\| R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(z_n)-R^{H(\cdot,\cdot)-\eta}_{\lambda,M}(z_{n-1})\|. \end{eqnarray} \noindent Since $g$ is $t$-strongly monotone and $\lambda_g$-Lipschitz continuous,by using technique of Noor {\cite{22}}, it follows that \begin{equation}\label{eq15} \|u_n-u_{n-1}-(g(u_n)-g(u_{n-1}))\| \leq\sqrt{1-2t+\lambda_g^2}\|u_n-u_{n-1}\|. \end{equation} \noindent Using equation (\ref{eq15}) and Lipschitz continuity of the resolvent operator $ R^{H(\cdot,\cdot)-\eta}_{\lambda,M}$, equation (\ref{eq14}) becomes \begin{eqnarray}\label{eq16} \|u_n-u_{n-1}\|&\leq& \sqrt{1-2t+\lambda_g^2}\|u_n-u_{n-1}\| + \theta \|z_n-z_{n-1}\|,\nonumber\\ \|u_n-u_{n-1}\|&\leq& \frac{\theta}{[1-\sqrt{1-2t+\lambda_g^2}]}\|z_n-z_{n-1}\|. \end{eqnarray} Combining (\ref{eq16}) with (\ref{eq13}), we obtain \begin{eqnarray}\label{eq17} \|z_{n+1}-z_n\|&\leq& \frac{\theta \theta_1}{[1-\sqrt{1-2t+\lambda_g^2}]}\|z_n-z_{n-1}\|,\nonumber\\ \|z_{n+1}-z_n\|&\leq& P(\theta)\|z_n-z_{n-1}\|, \end{eqnarray} \noindent where $P(\theta)=\frac{\theta \theta_1}{1-\sqrt{1-2t+\lambda_g^2}}$,\ \ $\theta=\displaystyle\frac{\sigma}{\left[\xi-m\lambda-(\mu\beta^2+\alpha\gamma^2)\right]}$\\ and $\theta_1$=$[(r_1+r_2)\lambda_g + \lambda(\lambda_{s_1}\delta_T +\lambda_{s_2}\delta_F)].$\\ From (\ref{eq9}) ,it follows that $P(\theta)<1.$ Consequently from (\ref{eq17}) it follows that $\{z_n\}$ is a cauchy sequence in $X$ and as $X$ is complete, $z_n\longrightarrow z$ as $n\longrightarrow \infty$. From (\ref{eq16}), it follows that $\{u_n\}$ is also a cauchy sequence in $X$ such that $u_n\longrightarrow u$ as $n\longrightarrow \infty$. From the $D$-Lipschitz continuous of $T$, $F$ and (ii) and (iii) of Algorithm \ref{algo1}, we know that $\{x_n\}$ and $\{y_n\}$ are also cauchy sequences in $X$ such that $x_n \longrightarrow x$ and $y_n \longrightarrow y$, as $n \longrightarrow \infty$. Further, we show that $x\in T(u)$, we have \begin{eqnarray*} d(x,T(u)) &\leq& \|x-x_n\| + d(x_n,T(u))\\ &\leq&\|x-x_n\| + D(T(u_n),T(u))\\ &\leq& \|x-x_n\| +\delta_T\|u_n-u_{n-1}\| \longrightarrow 0, \ \ as\ \ n\longrightarrow \infty. \end{eqnarray*} which implies that $d(x,T(u))=0$, it follows that $x\in T(u)$. Similarly we can show that $y\in F(u)$.\\ Since $H,A,B,g,T,F$ and $S$ all are continuous and by (iv) of Algorithm \ref{algo1}, it follows that $$z_{n+1}=H(A(g(u_n)),B(g(u_n)))-\lambda S(x_n,y_n),$$ \begin{equation}\label{eq18} \longrightarrow z=H(A(g(u)),B(g(u)))-\lambda S(x,y). \end{equation} \begin{equation}\label{eq19} Consequently, \ \ R_{\lambda,M}^{H(\cdot,\cdot)-\eta}(z_n)=g(u_n)\longrightarrow g(u)=R_{\lambda,M}^{H(\cdot,\cdot)-\eta}(z), as\ \ n\longrightarrow \infty. \end{equation} From (\ref{eq18}), (\ref{eq19}) and by Lemma \ref{lemma2}, the result follows. \begin{center} \begin{thebibliography}{99} % Enter references in alphabetical order and according to the following format. \bibitem{1} S. Adly and W.Oettli, Solvability of generalized nonlinear symmetric variational inequalities, \textit{University of Mannheim, Germany, (preprint)} (1997). \bibitem{2}R. Ahmad, M. Akram, Generalized monotone mapping with an application for solving a variational problem, \textit{J. Optim. 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Glowinski, Numerical Methods for Nonlinear Variational Problems, \textit{Springer-Verlag, Berlin,} (1984). \bibitem{19}M.A. Noor, On Variational Inequalities, \textit{Ph.D. Thesis, Brunel University, London, U.K.}, (1975). \bibitem{20}M.A. Noor, Quasi variational inequalities, \textit{Appl. Math. Lett.}, Vol. 1, No. 4, 367-370, (1988). \bibitem{21}M.A. Noor, sensitivity analysis for quasi variational inequalities, \textit{J. Optim. Theory Appl.}, Vol. 95, No. 2, 399-407, (1997). \bibitem{22}M.A. Noor, Set-Valued Mixed Quasi-Variational Inequalities and Implicit Resolvent Equations, \textit{J. Math. and Comp. Modelling}, Vol. 29 (1999). \bibitem{23}R.U. Verma, Approximation solvability of a class of nonlinear set-valued variational inclusions involving $(A, \eta)$-monotone mappings, \textit{J. Math. A.A.}, Vol. 337 (2008) 969-975. \end{thebibliography} \end{center} {\small \noindent{\bf Haider Abbas Rizvi} \noindent Department of Mathematics \noindent Ph. D. Research Scholar \noindent Aligarh Muslim University \noindent Aligarh-202002, India \noindent E-mail: haider.alig.abbas@gmail.com}\\ {\small \noindent{\bf Rais Ahmad } \noindent Department of Mathematics \noindent Professor of Mathematics \noindent Aligarh Muslim University \noindent Aligarh-202002, India \noindent E-mail: raisain\_123@rediffmail.com}\\ {\small \noindent{\bf Mijanur Rahaman} \noindent Department of Mathematics \noindent Ph. D. Research Scholar \noindent Aligarh Muslim University \noindent Aligarh-202002, India \noindent E-mail: mrahman96@yahoo.com}\\ \end{document}