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# A scheme for a solution of a variational inequality for a monotone mapping and a fixed point of a pseudocontractive mapping

## Abstract

We introduce an iterative process which converges strongly to a common point of the solution set of a variational inequality problem for a Lipschitzian monotone mapping and the fixed point set of a continuous pseudocontractive mapping in Hilbert spaces. In addition, a numerical example which supports our main result is presented. Our theorems improve and unify most of the results that have been proved for this important class of nonlinear operators.

## Introduction

Let C be a subset of a real Hilbert space H. A mapping $T:C\to H$ is called Lipschitzian if there exists $L>0$ such that $\|Tx-Ty\|\leq L\|x-y\|$, $\forall x,y\in C$. If $L=1$ then T is called nonexpansive and if $L\in(0,1)$ then T is called a contraction. The operator T is called pseudocontractive if for each $x,y\in C$ we have

\begin{aligned} \langle Tx-Ty, x-y\rangle\leq\|x-y\|^{2}. \end{aligned}
(1.1)

T is called strongly pseudocontractive if there exists $k\in(0,1)$ such that

$$\langle x-y, Tx-Ty \rangle\leq k\|x-y\|^{2}, \quad\mbox{for all }x,y\in C,$$

and T is said to be a k-strict pseudocontractive if there exists a constant $0\leq k <1$ such that

$$\langle x-y, Tx-Ty \rangle\leq\|x-y\|^{2}-k\bigl\| (I-T)x-(I-T)y \bigr\| ^{2}, \quad\mbox{for all }x,y\in C.$$

Observe that the class of pseudocontractive mappings is a more general class of mappings in the sense that it includes the classes of nonexpansive, strongly pseudocontractive, and k-strict pseudocontractive mappings.

Interest in pseudocontractive mappings stems mainly from their firm connection with the important class of nonlinear monotone mappings, where a mapping A with domain $D(A)$ and range $R(A)$ in H is called monotone if

$$\langle Ax-Ay,x-y\rangle\geq0, \quad\forall x, y \in D(A).$$

A mapping A is called α-inverse strongly monotone if there exists a positive real number α such that

$$\langle Ax-Ay,x-y\rangle\geq\alpha\|Ax-Ay\|^{2},\quad \forall x, y \in D(A).$$

A mapping A is called α-strongly monotone if there exists a positive real number α such that

$$\langle Ax-Ay,x-y\rangle\geq\alpha\|x-y\|^{2},\quad \forall x, y \in D(A).$$

It is obvious to see that the class of monotone mappings includes the class of α-inverse strongly monotone and α-strongly monotone mappings. Furthermore, we observe that any α-inverse strongly monotone mappings A is a monotone and $\frac{1}{\alpha}$-Lipschitzian mapping.

We observe that A is monotone if and only if $T:=I-A$ is pseudocontractive and thus a zero of A, $N(A):=\{x\in D(A):Ax=0\}$, is a fixed point of T, $F(T):=\{ x\in D(T):Tx=x\}$. It is now well known that if A is monotone then the solutions of the equation $Ax=0$ correspond to the equilibrium points of some evolution systems. Consequently, considerable research efforts have been devoted to iterative methods for approximating fixed points of T when T is nonexpansive or pseudocontractive (see, e.g.,  and the references therein).

Let C be a nonempty, closed, and convex subset of a real Hilbert space H. The classical variational inequality problem is to find a $u\in C$ such that $\langle v-u, Au\rangle\geq0$ for all $v\in C$, where A is a nonlinear mapping. The set of solutions of the variational inequality is denoted by $VI(C, A)$. In the context of the variational inequality problem, this implies that $u \in VI(C, A)$ if and only if $u = P_{C}(u - \lambda Au)$, $\forall\lambda>0$, where $P_{C}$ is a metric projection of H into C.

It is now well known that variational inequalities cover disciplines such as partial differential equations, optimal control, optimization, mathematical programming, mechanics and finance. See, for instance, .

Variational inequalities were introduced and studied by Stampacchia  in 1964. Since then, several numerical methods have been developed for solving variational inequalities; see, for instance, [12, 15, 1823] and the references therein.

In 2003, Takahashi and Toyoda  introduced the following iterative scheme under the assumption that a set $C\subset H$ is closed and convex, a mapping T of C into itself is nonexpansive, and a mapping A of C into H is α-inverse strongly monotone:

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} x_{0}\in C,\\ x_{n+1}=\alpha_{n} x_{n}+ (1-\alpha_{n}) TP_{C}(x_{n}-\lambda_{n}Ax_{n}), \quad n\geq 0, \end{array}\displaystyle \right . \end{aligned}
(1.2)

for all $n\geq0$, where $\{\alpha_{n}\}$ is a sequence in $(0, 1)$ and $\{\lambda_{n}\}$ is a sequence in $(0, 2\alpha)$. They proved that if $F(T)\cap VI(C,A)$ is nonempty, then the sequence $\{x_{n}\}$ generated by (1.2) converges weakly to some $z\in F(T)\cap VI(C,A)$.

In order to obtain a strong convergence theorem, Iiduka and Takahashi  reconsidered the common element problem via the following iterative algorithm:

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} x_{0}=x\in C,\\ x_{n+1}=\alpha_{n} x+ (1-\alpha_{n}) TP_{C}(x_{n}-\lambda_{n}Ax_{n}), \quad n\geq 0, \end{array}\displaystyle \right . \end{aligned}
(1.3)

for all $n\geq0$, where $T:C\to C$ is a nonexpansive mapping, $A:C\to H$ is a α-inverse strongly monotone mapping, $\{\alpha_{n}\}$ is a sequence in $(0, 1)$ and $\{\lambda_{n}\}$ is a sequence in $(0, 2\alpha)$. They proved that if $F(T)\cap VI(C,A)$ is nonempty, then the sequence $\{x_{n}\}$ generated by (1.3) converges strongly to some $z\in F(T)\cap VI(C,A)$.

In 2006, Nadezhkina and Takahashi  introduced the following hybrid method for finding an element of $F(S) \cap V I(C,A)$ and established the following strong convergence theorem for the sequence generated by this process.

### Theorem NT



Let C be a closed convex subset of a real Hilbert space H. Let A be a Lipschitzian monotone mapping of C into H with Lipschitz constant L and let S be a nonexpansive mapping of C into itself such that $F(S) \cap V I(C,A)\neq\emptyset$. Let $\{x_{n}\}$, $\{y_{n}\}$ and $\{z_{n}\}$ be sequences generated by

$$\left \{ \textstyle\begin{array}{@{}l} x_{0} = x \in C,\\ y_{n} = P_{C}(x_{n} -\lambda_{n} Ax_{n}),\\ z_{n} =\alpha_{n} x_{n}+(1-\alpha_{n}) SP_{C}(x_{n} - \lambda_{n} Ay_{n}),\\ C_{n} = \{z \in C : \|z_{n} - z\| \leq\|x_{n} - z\|\},\\ Q_{n} = \{z \in C : \langle x_{n}-z, x - x_{n}\rangle\geq0\},\\ x_{n+1} = P_{C_{n}\cap Q_{n}}x, \end{array}\displaystyle \right .$$

for every $n \geq0$, where $\{\lambda_{n}\} \subset [a, b]$ for some $a, b \in (0, \frac{1}{L})$ and $\{\alpha_{n}\}\subset [0, c]$ for some $c \in[0, 1)$. Then the sequences $\{x_{n}\}$, $\{y_{n}\}$ and $\{z_{n}\}$ converge strongly to the same element of $P_{F(S)\cap V I(C,A)}x$.

Our concern now is the following: can an approximation sequence $\{x_{n}\}$ be constructed which converges to a common point of the solution set of a variational inequality problem for a monotone mapping and the fixed point set of a continuous pseudocontractive mapping?

In this paper, it is our purpose to introduce an iterative scheme which converges strongly to a common element of the solution set of a variational inequality problem for Lipschitzian monotone mapping and the fixed point set of a continuous pseudocontractive mapping in Hilbert spaces. Our results provide an affirmative answers to our concern. In addition, a numerical example which supports our main result is presented. Our theorems will extend and unify most of the results that have been proved for this important class of nonlinear operators.

## Preliminaries

Let C be a nonempty, closed, and convex subset of a real Hilbert space H. It is well known that for every point $x\in H$, there exists a unique nearest point in C, denoted by $P_{C}x$, i.e.,

\begin{aligned} \|x -P_{C}x\|\leq\|x-y\| \quad\mbox{for all }y\in C. \end{aligned}
(2.1)

The mapping $P_{C}$ is called the metric projection of H onto C and characterized by the following properties (see, e.g., ):

\begin{aligned}& P_{C}x\in C \quad\mbox{and}\quad \langle x-P_{C}x,P_{C}x-y \rangle\geq0, \quad\mbox{for all }x\in H,y\in C\mbox{ and } \end{aligned}
(2.2)
\begin{aligned}& \|y-P_{C}x\|^{2}\leq\|x-y\|^{2}- \|x-P_{C}x\|^{2}, \quad\mbox{for all }x\in H, y\in C. \end{aligned}
(2.3)

In the sequel we shall make use of the following lemmas.

### Lemma 2.1



Let H be a real Hilbert space. Then, for all $x,y \in H$ and $\alpha \in[0,1]$ the following equality holds:

$$\bigl\| \alpha x+(1-\alpha)y\bigr\| ^{2}=\alpha\|x\|^{2}+(1-\alpha)\|y \|^{2} -\alpha (1-\alpha)\|x-y\|^{2}.$$

### Lemma 2.2

Let H be a real Hilbert space. Then for any given $x,y\in H$, the following inequality holds:

$$\|x+y\|^{2}\leq \|x\|^{2}+2\langle y,x+y\rangle.$$

### Lemma 2.3



Let $\{a_{n}\}$ be a sequence of nonnegative real numbers satisfying the following relation:

$$a_{n+1} \leq(1-\alpha_{n})a_{n} + \alpha_{n}\delta_{n} ,\quad n\geq n_{0},$$

where $\{\alpha_{n}\} \subset(0,1)$ and $\{\delta_{n}\}\subset \mathbb{R}$ satisfying the following conditions: $\lim_{n\to \infty}\alpha_{n}=0$, $\sum_{n=1}^{\infty} \alpha_{n}=\infty$, and $\limsup_{n\to\infty}\delta_{n}\leq0$. Then $\lim_{n\to\infty}a_{n}=0$.

### Lemma 2.4



Let $\{a_{n}\}$ be sequences of real numbers such that there exists a subsequence $\{n_{i}\}$ of $\{n\}$ such that $a_{n_{i}}< a_{{n_{i}}+1}$, for all $i\in \mathbb{N}$. Then there exists a nondecreasing sequence $\{m_{k}\} \subset \mathbb{N}$ such that $m_{k}\to\infty$ and the following properties are satisfied by all (sufficiently large) numbers $k\in \mathbb{N}$:

$$a_{m_{k}}\leq a_{{m_{k}}+1} \quad\textit{and}\quad a_{k}\leq a_{{m_{k}}+1}.$$

In fact, $m_{k}=\max\{j\leq k:a_{j}< a_{j+1}\}$.

### Lemma 2.5



Let C be a nonempty, closed, and convex subset of a real Hilbert space H. Let $T:C\to H$ be continuous pseudocontractive mapping. For $r>0$ and $x\in H$, define a mapping $F_{r}:H\to C$ as follows:

$$F_{r}x:=\biggl\{ z\in C:\langle y-z, Tz\rangle-\frac{1}{r}\bigl\langle y-z,(1+r)z-x\bigr\rangle \leq0, \forall y\in C\biggr\}$$

for all $x\in H$. Then the following hold:

1. (1)

$F_{r}$ is single-valued;

2. (2)

$F_{r}$ is firmly nonexpansive type mapping, i.e., for all $x,y\in H$,

$$\|F_{r}x-F_{r}y\|^{2} \leq\langle F_{r}x-F_{r}y,x-y\rangle;$$
3. (3)

$F(F_{r})=F(T)$;

4. (4)

$F(T)$ is closed and convex.

Let C be a nonempty, closed, and convex subset of a real Hilbert space H. Let $T:C\to H$, be a continuous pseudocontractive mapping. Then, in what follows, $T_{r_{n}}:H\to C$ are defined as follows: For $x\in H$ and $\{r_{n}\}\subset [e,\infty)$, for some $e>0$, define

$$T_{r_{n}}x:=\biggl\{ z\in C: \langle y-z, Tz\rangle-\frac{1}{r_{n}}\bigl\langle y-z,(1+r_{n})z-x\bigr\rangle \leq 0, \forall y\in C\biggr\} .$$

Now, we prove our main convergence theorem.

## Main result

### Theorem 3.1

Let C be a nonempty, closed, and convex subset of a real Hilbert space H. Let $T:C\to C$ be a continuous pseudocontractive mapping. Let $A: C\to H$ be a Lipschitzian monotone mapping with Lipschitz constant L. Assume that $\mathcal{F}=F(T)\cap VI(C,A)$ is nonempty. Let $\{x_{n}\}$ be a sequence generated from an arbitrary $x_{0},u\in C$ by

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} z_{n}=P_{C}[x_{n}-\gamma_{n}Ax_{n}],\\ x_{n+1}=\alpha_{n} u+ (1-\alpha_{n})( a_{n}x_{n} +b_{n}T_{r_{n}} x_{n} + c_{n}P_{C}[x_{n}-\gamma_{n} Az_{n}]), \end{array}\displaystyle \right . \end{aligned}
(3.1)

where $P_{C}$ is a metric projection from H onto C, $\gamma_{n}\subset [a,b]\subset (0,\frac{1}{L})$, and $\{a_{n}\}, \{b_{n}\}, \{c_{n}\} \subset(a,b)\subset (0,1)$, $\{\alpha_{n}\} \subset (0,c)\subset(0,1)$ satisfying the following conditions: (i) $a_{n}+b_{n}+c_{n}=1$, (ii) $\lim_{n\to\infty} \alpha_{n}=0$, $\sum\alpha _{n}=\infty$. Then $\{x_{n}\}$ converges strongly to the point $x^{*}$ of $\mathcal {F}$ nearest to u.

### Proof

Let $u_{n}=P_{C}(x_{n}-\gamma_{n}Az_{n})$ and $w_{n}=T_{r_{n}}x_{n}$ for all $n\geq 0$. Let $p\in\mathcal{F}$. Then from Lemma 2.5 we get $\|w_{n}-p\|\leq \|T_{r_{n}}x_{n}-T_{r_{n}}p\|\leq\|x_{n}-p\|$. In addition, from (2.3) we have

\begin{aligned} \|u_{n}-p\|^{2} \leq&\|x_{n}-\gamma_{n}Az_{n}-p \|^{2}-\|x_{n}-\gamma _{n}Az_{n}-u_{n} \|^{2} \\ =& \|x_{n}-p\|^{2}-\|x_{n}-u_{n} \|^{2}+2\gamma_{n}\langle Az_{n},p-u_{n} \rangle \\ =& \|x_{n}-p\|^{2}-\|x_{n}-u_{n} \|^{2}+2\gamma_{n} \bigl(\langle Az_{n}-Ap,p-z_{n} \rangle \\ &{} +\langle Ap,p-z_{n}\rangle+\langle Az_{n},z_{n}-u_{n} \rangle \bigr) \\ \leq& \|x_{n}-p\|^{2}-\|x_{n}-u_{n} \|^{2}+2\gamma_{n}\langle Az_{n}, z_{n}-u_{n}\rangle \\ \leq& \|x_{n}-p\|^{2}-\|x_{n}-z_{n} \|^{2}-2\langle x_{n}-z_{n}, z_{n}-u_{n} \rangle \\ &{} -\|z_{n}-u_{n}\|^{2}+2\gamma_{n} \langle Az_{n}, z_{n}-u_{n}\rangle \\ = & \|x_{n}-p\|^{2}-\|x_{n}-z_{n} \|^{2}-\|z_{n}-u_{n}\|^{2} \\ &{} + 2\langle x_{n}-\gamma_{n}Az_{n}-z_{n},u_{n}- z_{n}\rangle, \end{aligned}
(3.2)

and from (2.2), we obtain

\begin{aligned} \langle x_{n}-\gamma_{n}Az_{n}-z_{n},u_{n}- z_{n}\rangle =&\langle x_{n}-\gamma_{n}Ax_{n}-z_{n},u_{n}- z_{n}\rangle +\langle\gamma_{n}Ax_{n}-\gamma_{n}Az_{n},u_{n}- z_{n}\rangle \\ \leq& \langle\gamma_{n}Ax_{n}-\gamma_{n}Az_{n},u_{n}- z_{n}\rangle \\ \leq& \gamma_{n}L\|x_{n}-z_{n}\| \|u_{n}-z_{n}\|. \end{aligned}

Thus, we get

\begin{aligned} \|u_{n}-p\|^{2} \leq& \|x_{n}-p\|^{2}- \|x_{n}-z_{n}\|^{2}-\|z_{n}-u_{n} \|^{2} \\ &{}+2\gamma_{n}L\|x_{n}-z_{n}\| \|u_{n}-z_{n}\| \\ \leq& \|x_{n}-p\|^{2}-\|x_{n}-z_{n} \|^{2}-\|z_{n}-u_{n}\|^{2} \\ &{}+\gamma_{n}L \bigl[\|x_{n}-z_{n} \|^{2}+\|z_{n}-u_{n}\|^{2} \bigr] \\ \leq& \|x_{n}-p\|^{2}+(\gamma_{n}L-1) \bigl[ \|x_{n}-z_{n}\|^{2}+\|z_{n}-u_{n} \|^{2} \bigr] \end{aligned}
(3.3)
\begin{aligned} \leq& \|x_{n}-p\|^{2}. \end{aligned}
(3.4)

Furthermore, from (3.1) and Lemma 2.1 we have the following:

\begin{aligned} \|x_{n+1}-p\|^{2} =&\bigl\| \alpha_{n} u+ (1- \alpha_{n}) (a_{n}x_{n}+ b_{n}w_{n}+ c_{n} u_{n})-p\bigr\| ^{2} \\ \leq& \alpha_{n}\|u-p\|^{2}+ (1-\alpha_{n})\bigl\| a_{n}(x_{n}-p)+b_{n}(w_{n}-p) \\ &{}+ c_{n} (u_{n}-p) \bigr\| ^{2} \\ \leq& \alpha_{n}\|u-p\|^{2}+ (1-\alpha_{n}) \bigl[ a_{n}\|x_{n}-p\|^{2}+b_{n} \|w_{n}-p\|^{2} \\ &{} + c_{n} \|u_{n}-p \|^{2} \bigr] -(1- \alpha_{n})a_{n} b_{n} \|w_{n}-x_{n} \|^{2}\\ &{} -(1-\alpha_{n})a_{n} c_{n} \|u_{n}-x_{n}\|^{2} -(1-\alpha_{n})b_{n} c_{n} \|w_{n}-u_{n}\|^{2}, \end{aligned}

and using (3.3) we get

\begin{aligned} \|x_{n+1}-p\|^{2} \leq& \alpha_{n}\|u-p \|^{2}+ (1-\alpha_{n}) a_{n}\|x_{n}-p \|^{2}+(1-\alpha _{n})b_{n}\|x_{n}-p \|^{2} \\ &{} +(1-\alpha_{n}) c_{n} \bigl[\|x_{n}-p \|^{2} + (\gamma_{n}L-1) \bigl[\|x_{n}-z_{n} \|^{2}+\|z_{n}-u_{n}\|^{2} \bigr] \bigr] \\ &{} -(1-\alpha_{n})a_{n} b_{n} \|w_{n}-x_{n}\|^{2} -(1-\alpha_{n})a_{n} c_{n} \|u_{n}-x_{n}\|^{2} \\ &{} -(1-\alpha_{n})b_{n} c_{n} \|w_{n}-u_{n}\|^{2} \\ \leq& \alpha_{n}\|u-p\|^{2}+ (1-\alpha_{n}) \|x_{n}-p\|^{2} \\ &{} + (1-\alpha_{n})c_{n}(\gamma_{n}L-1) \bigl[ \|x_{n}-z_{n}\|^{2}+\|z_{n}-u_{n} \|^{2} \bigr] \\ &{} -(1-\alpha_{n})a_{n} b_{n} \|w_{n}-x_{n}\|^{2} -(1-\alpha_{n})a_{n} c_{n} \|u_{n}-x_{n}\|^{2} \\ &{}-(1-\alpha_{n})b_{n} c_{n} \|w_{n}-u_{n}\|^{2}. \end{aligned}
(3.5)

Since $\gamma_{n}L<1$, from (3.5) we get

\begin{aligned} \|x_{n+1}-p\|^{2} \leq & \alpha_{n}\|u-p \|^{2}+ (1-\alpha_{n})\|x_{n}-p\|^{2}. \end{aligned}
(3.6)

Thus, by induction,

$$\|x_{n+1}-p\|^{2} \leq\max\bigl\{ \|u-p\|^{2}, \|x_{0}-p\|^{2} \bigr\} , \quad\forall n\geq 0,$$

which implies that $\{x_{n}\}$ and $\{z_{n}\}$ are bounded.

Let $x^{*}= P_{\mathcal{F}}(u)$. Then, using (3.1), Lemma 2.2, and following the methods used to get (3.5) we obtain that

\begin{aligned} \bigl\| x_{n+1}-x^{*}\bigr\| ^{2} =& \bigl\| \alpha_{n}u+ (1- \alpha_{n}) ( a_{n}x_{n} +b_{n}w_{n} + c_{n} u_{n}) -x^{*}\bigr\| ^{2} \\ \leq& \bigl\| \alpha_{n} \bigl(u-x^{*}\bigr) + (1-\alpha_{n}) \bigl[ (a_{n}x_{n}+b_{n}w_{n} + c_{n} u_{n})-x^{*} \bigr]\bigr\| ^{2} \\ \leq& (1-\alpha_{n}) \bigl\| a_{n}x_{n}+ b_{n}w_{n} + c_{n} u_{n}-x^{*} \bigr\| ^{2} \\ &{} +2\alpha_{n}\bigl\langle u-x^{*}, x_{n+1}-x^{*}\bigr\rangle \\ \leq& (1-\alpha_{n})a_{n} \bigl\| x_{n}-x^{*} \bigr\| ^{2} +(1-\alpha _{n})b_{n}\bigl\| w_{n}-x^{*} \bigr\| ^{2}\\ &{} +(1-\alpha_{n})c_{n}\bigl\| u_{n}-x^{*} \bigr\| ^{2} -(1-\alpha_{n})b_{n}a_{n} \|w_{n}-x_{n}\|^{2}\\ &{} -(1-\alpha _{n})b_{n}c_{n} \|u_{n}-w_{n}\|^{2} -(1-\alpha_{n})a_{n}c_{n} \|x_{n}-u_{n}\|^{2}\\ &{} +2\alpha_{n}\bigl\langle u-x^{*}, x_{n+1}-x^{*}\bigr\rangle , \end{aligned}

which implies that

\begin{aligned} \bigl\| x_{n+1}-x^{*}\bigr\| ^{2} \leq& (1-\alpha_{n})a_{n} \bigl\| x_{n}-x^{*}\bigr\| ^{2} +(1-\alpha _{n})b_{n} \bigl\| x_{n}-x^{*}\bigr\| ^{2} \\ &{} +(1-\alpha_{n})c_{n} \bigl[ \bigl\| x_{n}-x^{*} \bigr\| ^{2} +(\gamma_{n}L-1) \bigl[\|x_{n}-z_{n} \|^{2}+\|z_{n}-u_{n}\|^{2} \bigr] \bigr] \\ &{} -(1-\alpha_{n})b_{n}a_{n}\|w_{n}-x_{n} \|^{2} -(1-\alpha_{n})b_{n}c_{n} \|u_{n}-w_{n}\|^{2} \\ &{}-(1-\alpha_{n})a_{n}c_{n}\|x_{n}-u_{n} \|^{2}+2\alpha_{n}\bigl\langle u-x^{*}, x_{n+1}-x^{*} \bigr\rangle \\ \leq& (1-\alpha_{n})\|x_{n}-x^{*}\|^{2} +(1- \alpha_{n})c_{n}(\gamma_{n}L-1) \bigl[ \|x_{n}-z_{n}\|^{2}+\|z_{n}-u_{n} \|^{2} \bigr] \\ &{} -(1-\alpha_{n})b_{n}a_{n}\|w_{n}-x_{n} \|^{2} -(1-\alpha _{n})b_{n}c_{n} \|u_{n}-w_{n}\|^{2} \\ &{} -(1-\alpha_{n})a_{n}c_{n}\|x_{n}-u_{n} \|^{2} +2\alpha_{n}\bigl\langle u-x^{*}, x_{n+1}-x^{*} \bigr\rangle \end{aligned}
(3.7)
\begin{aligned} \leq & (1-\alpha_{n}) \bigl\| x_{n}-x^{*}\bigr\| ^{2} +2 \alpha_{n}\bigl\langle u-x^{*}, x_{n+1}-x^{*}\bigr\rangle . \end{aligned}
(3.8)

Now, we consider two cases.

Case 1. Suppose that there exists $n_{0}\in \mathbb{N}$ such that $\{ \|x_{n}-x^{*}\|\}$ is decreasing for all $n\geq n_{0}$. Then we get $\{ \|x_{n}-x^{*}\|\}$ is convergent. Thus, from (3.7) we have

\begin{aligned} x_{n}-z_{n}\to0,\qquad z_{n}-u_{n}\to0 \quad\mbox{as }n\to\infty, \end{aligned}
(3.9)

and

\begin{aligned} w_{n}-x_{n}\to0,\qquad u_{n}-w_{n} \to0,\qquad x_{n}-u_{n}\to0 \quad\mbox{as }n\to\infty. \end{aligned}
(3.10)

Moreover, from the fact that $\alpha_{n}\to0$, as $n\to\infty$, (3.1), (3.9), and (3.10) we have

\begin{aligned} \|x_{n+1}-x_{n}\| = &\bigl\| \alpha_{n}(u-x_{n})+(1- \alpha_{n}) \bigl( b_{n}w_{n}+c_{n}u_{n} -(1-a_{n})x_{n}\bigr)\bigr\| \\ \leq &\alpha_{n}\|u-x_{n}\|+(1-\alpha_{n}) b_{n}\|w_{n}-x_{n}\|+(1-\alpha _{n})c_{n} \|u_{n}-x_{n}\|\to0, \end{aligned}
(3.11)

as $n\to\infty$.

Furthermore, since $\{x_{n+1} \}$ is bounded subset of H which is reflexive, we can choose a subsequence $\{x_{n_{i}+1}\}$ of $\{ x_{n+1}\}$ such that $x_{n_{i}+1}\rightharpoonup z$ and $\limsup_{n\to\infty}\langle u-x^{*}, x_{n+1}-x^{*}\rangle=\lim_{i\to\infty }\langle u-x^{*},x_{n_{i}+1}-x^{*}\rangle$. This implies from (3.11) that $x_{n_{i}}\rightharpoonup z$.

Now, we show that $z\in VI(C,A)$. But, since A is Lipschitz continuous, we have $Az_{n}-Au_{n}\to0$, as $n\to\infty$ and from (3.9) and (3.10) we have $u_{n_{i}}\rightharpoonup z$ and $z_{n_{i}}\rightharpoonup z$. Let

\begin{aligned} Tv= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} Av+N_{C}v, &\mbox{if }v\in C,\\ \emptyset,& \mbox{if }v\notin C. \end{array}\displaystyle \right . \end{aligned}
(3.12)

Then T is maximal monotone and $0\in Tv$ if and only if $v\in VI(C,A)$ (see, e.g. ). Let $(v,w)\in G(T)$. Then we have $w\in Tv=Av+N_{C}v$ and hence $w-Av\in N_{C}v$. So, we have $\langle v-u,w-Av\rangle\geq0$, for all $u\in C$. On the other hand, from $u_{n}=P_{C}(x_{n}-\gamma_{n}Az_{n})$ and $v\in C$, we have $\langle x_{n}-\gamma_{n}Az_{n}-u_{n}, u_{n}-v\rangle\geq0$, and hence, $\langle v-u_{n},(u_{n}-x_{n})/\gamma_{n}+Az_{n}\rangle\geq0$. Therefore, from $w-Av\in N_{C}v$ and $u_{n_{i}}\in C$ we have

\begin{aligned} \langle v-u_{n_{i}},w\rangle \geq&\langle v-u_{n_{i}},Av\rangle \geq \langle v-u_{n_{i}},Av\rangle-\bigl\langle v-u_{n_{i}},(u_{n_{i}}-x_{n_{i}})/ \gamma_{n_{i}}+Az_{n_{i}}\bigr\rangle \\ =& \langle v-u_{n_{i}},Av-Au_{n_{i}}\rangle+ \langle v-u_{n_{i}},Au_{n_{i}}-Az_{n_{i}}\rangle \\ &{} -\bigl\langle v-u_{n_{i}}, (u_{n_{i}}-x_{n_{i}})/ \gamma_{n_{i}}\bigr\rangle \\ \geq& \langle v-u_{n_{i}},Au_{n_{i}}-Az_{n_{i}}\rangle- \bigl\langle v-u_{n_{i}}, (u_{n_{i}}-x_{n_{i}})/ \gamma_{n_{i}}\bigr\rangle . \end{aligned}

Hence, we have $\langle v-z,w\rangle\geq0$, as $i\to\infty$. Since T is maximal monotone, we have $z\in T^{-1}(0)$ and hence $z\in VI(C,A)$.

Now, we show that $z\in F(T)$. Note that, from the definition of $w_{n_{i}}$, we have

\begin{aligned} \langle y-w_{n_{i}}, Tw_{n_{i}}\rangle- \frac{1}{r_{n_{i}}}\bigl\langle y-w_{n_{i}}, (r_{n_{i}}+1)w_{n_{i}}-x_{n_{i}} \bigr\rangle \leq 0, \quad\forall y\in C. \end{aligned}
(3.13)

Put $z_{t}=tv+(1-t)z$ for all $t\in(0,1]$ and $v\in C$. Consequently, we get $z_{t}\in C$. From (3.13) and pseudocontractivity of T it follows that

\begin{aligned} \langle w_{n_{i}}-z_{t}, Tz_{t}\rangle \geq&\langle w_{n_{i}}-z_{t}, Tz_{t}\rangle +\langle z_{t}-w_{n_{i}}, Tw_{n_{i}}\rangle-\frac{1}{r_{n_{i}}} \bigl\langle z_{t}-w_{n_{i}},(1+r_{n_{i}})w_{n_{i}}-x_{n_{i}} \bigr\rangle \\ = & -\langle z_{t}-w_{n_{i}}, Tz_{t}-Tw_{n_{i}} \rangle -\frac{1}{r_{n_{i}}}\langle z_{t}-w_{n_{i}}, w_{n_{i}}-x_{n_{i}}\rangle -\langle z_{t}-w_{n_{i}}, w_{n_{i}}\rangle \\ \geq & -\| z_{t}-w_{n_{i}}\|^{2} -\frac{1}{r_{n_{i}}} \langle z_{t}-w_{n_{i}}, w_{n_{i}}-x_{n_{i}} \rangle-\langle z_{t}-w_{n_{i}}, w_{n_{i}}\rangle \\ = & \langle w_{n_{i}}-z_{t}, z_{t}\rangle-\biggl\langle z_{t}-w_{n_{i}}, \frac{w_{n_{i}}-x_{n_{i}}}{r_{n_{i}}}\biggr\rangle . \end{aligned}

Then, since $w_{n}-x_{n}\to0$, as $n\to\infty$ we obtain $\frac{w_{n_{i}}-x_{n_{i}}}{r_{n_{i}}}\to0$ as $i\to\infty$. Thus, it follows that

$$\langle z-z_{t}, Tz_{t} \rangle\geq\langle z-z_{t},z_{t}\rangle\quad\mbox{as }i\to\infty,$$

and hence

$$-\langle v-z, Tz_{t}\rangle\geq-\langle v-z,z_{t}\rangle,\quad \forall v\in C.$$

Letting $t\to0$ and using the fact that T is continuous we obtain

$$-\langle v-z, Tz\rangle\geq-\langle v-z,z \rangle, \quad\forall v\in C.$$

Now, let $v=Tz$. Then we obtain $z=Tz$ and hence $z\in F(T)$. Therefore, by (2.2) we immediately obtain

\begin{aligned} \limsup_{n\to\infty}\bigl\langle u-x^{*},x_{n+1}-x^{*}\bigr\rangle =& \lim_{i\to \infty}\bigl\langle u-x^{*}, x_{n_{i}+1}-x^{*} \bigr\rangle \\ =&\bigl\langle u-x^{*}, z-x^{*}\bigr\rangle \leq0. \end{aligned}
(3.14)

Then it follows from (3.8), (3.14), and Lemma 2.3 that $\|x_{n}-x^{*}\|\to0$, as $n\to\infty$. Consequently, $\{x_{n}\}$ converges to the minimum norm point of $\mathcal{F}$.

Case 2. Suppose that there exists a subsequence $\{ n_{i}\}$ of $\{n\}$ such that

$$\bigl\| x_{n_{i}}-x^{*}\bigr\| < \bigl\| x_{n_{i}+1}-x^{*}\bigr\| ,$$

for all $i\in\mathbb{N}$. Then, by Lemma 2.4, there exists a nondecreasing sequence $\{m_{k}\}\subset\mathbb{N}$ such that $m_{k}\to\infty$, and

\begin{aligned} \bigl\| x_{m_{k}}-x^{*}\bigr\| \leq\bigl\| x_{m_{k}+1}-x^{*}\bigr\| \quad\mbox{and}\quad \bigl\| x_{k}-x^{*}\bigr\| \leq \bigl\| x_{m_{k}+1}-x^{*}\bigr\| , \end{aligned}
(3.15)

for all $k\in\mathbb{N}$. Now, from (3.7) we get

\begin{aligned} x_{m_{k}}-z_{m_{k}}\to0, \qquad z_{m_{k}}-u_{m_{k}}\to0 \quad\mbox{as }k\to\infty, \end{aligned}
(3.16)

and

\begin{aligned} w_{m_{k}}-x_{m_{k}}\to0,\qquad u_{m_{k}}-w_{m_{k}} \to0,\qquad x_{m_{k}}-u_{m_{k}}\to0 \quad\mbox{as }k\to\infty. \end{aligned}
(3.17)

Thus, like in Case 1, we obtain $x_{m_{k}+1}-x_{m_{k}}\to0$ and

\begin{aligned} \limsup_{k\to\infty}\bigl\langle u-x^{*}, x_{m_{k}+1}-x^{*}\bigr\rangle \leq0. \end{aligned}
(3.18)

Now, from (3.8) we have

\begin{aligned} \bigl\| x_{m_{k}+1}-x^{*}\bigr\| ^{2} \leq& (1-\alpha_{m_{k}}) \bigl\| x_{m_{k}}-x^{*}\bigr\| ^{2}+ 2\alpha_{m_{k}}\bigl\langle u-x^{*}, x_{m_{k}+1}-x^{*}\bigr\rangle , \end{aligned}
(3.19)

and hence (3.15) and (3.19) imply that

\begin{aligned} \alpha_{m_{k}}\bigl\| x_{m_{k}}-x^{*}\bigr\| ^{2} \leq & \bigl\| x_{m_{k}}-x^{*}\bigr\| ^{2}-\bigl\| x_{{m_{k}}+1}-x^{*}\bigr\| ^{2} +2 \alpha_{m_{k}}\bigl\langle u-x^{*}, x_{m_{k}+1}-x^{*}\bigr\rangle \\ \leq& +2\alpha_{m_{k}}\bigl\langle u-x^{*}, x_{m_{k}+1}-x^{*}\bigr\rangle . \end{aligned}

But since that $\alpha_{m_{k}}>0$, we obtain

$$\bigl\| x_{m_{k}}-x^{*}\bigr\| ^{2} \leq + 2\bigl\langle u-x^{*},x_{m_{k}+1}-x^{*}\bigr\rangle .$$

Then, using (3.18), we get $\| x_{m_{k}}-x^{*}\|\to0$, as $k\to\infty$. This together with (3.19) imply that $\| x_{{m_{k}}+1}-x^{*}\|\to0$, as $k\to\infty$. But $\| x_{k}-x^{*}\|\leq \| x_{{m_{k}}+1}-x^{*}\|$, for all $k\in\mathbb {N}$, thus we obtain $x_{k}\to x^{*}$. Therefore, from the above two cases, we can conclude that $\{x_{n}\}$ converges strongly to the point $x^{*}$ of $\mathcal{F}$ nearest to u. □

If, in Theorem 3.1, we assume that $T=I$, the identity mapping on C, we obtain the following corollary.

### Corollary 3.2

Let C be a nonempty, closed, and convex subset of a real Hilbert space H. Let $A: C\to H$ be a Lipschitzian monotone mapping with Lipschitz constant L. Assume that $VI(C,A)$ is nonempty. Let $\{x_{n}\}$ be a sequence generated from an arbitrary $x_{0},u\in C$ by

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} z_{n}=P_{C}[x_{n}-\gamma_{n}Ax_{n}],\\ x_{n+1}=\alpha_{n} u+ (1-\alpha_{n})( a_{n}x_{n} + (1-a_{n})P_{C}[x_{n}-\gamma_{n} Az_{n}]), \end{array}\displaystyle \right . \end{aligned}
(3.20)

where $P_{C}$ is a metric projection from H onto C, $\gamma_{n}\subset [a,b]\subset (0,\frac{1}{L})$, and $\{a_{n}\} \subset(a,b)\subset (0,1)$, $\{\alpha_{n}\} \subset(0,c)\subset(0,1)$ satisfying $\lim_{n\to\infty} \alpha_{n}=0$, $\sum\alpha_{n}=\infty$. Then $\{x_{n}\}$ converges strongly to the point $x^{*}=P_{VI(C,A)}(u)$.

If, in Theorem 3.1, we assume that $A=0$, we obtain the following corollary, which is Theorem 3.1 of .

### Corollary 3.3

Let C be a nonempty, closed, and convex subset of a real Hilbert space H. Let $T:C\to C$ be a continuous pseudocontractive mapping. Assume that $F(T)$ is nonempty. Let $\{x_{n}\}$ be a sequence generated from an arbitrary $x_{0},u\in C$ by

$$x_{n+1}=\alpha_{n} u+ (1-\alpha_{n}) \bigl( a_{n}x_{n} +(1-a_{n})T_{r_{n}} x_{n}\bigr),$$

where $\{a_{n}\} \subset(a,b)\subset (0,1)$, $\{\alpha_{n}\} \subset (0,c)\subset(0,1)$ satisfying $\lim_{n\to\infty} \alpha _{n}=0$, $\sum\alpha_{n}=\infty$. Then $\{x_{n}\}$ converges strongly to the point $x^{*}=P_{F(T)}(u)$.

If, in Theorem 3.1, we assume that A is α-inverse strongly monotone then A is Lipschitzian and we obtain the following corollary.

### Corollary 3.4

Let C be a nonempty, closed, and convex subset of a real Hilbert space H. Let $T:C\to C$ be a continuous pseudocontractive mapping. Let $A: C\to H$ an α-inverse strongly monotone mapping. Assume that $\mathcal{F}=F(T)\cap VI(C,A)$ is nonempty. Let $\{x_{n}\}$ be a sequence generated from an arbitrary $x_{0},u\in C$ by

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} z_{n}=P_{C}[x_{n}-\gamma_{n}Ax_{n}],\\ x_{n+1}=\alpha_{n} u+ (1-\alpha_{n})( a_{n}x_{n} +b_{n}T_{r_{n}} x_{n} + c_{n}P_{C}[x_{n}-\gamma_{n} Az_{n}]), \end{array}\displaystyle \right . \end{aligned}
(3.21)

where $P_{C}$ is a metric projection from H onto C, $\gamma_{n}\subset [a,b]\subset (0,\alpha)$, and $\{a_{n}\}, \{b_{n}\}, \{c_{n}\} \subset (a,b)\subset (0,1)$, $\{\alpha_{n}\} \subset(0,c)\subset(0,1)$ satisfying (i) $a_{n}+b_{n}+c_{n}=1$, (ii) $\lim_{n\to\infty} \alpha_{n}=0$, $\sum\alpha _{n}=\infty$. Then $\{x_{n}\}$ converges strongly to the point $x^{*}=P_{\mathcal{F}}(u)$.

If, in Theorem 3.1, we assume that $C=H$, a real Hilbert space, then $P_{C}$ becomes identity mapping and $VI(C,A)=A^{-1}(0)$, and hence we get the following corollary.

### Corollary 3.5

Let H be a real Hilbert space. Let $T:H\to H$ be a continuous pseudocontractive mapping. Let $A: H\to H$ be a Lipschitzian monotone mapping with Lipschitz constant L. Assume that $\mathcal{F}=F(T)\cap A^{-1}(0)$ is nonempty. Let $\{x_{n}\}$ be a sequence generated from an arbitrary $x_{0},u\in C$ by

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} z_{n}=x_{n}-\gamma_{n}Ax_{n},\\ x_{n+1}=\alpha_{n} u+ (1-\alpha_{n})( a_{n}x_{n} +b_{n}T_{r_{n}} x_{n} + c_{n}[x_{n}-\gamma_{n} Az_{n}]), \end{array}\displaystyle \right . \end{aligned}
(3.22)

where $\gamma_{n}\subset [a,b]\subset (0,\frac{1}{L})$ and $\{a_{n}\}, \{ b_{n}\}, \{c_{n}\} \subset(a,b)\subset (0,1)$, $\{\alpha_{n}\} \subset (0,c)\subset(0,1)$ satisfying the following conditions: (i) $a_{n}+b_{n}+c_{n}=1$, (ii) $\lim_{n\to\infty} \alpha_{n}=0$, $\sum\alpha _{n}=\infty$. Then $\{x_{n}\}$ converges strongly to the point $x^{*}$ of $\mathcal {F}$ nearest to u.

We also note that the method of proof of Theorem 3.1 provides the following theorem for approximating the common minimum-norm point of the solution set of a variational inequality problem for monotone mapping and fixed point set of a continuous pseudocontractive mapping.

### Theorem 3.6

Let C be a nonempty, closed, and convex subset of a real Hilbert space H. Let $T:C\to C$ be a continuous pseudocontractive mapping. Let $A: C\to H$ be a Lipschitzian monotone mapping with Lipschitz constant L. Assume that $\mathcal{F}=F(T)\cap VI(C,A)$ is nonempty. Let $\{x_{n}\}$ be a sequence generated from an arbitrary $x_{0},u\in C$ by

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} z_{n}=P_{C}[x_{n}-\gamma_{n}Ax_{n}],\\ x_{n+1}= P_{C} [(1-\alpha_{n})( a_{n}x_{n} +b_{n}T_{r_{n}} x_{n} + c_{n}P_{C}[x_{n}-\gamma_{n} Az_{n}]) ], \end{array}\displaystyle \right . \end{aligned}
(3.23)

where $P_{C}$ is a metric projection from H onto C, $\gamma_{n}\subset [a,b]\subset (0,\frac{1}{L})$, and $\{a_{n}\}, \{b_{n}\}, \{c_{n}\} \subset (a,b)\subset (0,1)$, $\{\alpha_{n}\} \subset(0,c)\subset(0,1)$ satisfying the following conditions: (i) $a_{n}+b_{n}+c_{n}=1$, (ii) $\lim_{n\to\infty} \alpha_{n}=0$, $\sum\alpha _{n}=\infty$. Then $\{x_{n}\}$ converges strongly to the minimum-norm point $x^{*}$ of  $\mathcal{F}$.

### Remark 3.7

Theorem 3.1 extends Theorem 3.1 of Takahashi and Toyoda  and Theorem 3.2 of Yao et al. , Theorem 3.1 of Iiduka and Takahashi  and the results of Nadezhkina and Takahashi  in the sense that our scheme provides a common point of the solution set of variational inequalities for a more general class of monotone mappings and/or the fixed point set of a more general class of continuous pseudocontractive mappings. Our results provide an affirmative answer to our concern.

## Applications to minimization problems

In this section, we study the problem of finding a minimizer of a continuously Fréchet differentiable convex functional in Hilbert spaces. Let f be a continuously Fréchet differentiable convex functionals of H into $(-\infty,\infty)$ such that the gradient of f, $(\bigtriangledown f)$ is continuous and monotone. For $\gamma>0$, and $x\in H$, let $T_{r_{n}}x:=\{z\in H:\langle y-z,(I- (\bigtriangledown f))z\rangle -\frac{1}{\gamma}\langle y-z, (1+\gamma)z-x\rangle\leq0, \forall y\in H\}$. Then the following theorem holds.

### Theorem 4.1

Let H be a real Hilbert space. Let f be a continuously Fréchet differentiable convex functionals of H into $(-\infty,\infty)$ such that the gradient of f, $(\bigtriangledown f)$ is continuous and monotone such that $\mathcal{N}:=\arg\min_{y\in C}f(y)\neq\emptyset$. Let $\{x_{n}\}$ be a sequence generated from an arbitrary $x_{0},u\in C$ by

$$x_{n+1}=\alpha_{n} u+ (1-\alpha_{n}) \bigl( a_{n}x_{n} +(1-a_{n})T_{r_{n}} x_{n}\bigr),$$

where $\{a_{n}\} \subset(a,b)\subset (0,1)$, $\{\alpha_{n}\} \subset (0,c)\subset(0,1)$ satisfying $\lim_{n\to\infty} \alpha_{n}=0$, $\sum\alpha_{n}=\infty$. Then $\{x_{n}\}$ converges strongly to the point $x^{*}\in\mathcal{N}$ nearest to u.

### Proof

We note that $T:=(I-\bigtriangledown f)$ is continuous pseudocontractive mapping with $F(T)= (\bigtriangledown f)^{-1}(0)$ and from the convexity and Frećhet differentiability of f we see that the zero of f is given by $\mathcal{N}=\arg\min_{y\in C}f(y)$. Thus, the conclusion follows from Corollary 3.3. □

## Numerical example

In this section, we give an example of a continuous pseudocontractive mapping T and a Lipschitzian monotone mapping with all the conditions of Theorem 3.1 and some numerical experiment results to explain the conclusion of the theorem.

### Example 5.1

Let $H=\mathbb{R}$ with Euclidean norm. Let $C=[-2,6]$ and $T:C\to\mathbb{R}$ be defined by

$$Tx:= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} -3x, & x\in[-2,0], \\ x, & (0,6], \end{array}\displaystyle \right .$$

and

\begin{aligned} Ax:= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} 0, & x\in[-2,\frac{1}{2}], \\ 3(x-\frac{1}{2})^{2},& x\in(\frac{1}{2},6]. \end{array}\displaystyle \right . \end{aligned}
(5.1)

Then we easily see that T is continuous pseudocontractive with $F(T)=[0,6]$.

In addition, we observe that A is monotone with $VI(C,A)= [-2,\frac{1}{2}]$. Next, we show that A it is Lipschitzian with $L=36$. If $x,y\in[-2,\frac{1}{2}]$ then

$$|Ax-Ay|=|0-0|\leq36|x-y|.$$

If $x,y\in(\frac{1}{2},6]$ then

\begin{aligned} |Ax-Ay| =& 3\biggl|\biggl(x-\frac{1}{2}\biggr)^{2}- \biggl(y- \frac{1}{2}\biggr)^{2}\biggr| \\ =& 3\biggl| \biggl(\biggl(x-\frac{1}{2}\biggr)+ \biggl(y-\frac{1}{2} \biggr) \biggr) \biggl(\biggl(x-\frac{1}{2}\biggr)- \biggl(y- \frac{1}{2}\biggr) \biggr)\biggr| \\ =& 3|x+y-1||x-y|\leq36|x-y|. \end{aligned}

If $x\in[-2,\frac{1}{2}]$ and $y\in(\frac{1}{2},6]$ then

\begin{aligned} |Ax-Ay| =&\biggl|0-3\biggl(y-\frac{1}{2}\biggr)^{2}\biggr| =3\biggl(y- \frac{1}{2}\biggr)^{2} \\ =&3\biggl|\biggl(y-\frac{1}{2}\biggr)^{2}-\biggl(x-\frac{1}{2} \biggr)^{2}+\biggl(x-\frac{1}{2}\biggr)^{2}\biggr| \\ \leq &3|x+y-1||x-y|+(x-y)^{2} \\ = &3 \bigl[|x+y-1|+|x+y| \bigr]|x-y| \\ \leq&36|x-y|. \end{aligned}

Thus, we see that A is a Lipschtzian mapping with $L=36$. It is also clear that $F(T)\cap VI(C,A)= [0,1]\cap[-2,\frac {1}{2}]=[0,\frac{1}{2}]$.

Furthermore, if $x\in(0,6]$, the inequality

\begin{aligned} T_{r}x=\biggl\{ z\in C: \langle y-z,Tz\rangle-\frac{1}{r}\bigl\langle y-z,(1+r)z-x\bigr\rangle \leq0, \forall y\in C\biggr\} , \end{aligned}
(5.2)

shows that we may take $T_{r}(x)=x$. If $x\in[-2, 0]$, inequality (5.2) gives that

$$r(y-z) (-3z)-(y-z) \bigl[(1+r)z-x \bigr]\leq0,\quad \forall y\in C,$$

which implies that $T_{r}(x)=z=\frac{x}{4r+1}$ and hence we get

$$T_{r}(x):= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} x, & x\in(0,6], \\ \frac{x}{4r+1}, & x\in[-2,0]. \end{array}\displaystyle \right .$$

Now, if we take, $\alpha_{n}=\frac{1}{n+100}$, $a_{n}=b_{n}=\frac{1}{n+100}+0.1$, $c_{n}= 0.8-\frac{2}{n+100}$; $r_{n}=10$, $\forall n\geq1$ and $\gamma_{n}=0.01+\frac{1}{n+100}$, we observe that the conditions of Theorem 3.1 are satisfied and Scheme (3.1) reduces to

\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} z_{n}=P_{C}[x_{n}-\gamma_{n}Ax_{n}],\\ x_{n+1}=\alpha_{n} u+ (1-\alpha_{n})( a_{n}x_{n} +b_{n}T_{r_{n}} x_{n} + c_{n}P_{C}[x_{n}-\gamma_{n} Az_{n}]). \end{array}\displaystyle \right . \end{aligned}
(5.3)

When $u=-0.1$ and $x_{0}=0.8$ we see that Scheme (5.3) converges strongly to $x^{*}=0.0$ as shown in Figure 1.

When $u=0.6$ and $x_{0}=-2.0$ we see that Scheme (5.3) converges strongly to $x^{*}=0.5$ as shown in Figure 2.

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## Acknowledgements

This project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, under grant no. (5-130-36-RG). The authors, therefore, acknowledge with thanks DSR technical and financial support. The authors are grateful to the reviewers for their meticulous reading of the paper.

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### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

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