A Parallel-Viscosity-Type Subgradient Extragradient-Line Method for Finding the Common Solution of Variational Inequality Problems Applied to Image Restoration Problems

Abstract

In this paper, we study a modified viscosity type subgradient extragradient-line method with a parallel monotone hybrid algorithm for approximating a common solution of variational inequality problems. Under suitable conditions in Hilbert spaces, the strong convergence theorem of the proposed algorithm to such a common solution is proved. We then give numerical examples in both finite and infinite dimensional spaces to justify our main theorem. Finally, we can show that our proposed algorithm is flexible and has good quality for use with common types of blur effects in image recovery.

1. Introduction

Let H be a real Hilbert space with the inner product $\langle .,.\rangle$ and the induced norm $\parallel .\parallel$. Let C be a nonempty closed and convex subset of H. A mapping $f:C\to C$ is said to be a strict contraction if there exists $k\in [0,1)$ such that

$$\begin{array}{c}\hfill \parallel fx-fy\parallel \le k\parallel x-y\parallel ,\forall x,y\in C.\end{array}$$

(1)

A mapping $A:C\to H$ is said to be

• Monotone if

  $$\begin{array}{c}\hfill \langle Ax-Ay,x-y\rangle \ge 0\mathrm{for}\mathrm{all}x,y\in C;\end{array}$$

  (2)
• Pseudo-monotone if

  $$\begin{array}{c}\hfill \langle Ay,x-y\rangle \ge 0\Rightarrow \langle Ax,x-y\rangle \ge 0\mathrm{for}\mathrm{all}x,y\in C;\end{array}$$

  (3)
• L-Lipschitz continuous if there exists a positive constant L such that

  $$\begin{array}{c}\hfill \parallel Ax-Ay\parallel \le L\parallel x-y\parallel \mathrm{for}\mathrm{all}x,y\in C.\end{array}$$

  (4)

In this paper, we study the following variational inequality problem (VIP) for the operator A to find $x^{*},\in C$ such that

$$\begin{array}{c}\hfill \langle A{x}^{*},y-x^{*}\rangle \ge 0,\forall y\in C.\end{array}$$

(5)

The set of solutions of VIP (5) is denoted by VI(A,C). The VIP was introduced and studied by Hartman and Stampacchia in 1966 [1]. The variational inequality theory is an important tool based on studying a wide class of problems—unilateral and equilibrium problems arising in structural analysis, economics, optimization, operations research, and engineering sciences (see [2,3,4,5,6,7] and the references therein). Several algorithms have been improved for solving variational inequality and related optimization problems (see [6,8,9,10,11,12,13,14,15] and the references therein). It is well known that x is the solution of the VIP (5) if and only if x is the fixed point of the mapping $P_C(I-rA)$, $r>0$ (see [4] for details)

$$x=P_C(x-\gamma Ax),\gamma >0 \mathrm{and} r_{\gamma }(x):=x-P_C(x-\gamma Ax)=0.$$

Therefore, we can find the fixed point of the mapping $P_C(I-rA)$ replaces finding the solution of VIP (5) (see [7,9]). For solving VIP (5), the projection on closed and convex sets have been used. The gradient method is the simplest projection method, in which only one projection on the feasible set needs to be computed. However, strongly monotone or inverse strongly monotone operators have been required to obtain the convergence result. In 1976, Korpelevich [16] proposed another projection method called the extragradient method for finding a saddle point, and then it was extended to finding a solution of VIP for Lipschitz continuous and monotone (even pseudomonotone) mappings $A.$ The extragradient method is designed as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{y}_n=P_C(x_n-\lambda A(x_n)),\hfill \\ x_{n+1}=P_C(x_n-\lambda A(y_n)),\hfill \end{array}\right.\end{array}$$

(6)

where $P_C$ is the metric projection onto C and $\lambda$ is a suitable parameter. When the structure of C is simple, the extragradient method is computable and very useful because the projection onto C can be found easily. However, the computation of a projection onto a closed convex subset is generally difficult, and two distance optimization problems in the extragradient method are solved to obtain the next approximation $x_{n+1}$ over each iteration. This can be precious and seriously affect the efficiency of the used method. In 2011, the subgradient extragradient method was proposed in [17] for solving VIPs in Hilbert spaces. A projection onto a closed convex subset is reduced into one step and a special half-space is constructed for the projection in the second step. The method is generated as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{y}_n=P_C(x_n-\lambda A(x_n)),\hfill \\ x_{n+1}=P_{T_n}(x_n-\lambda A(y_n)),\hfill \end{array}\right.\end{array}$$

(7)

where $T_n$ is a half-space whose bounding hyperplane is supported on C at $y_n$, that is,

$$\begin{array}{c}\hfill T_n=\{\upsilon \in H:\langle (x_n-\lambda A(x_n))-y_n,\upsilon -y_n\rangle \}.\end{array}$$

The authors in [17] proved that two sequences $\left\{x_n\right\},\left\{y_n\right\}$ generated by (7) converge weakly to a solution of the VIP.

Recently, Gibali [18] proposed a new subgradient extragradient method by using adopting Armijo-like searches, called the self-adaptive subgradient extragradient method. Under the assumption of pseudomonotonicity and continuity of the operator, it has been proven that the convergence result for VIP (5) is ${\mathbb{R}}^n$. Gibali [9] remarked that the Armijo-like searches can be viewed as a local approximation of the Lipschitz constant of A.

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{x}_0\in {\mathbb{R}}^n,\hfill \\ y_n=P_C(x_n-\alpha A(x_n)),\alpha \in \{{\alpha }_{n-1},{\alpha }_{n-1}\beta ,{\alpha }_{n-1}{\beta }^2,...\},\hfill \\ (\alpha \mathrm{satisfies}\alpha \langle x_n-y_n,A(x_n)-A(y_n)\rangle \le (1-\epsilon ){\parallel x_n-y_n\parallel }^2),\hfill \\ x_{n+1}=P_{T_n}(x_n-{\alpha }_{n}A(y_n)),n\ge 1,\hfill \end{array}\right.\end{array}$$

(8)

where $T_n=\{w\in {\mathbb{R}}^n\mid \langle x_n-{\alpha }_{n}A(x_n)-y_n,w-y_n\rangle \le 0\}$, ${\alpha }_1\in (0,\mathsf{\infty })$, and $\epsilon ,\beta \in (0,1)$.

Very recently, solving the VIP (5) when A is a Lipschitz continuous monotone mapping such that the Lipschitz constant is unknown in Hilbert spaces by using the following viscosity-type subgradient extragradient-like method was proposed by Shehu and Iyiola [19].

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{y}_n=P_C(x_n-{\lambda }_{n}A{x}_n),{\lambda }_n={\rho }^{l_n},\hfill \\ (l_n\mathrm{is}\mathrm{the}\mathrm{smallest}\mathrm{non}-\mathrm{negative}\mathrm{integer}l\mathrm{such}\mathrm{that}{\lambda }_n\parallel A{x}_n-A{y}_n\parallel \le \mu \parallel r_{{\rho }^{l_n}}(x_n)\parallel ),\hfill \\ z_n=P_{T_n}(x_n-{\lambda }_{n}Ayn),\hfill \\ x_{n+1}={\alpha }_{n}f(x_n)+(1-{\alpha }_n)z_n,n\ge 1,\hfill \end{array}\right.\end{array}$$

(9)

where $T_n=\{z\in H:\langle x_n-{\lambda }_{n}A{x}_n-y_n,z-y_n\rangle \le 0\}$ with $\rho ,\mu \in (0,1)$ and $\left\{{\alpha }_n\right\}\subseteq (0,1)$. It was proved that the sequence $\left\{x_n\right\}$ generated by (9) converges strongly to $x^{*}\in$ VI(C,A), where $x^{*}=P_{VI(C,A)}f(x^{*})$ is the unique solution of the variational inequality

$$\begin{array}{c}\hfill \langle (I-f)x^{*},x-x^{*}\rangle \ge 0,\forall x\in VI(C,A),\end{array}$$

(10)

where $f:H\to H$ is a strict contraction mapping such that constant $k\in [0,1)$ under the following conditions:

$${\displaystyle (C_1)\underset{n\to \mathsf{\infty }}{lim}{\alpha }_n=0 \mathrm{and} (C_2)\sum _{n=1}^{\mathsf{\infty }}{\alpha }_n=\mathsf{\infty }.}$$

Our interest in this paper is to study the finding of common solutions of variational inequality problems (CSVIPs). The CSVIP is stated as follows: Let C be a nonempty closed and convex subset of H. Let $A_i:H\to H$, $i=1,2,...,N$ be mappings. The CSVIP is to find $x^{*}\in C$ such that

$$\begin{array}{c}\hfill \langle A_i{x}^{*},x-x^{*}\rangle \ge 0,\forall x\in C,i=1,2,...,N.\end{array}$$

(11)

If $N=1$, CSVIP (11) becomes VIP (5).

The CSVIP has received a great deal of attention due to its applications in a large variety of problems arising in structural analysis, convex feasibility problems, common fixed point problems, common minimizer problems, common saddle-point problems, and common variational inequality problems [20]. These problems have practical applicabilities in signal processing, network resource allocation, image processing, and many other fields [21,22]. Recently, many mathematicians have been widely studying this problem both theoretically and algorithmically; see [23,24,25,26,27] and the references therein.

Very recently, Anh and Hieu [28,29] proposed an important method for finding a common fixed point of a finite family of quasi ∅-nonexpansive mappings ${\left\{S_i\right\}}_{i=1}^N$ in Banach spaces, which they called the parallel monotone hybrid algorithm. This algorithm is related to Hilbert spaces as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{x}_0\in C,\hfill \\ y_n^i={\alpha }_n{x}_n+(1-{\alpha }_n)S_i{x}_n,i=1,...,N,\hfill \\ i_n=\mathrm{argmax}\{\parallel y_n^i-x_n\parallel :i=1,...,N\},{\overline{y}}_n:=y_n^{i_n},\hfill \\ C_{n+1}=\{\upsilon \in C_n:\parallel \upsilon -{\overline{y}}_n^i\parallel \le \parallel \upsilon -x_n\parallel \},\hfill \\ x_{x+1}=P_{C_{n+1}}{x}_0,\hfill \end{array}\right.\end{array}$$

(12)

where $0<{\alpha }_n<1$, ${\displaystyle \underset{n\to \mathsf{\infty }}{lim}sup{\alpha }_n<1}$. We see that a parallel algorithm is an algorithm that can execute several directions simultaneously on different processing devices and then combine all the individual outputs.

Inspired and encouraged by the previous results, in this paper we introduce a modified parallel method with a viscosity-type subgradient extragradient-like method for finding a common solution of variational inequality problems. Numerical experiments are also conducted to illustrate the efficiency of the proposed algorithms. Moreover, the problem of multiblur effects in an image is solved by applying our proposed algorithm.

2. Preliminaries

In order to prove our main result, we recall some basic definitions and lemmas needed for further investigation. In a Hilbert space H, let C be a nonempty closed and convex subset of H. For every point $x\in H$, there exists a unique nearest point of C, denoted by $P_{C}x$, such that $\parallel x-P_{C}x\parallel \le \parallel x-y\parallel$ for all $y\in C$. Such a $P_C$ is called the metric projection from H onto C.

Lemma 1

([30]). Let C be a nonempty closed and convex subset of a real Hilbert space H and let $P_C$ be the metric projection of H onto C. Let $x\in H$ and $z\in C$. Then, $z=P_{C}x$ if and only if

$$\langle x-z,y-z\rangle \le 0,\forall y\in C.$$

Lemma 2

([30]). The following statements hold in any real Hilbert space H:

$$\begin{array}{cc}\hfill {(i)\parallel x+y\parallel }^2& ={\parallel x\parallel }^2+2\langle x,y\rangle +{\parallel y\parallel }^2,\forall x,y\in H.\hfill \\ \hfill {(ii)\parallel x+y\parallel }^2& \le {\parallel x\parallel }^2+2\langle y,x+y\rangle ,\forall x,y\in H.\hfill \end{array}$$

Lemma 3

(Xu, [31]). Let ${\left\{a_n\right\}}_{n=0}^{\mathsf{\infty }}$ be a sequence of non-negative real numbers satisfying the following relation:

$$\begin{array}{c}\hfill a_{n+1}\le (1-{\alpha }_n)a_n+{\alpha }_n{\sigma }_n+{\gamma }_n,n\ge 1,\end{array}$$

where

(i) 

${\displaystyle {\left\{{\alpha }_n\right\}}_{n=0}^{\mathsf{\infty }}\subset [0,1],{\mathsf{\Sigma }}_{n=1}^{\mathsf{\infty }}{\alpha }_n=\mathsf{\infty };}$

(ii) 

${\displaystyle \underset{n\to \mathsf{\infty }}{lim\; sup}{\sigma }_n\le 0;}$

(iii) 

${\displaystyle {\gamma }_n\ge 0(n\ge 1),{\mathsf{\Sigma }}_{n=1}^{\mathsf{\infty }}{\gamma }_n<\mathsf{\infty }.}$

Then, $a_n\to 0$ as $n\to \mathsf{\infty }$.

Lemma 4

([8]). Let C be a nonempty closed and convex subset of a real Hilbert space H and $P_C:H\to C$ is the metric projection from H onto C. Then, the following inequality holds:

$${\parallel x-y\parallel }^2\ge {\parallel x-P_{C}x\parallel }^2+{\parallel y-P_{C}x\parallel }^2\forall x\in H,\forall y\in C.$$

(13)

Lemma 5

([19]). There exists a nonnegative integer $l_n$ satisfying (9).

Lemma 6

([32]). For each $x_1,x_2,...,x_m\in H$ and ${\alpha }_1,{\alpha }_2,...,{\alpha }_m\in [0,1]$ with ${\displaystyle \sum _{i=1}^m}{\alpha }_i=1$, we have

$${\parallel {\alpha }_1{x}_1+....+{\alpha }_m{x}_m\parallel }^2=\sum _{i=1}^m{\alpha }_i{\parallel x_i\parallel }^2-\sum _{1\le i<j\le m}{\alpha }_i{\alpha }_j{\parallel x_i-x_j\parallel }^2.$$

3. Main Results

In this section, we propose the parallel method with the viscosity-type subgradient extragradient-like method modified for solving common variational inequality problems. Let C be a nonempty closed and convex subset of a real Hilbert space H. Let $A_i:C\to H$ be a monotone mapping and $L_i$-Lipschitz continuous on H but $L_i$ is unknown for all $i=1,2,...,N$ such that ${\displaystyle F:=\bigcap _{i=1}^{N}VI(C,A_i)\ne \varnothing }$. Let $f:C\to C$ be a strict contraction mapping with constant $k\in (0,1]$. Suppose ${\left\{x_n\right\}}_{n=1}^{\mathsf{\infty }}$ is generated in the following Algorithm 1:

Algorithm 1. Given $\rho \in (0,1),\mu \in (0,1)$. Let ${\left\{{\alpha }_n\right\}}_{n=1}^{\mathsf{\infty }}$ be a real sequence in $(0,1)$. Let $x_1\in H$ be arbitrary.

Step 1:   Compute $y_n^i$ for all $i=1,2,...,N$ by $$y_n^i=P_C(x_n-{\lambda }_n^i{A}_i{x}_n),{\forall }_n\ge 1,$$ where ${\lambda }_n^i={\rho }^{l_n^i}$ and $l_n^i$ is the smallest non-negative integer $l^i$ such that $${\lambda }_n^i\Vert A_i{x}_n-A_i{y}_n^i\Vert \le \mu \Vert r_{{\rho }^{l^i}}(x_n)\Vert .$$ (14) Step 2:   Compute $$z_n^i=P_{T_n^i}(x_n-{\lambda }_n^i{A}_i{y}_n^i),$$ where $T_n^i:=\{z\in H:\langle x_n-{\lambda }_n^i{A}_i{x}_n-y_n^i,z-y_n^i\rangle \le 0\}.$ Step 3:   Compute $${\displaystyle x_{n+1}={\alpha }_n^{0}f(x_n)+\sum _{i=1}^N{\alpha }_n^i{z}_n^i,n\ge 1.}$$ (15) Set $n+1\to n$ and go to Step 1.

Theorem 1.

Assume that

(a) 

${\displaystyle \underset{n\to \mathsf{\infty }}{lim}{\alpha }_n^0=0}$, ${\displaystyle \sum _{n=1}^{\mathsf{\infty }}{\alpha }_n^0=\mathsf{\infty }}$

(b) 

${\displaystyle \underset{n\to \mathsf{\infty }}{lim\; inf}{\alpha }_n^i>0,{\forall }_i=1,2,...,N}$.

Then the sequence ${\left\{x_n\right\}}_{n=1}^{\mathsf{\infty }}$ generated by Algorithm 1 strongly converges to $x^{*}\in F$, where $x^{*}=P_{F}f(x^{*})$ is the unique solution of the variational inequality:

$$\langle (I-f)x^{*},x-x^{*}\rangle \ge 0,\forall x\in F.$$

(16)

Proof. 

Let $x^{*}\in F$ and $u_n^i=x_n-{\lambda }_n^i{A}_i{y}_n^i,\forall n\ge 1,i=1,2,..,N$

$$\begin{array}{ccc}\hfill {\Vert z_n^i-x^{*}\Vert }^2& =& \langle P_{T_n^i}(u_n^i)-x^{*},P_{T_n^i}(u_n^i)-x^{*}\rangle \hfill \\ & =& {\Vert P_{T_n^i}(u_n^i)-u_n^i\Vert }^2+2\langle P_{T_n^i}(u_n^i)-u_n^i,u_n^i-x^{*}\rangle +{\Vert u_n^i-x^{*}\Vert }^2\hfill \\ & =& {\Vert u_n^i-x^{*}\Vert }^2+{\Vert u_n^i-P_{T_n^i}(u_n^i)\Vert }^2+2\langle P_{T_n^i}(u_n^i)-u_n^i,u_n^i-x^{*}\rangle .\hfill \end{array}$$

(17)

By the characterization of the metric projection $P_{T_n}^i$ and $x^{*}\in F\subseteq C\subseteq T_n^i$, we get

$$2{\Vert u_n^i-P_{T_n^i}(u_n^i)\Vert }^2+2\langle P_{T_n^i}(u_n^i)-u_n^i,u_n^i-x^{*}\rangle$$

$$=2\langle u_n^i-P_{T_n^i}(u_n^i),x^{*}-P_{T_n^i}(u_n^i)\rangle \le 0.$$

(18)

This implies that

$${\Vert u_n^i-P_{T_n^i}(u_n^i)\Vert }^2+2\langle P_{T_n^i}(u_n^i)-u_n^i,u_n^i-x^{*}\rangle \le -{\Vert u_n^i-P_{T_n^i}(u_n^i)\Vert }^2.$$

(19)

We then obtain by the definition of Algorithm 1 that

$$\begin{array}{ccc}\hfill {\Vert z_n^i-x^{*}\Vert }^2& \le & {\Vert u_n^i-x^{*}\Vert }^2-{\Vert u_n^i-z_n^i\Vert }^2\hfill \\ & =& {\Vert (x_n-{\lambda }_n^i{A}_i{y}_n^i)-x^{*}\Vert }^2-{\Vert (x_n-{\lambda }_n^i{A}_i{y}_n^i)-z_n^i\Vert }^2\hfill \\ & =& {\Vert x_n-x^{*}\Vert }^2-{\Vert x_n-z_n^i\Vert }^2+2{\lambda }_n^i\langle -x_n+x^{*},A_i{y}_n^i\rangle \hfill \\ & & +2{\lambda }_n^i\langle x_n-z_n^i,A_i{y}_n^i\rangle \hfill \\ & =& {\Vert x_n-x^{*}\Vert }^2-{\Vert x_n-z_n^i\Vert }^2+2{\lambda }_n^i\langle x^{*}-z_n^i,A_i{y}_n^i\rangle .\hfill \end{array}$$

(20)

By the monotonicity of the operator $A_i$, we have

$$\begin{array}{cc}\hfill 0& \le \langle A_i{y}_n^i-A_i{x}^{*},y_n^i-x^{*}\rangle \hfill \\ & =\langle A_i{y}_n^i,y_n^i-x^{*}\rangle -\langle A_i{x}^{*},y_n^i-x^{*}\rangle \hfill \\ & \le \langle A_i{y}_n^i,y_n^i-x^{*}\rangle \hfill \\ & =\langle A_i{y}_n^i,y_n^i-z_n^i\rangle +\langle A_i{y}_n^i,z_n^i-x^{*}\rangle .\hfill \end{array}$$

Thus

$$\langle x^{*}-z_n^i,A_i{y}_n^i\rangle \le \langle A_i{y}_n^i,y_n^i-z_n^i\rangle .$$

(21)

Using (20) in (21), we obtain

$$\begin{array}{ccc}\hfill {\Vert z_n^i-x^{*}\Vert }^2& \le & {\Vert x_n-x^{*}\Vert }^2-{\Vert x_n-z_n^i\Vert }^2+2{\lambda }_n^i\langle A_i{y}_n^i,y_n^i-z_n^i\rangle \hfill \\ & =& {\Vert x_n-x^{*}\Vert }^2+2\langle {\lambda }_n^i{A}_i{y}_n^i,y_n^i-z_n^i\rangle -2\langle x_n-y_n^i,y_n^i-z_n^i\rangle \hfill \\ & & -{\Vert x_n-y_n^i\Vert }^2-{\Vert y_n^i-z_n^i\Vert }^2\hfill \\ & =& {\Vert x_n-x^{*}\Vert }^2+2\langle -{\lambda }_n^i{A}_i{y}_n^i+x_n-y_n^i,z_n^i-y_n^i\rangle \hfill \\ & & -{\Vert x_n-y_n^i\Vert }^2-{\Vert y_n^i-z_n^i\Vert }^2\hfill \\ & =& {\Vert x_n-x^{*}\Vert }^2+2\langle x_n-{\lambda }_n^i{A}_i{y}_n^i-y_n^i,z_n^i-y_n^i\rangle \hfill \\ & & -{\Vert x_n-y_n^i\Vert }^2-{\Vert y_n^i-z_n^i\Vert }^2.\hfill \end{array}$$

(22)

Observe that

$$\begin{array}{ccc}\hfill \langle x_n-{\lambda }_n^i{A}_i{y}_n^i-y_n^i,z_n^i-y_n^i\rangle & =& \langle x_n-{\lambda }_n^i{A}_i{x}_n-y_n^i,z_n^i-y_n^i\rangle +\langle {\lambda }_n^i{A}_i{x}_n-{\lambda }_n^i{A}_i{y}_n^i,z_n^i-y_n^i\rangle \hfill \\ & & \le \langle {\lambda }_n^i{A}_i{x}_n-{\lambda }_n^i{A}_i{y}_n^i,z_n^i-y_n^i\rangle .\hfill \end{array}$$

Using the last inequality in (22) and Lemma 5, we have

$$\begin{array}{ccc}\hfill {\Vert z_n^i-x^{*}\Vert }^2& \le & {\Vert x_n-x^{*}\Vert }^2+2\langle {\lambda }_n^i{A}_i{x}_n-{\lambda }_n^i{A}_i{y}_n^i,z_n^i-y_n^i\rangle -{\Vert x_n-y_n^i\Vert }^2-{\Vert y_n^i-z_n^i\Vert }^2\hfill \\ & \le & {\Vert x_n-x^{*}\Vert }^2+2{\lambda }_n^i\Vert A_i{x}_n-A_i{y}_n^i\Vert \Vert z_n^i-y_n^i\Vert -{\Vert x_n-y_n^i\Vert }^2-{\Vert y_n^i-z_n^i\Vert }^2\hfill \\ & \le & {\Vert x_n-x^{*}\Vert }^2+2\mu \Vert x_n-y_n^i\Vert \Vert z_n^i-y_n^i\Vert -{\Vert x_n-y_n^i\Vert }^2-{\Vert y_n^i-z_n^i\Vert }^2\hfill \\ & \le & {\Vert x_n-x^{*}\Vert }^2+\mu ({\Vert x_n-y_n^i\Vert }^2+{\Vert z_n^i-y_n^i\Vert }^2)-{\Vert x_n-y_n^i\Vert }^2-{\Vert y_n^i-z_n^i\Vert }^2\hfill \\ & =& {\Vert x_n-x^{*}\Vert }^2-(1-\mu ){\Vert x_n-y_n^i\Vert }^2-(1-\mu ){\Vert y_n^i-z_n^i\Vert }^2.\hfill \end{array}$$

(23)

It follows from (15) and (23) that

$${\displaystyle \begin{array}{ccc}\hfill \Vert x_{n+1}-x^{*}\Vert & =& \Vert {\alpha }_n^{0}f(x_n)+\sum _{i=1}^N{\alpha }_n^i{z}_n^i-x^{*}\Vert \hfill \\ & \le & {\alpha }_n^0\Vert f(x_n)-x^{*}\Vert +\sum _{i=1}^N{\alpha }_n^i\Vert z_n^i-x^{*}\Vert \hfill \\ & \le & {\alpha }_n^0\Vert f(x_n)-f(x^{*})\Vert +{\alpha }_n^0\Vert f(x^{*})-x^{*}\Vert +\sum _{i=1}^N{\alpha }_n^i\Vert z_n^i-x^{*}\Vert \hfill \\ & \le & {\alpha }_n^{0}k\Vert x_n-x^{*}\Vert +\sum _{i=1}^N{\alpha }_n^i\Vert x_n-x^{*}\Vert +{\alpha }_n^0\Vert f(x^{*})-x^{*}\Vert \hfill \\ & =& (1-{\alpha }_n^0(1-k))\Vert x_n-x^{*}\Vert +{\alpha }_n^0\Vert f(x^{*})-x^{*}\Vert \hfill \\ & =& (1-{\alpha }_n^0(1-k))\Vert x_n-x^{*}\Vert +{\alpha }_n^0(1-k)\frac{\Vert f(x^{*})-x^{*}\Vert }{1-k}\hfill \\ & \le & \max \{\Vert x_n-x^{*}\Vert ,\frac{\Vert f(x^{*})-x^{*}\Vert }{1-k}\}\hfill \\ & ⋮& \\ & \le & \max \{\Vert x_1-x^{*}\Vert ,\frac{\Vert f(x^{*})-x^{*}\Vert }{1-k}\}.\hfill \end{array}}$$

This implies that $\left\{x_n\right\}$ is bounded. Consequently, {$f(x_n)$}, {$y_n^i$}, and {$z_n^i$} are also bounded.

Let $z=P_{F}f(z)$. From (15) and (23), we have

$$\begin{array}{ccc}\hfill {\Vert x_{n+1}-z\Vert }^2& =& {\displaystyle {\Vert {\alpha }_n^{0}f(x_n)-\sum _{i=1}^N{\alpha }_n^i{z}_n^i-z\Vert }^2}\hfill \\ & =& {\displaystyle {\Vert {\alpha }_n^0(f(x_n)-z)+\sum _{i=1}^N{\alpha }_n^i(z_n^i-z)\Vert }^2}\hfill \\ & \le & {\displaystyle {\alpha }_n^0{\Vert f(x_n)-z\Vert }^2+\sum _{i=1}^N{\alpha }_n^i{\Vert z_n^i-z\Vert }^2}\hfill \\ & \le & {\displaystyle {\alpha }_n{\Vert f(x_n)-z\Vert }^2+\sum _{i=1}^N{\alpha }_n^i({\Vert x_n-z\Vert }^2}\hfill \\ & & -(1-\mu ){\Vert x_n-y_n^i\Vert }^2-(1-\mu ){\Vert y_n^i-z_n^i\Vert }^2)\hfill \\ & =& {\displaystyle \sum _{i=1}^N{\alpha }_n^i{\Vert x_n-z\Vert }^2+{\alpha }_n^0{\Vert f(x_n)-z\Vert }^2}\hfill \\ & & {\displaystyle -\sum _{i=1}^N{\alpha }_n^i(1-\mu ){\Vert x_n-y_n^i\Vert }^2-\sum _{i=1}^N{\alpha }_n^i(1-\mu ){\Vert y_n^i-z_n^i\Vert }^2}\hfill \\ & \le & {\displaystyle {\Vert x_n-z\Vert }^2+{\alpha }_n^0{\Vert f(x_n)-z\Vert }^2-\sum _{i=1}^N{\alpha }_n^i(1-\mu ){\Vert x_n-y_n^i\Vert }^2}\hfill \\ & & {\displaystyle -\sum _{i=1}^N{\alpha }_n^i(1-\mu ){\Vert y_n^i-z_n^i\Vert }^2.}\hfill \end{array}$$

(24)

Furthermore, using Lemma 2 (ii) in (15), we obtain

$$\begin{array}{ccc}\hfill {\Vert x_{n+1}-z\Vert }^2& =& {\displaystyle {\Vert {\alpha }_n^{0}f(x_n)+\sum _{i=1}^N{\alpha }_n^i{z}_n^i-z\Vert }^2}\hfill \\ & =& {\displaystyle {\Vert {\alpha }_n^0(f(x_n)-z)+\sum _{i=1}^N{\alpha }_n^i(z_n^i-z)\Vert }^2}\hfill \\ & =& {\displaystyle {\Vert \sum _{i=1}^N{\alpha }_n^i(z_n^i-z)+{\alpha }_n^0(f(x_n)-z)\Vert }^2}\hfill \\ & \le & {(1-{\alpha }_n^0)}^2{\Vert x_n-z\Vert }^2+2{\alpha }_n^0\langle f(x_n)-z,x_{n+1}-z\rangle \hfill \\ & =& {(1-{\alpha }_n^0)}^2{\Vert x_n-z\Vert }^2+2{\alpha }_n^0\langle f(x_n)-f(z),x_{n+1}-z\rangle \hfill \\ & & +2{\alpha }_n^0\langle f(z)-z,x_{n+1}-z\rangle \hfill \\ & \le & {(1-{\alpha }_n^0)}^2{\Vert x_n-z\Vert }^2+2{\alpha }_n^{0}k\Vert x_n-z\Vert \Vert x_{n+1}-z\Vert \hfill \\ & & +2{\alpha }_n^0\langle f(z)-z,x_{n+1}-z\rangle \hfill \\ & \le & {(1-{\alpha }_n^0)}^2{\Vert x_n-z\Vert }^2+{\alpha }_n^{0}k({\Vert x_n-z\Vert }^2+{\Vert x_{n+1}-z\Vert }^2)\hfill \\ & & +2{\alpha }_n^0\langle f(z)-z,x_{n+1}-z\rangle \hfill \end{array}$$

(25)

which implies that for some $M>0$,

$$\begin{array}{ccc}\hfill {\Vert x_{n+1}-z\Vert }^2& \le & \frac{{(1-{\alpha }_n^0)}^2+{\alpha }_n^{0}k}{1-{\alpha }_n^{0}k}{\Vert x_n-z\Vert }^2+\frac{2{\alpha }_n^0}{1-{\alpha }_n^{0}k}\langle f(z)-z,x_{n+1}-z\rangle \hfill \\ & =& \frac{1-2{\alpha }_n^0+{\alpha }_n^{0}k}{1-{\alpha }_n^{0}k}{\Vert x_n-z\Vert }^2+\frac{{({\alpha }_n^0)}^2}{1-{\alpha }_n^{0}k}{\Vert x_n-z\Vert }^2+\frac{2{\alpha }_n^0}{1-{\alpha }_n^{0}k}\langle f(z)-z,x_{n+1}-z\rangle \hfill \\ & \le & (1-\frac{2(1-k){\alpha }_n^0}{(1-{\alpha }_n^0)k}){\Vert x_n-z\Vert }^2+\frac{2(1-k){\alpha }_n^0}{(1-{\alpha }_n^0)k}\hfill \\ & & \times [\frac{{\alpha }_n^0}{2(1-k)}{\Vert x_n-z\Vert }^2+\frac{1}{1-k}\langle f(z)-z,x_{n+1}-z\rangle ]\hfill \\ & \le & (1-\frac{2(1-k){\alpha }_n^0}{1-{\alpha }_n^{0}k}){\Vert x_n-z\Vert }^2+\frac{2(1-k){\alpha }_n^0}{1-{\alpha }_n^{0}k}\hfill \\ & & \times [\frac{{\alpha }_n^0}{2(1-k)}M+\frac{1}{1-k}\langle f(z)-z,x_{n+1}-z\rangle ].\hfill \end{array}$$

(26)

We will divide the next proof into two parts.

Case 1 Suppose that there exists $n_0\in \mathbb{N}$ such that ${\{\Vert x_n-z\Vert \}}_{n=n_0}^{\mathsf{\infty }}$ is non-increasing. Then, ${\{\Vert x_n-z\Vert \}}_{n=1}^{\mathsf{\infty }}$ converges and ${\Vert x_n-z\Vert }^2-{\Vert x_{n+1}-z\Vert }^2\to 0,n\to \mathsf{\infty }$. From (24), we have

$${\displaystyle \sum _{i=1}^N{\alpha }_n^i(1-\mu ){\Vert x_n-y_n^i\Vert }^2\le {\Vert x_n-z\Vert }^2-{\Vert x_{n+1}-z\Vert }^2+{\alpha }_n^0{\Vert f(x_n)-z\Vert }^2.}$$

(27)

It follows from our assumptions (i) and (ii) that

$${\displaystyle \underset{n\to \mathsf{\infty }}{lim}\Vert x_n-y_n^i\Vert =0,\forall i=1,2,...,N.}$$

(28)

Similarly, from (24), we obtain that

$${\displaystyle \underset{n\to \mathsf{\infty }}{lim}\Vert y_n^i-z_n^i\Vert =0,\forall i=1,2,...,N.}$$

(29)

Set ${\displaystyle t_n={\alpha }_n^0{x}_n+\sum _{i=1}^N{\alpha }_n^i{z}_n^i}$ and ${\displaystyle s_n={\alpha }_n^0{x}_n+\sum _{i=1}^N{\alpha }_n^i{y}_n^i}$. It follows from our assumption (i) and (29) that

$${\displaystyle \underset{n\to \mathsf{\infty }}{lim}\Vert x_{n+1}-t_n\Vert =\underset{n\to \mathsf{\infty }}{lim}{\alpha }_n^0\Vert f(x_n)-x_n\Vert =0}$$

(30)

and

$${\displaystyle \underset{n\to \mathsf{\infty }}{lim}\Vert t_n-s_n\Vert \le \underset{n\to \mathsf{\infty }}{lim}\sum _{i=1}^N{\alpha }_n^i\Vert z_n^i-y_n^i\Vert =0.}$$

(31)

It follows from (28) that

$${\displaystyle \underset{n\to \mathsf{\infty }}{lim}\Vert s_n-x_n\Vert \le \underset{n\to \mathsf{\infty }}{lim}\sum _{i=1}^N{\alpha }_n^i\Vert y_n^i-x_n\Vert =0.}$$

(32)

It follows from (30)–(32) that

$$\Vert x_{n+1}-x_n\Vert \le \Vert x_{n+1}-t_n\Vert +\Vert t_n-s_n\Vert +\Vert s_n-x_n\Vert \to 0.$$

Since $\left\{x_n\right\}$ is bounded, it has a subsequence $\left\{x_{n_j}\right\}$ such that $\left\{x_{n_j}\right\}$ converges weakly to some $\omega \in H$ and ${\displaystyle \underset{n\to \mathsf{\infty }}{lim\; sup}\langle f(z)-z,x_n-z\rangle =\underset{j\to \mathsf{\infty }}{lim}\langle f(z)-z,x_{n_j}-z\rangle }$. We show that $\omega \in F$. Now, $x_n-y_n^i\to 0$ implies that $y_{n_j}^i\rightharpoonup \omega$ and since $y_n^i\in C$, we then obtain $\omega \in C$. For all $x\in C$ and using the property of the projection $P_C$, we have (since $A_i$ is monotone):

$$\begin{array}{ccc}\hfill 0& \le & \langle y_{n_j}^i-x_{n_j}-{\lambda }_{n_j}^i{A}_i{x}_{n_j},x-y_{n_j}^i\rangle \hfill \\ & =& \langle y_{n_j}^i-x_{n_j},x-y_{n_j}^i\rangle +{\lambda }_n^i\langle A_i{x}_{n_j},x_{n_j}-y_{n_j}^i\rangle +{\lambda }_{n_j}^i\langle A_i{x}_{n_j},x-x_{n_j}^i\rangle \hfill \\ & \le & \langle y_{n_j}^i-x_{n_j},x-y_{n_j}^i\rangle +{\lambda }_{n_j}^i\langle A_i{x}_{n_j},x_{n_j}-y_{n_j}^i\rangle +{\lambda }_{n_j}^i\langle A_i{x}_{n_j},x-x_{n_j}^i\rangle .\hfill \end{array}$$

Taking $j\to \mathsf{\infty }$, we get (recall that ${\displaystyle \underset{n\ge 1}{inf}{\lambda }_{n_j}>0}$ by Remark 3.2 in [19])

$$\langle A_i\omega ,x-\omega \rangle \ge 0,\forall x\in C.$$

This implies that $\omega \in F$. Since $z=P_{F}f(z)$, we have

$${\displaystyle \begin{array}{ccc}\hfill \underset{n\to \mathsf{\infty }}{lim\; sup}\langle f(z)-z,x_n-z\rangle & =& \underset{j\to \mathsf{\infty }}{lim}\langle f(z)-z,x_{n_j}-z\rangle \hfill \\ & =& \langle f(z)-z,w-z\rangle \hfill \\ & \le & 0.\hfill \end{array}}$$

Since ${\displaystyle \underset{n\to \mathsf{\infty }}{lim}\Vert x_{n+1}-x_n\Vert =0}$, we have

$${\displaystyle \underset{n\to \mathsf{\infty }}{lim\; sup}\langle f(z)-z,x_{n+1}-z\rangle \le 0.}$$

In (26), let $a_n:={\Vert x_n-z\Vert }^2,{\beta }_n:=\frac{2(1-k){\alpha }_n^0}{1-{\alpha }_n^{0}k}$, and ${\sigma }_n:=\frac{{\alpha }_n^0}{2(1-k)}M+\frac{1}{1-k}\langle f(z)-z,x_{n+1}-z\rangle$. Then, we can write (26) as

$$a_{n+1}\le (1-{\beta }_n)a_n+{\beta }_n{\sigma }_n.$$

(33)

It is easy to see that ${\displaystyle \underset{n\to \mathsf{\infty }}{lim}{\beta }_n=0,\sum _{n=1}^{\mathsf{\infty }}{\beta }_n=\mathsf{\infty }}$, and ${\displaystyle \underset{n\to \mathsf{\infty }}{lim\; sup}{\sigma }_n\le 0}$.

Using Lemma 3 in (33), we obtain ${\displaystyle \underset{n\to \mathsf{\infty }}{lim}\Vert x_n-z\Vert =0}$. Thus, $x_n\to z,n\to \mathsf{\infty }$.

Case 2 Assume that $\{\Vert x_n-z\Vert \}$ is not a monotone and decreasing sequence. Set $F_n={\Vert x_n-z\Vert }^2$ and let $\tau :\mathbb{N}\to \mathbb{N}$ be a mapping defined for all $n\ge n_0$ (for some $n_0$ large enough) by

$${\tau }_{(n)}:=\max \{k\in \mathbb{N}:k\le n,F_k\le F_{k+1}\}.$$

It is clear that $\tau$ is a non-decreasing sequence such that ${\tau }_{(n)}\to \mathsf{\infty }$ as $n\to \mathsf{\infty }$ and

$$0\le F_{\tau (n)}\le F_{\tau (n)+1},{\forall }_n\ge n_0.$$

This implies that $\Vert x_{\tau (n)}-z\Vert \le \Vert x_{\tau (n)+1}-z\Vert ,{\forall }_n\ge n_0$. Thus, ${\displaystyle \underset{n\to \mathsf{\infty }}{lim}\Vert x_{\tau (n)}-z\Vert }$ exists. By (27), we obtain

$${\displaystyle \begin{array}{ccc}\hfill \sum _{i=1}^N{\alpha }_{\tau (n)}^i(1-\mu ){\Vert x_{\tau (n)}-y_{\tau (n)}^i\Vert }^2& \le & {\Vert x_{\tau (n)}-z\Vert }^2-{\Vert x_{\tau (n)+1}-z\Vert }^2\hfill \\ & & +{\alpha }_{\tau (n)}^0{\Vert f(x_{\tau (n)})-z\Vert }^2\to 0,\hfill \end{array}}$$

as $n\to \mathsf{\infty }$. Thus,

$$\Vert x_{\tau (n)}-y_{\tau (n)}^i\Vert \to 0$$

as $n\to \mathsf{\infty }$. As in Case 1, we can prove that

$${\displaystyle \underset{n\to \mathsf{\infty }}{lim}\Vert y_{\tau (n)}^i-z_{\tau (n)}^i\Vert =\underset{n\to \mathsf{\infty }}{lim}\Vert x_{\tau (n)+1}-z_{\tau (n)}^i\Vert =\underset{n\to \mathsf{\infty }}{lim}\Vert x_{\tau (n)+1}-x_{\tau (n)}\Vert =0.}$$

Since $\left\{x_{\tau (n)}\right\}$ is bounded, there exists a subsequence of $\left\{x_{\tau (n)}\right\}$ that converges weakly to $\omega$. Without loss of generality, we assume that $x_{\tau (n)}\rightharpoonup \omega$. Observe that since ${\displaystyle \underset{n\to \mathsf{\infty }}{lim}\Vert x_{\tau (n)}-y_{\tau (n)}^i\Vert =0}$ we also have $y_{\tau (n)}^i\rightharpoonup \omega$. By similar argument in Case 1, we can show that $\omega \in F$ and

$${\displaystyle \underset{n\to \mathsf{\infty }}{lim\; sup}\langle f(z)-z,x_{\tau (n)}-z\rangle \le 0.}$$

Observe that since ${\displaystyle \underset{n\to \mathsf{\infty }}{lim}\Vert x_{\tau (n)+1}-x_{\tau (n)}\Vert =0}$ and ${\displaystyle \underset{n\to \mathsf{\infty }}{lim\; sup}\langle f(z)-z,x_{\tau (n)}-z\rangle \le 0}$, this implies that

$${\displaystyle \underset{n\to \mathsf{\infty }}{lim\; sup}\langle f(z)-z,x_{\tau (n)+1}-z\rangle \le 0.}$$

By (26), we obtain that

$$\begin{array}{ccc}\hfill {\Vert x_{\tau (n)+1}-z\Vert }^2& \le & (1-\frac{2(1-\tau ){\alpha }_{\tau (n)}^0}{1-{\alpha }_{\tau (n)}^{0}k}){\Vert x_{\tau (n)}-z\Vert }^2+\frac{2(1-k){\alpha }_{\tau (n)}^0}{1-{\alpha }_n^{0}k}\hfill \\ & & \times [\frac{{\alpha }_{\tau (n)}^0}{2(1-k)}M+\frac{1}{1-k}\langle f(z)-z,x_{\tau (n)+1}-z\rangle ]\hfill \\ & & =(1-{\beta }_{\tau (n)}){\Vert x_{\tau (n)}-z\Vert }^2+{\beta }_{\tau (n)}(\frac{{\alpha }_{\tau (n)}^0}{2(1-k)}M+\frac{1}{1-k}<f(z)-z,\hfill \\ & & x_{\tau (n)+1}-z>)\hfill \end{array}$$

where ${\beta }_{\tau (n)}:=\frac{2(1-k){\alpha }_{\tau (n)}^0}{1-{\alpha }_{\tau (n)}^{0}k}.$ Hence, we have (since $F_{\tau (n)}\le F_{\tau (n)+1}$)

$$\begin{array}{ccc}\hfill {\beta }_{\tau (n)}{\Vert x_{\tau (n)}-z\Vert }^2& \le & {\Vert x_{\tau (n)}-z\Vert }^2-{\Vert x_{\tau (n)+1}-z\Vert }^2\hfill \\ & & +{\beta }_{\tau (n)}(\frac{{\alpha }_{\tau (n)}^0}{2(1-k)}M+\frac{1}{1-k}\langle f(z)-z,x_{\tau (n)+1}-z\rangle )\hfill \\ & & \le {\beta }_{\tau (n)}(\frac{{\alpha }_{\tau (n)}^0}{2(1-k)}M+\frac{1}{1-k}\langle f(z)-z,x_{\tau (n)+1}-z\rangle ).\hfill \end{array}$$

Since ${\alpha }_{\tau (n)}^0>0$ and $k\in [0,1)$, we have that ${\beta }_{\tau (n)}>0$. So, we get

$${\Vert x_{\tau (n)}-z\Vert }^2\le \frac{{\alpha }_{\tau (n)}^0}{2(1-\tau )}M+\frac{1}{1-k}\langle f(z)-z,x_{\tau (n)+1}-z\rangle ,$$

and this implies that

$${\displaystyle \underset{n\to \mathsf{\infty }}{lim\; sup}{\Vert x_{\tau (n)}-z\Vert }^2\le \underset{n\to \mathsf{\infty }}{lim\; sup}\frac{{\alpha }_{\tau (n)}^0}{2(1-\tau )}M+\frac{1}{1-k}\langle f(z)-z,x_{\tau (n)+1}-z\rangle \le 0.}$$

Therefore,

$${\displaystyle \underset{n\to \mathsf{\infty }}{lim}\Vert x_{\tau (n)}-z\Vert =0}$$

and

$${\displaystyle \underset{n\to \mathsf{\infty }}{lim}\Vert x_{\tau (n)+1}-z\Vert =0.}$$

Hence,

$${\displaystyle \underset{n\to \mathsf{\infty }}{lim}{F}_{\tau (n)}=\underset{n\to \mathsf{\infty }}{lim}{F}_{\tau (n)+1}=0.}$$

Furthermore, for $n\ge n_0$, it is easy to see that $F_{\tau (n)}\le F_{\tau (n)+1}$ if $n\ne \tau (n)$ (that is $\tau (n)<n$), because $F_j\ge F_{j+1}$ for $\tau (n)+1\le j\le n$. As a consequence, we obtain for all $n\ge n_0$,

$$0\le F_n\le max\{F_{\tau (n)},F_{\tau (n)+1}\}=F_{\tau (n)+1.}$$

Hence, $lim{F}_n=0$, that is , ${\displaystyle \underset{n\to \mathsf{\infty }}{lim}\Vert x_n-z\Vert =0}$. Hence, $\left\{x_n\right\}$ converges strongly to z.

This completes the proof. □

We now give an example in Euclidian space ${\mathbb{R}}^3$ where $\parallel .\parallel$ is ${\ell }_2$-norm defined by $\parallel x\parallel =\sqrt{x_1^2+x_2^2+x_3^2}$ where $x=(x_1,x_2,x_3)$ to support the main theorem.

Example 1.

Let $A_1,A_2,A_3:{\mathbb{R}}^3\to {\mathbb{R}}^3$ be defined by $A_{1}x=4x$, $A_{2}x=7x+(5,-2,1)$ and $A_{3}x=\left(\begin{array}{ccc}10& -5& 5\\ -5& 10& -5\\ 5& -5& 10\end{array}\right)+(4,2,1)$ for all $x=(x_1,x_2,x_3)$. Let $f:{\mathbb{R}}^3\to {\mathbb{R}}^3$ be defined by $f(x)=\frac{x}{2}$ for all $x\in {\mathbb{R}}^3$. Let $C=\{x\in {\mathbb{R}}^3|\parallel x{\parallel }^2\le 4\}$. We can choose ${\alpha }_n^0=\frac{1}{{(n+1)}^{0.3}}$, ${\alpha }_n^1=\frac{1}{2n}$, ${\alpha }_n^2=\frac{n}{n+1}$ and ${\alpha }_n^3=1-{\alpha }_n^0-{\alpha }_n^1-{\alpha }_n^2$. The stopping criterion is defined by $\parallel x_n-x_{n-1}\parallel <{10}^{-15}$ (See in Figure 1, Figure 2 and Figure 3). The different choices of $x_1$ are given in

Table 1. Comparison of the number of iterations in Example 1.

Inputting	${\mathit{x}}_1=(-3,-5,8)$		${\mathit{x}}_1=(-1,7,6)$		${\mathit{x}}_1=(6.13,-5.24,-1.19)$
	CPU Time	Iter No.	CPU Time	Iter No.	CPU Time	Iter No.
$A_1$	0.0000068	592	0.0000056	591	0.00001	589
$A_2$	0.0003795	230	0.0002848	230	0.0002887	229
$A_3$	0.0004619	230	0.0007852	230	0.000766	229
$A_1,A_2$	0.0002942	231	0.0002965	231	0.0002945	231
$A_1,A_3$	0.0008444	231	0.0009953	231	0.0009992	231
$A_2,A_3$	0.0011516	230	0.0009781	230	0.0007956	229
$A_1,A_2,A_3$	0.0007429	231	0.0007586	231	0.0007621	231

as follows in Example 1.

For infinitely dimensional space, we give an example in function space $L_2[0,1]$ such that $\parallel .\parallel$ is $L_2$-norm defined by $\parallel x(t)\parallel =\sqrt{{\int }_0^1{\left|x(t)\right|}^{2}dt}$ where $x(t)\in L_2[0,1]$.

Example 2.

Let $A_1,A_2,A_3:L_2[0,1]\to L_2[0,1]$ be defined by $A_{1}x(t)={\int }_0^{t}4x(s)ds$, $A_{2}x(t)={\int }_0^{t}tx(s)ds$ and $A_{3}x(t)={\int }_0^t(t^2-1)x(s)ds$ where $x(t)\in L_2[0,1]$. Let $f:L_2[0,1]\to L_2[0,1]$ be defined by $f(x(t))=\frac{x(t)}{2}$ where $x(t)\in L_2[0,1]$. Let $C=\{x(t)\in L_2[0,1]:{\int }_0^1(t^2+1)x(t)dt\}$. We can choose ${\alpha }_n^0=\frac{1}{{(n+1)}^{0.3}}$, ${\alpha }_n^1=\frac{1}{2n}$, ${\alpha }_n^2=\frac{n}{n+1}$ and ${\alpha }_n^3=1-{\alpha }_n^0-{\alpha }_n^1-{\alpha }_n^2$. The stopping criterion is defined by $\parallel x_n(t)-x_{n-1}(t)\parallel <{10}^{-5}$ (See in Figure 4, Figure 5 and Figure 6).

The different choices of $x_1(t)$ are given in

Table 2. Comparison of the number of iterations in Example 2.

Inputting	Bernstein Initial Data		Chebyshev Initial Data		Legendre Initial Data
	CPU Time	Iter. No.	CPU Time	Iter. No.	CPU Time	Iter. No.
$A_1$	2.20542	40	4.53568	40	2.66656	33
$A_2$	2.93440	35	1.53655	39	1.46195	33
$A_3$	2.699356	28	2.13809	38	1.38359	32
$A_1,A_2$	20.5162	36	33.9656	39	20.3007	33
$A_1,A_3$	11.0109	29	77.1907	38	44.6389	32
$A_2,A_3$	7.47927	28	52.5607	38	30.7733	32
$A_1,A_2,A_3$	6.20955	28	82.3549	38	45.8789	32

as follows:

Choice 1 

Bernstein initial data: $x_1(t)=-120t^7{(t-1)}^3$;

Choice 2 

Chebyshev initial data: $x_1(t)=64t^7-112t^5+56t^3-7t$;

Choice 3 

Legendre initial data: $x_1(t)=\frac{315}{128}t-\frac{1155}{32}{t}^3+\frac{9009}{64}{t}^5-\frac{6435}{32}{t}^7+\frac{12155}{128}{t}^9$.

From Table 1 and Table 2, we see that the advantage of the parallel viscosity type subgradient extragradient-line method Algorithm 1 when the common solution of two or more inputting $A_i$ gives the number of iterations smaller than one inputting.

4. Application to Image Restoration Problems

The image restoration problem can be modeled in one-dimensional vectors by the following linear equation system:

$$\begin{array}{c}\hfill b=Ax+\upsilon ,\end{array}$$

(34)

where $x\in {\mathbb{R}}^{n\times 1}$ is an original image, $b\in {\mathbb{R}}^{m\times 1}$ is the observed image, $\upsilon$ is additive noise, and $A\in {\mathbb{R}}^{m\times n}$ is the blurring matrix. For solving problem (34), we aim to approximate the original image, vector x, by minimizing the additive noise, which is called a least squares (LS) problem as follows:

$$\begin{array}{c}\hfill {\displaystyle \underset{x}{min}\frac{1}{2}{\parallel b-Ax\parallel }^2,}\end{array}$$

(35)

where $\parallel .\parallel$ is ${\ell }_2$-norm defined by $\parallel x\parallel =\sqrt{{\sum }_{i=1}^n{\left|x_i\right|}^2}$. The solution of the problem (35) can be approximated by many well-known iteration methods.

The Richardson iteration, which is often called the Landweber method [33,34,35,36], is generally used as an iterative regularization method to solve (35). The basic iteration takes the form:

$$\begin{array}{c}\hfill x_{n+1}=x_n+\tau A^T(b-A{x}_n).\end{array}$$

(36)

Here the step size $\tau$ remains constant for each iteration. The convergence can be proved under the step size $\tau$ such that $0<\tau <\frac{2}{{\sigma }_{max}^2}$ where ${\sigma }_{max}$ is the largest singular value of A.

The goal in image restoration is to deblur an image without knowing which one is the blurring operator. Thus, we focus on the following problem:

$$\begin{array}{c}\hfill {\displaystyle \underset{x\in {\mathbb{R}}^n}{min}\frac{1}{2}{\parallel A_{1}x-b_1\parallel }^2,\underset{x\in {\mathbb{R}}^n}{min}\frac{1}{2}{\parallel A_{2}x-b_2\parallel }^2,...,\underset{x\in {\mathbb{R}}^n}{min}\frac{1}{2}{\parallel A_{N}x-b_N\parallel }^2}\end{array}$$

(37)

where x is the original true image, $A_i$ is the blurred matrix, $b_i$ is the blurred image by the blurred matrix $A_i$ for all $i=1,2,...,N$. For solving this problem, we designed the following flowchart:

Where $\tilde{X}$ is the deblurred image or the common solutions of the problem (37) and as seen in Figure 7. We can apply the algorithm in Theorem 1 to solve the problem (37), and as a result, we know that $A_i^T(A_{i}x-b_i)$ is Lipschitz continuous for each $i=1,2,...,N$. Let $f:{\mathbb{R}}^n\to {\mathbb{R}}^n$ be a strict contraction mapping with constant $k\in (0,1]$. Suppose ${\left\{x_n\right\}}_{n=1}^{\mathsf{\infty }}$ is generated in the following Algorithm 2:

Algorithm 2. Given $\rho \in (0,1),\mu \in (0,1)$. Let ${\left\{{\alpha }_n\right\}}_{n=1}^{\mathsf{\infty }}$ be a real sequence in $(0,1)$. Let $x_1\in H$ be arbitrary.

Step 1:   Compute for all $i=1,2,...,N$ by $$y_n^i=P_C(x_n-{\lambda }_n^i{A}_i^T(A_i{x}_n-b_i)),{\forall }_n\ge 1,$$ where ${\lambda }_n^i={\rho }^{l_n^i}$ and $l_n^i$ is the smallest nonnegative integer $l^i$ such that $${\lambda }_n^i\Vert A_i{x}_n-A_i{y}_n^i\Vert \le \mu \Vert r_{{\rho }^{l^i}}(x_n)\Vert .$$ (38) Step 2:   Compute $$z_n^i=P_{T_n^i}(x_n-{\lambda }_n^i{A}_i^T(A_i{y}_n^i-b_i)),$$ where $T_n^i:=\{z\in H:\langle x_n-{\lambda }_n^i{A}_i{x}_n-y_n^i,z-y_n^i\rangle \le 0\}.$ Step 3:   Compute $${\displaystyle x_{n+1}={\alpha }_n^{0}f(x_n)+\sum _{i=1}^N{\alpha }_n^i{z}_n^i,n\ge 1.}$$ (39) Set $n+1\to n$ and go to Step 1.

We will present restoration of images corrupted by the following blur types:

(1)

Gaussian blur of filter size $9\times 9$ with standard deviation $\sigma =4$ (the original image was degraded by the blurring matrix $A_1$).

(2)

Out-of-focus blur (disk) with radius $r=6$ (the original image was degraded by the blurring matrix $A_2$).

(3)

Motion blur specifying with motion length of 21 pixels (len $=21$) and motion orientation ${11}^{\circ }$ ($\theta =11$) (the original image was degraded by the blurring matrix $A_3$).

The performance of the studied proposed Algorithm 2 with the following original grey and RGB images was tested, as can be seen in Figure 8 and Figure 9.

The parameter ${\alpha }_n^i$ on the implemented algorithm for solving the problem (VIP) was set as

$$\begin{array}{c}\hfill {\alpha }_n^i=\frac{n}{n+1},i=1,2,3.\end{array}$$

Three different types of blurred grey and RGB images degraded by the blurring matrices $A_1,A_2$ and $A_3$ are shown in Figure 10, Figure 11, Figure 12, Figure 13, Figure 14 and Figure 15.

We applied the proposed algorithm to obtain the solution of the deblurring problem (VIP) with ($N=1$) by inputting $A_1$, $A_2$, and $A_3$. The results of the proposed algorithm with 10,000 iterations for the following three cases:

Case I:

Inputting $A_1$ in the proposed algorithm;

Case II:

Inputting $A_2$ in the proposed algorithm; and

Case III:

Inputting $A_3$ in the proposed algorithm

are shown in Figure 16, Figure 17, Figure 18, Figure 19, Figure 20 and Figure 21 that are composed of the restored images and their peak signal-to-noise ratios (PSNRs).

Next found the common solutions of a deblurring problem (VIP) with ($N=2$) by using the proposed algorithm. So, we can consider the results of the proposed algorithm with 10,000 iterations in the following three cases:

Case I:

Inputting $A_1$ and $A_2$ in the proposed algorithm;

Case II:

Inputting $A_1$ and $A_3$ in the proposed algorithm; and

Case III:

Inputting $A_2$ and $A_3$ in the proposed algorithm.

It can be seen from Figure 22, Figure 23, Figure 24, Figure 25, Figure 26 and Figure 27 that the quality of restored images by using the proposed algorithm in solving the common solutions of the deblurring problem (VIP) with ($N=2$) was improved compared with the previous results in Figure 16, Figure 17, Figure 18, Figure 19, Figure 20 and Figure 21 .

Finally, the common solution of the deblurring problem (VIP) with ($N=3$) using the proposed algorithm was also tested (inputting $A_1$, $A_2$, and $A_3$ in the proposed algorithm).

Figure 28 and Figure 29 show the reconstructed grey and RGB images with $10,000$ iterations. The quality of the recovered grey and RGB images obtained by this algorithm were the highest compared to the previous two algorithms.

Moreover, the Cauchy error, the figure error, and the peak signal-to-noise ratio (PSNR) for recovering the degraded grey and RGB images by using the proposed method within the first 10,000 iterations are shown in Figure 30, Figure 31, Figure 32, Figure 33, Figure 34 and Figure 35.

The Cauchy error is defined as $\parallel x_n-x_{n-1}\parallel <{10}^{-8}$. The figure error is defined as $\parallel x_n-x\parallel$, where x is the solution of the problem (VIP). The performance of the proposed algorithm at $x_n$ in the image restoration process was measured quantitatively by the means of the peak signal-to-noise ratio (PSNR), which is defined by

$$\begin{array}{c}\hfill \mathrm{PSNR}(x_n)=20{\log }_{10}(\frac{{255}^2}{MSE}),\end{array}$$

where $\mathrm{MSE}={\parallel x_n-x\parallel }^2$, $\parallel x_n-x\parallel$ is the ${\ell }_2$-norm of vec($x_n-x$) and vec($x_n-x$) = A reshape matrix $x_n-x$ as vector.

The Cauchy error plot shows the validity of the proposed method, while the figure error plot confirms the convergence of the proposed method and the PSNR quality plot shows the measured quantitatively of the image. From Figure 30, Figure 31, Figure 32, Figure 33, Figure 34 and Figure 35, it is clearly seen that the common solution of the deblurring problem (VIP) with $(N\ge 2)$ obtained quality improvements in the reconstructed grey and RGB images. Another advantage of the proposed method when the common solution of two or more image deblurring problems was used to restore the image is that the received image is more consistent than usual (see Figure 36, Figure 37, Figure 38, Figure 39, Figure 40, Figure 41, Figure 42, Figure 43, Figure 44, Figure 45, Figure 46, Figure 47, Figure 48 and Figure 49). Figure 36, Figure 37, Figure 38, Figure 39, Figure 40, Figure 41, Figure 42, Figure 43, Figure 44, Figure 45, Figure 46, Figure 47, Figure 48 and Figure 49 show the reconstructed grey and RGB images using the proposed algorithm in obtaining the common solution of the following problem with the same PSNR.

(1)

Deblurring problem (VIP) with $(N=1)$ by inputting $A_1$, $A_2$, and $A_3$ in the proposed algorithm.

(2)

Deblurring problem (VIP) with $(N=2)$ by inputting $A_1$ and $A_2$, $A_1$ and $A_3$, and $A_2$ and $A_3$ in the proposed algorithm respectively.

(3)

Deblurring problem (VIP) with $(N=3)$ by inputting $A_1$, $A_2$, and $A_3$ in the proposed algorithm.

5. Conclusions

In this work, we considered the problem of finding a common solution of variational inequalities with monotonic and Lipschitz operators in a Hilbert space. Under some suitable conditions imposed on the parameters, we proved the strong convergence of the algorithm. Several numerical examples in both finite and infinite dimensional spaces were performed to illustrate the performance of the proposed algorithm (see Table 1 and Table 2 and Figure 1, Figure 2, Figure 3, Figure 4, Figure 5 and Figure 6). We applied our proposed algorithm to image recovery (2) under a situation without knowing the type of matrix blurs to demonstrate the computational performance (see Figure 10, Figure 11, Figure 12, Figure 13, Figure 14, Figure 15, Figure 16, Figure 17, Figure 18, Figure 19, Figure 20, Figure 21, Figure 22, Figure 23, Figure 24, Figure 25, Figure 26, Figure 27, Figure 28, Figure 29, Figure 30, Figure 31, Figure 32, Figure 33, Figure 34 and Figure 35). We found that the advantage of our proposed algorithm was its ability to restore two or more multiblur effects in an image, giving a restoration performance better than one (see Figure 36, Figure 37, Figure 38, Figure 39, Figure 40, Figure 41, Figure 42, Figure 43, Figure 44, Figure 45, Figure 46, Figure 47, Figure 48 and Figure 49).