A Modified Tseng’s Method for Solving the Modified Variational Inclusion Problems and Its Applications

Abstract

The goal of this study was to show how a modified variational inclusion problem can be solved based on Tseng’s method. In this study, we propose a modified Tseng’s method and increase the reliability of the proposed method. This method is to modify the relaxed inertial Tseng’s method by using certain conditions and the parallel technique. We also prove a weak convergence theorem under appropriate assumptions and some symmetry properties and then provide numerical experiments to demonstrate the convergence behavior of the proposed method. Moreover, the proposed method is used for image restoration technology, which takes a corrupt/noisy image and estimates the clean, original image. Finally, we show the signal-to-noise ratio (SNR) to guarantee image quality.

1. Introduction

As technology advances, several new and challenging real-world problems develop. These problems appear in a variety of fields, such as chemistry, engineering, biology, physics, and computer science. The problems can be formulated as optimization problems, i.e., to find x such that the following holds:

$$\underset{x\in {\mathbb{R}}^n}{min}f\left(x\right)$$

(1)

where $f:{\mathbb{R}}^n\to \mathbb{R}$ is continuously differentiable. In solving these problems, optimization and control methodologies and techniques are certainly required. Since symmetry appears in certain natural and engineering systems, there is some form of symmetry in many mathematical models and optimization problems. Therefore, researchers have to pay attention to adjusting certain constrained optimization problems, where certain variables appear symmetrically in the objective and constraint functions. One of the basic optimization concepts for finding minimization is the variational inclusion problem (VIP), which is to find x in a real Hilbert space H, such that the following holds:

$$0\in Ax+Bx,$$

(2)

where operators $A:H\to H$ and $B:H\to 2^H$ are, respectively, single-valued and multi-valued. The solution of the VIP can apply to real-world problems, such as engineering, economics, machine learning, equilibrium, image processing, and transportation problems [1,2,3,4,5,6,7]. An increasing number of researchers have investigated methods to solve the variational inclusion problem. A popular one is the forward–backward splitting method (see [8,9,10]), given by the following:

$$u_{n+1}=J_r^B(u_n-rA{u}_n),n\ge 1,$$

where $J_r^B={(I+rB)}^{-1}$ with $r>0$. Moreover, researchers have modified these methods not only with more versatility by using relaxation techniques (see [11,12]), but also with more acceleration by using the inertial techniques (see [13,14,15,16]). Later, Alvarez and Attouch [14,15] expanded the inertial idea and introduced the inertial forward–backward method given by the following:

$$\left.\begin{array}{c}\\ \end{array}\right\{\begin{array}{c}{s}_n=u_n+{\xi }_n(u_n-u_{n-1})\\ u_{n+1}={(I+rB)}^{-1}(I-rA)s_n,n\ge 1.\end{array}$$

The technique of speeding up this method consists in the term ${\xi }_n(u_n-u_{n-1})$. The inertial term with an extrapolation factor ${\xi }_n$ is well known (for more details, see [17,18,19,20]). For monotone inclusions/non-smooth convex minimization problems, they also proved the convergence theorem. Attouch and Cabot [21,22] established relaxed inertial proximal algorithms by using both techniques to increase the performance of algorithms to solve the previous problems. Later, many researchers focused on both techniques and introduced the relaxed inertial forward–backward algorithm [23], the inertial Douglas–Rachford algorithm [24], a Tseng extragradient method in Banach spaces [25], and the relaxed inertial Tseng’s type algorithm [26]. Among the well-known algorithms is the relaxed inertial Tseng’s type method developed by Abubakar et al. [26]. They solved the problem (2) by modifying Tseng’s forward–backward forward splitting approach by employing the relaxation parameter $\rho$ and the inertial extrapolation factor $\xi$. Their algorithm is given by the following:

$$\left.\begin{array}{c}\\ \\ \end{array}\right\{\begin{array}{c}{t}_n=u_n+\xi (u_n-u_{n-1})\\ s_n={(I+{\lambda }_{n}B)}^{-1}(I-{\lambda }_{n}A)t_n,n\ge 1\\ u_{n+1}=(1-\rho )t_n+\rho s_n+\rho {\lambda }_n(A{t}_n-A{s}_n),n\ge 1.\end{array}$$

A new simple step size rule was designed to self-adaptively update the step size $\lambda$. Furthermore, they proved a weak convergence theorem generated by their algorithm and applied the algorithm to solve the problem of image deblurring.

In 2014, Khuangsatung and Kangtunyakarn [27,28] presented the modified variational inclusion problem (MVIP), that is, to find $u\in H$ such that the following holds:

$$0\in \sum _{i=1}^N{a}_i{A}_{i}u+Bu,$$

(3)

where, for every $i=1,2,\dots ,N,a_i\in (0,1)$ with ${\sum }_{i=1}^N{a}_i=1$, operator $A_i:H\to H$ and $B:H\to 2^H$. Obviously, if $A_i\equiv A$ for every $i=1,2,\dots ,N$, then reduce from (3) to (2), and the iterative methods are used for solving a finite family of nonexpansive mappings of fixed point T and a finite family of variational inclusion problems in Hilbert spaces under the condition ${\sum }_{i=1}^n{a}_i={\sum }_{i=1}^n{\delta }_i=1.$ Their method is given by the following:   

$$\left.\begin{array}{c}\\ \end{array}\right\{\begin{array}{c}{z}_n^i=b_n{u}_n+(1-b_n)T_i{u}_n,\forall n\ge 1\\ u_{n+1}={\alpha }_{n}f\left(x_n\right)+{\beta }_n{\gamma }_n{J}_{M,\lambda }(I-\lambda \sum _{i=1}^N{\delta }_i{A}_i)u_n+{\gamma }_n\sum _{i=1}^N{a}_i{z}_n^i.\end{array}$$

Moreover, their iterative method provided a strong convergence theorem, which they proved. Apart from that, there is another technique that can reduce the overall computational effort under widely used conditions. That is the parallel technique (for more details, see [29,30,31]). Cholamjiak et al. recently proposed an inertial parallel monotone hybrid method that solves the common variational inclusion problems. Their method is given by the following:

$$\left.\begin{array}{c}\\ \\ \\ \\ \\ \end{array}\right\{\begin{array}{c}{s}_n=u_n+{\xi }_n(u_n-u_{n-1}),\\ z_n^i=(1-{\alpha }_n^i)s_n+{\alpha }_n^i{J}_{r_n}^B(I-r_n{A}_i)s_n,\\ i=argmax\{\parallel z_n^i-u_n\parallel :i=1,2,\dots ,N\},{\overline{z}}_n=z_n^i,\\ C_{n+1}=\{w\in C_n:\\ \parallel {\overline{z}}_n{-w\parallel }^2\le \parallel u_n{-w\parallel }^2+{\theta }_n^2\parallel u_n-u_{n-1}{\parallel }^2-2{\theta }_n\langle u_n-w,u_{n-1}-u_n\rangle \},\\ u_{n+1}=P_{C_{n+1}}{u}_1,n\ge 1,\end{array}$$

The idea of the parallel technique is to find the farthest element ${\overline{z}}_n=z_n^i$ from the previous approximation $u_n.$ They demonstrated that the method yielded strong convergence results and applied their results to solve image restoration problems.

Inspired by the ideas of [26,28,31], we provide a new method for determining the solution set of the modified variational inclusion problem in Hilbert spaces. For obtaining high performance, our method modifies the relaxed inertial Tseng’s type method by using the condition ${\sum }_{i=1}^n{\delta }_i=1$ and the parallel technique, which finds the farthest element $\overline{A}$. Furthermore, under appropriate conditions, we prove a weak convergence theorem and present numerical experiments to demonstrate convergence behavior. Finally, the image restoration problems are solved by our result.

The substance of this work is structured as follows: in Section 2, we gather basic definitions and lemmas. In Section 3, the proposed algorithms are explained. In Section 4, the numerical experiments are discussed. In the last section, the conclusion of this work is presented.

2. Preliminaries

Throughout the paper, we suppose that H is a real Hilbert space with norm $\parallel ·\parallel$ and the inner product $\langle ·,·\rangle$ and that C is a nonempty closed and convex subset of $H.$ In this section, we go over several fundamental concepts and lemmas that are utilized in the main result section.

Lemma 1.

[32] The following statements are accurate:

(i) 

for every ${\alpha ,\beta \in H,\parallel \alpha -\beta \parallel }^2\le {\parallel \alpha \parallel }^2-{\parallel \beta \parallel }^2-2\langle \alpha -\beta ,\beta \rangle$;

(ii) 

for every ${\alpha ,\beta \in H,\parallel \alpha +\beta \parallel }^2\le {\parallel \alpha \parallel }^2+2\langle \beta ,\alpha +\beta \rangle$;

(iii) 

for every $\gamma , \lambda \in [0,1]$ with $\gamma +\lambda ={1,\parallel \gamma \alpha +\lambda \beta \parallel }^2={\gamma \parallel \alpha \parallel }^2+{\lambda \parallel \lambda \parallel }^2-\gamma \lambda {\parallel \alpha -\beta \parallel }^2.$

Definition 1.

Suppose that $A:H\to H$. For any $x,y\in C$,

A is a monotone mapping if the following holds:

$$\langle Ax-Ay,x-y\rangle \ge 0;$$

A is L-Lipschitz continuous if there is a constant $L>0,$ as follows:

$$\parallel Ax-Ay\parallel \le L\parallel x-y\parallel .$$

We mention a mapping $B:H\to 2^H$. If $u\in Bx$ and $v\in By$ for every $x,y\in H$, then $\langle u-v,x-y\rangle \ge 0.$ If $B:H\to 2^H$ is monotone and every $(y,v)\in Graph\left(B\right)$ implies that $u\in Bx$ for every $(x,u)\in H,\langle u-v,x-y\rangle \ge 0$, it is called the maximal monotone.

Definition 2.

The resolvent operator associated with B is referred to as the following:

$$J_{\lambda }^B={(I+\lambda B)}^{-1}\left(u\right),\forall u\in H,$$

where $B:H\to 2^H$ is the maximal monotone, and I and λ are an identity mapping and a positive number, respectively.

Lemma 2.

[9] Let $A:H\to H$ be a monotone Lipschitz continuous mapping, and $B:H\to 2^H$ be a maximal monotone operator; then, $A+B$ is the maximal monotone.

Lemma 3.

[33] Let Γ be a non empty subset of H, and $\left\{x_n\right\}$ a sequence of elements of $H.$ Assume the following:

(i) 

For every $x\in C,{lim}_{n\to +\infty }\parallel x_n-x\parallel$ exists;

(ii) 

Every weak sequential limit point of $\left\{x_n\right\}$, as $n\to +\infty$, belongs to $C.$

Then $\left\{x_n\right\}$ converges weakly as $n\to +\infty$ to a point in $C.$

Lemma 4.

[34] Suppose that $\left\{{\zeta }_n\right\},\left\{{\nu }_n\right\}$ and $\left\{{\eta }_n\right\}$ are sequences in $[0,\infty )$. For every $n\ge 1,$ there is $\eta \in \mathbb{R}$ with $0\le {\eta }_n\le \eta \le 1$ and

$${\zeta }_{n+1}\le {\zeta }_n+{\eta }_n({\zeta }_n-{\zeta }_{n-1})+{\nu }_n,\sum {\nu }_n<\infty .$$

Then the following conditions are satisfied:

(i) 

$\sum {[{\zeta }_n-{\zeta }_{n-1}]}_{+}<\infty ,$ where ${\left[k\right]}_{+}=max\{k,0\}$;

(ii) 

there is $\overline{\zeta }\in [0,\infty )$ with $lim{\zeta }_n=\overline{\zeta }.$

3. Main Results

Two algorithms are presented in this section. First, we propose Algorithm 1, which is known as a modified Tseng method for solving the modified variational inclusion problem. After that, we prove a weak convergence theorem by using some symmetry properties. Assume the following assumptions to be correct for the purposes of the investigation.

Assumption 1.

The feasible set of the VIP is a nonempty closed and convex subset of $H.$

Assumption 2.

The solution set Γ of the MVIP is nonempty.

Assumption 3.

$A_i:H\to H$ is monotone and $L_i$-Lipschitz continuous on H for every i, and $B:H\to 2^H$ is the maximal monotone.

Algorithm 1 (Modified Tseng’s method for solving the MVIP)

Pick $u_0,u_1\in H,\mu \in (0,1),\rho >0,\xi \ge 0,{\lambda }_0>0.$ Iterative steps: Given iterates $u_{n-1}$ and $u_n$ in $H.$ Step 1. Set $t_n$ as $$t_n:=u_n+\xi (u_n-u_{n-1}).$$ (4) Step 2. Compute $$s_n={(1+{\lambda }_{n}B)}^{-1}(I-{\lambda }_n\sum _{i=1}^N{\delta }_i{A}_i)t_n.$$ (5) If $t_n=s_n,$ stop. $t_n$ is the solution of MVIP. Else, go to Step 3. Step 3. Compute $$u_{n+1}=(1-\rho )t_n+\rho s_n+\rho {\lambda }_n(A_{i_n}{t}_n-A_{i_n}{s}_n),$$ (6) where $A_{i_n}:=argmax\{\parallel A_i{t}_n-A_i{s}_n\parallel |i=1,2,3,\dots ,N\}$ for some $i_n\in \{1,..,N\}$, and the stepsize sequence $\left\{{\lambda }_n\right\}$ is updated as follows: $$\left.\begin{array}{c}\\ \\ \end{array}{\lambda }_{n+1}:=\right\{\begin{array}{c}min\{{\lambda }_n,\frac{\mu \parallel t_n-s_n\parallel }{\parallel A_i{t}_n-A_i{s}_n\parallel }\},\mathrm{if}{A}_i{t}_n\ne A_i{s}_n\mathrm{for}\mathrm{some}i\\ {\lambda }_n,\mathrm{otherwise}.\end{array}$$ (7) Set $n:=n+1$, and go back to Step 1.

Lemma 5.

The sequence $\left\{{\lambda }_n\right\}$ is bounded below.

Proof. 

Since $A_i$ is $L_i$-Lipschitz continuous for $i=1,2,\dots ,N$, we have

$$\parallel A_i{t}_n-A_i{s}_n\parallel \le L_i\parallel t_n-s_n\parallel$$

(8)

By (7), if $\parallel A_i{t}_n-A_i{s}_n\parallel \ne 0$ for $i=1,2,\dots ,N$, then the following holds:

$$\frac{\mu \parallel t_n-s_n\parallel }{\parallel A_i{t}_n-A_i{s}_n\parallel }\ge \frac{\mu }{L_i}.$$

(9)

Thus, the sequence $\left\{{\lambda }_n\right\}$ is bounded below by $\frac{\mu }{L_i}$ for $i=1,2,\dots ,N$.

In the case of $\parallel A_i{t}_n-A_i{s}_n\parallel =0$ for $i=1,2,\dots ,N$, it is obvious that the sequence $\left\{{\lambda }_n\right\}$ is bounded below by ${\lambda }_0$. We can conclude that $min\{\frac{\mu }{L_i},{\lambda }_0\}\le {\lambda }_n.$    □

Remark 1.

According to Lemma 5, it is easy to see that the sequence $\left\{{\lambda }_n\right\}$ is monotonically decreasing and bounded below by $min\{\frac{\mu }{L_i},{\lambda }_0\}.$ This implies that the update (7) is well defined and the following holds:

$${\lambda }_{n+1}\parallel A_i{t}_n-A_i{s}_n\parallel \le \mu \parallel t_n-s_n\parallel .$$

(10)

Lemma 6.

Let $\left\{t_n\right\}$ be a sequence generated by Algorithm 1. If there is a subsequence $\left\{t_{n_k}\right\}$ with weak convergence to q with ${lim}_{n\to \infty }\parallel t_n-s_n\parallel =0,$ then $q\in \Gamma .$

Proof. 

We assume that $(y,x)\in Graph(A+B)$, which means $x-Ay\in By.$ By $s_{n_k}=(I+{\lambda }_{n_k}B)(I-{\lambda }_{n_k}{\sum }_{i=1}^N{\delta }_i{A}_i)t_{n_k},$ we have the following:

$$(I-{\lambda }_{n_k}\sum _{i=1}^N{\delta }_i{A}_i)t_{n_k}\in (I+{\lambda }_{n_k}B)s_{n_k}.$$

Therefore, $\frac{1}{{\lambda }_{n_k}}(t_{n_k}-s_{n_k}-{\lambda }_{n_k}{\sum }_{i=1}^N{\delta }_i{A}_i{t}_{n_k})\in B{s}_{n_k}.$ Because B is the maximal monotone, we obtain the following:

$$\langle y-s_{n_k},x-Ay-\frac{1}{{\lambda }_{n_k}}(t_{n_k}-s_{n_k}{\lambda }_{n_k}\sum _{i=1}^N{\delta }_i{A}_i)\rangle \ge 0.$$

Hence,

$$\begin{array}{ccc}\hfill \langle y-s_{n_k},x\rangle & \ge & \langle y-s_{n_k},Ay+\frac{1}{{\lambda }_{n_k}}(t_{n_k}-s_{n_k}{\lambda }_{n_k}\sum _{i=1}^N{\delta }_i{A}_i{t}_{n_k})\rangle \hfill \\ & =& \langle y-s_{n_k},Ay-\sum _{i=1}^N{\delta }_i{A}_i{t}_{n_k}\rangle +\langle y-s_{n_k},\frac{1}{{\lambda }_{n_k}}(t_{n_k}-s_{n_k})\rangle \hfill \\ & =& \langle y-s_{n_k},Ay-\sum _{i=1}^N{\delta }_i{A}_i{s}_{n_k}\rangle +\langle y-s_{n_k},\sum _{i=1}^N{\delta }_i{A}_i{s}_{n_k}-\sum _{i=1}^N{\delta }_i{A}_i{t}_{n_k}\rangle \hfill \\ & & +\langle y-s_{n_k},\frac{1}{{\lambda }_{n_k}}(t_{n_k}-s_{n_k})\rangle \hfill \\ & \ge & \langle y-s_{n_k},\sum _{i=1}^N{\delta }_i{A}_i{s}_{n_k}-\sum _{i=1}^N{\delta }_i{A}_i{t}_{n_k}\rangle +\langle y-s_{n_k},\frac{1}{{\lambda }_{n_k}}(t_{n_k}-s_{n_k})\rangle .\hfill \end{array}$$

Since $A_i$ is $L_i$-Lipschitz continuous and ${lim}_{n\to \infty }\parallel t_n-s_n\parallel =0,$ it follows that ${lim}_{n\to \infty }\parallel {\sum }_{i=1}^N{\delta }_i{A}_i{t}_n-{\sum }_{i=1}^N{\delta }_i{A}_i{s}_n\parallel =0.$ If ${lim}_{n\to \infty }{\lambda }_n$ exists, it receives the following:

$$0\le \langle y-q,x\rangle =\underset{k\to \infty }{lim}\langle y-s_{n_k},x\rangle .$$

(11)

Based on the maximal monotonicity of $A+B$ and (11), a zero is in $(A+B)q.$ We can conclude that $q\in \Gamma .$    □

Theorem 1.

Assume the following:

$$0<\rho <\frac{\sqrt{1+8\omega }-1-2\omega }{2(1-\omega )}$$

(12)

with $\omega \in (0,\frac{1-\mu }{2-\rho (1-\mu )}),$ and ${\sum }_{i=1}^N{\delta }_i=1.$ If every $\overline{u}\in \Gamma \ne Ø,$ then $\left\{u_n\right\}$ generated by Algorithm 1 converges weakly to $\overline{u}$.

Proof. 

We know that the resolvent $J_{{\lambda }_n}^B$ is firmly non-expansive and $s_n={(I+{\lambda }_{n}B)}^{-1}(I-{\lambda }_n{\sum }_{i=1}^N{\delta }_i{A}_i)t_n=J_{{\lambda }_n}^B(I-{\lambda }_n{\sum }_{i=1}^N{\delta }_i{A}_i)t_n.$

Thus, we have the following:

$$\begin{array}{ccc}\hfill 〈s_n-\overline{u},t_n-s_n-{\lambda }_n\sum _{i=1}^N{\delta }_i{A}_i{t}_n〉& =& \langle J_{{\lambda }_n}^B(I-{\lambda }_n\sum _{i=1}^N{\delta }_i{A}_i)t_n-J_{{\lambda }_n}^B(I-{\lambda }_n\sum _{i=1}^N{\delta }_i{A}_i)\overline{u},\hfill \\ & & (I-{\lambda }_n\sum _{i=1}^N{\delta }_i{A}_i)t_n-(I-{\lambda }_n\sum _{i=1}^N{\delta }_i{A}_i)\overline{u}\hfill \\ & & +(I-{\lambda }_n\sum _{i=1}^N{\delta }_i{A}_i)\overline{u}-s_n\rangle \hfill \\ & \ge & \parallel s_n-\overline{u}{\parallel }^2+\langle s_n-\overline{u},\overline{u}-s_n\rangle -\langle s_n-\overline{u},{\lambda }_n\sum _{i=1}^N{\delta }_i{A}_i\overline{u}\rangle \hfill \\ & =& -\langle s_n-\overline{u},{\lambda }_n\sum _{i=1}^N{\delta }_i{A}_i\overline{u}\rangle .\hfill \end{array}$$

Obviously, $\langle s_n-\overline{u},t_n-s_n-{\lambda }_n{\sum }_{i=1}^N{\delta }_i(A_i{t}_n-A_i{s}_n)\rangle \ge 0.$ This implies that   

$$2\langle t_n-s_n,s_n-\overline{u}\rangle -2{\lambda }_n\sum _{i=1}^N{\delta }_i\langle A_i{t}_n-A_i{s}_n,s_n-\overline{u}\rangle \ge 0.$$

(13)

However, $2\langle t_n-s_n,s_n-\overline{u}\rangle = \parallel t_n-\overline{u}{\parallel }^2-\parallel t_n-s_n{\parallel }^2-{\parallel s_n-\overline{u}\parallel }^2.$ To replace the previous equation in (13), we obtain the following:

$$\parallel s_n-\overline{u}{\parallel }^2\le \parallel t_n-\overline{u}{\parallel }^2-{\parallel t_n-s_n\parallel }^2-2{\lambda }_n\sum _{i=1}^N{\delta }_i\langle A_i{t}_n-A_i{s}_n,s_n-\overline{u}\rangle .$$

(14)

The definition of $u_{n+1}$ implies the following:

$$\begin{array}{ccc}\hfill \parallel u_{n+1}-\overline{u}{\parallel }^2& =& \parallel (1-\rho )t_n+\rho s_n+\rho {\lambda }_n(A_{i_n}{t}_n-A_{i_n}{s}_n)-\overline{u}{\parallel }^2,\hfill \\ & =& \parallel (1-\rho )(t_n-\overline{u})+\rho (s_n-\overline{u})+\rho {\lambda }_n(A_{i_n}{t}_n-A_{i_n}{s}_n){\parallel }^2,\hfill \\ & =& {(1-\rho )}^2\parallel t_n-\overline{u}{\parallel }^2+{\rho }^2\parallel s_n-\overline{u}{\parallel }^2+{\rho }^2{\lambda }_n^2{\parallel A_i{t}_n-A_i{s}_n\parallel }^2\hfill \\ & & +2\rho (1-\rho )\langle t_n-\overline{u},s_n-\overline{u}\rangle +2{\lambda }_n\rho (1-\rho )\langle t_n-\overline{u},A_i{t}_n-A_i{s}_n\rangle \hfill \\ & & +2{\lambda }_n{\rho }^2\langle s_n-\overline{u},A_i{t}_n-A_i{s}_n\rangle ,\hfill \end{array}$$

(15)

for each $i=1,2,\dots ,N.$ Consider the following:

$$\parallel t_n-\overline{u}{\parallel }^2+\parallel s_n-\overline{u}{\parallel }^2-{\parallel t_n-s_n\parallel }^2=2\langle t_n-\overline{u},s_n-\overline{u}\rangle .$$

(16)

Substituting (16) in (15), we obtain the following:

$$\begin{array}{ccc}\hfill \parallel u_{n+1}-\overline{u}{\parallel }^2& =& {(1-\rho )}^2\parallel t_n-\overline{u}{\parallel }^2+{\rho }^2\parallel s_n-\overline{u}{\parallel }^2+{\rho }^2{\lambda }_n^2{\parallel A_i{t}_n-A_i{s}_n\parallel }^2\hfill \\ & & +\rho (1-\rho )(\parallel t_n-\overline{u}{\parallel }^2+\parallel s_n-\overline{u}{\parallel }^2-\parallel t_n-s_n{\parallel }^2)\hfill \\ & & +2{\lambda }_n\rho (1-\rho )\langle t_n-\overline{u},A_i{t}_n-A_i{s}_n\rangle +2{\lambda }_n{\rho }^2\langle s_n-\overline{u},A_i{t}_n-A_i{s}_n\rangle ,\hfill \\ & =& (1-\rho )\parallel t_n-\overline{u}{\parallel }^2+\rho \parallel s_n-\overline{u}{\parallel }^2-\rho (1-\rho ){\parallel t_n-s_n\parallel }^2\hfill \\ & & +{\lambda }_n^2{\rho }^2{\parallel A_i{t}_n-A_i{s}_n\parallel }^2+2{\lambda }_n\rho (1-\rho )\langle t_n-\overline{u},A_i{t}_n-A_i{s}_n\rangle \hfill \\ & & +2{\lambda }_n{\rho }^2\langle s_n-\overline{u},A_i{t}_n-A_i{s}_n\rangle ,\hfill \end{array}$$

(17)

for each $i=1,2,\dots ,N.$ Putting (14) in (17), we obtain that, for every $i=1,2,\dots ,N,$

$$\begin{array}{ccc}\hfill \parallel u_{n+1}-\overline{u}{\parallel }^2& \le & (1-\rho )\parallel t_n-\overline{u}{\parallel }^2+\rho (\parallel t_n-\overline{u}{\parallel }^2-{\parallel t_n-s_n\parallel }^2\hfill \\ & & -2{\lambda }_n\sum _{i=1}^N{\delta }_i\langle A_i{t}_n-A_i{s}_n,s_n-\overline{u}\rangle )-\rho (1-\rho ){\parallel t_n-s_n\parallel }^2\hfill \\ & & +{\lambda }_n^2{\rho }^2{\parallel A_i{t}_n-A_i{s}_n\parallel }^2+2{\lambda }_n\rho (1-\rho )\langle t_n-\overline{u},A_i{t}_n-A_i{s}_n\rangle \hfill \\ & & +2{\lambda }_n{\rho }^2\langle s_n-\overline{u},A_i{t}_n-A_i{s}_n\rangle ,\hfill \\ & =& \parallel t_n-\overline{u}{\parallel }^2-\rho (2-\rho ){\parallel t_n-s_n\parallel }^2-2\rho {\lambda }_n\sum _{i=1}^N{\delta }_i\langle A_i{t}_n-A_i{s}_n,s_n-\overline{u}\rangle \hfill \\ & & +{\rho }^2{\lambda }_n{\parallel A_i{t}_n-A_i{s}_n\parallel }^2+2{\lambda }_n\rho (1-\rho )\langle t_n-\overline{u},A_i{t}_n-A_i{s}_n\rangle \hfill \\ & & +2{\lambda }_n{\rho }^2\langle s_n-\overline{u},A_i{t}_n-A_i{s}_n\rangle \hfill \\ & \le & \parallel t_n-\overline{u}{\parallel }^2-\rho (2-\rho )\parallel t_n-s_n{\parallel }^2+{\rho }^2{\lambda }_n^2\frac{\mu }{{\lambda }_{n+1}^2}{\parallel t_n-s_n\parallel }^2\hfill \\ & & +2\rho {\lambda }_n(\sum _{i=1}^N{\delta }_i-\rho )\langle \overline{u}-s_n,A_i{t}_n-A_i{s}_n\rangle \hfill \\ & & +2\rho {\lambda }_n(1-\rho )\langle t_n-\overline{u},A_i{t}_n-A_i{s}_n\rangle \hfill \\ \hfill & \le & \parallel t_n-\overline{u}{\parallel }^2-\rho (2-\rho )\parallel t_n-s_n{\parallel }^2+{\rho }^2{\lambda }_n^2\frac{\mu }{{\lambda }_{n+1}^2}{\parallel t_n-s_n\parallel }^2\hfill \\ & & +2\rho {\lambda }_n(\sum _{i=1}^N{\delta }_i-\rho )\langle \overline{u}-s_n,A_i{t}_n-A_i{s}_n\rangle \hfill \\ & & +2\rho {\lambda }_n(1-\rho )\langle t_n-\overline{u},A_i{t}_n-A_i{s}_n\rangle \hfill \\ & \le & \parallel t_n-\overline{u}{\parallel }^2-\rho (2-\rho )\parallel t_n-s_n{\parallel }^2+{\rho }^2{\lambda }_n^2\frac{\mu }{{\lambda }_{n+1}^2}{\parallel t_n-s_n\parallel }^2\hfill \\ & & +2\rho {\lambda }_n(1-\rho )\langle t_n-s_n,A_i{t}_n-A_i{s}_n\rangle \hfill \\ & \le & \parallel t_n-\overline{u}{\parallel }^2-\rho (2-\rho )\parallel t_n-s_n{\parallel }^2+{\rho }^2{\lambda }_n^2\frac{\mu }{{\lambda }_{n+1}^2}{\parallel t_n-s_n\parallel }^2\hfill \\ & & +2\rho {\lambda }_n(1-\rho )\frac{\mu }{{\lambda }_{n+1}}{\parallel t_n-s_n\parallel }^2\hfill \\ & & \le \parallel t_n-\overline{u}{\parallel }^2-\rho (2-\rho -2\mu (1-\rho )\frac{{\lambda }_n}{{\lambda }_{n+1}}-\rho {\mu }^2\frac{{\lambda }_n^2}{{\lambda }_{n+1}^2}){\parallel t_n-s_n\parallel }^2.\hfill \\ & & =\parallel t_n-\overline{u}{\parallel }^2-\eta \rho {\parallel t_n-s_n\parallel }^2\hfill \end{array}$$

(18)

where $\eta =(2-\rho -2\mu (1-\rho )\frac{{\lambda }_n}{{\lambda }_{n+1}}-\rho {\mu }^2\frac{{\lambda }_n^2}{{\lambda }_{n+1}^2}).$ By the definition of $u_{n+1}$ and (10), it follows that

$$\begin{array}{ccc}\hfill \parallel u_{n+1}-s_n\parallel & =& \parallel (1-\rho )t_n+\rho s_n+\rho {\lambda }_n(A_{i_n}{t}_n-A_{i_n}{s}_n)-s_n\parallel \hfill \\ & \le & (1-\rho )\parallel t_n-s_n\parallel +\rho {\lambda }_n\parallel A_i{t}_n-A_i{s}_n\parallel \hfill \\ & \le & (1-\rho )\parallel t_n-s_n\parallel +\rho \mu \frac{{\lambda }_n}{{\lambda }_{n+1}}\parallel t_n-s_n\parallel \hfill \\ & =& (1-\rho (1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}}))\parallel t_n-s_n\parallel ,\hfill \end{array}$$

(19)

for every $i=1,2,\dots ,N.$ The following can be stated:

$$\parallel u_{n+1}-t_n\parallel \le \parallel u_{n+1}-s_n\parallel +\parallel s_n-t_n\parallel .$$

(20)

From (19), it follows that

$$\begin{array}{ccc}\hfill \parallel u_{n+1}-t_n\parallel & \le & (1-\rho (1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}}))\parallel t_n-s_n\parallel +\parallel s_n-t_n\parallel \hfill \\ & =& (2-\rho (1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}}))\parallel t_n-s_n\parallel .\hfill \end{array}$$

Therefore,

$$\frac{1}{2-\rho (1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})}\parallel u_{n+1}-t_n\parallel \le \parallel t_n-s_n\parallel .$$

(21)

By (18) and (21), we have the following:

$$\begin{array}{ccc}\hfill \parallel u_{n+1}-s_n{\parallel }^2& \le & \parallel t_n-\overline{u}{\parallel }^2-\eta \rho {\parallel t_n-s_n\parallel }^2\hfill \\ & \le & \parallel t_n-\overline{u}{\parallel }^2-\frac{\eta \rho }{2-\rho (1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})}\parallel u_{n+1}-t_n\parallel .\hfill \end{array}$$

(22)

We also can state the following:

$${\psi }_n=\rho (2-\rho -2\mu (1-\rho )\frac{{\lambda }_n}{{\lambda }_{n+1}}-\rho {\mu }^2\frac{{\lambda }_n^2}{{\lambda }_{n+1}^2}){(2-\rho (1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}}))}^{-2}.$$

Obviously, with ${\lambda }_n\to \lambda$ as $n\to \infty$, we have the following:

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}{\psi }_n& =& \rho (2-\rho -2\mu (1-\rho )-\rho {\mu }^2){(2-\rho (1-\mu ))}^{-2}\hfill \\ & =& (1-\mu )(2-\rho +\rho \mu ){(2-\rho (1-\mu ))}^{-2}>0.\hfill \end{array}$$

From (12), there is an $\omega$ such that

$$(2-\rho -2\mu (1-\rho )\frac{{\lambda }_n}{{\lambda }_{n+1}}-\rho {\mu }^2\frac{{\lambda }_n^2}{{\lambda }_{n+1}^2}){(2-\rho (1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}}))}^{-2}>\omega$$

(23)

for every $n\ge \mathbb{N}.$ The previous equation and (20) imply the following:

$$\parallel u_{n+1}-s_n{\parallel }^2\le \parallel t_n-\overline{u}{\parallel }^2-\omega \parallel u_{n+1}-t_n\parallel .$$

(24)

By the definition of $t_n$, we have the following:

$$\begin{array}{ccc}\hfill \parallel t_n-\overline{u}{\parallel }^2& =& \parallel u_n+\xi (u_n-u_{n-1})-\overline{u}{\parallel }^2\hfill \\ & =& \parallel (1-\xi )(u_n-\overline{u})-\xi (u_{n-1}-\overline{u}){\parallel }^2\hfill \\ & =& (1-\xi )\parallel u_n-\overline{u}{\parallel }^2-\xi \parallel u_{n-1}-\overline{u}{\parallel }^2+\xi (1+\xi ){\parallel u_n-u_{n-1}\parallel }^2.\hfill \end{array}$$

(25)

By the Cauchy–Schwartz inequality, it follows that

$$\begin{array}{ccc}\hfill \parallel u_{n+1}-t_n{\parallel }^2& =& \parallel u_{n+1}-u_n-\xi (u_n-u_{n-1}){\parallel }^2\hfill \\ & =& \parallel u_{n+1}-u_n{\parallel }^2+{\xi }^2{\parallel u_n-u_{n-1}\parallel }^2-2\xi \langle u_{n+1}-u_n,u_n-u_{n-1}\rangle \hfill \\ & \ge & \parallel u_{n+1}-u_n{\parallel }^2+{\xi }^2\parallel u_n-u_{n-1}{\parallel }^2-2\xi \parallel u_{n+1}-u_n\parallel \parallel u_n-u_{n-1}\parallel \hfill \\ & \ge & \parallel u_{n+1}-u_n{\parallel }^2+{\xi }^2\parallel u_n-u_{n-1}{\parallel }^2-\xi {\parallel u_{n+1}-u_n\parallel }^2\hfill \\ & & -\xi \parallel u_n-u_{n-1}{\parallel }^2\hfill \\ & \ge & (1-\xi )\parallel u_{n+1}-u_n{\parallel }^2+({\xi }^2-\xi ){\parallel u_n-u_{n-1}\parallel }^2.\hfill \end{array}$$

(26)

Thank to (24)–(26), we can see that

$$\begin{array}{ccc}\hfill \parallel u_{n+1}-\overline{u}{\parallel }^2& \le & (1+\xi )\parallel u_n-\overline{u}{\parallel }^2-\xi \parallel u_{n-1}-\overline{u}{\parallel }^2+\xi (1+\xi ){\parallel u_n-u_{n-1}\parallel }^2\hfill \\ & & -\omega ((1-\xi )\parallel u_{n+1}-u_n{\parallel }^2+({\xi }^2-\xi )\parallel u_n-u_{n-1}{\parallel }^2)\hfill \\ & \le & (1+\xi )\parallel u_n-\overline{u}{\parallel }^2-\xi \parallel u_{n-1}-\overline{u}{\parallel }^2-\omega (1-\xi ){\parallel u_{n+1}-u_n\parallel }^2\hfill \\ & & +(\xi (1+\xi )-\omega ({\xi }^2-\xi )){\parallel u_n-u_{n-1}\parallel }^2\hfill \\ & \le & (1+\xi )\parallel u_n{-p\parallel }^2-\xi \parallel u_{n-1}{-p\parallel }^2+{\beta }_n{\parallel u_{n+1}-u_n\parallel }^2\hfill \\ & & +{\gamma }_n{\parallel u_n-u_{n-1}\parallel }^2\hfill \end{array}$$

(27)

where ${\beta }_n={\omega }_n(1-\xi )$ and ${\gamma }_n=(\xi (1+\xi )-\omega ({\xi }^2-\xi )).$ It can be stated that

$${\kappa }_n=\parallel u_n-\overline{u}{\parallel }^2-\xi \parallel u_{n-1}-\overline{u}{\parallel }^2+{\beta }_n{\parallel u_n-u_{n-1}\parallel }^2.$$

(28)

Therefore,

$$\begin{array}{ccc}\hfill {\kappa }_{n+1}-{\kappa }_n& =& \parallel u_{n+1}-\overline{u}{\parallel }^2-\xi \parallel u_n-\overline{u}{\parallel }^2+{\beta }_{n+1}{\parallel u_{n+1}-u_n\parallel }^2\hfill \\ & & -\parallel u_n-\overline{u}{\parallel }^2+\xi \parallel u_{n-1}-\overline{u}{\parallel }^2-{\beta }_n{\parallel u_n-u_{n-1}\parallel }^2\hfill \\ & =& \parallel u_{n+1}{-p\parallel }^2-(1+\xi )\parallel u_n\overline{u}{\parallel }^2+\xi {\parallel u_{n-1}\overline{u}\parallel }^2\hfill \\ & & +{\beta }_{n+1}\parallel u_{n+1}-u_n{\parallel }^2-{\beta }_n{\parallel u_n-u_{n-1}\parallel }^2\hfill \\ & \le & \parallel u_{n+1}-\overline{u}{\parallel }^2-(1+\xi )\parallel u_n\overline{u}{\parallel }^2+\xi {\parallel u_{n-1}-\overline{u}\parallel }^2\hfill \\ & & +{\beta }_{n+1}\parallel u_{n+1}-u_n{\parallel }^2-{\beta }_n{\parallel u_n-u_{n-1}\parallel }^2\hfill \\ & & +{\beta }_{n+1}\parallel u_{n+1}-u_n{\parallel }^2-{\beta }_n{\parallel u_n-u_{n-1}\parallel }^2\hfill \\ & \le & -({\gamma }_n-{\beta }_{n+1}){\parallel u_{n+1}-u_n\parallel }^2.\hfill \end{array}$$

(29)

We note the following:

$$\begin{array}{ccc}\hfill {\gamma }_n-{\beta }_{n+1}& =& {\omega }_n(1-\xi )-(\xi (1+\xi )-\omega ({\xi }^2-\xi ))\hfill \\ & =& -(1-\omega ){\xi }^2-(1+2\omega )\xi +\omega .\hfill \end{array}$$

Thus,

$${\kappa }_{n+1}-{\kappa }_n\le -\varphi {\parallel u_{n+1}-u_n\parallel }^2$$

(30)

where $\varphi =(1-\omega ){\xi }^2-(1+2\omega )\xi +\omega .$ By the assumption of (12), we obtain $\varphi >0.$ This indicates that ${\kappa }_n$ is nonincreasing. Furthermore, we have the following:

$$\begin{array}{ccc}\hfill {\kappa }_{n+1}& =& \parallel u_{n+1}-\overline{u}{\parallel }^2-\xi \parallel u_n-\overline{u}{\parallel }^2+{\beta }_{n+1}{\parallel u_{n+1}-u_n\parallel }^2\hfill \\ & \ge & -\xi \parallel u_n-\overline{u}{\parallel }^2\hfill \end{array}$$

(31)

and

$$\begin{array}{ccc}\hfill {\kappa }_n& =& \parallel u_n-\overline{u}{\parallel }^2-\xi \parallel u_{n-1}-\overline{u}{\parallel }^2+{\beta }_n{\parallel u_n-u_{n-1}\parallel }^2\hfill \\ & \ge & \parallel u_n-\overline{u}{\parallel }^2-\xi {\parallel u_{n-1}-\overline{u}\parallel }^2.\hfill \end{array}$$

(32)

It follows that

$$\begin{array}{ccc}\hfill \parallel u_n-\overline{u}{\parallel }^2& \le & {\kappa }_n+\xi {\parallel u_{n-1}-\overline{u}\parallel }^2\hfill \\ & \le & {\kappa }_1+\xi {\parallel u_{n-1}-\overline{u}\parallel }^2\hfill \\ & \le & \dots \le {\kappa }_1({\xi }_{n-1}+\dots +1)+{\xi }_n{\parallel u_0-\overline{u}\parallel }^2\hfill \\ & \le & \frac{{\kappa }_1}{1-\xi }+{\xi }_n{\parallel u_0-\overline{u}\parallel }^2.\hfill \end{array}$$

(33)

Combining (31) and (33), we obtain the following:

$$-{\kappa }_{n+1}\le \xi \parallel u_n-\overline{u}{\parallel }^2\le \xi \frac{{\kappa }_1}{1-\xi }+{\kappa }_{n+1}{\parallel u_0-\overline{u}\parallel }^2.$$

(34)

According to (30) and (33), it follows that

$$\begin{array}{ccc}\hfill \varphi \sum _{n=1}^m\parallel u_{n+1}-u_n\parallel & \le & {\kappa }_1-{\kappa }_{m+1}\le {\kappa }_1+\xi \frac{{\kappa }_1}{1-\xi }+{\xi }_{m+1}{\parallel u_0-\overline{u}\parallel }^2\hfill \\ & \le & \frac{{\kappa }_1}{1-\xi }+{\parallel u_0-\overline{u}\parallel }^2.\hfill \end{array}$$

(35)

Taking $m\to \infty$ in the previous equation, we have ${\sum }_{n=1}^{\infty }\parallel u_{n+1}-u_n\parallel <+\infty .$ It can be concluded that ${lim}_{n\to \infty }\parallel u_{n+1}-u_n\parallel =0.$ We consider the following:

$$\parallel u_{n+1}-t_n\parallel =\parallel u_{n+1}-u_n{\parallel }^2+{\xi }^2{\parallel u_n-u_{n-1}\parallel }^2-2\xi \langle u_{n+1}-u_n,u_n-u_{n-1}\rangle .$$

Obviously, $\parallel u_{n+1}-t_n\parallel \to 0$ as $n\to \infty$. By means of (30) and Lemma 4, we obtain the following: ${lim}_{n\to \infty }{\parallel u_n-\overline{u}\parallel }^2=h,$ for some $h>0.$ From (25), we have ${lim}_{n\to \infty }{\parallel t_n-\overline{u}\parallel }^2=h.$ Furthermore,

$$0\le \parallel u_n-t_n\parallel \le \parallel u_n-u_{n+1}\parallel +\parallel u_{n+1}-t_n\parallel \to 0.$$

(36)

By (22), we see that $\parallel s_n-t_n\parallel \to 0.$ Because ${lim}_{n\to \infty }{\parallel u_n-\overline{u}\parallel }^2$ exists, it is implied that the sequence $\left\{u_n\right\}$ is bounded. We note that $\left\{u_{n_k}\right\}$ is a subsequence of $\left\{u_n\right\}$, and $u_{n_k}\rightharpoonup \widehat{u}$. It follows from (36) that $t_{n_k}\rightharpoonup \widehat{u}.$ Since ${lim}_{n\to \infty }\parallel t_n-s_n\parallel =0$ and because of Lemma 6, it follows that $\widehat{u}\in \Gamma .$ As a result of Lemma 3, we may deduce that the sequence $\left\{u_n\right\}$ converges weakly to the solution of the MVIP.    □

For solving the problem of (2), the following algorithm (Algorithm 2) is suggested when setting $A_i\equiv A$ in Algorithm 1

Algorithm 2 (Modified Tseng’s method for solving the VIP)

Pick $u_0,u_1\in H,\mu \in (0,1),\rho >0,\xi \ge 0,{\lambda }_0>0.$ Iterative steps: Given iterates $u_{n-1}$ and $u_n$ in $H.$ Step 1. Set $t_n$ as $$t_n:=u_n+\xi (u_n-u_{n-1}).$$ (37) Step 2. Compute $$s_n={(1+{\lambda }_{n}B)}^{-1}(I-{\lambda }_{n}A)t_n.$$ (38) If $t_n=s_n,$ stop. $t_n$ is the solution of the VIP. Else, go to Step 3. Step 3. Compute $$u_{n+1}=(1-\rho )t_n+\rho s_n+\rho {\lambda }_n(A{t}_n-A{s}_n).$$ The stepsize sequence $\left\{{\lambda }_n\right\}$ is updated as follows: $$\left.\begin{array}{c}\\ \\ \end{array}{\lambda }_{n+1}:=\right\{\begin{array}{c}min\{{\lambda }_n,\frac{\mu \parallel t_n-s_n\parallel }{\parallel A{t}_n-A{s}_n\parallel }\},\mathrm{if}A{t}_n\ne A{s}_n\\ {\lambda }_n,\mathrm{otherwise}.\end{array}$$ Set $n:=n+1$, and go back to Step 1.

Corollary 1.

If the solution set of the VIP $(\Gamma )$ is nonempty, then the sequence $\left\{u_n\right\}$ generated by Algorithm 2 converges weakly to $\overline{u}\in \Gamma$.

Proof. 

Proofs are similar by setting $A_i\equiv A$ in Lemma 5, 6, and Theorem 1. □

4. Numerical Experiments

We provide numerical examples to demonstrate the behavior of our algorithm in this section. We have broken it down into two examples.

The behavior of an error of the sequence $\left\{x_n\right\}$ generated by Algorithm 1 is demonstrated in Example 1. Furthermore, in Example 1, we demonstrate the behavior of the sequence $\left\{x_n\right\}$, which is generated by Algorithm 1 by separating the dimension of the sequence $\left\{x_n\right\}$.

In Example 2, we apply our main result to solve image deblurring problems. Image recovery problems could be explained using the inversion of the following observation model:

$$\begin{array}{c}\hfill b=Mx+\delta \end{array}$$

(39)

where M is a deblurring matrix, $x\in {\mathbb{R}}^n$ represents the original image, b is the deblurring image, and $\delta \in R^m$ is the Gaussian noise. The convex unconstrained optimization problem is known to be the same as solving (39):

$$\begin{array}{c}\hfill \underset{x\in {\mathbb{R}}^n}{min}\left\{{\displaystyle \frac{1}{2}}{\parallel Mx-b\parallel }_2^2+ϵ{\parallel x\parallel }_1^2\right\},\end{array}$$

(40)

as the regularization parameter $ϵ>0$. For solving (40), we suppose $A=\nabla F\left(x\right)$ and $B=\partial G$, where $F\left(x\right)={\displaystyle \frac{1}{2}}{\parallel Mx-b\parallel }_2^2$ and $G\left(x\right)={\parallel x\parallel }_1^2$. Therefore, $\nabla F\left(x\right)=M^t(Mx-b)$ is ${\displaystyle \frac{1}{{\parallel M\parallel }^2}}$-cocoercive. This implies that $(I-\tau \nabla F)$ is nonexpansive and that $0<\tau <{\displaystyle \frac{2}{{\parallel M\parallel }^2}}$. Because $\partial G$ is the maximal monotone, x is an (40) solution if and only if the following holds:

$$\begin{array}{c}\hfill 0\in \nabla F\left(x\right)+\partial G\left(x\right)\iff x=arg\underset{u\in {\mathbb{R}}^n}{min}\left\{G\left(u\right)+{\displaystyle \frac{1}{2ϵ}}{\parallel u-(I-ϵ\nabla F)\left(x\right)\parallel }^2\right\}.\end{array}$$

In addition, we might still formulate (40) as a convex constrained optimization problem for the split feasibility problem (SFP):

$$\begin{array}{c}\hfill \underset{x\in {\mathbb{R}}^n}{min}{\displaystyle \frac{1}{2}}{\parallel Mx-b\parallel }_2^2\\ \hfill {\mathrm{subject}\mathrm{to}\parallel x\parallel }_1\le k,\end{array}$$

(41)

where $k>0$ is a given constant and (41) is solved. Setting $Ax=\nabla F\left(x\right)$, we consider $C=\{x\in {\mathbb{R}}^n|\parallel x{\parallel }_1\le k\}$ and $Q=\left\{b\right\}.$ We use Algorithm 1 to solve this image deblurring problem. Figure 1 shows a flowchart of the technique for selecting a blurring function A by summing N blurring functions $A_1-A_N$.

Example 1.

Let $H={\mathbb{R}}^2$ be a two-dimensional space of real numbers. Define monotone operators $A_1\left(x\right)=2x$, $A_2\left(x\right)={\displaystyle \frac{1}{2}}x$ and maximal monotone operator $B\left(x\right)=4x$ for every $x\in {\mathbb{R}}^2$. Moreover, we set parameters $\xi =0.5,\rho =0.015,\omega =1.5,{\delta }_1=0.5,{\delta }_2=0.5$, and $\mu =0.5$ and choose the initial point as $x_1=(1,1)$ and $x_2=(2,0)$. For the time being, it is exciting to see that all of our parameters and operators are satisfied with our main result. By applying Algorithm 1, we can solve the problem of (3) and show the behavior of our algorithm.

After completion the experiment, we can observe that a sequence $\left\{x_n\right\}$ always converges to the solution point of the problem of (3), independently of the arbitrary starting point.

In Figure 2, we plot the error graph of a sequence $\left\{x_n\right\}$. This shows that the sequence achieves a small error when the number of iterations is higher. We are also confident that increasing the number of iterations will result in an inaccuracy of less than $ϵ<{10}^{-7}$.

In Figure 3, we plot the sequence $\left\{x_n\right\}$ by separating dimension of $x_n$. We can conclude that a sequence $\left\{x_n\right\}$ always converges to the solution of the interested problem. That is, $x_n=(x_n^1,x_n^2)\to (0,0)$ as $n\to \infty$.

Furthermore, the detailed analysis of the computations of our algorithm is shown in

Table 1. Detailed analysis of computational of Algorithm 1.

	Algorithm 1
$\mathit{n}$	$\mathit{E}\mathbf{\left(}{\mathit{x}}_{\mathit{n}}^{\mathbf{1}}\mathbf{\right)}$	$\mathit{E}\mathbf{\left(}{\mathit{x}}_{\mathit{n}}^{\mathbf{2}}\mathbf{\right)}$
1	0.100000000000001	0.100000000000000
2	0.000385156249999685	0.00945859374999999
3	0.0103723678649903	0.000402462457275377
4	0.0113151219476872	0.000500674227283338
5	0.0113532417283455	0.000588079127189875
⋮	⋮	⋮
29	0.00996291653487358	0.000524394399890646
30	0.00990842850392615	0.000521526442682538

for $E\left(x_n^1\right)=\parallel x_{n+1}^1-x_n^1\parallel$ and $E\left(x_n^2\right)=\parallel x_{n+1}^2-x_n^2\parallel$ for all $n\in \mathbb{N}$.

Example 2.

As the blurring function, we used MATLAB’s motion blur by using $A_1$ as fspecial (’motion’,9,40) and $A_2$ as fspecial (’gaussian’,9,2). The standard test images of a house $(256\times 256)$ and of boats $(512\times 512)$ are used in the comparison (see Figure 4). We compare our proposed algorithm (Algorithm 1) with the algorithm in [28]. The control parameters are set as follows: $\xi =0.5,\rho =0.015,\omega =1.5,{\delta }_1=0.5,{\delta }_2=0.5$, and $\mu =0.5$. The results of this experiment can be seen in Figure 5, Figure 6 and Figure 7. To analyze the image quality that is restored, we adopted the signal-to-noise ratio (SNR) as follows:

$$\begin{array}{c}\hfill \mathit{SNR}=20log{\displaystyle \frac{{\parallel x\parallel }_2^2}{\parallel x-x_n{\parallel }_2^2}}\end{array}$$

(42)

where x is the original image, and $x_n$ is the estimated image at iteration n. In Figure 7, the SNR values of the house and boat images restored by Algorithm 1 (Figure 5b and Figure 6b) are higher than those images restored by the algorithm in [28] (Figure 5c and Figure 6c). The superiority of the proposed algorithm in measures of SNR is shown.

5. Conclusions

Two modified Tseng’s methods are presented for solving the modified variational inclusion and the variational inclusion problems by using the condition for the modified variational inclusion problems and the parallel technique. Additionally, we demonstrate the behavior of our algorithm and use it to solve image deblurring problems.