Gradient Methods with Selection Technique for the Multiple-Sets Split Equality Problem

Abstract

The inverse problem is one of the four major problems in computational mathematics. There is an inverse problem in medical image reconstruction and radiotherapy that is called the multiple-sets split equality problem. The multiple-sets split equality problem is a unified form of the split feasibility problem, split equality problem, and split common fixed point problem. In this paper, we present two iterative algorithms for solving it. The suggested algorithms are based on the gradient method with a selection technique. Based on this technique, we only need to calculate one projection in each iteration.

1. Introduction

The inverse problem is one of the four major problems in computational mathematics. The rapid development of the inverse problem has been a feature of recent decades; it can be found in computer vision, machine learning, statistics, geography, medical imaging, remote sensing, ocean acoustics, tomography, aviation, physics, and other fields. There is an inverse problem in medical image reconstruction and radiotherapy that can be expressed as a split feasibility problem [1,2,3,4,5,6,7,8,9], split equality problem [10,11,12,13], and split common fixed point problem [14,15,16,17,18,19].

In this paper, we focus on a unified form of the split feasibility problem, split equality problem, and split common fixed point problem that is called the multiple-sets split equality problem.

Let $H_1,H_2,H_3$ be three real Hilbert spaces. $r,t$ are positive integers, ${\left\{C_i\right\}}_{i=1}^r$ and ${\left\{Q_j\right\}}_{j=1}^t$ are two families of closed and convex subsets of $H_1$ and $H_2$, respectively. $A:H_1\to H_3$, $B:H_2\to H_3$ are two bounded and linear operators. Then the multiple-sets split equality problem (MSSEP for short) can be formulated as

$$findingx\in \bigcap _{i=1}^r{C}_i,y\in \bigcap _{j=1}^t{Q}_{j}suchthatAx=By.$$

(1)

It reduces to the split equality problem if $r=t=1$; moreover, it is the split feasibility problem if $H_2=H_3$ and B is the identity operator on $H_2$. In addition, it is the split common fixed point problem if we take $x\in C_i$ to $x=P_{C_i}x$, $y\in Q_j$ to $y\in P_{Q_j}$ where $P_{C_i},P_{Q_j}$ are the metric projections on $C_i,Q_j$.

In the problem (1), without loss of generality, we may assume that $t\ge r$ and let $C_{r+1}=C_{r+2}=\cdots =C_t=H_1$. Then the problem (1) can be described equivalently as:

$$findingx\in \bigcap _{i=1}^t{C}_i,y\in \bigcap _{j=1}^t{Q}_{j}suchthatAx=By.$$

(2)

Let $S_i=C_i\times Q_i\subseteq H=H_1\times H_2$, $i\in \Lambda =\{1,2,\cdots ,t\}$, $G=[A,-B]:H\to H_3$, $G^{*}$ be the adjoint operator of G where $H_1\times H_2$ is the Cartesian product of $H_1$ and $H_2$. Then the original problem (1) can be modified as

$$findingw=(x,y)\in \bigcap _{i=1}^t{S}_{i}suchthatGw=0.$$

(3)

Assume the problem (3) is consistent and let $\Omega$ denote its solution set, that is, $\Omega$ is not empty. We consider the proximity function

$$f\left(w\right)=\frac{1}{2}{\Sigma }_{i=1}^t{\alpha }_i\parallel w-P_{S_i}{w\parallel }^2+\frac{1}{2}{\parallel Gw\parallel }^2,$$

where ${\alpha }_i,i=1,\cdots ,t$ are positive real numbers and $P_{S_i},i=1,\cdots ,t$ are metric projections from H onto $S_i$. Since $C_i$ and $Q_i$ are closed convex, so are $S_i$, and then $P_{S_i}$ are well defined. Then problem (3) can be transformed into the minimization problem

$$\underset{w\in {\bigcap }_{i=1}^t{S}_i}{min}f\left(w\right).$$

(4)

Note that the proximity function $f\left(w\right)$ is convex and differentiable with gradient

$$\nabla f\left(w\right)={\Sigma }_{i=1}^t{\alpha }_i(I-P_{S_i})w+G^{*}Gw,$$

where I is the identity operator on H. The gradient function $\nabla f\left(w\right)$ is L-Lipschitz continuous with Lipschitz constant [20]

$$L={\Sigma }_{i=1}^t{\alpha }_i+{\parallel G\parallel }^2.$$

To solve the minimization problem (4), a classical method is the gradient algorithm, which takes the iterative issue

$$w_{n+1}=w_n-{\gamma }_n\nabla f\left(w_n\right),$$

(5)

where ${\gamma }_n$ is the iterative step size in step n.

Note that in iteration (5), we need to calculate projections for t times in each step. On the other hand, notice that $w^{*}\in \Omega$ if and only if $g\left(w^{*}\right)=0$, where

$$g\left(w\right)=\frac{1}{2}\parallel w-P_{S_{i\left(n\right)}}{w\parallel }^2+\frac{1}{2}{\parallel Gw\parallel }^2,$$

in which

$$i\left(n\right)\in \{i|\underset{1\le i\le t}{max}\parallel w-P_{S_i}w\parallel \}.$$

Then we consider the iterative issue

$$w_{n+1}=w_n-{\gamma }_n\nabla g\left(w_n\right).$$

(6)

In iteration (6), we only need to implement a projection once in each step. Motivated by this point, we present Algorithms 1 and 2 in Section 3 to solve problem (3).

The general structure of this paper is as follows. In the next section, we go over some preliminaries. In Section 3, we present the main algorithms and their convergence. In Section 4, several numerical results are shown to confirm the effectiveness of the suggested algorithm. In the last section, there are some conclusions.

2. Preliminaries

For convenience, note that H is a real Hilbert space and I denotes the identity operator on H. By $x_n\to x^{*}$ and $x_n\rightharpoonup x^{*}$, the strong and weak convergence of sequence $\left\{x_n\right\}$ to a point $x^{*}$, respectively, and ${\omega }_w\left(x_n\right)$ denotes the set of weak cluster points of the sequence $\left\{x_n\right\}$. $P_S$ is the projection from H onto its closed and convex subset S.

Lemma 1

([21]). Let S be a closed, convex, and nonempty subset of H, then for any $x,y\in H$ and $z\in S$,

(i) $\langle x-P_{S}x,z-P_{S}x\rangle \le 0$;

(ii) $\parallel P_{S}x-P_S{y\parallel }^2\le \langle P_{S}x-P_{S}y,x-y\rangle$;

(iii) $\langle x-P_{S}x,x-z\rangle \ge {\parallel x-P_{S}x\parallel }^2$.

Lemma 2

([22]). Let $\left\{a_n\right\},\left\{{\alpha }_n\right\},\left\{u_n\right\}$ be sequences of non-negative real numbers with

$$\left\{{\alpha }_n\right\}\subset [0,1],{\Sigma }_{n=1}^{\infty }{\alpha }_n=\infty ,{\Sigma }_{n=1}^{\infty }{u}_n<\infty .$$

Let $\left\{t_n\right\}$ be a sequence of real numbers with ${lim\; sup}_{n\to \infty }{t}_n\le 0$ and

$$a_{n+1}\le (1-{\alpha }_n)a_n+{\alpha }_n{t}_n+u_n.$$

Then ${lim}_{n\to \infty }{a}_n=0$.

Lemma 3

([23]). Let S be a closed and convex subset of H, and $T:S\to S$ be non-expansive, and $\left\{x_n\right\}\subseteq S$. If $x_n\rightharpoonup x$ and ${lim}_{n\to \infty }\parallel x_n-T{x}_n\parallel =0$, then $Tx=x$.

Lemma 4

([24]). Let S be a closed, convex, and nonempty subset of H and $\left\{x_n\right\}$ be a sequence in H. If

(i) ${lim}_{n\to \infty }\parallel x_n-x\parallel$ exists for each $x\in S$;

(ii) ${\omega }_w\left(x_n\right)\subseteq S$;

then $\left\{x_n\right\}$ converges weakly to a point in S.

3. Main Results

Assume that the problem (3) is consistent and let $\Omega$ denote its solution set. That is, $\Omega$ is not empty and $\Omega :=\{w\in H:w\in {\bigcap }_{i=1}^t{S}_i,Gw=0\}$.

Remark 1.

$w_n$ is a solution of the problem (3) if and only if the equality (8) holds.

On the one hand, if $\parallel w_n+q_n-z_n\parallel =0$, then take $z\in \Omega$. We have

$$\begin{array}{ccc}\hfill 0& =& \langle w_n+q_n-z_n,w_n-z\rangle \hfill \\ & =& \langle w_n+G^{*}G{w}_n-P_{S_{i\left(n\right)}}{w}_n,w_n-z\rangle \hfill \\ & =& \langle w_n-P_{S_{i\left(n\right)}}{w}_n,w_n-z\rangle +\langle G{w}_n,G{w}_n-Gz\rangle \hfill \\ & \ge & \parallel w_n-P_{S_{i\left(n\right)}}{w}_n{\parallel }^2+{\parallel G{w}_n\parallel }^2.\hfill \end{array}$$

The first equality is from $\parallel w_n+q_n-z_n\parallel =0$, the second one is from the definition of $q_n$ and $z_n$, and the last inequality is from Lemma 1$\left(iii\right)$ and $Gz=0$. Then

$$\begin{array}{c}\hfill \parallel w_n-P_{S_{i\left(n\right)}}{w}_n\parallel =0and\parallel G{w}_n\parallel =0,\end{array}$$

which implies that

$$\begin{array}{c}\hfill \parallel w_n-P_{S_i}{w}_n\parallel =0,i\in \Lambda and\parallel G{w}_n\parallel =0.\end{array}$$

Hence $w_n\in {\bigcap }_{i=1}^t{S}_i$ and $G{w}_n=0$. Namely, $w_n\in \Omega$.

Conversely, if $w_n$ is a solution of the problem (3), that is $w_n\in {\bigcap }_{i=1}^t{S}_i$ and $G{w}_n=0$, then $q_n=G^{*}G{w}_n=0$ and $z_n=P_{S_{i\left(n\right)}}{w}_n=w_n$, so $w_n+q_n-z_n=0$. That is, $\parallel w_n+q_n-z_n\parallel =0$.

Next we discuss the convergence of the iterative sequence $\left\{w_n\right\}$ generated by Algorithm 1 if it does not terminate in finite steps.

Algorithm 1: Gradient method 1

Take $w_0\in H$ arbitrarily and compute $$\begin{array}{c}{z}_n=P_{S_{i\left(n\right)}}{w}_n,\hfill \\ q_n=G^{*}G{w}_n,\hfill \end{array}$$ (7) where $n\ge 0$ and $$i\left(n\right)\in \{i|\underset{i\in \Lambda }{max}\parallel w_n-P_{S_i}{w}_n\parallel ,\Lambda =\{1,2,\cdots ,t\}\}.$$ If $$\parallel w_n+q_n-z_n\parallel =0,$$ (8) then stop. $w_n$ is the solution (based on Remark 1). Otherwise, calculate $$w_{n+1}=w_n-{\tau }_n(w_n+q_n-z_n),$$ (9) where $${\tau }_n={\lambda }_n\frac{\parallel w_n-z_n{\parallel }^2+{\parallel G{w}_n\parallel }^2}{2\parallel w_n+q_n-z_n{\parallel }^2},$$ in which ${\lambda }_n\in (0,4)$.

Theorem 1.

If $0<{lim\; inf}_{n\to \infty }{\lambda }_n\le {lim\; sup}_{n\to \infty }{\lambda }_n<4$, taking initial point $w_0\in H$ arbitrarily, then the sequence $\left\{w_n\right\}$ generated by Algorithm 1 converges weakly to a solution of the problem (3).

Proof. 

First, we show the boundedness of $\left\{w_n\right\}$. Take $z\in \Omega$. Based on the inequality in the process of Remark 1, we get

$$\begin{array}{ccc}\hfill \parallel w_{n+1}{-z\parallel }^2& =& \parallel w_n-z-{\tau }_n(w_n+q_n-z_n){\parallel }^2\hfill \\ & =& \parallel w_n{-z\parallel }^2-2{\tau }_n\langle w_n+q_n-z_n,w_n-z\rangle +{\tau }_n^2{\parallel w_n+q_n-z_n\parallel }^2\hfill \\ & \le & \parallel w_n{-z\parallel }^2-{\lambda }_n\frac{(\parallel w_n-z_n{\parallel }^2+\parallel G{w}_n{{\parallel }^2)}^2}{\parallel w_n+q_n-z_n{\parallel }^2}\hfill \\ & & +\frac{{\lambda }_n^2}{4}\frac{(\parallel w_n-z_n{\parallel }^2+\parallel G{w}_n{{\parallel }^2)}^2}{\parallel w_n+q_n-z_n{\parallel }^2}\hfill \\ & =& \parallel w_n{-z\parallel }^2-{\lambda }_n(1-\frac{{\lambda }_n}{4})\frac{(\parallel w_n-z_n{\parallel }^2+\parallel G{w}_n{{\parallel }^2)}^2}{\parallel w_n+q_n-z_n{\parallel }^2}.\hfill \end{array}$$

(10)

This implies that ${lim}_{n\to \infty }\parallel w_n-z\parallel$ exists. Thus the sequence $\left\{w_n\right\}$ is bounded and so are the sequences $\left\{G{w}_n\right\}$ and $\left\{P_{S_i}{w}_n\right\}$, $i\in \Lambda$.

Next we show that ${\omega }_w\left(w_n\right)\subset \Omega$.

Since ${lim}_{n\to \infty }\parallel w_n-z\parallel$ exists and

$$\begin{array}{c}\hfill {\lambda }_n(1-\frac{{\lambda }_n}{4})\frac{(\parallel w_n-z_n{\parallel }^2+\parallel G{w}_n{{\parallel }^2)}^2}{\parallel w_n+q_n-z_n{\parallel }^2}\le \parallel w_n{-z\parallel }^2-{\parallel w_{n+1}-z\parallel }^2,\end{array}$$

together with the boundedness of the sequence $\{w_n+q_n-z_n\}$ and the definition of ${\lambda }_n$, it follows that

$$\begin{array}{c}\hfill \underset{n\to }{lim}\frac{(\parallel w_n-z_n{\parallel }^2+\parallel G{w}_n{{\parallel }^2)}^2}{\parallel w_n+q_n-z_n{\parallel }^2}=0,\end{array}$$

which implies that

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\parallel w_n-z_n\parallel =0and\underset{n\to \infty }{lim}\parallel G{w}_n\parallel =0.\end{array}$$

Hence,

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\parallel w_n-P_{S_i}{w}_n\parallel =0,i\in \Lambda and\underset{n\to \infty }{lim}\parallel G{w}_n\parallel =0.\end{array}$$

Since $\left\{w_n\right\}$ is bounded, let $w^{*}$ be a weak cluster point of $\left\{w_n\right\}$ with subsequence $\left\{w_{n_i}\right\}$ weakly convergent to it.

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\parallel w_{n_i}-P_{S_i}{w}_{n_i}\parallel =0,i\in \Lambda and\underset{n\to \infty }{lim}\parallel G{w}_{n_i}\parallel =0.\end{array}$$

By Lemma 3, we get $w^{*}\in \Omega$, and by the arbitrariness of $w^{*}\in {\omega }_w\left(w_n\right)$, we deduce that ${\omega }_w\left(w_n\right)\subset \Omega$. Moreover, the conditions in Lemma 4 have also been satisfied, and the sequence $\left\{w_n\right\}$ generated by the Algorithm 1 converges weakly to some solution of the problem (3). The proof is completed. □

There is only weak convergence in Theorem 1. Next, we show a strong convergence theorem for solving the problem (3).

Next, we discuss the convergence of the iterative sequence $\left\{w_n\right\}$ generated by Algorithm 2 if it does not terminate in finite steps.

Algorithm 2: Gradient method 2

Take $u\in H$ and initial point $w_0\in H$. Compute $$\begin{array}{c}{z}_n=P_{S_{i\left(n\right)}}{w}_n,\hfill \\ q_n=G^{*}G{w}_n,\hfill \end{array}$$ (11) where $$i\left(n\right)=\left\{i\right|\underset{i\in \Lambda }{max}\parallel w_n-P_{S_i}{w}_n\parallel ,\Lambda =\{1,2,\cdots ,t\}\}.$$ If $$\parallel w_n+q_n-z_n\parallel =0,$$ then stop. $w_n$ is the solution (by Remark 1). Otherwise, calculate $$w_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)(w_n-{\tau }_n(w_n+q_n-z_n)),$$ (12) where ${\alpha }_n\in (0,1)$, $n\ge 0$ and $${\tau }_n={\lambda }_n\frac{\parallel w_n-z_n{\parallel }^2+{\parallel G{w}_n\parallel }^2}{2\parallel w_n+q_n-z_n{\parallel }^2},$$ in which ${\lambda }_n\in (0,4)$.

Theorem 2.

Suppose that ${lim}_{n\to \infty }{\alpha }_n=0$, ${\Sigma }_{n=0}^{\infty }{\alpha }_n=\infty$, and $0<{lim\; inf}_{n\to \infty }{\lambda }_n\le {lim\; sup}_{n\to \infty }{\lambda }_n<4$. Taking $u\in H$ and initial point $w_0\in H$ arbitrarily, then the sequence $\left\{w_n\right\}$ generated by the Algorithm 2 converges strongly to $z=P_{\Omega }u$.

Proof. 

Let $u_n=w_n-{\tau }_n(w_n+q_n-z_n)$, for $n\ge 0$. From the process (10) in Theorem 1, we get

$$\parallel u_n{-z\parallel }^2\le {\parallel w_n-z\parallel }^2-{\lambda }_n(1-\frac{{\lambda }_n}{4})\frac{(\parallel w_n-z_n{\parallel }^2+\parallel G{w}_n{{\parallel }^2)}^2}{\parallel w_n+q_n-z_n{\parallel }^2}$$

(13)

by the definition of ${\lambda }_n$, that is, $\parallel u_n-z\parallel \le \parallel w_n-z\parallel$. Thus

$$\begin{array}{ccc}\hfill \parallel w_{n+1}-z\parallel & =& \parallel {\alpha }_{n}u+(1-{\alpha }_n)u_n-z\parallel \hfill \\ & \le & {\alpha }_n\parallel u-z\parallel +(1-{\alpha }_n)\parallel u_n-z\parallel \hfill \\ & \le & {\alpha }_n\parallel u-z\parallel +(1-{\alpha }_n)\parallel w_n-z\parallel \hfill \\ & \le & max\{\parallel w_n-z\parallel ,\parallel u-z\parallel \}.\hfill \end{array}$$

By induction, we derive

$$\parallel w_{n+1}-z\parallel \le max\{\parallel w_0-z\parallel ,\parallel u-z\parallel \},$$

which means that the sequence $\left\{w_n\right\}$ is bounded and so are the sequences $\left\{G{w}_n\right\}$ and $\left\{P_{S_i}{w}_n\right\}$, $i\in \Lambda$. By a simple derivation,

$$\begin{array}{ccc}\hfill \parallel w_{n+1}{-z\parallel }^2& =& \parallel {\alpha }_n(u-z)+(1-{\alpha }_n)(u_n-z){\parallel }^2\hfill \\ & \le & (1-{\alpha }_n){\parallel u_n-z\parallel }^2+2{\alpha }_n\langle u-z,w_{n+1}-z\rangle .\hfill \end{array}$$

Then by (13),

$$\begin{array}{ccc}\hfill \parallel w_{n+1}{-z\parallel }^2& \le & (1-{\alpha }_n){\parallel w_n-z\parallel }^2+2{\alpha }_n\langle u-z,w_{n+1}-z\rangle \hfill \\ & & -(1-{\alpha }_n){\lambda }_n(1-\frac{{\lambda }_n}{4})\frac{(\parallel w_n-z_n{\parallel }^2+\parallel G{w}_n{{\parallel }^2)}^2}{\parallel w_n+q_n-z_n{\parallel }^2}\hfill \\ & =& (1-{\alpha }_n)\parallel w_n-z{\parallel }^2+{\alpha }_n[2\langle u-z,w_{n+1}-z\rangle \hfill \\ & & -\frac{(1-{\alpha }_n)}{{\alpha }_n}{\lambda }_n(1-\frac{{\lambda }_n}{4})\frac{(\parallel w_n-z_n{\parallel }^2+\parallel G{w}_n{{\parallel }^2)}^2}{\parallel w_n+q_n-z_n{\parallel }^2}].\hfill \end{array}$$

(14)

Let

$${\theta }_n={\parallel w_n-z\parallel }^2,$$

$${\delta }_n=2\langle u-z,w_{n+1}-z\rangle -\frac{(1-{\alpha }_n)}{{\alpha }_n}{\lambda }_n(1-\frac{{\lambda }_n}{4})\frac{(\parallel w_n-z_n{\parallel }^2+\parallel G{w}_n{{\parallel }^2)}^2}{\parallel w_n+q_n-z_n{\parallel }^2}.$$

Then the inequality (14) equals

$${\theta }_{n+1}\le (1-{\alpha }_n){\theta }_n+{\alpha }_n{\delta }_n,$$

(15)

and also

$$0\le {\theta }_{n+1}\le (1-{\alpha }_n){\theta }_n+{\alpha }_n{\delta }_n,n\ge 0.$$

It follows that

$${\delta }_n\le 2\langle u-z,w_{n+1}-z\rangle \le 2\parallel u-z\parallel \parallel w_{n+1}-z\parallel .$$

So

$$\underset{n\to \infty }{lim\; sup}{\delta }_n<\infty .$$

Next, we show that ${lim\; sup}_{n\to \infty }{\delta }_n\ge -1$. Otherwise, if ${lim\; sup}_{n\to \infty }{\delta }_n<-1$, then by the definition of the supremum, there exists m such that ${\delta }_n\le -1$ for all $n\ge m$. It follows that for all $n\ge m$,

$$\begin{array}{ccc}\hfill {\theta }_{n+1}& \le & (1-{\alpha }_n){\theta }_n+{\alpha }_n{\delta }_n\hfill \\ & =& {\theta }_n+{\alpha }_n({\delta }_n-{\theta }_n)\hfill \\ & \le & {\theta }_n-{\alpha }_n.\hfill \end{array}$$

Thus

$${\theta }_{n+1}\le {\theta }_m-{\Sigma }_{i=m}^n{\alpha }_i.$$

Hence, taking lim sup as $n\to \infty$ in the above inequality, we obtain

$$0\le \underset{n\to \infty }{lim\; sup}{\theta }_{n+1}\le {\theta }_m-lim\; sup{\Sigma }_{i=m}^n{\alpha }_i=-\infty ,$$

which is a contradiction. Therefore, ${lim\; sup}_{n\to \infty }{\delta }_n\ge -1$, and it is finite. By the boundedness of $\left\{{\delta }_n\right\}$, we can take a subsequence $\left\{n_k\right\}$ of $\left\{n\right\}$ such that

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim\; sup}{\delta }_n& =& \underset{k\to \infty }{lim}{\delta }_{n_k}\hfill \\ & =& \underset{k\to \infty }{lim}[2\langle u-z,w_{n_k+1}-z\rangle \hfill \\ & & -\frac{(1-{\alpha }_{n_k})}{{\alpha }_{n_k}}{\lambda }_{n_k}(1-\frac{{\lambda }_{n_k}}{4})\frac{(\parallel w_{n_k}-z_{n_k}{\parallel }^2+\parallel G{w}_{n_k}{{\parallel }^2)}^2}{\parallel w_{n_k}+q_{n_k}-z_{n_k}{\parallel }^2}].\hfill \end{array}$$

Since the sequence $\left\{w_{n_k+1}\right\}$ is bounded, there exists a subsequence of $\left\{w_{n_k+1}\right\}$. Without loss of generality, we may assume it’s $\left\{w_{n_k+1}\right\}$ itself, such that ${lim}_{k\to \infty }\langle u-z,w_{n_k+1}-z\rangle$ exists. Consequently, the following limit exists:

$$\underset{k\to \infty }{lim}-\frac{(1-{\alpha }_{n_k})}{{\alpha }_{n_k}}{\lambda }_{n_k}(1-\frac{{\lambda }_{n_k}}{4})\frac{(\parallel w_{n_k}-z_{n_k}{\parallel }^2+\parallel G{w}_{n_k}{{\parallel }^2)}^2}{\parallel w_{n_k}+q_{n_k}-z_{n_k}{\parallel }^2}.$$

Together with the definitions of ${\alpha }_n$ and ${\lambda }_n$, it shows that

$$\underset{k\to \infty }{lim}\frac{(\parallel w_{n_k}-z_{n_k}{\parallel )}^2+{\parallel G{w}_{n_k}\parallel }^2}{\parallel w_{n_k}+q_{n_k}-z_{n_k}{\parallel }^2}=0,$$

which yields

$$\underset{k\to \infty }{lim}\parallel w_{n_k}-z_{n_k}\parallel =0and\underset{k\to \infty }{lim}\parallel G{w}_{n_k}\parallel =0.$$

Following the proof procedure of Theorem 1, we conclude that ${\omega }_w\left(w_{n_k}\right)\subset \Omega$. Since

$$\begin{array}{ccc}\hfill \parallel w_{n_k+1}-w_{n_k}\parallel & =& \parallel {\alpha }_{n_k}u+(1-{\alpha }_{n_k})u_{n_k}-w_{n_k}\parallel \hfill \\ & \le & {\alpha }_{n_k}\parallel u-w_{n_k}\parallel +(1-{\alpha }_{n_k})\parallel u_{n_k}-w_{n_k}\parallel \hfill \\ & =& {\alpha }_{n_k}\parallel u-w_{n_k}\parallel +(1-{\alpha }_{n_k}){\tau }_{n_k}\parallel w_{n_k}+q_{n_k}-z_{n_k}\parallel \hfill \\ & =& {\alpha }_{n_k}\parallel u-w_{n_k}\parallel +(1-{\alpha }_{n_k}){\lambda }_{n_k}\frac{\parallel w_{n_k}-z_{n_k}{\parallel }^2+{\parallel G{w}_{n_k}\parallel }^2}{\parallel w_{n_k}+q_{n_k}-z_{n_k}\parallel }\hfill \\ & \to & 0,\hfill \end{array}$$

assume that $w_{n_k+1}\rightharpoonup w^{*}\in \Omega$. Then

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim\; sup}{\delta }_n& =& \underset{k\to \infty }{lim}{\delta }_{n_k}\hfill \\ & =& \underset{k\to \infty }{lim}[2\langle u-z,w_{n_k+1}-z\rangle \hfill \\ & & -\frac{(1-{\alpha }_{n_k})}{{\alpha }_{n_k}}{\lambda }_{n_k}(1-\frac{{\lambda }_{n_k}}{4})\frac{(\parallel w_{n_k}-z_{n_k}{\parallel }^2+\parallel G{w}_{n_k}{{\parallel }^2)}^2}{\parallel w_{n_k}+q_{n_k}-z_{n_k}{\parallel }^2}]\hfill \\ & \le & \underset{k\to \infty }{lim}2\langle u-z,w_{n_k+1}-z\rangle \hfill \\ & =& 2\langle u-z,w^{*}-z\rangle \hfill \\ & \le & 0,\hfill \end{array}$$

due to the fact that $z=P_{\Omega }u$ and Lemma 1. Finally, applying Lemma 2 to (15), we conclude that $w_n\to z$. The proof is completed. □

4. Numerical Experiments

In this section, we provide several numerical results of the MSSEP (2) to confirm the effectiveness of the suggested Algorithm 1. The whole program was written in Wolfram Mathematica (version 9.0). All of the numerical results were carried out on a personal Lenovo computer with Intel(R)Core(TM) i5-6600 CPU 3.30 GHz and RAM 8.00 GB.

The MSSEP with $C_1=\{x\in R^2|\parallel x-(-1,1)\parallel \le 5\}$, $C_2=\{x\in R^2|\parallel x-(1,1)\parallel \le 5\}$, $C_3=\{x\in R^2|\parallel x-(0,-3)\parallel \le 5\}$, $Q_1=\{y\in R^3|\parallel y-(1,1,1)\parallel \le 5\}$, $Q_2=\{y\in R^3|\parallel y-(0,0,0)\parallel \le 5\}$, $Q_3=\{y\in R^3|\parallel y-(1,0,0)\parallel \le 5\}$, $A=\left(\begin{array}{cc}1& 2\\ 0& 3\\ 5& 2\end{array}\right)$, $B=\left(\begin{array}{ccc}2& 0& 1\\ 3& 2& 3\\ 1& 0& 0\end{array}\right)$, $\Lambda =\{1,2,3\}$, ${\lambda }_n=0.6$. We choose two initial values $x_0=(2,2)$, $y_0=(2,2,2)$ and $x_0=(20,20)$, $y_0=(10,10,10)$ and take the iterative steps n as the transverse axis and $\parallel Ax-By\parallel$ as the vertical axis in the figures below (Figure 1 and Figure 2). We considered using the Algorithm 1 to solve this MSSEP.

The figures above confirm the effectiveness of the proposed Algorithm 1 and also show that there is an approximately linear downward trend after finite steps, which means the convergence rate of the proposed Algorithm 1 may be fast enough.