An Iterative Method for Solving the Multiple-Sets Split Variational Inequality Problem

Abstract

In this work, we introduce a new algorithm for finding the minimum-norm solution of the multiple-sets split variational inequality problem in real Hilbert spaces. The strong convergence of the iterative sequence generated by the algorithm method is established under the condition that the mappings are monotone and Lipschitz continuous. We apply our main result to study the minimum-norm solution of the multiple-sets split feasibility problem and the split variational inequality problem. Finally, a numerical example is given to illustrate the proposed algorithm.

1 Introduction

Let $\mathcal {H}_1$ and $\mathcal {H}_2$ be two real Hilbert spaces, and let $A:\mathcal {H}_1\longrightarrow \mathcal {H}_2$ be a bounded linear operator. Given nonempty closed convex subsets $C_i\subseteq \mathcal {H}_1$, $Q_j\subseteq \mathcal {H}_2$ for $i=1,2,\ldots ,M$ and $j=1,2,\ldots , N$, respectively. Let $F_i: \mathcal {H}_1\longrightarrow \mathcal {H}_1$, $i=1,2,\ldots ,M$ and $G_j: \mathcal {H}_2\longrightarrow \mathcal {H}_2$, $j=1,2,\ldots , N$ be mappings. The multiple-sets split variational inequality problem (MSSVIP), introduced first by Censor et al. [18], can be formulated as follows:

such that

If the solution sets of variational inequality problems $VIP(C_i, F_i)$ and $VIP(Q_j, G_j)$ are denoted by $\hbox {Sol}(C_i, F_i)$ and $\hbox {Sol}(Q_j, G_j)$, respectively, then the MSSVIP becomes the problem of finding $x^*\in \displaystyle \bigcap \nolimits _{i=1}^{M}\hbox {Sol}(C_i, F_i)$ such that $Ax^*\in \displaystyle \bigcap \nolimits _{j=1}^{N}\hbox {Sol}(Q_j, G_j)$.

When $M=N=1$, the MSSVIP reduces to the split variational inequality problem, shortly SVIP,

such that

where C and Q are two nonempty closed convex subsets of $\mathcal {H}_1$ and $\mathcal {H}_2$, respectively, and $F: \mathcal {H}_1\longrightarrow \mathcal {H}_1$, $G: \mathcal {H}_2\longrightarrow \mathcal {H}_2$ are two mappings.

The SVIP was introduced and investigated by Censor et al. [18] in the case when F is $\alpha$-inverse strongly monotone on $\mathcal {H}_1$, G is $\beta$-inverse strongly monotone on $\mathcal {H}_2$, which starts from a given point $x^0\in \mathcal {H}_1$, for all $n\ge 0$, the next iterate is defined as

$$\begin{aligned} x^{n+1}=P_C^{F, \lambda }(x^n+\gamma A^*(P_Q^{G, \lambda }-I)(Ax^n)), \end{aligned}$$

where $\gamma \in \Big (0, \displaystyle \frac{1}{\Vert A\Vert ^2}\Big )$, $0\le \lambda \le 2\min \{\alpha , \beta \}$, $P_C^{F, \lambda }=P_C(I-\lambda F)$ and $P_Q^{G, \lambda }=P_Q(I-\lambda G)$. They showed that the sequence $\{x^n\}$ converges weakly to a solution of the split variational inequality problem, provided that the solution set of the SVIP is nonempty. If we consider only the problem VIP(C, F), then VIP(C, F) is a classical variational inequality problem, which was studied by many authors, see, for example, [1, 4, 5, 19, 20, 24,25,26, 28, 31, 34, 35, 38, 42, 43] and the references therein.

If $F_i=0$ for all $i=1,2,\ldots ,M$ and $G_j=0$ for all $j=1,2,\ldots ,N$, then the MSSVIP becomes the multiple-sets split feasibility problem (MSSFP):

$$\begin{aligned} \text {Find } x^*\in \displaystyle \bigcap \limits _{i=1}^{M}C_i\text { such that } Ax^*\in \displaystyle \bigcap \limits _{j=1}^{N}Q_j. \end{aligned}$$

The MSSFP, which has been first introduced and studied by Censor et al [17], is a general way to characterize various linear inverse problems which arises in many real-world application problems such as intensity-modulated radiation therapy [14, 15] and medical image reconstruction [22]. A special case of the MSSFP, when $M=N=1$, is the split feasibility problem (SFP), which is formulated as finding a point $x^*\in C$ such that $Ax^*\in Q$, where C and Q are two nonempty closed convex subsets of real Hilbert spaces $\mathcal {H}_1$ and $\mathcal {H}_2$, respectively. Recently, it has been found that the SFP has some practical applications in various disciplines such as signal processing, image restoration, intensity-modulated radiation therapy [15, 16, 21] and many other fields. Many iterative projection methods have been developed to solve the MSSFP, the SFP and their generalizations. See, for example, [2, 3, 6,7,8,9,10,11,12, 17, 23, 27, 29, 30, 32, 33, 36, 39,40,41, 44, 45, 47, 48] and the references cited there.

In 2012, Ceng et al. [13, Theorem 3.1] presented a method, which is called the relaxed extragradient method, for finding minimum-norm solution of the SFP. This method is given as: For $x^0\in \mathcal {H}_1$, the sequence $\{x^n\}$ is generated by

$$\begin{aligned} {\left\{ \begin{array}{ll} {y^n=P_C(x^n-\lambda _n\nabla f_{\alpha _n}(x^n))},\\ {x^{n+1}=\beta _nx^n+\gamma _ny^n+\delta _nP_C(x^n-\lambda _n\nabla f_{\alpha _n}(y^n))\;\;\forall n\ge 0}, \end{array}\right. } \end{aligned}$$

(1)

where $\nabla f_{\alpha _n}$ stands for $A^*(I-P_Q)A+\alpha _n I$. They proved that, under some conditions imposed on $\{\lambda _n\}$, $\{\alpha _n\}$, $\{\beta _n\}$, $\{\gamma _n\}$ and $\{\delta _n\}$, the sequence $\{x^n\}$ generated by (1) converges strongly to the minimum-norm solution of the SFP if the solution set $\varGamma$ of the SFP is nonempty.

Motivated and inspired by the above-mentioned works, in this paper, we present an iterative method for finding minimum-norm solution of the MSSVIP, which is to find a point $x^*\in \varOmega$ such that

$$\begin{aligned} \Vert x^*\Vert \le \Vert x\Vert \;\;\forall x\in \varOmega , \end{aligned}$$

where $\varOmega =\Big \{x^*\in \displaystyle \bigcap \limits _{i=1}^{M}\hbox {Sol}(C_i, F_i): Ax^*\in \displaystyle \bigcap \limits _{j=1}^{N}\hbox {Sol}(Q_j, G_j)\Big \}$ is the solution set of the MSSVIP and $A:\mathcal {H}_1\longrightarrow \mathcal {H}_2$ is a bounded linear operator between $\mathcal {H}_1$ and $\mathcal {H}_2$.

The paper is organized as follows. Section 2 deals with some definitions and lemmas that will be used in the third section, where we describe the algorithm and prove its strong convergence. We close Sect. 3 by considering some applications to the multiple-sets split feasibility problem and the split variational inequality problem. Finally, in the last section, we perform a numerical example to illustrate the convergence of the proposed algorithm.

2 Preliminaries

Let C be a nonempty closed convex subset of a real Hilbert space $\mathcal {H}$. It is well known that for all point $x\in \mathcal {H}$, there exists a unique point $P_C(x)\in C$ such that

$$\begin{aligned} \Vert x-P_C(x)\Vert =\min \{\Vert x-y\Vert : y\in C\}. \end{aligned}$$

(2)

The mapping $P_C:\mathcal {H}\longrightarrow C$ defined by (2) is called the metric projection of $\mathcal {H}$ onto C. It is known that $P_C$ is nonexpansive. Moreover, for all $x\in \mathcal {H}$ and $y\in C$, $y=P_C(x)$ if and only if

$$\begin{aligned} \langle x-y, z-y\rangle \le 0\;\;\forall z\in C. \end{aligned}$$

We denote the strong (weak) convergence of a sequence $\{x^n\}$ to x in $\mathcal {H}$ by $x^n\longrightarrow x$ ($x^n\rightharpoonup x$), respectively. In what follows, we recall some well-known definitions and lemmas which will be used in this paper.

Definition 1

Let $\mathcal {H}_1$ and $\mathcal {H}_2$ be two Hilbert spaces and let $A: \mathcal {H}_1\longrightarrow \mathcal {H}_2$ be a bounded linear operator. An operator $A^*: \mathcal {H}_2\longrightarrow \mathcal {H}_1$ with the property $\langle A(x), y\rangle =\langle x, A^*(y)\rangle$ for all $x\in \mathcal {H}_1$ and $y\in \mathcal {H}_2$, is called an adjoint operator.

The adjoint operator of a bounded linear operator A between Hilbert spaces $\mathcal {H}_1$, $\mathcal {H}_2$ always exists and is uniquely determined. Furthermore, $A^*$ is a bounded linear operator and $\Vert A^*\Vert =\Vert A\Vert$.

Definition 2

A mapping $F:\mathcal {H}\longrightarrow \mathcal {H}$ is said to be

(i) L-Lipschitz continuous on $\mathcal {H}$ if

$$\begin{aligned} \Vert F(x)-F(y)\Vert \le L\Vert x-y\Vert \;\;\forall x, y\in \mathcal {H}; \end{aligned}$$

(ii) monotone on $\mathcal {H}$ if

$$\begin{aligned} \langle F(x)-F(y), x-y\rangle \ge 0\;\;\forall x, y\in \mathcal {H}. \end{aligned}$$

Lemma 1

Let C be a nonempty closed convex subset of a real Hilbert space $\mathcal {H}$. Let $G: \mathcal {H}\longrightarrow \mathcal {H}$ be monotone and L-Lipschitz continuous on $\mathcal {H}$ such that $\hbox {Sol}(C, G)\ne \emptyset$. Let $x\in \mathcal {H}$, $0<\lambda <\dfrac{1}{L}$ and define

$$\begin{aligned} \begin{aligned} y&=P_C(x-\lambda G(x)),\\ z&=x-y-\lambda (G(x)-G(y)),\\ t&={\left\{ \begin{array}{ll} x-\dfrac{\langle x-y, z\rangle }{\Vert z\Vert ^2}z &{} \text{ if } y\ne x,\\ x &{} \text{ if } y=x. \end{array}\right. } \end{aligned} \end{aligned}$$

Then,

• (i)
  $$\begin{aligned} \langle x-y, z\rangle \ge (1-\lambda L)\Vert x-y\Vert ^2. \end{aligned}$$

  (3)
• (ii)
  $$\begin{aligned} \Vert x-y\Vert \le \dfrac{\sqrt{1+\lambda ^2L^2}}{1-\lambda L}\Vert x-t\Vert . \end{aligned}$$

  (4)

• (iii) For all $x^*\in \hbox {Sol}(C, G)$

  $$\begin{aligned} \Vert t-x^*\Vert ^2\le \Vert x-x^*\Vert ^2-\Vert x-t\Vert ^2. \end{aligned}$$

  (5)

Proof

(i) From the definition of z and the Lipschitz continuity of G, we have

$$\begin{aligned} \langle x-y, z\rangle&=\langle x-y, x-y-\lambda (G(x)-G(y))\rangle \\&=\Vert x-y\Vert ^2-\lambda \langle x-y, G(x)-G(y)\rangle \\&\ge \Vert x-y\Vert ^2-\lambda \Vert x-y\Vert \Vert G(x)-G(y)\Vert \\&\ge \Vert x-y\Vert ^2-\lambda L\Vert x-y\Vert ^2\\&=(1-\lambda L)\Vert x-y\Vert ^2. \end{aligned}$$

The inequality (3) holds. Since $1-\lambda L>0$, from (3) we have that $z\ne 0$ if $y\ne x$. So t is well defined. (ii) If $y=x$ then $t=x$, so (4) holds. Suppose that $y\ne x$, from the monotonicity and the L-Lipschitz continuity of G, we have

$$\begin{aligned} \begin{aligned} \Vert z\Vert ^2&=\Vert x-y-\lambda (G(x)-G(y))\Vert ^2\\&=\Vert x-y\Vert ^2+\lambda ^2\Vert G(x)-G(y)\Vert ^2-2\lambda \langle x-y, G(x)-G(y)\rangle \\&\le \Vert x-y\Vert ^2+\lambda ^2\Vert G(x)-G(y)\Vert ^2\\&\le \Vert x-y\Vert ^2+\lambda ^2L^2\Vert x-y\Vert ^2\\&=(1+\lambda ^2L^2)\Vert x-y\Vert ^2. \end{aligned} \end{aligned}$$

This implies that

$$\begin{aligned} \Vert x-y\Vert \ge \dfrac{\Vert z\Vert }{\sqrt{1+\lambda ^2L^2}}. \end{aligned}$$

So, by the definition of t and (3), we have

$$\begin{aligned} \Vert x-t\Vert =\bigg \Vert \dfrac{\langle x-y, z\rangle }{\Vert z\Vert ^2}z\bigg \Vert =\dfrac{\langle x-y, z\rangle }{\Vert z\Vert }\ge \dfrac{(1-\lambda L)\Vert x-y\Vert ^2}{\Vert z\Vert }\ge \dfrac{(1-\lambda L)\Vert x-y\Vert }{\sqrt{1+\lambda ^2L^2}} \end{aligned}$$

and hence

$$\begin{aligned} \Vert x-y\Vert \le \dfrac{\sqrt{1+\lambda ^2L^2}}{1-\lambda L}\Vert x-t\Vert . \end{aligned}$$

(iii) If $y=x$ then $t=x$. The inequality (5) holds. Suppose that $y\ne x$. Since G is monotone and $x^*\in \hbox {Sol}(C, G)$, we have

$$\begin{aligned} \langle G(y), y-x^*\rangle \ge \langle G(x^*), y-x^*\rangle \ge 0. \end{aligned}$$

(6)

From $y=P_C(x-\lambda G(x))$ and $x^*\in C$, we have

$$\begin{aligned} \langle x-\lambda G(x)-y, x^*-y\rangle \le 0. \end{aligned}$$

Combining the above inequality with $x-\lambda G(x)-y=z-\lambda G(y)$, we get

$$\begin{aligned} \langle z-\lambda G(y), x^*-y\rangle \le 0. \end{aligned}$$

(7)

It follows from (6) and (7) that

$$\begin{aligned} \langle z, x^*-y\rangle \le \lambda \langle G(y), x^*-y\rangle \le 0. \end{aligned}$$

(8)

By the definition of t, (3) and (8), we have

$$\begin{aligned} \Vert t-x^*\Vert ^2&=\Big \Vert (x-x^*)-\dfrac{\langle x-y, z\rangle }{\Vert z\Vert ^2}z\Big \Vert ^2\\&=\Vert x-x^*\Vert ^2-\dfrac{2\langle x-y, z\rangle }{\Vert z\Vert ^2}\langle x-x^*, z\rangle + \dfrac{(\langle x-y, z\rangle )^2}{\Vert z\Vert ^2}\\&=\Vert x-x^*\Vert ^2-\dfrac{(\langle x-y, z\rangle )^2}{\Vert z\Vert ^2}+\dfrac{2\langle x-y, z\rangle }{\Vert z\Vert ^2}\langle x^*-y, z\rangle \\&\le \Vert x-x^*\Vert ^2-\dfrac{(\langle x-y, z\rangle )^2}{\Vert z\Vert ^2}\\&=\Vert x-x^*\Vert ^2-\Vert x-t\Vert ^2. \end{aligned}$$

The proof is complete. $\square$

Lemma 2

([29, Lemma 3]) Let C be a nonempty closed convex subset of a real Hilbert space $\mathcal {H}$. Let $F: \mathcal {H}\longrightarrow \mathcal {H}$ be monotone and L-Lipschitz continuous on $\mathcal {H}$. Assume that $\lambda _n\ge a>0$ for all n, $\{x^n\}$ is a sequence in $\mathcal {H}$ satisfying $x^n\rightharpoonup \overline{x}$ and $\lim \limits _{n\longrightarrow \infty }\Vert x^n-y^n\Vert =0$, where $y^n=P_C(x^n-\lambda _n F(x^n))$ for all n. Then $\overline{x}\in \hbox {Sol}(C, F)$.

Lemma 3

([37, Remark 4.4]) Let $\{a_n\}$ be a sequence of nonnegative real numbers. Suppose that for any integer m, there exists an integer p such that $p\ge m$ and $a_p\le a_{p+1}$. Let $n_0$ be an integer such that $a_{n_0}\le a_{n_0+1}$ and define, for all integer $n\ge n_0$, by

$$\begin{aligned} \tau (n)=\max \{k\in \mathbb {N}: n_0\le k\le n, a_k\le a_{k+1}\}. \end{aligned}$$

Then, $\{\tau (n)\}_{n\ge n_0}$ is a nondecreasing sequence satisfying $\lim \limits _{n\longrightarrow \infty }\tau (n)=\infty$ and the following inequalities hold:

$$\begin{aligned} a_{\tau (n)}\le a_{\tau (n)+1}, \ a_n\le a_{\tau (n)+1} \ \forall n\ge n_0. \end{aligned}$$

Lemma 4

( [46, Lemma 2.5]) Assume $\{a_n\}$ is a sequence of nonnegative real numbers satisfying the condition

$$\begin{aligned} a_{n+1}\le (1-\alpha _n)a_n+\alpha _n\xi _n,\;\;\forall n\ge 0, \end{aligned}$$

where $\{\alpha _n\}$ is a sequence in (0, 1) and $\{\xi _n\}$ is a sequence in $\mathbb {R}$ such that

• (i) $\displaystyle \sum \limits _{n=0}^{\infty }\alpha _n=\infty$;

• (ii) $\limsup \limits _{n\longrightarrow \infty }\xi _n\le 0$. Then, $\lim \limits _{n\longrightarrow \infty }a_n=0$.

3 The Algorithm and Convergence Analysis

In this section, we propose a strong convergence algorithm for finding the minimum-norm solution of the multiple-sets split variational inequality problem. We impose the following assumptions on the mappings $F_i: \mathcal {H}_1\longrightarrow \mathcal {H}_1$ and $G_j: \mathcal {H}_2\longrightarrow \mathcal {H}_2$, $i=1,2,\ldots ,M$ and $j=1,2,\ldots , N$.

$(A_F)$: $F_i$ is monotone and $L_i$-Lipschitz continuous on $\mathcal {H}_1$ for each $i=1,2,\ldots ,M$.

$(A_G)$: $G_j$ is monotone and $\kappa _j$-Lipschitz continuous on $\mathcal {H}_2$ for each $j=1,2,\ldots ,N$.

The algorithm can be expressed as follows

Algorithm 1.

Step 0. Choose $\{\delta _n\}\subset [\underline{\delta }, \overline{\delta }]\subset \Big (0, \dfrac{1}{\Vert A\Vert ^2+1}\Big )$, $\{\lambda _n\}\subset [\underline{\lambda }, \overline{\lambda }]\subset \Big (0, \min \limits _{1\le i\le M}\Big \{\displaystyle \frac{1}{L_i}\Big \}\Big )$, $\{\mu _n\}\subset [\underline{\mu }, \overline{\mu }]\subset \Big (0, \min \limits _{1\le j\le N}\Big \{\displaystyle \frac{1}{\kappa _j}\Big \}\Big )$, $\{\alpha _n\}\subset (0, 1)$ such that $\lim \limits _{n\longrightarrow \infty }\alpha _n\!=\!0$, $\displaystyle \sum \limits _{n=0}^{\infty }\!\alpha _n\!=\!\infty$. Step1. Let $x^0\in \mathcal {H}_1$. Set $n:=0$.

Step 2. Compute, for all $j=1,2,\ldots ,N$

$$\begin{aligned} \begin{aligned} b^n&=A(x^n),\\ u_j^n&=P_{Q_j}(b^n-\mu _n G_j(b^n)),\\ v_j^n&=b^n-u_j^n-\mu _n(G_j(b^n)-G_j(u_j^n)),\\ w_j^n&= {\left\{ \begin{array}{ll} b^n-\dfrac{\langle b^n-u_j^n, v_j^n\rangle }{\Vert v_j^n\Vert ^2}v_j^n &{} \text{ if } u_j^n\ne b^n,\\ b^n &{} \text{ if } u_j^n=b^n. \end{array}\right. }\\ \end{aligned} \end{aligned}$$

Step 3. Find among $w_j^n$, $j=1,2,\ldots ,N$, the farthest element from $b^n$, i.e.,

$$\begin{aligned} j_n=\hbox {argmax}\{\Vert w_j^n-b^n\Vert : j=1,2,\ldots ,N\},\;r^n:=w_{j_n}^n. \end{aligned}$$

Step 4. Compute, for all $i=1,2,\ldots ,M$

$$\begin{aligned} \begin{aligned} a^n&=x^n+\delta _n A^*(r^n-b^n),\\ y_i^n&=P_{C_i}(a^n-\lambda _n F_i(a^n)),\\ z_i^n&=a^n-y_i^n-\lambda _n(F_i(a^n)-F_i(y_i^n)),\\ s_i^n&= {\left\{ \begin{array}{ll} a^n-\dfrac{\langle a^n-y_i^n, z_i^n\rangle }{\Vert z_i^n\Vert ^2}z_i^n &{} \text{ if } y_i^n\ne a^n,\\ a^n &{} \text{ if } y_i^n=a^n. \end{array}\right. } \end{aligned} \end{aligned}$$

Step 5. Find among $s_i^n$, $i=1,2,\ldots ,M$, the farthest element from $a^n$, i.e.,

$$\begin{aligned} i_n=\hbox {argmax}\{\Vert s_i^n-a^n\Vert : i=1,2,\ldots ,M\},\;t^n:=s_{i_n}^n. \end{aligned}$$

Step 6. Compute

$$\begin{aligned} x^{n+1}=(1-\alpha _n)t^n. \end{aligned}$$

Step 7. Set $n:=n+1$, and go to Step 2.

The following theorem shows validity and convergence of the algorithm.

Theorem 1

Suppose that the assumptions $(A_F),(A_G)$ hold. Then, the sequence $\{x^n\}$ generated by Algorithm 1 converges strongly to the minimum-norm solution of the multiple-sets split variational inequality problem.

Proof

The proof of the theorem is divided into several steps. Let $x^*=P_\varOmega (0)$. We will prove that $\{x^n\}$ converges in norm to $x^*$. We divide the proof into several steps. Step 1. For all $n\ge 0$, we have

$$\begin{aligned} \Vert x^{n+1}-x^*\Vert&\le (1-\alpha _n)\Vert t^n-x^*\Vert +\alpha _n\Vert x^*\Vert , \end{aligned}$$

(9)

$$\begin{aligned} \Vert x^{n+1}-x^*\Vert ^2&\le (1-\alpha _n)\Vert t^n-x^*\Vert ^2+2\alpha _n\langle x^*, x^*-t^n+\alpha _n t^n\rangle . \end{aligned}$$

(10)

In fact, it holds that

$$\begin{aligned} \Vert x^{n+1}-x^*\Vert&=\Vert (1-\alpha _n)t^n-x^*\Vert \\&=\Vert (1-\alpha _n)(t^n-x^*)-\alpha _nx^*\Vert \\&\le (1-\alpha _n)\Vert t^n-x^*\Vert +\alpha _n\Vert x^*\Vert , \end{aligned}$$

and

$$\begin{aligned} \Vert x^{n+1}-x^*\Vert ^2&=\Vert (1-\alpha _n)(t^n-x^*)-\alpha _nx^*\Vert ^2\\&=(1-\alpha _n)^2\Vert t^n-x^*\Vert ^2-2\alpha _n(1-\alpha _n)\langle x^*, t^n-x^*\rangle +\alpha _n^2\Vert x^*\Vert ^2\\&\le (1-\alpha _n)\Vert t^n-x^*\Vert ^2-2\alpha _n(1-\alpha _n)\langle x^*, t^n-x^*\rangle +2\alpha _n^2\Vert x^*\Vert ^2\\&=(1-\alpha _n)\Vert t^n-x^*\Vert ^2+2\alpha _n\langle x^*, x^*-t^n+\alpha _n t^n\rangle . \end{aligned}$$

Step 2. For all $n\ge 0$, we show that

$$\begin{aligned} \Vert a^n-x^*\Vert ^2\le \Vert x^n-x^*\Vert ^2-\delta _n(1-\delta _n\Vert A\Vert ^2)\Vert r^n-b^n\Vert ^2. \end{aligned}$$

(11)

Since $x^*\in \varOmega$, we have $x^*\in \hbox {Sol}(C_i, F_i)$, $Ax^*\in \hbox {Sol}(Q_j, G_j)$ for all $i=1,2,\ldots ,M$, $j=1,2,\ldots ,N$. From Lemma 1, we get, for all $n\ge 0$

$$\begin{aligned} \Vert r^n-Ax^*\Vert ^2&=\Vert w_{j_n}^n-Ax^*\Vert ^2\nonumber \\&\le \Vert b^n-Ax^*\Vert ^2-\Vert b^n-w_{j_n}^n\Vert ^2\nonumber \\&=\Vert b^n-Ax^*\Vert ^2-\Vert b^n-r^n\Vert ^2. \end{aligned}$$

(12)

Hence,

$$\begin{aligned} \Vert r^n-Ax^*\Vert \le \Vert b^n-Ax^*\Vert . \end{aligned}$$

(13)

From (13), we obtain

$$\begin{aligned} \Vert a^n-x^*\Vert ^2&=\Vert (x^n-x^*)+\delta _n A^*(r^n-b^n)\Vert ^2\\&=\Vert x^n-x^*\Vert ^2+\Vert \delta _n A^*(r^n-b^n)\Vert ^2+2\delta _n\langle x^n-x^*, A^*(r^n-b^n)\rangle \\&\le \Vert x^n-x^*\Vert ^2+\delta _n^2\Vert A^*\Vert ^2\Vert r^n-b^n\Vert ^2+2\delta _n\langle A(x^n-x^*), r^n-b^n\rangle \\&=\Vert x^n-x^*\Vert ^2+\delta _n^2\Vert A\Vert ^2\Vert r^n-b^n\Vert ^2+2\delta _n\langle b^n-Ax^*, r^n-b^n\rangle \\&=\Vert x^n-x^*\Vert ^2+\delta _n^2\Vert A\Vert ^2\Vert r^n-b^n\Vert ^2+2\delta _n[\langle r^n-Ax^*, r^n-b^n\rangle \\&\quad -\Vert r^n-b^n\Vert ^2]\\&=\Vert x^n-x^*\Vert ^2+\delta _n^2\Vert A\Vert ^2\Vert r^n-b^n\Vert ^2+\delta _n\big [\big (\Vert r^n-Ax^*\Vert ^2-\Vert b^n-Ax^*\Vert ^2\big )\\&\quad -\Vert r^n-b^n\Vert ^2\big ]\\&\le \Vert x^n-x^*\Vert ^2+\delta _n^2\Vert A\Vert ^2\Vert r^n-b^n\Vert ^2-\delta _n\Vert r^n-b^n\Vert ^2\\&=\Vert x^n-x^*\Vert ^2-\delta _n(1-\delta _n\Vert A\Vert ^2)\Vert r^n-b^n\Vert ^2. \end{aligned}$$

Step 3. We show that the sequences $\{x^n\}$, $\{a^n\}$ and $\{t^n\}$ are bounded. From Lemma 1, we have, for all $n\ge 0$

$$\begin{aligned} \Vert t^n-x^*\Vert ^2&=\Vert s_{i_n}^n-x^*\Vert ^2\nonumber \\&\le \Vert a^n-x^*\Vert ^2-\Vert a^n-s_{i_n}^n\Vert ^2\nonumber \\&=\Vert a^n-x^*\Vert ^2-\Vert a^n-t^n\Vert ^2. \end{aligned}$$

(14)

From (11), (14) and $\{\delta _n\}\subset {[}\underline{\delta }, \overline{\delta }]\subset \Big (0, \displaystyle \frac{1}{\Vert A\Vert ^2+1}\Big )$, we get

$$\begin{aligned} \Vert t^n-x^*\Vert \le \Vert a^n-x^*\Vert \le \Vert x^n-x^*\Vert \;\;\forall n\ge 0. \end{aligned}$$

(15)

Using (9) and (15), we have

$$\begin{aligned} \Vert x^{n+1}-x^*\Vert \le (1-\alpha _n)\Vert x^n-x^*\Vert +\alpha _n\Vert x^*\Vert \;\;\forall n\ge 0. \end{aligned}$$

This implies that

$$\begin{aligned} \Vert x^{n+1}-x^*\Vert \le \max \{\Vert x^n-x^*\Vert ,\Vert x^*\Vert \}\;\;\forall n\ge 0. \end{aligned}$$

So, by induction, we obtain, for every $n\ge 0$ that

$$\begin{aligned} \Vert x^n-x^*\Vert \le \max \{\Vert x^0-x^*\Vert ,\Vert x^*\Vert \}. \end{aligned}$$

Hence, the sequence $\{x^n\}$ is bounded and so are the sequences $\{a^n\}$ and $\{t^n\}$ thanks to (15). Step 4. We prove that $\{x^n\}$ converges strongly to $x^*=P_\varOmega (0)$. Let us consider two cases. Case 1. There exists $n_0$ such that the sequence $\{\Vert x^n-x^*\Vert \}$ is decreasing for $n\ge n_0$. In this case, the limit of $\{\Vert x^n-x^*\Vert \}$ exists. So, it follows from (10) and (15), for all $n\ge 0$, that

$$\begin{aligned} (\Vert x^{n+1}-x^*\Vert ^2-\Vert x^n-x^*\Vert ^2)-2\alpha _n\langle x^*, x^*-t^n+\alpha _n t^n\rangle&\le \Vert t^n-x^*\Vert ^2-\Vert x^n-x^*\Vert ^2\\&\le \Vert a^n-x^*\Vert ^2-\Vert x^n-x^*\Vert ^2\\&\le 0. \end{aligned}$$

Since the limit of $\{\Vert x^n-x^*\Vert \}$ exists, $\lim \limits _{n\longrightarrow \infty }\alpha _n=0$, $\{t^n\}$ is bounded, it follows from the above inequality that

$$\begin{aligned}&\lim \limits _{n\longrightarrow \infty }(\Vert t^n-x^*\Vert ^2-\Vert x^n-x^*\Vert ^2)=0, \end{aligned}$$

(16)

$$\begin{aligned}&\lim \limits _{n\longrightarrow \infty }(\Vert a^n-x^*\Vert ^2-\Vert x^n-x^*\Vert ^2)=0. \end{aligned}$$

(17)

Combining (16) and (17), we have

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty }(\Vert a^n-x^*\Vert ^2-\Vert t^n-x^*\Vert ^2)=0. \end{aligned}$$

(18)

From (14), we find

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty }\Vert a^n-t^n\Vert =0. \end{aligned}$$

(19)

From (11), (17) and $\{\delta _n\}\subset [\underline{\delta }, \overline{\delta }]\subset \Big (0, \displaystyle \frac{1}{\Vert A\Vert ^2+1}\Big )$, we get

$$\begin{aligned} \lim \nolimits _{n\longrightarrow \infty }\Vert r^n-b^n\Vert =0. \end{aligned}$$

(20)

On the other hand,

$$\begin{aligned} \Vert a^n-x^n\Vert&=\Vert \delta _n A^*(r^n-b^n)\Vert \le \delta _n\Vert A^*\Vert \Vert r^n-b^n\Vert \le \overline{\delta }\Vert A\Vert \Vert r^n-b^n\Vert , \end{aligned}$$

which together with (20) implies

$$\begin{aligned} \lim \nolimits _{n\longrightarrow \infty }\Vert a^n-x^n\Vert =0. \end{aligned}$$

(21)

From Lemma 1, we have, for all $i=1,2,\ldots ,M$

$$\begin{aligned} \Vert a^n-y_i^n\Vert&\le \dfrac{\sqrt{1+\lambda _n^2L_i^2}}{1-\lambda _n L_i}\Vert a^n-s_i^n\Vert \nonumber \\&\le \dfrac{\sqrt{1+\overline{\lambda }^2L_i^2}}{1-\overline{\lambda } L_i}\Vert a^n-s_{i_n}^n\Vert \nonumber \\&=\dfrac{\sqrt{1+\overline{\lambda }^2L_i^2}}{1-\overline{\lambda } L_i}\Vert a^n-t^n\Vert . \end{aligned}$$

(22)

From (19) and (22), we get

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty }\Vert a^n-y_i^n\Vert =0\;\;\forall i=1,2,\ldots ,M. \end{aligned}$$

(23)

From Lemma 1, we have

$$\begin{aligned} \Vert b^n-u_j^n\Vert&\le \dfrac{\sqrt{1+\mu _n^2\kappa _j^2}}{1-\mu _n \kappa _j}\Vert b^n-w_j^n\Vert \nonumber \\&\le \dfrac{\sqrt{1+\overline{\mu }^2\kappa _j^2}}{1-\overline{\mu }\kappa _j}\Vert b^n-w_{j_n}^n\Vert \nonumber \\&=\dfrac{\sqrt{1+\overline{\mu }^2\kappa _j^2}}{1-\overline{\mu }\kappa _j}\Vert b^n-r^n\Vert . \end{aligned}$$

(24)

From (20) and (24), we have

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty }\Vert b^n-u_j^n\Vert =0\;\;\forall j=1,2,\ldots ,N. \end{aligned}$$

(25)

Note that

$$\begin{aligned} \Vert x^n-t^n\Vert \le \Vert x^n-a^n\Vert +\Vert a^n-t^n\Vert . \end{aligned}$$

Using the above inequality, (19) and (21), we have

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty }\Vert x^n-t^n\Vert =0. \end{aligned}$$

(26)

Choose a subsequence $\{t^{n_k}\}$ of $\{t^n\}$ such that

$$\begin{aligned} \limsup \limits _{n\longrightarrow \infty }\langle x^*, x^*-t^n\rangle =\lim \limits _{k\longrightarrow \infty }\langle x^*, x^*-t^{n_k}\rangle . \end{aligned}$$

Since $\{t^{n_k}\}$ is bounded, we may assume without loss of generality that $t^{n_k}\rightharpoonup \overline{t}$. Therefore,

$$\begin{aligned} \limsup \limits _{n\longrightarrow \infty }\langle x^*, x^*-t^n\rangle =\langle x^*, x^*-\overline{t}\rangle . \end{aligned}$$

From $t^{n_k}\rightharpoonup \overline{t}$ and (19), we get $a^{n_k}\rightharpoonup \overline{t}$. From (23), we have $\lim \limits _{k\longrightarrow \infty }\Vert a^{n_k}-y_i^{n_k}\Vert =0$ for all $i=1,2,\ldots ,M$. Note that $y_i^{n_k}=P_{C_i}(a^{n_k}-\lambda _{n_k} F_i(a^{n_k}))$. From Lemma 2, we get $\overline{t}\in \hbox {Sol}(C_i, F_i)$ for all $i=1,2,\ldots ,M$, that is, $\overline{t}\in \displaystyle \bigcap \nolimits _{i=1}^{M}\hbox {Sol}(C_i, F_i)$.

From $t^{n_k}\rightharpoonup \overline{t}$ and (26), we get $x^{n_k}\rightharpoonup \overline{t}$. Thus, $b^{n_k}=A(x^{n_k})\rightharpoonup A(\overline{t})$. From (25), we have $\lim \limits _{k\longrightarrow \infty }\Vert b^{n_k}-u_j^{n_k}\Vert =0$ for all $j=1,2,\ldots ,N$. Note that $u_j^{n_k}=P_{Q_j}(b^{n_k}-\mu _{n_k} G_j(b^{n_k}))$. From Lemma 2, we obtain $A(\overline{t})\in \hbox {Sol}(Q_j, G_j)$ for all $j=1,2,\ldots ,N$, that is, $A(\overline{t})\in \displaystyle \bigcap \nolimits _{j=1}^{N}\hbox {Sol}(Q_j, G_j)$, which together with $\overline{t}\in \displaystyle \bigcap \nolimits _{i=1}^{M}\hbox {Sol}(C_i, F_i)$ implies $\overline{t}\in \varOmega$.

From (10) and (15), we get

$$\begin{aligned} \Vert x^{n+1}-x^*\Vert ^2\le (1-\alpha _n)\Vert x^n-x^*\Vert ^2+\alpha _n\xi _n, \end{aligned}$$

where

$$\begin{aligned} \xi _n=2\langle x^*, x^*-t^n+\alpha _n t^n\rangle . \end{aligned}$$

Since $x^*=P_\varOmega (0)$ and $\overline{t}\in \varOmega$, it follows that

$$\begin{aligned} \langle x^*, x^*-\overline{t}\rangle \le 0. \end{aligned}$$

Therefore, since $\{t^n\}$ is bounded, $\lim \limits _{n\longrightarrow \infty }\alpha ^n=0$, we get

$$\begin{aligned} \limsup \limits _{n\longrightarrow \infty }\xi _n&=\limsup \limits _{n\longrightarrow \infty }\big [2\langle x^*, x^*-t^n\rangle +2\alpha _n\langle x^*, t^n\rangle \big ]\\&=\limsup \limits _{n\longrightarrow \infty }2\langle x^*, x^*-t^n\rangle \\&=2\langle x^*, x^*-\overline{t}\rangle \\&\le 0. \end{aligned}$$

By Lemma 4, we have $\lim \limits _{n\longrightarrow \infty }\Vert x^n-x^*\Vert ^2=0$, that is, $x^n\longrightarrow x^*$. Case 2. Suppose that for any integer m, there exists an integer n such that $n\ge m$ and $\Vert x^n-x^*\Vert \le \Vert x^{n+1}-x^*\Vert$. According to Lemma 3, there exists a nondecreasing sequence $\{\tau (n)\}$ of $\mathbb N$ such that $\lim \limits _{n\longrightarrow \infty }\tau (n)=\infty$ and the following inequalities hold for all (sufficiently large) $n\in \mathbb N$.

$$\begin{aligned} \Vert x^{\tau (n)}-x^*\Vert \le \Vert x^{\tau (n)+1}-x^*\Vert ,\;\;\Vert x^n-x^*\Vert \le \Vert x^{\tau (n)+1}-x^*\Vert . \end{aligned}$$

(27)

From (9) and (15), we get

$$\begin{aligned} \begin{aligned} 0&\le \Vert a^{\tau (n)}-x^*\Vert -\Vert t^{\tau (n)}-x^*\Vert \\&\le \Vert x^{\tau (n)}-x^*\Vert -\Vert t^{\tau (n)}-x^*\Vert \\&\le \Vert x^{\tau (n)+1}-x^*\Vert -\Vert t^{\tau (n)}-x^*\Vert \\&\le \alpha _{\tau (n)}\Vert x^*\Vert -\alpha _{\tau (n)}\Vert t^{\tau (n)}-x^*\Vert . \end{aligned} \end{aligned}$$

(28)

From the boundedness of $\{t^n\}$, $\lim \limits _{n\longrightarrow \infty }\alpha _n=0$ and (28), we have

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty }(\Vert a^{\tau (n)}-x^*\Vert -\Vert t^{\tau (n)}-x^*\Vert )=0,\;\lim \limits _{n\longrightarrow \infty }(\Vert x^{\tau (n)}-x^*\Vert -\Vert t^{\tau (n)}-x^*\Vert )=0.\nonumber \\ \end{aligned}$$

(29)

Thus,

$$\begin{aligned} \lim \limits _{n\longrightarrow \infty }(\Vert a^{\tau (n)}-x^*\Vert -\Vert x^{\tau (n)}-x^*\Vert )=0. \end{aligned}$$

(30)

Thus, from (29), (30) and the boundedness of $\{x^n\}$, $\{a^n\}$, $\{t^n\}$, we obtain

$$\begin{aligned}&\lim \limits _{n\longrightarrow \infty }(\Vert a^{\tau (n)}-x^*\Vert ^2-\Vert t^{\tau (n)}-x^*\Vert ^2)=0, \\&\lim \limits _{n\longrightarrow \infty }(\Vert t^{\tau (n)}-x^*\Vert ^2-\Vert x^{\tau (n)}-x^*\Vert ^2)=0, \\&\lim \limits _{n\longrightarrow \infty }(\Vert a^{\tau (n)}-x^*\Vert ^2-\Vert x^{\tau (n)}-x^*\Vert ^2)=0. \end{aligned}$$

As proved in the first case, we obtain

$$\begin{aligned} \limsup \limits _{n\longrightarrow \infty }\langle x^*, x^*-t^{\tau (n)}\rangle \le 0. \end{aligned}$$

So, from the boundedness of $\{t^n\}$ and $\lim \limits _{n\longrightarrow \infty }\alpha _n=0$, we have

$$\begin{aligned} \limsup \limits _{n\longrightarrow \infty }\langle x^*, x^*-t^{\tau (n)}+\alpha _{\tau (n)}t^{\tau (n)}\rangle&=\limsup \limits _{n\longrightarrow \infty }\big [\langle x^*, x^*-t^{\tau (n)}\rangle +\alpha _{\tau (n)}\langle x^*, t^{\tau (n)}\rangle \big ]\nonumber \\&=\limsup \limits _{n\longrightarrow \infty }\langle x^*, x^*-t^{\tau (n)}\rangle \nonumber \\&\le 0. \end{aligned}$$

(31)

From (10), (15) and (27), we have

$$\begin{aligned} \Vert x^{\tau (n)+1}-x^*\Vert ^2\le (1-\alpha _{\tau (n)})\Vert x^{\tau (n)+1}-x^*\Vert ^2+2\alpha _{\tau (n)}\langle x^*, x^*-t^{\tau (n)}+\alpha _{\tau (n)} t^{\tau (n)}\rangle . \end{aligned}$$

Since $\alpha _{\tau (n)}>0$, it follows from above inequality that

$$\begin{aligned} \Vert x^{\tau (n)+1}-x^*\Vert ^2\le 2\langle x^*, x^*-t^{\tau (n)}+\alpha _{\tau (n)}t^{\tau (n)}\rangle . \end{aligned}$$

(32)

From (27) and (32), we obtain

$$\begin{aligned} \Vert x^n-x^*\Vert ^2\le 2\langle x^*, x^*-t^{\tau (n)}+\alpha _{\tau (n)}t^{\tau (n)}\rangle . \end{aligned}$$

(33)

Taking the limit in (33) as $n\longrightarrow \infty$, and using (31), we obtain that

$$\begin{aligned} \limsup \limits _{n\longrightarrow \infty }\Vert x^n-x^*\Vert ^2\le 0. \end{aligned}$$

Therefore, $x^n\longrightarrow x^*$. This completes the proof of Theorem. $\square$

From Algorithm 1 and Theorem 1, by taking $F_i=0$ for all $i=1,2,\ldots ,M$ and $G_j=0$ for all $j=1,2,\ldots ,N$, we get the following corollary:

Corollary 1

Assume that the solution set $\varOmega$ of the MSSFP is nonempty. Choose $x^0\in \mathcal {H}_1$, $\{\delta _n\}\subset [\underline{\delta }, \overline{\delta }]\subset \Big (0, \dfrac{1}{\Vert A\Vert ^2+1}\Big )$, $\{\alpha _n\}\subset (0, 1)$ such that $\lim \nolimits _{n\longrightarrow \infty }\alpha _n=0$ and $\displaystyle \sum \nolimits _{n=0}^{\infty }\alpha _n=\infty$.

For each iteration $n\ge 0$, compute

$$\begin{aligned} u^n=A(x^n),\;\;v^n:=P_{Q_{j_n}}(u^n), \end{aligned}$$

where

$$\begin{aligned} j_n=\hbox {argmax}\{\Vert P_{Q_j}(u^n)-u^n\Vert : j=1,2,\ldots ,N\}. \end{aligned}$$

Further, we compute

$$\begin{aligned} y^n=x^n+\delta _n A^*(v^n-u^n),\;\;z^n:=P_{C_{i_n}}(y^n), \end{aligned}$$

where

$$\begin{aligned} i_n=\hbox {argmax}\{\Vert P_{C_i}(y^n)-y^n\Vert : i=1,2,\ldots ,M\} \end{aligned}$$

and define the next iteration as

$$\begin{aligned} x^{n+1}=(1-\alpha _n)z^n. \end{aligned}$$

Then, the sequence $\{x^n\}$ converges strongly to the minimum-norm solution of the MSSFP.

Corollary 2

Let C and Q be two nonempty closed convex subset of two real Hilbert spaces $\mathcal {H}_1$ and $\mathcal {H}_2$, respectively. Let $F: \mathcal {H}_1\longrightarrow \mathcal {H}_1$ be monotone and L-Lipschitz continuous on $\mathcal {H}_1$, $G: \mathcal {H}_2\longrightarrow \mathcal {H}_2$ be monotone and $\kappa$-Lipschitz continuous on $\mathcal {H}_2$. Suppose that the sequences $\{\alpha _n\}$, $\{\delta _n\}$, $\{\lambda _n\}$ and $\{\mu _n\}$ satisfy the following restrictions:

$$\begin{aligned} {\left\{ \begin{array}{ll} {\{\alpha _n\}\subset (0, 1), \lim \limits _{n\longrightarrow \infty }\alpha _n=0, \displaystyle \sum \limits _{n=0}^{\infty }\alpha _n=\infty },\\ {\{\delta _n\}\subset [\underline{\delta }, \overline{\delta }]\subset \Big (0, \dfrac{1}{\Vert A\Vert ^2+1}\Big )},\\ {\{\lambda _n\}\subset [\underline{\lambda }, \overline{\lambda }]\subset \Big (0, \dfrac{1}{L}\Big )},\\ {\{\mu _n\}\subset [\underline{\mu }, \overline{\mu }]\subset \Big (0, \dfrac{1}{\kappa }\Big )}. \end{array}\right. } \end{aligned}$$

Choose $x^0\in \mathcal {H}_1$, and for each iteration $n\ge 0$, compute

$$\begin{aligned} \begin{aligned} b^n&=A(x^n),\\ u^n&=P_Q(b^n-\mu _n G(b^n)),\\ v^n&=b^n-u^n-\mu _n(G(b^n)-G(u^n)),\\ w^n&= {\left\{ \begin{array}{ll} b^n-\dfrac{\langle b^n-u^n, v^n\rangle }{\Vert v^n\Vert ^2}v^n &{} \text{ if } u^n\ne b^n,\\ b^n &{} \text{ if } u^n=b^n. \end{array}\right. }\\ \end{aligned} \end{aligned}$$

Further, we compute

$$\begin{aligned} \begin{aligned} a^n&=x^n+\delta _n A^*(w^n-b^n),\\ y^n&=P_C(a^n-\lambda _n F(a^n)),\\ z^n&=a^n-y^n-\lambda _n(F(a^n)-F(y^n)),\\ t^n&= {\left\{ \begin{array}{ll} a^n-\dfrac{\langle a^n-y^n, z^n\rangle }{\Vert z^n\Vert ^2}z^n &{} \text{ if } y^n\ne a^n,\\ a^n &{} \text{ if } y^n=a^n. \end{array}\right. } \end{aligned} \end{aligned}$$

and define the next iteration as

$$\begin{aligned} x^{n+1}=(1-\alpha _n)t^n. \end{aligned}$$

Then, the sequence $\{x^n\}$ converges strongly to the minimum-norm solution of the SVIP, provided that the solution set $\varOmega =\{x^*\in \hbox {Sol}(C, F): Ax^*\in \hbox {Sol}(Q, G)\}$ of the SVIP is nonempty.

4 Numerical Illustrations

Example 1

Let $\mathcal {H}_1=\mathbb {R}^4$ with the norm $\Vert x\Vert =(x_1^2+x_2^2+x_3^2+x_4^2)^{\frac{1}{2}}$ for $x=(x_1, x_2, x_3, x_4)^T\in \mathbb {R}^4$ and $\mathcal {H}_2=\mathbb {R}^2$ with the norm $\Vert y\Vert =(y_1^2+y_2^2)^{\frac{1}{2}}$ for all $y=(y_1, y_2)^T\in \mathbb {R}^2$.

Let $A(x)=(x_1+x_3+x_4, x_2+x_3-x_4)^T$ for all $x=(x_1, x_2, x_3, x_4)^T\in \mathbb {R}^4$. Then, A is a bounded linear operator from $\mathbb {R}^4$ into $\mathbb {R}^2$ with $\Vert A\Vert =\sqrt{3}$. For $y=(y_1, y_2)^T\in \mathbb {R}^2$, let $B(y)=(y_1, y_2, y_1+y_2, y_1-y_2)^T$. Then, B is a bounded linear operator from $\mathbb {R}^2$ into $\mathbb {R}^4$ with $\Vert B\Vert =\sqrt{3}$. Moreover, for any $x=(x_1, x_2, x_3, x_4)^T\in \mathbb {R}^4$ and $y=(y_1, y_2)^T\in \mathbb {R}^2$, $\langle A(x), y\rangle =\langle x, B(y)\rangle$, so $B=A^*$ is an adjoint operator of A.

Let

$$\begin{aligned} \begin{aligned} C_1&=\{x=(x_1, x_2, x_3, x_4)^T\in \mathbb {R}^4: 2x_1+x_2+3x_3+x_4\le 8\},\\ C_2&=\{x=(x_1, x_2, x_3, x_4)^T\in \mathbb {R}^4: x_1+x_2+2x_3\le 5\},\\ C_3&=\{x=(x_1, x_2, x_3, x_4)^T\in \mathbb {R}^4: x_1-x_2-4x_3\le 3\},\\ C_4&=\{x=(x_1, x_2, x_3, x_4)^T\in \mathbb {R}^4: 4x_1-x_2+x_3-x_4\le 18\},\\ C_5&=\{x=(x_1, x_2, x_3, x_4)^T\in \mathbb {R}^4: 3x_1-2x_2-5x_3-3x_4\ge 14\},\\ C_6&=\{x=(x_1, x_2, x_3, x_4)^T\in \mathbb {R}^4: -x_1+x_2+4x_3\le -3\} \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} Q_1&=\{(u_1, u_2)^T\in \mathbb {R}^2: 2u_1+u_2\ge 8\},\\ Q_2&=\{(u_1, u_2)^T\in \mathbb {R}^2: u_1+u_2\ge 5\}. \end{aligned} \end{aligned}$$

The solution set $\varOmega =\{x=(x_1, x_2, x_3, x_4)^T\in C_1\cap C_2\cap C_3\cap C_4\cap C_5\cap C_6: A(x)\in Q_1\cap Q_2\}$ of the MSSFP is given by

$$\begin{aligned} \varOmega ={\left\{ \begin{array}{ll} {2x_1+x_2+3x_3+x_4=8},\\ {x_1+x_2+2x_3=5},\\ {x_1-x_2-4x_3=3},\\ {4x_1-x_2+x_3-x_4\le 18},\\ {3x_1-2x_2-5x_3-3x_4\ge 14} \end{array}\right. } \end{aligned}$$

which can be rewritten as follows

$$\begin{aligned} \varOmega =\Big \{(a+4, 1-3a, a, -2a-1)^T: \dfrac{1}{10}\le a\le \dfrac{1}{5}\Big \}. \end{aligned}$$

Suppose $x=(a+4, 1-3a, a, -2a-1)^T\in \varOmega$, then

$$\begin{aligned} \begin{aligned} \Vert x\Vert&=\sqrt{(a+4)^2+(1-3a)^2+a^2+(-2a-1)^2}\\&=\sqrt{15a^2+6a+18}\\&=\sqrt{15\Big (\dfrac{1}{10}-a\Big )^2+9a+\dfrac{357}{20}}\\&\ge \sqrt{\dfrac{9}{10}+\dfrac{357}{20}}=\dfrac{\sqrt{75}}{2}. \end{aligned} \end{aligned}$$

The above equality holds if and only if $a=\dfrac{1}{10}$. So the minimum-norm solution $x^*$ of the MSSFP is $x^*=\Big (\dfrac{41}{10}, \dfrac{7}{10}, \dfrac{1}{10}, -\dfrac{6}{5}\Big )^T$. Select a random starting point $x^0=(-3,7,-4,10)^T\in \mathbb {R}^4$ for Algorithm 1. With $\delta _n=\dfrac{n+1}{5n+6}$, $\alpha _n=\dfrac{1}{n+2}$, $\varepsilon =10^{-5}$ and the stopping criterion $\Vert x^{n+1}-x^n\Vert \le \varepsilon$, we obtain an approximate solution obtained after 111490 iterations (with time 27.2783 seconds) is

$$\begin{aligned} x^{111490}=(4.09989912, 0.69989497, 0.10001402, -1.20001237)^T, \end{aligned}$$

which is a good approximation to the minimum-norm solution of the MSSFP $x^*=\Big (\dfrac{41}{10}, \dfrac{7}{10}, \dfrac{1}{10}, -\dfrac{6}{5}\Big )^T$.

We perform the iterative schemes in PYTHON 3.9 running on a laptop with Intel(R) Core(TM) i5-5200U CPU @ 2.20GHz, 4 GB RAM.

5 Conclusion

The paper has presented an iterative algorithm for finding minimum-norm solutions of the multiple-sets split variational inequality problem in real Hilbert spaces. Under some suitable conditions imposed on parameters, we have proved the strong convergence of the iterative sequence to the minimum-norm solution of the MSSVIP. As special cases, we apply the main theorem to find the minimum-norm solution of the MSSFP and the SVIP. A simple numerical example has been performed to illustrate the performance of the proposed algorithm.