1 Introduction

The split equality problem was first introduced and studied by Moudafi in [19]. This problem is formulated as follows: Let \(H, H_1\) and \(H_2\) be three real Hilbert spaces. Let C and Q be two nonempty closed and convex subsets of \(H_1\) and \(H_2\), respectively.

$$\begin{aligned} \text {Find }p\in C \text { and } q\in Q\text { such that } Ap-Bq=0, \end{aligned}$$
(1.1)

where \(A:\ H_1\rightarrow H\) and \(B:\ H_2\rightarrow H\), are two bounded linear operators.

The origins of Problem (1.1) can be found in [2]. The authors studied alternating minimization algorithms with costs-to-move. Next, a particular case of Problem (1.1) has been also studied by Attouch et al. in [3, 4] when C and Q are two minimizer point sets of two convex functions f and g, which are defined on \(H_1\) and \(H_2\), respectively.

We know that Problem (1.1) has some important applications in game theory, variational problems, partial differential equations, and intensity-modulated radiation therapy (see, e.g., [3, 9]). Thus, Problem (1.1) is an interesting topic of nonlinear analysis, which has attracted the attention of many mathematicians around the world. So, many algorithms have been presented for solving Problem (1.1) (see, e.g., [8, 11,12,13,14, 18, 20, 21, 25, 27,28,29,30,31]).

If \(H =H_2\) and \(B=I^{H_2}\), then Problem (1.1) becomes the split convex feasibility problem. Let C and Q be nonempty, closed and convex subsets of real Hilbert spaces \(H_1\) and \(H_2\), respectively. Let \(A:\ H_1\rightarrow H_2\) be a bounded linear operator. The split convex feasibility problem (SCFP for short) is formulated as follows:

$$\begin{aligned} \text {Find an element}\quad p\in C \text { such that }Ap\in Q. \end{aligned}$$
(1.2)

The SCFP was first introduced and studied by Censor and Elfving [10] for modeling certain inverse problems. It plays an important role in medical image reconstruction and in signal processing (see [6, 7]). Thus, all of the iterative methods or algorithms to approximate the solution to Problem (1.1) can be applied to find the solution to Problem (1.2).

We know that most applied problems are ill-posed problems in the sense, they do not satisfy at least one of the following three requirements: (1) The problem is solvable; (2) Its solution is unique; (3) The problem is stable in the sense, any small change in the input data leads to only small changes in the output data (the solution of the problem) (see, e.g., [17]). Thus SEP and its related problems are also ill-posed problems. One popular effective method for solving the ill-posed problems is the Tikhonov regularization method which is introduced by Tikhonov (see, e.g., [1, 26]).

In this paper, we study the system of the split equality problems. In order to find a solution to this problem, we first introduce a new implicit iterative method by using the Tikhonov regularization method. Next, we propose an explicit iterative regularization method and prove the strong convergence of it when the control parameters are chosen suitably.

2 Preliminaries

Let H be a real Hilbert space. We denote by \(\langle x,y\rangle _H\) the inner product of two elements \(x,y\in H\). The induced norm is denoted by \(\Vert \cdot \Vert _H\), that is, \(\Vert x\Vert _H=\sqrt{\langle x,x\rangle _H}\) for all \(x\in H\).

Let C be a nonempty, closed and convex subset of H. It is well known that for each \(x\in H\), there is unique point \(P^H_Cx \in C\) such that

$$\begin{aligned} \Vert x-P^H_Cx\Vert _H=\inf _{u\in C}\Vert x-u\Vert _H. \end{aligned}$$
(2.1)

The mapping \(P^H_C:\ H\rightarrow C\) defined by (2.1) is called the metric projection of H onto C. We also recall (see, for example, Section 3 in [16]) that

$$\begin{aligned} \langle x-P^H_Cx,y-P^H_Cx\rangle _H\le 0 \ \; \forall x \in H, \ \forall y \in C. \end{aligned}$$

Lemma 2.1

Let H be a real Hilbert space and C be a nonempty, closed and convex subset of H. Then for all \(x,y\in H\), we have

  1. i)

    \(\langle x-y,P_C^Hx-P_C^Hy\rangle _H \ge \Vert P_C^Hx-P_C^Hy\Vert ^2_H\);

  2. ii)

    \(\langle x-y, (I-P_C^H)x-(I-P_C^H)y\rangle _H \ge \Vert (I-P_C^H)x-(I-P_C^H)y\Vert ^2_H\).

Definition 2.2

Let \(f:\ H\longrightarrow (-\infty , \infty ]\) be a proper function. The subdifferential of f is the set-valued operator \(\partial f:\ H\rightarrow 2^H\) which is defined by

$$\begin{aligned} \partial f(x) := \{v\in H:\ f(y)-f(x)\ge \langle v,y-x\rangle _H \ \; \forall y\in H\}. \end{aligned}$$

Definition 2.3

An operator \(S:\ H\rightarrow H\) is called

  1. a)

    L-Lipschitz with Lipschitz constant \(L\ge 0\) if for all \(x,y\in H\), we have

    $$\begin{aligned} \Vert Sx-Sy\Vert _H\le L\Vert x-y\Vert _H. \end{aligned}$$

    In the above inequality, if \(L\in [0,1)\), then S is called a strict contraction and if \(L=1\), then it is said to be nonexpansive.

  2. b)

    \(\gamma \)-strongly monotone with constant \(\gamma >0\) if the inequality

    $$\begin{aligned} \langle x-y,Sx-Sy\rangle _H \ge \gamma \Vert x-y\Vert ^2_H \end{aligned}$$

    holds for all \(x,y\in H\).

  3. c)

    hemicontinuous at a point \(x_0\in H\) if \(S(x_0+t_nx)\rightharpoonup Sx_0\) as \(t_n \rightarrow 0^{+}\) where \(\{t_n\}\) is any sequence of positive real numbers. The operator S is called hemicontinuous if it is hemicontinuous at each element \(x\in H\).

  4. d)

    coercive if there exists a function c(t) defined for \(t\ge 0\) such that \(c(t)\rightarrow \infty \) as \(t\rightarrow \infty \), and the inequality

    $$\begin{aligned} \langle x, Sx\rangle _H \ge c(\Vert x\Vert _H)\Vert x\Vert _H \end{aligned}$$

    holds for all \(x\in H\).

  5. e)

    bounded on bounded sets if S(M) is a bounded set for each bounded set \(M\subset H\).

Remark 2.4

It follows from Lemma 2.1 that \(I^H-P_C^H\) is a nonexpansive mapping.

In the sequel, we are going to use the following lemmas in the proofs of the main results of this paper.

Lemma 2.5

Let H be a real Hilbert space. Suppose that \(F:\ H\rightarrow H\) is a L-Lipschitz and \(\gamma \)-strongly monotone mapping. Then for any \(\varepsilon \in (0,2\gamma /L^2)\), we have \(I^H-\varepsilon F\) is a strict contraction mapping with the contraction coefficient \(\tau =\sqrt{1-\varepsilon (2\gamma -\varepsilon L^2)}\).

Proof

For any \(x,y\in H\), we have

$$\begin{aligned} \Vert (I^H-\varepsilon F)x&-(I^H-\varepsilon F)y\Vert ^2_H\\&=\Vert (x-y)-\varepsilon (Fx-Fy)\Vert ^2_H\\&=\Vert x-y\Vert ^2_H-2\varepsilon \langle x-y, Fx-Fy\rangle _H +\varepsilon ^2 \Vert Fx-Fy\Vert ^2_H\\&\le \Vert x-y\Vert ^2_H-2\varepsilon \gamma \Vert x-y\Vert ^2_H +\varepsilon ^2 L^2\Vert x-y\Vert ^2_H\\&=[1-\varepsilon (2\gamma -\varepsilon L^2)]\Vert x-y\Vert ^2_H. \end{aligned}$$

Thus, it follows from the condition \(\varepsilon \in (0,2\gamma /L^2)\) that \(I^H-\varepsilon F\) is strict contraction mapping with the contraction coefficient \(\tau =\sqrt{1-\varepsilon (2\gamma -\varepsilon L^2)}\). \(\square \)

Lemma 2.6

[15] Let T be a nonexpansive self-mapping of a closed and convex subset C of a Hilbert space H. Then the mapping \(I^{H} - T\) is demiclosed, that is, whenever \(\{x_n\}\) is a sequence in C which weakly converges to some \(x \in C\) and the sequence \(\{(I-T)(x_n)\}\) strongly converges to some y, it follows that \((I^H -T)(x) = y\).

Lemma 2.7

[24]

Let \(\{\Gamma _n\}\) be a sequence of nonnegative numbers, \(\{b_n\}\) be a sequence in (0, 1) and let \(\{c_n\}\) be a sequence of real numbers satisfying the following two conditions:

  1. i)

    \(\Gamma _{n+1}\le (1-b_n)\Gamma _n+b_nc_n\);

  2. ii)

    \(\sum _{n=0}^{\infty }b_n=\infty ,\ \limsup _{n\rightarrow \infty }c_n\le 0.\)

Then \(\lim _{n\rightarrow \infty }\Gamma _n=0.\)

3 Main Results

Let \(H_1\), \(H_2\) and H, be three real Hilbert spaces. Let \(C_i\) and \(Q_i\), be nonempty closed convex subsets of \(H_1\) and \(H_2\), respectively, \(i=1,2,\ldots ,N\). Let  \(A_i:\ H_1\rightarrow H\) and \(B_i:\ H_2\rightarrow H\), \(i=1,2,\ldots ,N\), be bounded linear mappings and let \(~b_i\), \(i=1,2,\ldots ,N\), be N given elements in H. Suppose that

$$\begin{aligned} \Omega :=\{(x,y)\in \cap _{i=1}^N(C_i\times Q_i):\ A_ix-B_iy=b_i,\ i=1,2,\ldots ,N\}\ne \emptyset . \end{aligned}$$

Consider the problem of finding an element \((x^*,y^*)\in \Omega \), we denote this problem by (SSEP).

3.1 Implicit Iterative Method

First, we introduce a new Tikhonov regularization method type to approximate a solution of Problem (SSEP).

We define the sequences \(\{x_n\}\) and \(\{y_n\}\) by the following implicit iterative method:

$$\begin{aligned}&\sum _{i=1}^N\Big ((I^{H_1}-P^{H_1}_{C_{i}})x_n+A_{i}^*(A_{i}x_n-B_{i}y_n-b_{i})\Big )+\alpha _n Sx_n=0, \end{aligned}$$
(3.1)
$$\begin{aligned}&\sum _{i=1}^N \Big ((I^{H_2}-P^{H_2}_{Q_{i}})y_n-B_{i}^*(A_{i}x_n-B_{i}y_n-b_{i}) \Big ) +\alpha _n Ty_n=0, \end{aligned}$$
(3.2)

where \(\{\alpha _n\}\) is a sequence of positive real numbers, \(S:\ H_1\rightarrow H_1\) and \(T:\ H_2\rightarrow H_2\) are two bounded on bounded sets, hemicontinuous and strongly monotone mappings with the constants \(\gamma _S\) and \(\gamma _T\), respectively.

Theorem 3.1

  1. i)

    For each n, the system of equations (3.1)–(3.2) has a unique solution \((x_n,y_n)\).

  2. ii)

    If \(\lim _{n\rightarrow \infty }\alpha _n=0\), then \(x_n\rightarrow x^*\), \(y_n\rightarrow y^*\) with \((x^*,y^*)\in \Omega \) and \((x^*,y^*)\) is a unique solution to the following variational inequality

    $$\begin{aligned} \langle x-x^*, Sx^*\rangle _{H_1} + \langle y-y^*, Ty^*\rangle _{H_2} \ge 0,\ \forall (x,y)\in \Omega . \end{aligned}$$
    (3.3)

Proof

i) We rewrite the equations (3.1) and (3.2) in the following form

$$\begin{aligned} F_{\alpha _n}(x_n,y_n)=0, \end{aligned}$$
(3.4)

where \(F_{\alpha _n}:\ H_1\times H_2\rightarrow H_1\times H_2\) defined by

$$\begin{aligned} F_{\alpha _n}(a)=\Bigg (\sum _{i=1}^N\Big ((I^{H_1}&-P^{H_1}_{C_{i}})x+A_{i}^*(A_{i}x-B_{i}y-b_{i})\Big )+\alpha _n Sx,\\&\sum _{i=1}^N\Big ((I^{H_2}-P^{H_2}_{Q_{i}})y-B_{i}^*(A_{i}x-B_{i}y-b_{i})\Big )+\alpha _n Ty \Bigg ), \end{aligned}$$

for all \(a=(x,y)\in H_1\times H_2\).

We know that \(H_1\times H_2\) is a Hilbert spaces with the inner product

$$\begin{aligned} \langle (x_1,y_1), (x_2,y_2)\rangle _{H_1\times H_2}=\langle x_1, y_1\rangle _{H_1} +\langle x_2, y_2\rangle _{H_2} \end{aligned}$$

for all \((x_1,y_1)\) and \((x_2,y_2)\) in \(H_1\times H_2\) and the norm on \(H_1\times H_2\) is defined by

$$\begin{aligned} \Vert (x,y)\Vert _{H_1\times H_2}=\sqrt{\Vert x\Vert _{H_1}^2+\Vert y\Vert _{H_2}^2} \end{aligned}$$

for all \((x,y)\in H_1\times H_2\) (see, e.g., [22, Proposition 2.2], [23, Proposition 2.4]).

We first show that \(F_{\alpha _n}\) is a monotone mapping. Indeed, for any \(a=(x_1,y_1)\) and \(b=(x_2,y_2)\) in \(H_1\times H_2\), we have

$$\begin{aligned} \langle a-b,&F_{\alpha _n}(a)-F_{\alpha _n}(b)\rangle _{H_1\times H_2} \\&=\sum _{i=1}^N\langle x_1-x_2, (I^{H_1}-P^{H_1}_{C_{i}})x_1-(I^{H_1}-P^{H_1}_{C_{i}})x_2\rangle _{H_1}\\&\quad +\sum _{i=1}^N\langle x_1-x_2,A_{i}^*(A_{i}(x_1-x_2)-B_{i}(y_1-y_2))\rangle _{H_1}\\&\quad +\alpha _n \langle x_1-x_2, Sx_1-Sx_2\rangle _{H_1}\\&\quad +\sum _{i=1}^N\langle y_1-y_2, (I^{H_2}-P^{H_2}_{Q_{i}})y_1-(I^{H_2}-P^{H_2}_{Q_{i}})y_2\rangle _{H_2}\\&\quad -\sum _{i=1}^N\langle y_1-y_2,B_{i}^*(A_{i}(x_1-x_2)-B_{i}(y_1-y_2))\rangle _{H_2}\\&\quad +\alpha _n \langle y_1-y_2, Ty_1-Ty_2\rangle _{H_2}\\&=\sum _{i=1}^N\langle x_1-x_2, (I^{H_1}-P^{H_1}_{C_{i}})x_1-(I^{H_1}-P^{H_1}_{C_{i}})x_2\rangle _{H_1}\\&\quad +\sum _{i=1}^N\langle A_{i}(x_1-x_2),A_{i}(x_1-x_2)-B_{i}(y_1-y_2)\rangle _{H}\\&\quad +\alpha _n \langle x_1-x_2, Sx_1-Sx_2\rangle _{H_1}\\&\quad +\sum _{i=1}^N\langle y_1-y_2, (I^{H_2}-P^{H_2}_{Q_{i}})y_1-(I^{H_2}-P^{H_2}_{Q_{i}})y_2\rangle _{H_2}\\&\quad -\sum _{i=1}^N\langle B_{i}(y_1-y_2),A_{i}(x_1-x_2)-B_{i}(y_1-y_2)\rangle _{H}\\&\quad +\alpha _n \langle y_1-y_2, Ty_1-Ty_2\rangle _{H_2}\\&=\sum _{i=1}^N\langle x_1-x_2, (I^{H_1}-P^{H_1}_{C_{i}})x_1-(I^{H_1}-P^{H_1}_{C_{i}})x_2\rangle _{H_1}\\&\quad +\alpha _n \langle x_1-x_2, Sx_1-Sx_2\rangle _{H_1}\\&\quad +\sum _{i=1}^N\langle y_1-y_2, (I^{H_2}-P^{H_2}_{Q_{i}})y_1-(I^{H_2}-P^{H_2}_{Q_{i}})y_2\rangle _{H_2}\\&\quad +\sum _{i=1}^N\Vert A_{i}(x_1-x_2)-B_{i}(y_1-y_2)\Vert ^2 _{H}\\&\quad +\alpha _n \langle y_1-y_2, Ty_1-Ty_2\rangle _{H_2}. \end{aligned}$$

It follows from Lemma 2.1, the strongly monotone of S and T, and the above equality that

$$\begin{aligned} \langle a-b, F_{\alpha _n}(a)-F_{\alpha _n}(b)\rangle _{H_1\times H_2}&\ge \sum _{i=1}^N\Vert (I^{H_1}-P^{H_1}_{C_{i}})x_1-(I^{H_1}-P^{H_1}_{C_{i}})x_2\Vert _{H_1}^2\nonumber \nonumber \\&\quad +\sum _{i=1}^N\Vert (I^{H_2}-P^{H_2}_{Q_{i}})y_1-(I^{H_2}-P^{H_2}_{Q_{i}})y_2\Vert ^2_{H_2}\nonumber \\&\quad +\sum _{i=1}^N\Vert A_{i}(x_1-x_2)-B_{i}(y_1-y_2)\Vert ^2 _{H}\nonumber \\&\quad +\alpha _n (\gamma _S\Vert x_1-x_2\Vert ^2_{H_1} +\gamma _T\Vert y_1-y_2\Vert ^2_{H_2})\nonumber \\&\ge 0. \end{aligned}$$
(3.5)

This implies that \(F_{\alpha _n}\) is a monotone mapping. Moreover, we also have

$$\begin{aligned} \langle a-b, F_{\alpha _n}(a)-F_{\alpha _n}(b)\rangle _{H_1\times H_2}&\ge \alpha _n (\gamma _S\Vert x_1-x_2\Vert ^2_{H_1} +\gamma _T\Vert y_1-y_2\Vert ^2_{H_2})\\&\ge \alpha _n\min \{\gamma _S,\gamma _T\}(\Vert x_1-x_2\Vert ^2_{H_1} +\Vert y_1-y_2\Vert ^2_{H_2})\\&=c(\Vert a-b\Vert _{H_1\times H_2})\Vert a-b\Vert _{H_1\times H_2}, \end{aligned}$$

with \(c(t)= \alpha _n\min \{\gamma _S,\gamma _T\}t\) for all \(t\ge 0\). This implies that \(F_{\alpha _n}\) is a coercive mapping.

It follows from the hemicontinuity of S and T that \(F_{\alpha _n}\) is a hemicontinuous mapping. We conclude that \(F_{\alpha _n}\) is a single valued hemicontinuous monotone and coercive mapping on \(H_1\times H_2\), and hence \(R(F_{\alpha _n})=H_1\times H_2\) (see, e.g. [5, Proposition 1]). Thus Equation (3.4) is solvable.

Next, we prove the uniqueness of the solution \((x_n,y_n)\) of Equation (3.4). Indeed, suppose that \((u_n,v_n)\) is also another solution to Equation (3.4). Let \(a=(x_n,y_n)\) and \(b=(u_n,v_n)\), it follows from \(F_{\alpha _n}(a)=F_{\alpha _n}(b)=0\) and (3.5) that

$$\begin{aligned} \gamma _S \Vert x_n-u_n\Vert ^2_{H_1} +\gamma _T\Vert y_n-v_n\Vert ^2_{H_2}\le 0. \end{aligned}$$

This implies that \(x_n=u_n\) and \(y_n=v_n\). So, Equation (3.4) has a unique solution, that is, the system of equations (3.1)-(3.2) has unique solution \((x_n,y_n)\).

ii) Take any \(c=(\bar{x}, \bar{y})\in \Omega \), that is, \(F_{\alpha _n}(c)=\alpha _n (S\bar{x}, T\bar{y})\). From (3.5), we have

$$\begin{aligned}&\langle a-c, F_{\alpha _n}(a)-F_{\alpha _n}(c)\rangle _{H_1\times H_2}\ge \alpha _n (\gamma _S\Vert x_n-\bar{x}\Vert ^2_{H_1} +\gamma _T\Vert y_n-\bar{y}\Vert ^2_{H_2}). \end{aligned}$$

On the other hand, we also have

$$\begin{aligned} \langle a-c, F_{\alpha _n}(a)-F_{\alpha _n}(c)\rangle _{H_1\times H_2}=-\alpha _n (\langle x_n-\bar{x}, S\bar{x}\rangle _{H_1}+\langle y_n-\bar{y}, T\bar{y}\rangle _{H_2}). \end{aligned}$$

Thus, we obtain

$$\begin{aligned} \Vert x_n-\bar{x}\Vert ^2_{H_1}&+\Vert y_n-\bar{y}\Vert ^2_{H_2}\nonumber \\&\le - \dfrac{1}{\min \{\gamma _S,\gamma _T\}} (\langle x_n-\bar{x}, S\bar{x}\rangle _{H_1}+\langle y_n-\bar{y}, T\bar{y}\rangle _{H_2})\nonumber \\&\le \dfrac{1}{\min \{\gamma _S,\gamma _T\}}( \Vert x_n-\bar{x}\Vert _{H_1}\Vert S\bar{x}\Vert _{H_1} +\Vert y_n-\bar{y}\Vert _{H_2} \Vert T\bar{y}\Vert _{H_2})\nonumber \\&\le \dfrac{1}{\min \{\gamma _S,\gamma _T\}}\sqrt{ \Vert x_n-\bar{x}\Vert _{H_1}^2+\Vert y_n-\bar{y}\Vert _{H_2}^2}\sqrt{\Vert S\bar{x}\Vert _{H_1}^2+\Vert T\bar{y}\Vert _{H_2}^2}. \end{aligned}$$
(3.6)

This deduces that

$$\begin{aligned} \Vert x_n-\bar{x}\Vert ^2_{H_1}&+\Vert y_n-\bar{y}\Vert ^2_{H_2}\le \dfrac{1}{(\min \{\gamma _S,\gamma _T\})^2}(\Vert S\bar{x}\Vert _{H_1}^2+\Vert T\bar{y}\Vert _{H_2}^2). \end{aligned}$$

Thus we see that two sequences \(\{x_n\}\) and \(\{y_n\}\) are bounded. Hence, there are two subsequences \(\{x_{m_n}\}\) and \(\{y_{m_n}\}\) of \(\{x_n\}\) and \(\{y_n\}\), respectively, such that \(x_{m_n}\rightharpoonup x^*\) and \(y_{m_n}\rightharpoonup y^*\), as \(n\rightarrow \infty \).

It follows from Lemma 2.1, (3.1) and (3.2) that

$$\begin{aligned} -\alpha _n&(\langle x_n-\bar{x}, Sx_n\rangle _{H_1} +\langle y_n-\bar{y}, Ty_n\rangle _{H_2})\nonumber \\&=\sum _{i=1}^N\langle x_n-\bar{x},(I^{H_1}-P^{H_1}_{C_{i}})x_n-(I^{H_1}-P^{H_1}_{C_{i}})\bar{x}\rangle _{H_1}\nonumber \\&\quad +\sum _{i=1}^N\langle y_n-\bar{y}, (I^{H_2}-P^{H_2}_{Q_{i}})y_n-(I^{H_2}-P^{H_2}_{Q_{i}})\bar{y}\rangle _{H_2}\nonumber \\&\quad + \sum _{i=1}^N\langle x_n-\bar{x}, A_{i}^*(A_{i}x_n-B_{i}y_n-b_{i})\rangle _{H_1}\nonumber \\&\quad - \sum _{i=1}^N \langle y_n-\bar{y}, B_{i}^*(A_{i}x_n-B_{i}y_n-b_{i})\rangle _{H_2}\nonumber \\&=\sum _{i=1}^N\langle x_n-\bar{x},(I^{H_1}-P^{H_1}_{C_{i}})x_n-(I^{H_1}-P^{H_1}_{C_{i}})\bar{x}\rangle _{H_1}\nonumber \\&\quad +\sum _{i=1}^N\langle y_n-\bar{y}, (I^{H_2}-P^{H_2}_{Q_{i}})y_n-(I^{H_2}-P^{H_2}_{Q_{i}})\bar{y}\rangle _{H_2}\nonumber \\&\quad + \sum _{i=1}^N\langle A_{i}x_n-B_{i}y_n-A_{i} \bar{x} +B_{i}\bar{y}, A_{i}x_n-B_{i}y_n-b_{i}\rangle _{H}\nonumber \\&\ge \sum _{i=1}^N\Vert (I^{H_1}-P^{H_1}_{C_{i}})x_n\Vert _{H_1}^2+\sum _{i=1}^N \Vert (I^{H_2}-P^{H_2}_{Q_{i}})y_n\Vert _{H_2}^2\nonumber \\&\quad +\sum _{i=1}^N\Vert A_{i}x_n-B_{i}y_n-b_{i}\Vert ^2_{H}. \end{aligned}$$
(3.7)

Since \(\{x_n\}\), \(\{y_n\}\) are bounded and S, T are two bounded on bounded sets mappings, there is a positive real number K such that

$$\begin{aligned} \max \{\sup _n\Vert x_n\Vert _{H_1},\sup _n\Vert y_n\Vert _{H_2},\sup _n\Vert Sx_n\Vert _{H_1},\sup _n\Vert Ty_n\Vert _{H_2}\}\le K. \end{aligned}$$

From (3.7), we get

$$\begin{aligned} \sum _{i=1}^N\Vert (I^{H_1}-P^{H_1}_{C_{i}})x_n\Vert _{H_1}^2&+\sum _{i=1}^N \Vert (I^{H_2}-P^{H_2}_{Q_{i}})y_n\Vert _{H_2}^2\\&+\sum _{i=1}^N\Vert A_{i}x_n-B_{i}y_n-b_{i}\Vert ^2_{H}\\&\le K(2K+\Vert \bar{x}\Vert _{H_1} +\Vert \bar{y}\Vert _{H_2})\alpha _n. \end{aligned}$$

It follows from \(\lim _{n\rightarrow \infty }\alpha _n =0\) that

$$\begin{aligned}&\lim _{n\rightarrow \infty }(I^{H_1}-P^{H_1}_{C_{i}})x_{n}\Vert _{H_1}=0,\ \lim _{n\rightarrow \infty } \Vert (I^{H_2}-P^{H_2}_{Q_{i}})y_{n}\Vert _{H_2}=0, \end{aligned}$$
(3.8)
$$\begin{aligned}&\lim _{n\rightarrow \infty }\Vert A_{i}x_{n}-B_{i}y_{n}-b_{i}\Vert _{H}=0, \end{aligned}$$
(3.9)

for all \(i=1,2,\ldots ,N\). In particular, we have that

$$\begin{aligned}&\lim _{n\rightarrow \infty }(I^{H_1}-P^{H_1}_{C_{i}})x_{m_n}\Vert _{H_1}=0,\ \lim _{n\rightarrow \infty } \Vert (I^{H_2}-P^{H_2}_{Q_{i}})y_{m_n}\Vert _{H_2}=0, \end{aligned}$$
(3.10)
$$\begin{aligned}&\lim _{n\rightarrow \infty }\Vert A_{i}x_{m_n}-B_{i}y_{m_n}-b_{i}\Vert _{H}=0. \end{aligned}$$
(3.11)

From Lemma 2.6 and (3.10), we infer that \((x^*,y^*)\in C_i\times Q_i\). Since \(A_i\) and \(B_i\) are bounded linear mappings, and \(x_{m_n}\rightharpoonup x^*\), \(y_{m_n}\rightharpoonup y^*\), we obtain

$$\begin{aligned} A_{i}x_{m_n}-B_{i}y_{m_n}-b_{i}\rightharpoonup A_{i}x^*-B_{i}y^*-b_{i}. \end{aligned}$$

This combines with (3.11), one has \(A_{i}x^*-B_{i}y^*=b_{i}\). Thus, we have \((x^*,y^*)\in \Omega \).

Next, we show that \((x^*,y^*)\) is a unique solution to the variational inequality (3.3). It follows from (3.6) that

$$\begin{aligned} \langle x_n-\bar{x}, S\bar{x}\rangle _{H_1}+\langle y_n-\bar{y}, T\bar{y}\rangle _{H_2}\le 0. \end{aligned}$$

Replacing n by \(m_n\) and letting \(n\rightarrow \infty \), we get

$$\begin{aligned} \langle x^*-\bar{x}, S\bar{x}\rangle _{H_1}+\langle y^*-\bar{y}, T\bar{y}\rangle _{H_2}\le 0, \end{aligned}$$
(3.12)

for all \((\bar{x},\bar{y})\in \Omega \). Set \((x_t,y_t)=(x^*,y^*)+t(\bar{x}-x^*,\bar{y}-y^*)\) with \(t\in (0,1)\). Since \(\Omega \) is a closed convex set, and \((x^*,y^*), (\bar{x},\bar{y} )\in \Omega \), we have \((x_t,y_t)\in \Omega \). So, in (3.12), replacing \((\bar{x},\bar{y})\) by \((x_t,y_t)\), one has

$$\begin{aligned} \langle \bar{x}-x^*, S(x^*+t(\bar{x}-x^*))\rangle _{H_1}+\langle \bar{y}-y^*, T(y^*+t(\bar{y}-y^*))\rangle _{H_2}\ge 0 ,\ \forall t\in (0,1). \end{aligned}$$

Using the hemicontinuity of S, T and letting \(t\rightarrow 0^+\), we obtain

$$\begin{aligned} \langle \bar{x}-x^*, Sx^*\rangle _{H_1}+\langle \bar{y}-y^*, Ty^*\rangle _{H_2}\ge 0 ,\ \forall (\bar{x},\bar{y})\in \Omega , \end{aligned}$$
(3.13)

that is, \((x^*,y^*)\) is a solution to the variational inequality (3.3).

We now establish the uniqueness of \((x^*,y^*)\). Suppose that \((u^*,v^*)\) is also another solution to the variational inequality (3.3), that is,

$$\begin{aligned} \langle \bar{x}-u^*, Su^*\rangle _{H_1}+\langle \bar{y}-u^*, Tu^*\rangle _{H_2}\ge 0 ,\ \forall (\bar{x},\bar{y})\in \Omega . \end{aligned}$$
(3.14)

In (3.13) and (3.14), replacing \((\bar{x},\bar{y})\) by \((u^*,v^*)\) and \((x^*,y^*)\), and adding the resulting inequalities, we obtain

$$\begin{aligned} \langle x^*-u^*, Sx^*-Sy^*\rangle _{H_1}+\langle y^*-v^*, Ty^*-Tv^*\rangle _{H_2}\le 0. \end{aligned}$$

Since S and T are strongly monotone with the constants \(\gamma _S\) and \(\gamma _T\), it follows that

$$\begin{aligned} \gamma _S\Vert x^*-u^*\Vert ^2_{H_1}+\gamma _T\Vert y^*-v^*\Vert ^2_{H_2}\le 0, \end{aligned}$$

which implies that \(u^*=x^*\) and \(v^*=y^*\). Thus \((x^*,y^*)\) is unique solution to the variational inequality (3.3).

It now follows from the uniqueness of \((x^*,y^*)\) that \(x_n\rightharpoonup x^*\) and \(y_n \rightharpoonup y^*\). Finally, we show that \(x_n\rightarrow x^*\) and \(y_n\rightarrow y^*\). Indeed, in (3.6), replacing \((\bar{x},\bar{y})\) by \((x^*,y^*)\), one has

$$\begin{aligned} \Vert x_n- x^*\Vert ^2_{H_1} +\Vert y_n- y^*\Vert ^2_{H_2}\le - \dfrac{1}{\min \{\gamma _S,\gamma _T\}} (\langle x_n-x^*, S x^*\rangle _{H_1}+\langle y_n-y^*, T y^*\rangle _{H_2}). \end{aligned}$$

Letting \(n\rightarrow \infty \), we infer that \(x_n\rightarrow x^*\) and \(y_n\rightarrow y^*\).

This completes the proof. \(\square \)

We have the following theorem regarding to the distance between two solutions \((x_n,y_n)\) and \((x_m,y_m)\) of the system (3.1)–(3.2).

Theorem 3.2

Let \((x_n,y_n)\) and \((x_m,y_m)\) be two solutions of the system (3.1)-(3.2) corresponding to \(\alpha _n\) and \(\alpha _m\). Then we have the following estimate

$$\begin{aligned} \sqrt{\Vert x_n-x_m\Vert _{H_1}^2+\Vert y_n-y_m\Vert _{H_2}^2}\le K_1\dfrac{|\alpha _n-\alpha _m|}{\alpha _n}, \end{aligned}$$
(3.15)

where \(K_1=K\sqrt{2}/\min \{\gamma _S,\gamma _T\}\).

Proof

Let \(a=(x_n,y_n)\) and \(b=(x_m,y_m)\). It follows from \(F_{\alpha _n}(a)=F_{\alpha _m}(b)=0\) and (3.5) that

$$\begin{aligned} 0&=\langle a-b, F_{\alpha _n}(a)-F_{\alpha _m}(b)\rangle _{H_1\times H_2}\\&=\langle a-b, F_{\alpha _n}(a)-F_{\alpha _n}(b)\rangle _{H_1\times H_2} + \langle a-b, F_{\alpha _n}(b)-F_{\alpha _m}(b)\rangle _{H_1\times H_2}\\&=\langle a-b, F_{\alpha _n}(a)-F_{\alpha _n}(b)\rangle _{H_1\times H_2}\\&\quad +(\alpha _n-\alpha _m)(\langle x_n-x_m, Sx_m\rangle _{H_1}+\langle y_n-y_m, Ty_m\rangle _{H_2})\\&\ge \alpha _n( \gamma _S\Vert x_n-x_m\Vert _{H_1}^2 +\gamma _T\Vert y_n-y_m\Vert _{H_2}^2)\\&\quad +(\alpha _n-\alpha _m)(\langle x_n-x_m, Sx_m\rangle _{H_1}+\langle y_n-y_m, Ty_m\rangle _{H_2}). \end{aligned}$$

This implies that

$$\begin{aligned} \alpha _n(\gamma _S\Vert x_n-x_m\Vert _{H_1}^2&+\gamma _T\Vert y_n-y_m\Vert _{H_2}^2)\\&\le K|\alpha _n-\alpha _m| (\Vert x_n-x_m\Vert _{H_1}+\Vert y_n-y_m\Vert _{H_2}). \end{aligned}$$

And hence, we have

$$\begin{aligned} \Vert x_n-x_m\Vert _{H_1}^2&+\Vert y_n-y_m\Vert _{H_2}^2\\&\le (K/\min \{\gamma _S,\gamma _T\})\dfrac{|\alpha _n-\alpha _m|}{\alpha _n} (\Vert x_n-x_m\Vert _{H_1}+\Vert y_n-y_m\Vert _{H_2})\\&\le (K/\min \{\gamma _S,\gamma _T\})\dfrac{|\alpha _n-\alpha _m|}{\alpha _n} \sqrt{2(\Vert x_n-x_m\Vert _{H_1}^2+\Vert y_n-y_m\Vert _{H_2}^2)}, \end{aligned}$$

which shows that

$$\begin{aligned} \sqrt{\Vert x_n-x_m\Vert _{H_1}^2+\Vert y_n-y_m\Vert _{H_2}^2}\le K_1\dfrac{|\alpha _n-\alpha _m|}{\alpha _n}, \end{aligned}$$

This completes the proof. \(\square \)

3.2 Explicit Iterative Method

First, we have the following proposition.

Proposition 3.3

Let \(H_1\), \(H_2\) and H, be three real Hilbert spaces. Let \(C_i\) and \(Q_i\), be nonempty closed convex subsets of \(H_1\) and \(H_2\), respectively, \(i=1,2,\ldots ,N\). Let  \(A_i:\ H_1\rightarrow H\) and \(B_i:\ H_2\rightarrow H\), \(i=1,2,\ldots ,N\), be bounded linear mappings and let \(~b_i\), \(i=1,2,\ldots ,N\), be N given elements in H. Let \(S:\ H_1\rightarrow H_1\) and \(T:\ H_2\rightarrow H_2\) be two strongly monotone mappings with constants \(\gamma _S,\gamma _T\) and Lipschitz mappings with the constants \(L_S, L_T\), respectively. Then for each \(\alpha >0\), we have that

$$\begin{aligned} F_\alpha (x,y)&=\Bigg (\sum _{i=1}^N\Big ((I^{H_1}-P_{C_i}^{H_1})x+A_i^*(A_ix-B_iy-b_i)\Big )+\alpha Sx,\\&\quad \sum _{i=1}^N\Big ( (I^{H_2}-P_{Q_i}^{H_2})y-B_i^*(A_ix-B_iy-b_i)\Big )+\alpha Ty\Bigg ),\ \\&\quad \forall (x,y)\in H_1\times H_2 \end{aligned}$$

is a \(\gamma \)-strongly monotone and L-Lipschitz mapping on \(H_1\times H_2\) with \(\gamma =\min \{\gamma _S, \gamma _T\}\alpha \) and \(L=\sqrt{ [(N+\alpha L_{S,T})^2+\gamma _{A,B}^4] (1+4N^2)}\), where \(L_{S,T}=\max \{L_S, L_T\}\) and \(\gamma _{A,B}=\max _{i=12,\ldots ,N}\{\Vert A_i\Vert ,\Vert B_i\Vert \}\).

Proof

For any \(a=(x_1,y_1)\) and \(b=(x_2,y_2)\) in \(H_1\times H_2\), it follows from (3.5) that

$$\begin{aligned} \langle a-b, F_\alpha (a)-F_\alpha (b)\rangle _{H_1\times H_2}&\ge \alpha (\gamma _S\Vert x_1-x_2\Vert ^2_{H_1} +\gamma _T\Vert y_1-y_2\Vert ^2_{H_2})\\&\ge \alpha \min \{\gamma _S, \gamma _T\} \Vert a-b\Vert _{H_1\times H_2}^2. \end{aligned}$$

This implies that \(F_\alpha \) is \(\gamma \)-strongly monotone mapping on \(H_1\times H_2\) with \(\gamma =\alpha \min \{\gamma _S, \gamma _T\}\).

Next, we have

$$\begin{aligned} \Vert&F_\alpha (a)-F_\alpha (b)\Vert ^2_{H_1\times H_2}\\&=\Big \Vert \sum _{i=1}^N\Big ((I^{H_1}-P_{C_i}^{H_1})x_1-(I^{H_1}-P_{C_i}^{H_1})x_2\Big )\\&\quad +\sum _{i=1}^NA_i^*(A_i(x_1-x_2)-B_i(y_1-y_2))+\alpha (Sx_1-Sx_2)\Big \Vert ^2_{H_1}\\&+\Big \Vert \sum _{i=1}^N\Big ((I^{H_2}-P_{Q_i}^{H_2})y_1-(I^{H_2}-P_{Q_i}^{H_2})y_2\Big ) \\&\quad -\sum _{i=1}^N B_i^*(A_i(x_1-x_2)-B_i(y_1-y_2))+\alpha (Ty_1-Ty_2)\Big \Vert ^2_{H_2}\\&\le \Big [(N+\alpha L_S)\Vert x_1-x_2\Vert _{H_1} +\sum _{i=1}^N\Vert A_i\Vert (\Vert A_i\Vert \Vert x_1-x_2\Vert _{H_1}+\Vert B_i\Vert \Vert y_1-y_2\Vert _{H_2})\Big ]^2\\&\quad +\Big [(N+\alpha L_T)\Vert y_1-y_2\Vert _{H_1} +\sum _{i=1}^N\Vert B_i\Vert (\Vert A_i\Vert \Vert x_1-x_2\Vert _{H_1}+\Vert B_i\Vert \Vert y_1-y_2\Vert _{H_2})\Big ]^2\\&\le \Big [(N+\alpha L_S)\Vert x_1-x_2\Vert _{H_1} +\gamma _{A,B}^2\sum _{i=1}^N(\Vert x_1-x_2\Vert _{H_1}+\Vert y_1-y_2\Vert _{H_2})\Big ]^2\\&\quad +\Big [(N+\alpha L_T)\Vert y_1-y_2\Vert _{H_1} +\gamma _{A,B}^2\sum _{i=1}^N(\Vert x_1-x_2\Vert _{H_1}+\Vert y_1-y_2\Vert _{H_2})\Big ]^2\\&\le [(N+\alpha L_S)^2+\gamma _{A,B}^4]\Bigg [\Vert x_1-x_2\Vert _{H_1}^2+\Bigg (\sum _{i=1}^N(\Vert x_1-x_2\Vert _{H_1}+\Vert y_1-y_2\Vert _{H_2})\Bigg )^2\Bigg ]\\&\quad + [(N+\alpha L_T)^2+\gamma _{A,B}^4]\Bigg [\Vert y_1-y_2\Vert _{H_2}^2+\Bigg (\sum _{i=1}^N(\Vert x_1-x_2\Vert _{H_1}+\Vert y_1-y_2\Vert _{H_2})\Bigg )^2\Bigg ]\\&\le [(N+\alpha L_{S,T})^2+\gamma _{A,B}^4][\Vert a-b\Vert _{H_1\times H_2}^2 +2N^2(\Vert x_1-x_2\Vert _{H_1}+\Vert y_1-y_2\Vert _{H_2})^2]\\&\le [(N+\alpha L_{S,T})^2+\gamma _{A,B}^4][\Vert a-b\Vert _{H_1\times H_2}^2 +4N^2(\Vert x_1-x_2\Vert _{H_1}^2+\Vert y_1-y_2\Vert _{H_2}^2)]\\&\le [(N+\alpha L_{S,T})^2+\gamma _{A,B}^4] (1+4N^2)\Vert a-b\Vert _{H_1\times H_2}^2, \end{aligned}$$

where \(L_{S,T}=\max \{L_S, L_T\}\) and \(\gamma _{A,B}=\max \{\Vert A\Vert ,\Vert B\Vert \}\). This implies that \(F_\alpha \) is a Lipschitz mapping with the constant

$$\begin{aligned} L=\sqrt{ [(N+\alpha L_{S,T})^2+\gamma _{A,B}^4] (1+4N^2)}. \end{aligned}$$

This completes the proof. \(\square \)

We now introduce the following explicit iterative regularization method for finding an element in \(\Omega \). For any \((d_0,e_0)\in H_1\times H_2\), define the two sequences \(\{d_n\}\) and \(\{e_n\}\) as follows:

$$\begin{aligned}&d_{n+1}=d_n-\varepsilon _n \Big [\sum _{i=1}^N\Big ((I^{H_1}-P^{H_1}_{C_{i}})d_n+A_{i}^*(A_{i}d_n-B_{i}e_n-b_{i})\Big )+\alpha _n Sd_n\Big ], \end{aligned}$$
(3.16)
$$\begin{aligned}&e_{n+1}=e_n-\varepsilon _n \Big [\sum _{i=1}^N\Big ((I^{H_2}-P^{H_2}_{Q_{i}})e_n-B_{i}^*(A_{i}d_n-B_{i}e_n-b_{i})\Big )+\alpha _n Te_n\Big ], \end{aligned}$$
(3.17)

where \(\{\alpha _n\}\) and \(\{\varepsilon _n\}\) are two sequences of positive real numbers.

Remark 3.4

The iterative method (3.16)–(3.17) can be rewritten in the following form.

$$\begin{aligned} (d_{n+1},e_{n+1})=(I^{H_1\times H_2}-\varepsilon _n F_{\alpha _n})(d_n,e_n),\ \forall n\ge 0. \end{aligned}$$
(3.18)

The strong convergence of the sequences \(\{d_n\}\) and \(\{e_n\}\) generated by (3.16)-(3.17) are given in the following theorem.

Theorem 3.5

Suppose that the mappings \(S:\ H_1\rightarrow H_1\) and \(T:\ H_2\rightarrow H_2\) are \(L_S\) and \(L_T\) Lipschitz, and \(\gamma _S\) and \(\gamma _T\) strongly monotone, respectively. If

$$\begin{aligned} \varepsilon _n \in \bigg (0, \dfrac{2\min \{\gamma _{S},\gamma _T\}\alpha _n}{[(N+\alpha _n L_{S,T})^2+\gamma _{A,B}^4] (1+4N^2)}\bigg ),\ \forall n\ge 0\quad \quad ({C}), \end{aligned}$$

and the following conditions hold

  1. (C1)

    \(\lim _{n\rightarrow \infty }\alpha _n=0\);

  2. (C2)

    \(\sum _{n=1}^\infty \varepsilon _n\alpha _n=\infty \);

  3. (C3)

    \(\lim _{n\rightarrow \infty }\varepsilon _n/\alpha _n=0\);

  4. (C4)

    \(\lim _{n\rightarrow \infty }\dfrac{|\alpha _{n+1}-\alpha _n|}{\varepsilon _n\alpha _n^2}=0\),

then \(d_n\rightarrow x^*\), \(e_n\rightarrow y^*\) with \((x^*,y^*)\in \Omega \) and \((x^*,y^*)\) is a unique solution to the variational inequality (3.3).

Proof

We first show that the sequences \(\{d_n\}\) and \(\{e_n\}\) are bounded. Indeed, fixing \(\bar{w}=(\bar{x},\bar{y})\in \Omega \), i.e., \(F_{\alpha _n}(\bar{w})=\alpha _n (S\bar{x}, T\bar{y})\). Let \(w_n=(d_n,e_n)\). It follows from Lemma 2.5, Proposition 3.3 and the condition (C) that \(I^{H_1\times H_2}-\varepsilon _n F_{\alpha _n}\) is a strict contraction mapping with the contraction coefficient

$$\begin{aligned} \tau _n= \sqrt{1-\varepsilon _n (2\min \{\gamma _{S},\gamma _T\}\alpha _n-\varepsilon _n L)}, \end{aligned}$$

where \(L=\sqrt{[(N+\alpha L_{S,T})^2+\gamma _{A,B}^4] (1+4N^2)}\). Thus, from (3.18) we have

$$\begin{aligned} \Vert w_{n+1}-\bar{w}\Vert _{H_1\times H_2}&=\Vert (I^{H_1\times H_2}-\varepsilon _n F_{\alpha _n})(w_n)-\bar{w}\Vert _{H_1\times H_2}\nonumber \\&\le \Vert (I^{H_1\times H_2}-\varepsilon _n F_{\alpha _n})(w_n)-(I^{H_1\times H_2}-\varepsilon _n F_{\alpha _n})(\bar{w})\Vert _{H_1\times H_2}\nonumber \\&\quad +\varepsilon _n\Vert F_{\alpha _n}(\bar{w})\Vert _{H_1\times H_2}\nonumber \\&=\Vert (I^{H_1\times H_2}-\varepsilon _n F_{\alpha _n})(w_n)-(I^{H_1\times H_2}-\varepsilon _n F_{\alpha _n})(\bar{w})\Vert _{H_1\times H_2}\nonumber \\&\quad +\varepsilon _n\alpha _n\Vert (S\bar{x}, T\bar{y})\Vert _{H_1\times H_2}\nonumber \\&\le \tau _n \Vert w_n-\bar{w}\Vert _{H_1\times H_2}+(1-\tau _n)\dfrac{\varepsilon _n\alpha _n\Vert (S\bar{x}, T\bar{y})\Vert _{H_1\times H_2}}{1-\tau _n}. \end{aligned}$$
(3.19)

Next, since \(\lim _{n\rightarrow \infty }\varepsilon _n/\alpha _n=0\), it follows that

$$\begin{aligned} \dfrac{\varepsilon _n\alpha _n}{1-\tau _n}&=\dfrac{\varepsilon _n\alpha _n (1+\tau _n)}{\varepsilon _n (2\min \{\gamma _{S},\gamma _T\}\alpha _n-\varepsilon _n L)}\\&=\dfrac{1+\tau _n}{(2\min \{\gamma _{S},\gamma _T\}-L\varepsilon _n /\alpha _n)}\rightarrow 1/2\min \{\gamma _{S},\gamma _T\}. \end{aligned}$$

Thus, there is a positive real number \(K_2\) such that \(\sup _n \dfrac{\varepsilon _n\alpha _n}{1-\tau _n}\le K_2\). Hence, using (3.19), we get

$$\begin{aligned} \Vert w_{n+1}-\bar{w}\Vert _{H_1\times H_2}&\le \tau _n \Vert w_n-\bar{w}\Vert _{H_1\times H_2}+(1-\tau _n) K_2\Vert (S\bar{x}, T\bar{y})\Vert _{H_1\times H_2}\\&\le \max \{\Vert w_n-\bar{w}\Vert _{H_1\times H_2}, K_2\Vert (S\bar{x}, T\bar{y})\Vert _{H_1\times H_2}\}\\&\quad \vdots \\&\le \max \{\Vert w_0-\bar{w}\Vert _{H_1\times H_2}, K_2\Vert (S\bar{x}, T\bar{y})\Vert _{H_1\times H_2}\}. \end{aligned}$$

This implies that the sequence \(\{w_n\}\) is bounded, i.e., two sequences \(\{d_n\}\) and \(\{e_n\}\) are bounded.

Let \(h_n=(x_n,y_n)\) which defined by the system of equations (3.1)-(3.2). It is easy to see that two mappings S and T satisfy all conditions in Theorem 3.1 and hence \(x_n\rightarrow x^*\), \(y_n\rightarrow y^*\) with \((x^*,y^*)\in \Omega \) and \((x^*,y^*)\) is a unique solution to the variational inequality (3.3).

Now, in order to prove \(d_n\rightarrow x^*\), \(e_n\rightarrow y^*\), we will show that \(d_n-x_n\rightarrow 0\) and \(e_n-y_n\rightarrow 0\). Indeed, it follows from the inequality

$$\begin{aligned} \Vert x\Vert _{H_1\times H_2}^2\le \Vert y\Vert _{H_1\times H_2}^2+2\langle x,x-y\rangle _{H_1\times H_2} \end{aligned}$$

which holds for all \(x,y\in {H_1\times H_2}\), that

$$\begin{aligned}&\Vert w_{n+1}-h_{n+1}\Vert _{H_1\times H_2}^2\le \Vert w_{n+1}-h_n\Vert _{H_1\times H_2}^2+2\langle w_{n+1}-h_{n+1}, h_n-h_{n+1}\rangle _{H_1\times H_2},\\&\Vert w_{n+1}-h_n\Vert _{H_1\times H_2}^2\le \Vert w_n-h_n\Vert _{H_1\times H_2}^2 +2\langle w_{n+1}-h_n,w_{n+1}-w_n\rangle _{H_1\times H_2}. \end{aligned}$$

This leads to

$$\begin{aligned} \Vert w_{n+1}-h_{n+1}\Vert _{H_1\times H_2}^2&\le \Vert w_n-h_n\Vert _{H_1\times H_2}^2 +2\langle w_{n+1}-h_n,w_{n+1}-w_n\rangle _{H_1\times H_2}\nonumber \\&\quad +2\langle w_{n+1}-h_{n+1}, h_n-h_{n+1}\rangle _{H_1\times H_2}\nonumber \\&\le \Vert w_n-h_n\Vert _{H_1\times H_2}^2 \nonumber \\&\quad +2 (\Vert w_{n+1}\Vert _{H_1\times H_2}+\Vert h_{n+1}\Vert _{H_1\times H_2})\Vert h_{n+1}-h_n\Vert _{H_1\times H_2}\nonumber \\&\quad +2\langle w_{n+1}-h_n,w_{n+1}-w_n\rangle _{H_1\times H_2}. \end{aligned}$$
(3.20)

We now estimate the quantity \(\langle w_{n+1}-h_n,w_{n+1}-w_n\rangle _{H_1\times H_2}\). It follows from \(F_{\alpha _n} (h_n)=0\) and (3.5) that

$$\begin{aligned} \langle w_{n+1}-h_n,&w_{n+1}-w_n\rangle _{H_1\times H_2}\nonumber \\&=\langle w_n-\varepsilon _n F_{\alpha _n}(w_n) -h_n, w_n-\varepsilon _n F_{\alpha _n}(w_n) -w_n\rangle _{H_1\times H_2}\nonumber \\&=\langle w_n-\varepsilon _n F_{\alpha _n}(w_n) -h_n, -\varepsilon _n F_{\alpha _n}(w_n) \rangle _{H_1\times H_2}\nonumber \\&=-\varepsilon _n \langle w_n-h_n, F_{\alpha _n}(w_n) \rangle _{H_1\times H_2} +\varepsilon _n^2\Vert F_{\alpha _n}(w_n)\Vert ^2_{H_1\times H_2}\nonumber \\&=-\varepsilon _n \langle w_n-h_n, F_{\alpha _n}(w_n) -F_{\alpha _n}h_n \rangle _{H_1\times H_2} +\varepsilon _n^2\Vert F_{\alpha _n}(w_n)\Vert ^2_{H_1\times H_2}\nonumber \\&\le -\varepsilon _n\alpha _n (\gamma _S \Vert d_n-x_n\Vert ^2_{H_1} +\gamma _T \Vert e_n-y_n\Vert ^2_{H_2}) + \varepsilon _n^2\Vert F_{\alpha _n}(w_n)\Vert ^2_{H_1\times H_2}\nonumber \\&\le - \varepsilon _n\alpha _n\min \{\gamma _{S},\gamma _T\}\Vert w_n-h_n\Vert _{H_1\times H_2}^2 + \varepsilon _n^2\Vert F_{\alpha _n}(w_n)\Vert ^2_{H_1\times H_2}. \end{aligned}$$
(3.21)

From the boundedness of \(\{w_n\}\), it is easy to see that \(\{F_{\alpha _n}(w_n)\}\) is bounded too. Thus, there exists a positive real number \(K_3\) such that

$$\begin{aligned} \sup _n \{\Vert w_n\Vert _{H_1\times H_2}, \Vert F_{\alpha _n}(w_n)\Vert _{H_1\times H_2}\}\le K_3. \end{aligned}$$

This combines with (3.15), (3.20) and (3.21), we obtain

$$\begin{aligned} \Vert w_{n+1}-h_{n+1}\Vert _{H_1\times H_2}^2&\le (1- 2\min \{\gamma _{S},\gamma _T\}\varepsilon _n\alpha _n)\Vert w_n-h_n\Vert _{H_1\times H_2}^2\nonumber \\&\quad +2(K\sqrt{2}+K_3)K_1\dfrac{|\alpha _{n+1}-\alpha _n|}{\alpha _n}\nonumber \\&\quad +2K_3^2 \varepsilon _n^2. \end{aligned}$$
(3.22)

Letting

$$\begin{aligned}&\Gamma _n=\Vert w_n-h_n\Vert _{H_1\times H_2}^2,\\&b_n=2\min \{\gamma _{S},\gamma _T\}\varepsilon _n\alpha _n,\\&c_n=\dfrac{1}{2\min \{\gamma _{S},\gamma _T\}}\bigg (2(K\sqrt{2}+K_3)\dfrac{|\alpha _{n+1}-\alpha _n|}{\varepsilon _n\alpha _n^2}+2K_3^2\dfrac{\varepsilon _n}{\alpha _n}\bigg ). \end{aligned}$$

we can rewrite the above inequality as follows:

$$\begin{aligned} \Gamma _{n+1}\le (1-b_n)\Gamma _n + b_nc_n. \end{aligned}$$

It is not difficult to see that conditions (C1)–(C4) ensure that all the assumptions of Lemma 2.7 are satisfied. Therefore we immediately infer that \(\Gamma _n\rightarrow 0\), that is, \(\Vert w_n-h_n\Vert _{H_1\times H_2}^2\rightarrow 0\) or \(\Vert d_n-x_n\Vert _{H_1}^2+\Vert e_n-y_n\Vert _{H_2}^2\rightarrow 0\). This shows that \(\Vert d_n-x_n\Vert _{H_1}\rightarrow 0\), \(\Vert e_n-y_n\Vert _{H_2}\rightarrow 0\) and hence \(d_n\rightarrow x^*\), \(e_n\rightarrow y^*\).

This completes the proof. \(\square \)

Remark 3.6

It is easy to check that \(\alpha _n=1/\root 4 \of {n}\) and

$$\begin{aligned} \varepsilon _n= \dfrac{2\min \{\gamma _{S},\gamma _T\}\alpha _n^2}{[(N+\alpha _n L_{S,T})^2+\gamma _{A,B}^4] (1+4N^2)M}, \end{aligned}$$

with \(M > 1\), satisfy all conditions of Theorem 3.5.

4 Relaxed Iterative Methods

In this section we consider Problem (SSFP) when \(C_i\) and \(Q_i\), \(i=1,2,\ldots ,N\), are sublevel sets of the lower semicontinuous convex functions \(c_i:\ H_1\rightarrow \mathbb R\) and \(q_i:\ H_2\rightarrow \mathbb R\), \(i=1,2,\ldots ,N\), respectively. In other words,

$$\begin{aligned}&C_i=\{x\in H_1:\ c_i(x)\le 0\},\\&Q_i=\{y\in H_2:\ q_i(y)\le 0\},\ i=1,2,\ldots ,N. \end{aligned}$$

Assume that \(c_i\) and \(q_i\), \(i=1,2,\ldots ,N\), are subdifferentiable on \(H_1\) and \(H_2\), respectively, and that the subdifferentials \(\partial c_i\) and \(\partial q_i\), \(i=1,2,\ldots ,N\), are bounded (on bounded sets). At a point \(x_n\in H_1\) and \(y_n\in H_2\), we define the subsets \(C_{i,n}\) and \(Q_{i,n}\) as follows:

$$\begin{aligned}&C_{i,n} := \{x\in H_1:\ c_i(x_n)\le \langle x_n-x,\xi _{i,n}\rangle _{H_1}\},\\&Q_{i,n} := \{y\in H_2:\ q_i(y_n)\le \langle y_n-y,\eta _{i,n}\rangle _{H_2}\}, \end{aligned}$$

where \(\xi _{i,n}\in \partial c_i(x_n)\) and \(\eta _{i,n}\in \partial q_i(y_n)\) for all \(i=1,2,\ldots ,N\). The sets \(C_{i,n}\) and \(Q_{i,n}\) are called the relaxed sets of \(C_i\) and \(Q_i\), respectively. It is not difficult to see that \(C_{i,n} \) and \(Q_{i,n}\) are half-spaces of \(H_1\) and \(H_2\), and that \(C_i\subset C_{i,n}\), \(Q_i\subset Q_{i,n}\), for all \(i=1,2,\ldots ,N\).

It is well known that generally speaking, it is not easy to calculate the projections \(P_{C_i}^{H_1}x\) and \(P_{Q_i}^{H_2}y\). Therefore we introduce two relaxed iterative methods corresponding to the the iterative methods (3.1)–(3.2) and (3.16)–(3.17), where \(P_{C_i}^{H_1}\) and \(P_{Q_i}^{H_2}\) are replaced by the operators \(P_{C_{i,n}}^{H_1}\) and \(P_{Q_{i,n}}^{H_2}\), respectively, which are defined as follows:

$$\begin{aligned}&P_{C_{i,n}}^{H_1}x=x - \max \bigg \{\dfrac{\langle x,\xi _{i,n}\rangle _{H_1}-\langle x_n,\xi _{i,n}\rangle _{H_1}+c_i(x_n)}{\Vert \xi _{i,n}\Vert ^2_{H_1}},0\bigg \}\xi _{i,n},\\&P_{Q_{i,n}}^{H_2}y = y-\max \bigg \{\dfrac{\langle y,\eta _{i,n}\rangle _{H_2}-\langle y_n,\eta _{i,n}\rangle _{H_2}+q_i(y_n)}{\Vert \eta _{i,n}\Vert ^2_{H_2}},0\bigg \}\eta _{i,n}. \end{aligned}$$

We first state and prove the following theorem.

Theorem 4.1

Let \(S:\ H_1\rightarrow H_1\) and \(T:\ H_2\rightarrow H_2\) be two bounded on bounded sets, hemicontinuous and strongly monotone mappings with the constants \(\gamma _S\) and \(\gamma _T\), respectively. Let \(\{\alpha _n\}\) be a sequence of positive real numbers. Then the system of regularization equations

$$\begin{aligned}&\sum _{i=1}^N\Big ((I^{H_1}-P^{H_1}_{C_{i,n}})x_n+A_{i}^*(A_{i}x_n-B_{i}y_n-b_{i})\Big )+\alpha _n Sx_n=0,\ \end{aligned}$$
(4.1)
$$\begin{aligned}&\sum _{i=1}^N\Big ((I^{H_2}-P^{H_2}_{Q_{i,n}})y_n-B_{i}^*(A_{i}x_n-B_{i}y_n-b_{i})\Big )+\alpha _n Ty_n=0, \end{aligned}$$
(4.2)

has a unique solution \((x_n,y_n)\) for each \(n \ge 1\). Moreover, if \(\alpha _n\rightarrow 0\), then

\(x_n\rightarrow x^*\), \(y_n\rightarrow y^*\) with \((x^*,y^*)\in \Omega \) and \((x^*,y^*)\) is a unique solution to the variational inequality (3.3).

Proof

Following the proof of Theorem 3.1, the system (4.1)–(4.2) has unique solution \((x_n,y_n)\). Moreover, the sequences \(\{x_n\}\) and \(\{y_n\}\) are bounded.

We now prove that all weak subsequential limits of the sequence \(\{(x_n,y_n)\}\) belongs to \(\Omega \). Indeed, suppose that \((x^*,y^*)\) is a weak subsequential limit of \(\{(x_n,y_n)\}\). There are the subsequences \(\{x_{p_{n}}\}\) and \(\{y_{p_{n}}\}\) of \(\{x_n\}\) and \(\{y_n\}\), respectively, such that \(x_{p_n}\rightharpoonup x^*\) and \(y_{p_n}\rightharpoonup y^*\). Moreover, we also have

  1. (1)

    \(\lim _{n\rightarrow \infty }(I^{H_1}-P^{H_1}_{C_{i,n}})x_{p_n}\Vert _{H_1}=0,\ \lim _{n\rightarrow \infty } \Vert (I^{H_2}-P^{H_2}_{Q_{i,n}})y_{p_n}\Vert _{H_2}=0\),

  2. (2)

    \(\lim _{n\rightarrow \infty }\Vert A_{i}x_{p_n}-B_{i}y_{p_n}-b_{i}\Vert _{H}=0\),

for all \(i=1,2,\ldots ,N\).

Since the subdifferential \(\partial c_i\) is assumed to be bounded on bounded sets and the sequence \(\{x_n\}\) is bounded, there is a positive real number \(K_4\) such that \(\Vert \xi _{i,n}\Vert \le K_4\) for all \(n\ge 1\). It follows from \(P_{C_{i,n}}^{H_1}x_{p_n}\in C_{i,n}\) and the definition of \(C_{i,n}\) that

$$\begin{aligned} c_i(x_{p_n})\le \langle (I^{H_1}-P_{C_{i,n}}^{H_1})x_{p_n}, \xi _{i,p_n}\rangle _{H_1}\le K_4\Vert (I^{H_1}-P_{C_{i,n}}^{H_1})x_{p_n}\Vert _{H_1}\rightarrow 0. \end{aligned}$$

This implies that \(\liminf _{n\rightarrow \infty }c_i(x_{p_n}) \le 0\). By the lower semicontinuity of the function c, we have

$$\begin{aligned} c_i(x^*)\le \liminf _{n\rightarrow \infty } c_i(x_{p_n}) \le 0. \end{aligned}$$

This implies that \(x^*\in C_i\). By an argument similar to the one above, we also obtain that \(y^*\in Q_i\). Furthermore, from \(\lim _{n\rightarrow \infty }\Vert A_{i}x_{p_n}-B_{i}y_{p_n}-b_{i}\Vert _{H}=0\), it is easy to deduce that \(A_ix^*-B_iy^*=b_i\). Thus, we have \((x^*,y^*)\in \Omega \).

Using similar argument to the one employed in the proof of Theorem 3.1, we conclude that \((x^*,y^*)\) is the unique weak subsequential limit of \(\{(x_n,y_n)\}\), that it is the unique solution to the variational inequality (3.3) and that \(x_n\rightarrow x^*\) and \(y_n\rightarrow y^*\), as \(n\rightarrow \infty \).

This completes the proof. \(\square \)

Finally, by using a line of proof similar to the one in the proof of Theorem 3.5 and combining it with Theorem 4.1, we obtain the following theorem.

Theorem 4.2

Suppose that the mappings \(S:\ H_1\rightarrow H_1\) and \(T:\ H_2\rightarrow H_2\) are \(L_S\) and \(L_T\) lipschitz, and \(\gamma _S\) and \(\gamma _T\) strongly monotone, respectively. Let \(\{\varepsilon _n\}\) and \(\{\alpha _n\}\) be two sequences of positive real numbers. For any \((d_0,e_0)\in H_1\times H_2\), define the two sequences \(\{d_n\}\) and \(\{e_n\}\) as follows:

$$\begin{aligned}&d_{n+1}=d_n-\varepsilon _n \Big [\sum _{i=1}^N\Big ((I^{H_1}-P^{H_1}_{C_{i,n}})d_n+A_{i}^*(A_{i}d_n-B_{i}e_n-b_{i})\Big )+\alpha _n Sd_n\Big ],\\&e_{n+1}=e_n-\varepsilon _n \Big [\sum _{i=1}^N\Big ((I^{H_2}-P^{H_2}_{Q_{i,n}})e_n-B_{i}^*(A_{i}d_n-B_{i}e_n-b_{i})\Big )+\alpha _n Te_n\Big ], \end{aligned}$$

If conditions (C) and (C1)–(C4) of Theorem 3.5 hold, then \(d_n\rightarrow x^*\), \(e_n\rightarrow y^*\) with \((x^*,y^*)\in \Omega \) and \((x^*,y^*)\) is a unique solution to the variational inequality (3.3).