Split hierarchical variational inequality problems and related problems

Abstract

The main objective of this paper is to introduce a split hierarchical variational inequality problem. Several related problems are also considered. We propose an iterative method for finding a solution of our problem. The weak convergence of the sequence generated by the proposed method is studied.

MSC:47H09, 47H10, 47J25, 49J40, 65K10.

1 Introduction

In 1994, Censor and Elfving [1] introduced the following split feasibility problem (in short, SFP): find a point

$$x^{\ast }\in C\text{such that}A{x}^{\ast }\in Q,$$

(1.1)

where C and Q are nonempty closed convex subsets of ${\mathbb{R}}^n$ and ${\mathbb{R}}^m$, respectively, and A is an $m\times n$ matrix. They proposed an algorithm to find the solution of SFP. Their algorithm did not become popular, since it concerns the complicated matrix inverse computations and, subsequently, is considered in the case when $n=m$. Based on these observations, Byrne [2] applied the forward-backward method, a type of projected gradient method, presenting the so-called CQ-iterative procedure, which is defined by

$$x_{k+1}=P_C(x_k+\gamma A^{\mathrm{\top }}(P_Q-I)A{x}_k),\mathrm{\forall }k\ge 1,$$

where an initial $x_1\in {\mathbb{R}}^n$, $\gamma \in (0,2/{\parallel A\parallel }^2)$, and $P_C$ and $P_Q$ denote the metric projections onto C and Q, respectively. The convergence result of the sequence ${\{x_k\}}_{k=1}^{\mathrm{\infty }}$ to a solution of the considered split feasibility problem was presented. Further, Byrne also proposed an application to dynamic emission tomographic image reconstruction. A few years later, Censor et al. [3, 4] proposed an incredible application of the split feasibility problem to the inverse problem of intensity-modulated radiation therapy treatment planning. Recently, Xu [5] considered SFP in the setting of infinite-dimensional Hilbert spaces and established the following CQ-algorithm: Let $H_1$ and $H_2$ be real Hilbert spaces and $A:H_1\to H_2$ be a bounded linear operator. For a given $x_1\in H_1$, generated a sequence ${\{x_k\}}_{k=1}^{\mathrm{\infty }}$ by the following iterative scheme:

$$x_{k+1}=P_C(x_k+\gamma A^{\ast }(P_Q-I)A{x}_k),\mathrm{\forall }k\ge 1,$$

where $\gamma \in (0,2/{\parallel A\parallel }^2)$, and $A^{\ast }$ is the adjoint operator of A. He also proved the weak convergence of the sequence produced by the above procedure to a solution of the SFP. The details of the CQ-algorithm for SFP problem are given in [6]. A comprehensive literature, survey, and references on SFP can be found in [7].

In 2009, Censor and Segal [8] presented an important form of the split feasibility problem called the split common fixed point problem, which is to find a point

$$x^{\ast }\in Fix(T)\text{such that}A{x}^{\ast }\in Fix(S),$$

(1.2)

where T and S are some nonlinear operators on ${\mathbb{R}}^n$ and ${\mathbb{R}}^m$, respectively, and A is a real $m\times n$ matrix. Based on the properties of operators T and S, called cutter or directed operators, they presented the following algorithm for solving the split common fixed point problem:

$$x_{k+1}=T(x_k+\gamma A^{\mathrm{\top }}(S-I)A{x}_k),\mathrm{\forall }k\ge 1,$$

(1.3)

where an initial $x_1\in {\mathbb{R}}^n$, $\gamma \in (0,2/{\parallel A\parallel }^2)$. They also presented a convergence result for this algorithm. Moudafi [9] studied the split common fixed point problem in the context of the demicontractive operators T and S in the setting of infinite-dimensional Hilbert spaces. He established the weak convergence of the sequence generated by his scheme to a solution of the split common fixed point problem.

On the other hand, the theory of variational inequalities is well known and well developed because of its applications in different areas of science, social science, engineering, and management. There are several monographs on variational inequalities, but we mention here a few [10–12]. Let C be a nonempty closed convex subset of a real Hilbert space H and $f:H\to H$ be an operator. The variational inequality problem defined by C and f is to find $x^{\ast }\in C$ such that

($VIP(C;f)$)  $〈f\left(x^{\ast }\right),x-x^{\ast }〉\ge 0$, for all $x\in C$.

Another problem closely related to $VIP(C;f)$ is known as the Minty variational inequality problem: find $x^{\ast }\in C$ such that

($MVIP(C;f)$)  $〈f(x),x-x^{\ast }〉\ge 0$, for all $x\in C$.

The trivial unlikeness of two proposed problems is the linearity of variational inequalities. In fact, the Minty variational inequality $MVIP(C;f)$ is linear but the variational inequality $VIP(C;f)$ is not. However, under the (hemi)continuity and monotonicity of f, the solution sets of these problems are the same (see [[13], Lemma 1]).

If the constrained set C in variational inequality formulations is a set of fixed points of an operator, then the variational inequality problem is known as hierarchical variational inequality problem.

Let $T:H\to H$ be a nonlinear operator with the set of fixed points $Fix(T)\ne \mathrm{\varnothing }$ and $f:H\to H$ be an operator. The hierarchical variational inequality problem is to find $x^{\ast }\in Fix(T)$ such that

$$〈f\left(x^{\ast }\right),x-x^{\ast }〉\ge 0,\text{for all }x\in Fix(T).$$

(1.4)

A closely related problem to hierarchical variational inequality problem is the hierarchical Minty variational inequality problem: find $x^{\ast }\in Fix(T)$ such that

$$〈f(x),x-x^{\ast }〉\ge 0,\text{for all }x\in Fix(T).$$

(1.5)

In the recent past, several methods for solving hierarchical variational inequalities have been investigated in the literature; see, for example, [14–19] and the references therein.

The goal of this paper is to introduce a split-type problem, by combining a split fixed point problem and a hierarchical variational inequality problem. The considered problem can be applied to solve many existing problems. We present an iterative procedure for finding a solution of the proposed problem and show that under some suitable assumptions, the sequence generated by our algorithm converges weakly to a solution of the considered problem.

The rest of this paper is divided into four sections. In Section 2, we recall and state preliminaries on numerous nonlinear operators and their useful properties. Section 3 we divide in two subsections, that is, first, we present a split problem, called the split hierarchical Minty variational inequality problem. We subsequently propose an algorithm for solving our problem and establish a convergence result under some assumptions. Second, we present the split hierarchical variational inequality problem. In the last section, we investigate some related problems, where we can apply the considered problem.

2 Preliminaries

Let H be a real Hilbert space whose inner product and norm are denoted by $〈\cdot ,\cdot 〉$ and $\parallel \cdot \parallel$, respectively. The strong convergence and weak convergence of a sequence ${\{x_k\}}_{k=1}^{\mathrm{\infty }}$ to $x\in H$ are denoted by $x_k\to x$ and $x_k\rightharpoonup x$, respectively. Let $T:H\to H$ be an operator. We denote by $R(T)$ the range of T, and by $Fix(T)$ the set of all fixed points of T, that is, $Fix(T)=\{x\in H:x=Tx\}$. The operator T is said to be nonexpansive if for all $x,y\in H$, $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$; strongly nonexpansive [20, 21] if T is nonexpansive and for all bounded sequences ${\{x_k\}}_{k=1}^{\mathrm{\infty }}$, ${\{y_k\}}_{k=1}^{\mathrm{\infty }}$ in H, the condition ${lim}_{k\to \mathrm{\infty }}(\parallel x_k-y_k\parallel -\parallel T{x}_k-T{y}_k\parallel )=0$ implies ${lim}_{k\to \mathrm{\infty }}\parallel (x_k-y_k)-(T{x}_k-T{y}_k)\parallel =0$; averaged nonexpansive if for all $x,y\in H$, $T=(1-\alpha )I+\alpha S$ holds for a nonexpansive operator $S:H\to H$ and $\alpha \in (0,1)$; firmly nonexpansive if $2T-I$ is nonexpansive, or equivalently for all $x,y\in H$, ${\parallel Tx-Ty\parallel }^2\le 〈x-y,Tx-Ty〉$; cutter [21] if $〈x-Tx,z-Tx〉\le 0$ for all $x\in H$ and all $z\in Fix(T)$; monotone if $〈Tx-Ty,x-y〉\ge 0$, for all $x,y\in H$; α-inverse strongly monotone (or α-cocoercive) if there exists a positive real number α such that $〈Tx-Ty,x-y〉\ge \alpha {\parallel Tx-Ty\parallel }^2$, for all $x,y\in H$.

Let $B:H\to 2^H$ be a set-valued operator, we define a graph of B by $\{(x,y)\in H\times H:y\in B(x)\}$ and an inverse operator of B, denoted $B^{-1}$, by $\{(y,x)\in H\times H:x\in B(y)\}$. A set-valued operator $B:H\to 2^H$ is called monotone if for all $x,y\in H$, $u\in B(x)$ and $v\in B(y)$ such that $〈x-y,u-v〉\ge 0$. A monotone operator B is said to be maximal monotone if there exists no other monotone operator such that its graph properly contains the graph of B. For a maximal monotone operator B, we know that for each $x\in H$ and a positive real number σ, there is a unique $z\in H$ such that $x\in (I+\sigma B)z$. We define the resolvent of B with parameter σ by $J_{\sigma }^B:={(I+\sigma B)}^{-1}$. It is well known that the resolvent is a single-valued and firmly nonexpansive operator.

Remark 2.1 It can be seen that the class of averaged nonexpansive operators is a proper subclass of the class of strongly nonexpansive operators. Since any firmly nonexpansive operator is an averaged nonexpansive operator, it is clear that a class of firmly nonexpansive operators is contained in the class of strongly nonexpansive operators. Further, a firmly nonexpansive operator with fixed point is cutter. However, a nonexpansive cutter operator need not to be firmly nonexpansive; for further details, see [21].

The following well-known lemma is due to Opial [22].

Lemma 2.2 (Demiclosedness principle) [[22], Lemma 2]

Let C be a nonempty closed convex subset of a real Hilbert space H and $T:C\to H$ be a nonexpansive operator. If the sequence ${\{x_k\}}_{k=1}^{\mathrm{\infty }}\subseteq C$ converges weakly to an element $x\in C$ and the sequence ${\{x_k-T{x}_k\}}_{k=1}^{\mathrm{\infty }}$ converges strongly to 0, then x is a fixed point of the operator T.

To prove the main theorem of this paper, we need the following lemma.

Lemma 2.3 [[23], Section 2.2.1, Lemma 2]

Assume that ${\{a_k\}}_{k=1}^{\mathrm{\infty }}$ and ${\{b_k\}}_{k=1}^{\mathrm{\infty }}$ are nonnegative real sequences such that $a_{k+1}\le a_k+b_k$. If ${\sum }_{k=1}^{\mathrm{\infty }}{b}_k<\mathrm{\infty }$, then ${lim}_{k\to \mathrm{\infty }}{a}_k$ exists.

The following lemma can be immediately obtained by the properties of an inner product.

Lemma 2.4 Let H be a real Hilbert space H. Then, for all $x,y\in H$,

$${\parallel x-y\parallel }^2\le {\parallel x\parallel }^2+2〈y,y-x〉.$$

3 Split hierarchical variational inequality problems and convergent results

In this section we introduce split hierarchical variational inequality problems and discuss some related problems. Further, we propose an iterative method for finding a solution of the hierarchical variational inequality problem and prove the convergence result for the sequence generated by the proposed iterative method.

3.1 Split hierarchical Minty variational inequality problem

Let $H_1$ and $H_2$ be two real Hilbert spaces, $f,T:H_1\to H_1$ be operators such that $Fix(T)\ne \mathrm{\varnothing }$, and $h,S:H_2\to H_2$ be operators such that $Fix(S)\ne \mathrm{\varnothing }$. Let $A:H_1\to H_2$ be an operator with $R(A)\cap Fix(S)\ne \mathrm{\varnothing }$. The split hierarchical Minty variational inequality problem (in short, SHMVIP) is to find $x^{\ast }\in Fix(T)$ such that

$$〈f(x),x-x^{\ast }〉\ge 0,\text{for all }x\in Fix(T),$$

(3.1)

and such that $A{x}^{\ast }\in Fix(S)$ satisfies

$$〈h(y),y-A{x}^{\ast }〉\ge 0,\text{for all }y\in Fix(S).$$

(3.2)

The solution set of SHMVIP (3.1)-(3.2) is denoted by Γ, that is,

$$\mathrm{\Gamma }:=\{x\text{ which solves }(\text{3.1}):Ax\text{ solves }(\text{3.2})\}.$$

We note that the problems (3.1) and (3.2) are nothing but hierarchical Minty variational inequality problems. If f and h are zero operators, that is, $f\equiv h\equiv 0$, then SHMVIP (3.1) and (3.2) reduces to the split fixed point problem (1.2). Moreover, SHMVIP (3.1) and (3.2) can be applied to several existing split-type problems, which we will discuss in Section 4.

Inspired by the iterative scheme (1.3) for solving split common fixed point problem (1.2) and the existing algorithms, a generalization of the projected gradient method, for solving the hierarchical variational inequality problem of Yamada [19] and Iiduka [16], we now present an iterative algorithm for solving HMVIP (3.1)-(3.2).

Algorithm 3.1 Initialization: Choose ${\{{\alpha }_k\}}_{k=1}^{\mathrm{\infty }},{\{{\beta }_k\}}_{k=1}^{\mathrm{\infty }}\subset (0,+\mathrm{\infty })$. Take arbitrary $x_1\in H_1$.

Iterative Step: For a given current iterate $x_k\in H_1$, compute

$$y_k:=x_k-\gamma A^{\ast }(I-S(I-{\beta }_{k}h))A{x}_k,$$

where $\gamma \in (0,\frac{2}{{\parallel A\parallel }^2})$ and define $x_{k+1}\in H_1$ by

$$x_{k+1}:=T(I-{\alpha }_{k}f)y_k.$$

Update $k:=k+1$.

Rest of the section, unless otherwise specified, we assume that T is a strongly nonexpansive operator on $H_1$ with $Fix(T)\ne \mathrm{\varnothing }$ and S is a strongly nonexpansive cutter operator on $H_2$ with $Fix(S)\ne \mathrm{\varnothing }$, f (respectively, h) is a monotone and continuous operator on $H_1$ (respectively, $H_2$) and $A:H_1\to H_2$ is a bounded linear operator with $R(A)\cap Fix(S)\ne \mathrm{\varnothing }$.

Remark 3.2 (i) The Algorithm 3.1 can be applied to the iterative scheme (1.3) by setting the operators $f\equiv h\equiv 0$.

(ii) The iterative Algorithm 3.1 extends and develops the algorithms in [[24], Algorithm 6.1] and [[25], Algorithm (8)] in many aspects. Indeed, the metric projections $P_C$ and $P_Q$ in [[24], Algorithm 6.1] and the resolvent operators $J_{\lambda }^{B_1}$ and $J_{\lambda }^{B_2}$ in [[25], Algorithm (8)] are firmly nonexpansive operators which are special cases of the assumption on the operators T and S. Second, its involves control sequences ${\{{\alpha }_k\}}_{k=1}^{\mathrm{\infty }}$ and ${\{{\beta }_k\}}_{k=1}^{\mathrm{\infty }}$, while in [[24], Algorithm 6.1] and [[25], Algorithm (8)] they are constant. Furthermore, we assume $\gamma \in (0,\frac{2}{{\parallel A\parallel }^2})$, while in [[24], Algorithm 6.1] and [[25], Algorithm (8)], it was assumed to be in $(0,\frac{1}{{\parallel A\parallel }^2})$, which clearly is a more restrictive assumption.

(iii) By setting an operator A to be zero operator, the iterative Algorithm 3.1 relates to existing iterative schemes for solving the hierarchical variational inequality problem in, for example, [[15], Algorithm 3.1] and [[17], Algorithm 5].

The following theorem provides the weak convergence of the sequence generated by the Algorithm 3.1 to an element of Γ.

Theorem 3.3 Let a sequence ${\{x_k\}}_{k=1}^{\mathrm{\infty }}$ be generated by Algorithm 3.1 with $x_1\in H_1$, and let ${\{{\alpha }_k\}}_{k=1}^{\mathrm{\infty }}$ and ${\{{\beta }_k\}}_{k=1}^{\mathrm{\infty }}\subset (0,1)$ be sequences such that ${\sum }_{k=1}^{\mathrm{\infty }}{\alpha }_k<\mathrm{\infty }$ and ${lim}_{k\to \mathrm{\infty }}{\beta }_k=0$. If $\mathrm{\Gamma }\ne \mathrm{\varnothing }$, then the following statements hold.

• (i) If there exists a natural number $k_0$ such that

  $$\mathrm{\Gamma }\subset \bigcap _{k=k_0}^{\mathrm{\infty }}\{z\in H_1:〈h(A{x}_k),S(I-{\beta }_{k}h)A{x}_k-Az〉\ge 0\},$$

  then, for all $k\ge k_0$ and $z\in \mathrm{\Gamma }$, we have

  $${\parallel y_k-z\parallel }^2\le {\parallel x_k-z\parallel }^2-\gamma (2-\gamma {\parallel A\parallel }^2){\parallel (I-S(I-{\beta }_{k}h))A{x}_k\parallel }^2.$$

• (ii) If the sequence ${\{f(y_k)\}}_{k=1}^{\mathrm{\infty }}$ is bounded, then ${lim}_{k\to \mathrm{\infty }}\parallel x_k-z\parallel$ exists for all $z\in \mathrm{\Gamma }$.

• (iii) If the sequence ${\{h(A{x}_k)\}}_{k=1}^{\mathrm{\infty }}$ is bounded, then ${lim}_{k\to \mathrm{\infty }}\parallel x_{k+1}-x_k\parallel =0$, ${lim}_{k\to \mathrm{\infty }}\parallel x_k-T{x}_k\parallel =0$, and ${lim}_{k\to \mathrm{\infty }}\parallel A{x}_k-SA{x}_k\parallel =0$.

• (iv) If $\parallel x_{k+1}-x_k\parallel =o({\alpha }_k)$, and ${\alpha }_k=o({\beta }_k^2)$, then the sequence ${\{x_k\}}_{k=1}^{\mathrm{\infty }}$ converges weakly to an element of Γ.

Proof (i) Let $z\in \mathrm{\Gamma }$ be given. Then

$$\begin{array}{rcl}{\parallel y_k-z\parallel }^2& =& {\parallel x_k-\gamma A^{\ast }(I-S(I-{\beta }_{k}h))A{x}_k-z\parallel }^2\\ =& {\parallel x_k-z\parallel }^2+{\gamma }^2{\parallel A^{\ast }(I-S(I-{\beta }_{k}h))A{x}_k\parallel }^2\\ -2\gamma 〈x_k-z,A^{\ast }(I-S(I-{\beta }_{k}h))A{x}_k〉\\ \le & {\parallel x_k-z\parallel }^2+{\gamma }^2{\parallel A\parallel }^2{\parallel (I-S(I-{\beta }_{k}h))A{x}_k\parallel }^2\\ -2\gamma 〈x_k-z,A^{\ast }(I-S(I-{\beta }_{k}h))A{x}_k〉,\end{array}$$

(3.3)

for all $k\ge 1$. On the other hand, since S is a cutter operator, we have with

$$\begin{array}{r}〈x_k-z,A^{\ast }(S(I-{\beta }_{k}h)-I)A{x}_k〉\\ =〈A{x}_k-Az,(S(I-{\beta }_{k}h)-I)A{x}_k〉\\ =〈S(I-{\beta }_{k}h)(A{x}_k)-Az,S(I-{\beta }_{k}h)(A{x}_k)-A{x}_k〉\\ -{\parallel (S(I-{\beta }_{k}h)-I)A{x}_k\parallel }^2\\ =〈S(I-{\beta }_{k}h)(A{x}_k)-Az,S(I-{\beta }_{k}h)(A{x}_k)-(I-{\beta }_{k}h)A{x}_k〉\\ -{\beta }_k〈S(I-{\beta }_{k}h)(A{x}_k)-Az,h(A{x}_k)〉-{\parallel (S(I-{\beta }_{k}h)-I)A{x}_k\parallel }^2\\ \le -{\parallel (S(I-{\beta }_{k}h)-I)A{x}_k\parallel }^2,\end{array}$$

(3.4)

for all $k\ge k_0$. By using (3.4), the inequality (3.3) becomes

$$\begin{array}{rcl}{\parallel y_k-z\parallel }^2& \le & {\parallel x_k-z\parallel }^2+{\gamma }^2{\parallel A\parallel }^2{\parallel (S(I-{\beta }_{k}h)-I)A{x}_k\parallel }^2\\ -2\gamma {\parallel (S(I-{\beta }_{k}h)-I)A{x}_k\parallel }^2\\ =& {\parallel x_k-z\parallel }^2-\gamma (2-\gamma {\parallel A\parallel }^2){\parallel (S(I-{\beta }_{k}h)-I)A{x}_k\parallel }^2,\end{array}$$

for all $k\ge k_0$, as required. Furthermore, since $\gamma <\frac{2}{{\parallel A\parallel }^2}$, we observe that $\gamma (2-\gamma {\parallel A\parallel }^2)>0$, and hence

$$\parallel y_k-z\parallel \le \parallel x_k-z\parallel ,\mathrm{\forall }k\ge k_0.$$

(3.5)

(ii) Let $M_1:=sup\{\parallel f(y_k)\parallel :k\ge 1\}$. For $k\ge k_0$, we have $\parallel x_{k+1}-z\parallel \le \parallel x_k-z\parallel +{\alpha }_k{M}_1$. Since ${\sum }_{k=1}^{\mathrm{\infty }}{\alpha }_k<\mathrm{\infty }$, by using Lemma 2.3, we obtain the result that ${lim}_{k\to \mathrm{\infty }}\parallel x_k-z\parallel$ exists.

(iii) We first show that ${lim}_{k\to \mathrm{\infty }}\parallel x_k-x_{k+1}\parallel =0$.

Let $w_k:=y_k-{\alpha }_{k}f(y_k)$ for all $k\ge 1$. Note that ${\{w_k\}}_{k=1}^{\mathrm{\infty }}$ is a bounded sequence, since for all $k\ge k_0$ we know that $\parallel w_k-z\parallel \le \parallel y_k-z\parallel +{\alpha }_k\parallel f(y_k)\parallel$ and ${\{y_k\}}_{k=1}^{\mathrm{\infty }}$ is a bounded sequence. By using the nonexpansiveness of T and (3.5), we have

$$0\le \parallel w_k-z\parallel -\parallel T{w}_k-Tz\parallel \le \parallel x_k-z\parallel +{\alpha }_k\parallel f(y_k)\parallel -\parallel x_{k+1}-z\parallel ,\mathrm{\forall }k\ge k_0.$$

Subsequently, by the existence of ${lim}_{k\to \mathrm{\infty }}\parallel x_k-z\parallel$ and the fact that ${\alpha }_k\to 0$, it follows that

$$\underset{k\to \mathrm{\infty }}{lim}(\parallel w_k-z\parallel -\parallel T{w}_k-Tz\parallel )=0.$$

By the strong nonexpansiveness of T and the boundedness of ${\{w_k\}}_{k=1}^{\mathrm{\infty }}$, we have

$$\underset{k\to \mathrm{\infty }}{lim}\parallel w_k-T{w}_k\parallel =0.$$

On the other hand, by (i), we observe that

$$\begin{array}{rcl}{\parallel x_{k+1}-z\parallel }^2& \le & {\parallel y_k-{\alpha }_{k}f(y_k)-z\parallel }^2\\ \le & {\parallel y_k-z\parallel }^2+{\alpha }_k^2{\parallel f(y_k)\parallel }^2+2{\alpha }_k\parallel f(y_k)\parallel \parallel y_k-z\parallel \\ \le & {\parallel x_k-z\parallel }^2-\gamma (2-\gamma {\parallel A\parallel }^2){\parallel (I-S(I-{\beta }_{k}h))A{x}_k\parallel }^2\\ +{\alpha }_k{\parallel f(y_k)\parallel }^2+2{\alpha }_k\parallel f(y_k)\parallel \parallel y_k-z\parallel ,\end{array}$$

which is equivalent to

$$\begin{array}{rcl}\gamma (2-\gamma {\parallel A\parallel }^2){\parallel (I-S(I-{\beta }_{k}h))A{x}_k\parallel }^2& \le & {\parallel x_k-z\parallel }^2-{\parallel x_{k+1}-z\parallel }^2+{\alpha }_k{\parallel f(y_k)\parallel }^2\\ +2{\alpha }_k\parallel f(y_k)\parallel \parallel y_k-z\parallel ,\end{array}$$

(3.6)

for all $k\ge k_0$. Taking the limit as $k\to \mathrm{\infty }$, we get

$$\underset{k\to \mathrm{\infty }}{lim}\parallel (I-S(I-{\beta }_{k}h))A{x}_k\parallel =0,$$

(3.7)

which implies, by the definition of $y_k$, that

$$\underset{k\to \mathrm{\infty }}{lim}\parallel y_k-x_k\parallel =0,$$

(3.8)

and also

$$\underset{k\to \mathrm{\infty }}{lim}\parallel w_k-x_k\parallel =0.$$

These together imply that

$$\underset{k\to \mathrm{\infty }}{lim}\parallel x_k-x_{k+1}\parallel \le \underset{k\to \mathrm{\infty }}{lim}(\parallel x_k-w_k\parallel +\parallel w_k-T{w}_k\parallel )=0,$$

(3.9)

as desired.

We now show that ${lim}_{k\to \mathrm{\infty }}\parallel x_k-T{x}_k\parallel =0$.

Observe that $\parallel x_{k+1}-T{y}_k\parallel \le {\alpha }_k\parallel f(y_k)\parallel \to 0$ as $k\to \mathrm{\infty }$. Using this one together with (3.8), (3.9) and the nonexpansiveness of T, we obtain

$$\underset{k\to \mathrm{\infty }}{lim}\parallel x_k-T{x}_k\parallel =0.$$

(3.10)

Next, we show that ${lim}_{k\to \mathrm{\infty }}\parallel A{x}_k-SA{x}_k\parallel =0$.

Set $z_k:=A{x}_k-{\beta }_{k}h(A{x}_k)$ for all $k\ge 1$. For every $k\ge k_0$, we observe that

$$0\le \parallel z_k-Az\parallel -\parallel S{z}_k-SAz\parallel \le {\beta }_k\parallel h(A{x}_k)\parallel +\parallel A{x}_k-S{z}_k\parallel .$$

Thus, the boundedness of ${\{h(A{x}_k)\}}_{k=1}^{\mathrm{\infty }}$ and (3.7) yield

$$\underset{k\to \mathrm{\infty }}{lim}(\parallel z_k-Az\parallel -\parallel S{z}_k-SAz\parallel )=0,$$

and hence, by the fact that the sequence ${\{z_k\}}_{k=1}^{\mathrm{\infty }}$ is bounded and S is a strongly nonexpansive mapping, we have

$$\underset{k\to \mathrm{\infty }}{lim}\parallel z_k-S{z}_k\parallel =0.$$

(3.11)

We note that $\parallel A{x}_k-z_k\parallel ={\beta }_k\parallel h(A{x}_k)\parallel \to 0$ as $k\to \mathrm{\infty }$. Also, we observe that

$$\parallel z_k-SA{x}_k\parallel \le \parallel z_k-S{z}_k\parallel +\parallel S(A{x}_k-{\beta }_{k}h(A{x}_k))-SA{x}_k\parallel \le \parallel z_k-S{z}_k\parallel +{\beta }_k\parallel h(A{x}_k)\parallel ,$$

for all $k\ge k_0$. By using (3.11) and taking the limit as $k\to \mathrm{\infty }$, we get

$$\underset{k\to \mathrm{\infty }}{lim}\parallel z_k-SA{x}_k\parallel =0.$$

Hence, we get

$$\underset{k\to \mathrm{\infty }}{lim}\parallel A{x}_k-SA{x}_k\parallel =0,$$

(3.12)

as required.

• (iv) Since ${\{x_k\}}_{k=1}^{\mathrm{\infty }}$ is a bounded sequence, there exist a subsequence ${\{x_{k_j}\}}_{j=1}^{\mathrm{\infty }}$ of ${\{x_k\}}_{k=1}^{\mathrm{\infty }}$ and $q\in H_1$ such that $x_{k_j}\rightharpoonup q\in H_1$. By (iii) and the demiclosed principle of the nonexpansive operator T, we obtain $q\in Fix(T)$. We claim that $q\in Fix(T)$ solves (3.1).

In fact, for $k\ge k_0$, we have

$$\begin{array}{rcl}{\parallel x_{k+1}-z\parallel }^2& \le & {\parallel y_k-z\parallel }^2+2{\alpha }_k〈f(y_k),z-w_k〉\\ =& {\parallel x_k-z\parallel }^2+2{\alpha }_k〈f(y_k),z-y_k〉+2{\alpha }_k^2{\parallel f(y_k)\parallel }^2\\ =& {\parallel x_k-z\parallel }^2+2{\alpha }_k〈f(z),z-y_k〉\\ -2{\alpha }_k〈f(y_k)-f(z),y_k-z〉+2{\alpha }_k^2{\parallel f(y_k)\parallel }^2\\ \le & {\parallel x_k-z\parallel }^2+2{\alpha }_k〈f(z),z-y_k〉+2{\alpha }_k^2{\parallel f(y_k)\parallel }^2,\end{array}$$

(3.13)

and then

$$\begin{array}{rcl}2〈f(z),y_k-z〉& \le & \frac{1}{{\alpha }_k}({\parallel x_k-z\parallel }^2-{\parallel x_{k+1}-z\parallel }^2)+2{\alpha }_k{\parallel f(y_k)\parallel }^2\\ \le & \frac{1}{{\alpha }_k}(\parallel x_k-z\parallel +\parallel x_{k+1}-z\parallel )(\parallel x_k-z\parallel -\parallel x_{k+1}-z\parallel )+2{\alpha }_k{\parallel f(y_k)\parallel }^2\\ \le & \frac{\parallel x_k-x_{k+1}\parallel }{{\alpha }_k}M+2{\alpha }_k{\parallel f(y_k)\parallel }^2,\end{array}$$

where $M:=sup\{\parallel x_k-z\parallel +\parallel x_{k+1}-z\parallel :k\ge 1\}<\mathrm{\infty }$. Since $x_{k_j}\rightharpoonup q$, ${\alpha }_k\to 0$, and $\parallel x_{k+1}-x_k\parallel =o({\alpha }_k)$, by (3.8), we have $〈f(z),q-z〉\le 0$ for all $z\in Fix(T)$ which means that $q\in Fix(T)$ solves (3.1).

Next, by $A{x}_{k_j}\rightharpoonup Aq\in H_2$ together with (iii) and the demiclosedness of the nonexpansive operator S, we know that $Aq\in Fix(S)$. We show that such $Aq\in Fix(S)$ solves (3.2).

Since ${\alpha }_k=o({\beta }_k^2)$, we may assume that ${\alpha }_k\le {\beta }_k^2$ for all $k\ge k_0$. From (3.6), for all $k\ge k_0$, we have

$$\gamma (1-\gamma {\parallel A\parallel }^2){\parallel A{x}_k-S{z}_k\parallel }^2\le \parallel x_k-x_{k+1}\parallel M_2+{\alpha }_k{\parallel f(y_k)\parallel }^2+2{\alpha }_k\parallel f(y_k)\parallel \parallel y_k-z\parallel ,$$

where $M_2:=sup\{\parallel x_k-z\parallel +\parallel x_{k+1}-z\parallel :k\ge k_0\}$. This implies that, for $k\ge k_0$,

$$\gamma (2-\gamma {\parallel A\parallel }^2)\frac{{\parallel A{x}_k-S{z}_k\parallel }^2}{{\beta }_k^2}\le \frac{\parallel x_k-x_{k+1}\parallel }{{\beta }_k^2}{M}_3+\frac{{\alpha }_k}{{\beta }_k^2}{M}_3\le \frac{\parallel x_k-x_{k+1}\parallel }{{\alpha }_k}{M}_3+\frac{{\alpha }_k}{{\beta }_k^2}{M}_3,$$

where $M_3:=max\{M_2,{sup}_{k\ge 1}\{{\parallel f(y_k)\parallel }^2+2\parallel f(y_k)\parallel \parallel y_k-z\parallel \}\}$. Subsequently, since $\parallel x_{k+1}-x_k\parallel =o({\alpha }_k)$, ${\alpha }_k=o({\beta }_k^2)$, and $\gamma (2-\gamma {\parallel A\parallel }^2)>0$, we have

$$\underset{k\to \mathrm{\infty }}{lim}\frac{\parallel A{x}_k-S{z}_k\parallel }{{\beta }_k}=0.$$

(3.14)

For $k\ge k_0$, we compute

$$\begin{array}{rcl}{\parallel S{z}_k-SAz\parallel }^2& \le & {\parallel A{x}_k-{\beta }_{k}h(A{x}_k)-Az\parallel }^2\\ \le & {\parallel A{x}_k-Az\parallel }^2+2{\beta }_k〈h(A{x}_k),Az-z_k〉\\ =& {\parallel A{x}_k-Az\parallel }^2+2{\beta }_k〈h(A{x}_k),Az-A{x}_k〉+2{\beta }_k^2{\parallel h(A{x}_k)\parallel }^2\\ =& {\parallel A{x}_k-Az\parallel }^2-2{\beta }_k〈A{x}_k-Az,h(A{x}_k)-h(Az)〉\\ +2{\beta }_k〈h(Az),Az-A{x}_k〉+2{\beta }_k^2{\parallel h(A{x}_k)\parallel }^2\\ \le & {\parallel A{x}_k-Az\parallel }^2+{\beta }_k^2{\parallel h(A{x}_k)-h(Az)\parallel }^2+2{\beta }_k〈h(Az),Az-A{x}_k〉\\ +2{\beta }_k^2{\parallel h(A{x}_k)\parallel }^2.\end{array}$$

(3.15)

This gives

$$\begin{array}{rcl}2〈h(Az),A{x}_k-Az〉& \le & \frac{1}{{\beta }_k}({\parallel A{x}_k-Az\parallel }^2-{\parallel S{z}_k-SAz\parallel }^2)+{\beta }_k{\parallel h(A{x}_k)-h(Az)\parallel }^2\\ +2{\beta }_k{\parallel h(A{x}_k)\parallel }^2\\ \le & \frac{\parallel A{x}_k-S{z}_k\parallel }{{\beta }_k}{M}_4+{\beta }_k{\parallel h(A{x}_k)-h(Az)\parallel }^2+2{\beta }_k{\parallel h(A{x}_k)\parallel }^2,\end{array}$$

where $M_4:=sup\{\parallel A{x}_k-Az\parallel +\parallel S{z}_k-SAz\parallel :k\ge 1\}<\mathrm{\infty }$. Using this together with $A{x}_{k_j}\rightharpoonup Aq\in Fix(S)$ and (3.14), we obtain $〈h(Az),Aq-Az〉\le 0$ for all $Az\in Fix(S)$. Thus, $Aq\in Fix(S)$ solves (3.2), and therefore, $q\in \mathrm{\Gamma }$.

Finally, it remains to show that $x_k\rightharpoonup q\in \mathrm{\Gamma }$. By the boundedness of ${\{x_k\}}_{k=1}^{\mathrm{\infty }}$, it suffices to show that there is no subsequence ${\{x_{k_i}\}}_{i=1}^{\mathrm{\infty }}$ of ${\{x_k\}}_{k=1}^{\mathrm{\infty }}$ such that $x_{k_i}\rightharpoonup p\in H_1$ and $p\ne q$.

Indeed, if this is not true, the well-known Opial theorem would imply

$$\begin{array}{rcl}\underset{k\to \mathrm{\infty }}{lim}\parallel x_k-p\parallel & =& \underset{j\to \mathrm{\infty }}{lim}\parallel x_{k_j}-q\parallel \\ <& \underset{j\to \mathrm{\infty }}{lim}\parallel x_{k_j}-p\parallel \\ =& \underset{k\to \mathrm{\infty }}{lim}\parallel x_k-p\parallel \\ =& \underset{i\to \mathrm{\infty }}{lim}\parallel x_{k_i}-p\parallel \\ <& \underset{i\to \mathrm{\infty }}{lim}\parallel x_{k_i}-q\parallel \\ =& \underset{k\to \mathrm{\infty }}{lim}\parallel x_k-p\parallel ,\end{array}$$

which leads to a contradiction. Therefore, the sequence ${\{x_k\}}_{k=1}^{\mathrm{\infty }}$ converges weakly to a point $q\in \mathrm{\Gamma }$. □

3.2 Split hierarchical variational inequality problems

We consider another kind of split hierarchical variational inequality problem (in short, SHVIP) in which we consider a variational inequality formulation instead of the Minty variational inequality. More precisely, we consider the following split hierarchical variational inequality problem: find a point $x^{\ast }\in Fix(T)$ such that

$$〈f\left(x^{\ast }\right),x-x^{\ast }〉\ge 0,\text{for all }x\in Fix(T),$$

(3.16)

such that $A{x}^{\ast }\in Fix(S)$ and it satisfies

$$〈h\left(A{x}^{\ast }\right),y-A{x}^{\ast }〉\ge 0,\text{for all }y\in Fix(S).$$

(3.17)

The solution set of the SHVIP is denoted by Ω. Since $Fix(T)$ and $Fix(S)$ are nonempty closed convex and f and h are monotone and continuous, by the Minty lemma [[13], Lemma 1], SHVIP (3.16)-(3.17) and SHMVIP (3.1)-(3.2) are equivalent. Hence, Algorithm 3.1 and Theorem 3.3 are also applicable for SHVIP (3.16)-(3.17).

When $Fix(T)=C$ a closed convex subset of a Hilbert space $H_1$ and $Fix(S)=Q$ a closed convex subset of a Hilbert space $H_2$, then SHVIP (3.16)-(3.17) is considered and studied by Censor et al. [24].

Remark 3.4 It is worth to note that, in the context of SHVIP (3.16)-(3.17), if we assume either the finite dimensional settings of $H_1$ and $H_2$ or the compactness of spaces $H_1$ and $H_2$, then the monotonicity assumptions of operators f and h can be omitted.

In fact, as in the statement and the proof of Theorem 3.3, since we know that $x_{k_j}\rightharpoonup q\in Fix(T)$, we also have $x_{k_j}\to q\in Fix(T)$. Recalling the inequality (3.13), we have, for $k\ge k_0$,

$${\parallel x_{k+1}-z\parallel }^2\le {\parallel x_k-z\parallel }^2+2{\alpha }_k〈f(y_k),z-y_k〉+2{\alpha }_k^2{\parallel f(y_k)\parallel }^2,$$

which is equivalent to

$$2〈f(y_k),y_k-z〉\le \frac{\parallel x_k-x_{k+1}\parallel }{{\alpha }_k}M+2{\alpha }_k{\parallel f(y_k)\parallel }^2,$$

where $M:=sup\{\parallel x_k-z\parallel +\parallel x_{k+1}-z\parallel :k\ge 1\}<\mathrm{\infty }$. Since we know that $y_{k_j}\to q\in Fix(T)$ and f is continuous, by approaching j to infinity, we obtain $〈f(q),q-z〉\le 0$, for all $z\in Fix(T)$, which means that $q\in Fix(T)$ solves (3.16). Similarly, from the inequality (3.15), we note that, for $k\ge k_0$,

$${\parallel S{z}_k-SAz\parallel }^2\le {\parallel A{x}_k-Az\parallel }^2+2{\beta }_k〈h(A{x}_k),Az-A{x}_k〉+2{\beta }_k^2{\parallel h(A{x}_k)\parallel }^2,$$

which is equivalent to

$$2〈h(A{x}_k),A{x}_k-Az〉\le \frac{\parallel A{x}_k-S{z}_k\parallel }{{\beta }_k}{M}_4+2{\beta }_k{\parallel h(A{x}_k)\parallel }^2,$$

where $M_4:=sup\{\parallel A{x}_k-Az\parallel +\parallel S{z}_k-SAz\parallel :k\ge 1\}<\mathrm{\infty }$. Since $A{x}_{k_j}\to Aq\in Fix(S)$ and h is continuous, we also get $〈h(Aq),Aq-Az〉\le 0$, for all $Az\in Fix(S)$, which means that $Aq\in Fix(S)$ solves (3.17), and subsequently, $q\in \mathrm{\Omega }$.

Remark 3.5 The strong nonexpansiveness of operators T and S in Theorem 3.3 can be applied to the cases when the operators T and S are not only firmly nonexpansive, but with relaxation of being firmly nonexpansive and averaged nonexpansive but also strictly nonexpansive, that is, $\parallel Tx-Ty\parallel <\parallel x-y\parallel$ or $x-y=Tx-Ty$ for all x, y; regarding the compactness of the spaces $H_1$ and $H_2$, for more details, see [[21], Remark 2.3.3].

4 Some related problems

In this section, we present some split-type problems which are special cases of SHVIP (3.1) and (3.2) and can be solved by using Algorithm 3.1.

4.1 A split convex minimization problem

Let $\varphi :H_1\to \mathbb{R}$ and $\phi :H_2\to \mathbb{R}$ be convex continuously differentiable functions and $A:H_1\to H_2$ be a bounded linear operator such that $A(H_1)\cap Fix(S)\ne \mathrm{\varnothing }$. Consider the following split convex minimization problem (in short, SCMP): find

$$x^{\ast }\in Fix(T)\text{such that}{x}^{\ast }=\underset{x\in Fix(T)}{argmin}\varphi (x),$$

(4.1)

and such that

$$A{x}^{\ast }\in Fix(S)\text{such that}A{x}^{\ast }=\underset{y\in Fix(S)}{argmin}\phi (y).$$

(4.2)

We know that the SCMP (4.1)-(4.2) can be formulated as the following split hierarchical variational inequality problem: find a point

$$x^{\ast }\in Fix(T)\text{such that}〈\mathrm{\nabla }\varphi \left(x^{\ast }\right),x-x^{\ast }〉\ge 0,\text{for all }x\in Fix(T),$$

and such that the point

$$A{x}^{\ast }\in Fix(S)\text{such that}〈\mathrm{\nabla }\phi \left(A{x}^{\ast }\right),y-A{x}^{\ast }〉\ge 0,\text{for all }y\in Fix(S),$$

where ∇ϕ and ∇φ denote the gradient of ϕ and φ, respectively. Since ∇ϕ and ∇φ are monotone [[12], Proposition 5.3] and continuous, we can apply Algorithm 3.1 to obtain the solution of SCMP (4.1)-(4.2), and Theorem 3.3 will provide the convergence of the sequence ${\{x_k\}}_{k=1}^{\mathrm{\infty }}$ to a solution of SCMP (4.1)-(4.2). Furthermore, in the context of finite-dimensional cases, we can remove the convexity of ϕ and φ.

4.2 A split variational inequality problem over the solution set of monotone variational inclusion problem

Let $H_1$ and $H_2$ be two real Hilbert spaces, $\varphi :H_1\to H_1$, $\phi :H_2\to H_2$ be $L_1$, (respectively, $L_2$)-inverse strongly monotone operators, $B_1:H_1\to 2^{H_1}$, $B_2:H_2\to 2^{H_2}$ are set-valued maximal monotone operators. Let us consider the following monotone variational inclusion problem (in short, MVIP): find

$$x^{\ast }\in H_1\text{such that}0\in \varphi \left(x^{\ast }\right)+B_1\left(x^{\ast }\right).$$

We denote the solution set of MVIP by $SOL(\varphi ,B_1)$.

For $\sigma >0$, $\lambda \in (0,2L_1)$, we know that the operator $J_{\sigma }^{B_1}(I-\lambda \varphi )$ is an averaged nonexpansive operator and

$$Fix(J_{\sigma }^{B_1}(I-\lambda \varphi ))=SOL(\varphi ,B_1),$$

where $J_{\sigma }^{B_1}$ represents the resolvent of $B_1$ with parameter σ.

We consider the following split variational inequality problem: find $x^{\ast }\in SOL(\varphi ,B_1)$ such that

$$〈f(x),x-x^{\ast }〉\ge 0,\text{for all }x\in SOL(\varphi ,B_1),$$

(4.3)

and such that $A{x}^{\ast }\in SOL(\phi ,B_2)$ satisfies

$$〈h(y),y-A{x}^{\ast }〉\ge 0,\text{for all }y\in SOL(\phi ,B_2).$$

(4.4)

By using these facts and adding the assumption that $J_{\sigma }^{B_2}(I-\lambda \phi )$ is cutter, we can apply Algorithm 3.1 and Theorem 3.3 to obtain the solution of SVIP (4.3)-(4.4).

4.3 A split variational inequality problem over the solution set of equilibrium problem

Let $H_1$ and $H_2$ be real Hilbert spaces, $C\subseteq H_1$ and $Q\subset H_2$ be nonempty closed convex sets, $\varphi :C\times C\to \mathbb{R}$ and $\phi :Q\times Q\to \mathbb{R}$ be bifunctions. The equilibrium problem defined by C and ϕ is the problem of finding $x^{\ast }\in C$ such that

$$\varphi (x^{\ast },x)\ge 0,\text{for all }x\in C,$$

and we denote the solution set of equilibrium problem of C and ϕ by $EP(C,\varphi )$.

Recall that Blum and Oettli [26] showed that the bifunction ϕ satisfies the following conditions:

(A1) $\varphi (x,x)=0$ for all $x\in C$;

(A2) ϕ is monotone, that is, $\varphi (x,y)+\varphi (y,x)\le 0$ for all $x,y\in C$;

(A3) for all $x,y,z\in C$,

$$\underset{t\downarrow 0}{lim\hspace{0.17em}sup}\varphi (tz+(1-t)x,y)\le \varphi (x,y);$$

(A4) for all $x\in C$, $\varphi (x,\cdot )$ is a convex and lower semicontinuous function,

and let $r>0$ and $z\in H_1$, then the set

$$T_r(z):=\{x^{\ast }\in C:\varphi (x^{\ast },x)+\frac{1}{r}〈x-x^{\ast },x^{\ast }-x〉\ge 0\text{ for all }x\in C\}\ne \mathrm{\varnothing }.$$

Moreover, from [27], we know that

• (i) $T_r$ is single-valued;

• (ii) $T_r$ is firmly nonexpansive;

• (iii) $Fix(T_r)=EP(C;\varphi )$.

We now consider the following split variational inequality problem: find $x^{\ast }\in EP(C,\varphi )$ such that

$$〈f(x),x-x^{\ast }〉\ge 0,\text{for all }x\in EP(C,\varphi ),$$

(4.5)

and such that $A{x}^{\ast }\in EP(Q,\phi )$ satisfies

$$〈h(y),y-A{x}^{\ast }〉\ge 0,\text{for all }y\in EP(Q,\phi ).$$

(4.6)

Similarly, by assuming that conditions (A1)-(A4) in the context of the equilibrium problem defined by Q and φ hold, we see that, for all $u\in H_2$ and $s>0$, the set

$$T_s(u):=\{y^{\ast }\in Q:\phi (y^{\ast },y)+\frac{1}{s}〈y-y^{\ast },y^{\ast }-u〉\ge 0\text{ for all }y\in Q\}$$

is a nonempty set. Moreover, we also see that $T_s$ is single-valued and firmly nonexpansive; and $Fix(T_s)=EP(Q;\phi )$.

By using these facts, we obtain the result that the problem (4.5)-(4.6) forms a special case of the SHVIP (3.1)-(3.2).