Hybrid viscosity CQ method for finding a common solution of a variational inequality, a general system of variational inequalities, and a fixed point problem

Abstract

In the literature, various iterative methods have been proposed for finding a common solution of the classical variational inequality problem and a fixed point problem. Research along these lines is performed either by relaxing the assumptions on the mappings in the settings (for instance, commonly seen assumptions for the mapping involved in the fixed point problem are nonexpansive or strictly pseudocontractive) or by adding a general system of variational inequalities into the settings. In this paper, we consider both possible ways in our settings. Specifically, we propose an iterative method for finding a common solution of the classical variational inequality problem, a general system of variational inequalities and a fixed point problem of a uniformly continuous asymptotically strictly pseudocontractive mapping in the intermediate sense. Our iterative method is hybridized by utilizing the well-known extragradient method, the CQ method, the Mann-type iterative method and the viscosity approximation method. The iterates yielded by our method converge strongly to a common solution of these three problems. In addition, we propose a hybridized extragradient-like method to yield iterates converging weakly to a common solution of these three problems.

MSC:49J30, 47H09, 47J20.

1 Introduction

Let H be a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$, let C be a nonempty closed convex subset of H, and let $P_C$ be the metric projection of H onto C. Let $S:C\to C$ be a self-mapping on C. We denote by $Fix(S)$ the set of fixed points of S and by R the set of all real numbers. A mapping $A:C\to H$ is called L-Lipschitz continuous if there exists a constant $L\ge 0$ such that

$$\parallel Ax-Ay\parallel \le L\parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

In particular, if $L=1$, then A is called a nonexpansive mapping [1]; if $L\in [0,1)$, then A is called a contraction. Also, a mapping $A:C\to H$ is called monotone if $〈Ax-Ay,x-y〉\ge 0$ for all $x,y\in C$. A is called η-strongly monotone if there exists a constant $\eta >0$ such that

$$〈Ax-Ay,x-y〉\ge \eta {\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

A is called α-inverse-strongly monotone if there exists a constant $\alpha >0$ such that

$$〈Ax-Ay,x-y〉\ge \alpha {\parallel Ax-Ay\parallel }^2,\mathrm{\forall }x,y\in C.$$

It is obvious that if A is α-inverse-strongly monotone, then A is monotone and $\frac{1}{\alpha }$-Lipschitz continuous.

For a given nonlinear operator $A:C\to H$, we consider the variational inequality problem (VIP) of finding $x^{\ast }\in C$ such that

$$〈A{x}^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in C.$$

(1.1)

The solution set of VIP (1.1) is denoted by $VI(C,A)$. VIP (1.1) was first discussed by Lions [2] and now has many applications in computational mathematics, mathematical physics, operations research, mathematical economics, optimization theory, and other fields; see, e.g., [3–6]. It is well known that if A is a strongly monotone and Lipschitz-continuous mapping on C, then VIP (1.1) has a unique solution.

In the literature, there is a growing interest in studying how to find a common solution of $Fix(S)\cap VI(C,A)$. Under various assumptions imposed on A and S, iterative algorithms were derived to yield iterates which converge strongly or weakly to a common solution of these two problems.

1.1 Finding a common element and weak convergence

Consider that a set $C\subset H$ is nonempty, closed and convex, a mapping $S:C\to C$ is nonexpansive and a mapping $A:C\to H$ is α-inverse-strongly monotone. Takahashi and Toyoda [7] introduced the Mann-type iterative scheme:

$$\{\begin{array}{c}{x}_0=x\in C\text{chosen arbitrarily},\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)S{P}_C(x_n-{\lambda }_{n}A{x}_n),\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(1.2)

where $\{{\alpha }_n\}$ is a sequence in $(0,1)$ and $\{{\lambda }_n\}$ is a sequence in $(0,2\alpha )$. They proved that if $Fix(S)\cap VI(C,A)\ne \mathrm{\varnothing }$, then the sequence $\{x_n\}$ generated by (1.2) converges weakly to some $z\in Fix(S)\cap VI(C,A)$.

Motivated by Korpelevich’s extragradient method [8], Nadezhkina and Takahashi [9] proposed an extragradient iterative method and showed the iterates converge weakly to a common element of $Fix(S)\cap VI(C,A)$:

$$\{\begin{array}{c}{x}_0=x\in C\text{chosen arbitrarily},\hfill \\ y_n=P_C(x_n-{\lambda }_{n}A{x}_n),\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)S{P}_C(x_n-{\lambda }_{n}A{y}_n),\mathrm{\forall }n\ge 0,\hfill \end{array}$$

where $A:C\to H$ is a monotone, L-Lipschitz continuous mapping and $S:C\to C$ is a nonexpansive mapping and $\{{\lambda }_n\}\subset [a,b]$ for some $a,b\in (0,1/L)$ and $\{{\alpha }_n\}\subset [c,d]$ for some $c,d\in (0,1)$. See also Zeng and Yao [10], in which a hybridized iterative method was proposed to yield a new weak convergence result.

1.2 Finding a common element and strong convergence

Let $C\subset H$ be a nonempty closed convex subset, let $S:C\to C$ be a nonexpansive mapping, and let $A:C\to H$ be an α-inverse strongly monotone mapping. Iiduka and Takahashi [11] introduced the following hybrid method:

$$\{\begin{array}{c}{x}_0=x\in C\text{chosen arbitrarily},\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)S{P}_C(x_n-{\lambda }_{n}A{x}_n),\hfill \\ C_n=\{z\in C:\parallel y_n-z\parallel \le \parallel x_n-z\parallel \},\hfill \\ Q_n=\{z\in C:〈x_n-z,x-x_n〉\ge 0\},\hfill \\ x_{n+1}=P_{C_n\cap Q_n}x,\mathrm{\forall }n\ge 0,\hfill \end{array}$$

where $0\le {\alpha }_n\le c<1$ and $0<a\le {\lambda }_n\le b<2\alpha$. They showed that if $Fix(S)\cap VI(C,A)\ne \mathrm{\varnothing }$, then the sequence $\{x_n\}$, generated by this iterative process, converges strongly to $P_{Fix(S)\cap VI(C,A)}x$. Recently, the method proposed by Nadezhkina and Takahashi [12] also demonstrated the strong convergence result. However, note that they assumed that A is monotone and L-Lipschitz-continuous while S is nonexpansive. For another strong convergence result, see Ceng and Yao [13] whose method is based on the extragradient method and the viscosity approximation method.

As we have seen, most of the papers were based on the different assumptions imposed on A while the mapping S is nonexpansive. In the following, we shall relax the nonexpansive requirement on S (for instance, κ-strictly pseudocontractive, asymptotically κ-strictly pseudocontractive mapping in the intermediate sense, etc.). Furthermore, we also consider adding a general system of variational inequalities to our settings.

1.3 Relaxation on nonexpansive S

Definition 1.1 Let C be a nonempty subset of a normed space X, and let $S:C\to C$ be a self-mapping on C.

• (i) S is asymptotically nonexpansive (cf. [14]) if there exists a sequence $\{k_n\}$ of positive numbers satisfying the property ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ and

  $$\parallel S^{n}x-S^{n}y\parallel \le k_n\parallel x-y\parallel ,\mathrm{\forall }n\ge 1,\mathrm{\forall }x,y\in C;$$

• (ii) S is asymptotically nonexpansive in the intermediate sense [15] provided S is uniformly continuous and

  $$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{x,y\in C}{sup}(\parallel S^{n}x-S^{n}y\parallel -\parallel x-y\parallel )\le 0;$$

• (iii) S is uniformly Lipschitzian if there exists a constant $\mathcal{L}>0$ such that

  $$\parallel S^{n}x-S^{n}y\parallel \le \mathcal{L}\parallel x-y\parallel ,\mathrm{\forall }n\ge 1,\mathrm{\forall }x,y\in C.$$

It is clear that every nonexpansive mapping is asymptotically nonexpansive and every asymptotically nonexpansive mapping is uniformly Lipschitzian.

The class of asymptotically nonexpansive mappings was introduced by Goebel and Kirk [14] as an important generalization of the class of nonexpansive mappings. The existence of fixed points of asymptotically nonexpansive mappings was proved by Goebel and Kirk [14] as follows.

Theorem GK (see [[14], Theorem 1])

If C is a nonempty closed convex bounded subset of a uniformly convex Banach space, then every asymptotically nonexpansive mapping $S:C\to C$ has a fixed point in C.

Let C be a nonempty closed convex bounded subset of a Hilbert space H. An iterative method for the approximation of fixed points of an asymptotically nonexpansive mapping with sequence $\{k_n\}$ was developed by Schu [16] via the following Mann-type iterative scheme:

$$x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_n{S}^n{x}_n,\mathrm{\forall }n\ge 1,$$

(1.3)

where $\delta \le {\alpha }_n\le 1-\delta$ ($\mathrm{\forall }n\ge 1$) for some $\delta >0$. He proved the weak convergence of $\{x_n\}$ to a fixed point of S if ${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$. Moreover, iterative methods for approximation of fixed points of asymptotically nonexpansive mappings have been further studied by other authors (see, e.g., [16–18] and references therein).

The class of asymptotically nonexpansive mappings in the intermediate sense was introduced by Bruck et al. [15] and iterative methods for the approximation of fixed points of such types of non-Lipschitzian mappings were studied by Bruck et al. [15], Agarwal et al. [19], Chidume et al. [20], Kim and Kim [21] and many others.

Recently, Kim and Xu [22] introduced the concept of asymptotically κ-strictly pseudocontractive mappings in a Hilbert space as follows.

Definition 1.2 Let C be a nonempty subset of a Hilbert space H. A mapping $S:C\to C$ is said to be an asymptotically κ-strictly pseudocontractive mapping with sequence $\{{\gamma }_n\}$ if there exists a constant $\kappa \in [0,1)$ and a sequence $\{{\gamma }_n\}$ in $[0,\mathrm{\infty })$ with ${lim}_{n\to \mathrm{\infty }}{\gamma }_n=0$ such that

$${\parallel S^{n}x-S^{n}y\parallel }^2\le (1+{\gamma }_n){\parallel x-y\parallel }^2+\kappa {\parallel x-S^{n}x-(y-S^{n}y)\parallel }^2,\mathrm{\forall }n\ge 1,\mathrm{\forall }x,y\in C.$$

(1.4)

They studied weak and strong convergence theorems for this class of mappings. It is important to note that every asymptotically κ-strictly pseudocontractive mapping with sequence $\{{\gamma }_n\}$ is a uniformly ℒ-Lipschitzian mapping with $\mathcal{L}=sup\{\frac{\kappa +\sqrt{1+(1-\kappa ){\gamma }_n}}{1+\kappa }:n\ge 1\}$.

Very recently, Sahu et al. [23] considered the concept of asymptotically κ-strictly pseudocontractive mappings in the intermediate sense, which are not necessarily Lipschitzian.

Definition 1.3 Let C be a nonempty subset of a Hilbert space H. A mapping $S:C\to C$ is said to be an asymptotically κ-strictly pseudocontractive mapping in the intermediate sense with sequence $\{{\gamma }_n\}$ if there exists a constant $\kappa \in [0,1)$ and a sequence $\{{\gamma }_n\}$ in $[0,\mathrm{\infty })$ with ${lim}_{n\to \mathrm{\infty }}{\gamma }_n=0$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{x,y\in C}{sup}({\parallel S^{n}x-S^{n}y\parallel }^2-(1+{\gamma }_n){\parallel x-y\parallel }^2-\kappa {\parallel x-S^{n}x-(y-S^{n}y)\parallel }^2)\le 0.$$

(1.5)

Put $c_n:=max\{0,{sup}_{x,y\in C}({\parallel S^{n}x-S^{n}y\parallel }^2-(1+{\gamma }_n){\parallel x-y\parallel }^2-\kappa {\parallel x-S^{n}x-(y-S^{n}y)\parallel }^2)\}$. Then $c_n\ge 0$ ($\mathrm{\forall }n\ge 1$), $c_n\to 0$ ($n\to \mathrm{\infty }$) and (1.5) reduces to the relation

$${\parallel S^{n}x-S^{n}y\parallel }^2\le (1+{\gamma }_n){\parallel x-y\parallel }^2+\kappa {\parallel x-S^{n}x-(y-S^{n}y)\parallel }^2+c_n,\mathrm{\forall }n\ge 1,\mathrm{\forall }x,y\in C.$$

(1.6)

Whenever $c_n=0$ for all $n\ge 1$ in (1.6), then S is an asymptotically κ-strictly pseudocontractive mapping with sequence $\{{\gamma }_n\}$.

For S to be a uniformly continuous asymptotically κ-strictly pseudocontractive mapping in the intermediate sense with sequence $\{{\gamma }_n\}$ such that $Fix(S)$ is nonempty and bounded, Sahu et al. [23] proposed an iterative Mann-type CQ method in which the iterates converge strongly to a fixed point of S.

Theorem SXY (see [[23], Theorem 4.1])

Let C be a nonempty closed convex subset of a real Hilbert space H, and let $S:C\to C$ be a uniformly continuous asymptotically κ-strictly pseudocontractive mapping in the intermediate sense with sequence $\{{\gamma }_n\}$ such that $Fix(S)$ is nonempty and bounded. Let $\{{\alpha }_n\}$ be a sequence in $[0,1]$ such that $0<\delta \le {\alpha }_n\le 1-\kappa$ for all $n\ge 1$. Let $\{x_n\}$ be a sequence in C generated by the following (CQ) algorithm:

$$\{\begin{array}{c}{x}_1=x\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=(1-{\alpha }_n)x_n+{\alpha }_n{S}^n{x}_n,\hfill \\ C_n=\{z\in C:{\parallel y_n-z\parallel }^2\le {\parallel x_n-z\parallel }^2+{\theta }_n\},\hfill \\ Q_n=\{z\in C:〈x_n-z,x-x_n〉\ge 0\},\hfill \\ x_{n+1}=P_{C_n\cap Q_n}x,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(1.7)

where ${\theta }_n=c_n+{\gamma }_n{\Delta }_n$ and ${\Delta }_n=sup\{{\parallel x_n-z\parallel }^2:z\in Fix(S)\}<\mathrm{\infty }$. Then $\{x_n\}$ converges strongly to $P_{Fix(S)}x$.

1.4 Common solution of three problems

Let $B_1,B_2:C\to H$ be two mappings. Recently, Ceng et al. [24] introduced and considered the problem of finding $(x^{\ast },y^{\ast })\in C\times C$ such that

$$\{\begin{array}{c}〈{\mu }_1{B}_1{y}^{\ast }+x^{\ast }-y^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ 〈{\mu }_2{B}_2{x}^{\ast }+y^{\ast }-x^{\ast },x-y^{\ast }〉\ge 0,\mathrm{\forall }x\in C,\hfill \end{array}$$

(1.8)

which is called a general system of variational inequalities (GSVI), where ${\mu }_1>0$ and ${\mu }_2>0$ are two constants. The set of solutions of GSVI (1.8) is denoted by $GSVI(C,B_1,B_2)$. In particular, if $B_1=B_2$, then GSVI (1.8) reduces to the new system of variational inequalities (NSVI), introduced and studied by Verma [25]. Further, if $x^{\ast }=y^{\ast }$ additionally, then the NSVI reduces to VIP (1.1). Moreover, they transformed GSVI (1.8) into a fixed point problem in the following way.

Lemma CWY (see [24])

For given $\overline{x},\overline{y}\in C$, $(\overline{x},\overline{y})$ is a solution of GSVI (1.8) if and only if $\overline{x}$ is a fixed point of the mapping $G:C\to C$ defined by

$$G(x)=P_C[P_C(x-{\mu }_2{B}_{2}x)-{\mu }_1{B}_1{P}_C(x-{\mu }_2{B}_{2}x)],\mathrm{\forall }x\in C,$$

where $\overline{x}=P_C(\overline{x}-{\mu }_2{B}_2\overline{x})$.

In particular, if the mapping $B_i:C\to H$ is ${\beta }_i$-inverse strongly monotone for $i=1,2$, then the mapping G is nonexpansive provided ${\mu }_i\in (0,2{\beta }_i)$ for $i=1,2$.

Utilizing Lemma CWY, they introduced and studied a relaxed extragradient method for solving GSVI (1.8). Throughout this paper, the set of fixed points of the mapping G is denoted by Ξ. Based on the relaxed extragradient method and the viscosity approximation method, Yao et al. [26] proposed and analyzed an iterative algorithm for finding a common solution of GSVI (1.8), and the fixed point problem of a κ-strictly pseudocontractive mapping $S:C\to C$ (namely, there exists a constant $\kappa \in [0,1)$ such that ${\parallel Sx-Sy\parallel }^2\le {\parallel x-y\parallel }^2+\kappa {\parallel (I-S)x-(I-S)y\parallel }^2$ for all $x,y\in C$).

The main theme of this paper is to study the problem of finding a common element of the solution set of VIP (1.1), the solution set of GSVI (1.8) and the fixed point set of a self-mapping $S:C\to C$. Ceng et al. [27] analyzed this problem by assuming the mapping S to be strictly pseudocontractive as follows.

Theorem CGY (see [[27], Theorem 3.1])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $A:C\to H$ be α-inverse strongly monotone and $B_i:C\to H$ be ${\beta }_i$-inverse strongly monotone for $i=1,2$. Let $S:C\to C$ be a κ-strictly pseudocontractive mapping such that $Fix(S)\cap \Xi \cap VI(C,A)\ne \mathrm{\varnothing }$. Let $f:C\to C$ be a ρ-contraction with $\rho \in [0,\frac{1}{2})$. For given $x_0\in C$ arbitrarily, let the sequences $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ be generated iteratively by

$$\{\begin{array}{c}{z}_n=P_C(x_n-{\lambda }_{n}A{x}_n),\hfill \\ y_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)P_C[P_C(z_n-{\mu }_2{B}_2{z}_n)-{\mu }_1{B}_1{P}_C(z_n-{\mu }_2{B}_2{z}_n)],\hfill \\ x_{n+1}={\beta }_n{x}_n+{\gamma }_n{y}_n+{\delta }_{n}S{y}_n,\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(1.9)

where ${\mu }_i\in (0,2{\beta }_i)$ for $i=1,2$, $\{{\lambda }_n\}\subset (0,2\alpha ]$ and $\{{\alpha }_n\},\{{\beta }_n\},\{{\gamma }_n\},\{{\delta }_n\}\subset [0,1]$ such that

• (i) ${\beta }_n+{\gamma }_n+{\delta }_n=1$ and $({\gamma }_n+{\delta }_n)k\le {\gamma }_n$ for all $n\ge 0$;

• (ii) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (iii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$;

• (iv) ${lim}_{n\to \mathrm{\infty }}(\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\gamma }_n}{1-{\beta }_n})=0$;

• (v) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\lambda }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\lambda }_n<2\alpha$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}|{\lambda }_{n+1}-{\lambda }_n|=0$.

Then the sequence $\{x_n\}$ generated by (1.9) converges strongly to $\overline{x}=P_{Fix(S)\cap \Xi \cap VI(C,A)}Q\overline{x}$ and $(\overline{x},\overline{y})$ is a solution of GSVI (1.8), where $\overline{y}=P_C(\overline{x}-{\mu }_2{B}_2\overline{x})$.

In this paper, we study the problem of finding a common element of the solution set of VIP (1.1), the solution set of GSVI (1.8) and the fixed point set of a self-mapping $S:C\to C$, where the mapping S is assumed to be a uniformly continuous asymptotically κ-strictly pseudocontractive mapping in the intermediate sense with sequence $\{{\gamma }_n\}$ such that $Fix(S)\cap \Xi \cap VI(C,A)$ is nonempty and bounded. Not surprisingly, our main points of proof come from the ideas in [[23], Theorem 4.1] and [[27], Theorem 3.1]. Our major contribution ensures a strong convergence result to the extent of involving uniformly continuous asymptotically κ-strictly pseudocontractive mappings in the intermediate sense. Moreover, in Section 4 we extend Ceng, Hadjisavvas and Wong’s hybrid extragradient-like approximation method given in [[28], Theorem 5] to establish a new weak convergence theorem for finding a common solution of VIP (1.1), GSVI (1.8) and the fixed point problem of S.

2 Preliminaries

Let H be a real Hilbert space whose inner product and norm are denoted by $〈\cdot ,\cdot 〉$ and $\parallel \cdot \parallel$, respectively. Let C be a nonempty closed convex subset of H. We write $x_n\rightharpoonup x$ to indicate that the sequence $\{x_n\}$ converges weakly to x and $x_n\to x$ to indicate that the sequence $\{x_n\}$ converges strongly to x. Moreover, we use ${\omega }_w(x_n)$ to denote the weak ω-limit set of the sequence $\{x_n\}$, i.e.,

$${\omega }_w(x_n):=\{x\in H:x_{n_i}\rightharpoonup x\text{ for some subsequence }\{x_{n_i}\}\text{ of }\{x_n\}\}.$$

The metric (or nearest point) projection from H onto C is the mapping $P_C:H\to C$ which assigns to each point $x\in H$ the unique point $P_{C}x\in C$ satisfying the property

$$\parallel x-P_{C}x\parallel =\underset{y\in C}{inf}\parallel x-y\parallel =:d(x,C).$$

Some important properties of projections are gathered in the following proposition.

Proposition 2.1 For given $x\in H$ and $z\in C$:

• (i) $z=P_{C}x\iff 〈x-z,y-z〉\le 0$, $\mathrm{\forall }y\in C$;

• (ii) $z=P_{C}x\iff {\parallel x-z\parallel }^2\le {\parallel x-y\parallel }^2-{\parallel y-z\parallel }^2$, $\mathrm{\forall }y\in C$;

• (iii) $〈P_{C}x-P_{C}y,x-y〉\ge {\parallel P_{C}x-P_{C}y\parallel }^2$, $\mathrm{\forall }y\in H$.

Consequently, $P_C$ is nonexpansive and monotone.

We need some facts and tools which are listed as lemmas below.

Lemma 2.1 (see [[29], demiclosedness principle])

Let C be a nonempty closed and convex subset of a Hilbert space H, and let $S:C\to C$ be a nonexpansive mapping. Then the mapping $I-S$ is demiclosed on C. That is, whenever $\{x_n\}$ is a sequence in C such that $x_n\rightharpoonup x\in C$ and $(I-S)x_n\to y$, it follows that $(I-S)x=y$. Here I is the identity operator of H.

Lemma 2.2 ([[19], Proposition 2.4])

Let $\{x_n\}$ be a bounded sequence on a reflexive Banach space X. If ${\omega }_w(\{x_n\})=\{x\}$, then $x_n\rightharpoonup x$.

Lemma 2.3 Let $A:C\to H$ be a monotone mapping. In the context of the variational inequality problem, the characterization of the projection (see Proposition  2.1(i)) implies

$$u\in VI(C,A)\iff u=P_C(u-\lambda Au),\mathrm{\forall }\lambda >0.$$

Lemma 2.4 Let H be a real Hilbert space. Then the following hold:

• (a) ${\parallel x-y\parallel }^2={\parallel x\parallel }^2-{\parallel y\parallel }^2-2〈x-y,y〉$ for all $x,y\in H$;

• (b) ${\parallel (1-t)x+ty\parallel }^2=(1-t){\parallel x\parallel }^2+t{\parallel y\parallel }^2-t(1-t){\parallel x-y\parallel }^2$ for all $t\in [0,1]$ and for all $x,y\in H$;

• (c) If $\{x_n\}$ is a sequence in H such that $x_n\rightharpoonup x$, it follows that

  $$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-y\parallel }^2=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-x\parallel }^2+{\parallel x-y\parallel }^2,\mathrm{\forall }y\in H.$$

Lemma 2.5 ([[23], Lemma 2.5])

Let H be a real Hilbert space. Given a nonempty closed convex subset of H and points $x,y,z\in H$, and given also a real number $a\in \mathbf{R}$, the set

$$\{v\in C:{\parallel y-v\parallel }^2\le {\parallel x-v\parallel }^2+〈z,v〉+a\}$$

is convex (and closed).

Lemma 2.6 ([[23], Lemma 2.6])

Let C be a nonempty subset of a Hilbert space H, and let $S:C\to C$ be an asymptotically κ-strictly pseudocontractive mapping in the intermediate sense with sequence $\{{\gamma }_n\}$. Then

$$\parallel S^{n}x-S^{n}y\parallel \le \frac{1}{1-\kappa }(\kappa \parallel x-y\parallel +\sqrt{(1+(1-\kappa ){\gamma }_n){\parallel x-y\parallel }^2+(1-\kappa )c_n})$$

for all $x,y\in C$ and $n\ge 1$.

Lemma 2.7 ([[23], Lemma 2.7])

Let C be a nonempty subset of a Hilbert space H, and let $S:C\to C$ be a uniformly continuous asymptotically κ-strictly pseudocontractive mapping in the intermediate sense with sequence $\{{\gamma }_n\}$. Let $\{x_n\}$ be a sequence in C such that $\parallel x_n-x_{n+1}\parallel \to 0$ and $\parallel x_n-S^n{x}_n\parallel \to 0$ as $n\to \mathrm{\infty }$. Then $\parallel x_n-S{x}_n\parallel \to 0$ as $n\to \mathrm{\infty }$.

Lemma 2.8 (Demiclosedness principle [[23], Proposition 3.1])

Let C be a nonempty closed convex subset of a Hilbert space H, and let $S:C\to C$ be a continuous asymptotically κ-strictly pseudocontractive mapping in the intermediate sense with sequence $\{{\gamma }_n\}$. Then $I-S$ is demiclosed at zero in the sense that if $\{x_n\}$ is a sequence in C such that $x_n\rightharpoonup x\in C$ and ${lim\hspace{0.17em}sup}_{m\to \mathrm{\infty }}{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel x_n-S^m{x}_n\parallel =0$, then $(I-S)x=0$.

Lemma 2.9 ([[23], Proposition 3.2])

Let C be a nonempty closed convex subset of a Hilbert space H, and let $S:C\to C$ be a continuous asymptotically κ-strictly pseudocontractive mapping in the intermediate sense with sequence $\{{\gamma }_n\}$ such that $Fix(S)\ne \mathrm{\varnothing }$. Then $Fix(S)$ is closed and convex.

Remark 2.1 Lemmas 2.8 and 2.9 give some basic properties of an asymptotically κ-strictly pseudocontractive mapping in the intermediate sense with sequence $\{{\gamma }_n\}$. Moreover, Lemma 2.8 extends the demiclosedness principles studied for certain classes of nonlinear mappings in Kim and Xu [22], Gornicki [30], Marino and Xu [31] and Xu [32].

To prove a weak convergence theorem by the hybrid extragradient-like method for finding a common solution of VIP (1.1), GSVI (1.8) and the fixed point problem of an asymptotically κ-strictly pseudocontractive mapping in the intermediate sense, we need the following lemma by Osilike et al. [33].

Lemma 2.10 ([[33], p.80])

Let ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{b_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\delta }_n\}}_{n=1}^{\mathrm{\infty }}$ be sequences of nonnegative real numbers satisfying the inequality

$$a_{n+1}\le (1+{\delta }_n)a_n+b_n,\mathrm{\forall }n\ge 1.$$

If ${\sum }_{n=1}^{\mathrm{\infty }}{\delta }_n<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{b}_n<\mathrm{\infty }$, then ${lim}_{n\to \mathrm{\infty }}{a}_n$ exists. If, in addition, ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$ has a subsequence which converges to zero, then ${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

Corollary 2.1 ([[34], p.303])

Let ${\{a_n\}}_{n=0}^{\mathrm{\infty }}$ and ${\{b_n\}}_{n=0}^{\mathrm{\infty }}$ be two sequences of nonnegative real numbers satisfying the inequality

$$a_{n+1}\le a_n+b_n,\mathrm{\forall }n\ge 0.$$

If ${\sum }_{n=0}^{\mathrm{\infty }}{b}_n$ converges, then ${lim}_{n\to \mathrm{\infty }}{a}_n$ exists.

We need a technique lemma in the sequel, whose proof is an immediate consequence of Opial’s property [35] of a Hilbert space and is hence omitted.

Lemma 2.11 Let K be a nonempty closed and convex subset of a real Hilbert space H. Let ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence in H satisfying the properties:

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

• (ii) ${\omega }_w(x_n)\subset K$.

Then ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is weakly convergent to a point in K.

A set-valued mapping $T:H\to 2^H$ is called monotone if for all $x,y\in H$, $f\in Tx$ and $g\in Ty$ imply $〈x-y,f-g〉\ge 0$. A monotone mapping $T:H\to 2^H$ is maximal if its graph $Gph(T)$ is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping T is maximal if and only if for $(x,f)\in H\times H$, $〈x-y,f-g〉\ge 0$ for all $(y,g)\in Gph(T)$ implies $f\in Tx$. Let $A:C\to H$ be a monotone and Lipschitzian mapping, and let $N_{C}v$ be the normal cone to C at $v\in C$, i.e., $N_{C}v=\{w\in H:〈v-u,w〉\ge 0,\mathrm{\forall }u\in C\}$. Define

$$Tv=\{\begin{array}{cc}Av+N_{C}v\hfill & \text{if }v\in C,\hfill \\ \mathrm{\varnothing }\hfill & \text{if }v\notin C.\hfill \end{array}$$

It is known that in this case T is maximal monotone, and $0\in Tv$ if and only if $v\in \Omega$; see [36].

3 Strong convergence theorem

In this section, we prove a strong convergence theorem for a hybrid viscosity CQ iterative algorithm for finding a common solution of VIP (1.1), GSVI (1.8) and the fixed point problem of a uniformly continuous asymptotically κ-strictly pseudocontractive mapping $S:C\to C$ in the intermediate sense. This iterative algorithm is based on the extragradient method, the CQ method, the Mann-type iterative method and the viscosity approximation method.

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $A:C\to H$ be α-inverse strongly monotone, and let $B_i:C\to H$ be ${\beta }_i$-inverse strongly monotone for $i=1,2$. Let $f:C\to C$ be a ρ-contraction with $\rho \in [0,1)$, and let $S:C\to C$ be a uniformly continuous asymptotically κ-strictly pseudocontractive mapping in the intermediate sense with sequence $\{{\gamma }_n\}$ such that $Fix(S)\cap \Xi \cap VI(C,A)$ is nonempty and bounded. Let $\{{\gamma }_n\}$ and $\{c_n\}$ be defined as in (1.6). Let $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ be the sequences generated by

$$\{\begin{array}{c}{x}_1=x\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=P_C(x_n-{\lambda }_{n}A{x}_n),\hfill \\ t_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)P_C[P_C(y_n-{\mu }_2{B}_2{y}_n)-{\mu }_1{B}_1{P}_C(y_n-{\mu }_2{B}_2{y}_n)],\hfill \\ z_n=(1-{\mu }_n-{\nu }_n)x_n+{\mu }_n{t}_n+{\nu }_n{S}^n{t}_n,\hfill \\ C_n=\{z\in C:{\parallel z_n-z\parallel }^2\le {\parallel x_n-z\parallel }^2+{\theta }_n\},\hfill \\ Q_n=\{z\in C:〈x_n-z,x-x_n〉\ge 0\},\hfill \\ x_{n+1}=P_{C_n\cap Q_n}x,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(3.1)

where ${\mu }_i\in (0,2{\beta }_i)$ for $i=1,2$, ${\theta }_n=c_n+({\alpha }_n+{\gamma }_n){\Delta }_n$,

$${\Delta }_n=sup\{{\parallel x_n-z\parallel }^2+\frac{1+{\gamma }_n}{1-\rho }{\parallel (I-f)z\parallel }^2:z\in Fix(S)\cap \Xi \cap VI(C,A)\}<\mathrm{\infty },$$

$\{{\lambda }_n\}$ is a sequence in $(0,2\alpha )$ and $\{{\alpha }_n\}$, $\{{\mu }_n\}$, $\{{\nu }_n\}$ are three sequences in $[0,1]$ such that ${\mu }_n+{\nu }_n\le 1$ for all $n\ge 1$. Assume that the following conditions hold:

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

• (ii) $\{{\lambda }_n\}\subset [a,b]$ for some $a,b\in (0,2\alpha )$;

• (iii) $\kappa \le {\mu }_n$ for all $n\ge 1$;

• (iv) $0<\sigma \le {\nu }_n$ for all $n\ge 1$.

Then the sequences $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ converge strongly to $P_{Fix(S)\cap \Xi \cap VI(C,A)}x$.

Proof It is obvious that $C_n$ is closed and $Q_n$ is closed and convex for every $n=1,2,\dots$ . As the defining inequality in $C_n$ is equivalent to the inequality

$$〈2(x_n-z_n),z〉\le {\parallel x_n\parallel }^2-{\parallel z_n\parallel }^2+{\theta }_n,$$

by Lemma 2.5 we also have that $C_n$ is convex for every $n=1,2,\dots$ . As $Q_n=\{z\in C:〈x_n-z,x-x_n〉\ge 0\}$, we have $〈x_n-z,x-x_n〉\ge 0$ for all $z\in Q_n$ and, by Proposition 2.1(i), we get $x_n=P_{Q_n}x$.

Next, we divide the rest of the proof into several steps.

Step 1. $Fix(S)\cap \Xi \cap VI(C,A)\subset C_n\cap Q_n$ for all $n\ge 1$.

Indeed, take $x^{\ast }\in Fix(S)\cap \Xi \cap VI(C,A)$ arbitrarily. Then $S{x}^{\ast }=x^{\ast }$, $x^{\ast }=P_C(x^{\ast }-{\lambda }_{n}A{x}^{\ast })$ and

$$x^{\ast }=P_C[P_C(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })-{\mu }_1{B}_1{P}_C(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })].$$

(3.2)

Since $A:C\to H$ is α-inverse strongly monotone and $0<{\lambda }_n\le 2\alpha$, we have, for all $n\ge 1$,

$$\begin{array}{rcl}{\parallel y_n-x^{\ast }\parallel }^2& =& {\parallel P_C(x_n-{\lambda }_{n}A{x}_n)-P_C(x^{\ast }-{\lambda }_{n}A{x}^{\ast })\parallel }^2\\ \le & {\parallel (x_n-{\lambda }_{n}A{x}_n)-(x^{\ast }-{\lambda }_{n}A{x}^{\ast })\parallel }^2\\ =& {\parallel (x_n-x^{\ast })-{\lambda }_n(A{x}_n-A{x}^{\ast })\parallel }^2\\ \le & {\parallel x_n-x^{\ast }\parallel }^2-{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{x}_n-A{x}^{\ast }\parallel }^2\\ \le & {\parallel x_n-x^{\ast }\parallel }^2.\end{array}$$

(3.3)

For simplicity, we write $y^{\ast }=P_C(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })$, $u_n=P_C(y_n-{\mu }_2{B}_2{y}_n)$ and

$$v_n:=P_C(u_n-{\mu }_1{B}_1{u}_n)=P_C[P_C(y_n-{\mu }_2{B}_2{y}_n)-{\mu }_1{B}_1{P}_C(y_n-{\mu }_2{B}_2{y}_n)]$$

for all $n\ge 1$. Since $B_i:C\to H$ is ${\beta }_i$-inverse strongly monotone and $0<{\mu }_i<2{\beta }_i$ for $i=1,2$, we know that for all $n\ge 1$,

$$\begin{array}{c}\parallel v_n-x^{\ast }\parallel \hfill \\ ={\parallel P_C[P_C(y_n-{\mu }_2{B}_2{y}_n)-{\mu }_1{B}_1{P}_C(y_n-{\mu }_2{B}_2{y}_n)]-x^{\ast }\parallel }^2\hfill \\ =\parallel P_C[P_C(y_n-{\mu }_2{B}_2{y}_n)-{\mu }_1{B}_1{P}_C(y_n-{\mu }_2{B}_2{y}_n)]\hfill \\ -P_C[P_C(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })-{\mu }_1{B}_1{P}_C(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })]{\parallel }^2\hfill \\ \le \parallel [P_C(y_n-{\mu }_2{B}_2{y}_n)-{\mu }_1{B}_1{P}_C(y_n-{\mu }_2{B}_2{y}_n)]\hfill \\ -[P_C(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })-{\mu }_1{B}_1{P}_C(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })]{\parallel }^2\hfill \\ =\parallel [P_C(y_n-{\mu }_2{B}_2{y}_n)-P_C(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })]\hfill \\ -{\mu }_1[B_1{P}_C(y_n-{\mu }_2{B}_2{y}_n)-B_1{P}_C(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })]{\parallel }^2\hfill \\ \le {\parallel P_C(y_n-{\mu }_2{B}_2{y}_n)-P_C(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })\parallel }^2\hfill \\ -{\mu }_1(2{\beta }_1-{\mu }_1){\parallel B_1{P}_C(y_n-{\mu }_2{B}_2{y}_n)-B_1{P}_C(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })\parallel }^2\hfill \\ \le {\parallel (y_n-{\mu }_2{B}_2{y}_n)-(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })\parallel }^2-{\mu }_1(2{\beta }_1-{\mu }_1){\parallel B_1{u}_n-B_1{y}^{\ast }\parallel }^2\hfill \\ ={\parallel (y_n-x^{\ast })-{\mu }_2(B_2{y}_n-B_2{x}^{\ast })\parallel }^2-{\mu }_1(2{\beta }_1-{\mu }_1){\parallel B_1{u}_n-B_1{y}^{\ast }\parallel }^2\hfill \\ \le {\parallel y_n-x^{\ast }\parallel }^2-{\mu }_2(2{\beta }_2-{\mu }_2){\parallel B_2{y}_n-B_2{x}^{\ast }\parallel }^2-{\mu }_1(2{\beta }_1-{\mu }_1){\parallel B_1{u}_n-B_1{y}^{\ast }\parallel }^2\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2-{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{x}_n-A{x}^{\ast }\parallel }^2\hfill \\ -{\mu }_2(2{\beta }_2-{\mu }_2){\parallel B_2{y}_n-B_2{x}^{\ast }\parallel }^2-{\mu }_1(2{\beta }_1-{\mu }_1){\parallel B_1{u}_n-B_1{y}^{\ast }\parallel }^2\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2.\hfill \end{array}$$

(3.4)

Hence we get

$$\begin{array}{c}{\parallel t_n-x^{\ast }\parallel }^2\hfill \\ ={\parallel {\alpha }_n(f(x_n)-x^{\ast })+(1-{\alpha }_n)(P_C[P_C(y_n-{\mu }_2{B}_2{y}_n)-{\mu }_1{B}_1{P}_C(y_n-{\mu }_2{B}_2{y}_n)]-x^{\ast })\parallel }^2\hfill \\ \le {[{\alpha }_n\parallel f(x_n)-x^{\ast }\parallel +(1-{\alpha }_n)\parallel P_C[P_C(y_n-{\mu }_2{B}_2{y}_n)-{\mu }_1{B}_1{P}_C(y_n-{\mu }_2{B}_2{y}_n)]-x^{\ast }\parallel ]}^2\hfill \\ \le {[{\alpha }_n(\parallel f(x_n)-f\left(x^{\ast }\right)\parallel +\parallel f\left(x^{\ast }\right)-x^{\ast }\parallel )+(1-{\alpha }_n)\parallel x_n-x^{\ast }\parallel ]}^2\hfill \\ \le {[{\alpha }_n(\rho \parallel x_n-x^{\ast }\parallel +\parallel f\left(x^{\ast }\right)-x^{\ast }\parallel )+(1-{\alpha }_n)\parallel x_n-x^{\ast }\parallel ]}^2\hfill \\ ={[(1-(1-\rho ){\alpha }_n)\parallel x_n-x^{\ast }\parallel +(1-\rho ){\alpha }_n\frac{\parallel f(x^{\ast })-x^{\ast }\parallel }{1-\rho }]}^2\hfill \\ \le (1-(1-\rho ){\alpha }_n){\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n\frac{{\parallel f(x^{\ast })-x^{\ast }\parallel }^2}{1-\rho }\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n\frac{{\parallel f(x^{\ast })-x^{\ast }\parallel }^2}{1-\rho }.\hfill \end{array}$$

(3.5)

Therefore, from (3.5), $z_n=(1-{\mu }_n-{\nu }_n)x_n+{\mu }_n{t}_n+{\nu }_n{S}^n{t}_n$, and $x^{\ast }=S{x}^{\ast }$, we have

$$\begin{array}{c}{\parallel z_n-x^{\ast }\parallel }^2\hfill \\ ={\parallel (1-{\mu }_n-{\nu }_n)x_n+{\mu }_n{t}_n+{\nu }_n{S}^n{t}_n-x^{\ast }\parallel }^2\hfill \\ ={\parallel (1-{\mu }_n-{\nu }_n)(x_n-x^{\ast })+{\mu }_n(t_n-x^{\ast })+{\nu }_n(S^n{t}_n-x^{\ast })\parallel }^2\hfill \\ \le (1-{\mu }_n-{\nu }_n){\parallel x_n-x^{\ast }\parallel }^2+({\mu }_n+{\nu }_n){\parallel \frac{{\mu }_n}{{\mu }_n+{\nu }_n}(t_n-x^{\ast })+\frac{{\nu }_n}{{\mu }_n+{\nu }_n}(S^n{t}_n-x^{\ast })\parallel }^2\hfill \\ =(1-{\mu }_n-{\nu }_n){\parallel x_n-x^{\ast }\parallel }^2+({\mu }_n+{\nu }_n)\{\frac{{\mu }_n}{{\mu }_n+{\nu }_n}{\parallel t_n-x^{\ast }\parallel }^2\hfill \\ +\frac{{\nu }_n}{{\mu }_n+{\nu }_n}{\parallel S^n{t}_n-x^{\ast }\parallel }^2-\frac{{\mu }_n{\nu }_n}{{({\mu }_n+{\nu }_n)}^2}{\parallel t_n-S^n{t}_n\parallel }^2\}\hfill \\ \le (1-{\mu }_n-{\nu }_n){\parallel x_n-x^{\ast }\parallel }^2+({\mu }_n+{\nu }_n)\{\frac{{\mu }_n}{{\mu }_n+{\nu }_n}{\parallel t_n-x^{\ast }\parallel }^2\hfill \\ +\frac{{\nu }_n}{{\mu }_n+{\nu }_n}[(1+{\gamma }_n){\parallel t_n-x^{\ast }\parallel }^2+\kappa {\parallel t_n-S^n{t}_n\parallel }^2+c_n]\hfill \\ -\frac{{\mu }_n{\nu }_n}{{({\mu }_n+{\nu }_n)}^2}{\parallel t_n-S^n{t}_n\parallel }^2\}\hfill \\ \le (1-{\mu }_n-{\nu }_n){\parallel x_n-x^{\ast }\parallel }^2+({\mu }_n+{\nu }_n)\{(1+{\gamma }_n){\parallel t_n-x^{\ast }\parallel }^2\hfill \\ +\frac{{\nu }_n}{{\mu }_n+{\nu }_n}(\kappa -\frac{{\mu }_n}{{\mu }_n+{\nu }_n}){\parallel t_n-S^n{t}_n\parallel }^2+\frac{{\nu }_n{c}_n}{{\mu }_n+{\nu }_n}\}\hfill \\ \le (1-{\mu }_n-{\nu }_n){\parallel x_n-x^{\ast }\parallel }^2+({\mu }_n+{\nu }_n)\{(1+{\gamma }_n){\parallel t_n-x^{\ast }\parallel }^2+\frac{{\nu }_n{c}_n}{{\mu }_n+{\nu }_n}\}\hfill \\ \le (1-{\mu }_n-{\nu }_n){\parallel x_n-x^{\ast }\parallel }^2+({\mu }_n+{\nu }_n)\{(1+{\gamma }_n)[{\parallel x_n-x^{\ast }\parallel }^2\hfill \\ +{\alpha }_n\frac{{\parallel f(x^{\ast })-x^{\ast }\parallel }^2}{1-\rho }]+\frac{{\nu }_n{c}_n}{{\mu }_n+{\nu }_n}\}\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n(1+{\gamma }_n)\frac{{\parallel f(x^{\ast })-x^{\ast }\parallel }^2}{1-\rho }+c_n\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2+({\gamma }_n+{\alpha }_n)({\parallel x_n-x^{\ast }\parallel }^2+\frac{1+{\gamma }_n}{1-\rho }{\parallel (I-f)x^{\ast }\parallel }^2)+c_n\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2+c_n+({\alpha }_n+{\gamma }_n){\Delta }_n\hfill \end{array}$$

(3.6)

for every $n=1,2,\dots$ , and hence $x^{\ast }\in C_n$. So, $Fix(S)\cap \Xi \cap VI(C,A)\subset C_n$ for every $n=1,2,\dots$ . Now, let us show by mathematical induction that $\{x_n\}$ is well defined and $Fix(S)\cap \Xi \cap VI(C,A)\subset C_n\cap Q_n$ for every $n=1,2,\dots$ . For $n=1$, we have $Q_1=C$. Hence we obtain $Fix(S)\cap \Xi \cap VI(C,A)\subset C_1\cap Q_1$. Suppose that $x_k$ is given and $Fix(S)\cap \Xi \cap VI(C,A)\subset C_k\cap Q_k$ for some integer $k\ge 1$. Since $Fix(S)\cap \Xi \cap VI(C,A)$ is nonempty, $C_k\cap Q_k$ is a nonempty closed convex subset of C. So, there exists a unique element $x_{k+1}\in C_k\cap Q_k$ such that $x_{k+1}=P_{C_k\cap Q_k}x$. It is also obvious that there holds $〈x_{k+1}-z,x-x_{k+1}〉\ge 0$ for every $z\in C_k\cap Q_k$. Since $Fix(S)\cap \Xi \cap VI(C,A)\subset C_k\cap Q_k$, we have $〈x_{k+1}-z,x-x_{k+1}〉\ge 0$ for every $z\in Fix(S)\cap \Xi \cap VI(C,A)$, and hence $Fix(S)\cap \Xi \cap VI(C,A)\subset Q_{k+1}$. Therefore, we obtain $Fix(S)\cap \Xi \cap VI(C,A)\subset C_{k+1}\cap Q_{k+1}$.

Step 2. $\{x_n\}$ is bounded and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x_{n+1}\parallel ={lim}_{n\to \mathrm{\infty }}\parallel x_n-z_n\parallel =0$.

Indeed, let $q=P_{Fix(S)\cap \Xi \cap VI(C,A)}x$. From $x_{n+1}=P_{C_n\cap Q_n}x$ and $q\in Fix(S)\cap \Xi \cap VI(C,A)\subset C_n\cap Q_n$, we have

$$\parallel x_{n+1}-x\parallel \le \parallel q-x\parallel$$

(3.7)

for every $n=1,2,\dots$ . Therefore, $\{x_n\}$ is bounded. From (3.3)-(3.6) we also obtain that $\{u_n\}$, $\{v_n\}$, $\{y_n\}$, $\{z_n\}$ and $\{t_n\}$ are bounded. Since $x_{n+1}\in C_n\cap Q_n\subset Q_n$ and $x_n=P_{Q_n}x$, we have

$$\parallel x_n-x\parallel \le \parallel x_{n+1}-x\parallel$$

for every $n=1,2,\dots$ . Therefore, there exists ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x\parallel$. Since $x_n=P_{Q_n}x$ and $x_{n+1}\in Q_n$, using Proposition 2.1(ii), we have

$${\parallel x_{n+1}-x_n\parallel }^2\le {\parallel x_{n+1}-x\parallel }^2-{\parallel x_n-x\parallel }^2$$

for every $n=1,2,\dots$ . This implies that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =0.$$

Since $x_{n+1}\in C_n$, we have

$${\parallel z_n-x_{n+1}\parallel }^2\le {\parallel x_n-x_{n+1}\parallel }^2+{\theta }_n,$$

which implies that

$$\parallel z_n-x_{n+1}\parallel \le \parallel x_n-x_{n+1}\parallel +\sqrt{{\theta }_n}.$$

Hence we get

$$\parallel x_n-z_n\parallel \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-z_n\parallel \le 2\parallel x_{n+1}-x_n\parallel +\sqrt{{\theta }_n}$$

for every $n=1,2,\dots$ . From $\parallel x_{n+1}-x_n\parallel \to 0$ and ${\theta }_n\to 0$, we have $\parallel x_n-z_n\parallel \to 0$.

Step 3. ${lim}_{n\to \mathrm{\infty }}\parallel A{x}_n-A{x}^{\ast }\parallel ={lim}_{n\to \mathrm{\infty }}\parallel B_2{y}_n-B_2{x}^{\ast }\parallel ={lim}_{n\to \mathrm{\infty }}\parallel B_1{u}_n-B_1{y}^{\ast }\parallel =0$.

Indeed, from (3.1), (3.4) and (3.6) it follows that

$$\begin{array}{c}{\parallel z_n-x^{\ast }\parallel }^2\hfill \\ \le (1-{\mu }_n-{\nu }_n){\parallel x_n-x^{\ast }\parallel }^2+({\mu }_n+{\nu }_n)\{(1+{\gamma }_n){\parallel t_n-x^{\ast }\parallel }^2+\frac{{\nu }_n{c}_n}{{\mu }_n+{\nu }_n}\}\hfill \\ \le (1-{\mu }_n-{\nu }_n){\parallel x_n-x^{\ast }\parallel }^2+({\mu }_n+{\nu }_n)\{(1+{\gamma }_n)[{\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2\hfill \\ +(1-{\alpha }_n){\parallel P_C[P_C(y_n-{\mu }_2{B}_2{y}_n)-{\mu }_1{B}_1{P}_C(y_n-{\mu }_2{B}_2{y}_n)]-x^{\ast }\parallel }^2]+\frac{{\nu }_n{c}_n}{{\mu }_n+{\nu }_n}\}\hfill \\ \le (1-{\mu }_n-{\nu }_n){\parallel x_n-x^{\ast }\parallel }^2+({\mu }_n+{\nu }_n)\{(1+{\gamma }_n)[{\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2\hfill \\ +{\parallel P_C[P_C(y_n-{\mu }_2{B}_2{y}_n)-{\mu }_1{B}_1{P}_C(y_n-{\mu }_2{B}_2{y}_n)]-x^{\ast }\parallel }^2]+\frac{{\nu }_n{c}_n}{{\mu }_n+{\nu }_n}\}\hfill \\ \le (1-{\mu }_n-{\nu }_n){\parallel x_n-x^{\ast }\parallel }^2+({\mu }_n+{\nu }_n)\{(1+{\gamma }_n)[{\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2\hfill \\ +{\parallel x_n-x^{\ast }\parallel }^2-{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{x}_n-A{x}^{\ast }\parallel }^2\hfill \\ -{\mu }_2(2{\beta }_2-{\mu }_2){\parallel B_2{y}_n-B_2{x}^{\ast }\parallel }^2-{\mu }_1(2{\beta }_1-{\mu }_1){\parallel B_1{u}_n-B_1{y}^{\ast }\parallel }^2]+\frac{{\nu }_n{c}_n}{{\mu }_n+{\nu }_n}\}\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2+c_n+{\alpha }_n(1+{\gamma }_n){\parallel f(x_n)-x^{\ast }\parallel }^2\hfill \\ -({\mu }_n+{\nu }_n)(1+{\gamma }_n)[{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{x}_n-A{x}^{\ast }\parallel }^2\hfill \\ +{\mu }_2(2{\beta }_2-{\mu }_2){\parallel B_2{y}_n-B_2{x}^{\ast }\parallel }^2+{\mu }_1(2{\beta }_1-{\mu }_1){\parallel B_1{u}_n-B_1{y}^{\ast }\parallel }^2],\hfill \end{array}$$

(3.8)

which hence implies that

$$\begin{array}{c}(\kappa +\sigma )(1+{\gamma }_n)[a(2\alpha -b){\parallel A{x}_n-A{x}^{\ast }\parallel }^2\hfill \\ +{\mu }_2(2{\beta }_2-{\mu }_2){\parallel B_2{y}_n-B_2{x}^{\ast }\parallel }^2+{\mu }_1(2{\beta }_1-{\mu }_1){\parallel B_1{u}_n-B_1{y}^{\ast }\parallel }^2]\hfill \\ \le ({\mu }_n+{\nu }_n)(1+{\gamma }_n)[{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{x}_n-A{x}^{\ast }\parallel }^2\hfill \\ +{\mu }_2(2{\beta }_2-{\mu }_2){\parallel B_2{y}_n-B_2{x}^{\ast }\parallel }^2+{\mu }_1(2{\beta }_1-{\mu }_1){\parallel B_1{u}_n-B_1{y}^{\ast }\parallel }^2]\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel z_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2+c_n+{\alpha }_n(1+{\gamma }_n){\parallel f(x_n)-x^{\ast }\parallel }^2\hfill \\ \le (\parallel x_n-x^{\ast }\parallel +\parallel z_n-x^{\ast }\parallel )\parallel x_n-z_n\parallel +{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2+c_n+{\alpha }_n(1+{\gamma }_n){\parallel f(x_n)-x^{\ast }\parallel }^2.\hfill \end{array}$$

Since ${\alpha }_n\to 0$, ${\gamma }_n\to 0$, $c_n\to 0$ and $\parallel x_n-z_n\parallel \to 0$, from the boundedness of $\{x_n\}$ and $\{z_n\}$ we obtain that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel A{x}_n-A{x}^{\ast }\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel B_2{y}_n-B_2{x}^{\ast }\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel B_1{u}_n-B_1{y}^{\ast }\parallel =0.$$

(3.9)

Step 4. ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$.

Indeed, utilizing Proposition 2.1(iii), we deduce from (3.1) that

$$\begin{array}{rcl}{\parallel y_n-x^{\ast }\parallel }^2& =& {\parallel P_C(x_n-{\lambda }_{n}A{x}_n)-P_C(x^{\ast }-{\lambda }_{n}A{x}^{\ast })\parallel }^2\\ \le & 〈(x_n-{\lambda }_{n}A{x}_n)-(x^{\ast }-{\lambda }_{n}A{x}^{\ast }),y_n-x^{\ast }〉\\ =& \frac{1}{2}[{\parallel x_n-x^{\ast }-{\lambda }_n(A{x}_n-A{x}^{\ast })\parallel }^2+{\parallel y_n-x^{\ast }\parallel }^2\\ -{\parallel (x_n-x^{\ast })-{\lambda }_n(A{x}_n-A{x}^{\ast })-(y_n-x^{\ast })\parallel }^2]\\ \le & \frac{1}{2}[{\parallel x_n-x^{\ast }\parallel }^2+{\parallel y_n-x^{\ast }\parallel }^2-{\parallel (x_n-y_n)-{\lambda }_n(A{x}_n-A{x}^{\ast })\parallel }^2]\\ =& \frac{1}{2}[{\parallel x_n-x^{\ast }\parallel }^2+{\parallel y_n-x^{\ast }\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n〈x_n-y_n,A{x}_n-A{x}^{\ast }〉\\ -{\lambda }_n^2{\parallel A{x}_n-A{x}^{\ast }\parallel }^2]\\ \le & \frac{1}{2}[{\parallel x_n-x^{\ast }\parallel }^2+{\parallel y_n-x^{\ast }\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n\parallel x_n-y_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel ].\end{array}$$

Thus,

$${\parallel y_n-x^{\ast }\parallel }^2\le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n\parallel x_n-y_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel .$$

(3.10)

Similarly to the above argument, utilizing Proposition 2.1(iii), we conclude from $u_n=P_C(y_n-{\mu }_2{B}_2{y}_n)$ that

$$\begin{array}{c}{\parallel u_n-y^{\ast }\parallel }^2\hfill \\ ={\parallel P_C(y_n-{\mu }_2{B}_2{y}_n)-P_C(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })\parallel }^2\hfill \\ \le 〈(y_n-{\mu }_2{B}_2{y}_n)-(x^{\ast }-{\mu }_2{B}_2{x}^{\ast }),u_n-y^{\ast }〉\hfill \\ =\frac{1}{2}[{\parallel y_n-x^{\ast }-{\mu }_2(B_2{y}_n-B_2{x}^{\ast })\parallel }^2+{\parallel u_n-y^{\ast }\parallel }^2\hfill \\ -{\parallel (y_n-x^{\ast })-{\mu }_2(B_2{y}_n-B_2{x}^{\ast })-(u_n-y^{\ast })\parallel }^2]\hfill \\ \le \frac{1}{2}[{\parallel y_n-x^{\ast }\parallel }^2+{\parallel u_n-y^{\ast }\parallel }^2-{\parallel (y_n-u_n)-{\mu }_2(B_2{y}_n-B_2{x}^{\ast })-(x^{\ast }-y^{\ast })\parallel }^2]\hfill \\ =\frac{1}{2}[{\parallel y_n-x^{\ast }\parallel }^2+{\parallel u_n-y^{\ast }\parallel }^2-{\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel }^2\hfill \\ +2{\mu }_2〈y_n-u_n-(x^{\ast }-y^{\ast }),B_2{y}_n-B_2{x}^{\ast }〉-{\mu }_2^2{\parallel B_2{y}_n-B_2{x}^{\ast }\parallel }^2],\hfill \end{array}$$

that is,

$$\begin{array}{rcl}{\parallel u_n-y^{\ast }\parallel }^2& \le & {\parallel y_n-x^{\ast }\parallel }^2-{\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel }^2\\ +2{\mu }_2\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{y}_n-B_2{x}^{\ast }\parallel .\end{array}$$

(3.11)

Substituting (3.10) in (3.11), we have

$$\begin{array}{c}{\parallel u_n-y^{\ast }\parallel }^2\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n\parallel x_n-y_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel \hfill \\ -{\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel }^2+2{\mu }_2\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{y}_n-B_2{x}^{\ast }\parallel .\hfill \end{array}$$

(3.12)

Similarly to the above argument, utilizing Proposition 2.1(iii), we conclude from $v_n=P_C(u_n-{\mu }_1{B}_1{u}_n)$ that

$$\begin{array}{c}{\parallel v_n-x^{\ast }\parallel }^2\hfill \\ ={\parallel P_C(u_n-{\mu }_1{B}_1{u}_n)-P_C(y^{\ast }-{\mu }_1{B}_1{y}^{\ast })\parallel }^2\hfill \\ \le 〈(u_n-{\mu }_1{B}_1{u}_n)-(y^{\ast }-{\mu }_1{B}_1{y}^{\ast }),v_n-x^{\ast }〉\hfill \\ =\frac{1}{2}[{\parallel u_n-y^{\ast }-{\mu }_1(B_1{u}_n-B_1{y}^{\ast })\parallel }^2+{\parallel v_n-x^{\ast }\parallel }^2\hfill \\ -{\parallel (u_n-y^{\ast })-{\mu }_1(B_1{u}_n-B_1{y}^{\ast })-(v_n-x^{\ast })\parallel }^2]\hfill \\ \le \frac{1}{2}[{\parallel u_n-y^{\ast }\parallel }^2+{\parallel v_n-x^{\ast }\parallel }^2-{\parallel (u_n-v_n)-{\mu }_1(B_1{u}_n-B_1{y}^{\ast })-(y^{\ast }-x^{\ast })\parallel }^2]\hfill \\ =\frac{1}{2}[{\parallel u_n-y^{\ast }\parallel }^2+{\parallel v_n-x^{\ast }\parallel }^2-{\parallel u_n-v_n-(y^{\ast }-x^{\ast })\parallel }^2\hfill \\ +2{\mu }_1〈u_n-v_n-(y^{\ast }-x^{\ast }),B_1{u}_n-B_1{y}^{\ast }〉-{\mu }_1^2{\parallel B_1{u}_n-B_1{y}^{\ast }\parallel }^2],\hfill \end{array}$$

that is,

$$\begin{array}{rcl}{\parallel v_n-x^{\ast }\parallel }^2& \le & {\parallel u_n-y^{\ast }\parallel }^2-{\parallel u_n-v_n-(y^{\ast }-x^{\ast })\parallel }^2\\ +2{\mu }_1\parallel u_n-v_n-(y^{\ast }-x^{\ast })\parallel \parallel B_1{u}_n-B_1{y}^{\ast }\parallel .\end{array}$$

(3.13)

Substituting (3.12) in (3.13), we have

$$\begin{array}{rcl}{\parallel v_n-x^{\ast }\parallel }^2& \le & {\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n\parallel x_n-y_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel \\ -{\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel }^2+2{\mu }_2\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{y}_n-B_2{x}^{\ast }\parallel \\ -{\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel }^2+2{\mu }_1\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel \parallel B_1{u}_n-B_1{y}^{\ast }\parallel .\end{array}$$

This together with (3.4) and (3.8) implies that

$$\begin{array}{c}{\parallel z_n-x^{\ast }\parallel }^2\hfill \\ \le (1-{\mu }_n-{\nu }_n){\parallel x_n-x^{\ast }\parallel }^2+({\mu }_n+{\nu }_n)\{(1+{\gamma }_n)[{\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2\hfill \\ +{\parallel P_C[P_C(y_n-{\mu }_2{B}_2{y}_n)-{\mu }_1{B}_1{P}_C(y_n-{\mu }_2{B}_2{y}_n)]-x^{\ast }\parallel }^2]+\frac{{\nu }_n{c}_n}{{\mu }_n+{\nu }_n}\}\hfill \\ \le (1-{\mu }_n-{\nu }_n){\parallel x_n-x^{\ast }\parallel }^2+({\mu }_n+{\nu }_n)\{(1+{\gamma }_n)[{\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2\hfill \\ +{\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n\parallel x_n-y_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel \hfill \\ -{\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel }^2+2{\mu }_2\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{y}_n-B_2{x}^{\ast }\parallel \hfill \\ -{\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel }^2+2{\mu }_1\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel \parallel B_1{u}_n-B_1{y}^{\ast }\parallel ]+\frac{{\nu }_n{c}_n}{{\mu }_n+{\nu }_n}\}\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2+c_n+{\alpha }_n(1+{\gamma }_n){\parallel f(x_n)-x^{\ast }\parallel }^2\hfill \\ +2(1+{\gamma }_n)[{\lambda }_n\parallel x_n-y_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel +{\mu }_2\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{y}_n-B_2{x}^{\ast }\parallel \hfill \\ +{\mu }_1\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel \parallel B_1{u}_n-B_1{y}^{\ast }\parallel ]-({\mu }_n+{\nu }_n)(1+{\gamma }_n)[{\parallel x_n-y_n\parallel }^2\hfill \\ +{\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel }^2+{\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel }^2].\hfill \end{array}$$

(3.14)

So, we have

$$\begin{array}{c}(\kappa +\sigma )(1+{\gamma }_n)[{\parallel x_n-y_n\parallel }^2+{\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel }^2+{\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel }^2]\hfill \\ \le ({\mu }_n+{\nu }_n)(1+{\gamma }_n)[{\parallel x_n-y_n\parallel }^2+{\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel }^2+{\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel }^2]\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel z_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2+c_n+{\alpha }_n(1+{\gamma }_n){\parallel f(x_n)-x^{\ast }\parallel }^2\hfill \\ +2(1+{\gamma }_n)[{\lambda }_n\parallel x_n-y_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel +{\mu }_2\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{y}_n-B_2{x}^{\ast }\parallel \hfill \\ +{\mu }_1\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel \parallel B_1{u}_n-B_1{y}^{\ast }\parallel ]\hfill \\ \le (\parallel x_n-x^{\ast }\parallel +\parallel z_n-x^{\ast }\parallel )\parallel x_n-z_n\parallel +{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2+c_n+{\alpha }_n(1+{\gamma }_n){\parallel f(x_n)-x^{\ast }\parallel }^2\hfill \\ +2(1+{\gamma }_n)[{\lambda }_n\parallel x_n-y_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel +{\mu }_2\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{y}_n-B_2{x}^{\ast }\parallel \hfill \\ +{\mu }_1\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel \parallel B_1{u}_n-B_1{y}^{\ast }\parallel ].\hfill \end{array}$$

Since ${\alpha }_n\to 0$, ${\gamma }_n\to 0$, $c_n\to 0$, $\parallel x_n-z_n\parallel \to 0$, $\parallel A{x}_n-A{x}^{\ast }\parallel \to 0$, $\parallel B_2{y}_n-B_2{x}^{\ast }\parallel \to 0$ and $\parallel B_1{u}_n-B_1{y}^{\ast }\parallel \to 0$, from the boundedness of $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ we obtain that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-y_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel =0,$$

and hence

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-v_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-v_n\parallel =0.$$

(3.15)

Step 5. ${lim}_{n\to \mathrm{\infty }}\parallel x_n-t_n\parallel ={lim}_{n\to \mathrm{\infty }}\parallel x_n-S{x}_n\parallel =0$.

Indeed, it follows from (3.1) that

$$\parallel t_n-y_n\parallel \le {\alpha }_n\parallel f(x_n)-y_n\parallel +(1-{\alpha }_n)\parallel v_n-y_n\parallel \le {\alpha }_n\parallel f(x_n)-y_n\parallel +\parallel v_n-y_n\parallel .$$

Since ${\alpha }_n\to 0$ and $\parallel v_n-y_n\parallel \to 0$, from the boundedness of $\{x_n\}$ and $\{y_n\}$ we know that $\parallel t_n-y_n\parallel \to 0$ as $n\to \mathrm{\infty }$. Also, from $\parallel x_n-t_n\parallel \le \parallel x_n-y_n\parallel +\parallel y_n-t_n\parallel$ we also have $\parallel x_n-t_n\parallel \to 0$. Since $z_n=(1-{\mu }_n-{\nu }_n)x_n+{\mu }_n{t}_n+{\nu }_n{S}^n{t}_n$, we have ${\nu }_n(S^n{t}_n-t_n)=(1-{\mu }_n-{\nu }_n)(t_n-x_n)+(z_n-t_n)$. Then

$$\begin{array}{rcl}\sigma \parallel S^n{t}_n-t_n\parallel & \le & {\nu }_n\parallel S^n{t}_n-t_n\parallel \\ \le & (1-{\mu }_n-{\nu }_n)\parallel t_n-x_n\parallel +\parallel z_n-t_n\parallel \\ \le & (1-{\mu }_n-{\nu }_n)\parallel t_n-x_n\parallel +\parallel z_n-x_n\parallel +\parallel x_n-t_n\parallel \\ \le & 2\parallel t_n-x_n\parallel +\parallel z_n-x_n\parallel ,\end{array}$$

and hence $\parallel t_n-S^n{t}_n\parallel \to 0$. Furthermore, observe that

$$\parallel x_n-S^n{x}_n\parallel \le \parallel x_n-t_n\parallel +\parallel t_n-S^n{t}_n\parallel +\parallel S^n{t}_n-S^n{x}_n\parallel .$$

(3.16)

Utilizing Lemma 2.6, we have

$$\parallel S^n{t}_n-S^n{x}_n\parallel \le \frac{1}{1-\kappa }(\kappa \parallel t_n-x_n\parallel +\sqrt{(1+(1-\kappa ){\gamma }_n){\parallel t_n-x_n\parallel }^2+(1-\kappa )c_n})$$

for every $n=1,2,\dots$ . Hence it follows from $\parallel x_n-t_n\parallel \to 0$ that $\parallel S^n{t}_n-S^n{x}_n\parallel \to 0$. Thus from (3.16) and $\parallel t_n-S^n{t}_n\parallel \to 0$ we get $\parallel x_n-S^n{x}_n\parallel \to 0$. Since $\parallel x_{n+1}-x_n\parallel \to 0$, $\parallel x_n-S^n{x}_n\parallel \to 0$ as $n\to \mathrm{\infty }$ and S is uniformly continuous, we obtain from Lemma 2.7 that $\parallel x_n-S{x}_n\parallel \to 0$ as $n\to \mathrm{\infty }$.

Step 6. ${\omega }_w(\{x_n\})\subset Fix(S)\cap \Xi \cap VI(C,A)$.

Indeed, by the boundedness of $\{x_n\}$, we know that ${\omega }_w(\{x_n\})\ne \mathrm{\varnothing }$. Take $\hat{x}\in {\omega }_w(\{x_n\})$ arbitrarily. Then there exists a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that $\{x_{n_i}\}$ converges weakly to $\hat{x}$. We can assert that $\hat{x}\in Fix(S)\cap \Xi \cap VI(C,A)$. First, note that S is uniformly continuous and $\parallel x_n-S{x}_n\parallel \to 0$. Hence it is easy to see that $\parallel x_n-S^m{x}_n\parallel \to 0$ for all $m\ge 1$. By Lemma 2.8, we obtain $\hat{x}\in Fix(S)$. Now let us show that $\hat{x}\in \Xi$. We note that

$$\begin{array}{c}\parallel t_n-G(t_n)\parallel \hfill \\ \le {\alpha }_n\parallel f(x_n)-G(t_n)\parallel \hfill \\ +(1-{\alpha }_n)\parallel P_C[P_C(y_n-{\mu }_2{B}_2{y}_n)-{\mu }_1{B}_1{P}_C(y_n-{\mu }_2{B}_2{y}_n)]-G(t_n)\parallel \hfill \\ ={\alpha }_n\parallel f(x_n)-G(t_n)\parallel +(1-{\alpha }_n)\parallel G(y_n)-G(t_n)\parallel \hfill \\ \le {\alpha }_n\parallel f(x_n)-G(t_n)\parallel +(1-{\alpha }_n)\parallel y_n-t_n\parallel \to 0.\hfill \end{array}$$

(3.17)

Since $x_{n_i}\rightharpoonup \hat{x}$ and $\parallel x_n-t_n\parallel \to 0$, it follows that $t_{n_i}\rightharpoonup \hat{x}$. Thus, according to Lemma 2.1 we get $\hat{x}\in \Xi$. Furthermore, we show $\hat{x}\in VI(C,A)$. Since $x_n-y_n\to 0$ and $x_n-t_n\to 0$, we have $y_{n_i}\rightharpoonup \hat{x}$ and $t_{n_i}\rightharpoonup \hat{x}$. Let

$$Tv=\{\begin{array}{cc}Av+N_{C}v\hfill & \text{if }v\in C,\hfill \\ \mathrm{\varnothing }\hfill & \text{if }v\notin C,\hfill \end{array}$$

where $N_{C}v$ is the normal cone to C at $v\in C$. We have already mentioned that in this case the mapping T is maximal monotone, and $0\in Tv$ if and only if $v\in VI(C,A)$; see [36] for more details. Let $Gph(T)$ be the graph of T, and let $(v,w)\in Gph(T)$. Then we have $w\in Tv=Av+N_{C}v$, and hence $w-Av\in N_{C}v$. So, we have $〈v-t,w-Av〉\ge 0$ for all $t\in C$. On the other hand, from $y_n=P_C(x_n-{\lambda }_{n}A{x}_n)$ and $v\in C$ we have

$$〈x_n-{\lambda }_{n}A{x}_n-y_n,y_n-v〉\ge 0,$$

and hence

$$〈v-y_n,\frac{y_n-x_n}{{\lambda }_n}+A{x}_n〉\ge 0.$$

Therefore, from $〈v-t,w-Av〉\ge 0$ for all $t\in C$ and $y_{n_i}\in C$, we have

$$\begin{array}{rcl}〈v-y_{n_i},w〉& \ge & 〈v-y_{n_i},Av〉\\ \ge & 〈v-y_{n_i},Av〉-〈v-y_{n_i},\frac{y_{n_i}-x_{n_i}}{{\lambda }_{n_i}}+A{x}_{n_i}〉\\ =& 〈v-y_{n_i},Av-A{y}_{n_i}〉+〈v-y_{n_i},A{y}_{n_i}-A{x}_{n_i}〉-〈v-y_{n_i},\frac{y_{n_i}-x_{n_i}}{{\lambda }_{n_i}}〉\\ \ge & 〈v-y_{n_i},A{y}_{n_i}-A{x}_{n_i}〉-〈v-y_{n_i},\frac{y_{n_i}-x_{n_i}}{{\lambda }_{n_i}}〉.\end{array}$$

Thus, we obtain $〈v-\hat{x},w〉\ge 0$ as $i\to \mathrm{\infty }$. Since T is maximal monotone, we have $\hat{x}\in T^{-1}0$ and hence $\hat{x}\in VI(C,A)$. Consequently, $\hat{x}\in Fix(S)\cap \Xi \cap VI(C,A)$. This implies that ${\omega }_w(\{x_n\})\subset Fix(S)\cap \Xi \cap VI(C,A)$.

Step 7. $x_n\to q=P_{Fix(S)\cap \Xi \cap VI(C,A)}x$.

Indeed, from $q=P_{Fix(S)\cap \Xi \cap VI(C,A)}x$, $\hat{x}\in Fix(S)\cap \Xi \cap VI(C,A)$ and (3.7), we have

$$\parallel q-x\parallel \le \parallel \hat{x}-x\parallel \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_i}-x\parallel \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_i}-x\parallel \le \parallel q-x\parallel .$$

So, we obtain

$$\underset{i\to \mathrm{\infty }}{lim}\parallel x_{n_i}-x\parallel =\parallel \hat{x}-x\parallel .$$

From $x_{n_i}-x\rightharpoonup \hat{x}-x$ we have $x_{n_i}-x\to \hat{x}-x$ due to the Kadec-Klee property of a real Hilbert space [29]. So, it is clear that $x_{n_i}\to \hat{x}$. Since $x_n=P_{Q_n}x$ and $q\in Fix(S)\cap \Xi \cap VI(C,A)\subset C_n\cap Q_n\subset Q_n$, we have

$$-{\parallel q-x_{n_i}\parallel }^2=〈q-x_{n_i},x_{n_i}-x〉+〈q-x_{n_i},x-q〉\ge 〈q-x_{n_i},x-q〉.$$

As $i\to \mathrm{\infty }$, we obtain $-{\parallel q-\hat{x}\parallel }^2\ge 〈q-\hat{x},x-q〉\ge 0$ by $q=P_{Fix(S)\cap \Xi \cap VI(C,A)}x$ and $\hat{x}\in Fix(S)\cap \Xi \cap VI(C,A)$. Hence we have $\hat{x}=q$. This implies that $x_n\to q$. It is easy to see that $y_n\to q$ and $z_n\to q$. This completes the proof. □

4 Weak convergence theorem

In this section, we prove a new weak convergence theorem by the hybrid extragradient-like method for finding a common element of the solution set of VIP (1.1), the solution set of GSVI (1.8) and the fixed point set of a uniformly continuous asymptotically κ-strictly pseudocontractive mapping $S:C\to C$ in the intermediate sense.

Theorem 4.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $A:C\to H$ be α-inverse strongly monotone, let $B_i:C\to H$ be ${\beta }_i$-inverse strongly monotone for $i=1,2$, let $f:C\to C$ be a ρ-contraction with $\rho \in [0,1)$, and let $S:C\to C$ be a uniformly continuous asymptotically κ-strictly pseudocontractive mapping in the intermediate sense with sequence $\{{\gamma }_n\}$ such that $Fix(S)\cap \Xi \cap VI(C,A)$ is nonempty and bounded. Let $\{{\gamma }_n\}$ and $\{c_n\}$ be defined as in (1.6). Let $\{x_n\}$ and $\{y_n\}$ be the sequences generated by

$$\{\begin{array}{c}{x}_1=x\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=P_C(x_n-{\lambda }_{n}A{x}_n),\hfill \\ t_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)P_C[P_C(y_n-{\mu }_2{B}_2{y}_n)-{\mu }_1{B}_1{P}_C(y_n-{\mu }_2{B}_2{y}_n)],\hfill \\ x_{n+1}=(1-{\mu }_n-{\nu }_n)x_n+{\mu }_n{t}_n+{\nu }_n{S}^n{t}_n,\hfill \end{array}$$

(4.1)

where ${\mu }_i\in (0,2{\beta }_i)$ for $i=1,2$, $\{{\lambda }_n\}$ is a sequence in $(0,2\alpha )$ and $\{{\alpha }_n\}$, $\{{\mu }_n\}$, $\{{\nu }_n\}$ are three sequences in $[0,1]$ such that ${\mu }_n+{\nu }_n\le 1$ for all $n\ge 1$. Assume that the following conditions hold:

• (i) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n<\mathrm{\infty }$;

• (ii) $\{{\lambda }_n\}\subset [a,b]$ for some $a,b\in (0,2\alpha )$;

• (iii) ${\sum }_{n=1}^{\mathrm{\infty }}{c}_n<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<\mathrm{\infty }$;

• (iv) and for all $n\ge 1$, $\kappa \le {\mu }_n$, $\sigma \le {\nu }_n$ and ${\mu }_n+{\nu }_n\le c$ for some $c,\sigma \in (0,1)$.

Then the sequences $\{x_n\}$ and $\{y_n\}$ converge weakly to an element of $Fix(S)\cap \Xi \cap VI(C,A)$.

Proof First of all, take $x^{\ast }\in Fix(S)\cap \Xi \cap VI(C,A)$ arbitrarily. Then, repeating the same arguments as in (3.3) and (3.5), we deduce from (4.1) that

$$\parallel y_n-x^{\ast }\parallel \le \parallel x_n-x^{\ast }\parallel ,$$

(4.2)

and

$${\parallel t_n-x^{\ast }\parallel }^2\le {\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n\frac{{\parallel f(x^{\ast })-x^{\ast }\parallel }^2}{1-\rho }.$$

(4.3)

Repeating the same arguments as in (3.6), we can obtain that

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & {\parallel x_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n(1+{\gamma }_n)\frac{{\parallel f(x^{\ast })-x^{\ast }\parallel }^2}{1-\rho }+c_n\\ =& (1+{\gamma }_n){\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n(1+{\gamma }_n)\frac{{\parallel f(x^{\ast })-x^{\ast }\parallel }^2}{1-\rho }+c_n.\end{array}$$

(4.4)

Since ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{c}_n<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n<\mathrm{\infty }$ it follows that

$$\sum _{n=1}^{\mathrm{\infty }}\{{\alpha }_n(1+{\gamma }_n)\frac{{\parallel f(x^{\ast })-x^{\ast }\parallel }^2}{1-\rho }+c_n\}<\mathrm{\infty }.$$

So, by Lemma 2.10 we know that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-x^{\ast }\parallel \text{exists for all }{x}^{\ast }\in Fix(S)\cap \Xi \cap VI(C,A).$$

This implies that $\{x_n\}$ is bounded and hence $\{t_n\}$, $\{y_n\}$ are bounded due to (4.2) and (4.3).

Repeating the same arguments as in (3.8), we can conclude that

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & {\parallel x_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2+c_n+{\alpha }_n(1+{\gamma }_n){\parallel f(x_n)-x^{\ast }\parallel }^2\\ -({\mu }_n+{\nu }_n)(1+{\gamma }_n)[{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{x}_n-A{x}^{\ast }\parallel }^2\\ +{\mu }_2(2{\beta }_2-{\mu }_2){\parallel B_2{y}_n-B_2{x}^{\ast }\parallel }^2+{\mu }_1(2{\beta }_1-{\mu }_1){\parallel B_1{u}_n-B_1{y}^{\ast }\parallel }^2],\end{array}$$

(4.5)

which hence implies that

$$\begin{array}{c}(\kappa +\sigma )(1+{\gamma }_n)[a(2\alpha -b){\parallel A{x}_n-A{x}^{\ast }\parallel }^2\hfill \\ +{\mu }_2(2{\beta }_2-{\mu }_2){\parallel B_2{y}_n-B_2{x}^{\ast }\parallel }^2+{\mu }_1(2{\beta }_1-{\mu }_1){\parallel B_1{u}_n-B_1{y}^{\ast }\parallel }^2]\hfill \\ \le ({\mu }_n+{\nu }_n)(1+{\gamma }_n)[{\lambda }_n(2\alpha -{\lambda }_n){\parallel A{x}_n-A{x}^{\ast }\parallel }^2\hfill \\ +{\mu }_2(2{\beta }_2-{\mu }_2){\parallel B_2{y}_n-B_2{x}^{\ast }\parallel }^2+{\mu }_1(2{\beta }_1-{\mu }_1){\parallel B_1{u}_n-B_1{y}^{\ast }\parallel }^2]\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2+c_n+{\alpha }_n(1+{\gamma }_n){\parallel f(x_n)-x^{\ast }\parallel }^2.\hfill \end{array}$$

Since ${\alpha }_n\to 0$, ${\gamma }_n\to 0$, $c_n\to 0$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x^{\ast }\parallel$ exists, from the boundedness of $\{x_n\}$ we conclude that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel A{x}_n-A{x}^{\ast }\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel B_2{y}_n-B_2{x}^{\ast }\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel B_1{u}_n-B_1{y}^{\ast }\parallel =0.$$

(4.6)

Repeating the same arguments as in (3.14), we can conclude that

$$\begin{array}{c}{\parallel x_{n+1}-x^{\ast }\parallel }^2\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2+c_n+{\alpha }_n(1+{\gamma }_n){\parallel f(x_n)-x^{\ast }\parallel }^2\hfill \\ +2(1+{\gamma }_n)[{\lambda }_n\parallel x_n-y_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel +{\mu }_2\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{y}_n-B_2{x}^{\ast }\parallel \hfill \\ +{\mu }_1\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel \parallel B_1{u}_n-B_1{y}^{\ast }\parallel ]-({\mu }_n+{\nu }_n)(1+{\gamma }_n)[{\parallel x_n-y_n\parallel }^2\hfill \\ +{\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel }^2+{\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel }^2],\hfill \end{array}$$

which hence implies that

$$\begin{array}{c}(\kappa +\sigma )(1+{\gamma }_n)[{\parallel x_n-y_n\parallel }^2+{\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel }^2+{\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel }^2]\hfill \\ \le ({\mu }_n+{\nu }_n)(1+{\gamma }_n)[{\parallel x_n-y_n\parallel }^2+{\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel }^2+{\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel }^2]\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2+c_n+{\alpha }_n(1+{\gamma }_n){\parallel f(x_n)-x^{\ast }\parallel }^2\hfill \\ +2(1+{\gamma }_n)[{\lambda }_n\parallel x_n-y_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel +{\mu }_2\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel \parallel B_2{y}_n-B_2{x}^{\ast }\parallel \hfill \\ +{\mu }_1\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel \parallel B_1{u}_n-B_1{y}^{\ast }\parallel ].\hfill \end{array}$$

Since ${\alpha }_n\to 0$, ${\gamma }_n\to 0$, $c_n\to 0$, $\parallel A{x}_n-A{x}^{\ast }\parallel \to 0$, $\parallel B_2{y}_n-B_2{x}^{\ast }\parallel \to 0$, $\parallel B_1{u}_n-B_1{y}^{\ast }\parallel \to 0$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x^{\ast }\parallel$ exists, from the boundedness of $\{x_n\}$ and $\{y_n\}$ we obtain that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-y_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-u_n-(x^{\ast }-y^{\ast })\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-v_n+(x^{\ast }-y^{\ast })\parallel =0,$$

and hence

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-v_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-v_n\parallel =0.$$

(4.7)

On the other hand, it follows from (4.1) that

$$\parallel t_n-y_n\parallel \le {\alpha }_n\parallel f(x_n)-y_n\parallel +(1-{\alpha }_n)\parallel v_n-y_n\parallel \le {\alpha }_n\parallel f(x_n)-y_n\parallel +\parallel v_n-y_n\parallel .$$

Since ${\alpha }_n\to 0$ and $\parallel v_n-y_n\parallel \to 0$, from the boundedness of $\{x_n\}$ and $\{y_n\}$ we know that $\parallel t_n-y_n\parallel \to 0$ as $n\to \mathrm{\infty }$. Also, from $\parallel x_n-t_n\parallel \le \parallel x_n-y_n\parallel +\parallel y_n-t_n\parallel$ we also have $\parallel x_n-t_n\parallel \to 0$. Repeating the same arguments as in (3.6), we have

$$\begin{array}{c}{\parallel x_{n+1}-x^{\ast }\parallel }^2\hfill \\ ={\parallel (1-{\mu }_n-{\nu }_n)(x_n-x^{\ast })+{\mu }_n(t_n-x^{\ast })+{\nu }_n(S^n{t}_n-x^{\ast })\parallel }^2\hfill \\ =(1-{\mu }_n-{\nu }_n){\parallel x_n-x^{\ast }\parallel }^2+({\mu }_n+{\nu }_n){\parallel \frac{{\mu }_n}{{\mu }_n+{\nu }_n}(t_n-x^{\ast })+\frac{{\nu }_n}{{\mu }_n+{\nu }_n}(S^n{t}_n-x^{\ast })\parallel }^2\hfill \\ -(1-{\mu }_n-{\nu }_n)({\mu }_n+{\nu }_n){\parallel \frac{{\mu }_n}{{\mu }_n+{\nu }_n}(t_n-x_n)+\frac{{\nu }_n}{{\mu }_n+{\nu }_n}(S^n{t}_n-x_n)\parallel }^2\hfill \\ =(1-{\mu }_n-{\nu }_n){\parallel x_n-x^{\ast }\parallel }^2+({\mu }_n+{\nu }_n){\parallel \frac{{\mu }_n}{{\mu }_n+{\nu }_n}(t_n-x^{\ast })+\frac{{\nu }_n}{{\mu }_n+{\nu }_n}(S^n{t}_n-x^{\ast })\parallel }^2\hfill \\ -(1-{\mu }_n-{\nu }_n)({\mu }_n+{\nu }_n){\parallel \frac{1}{{\mu }_n+{\nu }_n}(x_{n+1}-x_n)\parallel }^2\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n(1+{\gamma }_n)\frac{{\parallel f(x^{\ast })-x^{\ast }\parallel }^2}{1-\rho }\hfill \\ +c_n-\frac{(1-{\mu }_n-{\nu }_n)}{{\mu }_n+{\nu }_n}{\parallel x_{n+1}-x_n\parallel }^2,\hfill \end{array}$$

which hence implies that

$$\begin{array}{c}\frac{(1-c)}{c}{\parallel x_{n+1}-x_n\parallel }^2\hfill \\ \le \frac{(1-{\mu }_n-{\nu }_n)}{{\mu }_n+{\nu }_n}{\parallel x_{n+1}-x_n\parallel }^2\hfill \\ \le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n(1+{\gamma }_n)\frac{{\parallel f(x^{\ast })-x^{\ast }\parallel }^2}{1-\rho }+c_n.\hfill \end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x^{\ast }\parallel$ exists, ${\alpha }_n\to 0$, ${\gamma }_n\to 0$, $c_n\to 0$ and the sequence $\{x_n\}$ is bounded, we obtain that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =0.$$

Also, since $x_{n+1}=(1-{\mu }_n-{\nu }_n)x_n+{\mu }_n{t}_n+{\nu }_n{S}^n{t}_n$, we have ${\nu }_n(S^n{t}_n-t_n)=(1-{\mu }_n-{\nu }_n)(t_n-x_n)+(x_{n+1}-t_n)$. Then

$$\begin{array}{rcl}\sigma \parallel S^n{t}_n-t_n\parallel & \le & {\nu }_n\parallel S^n{t}_n-t_n\parallel \\ \le & (1-{\mu }_n-{\nu }_n)\parallel t_n-x_n\parallel +\parallel x_{n+1}-t_n\parallel \\ \le & (1-{\mu }_n-{\nu }_n)\parallel t_n-x_n\parallel +\parallel x_{n+1}-x_n\parallel +\parallel x_n-t_n\parallel \\ \le & 2\parallel t_n-x_n\parallel +\parallel x_{n+1}-x_n\parallel ,\end{array}$$

and hence $\parallel t_n-S^n{t}_n\parallel \to 0$. Furthermore, observe that

$$\parallel x_n-S^n{x}_n\parallel \le \parallel x_n-t_n\parallel +\parallel t_n-S^n{t}_n\parallel +\parallel S^n{t}_n-S^n{x}_n\parallel .$$

(4.8)

Utilizing Lemma 2.6, we have

$$\parallel S^n{t}_n-S^n{x}_n\parallel \le \frac{1}{1-\kappa }(\kappa \parallel t_n-x_n\parallel +\sqrt{(1+(1-\kappa ){\gamma }_n){\parallel t_n-x_n\parallel }^2+(1-\kappa )c_n})$$

for every $n=1,2,\dots$ . Hence it follows from $\parallel x_n-t_n\parallel \to 0$ that $\parallel S^n{t}_n-S^n{x}_n\parallel \to 0$. Thus from (4.8) and $\parallel t_n-S^n{t}_n\parallel \to 0$ we get $\parallel x_n-S^n{x}_n\parallel \to 0$. Since $\parallel x_{n+1}-x_n\parallel \to 0$, $\parallel x_n-S^n{x}_n\parallel \to 0$ as $n\to \mathrm{\infty }$ and S is uniformly continuous, we obtain from Lemma 2.7 that $\parallel x_n-S{x}_n\parallel \to 0$ as $n\to \mathrm{\infty }$.

Further, repeating the same arguments as in the proof of Theorem 3.1, we can derive that ${\omega }_w(\{x_n\})\subset Fix(S)\cap \Xi \cap VI(C,A)$. Utilizing Lemma 2.11, from the existence of ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x^{\ast }\parallel$ for each $x^{\ast }\in Fix(S)\cap \Xi \cap VI(C,A)$, we infer that $\{x_n\}$ converges weakly to an element $\hat{x}\in Fix(S)\cap \Xi \cap VI(C,A)$. Since $\parallel x_n-y_n\parallel \to 0$ as $n\to \mathrm{\infty }$, it is clear that $\{y_n\}$ converges weakly to $\hat{x}\in Fix(S)\cap \Xi \cap VI(C,A)$. □

In the following, we present a numerical example to illustrate how Theorem 4.1 works.

Example 4.1 Let $H={\mathbf{R}}^2$ with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$ which are defined by

$$〈x,y〉=ac+bd\text{and}\parallel x\parallel =\sqrt{a^2+b^2}$$

for all $x,y\in {\mathbf{R}}^2$ with $x=(a,b)$ and $y=(c,d)$. Let $C=\{(a,a):a\in \mathbf{R}\}$. Clearly, C is a nonempty closed convex subset of a real Hilbert space $H={\mathbf{R}}^2$. Let $A:C\to H$ be α-inverse strongly monotone, let $B_i:C\to H$ be ${\beta }_i$-inverse strongly monotone for $i=1,2$, let $f:C\to C$ be a ρ-contraction with $\rho \in [0,1)$, and let $S:C\to C$ be a uniformly continuous asymptotically κ-strictly pseudocontractive mapping in the intermediate sense with sequence $\{{\gamma }_n\}$ such that $Fix(S)\cap \Xi \cap VI(C,A)$ is nonempty bounded; for instance, putting $A=\left[\begin{array}{cc}\frac{3}{5}& \frac{2}{5}\\ \frac{2}{5}& \frac{3}{5}\end{array}\right]$, $S=\left[\begin{array}{cc}\frac{2}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{2}{3}\end{array}\right]$, $B_1=I-A=\left[\begin{array}{cc}\frac{2}{5}& -\frac{2}{5}\\ -\frac{2}{5}& \frac{2}{5}\end{array}\right]$, $B_2=I-S=\left[\begin{array}{cc}\frac{1}{3}& -\frac{1}{3}\\ -\frac{1}{3}& \frac{1}{3}\end{array}\right]$ and $f=\frac{1}{2}S=\left[\begin{array}{cc}\frac{2}{6}& \frac{1}{6}\\ \frac{1}{6}& \frac{2}{6}\end{array}\right]$. It is easy to see that $\parallel A\parallel =\parallel S\parallel =1$, $\parallel f\parallel =\frac{1}{2}\parallel S\parallel =\frac{1}{2}$, and that A is α-inverse strongly monotone with $\alpha =\frac{1}{2}$, that $B_1$ and $B_2$ are $\frac{1}{2}$-inverse strongly monotone, f is a $\frac{1}{2}$-contraction, S is a nonexpansive mapping, i.e., a uniformly continuous asymptotically 0-strictly pseudocontractive mapping in the intermediate sense with sequences $\{{\gamma }_n\}$ (${\gamma }_n\equiv 0$) and $\{c_n\}$ ($c_n\equiv 0$). Moreover, it is clear that $Fix(S)=C$, $VI(C,A)=\{0\}$ and $\Xi =C$. Hence, $Fix(S)\cap \Xi \cap VI(C,A)=\{0\}$. In this case, from iterative scheme (4.1) in Theorem 4.1, we obtain that for any given $x_1\in C$,

$$\{\begin{array}{c}{y}_n=P_C(x_n-{\lambda }_{n}A{x}_n)=(1-{\lambda }_n)x_n,\hfill \\ t_n={\alpha }_{n}f(x_n)+(1-{\alpha }_n)P_C(I-{\mu }_1{B}_1)P_C(I-{\mu }_2{B}_2)y_n\hfill \\ \phantom{t_n}=\frac{1}{2}{\alpha }_{n}S{x}_n+(1-{\alpha }_n)P_C(I-{\mu }_1{B}_1)P_C(I-{\mu }_2{B}_2)(1-{\lambda }_n)x_n\hfill \\ \phantom{t_n}=\frac{1}{2}{\alpha }_n{x}_n+(1-{\alpha }_n)P_C(I-{\mu }_1{B}_1)(1-{\lambda }_n)x_n\hfill \\ \phantom{t_n}=\frac{1}{2}{\alpha }_n{x}_n+(1-{\alpha }_n)(1-{\lambda }_n)x_n\hfill \\ \phantom{t_n}=[\frac{1}{2}{\alpha }_n+(1-{\alpha }_n)(1-{\lambda }_n)]x_n,\hfill \\ x_{n+1}=(1-{\mu }_n-{\nu }_n)x_n+{\mu }_n{t}_n+{\nu }_n{S}^n{t}_n\hfill \\ \phantom{x_{n+1}}=(1-{\mu }_n-{\nu }_n)x_n+{\mu }_n[\frac{1}{2}{\alpha }_n+(1-{\alpha }_n)(1-{\lambda }_n)]x_n\hfill \\ \phantom{x_{n+1}=}+{\nu }_n{S}^n[\frac{1}{2}{\alpha }_n+(1-{\alpha }_n)(1-{\lambda }_n)]x_n\hfill \\ \phantom{x_{n+1}}=\{(1-{\mu }_n-{\nu }_n)+({\mu }_n+{\nu }_n)[\frac{1}{2}{\alpha }_n+(1-{\alpha }_n)(1-{\lambda }_n)]\}{x}_n\hfill \\ \phantom{x_{n+1}}=[1-({\mu }_n+{\nu }_n)(\frac{1}{2}{\alpha }_n+{\lambda }_n(1-{\alpha }_n))]x_n.\hfill \end{array}$$

Whenever ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n<\mathrm{\infty }$, $\{{\lambda }_n\}\subset [a,b]\subset (0,1)$, and $\{{\mu }_n+{\nu }_n\}\subset [\sigma ,c]\subset (0,1)$, we have

$$\begin{array}{rcl}\parallel x_{n+1}\parallel & =& [1-({\mu }_n+{\nu }_n)(\frac{1}{2}{\alpha }_n+{\lambda }_n(1-{\alpha }_n))]\parallel x_n\parallel \\ \le & [1-({\mu }_n+{\nu }_n)(\frac{1}{2}{\lambda }_n{\alpha }_n+{\lambda }_n(1-{\alpha }_n))]\parallel x_n\parallel \\ =& [1-({\mu }_n+{\nu }_n){\lambda }_n(1-\frac{1}{2}{\alpha }_n)]\parallel x_n\parallel \\ \le & [1-\sigma a(1-\frac{1}{2}{\alpha }_n)]\parallel x_n\parallel \\ \le & \parallel x_n\parallel exp\{-\sigma a(1-\frac{1}{2}{\alpha }_n)\}\\ \cdots \\ \le & \parallel x_n\parallel exp\{-n\sigma a+\frac{1}{2}\sigma a\sum _{i=1}^n{\alpha }_i\}\\ \to & 0\text{as }n\to \mathrm{\infty }.\end{array}$$

This shows that $\{x_n\}$ converges to the unique element 0 of $Fix(S)\cap \Xi \cap VI(C,A)$. Note that as $n\to \mathrm{\infty }$,

$$\parallel x_n-y_n\parallel ={\lambda }_n\parallel x_n\parallel \le \parallel x_n\parallel \to 0.$$

Hence, $\{y_n\}$ also converges to the unique element 0 of $Fix(S)\cap \Xi \cap VI(C,A)$.