The hybrid steepest descent method for solving variational inequality over triple hierarchical problems

Abstract

An explicit algorithm is introduced to solve the monotone variational inequality over a triple hierarchical problem. The strong convergence for the proposed algorithm to the solution is guaranteed under some assumptions. Our results extend those of Iiduka (Nonlinear Anal. 71:e1292-e1297, 2009), Marino and Xu (J. Optim. Theory Appl. 149(1):61-78, 2011), Yao et al. (Fixed Point Theory Appl. 2011:794203, 2011) and other authors.

MSC:46C05, 47H06, 47H09, 47H10, 47J20, 47J25, 65J15.

1 Introduction

Let C be a closed convex subset of a real Hilbert space H with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$. We denote weak convergence and strong convergence by notations ⇀ and →, respectively. A mapping A is a nonlinear mapping. The Hartmann-Stampacchia variational inequality [1] for finding $x\in C$ such that

$$〈Ax,y-x〉\ge 0,\mathrm{\forall }y\in C.$$

(1.1)

The set of solutions of (1.1) is denoted by $\mathit{VI}(C,A)$. The variational inequality has been extensively studied in the literature [2, 3].

Let $f:C\to C$ be called a ρ-contraction if there exists a constant $\rho \in [0,1)$ such that

$$\parallel f(x)-f(y)\parallel \le \rho \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

A mapping $T:C\to C$ is said to be nonexpansive if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

A point $x\in C$ is a fixed point of T provided $Tx=x$. Denote by $F(T)$ the set of fixed points of T; that is, $F(T)=\{x\in C:Tx=x\}$. If C is bounded closed convex and T is a nonexpansive mapping of C into itself, then $F(T)$ is nonempty [4].

We discuss the following variational inequality problem over the fixed point set of a nonexpansive mapping (see [5–13]), which is called a hierarchical problem. Consider a monotone, continuous mapping $A:H\to H$ and a nonexpansive mapping $T:H\to H$. Find $x^{\ast }\in \mathit{VI}(F(T),A)=\{x^{\ast }\in F(T):〈A{x}^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in F(T)\}$, where $F(T)\ne \mathrm{\varnothing }$. This solution set is denoted by Ξ.

We introduce the following variational inequality problem over the solution set of the variational inequality problem over the fixed point set of a nonexpansive mapping (see [14–18]), which is called a triple hierarchical problem. Consider an inverse-strongly monotone $A:H\to H$, a strongly monotone and Lipschitz continuous $B:H\to H$ and a nonexpansive mapping $T:H\to H$. Find $x^{\ast }\in \mathit{VI}(\mathrm{\Xi },B)=\{x^{\ast }\in \mathrm{\Xi }:〈B{x}^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in \mathrm{\Xi }\}$, where $\mathrm{\Xi }:=\mathit{VI}(F(T),A)\ne \mathrm{\varnothing }$.

A mapping $A:H\to H$ is said to be monotone if

$$〈Ax-Ay,x-y〉\ge 0,\mathrm{\forall }x,y\in H.$$

A mapping $A:H\to H$ is said to be α-strongly monotone if there exists a positive real number α such that

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

A mapping $A:H\to H$ is said to be β-inverse-strongly monotone if there exists a positive real number β such that

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

A mapping $A:H\to H$ is said to be L-Lipschitz continuous if there exists a positive real number L such that

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

A linear bounded operator A is said to be strongly positive on H if there exists a constant $\overline{\gamma }>0$ with the property

$$〈Ax,x〉\ge \overline{\gamma }{\parallel x\parallel }^2,\mathrm{\forall }x\in H.$$

In 2009, Iiduka [14] introduced an iterative algorithm for the following triple hierarchical constrained optimization problem. The sequence $\{x_n\}$ defined by the iterative method below, with the initial guess $x_1\in H$, is chosen arbitrarily,

$$\{\begin{array}{c}{y}_n=T(x_n-{\lambda }_n{A}_1{x}_n),\hfill \\ x_{n+1}=y_n-\mu {\alpha }_n{A}_2{y}_n,\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(1.2)

where ${\alpha }_n\in (0,1]$ and ${\lambda }_n\in (0,2\alpha ]$ satisfy certain conditions. Let $A_1:H\to H$ be an inverse-strongly monotone, $A_2:H\to H$ be a strongly monotone and Lipschitz continuous and $T:H\to H$ be a nonexpansive mapping, then the sequence converges strongly to the set solution of (1.2).

In 2011, Yao et al. [19] studied new algorithms. For $x_0\in C$ chosen arbitrarily, let the sequence $\{x_n\}$ be generated iteratively by

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)T{P}_C[I-{\alpha }_n(A-\gamma f)]x_n,\mathrm{\forall }n\ge 0,$$

where the sequences $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are two sequences in $[0,1]$. Then $\{x_n\}$ converges strongly to $x^{\ast }\in F(T)$ which is the unique solution of the variational inequality:

$$\text{Find }{x}^{\ast }\in F(T)\text{ such that }〈(A-\gamma f)x^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in F(T).$$

(1.3)

Let $A:C\to H$ be a strongly positive linear bounded operator, $f:C\to H$ be a ρ-contraction and $T:C\to C$ be a nonexpansive mapping satisfying some conditions. The solution set of (1.3) is denoted by $\mathrm{\Omega }:=\mathit{VI}(F(T),A-\gamma f)$.

Very recently, Marino and Xu [20] generated a sequence $\{x_n\}$ through the recursive formula

$$x_{n+1}={\lambda }_{n}f{x}_n+(1-{\lambda }_n)({\alpha }_{n}V{x}_n+(1-{\alpha }_n)T{x}_n),\mathrm{\forall }n\ge 0,$$

where f is a contraction on C, the initial guess $x_0\in C$ is arbitrary and $\{{\lambda }_n\}$, $\{{\alpha }_n\}$ are two sequences in $(0,1)$ and $T,V:C\to C$ are two nonexpansive self mappings. Strong convergence of the algorithm is proved under different circumstances of parameter selection.

In this paper, we introduce iterative algorithms and prove a strong convergence theorem for the following variational inequality over the triple hierarchical problem (1.4) below. Let $B:C\to C$ be a β-strongly monotone and L-Lipschitz continuous. Find $x^{\ast }\in \mathrm{\Omega }$ such that

$$〈B{x}^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in \mathrm{\Omega },$$

(1.4)

where $\mathrm{\Omega }:=\mathit{VI}(F(T),A-\gamma f)\ne \mathrm{\varnothing }$, T is a nonexpansive mapping, $A:C\to H$ is a strongly positive linear bounded operator and $f:C\to H$ is a ρ-contraction. This solution set of (1.4) is denoted by $\mathrm{\Upsilon }:=\mathit{VI}(\mathrm{\Omega },B):=\mathit{VI}(\mathit{VI}(F(T),A-\gamma f),B)$. Then the sequence $\{x_n\}$ strongly converges to the unique solution of (3.2) in Section 3.17 and we shall denote the set of such solutions by $\mathrm{\Theta }:=\mathit{VI}(\mathrm{\Upsilon },I-\varphi )$.

2 Preliminaries

Let H be a real Hilbert space and C be a nonempty closed convex subset of H. Recall that the (nearest point) projection $P_C$ from H onto C assigns, to each $x\in H$, the unique point in $P_{C}x\in C$ satisfying the property

$$\parallel x-P_{C}x\parallel =\underset{y\in C}{min}\parallel x-y\parallel .$$

The following characterizes the projection $P_C$. We recall some lemmas which will be needed in the rest of this paper.

Lemma 2.1 The function $u\in C$ is a solution of the variational inequality (1.1) if and only if $u\in C$ satisfies the relation $u=P_C(u-\lambda Au)$ for all $\lambda >0$.

Lemma 2.2 For a given $z\in H$, $u\in C$, $u=P_{C}z\iff 〈u-z,v-u〉\ge 0$, $\mathrm{\forall }v\in C$.

It is well known that $P_C$ is a firmly nonexpansive mapping of H onto C and satisfies

$${\parallel P_{C}x-P_{C}y\parallel }^2\le 〈P_{C}x-P_{C}y,x-y〉,\mathrm{\forall }x,y\in H.$$

(2.1)

Moreover, $P_{C}x$ is characterized by the following properties: $P_{C}x\in C$ and for all $x\in H$, $y\in C$,

$$〈x-P_{C}x,y-P_{C}x〉\le 0.$$

(2.2)

Lemma 2.3 The following inequality holds in an inner product space H:

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,x+y〉,\mathrm{\forall }x,y\in H.$$

Lemma 2.4 [21]

Let C be a closed convex subset of a real Hilbert space H and let $T:C\to C$ be a nonexpansive mapping. Then $I-T$ is demiclosed at zero, that is, $x_n\rightharpoonup x$ and $x_n-T{x}_n\to 0$ imply $x=Tx$.

Lemma 2.5 [22]

Assume A is a strongly positive linear bounded operator on a Hilbert space H with the coefficient $\overline{\gamma }>0$ and $0<\rho \le {\parallel A\parallel }^{-1}$, then $\parallel I-\rho A\parallel \le 1-\rho \overline{\gamma }$.

Lemma 2.6 [23]

Each Hilbert space H satisfies Opial’s condition, that is, for any sequence $\{x_n\}\subset H$ with $x_n\rightharpoonup x$, the inequality

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-x\parallel <\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-y\parallel$$

holds for each $y\in H$ with $y\ne x$.

Lemma 2.7 [24]

Let $\{x_n\}$ and $\{y_n\}$ be bounded sequences in a Banach space X and let $\{{\beta }_n\}$ be a sequence in $[0,1]$ with $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$. Suppose $x_{n+1}=(1-{\beta }_n)y_n+{\beta }_n{x}_n$ for all integers $n\ge 0$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}(\parallel y_{n+1}-y_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$. Then ${lim}_{n\to \mathrm{\infty }}\parallel y_n-x_n\parallel =0$.

Lemma 2.8 [10]

Let $B:H\to H$ be β-strongly monotone and L-Lipschitz continuous and let $\mu \in (0,\frac{2\beta }{L^2})$. For $\lambda \in [0,1]$, define $T_{\lambda }:H\to H$ by $T_{\lambda }(x):=x-\lambda \mu B(x)$ for all $x\in H$. Then, for all $x,y\in H$,

$$\parallel T_{\lambda }(x)-T_{\lambda }(y)\parallel \le (1-\lambda \tau )\parallel x-y\parallel$$

holds, where $\tau :=1-\sqrt{1-\mu (2\beta -\mu L^2)}\in (0,1]$.

Lemma 2.9 [25]

Assume $\{a_n\}$ is a sequence of nonnegative real numbers such that

$$a_{n+1}\le (1-{\gamma }_n)a_n+{\delta }_n,\mathrm{\forall }n\ge 0,$$

where $\{{\gamma }_n\}\subset (0,1)$ and $\{{\delta }_n\}$ is a sequence in ℛ such that

• (i) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }$,

• (ii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{{\delta }_n}{{\gamma }_n}\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_n|<\mathrm{\infty }$.

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

3 Strong convergence theorem

In this section, we introduce an iterative algorithm for solving the monotone variational inequality over a triple hierarchical problem.

Theorem 3.1 Let H be a real Hilbert space, C be a closed convex subset of H. Let $A:C\to H$ be a strongly positive linear bounded operator, $f:C\to H$ be a ρ-contraction, γ be a positive real number such that $\frac{\overline{\gamma }-1}{\rho }<\gamma <\frac{\overline{\gamma }}{\rho }$. Let $T:C\to C$ be a nonexpansive mapping, $B:C\to C$ be a β-strongly monotone and L-Lipschitz continuous. Let $\varphi :C\to C$ be a k-contraction mapping with $k\in [0,1)$. Assume that $\mathrm{\Upsilon }:=\mathit{VI}(\mathit{VI}(F(T),A-\gamma f),B)$ is nonempty set. Suppose $\{x_n\}$ is a sequence generated by the following algorithm $x_0\in C$ arbitrarily:

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

(3.1)

where $\{{\alpha }_n\},\{{\delta }_n\}\subset [0,1]$. If $\mu \in (0,\frac{2\beta }{L^2})$ is used and if $\{{\beta }_n\}\subset (0,1]$ satisfies the following conditions:

(C1): ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_{n+1}-{\delta }_n|<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{\delta }_n=\mathrm{\infty }$;

(C2): ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n+1}-{\beta }_n|<\mathrm{\infty }$;

(C3): ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty }$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

(C4): ${\delta }_n\le {\beta }_n$ and ${\beta }_n\le {\alpha }_n$.

Then $\{x_n\}$ converges strongly to $x^{\ast }\in \mathrm{\Upsilon }$, which is the unique solution of the variational inequality:

$$\mathit{\text{Find}}{x}^{\ast }\in \mathrm{\Upsilon }\mathit{\text{such that}}〈(I-\varphi )x^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in \mathrm{\Upsilon }.$$

(3.2)

Proof We will divide the proof into four steps.

Step 1. We will show $\{x_n\}$ is bounded. For any $q\in \mathrm{\Upsilon }$, we have

$$\begin{array}{rcl}\parallel z_n-q\parallel & =& \parallel T{P}_C[I-{\delta }_n(A-\gamma f)]x_n-T{P}_{C}q\parallel \\ \le & \parallel [I-{\delta }_n(A-\gamma f)]x_n-q\parallel \\ \le & {\delta }_n\parallel \gamma f(x_n)-\gamma f(q)\parallel +{\delta }_n\parallel \gamma f(q)-Aq\parallel +\parallel I-{\delta }_{n}A\parallel \parallel x_n-q\parallel \\ \le & {\delta }_n\gamma \rho \parallel x_n-q\parallel +{\delta }_n\parallel \gamma f(q)-Aq\parallel +(1-{\delta }_n\overline{\gamma })\parallel x_n-q\parallel \\ =& [1-(\overline{\gamma }-\gamma \rho ){\delta }_n]\parallel x_n-q\parallel +{\delta }_n\parallel \gamma f(q)-Aq\parallel .\end{array}$$

(3.3)

By Lemma 2.8, it is found that

$$\begin{array}{rcl}\parallel y_n-q\parallel & =& \parallel (I-\mu {\beta }_{n}B)z_n-(I-\mu {\beta }_{n}B)q\parallel \\ \le & (1-{\beta }_n\tau )\parallel z_n-q\parallel \\ \le & (1-{\beta }_n\tau )\{[1-(\overline{\gamma }-\gamma \rho ){\delta }_n]\parallel x_n-q\parallel +{\delta }_n\parallel \gamma f(q)-Aq\parallel \}.\end{array}$$

(3.4)

From (3.1), we get

$$\begin{array}{rcl}\parallel x_{n+1}-q\parallel & \le & {\alpha }_n\parallel \varphi (x_n)-\varphi (q)\parallel +{\alpha }_n\parallel \varphi (q)-q\parallel +(1-{\alpha }_n)\parallel y_n-q\parallel \\ \le & {\alpha }_{n}k\parallel x_n-q\parallel +{\alpha }_n\parallel \varphi (q)-\varphi (q)\parallel +(1-{\alpha }_n)\parallel y_n-q\parallel \\ \le & {\alpha }_{n}k\parallel x_n-q\parallel +(1-{\alpha }_n)(1-{\beta }_n\tau )\\ \times \{[1-(\overline{\gamma }-\gamma \rho ){\delta }_n]\parallel x_n-q\parallel +{\delta }_n\parallel \gamma f(q)-Aq\parallel \}\\ \le & {\alpha }_n\parallel x_n-q\parallel +(1-{\alpha }_n)(1-{\beta }_n\tau )[1-(\overline{\gamma }-\gamma \rho ){\delta }_n]\parallel x_n-q\parallel \\ +(1-{\alpha }_n)(1-{\beta }_n\tau ){\delta }_n\parallel \gamma f(q)-Aq\parallel \\ =& {\alpha }_n\parallel x_n-q\parallel +(1-{\alpha }_n)[1-(\overline{\gamma }-\gamma \rho ){\delta }_n-{\beta }_n\tau +{\beta }_n\tau (\overline{\gamma }-\gamma \rho ){\delta }_n]\parallel x_n-q\parallel \\ +(1-{\alpha }_n)(1-{\beta }_n\tau ){\delta }_n\parallel \gamma f(q)-Aq\parallel \\ =& {\alpha }_n\parallel x_n-q\parallel +(1-{\alpha }_n)[1-\{(\overline{\gamma }-\gamma \rho ){\delta }_n+{\beta }_n\tau -{\beta }_n\tau (\overline{\gamma }-\gamma \rho ){\delta }_n\}]\parallel x_n-q\parallel \\ +(1-{\alpha }_n)(1-{\beta }_n\tau ){\delta }_n\parallel \gamma f(q)-Aq\parallel \\ =& {\alpha }_n\parallel x_n-q\parallel \\ +[1-{\alpha }_n-\{(\overline{\gamma }-\gamma \rho ){\delta }_n+{\beta }_n\tau -{\beta }_n\tau (\overline{\gamma }-\gamma \rho ){\delta }_n\}(1-{\alpha }_n)]\parallel x_n-q\parallel \\ +(1-{\alpha }_n)(1-{\beta }_n\tau ){\delta }_n\parallel \gamma f(q)-Aq\parallel \\ =& [1-(1-{\alpha }_n)\{(\overline{\gamma }-\gamma \rho ){\delta }_n+{\beta }_n\tau -{\beta }_n\tau (\overline{\gamma }-\gamma \rho ){\delta }_n\}]\parallel x_n-q\parallel \\ +(1-{\alpha }_n)(1-{\beta }_n\tau ){\delta }_n\parallel \gamma f(q)-Aq\parallel \\ =& [1-(1-{\alpha }_n)\{(\overline{\gamma }-\gamma \rho ){\delta }_n(1-{\beta }_n\tau )+{\beta }_n\tau \}]\parallel x_n-q\parallel \\ +(1-{\alpha }_n)(1-{\beta }_n\tau ){\delta }_n\parallel \gamma f(q)-Aq\parallel \\ =& [1-(1-{\alpha }_n)(\overline{\gamma }-\gamma \rho ){\delta }_n(1-{\beta }_n\tau )-(1-{\alpha }_n){\beta }_n\tau ]\parallel x_n-q\parallel \\ +(1-{\alpha }_n)(1-{\beta }_n\tau ){\delta }_n\parallel \gamma f(q)-Aq\parallel \\ =& [1-(1-{\alpha }_n)(\overline{\gamma }-\gamma \rho ){\delta }_n(1-{\beta }_n\tau )]\parallel x_n-q\parallel -(1-{\alpha }_n){\beta }_n\tau \parallel x_n-q\parallel \\ +(1-{\alpha }_n)(1-{\beta }_n\tau ){\delta }_n\parallel \gamma f(q)-Aq\parallel \\ \le & [1-(\overline{\gamma }-\gamma \rho )(1-{\alpha }_n)(1-{\beta }_n\tau ){\delta }_n]\parallel x_n-q\parallel \\ +(\overline{\gamma }-\gamma \rho )(1-{\alpha }_n)(1-{\beta }_n\tau ){\delta }_n\frac{\parallel \gamma f(q)-Aq\parallel }{\overline{\gamma }-\gamma \rho }\\ =& (1-{\sigma }_n)\parallel x_n-q\parallel +{\sigma }_n\frac{\parallel \gamma f(q)-Aq\parallel }{\overline{\gamma }-\gamma \rho },\end{array}$$

where ${\sigma }_n:=(\overline{\gamma }-\gamma \rho )(1-{\alpha }_n)(1-{\beta }_n\tau ){\delta }_n$. Then the mathematical induction implies that

$$\parallel x_n-q\parallel \le max\{\parallel x_0-q\parallel ,\frac{\parallel \gamma f(q)-Aq\parallel }{\overline{\gamma }-\gamma \rho }\},\mathrm{\forall }n\ge 0.$$

Therefore, $\{x_n\}$ is bounded and so are $\{y_n\}$, $\{z_n\}$, $\{A{x}_n\}$, $\{B{x}_n\}$, $\{\varphi (x_n)\}$ and $\{f(x_n)\}$.

Step 2. We claim that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$. From (3.1), we have

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel & =& \parallel T{P}_C[I-{\delta }_{n+1}(A-\gamma f)]x_{n+1}-T{P}_C[I-{\delta }_n(A-\gamma f)]x_n\parallel \\ \le & \parallel P_C[I-{\delta }_{n+1}(A-\gamma f)]x_{n+1}-P_C[I-{\delta }_n(A-\gamma f)]x_n\parallel \\ \le & \parallel [I-{\delta }_{n+1}(A-\gamma f)]x_{n+1}-[I-{\delta }_n(A-\gamma f)]x_n\parallel \\ =& \parallel {\delta }_{n+1}(\gamma f(x_{n+1})-\gamma f(x_n))+({\delta }_{n+1}-{\delta }_n)\gamma f(x_n)+(I-{\delta }_{n+1}A)(x_{n+1}-x_n)\\ +({\delta }_n-{\delta }_{n+1})A{x}_n\parallel \\ \le & {\delta }_{n+1}\gamma \parallel f(x_{n+1})-f(x_n)\parallel +(1-{\delta }_{n+1}\overline{\gamma })\parallel x_{n+1}-x_n\parallel \\ +|{\delta }_{n+1}-{\delta }_n|(\parallel \gamma f(x_n)\parallel +\parallel A{x}_n\parallel )\\ \le & {\delta }_{n+1}\gamma \rho \parallel x_{n+1}-x_n\parallel +(1-{\delta }_{n+1}\overline{\gamma })\parallel x_{n+1}-x_n\parallel \\ +|{\delta }_{n+1}-{\delta }_n|(\parallel \gamma f(x_n)\parallel +\parallel A{x}_n\parallel )\\ =& [1-(\overline{\gamma }-\gamma \rho ){\delta }_{n+1}]\parallel x_{n+1}-x_n\parallel +|{\delta }_{n+1}-{\delta }_n|(\parallel \gamma f(x_n)\parallel +\parallel A{x}_n\parallel )\end{array}$$

(3.5)

and

$$\begin{array}{rcl}\parallel y_{n+1}-y_n\parallel & =& \parallel (I-\mu {\beta }_{n+1}B)z_{n+1}-(I-\mu {\beta }_{n}B)z_n\parallel \\ \le & \parallel (I-\mu {\beta }_{n+1}B)z_{n+1}-(I-\mu {\beta }_{n+1}B)z_n\parallel \\ +\parallel (I-\mu {\beta }_{n+1}B)z_n-(I-\mu {\beta }_{n}B)z_n\parallel \\ \le & (1-{\beta }_n\tau )\parallel z_{n+1}-z_n\parallel +\mu |{\beta }_{n+1}-{\beta }_n|\parallel B{z}_n\parallel .\end{array}$$

(3.6)

Using (3.5) and (3.6), we get

$$\begin{array}{rcl}\parallel x_{n+2}-x_{n+1}\parallel & =& \parallel {\alpha }_{n+1}\varphi (x_{n+1})+(1-{\alpha }_{n+1})y_{n+1}-{\alpha }_n\varphi (x_n)-(1-{\alpha }_n)y_n\parallel \\ \le & {\alpha }_{n+1}\parallel \varphi (x_{n+1})-\varphi (x_n)\parallel +|{\alpha }_{n+1}-{\alpha }_n|\parallel \varphi (x_{n+1})\parallel +(1-{\alpha }_{n+1})\parallel y_{n+1}-y_n\parallel \\ +|{\alpha }_{n+1}-{\alpha }_n|\parallel y_n\parallel \\ \le & {\alpha }_{n+1}k\parallel x_{n+1}-x_n\parallel +|{\alpha }_{n+1}-{\alpha }_n|(\parallel \varphi (x_{n+1})\parallel +\parallel y_n\parallel )\\ +(1-{\alpha }_{n+1})\parallel y_{n+1}-y_n\parallel \\ \le & {\alpha }_{n+1}k\parallel x_{n+1}-x_n\parallel +|{\alpha }_{n+1}-{\alpha }_n|(\parallel \varphi (x_{n+1})\parallel +\parallel y_n\parallel )\\ +(1-{\alpha }_{n+1})\{(1-{\beta }_n\tau )\parallel z_{n+1}-z_n\parallel +\mu |{\beta }_{n+1}-{\beta }_n|\parallel B{z}_n\parallel \}\\ =& {\alpha }_{n+1}k\parallel x_{n+1}-x_n\parallel +|{\alpha }_{n+1}-{\alpha }_n|(\parallel \varphi (x_{n+1})\parallel +\parallel y_n\parallel )\\ +(1-{\alpha }_{n+1})(1-{\beta }_n\tau )\parallel z_{n+1}-z_n\parallel +(1-{\alpha }_{n+1})\mu |{\beta }_{n+1}-{\beta }_n|\parallel B{z}_n\parallel \\ \le & {\alpha }_{n+1}\parallel x_{n+1}-x_n\parallel +|{\alpha }_{n+1}-{\alpha }_n|(\parallel \varphi (x_{n+1})\parallel +\parallel y_n\parallel )\\ +(1-{\alpha }_{n+1})\mu |{\beta }_{n+1}-{\beta }_n|\parallel B{z}_n\parallel \\ +(1-{\alpha }_{n+1})\{[1-(\overline{\gamma }-\gamma \rho ){\delta }_{n+1}]\parallel x_{n+1}-x_n\parallel \\ +|{\delta }_{n+1}-{\delta }_n|(\parallel \gamma f(x_n)\parallel +\parallel A{x}_n\parallel )\}\\ \le & {\alpha }_{n+1}\parallel x_{n+1}-x_n\parallel +|{\alpha }_{n+1}-{\alpha }_n|(\parallel \varphi (x_{n+1})\parallel +\parallel y_n\parallel )\\ +(1-{\alpha }_{n+1})\mu |{\beta }_{n+1}-{\beta }_n|\parallel B{z}_n\parallel \\ +(1-{\alpha }_{n+1})[1-(\overline{\gamma }-\gamma \rho ){\delta }_{n+1}]\parallel x_{n+1}-x_n\parallel \\ +(1-{\alpha }_{n+1})|{\delta }_{n+1}-{\delta }_n|(\parallel \gamma f(x_n)\parallel +\parallel A{x}_n\parallel )\\ \le & [1-(\overline{\gamma }-\gamma \rho ){\delta }_{n+1}(1-{\alpha }_{n+1})]\parallel x_{n+1}-x_n\parallel \\ +|{\alpha }_{n+1}-{\alpha }_n|(\parallel \varphi (x_{n+1})\parallel +\parallel y_n\parallel )\\ +\mu |{\beta }_{n+1}-{\beta }_n|\parallel B{z}_n\parallel +|{\delta }_{n+1}-{\delta }_n|(\parallel \gamma f(x_n)\parallel +\parallel A{x}_n\parallel )\\ \le & [1-(\overline{\gamma }-\gamma \rho ){\delta }_{n+1}(1-{\alpha }_{n+1})]\parallel x_{n+1}-x_n\parallel \\ +\{|{\alpha }_{n+1}-{\alpha }_n|+|{\beta }_{n+1}-{\beta }_n|+|{\delta }_{n+1}-{\delta }_n|\}M,\end{array}$$

where M is some constant such that

$$\underset{n\ge 0}{sup}\{\parallel \varphi (x_n)\parallel +\parallel y_n\parallel ,\mu \parallel B{z}_n\parallel ,\parallel \gamma f(x_n)\parallel +\parallel A{x}_n\parallel \}\le M.$$

From (C1)-(C3) and the boundedness of $\{x_n\}$, $\{y_n\}$, $\{A{x}_n\}$, $\{B{z}_n\}$, $\{\varphi (x_n)\}$ and $\{f(x_n)\}$, by Lemma 2.9, we have

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

(3.7)

On the other hand, we note that

$$\begin{array}{rcl}\parallel z_n-T{x}_n\parallel & =& \parallel T{P}_C[I-{\delta }_n(A-\gamma f)]x_n-T{x}_n\parallel \\ =& \parallel T{P}_C[I-{\delta }_n(A-\gamma f)]x_n-T{P}_C{x}_n\parallel \\ \le & \parallel [I-{\delta }_n(A-\gamma f)]x_n-x_n\parallel \\ \le & {\delta }_n\parallel (A-\gamma f)]x_n\parallel ,\end{array}$$

by (C3)-(C4) and it follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel z_n-T{x}_n\parallel =0.$$

(3.8)

From (3.1), we compute

$$\begin{array}{rcl}\parallel x_{n+1}-z_n\parallel & =& \parallel {\alpha }_n\varphi (x_n)+(1-{\alpha }_n)y_n-z_n\parallel \\ =& \parallel {\alpha }_n\varphi (x_n)+(1-{\alpha }_n)(I-\mu {\beta }_{n}B)z_n-z_n\parallel \\ \le & {\alpha }_n\parallel \varphi (x_n)-z_n\parallel +(1-{\alpha }_n)\parallel (I-\mu {\beta }_{n}B)z_n-z_n\parallel \\ \le & {\alpha }_{n}k\parallel x_n-z_n\parallel +{\alpha }_n\parallel \varphi (z_n)-z_n\parallel +(1-{\alpha }_n)\mu {\beta }_n\parallel B{z}_n\parallel .\end{array}$$

By (C3) and (C4), it follows that

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

(3.9)

Since

$$\parallel x_n-T{x}_n\parallel \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-z_n\parallel +\parallel z_n-T{x}_n\parallel .$$

By (3.7), (3.8) and (3.9), we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T{x}_n\parallel =0.$$

(3.10)

From (3.1), we compute

$$\begin{array}{rcl}\parallel x_{n+1}-y_n\parallel & =& \parallel {\alpha }_n\varphi (x_n)+(1-{\alpha }_n)y_n-y_n\parallel \\ =& \parallel {\alpha }_n\varphi (x_n)+y_n-{\alpha }_n{y}_n-y_n\parallel \\ \le & {\alpha }_n\parallel \varphi (x_n)-y_n\parallel .\end{array}$$

(3.11)

By (C3), it follows that

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

(3.12)

Since

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

From (3.7) and (3.12), we obtain

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

(3.13)

Step 3. First, ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈u_n-x^{\ast },\gamma f(x^{\ast })-A{x}^{\ast }〉\le 0$ is proven. Choose a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x_n-x^{\ast },\gamma f\left(x^{\ast }\right)-A{x}^{\ast }〉=\underset{i\to \mathrm{\infty }}{lim}〈x_{n_i}-x^{\ast },\gamma f\left(x^{\ast }\right)-A{x}^{\ast }〉.$$

The boundedness of $\{x_{n_i}\}$ implies the existences of a subsequence $\{x_{n_{i_j}}\}$ of $\{x_{n_i}\}$ and a point $\hat{x}\in H$ such that $\{x_{n_{i_j}}\}$ converges weakly to $\hat{x}$. We may assume, without loss of generality, that ${lim}_{i\to \mathrm{\infty }}〈x_{n_i},w〉=〈\hat{x},w〉$, $w\in H$. Assume $\hat{x}\ne T(\hat{x})$. ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$ with $F(T)\ne \mathrm{\varnothing }$ guarantees that

which is a contradiction. Therefore, $\hat{x}\in F(T)$. From $x^{\ast }\in \mathit{VI}(F(T),A-\gamma f)$, we find

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x_n-x^{\ast },\gamma f\left(x^{\ast }\right)-A{x}^{\ast }〉& =& \underset{i\to \mathrm{\infty }}{lim}〈x_{n_i}-x^{\ast },\gamma f\left(x^{\ast }\right)-A{x}^{\ast }〉\\ =& 〈\hat{x}-x^{\ast },\gamma f\left(x^{\ast }\right)-A{x}^{\ast }〉\\ \le & 0.\end{array}$$

Setting $u_n=[I-{\delta }_n(A-\gamma f)]x_n$, by (C3)-(C4), we notice that

$$\parallel u_n-x_n\parallel \le {\delta }_n\parallel (A-\gamma f)\parallel \to 0.$$

Hence, we get

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u_n-x^{\ast },\gamma f\left(x^{\ast }\right)-A{x}^{\ast }〉\le 0.$$

(3.14)

Second, ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈x^{\ast }-x_{n+1},B{x}^{\ast }〉\le 0$ is proven. ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$ guarantees the existences of a subsequence $\{x_{n_k+1}\}$ of $\{x_{n_k}\}$ and a point $\overline{x}\in H$ such that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈x^{\ast }-x_{n+1},B{x}^{\ast }〉={lim}_{k\to \mathrm{\infty }}〈x^{\ast }-x_{n_k+1},B{x}^{\ast }〉$ and ${lim}_{k\to \mathrm{\infty }}〈x_{n_k},w〉={lim}_{k\to \mathrm{\infty }}〈x_{n_k+1},w〉=〈\overline{x},w〉$, $w\in H$. By the same discussion as in the proof of $\hat{x}\in F(T)$, we have $\overline{x}\in F(T)$. Let $y\in F(T)$ be fixed arbitrarily. Then it follows that $T:C\to C$ is a nonexpansive mapping with $F(T)\ne \mathrm{\varnothing }$, $A:C\to H$ is a strongly positive linear bounded operator and $f:C\to H$ is a contraction for all $n\in \mathbb{N}$. From (3.1),

$$\begin{array}{rcl}\parallel z_n-y\parallel & =& \parallel T{P}_C{u}_n-T{P}_{C}y\parallel \\ \le & \parallel u_n-y\parallel .\end{array}$$

(3.15)

By (C3)-(C4), we observe that

$$\begin{array}{rcl}\parallel u_n-y\parallel & =& \parallel [I-{\delta }_n(A-\gamma f)]x_n-y\parallel \\ \le & \parallel x_n-y\parallel +{\delta }_n\parallel (A-\gamma f)x_n\parallel \\ \le & \parallel x_n-y\parallel .\end{array}$$

(3.16)

Using (3.15) and (3.16),

$$\begin{array}{rcl}{\parallel u_n-y\parallel }^2& =& {\parallel [I-{\delta }_n(A-\gamma f)]x_n-y\parallel }^2\\ =& {\parallel {\delta }_n(\gamma f(x_n)-Ay)+(I-{\delta }_{n}A)(x_n-y)\parallel }^2\\ \le & {(1-{\delta }_n\overline{\gamma })}^2{\parallel x_n-y\parallel }^2+2{\delta }_n〈\gamma f(x_n)-Ay,u_n-y〉\\ \le & (1-2{\delta }_n\overline{\gamma }+{\delta }_n^2{\overline{\gamma }}^2){\parallel x_n-y\parallel }^2+2{\delta }_n\gamma \rho \parallel x_n-y\parallel \parallel u_n-y\parallel \\ +2{\delta }_n〈\gamma f(y)-Ay,u_n-y〉\\ \le & (1-2{\delta }_n\overline{\gamma }+{\delta }_n^2{\overline{\gamma }}^2){\parallel x_n-y\parallel }^2+2{\delta }_n\gamma \rho {\parallel x_n-y\parallel }^2+2{\delta }_n〈\gamma f(y)-Ay,u_n-y〉\\ =& [1-2{\delta }_n(\overline{\gamma }-\gamma \rho )]{\parallel x_n-y\parallel }^2+{\delta }_n^2{\overline{\gamma }}^2{\parallel x_n-y\parallel }^2+2{\delta }_n〈\gamma f(y)-Ay,u_n-y〉,\end{array}$$

which implies that

$$\begin{array}{rcl}0& \le & ({\parallel x_n-y\parallel }^2-{\parallel u_n-y\parallel }^2)-2{\delta }_n(\overline{\gamma }-\gamma \rho ){\parallel x_n-y\parallel }^2+{\delta }_n^2{\overline{\gamma }}^2{\parallel x_n-y\parallel }^2\\ +2{\delta }_n〈\gamma f(y)-Ay,u_n-y〉\\ =& (\parallel x_n-y\parallel +\parallel u_n-y\parallel )(\parallel x_n-y\parallel -\parallel u_n-y\parallel )\\ -2{\delta }_n(\overline{\gamma }-\gamma \rho ){\parallel x_n-y\parallel }^2+{\delta }_n^2{\overline{\gamma }}^2{\parallel x_n-y\parallel }^2\\ +2{\delta }_n〈\gamma f(y)-Ay,u_n-y〉\\ \le & M_2\parallel x_n-u_n\parallel -2{\delta }_n(\overline{\gamma }-\gamma \rho ){\parallel x_n-y\parallel }^2+{\delta }_n^2{\overline{\gamma }}^2{\parallel x_n-y\parallel }^2+2{\delta }_n〈\gamma f(y)-Ay,u_n-y〉,\end{array}$$

where $M_2:=sup\{\parallel x_n-y\parallel +\parallel u_n-y\parallel :n\in \mathbb{N}\}<\mathrm{\infty }$ for every $n\in \mathbb{N}$. By the weak convergence of $\{u_{n_i}\}$ to $\overline{x}\in F(T)$, ${lim}_{n\to \mathrm{\infty }}\parallel u_n-x_n\parallel =0$ and (C3)-(C4), we get $〈(\gamma f-A)y,\overline{x}-y〉\le 0$ for all $y\in F(T)$. A mapping A being a strongly positive linear bounded operator and f being a contraction ensure $〈(\gamma f-A)y,\overline{x}-y〉\le 0$ for all $y\in F(T)$, that is, $\overline{x}\in \mathit{VI}(F(T),A-\gamma f)$. Thus, $x^{\ast }\in \mathit{VI}(\mathit{VI}(F(T),A-\gamma f),B)$, we have

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x^{\ast }-x_n,B{x}^{\ast }〉& =& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x^{\ast }-x_{n_i},B{x}^{\ast }〉\\ =& 〈x^{\ast }-\overline{x},B{x}^{\ast }〉\\ \le & 0.\end{array}$$

From (3.13), we notice that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x^{\ast }-y_n,B{x}^{\ast }〉\le 0.$$

(3.17)

Third, ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈x_n-x^{\ast },\varphi (x^{\ast })-x^{\ast }〉\le 0$ is proven. Choose a subsequence $\{x_{n_g}\}$ of $\{x_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x_n-x^{\ast },\varphi \left(x^{\ast }\right)-x^{\ast }〉=\underset{g\to \mathrm{\infty }}{lim}〈x_{n_g}-x^{\ast },\varphi \left(x^{\ast }\right)-x^{\ast }〉.$$

The boundedness of $\{x_{n_g}\}$ implies the existence of a subsequence $\{x_{n_{g_h}}\}$ of $\{x_{n_g}\}$ and a point $\tilde{x}\in H$ such that $\{x_{n_{g_h}}\}$ converges weakly to $\tilde{x}$. By ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, we have ${lim}_{h\to \mathrm{\infty }}〈x_{n_{g_h}+1},w〉=〈\tilde{x},w〉$, $w\in H$. We may assume, without loss of generality, that ${lim}_{i\to \mathrm{\infty }}〈x_{n_g},w〉=〈\tilde{x},w〉$, $w\in H$. Assume $\tilde{x}\ne T(\tilde{x})$. ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$ with $F(T)\ne \mathrm{\varnothing }$ guarantees that

$$\begin{array}{rcl}\underset{g\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_g}-\tilde{x}\parallel & <& \underset{g\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_g}-T(\tilde{x})\parallel \\ =& \underset{g\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_g}-T(x_{n_g})+T(x_{n_g})-T(\tilde{x})\parallel \\ =& \underset{g\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel T(x_{n_g})-T(\tilde{x})\parallel \\ \le & \underset{g\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_g}-\tilde{x}\parallel .\end{array}$$

This is a contradiction, that is, $\tilde{x}\in F(T)$. From $x^{\ast }\in \mathit{VI}(\mathit{VI}(\mathit{VI}(F(T),A-\gamma f),B),I-\varphi )$, we find

(3.18)

Step 4. Finally, we prove ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x^{\ast }\parallel =0$. By Lemma 2.8, we compute

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& =& {\parallel {\alpha }_n\varphi (x_n)+(1-{\alpha }_n)y_n-x^{\ast }\parallel }^2\\ =& {\parallel {\alpha }_n(\varphi (x_n)-\varphi \left(x^{\ast }\right))+{\alpha }_n(\varphi \left(x^{\ast }\right)-x^{\ast })+(1-{\alpha }_n)(y_n-x^{\ast })\parallel }^2\\ \le & {\alpha }_n{\parallel \varphi (x_n)-\varphi \left(x^{\ast }\right)\parallel }^2+(1-{\alpha }_n){\parallel y_n-x^{\ast }\parallel }^2+2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ \le & {\alpha }_n{k}^2{\parallel x_n-x^{\ast }\parallel }^2+(1-{\alpha }_n){\parallel (I-\mu {\beta }_{n}B)z_n-x^{\ast }\parallel }^2\\ +2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ =& {\alpha }_n{k}^2{\parallel x_n-x^{\ast }\parallel }^2+(1-{\alpha }_n){\parallel (z_n-\mu {\beta }_{n}B{z}_n)-(x^{\ast }-\mu {\beta }_{n}B{x}^{\ast })-\mu {\beta }_{n}B{x}^{\ast }\parallel }^2\\ +2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ \le & {\alpha }_n{k}^2{\parallel x_n-x^{\ast }\parallel }^2+(1-{\alpha }_n)\\ \times \{{\parallel (z_n-\mu {\beta }_{n}B{z}_n)-(x^{\ast }-\mu {\beta }_{n}B{x}^{\ast })\parallel }^2+2\mu {\beta }_n〈x^{\ast }-y_n,B{x}^{\ast }〉\}\\ +2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ \le & {\alpha }_n{k}^2{\parallel x_n-x^{\ast }\parallel }^2+(1-{\alpha }_n){(1-\tau {\beta }_n)}^2{\parallel z_n-x^{\ast }\parallel }^2+2\mu {\beta }_n〈x^{\ast }-y_n,B{x}^{\ast }〉\\ +2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ \le & {\alpha }_n{k}^2{\parallel x_n-x^{\ast }\parallel }^2+(1-{\alpha }_n)(1-\tau {\beta }_n){\parallel u_n-x^{\ast }\parallel }^2+2\mu {\beta }_n〈x^{\ast }-y_n,B{x}^{\ast }〉\\ +2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ =& {\alpha }_n{k}^2{\parallel x_n-x^{\ast }\parallel }^2+(1-{\alpha }_n)(1-\tau {\beta }_n){\parallel [I-{\delta }_n(A-\gamma f)]x_n-x^{\ast }\parallel }^2\\ +2\mu {\beta }_n〈x^{\ast }-y_n,B{x}^{\ast }〉+2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ =& {\alpha }_n{k}^2{\parallel x_n-x^{\ast }\parallel }^2+(1-{\alpha }_n)(1-\tau {\beta }_n)\\ \times {\parallel (I-{\delta }_{n}A)(x_n-x^{\ast })+{\delta }_n(\gamma f(x_n)-A{x}^{\ast })\parallel }^2\\ +2\mu {\beta }_n〈x^{\ast }-y_n,B{x}^{\ast }〉+2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ \le & {\alpha }_n{k}^2{\parallel x_n-x^{\ast }\parallel }^2+(1-{\alpha }_n)(1-\tau {\beta }_n)\\ \times \{{(1-{\delta }_n\overline{\gamma })}^2{\parallel x_n-x^{\ast }\parallel }^2+2{\delta }_n〈\gamma f(x_n)-A{x}^{\ast },u_n-x^{\ast }〉\}\\ +2\mu {\beta }_n〈x^{\ast }-y_n,B{x}^{\ast }〉+2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ \le & {\alpha }_{n}k{\parallel x_n-x^{\ast }\parallel }^2+(1-{\alpha }_n)(1-\tau {\beta }_n)\{(1-2{\delta }_n\overline{\gamma }+{\delta }_n^2{\overline{\gamma }}^2){\parallel x_n-x^{\ast }\parallel }^2\\ +2{\delta }_n〈\gamma f(x_n)-\gamma f\left(x^{\ast }\right),u_n-x^{\ast }〉+2{\delta }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },u_n-x^{\ast }〉\}\\ +2\mu {\beta }_n〈x^{\ast }-y_n,B{x}^{\ast }〉+2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ \le & {\alpha }_{n}k{\parallel x_n-x^{\ast }\parallel }^2+(1-{\alpha }_n)(1-\tau {\beta }_n)\{(1-2{\delta }_n\overline{\gamma }){\parallel x_n-x^{\ast }\parallel }^2\\ +{\delta }_n^2{\overline{\gamma }}^2{\parallel x_n-x^{\ast }\parallel }^2+2{\delta }_n\gamma \rho \parallel x_n-x^{\ast }\parallel \parallel u_n-x^{\ast }\parallel \\ +2{\delta }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },u_n-x^{\ast }〉\}\\ +2\mu {\beta }_n〈x^{\ast }-y_n,B{x}^{\ast }〉+2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ \le & {\alpha }_n{\parallel x_n-x^{\ast }\parallel }^2+(1-{\alpha }_n)(1-\tau {\beta }_n)[1-2{\delta }_n(\overline{\gamma }-\gamma \rho )]{\parallel x_n-x^{\ast }\parallel }^2\\ +(1-{\alpha }_n)(1-\tau {\beta }_n){\delta }_n^2{\overline{\gamma }}^2{\parallel x_n-x^{\ast }\parallel }^2+2{\delta }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },u_n-x^{\ast }〉\\ +2\mu {\beta }_n〈x^{\ast }-y_n,B{x}^{\ast }〉+2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ =& {\alpha }_n{\parallel x_n-x^{\ast }\parallel }^2\\ +(1-{\alpha }_n)[1-2{\delta }_n(\overline{\gamma }-\gamma \rho )-\tau {\beta }_n+\tau {\beta }_{n}2{\delta }_n(\overline{\gamma }-\gamma \rho )]{\parallel x_n-x^{\ast }\parallel }^2\\ +(1-{\alpha }_n)(1-\tau {\beta }_n){\delta }_n^2{\overline{\gamma }}^2{\parallel x_n-x^{\ast }\parallel }^2+2{\delta }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },u_n-x^{\ast }〉\\ +2\mu {\beta }_n〈x^{\ast }-y_n,B{x}^{\ast }〉+2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ =& {\alpha }_n{\parallel x_n-x^{\ast }\parallel }^2\\ +(1-{\alpha }_n)[1-\{2{\delta }_n(\overline{\gamma }-\gamma \rho )+\tau {\beta }_n-\tau {\beta }_{n}2{\delta }_n(\overline{\gamma }-\gamma \rho )\}]{\parallel x_n-x^{\ast }\parallel }^2\\ +(1-{\alpha }_n)(1-\tau {\beta }_n){\delta }_n^2{\overline{\gamma }}^2{\parallel x_n-x^{\ast }\parallel }^2+2{\delta }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },u_n-x^{\ast }〉\\ +2\mu {\beta }_n〈x^{\ast }-y_n,B{x}^{\ast }〉+2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ =& [1-(1-{\alpha }_n)\{2{\delta }_n(\overline{\gamma }-\gamma \rho )+\tau {\beta }_n-\tau {\beta }_{n}2{\delta }_n(\overline{\gamma }-\gamma \rho )\}]{\parallel x_n-x^{\ast }\parallel }^2\\ +(1-{\alpha }_n)(1-\tau {\beta }_n){\delta }_n^2{\overline{\gamma }}^2{\parallel x_n-x^{\ast }\parallel }^2+2{\delta }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },u_n-x^{\ast }〉\\ +2\mu {\beta }_n〈x^{\ast }-y_n,B{x}^{\ast }〉+2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ =& [1-(1-{\alpha }_n)\{2{\delta }_n(\overline{\gamma }-\gamma \rho )(1-\tau {\beta }_n)+\tau {\beta }_n\}]{\parallel x_n-x^{\ast }\parallel }^2\\ +(1-{\alpha }_n)(1-\tau {\beta }_n){\delta }_n^2{\overline{\gamma }}^2{\parallel x_n-x^{\ast }\parallel }^2+2{\delta }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },u_n-x^{\ast }〉\\ +2\mu {\beta }_n〈x^{\ast }-y_n,B{x}^{\ast }〉+2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ =& [1-(1-{\alpha }_n)2{\delta }_n(\overline{\gamma }-\gamma \rho )(1-\tau {\beta }_n)]{\parallel x_n-x^{\ast }\parallel }^2-(1-{\alpha }_n)\tau {\beta }_n{\parallel x_n-x^{\ast }\parallel }^2\\ +(1-{\alpha }_n)(1-\tau {\beta }_n){\delta }_n^2{\overline{\gamma }}^2{\parallel x_n-x^{\ast }\parallel }^2+2{\delta }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },u_n-x^{\ast }〉\\ +2\mu {\beta }_n〈x^{\ast }-y_n,B{x}^{\ast }〉+2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉\\ \le & [1-2(\overline{\gamma }-\gamma \rho )(1-{\alpha }_n)(1-\tau {\beta }_n){\delta }_n]{\parallel x_n-x^{\ast }\parallel }^2\\ +(1-{\alpha }_n)(1-\tau {\beta }_n){\delta }_n^2{\overline{\gamma }}^2{\parallel x_n-x^{\ast }\parallel }^2+2{\delta }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },u_n-x^{\ast }〉\\ +2\mu {\beta }_n〈x^{\ast }-y_n,B{x}^{\ast }〉+2{\alpha }_n〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉.\end{array}$$

(3.19)

Since $\{x_n\}$, $\{A{x}_n\}$, $\{B{x}_n\}$, $\{\varphi (x_n)\}$ and $\{f(x_n)\}$ are all bounded, we can choose a constant $M_1>0$ such that

$$\underset{n}{sup}\frac{1}{\overline{\gamma }-\gamma \rho }\left\{\frac{{\delta }_n{\overline{\gamma }}^2}{2}{\parallel x_n-x^{\ast }\parallel }^2\right\}\le M_1.$$

It follows that

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & [1-2(\overline{\gamma }-\gamma \rho )(1-{\alpha }_n)(1-\tau {\beta }_n){\delta }_n]{\parallel x_n-x^{\ast }\parallel }^2\\ +2(\overline{\gamma }-\gamma \rho )(1-{\alpha }_n)(1-\tau {\beta }_n){\delta }_n{\varsigma }_n,\end{array}$$

(3.20)

where

$$\begin{array}{rcl}{\varsigma }_n& =& {\delta }_n{M}_1+\frac{1}{(\overline{\gamma }-\gamma \rho )(1-{\alpha }_n)(1-\tau {\beta }_n)}〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },u_n-x^{\ast }〉\\ +\frac{\mu {\beta }_n}{(\overline{\gamma }-\gamma \rho )(1-{\alpha }_n)(1-\tau {\beta }_n){\delta }_n}〈x^{\ast }-y_n,B{x}^{\ast }〉\\ +\frac{{\alpha }_n}{(\overline{\gamma }-\gamma \rho )(1-{\alpha }_n)(1-\tau {\beta }_n){\delta }_n}〈\varphi \left(x^{\ast }\right)-x^{\ast },x_{n+1}-x^{\ast }〉.\end{array}$$

By (3.14), (3.17), (3.18) and (C3)-(C4), we get ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\varsigma }_n\le 0$. Applying Lemma 2.9, we can conclude that $x_n\to x^{\ast }$. This completes the proof. □

4 An example

Next, the following example shows that all the conditions of Theorem 3.1 are satisfied.

Example 4.1 For instance, let ${\alpha }_n=\frac{1}{n}$, ${\beta }_n=\frac{1}{2n}$ and ${\delta }_n=\frac{1}{3n}$. We will show that the condition (C1) is achieved. Then, clearly, the sequence $\{{\delta }_n\}$

$$\sum _{n=1}^{\mathrm{\infty }}{\delta }_n=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{3n}=\mathrm{\infty }$$

and

$$\begin{array}{rcl}\sum _{n=1}^{\mathrm{\infty }}|{\delta }_{n+1}-{\delta }_n|& =& \sum _{n=1}^{\mathrm{\infty }}|\frac{1}{3(n+1)}-\frac{1}{3n}|\\ \le & |\frac{1}{3\cdot 1}-\frac{1}{3\cdot 2}|+|\frac{1}{3\cdot 2}-\frac{1}{3\cdot 3}|+|\frac{1}{3\cdot 3}-\frac{1}{3\cdot 4}|+\cdots \\ =& \frac{1}{3}.\end{array}$$

The sequence $\{{\delta }_n\}$ satisfies the condition (C1).

Next, we will show that the condition (C2) is achieved. We compute

$$\begin{array}{rcl}\sum _{n=1}^{\mathrm{\infty }}|{\beta }_{n+1}-{\beta }_n|& =& \sum _{n=1}^{\mathrm{\infty }}|\frac{1}{2(n+1)}-\frac{1}{2n}|\\ \le & |\frac{1}{2\cdot 1}-\frac{1}{2\cdot 2}|+|\frac{1}{2\cdot 2}-\frac{1}{2\cdot 3}|+|\frac{1}{2\cdot 3}-\frac{1}{2\cdot 4}|+\cdots \\ =& \frac{1}{2}.\end{array}$$

The sequence $\{{\beta }_n\}$ satisfies the condition (C2).

Next, we will show that the condition (C3) is achieved. We compute

$$\begin{array}{rcl}\sum _{n=1}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|& =& \sum _{n=1}^{\mathrm{\infty }}|\frac{1}{n+1}-\frac{1}{n}|\\ \le & |\frac{1}{1}-\frac{1}{2}|+|\frac{1}{2}-\frac{1}{3}|+|\frac{1}{3}-\frac{1}{4}|+\cdots \\ =& 1\end{array}$$

and

$$\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0.$$

The sequence $\{{\alpha }_n\}$ satisfies the condition (C3).

Finally, we will show that the condition (C4) is achieved.

$$\frac{1}{3n}<\frac{1}{2n}\text{and}\frac{1}{2n}<\frac{1}{n}.$$

Corollary 4.2 Let H be a real Hilbert space, C be a closed convex subset of H. Let $A:C\to H$ be inverse-strongly monotone. Let $T:C\to C$ be a nonexpansive mapping. Let $B:C\to C$ be β-strongly monotone and L-Lipschitz continuous. Assume that $\mathit{VI}(F(T),A)$ is nonempty set. Suppose $\{x_n\}$ is a sequence generated by the following algorithm $x_0\in C$ arbitrarily:

$$\{\begin{array}{c}{z}_n=T(I-{\delta }_{n}A)x_n,\hfill \\ y_n=(I-\mu {\beta }_{n}B)z_n,\hfill \\ x_{n+1}=(1-{\alpha }_n)y_n,\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(4.1)

$\{{\alpha }_n\},\{{\delta }_n\}\subset [0,1]$. If $\mu \in (0,\frac{2\beta }{L^2})$ is used and if $\{{\beta }_n\}\subset (0,1]$ satisfies the following conditions:

(C1): ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_{n+1}-{\delta }_n|<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{\delta }_n=\mathrm{\infty }$;

(C2): ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n+1}-{\beta }_n|<\mathrm{\infty }$;

(C3): ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty }$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

(C4): ${\delta }_n\le {\beta }_n$ and ${\beta }_n\le {\alpha }_n$.

Then $\{x_n\}$ converges strongly to $x^{\ast }\in \mathit{VI}(F(T),A)$, which is the unique solution of the variational inequality:

$$\mathit{\text{Find}}{x}^{\ast }\in \mathit{VI}(F(T),A)\mathit{\text{such that}}〈B{x}^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in \mathit{VI}(F(T),A).$$

(4.2)

Proof Putting $P_C$ is the identity and $f,\varphi \equiv 0$ in Theorem 3.1, we can obtain the desired conclusion immediately. □

Remark 4.3 Corollary 4.2 generalizes and improves the results of Iiduka [14].

Corollary 4.4 Let H be a real Hilbert space, C be a closed convex subset of H. Let $A:C\to H$ be a strongly positive linear bounded operator, $f:C\to H$ be a ρ-contraction, γ be a positive real number such that $\frac{\overline{\gamma }-1}{\rho }<\gamma <\frac{\overline{\gamma }}{\rho }$. Let $T:C\to C$ be a nonexpansive mapping. Assume that Ω is nonempty set. Suppose $\{x_n\}$ is a sequence generated by the following algorithm $x_0\in C$ arbitrarily:

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

(4.3)

where $\{{\alpha }_n\},\{{\delta }_n\}\subset [0,1]$. If $\mu \in (0,\frac{2\beta }{L^2})$ is used and if $\{{\beta }_n\}\subset (0,1]$ satisfies the following conditions:

(C1): ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_{n+1}-{\delta }_n|<\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{\delta }_n=\mathrm{\infty }$;

(C2): ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n+1}-{\beta }_n|<\mathrm{\infty }$;

(C3): ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty }$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

(C4): ${\delta }_n\le {\beta }_n$ and ${\beta }_n\le {\alpha }_n$.

Then $\{x_n\}$ converges strongly to $x^{\ast }\in \mathrm{\Omega }$, which is the unique solution of the variational inequality:

$$\mathit{\text{Find}}{x}^{\ast }\in \mathrm{\Omega }\mathit{\text{such that}}〈B{x}^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in \mathrm{\Omega }.$$

(4.4)

Proof Putting ϕ is the identity in Theorem 3.1, we can obtain the desired conclusion immediately. □

Remark 4.5 Corollary 4.4 generalizes and improves the results of Marino and Xu [20].