Convergence of a regularization algorithm for nonexpansive and monotone operators in Hilbert spaces

Abstract

Variational inequality, fixed point and generalized equilibrium problems are investigated via a regularization algorithm. It is proved that the sequence generated in the regularization algorithm converges strongly to a common solution of the three problems in the framework of Hilbert spaces. The results presented in this paper improve and extend the corresponding ones announced by many authors.

1 Introduction and preliminaries

In this paper, we always assume that H is a real Hilbert space with inner product $〈x,y〉$ and induced norm $\parallel x\parallel =\sqrt{〈x,x〉}$ for $x,y\in H$. Let C be a nonempty, closed, and convex subset of H.

Let $A:C\to H$ be a mapping. Recall that A is said to be monotone iff

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

Recall that A is said to be strongly monotone iff there exists a constant $\alpha >0$ such that

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

For such a case, A is also said to be α-strongly monotone. Recall that A is said to be inverse-strongly monotone iff there exists a constant $\alpha >0$ such that

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

For such a case, A is also said to be α-inverse-strongly monotone.

Recall that the classical variational inequality is to find an $x\in C$ such that

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

(1.1)

In this paper, we always use $VI(C,A)$ to denote the solution set of (1.1) and use $P_C$ denote the metric projection from H onto C. It is well known that $x\in C$ is a solution of (1.1) iff x is a fixed point of the mapping $P_C(I-rA)$, where $r>0$ is a constant, I stands for the identity mapping. If A is strongly monotone and Lipschitz continuous, the existence and uniqueness of solutions of equilibrium (1.1) is guaranteed by the Banach contraction principle.

Recall that a set-valued mapping $M:H\rightrightarrows H$ is said to be monotone iff, for all $x,y\in H$, $f\in Mx$, and $g\in My$ imply $〈x-y,f-g〉>0$. M is maximal iff the graph $Graph(M)$ of R is not properly contained in the graph of any other monotone mapping. It is well known that a monotone mapping M is maximal if and only if, for any $(x,f)\in H\times H$, $〈x-y,f-g〉\ge 0$, for all $(y,g)\in Graph(M)$ implies $f\in Rx$.

Let $S:C\to C$ be a mapping. $F(S)$ stands for the fixed point set of S; that is, $F(S):=\{x\in C:x=Sx\}$.

Recall that S is said to be contractive iff there exists a constant $\alpha \in (0,1)$ such that

$$\parallel Sx-Sy\parallel \le \alpha \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

For such a case, S is also said to be α-contractive. We know that the mapping enjoys a unique fixed point and Picard’s algorithm can be employed to approximate its unique fixed point.

Recall that S is said to be nonexpansive iff

$$\parallel Sx-Sy\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

If C is a closed, bounded and convex subset of H, then $F(S)$ is not empty; see [1].

Let $\{S_i:C\to C\}$ be a family of infinitely nonexpansive mappings and $\{{\gamma }_i\}$ be a nonnegative real sequence with $0\le {\gamma }_i<1$, $\mathrm{\forall }i\ge 1$. For $n\ge 1$, define a mapping $W_n:C\to C$ as follows:

$$\begin{array}{r}{U}_{n,n+1}=I,\\ U_{n,n}={\gamma }_n{S}_n{U}_{n,n+1}+(1-{\gamma }_n)I,\\ U_{n,n-1}={\gamma }_{n-1}{S}_{n-1}{U}_{n,n}+(1-{\gamma }_{n-1})I,\\ ⋮\\ U_{n,k}={\gamma }_k{S}_k{U}_{n,k+1}+(1-{\gamma }_k)I,\\ U_{n,k-1}={\gamma }_{k-1}{S}_{k-1}{U}_{n,k}+(1-{\gamma }_{k-1})I,\\ ⋮\\ U_{n,2}={\gamma }_2{S}_2{U}_{n,3}+(1-{\gamma }_2)I,\\ W_n=U_{n,1}={\gamma }_1{S}_1{U}_{n,2}+(1-{\gamma }_1)I.\end{array}$$

(1.2)

Such a mapping $W_n$ is nonexpansive from C to C and it is called a W-mapping generated by $S_n,S_{n-1},\dots ,S_1$ and ${\gamma }_n,{\gamma }_{n-1},\dots ,{\gamma }_1$; see [2] and the references therein.

Let $T:C\to H$ be a monotone mapping and let F be a bifunction of $C\times C$ into ℝ, where ℝ denotes the set of real numbers. We consider the following generalized equilibrium problem:

$$\text{Find }x\in C\text{ such that }F(x,y)+〈Tx,y-x〉\ge 0,\mathrm{\forall }y\in C.$$

(1.3)

In this paper, the set of such $x\in C$ is denoted by $EP(F,T)$, i.e.,

$$GEP(F,A)=\{x\in C:F(x,y)+〈Tx,y-x〉\ge 0,\mathrm{\forall }y\in C\}.$$

If $T\equiv 0$, the zero mapping, then the problem (1.3) is reduced to the following equilibrium problem [3]:

$$\text{Find }x\in C\text{ such that }F(x,y)\ge 0,\mathrm{\forall }y\in C.$$

(1.4)

In this paper, the set of such an $x\in C$ is denoted by $EP(F)$.

If $F\equiv 0$, then the problem (1.3) is reduced to the classical variational inequality (1.1).

To study equilibrium problems (1.3) and (1.4), we may assume that F satisfies the following conditions:

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

(A2) F is monotone, i.e., $F(x,y)+F(y,x)\le 0$ for all $x,y\in C$;

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

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

(A4) for each $x\in C$, $y\mapsto F(x,y)$ is convex and weakly lower semi-continuous.

Many important problems have reformulations which require finding solutions of equilibriums (1.3) and (1.4), for instance, image recovery, inverse problems, network allocation, transportation problems and optimization problems; see [3–11] and the references therein. For solving solutions of equilibriums (1.3) and (1.4), regularization methods recently have been extensively studied; see [11–28] and the references therein.

In this paper, motivated and inspired by the research going on in this direction, we study the variational inequality (1.1), and the fixed point and equilibrium problem (1.3) based on a regularization algorithm. It is proved that the sequence generated in the regularization algorithm converges strongly to a common solutions of the three problems in the framework of Hilbert spaces. The results presented in this paper improve and extend the corresponding results in Chang et al. [11], Takahashi and Takahashi [13] and Hao [29].

The following lemmas play an important role in our paper.

Lemma 1.1 [3]

Let $F:C\times C\to \mathbb{R}$ be a bifunction satisfying (A1)-(A4). Then, for any $r>0$ and $x\in H$, there exists $z\in C$ such that

$$F(z,y)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C.$$

Define a mapping $T_r:H\to C$ as follows:

$$T_{r}x=\{z\in C:F(z,y)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\},x\in H,$$

then the following conclusions hold:

• (1) $T_r$ is single-valued;

• (2) $T_r$ is firmly nonexpansive, i.e., for any $x,y\in H$,

  $${\parallel T_{r}x-T_{r}y\parallel }^2\le 〈T_{r}x-T_{r}y,x-y〉;$$

• (3) $F(T_r)=EP(F)$;

• (4) $EP(F)$ is closed and convex.

Lemma 1.2 [30]

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

$${\alpha }_{n+1}\le (1-{\gamma }_n){\alpha }_n+{\delta }_n,$$

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

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

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

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

Lemma 1.3 [2]

Let $\{S_i:C\to C\}$ be a family of infinitely nonexpansive mappings with a nonempty common fixed point set and let $\{{\gamma }_i\}$ be a real sequence such that $0<{\gamma }_i\le l<1$, where l is some real number, $\mathrm{\forall }i\ge 1$. Then

• (1) $W_n$ is nonexpansive and $F(W_n)={\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)$, for each $n\ge 1$;

• (2) for each $x\in C$ and for each positive integer k, the limit ${lim}_{n\to \mathrm{\infty }}{U}_{n,k}$ exists.

• (3) the mapping $W:C\to C$ defined by

  $$Wx:=\underset{n\to \mathrm{\infty }}{lim}{W}_{n}x=\underset{n\to \mathrm{\infty }}{lim}{U}_{n,1}x,x\in C,$$

  (1.5)

is a nonexpansive mapping satisfying $F(W)={\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)$ and it is called the W-mapping generated by $S_1,S_2,\dots$ and ${\gamma }_1,{\gamma }_2,\dots$ .

Lemma 1.4 [31]

Let $\{x_n\}$ and $\{y_n\}$ be bounded sequences in H 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 that $x_{n+1}=(1-{\beta }_n)y_n+{\beta }_n{x}_n$ for all $n\ge 0$ and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel y_{n+1}-y_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

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

Lemma 1.5 [11]

Let $\{S_i:C\to C\}$ be a family of infinitely nonexpansive mappings with a nonempty common fixed point set and let $\{{\gamma }_i\}$ be a real sequence such that $0<{\gamma }_i\le l<1$, $\mathrm{\forall }i\ge 1$. If K is any bounded subset of C, then

$$\underset{n\to \mathrm{\infty }}{lim}\underset{x\in K}{sup}\parallel Wx-W_{n}x\parallel =0.$$

Throughout this paper, we always assume that $0<{\gamma }_i\le l<1$, $\mathrm{\forall }i\ge 1$.

Lemma 1.6 [10]

Let $A:C\to H$ a Lipschitz monotone mapping and let $N_{C}x$ be the normal cone to C at $x\in C$; that is, $N_{C}x=\{y\in H:〈x-u,y〉,\mathrm{\forall }u\in C\}$. Define

$$Dx=\{\begin{array}{cc}Ax+N_{C}x,\hfill & x\in C,\hfill \\ \mathrm{\varnothing }\hfill & x\notin C.\hfill \end{array}$$

Then D is maximal monotone and $0\in Dx$ if and only if $x\in VI(C,A)$.

2 Main results

Theorem 2.1 Let C be a nonempty closed convex subset of H. Let F be a bifunction from $C\times C$ to ℝ which satisfies (A1)-(A4) and let $f:C\to C$ be a κ-contraction. Let $A:C\to H$ be an α-inverse-strongly monotone mapping and let $B:C\to H$ be a β-inverse-strongly monotone mapping. Let $T:C\to H$ be a τ-inverse-strongly monotone mapping. Let $\{S_i:C\to C\}$ be a family of infinitely nonexpansive mappings. Assume that $\mathrm{\Sigma }={\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)\cap GEP(F,T)\cap VI(C,A)\cap VI(C,B)$ is not empty. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ be sequences in $(0,1)$ such that ${\alpha }_n+{\beta }_n+{\gamma }_n=1$. Let $\{r_n\}$, $\{s_n\}$, and $\{{\lambda }_n\}$ be positive number sequences. Let $x_1\in C$ and let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{c}{y}_n=P_C(u_n-s_{n}B{u}_n),\hfill \\ x_{n+1}={\alpha }_{n}f(y_n)+{\beta }_n{W}_n{P}_C(y_n-r_{n}A{y}_n)+{\gamma }_n{x}_n,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(2.1)

where $\{u_n\}$ is such that $F(u_n,y)+〈T{x}_n,y-u_n〉+\frac{1}{{\lambda }_n}〈y-u_n,u_n-x_n〉\ge 0$, $\mathrm{\forall }y\in C$, and $\{W_n\}$ is the sequence generated in (1.5). Assume that the following restrictions hold:

• (a) $0<a\le {\lambda }_n\le b<2\tau$ and ${lim}_{n\to \mathrm{\infty }}|{\lambda }_{n+1}-{\lambda }_n|=0$,

• (b) $0<a^{\prime }\le r_n\le b^{\prime }<2\alpha$ and ${lim}_{n\to \mathrm{\infty }}|r_{n+1}-r_n|=0$,

• (c) $0<a^{″}\le s_n\le b^{″}<2\beta$ and ${lim}_{n\to \mathrm{\infty }}|s_{n+1}-s_n|=0$,

• (d) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$,

• (e) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\gamma }_n<1$,

where a, $a^{\prime }$, $a^{″}$, b, $b^{\prime }$, and $b^{″}$ are real constants. Then $\{x_n\}$ converges strongly to $\overline{x}\in \mathrm{\Sigma }$, which solves uniquely the following variational inequality:

$$〈\overline{x}-f(\overline{x}),\overline{x}-x〉\le 0,\mathrm{\forall }x\in \mathrm{\Sigma }.$$

Proof Since A is inverse-strongly monotone, we see from restriction (b) that

$$\begin{array}{r}{\parallel (I-r_{n}A)x-(I-r_{n}A)y\parallel }^2\\ ={\parallel x-y\parallel }^2-2r_n〈x-y,Ax-Ay〉+r_n^2{\parallel Ax-Ay\parallel }^2\\ \le {\parallel x-y\parallel }^2-2r_n\alpha {\parallel Ax-Ay\parallel }^2+r_n^2{\parallel Ax-Ay\parallel }^2\\ ={\parallel x-y\parallel }^2+r_n(r_n-2\alpha ){\parallel Ax-Ay\parallel }^2\\ \le {\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.\end{array}$$

This shows that $I-r_{n}A$ is nonexpansive. In the same way, we find that $I-s_{n}B$ and $I-{\lambda }_{n}T$ are nonexpansive. Note that $u_n$ can be re-written as $u_n=T_{{\lambda }_n}(I-{\lambda }_{n}T)x_n$. Let $x^{\ast }\in \mathrm{\Sigma }$. It follows that

$$\parallel u_n-x^{\ast }\parallel \le \parallel (I-{\lambda }_{n}T)x_n-(I-{\lambda }_{n}T)x^{\ast }\parallel \le \parallel x_n-x^{\ast }\parallel .$$

Putting $z_n=P_C(y_n-r_n)A{y}_n$, we see that $\parallel z_n-x^{\ast }\parallel \le \parallel y_n-x^{\ast }\parallel \le \parallel x_n-x^{\ast }\parallel$.

$$\begin{array}{rl}\parallel x_{n+1}-x^{\ast }\parallel & \le {\alpha }_n\parallel f(y_n)-x^{\ast }\parallel +{\beta }_n\parallel W_n{z}_n-x^{\ast }\parallel +{\gamma }_n\parallel x_n-x^{\ast }\parallel \\ \le {\alpha }_n\kappa \parallel y_n-x^{\ast }\parallel +{\alpha }_n\parallel f\left(x^{\ast }\right)-x^{\ast }\parallel +{\beta }_n\parallel z_n-x^{\ast }\parallel +{\gamma }_n\parallel x_n-x^{\ast }\parallel \\ \le (1-{\alpha }_n(1-\kappa ))\parallel x_n-x^{\ast }\parallel +{\alpha }_n(1-\kappa )\frac{\parallel f(x^{\ast })-x^{\ast }\parallel }{1-\kappa }.\end{array}$$

This implies that

$$\parallel x_n-x^{\ast }\parallel \le max\{\parallel x_1-x^{\ast }\parallel ,\frac{\parallel f(x^{\ast })-x^{\ast }\parallel }{1-\kappa }\}<\mathrm{\infty }.$$

This yields the result that the sequence $\{x_n\}$ is bounded, and so are $\{y_n\}$, $\{z_n\}$, and $\{u_n\}$. Without loss of generality, we can assume that there exists a bounded set $K\subset C$ such that $x_n,y_n,z_n,u_n\in K$. Since $u_n=T_{{\lambda }_n}(I-{\lambda }_n)x_n$, we find that

$$F(u_{n+1},y)+\frac{1}{{\lambda }_{n+1}}〈y-u_{n+1},u_{n+1}-(I-r_{n+1}T)x_{n+1}〉\ge 0,\mathrm{\forall }y\in C,$$

(2.2)

and

$$F(u_n,y)+\frac{1}{{\lambda }_n}〈y-u_n,u_n-(I-{\lambda }_{n}T)x_n〉\ge 0,\mathrm{\forall }y\in C.$$

(2.3)

Let $y=u_n$ in (2.2) and $y=u_{n+1}$ in (2.3). By adding up these two inequalities, we obtain

$$〈u_{n+1}-u_n,u_n-u_{n+1}+u_{n+1}-(I-{\lambda }_n{A}_3)x_n-\frac{{\lambda }_n}{{\lambda }_{n+1}}(u_{n+1}-(I-{\lambda }_{n+1}{A}_3)x_{n+1})〉\ge 0.$$

This implies that

$$\begin{array}{rl}{\parallel u_{n+1}-u_n\parallel }^2\le & 〈u_{n+1}-u_n,(I-{\lambda }_{n+1}T)x_{n+1}-(I-{\lambda }_{n}T)x_n\\ +(1-\frac{{\lambda }_n}{{\lambda }_{n+1}})(u_{n+1}-(I-{\lambda }_{n+1}T)x_{n+1})〉\\ \le & \parallel u_{n+1}-u_n\parallel (\parallel (I-{\lambda }_{n+1}T)x_{n+1}-(I-{\lambda }_{n}T)x_n\parallel \\ +|1-\frac{{\lambda }_n}{{\lambda }_{n+1}}|\parallel u_{n+1}-(I-{\lambda }_{n+1}T)x_{n+1}\parallel ).\end{array}$$

It follows that

$$\begin{array}{rl}\parallel u_{n+1}-u_n\parallel \le & \parallel (I-{\lambda }_{n+1}T)x_{n+1}-(I-{\lambda }_{n}T)x_n\parallel \\ +\frac{|{\lambda }_{n+1}-{\lambda }_n|}{{\lambda }_{n+1}}\parallel u_{n+1}-(I-{\lambda }_{n+1}T)x_{n+1}\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +|{\lambda }_{n+1}-{\lambda }_n|M_1,\end{array}$$

(2.4)

where $M_1$ is an appropriate constant such that

$$M_1=\underset{n\ge 1}{sup}\{\parallel T{x}_n\parallel +\frac{\parallel u_{n+1}-(I-{\lambda }_{n+1}T)x_{n+1}\parallel }{a}\}.$$

It follows from (2.4) that

$$\begin{array}{rl}\parallel y_{n+1}-y_n\parallel \le & \parallel P_C(u_{n+1}-s_{n+1}B{u}_{n+1})-P_C(u_n-s_{n+1}B{u}_n)\parallel \\ +\parallel P_C(u_n-s_{n+1}B{u}_n)-P_C(u_n-s_{n}B{u}_n)\parallel \\ \le & \parallel u_{n+1}-u_n\parallel +|s_{n+1}-s_n|\parallel B{u}_n\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +|{\lambda }_{n+1}-{\lambda }_n|M_1+|s_{n+1}-s_n|\parallel B{u}_n\parallel .\end{array}$$

Hence, we have

$$\begin{array}{rl}\parallel z_{n+1}-z_n\parallel \le & \parallel P_C(y_{n+1}-r_{n+1}A{y}_{n+1})-P_C(y_n-r_{n+1}A{y}_n)\parallel \\ +\parallel P_C(y_n-r_{n+1}A{y}_n)-P_C(y_n-r_{n}A{y}_n)\parallel \\ \le & \parallel y_{n+1}-y_n\parallel +|r_{n+1}-r_n|\parallel A{y}_n\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +M_2(|{\lambda }_{n+1}-{\lambda }_n|+|s_{n+1}-s_n|+|r_{n+1}-r_n|),\end{array}$$

(2.5)

where $M_2=max\{M_1,{sup}_{n\ge 1}\{A{y}_n\},{sup}_{n\ge 1}\{B{u}_n\}\}$. This implies from (2.5) that

$$\begin{array}{r}\parallel W_{n+1}{z}_{n+1}-W_n{z}_n\parallel \\ \le \parallel W_{n+1}{z}_{n+1}-W{z}_{n+1}\parallel +\parallel W{z}_{n+1}-W{z}_n\parallel +\parallel W{z}_n-W_n{z}_n\parallel \\ \le \underset{x\in K}{sup}\{\parallel W_{n+1}x-Wx\parallel +\parallel Wx-W_{n}x\parallel \}+\parallel x_{n+1}-x_n\parallel \\ +M_2(|{\lambda }_{n+1}-{\lambda }_n|+|s_{n+1}-s_n|+|r_{n+1}-r_n|),\end{array}$$

(2.6)

where K is the bounded subset of C defined above. Let $x_{n+1}=(1-{\gamma }_n)q_n+{\gamma }_n{x}_n$. It follows that

$$\begin{array}{rl}{q}_{n+1}-q_n=& \frac{{\alpha }_{n+1}f(y_{n+1})+{\beta }_{n+1}{W}_{n+1}{z}_{n+1}}{1-{\gamma }_{n+1}}-\frac{{\alpha }_{n}f(y_n)+{\beta }_n{W}_n{z}_n}{1-{\gamma }_n}\\ =& \frac{{\alpha }_{n+1}}{1-{\gamma }_{n+1}}f(y_{n+1})+\frac{1-{\alpha }_{n+1}-{\gamma }_{n+1}}{1-{\gamma }_{n+1}}{W}_{n+1}{z}_{n+1}\\ -(\frac{{\alpha }_n}{1-{\gamma }_n}f(y_n)+\frac{1-{\alpha }_n-{\gamma }_n}{1-{\gamma }_n}{W}_n{z}_n)\\ =& \frac{{\alpha }_{n+1}}{1-{\gamma }_{n+1}}(f(y_{n+1})-W_{n+1}{z}_{n+1})-\frac{{\alpha }_n}{1-{\gamma }_n}(f(y_n)-W_n{z}_n)\\ +W_{n+1}{z}_{n+1}-W_n{z}_n.\end{array}$$

By use of (2.6), we find that

$$\begin{array}{rl}\parallel q_{n+1}-q_n\parallel \le & \frac{{\alpha }_{n+1}}{1-{\gamma }_{n+1}}\parallel f(y_{n+1})-W_{n+1}{z}_{n+1}\parallel +\frac{{\alpha }_n}{1-{\gamma }_n}\parallel f(y_n)-W_n{z}_n\parallel \\ +\parallel W_{n+1}{z}_{n+1}-W_n{z}_n\parallel \\ \le & \frac{{\alpha }_{n+1}}{1-{\gamma }_{n+1}}\parallel f(y_{n+1})-W_{n+1}{z}_{n+1}\parallel +\frac{{\alpha }_n}{1-{\gamma }_n}\parallel f(y_n)-W_n{z}_n\parallel \\ +\underset{x\in K}{sup}\{\parallel W_{n+1}x-Wx\parallel +\parallel Wx-W_{n}x\parallel \}+\parallel x_{n+1}-x_n\parallel \\ +M_2(|{\lambda }_{n+1}-{\lambda }_n|+|s_{n+1}-s_n|+|r_{n+1}-r_n|).\end{array}$$

This implies that

$$\begin{array}{r}\parallel q_{n+1}-q_n\parallel -\parallel x_{n+1}-x_n\parallel \\ \le \frac{{\alpha }_{n+1}}{1-{\gamma }_{n+1}}\parallel f(y_{n+1})-W_{n+1}{z}_{n+1}\parallel +\frac{{\alpha }_n}{1-{\gamma }_n}\parallel f(y_n)-W_n{z}_n\parallel \\ +\underset{x\in K}{sup}\{\parallel W_{n+1}x-Wx\parallel +\parallel Wx-W_{n}x\parallel \}\\ +M_2(|{\lambda }_{n+1}-{\lambda }_n|+|s_{n+1}-s_n|+|r_{n+1}-r_n|).\end{array}$$

It follows from restrictions (a)-(e) that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel q_{n+1}-q_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

This implies from Lemma 1.4 that ${lim}_{n\to \mathrm{\infty }}\parallel q_n-x_n\parallel =0$. It follows that

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

(2.7)

Since A is inverse-strongly monotone, we find that

$$\begin{array}{rl}{\parallel z_n-x^{\ast }\parallel }^2& \le {\parallel (I-r_{n}A)y_n-(I-r_{n}A)x^{\ast }\parallel }^2\\ \le {\parallel y_n-x^{\ast }\parallel }^2-2r_n\alpha {\parallel A{y}_n-A{x}^{\ast }\parallel }^2+r_n^2{\parallel A{y}_n-A{x}^{\ast }\parallel }^2\\ \le {\parallel x_n-x^{\ast }\parallel }^2+r_n(r_n-2\alpha ){\parallel A{y}_n-A{x}^{\ast }\parallel }^2.\end{array}$$

It follows that

$$\begin{array}{rl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel W_n{z}_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2\\ \le {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel z_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2\\ \le {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+r_n(r_n-2\alpha ){\beta }_n{\parallel A{y}_n-A{x}^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2.\end{array}$$

Hence, we have

$$\begin{array}{r}{r}_n(2\alpha -r_n){\beta }_n{\parallel A{y}_n-A{x}^{\ast }\parallel }^2\\ \le {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+(\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\parallel x_n-x_{n+1}\parallel .\end{array}$$

By use of the restrictions (a), (d), and (e), we obtain from (2.7)

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

(2.8)

Since the metric projection is firmly nonexpansive, we find that

$$\begin{array}{rl}{\parallel z_n-x^{\ast }\parallel }^2\le & 〈(I-r_{n}A)y_n-(I-r_{n}A)x^{\ast },z_n-x^{\ast }〉\\ =& \frac{1}{2}\{{\parallel (I-r_{n}A)y_n-(I-r_{n}A)x^{\ast }\parallel }^2+{\parallel z_n-x^{\ast }\parallel }^2\\ -{\parallel (I-r_{n}A)y_n-(I-r_{n}A)x^{\ast }-(z_n-x^{\ast })\parallel }^2\}\\ \le & \frac{1}{2}\{{\parallel y_n-x^{\ast }\parallel }^2+{\parallel z_n-x^{\ast }\parallel }^2-{\parallel y_n-z_n-r_n(A{y}_n-A{x}^{\ast })\parallel }^2\}\\ \le & \frac{1}{2}\{{\parallel x_n-x^{\ast }\parallel }^2+{\parallel z_n-x^{\ast }\parallel }^2-{\parallel y_n-z_n\parallel }^2\\ +2r_n\parallel y_n-z_n\parallel \parallel A{y}_n-A{x}^{\ast }\parallel \}.\end{array}$$

Hence, we have

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

This further implies that

$$\begin{array}{rl}{\parallel x_{n+1}-x^{\ast }\parallel }^2\le & {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel W_n{z}_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel z_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2-{\beta }_n{\parallel y_n-z_n\parallel }^2+2r_n\parallel y_n-z_n\parallel \parallel A{y}_n-A{x}^{\ast }\parallel \\ +{\parallel x_n-x^{\ast }\parallel }^2,\end{array}$$

which yields

$$\begin{array}{rl}{\beta }_n{\parallel y_n-z_n\parallel }^2\le & {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+2r_n\parallel y_n-z_n\parallel \parallel A{y}_n-A{x}^{\ast }\parallel \\ +{\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+2r_n\parallel y_n-z_n\parallel \parallel A{y}_n-A{x}^{\ast }\parallel \\ +(\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\parallel x_{n+1}-x_n\parallel .\end{array}$$

By use of restrictions (b), (d), and (e), we find from (2.7) that

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

(2.9)

Since B is inverse-strongly monotone, we find that

$$\begin{array}{rl}{\parallel y_n-x^{\ast }\parallel }^2& \le {\parallel (I-s_{n}B)u_n-(I-s_{n}B)x^{\ast }\parallel }^2\\ \le {\parallel u_n-x^{\ast }\parallel }^2-2s_n\beta {\parallel B{u}_n-B{x}^{\ast }\parallel }^2+s_n^2{\parallel B{u}_n-B{x}^{\ast }\parallel }^2\\ \le {\parallel x_n-x^{\ast }\parallel }^2+s_n(s_n-2\beta ){\parallel B{u}_n-B{x}^{\ast }\parallel }^2.\end{array}$$

It follows that

$$\begin{array}{rl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel W_n{z}_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2\\ \le {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel y_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2\\ \le {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+s_n(s_n-2\beta ){\beta }_n{\parallel B{u}_n-B{x}^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2.\end{array}$$

Hence, we have

$$\begin{array}{r}{s}_n(2\beta -s_n){\beta }_n{\parallel B{u}_n-B{x}^{\ast }\parallel }^2\\ \le {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+(\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\parallel x_n-x_{n+1}\parallel .\end{array}$$

By use of the restrictions (c), (d), and (e), we obtain from (2.7)

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

(2.10)

Since the metric projection is firmly nonexpansive, we find that

$$\begin{array}{rl}{\parallel y_n-x^{\ast }\parallel }^2\le & 〈(I-s_{n}B)u_n-(I-s_{n}B)x^{\ast },y_n-x^{\ast }〉\\ =& \frac{1}{2}\{{\parallel (I-s_{n}B)u_n-(I-s_{n}B)x^{\ast }\parallel }^2+{\parallel y_n-x^{\ast }\parallel }^2\\ -{\parallel (I-s_{n}B)y_n-(I-s_{n}B)x^{\ast }-(y_n-x^{\ast })\parallel }^2\}\\ \le & \frac{1}{2}\{{\parallel u_n-x^{\ast }\parallel }^2+{\parallel y_n-x^{\ast }\parallel }^2-{\parallel u_n-y_n-s_n(B{u}_n-B{x}^{\ast })\parallel }^2\}\\ \le & \frac{1}{2}\{{\parallel x_n-x^{\ast }\parallel }^2+{\parallel y_n-x^{\ast }\parallel }^2-{\parallel u_n-y_n\parallel }^2\\ +2s_n\parallel u_n-y_n\parallel \parallel B{u}_n-B{x}^{\ast }\parallel \}.\end{array}$$

Hence, we have

$${\parallel y_n-x^{\ast }\parallel }^2\le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel u_n-y_n\parallel }^2+2s_n\parallel u_n-y_n\parallel \parallel B{u}_n-B{x}^{\ast }\parallel .$$

This further implies that

$$\begin{array}{rl}{\parallel x_{n+1}-x^{\ast }\parallel }^2\le & {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel W_n{z}_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel y_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2-{\beta }_n{\parallel u_n-y_n\parallel }^2+2s_n\parallel u_n-y_n\parallel \parallel B{u}_n-B{x}^{\ast }\parallel \\ +{\parallel x_n-x^{\ast }\parallel }^2,\end{array}$$

which yields

$$\begin{array}{rl}{\beta }_n{\parallel u_n-y_n\parallel }^2\le & {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+2s_n\parallel u_n-y_n\parallel \parallel B{u}_n-B{x}^{\ast }\parallel \\ +{\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+2s_n\parallel u_n-y_n\parallel \parallel B{u}_n-B{x}^{\ast }\parallel \\ +(\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\parallel x_{n+1}-x_n\parallel .\end{array}$$

By use of restrictions (c), (d), and (e), we find from (2.7) that

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

(2.11)

Since T is inverse-strongly monotone, we find that

$$\begin{array}{rl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel u_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2\\ \le {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel x_n-x^{\ast }-{\lambda }_n(T{x}_n-T{x}^{\ast })\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2\\ \le {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2-{\lambda }_n{\beta }_n(2\tau -{\lambda }_n){\parallel T{x}_n-T{x}^{\ast }\parallel }^2.\end{array}$$

This implies that

$$\begin{array}{r}{\lambda }_n{\beta }_n(2\tau -{\lambda }_n){\parallel T{x}_n-T{x}^{\ast }\parallel }^2\\ \le {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+(\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\parallel x_n-x_{n+1}\parallel .\end{array}$$

In view of the restrictions (a), (d), and (e), we see from (2.7) that

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

(2.12)

Since $T_{{\lambda }_n}$ is firmly nonexpansive, we find that

$$\begin{array}{rl}{\parallel u_n-x^{\ast }\parallel }^2=& {\parallel T_{{\lambda }_n}(I-{\lambda }_{n}T)x_n-T_{{\lambda }_n}(I-{\lambda }_{n}T)x^{\ast }\parallel }^2\\ \le & 〈(I-{\lambda }_{n}T)x_n-(I-{\lambda }_{n}T)x^{\ast },u_n-x^{\ast }〉\\ \le & \frac{1}{2}({\parallel x_n-x^{\ast }\parallel }^2+{\parallel u_n-x^{\ast }\parallel }^2-{\parallel x_n-u_n\parallel }^2\\ +2{\lambda }_n\parallel T{x}_n-T{x}^{\ast }\parallel \parallel x_n-u_n\parallel ).\end{array}$$

This in turn implies that

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

It follows that

$$\begin{array}{rl}{\parallel x_{n+1}-x^{\ast }\parallel }^2\le & {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel u_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2\\ \le & {\alpha }_n{\parallel f(y_n)-x^{\ast }\parallel }^2-{\beta }_n{\parallel x_n-u_n\parallel }^2\\ +2{\lambda }_n\parallel T{x}_n-T{x}^{\ast }\parallel \parallel x_n-u_n\parallel +{\parallel x_n-x^{\ast }\parallel }^2.\end{array}$$

By use of restrictions (a), (d), and (e), we see from (2.7) and (2.12) that

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

(2.13)

Next, we prove that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(\overline{x})-\overline{x},x_n-\overline{x}〉\le 0,$$

where $\overline{x}=P_{\mathrm{\Sigma }}f(\overline{x})$. To see this, we choose a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(f-I)\overline{x},x_n-\overline{x}〉=\underset{i\to \mathrm{\infty }}{lim}〈(f-I)\overline{x},x_{n_i}-\overline{x}〉.$$

Since $\{x_{n_i}\}$ is bounded, there exists a subsequence $\{x_{n_{i_j}}\}$ of $\{x_{n_i}\}$ which converges weakly to w. Without loss of generality, we may assume that $x_{n_i}\rightharpoonup w$. Since

$${\beta }_n\parallel W_n{z}_n-x_n\parallel \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f(y_n)-x_n\parallel .$$

In view of the restrictions (d) and (e), we obtain from (2.7)

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

(2.14)

Note that

$$\parallel W_n{z}_n-z_n\parallel \le \parallel W_n{z}_n-x_n\parallel +\parallel x_n-u_n\parallel +\parallel u_n-y_n\parallel +\parallel y_n-z_n\parallel .$$

In view of (2.8), (2.11), (2.13), and (2.14), we find that

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

(2.15)

Suppose the contrary, $w\notin {\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)$, i.e., $Ww\ne w$. Since $y_{n_i}\rightharpoonup w$, we find from Opial’s condition [32] that

$$\begin{array}{rl}\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel z_{n_i}-w\parallel & <\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel z_{n_i}-Ww\parallel \\ \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{\parallel z_{n_i}-W{z}_{n_i}\parallel +\parallel W{z}_{n_i}-Ww\parallel \}\\ \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{\parallel z_{n_i}-W{z}_{n_i}\parallel +\parallel z_{n_i}-w\parallel \}.\end{array}$$

On the other hand, we have

$$\begin{array}{rl}\parallel W{z}_n-z_n\parallel & \le \parallel W{z}_n-W_n{z}_n\parallel +\parallel W_n{z}_n-y_n\parallel \\ \le \underset{x\in K}{sup}\parallel Wx-W_{n}x\parallel +\parallel W_n{z}_n-z_n\parallel .\end{array}$$

In view of Lemma 1.5, we obtain from (2.15) ${lim}_{n\to \mathrm{\infty }}\parallel W{z}_n-z_n\parallel =0$. It follows that ${lim\hspace{0.17em}inf}_{i\to \mathrm{\infty }}\parallel z_{n_i}-w\parallel <{lim\hspace{0.17em}inf}_{i\to \mathrm{\infty }}\parallel z_{n_i}-w\parallel$. Thus one derives a contradiction. Thus, we have $w\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)$.

Next, we show that $\overline{x}\in VI(C,A)$. Let T be the maximal monotone mapping defined by

$$Dx=\{\begin{array}{cc}Bx+N_{C}x,\hfill & x\in C,\hfill \\ \mathrm{\varnothing },\hfill & x\notin C.\hfill \end{array}$$

For any given $(x,y)\in Graph(D)$, we have $y-Bx\in N_{C}x$. Since $y_n\in C$, by the definition of $N_C$, we have $〈x-y_n,y-Bx〉\ge 0$. Since $y_n=P_C(u_n-s_{n}B{u}_n)$, we see that $〈x-y_n,\frac{y_n-u_n}{s_n}+B{u}_n〉\ge 0$. It follows that

$$\begin{array}{rl}〈x-y_{n_i},y〉\ge & 〈x-y_{n_i},Bx〉\\ \ge & 〈x-y_{n_i},Bx〉-〈x-y_{n_i},\frac{y_{n_i}-u_{n_i}}{s_{n_i}}+B{u}_{n_i}〉\\ =& 〈x-y_{n_i},Bx-B{y}_{n_i}〉+〈x-y_{n_i},B{y}_{n_i}-B{u}_{n_i}〉-〈x-y_{n_i},\frac{y_{n_i}-u_{n_i}}{s_{n_i}}〉\\ \ge & 〈x-y_{n_i},B{y}_{n_i}-B{u}_{n_i}〉-〈x-y_{n_i},\frac{y_{n_i}-u_{n_i}}{s_{n_i}}〉.\end{array}$$

Since B is Lipschitz continuous, we see that $〈x-w,y〉\ge 0$. Notice that D is maximal monotone and hence $0\in Tw$. This shows that $w\in VI(C,A)$. In the same way, we find that $w\in VI(C,B)$.

Next, we show that $w\in GEP(F,T)$. Since $u_n=T_{{\lambda }_n}(I-{\lambda }_{n}T)x_n$, for any $y\in C$, we find from (A2) that

$$〈T{x}_{n_i},y-u_{n_i}〉+〈y-u_{n_i},\frac{u_{n_i}-x_{n_i}}{{\lambda }_{n_i}}〉\ge F(y,u_{n_i}),\mathrm{\forall }y\in C.$$

(2.16)

Putting $y_t=ty+(1-t)w$ for any $t\in (0,1]$ and $y\in C$, we see that $y_t\in C$. It follows from (2.16) that

$$\begin{array}{r}〈y_t-u_{n_i},T{y}_t〉\\ \ge 〈y_t-u_{n_i},T{y}_t〉-〈T{x}_{n_i},y_t-u_{n_i}〉-〈y_t-u_{n_i},\frac{u_{n_i}-x_{n_i}}{{\lambda }_{n_i}}〉+F(y_t,u_{n_i})\\ =〈y_t-u_{n_i},T{y}_t-T{u}_{n_i}〉+〈y_t-u_{n_i},T{u}_{n_i}-T{x}_{n_i}〉-〈y_t-u_{n_i},\frac{u_{n_i}-x_{n_i}}{{\lambda }_{n_i}}〉+F(y_t,u_{n_i}).\end{array}$$

In view of the monotonicity of T, and the restriction (a), we obtain from (A4)

$$〈y_t-w,T{y}_t〉\ge F(y_t,q).$$

(2.17)

From (A1) and (A4), we see that

$$\begin{array}{rl}0& =F(y_t,y_t)\le tF(y_t,y)+(1-t)F(y_t,w)\\ \le tF(y_t,y)+(1-t)〈y_t-w,T{y}_t〉\\ =tF(y_t,y)+(1-t)t〈y-w,T{y}_t〉.\end{array}$$

It follows from (A3) that $w\in GEP(F,T)$. This proves that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈f(\overline{x})-\overline{x},x_n-\overline{x}〉\le 0$.

Finally, we show that $x_n\to \overline{x}$, as $n\to \mathrm{\infty }$. Note that

$$\begin{array}{r}{\parallel x_{n+1}-\overline{x}\parallel }^2\\ ={\alpha }_n〈f(y_n)-\overline{x},x_{n+1}-\overline{x}〉+{\beta }_n〈W_n{z}_n-\overline{x},x_{n+1}-\overline{x}〉+{\gamma }_n〈x_n-\overline{x},x_{n+1}-\overline{x}〉\\ \le {\alpha }_n\kappa \parallel y_n-\overline{x}\parallel \parallel x_{n+1}-\overline{x}\parallel +{\alpha }_n〈f(\overline{x})-\overline{x},x_{n+1}-\overline{x}〉+{\beta }_n\parallel z_n-\overline{x}\parallel \parallel x_{n+1}-\overline{x}\parallel \\ +{\gamma }_n\parallel x_n-\overline{x}\parallel \parallel x_{n+1}-\overline{x}\parallel \\ \le (1-{\alpha }_n(1-\kappa ))\parallel x_n-\overline{x}\parallel \parallel x_{n+1}-\overline{x}\parallel +{\alpha }_n〈f(\overline{x})-\overline{x},x_{n+1}-\overline{x}〉.\end{array}$$

This implies that

$${\parallel x_{n+1}-\overline{x}\parallel }^2\le (1-{\alpha }_n(1-\kappa )){\parallel x_n-\overline{x}\parallel }^2+2{\alpha }_n〈f(\overline{x})-\overline{x},x_{n+1}-\overline{x}〉.$$

From the restriction (d), we obtain from Lemma 1.2 ${lim}_{n\to \mathrm{\infty }}\parallel x_n-\overline{x}\parallel =0$. This completes the proof. □

Corollary 2.2 Let C be a nonempty closed convex subset of H. Let F be a bifunction from $C\times C$ to ℝ which satisfies (A1)-(A4) and let $f:C\to C$ be a κ-contraction. Let $T:C\to H$ be a τ-inverse-strongly monotone mapping. Let $\{S_i:C\to C\}$ be a family of infinitely nonexpansive mappings. Assume that $\mathrm{\Sigma }={\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)\cap GEP(F,T)$ is not empty. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ be sequences in $(0,1)$ such that ${\alpha }_n+{\beta }_n+{\gamma }_n=1$. Let $\{{\lambda }_n\}$ be a positive number sequence. Let $x_1\in C$ and let $\{x_n\}$ be a sequence generated by

$$x_{n+1}={\alpha }_{n}f(y_n)+{\beta }_n{W}_n{u}_n+{\gamma }_n{x}_n,\mathrm{\forall }n\ge 1,$$

where $\{u_n\}$ is such that $F(u_n,y)+〈T{x}_n,y-u_n〉+\frac{1}{{\lambda }_n}〈y-u_n,u_n-x_n〉\ge 0$, $\mathrm{\forall }y\in C$, and $\{W_n\}$ is the sequence generated in (1.5). Assume that the following restrictions hold:

• (a) $0<a\le {\lambda }_n\le b<2\tau$ and ${lim}_{n\to \mathrm{\infty }}|{\lambda }_{n+1}-{\lambda }_n|=0$,

• (b) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$,

• (c) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\gamma }_n<1$,

where a and b are real constants. Then $\{x_n\}$ converges strongly to $\overline{x}\in \mathrm{\Sigma }$, which solves uniquely the following variational inequality:

$$〈\overline{x}-f(\overline{x}),\overline{x}-x〉\le 0,\mathrm{\forall }x\in \mathrm{\Sigma }.$$

Corollary 2.3 Let C be a nonempty closed convex subset of H. Let F be a bifunction from $C\times C$ to ℝ which satisfies (A1)-(A4) and let $f:C\to C$ be a κ-contraction. Let $B:C\to H$ be a β-inverse-strongly monotone mapping. Let $T:C\to H$ be a τ-inverse-strongly monotone mapping. Let $\{S_i:C\to C\}$ be a family of infinitely nonexpansive mappings. Assume that $\mathrm{\Sigma }={\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)\cap GEP(F,T)\cap VI(C,B)$ is not empty. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ be sequences in $(0,1)$ such that ${\alpha }_n+{\beta }_n+{\gamma }_n=1$. Let $\{s_n\}$ and $\{{\lambda }_n\}$ be positive number sequences. Let $x_1\in C$ and let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{c}{y}_n=P_C(u_n-s_{n}B{u}_n),\hfill \\ x_{n+1}={\alpha }_{n}f(y_n)+{\beta }_n{W}_n{y}_n+{\gamma }_n{x}_n,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

where $\{u_n\}$ is such that $F(u_n,y)+〈T{x}_n,y-u_n〉+\frac{1}{{\lambda }_n}〈y-u_n,u_n-x_n〉\ge 0$, $\mathrm{\forall }y\in C$, and $\{W_n\}$ is the sequence generated in (1.5). Assume that the following restrictions hold:

• (a) $0<a\le {\lambda }_n\le b<2\tau$ and ${lim}_{n\to \mathrm{\infty }}|{\lambda }_{n+1}-{\lambda }_n|=0$,

• (b) $0<a^{″}\le s_n\le b^{″}<2\beta$ and ${lim}_{n\to \mathrm{\infty }}|s_{n+1}-s_n|=0$,

• (c) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$,

• (d) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\gamma }_n<1$,

where a, $a^{\prime }$, b, and $b^{\prime }$ are real constants. Then $\{x_n\}$ converges strongly to $\overline{x}\in \mathrm{\Sigma }$, which solves uniquely the following variational inequality:

$$〈\overline{x}-f(\overline{x}),\overline{x}-x〉\le 0,\mathrm{\forall }x\in \mathrm{\Sigma }.$$

Corollary 2.4 Let C be a nonempty closed convex subset of H. Let F be a bifunction from $C\times C$ to ℝ which satisfies (A1)-(A4) and let $f:C\to C$ be a κ-contraction. Let $A:C\to H$ be an α-inverse-strongly monotone mapping and let $B:C\to H$ be a β-inverse-strongly monotone mapping. Let $\{S_i:C\to C\}$ be a family of infinitely nonexpansive mappings. Assume that $\mathrm{\Sigma }={\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)\cap EP(F)\cap VI(C,A)\cap VI(C,B)$ is not empty. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ be sequences in $(0,1)$ such that ${\alpha }_n+{\beta }_n+{\gamma }_n=1$. Let $\{r_n\}$, $\{s_n\}$, and $\{{\lambda }_n\}$ be positive number sequences. Let $x_1\in C$ and let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{c}{y}_n=P_C(u_n-s_{n}B{u}_n),\hfill \\ x_{n+1}={\alpha }_{n}f(y_n)+{\beta }_n{W}_n{P}_C(y_n-r_{n}A{y}_n)+{\gamma }_n{x}_n,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

where $\{u_n\}$ is such that $F(u_n,y)+\frac{1}{{\lambda }_n}〈y-u_n,u_n-x_n〉\ge 0$, $\mathrm{\forall }y\in C$, and $\{W_n\}$ is the sequence generated in (1.5). Assume that the following restrictions hold:

• (a) $0<a\le {\lambda }_n\le b<2\tau$ and ${lim}_{n\to \mathrm{\infty }}|{\lambda }_{n+1}-{\lambda }_n|=0$,

• (b) $0<a^{\prime }\le r_n\le b^{\prime }<2\alpha$ and ${lim}_{n\to \mathrm{\infty }}|r_{n+1}-r_n|=0$,

• (c) $0<a^{″}\le s_n\le b^{″}<2\beta$ and ${lim}_{n\to \mathrm{\infty }}|s_{n+1}-s_n|=0$,

• (d) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$,

• (e) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\gamma }_n<1$,

where a, $a^{\prime }$, $a^{″}$, b, $b^{\prime }$, and $b^{″}$ are real constants. Then $\{x_n\}$ converges strongly to $\overline{x}\in \mathrm{\Sigma }$, which solves uniquely the following variational inequality:

$$〈\overline{x}-f(\overline{x}),\overline{x}-x〉\le 0,\mathrm{\forall }x\in \mathrm{\Sigma }.$$

Proof In Theorem 2.1, put $T=0$. Then, for all $\tau \in (0,\mathrm{\infty })$, we have

$$〈x,y,Tx-Ty〉\ge \tau {\parallel Tx-Ty\parallel }^2,\mathrm{\forall }x,y\in C.$$

Taking $a,b\in (0,\mathrm{\infty })$ with $0<a<b<\mathrm{\infty }$ and choosing a sequence $\{{\lambda }_n\}$ of real numbers with $a\le {\lambda }_n\le b$, we obtain the desired result by Theorem 2.1. □

Corollary 2.5 Let C be a nonempty closed convex subset of H. Let $f:C\to C$ be a κ-contraction and let $T:C\to H$ be a τ-inverse-strongly monotone mapping. Let $A:C\to H$ be an α-inverse-strongly monotone mapping and let $B:C\to H$ be a β-inverse-strongly monotone mapping. Let $\{S_i:C\to C\}$ be a family of infinitely nonexpansive mappings. Assume that $\mathrm{\Sigma }={\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)\cap VI(C,T)\cap VI(C,A)\cap VI(C,B)$ is not empty. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ be sequences in $(0,1)$ such that ${\alpha }_n+{\beta }_n+{\gamma }_n=1$. Let $\{r_n\}$, $\{s_n\}$, and $\{{\lambda }_n\}$ be positive number sequences. Let $x_1\in C$ and let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{c}{u}_n=P_C(x_n-{\lambda }_{n}T{x}_n),\hfill \\ y_n=P_C(u_n-s_{n}B{u}_n),\hfill \\ x_{n+1}={\alpha }_{n}f(y_n)+{\beta }_n{W}_n{P}_C(y_n-r_{n}A{y}_n)+{\gamma }_n{x}_n,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

where $\{W_n\}$ is the sequence generated in (1.5). Assume that the following restrictions hold:

• (a) $0<a\le {\lambda }_n\le b<2\tau$ and ${lim}_{n\to \mathrm{\infty }}|{\lambda }_{n+1}-{\lambda }_n|=0$,

• (b) $0<a^{\prime }\le r_n\le b^{\prime }<2\alpha$ and ${lim}_{n\to \mathrm{\infty }}|r_{n+1}-r_n|=0$,

• (c) $0<a^{″}\le s_n\le b^{″}<2\beta$ and ${lim}_{n\to \mathrm{\infty }}|s_{n+1}-s_n|=0$,

• (d) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$,

• (e) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\gamma }_n<1$,

where a, $a^{\prime }$, $a^{″}$, b, $b^{\prime }$, and $b^{″}$ are real constants. Then $\{x_n\}$ converges strongly to $\overline{x}\in \mathrm{\Sigma }$, which solves uniquely the following variational inequality:

$$〈\overline{x}-f(\overline{x}),\overline{x}-x〉\le 0,\mathrm{\forall }x\in \mathrm{\Sigma }.$$

Proof Putting $F=0$, we find that

$$〈T{x}_n,y-u_n〉+\frac{1}{{\lambda }_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,$$

is equivalent to

$$〈y-u_n,x_n-{\lambda }_{n}T{x}_n-u_n〉\le 0,\mathrm{\forall }y\in C,$$

that is, $u_n=P_C(x_n-{\lambda }_{n}T{x}_n)$. This completes the proof. □