Strong convergence of a Halpern-type algorithm for common solutions of fixed point and equilibrium problems

Abstract

In this article, fixed points of nonexpansive mappings and equilibrium problems based on a Halpern-type algorithm are investigated. Strong convergence theorems for common solutions of the two problems are obtained in the framework of real Hilbert spaces.

1 Introduction

The study of equilibrium problems is an important branch of optimization theory and nonlinear functional analysis. Numerous problems in physics, optimization, transportation, signal processing, and economics are reduced to find a solution to equilibrium problems, which cover fixed point problems, variational inequalities, saddle problems, inclusion problems, and so on. A closely related subject of current interest is the problem of finding common elements in the fixed point set of nonlinear operators and in the solution set of monotone variational inequalities; see [1–15] and the references therein. The motivation for this subject is mainly due to its possible applications to mathematical modeling of concrete complex problems. The aim of this paper is to investigate a common element problem based on a Halpern-type algorithm. Strong convergence of the algorithm is obtained in the framework of real Hilbert spaces. The organization of this paper is as follows. In Section 2, we provide some necessary preliminaries. In Section 3, a Halpern-type algorithm is proposed and analyzed. Strong convergence theorems for common solutions of two problems are established in the framework of Hilbert spaces. In Section 4, applications of the main results are provided.

2 Preliminaries

Let H be a real Hilbert space with inner product $〈\cdot ,\cdot 〉$ and norm $\parallel \cdot \parallel$. Let C be a nonempty, closed, and convex subset of H and let ${Proj}_C$ be the metric projection from H onto C.

Let $T:C\to C$ be a mapping. In this paper, we use $F(T)$ to denote the fixed point set of T. Recall that T is said to be contractive iff there exists a constant $\alpha \in (0,1)$ such that

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

For such a case, T is also said to be α-contractive. Recall that T is said to be nonexpansive iff

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

It is well known that the fixed point set of nonexpansive mappings is nonempty provided that the subset C is bounded, convex, and closed.

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 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

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

(2.1)

In this paper, we use $VI(C,A)$ to denote the solution set of (2.1). It is well known that $x\in C$ is a solution of the variational inequality (2.1) iff x is a solution of the fixed point equation $P_C(I-rA)x=x$, where $r>0$ is a constant.

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 $f\in Rx$.

For a maximal monotone operator M on H, and $r>0$, we may define the single-valued resolvent $J_r:H\to D(M)$, where $D(M)$ denotes the domain of M. It is known that $J_r$ is firmly nonexpansive, and $M^{-1}(0)=F(J_r)$, where $F(J_r):=\{x\in D(M):x=J_{r}x\}$, and $M^{-1}(0):=\{x\in H:0\in Mx\}$.

Let $A:C\to H$ 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:

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

(2.2)

In this paper, we use $EP(F,A)$ to denote the solution set of the generalized equilibrium problem (2.2).

Next, we give some special cases of the generalized equilibrium problem (2.2).

• (I) If $F\equiv 0$, then problem (2.2) is reduced to the classical variational inequality (2.1).

• (II) If $A\equiv 0$, the zero mapping, then problem (2.2) is reduced to the following equilibrium problem:

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

  (2.3)

In this paper, we use $EP(F)$ to denote the solution set of the equilibrium problem (2.3).

To study the equilibrium problems, 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.

Recently, many authors have studied fixed point problems of nonexpansive mappings and solution problems of the equilibrium problems (2.2) and (2.3); for more details, see [16–25] and the references therein. In this paper, motivated and inspired by the research going on in this direction, we consider common element problems based on a mean iterative process. Strong convergence of the iterative process is obtained in the framework of real Hilbert spaces. The results presented in this paper improve and extend the corresponding results in Hao [1], Qin et al. [24], Chang et al. [25].

In order to prove our main results, we need the following lemmas.

Lemma 2.1 [26]

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 2.2 [27]

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.

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:

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

(2.4)

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 [28] and the references therein.

Lemma 2.3 [28]

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

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

  (2.5)

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 2.4 [29]

Let $B:C\to H$ be a mapping and let $M:H\rightrightarrows H$ be a maximal monotone operator. Then $F(J_r(I-rB))={(B+M)}^{-1}(0)$, where r is some positive real number.

Lemma 2.5 [25]

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 2.6 [30]

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

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

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

Lemma 2.7 [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 ${lim}_{n\to \mathrm{\infty }}\parallel y_n-x_n\parallel =0$.

3 Main results

Theorem 3.1 Let C be a nonempty closed convex subset of a Hilbert space H and let F be a bifunction from $C\times C$ to ℝ which satisfies (A1)-(A4). 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 $M:H\rightrightarrows H$ be a maximal monotone operator. Let $\{S_i:C\to C\}$ be a family of infinitely nonexpansive mappings. Assume that $F:={\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)\cap EP(F,B)\cap {(A+M)}^{-1}(0)\ne \mathrm{\varnothing }$. Let $f:C\to C$ be a κ-contraction. Let $\{x_n\}$ be a sequence generated by the process: $x_1\in C$ and

$$\{\begin{array}{c}F(u_n,y)+〈B{x}_n,y-u_n〉+\frac{1}{s_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_n{W}_n{Proj}_C{J}_{r_n}(u_n-r_{n}A{u}_n),\mathrm{\forall }n\ge 1,\hfill \end{array}$$

where $\{W_n:C\to C\}$ is the sequence generated in (2.4), $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ are sequences in $(0,1)$ such that ${\alpha }_n+{\beta }_n+{\gamma }_n=1$ and $\{r_n\}$ and $\{s_n\}$ are positive number sequences. Assume that the above control sequences satisfy the following restrictions:

• (a) $0<r\le r_n\le r^{\prime }<2\alpha$, $0<r^{″}\le s_n\le r^{‴}<2\beta$;

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

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

• (d) ${lim}_{n\to \mathrm{\infty }}|s_n-s_{n+1}|={lim}_{n\to \mathrm{\infty }}|r_n-r_{n+1}|=0$.

Then the sequence $\{x_n\}$ converges strongly to $\overline{x}={Proj}_{F}f(\overline{x})$.

Proof First, we show that $I-r_{n}A$ is nonexpansive. For $\mathrm{\forall }x,y\in C$, we have

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

Using restriction (a), we have $I-r_{n}A$ is nonexpansive, so is $I-s_{n}B$. Fix $x^{\ast }\in F$. It follows that $\parallel u_n-x^{\ast }\parallel \le \parallel (I-s_{n}B)x_n-(I-s_{n}B)x^{\ast }\parallel \le \parallel x_n-x^{\ast }\parallel$. Putting $y_n=J_{r_n}(u_n-r_{n}A{u}_n)$, one finds that $\parallel y_n-x^{\ast }\parallel \le \parallel u_n-x^{\ast }\parallel \le \parallel x_n-x^{\ast }\parallel$. It follows that

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

Hence, we have $\parallel x_n-x^{\ast }\parallel \le max\{\parallel x_1-x^{\ast }\parallel ,\frac{\parallel f(x^{\ast })-x^{\ast }\parallel }{1-\alpha }\}$. This yields the result that $\{x_n\}$ is bounded. Therefore, both $\{y_n\}$ and $\{u_n\}$ are also bounded. Next, without loss of generality, we assume that there exists a bounded set $K\subset C$ such that $x_n,y_n,u_n\in K$. Notice that $F(u_{n+1},y)+\frac{1}{s_{n+1}}〈y-u_{n+1},u_{n+1}-(I-s_{n+1}B)x_{n+1}〉\ge 0$, $\mathrm{\forall }y\in C$, and $F(u_n,y)+\frac{1}{s_n}〈y-u_n,u_n-(I-s_{n}B)x_n〉\ge 0$, $\mathrm{\forall }y\in C$. It follows that

$$〈u_{n+1}-u_n,\frac{u_n-(I-s_{n}B)x_n}{s_n}-\frac{u_{n+1}-(I-s_{n+1}B)x_{n+1}}{s_{n+1}}〉\ge 0.$$

Hence, we have

$$\begin{array}{rcl}{\parallel u_{n+1}-u_n\parallel }^2& \le & 〈u_{n+1}-u_n,(I-s_{n+1}B)x_{n+1}-(I-s_{n}B)x_n\\ +(1-\frac{s_n}{s_{n+1}})(u_{n+1}-(I-s_{n+1}B)x_{n+1})〉\\ \le & \parallel u_{n+1}-u_n\parallel (\parallel (I-s_{n+1}B)x_{n+1}-(I-s_{n}B)x_n\parallel \\ +|1-\frac{s_n}{s_{n+1}}|\parallel u_{n+1}-(I-s_{n+1}B)x_{n+1}\parallel ).\end{array}$$

This yields the result that

$$\begin{array}{rcl}\parallel u_{n+1}-u_n\parallel & \le & \parallel (I-s_{n+1}B)x_{n+1}-(I-s_{n}B)x_n\parallel \\ +\frac{|s_{n+1}-s_n|}{s_{n+1}}\parallel u_{n+1}-(I-s_{n+1}B)x_{n+1}\parallel \\ =& \parallel (I-s_{n+1}B)x_{n+1}-(I-s_{n+1}B)x_n+(I-s_{n+1}B)x_n-(I-s_{n}B)x_n\parallel \\ +\frac{|s_{n+1}-s_n|}{s_{n+1}}\parallel u_{n+1}-(I-s_{n+1}B)x_{n+1}\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +|s_{n+1}-s_n|M_1,\end{array}$$

(3.1)

where $M_1$ is an appropriate constant such that

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

Since $J_{r_n}$ is firmly nonexpansive, one sees that

$$\begin{array}{c}\parallel y_{n+1}-y_n\parallel \hfill \\ =\parallel J_{r_n}(u_{n+1}-r_{n+1}A{u}_{n+1})-J_{r_n}(u_n-r_{n}A{u}_n)\parallel \hfill \\ \le \parallel u_{n+1}-r_{n+1}A{u}_{n+1}-(u_n-r_{n}A{u}_n)\parallel \hfill \\ =\parallel (I-r_{n+1}A)u_{n+1}-(I-r_{n+1}A)u_n+(r_n-r_{n+1})A{u}_n\parallel \hfill \\ \le \parallel u_{n+1}-u_n\parallel +|r_n-r_{n+1}|\parallel A{u}_n\parallel .\hfill \end{array}$$

(3.2)

Substituting (3.1) into (3.2), one finds that

$$\parallel y_{n+1}-y_n\parallel \le \parallel x_{n+1}-x_n\parallel +(|s_{n+1}-s_n|+|r_n-r_{n+1}|)M_2,$$

(3.3)

where $M_2$ is an appropriate constant such that $M_2=max\{{sup}_{n\ge 1}\{\parallel A{u}_n\parallel \},M_1\}$. On the other hand, one has

$$\begin{array}{c}\parallel W_{n+1}{Proj}_C{y}_{n+1}-W_n{Proj}_C{y}_n\parallel \hfill \\ \le \parallel W_{n+1}{y}_{n+1}-W_n{y}_n\parallel \hfill \\ \le \parallel W_{n+1}{y}_{n+1}-W{y}_{n+1}\parallel +\parallel W{y}_{n+1}-W{y}_n\parallel +\parallel W{y}_n-W_n{y}_n\parallel \hfill \\ \le \underset{x\in K}{sup}\{\parallel W_{n+1}x-Wx\parallel +\parallel Wx-W_{n}x\parallel \}+\parallel y_{n+1}-y_n\parallel ,\hfill \end{array}$$

(3.4)

where K is the bounded subset of C defined above. Combining (3.3) with (3.4), one finds

$$\begin{array}{c}\parallel W_{n+1}{Proj}_C{y}_{n+1}-W_n{Proj}_C{y}_n\parallel \hfill \\ \le \underset{x\in K}{sup}\{\parallel W_{n+1}x-Wx\parallel +\parallel Wx-W_{n}x\parallel \}+\parallel x_{n+1}-x_n\parallel \hfill \\ +(|r_{n+1}-r_n|+|s_n-s_{n+1}|)M_2.\hfill \end{array}$$

(3.5)

Letting $x_{n+1}=(1-{\beta }_n)z_n+{\beta }_n{x}_n$ we see that

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel & \le & \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel f(x_{n+1})-W_{n+1}{y}_{n+1}\parallel +\frac{{\alpha }_n}{1-{\beta }_n}\parallel f(x_n)-W_n{y}_n\parallel \\ +\parallel W_{n+1}{Proj}_C{y}_{n+1}-W_n{Proj}_C{y}_n\parallel .\end{array}$$

(3.6)

Substituting (3.5) into (3.6), we see that

$$\begin{array}{c}\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel \hfill \\ \le \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel f(x_{n+1})-W_{n+1}{y}_{n+1}\parallel +\frac{{\alpha }_n}{1-{\beta }_n}\parallel f(x_n)-W_n{y}_n\parallel \hfill \\ +\underset{x\in K}{sup}\{\parallel W_{n+1}x-Wx\parallel +\parallel Wx-W_{n}x\parallel \}\hfill \\ +(|r_{n+1}-r_n|+|s_n-s_{n+1}|)M_2.\hfill \end{array}$$

It follows from restrictions (a), (c), and (d) that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}(\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$. Using Lemma 2.7, we find that ${lim}_{n\to \mathrm{\infty }}\parallel z_n-x_n\parallel =0$. It follows that

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

(3.7)

For any $x^{\ast }\in F$, we see that

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel x_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel W_n{Proj}_C{y}_n-x^{\ast }\parallel }^2\\ \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel x_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel y_n-x^{\ast }\parallel }^2.\end{array}$$

(3.8)

Since

$$\begin{array}{rcl}{\parallel y_n-x^{\ast }\parallel }^2& =& {\parallel J_{r_n}(u_n-r_{n}A{u}_n)-x^{\ast }\parallel }^2\\ \le & {\parallel (I-r_{n}A)u_n-(I-r_{n}A)x^{\ast }\parallel }^2\\ =& {\parallel u_n-x^{\ast }\parallel }^2-2r_n〈u_n-x^{\ast },A{u}_n-A{x}^{\ast }〉+r_n^2{\parallel A{u}_n-A{x}^{\ast }\parallel }^2\\ \le & {\parallel x_n-x^{\ast }\parallel }^2+r_n(r_n-2\alpha ){\parallel A{u}_n-A{x}^{\ast }\parallel }^2,\end{array}$$

we find from (3.8) that

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

(3.9)

It also follows from (3.8) that

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel x_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }-s_n(B{x}_n-B{x}^{\ast })\parallel }^2\\ \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel x_n-x^{\ast }\parallel }^2\\ +{\gamma }_n({\parallel x_n-x^{\ast }\parallel }^2+s_n^2{\parallel B{x}_n-B{x}^{\ast }\parallel }^2-2s_n〈B{x}_n-B{x}^{\ast },x_n-x^{\ast }〉)\\ \le & {\alpha }_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\beta }_n{\parallel x_n-x^{\ast }\parallel }^2+{\gamma }_n{\parallel x_n-x^{\ast }\parallel }^2\\ -s_n{\gamma }_n(2\beta -s_n){\parallel B{x}_n-B{x}^{\ast }\parallel }^2.\end{array}$$

Using (3.7), one arrives at

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

(3.10)

Since $T_{s_n}$ is firmly nonexpansive, we find that

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

which implies that ${\parallel u_n-x^{\ast }\parallel }^2\le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_n-u_n\parallel }^2+2s_n\parallel B{x}_n-B{x}^{\ast }\parallel \parallel x_n-u_n\parallel$. Hence

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

Using (3.7) and (3.10), one has

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

(3.11)

Similarly, one also has

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

(3.12)

Since

$$\parallel W_n{y}_n-y_n\parallel \le \parallel y_n-u_n\parallel +\parallel u_n-x_n\parallel +\parallel x_n-W_n{y}_n\parallel ,$$

we find from (3.11) and (3.12) that

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

(3.13)

Now, we are in a position to show ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈f(\overline{x})-\overline{x},x_n-z〉\le 0$, where $\overline{x}={Proj}_{F}f(\overline{x})$. To prove this, we choose a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that

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

(3.14)

Since $\{x_{n_i}\}$ is bounded, without loss of generality, we may assume that $x_{n_i}\rightharpoonup q$. Using (3.11) and (3.12), we have ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$. Therefore, we see that $y_{n_i}\rightharpoonup q$. Now, we are in a position to prove $q\in {(A+M)}^{-1}(0)$. Notice that $\frac{u_n-y_n}{r_n}-A{u}_n\in M{y}_n$. Let $\mu \in M\nu$. Since M is monotone, we find that $〈\frac{u_n-y_n}{r_n}-A{u}_n-\mu ,y_n-\nu 〉\ge 0$. This implies that $〈-Aq-\mu ,q-\nu 〉\ge 0$. This implies that $-Aq\in Mq$, that is, $q\in {(A+M)}^{-1}(0)$. Next, we show that $q\in EP(F,B)$. Since $u_n=T_{s_n}(I-s_{n}B)x_n$, we find from (A2) that

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

(3.15)

Putting $y_t=ty+(1-t)q$ for any $t\in (0,1]$ and $y\in C$, we see that $y_t\in C$. Using (3.15), we find that

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

Since B is monotone, we obtain from (A4) that $〈y_t-w,B{y}_t〉\ge F(y_t,w)$. Using (A1) and (A4), we find that

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

Hence, $0\le F(y_t,y)+(1-t)〈y-w,B{y}_t〉$, $\mathrm{\forall }y\in C$. It follows from (A3) that $w\in EP(F,B)$. Next, we prove that $q\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)$. Suppose to the contrary, $q\notin {\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)$, i.e., $Wq\ne q$. Since $y_{n_i}\rightharpoonup q$ and the space satisfies Opial’s condition, one has

$$\begin{array}{rcl}\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel y_{n_i}-q\parallel & <& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel y_{n_i}-Wq\parallel \\ \le & \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{\parallel y_{n_i}-W{y}_{n_i}\parallel +\parallel W{y}_{n_i}-Wq\parallel \}\\ \le & \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{\parallel y_{n_i}-W{y}_{n_i}\parallel +\parallel y_{n_i}-q\parallel \}.\end{array}$$

(3.16)

Since $\parallel W{y}_n-y_n\parallel \le {sup}_{x\in K}\parallel Wx-W_{n}x\parallel +\parallel W_n{y}_n-y_n\parallel$, we find from Lemma 2.5 that ${lim}_{n\to \mathrm{\infty }}\parallel W{y}_n-y_n\parallel =0$. It follows that ${lim\hspace{0.17em}inf}_{i\to \mathrm{\infty }}\parallel y_{n_i}-q\parallel <{lim\hspace{0.17em}inf}_{i\to \mathrm{\infty }}\parallel y_{n_i}-q\parallel$. This leads to a contradiction. Thus, we have $q\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)$. This proves that $q\in F$. Therefore, one has

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

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

$$\begin{array}{c}{\parallel x_{n+1}-\overline{x}\parallel }^2\hfill \\ \le {\alpha }_n〈f(x_n)-f(\overline{x}),x_{n+1}-\overline{x}〉+{\alpha }_n〈f(\overline{x})-\overline{x},x_{n+1}-\overline{x}〉\hfill \\ +{\beta }_n\parallel x_n-\overline{x}\parallel \parallel x_{n+1}-\overline{x}\parallel +{\gamma }_n\parallel y_n-\overline{x}\parallel \parallel x_{n+1}-\overline{x}\parallel \hfill \\ \le \frac{\kappa }{2}{\alpha }_n({\parallel x_n-\overline{x}\parallel }^2+{\parallel x_{n+1}-\overline{x}\parallel }^2)+{\alpha }_n〈f(\overline{x})-\overline{x},x_{n+1}-\overline{x}〉\hfill \\ +(1-{\alpha }_n)\parallel x_n-\overline{x}\parallel \parallel x_{n+1}-\overline{x}\parallel \hfill \\ \le \frac{1-{\alpha }_n(1-\kappa )}{2}{\parallel x_n-\overline{x}\parallel }^2+\frac{1}{2}{\parallel x_{n+1}-\overline{x}\parallel }^2+{\alpha }_n〈f(\overline{x})-\overline{x},x_{n+1}-\overline{x}〉,\hfill \end{array}$$

which implies that

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

Using Lemma 2.1, we find that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-\overline{x}\parallel =0$. This completes the proof. □

4 Applications

Recall that a mapping $T:C\to C$ is said to be a k-strict pseudo-contraction if there exists a constant $k\in [0,1)$ such that

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+k{\parallel (I-T)x-(I-T)y\parallel }^2$$

for all $x,y\in C$. Note that the class of k-strict pseudo-contractions strictly includes the class of nonexpansive mappings. Put $A=I-T$, where $T:C\to C$ is a k-strict pseudo-contraction. Then A is $\frac{1-k}{2}$-inverse-strongly monotone. Now, we are in a position to state a results on fixed points of strict pseudo-contractions.

Theorem 4.1 Let C be a nonempty closed convex subset of a Hilbert space H and let F be a bifunction from $C\times C$ to ℝ which satisfies (A1)-(A4). Let $T:C\to H$ be a k-strict pseudo-contraction, $B:C\to H$ be a β-inverse-strongly monotone mapping, and $\{S_i:C\to C\}$ be a family of infinitely nonexpansive mappings. Assume that $F:={\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)\cap EP(F,B)\cap F(T)\ne \mathrm{\varnothing }$. Let $f:C\to C$ be a κ-contraction. Let $\{x_n\}$ be a sequence generated by $x_1\in C$ and

$$\{\begin{array}{c}F(u_n,y)+〈B{x}_n,y-u_n〉+\frac{1}{s_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ y_n=(1-r_n)u_n+r_{n}T{u}_n,\hfill \\ x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_n{W}_n{y}_n,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

where $\{W_n:C\to C\}$ is the sequence generated in (2.4), $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ are sequences in $(0,1)$ such that ${\alpha }_n+{\beta }_n+{\gamma }_n=1$ and $\{r_n\}$, and $\{s_n\}$ are positive number sequences. Assume that the above control sequences satisfy the following restrictions:

• (a) $0<r\le s_n\le r^{\prime }<2\beta$, $0<r^{″}\le r_n\le r^{‴}<1-k$;

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

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

• (d) ${lim}_{n\to \mathrm{\infty }}|s_n-s_{n+1}|={lim}_{n\to \mathrm{\infty }}|r_n-r_{n+1}|=0$.

Then the sequence $\{x_n\}$ converges strongly to $\overline{x}={Proj}_{F}f(\overline{x})$.

Proof Taking $A=I-T$, wee see that $A:C\to H$ is a α-strict pseudo-contraction with $\alpha =\frac{1-k}{2}$ and $F(T)=VI(C,A)$. Using Theorem 3.1, we find the desired conclusion immediately. □

Let $g:H\to (-\mathrm{\infty },+\mathrm{\infty }]$ be a proper convex lower semi-continuous function. Then the subdifferential ∂g of g is defined as follows:

$$\partial fg(x)=\{y\in H:g(z)\ge g(x)+〈z-x,y〉,z\in H\},\mathrm{\forall }x\in H.$$

From Rockafellar [30], we know that ∂g is maximal monotone. It is not hard to verify that $0\in \partial g(x)$ if and only if $g(x)={min}_{y\in H}g(y)$.

Let $I_C$ be the indicator function of C, i.e.,

$$I_C(x)=\{\begin{array}{cc}0,\hfill & x\in C,\hfill \\ +\mathrm{\infty },\hfill & x\notin C.\hfill \end{array}$$

Since $I_C$ is a proper lower semi-continuous convex function on H, we see that the subdifferential $\partial I_C$ of $I_C$ is a maximal monotone operator. It is clear that $J_{r}x=P_{C}x$, $\mathrm{\forall }x\in H$. Notice that ${(A+\partial I_C)}^{-1}(0)=VI(C,A_1)$. Now, we are in a position to state the result on variational inequalities.

Theorem 4.2 Let C be a nonempty closed convex subset of a Hilbert space H and let F be a bifunction from $C\times C$ to ℝ which satisfies (A1)-(A4). Let $A:C\to H$ be an α-inverse-strongly monotone mapping, $B:C\to H$ be a β-inverse-strongly monotone mapping, and $\{S_i:C\to C\}$ be a family of infinitely nonexpansive mappings. Assume that $F:={\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)\cap EP(F,B)\cap VI(C,A)\ne \mathrm{\varnothing }$. Let $f:C\to C$ be a κ-contraction. Let $\{x_n\}$ be a sequence generated by $x_1\in C$ and

$$\{\begin{array}{c}F(u_n,y)+〈B{x}_n,y-u_n〉+\frac{1}{s_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_n{W}_n{P}_C(u_n-s_{n}A{u}_n),\mathrm{\forall }n\ge 1,\hfill \end{array}$$

where $\{W_n:C\to C\}$ is the sequence generated in (2.4), $\{{\alpha }_n\}$, $\{{\beta }_n\}$, and $\{{\gamma }_n\}$ are sequences in $(0,1)$ such that ${\alpha }_n+{\beta }_n+{\gamma }_n=1$ and $\{r_n\}$, and $\{s_n\}$ are positive number sequences. Assume that the above control sequences satisfy the following restrictions:

• (a) $0<r\le s_n\le r^{\prime }<2\beta$, $0<r^{″}\le r_n\le r^{‴}<2\alpha$;

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

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

• (d) ${lim}_{n\to \mathrm{\infty }}|s_n-s_{n+1}|={lim}_{n\to \mathrm{\infty }}|r_n-r_{n+1}|=0$.

Then the sequence $\{x_n\}$ converges strongly to $\overline{x}={Proj}_{F}f(\overline{x})$.