Strong convergence theorems for equilibrium problems involving a family of nonexpansive mappings

Abstract

We give new hybrid variants of extragradient methods for finding a common solution of an equilibrium problem and a family of nonexpansive mappings. We present a scheme that combines the idea of an extragradient method and a successive iteration method as a hybrid variant. Then, this scheme is modified by projecting on a suitable convex set to get a better convergence property under certain assumptions in a real Hilbert space.

MSC:65K10, 65K15, 90C25, 90C33.

1 Introduction

In this paper, we always assume that ℋ is a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the induced norm $\parallel \cdot \parallel$. Let C be a nonempty closed convex subset of ℋ and the bifunction $f:C\times C\to \mathcal{R}$. Then f is called strongly monotone on C with $\beta >0$ iff

$$f(x,y)+f(y,x)\le -\beta {\parallel x-y\parallel }^2\mathrm{\forall }x,y\in C;$$

monotone on C iff

$$f(x,y)+f(y,x)\le 0\mathrm{\forall }x,y\in C;$$

pseudomonotone on C iff

$$f(x,y)\ge 0\text{implies}f(y,x)\le 0\mathrm{\forall }x,y\in C;$$

Lipschitz-type continuous on C in the sense of Mastroeni [1] iff there exist positive constants $c_1>0$, $c_2>0$ such that

$$f(x,y)+f(y,z)\ge f(x,z)-c_1{\parallel x-y\parallel }^2-c_2{\parallel y-z\parallel }^2\mathrm{\forall }x,y,z\in C.$$

An equilibrium problem, shortly $EP(f,C)$, is to find a point in

$$Sol(f,C)=\{x^{\ast }\in C:f(x^{\ast },y)\ge 0\mathrm{\forall }y\in C\}.$$

Let a mapping T of C into itself. Then T is called contractive with constant $\delta \in (0,1)$ iff

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

The mapping T is called strictly pseudocontractive iff there exists a constant $k\in [0,1)$ such that

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

In the case $k=0$, the mapping T is called nonexpansive on C. We denote by $Fix(T)$ the set of fixed points of T.

Let $T_i:C\to C$, $i\in \mathrm{\Gamma }$, be a family of nonexpansive mappings where Γ stands for an index set. In this paper, we are interested in the problem of finding a common element of the solution set of problem $EP(f,C)$ and the set of fixed points $F={\bigcap }_{i\in \mathrm{\Gamma }}Fix(T_i)$, namely:

$$\text{Find }{x}^{\ast }\in F\cap Sol(f,C),$$

(1.1)

where the function f and the mappings $T_i$, $i\in \mathrm{\Gamma }$, satisfy the following conditions:

(A₁) $f(x,x)=0$ for all $x\in C$ and f is pseudomonotone on C,

(A₂) f is Lipschitz-type continuous on C with constants $c_1>0$ and $c_2>0$,

(A₃) f is upper semicontinuous on C,

(A₄) For each $x\in C$, $f(x,\cdot )$ is convex and subdifferentiable on C,

(A₅) $F\cap Sol(f,C)\ne \mathrm{\varnothing }$.

Under these assumptions, for each $r>0$ and $x\in C$, there exists a unique element $z\in C$ such that

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

(1.2)

Problem (1.1) is very general in the sense that it includes, as special cases, optimization problems, variational inequalities, minimax problems, equilibrium equilibriums, fixed point problems (see, e.g., [2–7]). Recently, it has become an attractive field for many researchers in both theory and its solution methods (see, e.g., [3, 4, 8–12] and the references therein). Most of these algorithms are based on inequality (1.2) for solving the underlying equilibrium problem when $F\cap Sol(f,C)\ne \mathrm{\varnothing }$. Motivated by this idea for finding a common point of $Sol(f,C)$ and the fixed point set $Fix(T)$ of a nonexpansive mapping T, Takahashi and Takahashi [13] first introduced an iterative scheme by the viscosity approximation method. The sequence $\{x^n\}$ is defined by

$$\{\begin{array}{c}{x}^0\in C,\hfill \\ f(u^n,y)+\frac{1}{r_n}〈y-u^n,u^n-x^n〉\ge 0\mathrm{\forall }y\in C,\hfill \\ x^{n+1}={\alpha }_{n}g(x^n)+(1-{\alpha }_n)T(u^n)\mathrm{\forall }n\ge 0,\hfill \end{array}$$

where $g:C\to C$ is contractive. Under certain conditions over the parameters $\{{\alpha }_n\}$ and $\{r_n\}$, they showed that the sequences $\{x^n\}$ and $\{u^n\}$ strongly converge to $z={Pr}_{Fix(T)\cap Sol(f,C)}g(z)$, where ${Pr}_C$ denotes the projection on C. At each iteration n in all of these algorithms, it requires to solve approximation auxiliary equilibrium problems for finding a common solution of an equilibrium problem and a fixed point problem. In order to avoid this requirement, Anh [14] recently proposed a hybrid extragradient algorithm for finding a common point of the set $Fix(T)\cap Sol(f,C)$. Starting with an arbitrary initial point $x^0\in C$, iteration sequences are defined by

$$\{\begin{array}{c}{y}^k=argmin\{{\lambda }_{k}f(x^k,y)+\frac{1}{2}{\parallel y-x^k\parallel }^2:y\in C\},\hfill \\ t^k=argmin\{{\lambda }_{k}f(y^k,t)+\frac{1}{2}{\parallel t-x^k\parallel }^2:t\in C\},\hfill \\ x^{k+1}={\alpha }_k{x}^0+(1-{\alpha }_k)T(x^k).\hfill \end{array}$$

(1.3)

Under certain conditions onto parameters $\{{\lambda }_k\}$ and $\{{\alpha }_k\}$, he showed that the sequences $\{x^k\}$, $\{y^k\}$ and $\{t^k\}$ weakly converge to the point $x\in Fix(T)\cap Sol(f,C)$ in a real Hilbert space. At each main iteration n of the scheme, he only solved strongly convex problems on C, but the proof of convergence was still done under the assumptions that $x^{n+1}-x^n\to 0$.

For finding a common point of a family of nonexpansive mappings $T_i$ ($i\in \mathrm{\Gamma }$), as a corollary of Theorem 2.1 in [15], Zhou proposed the following iteration scheme:

$$\{\begin{array}{c}{x}^0\in \mathcal{H}\text{ chosen arbitrarily,}\hfill \\ C_{1,i}=C,C_1={\bigcap }_{i\in \mathrm{\Gamma }}{C}_{1,i},\hfill \\ x^1={Pr}_{C_1}(x^0),\hfill \\ y^{n,i}=(1-{\alpha }_{n,i})x^n+{\alpha }_{n,i}{T}_i(x^n),\hfill \\ C_{n+1,i}=\{z\in C_{n,i}:{\alpha }_{n,i}(1-2{\alpha }_{n,i}){\parallel x^n-T_i(x^n)\parallel }^2\le 〈x^n-z,y^{n,i}-T_i(y^{n,i})〉\},\hfill \\ C_{n+1}={\bigcap }_{i\in \mathrm{\Gamma }}{C}_{n+1,i},\hfill \\ x^{n+1}={Pr}_{C_{n+1}}(x^0).\hfill \end{array}$$

(1.4)

Under the restrictions of the control sequences $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_{n,i}\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_{n,i}\le a_i<\frac{1}{2}$, he showed that the sequence $\{x^n\}$ defined by (1.4) strongly converges to $x^{\ast }={Pr}_F(x^0)$ in a real Hilbert space ℋ, where $F={\bigcap }_{i\in \mathrm{\Gamma }}Fix(T_i)$.

In this paper, motivated by Ceng et al. [16, 17], Wang and Guo [18], Zhou [15], Nadezhkina and Takahashi [10], Cho et al. [19], Takahashi and Takahashi [13], Anh [6, 12] and Anh et al. [20, 21], we introduce several modified hybrid extragradient schemes to modify the iteration schemes (1.3) and (1.4) to obtain new strong convergence theorems for a family of nonexpansive mappings and the equilibrium problem $EP(f,C)$ in the framework of a real Hilbert space ℋ.

To investigate the convergence of this scheme, we recall the following technical lemmas which will be used in the sequel.

Lemma 1.1 ([14], Lemma 3.1)

Let C be a nonempty closed convex subset of a real Hilbert space ℋ. Let $f:C\times C\to \mathcal{R}$ be a pseudomonotone and Lipschitz-type continuous bifunction. For each $x\in C$, let $f(x,\cdot )$ be convex and subdifferentiable on C. Suppose that the sequences $\{x^n\}$, $\{y^n\}$, $\{t^n\}$ are generated by scheme (1.3) and $x^{\ast }\in Sol(f,C)$. Then

$${\parallel t^n-x^{\ast }\parallel }^2\le {\parallel x^n-x^{\ast }\parallel }^2-(1-2{\lambda }_n{c}_1){\parallel x^n-y^n\parallel }^2-(1-2{\lambda }_n{c}_2){\parallel y^n-t^n\parallel }^2\mathrm{\forall }n\ge 0.$$

Lemma 1.2 Let C be a closed convex subset of a real Hilbert space ℋ, and let ${Pr}_C$ be the metric projection from ℋ on to C (i.e., for $x\in \mathcal{H}$, ${Pr}_C$ is the only point in C such that $\parallel x-{Pr}_{C}x\parallel =inf\{\parallel x-z\parallel :z\in C\}$). Given $x\in \mathcal{H}$ and $z\in C$. Then $z={Pr}_{C}x$ if only if there holds the relation $〈x-z,y-z〉\le 0$ for all $y\in C$.

Lemma 1.3 Let ℋ be a real Hilbert space. Then the following equations hold:

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

• (ii) ${\parallel tx+(1-t)y\parallel }^2=t{\parallel x\parallel }^2+(1-t){\parallel y\parallel }^2-t(1-t){\parallel x-y\parallel }^2$ for all $t\in [0,1]$ and $x,y\in \mathcal{H}$.

2 Convergence theorems

Now, we prove the main convergence theorem.

Theorem 2.1 Let C be a nonempty closed convex subset of a real Hilbert space ℋ. Suppose that assumptions (A₁)-(A₅) are satisfied and ${\{T_i\}}_{i\in \mathrm{\Gamma }}$ is a family of nonexpansive mappings from C into itself and a nonempty common fixed points set F. Let $\{x^n\}$ be a sequence generated by the following scheme:

$$\{\begin{array}{c}{x}^0\in \mathcal{H}\mathit{\text{ chosen arbitrarily}},\hfill \\ C_{1,i}=D_{1,i}=C,C_1={\bigcap }_{i\in \mathrm{\Gamma }}{C}_{1,i},D_1={\bigcap }_{i\in \mathrm{\Gamma }}{D}_{1,i},\hfill \\ x^1={Pr}_{C_1\cap D_1}{x}^0,\hfill \\ y^n=argmin\{{\lambda }_{n}f(x^n,y)+\frac{1}{2}{\parallel y-x^n\parallel }^2:y\in C\},\hfill \\ z^n=argmin\{{\lambda }_{n}f(y^n,y)+\frac{1}{2}{\parallel z-x^n\parallel }^2:z\in C\},\hfill \\ y^{n,i}=(1-{\alpha }_{n,i})z^n+{\alpha }_{n,i}{T}_i{z}^n,\hfill \\ C_{n+1,i}=\{z\in C_{n,i}:{\alpha }_{n,i}(1-2{\alpha }_{n,i}){\parallel z^n-T_i{z}^n\parallel }^2\le 〈z^n-z,y^{n,i}-T_i{y}^{n,i}〉\},\hfill \\ C_{n+1}={\bigcap }_{i\in \mathrm{\Gamma }}{C}_{n+1,i},\hfill \\ D_{n+1,i}=\{z\in D_{n,i}:\parallel y^{n,i}-z\parallel \le \parallel x^n-z\parallel \},\hfill \\ D_{n+1}={\bigcap }_{i\in \mathrm{\Gamma }}{D}_{n+1,i},\hfill \\ x^{n+1}={Pr}_{C_{n+1}\cap D_{n+1}}{x}^0,\hfill \\ 0<lim\hspace{0.17em}inf{\alpha }_{n,i}\le lim\hspace{0.17em}sup{\alpha }_{n,i}<1,\hfill \\ \{{\lambda }_n\}\subset [a,b]\mathit{\text{ for some }}a,b\in (0,\frac{1}{L}),\mathit{\text{ where }}L=max\{2c_1,2c_2\}.\hfill \end{array}$$

Then the sequences $\{x^n\}$, $\{y^n\}$ and $\{z^n\}$ strongly converge to the same point ${Pr}_{F\cap Sol(f,C)}{x}^0$.

Proof The proof of this theorem is divided into several steps.

Step 1. Claim that $C_n$ and $D_n$ are closed and convex for all $n\ge 0$.

We have to show that for any fixed point but arbitrary $i\in \mathrm{\Gamma }$, $C_{n,i}$ is closed and convex for every $n\ge 0$. This can be proved by induction on n. It is obvious that $C_{1,i}=C$ is closed and convex. Assume that $C_{n,i}$ is closed and convex for some $n\in {\mathcal{N}}^{\ast }=\{1,2,\dots \}$. We have that the set

$$A=\{z\in C:{\alpha }_{n,i}(1-2{\alpha }_{n,i}){\parallel z^n-T_i{z}^n\parallel }^2\le 〈z^n-z,y^{n,i}-T_i{y}^{n,i}〉\}$$

is closed and convex, and $C_{n+1,i}=C_{n,i}\cap A$, hence $C_{n+1,i}$ is closed and convex. Then $C_n$ is closed and convex for all $n\ge 0$. We can write $D_{n+1,i}$ under the form

$$D_{n+1,i}=\{z\in D_{n,i}:{\parallel y^{n,i}-x^n\parallel }^2+2〈y^{n,i}-x^n,x^n-z〉\le 0\}.$$

Then $D_{n+1,i}$ is closed and convex. Thus, $D_n$ is closed and convex.

Step 2. Claim that $F\cap Sol(f,C)\subseteq C_n\cap D_n$ for all $n\in {\mathcal{N}}^{\ast }$.

First, we show that $F\subseteq C_n$ by induction on n. It suffices to show that $F\subseteq C_{n,i}$.

We have $F\subseteq C=C_{1,i}$ is obvious. Suppose $F\subseteq C_{n,i}$ for some $n\in \mathcal{N}$. We have to show that $F\subseteq C_{n+1,i}$. Indeed, let $w\in F$, by inductive hypothesis, we have $w\in C_{n,i}$ and

$$\begin{array}{rl}{\parallel z^n-T_i{z}^n\parallel }^2=& 〈z^n-T_i{z}^n,z^n-T_i{z}^n〉\\ =& \frac{1}{{\alpha }_{n,i}}〈z^n-y^{n,i},z^n-T_i{z}^n〉\\ =& \frac{1}{{\alpha }_{n,i}}〈z^n-y^{n,i},z^n-T_i{z}^n-(y^{n,i}-T_i{y}^{n,i})〉+\frac{1}{{\alpha }_{n,i}}〈z^n-y^{n,i},y^{n,i}-T_i{y}^{n,i}〉\\ =& \frac{1}{{\alpha }_{n,i}}〈z^n-y^{n,i},z^n-T_i{z}^n-(y^{n,i}-T_i{y}^{n,i})〉\\ +\frac{1}{{\alpha }_{n,i}}〈z^n-w+w-y^{n,i},y^{n,i}-T_i{y}^{n,i}〉\\ =& \frac{1}{{\alpha }_{n,i}}〈z^n-y^{n,i},z^n-y^{n,i}〉+\frac{1}{{\alpha }_{n,i}}〈z^n-y^{n,i},T_i{y}^{n,i}-T_i{z}^n〉\\ +\frac{1}{{\alpha }_{n,i}}〈z^n-w,y^{n,i}-T_i{y}^{n,i}〉+\frac{1}{{\alpha }_{n,i}}〈w-y^{n,i},y^{n,i}-T_i{y}^{n,i}〉\\ \le & \frac{2}{{\alpha }_{n,i}}{\parallel z^n-y^{n,i}\parallel }^2+\frac{1}{{\alpha }_{n,i}}〈z^n-w,y^{n,i}-T_i{y}^{n,i}〉\\ +\frac{1}{{\alpha }_{n,i}}〈w-y^{n,i},y^{n,i}-T_i{y}^{n,i}〉.\end{array}$$

(2.1)

On the other hand, for all $w\in F$ and $y^{n,i}\in C$, we have

$$\begin{array}{rl}{\parallel w-y^{n,i}\parallel }^2& \ge 〈T_{i}w-T_i{y}^{n,i},w-y^{n,i}〉\\ =〈w-T_i{y}^{n,i},w-y^{n,i}〉\\ =〈w-y^{n,i}+y^{n,i}-T_i{y}^{n,i},w-y^{n,i}〉\\ ={\parallel w-y^{n,i}\parallel }^2+〈y^{n,i}-T_i{y}^{n,i},w-y^{n,i}〉,\end{array}$$

and hence

$$〈w-y^{n,i},y^{n,i}-T_i{y}^{n,i}〉\le 0.$$

Combining this with (2.1), we obtain

$$\begin{array}{rl}{\parallel z^n-T_i{z}^n\parallel }^2& \le \frac{2}{{\alpha }_{n,i}}{\parallel z^n-y^{n,i}\parallel }^2+\frac{1}{{\alpha }_{n,i}}〈z^n-w,y^{n,i}-T_i{y}^{n,i}〉\\ \le 2{\alpha }_{n,i}{\parallel z^n-T_i{z}^n\parallel }^2+\frac{1}{{\alpha }_{n,i}}〈z^n-w,y^{n,i}-T_i{y}^{n,i}〉.\end{array}$$

This follows that

$${\alpha }_{n,i}(1-2{\alpha }_{n,i}){\parallel z^n-T_i{z}^n\parallel }^2\le 〈z^n-w,y^{n,i}-T_i{y}^{n,i}〉.$$

By the definition of $C_{n+1,i}$, we have $w\in C_{n+1,i}$, and so $F\subseteq C_{n+1,i}$ for all $i\in \mathrm{\Gamma }$, which deduces that $F\subseteq C_n$. This shows that $F\cap Sol(f,C)\subseteq C_n$ for all $n\in {\mathcal{N}}^{\ast }$.

Next, we will prove $F\cap Sol(f,C)\subseteq D_n$ by induction on $n\in {\mathcal{N}}^{\ast }$. It suffices to show that $F\cap Sol(f,C)\subseteq D_{n,i}$. Indeed, $F\subseteq C=D_{1,i}$ so $F\cap Sol(f,C)\subseteq D_{1,i}$. Suppose that $F\cap Sol(f,C)\subseteq D_{n,i}$. Let $x^{\ast }\in F\cap Sol(f,C)$, then $x^{\ast }\in D_{n,i}$. Using Lemma 1.1, we get

$$\begin{array}{rl}{\parallel y^{n,i}-x^{\ast }\parallel }^2& ={\parallel (1-{\alpha }_{n,i})z^n+{\alpha }_{n,i}{T}_i{z}^n-x^{\ast }\parallel }^2\\ \le (1-{\alpha }_{n,i}){\parallel z^n-x^{\ast }\parallel }^2+{\alpha }_{n,i}{\parallel T_i{z}^n-T_i{x}^{\ast }\parallel }^2\\ \le {\parallel z^n-x^{\ast }\parallel }^2\\ \le {\parallel x^n-x^{\ast }\parallel }^2-(1-2{\lambda }_n{c}_1){\parallel x^n-y^n\parallel }^2-(1-2{\lambda }_n{c}_2){\parallel y^n-z^n\parallel }^2\\ \le {\parallel x^n-x^{\ast }\parallel }^2.\end{array}$$

(2.2)

Then we have $x^{\ast }\in D_{n+1,i}$ and hence $F\cap Sol(f,C)\subseteq D_{n+1,i}$. This shows that $F\cap Sol(f,C)\subseteq D_n$, which yields that $F\cap Sol(f,C)\subseteq C_n\cap D_n$ for all $n\in {\mathcal{N}}^{\ast }$.

Step 3. Claim that the sequence $\{x^n\}$ is bounded and there exists the limit ${lim}_{n\to \mathrm{\infty }}\parallel x^n-x^0\parallel =c$.

From $x^n={Pr}_{C_n\cap D_n}{x}^0$, it follows that

$$〈x^0-x^n,x^n-y〉\ge 0\mathrm{\forall }y\in C_n\cap D_n.$$

(2.3)

Then, using Step 2, we have $F\cap Sol(f,C)\subseteq C_n\cap D_n$ and

$$〈x^0-x^n,x^n-w〉\ge 0\mathrm{\forall }w\in F\cap Sol(f,C).$$

(2.4)

Combining this and assumption (A₅), the projection ${Pr}_{F\cap Sol(f,C)}{x}^0$ is well defined and there exits a unique point p such that $p={Pr}_{F\cap Sol(f,C)}{x}^0$. So, we have

$$\begin{array}{rl}0& \le 〈x^0-x^n,x^n-p〉=〈x^0-x^n,x^n-x^0+x^0-p〉\\ \le -{\parallel x^0-x^n\parallel }^2+\parallel x^0-x^n\parallel \parallel x^0-p\parallel ,\end{array}$$

and hence

$$\parallel x^0-x^n\parallel \le \parallel x^0-p\parallel .$$

Then the sequence $\{x^n\}$ is bounded. So, the sequences $\{y^n\}$, $\{z^n\}$, $\{y^{n,i}\}$, $\{T_i{y}^{n,i}\}$ also are bounded. Since $x^{n+1}\in C_{n+1}\cap D_{n+1}\subset C_n\cap D_n$ and (2.3), we have

$$\begin{array}{rl}0& \le 〈x^0-x^n,x^n-x^{n+1}〉=〈x^0-x^n,x^n-x^0+x^0-x^{n+1}〉\\ \le -{\parallel x^0-x^n\parallel }^2+\parallel x^0-x^n\parallel \parallel x^0-x^{n+1}\parallel ,\end{array}$$

and hence $\parallel x^0-x^n\parallel \le \parallel x^0-x^{n+1}\parallel$. This together with the boundedness of $\{x^n\}$ implies that the limit ${lim}_{n\to \mathrm{\infty }}\parallel x^n-x^0\parallel =c$ exists.

Step 4. We claim that ${lim}_{n\to \mathrm{\infty }}{x}^n=q\in C$.

Since $C_m\cap D_m\subseteq C_n\cap D_n$, $x^m={Pr}_{C_m\cap D_m}{x}^0\in C_n\cap D_n$ for any positive integer $m\ge n$ and (2.3), we have

$$〈x^0-x^n,x^n-x^{n+m}〉\ge 0.$$

Then

$$\begin{array}{rl}{\parallel x^n-x^{n+m}\parallel }^2& ={\parallel x^n-x^0+x^0-x^{n+m}\parallel }^2\\ ={\parallel x^n-x^0\parallel }^2+{\parallel x^0-x^{n+m}\parallel }^2-2〈x^0-x^n,x^0-x^{n+m}〉\\ \le {\parallel x^0-x^{n+m}\parallel }^2-{\parallel x^n-x^0\parallel }^2-2〈x^0-x^n,x^n-x^{n+m}〉\\ \le {\parallel x^0-x^{n+m}\parallel }^2-{\parallel x^n-x^0\parallel }^2.\end{array}$$

(2.5)

Passing the limit in (2.5) as $n\to \mathrm{\infty }$, we get ${lim}_{n\to \mathrm{\infty }}\parallel x^n-x^{n+m}\parallel =0$ $\mathrm{\forall }m\in {\mathcal{N}}^{\ast }$. Hence, $\{x^n\}$ is a Cauchy sequence in a real Hilbert space ℋ and so ${lim}_{n\to \mathrm{\infty }}{x}^n=q\in C$.

Step 5. We claim that $q={Pr}_{F\cap Sol(f,C)}{x}^0$, where $q={lim}_{n\to \mathrm{\infty }}{x}^n$.

First we show that $q\in F\cap Sol(f,C)$. Since $x^{n+1}={Pr}_{C_{n+1}\cap D_{n+1}}{x}^0$, we have $x^{n+1}\in D_{n+1}$. Then $x^{n+1}\in D_{n+1,i}$ and

$$\parallel y^{n,i}-x^{n+1}\parallel \le \parallel x^n-x^{n+1}\parallel ,$$

which yields that

$$\begin{array}{rl}\parallel x^n-y^{n,i}\parallel & \le \parallel x^n-x^{n+1}\parallel +\parallel x^{n+1}-y^{n,i}\parallel \\ \le 2\parallel x^n-x^{n+1}\parallel .\end{array}$$

Combining this and ${lim}_{n\to \mathrm{\infty }}\parallel x^n-x^m\parallel =0$ for all $m\in {\mathcal{N}}^{\ast }$, we get

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x^n-y^{n,i}\parallel =0.$$

(2.6)

For each $x^{\ast }\in Sol(f,C)\cap F$, by (2.2) we have

$$\begin{array}{rl}(1-2b{c}_1){\parallel x^n-y^n\parallel }^2& \le (1-2{\lambda }_n{c}_1){\parallel x^n-y^n\parallel }^2\\ \le {\parallel x^n-x^{\ast }\parallel }^2-{\parallel y^{n,i}-x^{\ast }\parallel }^2\\ =(\parallel x^n-x^{\ast }\parallel +\parallel y^{n,i}-x^{\ast }\parallel )(\parallel x^n-x^{\ast }\parallel -\parallel y^{n,i}-x^{\ast }\parallel )\\ \le (\parallel x^n-x^{\ast }\parallel +\parallel y^{n,i}-x^{\ast }\parallel )\left(\parallel x^n-y^{n,i}\parallel \right).\end{array}$$

Using this, the boundedness of sequences $\{x^n\}$, $\{y^{n,i}\}$ and (2.6), we obtain

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

(2.7)

By a similar way, we also have ${lim}_{n\to \mathrm{\infty }}\parallel z^n-y^n\parallel =0$. Then it follows from the inequality

$$\parallel x^n-z^n\parallel \le \parallel x^n-y^n\parallel +\parallel y^n-z^n\parallel$$

that

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

(2.8)

On the other hand, we have

$$\parallel y^{n,i}-z^n\parallel \le \parallel y^{n,i}-x^n\parallel +\parallel x^n-z^n\parallel .$$

Combining this, (2.6) and (2.8), we obtain ${lim}_{n\to \mathrm{\infty }}\parallel y^{n,i}-z^n\parallel =0$. By the definition of the sequence $\{y^{n,i}\}$, we have

$$\parallel y^{n,i}-z^n\parallel ={\alpha }_{n,i}\parallel T_i{z}^n-z^n\parallel ,$$

and hence

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T_i{z}^n-z^n\parallel =0,$$

which yields that

$$\begin{array}{rl}\parallel T_i{x}^n-x^n\parallel & \le \parallel T_i{x}^n-T_i{z}^n\parallel +\parallel T_i{z}^n-z^n\parallel +\parallel x^n-z^n\parallel \\ \le 2\parallel x^n-z^n\parallel +\parallel T_i{z}^n-z^n\parallel \\ \to 0\text{as }n\to \mathrm{\infty }\end{array}$$

and

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

It follows from Step 4 that ${lim}_{n\to \mathrm{\infty }}{T}_i{x}^n=q$. Hence $q\in F$.

Now we show that $q\in Sol(f,C)$. By Step 5, we have $y^n\to q$ as $n\to \mathrm{\infty }$.

Since $y^n$ is the unique solution of the strongly convex problem

$$min\{\frac{1}{2}{\parallel y-x^n\parallel }^2+{\lambda }_{n}f(x^n,y):y\in C\},$$

we get

$$0\in {\partial }_2({\lambda }_{n}f(x^n,y)+\frac{1}{2}{\parallel y-x^n\parallel }^2)\left(y^n\right)+N_C\left(y^n\right).$$

From this it follows that

$$0={\lambda }_{n}w+y^n-x^n+\overline{w},$$

where $w\in {\partial }_{2}f(x^n,\cdot )(y^n)$ and $\overline{w}\in N_C(y^n)$. By the definition of the normal cone $N_C$, we have

$$〈y^n-x^n,y-y^n〉\ge {\lambda }_n〈w,y^n-y〉\mathrm{\forall }y\in C.$$

(2.9)

On the other hand, since $f(x^n,\cdot )$ is subdifferentiable on C, by the well-known Moreau-Rockafellar theorem, there exists $w\in {\partial }_{2}f(x^n,\cdot )(y^n)$ such that

$$f(x^n,y)-f(x^n,y^n)\ge 〈w,y-y^n〉\mathrm{\forall }y\in C.$$

Combining this with (2.9), we have

$${\lambda }_n(f(x^n,y)-f(x^n,y^n))\ge 〈y^n-x^n,y^n-y〉\mathrm{\forall }y\in C.$$

Then, using $\{{\lambda }_n\}\subset [a,b]\subset (0,\frac{1}{L})$, (2.7), $x^n\to q$, $y^n\to q$ as $n\to \mathrm{\infty }$ and the upper semicontinuity of f, we have

$$f(q,y)\ge 0\mathrm{\forall }q\in C.$$

This means that $q\in Sol(f,C)$. By taking the limit in (2.4), we have

$$〈x^0-q,q-w〉\ge 0\mathrm{\forall }w\in F\cap Sol(f,C),$$

which implies that $q={Pr}_{F\cap Sol(f,C)}{x}^0$. Thus, the subsequences $\{x^n\}$, $\{y^n\}$, $\{z^n\}$ strongly converge to the same point $q={Pr}_{F\cap Sol(f,C)}{x}^0$. This completes the proof. □

Now, notice that $\mathrm{\forall }w\in F$

$$\begin{array}{rl}{\parallel z^n-T_i{z}^n\parallel }^2& ={\parallel z^n-w+w-T_i{z}^n\parallel }^2\\ ={\parallel z^n-w\parallel }^2+{\parallel w-T_i{z}^n\parallel }^2+2〈z^n-w,w-T_i{z}^n〉\\ \le 2{\parallel z^n-w\parallel }^2+2〈z^n-w,w-z^n+z^n-T_i{z}^n〉\\ =2{\parallel z^n-w\parallel }^2-2{\parallel z^n-w\parallel }^2+2〈z^n-w,z^n-T_i{z}^n〉\\ =2〈z^n-w,z^n-T_i{z}^n〉.\end{array}$$

Hence

$$\begin{array}{rl}{\parallel y^{n,i}-w\parallel }^2=& {\parallel (1-{\alpha }_{n,i})(z^n-w)+{\alpha }_{n,i}(T_i{z}^n-w)\parallel }^2\\ =& (1-{\alpha }_{n,i}){\parallel z^n-w\parallel }^2+{\alpha }_{n,i}{\parallel T_i{z}^n-w\parallel }^2-{\alpha }_{n,i}(1-{\alpha }_{n,i}){\parallel T_i{z}^n-z^n\parallel }^2\\ =& (1-{\alpha }_{n,i}){\parallel z^n-w\parallel }^2+{\alpha }_{n,i}{\parallel T_i{z}^n-z^n+z^n-w\parallel }^2\\ -{\alpha }_{n,i}(1-{\alpha }_{n,i}){\parallel T_i{z}^n-z^n\parallel }^2\\ =& (1-{\alpha }_{n,i}){\parallel z^n-w\parallel }^2+{\alpha }_{n,i}{\parallel T_i{z}^n-z^n\parallel }^2+{\alpha }_{n,i}{\parallel z^n-w\parallel }^2\\ +2{\alpha }_{n,i}〈T_i{z}^n-z^n,z^n-w〉-{\alpha }_{n,i}(1-{\alpha }_{n,i}){\parallel T_i{z}^n-z^n\parallel }^2\\ \le & {\parallel z^n-w\parallel }^2+2{\alpha }_{n,i}〈z^n-w,z^n-T_i{z}^n〉+2{\alpha }_{n,i}〈T_i{z}^n-z^n,z^n-w〉\\ -{\alpha }_{n,i}(1-{\alpha }_{n,i}){\parallel T_i{z}^n-z^n\parallel }^2\\ =& {\parallel z^n-w\parallel }^2-{\alpha }_{n,i}(1-{\alpha }_{n,i}){\parallel T_i{z}^n-z^n\parallel }^2.\end{array}$$

(2.10)

From (2.10) and using the methods in Theorem 2.1, we can get the following convergence result.

Theorem 2.2 Let C be a nonempty closed convex subset of a real Hilbert space ℋ. Suppose that assumptions (A₁)-(A₅) are satisfied and ${\{T_i\}}_{i\in \mathrm{\Gamma }}$ is a family of nonexpansive mappings from C into itself and a nonempty common fixed points set F. Let $\{x^n\}$ be a sequence generated by the following scheme:

$$\{\begin{array}{c}{x}^0\in \mathcal{H}\mathit{\text{ chosen arbitrarily}},\hfill \\ C_{1,i}=D_{1,i}=C,C_1={\bigcap }_{i\in \mathrm{\Gamma }}{C}_{1,i},D_1={\bigcap }_{i\in \mathrm{\Gamma }}{D}_{1,i},\hfill \\ x^1={Pr}_{C_1\cap D_1}{x}^0,\hfill \\ y^n=argmin\{{\lambda }_{n}f(x^n,y)+\frac{1}{2}{\parallel y-x^n\parallel }^2:y\in C\},\hfill \\ z^n=argmin\{{\lambda }_{n}f(y^n,y)+\frac{1}{2}{\parallel z-x^n\parallel }^2:z\in C\},\hfill \\ y^{n,i}=(1-{\alpha }_{n,i})z^n+{\alpha }_{n,i}{T}_i{z}^n,\hfill \\ C_{n+1,i}=\{z\in C_{n,i}:{\parallel y^{n,i}-z\parallel }^2\le {\parallel z^n-z\parallel }^2-{\alpha }_{n,i}(1-{\alpha }_{n,i}){\parallel z^n-T_i{z}^n\parallel }^2\},\hfill \\ C_{n+1}={\bigcap }_{i\in \mathrm{\Gamma }}{C}_{n+1,i},\hfill \\ D_{n+1,i}=\{z\in D_{n,i}:\parallel y^{n,i}-z\parallel \le \parallel x^n-z\parallel \},\hfill \\ D_{n+1}={\bigcap }_{i\in \mathrm{\Gamma }}{D}_{n+1,i},\hfill \\ x^{n+1}={Pr}_{C_{n+1}\cap D_{n+1}}{x}^0,\hfill \\ 0<lim\hspace{0.17em}inf{\alpha }_{n,i}\le lim\hspace{0.17em}sup{\alpha }_{n,i}<1,\hfill \\ \{{\lambda }_n\}\subset [a,b]\mathit{\text{ for some }}a,b\in (0,\frac{1}{L}),\mathit{\text{ where }}L=max\{2c_1,2c_2\}.\hfill \end{array}$$

Then the sequences $\{x^n\}$, $\{y^n\}$ and $\{z^n\}$ converge strongly to the same point ${Pr}_{F\cap Sol(f,C)}{x}^0$.