Iterative approximation methods for mixed equilibrium problems for a countable family of quasi-ϕ-asymptotically nonexpansive multivalued mappings in Banach spaces

Abstract

In this paper, we prove the existence of a solution of the mixed equilibrium problem $\mathit{MEP}(f,\phi ,C)$ by using the KKM mapping in a Banach space setting. Then, by virtue of this result, we introduce a hybrid iterative scheme for finding a common element of the set of solutions of $\mathit{MEP}(f,\phi ,C)$ and the set of common fixed points of a countable family of quasi-ϕ-asymptotically nonexpansive multivalued mappings. Furthermore, we prove that the sequences generated by the hybrid iterative scheme converge strongly to a common element of the set of solutions of $\mathit{MEP}(f,\phi ,C)$ and the set of common fixed points of a countable family of quasi-ϕ-asymptotically nonexpansive multivalued mappings.

1 Introduction

Let E be a Banach space with the norm $\parallel \cdot \parallel$ and the dual $E^{\ast }$. We denote by ℕ and ℝ the sets of positive integers and real numbers, respectively. Also, we denote the normalized duality mapping from E to $2^{E^{\ast }}$ by J defined by

$$Jx=\{x^{\ast }\in E^{\ast }:〈x,x^{\ast }〉={\parallel x\parallel }^2={\parallel x^{\ast }\parallel }^2\},\mathrm{\forall }x\in E,$$

where $〈\cdot ,\cdot 〉$ denotes the generalized duality pairing. Let C be a nonempty, closed and convex subset of a Banach space E. The bifunction $f:C\times C\to \mathbb{R}$ is said to be relaxed ξ-monotone if there exists a function $\xi :E\to \mathbb{R}$ positively homogeneous of degree p, that is, $\xi (tz)=t^p\xi (z)$ for all $t>0$ and $z\in E$, where $p>1$ is a constant such that

$$f(x,y)+f(y,x)\le \xi (y-x)\text{for all }x,y\in C.$$

(1.1)

If $\xi =0$, then f is monotone, i.e.,

$$f(x,y)+f(y,x)\le 0\text{for all }x,y\in C.$$

Example 1.1 Let $E=\mathbb{R}$ and $C=[0,1]$. A bifunction $f:C\times C\to \mathbb{R}$ is defined by

$$f(x,y)=y(y-x),\mathrm{\forall }x,y\in [0,1].$$

It is easy to see that $f(x,y)+f(y,x)={(x-y)}^2\ge 0$. If we put $\xi (x)=x^2$ for all $x\in \mathbb{R}$, then f is relaxed ξ-monotone, but not monotone.

Let us consider the bifunction $f:C\times C\to \mathbb{R}$ and a function $\phi :C\to \mathbb{R}$. The mixed equilibrium problem (in short, $\mathit{MEP}(f,\phi ,C)$) is to find $\tilde{x}\in C$ such that

$$f(\tilde{x},y)+\phi (y)-\phi (\tilde{x})\ge 0,\mathrm{\forall }y\in C.$$

(1.2)

Problem (1.2) was studied by Ceng and Yao [1], Cholamjiak and Suantai [2] in Hilbert spaces and Banach spaces, resp. By using the well-known KKM technique, they gave the existence and uniqueness of solutions of the $\mathit{MEP}(f,\phi ,C)$ when f is monotone.

In the case of $f(x,y)=〈Ax,\eta (y,x)〉$, where $A:C\to E^{\ast }$ and $\eta :C\times C\to E$, problem (1.2) is reduced to the following variational-like inequality problem (in short, $\mathit{VLIP}(A,\phi ,C)$), which is to find $\tilde{x}\in C$ such that

$$〈A(\tilde{x}),\eta (y,\tilde{x})〉+\phi (y)-\phi (\tilde{x})\ge 0,\mathrm{\forall }y\in C.$$

(1.3)

Problem (1.3) was studied by Fang and Huang [3]. By using the KKM technique and η-ξ monotonicity of the mapping A, they obtained the existence of solutions of the variational-like inequality problem (1.3) in a real reflexive Banach space.

In particular, if $\phi \equiv 0$, problem (1.2) is reduced to the well-known equilibrium problem (in short, $\mathit{EP}(f,C)$), which is to find $\tilde{x}\in C$ such that

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

(1.4)

The $\mathit{EP}(f,C)$ includes fixed point problems, optimization problems, variational inequality problems and Nash equilibrium problems as special cases (see also [4–6]).

Let S be a mapping from C into itself. A mapping S is said to be Lipschitz if there exists a constant $L>0$ such that $\parallel Sx-Sy\parallel \le L\parallel x-y\parallel$ for all $x,y\in C$. S is said to be nonexpansive, when $L=1$. A point $p\in C$ is said to be a fixed point of S if $Sp=p$. A point $p\in C$ is said to be an asymptotic fixed point of S if there exists a sequence $\{x_n\}\subset C$ such that $x_n\rightharpoonup x\in E$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-S{x}_n\parallel =0$. Denote the set of all fixed points of S and the set of all asymptotic fixed points of S by $F(S)$ and $\hat{F}(S)$, respectively.

Now, we assume that E is a smooth, strictly convex and reflexive Banach space. The Lyapunov function $\varphi :E\times E\to {\mathbb{R}}^{+}$ is defined by

$$\varphi (x,y)={\parallel x\parallel }^2-2〈x,Jy〉+{\parallel y\parallel }^2,\mathrm{\forall }x,y\in E.$$

A mapping $S:C\to C$ is said to be relatively nonexpansive (see also [7–9]) if the following conditions are satisfied:

• (1) $F(S)$ is nonempty;

• (2) $\varphi (p,Sx)\le \varphi (p,x)$ for all $x\in C$, $p\in F(S)$;

• (3) $\hat{F}(S)=F(S)$.

Lemma 1.2 [8]

Let C be a nonempty, closed and convex subset of a smooth, strictly convex and reflexive Banach space E, and let T be a relatively nonexpansive mapping from C into itself. Then $F(T)$ is closed and convex.

In 2008, Takahashi and Zembayashi [10] introduced the following iterative scheme which is called the shrinking projection method:

$$\{\begin{array}{l}{x}_0\in C,C_0=C,\\ y_n=J^{-1}[{\alpha }_{n}J{x}_0+(1-{\alpha }_n)JS{x}_n],\\ u_n\in C\text{such that }\\ f(u_n,y)+\frac{1}{r_n}〈y-u_n,J{u}_n-J{w}_n〉\ge 0,\mathrm{\forall }y\in C,\\ C_{n+1}=\{v\in C_n:\varphi (v,u_n)\le \varphi (v,x_n)\},\\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_0,n\ge 0,\end{array}$$

(1.5)

where J is the duality mapping on E, ${\mathrm{\Pi }}_C$ is the generalized projection from E onto C and S is relatively nonexpansive single-valued mapping. They proved that the sequence $\{x_n\}$ converges strongly to a common element of the set of fixed points of a relatively nonexpansive single-valued mapping and the set of solutions of an equilibrium problem under appropriate conditions in a uniformly smooth and uniformly convex Banach space.

In the case of single-valued mapping, many authors such as in references [8, 9, 11–19] studied the problem of finding a common element of the set of fixed points of a nonexpansive or relatively nonexpansive mapping and the set of solutions of an equilibrium problem in the framework of Hilbert spaces and Banach spaces.

Let $N(C)$ and $\mathit{CB}(C)$ denote the family of nonempty subsets and nonempty closed bounded subsets of C, respectively. The Hausdorff metric on $\mathit{CB}(C)$ is defined by

$$H(A,B)=max\{\underset{a\in A}{sup}d(a,B),\underset{b\in B}{sup}d(A,b)\}$$

for all $A,B\in \mathit{CB}(C)$, where $d(x,D)=inf\{\parallel x-y\parallel ,y\in D\}$. Homaeipour and Razani [20] have defined the relatively nonexpansive multivalued mapping as follows.

A multivalued mapping $S:C\to N(C)$ is said to be relatively nonexpansive if the following conditions are satisfied:

• (1) $F(S)$ is nonempty;

• (2) $\varphi (p,z)\le \varphi (p,x)$ for all $x\in C$, $p\in F(S)$, $z\in Sx$;

• (3) $\hat{F}(S)=F(S)$.

It is obvious that the class of relatively nonexpansive multivalued mappings contains properly the class of relatively nonexpansive multivalued mappings (for example, see [20]). They introduced the following iterative sequence for finding a fixed point of the relatively nonexpansive multivalued mapping $S:C\to N(C)$:

$$\{\begin{array}{l}{x}_1\in C,\\ x_{n+1}={\mathrm{\Pi }}_C({\alpha }_{n}J(x_n)+(1-{\alpha }_n)J(z_n)),z_n\in S{x}_n,\mathrm{\forall }n\ge 1,\end{array}$$

(1.6)

where $\{{\alpha }_n\}\subset [0,1]$. Moreover, they proved weak and strong convergence theorems for a single relatively nonexpansive multivalued mapping in a uniformly convex and uniformly smooth Banach space E.

A multivalued mapping $S:C\to \mathit{CB}(C)$ is said to be quasi-ϕ-nonexpansive [21, 22] if $F(S)\ne \mathrm{\varnothing }$ and

$$\varphi (p,z)\le \varphi (p,x),\mathrm{\forall }x\in C,p\in F(S),z\in Sx.$$

$S:C\to \mathit{CB}(C)$ is said to be quasi-ϕ-asymptotically nonexpansive [21, 22] if $F(S)\ne \mathrm{\varnothing }$ and there exists a real sequence $l_n\subset [1,+\mathrm{\infty })$ with $l_n\to 1$ such that

$$\varphi (p,z_n)\le l_n\varphi (p,x),\mathrm{\forall }x\in C,p\in F(S),z_n\in S^{n}x.$$

We remark from the above definitions that the class of quasi-ϕ-asymptotically nonexpansive multivalued mappings contains properly the class of quasi-ϕ-nonexpansive multivalued mappings as a subclass, and the class of quasi-ϕ-nonexpansive multivalued mappings contains properly the class of relatively nonexpansive mappings as a subclass.

Recently, Yi [21] introduced the following modifying Halpern iterative sequence for a quasi-ϕ-asymptotically nonexpansive multivalued mapping S. Let $\{x_n\}$ be the sequence generated by

$$\{\begin{array}{l}{x}_0\in C,C_0=C,\\ y_n=J^{-1}[{\alpha }_{n}J{x}_0+(1-{\alpha }_n)J{z}_n],z_n\in S^n{x}_n,\\ C_{n+1}=\{v\in C_n:\varphi (v,y_n)\le {\alpha }_n\varphi (v,x_0)+(1-{\alpha }_n)\varphi (v,x_n)+{\zeta }_n\},\\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_0,n\ge 0,\end{array}$$

(1.7)

where ${\zeta }_n=(l_n-1){sup}_{p\in F(S)}\varphi (p,x_n)$ and $\{{\alpha }_n\}\subset (0,1)$. Under suitable conditions, the author proved that $\{x_n\}$ converges strongly to ${\mathrm{\Pi }}_{F(S)}{x}_0$.

Motivated and inspired by the above results, we investigate the problem of finding a common element of the set of solutions of $\mathit{MEP}(f,\phi ,C)$ and the set of common fixed points of a countable family of quasi-ϕ-asymptotically nonexpansive multivalued mappings in Banach spaces.

The rest of the paper is organized as follows. In Section 2, we provide necessary concepts and lemmas. In Section 3, we derive the existence and uniqueness of solutions of the auxiliary problems for the $\mathit{MEP}(f,\phi ,C)$ when the bifunction f is relaxed ξ-monotone by using the well-known KKM technique. In Section 4, we prove that our proposed hybrid iterative scheme converges strongly to a common element of the set of $\mathit{MEP}(f,\phi ,C)$ and the set of common fixed points of a countable family of quasi-ϕ-asymptotically nonexpansive multivalued mappings. In Section 5, we give an application for a system of mixed equilibrium problems. The last section is the conclusions.

2 Preliminaries

Throughout this paper, let E be a real Banach space and $E^{\ast }$ be its dual space. We write $x_n\rightharpoonup x$ (respectively $x_n{\rightharpoonup }^{\ast }x$) to indicate that the sequence $\{x_n\}$ weakly (respectively weak^∗) converges to x; as usual $x_n\to x$ will symbolize strong convergence. Let $S(E)=\{x\in E:\parallel x\parallel =1\}$ denote the unit sphere of a Banach space E.

A Banach space E is said to have a Gâteaux differentiable norm (we also say that E is smooth) if the limit

$$\underset{t\to 0}{lim}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$$

(2.1)

exists for each $x,y\in S(E)$. A Banach space E is said to have a uniformly Gâteaux differentiable norm if for each y in $S(E)$, the limit (2.1) is uniformly attained for $x\in S(E)$; a Fréchet differentiable norm if for each $x\in S(E)$, the limit (2.1) is attained uniformly for $y\in S(E)$; a uniformly Fréchet differentiable norm (we also say that E is uniformly smooth) if the limit (2.1) is attained uniformly for $(x,y)\in S(E)\times S(E)$.

A Banach space E is said to strictly convex if $\frac{\parallel x+y\parallel }{2}<1$ for $x,y\in S(E)$, $x\ne y$; uniformly convex if, for any $\epsilon \in (0,2]$, there exists $\delta >0$ such that for any $x,y\in S(E)$, $\parallel x-y\parallel \ge \epsilon$ implies $\parallel \frac{x+y}{2}\parallel \le 1-\delta$.

Remark 2.1 The following basic properties for the Banach space E and for the normalized duality mapping J can be found in Cioranescu [23].

• (i) E ($E^{\ast }$, resp.) is uniformly convex if and only if E ($E^{\ast }$, resp.) is uniformly smooth.

• (ii) If E is smooth, then J is single-valued and norm-to-weak continuous.

• (iii) If E is reflexive, then J is onto.

• (iv) If E is strictly convex, then $Jx\cap Jy\ne \mathrm{\varnothing }$ for all $x,y\in E$.

• (v) Each uniformly convex Banach space E has the Kadec-Klee property, that is, for any sequence $\{x_n\}\subset E$, if $x_n\rightharpoonup x\in E$ and $\parallel x_n\parallel \to \parallel x\parallel$, then $x_n\to x\in E$.

• (vi) If E is a strictly convex reflexive Banach space, then $J^{-1}$ is hemicontinuous, that is, $J^{-1}$ is norm-to-weak^∗-continuous.

Recall the Lyapunov function ϕ defined by

$$\varphi (x,y)={\parallel x\parallel }^2-2〈x,Jy〉+{\parallel y\parallel }^2,\mathrm{\forall }x,y\in E.$$

It follows from the definition of the function ϕ that

$$\varphi (x,J^{-1}(\lambda Jy+(1-\lambda )Jz))\le \lambda \varphi (x,y)+(1-\lambda )\varphi (x,z),\mathrm{\forall }\lambda \in [0,1]\text{ and }x,y,z\in E.$$

(2.2)

Following Alber [24], the generalized projection ${\mathrm{\Pi }}_C$ from E onto C is defined by

$${\mathrm{\Pi }}_{C}x=arg\underset{y\in C}{min}\varphi (y,x),\mathrm{\forall }x\in C.$$

If E is a Hilbert space, then $\varphi (y,x)={\parallel y-x\parallel }^2$ and ${\mathrm{\Pi }}_C$ is the metric projection of H onto C. We known the following lemma for generalized projection.

Lemma 2.2 [24]

Let E be a smooth, strictly convex and reflexive Banach space, and let C be a nonempty, closed and convex subset of E. Then the following conclusions hold:

• (i) $\varphi (x,y)=0$ if and only if $x=y$ for all $x,y\in E$;

• (ii) $\varphi (x,{\mathrm{\Pi }}_{C}y)+\varphi ({\mathrm{\Pi }}_{C}y,y)\le \varphi (x,y)$ for all $x\in C$, for all $y\in E$;

• (iii) if $x\in E$ and $z\in C$, then $z={\mathrm{\Pi }}_{C}x\iff 〈z-y,Jx-Jz〉\ge 0$ for all $y\in C$.

Now, let us recall the following useful concepts and results.

Definition 2.3 Let B be a subset of a topological vector space X. A mapping $G:B\to 2^X$ is called a KKM mapping if $co\{x_1,x_2,\dots ,x_m\}\subset {\bigcup }_{i=1}^{m}G(x_i)$ for $x_i\in B$ and $i=1,2,\dots ,m$, where coA denotes the convex hull of the set A.

Lemma 2.4 [25]

Let B be a nonempty subset of a Hausdorff topological vector space X, and let $G:B\to 2^X$ be a KKM mapping. If $G(x)$ is closed for all $x\in B$ and is compact for at least one $x\in B$, then ${\bigcap }_{x\in B}G(x)\ne \mathrm{\varnothing }$.

Definition 2.5 A multivalued mapping $S:C\to \mathit{CB}(C)$ is said to be closed if for any sequence $\{x_n\}\subset C$ with $x_n\to x\in C$ and $d(y,S(x_n))\to 0$, $d(y,S(x))=0$.

The following lemmas can be found in [21].

Lemma 2.6 [[21], Lemma 2.8]

Let E be a real uniformly smooth and strictly convex Banach space with the Kadec-Klee property, and let C be a nonempty, closed and convex subset of E. Let $\{x_n\}$ and $\{y_n\}$ be two sequences in C such that $x_n\to p$ and $\varphi (x_n,y_n)\to 0$, then $y_n\to p$.

Lemma 2.7 [[21], Lemma 2.9]

Let E be a real uniformly smooth and strictly convex Banach space with the Kadec-Klee property, and let C be a nonempty, closed and convex subset of E. Let $S:C\to \mathit{CB}(C)$ be a closed and quasi-ϕ-asymptotically nonexpansive multivalued mapping with nonnegative real sequences $\{l_n\}\subset [1,\mathrm{\infty })$, if $l_n\to 1$, then the fixed point set $F(S)$ of S is a closed and convex subset of C.

3 Existence results of mixed equilibrium problem

In this chapter, we prove the existence theorem for $\mathit{MEP}(f,\phi ,C)$ by using the KKM technique.

Before solving mixed equilibrium problem (1.2), let us assume the following conditions for a bifunction $f:C\times C\to \mathbb{R}$:

• (A1) $f(x,x)=0$ for all $x\in C$;

• (A2) f is relaxed ξ-monotone, i.e.,

  $$f(x,y)+f(y,x)\le \xi (y-x);$$

• (A3) for all $y\in C$, $f(\cdot ,y)$ is upper hemicontinuous, i.e., for all $x,y,z\in C$,

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

• (A4) for all $x\in C$, $f(x,\cdot )$ is convex and lower-semicontinuous.

Lemma 3.1 Let C be a nonempty, closed and convex subset of a real smooth, strictly convex and reflexive Banach space E, let f be a bifunction from $C\times C$ to ℝ satisfying (A1)-(A4), and let φ be a lower semicontinuous and convex function from C to ℝ. Let $z\in C$. Then the following problems (3.1) and (3.2) are equivalent:

$$\begin{array}{c}\mathit{\text{Find }}\tilde{x}\in C\mathit{\text{ such that }}f(\tilde{x},y)+\phi (y)+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2-〈y-\tilde{x},Jz〉\ge \phi (\tilde{x}),\hfill \\ \mathrm{\forall }y\in C.\hfill \end{array}$$

(3.1)

$$\begin{array}{c}\mathit{\text{Find }}\tilde{x}\in C\mathit{\text{ such that }}\phi (y)+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2-〈y-\tilde{x},Jz〉+\xi (y-x)\ge f(y,\tilde{x})+\phi (\tilde{x}),\hfill \\ \mathrm{\forall }y\in C.\hfill \end{array}$$

(3.2)

Proof Let $\tilde{x}\in C$ be a solution of problem (3.1). It follows from (A2) that

$$\begin{array}{rcl}0& \le & f(\tilde{x},y)+\phi (y)-\phi (\tilde{x})+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2-〈y-\tilde{x},Jz〉\\ =& f(\tilde{x},y)+f(y,\tilde{x})-f(y,\tilde{x})+\phi (y)-\phi (\tilde{x})+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2-〈y-\tilde{x},Jz〉\\ \le & \xi (y-\tilde{x})-f(y,\tilde{x})+\phi (y)-\phi (\tilde{x})+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2-〈y-\tilde{x},Jz〉,\mathrm{\forall }y\in C.\end{array}$$

(3.3)

Thus $\tilde{x}\in C$ is a solution of problem (3.2).

Conversely, let $\tilde{x}\in C$ be a solution of problem (3.2). For any $t\in (0,1]$, we put

$$y_t=(1-t)\tilde{x}+ty.$$

(3.4)

Then $y_t\in C$, because of the convexity of C. Since $\tilde{x}\in C$ is a solution of problem (3.2), it follows that

$$\begin{array}{rcl}f(y_t,\tilde{x})& \le & \phi (y_t)-\phi (\tilde{x})+\frac{1}{2}{\parallel y_t\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2-〈y_t-\tilde{x},Jz〉+\xi (y_t-x)\\ =& \phi (y_t)-\phi (\tilde{x})+\frac{1}{2}{\parallel y_t\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2-〈y_t-\tilde{x},Jz〉+t^p\xi (y-x).\end{array}$$

(3.5)

Using (A1) and (A4), we have

$$0=f(y_t,y_t)\le (1-t)f(y_t,\tilde{x})+tf(y_t,y),$$

and so

$$t(f(y_t,\tilde{x})-f(y_t,y))\le f(y_t,\tilde{x}).$$

(3.6)

The convexity of the function φ implies that

$$\phi (y_t)=\phi ((1-t)\tilde{x}+ty)\le (1-t)\phi (\tilde{x})+t\phi (y).$$

(3.7)

It follows from (3.5)-(3.7) and the convexity of $\frac{1}{2}{\parallel \cdot \parallel }^2$ that

$$\begin{array}{rcl}0& =& f(y_t,y_t)\\ \le & (1-t)f(y_t,\tilde{x})+tf(y_t,y)\\ \le & (1-t)[\phi (y_t)-\phi (\tilde{x})+\frac{1}{2}{\parallel y_t\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2-〈y_t-\tilde{x},Jz〉+t^p\xi (y-\tilde{x})]+tf(y_t,y)\\ \le & tf(y_t,y)+(1-t)[(1-t)\phi (\tilde{x})+t\phi (y)-\phi (\tilde{x})+\frac{(1-t)}{2}{\parallel \tilde{x}\parallel }^2+\frac{t}{2}{\parallel y\parallel }^2\\ -\frac{1}{2}{\parallel \tilde{x}\parallel }^2-t〈y-\tilde{x},Jz〉+t^p\xi (y-\tilde{x})]\\ \le & tf(y_t,y)+(1-t)t[\phi (y)-\phi (\tilde{x})+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2-〈y-\tilde{x},Jz〉+t^{p-1}\xi (y-\tilde{x})].\end{array}$$

(3.8)

This implies that

$$0\le f(y_t,y)+(1-t)[\phi (y)-\phi (\tilde{x})+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2-〈y-\tilde{x},Jz〉+t^{p-1}\xi (y-\tilde{x})].$$

(3.9)

Taking the upper limit $t\to 0$ in (3.9), by (A3) and $p>1$, we get that

$$\phi (\tilde{x})\le f(\tilde{x},y)+\phi (y)+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2-〈y-\tilde{x},Jz〉,\mathrm{\forall }y\in C.$$

Therefore, $\tilde{x}$ is also a solution of problem (3.1). This completes the proof. □

Lemma 3.2 Let C be a nonempty, bounded, closed and convex subset of a real smooth, strictly convex and reflexive Banach space E, let f be a bifunction from $C\times C$ to ℝ satisfying (A1) and (A4), and let φ be a lower semicontinuous and convex function from C to ℝ. Let $z\in C$. Assume that

• (i) $\xi :E\to \mathbb{R}$ is weakly upper semicontinuous; that is, for any net $\{x_{\beta }\}$, $x_{\beta }$ converges to x in $\sigma (E,E^{\ast })$, which implies that $\xi (x)\le lim\hspace{0.17em}inf\xi (x_{\beta })$.

Then the solution set of problem (3.1) is nonempty; that is, there exists $\tilde{x}\in C$ such that

$$f(\tilde{x},y)+\phi (y)+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2-〈y-\tilde{x},Jz〉\ge \phi (\tilde{x}),\mathrm{\forall }y\in C.$$

(3.10)

Proof Let $z\in C$. Define two set-valued mappings $F_z,G_z:C\to 2^C$ as follows:

$$F_z(y)=\{x\in C:f(x,y)+\phi (y)+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel x\parallel }^2-〈y-x,Jz〉\ge \phi (x)\}$$

and

$$G_z(y)=\{x\in C:\phi (y)+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel x\parallel }^2-〈y-x,Jz〉+\xi (y-x)\ge f(y,x)+\phi (x)\}$$

for every $y\in C$. It is easily seen that $y\in F_z(y)$ and $y\in G_z(y)$, and hence $F_z(y)$ and $G_z(y)$ are nonempty.

(a) We claim that $F_z$ is a KKM mapping. If $F_z$ is not a KKM mapping, then there exist $\{y_1,\dots ,y_n\}\subset K$ and ${\mu }_i>0$, $i=1,\dots ,n$, such that

$$\sum _{i=1}^n{\mu }_i=1,y_0=\sum _{i=1}^n{\mu }_i{y}_i\notin \bigcup _{i=1}^n{F}_z(y_i).$$

(3.11)

By the definition of $F_z$, we have

$$f(y_0,y_i)+\phi (y_i)-\phi (y_0)+\frac{1}{2}{\parallel y_i\parallel }^2-\frac{1}{2}{\parallel y_0\parallel }^2-〈y_i-y_0,Jz〉<0$$

(3.12)

for all $i=1,\dots ,n$. It follows from (A1), (A4), the convexity of φ and $\frac{1}{2}{\parallel \cdot \parallel }^2$ that

$$\begin{array}{rcl}0& =& f(y_0,y_0)+\phi (y_0)-\phi (y_0)+\frac{1}{2}{\parallel y_0\parallel }^2-\frac{1}{2}{\parallel y_0\parallel }^2-〈y_0-y_0,Jz〉\\ \le & \sum _{i=1}^n{\mu }_i(f(y_0,y_i)+\phi (y_i)-\phi (y_0)+\frac{1}{2}{\parallel y_i\parallel }^2-\frac{1}{2}{\parallel y_0\parallel }^2-〈y_i-y_0,Jz〉)<0,\end{array}$$

(3.13)

which is a contradiction. This implies that $F_z$ is a KKM mapping.

(b) We claim that $G_z$ is a KKM mapping. It is sufficient to show that

$$F_z(y)\subset G_z(y),\mathrm{\forall }y\in C.$$

For any given $y\in C$, taking $x\in F_z(y)$, then

$$f(x,y)+\phi (y)-\phi (x)+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel x\parallel }^2-〈y-x,Jz〉\ge 0.$$

(3.14)

It follows from the relaxed ξ-monotonicity of f that

$$\phi (y)+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel x\parallel }^2-〈y-x,Jz〉+\xi (y-x)\ge f(y,x)+\phi (x).$$

It follows that $x\in G_z(y)$ and so

$$F_z(y)\subset G_z(y),\mathrm{\forall }y\in C.$$

This implies that $G_z$ is also a KKM mapping.

(c) We show that $G_z(y)$ is weakly closed for all $y\in C$. Let $\{u_n\}$ be a sequence in $G_z(y)$ such that $u_n\rightharpoonup u$ as $n\to \mathrm{\infty }$. It then follows from $u_n\in G_z(y)$ that

$$\phi (y)+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel u_n\parallel }^2-〈y-u_n,Jz〉+\xi (y-u_n)\ge f(y,u_n)+\phi (u_n),\mathrm{\forall }y\in C.$$

(3.15)

By (A4), the weak lower semicontinuity of φ and ${\parallel \cdot \parallel }^2$, and the weak upper semicontinuity of ξ, we obtain from (3.15) that

$$\begin{array}{rcl}f(y,u)+\phi (u)& \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}f(y,u_n)+\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\phi (u_n)\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{f(y,u_n)+\phi (u_n)\}\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\{\phi (y)+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel u_n\parallel }^2-〈y-u_n,Jz〉+\xi (y-u_n)\}\\ \le & \phi (y)+\frac{1}{2}{\parallel y\parallel }^2-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{1}{2}{\parallel u_n\parallel }^2-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}〈y-u_n,Jz〉\\ +\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\xi (y-u_n)\\ =& \phi (y)+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel u\parallel }^2-〈y-u,Jz〉+\xi (y-u).\end{array}$$

This shows that $u\in G_z(y)$ and hence $G_z(y)$ is weakly closed for all $y\in C$.

• (d) We prove that $G_z(y)$ is weakly compact. Since C is a closed, bounded and convex subset of a reflexive Banach space E, it is weakly compact. Again, since $G_z(y)$ is a weakly closed subset of C, we also have $G_z(y)$ is weakly compact.

By using (a)-(d) and Lemma 2.4 and Lemma 3.1 that

$$\bigcap _{y\in C}{F}_z(y)=\bigcap _{y\in C}{G}_z(y)\ne \mathrm{\varnothing }.$$

Hence, there exists $x_0\in C$ satisfying inequality (3.10). This completes the proof. □

Proposition 3.3 Let C be a nonempty, closed and convex subset of a real smooth, strictly convex and reflexive Banach space E, $\psi :C\to \mathbb{R}$ be a convex function. The following two inequalities are equivalent:

$$\mathit{\text{there exists }}\tilde{x}\in C\mathit{\text{ such that }}\psi (\tilde{x})=0\mathit{\text{ and }}\psi (y)+〈y-\tilde{x},J\tilde{x}〉\ge 0,\mathrm{\forall }y\in C$$

(3.16)

and

$$\mathit{\text{there exists }}\tilde{x}\in C\mathit{\text{ such that }}\psi (\tilde{x})=0\mathit{\text{ and }}\psi (y)+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2\ge 0,\mathrm{\forall }y\in C.$$

(3.17)

Proof Let $\tilde{x}$ satisfy (3.16). It is well known that

$$〈y-x,Jx〉\le \frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel x\parallel }^2,\mathrm{\forall }x,y\in E.$$

Then

$$\begin{array}{rcl}0& \le & \psi (y)+〈y-\tilde{x},J\tilde{x}〉\\ =& \psi (y)+〈y-\tilde{x},J\tilde{x}〉\\ \le & \psi (y)+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2.\end{array}$$

Let $\tilde{x}$ satisfy (3.17). For any $t\in (0,1]$, let $y_t=(1-t)\tilde{x}+ty$. Then $y_t\in C$ because of the convexity of C, and so

$$\psi (y_t)+\frac{1}{2}{\parallel y_t\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2-〈y_t-\tilde{x},Jz〉\ge 0.$$

Notice that in a real smooth, strictly convex reflexive Banach space E, the duality mapping J is single-valued, 1-1, and onto. Since $\frac{1}{2}{\parallel \cdot \parallel }^2$ is continuous and Gâteaux differentiable, from the mean value theorem, there exists ${\alpha }_t\in (0,1)$ such that

$$\frac{1}{2}{\parallel y_t\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2=〈y_t-\tilde{x},J{w}_{{\alpha }_t}〉,$$

where $w_{{\alpha }_t}={\alpha }_t\tilde{x}+(1-{\alpha }_t)y_t$. Hence

$$\begin{array}{rcl}0& \le & \psi (y)+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2\\ =& \psi (y_t)+〈y_t-\tilde{x},J{w}_{{\alpha }_t}〉\\ \le & (1-t)\psi (\tilde{x})+t\psi (y)+t〈y-\tilde{x},J{w}_{{\alpha }_t}〉\\ =& t\psi (y)+t〈y-\tilde{x},J{w}_{{\alpha }_t}〉.\end{array}$$

Dividing t in the above inequality, we get

$$0\le \psi (y)+〈y-\tilde{x},J{w}_{{\alpha }_t}〉.$$

By the existence of $w_{{\alpha }_t}$, $w_{{\alpha }_t}\to \tilde{x}$ as $t\to 0$. Since J is norm-to-weak^∗ continuous, we have that

$$0\le \psi (y)+〈y-\tilde{x},J\tilde{x}〉.$$

This completes the proof. □

Remark 3.4 If $\tilde{x}\in C$ is a solution of (3.1), then

$$f(\tilde{x},y)+\phi (y)+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2-〈y-\tilde{x},Jz〉\ge \phi (\tilde{x}),\mathrm{\forall }y\in C.$$

For any fixed $z\in E$, we put ${\psi }_{\tilde{x}}(y)=f(\tilde{x},y)+\phi (y)-\phi (\tilde{x})-〈y-\tilde{x},Jz〉$ for all $y\in C$. Obviously, ${\psi }_{\tilde{x}}(\tilde{x})=0$ and ${\psi }_{\tilde{x}}$ is a convex function, since the linearity of a duality mapping and the convexity of $f(\tilde{x},\cdot )$ and φ. Hence

$${\psi }_{\tilde{x}}(y)+\frac{1}{2}{\parallel y\parallel }^2-\frac{1}{2}{\parallel \tilde{x}\parallel }^2\ge 0,\mathrm{\forall }y\in C.$$

It then follows from Proposition 3.3 that $\tilde{x}$ is a solution of the following problem:

$$f(\tilde{x},y)+\phi (y)+〈y-\tilde{x},J\tilde{x}-Jz〉\ge \phi (\tilde{x}),\mathrm{\forall }y\in C.$$

The existence result relaxes the results of Ceng and Yao [1] and Cholamjiak and Suantai [2], because of the ξ-monotonicity of f.

Lemma 3.5 Let C be a closed, bounded and convex subset of a uniformly smooth, strictly convex Banach space E, let f be a bifunction from $C\times C$ to ℝ satisfying (A1)-(A4), and let φ be a lower semicontinuous and convex function from C to ℝ. Suppose further that

• (i) $\xi :E\to \mathbb{R}$ is weakly upper semicontinuous;

• (ii) $\xi (y-x)+\xi (x-y)\le 0$.

Define a mapping $T_r(z):E\to C$ as follows:

$$T_r(z)=\{x\in C:f(x,y)+\phi (y)+\frac{1}{r}〈y-x,Jx-Jz〉\ge \phi (x),\mathrm{\forall }y\in C\}.$$

(3.18)

Then

• (1) $T_r$ is single-valued;

• (2) $T_r$ is a firmly nonexpansive mapping, i.e., for all $x,y\in E$,

  $$〈T_r(x)-T_r(y),J{T}_r(x)-J{T}_r(y)〉\le 〈T_r(x)-T_r(y),Jx-Jy〉;$$

• (3) $F(T_r)=\mathit{MEP}(f,\phi ,C)$;

• (4) $\mathit{MEP}(f,\phi ,C)$ is closed and convex;

• (5) $\varphi (p,T_{r}x)+\varphi (T_{r}x,x)\le \varphi (p,x)$.

Proof If $f(x,y):C\times C\to \mathbb{R}$ satisfies (A1)-(A4) and $\phi :C\to \mathbb{R}$ is convex lower semicontinuous, then for any $r>0$, we known that $rf(x,y)$ and rφ are also. We have from Lemma 3.2 and Remark 3.4 that $T_{r}z$ is nonempty for all $z\in E$.

(1) For each $z\in E$, let $x_1,x_2\in T_r(z)$. Then

$$f(x_1,x_2)+\phi (x_2)+\frac{1}{r}〈x_2-x_1,J{x}_1-Jz〉\ge \phi (x_1)$$

(3.19)

and

$$f(x_2,x_1)+\phi (x_1)+\frac{1}{r}〈x_1-x_2,J{x}_2-Jz〉\ge \phi (x_2).$$

(3.20)

Adding (3.19) and (3.20), we have

$$f(x_2,x_1)+f(x_1,x_2)+\frac{1}{r}〈x_1-x_2,J{x}_2-Jz〉+\frac{1}{r}〈x_2-x_1,J{x}_1-Jz〉\ge 0,$$

and so

$$f(x_2,x_1)+f(x_1,x_2)+\frac{1}{r}〈x_2-x_1,J{x}_1-J{x}_2〉\ge 0.$$

(3.21)

By the relaxed ξ-monotonicity of f, we know that

$$〈x_2-x_1,J{x}_1-J{x}_2〉\ge -r\xi (y-x).$$

(3.22)

In (3.21) exchanging the position of $x_1$ and $x_2$, we get

$$f(x_1,x_2)+f(x_2,x_1)+\frac{1}{r}〈x_1-x_2,J{x}_2-J{x}_1〉\ge 0,$$

(3.23)

and so

$$〈x_1-x_2,J{x}_2-J{x}_1〉\ge -r\xi (y-x).$$

(3.24)

Now, adding inequalities (3.22) and (3.24), by using (ii), we have

$$〈x_2-x_1,J{x}_1-J{x}_2〉\ge -\frac{r}{2}(\xi (y-x)+\xi (y-x))\ge 0.$$

Since J is monotone and E is strictly convex, we have $x_1=x_2$, and so $T_r(z)$ is single-valued.

(2) For $x,y\in C$, we have that

$$f(T_{r}x,T_{r}y)+\phi (T_{r}y)-\phi (T_{r}x)+\frac{1}{r}〈T_{r}y-T_{r}x,J{T}_{r}x-Jx〉\ge 0$$

and

$$f(T_{r}y,T_{r}x)+\phi (T_{r}x)-\phi (T_{r}y)+\frac{1}{r}〈T_{r}x-T_{r}y,J{T}_{r}y-Jy〉\ge 0.$$

Adding the two inequalities above, we get that

$$f(T_{r}x,T_{r}y)+f(T_{r}y,T_{r}x)+\frac{1}{r}〈T_{r}y-T_{r}x,J{T}_{r}x-J{T}_{r}y-Jx+Jy〉\ge 0.$$

From (A2), we obtain that

$$\xi (T_{r}y-T_{r}x)+\frac{1}{r}〈T_{r}y-T_{r}x,J{T}_{r}x-J{T}_{r}y-Jx+Jy〉\ge 0.$$

(3.25)

In (3.25), interchanging the position of $T_{r}x$ and $T_{r}y$, we get

$$\xi (T_{r}x-T_{r}y)+\frac{1}{r}〈T_{r}x-T_{r}y,J{T}_{r}y-J{T}_{r}x-Jy+Jx〉\ge 0.$$

(3.26)

Again adding (3.24) and (3.25), we get

$$\xi (T_{r}y-T_{r}x)+\xi (T_{r}x-T_{r}y)+\frac{2}{r}〈T_{r}y-T_{r}x,J{T}_{r}x-J{T}_{r}y-Jx+Jy〉\ge 0.$$

It follows from (ii) and $r>0$ that

$$〈T_{r}x-T_{r}y,J{T}_{r}x-J{T}_{r}y〉\le 〈T_{r}y-T_{r}x,Jx-Jy〉.$$

(3) Indeed, we have the following:

$$\begin{array}{rcl}p\in F(T_r)& \iff & p=T_{r}p\\ \iff & f(p,y)+\phi (y)-\phi (p)+\frac{1}{r}〈y-p,Jp-Jp〉\ge 0,\mathrm{\forall }y\in C\\ \iff & f(p,y)+\phi (y)-\phi (p)\ge 0,\mathrm{\forall }y\in C\\ \iff & p\in \mathit{MEP}(f,\phi ,C).\end{array}$$

(4) We claim that $\mathit{MEP}(f,\phi ,C)$ is closed and convex. Indeed, from (3) we have $\mathit{MEP}(f,\phi ,C)=F(T_r)$. From (2) we have, for all $x,y\in C$,

$$〈T_r(x)-T_r(y),J{T}_r(x)-J{T}_r(y)〉\le 〈T_r(x)-T_r(y),Jx-Jy〉.$$

Moreover, we have

$$\begin{array}{rcl}\varphi (T_{r}x,T_{r}y)+\varphi (T_{r}y,T_{r}x)& =& 2{\parallel T_{r}x\parallel }^2-〈T_{r}x,J{T}_{r}y〉-〈T_{r}y,J{T}_{r}x〉+2{\parallel T_{r}y\parallel }^2\\ =& 2〈T_{r}x,J{T}_{r}x-J{T}_{r}y〉+2〈T_{r}y,T_{r}y-J{T}_{r}x〉\\ =& 2〈T_{r}x-T_{r}y,J{T}_{r}x-J{T}_{r}y〉\end{array}$$

and

$$\begin{array}{r}\varphi (T_{r}x,y)+\varphi (T_{r}y,x)-\varphi (T_{r}x,x)-\varphi (T_{r}y,y)\\ ={\parallel T_{r}x\parallel }^2-2〈T_{r}x,Jy〉+{\parallel y\parallel }^2+{\parallel T_{r}y\parallel }^2-〈T_{r}y,Jx〉+{\parallel x\parallel }^2\\ =-{\parallel T_{r}x\parallel }^2+2〈T_{r}x,Jx〉-{\parallel x\parallel }^2-{\parallel T_{r}y\parallel }^2+〈T_{r}y,Jy〉-{\parallel y\parallel }^2\\ =2〈T_{r}x,Jx-Jy〉-2〈T_{r}y,Jx-Jy〉\\ =2〈T_{r}x-T_{r}y,Jx-Jy〉.\end{array}$$

Hence, we have

$$\varphi (T_{r}x,T_{r}y)+\varphi (T_{r}y,T_{r}x)\le \varphi (T_{r}x,y)+\varphi (T_{r}y,x)-\varphi (T_{r}x,x)-\varphi (T_{r}y,y).$$

(3.27)

So, we have, for any $x,y\in C$,

$$\varphi (T_{r}x,T_{r}y)+\varphi (T_{r}y,T_{r}x)\le \varphi (T_{r}x,y)+\varphi (T_{r}y,x).$$

Next, we show that $\hat{F}(T_r)=\mathit{MEP}(f,\phi ,C)$. Let $p\in \hat{F}(T_r)$. Then there exists $\{u_n\}\subset E$ such that $z_n\rightharpoonup p$ and ${lim}_{n\to \mathrm{\infty }}(u_n-T_r{u}_n)=0$. Moreover, we get that $T_r{u}_n\rightharpoonup p$. Since the duality mapping J is uniformly continuous on a bounded set, we get

$$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{r}\parallel J{u}_n-J{T}_r{u}_n\parallel =0.$$

From the definition of $T_r$, we have

$$f(T_r{u}_n,y)+\frac{1}{r}〈y-T_r{u}_n,J{T}_r{u}_n-J{u}_n〉\ge \phi (T_r{u}_n).$$

Since

$$\begin{array}{c}\frac{1}{r}〈y-T_r{u}_n,J{T}_r{u}_n-J{u}_n〉\hfill \\ \ge \phi (T_r{u}_n)-\phi (y)-f(T_r{u}_n,y)\hfill \\ \ge \phi (T_r{u}_n)-\phi (y)+f(y,T_r{u}_n)-\xi (y-T_r{u}_n).\hfill \end{array}$$

By (A4), the convexity and lower semicontinuity of $\phi (x)$ and the weak upper semicontinuity of ξ, we can obtain that

$$\begin{array}{rcl}0& \ge & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{\phi (T_r{u}_n)-\phi (y)+f(y,T_r{u}_n)-\xi (y-T_r{u}_n)\}\\ =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\phi (T_r{u}_n)-\phi (y)+\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}f(y,T_r{u}_n)-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\xi (y-T_r{u}_n)\\ \ge & \phi (p)-\phi (y)+f(y,p)-\xi (y-p).\end{array}$$

This implies that

$$\phi (p)-\phi (y)+f(y,p)-\xi (y-p)\le 0,\mathrm{\forall }y\in C.$$

(3.28)

Let $y\in K$ and set $p_t=ty+(1-t)p$ for $t\in (0,1]$. It follows from (A1) and (A4) that

$$\begin{array}{rl}0& =f(p_t,p_t)\\ \le (1-t)f(p_t,p)+tf(p_t,y).\end{array}$$

(3.29)

The convexity of the function φ implies that

$$\phi (p_t)=\phi ((1-t)p+ty)\le (1-t)\phi (p)+t\phi (y).$$

(3.30)

It follows from (3.28)-(3.30) that

$$0=f(p_t,p_t)$$

(3.31)

$$\begin{array}{rc}\le & (1-t)f(p_t,p)+tf(p_t,y)\\ \le & tf(p_t,y)+(1-t)[\phi (p_t)-\phi (p)+t^p\xi (y-p)]\\ \le & tf(p_t,y)+(1-t)[(1-t)\phi (p)+t\phi (y)-\phi (p)+t^p\xi (y-p)]\\ =& tf(p_t,y)+(1-t)t[\phi (y)-\phi (p)+t^{p-1}\xi (y-p)].\end{array}$$

(3.32)

Dividing by t, we have

$$0\le f(p_t,y)+(1-t)[\phi (y)-\phi (p)+t^{p-1}\xi (y-p)].$$

(3.33)

By (A3) and $p>1$, taking the upper limit $t\to 0$ in (3.33), we get

$$f(p,y)+\phi (y)\ge \phi (p),\mathrm{\forall }y\in C.$$

Hence, $p\in \mathit{MEP}(f,\phi ,C)$, and so $F(T_r)=\mathit{MEP}(f,\phi ,C)=\hat{F}(T_r)$. Therefore, we have $T_r$ is a relatively nonexpansive mapping. From Lemma 1.2, $\mathit{MEP}(f,\phi ,C)=F(T_r)$ is closed and convex.

(5) From (3.27) we have, for all $x,y\in E$,

$$\varphi (T_{r}x,T_{r}y)+\varphi (T_{r}y,T_{r}x)\le \varphi (T_{r}x,y)+\varphi (T_{r}y,x)-\varphi (T_{r}x,x)-\varphi (T_{r}y,y).$$

Letting $y=p\in F(T_r)$, we have

$$\varphi (p,T_{r}x)+\varphi (T_{r}x,x)\le \varphi (p,x).$$

This completes the proof. □

4 Strong convergence theorems

Before proving the convergence theorem, we recall some definitions of a countable family of multivalued mappings ${\{S_i:C\to \mathit{CB}(C)\}}_{i=1}^{\mathrm{\infty }}$.

Definition 4.1 ${\{S_i\}}_{i=1}^{\mathrm{\infty }}$ is said to be a family of uniformly quasi-ϕ-asymptotically nonexpansive mappings [15] if $\mathcal{F}:={\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)\ne \mathrm{\varnothing }$ and there exists a sequence $\{l_n\}\subset [1,\mathrm{\infty })$ with $l_n\to 1$ such that for each $i\ge 1$,

$$\varphi (p,z_{n,i})\le l_n\varphi (p,x),\mathrm{\forall }p\in \mathcal{F},z_{n,i}\in S_i^{n}x,\mathrm{\forall }n\ge 1.$$

Definition 4.2 A mapping $S:C\to \mathit{CB}(C)$ is said to be uniformly L-Lipschitz continuous [15] if there exists a constant $L>0$ such that $\parallel x_n-y_n\parallel \le L\parallel x-y\parallel$ for all $x,y\in C$, $x_n\in T^{n}x$, $y_n\in T^{n}y$.

Theorem 4.3 Let E be a real uniformly smooth and strictly convex Banach space with the Kadec-Klee property, and let C be a nonempty, bounded, closed and convex subset of E. Let f be a bifunction from $C\times C$ to ℝ satisfying (A1)-(A4), and let φ be a lower semicontinuous and convex function from C to ℝ. Let ${\{S_i:C\to \mathit{CB}(C)\}}_{i=1}^{\mathrm{\infty }}$ be a family of closed and uniformly quasi-ϕ-asymptotically nonexpansive multivalued mappings with a sequence $\{l_n\}\subset [1,\mathrm{\infty })$, $l_n\to 1$. Suppose that for each $i\ge 1$, $S_i$ is uniformly $L_i$-Lipschitz continuous and $\mathcal{F}\cap \mathit{MEP}(f,\phi ,C)\ne \mathrm{\varnothing }$.

Let $\{x_n\}$ be a sequence in C generated by

$$\{\begin{array}{l}{x}_0\in C,C_0=C,\\ w_n=J^{-1}({\alpha }_{n,0}J{x}_0+{\sum }_{i=1}^{\mathrm{\infty }}{\alpha }_{n,i}J{z}_{n,i}),z_{n,i}\in S_i^n{x}_n,\\ u_n\in K\mathit{\text{such that }}\\ f(u_n,y)+\phi (y)+\frac{1}{r_n}〈y-u_n,J{u}_n-J{w}_n〉\ge \phi (u_n),\mathrm{\forall }y\in C,\\ C_{n+1}=\{v\in C_n:\varphi (v,u_n)\le {\alpha }_{n,0}\varphi (v,x_0)+(1-{\alpha }_{n,0})\varphi (v,x_n)+{\zeta }_n\},\\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_0,n\ge 0,\end{array}$$

(4.1)

where for each ${\{r_n\}}_{n=0}^{\mathrm{\infty }}\subset [a,\mathrm{\infty })$ for some $a>0$, ${\{{\alpha }_{n,i}\}}_{n=0,i=0}^{\mathrm{\infty }}\subset [0,1]$, and ${\zeta }_n={sup}_{u\in \mathcal{F}}(l_n-1)\varphi (u,x_n)$. If ${\sum }_{i=0}^{\mathrm{\infty }}{\alpha }_{n,i}=1$, $\mathrm{\forall }n\ge 0$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n,0}=0$, then $\{x_n\}$ converges strongly as $n\to \mathrm{\infty }$ to ${\mathrm{\Pi }}_{\mathrm{\Omega }}{x}_0$, where $\mathrm{\Omega }:=\mathcal{F}\cap \mathit{MEP}(f,\phi ,C)$.

Proof We divide the proof into five steps. Firstly, we rewrite algorithm (4.1) as follows:

$$\{\begin{array}{l}{x}_0\in C,C_0=C,\\ w_n=J^{-1}({\alpha }_{n,0}J{x}_0+{\sum }_{i=1}^{\mathrm{\infty }}{\alpha }_{n,i}J{z}_{n,i}),z_{n,i}\in S_i^n{x}_n,\\ C_{n+1}=\{v\in C_n:\varphi (v,T_{r_n}{w}_n)\le {\alpha }_{n,0}\varphi (v,x_0)+(1-{\alpha }_{n,0})\varphi (v,x_n)+{\zeta }_n\},\\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_0,n\ge 0,\end{array}$$

(4.2)

where $T_r$ is the mapping defined by (3.18) for all $r>0$.

Step 1. We first show that the sequence $\{x_n\}$ is well defined. It suffices to prove that $C_n$ is closed and convex and that $\mathrm{\Omega }\subset C_n$ for all $n\ge 0$. Suppose that $C_n$ is closed and convex for some $n\ge 0$. By the definition of ϕ, we have

$$\begin{array}{rcl}{C}_{n+1}& =& \{v\in C_n:\varphi (v,u_n)\le {\alpha }_{n,0}\varphi (v,x_0)+(1-{\alpha }_{n,0})\varphi (v,x_n)+{\zeta }_n\}\\ =& \{v\in C:\varphi (v,u_n)\le {\alpha }_{n,0}\varphi (v,x_0)+(1-{\alpha }_{n,0})\varphi (v,x_n)+{\zeta }_n\}\cap C_n\\ =& \{v\in C:2{\alpha }_{n,0}〈v,J{x}_0〉+2(1-{\alpha }_{n,0})〈v,J{x}_n〉-2〈v,J{u}_n〉\\ \le {\alpha }_{n,0}{\parallel x_0\parallel }^2+(1-{\alpha }_{n,0}){\parallel x_n\parallel }^2-{\parallel y_n\parallel }^2+{\zeta }_n\}\cap C_n.\end{array}$$

This shows that $C_{n+1}$ is closed and convex. The conclusions are proved.

Step 2. Next, we prove that $\mathrm{\Omega }\subset C_n$ for all $n\ge 0$. In fact, it is obvious that $\mathrm{\Omega }\subset C_0:=C$. Suppose $\mathrm{\Omega }\subset C_n$ for some $n\ge 0$. Hence, for any $p\in \mathrm{\Omega }$, it follows from Lemma 3.5(5) that

$$\begin{array}{rcl}\varphi (p,u_n)& =& \varphi (p,T_{r_n}{w}_n)\le \varphi (p,w_n)\\ =& \varphi (p,J^{-1}({\alpha }_{n,0}J{x}_0+\sum _{i=1}^{\mathrm{\infty }}{\alpha }_{n,i}J{z}_{n,i}))\\ \le & {\alpha }_{n,0}\varphi (p,x_0)+\sum _{i=1}^{\mathrm{\infty }}{\alpha }_{n,i}\varphi (p,z_{n,i})\\ \le & {\alpha }_{n,0}\varphi (p,x_0)+(1-{\alpha }_{n,0})l_n\varphi (p,x_n)\\ \le & {\alpha }_{n,0}\varphi (p,x_0)+(1-{\alpha }_{n,0})\{\varphi (p,x_n)+(l_n-1)\varphi (p,x_n)\}\\ \le & {\alpha }_{n,0}\varphi (p,x_0)+(1-{\alpha }_{n,0})\{\varphi (p,x_n)+(l_n-1)\underset{p\in \mathcal{F}}{sup}\varphi (p,x_n)\}\\ =& {\alpha }_{n,0}\varphi (p,x_0)+(1-{\alpha }_{n,0})\varphi (p,x_n)+{\zeta }_n.\end{array}$$

(4.3)

This shows that $p\in \mathrm{\Omega }\subset C_{n+1}$. Hence $\mathrm{\Omega }\subset C_n$ for all $n\ge 0$.

Step 3. We show that $\{x_n\}$ converges strongly to some point $\tilde{x}$. Since $x_n={\mathrm{\Pi }}_{C_n}{x}_0$, from Lemma 2.2(iii), we have that

$$〈x_n-y,J{x}_0-J{x}_n〉\ge 0,\mathrm{\forall }y\in C_n.$$

Also, since $\mathcal{F}\subset C_n$, we have

$$〈x_n-p,J{x}_0-J{x}_n〉\ge 0,\mathrm{\forall }p\in \mathcal{F}.$$

It follows from Lemma 2.2(ii) that for each $p\in \mathcal{F}$ and for each $n\ge 1$,

$$\varphi (x_n,x_0)=\varphi ({\mathrm{\Pi }}_{C_n}{x}_0,x_0)\le \varphi (p,x_0)-\varphi (p,x_n)\le \varphi (p,x_0).$$

Therefore, $\{\varphi (x_n,x_0)\}$ is bounded, and so is $\{x_n\}$. Since $x_n={\mathrm{\Pi }}_{C_n}{x}_0$ and $x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_0\in C_{n+1}\subset C_n$, we have

$$\varphi (x_n,p)\le \varphi (x_{n+1},p),\mathrm{\forall }n\ge 0.$$

That is, $\{\varphi (x_n,p)\}$ is a nondecreasing sequence, and so ${lim}_{n\to \mathrm{\infty }}\varphi (x_n,p)$ exists. Since E is reflexive, there exists a subsequence $\{x_{n_i}\}\subset \{x_n\}$ such that $x_{n_i}\rightharpoonup \tilde{x}\in C$. Since $C_n$ is closed, convex and $C_{n+1}\subset C_n$, we get that $C_n$ is weakly closed and $\tilde{x}\in C_n$ for all $n\ge 0$. Since $x_{n_i}={\mathrm{\Pi }}_{C_{n_i}}{x}_0$, we have

$$\varphi (x_{n_i},x_0)\le \varphi (\tilde{x},x_0),\mathrm{\forall }{n}_i\ge 0.$$

By the weak lower semicontinuity of $\parallel \cdot \parallel$, we have

$$\begin{array}{rcl}\underset{n_i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\varphi (x_{n_i},x_0)& =& \underset{n_i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}({\parallel x_{n_i}\parallel }^2-2〈x_{n_i},J{x}_0〉+{\parallel x_0\parallel }^2)\\ \ge & {\parallel \tilde{x}\parallel }^2-2〈\tilde{x},J{x}_0〉+{\parallel x_0\parallel }^2\\ =& \varphi (\tilde{x},x_0),\end{array}$$

and so

$$\varphi (\tilde{x},x_0)\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\varphi (x_{n_i},x_0)\le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\varphi (x_{n_i},x_0)\le \varphi (\tilde{x},x_0).$$

Therefore ${lim}_{i\to \mathrm{\infty }}\varphi (x_{n_i},x_0)=(\tilde{x},x_0)$, and so $\parallel x_{n_i}\parallel \to \parallel \tilde{x}\parallel$. Since E has the Kadec-Klee property and $x_{n_i}\rightharpoonup \tilde{x}$, we obtain that $x_{n_i}\to \tilde{x}$. Since the limit of $\{\varphi (x_n,x_0)\}$ exists, this together with ${lim}_{n\to \mathrm{\infty }}\varphi (x_{n_i},x_0)=\varphi (\tilde{x},x_0)$ implies that ${lim}_{n\to \mathrm{\infty }}\varphi (x_n,x_0)=\varphi (\tilde{x},x_0)$. If there exists a subsequence $\{x_{n_j}\}\subset \{x_n\}$ such that $x_{n_j}\rightharpoonup \hat{x}$. Then by Lemma 2.2(ii) we have that

$$\begin{array}{rcl}\varphi (\tilde{x},\hat{x})& =& \underset{i,j\to \mathrm{\infty }}{lim}\varphi (x_{n_i},x_{n_j})=\underset{i,j\to \mathrm{\infty }}{lim}\varphi (x_{n_i},{\mathrm{\Pi }}_{C_{n_j}}{x}_0)\\ \le & \underset{i,j\to \mathrm{\infty }}{lim}\{\varphi (x_{n_i},x_0)-\varphi ({\mathrm{\Pi }}_{C_{n_j}}{x}_0,x_0)\}=\underset{i,j\to \mathrm{\infty }}{lim}\{\varphi (x_{n_i},x_0)-\varphi (x_{n_j},x_0)\}\\ =& \varphi (\tilde{x},x_0)-\varphi (\tilde{x},x_0).\end{array}$$

Hence $\tilde{x}=\hat{x}$, and so

$$\underset{n\to \mathrm{\infty }}{lim}{x}_n=\tilde{x}.$$

(4.4)

This implies that

$${\zeta }_n=(l_n-1)\underset{p\in \mathcal{F}}{sup}\varphi (p,x_n)\to 0.$$

(4.5)

Step 4. We prove that $\tilde{x}\in \mathcal{F}$. Since $x_n={\mathrm{\Pi }}_{C_n}{x}_0$, by Lemma 2.2(ii) we get that

$$\varphi (x,x_n)\le \varphi (x,x_0)-\varphi (x_n,x_0),\mathrm{\forall }x\in C_n.$$

Since $x_{n+1}\in C_{n+1}\subset C_n$, we get that

$$\varphi (x_{n+1},x_n)\le \varphi (x_{n+1},x_0)-\varphi (x_n,x_0).$$

Since ${lim}_{n\to \mathrm{\infty }}\varphi (x_n,x_0)=\varphi (\tilde{x},x_0)$, we get that

$$\underset{n\to \mathrm{\infty }}{lim}\varphi (x_{n+1},x_n)=0.$$

(4.6)

Since $x_{n+1}\in C_{n+1}$, we have from (4.1) that

$$\varphi (x_{n+1},u_n)\le {\alpha }_{n,0}\varphi (v,x_0)+(1-{\alpha }_{n,0})\varphi (x_{n+1},x_n)+{\zeta }_n.$$

(4.7)

By ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n,0}=0$, (4.5) and (4.6), it implies that

$$\underset{n\to \mathrm{\infty }}{lim}\varphi (x_{n+1},u_n)=0,$$

(4.8)

which together with (4.4) and Lemma 2.6 give that

$$\underset{n\to \mathrm{\infty }}{lim}{u}_n=\tilde{x}.$$

(4.9)

Recall from (4.3) that

$$\varphi (p,u_n)\le \varphi (p,w_n)\le {\alpha }_{n,0}\varphi (p,x_0)+(1-{\alpha }_{n,0})\varphi (p,x_n)+{\zeta }_n.$$

It follows from ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n,0}=0$, (4.4), (4.5) and (4.9) that

$$\underset{n\to \mathrm{\infty }}{lim}\varphi (p,w_n)=\varphi (p,\tilde{x}).$$

By Lemma 3.5, we get that

$$\varphi (u_n,w_n)\le \varphi (p,w_n)-\varphi (p,u_n)\to 0\text{as }n\to \mathrm{\infty }.$$

(4.10)

It follows from Lemma 2.6 and ${lim}_{n\to \mathrm{\infty }}{u}_n=\tilde{x}$ that

$$\underset{n\to \mathrm{\infty }}{lim}{w}_n=\tilde{x}.$$

(4.11)

From (4.1), we get that

$$\parallel J{w}_n-J{z}_{n,i}\parallel ={\alpha }_{n,0}\parallel J{x}_0-J{z}_{n,i}\parallel ,\mathrm{\forall }i\ge 1.$$

Since ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n,0}=0$ and the boundedness of $\{z_{n,i}\}$,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{w}_n-J{z}_{n,i}\parallel =0,\mathrm{\forall }i\ge 1.$$

(4.12)

Since $J{w}_n\to J\tilde{x}$, we have from the uniform continuity of J that $J{z}_{n_i}\to J\tilde{x}$ for all $i\ge 1$. Remark 2.1(vi) gives that

$$z_{n,i}\rightharpoonup \tilde{x},\mathrm{\forall }i\ge 1.$$

(4.13)

Again, since

$$|\parallel z_{n,i}\parallel -\parallel \tilde{x}\parallel |=|\parallel J{z}_{n,i}\parallel -\parallel J\tilde{x}\parallel |\le \parallel J{z}_{n,i}-J\tilde{x}\parallel \to 0\text{as }n\to \mathrm{\infty },\mathrm{\forall }i\ge 1,$$

(4.14)

it follows from (4.13), (4.14) and the Kadec-Klee property of E that

$$\underset{n\to \mathrm{\infty }}{lim}{z}_{n,i}=\tilde{x},\mathrm{\forall }i\ge 1.$$

(4.15)

For all $i\ge 1$, we consider

$$\begin{array}{rcl}d(S_i{z}_{n,i},z_{n,i})& \le & d(S_i{z}_{n,i},z_{n+1,i})+\parallel z_{n+1,i}-x_{n+1}\parallel +\parallel x_{n+1}-x_n\parallel +\parallel x_n-z_{n,i}\parallel \\ \le & d(S_i^{n+1}{x}_n,S_i^{n+1}{x}_{n+1})+\parallel z_{n+1,i}-x_{n+1}\parallel +\parallel x_{n+1}-x_n\parallel +\parallel x_n-z_{n,i}\parallel \\ \le & L_i\parallel x_n-x_{n+1}\parallel +\parallel z_{n+1,i}-x_{n+1}\parallel +\parallel x_{n+1}-x_n\parallel +\parallel x_n-z_{n,i}\parallel \\ =& (L_i+1)\parallel x_n-x_{n+1}\parallel +\parallel z_{n+1,i}-x_{n+1}\parallel +\parallel x_n-z_{n,i}\parallel .\end{array}$$

From (4.4) and (4.15), we get that $d(S_i{z}_{n,i},z_{n,i})\to 0$ as $n\to \mathrm{\infty }$ for all $i\ge 1$. By (4.15) and the closedness of $S_i$, we have $\tilde{x}\in S_i\tilde{x}$ for all $i\ge 1$, and so $\tilde{x}\in \mathcal{F}$.

Step 5. We show that $\tilde{x}\in \mathit{MEP}(f,\phi ,C)$. Since $u_n=T_{r_n}{w}_n$, we derive

$$f(u_n,y)+\phi (y)+\frac{1}{r_n}〈y-u_n,J{u}_n-J{w}_n〉\ge \phi (u_n),\mathrm{\forall }y\in C.$$

From relaxed ξ-monotonicity of f, we have

$$\frac{1}{r_n}〈y-u_n,J{u}_n-J{w}_n〉+\phi (y)-\phi (u_n)\ge -f(u_n,y)\ge f(y,u_n)-\xi (y-u_n),\mathrm{\forall }y\in C.$$

Since $\frac{J{u}_n-J{w}_n}{r_n}\to 0$ and $u_n\to \tilde{x}$. By (A4), the lower semicontinuity of φ and the upper semicontinuity of ξ, we have

$$\begin{array}{rcl}f(y,\tilde{x})-\xi (y-\tilde{x})& \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}f(y,u_n)-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\xi (y-u_n)\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{f(y,u_n)-\xi (y-u_n)\}\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{\frac{1}{r_n}〈y-u_n,J{u}_n-J{w}_n〉+\phi (y)-\phi (u_n)\}\\ =& \phi (y)-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\phi (u_n)\\ =& \phi (y)-\phi (\tilde{x}),\mathrm{\forall }y\in C.\end{array}$$

For all $y\in C$ and $t\in (0,1]$, let $y_t=ty+(1-t)\tilde{x}$. Since C is convex, we have $y_t\in C$, and then

$$f(y_t,\tilde{x})-\phi (y_t)+\phi (\tilde{x})\le \xi (y_t-\tilde{x}).$$

Using (A1) and (A4), we have

$$0=f(y_t,y_t)\le (1-t)f(y_t,\tilde{x})+tf(y_t,y).$$

(4.16)

The convexity of the function φ implies that

$$\phi (y_t)=\phi ((1-t)\tilde{x}+ty)\le (1-t)\phi (\tilde{x})+t\phi (y).$$

(4.17)

This implies that

$$\begin{array}{rcl}0& =& f(y_t,y_t)\\ \le & (1-t)f(y_t,\tilde{x})+tf(y_t,y)\\ \le & tf(y_t,y)+(1-t)[\phi (y_t)-\phi (\tilde{x})+\xi (y_t-\tilde{x})]\\ \le & tf(y_t,y)+(1-t)t[\phi (y)-\phi (\tilde{x})+t^{p-1}\xi (y-\tilde{x})].\end{array}$$

This implies that

$$0\le f(y_t,y)+(1-t)[\phi (y)-\phi (\tilde{x})+t^{p-1}\xi (y-\tilde{x})].$$

(4.18)

By (A3) and $p>1$, taking the upper limit $t\to 0$ in (4.18), we get

$$f(\tilde{x},y)+\phi (y)-\phi (\tilde{x})\ge 0,\mathrm{\forall }y\in C.$$

Therefore, $\tilde{x}\in \mathit{MEP}(f,\phi ,C)$.

Step 6. Finally, we prove that $\tilde{x}={\mathrm{\Pi }}_{\mathrm{\Omega }}{x}_0$. Let $w={\mathrm{\Pi }}_{\mathrm{\Omega }}{x}_0$. Then $\varphi (w,x_0)\le \varphi (\tilde{x},x_0)$.

Since $w\in \mathrm{\Omega }\subset C_n$, for all $n\ge 0$ and $x_n\in {\mathrm{\Pi }}_{C_n}{x}_0$, we have

$$\varphi (x_n,x_0)\le \varphi (w,x_0),\mathrm{\forall }n\ge 0.$$

Since $x_n\to \tilde{x}$, we get that

$$\begin{array}{rcl}\varphi (\tilde{x},x_0)& =& \parallel \tilde{x}\parallel -2〈\tilde{x},J{x}_0〉+{\parallel x_0\parallel }^2\\ =& \underset{n\to \mathrm{\infty }}{lim}\{\parallel x_n\parallel -2〈x_n,J{x}_0〉+{\parallel x_0\parallel }^2\}\\ =& \underset{n\to \mathrm{\infty }}{lim}\varphi (x_n,x_0)\le \varphi (w,x_0).\end{array}$$

Hence $x_n\to \tilde{x}={\mathrm{\Pi }}_{\mathrm{\Omega }}{x}_0$. □

In the case of $f\equiv 0$ and $\phi \equiv 0$ in Theorem 4.3, we have the following corollary.

Corollary 4.4 Let E be a real uniformly smooth and strictly convex Banach space with the Kadec-Klee property, and let C be a nonempty, bounded, closed and convex subset of E. Let ${\{S_i:C\to \mathit{CB}(C)\}}_{i=1}^{\mathrm{\infty }}$ be a family of closed and uniformly quasi-ϕ-asymptotically nonexpansive multivalued mappings with a sequence $\{l_n\}\subset [1,\mathrm{\infty })$, $l_n\to 1$. Suppose that for each $i\ge 1$, $S_i$ is uniformly $L_i$-Lipschitz continuous and $\mathcal{F}\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence in C generated by

$$\{\begin{array}{l}{x}_0\in C,C_0=C,\\ w_n=J^{-1}({\alpha }_{n,0}J{x}_0+{\sum }_{i=1}^{\mathrm{\infty }}{\alpha }_{n,i}J{z}_{n,i}),z_{n,i}\in S_i^n{x}_n,\\ C_{n+1}=\{v\in C_n:\varphi (v,w_n)\le {\alpha }_{n,0}\varphi (v,x_0)+(1-{\alpha }_{n,0})\varphi (v,x_n)+{\zeta }_n\},\\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_0,n\ge 0,\end{array}$$

(4.19)

where $\{{\alpha }_n\}\subset [0,1]$, and ${\zeta }_n={sup}_{u\in F(S)}(l_n-1)\varphi (u,x_n)$. If ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, then $\{x_n\}$ converges strongly as $n\to \mathrm{\infty }$ to ${\mathrm{\Pi }}_{\mathrm{\Omega }}{x}_0$, where $\mathrm{\Omega }:=F(S)\cap \mathit{MEP}(f,\phi ,K)$.

If $S_i=S$, for all $i\ge 1$, then the following corollary follows from Theorem 4.3.

Corollary 4.5 Let E be a real uniformly smooth and strictly convex Banach space with the Kadec-Klee property, and let C be a nonempty, bounded, closed and convex subset of E. Let f be a bifunction from $C\times C$ to ℝ satisfying (A1)-(A4), and let φ be a lower semicontinuous and convex function from C to ℝ. Let $S:C\to \mathit{CB}(C)$ be a closed and uniformly L-Lipschitz continuous quasi-ϕ-asymptotically nonexpansive multivalued mapping with a sequence $\{l_n\}\subset [1,\mathrm{\infty })$, $l_n\to 1$ with $F(S)\cap \mathit{MEP}(f,\phi ,C)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence in C generated by

$$\{\begin{array}{l}{x}_0\in C,C_0=C,\\ w_n=J^{-1}[{\alpha }_{n}J{x}_0+(1-{\alpha }_n)J{z}_n],z_n\in S^n{x}_n,\\ u_n\in C\mathit{\text{such that }}\\ f(u_n,y)+\phi (y)+\frac{1}{r_n}〈y-u_n,J{u}_n-J{w}_n〉\ge \phi (u_n),\mathrm{\forall }y\in C,\\ C_{n+1}=\{v\in C_n:\varphi (v,u_n)\le {\alpha }_{n,0}\varphi (v,x_0)+(1-{\alpha }_{n,0})\varphi (v,x_n)+{\zeta }_n\},\\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_0,n\ge 0,\end{array}$$

(4.20)

where for each ${\{r_n\}}_{n=0}^{\mathrm{\infty }}\subset [a,\mathrm{\infty })$ for some $a>0$, $\{{\alpha }_n\}\subset [0,1]$, and ${\zeta }_n={sup}_{u\in F(S)}(l_n-1)\varphi (u,x_n)$. If ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, then $\{x_n\}$ converges strongly as $n\to \mathrm{\infty }$ to ${\mathrm{\Pi }}_{\mathrm{\Omega }}{x}_0$, where $\mathrm{\Omega }:=F(S)\cap \mathit{MEP}(f,\phi ,K)$.

In the case of $f\equiv 0$ and $\phi \equiv 0$ in Corollary 4.4, we can omit the boundedness of C (necessary for the existence of a mapping $T_r$), so we have the following corollary.

Corollary 4.6 [[21], Theorem 3.1]

Let E be a real uniformly smooth and strictly convex Banach space with the Kadec-Klee property, let C be a nonempty, closed and convex subset of E, and let $S:C\to \mathit{CB}(C)$ be a closed and uniformly L-Lipschitz continuous quasi-ϕ-asymptotically nonexpansive multivalued mapping with nonnegative real sequences $\{l_n\}\subset [1,\mathrm{\infty })$ and $l_n\to 1$ with $F(S)$ being a nonempty bounded subset. Let $\{{\alpha }_n\}$ be a sequence in $(0,1)$. Let $\{x_n\}$ be the sequence generated by

$$\{\begin{array}{l}{x}_0\in C,C_0=C,\\ y_n=J^{-1}[{\alpha }_{n}J{x}_0+(1-{\alpha }_n)J{z}_n],z_n\in S^n{x}_n,\\ C_{n+1}=\{v\in C_n:\varphi (v,y_n)\le {\alpha }_n\varphi (v,x_0)+(1-{\alpha }_n)\varphi (v,x_n)+{\zeta }_n\},\\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_0,n\ge 0,\end{array}$$

(4.21)

where ${\zeta }_n=(l_n-1){sup}_{p\in F(S)}\varphi (p,x_n)$. If ${\alpha }_n\to 0$, then $\{x_n\}$ converges strongly to ${\mathrm{\Pi }}_{F(S)}{x}_0$.

5 Application

We utilize Theorem 4.3 to study a modified Halpern iterative algorithm for a system of mixed equilibrium problems.

Theorem 5.1 Let E be a real uniformly smooth and strictly convex Banach space with the Kadec-Klee property, and let C be a nonempty, bounded, closed and convex subset of E. Let $f_i:C\times C\to \mathbb{R}$, $i=1,2,\dots$ , be a countable family of bifunctions satisfying conditions (A1)-(A4), and let ${\phi }_i:C\to \mathbb{R}$, $i=1,2,\dots$ , be a countable family of lower semicontinuous and convex functions satisfying conditions (i)-(ii) as in Lemma  3.5 with $\mathrm{\Gamma }={\bigcap }_{i=1}^{\mathrm{\infty }}\mathit{MEP}(f_i,{\phi }_i,C)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence in C generated by

$$\{\begin{array}{l}{x}_0\in C,C_0=C,\\ u_{n,i}\in C\mathit{\text{such that }}\\ f_i(u_{n,i},y)+{\phi }_i(y)+\frac{1}{r}〈y-u_{n,i},J{u}_{n,i}-J{x}_{n,i}〉\ge \phi (u_{n,i}),\mathrm{\forall }y\in C,r>0,i\ge 1,\\ y_n=J^{-1}({\alpha }_{n,0}J{x}_0+{\sum }_{i=1}^{\mathrm{\infty }}{\alpha }_{n,i}J{u}_{n,i}),\\ C_{n+1}=\{v\in C_n:\varphi (v,y_n)\le {\alpha }_{n,0}\varphi (v,x_0)+(1-{\alpha }_{n,0})\varphi (v,x_n)+{\zeta }_n\},\\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_0,n\ge 0,\end{array}$$

(5.1)

where $\{{\alpha }_n\}\subset [0,1]$, and ${\zeta }_n={sup}_{u\in F(S)}(l_n-1)\varphi (u,x_n)$. If ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, then $\{x_n\}$ converges strongly as $n\to \mathrm{\infty }$ to ${\mathrm{\Pi }}_{\mathrm{\Gamma }}{x}_0$, which is a common solution of the system of mixed equilibrium problems.

Proof For any $i=1,2,\dots$ , we define a mapping $T_{r,i}(z):E\to C$ as follows:

$$T_{r,i}(z)=\{x\in C:f_i(x,y)+{\phi }_i(y)+\frac{1}{r}〈y-x,Jx-Jz〉\ge {\phi }_i(x),\mathrm{\forall }y\in C\}.$$

From Lemma 3.5, we get that $u_{n,i}=T_{r,i}{x}_n$ and $F(T_{r,i})=\mathit{MEP}(f_i,{\phi }_i,C)$ for all $i=1,2,\dots$ , and $T_{r,i}$ is a countable family of closed quasi-ϕ-nonexpansive mappings. Thus, (5.1) can be written as follows:

$$\{\begin{array}{l}{x}_0\in C,C_0=C,\\ y_n=J^{-1}({\alpha }_{n,0}J{x}_0+{\sum }_{i=1}^{\mathrm{\infty }}{\alpha }_{n,i}J{T}_{r,i}{x}_n),\\ C_{n+1}=\{v\in C_n:\varphi (v,y_n)\le {\alpha }_{n,0}\varphi (v,x_0)+(1-{\alpha }_{n,0})\varphi (v,x_n)+{\zeta }_n\},\\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_0,n\ge 0.\end{array}$$

Therefore, the result can be obtained from Corollary 4.4. □

6 Conclusion

In this paper, we establish the existence of a solution of the mixed equilibrium problem $\mathit{MEP}(f,\phi ,C)$ by using the KKM mapping in a Banach space setting, when f is relaxed ξ-monotone. Then, by virtue of this result, we introduce a hybrid iterative scheme and prove that our proposed iterative scheme converges strongly to a common element of the set of solutions of $\mathit{MEP}(f,\phi ,C)$ and the set of common fixed points of a countable family of quasi-ϕ-asymptotically nonexpansive multivalued mappings.