Existence and approximation for a solution of a generalized equilibrium problem on the dual space of a Banach space

Abstract

In this paper, we first prove the existence of a solution for a generalized equilibrium problem with a bifunction defined on the dual space in a Banach space setting. Second, by the virtue of this result, we construct the hybrid projection method for solving a solution of a generalized equilibrium problem. Consequently, we establish the strong convergence theorem by using sunny generalized nonexpansive retraction in the dual of Banach spaces.

MSC:47H09, 47H10, 47J25.

Dedication

Dedicated to Prof. W Takahashi on the occasion of his 70th birthday

1 Introduction

Let ℝ be the set of real numbers. Let E be a real Banach space with the norm $\parallel \cdot \parallel$, and $〈\cdot ,\cdot 〉$ is the dual pair between E and $E^{\ast }$, where $E^{\ast }$ is the dual space of E. Let C be a closed and convex subset of a real Banach space E with the dual space $E^{\ast }$, and let $C^{\ast }$ be a closed and convex subset of $E^{\ast }$. We recall the following definitions:

• (1) A mapping $A:C\to E^{\ast }$ is said to be monotone if for each $x,y\in C$ such that

  $$〈x-y,Ax-Ay〉\ge 0.$$

• (2) A mapping $A:C\to E^{\ast }$ is said to be δ-strongly monotone if there exists a constant $\delta >0$ such that

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

• (3) A mapping $A:C\to E^{\ast }$ is said to be δ-inverse strongly monotone if there exists a constant $\delta >0$ such that

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

• (4) A mapping $A:C^{\ast }\to E$ is said to be skew monotone if for each $x^{\ast },y^{\ast }\in C^{\ast }$ such that

  $$〈A{x}^{\ast }-A{y}^{\ast },x^{\ast }-y^{\ast }〉\ge 0.$$

• (5) A mapping $A:D(A)\subset E^{\ast }\to E$ is said to be α-inverse strongly skew monotone if there exists a constant $\alpha >0$ such that

  $$〈A{x}^{\ast }-A{y}^{\ast },x^{\ast }-y^{\ast }〉\ge \alpha {\parallel A{x}^{\ast }-A{y}^{\ast }\parallel }^2,\mathrm{\forall }{x}^{\ast },y^{\ast }\in D(A).$$

• (6) A mapping $A:D(A)\subset E^{\ast }\to E$ is said to be hemicontinuous if for all $x^{\ast },y^{\ast }\in D(A)$, the mapping f of $[0,1]$ into E defined by $f(t)=A(t{x}^{\ast }+(1-t)y^{\ast })$ is continuous.

Let C be a nonempty, closed and convex subset of E, and let J be the duality mapping from E into $E^{\ast }$ such that $J(C)$ is closed and convex of $E^{\ast }$, let us assume that a bifunction $F:J(C)\times J(C)\to \mathbb{R}$ satisfies suitable conditions, and $A:C^{\ast }\to E$ is a skew monotone operator from $J(C)$ into E.

The generalized equilibrium problem is to find $\hat{z}\in C$ such that

$$F(J\hat{z},Jy)+〈AJ\hat{z},Jy-J\hat{z}〉\ge 0,\mathrm{\forall }y\in C.$$

(1.1)

The set of solutions of (1.1) is denoted by $GEP(F,A)$, that is,

$$GEP(F,A)=\{\hat{z}\in C:F(J\hat{z},Jy)+〈AJ\hat{z},Jy-J\hat{z}〉\ge 0,\mathrm{\forall }y\in C\}.$$

(1.2)

If $A\equiv 0$, then problem (1.1) reduces to the equilibrium problem, which is to find $\hat{z}\in C$ such that

$$F(J\hat{z},Jy)\ge 0,\mathrm{\forall }y\in C.$$

(1.3)

The set of solutions of problem (1.3) is denoted by $EP(F)$, that is,

$$EP(F)=\{\hat{z}\in C:F(J\hat{z},Jy)\ge 0,\mathrm{\forall }y\in C\}.$$

(1.4)

The above formulation (1.3) was considered in Takahashi and Zembayashi [1], and they proved a strong convergence theorem for finding a solution of the equilibrium problem (1.3) in Banach spaces.

If $F\equiv 0$, then problem (1.1) reduces to variational inequality, which is to find $\hat{z}\in C$ such that

$$〈AJ\hat{z},Jy-J\hat{z}〉\ge 0,\mathrm{\forall }y\in C.$$

(1.5)

The set of solutions of problem (1.5) is denoted by $VI(J(C),A)$, that is,

$$VI(J(C),A)=\{\hat{z}\in C:〈AJ\hat{z},Jy-J\hat{z}〉\ge 0,\mathrm{\forall }y\in C\}.$$

(1.6)

In the sequel, let $T:C\to C$ be a mapping, we denote by $Fix(T)$ the set of fixed points of T, that is,

$$Fix(T)=\{x\in C:Tx=x\}.$$

(1.7)

We denote the strong convergence, weak convergence $\{x_n\}$ by $x_n\to x$, $x_n\rightharpoonup x$, respectively.

Let C be a nonempty, closed subset of a smooth, strictly convex and reflexive Banach space E such that $J(C)$ is closed and convex. For solving the equilibrium problem, let us assume that a bifunction $F:J(C)\times J(C)\to \mathbb{R}$ satisfies the following conditions:

• (DA1) $F(x^{\ast },x^{\ast })=0$ for all $x^{\ast }\in J(C)$;

• (DA2) F is monotone, i.e., $F(x^{\ast },y^{\ast })+F(y^{\ast },x^{\ast })\le 0$ for all $x^{\ast },y^{\ast }\in J(C)$;

• (DA3) for all $x^{\ast },y^{\ast },z^{\ast }\in J(C)$,

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

• (DA4) for all $x^{\ast }\in J(C)$, $F(x^{\ast },\cdot )$ is convex and lower semicontinuous.

The following result is in Blum and Oettli [2], and see the proof in [3].

Definition 1.1 Let E be a Banach space. Then,

• (1) E is said to be strictly convex if $\frac{\parallel x+y\parallel }{2}<1$ for all $x,y\in U_E=\{z\in E:\parallel z\parallel =1\}$ with $x\ne y$.

• (2) E is said to be uniformly convex if for each $ϵ\in (0,2]$, there exists $\delta >0$ such that $\frac{\parallel x+y\parallel }{2}\le 1-\delta$ for all $x,y\in U_E$ with $\parallel x-y\parallel >ϵ$.

• (3) E is said to be smooth if the limit (1.8)

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

  (1.8)

exists for each $x,y\in U_E$.

• (4) E is said to be uniformly smooth if the limit (1.8) is attained uniformly for all $x,y\in U_E$.

• (5) E is said to have uniformly Gâteaux differentiable norm if for all $y\in U(E)$, the limit (1.8) converges uniformly for $x\in U_E$.

Definition 1.2 Let E be a Banach space. Then a function ${\rho }_E:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is said to be the modulus of smoothness of E if

$${\rho }_E(t)=sup\{\frac{\parallel x+y\parallel +\parallel x-y\parallel }{2}-1:\parallel x\parallel =1,\parallel y\parallel =t\}.$$

• (1) E is said to be smooth if ${\rho }_E(t)>0$, $\mathrm{\forall }t>0$.

• (2) E is said to be uniformly smooth if and only if ${lim}_{t\to 0^{+}}\frac{{\rho }_E(t)}{t}=0$.

Definition 1.3 Let E be a Banach space. Then the modulus of convexity of E is the function ${\delta }_E:[0,2]\to [0,1]$ defined by

$${\delta }_E(ϵ)=inf\{1-\parallel \frac{x+y}{2}\parallel :\parallel x\parallel \le 1,\parallel y\parallel \le 1;\parallel x-y\parallel \ge ϵ\}.$$

• (1) E is said to be uniformly convex if and only if ${\delta }_E(ϵ)>0$ for all $ϵ\in (0,2]$.

• (2) Let p be a fixed real number $p>1$. Then E is said to be p-uniformly convex if there exists a constant $c>0$ such that ${\delta }_E(ϵ)\ge c{ϵ}^p$ for all $ϵ\in [0,2]$.

Observe that every p-uniformly convex is uniformly convex. One should note that no Banach space is p-uniformly convex for $1<p<2$. It is well known that a Hilbert space is 2-uniformly convex and uniformly smooth.

For any $p>1$, the generalized duality mapping $J_p:E\to 2^{E^{\ast }}$ is defined by

$$J_{p}x=\{f^{\ast }\in E^{\ast }:〈x,f^{\ast }〉={\parallel x\parallel }^p,\parallel f^{\ast }\parallel ={\parallel x\parallel }^{p-1}\},\mathrm{\forall }x\in E.$$

(1.9)

In particular, $J=J_2$ is called the normalized duality mapping. If E is a Hilbert space, then $J=I$, where I is the identity mapping. That is,

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

(1.10)

Remark 1.4 The basic properties below hold, see [4–6].

• (1) If E is a uniformly smooth real Banach space, then J is uniformly continuous on each bounded subset of E.

• (2) If E is a uniformly smooth real Banach space, then $J^{\ast }:E^{\ast }\to 2^E$ is a normalized duality mapping on $E^{\ast }$, then $J^{-1}=J^{\ast }$, $(J^{\ast })J=I_E$ and $J(J^{\ast })=I_{E^{\ast }}$, where on $I_E$ and $I_{E^{\ast }}$ are the identity mappings on E and $E^{\ast }$, respectively.

• (3) Let E be a smooth, strictly convex reflexive Banach space and J be the duality mapping from E into $E^{\ast }$. Then $J^{-1}$ is also single-valued, one-to-one, onto, and it is also the duality mapping from $E^{\ast }$ into E.

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

• (5) If E is a reflexive, smooth and strictly convex Banach space, then J is single-valued, one-to-one, and onto.

• (6) A Banach space E is uniformly smooth if and only if $E^{\ast }$ is uniformly convex.

• (7) 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 $x_n\to x$.

• (8) A Banach space E is strictly convex if and only if J is strictly monotone, that is,

  $$〈x-y,x^{\ast }-y^{\ast }〉>0,\text{whenever }x,y\in E,x\ne y,\text{ and }{x}^{\ast }\in Jx,y^{\ast }\in Jy.$$

• (9) Both, uniformly smooth Banach space and uniformly convex Banach space, are reflexive.

• (10) If $E^{\ast }$ is uniformly convex, then J is uniformly norm-to-norm continuous on each bounded subset of E.

• (11) If $E^{\ast }$ is a strictly convex Banach space, then J is one-to-one, that is, $x\ne y$ implies that $Jx\cap Jy\ne \mathrm{\varnothing }$.

Let J be the normalized duality mapping, then J is said to be weakly sequentially continuous if the strong convergence of a sequence $\{x_n\}$ to $x\in E$ implies the weak^∗ convergence of a sequence $\{J{x}_n\}$ to Jx in $E^{\ast }$.

Let E be a smooth and strictly convex reflexive Banach space, and let C be a nonempty, closed and convex subset of E. We assume that the Lyapunov functional $\varphi :E\times E\to {\mathbb{R}}^{+}$ is defined by [6, 7]

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

Let C be a nonempty, closed and convex subset of a Banach space E. The generalized projection [6]${\mathrm{\Pi }}_C:E\to C$ is defined by, for each $x\in E$,

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

Remark 1.5 From the definition of ϕ. It is easy to see that

• (1) ${(\parallel x\parallel -\parallel y\parallel )}^2\le \varphi (x,y)\le {(\parallel x\parallel +\parallel y\parallel )}^2$ for all $x,y\in E$.

• (2) $\varphi (x,y)=\varphi (x,z)+\varphi (z,y)+2〈x-z,Jz-Jy〉$ for all $x,y,z\in E$.

• (3) $\varphi (x,y)=〈x,Jx-Jy〉+〈y-x,Jy〉\le \parallel x\parallel \parallel Jx-Jy\parallel +\parallel y-x\parallel \parallel y\parallel$ for all $x,y,z\in E$.

• (4) If E is a real Hilbert space H, then $\varphi (x,y)={\parallel x-y\parallel }^2$, and ${\mathrm{\Pi }}_C=P_C$ (the metric projection of H onto C).

Lemma 1.6 [6, 7]

If C is a nonempty, closed and convex subset of a smooth and strictly convex reflexive real Banach space E, then

• (1) for $x\in E$, and $u\in C$, one has

  $$u={\mathrm{\Pi }}_C(x)\iff 〈u-y,Jx-Ju〉\ge 0,\mathrm{\forall }y\in C.$$

• (2) $\varphi (x,{\mathrm{\Pi }}_C(y))+\varphi ({\mathrm{\Pi }}_C(y),y)\le \varphi (x,y)$, $\mathrm{\forall }x\in C$, $y\in E$.

• (3) $\varphi (x,y)=0$ if and only if $x=y$, $\mathrm{\forall }x,y\in E$.

In 2007, Takahashi and Zembayashi [1] introduced an iterative algorithm for finding a solution of an equilibrium problem with a bifunction defined on the dual space of a Banach space by using the shrinking projection method, and they established the strong convergence as follows.

Theorem TZ Let E be a uniformly convex Banach space whose norm is uniformly Gâteaux differentiable, and let C be a nonempty, closed and convex subset of E such that $J(C)$ is closed and convex of $E^{\ast }$. Assume that a mapping $F:J(C)\times J(C)\to \mathbb{R}$ satisfies the condition (DA1)-(DA4) such that $EP(F)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence generated by the following algorithm:

$$\{\begin{array}{c}{x}_0\in C\mathit{\text{chosen arbitrary}},C_0=C,\hfill \\ u_n\in C\mathit{\text{such that}}F(J{u}_n,Jy)+\frac{1}{r_n}〈u_n-x_n,Jy-J{u}_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)u_n,\hfill \\ C_{n+1}=\{z\in C_n:\varphi (y_n,z)\le \varphi (x_n,z)\},\hfill \\ x_{n+1}=R_{C_{n+1}}(x_0),\mathrm{\forall }n\in \mathbb{N}\cup \{0\},\hfill \end{array}$$

(1.11)

where J is the duality mapping on E, the sequence $\{{\alpha }_n\}\subset [0,1]$ such that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$, $r_n\subset [a,\mathrm{\infty })$ for some $a>0$, and $R_{C_{n+1}}$ is the sunny generalized nonexpansive retraction from E onto $C_{n+1}$.

Then the sequence $\{x_n\}$ converges strongly to some point $p=R_{EP(F)}(x_0)$, where $R_{EP(F)}$ is the sunny generalized nonexpansive retraction from E to $EP(F)$.

In 2010, Plubtieng and Sriprad [8] proved the existence theorem of the variational inequality problem for skew monotone operator defined on the dual space of a smooth Banach space, and they established a weak convergence theorem for finding a solution of the variational inequality problem using the projection algorithm method with a new projection, which was introduced by Ibaraki and Takahashi [9] and Iiduka and Takahashi [10] in Banach spaces.

Theorem PS Let E be a uniformly convex and 2-uniformly smooth Banach space whose duality mapping J is weakly sequentially continuous. Let C be a nonempty, closed and convex subset of E such that $J(C)$ is closed and convex, and let A be an α-inverse-strongly-skew-monotone operator of $J(C)$ into E such that $VI(J(C),A)\ne \mathrm{\varnothing }$ and $\parallel AJz\parallel \le \parallel AJy-AJz\parallel$, for all $y\in C$ and $z\in VI(J(C),A)$. Let $\{x_n\}$ be a sequence defined by $x_1=x\in C$ and

$$x_{n+1}=R_C(x_n-{\lambda }_{n}AJ{x}_n)$$

(1.12)

for every $n=1,2,3,\dots$ , where $R_C$ is the sunny generalized nonexpansive retraction of E into C, $\{{\alpha }_n\}\subset [a,b]$, for some a, b with $0<a<b<\frac{\alpha }{c}$, where $c>0$ is a constant satisfying $\parallel Jx-Jy\parallel \le c\parallel x-y\parallel$ for all $x,y\in C$. Then the sequence $\{x_n\}$ converges weakly to some element $z\in VI(J(C),A)$. Further $z={lim}_{n\to \mathrm{\infty }}{R}_{VI(J(C),A)}(x_n)$.

In this paper, motivated and inspired by the previously mentioned results, we study the existence theorem for a generalized equilibrium problem with a bifunction defined on the dual space of a Banach space, and we also construct an iterative procedure generated by a hybrid method for solving the solution of a generalized equilibrium problem by using the sunny generalized nonexpansive retraction. Under some suitable assumptions, the strong convergence theorems are established in Banach spaces. The results obtained in this paper extend and improve several recent results in this area.

2 Preliminaries

Definition 2.1 Let C be a nonempty and closed subset of a smooth Banach space.

• (1) A mapping $T:C\to C$ is said to be closed if for each $\{x_n\}\subset C$, $x_n\to x$ and $T{x}_n\to y$ imply that $Tx=y$.

• (2) A mapping $T:C\to C$ is said to be nonexpansive if

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

• (3) A mapping $T:C\to C$ is said to be ϕ-nonexpansive if $Fix(T)\ne \mathrm{\varnothing }$, and

  $$\varphi (Tx,Ty)\le \varphi (x,y),\mathrm{\forall }x,y\in C.$$

• (4) A mapping $T:C\to C$ is said to be generalized nonexpansive [11] if $Fix(T)\ne \mathrm{\varnothing }$, and

  $$\varphi (Tx,p)\le \varphi (x,p),\mathrm{\forall }x\in C,p\in Fix(T).$$

Definition 2.2 [11]

Let C be a nonempty and closed subset of a smooth Banach space E. A mapping $R:E\to C$ is called

• (1) a retraction if $R^2=R$;

• (2) a sunny if $R(Rx+t(x-Rx))=Rx$ for all $x\in E$ and $t\ge 0$.

We also know that if E is a smooth, strictly convex and reflexive Banach space, and C is nonempty, closed and convex subset of E, then there exists a sunny generalized nonexpansive retraction $R_C$ of E onto C if and only if $J(C)$ is closed and convex. In this case, $R_C$ is given by

$$R_C=J^{-1}{\mathrm{\Pi }}_{J(C)}J.$$

Definition 2.3 [11]

Let C be a nonempty and closed subset of a smooth Banach space E. The set C is called a sunny generalized nonexpansive retraction of E if there exists a sunny generalized nonexpansive R from E onto C.

Lemma 2.4 [9]

Let C be a nonempty and closed subset of a smooth and strictly convex Banach space E, and let R be a retraction from E onto C. Then the following are equivalent:

• (1) R is sunny generalized nonexpansive;

• (2) $〈x-Rx,Jy-JRx〉\le 0$ for all $x\in E$ and $y\in C$.

Lemma 2.5 [9]

Let C be a nonempty, closed and sunny generalized nonexpansive retraction of a smooth and strictly convex Banach space E. Then the sunny generalized nonexpansive retraction from E onto C is uniquely determined.

Lemma 2.6 [9]

Let C be a nonempty and closed subset of a smooth and strictly convex Banach space E such that there exists a sunny generalized nonexpansive retraction R from E onto C. Let $x\in E$ and $z\in C$. Then the following hold:

• (1) $z=Rx$ if and only if $〈x-z,Jy-Jz〉\le 0$ for all $y\in C$;

• (2) $\varphi (x,Rx)+\varphi (Rx,z)\le \varphi (x,z)$.

Lemma 2.7 [12]

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

• (1) C is a sunny generalized nonexpansive retraction of E;

• (2) $J(C)$ is closed and convex.

Remark 2.8 From Lemmas 2.5 and 2.7. If E is a Hilbert space, then a sunny generalized nonexpansive retraction from E onto C reduces to a metric projection operator P from E onto C.

Lemma 2.9 [12]

Let E be a smooth, strictly convex and reflexive Banach space, let C be a nonempty, closed and sunny generalized nonexpansive retraction of E, and let R be the sunny generalized nonexpansive retraction from E onto C. Let $x\in E$ and $z\in C$. Then the following are equivalent:

• (1) $z=Rx$;

• (2) $\varphi (x,z)={min}_{y\in C}\varphi (x,y)$.

Lemma 2.10 [7]

Let E be a uniformly smooth and strictly convex real Banach space, and let $\{x_n\}$ and $\{y_n\}$ be two sequences of E. If $\varphi (x_n,y_n)\to 0$ and either $\{x_n\}$ or $\{y_n\}$ is bounded, then $\parallel x_n-y_n\parallel \to 0$.

Lemma 2.11 [13]

Let E be a uniformly smooth and strictly convex real 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 and $p\in E$. If $x_n\to p$ and $\varphi (x_n,y_n)\to 0$, then $y_n\to p$.

Lemma 2.12 [14]

Let $\{a_n\}$ and $\{b_n\}$ be two sequences of nonnegative real numbers satisfying the inequality

$$a_{n+1}\le a_n+b_n,\mathrm{\forall }n\ge 1.$$

If ${\sum }_{n=0}^{\mathrm{\infty }}{b}_n<\mathrm{\infty }$, then ${lim}_{n\to \mathrm{\infty }}{a}_n$ exists.

Now, let us recall the following well-known concept and the result.

Definition 2.13 [15]

Let B be a subset of a topological vector space X. A mapping $G:B\to 2^X$ is called a KKM mapping if $conv\{x_1,x_2,x_3,\dots ,x_m\}\subset {\bigcup }_{i=1}^{m}G(x_i)$, for $x_i\in B$ and $i=1,2,3,\dots ,m$, where convA denotes the convex hull of the set A.

In [16], Ky Fan gave the following famous infinite-dimensional generalization of Knaster, Kuratowski and Mazurkiewicz’s classical finite-dimensional result.

Lemma 2.14 [16]

Let B be a 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 ${\bigcap }_{x\in B}G(x)\ne \mathrm{\varnothing }$.

3 Existence theorem

In this section, we prove the existence theorem of a solution for a generalized equilibrium problem with a bifunction defined on the dual space of a Banach space. Now, we use the concept of KKM mapping to prove the lemma for our main result.

Lemma 3.1 Let C be a nonempty, compact and convex subset of a uniformly smooth, strictly convex and reflexive Banach space E, and let J be the duality mapping from E into $E^{\ast }$ such that $J(C)$ is closed and convex, let us assume that a bifunction $F:J(C)\times J(C)\to \mathbb{R}$ satisfies the following conditions (DA1)-(DA4), let $C^{\ast }$ be a nonempty, closed and convex subset of $E^{\ast }$, and let $A:C^{\ast }\to E$ be an α-inverse strongly skew monotone. Let any $r>0$ be a given real number, and let $x\in E$ be any point. Then there exists $z\in C$ such that

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

Proof Let $x_0$ be any point in E. For each $y\in C$, we define the mapping $H:C\to 2^E$ as follows:

$$H(y)=\{z\in C:F(Jz,Jy)+〈AJz,Jy-Jz〉+\frac{1}{r}〈z-x_0,J(y-z)〉\ge 0,\mathrm{\forall }y\in C\}.$$

It is easy to see that $y\in H(y)$, and hence $H(y)\ne \mathrm{\varnothing }$.

(a) First, we show that H is a KKM mapping.

Suppose that H is not a KKM mapping. Then there exists a finite subset $\{y_1,y_2,\dots ,y_m\}$ of C and ${\alpha }_i>0$ with ${\sum }_{i=1}^m{\alpha }_i=1$ such that $\hat{x}={\sum }_{i=1}^m{\alpha }_i{y}_i\notin {\bigcup }_{i=1}^{m}H(y_i)$ for all $i=1,2,3,\dots ,m$.

It follows from the definition of a mapping H that

$$F(J\hat{x},J{y}_i)+〈AJ\hat{x},J{y}_i-J\hat{x}〉+\frac{1}{r}〈\hat{x}-x_0,J(y_i-\hat{x})〉<0\text{for all }i=1,2,3,\dots ,m.$$

By the assumptions of (DA1) and (DA4), we get

$$\begin{array}{rcl}0& =& F(J\hat{x},J\hat{x})+〈AJ\hat{x},J\hat{x}-J\hat{x}〉+\frac{1}{r}〈\hat{x}-x_0,J(\hat{x}-\hat{x})〉\\ \le & \sum _{i=1}^m{\alpha }_i(F(J\hat{x},J{y}_i)+〈AJ\hat{x},J{y}_i-J\hat{x}〉+\frac{1}{r}〈\hat{x}-x_0,J(y_i-\hat{x})〉)\\ <& 0,\end{array}$$

which is a contradiction. Thus, H is a KKM mapping on C.

(b) Next, we show that $H(y)$ is closed for all $y\in C$.

Let $\{z_n\}$ be a sequence in $H(y)$ such that $z_n\to z_0$, as $n\to \mathrm{\infty }$.

It then follows from $z_n\in H(y)$ that

$$F(J{z}_n,Jy)+〈AJ{z}_n,Jy-J{z}_n〉+\frac{1}{r}〈z_n-x_0,J(y-z_n)〉\ge 0.$$

(3.1)

By assumption (DA3), the continuity of J and the lower semicontinuity of ${\parallel \cdot \parallel }^2$, we obtain from (3.1) that

$$\begin{array}{rcl}0& \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(F(J{z}_n,Jy)+〈AJ{z}_n,Jy-J{z}_n〉+\frac{1}{r}〈z_n-x_0,J(y-z_n)〉)\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(F(J{z}_n,Jy)+〈AJ{z}_n,Jy-J{z}_n〉+\frac{1}{r}〈z_n-x_0,J(y-z_n)〉)\\ =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(F(J{z}_n,Jy)+〈AJ{z}_n,Jy-J{z}_n〉+\frac{1}{r}〈(z_n-y)+(y-x_0),J(y-z_n)〉)\\ =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(F(J{z}_n,Jy)+〈AJ{z}_n,Jy-J{z}_n〉+\frac{1}{r}〈z_n-y,J(y-z_n)〉+\frac{1}{r}〈y-x_0,J(y-z_n)〉)\\ =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(F(J{z}_n,Jy)+〈AJ{z}_n,Jy-J{z}_n〉+\frac{1}{r}〈y-x_0,J(y-z_n)〉-\frac{1}{r}〈y-z_n,J(y-z_n)〉)\\ =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(F(J{z}_n,Jy)+〈AJ{z}_n,Jy-J{z}_n〉+\frac{1}{r}〈y-x_0,J(y-z_n)〉-\frac{1}{r}{\parallel y-z_n\parallel }^2)\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}F(J{z}_n,Jy)+\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈AJ{z}_n,Jy-J{z}_n〉\\ +\frac{1}{r}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈y-x_0,J(y-z_n)〉-\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{1}{r}{\parallel y-z_n\parallel }^2\\ \le & F(J{z}_0,Jy)+〈AJ{z}_0,Jy-J{z}_0〉+\frac{1}{r}〈y-x_0,J(y-z_0)〉-\frac{1}{r}{\parallel y-z_0\parallel }^2\\ =& F(J{z}_0,Jy)+〈AJ{z}_0,Jy-J{z}_0〉+\frac{1}{r}〈y-x_0,J(y-z_0)〉-\frac{1}{r}〈y-z_0,J(y-z_0)〉\\ =& F(J{z}_0,Jy)+〈AJ{z}_0,Jy-J{z}_0〉+\frac{1}{r}〈y-x_0,J(y-z_0)〉+\frac{1}{r}〈z_0-y,J(y-z_0)〉\\ =& F(J{z}_0,Jy)+〈AJ{z}_0,Jy-J{z}_0〉+\frac{1}{r}〈(y-x_0)+(z_0-y),J(y-z_0)〉\\ =& F(J{z}_0,Jy)+〈AJ{z}_0,Jy-J{z}_0〉+\frac{1}{r}〈z_0-x_0,J(y-z_0)〉.\end{array}$$

Now, we get

$$F(J{z}_0,Jy)+〈AJ{z}_0,Jy-J{z}_0〉+\frac{1}{r}〈z_0-x_0,J(y-z_0)〉\ge 0.$$

Therefore, $z_0\in H(y)$, and so $H(y)$ is closed for all $y\in C$.

(c) We show that $H(y)$ is weakly compact.

Now, we know that $H(y)$ is closed and subset of C.

Since C is compact. Therefore, $H(y)$ is compact, and then $H(y)$ is weakly compact.

By using (a), (b), (c) and Lemma 2.14, we can conclude that ${\bigcap }_{y\in C}H(y)\ne \mathrm{\varnothing }$.

Therefore, there exists $z\in C$ such that

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

 □

Theorem 3.2 Let C be a nonempty, closed and convex subset of a uniformly smooth, strictly convex and reflexive Banach space E, and let J be the duality mapping from E into $E^{\ast }$ such that $J(C)$ is closed and convex, let us assume that a bifunction $F:J(C)\times J(C)\to \mathbb{R}$ satisfies the following conditions (DA1)-(DA4), let $C^{\ast }$ be a nonempty, closed and convex subset of $E^{\ast }$, and let $A:C^{\ast }\to E$ be an α-inverse strongly skew monotone and hemicontinuous. Let any $r>0$ be a given real number, and let $x\in E$ be any point. We define a mapping $T_r:E\to C$ as follows:

$$T_r(x)=\{z\in C:F(Jz,Jy)+〈AJz,Jy-Jz〉+\frac{1}{r}〈z-x,J(y-z)〉\ge 0,\mathrm{\forall }y\in C\}.$$

(3.2)

Then the following conclusions hold:

• (1) $T_r$ is single-valued;

• (2) $〈T_{r}x-T_{r}y,J(T_{r}x-T_{r}y)〉\le 〈x-y,J(T_{r}x-T_{r}y)〉$, $\mathrm{\forall }x,y\in E$;

• (3) $Fix(T_r)=GEP(F,A)$;

• (4) $J(GEP(F,A))$ is closed and convex.

Proof We complete this proof by four items below.

(1) We show that $T_r$ is single-valued.

From the definition of $T_r(x)$, it is easy to see that

$$F(Jy,Jy)+〈AJy,Jy-Jy〉+\frac{1}{r}〈z-y,J(y-y)〉=0\ge 0.$$

Therefore, $y\in T_r(x)$. Hence, $T_r(x)\ne \mathrm{\varnothing }$.

Indeed, for any $x\in E$ and $r>0$, let $z_1,z_2\in T_r(x)$. Then

$$F(J{z}_1,J{z}_2)+〈AJ{z}_1,J{z}_2-J{z}_1〉+\frac{1}{r}〈z_1-x,J(z_2-z_1)〉\ge 0,$$

and

$$F(J{z}_2,J{z}_1)+〈AJ{z}_2,J{z}_1-J{z}_2〉+\frac{1}{r}〈z_2-x,J(z_1-z_2)〉\ge 0.$$

Adding the two inequalities, we have

$$\begin{array}{rcl}0& \le & F(J{z}_1,J{z}_2)+F(J{z}_2,J{z}_1)+〈AJ{z}_1,J{z}_2-J{z}_1〉+〈AJ{z}_2,J{z}_1-J{z}_2〉\\ +\frac{1}{r}〈z_1-x,J(z_2-z_1)〉+\frac{1}{r}〈z_2-x,J(z_1-z_2)〉\\ =& F(J{z}_1,J{z}_2)+F(J{z}_2,J{z}_1)+〈AJ{z}_1,J{z}_2-J{z}_1〉-〈AJ{z}_2,J{z}_2-J{z}_1〉\\ +\frac{1}{r}〈z_1-x,J(z_2-z_1)〉-\frac{1}{r}〈z_2-x,J(z_2-z_1)〉\\ =& F(J{z}_1,J{z}_2)+F(J{z}_2,J{z}_1)+〈AJ{z}_1-AJ{z}_2,J{z}_2-J{z}_1〉+\frac{1}{r}〈(z_1-x)-(z_2-x),J(z_2-z_1)〉\\ =& F(J{z}_1,J{z}_2)+F(J{z}_2,J{z}_1)+〈AJ{z}_1-AJ{z}_2,J{z}_2-J{z}_1〉+\frac{1}{r}〈z_1-z_2,J(z_2-z_1)〉\\ =& F(J{z}_1,J{z}_2)+F(J{z}_2,J{z}_1)-〈AJ{z}_1-AJ{z}_2,J{z}_1-J{z}_2〉+\frac{1}{r}〈z_1-z_2,J(z_2-z_1)〉.\end{array}$$

Therefore, we obtain

$$F(J{z}_1,J{z}_2)+F(J{z}_2,J{z}_1)-〈AJ{z}_1-AJ{z}_2,J{z}_1-J{z}_2〉+\frac{1}{r}〈z_1-z_2,J(z_2-z_1)〉\ge 0.$$

(3.3)

From condition (DA2) and the fact that A is an α-inverse strongly skew monotone, we have

$$\begin{array}{rcl}0& \le & F(J{z}_1,J{z}_2)+F(J{z}_2,J{z}_1)-〈AJ{z}_1-AJ{z}_2,J{z}_1-J{z}_2〉+\frac{1}{r}〈z_1-z_2,J(z_2-z_1)〉\\ \le & -\alpha {\parallel AJ{z}_1-AJ{z}_2\parallel }^2+\frac{1}{r}〈z_1-z_2,J(z_2-z_1)〉\\ \le & \frac{1}{r}〈z_1-z_2,J(z_2-z_1)〉.\end{array}$$

Since $r>0$, J is monotone, and E is strictly convex, we obtain

$$z_1=z_2.$$

This implies that $T_r$ is single-valued.

(2) We show that $〈T_{r}x-T_{r}y,J(T_{r}x-T_{r}y)〉\le 〈x-y,J(T_{r}x-T_{r}y)〉$ for all $x,y\in E$.

Indeed, for any $x,y\in C$ and $r>0$, we have

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

and

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

Adding the two inequalities, we have

$$\begin{array}{rcl}0& \le & F(J{T}_{r}x,J{T}_{r}y)+F(J{T}_{r}y,J{T}_{r}x)+〈AJ{T}_{r}x,J{T}_{r}y-J{T}_{r}x〉+〈AJ{T}_{r}y,J{T}_{r}x-J{T}_{r}y〉\\ +\frac{1}{r}〈T_{r}x-x,J(T_{r}y-T_{r}x)〉+\frac{1}{r}〈T_{r}y-y,J(T_{r}x-T_{r}y)〉\\ =& F(J{T}_{r}x,J{T}_{r}y)+F(J{T}_{r}y,J{T}_{r}x)+〈AJ{T}_{r}x,J{T}_{r}y-J{T}_{r}x〉-〈AJ{T}_{r}y,J{T}_{r}y-J{T}_{r}x〉\\ +\frac{1}{r}〈T_{r}x-x,J(T_{r}y-T_{r}x)〉-\frac{1}{r}〈T_{r}y-y,J(T_{r}y-T_{r}x)〉\\ =& F(J{T}_{r}x,J{T}_{r}y)+F(J{T}_{r}y,J{T}_{r}x)+〈AJ{T}_{r}x-AJ{T}_{r}y,J{T}_{r}y-J{T}_{r}x〉\\ +\frac{1}{r}〈(T_{r}x-x)-(T_{r}y-y),J(T_{r}y-T_{r}x)〉\\ =& F(J{T}_{r}x,J{T}_{r}y)+F(J{T}_{r}y,J{T}_{r}x)+〈AJ{T}_{r}x-AJ{T}_{r}y,J{T}_{r}y-J{T}_{r}x〉\\ +\frac{1}{r}〈(T_{r}x-T_{r}y)-(x-y),J(T_{r}y-T_{r}x)〉\\ =& F(J{T}_{r}x,J{T}_{r}y)+F(J{T}_{r}y,J{T}_{r}x)-〈AJ{T}_{r}x-AJ{T}_{r}y,J{T}_{r}x-J{T}_{r}y〉\\ +\frac{1}{r}〈(T_{r}x-T_{r}y)-(x-y),J(T_{r}y-T_{r}x)〉.\end{array}$$

(3.4)

It follows that

$$\begin{array}{r}F(J{T}_{r}x,J{T}_{r}y)+F(J{T}_{r}y,J{T}_{r}x)-〈AJ{T}_{r}x-AJ{T}_{r}y,J{T}_{r}x-J{T}_{r}y〉\\ +\frac{1}{r}〈(T_{r}x-T_{r}y)-(x-y),J(T_{r}y-T_{r}x)〉\ge 0.\end{array}$$

(3.5)

From condition (DA2) and the fact that A is an α-inverse strongly skew monotone, we get

$$\begin{array}{rcl}0& \le & F(J{T}_{r}x,J{T}_{r}y)+F(J{T}_{r}y,J{T}_{r}x)-〈AJ{T}_{r}x-AJ{T}_{r}y,J{T}_{r}x-J{T}_{r}y〉\\ +\frac{1}{r}〈(T_{r}x-T_{r}y)-(x-y),J(T_{r}y-T_{r}x)〉\\ \le & -\alpha {\parallel AJ{T}_{r}x-AJ{T}_{r}y\parallel }^2+\frac{1}{r}〈(T_{r}x-T_{r}y)-(x-y),J(T_{r}y-T_{r}x)〉\\ \le & \frac{1}{r}〈(T_{r}x-T_{r}y)-(x-y),J(T_{r}y-T_{r}x)〉\\ \le & -\frac{1}{r}〈(T_{r}x-T_{r}y)-(x-y),J(T_{r}x-T_{r}y)〉.\end{array}$$

Since $r>0$, we have

$$〈(T_{r}x-T_{r}y)-(x-y),J(T_{r}x-T_{r}y)〉\le 0.$$

Therefore, we also have

$$〈T_{r}x-T_{r}y,J(T_{r}x-T_{r}y)〉-〈x-y,J(T_{r}x-T_{r}y)〉\le 0.$$

This implies that

$$〈T_{r}x-T_{r}y,J(T_{r}x-T_{r}y)〉\le 〈x-y,J(T_{r}x-T_{r}y)〉.$$

(3) We show that $Fix(T_r)=GEP(F,A)$.

It is easy to see that

$$\begin{array}{rcl}z\in Fix(T_r)& \iff & z=T_{r}z\\ \iff & F(Jz,Jy)+〈AJz,Jy-Jz〉+\frac{1}{r}〈z-z,J(y-z)〉\ge 0\\ \iff & F(Jz,Jy)+〈AJz,Jy-Jz〉\ge 0\\ \iff & z\in GEP(F,A).\end{array}$$

This implies that $Fix(T_r)=GEP(F,A)$.

(4) We show that $J(GEP(F,A))$ is closed and convex.

For each $y\in C$, we define the mapping $G:C\to 2^E$ as follows:

$$G(y)=\{z\in C:F(Jz,Jy)+〈AJz,Jy-Jz〉\ge 0\}.$$

It is easy to see that $y\in G(y)$, so that $G(y)\ne \mathrm{\varnothing }$.

Next, we show that G is a KKM mapping.

Suppose that G is not a KKM mapping. Then there exists a finite subset $\{z_1,z_2,\dots ,z_m\}$ of C, and ${\beta }_i>0$ with ${\sum }_{i=1}^m{\beta }_i=1$ such that $\hat{z}={\sum }_{i=1}^m{\beta }_i{z}_i\notin G(z_i)$ for all $i=1,2,3,\dots ,m$. Then we have

$$F(J\hat{z},J{z}_i)+〈AJ\hat{z},J{z}_i-J\hat{z}〉<0,i=1,2,3,\dots ,m.$$

It follows from (DA1) and (DA4) that

$$0=F(J\hat{z},J\hat{z})+〈AJ\hat{z},J\hat{z}-J\hat{z}〉\le \sum _{i=1}^m{\beta }_i(F(J\hat{z},J{z}_i)+〈AJ\hat{z},J{z}_i-J\hat{z}〉)<0,$$

which is the contradiction. Hence, G is a KKM mapping on C.

(4.1) Next, we show that $G(y)$ is closed for each $y\in C$.

For any $y\in C$, let $\{z_n\}$ be any sequence in $G(y)$ such that $z_n\to z_0$, as $n\to \mathrm{\infty }$.

Hence, $z_n-x_0\to z_0-x_0$, as $n\to \mathrm{\infty }$.

Next, we show that $z_0\in G(y)$. Then for each $y\in C$, we have

$$F(J{z}_n,Jy)+〈AJ{z}_n,Jy-J{z}_n〉\ge 0.$$

(3.6)

It follows from assumption (DA3) that

$$F(J{z}_0,Jy)+〈AJ{z}_0,Jy-J{z}_0〉\ge \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}F(J{z}_n,Jy)+\underset{n\to \mathrm{\infty }}{lim}〈AJ{z}_n,Jy-J{z}_n〉\ge 0.$$

This implies that $z_0\in G(y)$, and hence $G(y)$ is closed for each $y\in C$.

Since J is continuous. Therefore, ${\bigcap }_{y\in C}G(y)=J(GEP(F,A))$ is closed.

(4.2) Next, we show that $J(GEP(F,A))$ is convex.

Let $z_1^{\ast },z_2^{\ast }\in J(GEP(F,A))$, then we have $z_1^{\ast }=J{z}_1\in J(C)$ and $z_2^{\ast }=J{z}_2\in J(C)$, where $z_1,z_2\in C$.

For $k,t\in (0,1)$, let $z^{\ast }=k{z}_1^{\ast }+(1-k)z_2^{\ast }$ and for any $y\in C$, we set $x_t^{\ast }=tJy+(1-t)z^{\ast }$.

It follows from (DA1) and (DA4) that

$$\begin{array}{rl}0=& F(x_t^{\ast },x_t^{\ast })\\ \le & F(x_t^{\ast },tJy+(1-t)z^{\ast })\\ \le & tF(x_t^{\ast },Jy)+(1-t)F(x_t^{\ast },z^{\ast })\\ \le & tF(x_t^{\ast },Jy),\end{array}$$

(3.7)

and

$$\begin{array}{rcl}0& =& 〈A{x}_t^{\ast },x_t^{\ast }-x_t^{\ast }〉\\ =& 〈A{x}_t^{\ast },(x_t^{\ast }-Jy)+(Jy-x_t^{\ast })〉\\ =& 〈A{x}_t^{\ast },x_t^{\ast }-Jy〉+〈A{x}_t^{\ast },Jy-x_t^{\ast }〉\\ =& 〈A{x}_t^{\ast },x_t^{\ast }-Jy〉-〈A{x}_t^{\ast },x_t^{\ast }-Jy〉\\ \le & 〈A{x}_t^{\ast },x_t^{\ast }-Jy〉\\ =& 〈A{x}_t^{\ast },tJy+(1-t)z^{\ast }-Jy〉\\ =& 〈A{x}_t^{\ast },tJy+(1-t)z^{\ast }-(t+(1-t))Jy〉\\ =& 〈A{x}_t^{\ast },(1-t)(z^{\ast }-Jy)〉\\ =& (1-t)〈A{x}_t^{\ast },z^{\ast }-Jy〉\\ =& (t-1)〈A{x}_t^{\ast },Jy-z^{\ast }〉\\ \le & t〈A{x}_t^{\ast },Jy-z^{\ast }〉.\end{array}$$

(3.8)

Adding two inequalities (3.7) and (3.8) and dividing by $t>0$, we get

$$\begin{array}{rcl}0& \le & tF(x_t^{\ast },Jy)+t〈A{x}_t^{\ast },Jy-z^{\ast }〉\\ \le & F(x_t^{\ast },Jy)+〈A{x}_t^{\ast },Jy-z^{\ast }〉.\end{array}$$

Letting t to 0 by (DA3) and the hemicontinuous of A, we obtain

$$F(z^{\ast },Jy)+〈A{z}^{\ast },Jy-z^{\ast }〉\ge 0\text{for all }y\in C.$$

Hence, $z^{\ast }\in J(GEP(F,A))$, and thus, $J(GEP(F,A))$ is convex.

This completes the proof. □

4 Convergence theorem

In this section, we use the hybrid projection method for finding a solution of a generalized equilibrium problem in the dual space of Banach spaces.

Theorem 4.1 Let E be a uniformly smooth, strictly convex and reflexive real Banach space, which has a Kadec-Klee property, let C be a nonempty, closed and convex subset of E, and let J be the duality mapping from E into $E^{\ast }$ such that $J(C)$ is closed and convex of $E^{\ast }$, let us assume that a bifunction $F:J(C)\times J(C)\to \mathbb{R}$ satisfies the following conditions (DA1)-(DA4), let $C^{\ast }$ be a nonempty, closed and convex subset of $E^{\ast }$, and let $A:C^{\ast }\to E$ be an α-inverse strongly skew monotone. For $x\in E$, we define a mapping $T_{r_n}:E\to C$ as follows:

$$T_{r_n}(x)=\{z\in C:F(Jz,Jy)+〈AJz,Jy-Jz〉+\frac{1}{r_n}〈z-x,J(y-z)〉\ge 0,\mathrm{\forall }y\in C\}.$$

(4.1)

Suppose that $GEP(F,A)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{c}{x}_0\in C\mathit{\text{chosen arbitrary}},C_0=C,\hfill \\ u_n=T_{r_n}{x}_n,\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)u_n,\hfill \\ C_{n+1}=\{z\in C_n:\varphi (y_n,z)\le \varphi (x_n,z)\},\hfill \\ x_{n+1}=R_{C_{n+1}}(x_0),\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(4.2)

where J is the duality mapping on E, $\{{\alpha }_n\}$ is the sequence in $[0,1]$ such that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$, $\{r_n\}\subset [a,\mathrm{\infty })$ for some $a>0$ and $R_{C_{n+1}}$ is the sunny generalized nonexpansive retraction from E onto $C_{n+1}$. Then the sequence $\{x_n\}$ converges strongly to $R_{GEP(F,A)}(x_0)$, where $R_{GEP(F,A)}$ is the sunny generalized nonexpansive retraction from E onto $GEP(F,A)$.

Proof We complete this proof by seven steps below.

Step 1. We show that $J(C_n)$ are closed and convex subsets of E for each $n\in \mathbb{N}\cup \{0\}$.

It is obvious that $J(C_0)$ is closed and convex. Suppose that $J(C_k)$ is closed and convex for some $k\in \mathbb{N}\cup \{0\}$.

For each $z\in C_n$, we see that

$$\begin{array}{rcl}\varphi (y_n,z)\le \varphi (x_n,z)& \iff & {\parallel y_n\parallel }^2-2〈y_n,Jz〉+{\parallel z\parallel }^2\le {\parallel x_n\parallel }^2-2〈x_n,Jz〉+{\parallel z\parallel }^2\\ \iff & {\parallel y_n\parallel }^2-{\parallel x_n\parallel }^2-2〈y_n,Jz〉+2〈x_n,Jz〉\le 0\\ \iff & {\parallel y_n\parallel }^2-{\parallel x_n\parallel }^2-2〈y_n-x_n,Jz〉\le 0.\end{array}$$

Hence, $J(C_{k+1})$ is closed and convex. Therefore, $J(C_n)$ are closed and convex subsets of E for each $n\in \mathbb{N}\cup \{0\}$.

Step 2. We show that $GEP(F,A)\subset C_n$ for all $n\in \mathbb{N}\cup \{0\}$.

Note that $u_n=T_{r_n}{x}_n$ and $T_{r_n}$ is generalized nonexpansive.

From $C_0=C$, we have $GEP(F,A)\subset C_0$. Suppose that $GEP(F,A)\subset C_k$, for some $k\in \mathbb{N}\cup \{0\}$.

For any $p\in GEP(F,A)$, from algorithm (4.2) and the fact that $T_{r_k}$ is generalized nonexpansive, we compute

$$\begin{array}{rcl}\varphi (y_k,p)& =& \varphi ({\alpha }_n{x}_k+(1-{\alpha }_n)u_k,p)\\ =& {\parallel {\alpha }_k{x}_k+(1-{\alpha }_k)u_k\parallel }^2-2〈{\alpha }_k{x}_k+(1-{\alpha }_k)u_k,Jp〉+{\parallel p\parallel }^2\\ \le & {\alpha }_k{\parallel x_k\parallel }^2+(1-{\alpha }_k){\parallel u_k\parallel }^2-2{\alpha }_k〈x_k,Jp〉-2(1-{\alpha }_k)〈u_k,Jp〉+{\parallel p\parallel }^2\\ =& {\alpha }_k{\parallel x_k\parallel }^2+(1-{\alpha }_k){\parallel u_k\parallel }^2-2{\alpha }_k〈x_k,Jp〉\\ -2(1-{\alpha }_k)〈u_k,Jp〉+({\alpha }_k+(1-{\alpha }_k)){\parallel p\parallel }^2\\ =& {\alpha }_k{\parallel x_k\parallel }^2-2{\alpha }_k〈x_k,Jp〉+{\alpha }_k{\parallel p\parallel }^2+(1-{\alpha }_k){\parallel u_k\parallel }^2\\ -2(1-{\alpha }_k)〈u_k,Jp〉+(1-{\alpha }_k){\parallel p\parallel }^2\\ =& {\alpha }_k\varphi (x_k,p)+(1-{\alpha }_k)\varphi (u_k,p)\\ =& {\alpha }_k\varphi (x_k,p)+(1-{\alpha }_k)\varphi (T_{r_k}{x}_k,p)\\ \le & {\alpha }_k\varphi (x_k,p)+(1-{\alpha }_k)\varphi (x_k,p)\\ =& \varphi (x_k,p).\end{array}$$

Therefore, $p\in C_k$. Hence, we get $p\in C_{k+1}$.

This implies that $GEP(F,A)\subset C_n$, $\mathrm{\forall }n\in \mathbb{N}\cup \{0\}$, and also the sequence $\{x_n\}$ is well defined.

Step 3. We show that $\{x_n\}$ is bounded.

From the definition of $\{x_n\}$, we know that $x_n=R_{C_n}(x_0)$ and $x_{n+1}=R_{C_{n+1}}(x_0)\in C_{n+1}\subset C_n$. For all $p\in GEP(F,A)\subset C_n$, we have

$$\begin{array}{rcl}\varphi (x_n,x_0)& =& \varphi (R_{C_n}(x_0),x_0)\\ \le & \varphi (p,x_0)-\varphi (p,R_{C_n}(x_0))\\ \le & \varphi (p,x_0).\end{array}$$

Then $\varphi (x_0,x_n)$ is bounded. Therefore, $\{x_n\}$ is bounded, and also $\{T_{r_n}{x}_n\}$ is bounded.

Step 4. We show that there exists $\hat{p}\in C$ such that $x_n\to \hat{p}$, as $n\to \mathrm{\infty }$.

Since $x_n=R_{C_n}(x_0)$ and $x_{n+1}=R_{C_{n+1}}(x_0)\in C_{n+1}\subset C_n$, we have

$$\varphi (x_n,x_0)\le \varphi (x_{n+1},x_0),\mathrm{\forall }n\in \mathbb{N}\cup \{0\}.$$

Therefore, the sequence $\{\varphi (x_n,x_0)\}$ is nondecreasing. Hence, ${lim}_{n\to \mathrm{\infty }}\varphi (x_n,x_0)$ exists.

By the definition of $C_n$, one has that $C_m\subset C_n$ and $x_m=R_{C_m}(x_0)\in C_n$ for any positive integer $m\ge n$. It follows that

$$\begin{array}{rcl}\varphi (x_m,x_n)& =& \varphi (x_m,R_{C_n}(x_0))\\ \le & \varphi (x_m,x_0)-\varphi (R_{C_n}(x_0),x_0)\\ \le & \varphi (x_m,x_0)-\varphi (x_n,x_0)\\ \to & 0,\text{as }m,n\to \mathrm{\infty }.\end{array}$$

It follows from Lemma 2.10 that $\parallel x_m-x_n\parallel \to 0$, as $m,n\to \mathrm{\infty }$.

Thus, the sequence $\{x_n\}$ is a Cauchy sequence.

Without loss of generality, we can assume that $x_n\rightharpoonup \hat{p}$. Since $\{x_n\}$ is bounded, and E is reflexive.

We know that $C_{n+1}\subset C_n$, and $C_n$ is closed and convex, for $\hat{p}\in C_n$, we have

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

It follows that

$$\varphi (\hat{p},x_0)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\varphi (x_n,x_0)\le \varphi (\hat{p},x_0).$$

This implies that

$$\underset{n\to \mathrm{\infty }}{lim}\varphi (x_n,x_0)=\varphi (\hat{p},x_0).$$

So $\parallel x_n\parallel \to \parallel \hat{p}\parallel$. Since $x_n\rightharpoonup \hat{p}$. By the Kadec-Klee property of E, we obtain that

$$x_n\to \hat{p},\text{as }n\to \mathrm{\infty }.$$

(4.3)

From J is uniformly norm-to-norm continuous on bounded subset of E, we also have

$$J{x}_n\to J\hat{p},\text{as }n\to \mathrm{\infty }.$$

(4.4)

Step 5. We show that $\parallel x_n-y_n\parallel \to 0$ and $\parallel x_n-u_n\parallel \to 0$, as $n\to \mathrm{\infty }$.

Since $x_n=R_{C_n}(x_0)$ and ${lim}_{n\to \mathrm{\infty }}\varphi (x_n,x_0)$ exists. We get

$$\begin{array}{rcl}\varphi (x_n,x_{n+1})& =& \varphi (R_{C_n}(x_0),x_{n+1})\\ \le & \varphi (x_0,x_{n+1})-\varphi (x_0,R_{C_n}(x_0))\\ =& \varphi (x_0,x_{n+1})-\varphi (x_0,x_n)\\ \to & 0,\text{as }n\to \mathrm{\infty }.\end{array}$$

From $x_{n+1}=R_{C_{n+1}}(x_0)\in C_{n+1}\subset C_n$, we have

$$\varphi (y_n,x_{n+1})\le \varphi (x_n,x_{n+1})\to 0,\text{as }n\to \mathrm{\infty }.$$

By Lemma 2.10, we obtain

$$\parallel x_n-x_{n+1}\parallel \to 0\text{and}\parallel y_n-x_{n+1}\parallel \to 0,\text{as }n\to \mathrm{\infty }.$$

Therefore,

$$\parallel x_n-y_n\parallel \to 0,\text{as }n\to \mathrm{\infty }.$$

(4.5)

Since

$$\begin{array}{rcl}0& =& \underset{n\to \mathrm{\infty }}{lim}\parallel x_n-y_n\parallel \\ =& \underset{n\to \mathrm{\infty }}{lim}\parallel x_n-({\alpha }_n{x}_n+(1-{\alpha }_n)u_n)\parallel \\ =& \underset{n\to \mathrm{\infty }}{lim}(1-{\alpha }_n)\parallel x_n-u_n\parallel .\end{array}$$

By the assumption, we have ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$, we obtain

$$\parallel x_n-u_n\parallel \to 0,\text{as }n\to \mathrm{\infty }.$$

(4.6)

Step 6. We show that $\hat{p}\in GEP(F,A)$, that is, $J\hat{p}\in J(GEP(F,A))$.

From (4.3), (4.5) and (4.6), it follows that

$$u_n\to \hat{p}\text{and}{y}_n\to \hat{p},\text{as }n\to \mathrm{\infty }.$$

(4.7)

From J is uniformly norm-to-norm continuous on bounded subset of E, we also have

$$J{u}_n\to J\hat{p}\text{and}J{y}_n\to J\hat{p},\text{as }n\to \mathrm{\infty }.$$

(4.8)

It follows from (4.5) and (4.6) and the property of J that

$$\parallel J{x}_n-J{y}_n\parallel \to 0\text{and}\parallel J{x}_n-J{u}_n\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

(4.9)

Since $\{x_n\}$ is bounded. Therefore, $\{u_n\}$ and $\{y_n\}$ are bounded.

Hence, $\{J{x}_n\}$, $\{J{u}_n\}$ and $\{J{y}_n\}$ are also bounded.

So, there exists a subsequence $\{J{x}_{n_k}\}$ of $\{J{x}_n\}$ such that $J{x}_{n_k}\rightharpoonup {\hat{p}}^{\ast }$, and there exists a subsequence $\{J{u}_{n_k}\}$ of $\{J{u}_n\}$ such that $J{u}_{n_k}\rightharpoonup {\hat{p}}^{\ast }$.

From (4.6) and $r_n\subset [a,\mathrm{\infty })$, we have

$$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{r_n}\parallel x_n-u_n\parallel =0.$$

Since $u_n=T_{r_n}{x}_n$, we have

$$F(J{u}_n,Jy)+〈AJ{u}_n,Jy-J{u}_n〉+\frac{1}{r_n}〈u_n-x_n,J(y-u_n)〉\ge 0,\mathrm{\forall }y\in C.$$

It follows from (DA2) that

$$〈AJ{u}_n,Jy-J{u}_n〉+\frac{1}{r_n}〈u_n-x_n,J(y-u_n)〉\ge -F(J{u}_n,Jy)\ge F(Jy,J{u}_n),\mathrm{\forall }y\in C.$$

From $x_n\to \hat{p}$ and $u_n\to \hat{p}$, we get

$$〈AJ\hat{p},Jy-J\hat{p}〉+\frac{1}{r_n}〈\hat{p}-\hat{p},J(y-\hat{p})〉\ge F(Jy,J\hat{p}),\mathrm{\forall }y\in C.$$

Therefore,

$$〈AJ\hat{p},Jy-J\hat{p}〉\ge F(Jy,J\hat{p}),\mathrm{\forall }y\in C.$$

For any $0<t<1$, $y\in C$ and setting $y_t^{\ast }=tJy+(1-t)J\hat{p}$. Then we get $y_t^{\ast }\in J(C)$, and so

$$〈AJ\hat{p},y_t^{\ast }-J\hat{p}〉\ge F(y_t^{\ast },J\hat{p}),\mathrm{\forall }y\in C.$$

It follows from (DA1) and (DA4) that

$$\begin{array}{rl}0& =F(y_t^{\ast },y_t^{\ast })\\ \le tF(y_t^{\ast },Jy)+(1-t)F(y_t^{\ast },J\hat{p})\\ \le tF(y_t^{\ast },Jy)+(1-t)〈AJ\hat{p},y_t^{\ast }-J\hat{p}〉.\end{array}$$

Letting $t\downarrow 0$, we obtain from assumption (DA3) that

$$F(J\hat{p},Jy)+〈AJ\hat{p},y_t^{\ast }-J\hat{p}〉\ge 0.$$

This implies that

$$J\hat{p}\in J(GEP(F,A)).$$

Step 7. We show that the sequence $\{x_n\}$ converges strongly to $R_{GEP(F,A)}(x_0)$.

We know that $x_n=R_{C_n}(x_0)\in GEP(F,A)\subset C_n$.

Let $w=R_{GEP(F,A)}(x_0)$. It follows from Lemma 2.9 that

$$\varphi (x_0,x_n)\le \varphi (x_0,w).$$

Since the norm is weakly lower semicontinuous, and from (DA4), we have

$$\begin{array}{rcl}\varphi (x_0,J^{-1}{\hat{p}}^{\ast })& =& {\parallel x_0\parallel }^2-2〈x_0,J\left(J^{-1}{\hat{p}}^{\ast }\right)〉+{\parallel \hat{p}\parallel }^2\\ =& {\parallel x_0\parallel }^2-2〈x_0,{\hat{p}}^{\ast }〉+{\parallel \hat{p}\parallel }^2\\ \le & \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}({\parallel x_0\parallel }^2-2〈x_0,J{x}_{n_k}〉+{\parallel x_{n_k}\parallel }^2)\\ =& \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\varphi (x_0,x_{n_k})\\ \le & \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\varphi (x_0,x_{n_k})\\ \le & \varphi (x_0,w).\end{array}$$

From the definition of $R_{GEP(F,A)}$, we have $J^{-1}{\hat{p}}^{\ast }=w$.

Finally, we show that $x_n\to w$, where $w=R_{GEP(F,A)}(x_0)$.

Now, we have

$$\varphi (x_0,x_n)\le \varphi (x_0,w)=\varphi (x_n,R_{GEP(F,A)}(x_0)),$$

and from Remark 1.5(2), we get

$$\varphi (w,x_n)=\varphi (w,x_0)+\varphi (x_0,x_n)+2〈w-x_0,J{x}_0-J{x}_n〉.$$

Therefore,

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\varphi (w,x_n)& =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\varphi (w,x_0)+\varphi (x_0,x_n)+2〈w-x_0,J{x}_0-J{x}_n〉)\\ \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\varphi (w,x_0)+\varphi (x_0,w)+2〈w-x_0,J{x}_0-J{x}_n〉)\\ =& \varphi (w,x_0)+\varphi (x_0,w)+2〈w-x_0,J{x}_0-Jw〉\\ =& \varphi (w,w)=0.\end{array}$$

This implies that

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

By Lemma 2.10, we get

$$\underset{n\to \mathrm{\infty }}{lim}\parallel w-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel R_{GEP(F,A)}(x_0)-x_n\parallel =0.$$

Hence,

$$x_n\to R_{GEP(F,A)}(x_0),\text{as }n\to \mathrm{\infty }.$$

Therefore, the sequence $\{x_n\}$ converges strongly to $R_{GEP(F,A)}(x_0)$. This completes the proof. □

If we substitute $x=x_n$ and $z=u_n$ in equation (4.1), then we obtain the following result which extends the following results by Takahashi and Zembayashi [1] from an equilibrium problem to a generalized equilibrium problem.

Corollary 4.2 Let E be a uniformly smooth, strictly convex and reflexive real Banach space which has a Kadec-Klee property, let C be a nonempty, closed and convex subset of E, and let J be the duality mapping from E into $E^{\ast }$ such that $J(C)$ is closed and convex of $E^{\ast }$, let us assume that a bifunction $F:J(C)\times J(C)\to \mathbb{R}$ satisfies the following conditions (DA1)-(DA4), let $C^{\ast }$ be a nonempty, closed and convex subset of $E^{\ast }$, and let $A:C^{\ast }\to E$ be an α-inverse strongly skew monotone. Suppose that $GEP(F,A)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{c}{x}_0\in C\mathit{\text{chosen arbitrary}},C_0=C,\hfill \\ u_n\in C\mathit{\text{such that}}\hfill \\ F(J{u}_n,Jy)+〈AJ{u}_n,Jy-J{u}_n〉+\frac{1}{r_n}〈u_n-x_n,J(y-u_n)〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)u_n,\hfill \\ C_{n+1}=\{z\in C_n:\varphi (y_n,z)\le \varphi (x_n,z)\},\hfill \\ x_{n+1}=R_{C_{n+1}}(x_0),\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(4.10)

where J is the duality mapping on E, $\{{\alpha }_n\}$ is the sequence in $[0,1]$ such that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$, $\{r_n\}\subset [a,\mathrm{\infty })$ for some $a>0$ and $R_{C_{n+1}}$ is the sunny generalized nonexpansive retraction from E onto $C_{n+1}$. Then the sequence $\{x_n\}$ converges strongly to $R_{GEP(F,A)}(x_0)$, where $R_{GEP(F,A)}$ is the sunny generalized nonexpansive retraction from E onto $GEP(F,A)$.

If we set $A\equiv 0$ in Corollary 4.2, then we obtain the following result which extends the following results by Takahashi and Zembayashi [1].

Corollary 4.3 Let E be a uniformly smooth, strictly convex and reflexive real Banach space which has Kadec-Klee property, let C be a nonempty, closed and convex subset of E, and let J be the duality mapping from E into $E^{\ast }$ such that $J(C)$ is closed and convex of $E^{\ast }$, let us assume that a bifunction $F:J(C)\times J(C)\to \mathbb{R}$ satisfies the following conditions (DA1)-(DA4). Suppose that $EP(F)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{c}{x}_0\in C\mathit{\text{chosen arbitrary}},C_0=C,\hfill \\ u_n\in C\mathit{\text{such that}}F(J{u}_n,Jy)+\frac{1}{r_n}〈u_n-x_n,J(y-u_n)〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)u_n,\hfill \\ C_{n+1}=\{z\in C_n:\varphi (y_n,z)\le \varphi (x_n,z)\},\hfill \\ x_{n+1}=R_{C_{n+1}}(x_0),\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(4.11)

where J is the duality mapping on E, $\{{\alpha }_n\}$ is the sequence in $[0,1]$ such that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$, $\{r_n\}\subset [a,\mathrm{\infty })$ for some $a>0$ and $R_{C_{n+1}}$ is the sunny generalized nonexpansive retraction from E onto $C_{n+1}$. Then the sequence $\{x_n\}$ converges strongly to $R_{EP(F)}(x_0)$, where $R_{EP(F)}$ is the sunny generalized nonexpansive retraction from E onto $EP(F)$.

If we set $F\equiv 0$ in Corollary 4.2, then Corollary 4.2 is reduced to the following corollary.

Corollary 4.4 Let E be a uniformly smooth, strictly convex and reflexive real Banach space which has a Kadec-Klee property, let C be a nonempty, closed and convex subset of E such that $J(C)$ is closed and convex of $E^{\ast }$, let $C^{\ast }$ be a nonempty, closed and convex subset of $E^{\ast }$ and $A:C^{\ast }\to E$ be an α-inverse strongly skew monotone. Suppose that $VI(J(C),A)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{c}{x}_0\in C\mathit{\text{chosen arbitrary}},C_0=C,\hfill \\ u_n\in C\mathit{\text{such that}}〈AJ{u}_n,Jy-J{u}_n〉+\frac{1}{r_n}〈u_n-x_n,J(y-u_n)〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)u_n,\hfill \\ C_{n+1}=\{z\in C_n:\varphi (y_n,z)\le \varphi (x_n,z)\},\hfill \\ x_{n+1}=R_{C_{n+1}}(x_0),\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(4.12)

where J is the duality mapping on E, $\{{\alpha }_n\}$ is the sequence in $[0,1]$ such that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$, $\{r_n\}\subset [a,\mathrm{\infty })$ for some $a>0$, and $R_{C_{n+1}}$ is the sunny generalized nonexpansive retraction from E onto $C_{n+1}$. Then the sequence $\{x_n\}$ converges strongly to $R_{VI(J(C),A)}(x_0)$, where $R_{VI(J(C),A)}$ is the sunny generalized nonexpansive retraction from E onto $VI(J(C),A)$.