New feasible iterative algorithms and strong convergence theorems for bilevel split equilibrium problems

Abstract

In this paper, we first introduce and investigate a bilevel split equilibrium problem (BSEP) which can be regarded as a new development in the field of equilibrium problems. We provide some new feasible iterative algorithms for BSEP and establish strong convergence theorems for these iterative algorithms in different spaces.

MSC:47J25, 47H09, 65K10.

1 Introduction and preliminaries

Let K be a closed convex subset of a real Hilbert space H. Let $〈\cdot ,\cdot 〉$ and $\parallel \cdot \parallel$ denote the inner product of H and the norm of H, respectively. For each point $x\in H$, there exists a unique nearest point in K, denoted by $P_{K}x$, such that

$$\parallel x-P_{K}x\parallel \le \parallel x-y\parallel \text{for all }y\in K.$$

The mapping $P_K$ is called the metric projection from H onto K. It is well known that $P_K$ has the following properties:

• (i) $〈x-y,P_{K}x-P_{K}y〉\ge {\parallel P_{K}x-P_{K}y\parallel }^2$ for every $x,y\in H$.

• (ii) For $x\in H$ and $z\in K$, $z=P_K(x)\iff 〈x-z,z-y〉\ge 0$ for all $y\in K$.

• (iii) For $x\in H$ and $y\in K$,

  $${\parallel y-P_K(x)\parallel }^2+{\parallel x-P_K(x)\parallel }^2\le {\parallel x-y\parallel }^2.$$

  (1.1)

Let $H_1$ and $H_2$ be two Hilbert spaces. Let $A:H_1\to H_2$ and $A^{\ast }:H_2\to H_1$ be two bounded linear operators. $A^{\ast }$ is called the adjoint operator (or adjoint) of A if

$$〈Az,w〉=〈z,A^{\ast }w〉\text{for all }z\in H_1\text{ and }w\in H_2.$$

It is known that the adjoint operator of a bounded linear operator on a Hilbert space always exists and is bounded linear and unique. Moreover, it is not hard to show that if $A^{\ast }$ is an adjoint operator of A, then $\parallel A\parallel =\parallel A^{\ast }\parallel$. The symbols ℕ and ℝ are used to denote the sets of positive integers and real numbers, respectively.

Example 1.1 ([1])

Let $H_2=\mathbb{R}$ with the standard norm $|\cdot |$ and $H_1={\mathbb{R}}^2$ with the norm $\parallel \alpha \parallel ={(a_1^2+a_2^2)}^{\frac{1}{2}}$ for some $\alpha =(a_1,a_2)\in {\mathbb{R}}^2$. $〈x,y〉=xy$ denotes the inner product of $H_2$ for some $x,y\in H_2$, and $〈\alpha ,\beta 〉={\sum }_{i=1}^2{a}_i{b}_i$ denotes the inner product of $H_1$ for some $\alpha =(a_1,a_2),\beta =(b_1,b_2)\in H_1$. Let $A\alpha =a_2-a_1$ for $\alpha =(a_1,a_2)\in H_1$ and $Bx=(-x,x)$ for $x\in H_2$, then B is an adjoint operator of A.

Example 1.2 ([1])

Let $H_1={\mathbb{R}}^2$ with the norm $\parallel \alpha \parallel ={(a_1^2+a_2^2)}^{\frac{1}{2}}$ for some $\alpha =(a_1,a_2)\in {\mathbb{R}}^2$ and $H_2={\mathbb{R}}^3$ with the norm $\parallel \gamma \parallel ={(c_1^2+c_2^2+c_3^2)}^{\frac{1}{2}}$ for some $\gamma =(c_1,c_2,c_3)\in {\mathbb{R}}^3$. Let $〈\alpha ,\beta 〉={\sum }_{i=1}^2{a}_i{b}_i$ and $〈\gamma ,\eta 〉={\sum }_{i=1}^3{c}_i{d}_i$ denote the inner product of $H_1$ and $H_2$, respectively, where $\alpha =(a_1,a_2),\beta =(b_1,b_2)\in H_1$, $\gamma =(c_1,c_2,c_3),\eta =(d_1,d_2,d_3)\in H_2$. Let $A\alpha =(a_2,a_1,a_1-a_2)$ for $\alpha =(a_1,a_2)\in H_1$ and $B\gamma =(c_2+c_3,c_1-c_3)$ for $\gamma =(c_1,c_2,c_3)\in H_2$. Obviously, B is an adjoint operator of A.

Let f be a bi-function from $C\times C$ to ℝ. The classical equilibrium problem (EP, for short) is defined as follows.

(EP) Find $p\in C$ such that $f(p,y)\ge 0$, $\mathrm{\forall }y\in C$.

The set of such solutions is denoted by $EP(f)$, that is, $EP(f)=\{u\in C:f(u,v)\ge 0,\mathrm{\forall }v\in C\}$. In fact, equilibrium problem has an important relationship with variational inequality problem. For example, let $T:C\to H$ be a nonlinear mapping satisfying $〈Tx,y-x〉\ge 0$ for all $x,y\in C$. Then $x\in EP(f)$ if and only if $x\in C$ is a solution of the variational inequality $〈Tx,y-x〉\ge 0$ for all $y\in C$. It is known that the EP is an important mathematical model for nonlinear analysis and applied sciences which is generalized to many new mathematical models and includes many important problems arising in physics, engineering, science optimization, economics, network, game theory, complementary problems, variational inequalities problems, saddle point problems, fixed point problems and others; for details, one can refer to [2–8] and references therein. Many authors have proposed some useful methods to solve the EP; see, for instance, [2–5, 9–17] and references therein.

Recent investigations and developments in equilibrium theory as well as optimization theory have been applied to connect fundamental sciences with the real world. According to our experience, useful methods of real world problems often need to be used to solve several problems arising in different spaces. In view of this, recent studies focus on split problems which are more closed in the real world applications; see, for instance, [1, 18–24] and the references therein. Recently, He [1] considered the following split equilibrium problem. Let $H_1$ and $H_2$ be two real Hilbert spaces. Let C be a closed convex subset of $H_1$ and K be a closed convex subset of $H_2$. Let $f:C\times C\to \mathbb{R}$ and $g:K\times K\to \mathbb{R}$ be two bi-functions, and $A:H_1\to H_2$ be a bounded linear operator. The split equilibrium problem (SEP, in short) is defined as follows:

(SEP) Find $p\in C$ such that $f(p,y)\ge 0$, $\mathrm{\forall }y\in C$, and $u:=Ap$ satisfying $g(u,v)\ge 0$, $\mathrm{\forall }v\in K$.

In [1], the author established weak convergence algorithms and strong convergence algorithms for SEP (see [1] for more details).

Motivated and inspired by the works mentioned above, in this paper we shall introduce and investigate the following new problem. Let $H_1$, $H_2$ and $H_3$ be three real Hilbert spaces. Let C be a closed convex subset of $H_1$, Q be a closed convex subset of $H_2$ and K be a closed convex subset of $H_3$. Let $f:C\times C\to \mathbb{R}$, $g:Q\times Q\to \mathbb{R}$ and $h:K\times K\to \mathbb{R}$ be three bi-functions. Let $A:H_1\to H_3$ and $B:H_2\to H_3$ be two bounded linear operators with theirs adjoint operators $A^{\ast }$ and $B^{\ast }$, respectively. The mathematical model about bilevel split equilibrium problem (BSEP, in short) is defined as follows:

(BSEP) Find $p\in C$ and $q\in Q$ such that

• (i) $f(p,x)\ge 0$ and $g(q,y)\ge 0$ for all $x\in C$ and $y\in Q$;

• (ii) $Ap=Bq:=u$;

• (iii) $h(u,z)\ge 0$ for all $z\in K$.

In fact, BSEP can be regarded as a new development in the field of equilibrium problems and contains several important problems as special cases. It was profoundly believed that BSEP will motivate and inspire further scientific activities in the fields of equilibrium problems, optimization problems, game problems, complementary problems, variational inequalities problems, fixed point problems and their applications.

Example A Let $H_1$, $H_2$ and $H_3$ be three real Hilbert spaces. Let $C\subset H_1$, $Q\subset H_2$ and $K\subset H_3$ be three closed convex sets. Let $f^{\ast }:C\to \mathbb{R}$, $g^{\ast }:Q\to \mathbb{R}$ and $h^{\ast }:K\to \mathbb{R}$ be three convex functions. Let $A:H_1\to H_3$ and $B:H_2\to H_3$ be two bounded linear operators with their adjoint operators $A^{\ast }$ and $B^{\ast }$, respectively. Let

$$\begin{array}{r}f(x,\alpha )=f^{\ast }(x)-f^{\ast }(\alpha )\text{for }x,\alpha \in C,\\ g(y,\beta )=g^{\ast }(y)-g^{\ast }(\beta )\text{for }y,\beta \in Q,\end{array}$$

and

$$h(z,\eta )=h^{\ast }(z)-h^{\ast }(\eta )\text{for }z,\eta \in K.$$

Then BSEP reduces the bilevel convex optimization problem (BCOP):

(BCOP) Find $p\in C$ and $q\in Q$ such that $u:=Ap=Bq\in K$, $f^{\ast }(x)\ge f^{\ast }(p)$, $g^{\ast }(y)\ge g^{\ast }(q)$ and $h^{\ast }(z)\ge h^{\ast }(u)$ for all $x\in C$, $y\in Q$ and $z\in K$.

Example B Let $H_1$, $H_2$ and $H_3$ be three real Hilbert spaces. Let $C\subset H_1$, $Q\subset H_2$ and $K\subset H_3$ be three closed convex sets. Let $T:C\to H_1$, $S:Q\to H_2$ and $G:K\to H_3$ be three nonlinear operators. Let $A:H_1\to H_3$ and $B:H_2\to H_3$ be two bounded linear operators with their adjoint operators $A^{\ast }$ and $B^{\ast }$, respectively. If $f(p,x)=〈Tp,x-p〉$, $g(q,y)=〈Sq,y-q〉$ and $h(u,z)=〈Tu,z-u〉$, then BSEP reduces to the bilevel split variational inequality problem (BSVI):

(BSVI) Find $p\in C$ and $q\in Q$ such that $u:=Ap=Bq\in K$ satisfying $〈Tp,x-p〉\ge 0$, $〈Sq,y-q〉\ge 0$ and $〈Tu,z-u〉\ge 0$ for all $x\in C$, $y\in Q$ and $z\in K$.

Example C Let $H_1$ and $H_2$ be two real Hilbert spaces and $B:H_1\to H_2$ be a bounded linear operator with its adjoint operator $B^{\ast }$. Let $C\subset H_1$, $Q\subset H_1$ and $K\subset H_2$ be three closed convex sets. If $H_1=H_2$ and $A=B$, then BSEP reduces to the following split equilibrium problem (1) (SEP${(1)}^{}$):

(${\mathrm{SEP}}^{(1)}$) Find $p\in C$ and $q\in Q$ such that $u:=Bp=Bq\in K$ satisfying $f(p,x)\ge 0$, $g(q,y)\ge 0$ and $h(u,z)\ge 0$ for all $x\in C$, $y\in Q$ and $z\in K$.

Example D Let $H_1$ and $H_2$ be two real Hilbert spaces and $A:H_1\to H_2$ be a bounded linear operator with its adjoint operator $A^{\ast }$. Let $C\subset H_1$, $Q\subset H_2$ and $K\subset H_2$ be three closed convex sets with $Q\cap K\ne \mathrm{\varnothing }$. If $H_2=H_3$ and $B=I$ (identity operator), then BSEP reduces to the following split equilibrium problem (2) (SEP${(2)}^{}$):

(${\mathrm{SEP}}^{(2)}$) Find $p\in C$ such that $u:=Ap\in Q\cap K$ satisfying $f(p,x)\ge 0$, $g(u,y)\ge 0$ and $h(u,z)\ge 0$ for each $x\in C$, $y\in Q$ and $z\in K$.

Especially, if $g(p,y)\equiv 0$ for all $p,y\in Q$, then (SEP${(1)}^{}$) reduces to finding $p\in C$ such that $u:=Ap\in K$ satisfying $f(p,x)\ge 0$ and $h(u,z)\ge 0$ for all $x\in C$ and $z\in K$, which was studied in [1].

Example E Let $H_1$ and $H_2$ be two real Hilbert spaces and $B:H_1\to H_2$ be a bounded linear operator with its adjoint operator $A^{\ast }$. Let $C\subset H_1$, $Q\subset H_1$ and $K\subset H_2$ be three closed convex sets with $C\cap Q\ne \mathrm{\varnothing }$. If $H_1=H_2$ and $A=I$ (identity operator), then BSEP reduces to the following split equilibrium problems (3) (SEP${(3)}^{}$):

(${\mathrm{SEP}}^{(3)}$) Find $p\in C\cap Q$ such that $u:=Bp\in K$ satisfying $f(p,x)\ge 0$, $g(p,y)\ge 0$ and $h(u,z)\ge 0$ for all $x\in C$, $y\in Q$ and $z\in K$.

Example F In Example A, if $H_1=H_2=H_3:=H$, $C=Q=K\subset H$ and $A=B=I$ (identity operator), then BSEP reduces to the common solution of equilibrium problems (CEP):

(CEP) Find $p\in C$ such that $f(p,x)\ge 0$, $g(p,y)\ge 0$ and $h(p,z)\ge 0$ for each $x,y,z\in C$.

The paper is divided into four sections. In Sections 1 and 2, we first introduce and investigate a bilevel split equilibrium problem (BSEP) and then provide some new feasible iterative algorithms for BSEP and establish strong convergence theorems for these iterative algorithms in different spaces. In Section 3, we give the proof of the main result Theorem 2.1 in detail. Finally, an example illustrating Theorem 2.1 is given in Section 4.

2 Feasible iterative algorithms for BSEP and their strong convergence theorems

In 1994, Blum and Oettli [2] established the following important existence theorem which plays a key role in solving equilibrium problems, variational inequality problems and optimization problems.

Lemma 2.1 (Blum and Oettli [2])

Let K be a nonempty closed convex subset of H and F be a bi-function of $K\times K$ into ℝ satisfying the following conditions.

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

(A2) F is monotone, that is, $F(x,y)+F(y,x)\le 0$ for all $x,y\in K$;

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

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

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

Let $r>0$ and $x\in H$. Then there exists $z\in K$ such that

$$F(z,y)+\frac{1}{r}〈y-z,z-x〉\ge 0\mathit{\text{for all}}y\in K.$$

In this paper, we first introduce a new iterative algorithm for BSEP and establish a strong convergence theorem for this iterative algorithm. Here, the space $H_1\times H_2$ denotes the product space of two real Hilbert spaces $H_1$ and $H_2$, which is endowed with the usual linear operation and norm, namely, for $(x,y),(\overline{x},\overline{y})\in H_1\times H_2$ and $a,b\in \mathbb{R}$,

$$a(x,y)+b(\overline{x},\overline{y})=(ax+b\overline{x},ay+b\overline{y})$$

and

$$\parallel (x,y)\parallel =\parallel x\parallel +\parallel y\parallel .$$

Theorem 2.1 Let $H_1$, $H_2$ and $H_3$ be three real Hilbert spaces. Let C be a closed convex subset of $H_1$, Q be a closed convex subset of $H_2$ and K be a closed convex subset of $H_3$. Let $f:C\times C\to \mathbb{R}$, $g:Q\times Q\to \mathbb{R}$ and $h:K\times K\to \mathbb{R}$ be three bi-functions. $A:H_1\to H_3$ and $B:H_2\to H_3$ are two bounded linear operators with their adjoint operators $A^{\ast }$ and $B^{\ast }$, respectively. Suppose that all the bi-functions f, g and h satisfy conditions (A1)-(A4). Let $x_1\in C$, $y_1\in Q$, $C_1=C$, $Q_1=Q$, $\{x_n\}$, $\{y_n\}$, $\{u_n\}$, $\{v_n\}$ and $\{w_n\}$ be sequences generated by

$$\{\begin{array}{c}{u}_n=T_{r_n}^f{x}_n,v_n=T_{r_n}^g{y}_n,w_n=T_{r_n}^h(\frac{1}{2}A{u}_n+\frac{1}{2}B{v}_n),\hfill \\ l_n=P_C(u_n-\xi A^{\ast }(A{u}_n-w_n)),k_n=P_Q(v_n-\xi B^{\ast }(B{v}_n-w_n)),\hfill \\ C_{n+1}\times Q_{n+1}=\{(x,y)\in C_n\times Q_n:{\parallel l_n-x\parallel }^2+{\parallel k_n-y\parallel }^2\hfill \\ \phantom{C_{n+1}\times Q_{n+1}=\{}\le {\parallel u_n-x\parallel }^2+{\parallel v_n-y\parallel }^2\le {\parallel x_n-x\parallel }^2+{\parallel y_n-y\parallel }^2\},\hfill \\ x_{n+1}=P_{C_{n+1}}(x_1),\hfill \\ y_{n+1}=P_{Q_{n+1}}(y_1),\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$

(2.1)

where $\xi \in (0,min(\frac{1}{{\parallel A\parallel }^2},\frac{1}{{\parallel B\parallel }^2}))$ and $\{r_n\}\subset (0,+\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{r}_n>0$, $P_C$ and $P_Q$ are two projection operators from $H_1$ into C and from $H_2$ into Q, respectively. Suppose that

$$\mathrm{\Omega }=\{(p,q)\in EP(f)\times EP(g):Ap=Bq\in EP(h)\}\ne \mathrm{\varnothing }.$$

Then there exists $(p,q)\in \mathrm{\Omega }$ such that

• (a) $(x_n,y_n)\to (p,q)$ as $n\to \mathrm{\infty }$;

• (b) $(u_n,v_n)\to (p,q)$ as $n\to \mathrm{\infty }$;

• (c) $w_n\to w^{\ast }:=Ap=Bq\in EP(h)$ as $n\to \mathrm{\infty }$.

The following conclusion is immediate from Theorem 2.1 by putting $A=B$.

Corollary 2.1 Let $H_1$ and $H_2$ be two real Hilbert spaces. Let C and Q be two closed convex subsets of $H_1$ and K be a closed convex subset of $H_2$. Let $f:C\times C\to \mathbb{R}$, $g:Q\times Q\to \mathbb{R}$ and $h:K\times K\to \mathbb{R}$ be three bi-functions. $B:H_1\to H_2$ is a bounded linear operator with its adjoint operator $B^{\ast }$. Suppose that all the bi-functions f, g and h satisfy conditions (A1)-(A4). Let $x_1\in C$, $y_1\in Q$, $C_1=C$, $Q_1=Q$, $\{x_n\}$, $\{y_n\}$, $\{u_n\}$, $\{v_n\}$ and $\{w_n\}$ be sequences generated by

$$\{\begin{array}{c}{u}_n=T_{r_n}^f{x}_n,v_n=T_{r_n}^g{y}_n,w_n=T_{r_n}^h(\frac{1}{2}B{u}_n+\frac{1}{2}B{v}_n),\hfill \\ l_n=P_C(u_n-\xi B^{\ast }(B{u}_n-w_n)),k_n=P_Q(v_n-\xi B^{\ast }(B{v}_n-w_n)),\hfill \\ C_{n+1}\times Q_{n+1}=\{(x,y)\in C_n\times Q_n:{\parallel l_n-x\parallel }^2+{\parallel k_n-y\parallel }^2\le {\parallel u_n-x\parallel }^2+{\parallel v_n-y\parallel }^2\hfill \\ \phantom{C_{n+1}\times Q_{n+1}=\{}\le {\parallel x_n-x\parallel }^2+{\parallel y_n-y\parallel }^2\},\hfill \\ x_{n+1}=P_{C_{n+1}}(x_1),\hfill \\ y_{n+1}=P_{Q_{n+1}}(y_1),\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$

where $\xi \in (0,\frac{1}{{\parallel B\parallel }^2})$ and $\{r_n\}\subset (0,+\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{r}_n>0$, $P_C$ and $P_Q$ are two projection operators from $H_1$ into C and from $H_1$ into Q, respectively. Suppose that

$$\mathrm{\Omega }=\{(p,q)\in EP(f)\times EP(g):Bp=Bq\in EP(h)\}\ne \mathrm{\varnothing }.$$

Then there exists $(p,q)\in \mathrm{\Omega }$ such that

• (a) $(x_n,y_n)\to (p,q)$ as $n\to \mathrm{\infty }$;

• (b) $(u_n,v_n)\to (p,q)$ as $n\to \mathrm{\infty }$;

• (c) $w_n\to w^{\ast }:=Bp=Bq\in EP(h)$ as $n\to \mathrm{\infty }$.

If $H_2=H_3$ and $B=I$, then Theorem 2.1 reduces to the following corollary.

Corollary 2.2 Let $H_1$ and $H_2$ be two real Hilbert spaces. Let $C\subset H_1$ and $Q,K\subset H_2$ be three closed convex sets. Let $f:C\times C\to \mathbb{R}$, $g:Q\times Q\to \mathbb{R}$ and $h:K\times K\to \mathbb{R}$ be three bi-functions. $A:H_1\to H_2$ is a bounded linear operator with its adjoint operator $A^{\ast }$. Suppose that all the bi-functions f, g and h satisfy conditions (A1)-(A4). Let $x_1\in C$, $y_1\in Q$, $C_1=C$, $Q_1=Q$, $\{x_n\}$, $\{y_n\}$, $\{u_n\}$, $\{v_n\}$ and $\{w_n\}$ be sequences generated by

$$\{\begin{array}{c}{u}_n=T_{r_n}^f{x}_n,v_n=T_{r_n}^g{y}_n,w_n=T_{r_n}^h(\frac{1}{2}A{u}_n+\frac{1}{2}{v}_n),\hfill \\ l_n=P_C(u_n-\xi A^{\ast }(A{u}_n-w_n)),k_n=v_n-\xi (v_n-w_n),\hfill \\ C_{n+1}\times Q_{n+1}=\{(x,y)\in C_n\times Q_n:{\parallel l_n-x\parallel }^2+{\parallel k_n-y\parallel }^2\le {\parallel u_n-x\parallel }^2+{\parallel v_n-y\parallel }^2\hfill \\ \phantom{C_{n+1}\times Q_{n+1}=\{}\le {\parallel x_n-x\parallel }^2+{\parallel y_n-y\parallel }^2\},\hfill \\ x_{n+1}=P_{C_{n+1}}(x_1),\hfill \\ y_{n+1}=P_{Q_{n+1}}(y_1),\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$

where $\xi \in (0,min\{1,\frac{1}{{\parallel A\parallel }^2}\})$ and $\{r_n\}\subset (0,+\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{r}_n>0$, $P_C$ and $P_Q$ are two projection operators from $H_1$ into C and from $H_2$ into Q, respectively. Suppose that

$$\mathrm{\Omega }=\{p\in EP(f):Ap\in EP(g)\cap EP(h)\}\ne \mathrm{\varnothing }.$$

Then there exists $p\in \mathrm{\Omega }$ such that

• (a) $x_n\to p$ as $n\to \mathrm{\infty }$;

• (b) $u_n\to p$ as $n\to \mathrm{\infty }$;

• (c) $v_n,y_n,w_n\to w^{\ast }:=Ap$ as $n\to \mathrm{\infty }$.

If $H_1=H_2$ and $A=I$, then Theorem 2.1 reduces to the following corollary.

Corollary 2.3 Let $H_1$ and $H_2$ be two real Hilbert spaces. Let $C,Q\subset H_1$ and $K\subset H_2$ be three closed convex sets. Let $f:C\times C\to \mathbb{R}$, $g:Q\times Q\to \mathbb{R}$ and $h:K\times K\to \mathbb{R}$ be three bi-functions. $B:H_1\to H_2$ is a bounded linear operator with its adjoint operator $B^{\ast }$. Suppose that all the bi-functions f, g and h satisfy conditions (A1)-(A4). Let $x_1\in C$, $y_1\in Q$, $C_1=C$, $Q_1=Q$, $\{x_n\}$, $\{y_n\}$, $\{u_n\}$, $\{v_n\}$ and $\{w_n\}$ be sequences generated by

$$\{\begin{array}{c}{u}_n=T_{r_n}^f{x}_n,v_n=T_{r_n}^g{y}_n,w_n=T_{r_n}^h(\frac{1}{2}{u}_n+\frac{1}{2}B{v}_n),\hfill \\ l_n=u_n-\xi (u_n-w_n),k_n=P_q(v_n-\xi B^{\ast }(B{v}_n-w_n)),\hfill \\ C_{n+1}\times Q_{n+1}=\{(x,y)\in C_n\times Q_n:{\parallel l_n-x\parallel }^2+{\parallel k_n-y\parallel }^2\le {\parallel u_n-x\parallel }^2+{\parallel v_n-y\parallel }^2\hfill \\ \phantom{C_{n+1}\times Q_{n+1}=\{}\le {\parallel x_n-x\parallel }^2+{\parallel y_n-y\parallel }^2\},\hfill \\ x_{n+1}=P_{C_{n+1}}(x_1),\hfill \\ y_{n+1}=P_{Q_{n+1}}(y_1),\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$

where $\xi \in (0,min\{1,\frac{1}{{\parallel B\parallel }^2}\})$ and $\{r_n\}\subset (0,+\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{r}_n>0$, $P_C$ and $P_Q$ are two projection operators from $H_1$ into C and from $H_2$ into Q, respectively. Suppose that

$$\mathrm{\Omega }=\{p\in EP(f)\cap EP(g):Ap\in EP(h)\}\ne \mathrm{\varnothing }.$$

Then there exists $p\in \mathrm{\Omega }$ such that

• (a) $x_n,u_n\to p$ as $n\to \mathrm{\infty }$;

• (b) $y_n,v_n\to p$ as $n\to \mathrm{\infty }$;

• (c) $w_n\to w^{\ast }:=Ap$ as $n\to \mathrm{\infty }$.

Putting $A=B=I$ (identical operator), $H_1=H_2=H_3=H$ and $C=Q=K$, then we have the following result.

Corollary 2.4 Let H be a real Hilbert space. Let C be a closed convex subset of H. Let $f,g,h:C\times C\to \mathbb{R}$ be three bi-functions. Suppose that all the bi-functions f, g and h satisfy conditions (A1)-(A4). Let $x_1,y_1\in C$, $C_1=C$, $\{x_n\}$, $\{y_n\}$, $\{u_n\}$, $\{v_n\}$ and $\{w_n\}$ be sequences generated by

$$\{\begin{array}{c}{u}_n=T_{r_n}^f{x}_n,v_n=T_{r_n}^g{y}_n,w_n=T_{r_n}^h(\frac{1}{2}{u}_n+\frac{1}{2}{v}_n),\hfill \\ l_n=u_n-\xi (u_n-w_n),k_n=v_n-\xi (v_n-w_n),\hfill \\ C_{n+1}\times C_{n+1}=\{(x,y)\in C_n\times C_n:{\parallel l_n-x\parallel }^2+{\parallel k_n-y\parallel }^2\le {\parallel u_n-x\parallel }^2+{\parallel v_n-y\parallel }^2\hfill \\ \phantom{C_{n+1}\times C_{n+1}=\{}\le {\parallel x_n-x\parallel }^2+{\parallel y_n-y\parallel }^2\},\hfill \\ x_{n+1}=P_{C_{n+1}}(x_1),\hfill \\ y_{n+1}=P_{C_{n+1}}(y_1),\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$

where $\xi \in (0,1)$ and $\{r_n\}\subset (0,+\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{r}_n>0$, $P_C$ is a projection operator from H into C. Suppose that

$$\mathrm{\Omega }=\{(p,q)\in EP(f)\times EP(g):p=q\in EP(h)\}\ne \mathrm{\varnothing }.$$

Then there exists $(p,q)\in \mathrm{\Omega }$ such that

• (a) $(x_n,y_n)\to (p,q)$ as $n\to \mathrm{\infty }$;

• (b) $(u_n,v_n)\to (p,q)$ as $n\to \mathrm{\infty }$;

• (c) $w_n\to w^{\ast }:=p=q\in EP(h)$ as $n\to \mathrm{\infty }$.

Remark 2.1 In Corollary 2.2, it is obvious that

$$\mathrm{\Omega }=\{(p,q)\in EP(f)\times EP(g):p=q\in EP(h)\}\ne \mathrm{\varnothing }$$

implies

$$\mathrm{\Omega }=\{p\in EP(f)\cap EP(g)\cap EP(h)\}\ne \mathrm{\varnothing }.$$

Hence, the problem studied in Corollary 2.4 is still the study of a common solution of three equilibrium problems in essence.

If C, Q, K are linear subspaces of a real Hilbert space, then we have the following corollaries from Theorem 2.1 and Corollary 2.1.

Corollary 2.5 Let $H_1$, $H_2$ and $H_3$ be three real Hilbert spaces. Let $C\subset H_1$, $Q\subset H_2$ and $K\subset H_3$ be three linear subspaces. Let $f:C\times C\to \mathbb{R}$, $g:Q\times Q\to \mathbb{R}$ and $h:K\times K\to \mathbb{R}$ be three bi-functions. $A:H_1\to H_3$ and $B:H_2\to H_3$ are two bounded linear operators with their adjoint operators $A^{\ast }$ and $B^{\ast }$, respectively. Suppose that all the bi-functions f, g and h satisfy conditions (A1)-(A4). Let $x_1\in C$, $y_1\in Q$, $C_1=C$, $Q_1=Q$, $\{x_n\}$, $\{y_n\}$, $\{u_n\}$, $\{v_n\}$ and $\{w_n\}$ be sequences generated by

$$\{\begin{array}{c}{u}_n=T_{r_n}^f{x}_n,v_n=T_{r_n}^g{y}_n,w_n=T_{r_n}^h(\frac{1}{2}A{u}_n+\frac{1}{2}B{v}_n),\hfill \\ l_n=u_n-\xi A^{\ast }(A{u}_n-w_n),k_n=v_n-\xi B^{\ast }(B{v}_n-w_n),\hfill \\ C_{n+1}\times Q_{n+1}=\{(x,y)\in C_n\times Q_n:{\parallel l_n-x\parallel }^2+{\parallel k_n-y\parallel }^2\le {\parallel u_n-x\parallel }^2+{\parallel v_n-y\parallel }^2\hfill \\ \phantom{C_{n+1}\times Q_{n+1}=\{}\le {\parallel x_n-x\parallel }^2+{\parallel y_n-y\parallel }^2\},\hfill \\ x_{n+1}=P_{C_{n+1}}(x_1),\hfill \\ y_{n+1}=P_{Q_{n+1}}(y_1),\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$

where $\xi \in (0,min(\frac{1}{{\parallel A\parallel }^2},\frac{1}{{\parallel B\parallel }^2}))$ and $\{r_n\}\subset (0,+\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{r}_n>0$, $P_C$ and $P_Q$ are two projection operators from $H_1$ into C and from $H_2$ into Q, respectively. Suppose that

$$\mathrm{\Omega }=\{(p,q)\in EP(f)\times EP(g):Ap=Bq\in EP(h)\}\ne \mathrm{\varnothing }.$$

Then there exists $(p,q)\in \mathrm{\Omega }$ such that

• (a) $(x_n,y_n)\to (p,q)$ as $n\to \mathrm{\infty }$;

• (b) $(u_n,v_n)\to (p,q)$ as $n\to \mathrm{\infty }$;

• (c) $w_n\to w^{\ast }:=Ap=Bq\in EP(h)$ as $n\to \mathrm{\infty }$.

Corollary 2.6 Let $H_1$ and $H_2$ be two real Hilbert spaces. Let $C\subset H_1$, $Q\subset H_1$ and $K\subset H_2$ be three linear subspaces. Let $f:C\times C\to \mathbb{R}$, $g:Q\times Q\to \mathbb{R}$ and $h:K\times K\to \mathbb{R}$ be three bi-functions. $B:H_1\to H_2$ is a bounded linear operator with its adjoint operator $B^{\ast }$. Suppose that all the bi-functions f, g and h satisfy conditions (A1)-(A4). Let $x_1\in C$, $y_1\in Q$, $C_1=C$, $Q_1=Q$, $\{x_n\}$, $\{y_n\}$, $\{u_n\}$, $\{v_n\}$ and $\{w_n\}$ be sequences generated by

$$\{\begin{array}{c}{u}_n=T_{r_n}^f{x}_n,v_n=T_{r_n}^g{y}_n,w_n=T_{r_n}^h(\frac{1}{2}B{u}_n+\frac{1}{2}B{v}_n),\hfill \\ l_n=u_n-\xi B^{\ast }(B{u}_n-w_n),k_n=v_n-\xi B^{\ast }(B{v}_n-w_n),\hfill \\ C_{n+1}\times Q_{n+1}=\{(x,y)\in C_n\times Q_n:{\parallel l_n-x\parallel }^2+{\parallel k_n-y\parallel }^2\le {\parallel u_n-x\parallel }^2+{\parallel v_n-y\parallel }^2\hfill \\ \phantom{C_{n+1}\times Q_{n+1}=\{}\le {\parallel x_n-x\parallel }^2+{\parallel y_n-y\parallel }^2\},\hfill \\ x_{n+1}=P_{C_{n+1}}(x_1),\hfill \\ y_{n+1}=P_{Q_{n+1}}(y_1),\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$

where $\xi \in (0,\frac{1}{{\parallel B\parallel }^2})$ and $\{r_n\}\subset (0,+\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{r}_n>0$, $P_C$ and $P_Q$ are two projection operators from $H_1$ into C and from $H_1$ into Q, respectively. Suppose that

$$\mathrm{\Omega }=\{(p,q)\in EP(f)\times EP(g):Bp=Bq\in EP(h)\}\ne \mathrm{\varnothing }.$$

Then there exists $(p,q)\in \mathrm{\Omega }$ such that

• (a) $(x_n,y_n)\to (p,q)$ as $n\to \mathrm{\infty }$;

• (b) $(u_n,v_n)\to (p,q)$ as $n\to \mathrm{\infty }$;

• (c) $w_n\to w^{\ast }:=Bp=Bq\in EP(h)$ as $n\to \mathrm{\infty }$.

Remark 2.2 In fact, the problem studied by Corollaries 2.1-2.3 and Corollary 2.6 is (SEP).

In order to prove Theorem 2.1, we need the following crucial known results.

Lemma 2.2 (see [10])

Let K be a nonempty closed convex subset of H, and let F be a bi-function of $K\times K$ into ℝ satisfying (A1)-(A4). For $r>0$, define a mapping $T_r^F:H\to K$ as follows:

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

(2.2)

for all $x\in H$. Then the following hold:

• (i) $T_r^F$ is single-valued and $\mathcal{F}(T_r^F)=EP(F)$ for $\mathrm{\forall }r>0$ and $EP(F)$ is closed and convex;

• (ii) $T_r^F$ is firmly nonexpansive, that is, for any $x,y\in H$, ${\parallel T_r^{F}x-T_r^{F}y\parallel }^2\le 〈T_r^{F}x-T_r^{F}y,x-y〉$.

Lemma 2.3 ([20])

Let $F_r^F$ be the same as in Lemma  2.2. If $\mathcal{F}(T_r^F)=EP(F)\ne \mathrm{\varnothing }$, then, for any $x\in H$ and $x^{\ast }\in \mathcal{F}(T_r^F)$, ${\parallel T_r^{F}x-x\parallel }^2\le {\parallel x-x^{\ast }\parallel }^2-{\parallel T_r^{F}x-x^{\ast }\parallel }^2$.

Lemma 2.4 ([1, 19])

Let the mapping $T_r^F$ be defined as in Lemma  2.2. Then, for $r,s>0$ and $x,y\in H$,

$$\parallel T_r^F(x)-T_s^F(y)\parallel \le \parallel x-y\parallel +\frac{|s-r|}{s}\parallel T_s^F(y)-y\parallel .$$

In particular, $\parallel T_r^F(x)-T_r^F(y)\parallel \le \parallel x-y\parallel$ for any $r>0$ and $x,y\in H$, that is, $T_r^F$ is nonexpansive for any $r>0$.

Lemma 2.5 (see, e.g., [25])

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

• (a) ${\parallel x-y\parallel }^2={\parallel x\parallel }^2+{\parallel y\parallel }^2-2〈x,y〉$ for all $x,y\in H$;

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

3 Proof of Theorem 2.1

Applying Lemmas 2.1 and 2.2, we know that $\{u_n\}$, $\{v_n\}$ and $\{w_n\}$ are all well defined. It is also easy to verify that $C_n$, $Q_n$, $C_n\times Q_n$ are closed convex sets for $n\in \mathbb{N}$.

Now, we claim $C_n\times Q_n\ne \mathrm{\varnothing }$ for all $n\in \mathbb{N}$. Indeed, it suffices to prove that $\mathrm{\Omega }\subset C_n\times Q_n$ for all $n\in \mathbb{N}$. Let $(x^{\ast },y^{\ast })\in \mathrm{\Omega }$. Then $x^{\ast }\in EP(f)$, $y^{\ast }\in EP(g)$ and

$$w^{\ast }:=A{x}^{\ast }=B{y}^{\ast }\in EP(h).$$

Let $n\in \mathbb{N}$ be given. By Lemma 2.3, we have

$$\begin{array}{r}\parallel u_n-x^{\ast }\parallel \le \parallel x_n-x^{\ast }\parallel ,\parallel v_n-y^{\ast }\parallel \le \parallel y_n-y^{\ast }\parallel ,\\ \parallel w_n-w^{\ast }\parallel \le \parallel \frac{A{u}_n+B{v}_n}{2}-w^{\ast }\parallel ,\\ {\parallel w_n-w^{\ast }\parallel }^2\le \frac{1}{2}{\parallel A{u}_n-w^{\ast }\parallel }^2+\frac{1}{2}{\parallel B{v}_n-w^{\ast }\parallel }^2(\text{by Lemma }\text{2.5}).\end{array}$$

(3.1)

From (2.1), (3.1) and Lemma 2.5, we obtain

$$\begin{array}{rl}{\parallel l_n-x^{\ast }\parallel }^2=& {\parallel P_C(u_n-\xi A^{\ast }(A{u}_n-w_n))-x^{\ast }\parallel }^2\le {\parallel u_n-x^{\ast }-\xi A^{\ast }(A{u}_n-w_n)\parallel }^2\\ =& {\parallel u_n-x^{\ast }\parallel }^2+{\parallel \xi A^{\ast }(A{u}_n-w_n)\parallel }^2-2\xi 〈u_n-x^{\ast },A^{\ast }(A{u}_n-w_n)〉\\ =& {\parallel u_n-x^{\ast }\parallel }^2+{\parallel \xi A^{\ast }(A{u}_n-w_n)\parallel }^2-2\xi 〈A{u}_n-A{x}^{\ast },A{u}_n-w_n〉\\ =& {\parallel u_n-x^{\ast }\parallel }^2+{\parallel \xi A^{\ast }(A{u}_n-w_n)\parallel }^2-2\xi 〈A{u}_n-w^{\ast },A{u}_n-w_n〉\\ =& {\parallel u_n-x^{\ast }\parallel }^2+{\parallel \xi A^{\ast }(A{u}_n-w_n)\parallel }^2-\xi {\parallel A{u}_n-w^{\ast }\parallel }^2\\ -\xi {\parallel A{u}_n-w_n\parallel }^2+\xi {\parallel w_n-w^{\ast }\parallel }^2\\ \le & {\parallel u_n-x^{\ast }\parallel }^2-\xi (1-\xi {\parallel A^{\ast }\parallel }^2){\parallel A{u}_n-w_n\parallel }^2\\ -\xi {\parallel A{u}_n-w^{\ast }\parallel }^2+\xi {\parallel w_n-w^{\ast }\parallel }^2\\ \le & {\parallel u_n-x^{\ast }\parallel }^2-\xi (1-\xi {\parallel A^{\ast }\parallel }^2){\parallel A{u}_n-w_n\parallel }^2-\xi {\parallel A{u}_n-w^{\ast }\parallel }^2\\ +\frac{\xi }{2}{\parallel A{u}_n-w^{\ast }\parallel }^2+\frac{\xi }{2}{\parallel B{v}_n-w^{\ast }\parallel }^2\\ =& {\parallel u_n-x^{\ast }\parallel }^2-\xi (1-\xi {\parallel A^{\ast }\parallel }^2){\parallel A{u}_n-w_n\parallel }^2\\ -\frac{\xi }{2}{\parallel A{u}_n-w^{\ast }\parallel }^2+\frac{\xi }{2}{\parallel B{v}_n-w^{\ast }\parallel }^2\end{array}$$

(3.2)

and

$$\begin{array}{rl}{\parallel k_n-y^{\ast }\parallel }^2=& {\parallel P_Q(v_n-\xi B^{\ast }(B{v}_n-w_n))-y^{\ast }\parallel }^2\le {\parallel v_n-y^{\ast }-\xi B^{\ast }(B{v}_n-w_n)\parallel }^2\\ =& {\parallel v_n-y^{\ast }\parallel }^2+{\parallel \xi B^{\ast }(B{v}_n-w_n)\parallel }^2-2\xi 〈v_n-y^{\ast },B^{\ast }(B{v}_n-w_n)〉\\ =& {\parallel v_n-y^{\ast }\parallel }^2+{\parallel \xi B^{\ast }(B{v}_n-w_n)\parallel }^2-2\xi 〈B{v}_n-B{y}^{\ast },B{v}_n-w_n〉\\ =& {\parallel v_n-y^{\ast }\parallel }^2+{\parallel \xi B^{\ast }(B{v}_n-w_n)\parallel }^2-2\xi 〈B{v}_n-w^{\ast },B{v}_n-w_n〉\\ =& {\parallel v_n-y^{\ast }\parallel }^2+{\parallel \xi B^{\ast }(B{v}_n-w_n)\parallel }^2-\xi {\parallel B{v}_n-w^{\ast }\parallel }^2\\ -\xi {\parallel B{v}_n-w_n\parallel }^2+\xi {\parallel w_n-w^{\ast }\parallel }^2\\ \le & {\parallel v_n-y^{\ast }\parallel }^2-\xi (1-\xi {\parallel B^{\ast }\parallel }^2){\parallel B{v}_n-w_n\parallel }^2-\xi {\parallel B{v}_n-w^{\ast }\parallel }^2+\xi {\parallel w_n-w^{\ast }\parallel }^2\\ \le & {\parallel v_n-y^{\ast }\parallel }^2-\xi (1-\xi {\parallel B^{\ast }\parallel }^2){\parallel B{v}_n-w_n\parallel }^2-\xi {\parallel B{v}_n-w^{\ast }\parallel }^2\\ +\frac{\xi }{2}{\parallel A{u}_n-w^{\ast }\parallel }^2+\frac{\xi }{2}{\parallel B{v}_n-w^{\ast }\parallel }^2\\ =& {\parallel v_n-y^{\ast }\parallel }^2-\xi (1-\xi {\parallel B^{\ast }\parallel }^2){\parallel B{v}_n-w_n\parallel }^2\\ -\frac{\xi }{2}{\parallel B{v}_n-w^{\ast }\parallel }^2+\frac{\xi }{2}{\parallel A{u}_n-w^{\ast }\parallel }^2.\end{array}$$

(3.3)

By taking into account inequalities (3.1), (3.2) and (3.3), we obtain

$$\begin{array}{rl}{\parallel l_n-x^{\ast }\parallel }^2+{\parallel k_n-y^{\ast }\parallel }^2\le & {\parallel u_n-x^{\ast }\parallel }^2+{\parallel v_n-y^{\ast }\parallel }^2-\xi (1-\xi {\parallel A^{\ast }\parallel }^2){\parallel A{u}_n-w_n\parallel }^2\\ -\xi (1-\xi {\parallel B^{\ast }\parallel }^2){\parallel B{v}_n-w_n\parallel }^2\\ \le & {\parallel x_n-x^{\ast }\parallel }^2+{\parallel y_n-y^{\ast }\parallel }^2-\xi (1-\xi {\parallel A^{\ast }\parallel }^2){\parallel A{u}_n-w_n\parallel }^2\\ -\xi (1-\xi {\parallel B^{\ast }\parallel }^2){\parallel B{v}_n-w_n\parallel }^2,\end{array}$$

(3.4)

which implies

$${\parallel l_n-x^{\ast }\parallel }^2+{\parallel k_n-y^{\ast }\parallel }^2\le {\parallel u_n-x^{\ast }\parallel }^2+{\parallel v_n-y^{\ast }\parallel }^2\le {\parallel x_n-x^{\ast }\parallel }^2+{\parallel y_n-y^{\ast }\parallel }^2.$$

(3.5)

Inequality (3.5) shows that $(x^{\ast },y^{\ast })\in C_n\times Q_n$. Hence $\mathrm{\Omega }\subset C_n\times Q_n$ and $C_n\times Q_n\ne \mathrm{\varnothing }$ for all $n\in \mathbb{N}$. For each $n\in \mathbb{N}$, since $\mathrm{\Omega }\subset C_n\times Q_n$, $C_{n+1}\subset C_n$, we have

$$x_{n+1}=P_{C_{n+1}}(x_1)\subset C_n.$$

Similarly, since $Q_{n+1}\subset Q_n$, we have

$$y_{n+1}=P_{Q_{n+1}}(y_1)\subset Q_n.$$

So, for any $(x^{\ast },y^{\ast })\in \mathrm{\Omega }$, we get

$$\parallel x_{n+1}-x_1\parallel \le \parallel x^{\ast }-x_1\parallel$$

and

$$\parallel y_{n+1}-y_1\parallel \le \parallel y^{\ast }-y_1\parallel .$$

The last inequalities deduce that $\{x_n\}$ and $\{y_n\}$ are bounded and hence show that $\{k_n\}$, $\{l_n\}$, $\{u_n\}$ and $\{v_n\}$ are all bounded. For some $n\in \mathbb{N}$ with $n>1$, from $x_n=P_{C_n}(x_1)\subset C_n$, $y_n=P_{Q_n}(y_1)\subset Q_n$ and (1.1), we have

$$\begin{array}{r}{\parallel x_{n+1}-x_n\parallel }^2+{\parallel x_1-x_n\parallel }^2={\parallel x_{n+1}-P_{C_n}(x_1)\parallel }^2+{\parallel x_1-P_{C_n}(x_1)\parallel }^2\le {\parallel x_{n+1}-x_1\parallel }^2,\\ {\parallel y_{n+1}-y_n\parallel }^2+{\parallel y_1-y_n\parallel }^2={\parallel y_{n+1}-P_{C_n}(y_1)\parallel }^2+{\parallel y_1-P_{C_n}(y_1)\parallel }^2\le {\parallel y_{n+1}-y_1\parallel }^2,\end{array}$$

which yields that

$$\begin{array}{l}\parallel x_1-x_n\parallel \le \parallel x_{n+1}-x_1\parallel ,\parallel y_1-y_n\parallel \le \parallel y_{n+1}-y_1\parallel .\end{array}$$

Together with the boundedness of $\{x_n\}$ and $\{y_n\}$, we know ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x_1\parallel$ and ${lim}_{n\to \mathrm{\infty }}\parallel y_n-y_1\parallel$ exist. For some $k,n\in \mathbb{N}$ with $k>n>1$, due to $x_k=P_{C_k}(x_1)\subset C_n$, $y_k=P_{Q_k}(y_1)\subset Q_n$ and (1.1), we have

$$\begin{array}{r}{\parallel x_k-x_n\parallel }^2+{\parallel x_1-x_n\parallel }^2={\parallel x_k-P_{C_n}(x_1)\parallel }^2+{\parallel x_1-P_{C_n}(x_1)\parallel }^2\le {\parallel x_k-x_1\parallel }^2,\\ {\parallel y_k-y_n\parallel }^2+{\parallel y_1-y_n\parallel }^2={\parallel y_k-P_{Q_n}(y_1)\parallel }^2+{\parallel y_1-P_{Q_n}(y_1)\parallel }^2\le {\parallel y_k-y_1\parallel }^2.\end{array}$$

(3.6)

By (3.6), we have ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x_k\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel y_n-y_k\parallel =0$. Hence $\{x_n\}$ and $\{y_n\}$ are all Cauchy sequences. Let $x_n\to p$ and $y_n\to q$ for some $(p,q)\in C\times Q$. We want to prove that $(p,q)\in \mathrm{\Omega }$. For any $n\in \mathbb{N}$, since

$$(x_{n+1},y_{n+1})\in C_{n+1}\times Q_{n+1}\subset C_n\times Q_n,$$

from (2.1), we have

$$\begin{array}{rl}{\parallel l_n-x_{n+1}\parallel }^2+{\parallel k_n-y_{n+1}\parallel }^2& \le {\parallel u_n-x_{n+1}\parallel }^2+{\parallel v_n-y_{n+1}\parallel }^2\\ \le {\parallel x_n-x_{n+1}\parallel }^2+{\parallel y_n-y_{n+1}\parallel }^2.\end{array}$$

(3.7)

By taking the limit from both sides of (3.7), we obtain

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}\parallel l_n-x_{n+1}\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel k_n-y_{n+1}\parallel =0,\\ \underset{n\to \mathrm{\infty }}{lim}\parallel u_n-x_{n+1}\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel v_n-y_{n+1}\parallel =0.\end{array}$$

(3.8)

Moreover, by (3.8), we get

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}\parallel l_n-u_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel l_n-x_n\parallel =0,\\ \underset{n\to \mathrm{\infty }}{lim}\parallel k_n-v_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel v_n-y_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel k_n-y_n\parallel =0.\end{array}$$

(3.9)

Since ${lim}_{n\to \mathrm{\infty }}\parallel u_n-x_n\parallel ={lim}_{n\to \mathrm{\infty }}\parallel v_n-y_n\parallel =0$, we have $u_n\to p$ and $v_n\to q$ as $n\to \mathrm{\infty }$. Moreover, we obtain $A{u}_n\to Ap$ and $B{v}_n\to Bq$ as $n\to \mathrm{\infty }$.

Now, we claim $p\in EP(f)$ and $q\in EP(g)$. In fact, for $r>0$, by Lemma 2.4, we have

$$\begin{array}{rcl}\parallel T_r^{f}p-p\parallel & =& \parallel T_r^{f}p-T_{r_n}^f{x}_n+T_{r_n}^f{x}_n-x_n+x_n-p\parallel \\ \le & \parallel x_n-p\parallel +\frac{|r_n-r|}{r_n}\parallel T_{r_n}^f{x}_n-x_n\parallel +\parallel T_{r_n}^f{x}_n-x_n\parallel +\parallel x_n-p\parallel \\ =& \parallel x_n-p\parallel +\frac{|r_n-r|}{r_n}\parallel u_n-x_n\parallel +\parallel u_n-x_n\parallel +\parallel x_n-p\parallel \to 0\end{array}$$

and

$$\begin{array}{rcl}\parallel T_r^{g}q-q\parallel & \le & \parallel T_r^{g}q-T_{r_n}^g{y}_n+T_{r_n}^g{y}_n-y_n+y_n-q\parallel \\ \le & \parallel y_n-q\parallel +\frac{|r_n-r|}{r_n}\parallel T_{r_n}^g{y}_n-y_n\parallel +\parallel T_{r_n}^g{y}_n-y_n\parallel +\parallel y_n-q\parallel \\ =& \parallel y_n-q\parallel +\frac{|r_n-r|}{r_n}\parallel v_n-y_n\parallel +\parallel v_n-y_n\parallel +\parallel y_n-q\parallel \to 0.\end{array}$$

So, $p\in EP(f)$ and $q\in EP(g)$.

Finally, we prove $Ap=Bq\in EP(h)$. Setting

$$\theta =min\{\xi (1-\xi {\parallel A^{\ast }\parallel }^2),\xi (1-\xi {\parallel B^{\ast }\parallel }^2)\}.$$

Then, for any $n\in \mathbb{N}$, by (3.4) and (3.9), we have

$$\begin{array}{r}\theta {\parallel A{u}_n-w_n\parallel }^2+\theta {\parallel B{v}_n-w_n\parallel }^2\\ \le {\parallel x_n-x^{\ast }\parallel }^2+{\parallel y_n-y^{\ast }\parallel }^2-{\parallel l_n-x^{\ast }\parallel }^2-{\parallel k_n-y^{\ast }\parallel }^2\\ =\{\parallel x_n-x^{\ast }\parallel -\parallel l_n-x^{\ast }\parallel \}\{\parallel x_n-x^{\ast }\parallel +\parallel l_n-x^{\ast }\parallel \}\\ +\{\parallel y_n-y^{\ast }\parallel -\parallel k_n-y^{\ast }\parallel \}\{\parallel y_n-y^{\ast }\parallel +\parallel k_n-y^{\ast }\parallel \}\\ \le \parallel l_n-x_n\parallel \{\parallel x_n-x^{\ast }\parallel +\parallel l_n-x^{\ast }\parallel \}+\parallel k_n-y_n\parallel \{\parallel y_n-y^{\ast }\parallel +\parallel k_n-y^{\ast }\parallel \}\\ \to 0.\end{array}$$

(3.10)

Hence (3.10) implies

$$\underset{n\to \mathrm{\infty }}{lim}\parallel A{u}_n-w_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel B{v}_n-w_n\parallel =0,\text{and}\underset{n\to \mathrm{\infty }}{lim}\parallel A{u}_n-B{v}_n\parallel =0.$$

(3.11)

Since $A{u}_n\to Ap$, $B{v}_n\to Bq$ and (3.11), we obtain $Ap=Bq$ and $w_n\to w^{\ast }$, where $w^{\ast }:=Ap=Bq$. On the other hand, for $r>0$, by Lemma 2.4 again, we have

$$\begin{array}{r}\parallel T_r^h{w}^{\ast }-w^{\ast }\parallel \\ =\parallel T_r^h{w}^{\ast }-T_{r_n}^h\frac{A{u}_n+B{v}_n}{2}+T_{r_n}^h\frac{A{u}_n+B{v}_n}{2}-\frac{A{u}_n+B{v}_n}{2}+\frac{A{u}_n+B{v}_n}{2}-w^{\ast }\parallel \\ \le \parallel \frac{A{u}_n+B{v}_n}{2}-w^{\ast }\parallel +\frac{|r_n-r|}{r_n}\parallel T_{r_n}^h\frac{A{u}_n+B{v}_n}{2}-\frac{A{u}_n+B{v}_n}{2}\parallel \\ +\parallel T_{r_n}^h\frac{A{u}_n+B{v}_n}{2}-\frac{A{u}_n+B{v}_n}{2}\parallel +\parallel \frac{A{u}_n+B{v}_n}{2}-w^{\ast }\parallel \\ =2\parallel \frac{A{u}_n+B{v}_n}{2}-w^{\ast }\parallel +\frac{|r_n-r|}{r_n}\parallel w_n-\frac{A{u}_n+B{v}_n}{2}\parallel +\parallel w_n-\frac{A{u}_n+B{v}_n}{2}\parallel \to 0.\end{array}$$

Hence $w^{\ast }\in EP(h)$, namely $Ap=Bq\in EP(h)$. Therefore, conclusions (a), (b) and (c) are all proved. The proof is completed.

4 An example of Theorem 2.1

In this section, we give an example illustrating Theorem 2.1.

Example 4.1 Let $H_1=R^2$, $H_2=R^3$ and $H_3=R^4$ be three real Hilbert spaces with the standard norm and inner product. For each $\alpha =({\alpha }_1,{\alpha }_2)\in R^2$ and $\nu =(z_1,z_2,z_3,z_4)\in R^4$, define

$$A\alpha =({\alpha }_1,{\alpha }_2,{\alpha }_1+{\alpha }_2,{\alpha }_1-{\alpha }_2)$$

and

$$A^{\ast }\nu =(z_1+z_3+z_4,z_2+z_3-z_4).$$

Then A is a bounded linear operator from $R^2$ into $R^4$ with $\parallel A\parallel =\sqrt{3}$, and $A^{\ast }$ is an adjoint operator of A with $\parallel A^{\ast }\parallel =\sqrt{3}$. For each $\beta =({\beta }_1,{\beta }_2,{\beta }_3)\in R^3$ and $\nu =(z_1,z_2,z_3,z_4)\in R^4$, let

$$B\beta =({\beta }_1,{\beta }_2,{\beta }_3,{\beta }_1-{\beta }_2)$$

and

$$B^{\ast }\nu =(z_1+z_4,z_2-z_4,z_3).$$

Then B is a bounded linear operator from $R^3$ into $R^4$ with $\parallel B\parallel =\sqrt{3}$, and $B^{\ast }$ is an adjoint operator of B with $\parallel B^{\ast }\parallel =\sqrt{3}$. Put

$$\begin{array}{r}C:=\{\alpha =({\alpha }_1,{\alpha }_2)\in R^2:-1\le {\alpha }_1\le 2,3\le {\alpha }_2\le 4\},\\ Q:=\{\beta =({\beta }_1,{\beta }_2,{\beta }_3)\in R^3:-1\le {\beta }_1\le 1,3\le {\beta }_2\le 4,3\le {\beta }_3\le 5\}\end{array}$$

and

$$K:=\{z=(z_1,z_2,z_3,z_4)\in R^4:0\le z_1\le 1,3\le z_2\le 6,3\le z_3\le 5,-5\le z_4\le -3\}.$$

For each $\alpha =({\alpha }_1,{\alpha }_2)\in C$, $\beta =({\beta }_1,{\beta }_2,{\beta }_3)\in Q$ and $z=(z_1,z_2,z_3,z_4)\in K$, define

$$\begin{array}{r}{f}^{\ast }(\alpha )={\alpha }_1^2+{\alpha }_2^2,\\ g^{\ast }(\beta )={\beta }_1^2+{\beta }_2^2+{\beta }_3^2\end{array}$$

and

$$h^{\ast }(z)=z_1^2+z_2^2+z_3^2+z_4^2.$$

For each $\alpha ,x\in C$, let

$$f(\alpha ,x)=f^{\ast }(x)-f^{\ast }(\alpha ).$$

For each $\beta ,y\in Q$, let

$$g(\beta ,y)=g^{\ast }(y)-g^{\ast }(\beta ).$$

For each $\eta ,z\in K$, let

$$h(\eta ,z)=h^{\ast }(z)-h^{\ast }(\eta ).$$

It is not hard to verify that f, g and h satisfy conditions (A1)-(A4) with $EP(f)=\{p=(0,3)\}$, $EP(g)=\{q=(0,3,3)\}$, $EP(h)=\{(0,3,3,-3)\}$ and

$$\mathrm{\Omega }=\{(p,q)\in EP(f)\times EP(g):Ap=Bq\in EP(h)\}\ne \mathrm{\varnothing }.$$

Let $C_1=C$, $Q_1=Q$, $\xi =\frac{1}{6}$ and $r_n\equiv 1$ for $n\in \mathbb{N}$. Thus, for each $\overline{x}=(a,b)\in C$ and $\overline{y}=(c,d,e)\in Q$ with $c>0$, we have the following:

• $u=(\frac{a}{3},3)=T_{r_n}^f\overline{x}$,

• $v=(\frac{c}{3},3,3)=T_{r_n}^g\overline{y}$,

• $w=(\frac{a+c}{6},3,3,-3)=T_{r_n}^h(\frac{1}{2}A\overline{x}+\frac{1}{2}B\overline{y})$,

• $l=P_C(u-\frac{1}{6}{A}^{\ast }(Au-w))=(\frac{7a+c}{36},3)$,

• $k=P_Q(v-\frac{1}{6}{B}^{\ast }(Bv-w))=(\frac{9c+a}{36},3+\frac{c}{18},3)$.

For $x_1=(a_1,b_1)\in C$ and $y_1=(c_1,d_1,e_1)\in Q$ with $15c_1\ge a_1>0$, $17a_1\ge c_1$ and $d_1>3+\frac{c_1}{18}$, we obtain the following:

• $u_1=(\frac{a_1}{3},3)$,

• $v_1=(\frac{c_1}{3},3,3)$,

• $w_1=(\frac{a_1+c_1}{6},3,3,-3)$,

• $l_1=(\frac{7a_1+c_1}{36},3)$,

• $k_1=(\frac{9c_1+a_1}{36},3+\frac{c_1}{18},3)$,

• $C_2=\{\alpha =({\alpha }_1,{\alpha }_2)\in C_1:-1\le {\alpha }_1\le \frac{19a_1+c_1}{72},3\le {\alpha }_2\le \frac{b_1+3}{2}\}$,

• $x_2=P_{C_2}(x_1)=(\frac{19a_1+c_1}{72},\frac{b_1+3}{2}):=(a_2,b_2)$,

• $Q_2=\{\beta =({\beta }_1,{\beta }_2,{\beta }_3)\in Q_1:-1\le {\beta }_1\le \frac{21c_1+a_1}{72},3+\frac{c_1}{36}\le {\beta }_2\le \frac{d_1+3}{2},3\le {\beta }_3\le \frac{e_1+3}{2}\}$,

• $y_2=P_{Q_2}(y_1)=(\frac{21c_1+a_1}{72},\frac{d_1+3}{2},\frac{e_1+3}{2}):=(c_2,d_2,e_2)$.

From $x_2$, $y_2$, we have $u_2=(\frac{1}{3}{a}_2,3)$, $v_2=(\frac{1}{3}{c}_2,3,3)$ and $w_2=(\frac{1}{6}(a_2+b_2),3,3,-3)$. Since

$$\begin{array}{r}{d}_2>3+\frac{c_2}{18},\\ 15c_2\ge a_2>0,\end{array}$$

and

$$17a_2\ge c_2,$$

we get the following:

• $C_3=\{\alpha =({\alpha }_1,{\alpha }_2)\in C_2:-1\le {\alpha }_1\le \frac{19a_2+c_2}{72},3\le {\alpha }_2\le \frac{1}{2}(b_2+3)\}$,

• $x_3=(\frac{19a_2+c_2}{72},\frac{1}{2}(b_2+3)):=(a_3,b_3)$,

• $Q_3=\{\beta =({\beta }_1,{\beta }_2,{\beta }_3)\in Q_1:-1\le {\beta }_1\le \frac{21c_2+a_2}{72},3+\frac{c_2}{36}\le {\beta }_2\le \frac{d_2+3}{2},3\le {\beta }_3\le \frac{e_2+3}{2}\}$,

• $y_3=(\frac{21c_2+a_2}{72},\frac{d_2+3}{2},\frac{e_2+3}{2}):=(c_3,d_3,e_3)$.

Similarly, for $n\in \mathbb{N}$ with $n>1$, we obtain

• $C_{n+1}=\{\alpha =({\alpha }_1,{\alpha }_2)\in C_n:-1\le {\alpha }_1\le \frac{19a_{n-1}+c_{n-1}}{72},3\le {\alpha }_2\le \frac{1}{2}(b_{n-1}+3)\}$,

• $x_{n+1}=(\frac{19a_{n-1}+c_{n-1}}{72},\frac{1}{2}(b_{n-1}+3))$,

• $Q_{n+1}=\{\beta =({\beta }_1,{\beta }_2,{\beta }_3)\in Q_n:-1\le {\beta }_1\le \frac{21c_{n-1}+a_{n-1}}{72},3+\frac{c_{n-1}}{36}\le {\beta }_2\le \frac{d_{n-1}+3}{2},3\le {\beta }_3\le \frac{e_{n-1}+3}{2}\}$,

• $y_{n+1}=(\frac{21c_{n-1}+a_{n-1}}{72},\frac{d_{n-1}+3}{2},\frac{e_{n-1}+3}{2})$,

• $u_{n+1}=(\frac{1}{3}{a}_{n-1},3)$,

• $v_{n+1}=(\frac{1}{3}{c}_{n-1},3,3)$,

• $w_{n+1}=(\frac{c_{n-1}+a_{n-1}}{6},3,3,-3)$.

By mathematical induction, we know that $\{a_n\}$, $\{b_n\}$, $\{c_n\}$, $\{d_n\}$ and $\{e_n\}$ all are decreasing sequences. Moreover, $a_n\to 0$, $b_n\to 3$, $c_n\to 0$, $d_n\to 3$ and $e_n\to 3$ as $n\to \mathrm{\infty }$. So, we have $u_n\to (0,3)$, $v_n\to (0,3,3)$, $w_n\to (0,3,3,-3)$, $x_n\to (0,3)$ and $y_n\to (0,3,3)$ as $n\to \mathrm{\infty }$.

5 Conclusion

In this paper, we first introduce and investigate BSEP which can be regarded as a new development in the field of equilibrium problems. We provide some new iterative algorithms for BSEP and establish strong convergence theorems for these iterative algorithms in different spaces. An example illustrating Theorem 2.1 is also given.