Open Access

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

Fixed Point Theory and Applications20142014:187

https://doi.org/10.1186/1687-1812-2014-187

Received: 1 June 2014

Accepted: 18 August 2014

Published: 2 September 2014

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.

Keywords

metric projection adjoint operator equilibrium problem (EP) split equilibrium problem (SEP) bilevel split equilibrium problem (BSEP) bilevel convex optimization problem (BCOP) the common solution of equilibrium problems (CEP) feasible iterative algorithm strong convergence theorem

1 Introduction and preliminaries

Let K be a closed convex subset of a real Hilbert space H. Let , and denote the inner product of H and the norm of H, respectively. For each point x H , there exists a unique nearest point in K, denoted by P K x , such that
x P K x x y for all  y 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:
  1. (i)

    x y , P K x P K y P K x P K y 2 for every x , y H .

     
  2. (ii)

    For x H and z K , z = P K ( x ) x z , z y 0 for all y K .

     
  3. (iii)
    For x H and y K ,
    y P K ( x ) 2 + x P K ( x ) 2 x y 2 .
    (1.1)
     
Let H 1 and H 2 be two Hilbert spaces. Let A : H 1 H 2 and A : H 2 H 1 be two bounded linear operators. A is called the adjoint operator (or adjoint) of A if
A z , w = z , A w for all  z H 1  and  w 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 is an adjoint operator of A, then A = A . The symbols and are used to denote the sets of positive integers and real numbers, respectively.

Example 1.1 ([1])

Let H 2 = R with the standard norm | | and H 1 = R 2 with the norm α = ( a 1 2 + a 2 2 ) 1 2 for some α = ( a 1 , a 2 ) R 2 . x , y = x y denotes the inner product of H 2 for some x , y H 2 , and α , β = i = 1 2 a i b i denotes the inner product of H 1 for some α = ( a 1 , a 2 ) , β = ( b 1 , b 2 ) H 1 . Let A α = a 2 a 1 for α = ( a 1 , a 2 ) H 1 and B x = ( x , x ) for x H 2 , then B is an adjoint operator of A.

Example 1.2 ([1])

Let H 1 = R 2 with the norm α = ( a 1 2 + a 2 2 ) 1 2 for some α = ( a 1 , a 2 ) R 2 and H 2 = R 3 with the norm γ = ( c 1 2 + c 2 2 + c 3 2 ) 1 2 for some γ = ( c 1 , c 2 , c 3 ) R 3 . Let α , β = i = 1 2 a i b i and γ , η = i = 1 3 c i d i denote the inner product of H 1 and H 2 , respectively, where α = ( a 1 , a 2 ) , β = ( b 1 , b 2 ) H 1 , γ = ( c 1 , c 2 , c 3 ) , η = ( d 1 , d 2 , d 3 ) H 2 . Let A α = ( a 2 , a 1 , a 1 a 2 ) for α = ( a 1 , a 2 ) H 1 and B γ = ( c 2 + c 3 , c 1 c 3 ) for γ = ( c 1 , c 2 , c 3 ) H 2 . Obviously, B is an adjoint operator of A.

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

(EP) Find p C such that f ( p , y ) 0 , y C .

The set of such solutions is denoted by EP ( f ) , that is, EP ( f ) = { u C : f ( u , v ) 0 , v C } . In fact, equilibrium problem has an important relationship with variational inequality problem. For example, let T : C H be a nonlinear mapping satisfying T x , y x 0 for all x , y C . Then x EP ( f ) if and only if x C is a solution of the variational inequality T x , y x 0 for all y 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 [28] and references therein. Many authors have proposed some useful methods to solve the EP; see, for instance, [25, 917] 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, 1824] 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 × C R and g : K × K R be two bi-functions, and A : H 1 H 2 be a bounded linear operator. The split equilibrium problem (SEP, in short) is defined as follows:

(SEP) Find p C such that f ( p , y ) 0 , y C , and u : = A p satisfying g ( u , v ) 0 , v 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 × C R , g : Q × Q R and h : K × K R be three bi-functions. Let A : H 1 H 3 and B : H 2 H 3 be two bounded linear operators with theirs adjoint operators A and B , respectively. The mathematical model about bilevel split equilibrium problem (BSEP, in short) is defined as follows:

(BSEP) Find p C and q Q such that
  1. (i)

    f ( p , x ) 0 and g ( q , y ) 0 for all x C and y Q ;

     
  2. (ii)

    A p = B q : = u ;

     
  3. (iii)

    h ( u , z ) 0 for all z 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 H 1 , Q H 2 and K H 3 be three closed convex sets. Let f : C R , g : Q R and h : K R be three convex functions. Let A : H 1 H 3 and B : H 2 H 3 be two bounded linear operators with their adjoint operators A and B , respectively. Let
f ( x , α ) = f ( x ) f ( α ) for  x , α C , g ( y , β ) = g ( y ) g ( β ) for  y , β Q ,
and
h ( z , η ) = h ( z ) h ( η ) for  z , η K .

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

(BCOP) Find p C and q Q such that u : = A p = B q K , f ( x ) f ( p ) , g ( y ) g ( q ) and h ( z ) h ( u ) for all x C , y Q and z K .

Example B Let H 1 , H 2 and H 3 be three real Hilbert spaces. Let C H 1 , Q H 2 and K H 3 be three closed convex sets. Let T : C H 1 , S : Q H 2 and G : K H 3 be three nonlinear operators. Let A : H 1 H 3 and B : H 2 H 3 be two bounded linear operators with their adjoint operators A and B , respectively. If f ( p , x ) = T p , x p , g ( q , y ) = S q , y q and h ( u , z ) = T u , z u , then BSEP reduces to the bilevel split variational inequality problem (BSVI):

(BSVI) Find p C and q Q such that u : = A p = B q K satisfying T p , x p 0 , S q , y q 0 and T u , z u 0 for all x C , y Q and z K .

Example C Let H 1 and H 2 be two real Hilbert spaces and B : H 1 H 2 be a bounded linear operator with its adjoint operator B . Let C H 1 , Q H 1 and K 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 ) ):

( SEP ( 1 ) ) Find p C and q Q such that u : = B p = B q K satisfying f ( p , x ) 0 , g ( q , y ) 0 and h ( u , z ) 0 for all x C , y Q and z K .

Example D Let H 1 and H 2 be two real Hilbert spaces and A : H 1 H 2 be a bounded linear operator with its adjoint operator A . Let C H 1 , Q H 2 and K H 2 be three closed convex sets with Q K . If H 2 = H 3 and B = I (identity operator), then BSEP reduces to the following split equilibrium problem (2) (SEP ( 2 ) ):

( SEP ( 2 ) ) Find p C such that u : = A p Q K satisfying f ( p , x ) 0 , g ( u , y ) 0 and h ( u , z ) 0 for each x C , y Q and z K .

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

Example E Let H 1 and H 2 be two real Hilbert spaces and B : H 1 H 2 be a bounded linear operator with its adjoint operator A . Let C H 1 , Q H 1 and K H 2 be three closed convex sets with C Q . If H 1 = H 2 and A = I (identity operator), then BSEP reduces to the following split equilibrium problems (3) (SEP ( 3 ) ):

( SEP ( 3 ) ) Find p C Q such that u : = B p K satisfying f ( p , x ) 0 , g ( p , y ) 0 and h ( u , z ) 0 for all x C , y Q and z K .

Example F In Example A, if H 1 = H 2 = H 3 : = H , C = Q = K H and A = B = I (identity operator), then BSEP reduces to the common solution of equilibrium problems (CEP):

(CEP) Find p C such that f ( p , x ) 0 , g ( p , y ) 0 and h ( p , z ) 0 for each x , y , z 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 × K into satisfying the following conditions.

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

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

(A3) for each x , y , z K ,
lim sup t 0 + F ( t z + ( 1 t ) x , y ) F ( x , y ) ;

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

Let r > 0 and x H . Then there exists z K such that
F ( z , y ) + 1 r y z , z x 0 for all y 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 × 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 ) , ( x ¯ , y ¯ ) H 1 × H 2 and a , b R ,
a ( x , y ) + b ( x ¯ , y ¯ ) = ( a x + b x ¯ , a y + b y ¯ )
and
( x , y ) = x + y .
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 × C R , g : Q × Q R and h : K × K R be three bi-functions. A : H 1 H 3 and B : H 2 H 3 are two bounded linear operators with their adjoint operators A and B , respectively. Suppose that all the bi-functions f, g and h satisfy conditions (A1)-(A4). Let x 1 C , y 1 Q , C 1 = C , Q 1 = Q , { x n } , { y n } , { u n } , { v n } and { w n } be sequences generated by
{ u n = T r n f x n , v n = T r n g y n , w n = T r n h ( 1 2 A u n + 1 2 B v n ) , l n = P C ( u n ξ A ( A u n w n ) ) , k n = P Q ( v n ξ B ( B v n w n ) ) , C n + 1 × Q n + 1 = { ( x , y ) C n × Q n : l n x 2 + k n y 2 C n + 1 × Q n + 1 = { u n x 2 + v n y 2 x n x 2 + y n y 2 } , x n + 1 = P C n + 1 ( x 1 ) , y n + 1 = P Q n + 1 ( y 1 ) , n N ,
(2.1)
where ξ ( 0 , min ( 1 A 2 , 1 B 2 ) ) and { r n } ( 0 , + ) with lim inf n + 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
Ω = { ( p , q ) EP ( f ) × EP ( g ) : A p = B q EP ( h ) } .
Then there exists ( p , q ) Ω such that
  1. (a)

    ( x n , y n ) ( p , q ) as n ;

     
  2. (b)

    ( u n , v n ) ( p , q ) as n ;

     
  3. (c)

    w n w : = A p = B q EP ( h ) as n .

     

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 × C R , g : Q × Q R and h : K × K R be three bi-functions. B : H 1 H 2 is a bounded linear operator with its adjoint operator B . Suppose that all the bi-functions f, g and h satisfy conditions (A1)-(A4). Let x 1 C , y 1 Q , C 1 = C , Q 1 = Q , { x n } , { y n } , { u n } , { v n } and { w n } be sequences generated by
{ u n = T r n f x n , v n = T r n g y n , w n = T r n h ( 1 2 B u n + 1 2 B v n ) , l n = P C ( u n ξ B ( B u n w n ) ) , k n = P Q ( v n ξ B ( B v n w n ) ) , C n + 1 × Q n + 1 = { ( x , y ) C n × Q n : l n x 2 + k n y 2 u n x 2 + v n y 2 C n + 1 × Q n + 1 = { x n x 2 + y n y 2 } , x n + 1 = P C n + 1 ( x 1 ) , y n + 1 = P Q n + 1 ( y 1 ) , n N ,
where ξ ( 0 , 1 B 2 ) and { r n } ( 0 , + ) with lim inf n + 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
Ω = { ( p , q ) EP ( f ) × EP ( g ) : B p = B q EP ( h ) } .
Then there exists ( p , q ) Ω such that
  1. (a)

    ( x n , y n ) ( p , q ) as n ;

     
  2. (b)

    ( u n , v n ) ( p , q ) as n ;

     
  3. (c)

    w n w : = B p = B q EP ( h ) as n .

     

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 H 1 and Q , K H 2 be three closed convex sets. Let f : C × C R , g : Q × Q R and h : K × K R be three bi-functions. A : H 1 H 2 is a bounded linear operator with its adjoint operator A . Suppose that all the bi-functions f, g and h satisfy conditions (A1)-(A4). Let x 1 C , y 1 Q , C 1 = C , Q 1 = Q , { x n } , { y n } , { u n } , { v n } and { w n } be sequences generated by
{ u n = T r n f x n , v n = T r n g y n , w n = T r n h ( 1 2 A u n + 1 2 v n ) , l n = P C ( u n ξ A ( A u n w n ) ) , k n = v n ξ ( v n w n ) , C n + 1 × Q n + 1 = { ( x , y ) C n × Q n : l n x 2 + k n y 2 u n x 2 + v n y 2 C n + 1 × Q n + 1 = { x n x 2 + y n y 2 } , x n + 1 = P C n + 1 ( x 1 ) , y n + 1 = P Q n + 1 ( y 1 ) , n N ,
where ξ ( 0 , min { 1 , 1 A 2 } ) and { r n } ( 0 , + ) with lim inf n + 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
Ω = { p EP ( f ) : A p EP ( g ) EP ( h ) } .
Then there exists p Ω such that
  1. (a)

    x n p as n ;

     
  2. (b)

    u n p as n ;

     
  3. (c)

    v n , y n , w n w : = A p as n .

     

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 H 1 and K H 2 be three closed convex sets. Let f : C × C R , g : Q × Q R and h : K × K R be three bi-functions. B : H 1 H 2 is a bounded linear operator with its adjoint operator B . Suppose that all the bi-functions f, g and h satisfy conditions (A1)-(A4). Let x 1 C , y 1 Q , C 1 = C , Q 1 = Q , { x n } , { y n } , { u n } , { v n } and { w n } be sequences generated by
{ u n = T r n f x n , v n = T r n g y n , w n = T r n h ( 1 2 u n + 1 2 B v n ) , l n = u n ξ ( u n w n ) , k n = P q ( v n ξ B ( B v n w n ) ) , C n + 1 × Q n + 1 = { ( x , y ) C n × Q n : l n x 2 + k n y 2 u n x 2 + v n y 2 C n + 1 × Q n + 1 = { x n x 2 + y n y 2 } , x n + 1 = P C n + 1 ( x 1 ) , y n + 1 = P Q n + 1 ( y 1 ) , n N ,
where ξ ( 0 , min { 1 , 1 B 2 } ) and { r n } ( 0 , + ) with lim inf n + 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
Ω = { p EP ( f ) EP ( g ) : A p EP ( h ) } .
Then there exists p Ω such that
  1. (a)

    x n , u n p as n ;

     
  2. (b)

    y n , v n p as n ;

     
  3. (c)

    w n w : = A p as n .

     

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 × C R be three bi-functions. Suppose that all the bi-functions f, g and h satisfy conditions (A1)-(A4). Let x 1 , y 1 C , C 1 = C , { x n } , { y n } , { u n } , { v n } and { w n } be sequences generated by
{ u n = T r n f x n , v n = T r n g y n , w n = T r n h ( 1 2 u n + 1 2 v n ) , l n = u n ξ ( u n w n ) , k n = v n ξ ( v n w n ) , C n + 1 × C n + 1 = { ( x , y ) C n × C n : l n x 2 + k n y 2 u n x 2 + v n y 2 C n + 1 × C n + 1 = { x n x 2 + y n y 2 } , x n + 1 = P C n + 1 ( x 1 ) , y n + 1 = P C n + 1 ( y 1 ) , n N ,
where ξ ( 0 , 1 ) and { r n } ( 0 , + ) with lim inf n + r n > 0 , P C is a projection operator from H into C. Suppose that
Ω = { ( p , q ) EP ( f ) × EP ( g ) : p = q EP ( h ) } .
Then there exists ( p , q ) Ω such that
  1. (a)

    ( x n , y n ) ( p , q ) as n ;

     
  2. (b)

    ( u n , v n ) ( p , q ) as n ;

     
  3. (c)

    w n w : = p = q EP ( h ) as n .

     
Remark 2.1 In Corollary 2.2, it is obvious that
Ω = { ( p , q ) EP ( f ) × EP ( g ) : p = q EP ( h ) }
implies
Ω = { p EP ( f ) EP ( g ) EP ( h ) } .

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 H 1 , Q H 2 and K H 3 be three linear subspaces. Let f : C × C R , g : Q × Q R and h : K × K R be three bi-functions. A : H 1 H 3 and B : H 2 H 3 are two bounded linear operators with their adjoint operators A and B , respectively. Suppose that all the bi-functions f, g and h satisfy conditions (A1)-(A4). Let x 1 C , y 1 Q , C 1 = C , Q 1 = Q , { x n } , { y n } , { u n } , { v n } and { w n } be sequences generated by
{ u n = T r n f x n , v n = T r n g y n , w n = T r n h ( 1 2 A u n + 1 2 B v n ) , l n = u n ξ A ( A u n w n ) , k n = v n ξ B ( B v n w n ) , C n + 1 × Q n + 1 = { ( x , y ) C n × Q n : l n x 2 + k n y 2 u n x 2 + v n y 2 C n + 1 × Q n + 1 = { x n x 2 + y n y 2 } , x n + 1 = P C n + 1 ( x 1 ) , y n + 1 = P Q n + 1 ( y 1 ) , n N ,
where ξ ( 0 , min ( 1 A 2 , 1 B 2 ) ) and { r n } ( 0 , + ) with lim inf n + 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
Ω = { ( p , q ) EP ( f ) × EP ( g ) : A p = B q EP ( h ) } .
Then there exists ( p , q ) Ω such that
  1. (a)

    ( x n , y n ) ( p , q ) as n ;

     
  2. (b)

    ( u n , v n ) ( p , q ) as n ;

     
  3. (c)

    w n w : = A p = B q EP ( h ) as n .

     
Corollary 2.6 Let H 1 and H 2 be two real Hilbert spaces. Let C H 1 , Q H 1 and K H 2 be three linear subspaces. Let f : C × C R , g : Q × Q R and h : K × K R be three bi-functions. B : H 1 H 2 is a bounded linear operator with its adjoint operator B . Suppose that all the bi-functions f, g and h satisfy conditions (A1)-(A4). Let x 1 C , y 1 Q , C 1 = C , Q 1 = Q , { x n } , { y n } , { u n } , { v n } and { w n } be sequences generated by
{ u n = T r n f x n , v n = T r n g y n , w n = T r n h ( 1 2 B u n + 1 2 B v n ) , l n = u n ξ B ( B u n w n ) , k n = v n ξ B ( B v n w n ) , C n + 1 × Q n + 1 = { ( x , y ) C n × Q n : l n x 2 + k n y 2 u n x 2 + v n y 2 C n + 1 × Q n + 1 = { x n x 2 + y n y 2 } , x n + 1 = P C n + 1 ( x 1 ) , y n + 1 = P Q n + 1 ( y 1 ) , n N ,
where ξ ( 0 , 1 B 2 ) and { r n } ( 0 , + ) with lim inf n + 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
Ω = { ( p , q ) EP ( f ) × EP ( g ) : B p = B q EP ( h ) } .
Then there exists ( p , q ) Ω such that
  1. (a)

    ( x n , y n ) ( p , q ) as n ;

     
  2. (b)

    ( u n , v n ) ( p , q ) as n ;

     
  3. (c)

    w n w : = B p = B q EP ( h ) as n .

     

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 × K into satisfying (A1)-(A4). For r > 0 , define a mapping T r F : H K as follows:
T r F ( x ) = { z K : F ( z , y ) + 1 r y z , z x 0 , y K }
(2.2)
for all x H . Then the following hold:
  1. (i)

    T r F is single-valued and F ( T r F ) = EP ( F ) for r > 0 and EP ( F ) is closed and convex;

     
  2. (ii)

    T r F is firmly nonexpansive, that is, for any x , y H , T r F x T r F y 2 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 F ( T r F ) = EP ( F ) , then, for any x H and x F ( T r F ) , T r F x x 2 x x 2 T r F x x 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 H ,
T r F ( x ) T s F ( y ) x y + | s r | s T s F ( y ) y .

In particular, T r F ( x ) T r F ( y ) x y for any r > 0 and x , y 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:
  1. (a)

    x y 2 = x 2 + y 2 2 x , y for all x , y H ;

     
  2. (b)

    α x + ( 1 α ) y 2 = α x 2 + ( 1 α ) y 2 α ( 1 α ) x y 2 for all x , y H and α [ 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 × Q n are closed convex sets for n N .

Now, we claim C n × Q n for all n N . Indeed, it suffices to prove that Ω C n × Q n for all n N . Let ( x , y ) Ω . Then x EP ( f ) , y EP ( g ) and
w : = A x = B y EP ( h ) .
Let n N be given. By Lemma 2.3, we have
u n x x n x , v n y y n y , w n w A u n + B v n 2 w , w n w 2 1 2 A u n w 2 + 1 2 B v n w 2 ( by Lemma  2.5 ) .
(3.1)
From (2.1), (3.1) and Lemma 2.5, we obtain
l n x 2 = P C ( u n ξ A ( A u n w n ) ) x 2 u n x ξ A ( A u n w n ) 2 = u n x 2 + ξ A ( A u n w n ) 2 2 ξ u n x , A ( A u n w n ) = u n x 2 + ξ A ( A u n w n ) 2 2 ξ A u n A x , A u n w n = u n x 2 + ξ A ( A u n w n ) 2 2 ξ A u n w , A u n w n = u n x 2 + ξ A ( A u n w n ) 2 ξ A u n w 2 ξ A u n w n 2 + ξ w n w 2 u n x 2 ξ ( 1 ξ A 2 ) A u n w n 2 ξ A u n w 2 + ξ w n w 2 u n x 2 ξ ( 1 ξ A 2 ) A u n w n 2 ξ A u n w 2 + ξ 2 A u n w 2 + ξ 2 B v n w 2 = u n x 2 ξ ( 1 ξ A 2 ) A u n w n 2 ξ 2 A u n w 2 + ξ 2 B v n w 2
(3.2)
and
k n y 2 = P Q ( v n ξ B ( B v n w n ) ) y 2 v n y ξ B ( B v n w n ) 2 = v n y 2 + ξ B ( B v n w n ) 2 2 ξ v n y , B ( B v n w n ) = v n y 2 + ξ B ( B v n w n ) 2 2 ξ B v n B y , B v n w n = v n y 2 + ξ B ( B v n w n ) 2 2 ξ B v n w , B v n w n = v n y 2 + ξ B ( B v n w n ) 2 ξ B v n w 2 ξ B v n w n 2 + ξ w n w 2 v n y 2 ξ ( 1 ξ B 2 ) B v n w n 2 ξ B v n w 2 + ξ w n w 2 v n y 2 ξ ( 1 ξ B 2 ) B v n w n 2 ξ B v n w 2 + ξ 2 A u n w 2 + ξ 2 B v n w 2 = v n y 2 ξ ( 1 ξ B 2 ) B v n w n 2 ξ 2 B v n w 2 + ξ 2 A u n w 2 .
(3.3)
By taking into account inequalities (3.1), (3.2) and (3.3), we obtain
l n x 2 + k n y 2 u n x 2 + v n y 2 ξ ( 1 ξ A 2 ) A u n w n 2 ξ ( 1 ξ B 2 ) B v n w n 2 x n x 2 + y n y 2 ξ ( 1 ξ A 2 ) A u n w n 2 ξ ( 1 ξ B 2 ) B v n w n 2 ,
(3.4)
which implies
l n x 2 + k n y 2 u n x 2 + v n y 2 x n x 2 + y n y 2 .
(3.5)
Inequality (3.5) shows that ( x , y ) C n × Q n . Hence Ω C n × Q n and C n × Q n for all n N . For each n N , since Ω C n × Q n , C n + 1 C n , we have
x n + 1 = P C n + 1 ( x 1 ) C n .
Similarly, since Q n + 1 Q n , we have
y n + 1 = P Q n + 1 ( y 1 ) Q n .
So, for any ( x , y ) Ω , we get
x n + 1 x 1 x x 1
and
y n + 1 y 1 y y 1 .
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 N with n > 1 , from x n = P C n ( x 1 ) C n , y n = P Q n ( y 1 ) Q n and (1.1), we have
x n + 1 x n 2 + x 1 x n 2 = x n + 1 P C n ( x 1 ) 2 + x 1 P C n ( x 1 ) 2 x n + 1 x 1 2 , y n + 1 y n 2 + y 1 y n 2 = y n + 1 P C n ( y 1 ) 2 + y 1 P C n ( y 1 ) 2 y n + 1 y 1 2 ,
which yields that
x 1 x n x n + 1 x 1 , y 1 y n y n + 1 y 1 .
Together with the boundedness of { x n } and { y n } , we know lim n x n x 1 and lim n y n y 1 exist. For some k , n N with k > n > 1 , due to x k = P C k ( x 1 ) C n , y k = P Q k ( y 1 ) Q n and (1.1), we have
x k x n 2 + x 1 x n 2 = x k P C n ( x 1 ) 2 + x 1 P C n ( x 1 ) 2 x k x 1 2 , y k y n 2 + y 1 y n 2 = y k P Q n ( y 1 ) 2 + y 1 P Q n ( y 1 ) 2 y k y 1 2 .
(3.6)
By (3.6), we have lim n x n x k = 0 and lim n y n y k = 0 . Hence { x n } and { y n } are all Cauchy sequences. Let x n p and y n q for some ( p , q ) C × Q . We want to prove that ( p , q ) Ω . For any n N , since
( x n + 1 , y n + 1 ) C n + 1 × Q n + 1 C n × Q n ,
from (2.1), we have
l n x n + 1 2 + k n y n + 1 2 u n x n + 1 2 + v n y n + 1 2 x n x n + 1 2 + y n y n + 1 2 .
(3.7)
By taking the limit from both sides of (3.7), we obtain
lim n l n x n + 1 = lim n k n y n + 1 = 0 , lim n u n x n + 1 = lim n v n y n + 1 = 0 .
(3.8)
Moreover, by (3.8), we get
lim n l n u n = lim n u n x n = lim n l n x n = 0 , lim n k n v n = lim n v n y n = lim n k n y n = 0 .
(3.9)

Since lim n u n x n = lim n v n y n = 0 , we have u n p and v n q as n . Moreover, we obtain A u n A p and B v n B q as n .

Now, we claim p EP ( f ) and q EP ( g ) . In fact, for r > 0 , by Lemma 2.4, we have
T r f p p = T r f p T r n f x n + T r n f x n x n + x n p x n p + | r n r | r n T r n f x n x n + T r n f x n x n + x n p = x n p + | r n r | r n u n x n + u n x n + x n p 0
and
T r g q q T r g q T r n g y n + T r n g y n y n + y n q y n q + | r n r | r n T r n g y n y n + T r n g y n y n + y n q = y n q + | r n r | r n v n y n + v n y n + y n q 0 .

So, p EP ( f ) and q EP ( g ) .

Finally, we prove A p = B q EP ( h ) . Setting
θ = min { ξ ( 1 ξ A 2 ) , ξ ( 1 ξ B 2 ) } .
Then, for any n N , by (3.4) and (3.9), we have
θ A u n w n 2 + θ B v n w n 2 x n x 2 + y n y 2 l n x 2 k n y 2 = { x n x l n x } { x n x + l n x } + { y n y k n y } { y n y + k n y } l n x n { x n x + l n x } + k n y n { y n y + k n y } 0 .
(3.10)
Hence (3.10) implies
lim n A u n w n = lim n B v n w n = 0 , and lim n A u n B v n = 0 .
(3.11)
Since A u n A p , B v n B q and (3.11), we obtain A p = B q and w n w , where w : = A p = B q . On the other hand, for r > 0 , by Lemma 2.4 again, we have
T r h w w = T r h w T r n h A u n + B v n 2 + T r n h A u n + B v n 2 A u n + B v n 2 + A u n + B v n 2 w A u n + B v n 2 w + | r n r | r n T r n h A u n + B v n 2 A u n + B v n 2 + T r n h A u n + B v n 2 A u n + B v n 2 + A u n + B v n 2 w = 2 A u n + B v n 2 w + | r n r | r n w n A u n + B v n 2 + w n A u n + B v n 2 0 .

Hence w EP ( h ) , namely A p = B q 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 α = ( α 1 , α 2 ) R 2 and ν = ( z 1 , z 2 , z 3 , z 4 ) R 4 , define
A α = ( α 1 , α 2 , α 1 + α 2 , α 1 α 2 )
and
A ν = ( 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 A = 3 , and A is an adjoint operator of A with A = 3 . For each β = ( β 1 , β 2 , β 3 ) R 3 and ν = ( z 1 , z 2 , z 3 , z 4 ) R 4 , let
B β = ( β 1 , β 2 , β 3 , β 1 β 2 )
and
B ν = ( z 1 + z 4 , z 2 z 4 , z 3 ) .
Then B is a bounded linear operator from R 3 into R 4 with B = 3 , and B is an adjoint operator of B with B = 3 . Put
C : = { α = ( α 1 , α 2 ) R 2 : 1 α 1 2 , 3 α 2 4 } , Q : = { β = ( β 1 , β 2 , β 3 ) R 3 : 1 β 1 1 , 3 β 2 4 , 3 β 3 5 }
and
K : = { z = ( z 1 , z 2 , z 3 , z 4 ) R 4 : 0 z 1 1 , 3 z 2 6 , 3 z 3 5 , 5 z 4 3 } .
For each α = ( α 1 , α 2 ) C , β = ( β 1 , β 2 , β 3 ) Q and z = ( z 1 , z 2 , z 3 , z 4 ) K , define
f ( α ) = α 1 2 + α 2 2 , g ( β ) = β 1 2 + β 2 2 + β 3 2
and
h ( z ) = z 1 2 + z 2 2 + z 3 2 + z 4 2 .
For each α , x C , let
f ( α , x ) = f ( x ) f ( α ) .
For each β , y Q , let
g ( β , y ) = g ( y ) g ( β ) .
For each η , z K , let
h ( η , z ) = h ( z ) h ( η ) .
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
Ω = { ( p , q ) EP ( f ) × EP ( g ) : A p = B q EP ( h ) } .

Let C 1 = C , Q 1 = Q , ξ = 1 6 and r n 1 for n N . Thus, for each x ¯ = ( a , b ) C and y ¯ = ( c , d , e ) Q with c > 0 , we have the following:

  • u = ( a 3 , 3 ) = T r n f x ¯ ,

  • v = ( c 3 , 3 , 3 ) = T r n g y ¯ ,

  • w = ( a + c 6 , 3 , 3 , 3 ) = T r n h ( 1 2 A x ¯ + 1 2 B y ¯ ) ,

  • l = P C ( u 1 6 A ( A u w ) ) = ( 7 a + c 36 , 3 ) ,

  • k = P Q ( v 1 6 B ( B v w ) ) = ( 9 c + a 36 , 3 + c 18 , 3 ) .

For x 1 = ( a 1 , b 1 ) C and y 1 = ( c 1 , d 1 , e 1 ) Q with 15 c 1 a 1 > 0 , 17 a 1 c 1 and d 1 > 3 + c 1 18 , we obtain the following:

  • u 1 = ( a 1 3 , 3 ) ,

  • v 1 = ( c 1 3 , 3 , 3 ) ,

  • w 1 = ( a 1 + c 1 6 , 3 , 3 , 3 ) ,

  • l 1 = ( 7 a 1 + c 1 36 , 3 ) ,

  • k 1 = ( 9 c 1 + a 1 36 , 3 + c 1 18 , 3 ) ,

  • C 2 = { α = ( α 1 , α 2 ) C 1 : 1 α 1 19 a 1 + c 1 72 , 3 α 2 b 1 + 3 2 } ,

  • x 2 = P C 2 ( x 1 ) = ( 19 a 1 + c 1 72 , b 1 + 3 2 ) : = ( a 2 , b 2 ) ,

  • Q 2 = { β = ( β 1 , β 2 , β 3 ) Q 1 : 1 β 1 21 c 1 + a 1 72 , 3 + c 1 36 β 2 d 1 + 3 2 , 3 β 3 e 1 + 3 2 } ,

  • y 2 = P Q 2 ( y 1 ) = ( 21 c 1 + a 1 72 , d 1 + 3 2 , e 1 + 3 2 ) : = ( c 2 , d 2 , e 2 ) .

From x 2 , y 2 , we have u 2 = ( 1 3 a 2 , 3 ) , v 2 = ( 1 3 c 2 , 3 , 3 ) and w 2 = ( 1 6 ( a 2 + b 2 ) , 3 , 3 , 3 ) . Since
d 2 > 3 + c 2 18 , 15 c 2 a 2 > 0 ,
and
17 a 2 c 2 ,

we get the following:

  • C 3 = { α = ( α 1 , α 2 ) C 2 : 1 α 1 19 a 2 + c 2 72 , 3 α 2 1 2 ( b 2 + 3 ) } ,

  • x 3 = ( 19 a 2 + c 2 72 , 1 2 ( b 2 + 3 ) ) : = ( a 3 , b 3 ) ,

  • Q 3 = { β = ( β 1 , β 2 , β 3 ) Q 1 : 1 β 1 21 c 2 + a 2 72 , 3 + c 2 36 β 2 d 2 + 3 2 , 3 β 3 e 2 + 3 2 } ,

  • y 3 = ( 21 c 2 + a 2 72 , d 2 + 3 2 , e 2 + 3 2 ) : = ( c 3 , d 3 , e 3 ) .

Similarly, for n N with n > 1 , we obtain

  • C n + 1 = { α = ( α 1 , α 2 ) C n : 1 α 1 19 a n 1 + c n 1 72 , 3 α 2 1 2 ( b n 1 + 3 ) } ,

  • x n + 1 = ( 19 a n 1 + c n 1 72 , 1 2 ( b n 1 + 3 ) ) ,

  • Q n + 1 = { β = ( β 1 , β 2 , β 3 ) Q n : 1 β 1 21 c n 1 + a n 1 72 , 3 + c n 1 36 β 2 d n 1 + 3 2 , 3 β 3 e n 1 + 3 2 } ,

  • y n + 1 = ( 21 c n 1 + a n 1 72 , d n 1 + 3 2 , e n 1 + 3 2 ) ,

  • u n + 1 = ( 1 3 a n 1 , 3 ) ,

  • v n + 1 = ( 1 3 c n 1 , 3 , 3 ) ,

  • w n + 1 = ( 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 0 , b n 3 , c n 0 , d n 3 and e n 3 as n . So, we have u n ( 0 , 3 ) , v n ( 0 , 3 , 3 ) , w n ( 0 , 3 , 3 , 3 ) , x n ( 0 , 3 ) and y n ( 0 , 3 , 3 ) as n .

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.

Declarations

Acknowledgements

The first author was supported by the Natural Science Foundation of Yunnan Province (201401CA00262) and the Candidate Foundation of Youth Academic Experts at Honghe University (2014HB0206); the second author was supported by Grant No. MOST 103-2115-M-017-001 of the Ministry of Science and Technology of the Republic of China.

Authors’ Affiliations

(1)
Department of Mathematics, Honghe University
(2)
Department of Mathematics, National Kaohsiung Normal University

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© He and Du; licensee Springer. 2014

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