Open Access

The split equilibrium problem and its convergence algorithms

Journal of Inequalities and Applications20122012:162

https://doi.org/10.1186/1029-242X-2012-162

Received: 1 April 2012

Accepted: 2 July 2012

Published: 20 July 2012

Abstract

The purpose of this paper is to introduce a split equilibrium problem (SEP) and find a solution of the equilibrium problem such that its image under a given bounded linear operator is a solution of another equilibrium problem. By using the iterative method, we construct some iterative algorithms to solve such problem in real Hilbert spaces and obtain some strong and weak convergence theorems. Finally, we point out that there exist many SEPs which need the use of new methods to solve them. Some examples are given to illustrate our results.

MSC:47J25, 47H09, 65K10.

Keywords

iterative algorithm split equilibrium problem strong and weak convergence

1 Introduction

Throughout this paper, the symbols N and R are used to denote the sets of positive integers and real numbers, respectively.

In this paper, we propose a new equilibrium problem, which is called a split equilibrium problem (SEP). Let E 1 and E 2 be two real Banach spaces. Let C be a closed convex subset of E 1 , K a closed convex subset of E 2 , and A : E 1 E 2 a bounded linear operator. f is a bi-function from C × C into R and g is a bi-function from K × K into R . The SEP is
to find an element p C such that f ( p , y ) 0 , y C ,
(1.1)
and
such that u : = A p K solves g ( u , v ) 0 , v K .
(1.2)

If we consider only the problem (1.1), then (1.1) is a classical equilibrium problem. From (1.1) and (1.2), we can see that the SEP contains two equilibrium problems, and the image of a solution of one equilibrium problem under a given bounded linear operator is a solution of another equilibrium problem. Since many problems coming from physics, optimization, and economics reduce to find a solution of the equilibrium problem (1.1) (see, for instance, [1, 2]), the equilibrium problem (1.1) is very important in the field of applied mathematics. Some authors have proposed some methods to find the solution of the equilibrium problem (1.1). As a generalization of the equilibrium problem (1.1), when finding a common solution for some equilibrium problems, it has been considered in the same subset of the same space; see [35]. However, in general, some equilibrium problems always belong to different subsets of spaces, so the SEP is important and quite general. The SEP should enable us to split the solution between two different subsets of spaces so that the image of a solution point of one problem, under a given bounded linear operator, is a solution point of another problem. A special case of the SEP is the split variational inequality problem (SVIP); see [6].

For convenience, in this paper let EP ( f ) , EP ( g ) and Ω = { p EP ( f ) : A p EP ( g ) } denote the solution set of (1.1), (1.2) and the SEP, respectively.

Example 1.1 Let E 1 = E 2 = R , C : = [ 1 , + ) and K : = ( , 4 ] . Let A ( x ) = 4 x for all x R , then A is a bounded linear operator. Let f : C × C R , and g : K × K R be defined by f ( x , y ) = y x , g ( u , v ) = 2 ( u v ) , respectively. Clearly, EP ( f ) = { 1 } and A ( 1 ) = 4 EP ( g ) . So Ω = { p EP ( f ) : A p EP ( g ) } .

Example 1.2 Let E 2 = R with the standard norm | | and E 1 = R 2 with the norm α = ( a 1 2 + a 2 2 ) 1 2 for some α = ( a 1 , a 2 ) R 2 . K : = [ 1 , + ) and C : = { α = ( a 1 , a 2 ) R 2 | a 2 a 1 1 } . Define a bi-function f ( w , α ) = w 1 w 2 + a 2 a 1 , where w = ( w 1 , w 2 ) , α = ( a 1 , a 2 ) C , then f is a bi-function from C × C into R with EP ( f ) = { p = ( p 1 , p 2 ) | p 2 p 1 = 1 } . For each α = ( a 1 , a 2 ) E 1 , let A α = a 2 a 1 , then A is a bounded linear operator from E 1 into E 2 . In fact, it is also easy to verify that A ( a α 1 + b α 2 ) = a A ( α 1 ) + b A ( α 2 ) and A = 2 for some α 1 , α 2 E 1 and a , b R . Now define another bi-function g as follows: g ( u , v ) = v u for all u , v K . Then g is a bi-function from K × K into R with EP ( g ) = { 1 } .

Clearly, when p EP ( f ) , we have A p = 1 EP ( g ) . So Ω = { p EP ( f ) : A p EP ( g ) } .

Remark 1.1 The SEP in Example 1.1 lies in two different subsets of the same space. While the SEP in Example 1.2 lies in two different subsets of the different space.

In this paper, we construct some iterative algorithms to solve the SEP. Some strong and weak convergence theorems are established. The results obtained in this paper can be reckoned as the new development of the equilibrium problem (1.1). Finally, we point out that there exist many SEPs which need the use of new methods to solve them. Some examples are given to illustrate our results.

2 Preliminaries

We assume that H is a real Hilbert space with zero vector θ whose inner product and norm are denoted by , and , respectively; and we use symbols → and to denote strong and weak convergence, respectively.

Let H 1 and H 2 be two Hilbert spaces. The operator A from H 1 into H 2 and the operator B from H 2 into H 1 are two bounded linear operators. B is called the adjoint operator of A, if for all z H 1 , w H 2 , B satisfies A z , w = z , B w . Especially, if H 1 = H 2 , then B reduces to the well-known adjoint operator of A.

Remark 2.1 It is easy to verify that the operator B, an adjoint operator of A, has the following characters:
  1. (i)

    B = A ; (ii) B is a unique adjoint operator of A.

     

Example 2.1 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 = a 1 b 1 + a 2 b 2 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 , then A is a bounded linear operator from H 1 into H 2 with A = 2 . For x H 2 , let B x = ( x , x ) , then B is a bounded linear operator from H 2 into H 1 with B = 2 . Moreover, for any α = ( a 1 , a 2 ) H 1 and x H 2 , A α , x = α , B x , so B is an adjoint operator of A.

Example 2.2 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 = a 1 b 1 + a 2 b 2 and γ , η = c 1 d 1 + c 2 d 2 + c 3 d 3 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 ) R 3 . Let A α = ( a 2 , a 1 , a 1 a 2 ) for α = ( a 1 , a 2 ) H 1 , then A is a bounded linear operator from H 1 into H 2 with A = 3 because ( 2 2 , 2 2 , 2 ) sup α = 1 A α 3 . For γ = ( c 1 , c 2 , c 3 ) H 2 , let B γ = ( c 2 + c 3 , c 1 c 3 ) , then B is a bounded linear operator from H 2 into H 1 with B = 3 . Moreover, for any α = ( a 1 , a 2 ) H 1 and γ = ( c 1 , c 2 , c 3 ) H 2 , A α , γ = α , B γ , so B is an adjoint operator of A.

Let K be a closed convex subset of a real Hilbert space H. 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 , 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 characters:
  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 , y K .

     
  3. (iii)
    For x H and y K ,
    y P K ( x ) 2 + x P K ( x ) 2 x y 2 .
    (2.1)
     
A Banach space ( X , ) is said to satisfy Opial’s condition if, for each sequence { x n } in X which converges weakly to a point x X , we have
lim inf n x n x < lim inf n x n y , y X , y x .

It is well known that each Hilbert space satisfies Opial’s condition.

The following results are crucial to our main results.

Lemma 2.1 (see [1])

Let K be a nonempty closed convex subset of H and F be a bi-function of K × K into R 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 .

Lemma 2.2 (see [7])

Let K be a nonempty closed convex subset of H and let F be a bi-function of K × K into R satisfying (A 1)-(A 4). For r > 0 and x H , 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;

     
  2. (ii)
    T r F is firmly non-expansive, 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 ;
     
  3. (iii)

    F ( T r F ) = EP ( F ) for r > 0 ;

     
  4. (iv)

    EP ( F ) is closed and convex.

     

Lemma 2.3 (see [3])

Let H be a real Hilbert space. Then for any x 1 , x 2 , , x k H and a 1 , a 2 , , a k [ 0 , 1 ] with i = 1 k a i = 1 , k N , we have
i = 1 k a i x i 2 = i = 1 k a i x i 2 i = 1 k 1 j = i + 1 k a i a j x i x j 2 .
In particular, we have
  1. (1)

    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 ] ;

     
  2. (2)

    the map f : H R defined by f ( x ) = x 2 is convex.

     

Lemma 2.4 (see, e.g., [8])

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

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

     
  2. (b)

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

     
Lemma 2.5 Let K be a nonempty closed convex subset of H. For x H , let the mapping T r F ( x ) be the same as in Lemma 2.2. Then for r , s > 0 and x , y H ,
T r F ( x ) T s F ( y ) y x + | s r | s T s F ( y ) y .
Proof For r , s > 0 and x , y H , by (i) of Lemma 2.2, we can let z 1 = T r F ( x ) and z 2 = T s F ( y ) . By the definition of T r F , we have
F ( z 1 , u ) + 1 r u z 1 , z 1 x 0 , u K
(2.3)
and
F ( z 2 , u ) + 1 s u z 2 , z 2 y 0 , u K .
(2.4)
Taking u = z 2 in (2.3) and u = z 1 in (2.4), we have
F ( z 1 , z 2 ) + 1 r z 2 z 1 , z 1 x 0 ,
(2.5)
and
F ( z 2 , z 1 ) + 1 s z 1 z 2 , z 2 y 0 .
(2.6)
Since the bi-function F satisfies the condition (A2), from (2.5) and (2.6) we have
1 r z 2 z 1 , z 1 x + 1 s z 1 z 2 , z 2 y 0 ,
which implies that
z 2 z 1 , z 1 x z 2 z 1 , r z 2 y s 0 .
Thus, we have
z 2 z 1 , z 1 z 2 + z 2 x r s ( z 2 y ) 0 ,
this implies that
z 2 z 1 2 z 2 z 1 , z 2 x r s ( z 2 y ) z 2 z 1 z 2 x r s ( z 2 y ) ,
so
z 2 z 1 z 2 x r s ( z 2 y ) y x + ( 1 r s ) ( z 2 y ) = y x + | s r | s T s F ( y ) y ,
namely,
T r F ( x ) T s F ( y ) y x + | s r | s T s F ( y ) y .

The proof is completed. □

3 Main results

In this section, we will solve the SEP which satisfies the conditions (A1)-(A4).

Theorem 3.1 (Weak convergence theorem)

Let C be a nonempty closed convex subset of H 1 and K a nonempty closed convex subset of H 2 , where H 1 and H 2 are two real Hilbert spaces. : = { 1 , 2 , , k } denotes a finite index set. For any i , f i : C × C R is a bi-function with i = 1 k EP ( f i ) . Let A : H 1 H 2 be a bounded linear operator with the adjoint B and g : K × K R a bi-function with EP ( g ) . Let { x n } and { u n i } ( i ) be sequences generated by
{ x 1 C , f i ( u n i , y ) + 1 r n y u n i , u n i x n 0 , y C , i , τ n = u n 1 + + u n k k , g ( w n , z ) + 1 r n z w n , w n A τ n 0 , z K , x n + 1 = P C ( τ n + μ B ( w n A τ n ) ) , n N ,
(3.1)

where { r n } ( 0 , + ) with lim inf n r n > 0 , P C is a projection operator from H 1 into C and μ ( 0 , 1 B 2 ) is a constant. Suppose that Ω = { p i = 1 k EP ( f i ) : A p EP ( g ) } , then the sequences { x n } and { u n i } ( i ) converge weakly to an element p Γ , while { w n } converges weakly to A p EP ( g ) .

Proof For each i and each r > 0 , let T r f i : H 1 C be defined by (2.2), then u n i = T r n f i x n for all n N by Lemma 2.2. Again let T r g : H 2 K be defined by (2.2), then w n = T r n g A τ n for all n N . So (3.1) can be rewritten as follows:
{ x 1 C , u n i = T r n f i x n , i , τ n = u n 1 + + u n k k , w n = T r n g A τ n , x n + 1 = P C ( τ n + μ B ( T r n g I ) A τ n ) , n N .
(3.2)
Let x C be a point such that x i = 1 k EP ( f i ) and A x EP ( g ) , namely, x Ω . By Lemma 2.2 and Lemma 2.4, it follows that
u n i x 2 = T r n f i x n T r n f i x 2 T r n f i x n T r n f i x , x n x = 1 2 { u n i x 2 + x n x 2 u n i x n 2 } ,
hence
u n i x 2 x n x 2 u n i x n 2 .
(3.3)
Applying Lemma 2.3, we get
τ n x 2 1 k i = 1 k u n i x 2 .
(3.4)
(3.3) and (3.4) imply that
τ n x 2 x n x 2 1 k i = 1 k u n i x n 2 .
(3.5)
Again from Lemma 2.2, we have
w n A x = T r n g A τ n A x A τ n A x for each n N .
(3.6)
By (b) of Lemma 2.4 and (3.6), for each n N , we have
(3.7)
We also have
B ( T r n g I ) A τ n 2 B 2 ( T r n g I ) A τ n 2 .
(3.8)
From (3.2), (3.5)-(3.8), we have
x n + 1 x 2 = P C ( τ n + μ B ( T r n g I ) A τ n ) P C x 2 τ n + μ B ( T r n g I ) A τ n ) x 2 = τ n x 2 + μ B ( T r n g I ) A τ n 2 + 2 μ τ n x , B ( T r n g I ) A τ n τ n x 2 + μ 2 B 2 ( T r n g I ) A τ n 2 μ ( T r n g I ) A τ n 2 = τ n x 2 μ ( 1 μ B 2 ) ( T r n g I ) A τ n 2 x n x 2 μ ( 1 μ B 2 ) ( T r n g I ) A τ n 2 .
(3.9)
Notice μ ( 0 , 1 B 2 ) , μ ( 1 μ B 2 ) > 0 . It follows from (3.9) that
x n + 1 x τ n x x n x
(3.10)
and
μ ( 1 μ B 2 ) ( T r n g I ) A τ n 2 x n x 2 x n + 1 x 2 .
(3.11)
(3.10) implies lim n x n x exists. Further, from (3.10)-(3.11),
lim n x n x = lim n τ n x , lim n ( T r n g I ) A τ n = 0 .
(3.12)
Again from (3.5), we have
lim n u n i x n = 0 , i ,
(3.13)
which yields that
lim n ( T r n f i I ) x n = lim n T r n f i x n x n = lim n u n i x n = 0 , i ,
(3.14)
and
τ n x n u n 1 x n + + u n k x n 0 as n .
(3.15)

Because lim n x n x exists, which implies { x n } is bounded, hence { x n } has a weakly convergence subsequence { x n j } . Assume that x n j p for some p C . Then u n j i p , τ n j p and A τ n j A p K by (3.13) and (3.15).

Now we prove p Ω or, to be more precise, we prove p i = 1 k EP ( f i ) and A p EP ( g ) . By Lemma 2.2, for any r > 0 , EP ( f i ) = F ( T r f i ) , i , and EP ( g ) = F ( T r g ) . For i , since ( I T r n f i ) x n 0 by (3.14), we have T r f i p = p for r > 0 . Otherwise, if T r f i p p for all i , then by Opial’s condition, we have
lim inf j x n j p < lim inf j x n j T r f i p = lim inf j x n j T r n j f i x n j + T r n j f i x n j T r f i p lim inf j { x n j T r n j f i x n j + T r n j f i x n j T r f i p } = lim inf j T r n j f i x n j T r f i p = lim inf j T r f i p T r n j f i x n j lim inf j ( x n j p + | r n j r | r n j T r n j f i x n j x n j ) (by Lemma 2.5) = lim inf j x n j p ,

this is a contradiction. So this shows that p i = 1 k F ( T r f i ) = i = 1 k EP ( f i ) . Similarly, we can prove A p EP ( g ) .

Finally, we prove { x n } and { u n i } converge weakly to p Ω , while { w n } converges weakly to A p EP ( g ) . Firstly, if there exists other subsequence of { x n } which is denoted by { x n t } such that x n t q Ω with q p , then, by Opial’s condition,
lim inf t x n t q < lim inf t x n t p < lim inf t x n t q .

This is a contradiction. Hence { x n } and { u n i } converge weakly to p Ω , respectively.

On the other hand, by (3.15) we also have τ n p . Notice that w n A τ n = ( T r n g I ) A τ n 0 by (3.12), so we have τ n A p and w n A p . We obtain the desired result. □

Corollary 3.1 Let C be a nonempty closed convex subset of H 1 and K a nonempty closed convex subset of H 2 , where H 1 and H 2 are two real Hilbert spaces. f : C × C R is a bi-function with EP ( f ) . Let A : H 1 H 2 be a bounded linear operator with the adjoint B and g : K × K R a bi-function with EP ( g ) . Let { x n } and { u n } be sequences generated by
{ x 1 C , f ( u n , y ) + 1 r n y u n , u n x n 0 , y C , g ( w n , z ) + 1 r n z w n , w n A u n 0 , z K , x n + 1 = P C ( u n + μ B ( w n A u n ) ) , n N ,
(3.16)

where { r n } ( 0 , + ) with lim inf n r n > 0 , P C is a projection operator from H 1 into C and μ ( 0 , 1 B 2 ) is a constant. Suppose that Ω = { p EP ( f ) : A p EP ( g ) } , then the sequences { x n } and { u n } converge weakly to an element p Ω , while { w n } converges weakly to A p EP ( g ) .

Theorem 3.2 (Strong convergence theorem)

Let C be a nonempty closed convex subset of H 1 and K a nonempty closed convex subset of H 2 , where H 1 and H 2 are two real Hilbert spaces. : = { 1 , 2 , , k } denotes a finite index set. For any i , f i : C × C R is a bi-function with i = 1 k EP ( f i ) . Let A : H 1 H 2 be a bounded linear operator with the adjoint B and g : K × K R a bi-function with EP ( g ) . Let C 1 = C and { x n } and { u n i } ( i ) be sequences generated by
{ x 1 C , f i ( u n i , y ) + 1 r n y u n i , u n i x n 0 , y C , i , τ n = u n 1 + + u n k k , g ( w n , z ) + 1 r n z w n , w n A τ n 0 , z K , y n = P C ( τ n + μ B ( w n A τ n ) ) , C n + 1 = { v C n : y n v τ n v x n v } , x n + 1 = P C n + 1 ( x 1 ) , n N ,
(3.17)

where { r n } ( 0 , + ) with lim inf n r n > 0 , P C is a projection operator from H 1 into C and μ ( 0 , 1 B 2 ) is a constant. Suppose that Ω = { p i = 1 k EP ( f i ) : A p EP ( g ) } , then the sequences { x n } and { u n i } ( i ) converge strongly to an element x Ω , while { w n } converges strongly to A x EP ( g ) .

Proof By Lemma 2.2, u n i = T r n f i x n for all i , n N and w n = T r n g A τ n for all n N . We claim C n for n N . In fact Ω C n for n N . Indeed, let p Ω , it follows from (3.7) and (3.8) that
2 μ τ n p , B ( T r n g I ) A τ n μ ( T r n g I ) A τ n 2 ,
(3.18)
and
B ( T r n g I ) A τ n 2 B 2 ( T r n g I ) A τ n 2 .
(3.19)
From (3.17)-(3.19) we have
y n p 2 τ n + μ B ( T r n g I ) A τ n p 2 = τ n p 2 + μ B ( T r n g I ) A τ n 2 + 2 μ τ n p , B ( T r n g I ) A τ n τ n p 2 + μ 2 B 2 ( T r n g I ) A τ n 2 μ ( T r n g I ) A τ n 2 = τ n p 2 μ ( 1 μ B 2 ) ( T r n g I ) A τ n 2 x n p 2 μ ( 1 μ B 2 ) ( T r n g I ) A τ n 2 .
(3.20)
Notice μ ( 0 , 1 B 2 ) , μ ( 1 μ B 2 ) > 0 . It follows from (3.20) that
y n p τ n p x n p for all n N ,
(3.21)

this shows p C n for all n N , so Ω C n and C n for n N .

We want to prove C n is a closed convex set for n N . It is easy to verify that C n is closed for n N , so it suffices to verify C n is convex for n N . In fact, let v 1 , v 2 C n + 1 , for each λ ( 0 , 1 ) , we have
y n ( λ v 1 + ( 1 λ ) v 2 ) 2 = λ ( y n v 1 ) + ( 1 λ ) ( y n v 2 ) 2 = λ y n v 1 2 + ( 1 λ ) y n v 2 2 λ ( 1 λ ) v 1 v 2 2 λ τ n v 1 2 + ( 1 λ ) τ n v 2 2 λ ( 1 λ ) v 1 v 2 2 = τ n ( λ v 1 + ( 1 λ ) v 2 ) 2 ,

namely, y n ( λ v 1 + ( 1 λ ) v 2 ) τ n ( λ v 1 + ( 1 λ ) v 2 ) . Similarly, we have τ n ( λ v 1 + ( 1 λ ) v 2 ) x n ( λ v 1 + ( 1 λ ) v 2 ) , this shows λ v 1 + ( 1 λ ) v 2 C n + 1 and C n + 1 is a convex set for n N .

By (iv) of Lemma 2.2, Ω is a closed convex set, so there exists a unique element q = P Ω ( x 1 ) Ω C n . Since x n = P C n ( x 1 ) , we have x n x 1 q x 1 , which shows that { x n } is bounded. So are { τ n } and { y n } . Notice that C n + 1 C n and x n + 1 = P C n + 1 ( x 1 ) C n , then
x n + 1 x 1 x n x 1 , n 2 .
(3.22)

It follows that lim n x n x 0 exists.

For some m , n N with m > n , from x m = P C m ( x 1 ) C n and (2.1), we have
x n x m 2 + x 1 x m 2 = x n P C m ( x 1 ) 2 + x 1 P C m ( x 0 ) 2 x n x 1 2 .
(3.23)

By (3.22)-(3.23) we have lim n x n x m = 0 , so { x n } is a Cauchy sequence. Let x n x .

Next we prove x Ω . Firstly, by x n + 1 = P C n + 1 ( x 1 ) C n + 1 C n , from (3.17) we have
(3.24)
Again from (3.20), we have
( T r n g I ) A τ n 2 1 μ ( 1 μ B 2 ) { x n p 2 y n p 2 } 1 μ ( 1 μ B 2 ) x n y n { x n p + y n p } 0 .
(3.25)
So
lim n ( T r n g I ) A τ n = 0 .
(3.26)
Notice τ n = u n 1 + + u n k k , hence from (3.24) and (3.5), we obtain
lim n T r n f i x n x n = lim n u n i x n = 0 , i .
(3.27)
Since x n x , (3.27) and Lemma 2.5 imply that for r > 0 ,
T r f i x x T r f i x T r n f i x n + T r n f i x n x n + x n x x n x + | r n r | r n T r n f i x n x n + T r n f i x n x n + x n x 0 ,
which yields that x F ( T r f i ) for all i , further x i = 1 k EP ( f i ) . Since A is a bounded linear operator, A x n A x 0 by x n x . Then for r > 0 , by (3.26) and Lemma 2.5 we have
T r g A x A x T r g A x T r n g A x n + T r n g A x n A x n + A x n A x A x n A x + | r n r | r n T r n g A x n A x n + T r n g A x n A x n + A x n A x 0 ,

hence, A x F ( T r g ) = EP ( g ) for r > 0 . Thus we have proved x Ω , namely, { x n } converges strongly to x Ω . Notice (3.27), we also have { u n i } ( i ) converges strongly to x Ω .

Since τ n x n 0 by (3.24), we have τ n x by x n x . Again from (3.26) we have
lim n w n A τ n = lim n ( T r n g I ) A τ n = 0 ,

hence w n A x EP ( g ) . The proof is completed. □

Corollary 3.2 Let C be a nonempty closed convex subset of H 1 and K a nonempty closed convex subset of H 2 , where H 1 and H 2 are two real Hilbert spaces. f : C × C R is a bi-function with EP ( f ) . Let A : H 1 H 2 be a bounded linear operator with the adjoint B and g : K × K R a bi-function with EP ( g ) . Let C 1 = C and { x n } and { u n } be sequences generated by
{ x 1 C , f ( u n , y ) + 1 r n y u n , u n x n 0 , y C , g ( w n , z ) + 1 r n z w n , w n A u n 0 , z K , y n = P C ( u n + μ B ( w n A u n ) ) , C n + 1 = { v C n : y n v u n v x n v } , x n + 1 = P C n + 1 ( x 1 ) , n N ,
(3.28)

where { r n } ( 0 , + ) with lim inf n r n > 0 , μ ( 0 , 1 B 2 ) is a constant. Suppose that Ω = { p EP ( f ) : A p EP ( g ) } . Then the sequences { x n } and { u n } converge strongly to an element x Ω , while { w n } converges strongly to A x EP ( g ) .

If C = H 1 and K = H 2 in Theorem 3.1 and Theorem 3.2, we have the following corollaries.

Corollary 3.3 Let H 1 and H 2 be two real Hilbert spaces. : = { 1 , 2 , , k } denotes a finite index set. For any i , f i : H 1 × H 1 R is a bi-function with i = 1 k EP ( f i ) . Let A : H 1 H 2 be a bounded linear operator with the adjoint B and g : H 2 × H 2 R a bi-function with EP ( g ) . Let { x n } and { u n i } be sequences generated by
{ x 1 H 1 , f i ( u n i , y ) + 1 r n y u n i , u n i x n 0 , y H 1 , i , τ n = u n 1 + + u n k k , g ( w n , z ) + 1 r n z w n , w n A τ n 0 , z H 2 , x n + 1 = τ n + μ B ( w n A τ n ) , n N ,
(3.29)

where { r n } ( 0 , + ) with lim inf n r n > 0 , μ ( 0 , 1 B 2 ) is a constant. Suppose that Ω = { p i = 1 k EP ( f i ) : A p EP ( g ) } . Then the sequences { x n } and { u n i } ( i ) converge weakly to an element p Ω , while { w n } converges weakly to A p EP ( g ) .

Corollary 3.4 Let H 1 and H 2 be two real Hilbert spaces. f : H 1 × H 1 R is a bi-function with EP ( f ) . Let A : H 1 H 2 be a bounded linear operator with the adjoint B and g : H 2 × H 2 R a bi-function with EP ( g ) . Let { x n } and { u n } be sequences generated by
{ x 1 H 1 , f ( u n , y ) + 1 r n y u n , u n x n 0 , y H 1 , g ( w n , z ) + 1 r n z w n , w n A u n 0 , z H 2 , x n + 1 = u n + μ B ( w n A u n ) , n N ,
(3.30)

where { r n } ( 0 , + ) with lim inf n r n > 0 , μ ( 0 , 1 B 2 ) is a constant. Suppose that Ω = { p EP ( f ) : A p EP ( g ) } . Then the sequences { x n } and { u n } converge weakly to an element p Ω , while { w n } converges weakly to A p EP ( g ) .

Corollary 3.5 Let H 1 and H 2 be two real Hilbert spaces. : = { 1 , 2 , , k } denotes a finite index set. For any i , f i : H 1 × H 1 R is a bi-function with i = 1 k EP ( f i ) . Let A : H 1 H 2 be a bounded linear operator with the adjoint B and g : H 2 × H 2 R a bi-function with EP ( g ) . Let C 1 = H 1 and { x n } and { u n i } ( i ) be sequences generated by
{ x 1 H 1 , f i ( u n i , y ) + 1 r n y u n i , u n i x n 0 , y H 1 , i , τ n = u n 1 + + u n k k , g ( w n , z ) + 1 r n z w n , w n A τ n 0 , z H 2 , y n = τ n + μ B ( w n A τ n ) , C n + 1 = { v C n : y n v τ n v x n v } , x n + 1 = P C n + 1 ( x 1 ) , n N ,
(3.31)

where { r n } ( 0 , + ) with lim inf n r n > 0 , μ ( 0 , 1 B 2 ) is a constant. Suppose that Ω = { p i = 1 k EP ( f i ) : A p EP ( g ) } . Then the sequences { x n } and { u n i } ( i ) converge strongly to an element p Ω , while { w n } converges strongly to A x EP ( g ) .

Corollary 3.6 Let H 1 and H 2 be two real Hilbert spaces. f : H 1 × H 1 R is a bi-function with EP ( f ) . Let A : H 1 H 2 be a bounded linear operator with the adjoint B and g : H 2 × H 2 R a bi-function with EP ( g ) . Let C 1 = H 1 and { x n } and { u n } be sequences generated by
{ x 1 H 1 , f ( u n , y ) + 1 r n y u n , u n x n 0 , y H 1 , g ( w n , z ) + 1 r n z w n , w n A u n 0 , z H 2 , y n = u n + μ B ( w n A u n ) , C n + 1 = { v C n : y n v u n v x n v } , x n + 1 = P C n + 1 ( x 1 ) , n N ,
(3.32)

where { r n } ( 0 , + ) with lim inf n r n > 0 , μ ( 0 , 1 B 2 ) is a constant. Suppose that Ω = { p EP ( f ) : A p EP ( g ) } . Then the sequences { x n } and { u n } converge strongly to an element x Ω , while { w n } converges strongly to A x EP ( g ) .

Remark 3.1 Since Example 1.1 and Example 1.2 satisfy the conditions of Corollary 3.1 and Corollary 3.2, the SEPs in Example 1.1 and Example 1.2 can be solved by the algorithm (3.16) and (3.28).

Remark 3.2 The results of this paper provide some solution algorithms for some SEPs; however, there are still some SEPs which cannot be solved by the results of this paper. The following examples belong to the case.

Example 3.1 Let H 2 = R and H 1 = R 2 with the norm z = ( x 2 + y 2 ) 1 2 for some z = ( x , y ) R 2 . K : = [ 1 , + ) and C : = { z = ( x , y ) R 2 | y x 1 } . Define a bi-function f ( w , z ) = x 1 + y 1 + y 2 x 2 , where w = ( x 1 , y 1 ) , z = ( x 2 , y 2 ) C , then f is a bi-function from C × C into R with EP ( f ) = { w = ( x , y ) | y x 1 , x + y 1 } . For each z = ( x , y ) H 1 , let A z = y x , then A is a bounded linear operator from H 1 into H 2 . Now define another bi-function g as follows: g ( u , v ) = v u for all u , v K . Then g is a bi-function from K × K into R with EP ( g ) = { 1 } .

Clearly, when p = ( x , y ) EP ( f ) with y x = 1 and x + y 1 , we have A p = 1 EP ( g ) . So Γ = { p EP ( f ) : A p EP ( g ) } . However, because the bi-function f does not satisfy the conditions (A1)-(A4), the SEP in this example cannot be solved by Corollary 3.1 or Corollary 3.2.

Example 3.2 Let H 1 , H 2 , A and B be the same as Example 2.2. Let C : = { α = ( a 1 , a 2 ) H 1 | a 2 a 1 1 } and K : = { γ = ( c 1 , c 2 , c 3 ) H 2 | γ 2 } . Define a bi-function f ( α , β ) = ( b 2 b 1 ) 2 ( a 1 + a 2 ) 2 , where α = ( a 1 , a 2 ) , β = ( b 1 , b 2 ) C , then f is a bi-function from C × C into R with EP ( f ) = { p = ( p 1 , p 2 ) | p 2 p 1 1 p 1 + p 2 1 } . Define another bi-function g ( γ , η ) = c 2 2 + c 3 2 ( c 1 2 + d 1 2 + d 2 2 + d 3 2 ) , where γ = ( c 1 , c 2 , c 3 ) , η = ( d 1 , d 2 , d 3 ) K , then EP ( g ) = { u = ( 0 , u 2 , u 3 ) K | u 2 2 + u 3 2 = 2 } .

Clearly, when p = ( 1 , 0 ) EP ( f ) , we have A p = ( 0 , 1 , 1 ) EP ( g ) . So Γ = { p EP ( f ) : A p EP ( g ) } . However, since all f and g do not satisfy the conditions (A1)-(A4), we cannot use the results obtained in this paper to solve the SEP in this example.

4 Conclusion

There are still many SEPs which do not satisfy the conditions (A1)-(A4), so we need to develop some new methods to solve these problems in the future.

Declarations

Acknowledgement

The author was supported by the Natural Science Foundation of Yunnan Province (2010ZC152).

Authors’ Affiliations

(1)
Department of Mathematics, Honghe University

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