Open Access

On hybrid split problem and its nonlinear algorithms

Fixed Point Theory and Applications20132013:47

https://doi.org/10.1186/1687-1812-2013-47

Received: 23 August 2012

Accepted: 18 February 2013

Published: 7 March 2013

Abstract

In this paper, we study a hybrid split problem (HSP for short) for equilibrium problems and fixed point problems of nonlinear operators. Some strong and weak convergence theorems are established.

MSC:47J25, 47H09, 65K10.

Keywords

fixed point problem equilibrium problem hybrid split problem iterative algorithm strong (weak) convergence theorem

1 Introduction

Throughout this paper, we assume that H is a real Hilbert space with zero vector θ, whose inner product and norm are denoted by , and , respectively. Let C be a nonempty subset of H and T : C H be a mapping. Denote by F ( T ) the set of fixed points of T. The symbols and are used to denote the sets of positive integers and real numbers, respectively.

Let H be a Hilbert space and C be a closed convex subset of H. Let f : C × C R be a bi-function. The classical equilibrium problem (EP for short) is defined as follows.
Find  p C  such that  f ( p , y ) 0 , y C .
(EP)
The symbol E P ( f ) is used to denote the set of all solutions of the problem (EP), that is,
E P ( f ) = { u K : f ( u , v ) 0 , v K } .

It is known that the problem (EP) contains optimization problems, complementary problems, variational inequalities problems, saddle point problems, fixed point problems, bilevel problems, semi-infinite problems and others as special cases and have many applications in physics and economics problems; for detail, one can refer to [13] and references therein.

In last ten years or so, the problem (EP) has been generalized and improved to find a common element of the set of fixed points of a nonlinear operator and the set of solutions of the problem (EP). More precisely, many authors have studied the following problem (FTEP) (see, for instance, [49]):
Find  p C  such that  T p = p  and  f ( p , y ) 0 , y C ,
(FTEP)

where C is a closed convex subset of a Hilbert space H, f : C × C R is a bi-function and T : C C is a nonlinear operator.

In this paper, motivated by the problems (EP) and (FTEP), we study the following mathematical model about a hybrid split problem for equilibrium problems and fixed point problems of nonlinear operators (HSP for short).

Let E 1 and E 2 be two real Banach spaces. Let C be a closed convex subset of E 1 and K be a closed convex subset of E 2 . Let f : C × C R and g : K × K R be two bi-functions, A : E 1 E 2 be a bounded linear operator. Let T : C C and S : K K be two nonlinear operators with F ( T ) and F ( S ) . The mathematical model about a hybrid split problem for equilibrium problems and fixed point problems of nonlinear operators (HSP for short) is defined as follows:
Find  p C  such that  T p = p , f ( p , y ) 0 , y C , and u : = A p  satisfying  S u = u K , g ( u , v ) 0 , v K .
(HSP)

In fact, (HSP) contains several important problems as special cases. We give some examples to explain about it.

Example A If T is an identity operator on C, then (HSP) will reduce to the following problem (P1):

(P1) Find p C such that f ( p , y ) 0 , y C , and u : = A p satisfying S u = u K , g ( u , v ) 0 , v K .

Example B If S is an identity operator on K, then (HSP) will reduce to the following problem (P2):

(P2) Find p C such that T p = p , f ( p , y ) 0 , y C , and u : = A p K satisfying g ( u , v ) 0 , v K .

Example C If T, S are all identity operators, then (HSP) will reduce to the following split equilibrium problem (P3) which has been considered in [10]:

(P3) Find p C such that f ( p , y ) 0 , y C , and u : = A p K satisfying g ( u , v ) 0 , v K .

Example D If S is an identity operator and f ( x , y ) 0 for all ( x , y ) C × C , then (HSP) will reduce to the following problem (P4) which has been studied in [11]:

(P4) Find p C such that T p = p and u : = A p K satisfying g ( u , v ) 0 , v K .

In this paper, we introduce some new iterative algorithms for (HSP) and some strong and weak convergence theorems for (HSP) will be established.

2 Preliminaries

In what follows, the symbols and → will symbolize weak convergence and strong convergence as usual, respectively. 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 any Hilbert space satisfies Opial’s condition. Let K be a nonempty subset of real Hilbert spaces H. Recall that a mapping T : K K is said to be nonexpansive if T x T y x y for all x , y K .

Let H 1 and H 2 be two Hilbert spaces. Let A : H 1 H 2 and B : H 2 H 1 be two bounded linear operators. B is called the adjoint operator (or adjoint) of A if for all z H 1 , w H 2 , B satisfies A z , w = z , B w . 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 B is an adjoint operator of A, then A = B .

Example 2.1 ([10])

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 2.2 ([10])

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 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 characteristics:
  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)
     

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 satisfying the following conditions:
  1. (A1)

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

     
  2. (A2)

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

     
  3. (A3)

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

     
  4. (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 [12])

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 ) = E P ( F ) for r > 0 and E P ( F ) is closed and convex;

     
  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 .

     

Lemma 2.3 (see, e.g., [6])

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

    x + y 2 y 2 + 2 x , x + y ;

     
  2. (b)

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

     
  3. (c)

    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 ] .

     

Lemma 2.4 Let F r F be the same as in Lemma  2.2. If F ( T r F ) = E P ( 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 .

Proof

By (ii) of Lemma 2.2 and (b) of Lemma 2.3,
T r F x x 2 T r F x x , x x = 1 2 ( T r F x x 2 + x x 2 T r F x x 2 ) ,

which shows that T r F x x 2 x x 2 T r F x x 2 . □

Lemma 2.5 ([10, 11])

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 .

Remark 2.1 In fact, Lemma 2.5 is motivated by a proof of [[5], Theorem 3.2]. In order to the sake of convenience for proving, we restated the fact and gave its proof in Lemma 2.5 [10, 11].

Lemma 2.6 ([13])

Let { a n } be a nonnegative real sequence satisfying the following condition:
a n + 1 ( 1 λ n ) a n + λ n b n , n n 0 ,
where n 0 is some nonnegative integer, { λ n } is a sequence in ( 0 , 1 ) and { b n } is a sequence in  R such that
  1. (i)

    n = 0 λ n = ;

     
  2. (ii)

    lim sup n b n 0 or n = 0 λ n b n is convergent. Then lim n a n = 0 .

     

Lemma 2.7 ([14])

Let { x n } and { y n } be bounded sequences in a Banach space E and let { β n } be a sequence in [ 0 , 1 ] with 0 < lim inf β n lim sup β n < 1 . Suppose x n + 1 = β n y n + ( 1 β n ) x n for all integers n 0 and lim sup n ( y n + 1 y n x n + 1 x n ) 0 , then lim n y n x n = 0 .

3 Weak convergence iterative algorithms for (HSP)

In this section, we will introduce some weak convergence iterative algorithms for the hybrid split problem.

Theorem 3.1 Let H 1 and H 2 be two real Hilbert spaces. Let C H 1 and K H 2 be two nonempty closed convex sets. Let T : C C and S : K K be non-expansive mappings and f : C × C R and g : K × K R be bi-functions satisfying the conditions (A1)-(A4). Let A : H 1 H 2 be a bounded linear operator with its adjoint B. Let x 1 C , { x n } and { u n } be sequences generated by
{ u n = T r n f x n , y n = ( 1 α ) u n + α T u n , w n = T r n g A y n , x n + 1 = P C ( y n + ξ B ( S w n A y n ) ) , n N ,
(3.1)

where α ( 0 , 1 ) , ξ ( 0 , 1 B 2 ) and { r n } ( 0 , + ) with lim inf n + r n > 0 , P C is a projection operator from H 1 into C. Suppose that Ω = { p F ( T ) E P ( f ) : A p F ( S ) E P ( g ) } , then x n , u n q Ω and w n A q F ( S ) E P ( g ) .

Proof Let p Ω , the following several inequalities can be proved easily:
y n p u n p x n p , w n A p A y n A p .
(3.2)
By Lemma 2.4, T r n g A y n A y n 2 A y n A p 2 T r n g A y n A p 2 , hence
S w n A p 2 = S T r n g A y n A p 2 T r n g A y n A p 2 A y n A p 2 T r n g A y n A y n 2 .
(3.3)
By (b) of Lemma 2.3 and (3.3), for each n N , we have
(3.4)
On the other hand, B ( S T r n g I ) A y n 2 B 2 ( S T r n g I ) A y n 2 , so from (3.1)-(3.4), we have
x n + 1 p 2 = P C ( y n + ξ B ( S T r n g I ) A y n ) p 2 y n + ξ B ( S T r n g I ) A y n p 2 = y n p 2 + ξ B ( S T r n g I ) A y n 2 + 2 ξ y n p , B ( S T r n g I ) A y n y n p 2 + ξ 2 B 2 ( S T r n g I ) A y n 2 ξ ( S T r n g I ) A y n 2 ξ ( T r n g I ) A y n 2 = y n p 2 ξ ( 1 ξ B 2 ) ( S T r n f I ) A y n 2 ξ ( T r n g I ) A y n 2 x n p 2 ξ ( 1 ξ B 2 ) ( S T r n f I ) A y n 2 ξ ( T r n g I ) A y n 2 .
(3.5)
Since ξ ( 0 , 1 B 2 ) , ξ ( 1 ξ B 2 ) > 0 , by (3.2) and (3.5), we have
x n + 1 p y n p u n p x n p
(3.6)
and
ξ ( 1 ξ B 2 ) ( S T r n g I ) A y n 2 + ξ ( T r n g I ) A y n 2 x n p 2 x n + 1 p 2 .
(3.7)
The inequality (3.6) implies lim n x n p exists. Further, from (3.6) and (3.7), we get
lim n x n p = lim n y n p = lim n u n p , lim n ( S T r n g I ) A y n = lim n ( T r n g I ) A y n = lim n w n A y n = 0 .
(3.8)
The inequality (3.8) also implies that
lim n S w n w n = 0 .
(3.9)
Using Lemma 2.4 and (3.8), we have
u n x n 2 = T r n f x n x n 2 x n p 2 T r n f x n p 2 = x n p 2 u n p 2 0 .
(3.10)
Notice that
y n p 2 = ( 1 α ) u n p 2 + α T u n p 2 α ( 1 α ) T u n u n 2 u n p 2 α ( 1 α ) T u n u n 2 ,
hence,
lim n T u n u n = 0 .
(3.11)
From (3.10) and (3.11), we also have
y n x n y n u n + u n x n = α T u n u n + u n x n 0 as  n .
(3.12)

The existence of lim n x n p implies that { x n } is bounded, hence { x n } has a weak convergence subsequence { x n j } . Assume that x n j q for some q C , then y n j q , A y n j A q K and w n j = T r n j g A y n j A q by (3.12) and (3.8).

We say q Ω , in other words, q F ( T ) E P ( f ) and A q F ( S ) E P ( g ) . By (3.10), we also obtain u n j q . If T q q , then, by Opial’s condition and (3.11), we get
lim inf j u n j q < lim inf j u n j T q lim inf j u n j T u n j + T u n j T q lim inf j u n j q ,
which is a contradiction. Hence T q = q or q F ( T ) . On the other hand, from Lemma 2.2, we know E P ( f ) = F ( T r f ) for any r > 0 . Hence, if T r f q q for r > 0 , then by Opial’s condition and (3.10) and Lemma 2.5, we have
lim inf j x n j q < lim inf j x n j T r f q = lim inf j x n j T r n j f x n j + T r n j f x n j T r f q lim inf j { x n j T r n j f x n j + T r f q T r n j f x n j } lim inf j { x n j T r n j f x n j + x n j q + | r r n j | r n j T r n j f x n j x n j } = lim inf j x n j q ,

which is also a contradiction. So, for each r > 0 , T r f q = q , namely q E P ( f ) . Thus, we have proved q F ( T ) E P ( f ) . Similarly, we can also prove A q F ( S ) E P ( g ) . Hence, q Ω .

Finally, we prove { x n } converges weakly to q Ω . Otherwise, if there exists another subsequence of { x n } , which is denoted by { x n l } , such that x n l x ¯ Ω with x ¯ q , then by Opial’s condition,
lim inf l x n l x ¯ < lim inf l x n l q = lim inf j x n j q < lim inf l x n l x ¯ .

This is a contradiction. Hence { x n } converges weakly to an element q Ω . Together with u n x n 0 (see (3.10)), we also get u n q .

Finally, we prove { w n = T r n g A y n } converges weakly to A q F ( S ) E P ( g ) . From (3.12), we have y n q , so A y n A q . Thus, from (3.8) we have w n = T r n g A y n A q F ( S ) E P ( g ) . The proof is completed. □

If T = I or S = I , where I denotes an identity operator, then the following corollaries follow from Theorem 3.1.

Corollary 3.1 Let H 1 and H 2 be two real Hilbert spaces. Let C H 1 and K H 2 be two nonempty closed convex sets. Let S : K K be a non-expansive mapping and f : C × C R and g : K × K R be bi-functions satisfying the conditions (A1)-(A4). Let A : H 1 H 2 be a bounded linear operator with its adjoint B. Let x 1 C , { x n } and { u n } be sequences generated by
{ u n = T r n f x n , w n = T r n g A u n , x n + 1 = P C ( u n + ξ B ( S w n A u n ) ) , n N ,

where ξ ( 0 , 1 B 2 ) and { r n } ( 0 , + ) with lim inf n + r n > 0 , P C is a projection operator from H 1 into C. Suppose that Ω = { p E P ( f ) : A p F ( S ) E P ( g ) } , then x n , u n q Ω and w n A q F ( S ) E P ( g ) .

Corollary 3.2 Let H 1 and H 2 be two real Hilbert spaces. Let C H 1 and K H 2 be two nonempty closed convex sets. Let T : C C be a non-expansive mapping and f : C × C R and g : K × K R be bi-functions satisfying the conditions (A1)-(A4). Let A : H 1 H 2 be a bounded linear operator with its adjoint B. Let x 1 C , { x n } and { u n } be sequences generated by
{ u n = T r n f x n , y n = ( 1 α ) u n + α T u n , w n = T r n g A y n , x n + 1 = P C ( y n + ξ B ( w n A y n ) ) , n N ,

where α ( 0 , 1 ) , ξ ( 0 , 1 B 2 ) and { r n } ( 0 , + ) with lim inf n + r n > 0 , P C is a projection operator from H 1 into C. Suppose that Ω = { p F ( T ) E P ( f ) : A p E P ( g ) } , then x n , u n q Ω and w n A q E P ( g ) .

Corollary 3.3 Let C H 1 and K H 2 be two nonempty closed convex sets. Let f : C × C R and g : K × K R be bi-functions satisfying the conditions (A1)-(A4). Let A : H 1 H 2 be a bounded linear operator with its adjoint B. Let x 1 C , { x n } and { u n } be sequences generated by
{ u n = T r n f x n , w n = T r n g A u n , x n + 1 = P C ( u n + ξ B ( w n A u n ) ) , n N ,

where ξ ( 0 , 1 B 2 ) and { r n } ( 0 , + ) with lim inf n + r n > 0 , P C is a projection operator from H 1 into C. Suppose that Ω = { p E P ( f ) : A p E P ( g ) } , then x n , u n q Ω and w n A q E P ( g ) .

4 Strong convergence iterative algorithms for (HSP)

In this section, we introduce two strong convergence algorithms for (HSP); see Theorem 4.1 and Theorem 4.2.

Theorem 4.1 Let H 1 and H 2 be two real Hilbert spaces. Let C H 1 and K H 2 be two nonempty closed convex sets. Let T : C C and S : K K be non-expansive mappings and f : C × C R and g : K × K R be bi-functions satisfying the conditions (A1)-(A4). Let A : H 1 H 2 be a bounded linear operator with its adjoint B. Let x 1 C 1 : = C , { x n } and { u n } be sequences generated by
{ u n = T r n f x n , y n = ( 1 α ) u n + α T u n , w n = T r n g A y n , z n = P C ( y n + ξ B ( S w n A y n ) ) , C n + 1 = { v C n : z n v y n v x n v } , x n + 1 = P C n + 1 ( x 1 ) , n N ,
(4.1)

where α ( 0 , 1 ) , ξ ( 0 , 1 B 2 ) and { r n } ( 0 , + ) with lim inf n + r n > 0 , P C is a projection operator from H 1 into C. Suppose that Ω = { p F ( T ) E P ( f ) : A p F ( S ) E P ( g ) } , then x n , u n q Ω and w n A q F ( S ) E P ( g ) .

Proof We claim that C n is a nonempty closed convex set for n N . In fact, let p Ω , it follows from (3.4) that
2 ξ y n p , B ( S w n A y n ) ξ ( T r n g I ) A x n 2 ξ S w n A y n 2 .
(4.2)
By (3.2), (4.1) and (4.2), we obtain
z n p 2 y n + ξ B ( S w n A y n ) p 2 = y n p 2 + ξ B ( S w n A y n ) 2 + 2 ξ y n p , B ( S w n A y n ) y n p 2 + ξ 2 B 2 S w n A y n 2 ξ ( T r n g I ) A y n 2 ξ S w n A y n 2 = y n p 2 ξ ( 1 ξ B 2 ) ( S T r n g I ) A y n 2 ξ ( T r n g I ) A y n 2 u n p 2 ( 1 α ) α u n T u n 2 ξ ( 1 ξ B 2 ) ( S T r n g I ) A y n 2 ξ ( T r n g I ) A y n 2 x n p 2 ξ ( 1 ξ B 2 ) ( S T r n g I ) A y n 2 ξ ( T r n g I ) A y n 2 ( 1 α ) α u n T u n 2 .
(4.3)
Notice ξ ( 0 , 1 B 2 ) , ξ ( 1 ξ B 2 ) > 0 . It follows from (4.3) that
z n p y n p u n p x n p for all  n N ,

hence p C n , which yields that Ω C n and C n for n N .

It is not hard to verify that C n is closed for n N , so it suffices to verify C n is convex for n N . Indeed, let w 1 , w 2 C n + 1 and γ [ 0 , 1 ] , we have

namely z n ( γ w 1 + ( 1 γ ) w 2 ) y n ( γ w 1 + ( 1 γ ) w 2 ) . Similarly, y n ( γ w 1 + ( 1 γ ) w 2 ) x n ( γ w 1 + ( 1 γ ) w 2 ) , which implies γ w 1 + ( 1 γ ) w 2 C n + 1 and C n + 1 is a convex set, n 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 for n > 1 . It follows that lim n x n x 1 exists. Hence { x n } is bounded, which yields that { z n } and { y n } are bounded. For some k , n N with k > n > 1 , from x k = P C k ( x 1 ) C n and (2.1), we have
x n x k 2 + x 1 x k 2 = x n P C k ( x 1 ) 2 + x 1 P C k ( x 1 ) 2 x n x 1 2 .
(4.4)

By lim n x n x 1 exists and (4.4), we have lim n x n x k = 0 , so { x n } is a Cauchy sequence.

Let x n q , then q Ω . Firstly, by x n + 1 = P C n + 1 ( x 1 ) C n + 1 C n , from (4.1) we have
z n x n z n x n + 1 + x n + 1 x n 2 x n + 1 x n 0 , y n x n y n x n + 1 + x n + 1 x n 2 x n + 1 x n 0 .
(4.5)
Setting ρ = ξ ( 1 ξ B 2 ) , by (4.3) again, we have
(4.6)
So,
lim n T u n u n = 0 , lim n w n A y n = lim n ( T r n g I ) A y n = 0 , lim n S w n A y n = lim n ( S T r n g I ) A y n = 0 , lim n S w n w n = 0 .
(4.7)
Notice that lim n T u n u n = 0 and y n u n = α T u n u n , so
lim n y n u n = 0 .
(4.8)
Further, from (4.5) and (4.8),
lim n x n u n = 0 .
(4.9)
Since x n q , we have u n q by (4.9). Thus
T q q T q T u n + T u n u n + u n q 0 ,
namely T q = q and q F ( T ) . On the other hand, for r > 0 , by Lemma 2.5, we have
T r f q q T r f q T r n f x n + T r n f x n x n + x n q x n q + | r n r | r n T r n f x n x n + T r n f x n x n + x n q 0 ,

which yields q F ( T r f ) = E P ( f ) . We have verified q F ( T ) E P ( f ) .

Next, we prove A q F ( S ) E P ( g ) . Since x n q and x n y n 0 by (4.8) and (4.9) and w n A y n 0 by (4.7), we have y n q and A y n A q and w n A q . So,
S A q A q S A q S w n + S w n w n + w n A q 0 ,
namely S A q = A q and A q F ( S ) . On the other hand, for r > 0 , by Lemma 2.5 again, we have
T r g A q A q T r g A q T r n g A y n + T r n g A y n A y n + A y n A q A y n A q + | r n r | r n T r n g A y n A y n + T r n g A y n A y n + A y n A q 0 ,

which implies that A q F ( T r g ) = E P ( g ) . We have verified A q F ( S ) E P ( g ) .

So, we have obtained q Ω and x n , u n q and w n A q , the proof is completed. □

If T = I or S = I , where I denotes an identity operator, then the following corollaries follow from Theorem 4.1.

Corollary 4.1 Let H 1 and H 2 be two real Hilbert spaces. Let C H 1 and K H 2 be two nonempty closed convex sets. Let f : C × C R and g : K × K R be bi-functions satisfying the conditions (A1)-(A4) and S : K K be a non-expansive mapping. Let A : H 1 H 2 be a bounded linear operator with its adjoint B. Let x 1 C 1 : = C , { x n } and { u n } be sequences generated by
{ u n = T r n f x n , w n = T r n g A u n , z n = P C ( u n + ξ B ( S w n A u n ) ) , C n + 1 = { v C n : z n v u n v x n v } , x n + 1 = P C n + 1 ( x 1 ) , n N ,

where ξ ( 0 , 1 B 2 ) and { r n } ( 0 , + ) with lim inf n + r n > 0 , P C is a projection operator from H 1 into C. Suppose that Ω = { p E P ( f ) : A p F ( S ) E P ( g ) } , then x n , u n q Ω and w n A q F ( S ) E P ( g ) .

Corollary 4.2 Let H 1 and H 2 be two real Hilbert spaces. Let C H 1 and K H 2 be two nonempty closed convex sets. Let T : C C be a non-expansive mapping and f : C × C R and g : K × K R be bi-functions satisfying the conditions (A1)-(A4). Let A : H 1 H 2 be a bounded linear operator with its adjoint B. Let x 1 C 1 : = C , { x n } and { u n } be sequences generated by
{ u n = T r n f x n , y n = ( 1 α ) u n + α T u n , w n = T r n g A y n , z n = P C ( y n + ξ B ( w n A y n ) ) , C n + 1 = { v C n : z n v y n v x n v } , x n + 1 = P C n + 1 ( x 1 ) , n N ,

where, α ( 0 , 1 ) , ξ ( 0 , 1 B 2 ) and { r n } ( 0 , + ) with lim inf n + r n > 0 , P C is a projection operator from H 1 into C. Suppose that Ω = { p F ( T ) E P ( f ) : A p E P ( g ) } , then x n , u n q Ω and w n A q E P ( g ) .

Corollary 4.3 Let H 1 and H 2 be two real Hilbert spaces. Let C H 1 and K H 2 be two nonempty closed convex sets. Let f : C × C R and g : K × K R be bi-functions satisfying the conditions (A1)-(A4). Let A : H 1 H 2 be a bounded linear operator with its adjoint B. Let x 1 C 1 : = C , { x n } and { u n } be sequences generated by
{ u n = T r n f x n , w n = T r n g A u n , z n = P C ( y n + ξ B ( w n A u n ) ) , C n + 1 = { v C n : z n v u n v x n v } , x n + 1 = P C n + 1 ( x 1 ) , n N ,

where ξ ( 0 , 1 B 2 ) and { r n } ( 0 , + ) with lim inf n + r n > 0 , P C is a projection operator from H 1 into C. Suppose that Ω = { p E P ( f ) : A p E P ( g ) } , then x n , u n q Ω and w n A q E P ( g ) .

It is well known that the viscosity iterative method is always applied to study the iterative solution for the fixed point problem of nonlinear operators, for example, [5, 6, 8, 15, 16]. Similarly, the viscosity iterative method can also be used to study the hybrid split problem (HSP). So, at the end of this paper, we introduce a viscosity iterative algorithm which can converge strongly to a solution of (HSP).

Theorem 4.2 Let H 1 and H 2 be two real Hilbert spaces. Let C H 1 and K H 2 be two nonempty closed convex sets. Let h : C C be a α-contraction mapping, T : C C and S : K K be non-expansive mappings and f : C × C R and g : K × K R be bi-functions satisfying the conditions (A1)-(A4). Let A : H 1 H 2 be a bounded linear operator with its adjoint B. Let x 1 C , { x n } and { u n } be sequences generated by
{ u n = T r n f x n , w n = T r n g A u n , y n = P C ( u n + ξ B ( S w n A u n ) ) , z n = ( 1 r ) x n + r T y n , x n + 1 = α n h ( x n ) + ( 1 α n ) z n , n N ,
(4.10)
where r ( 0 , 1 ) , ξ ( 0 , 1 B 2 ) and { r n } ( 0 , + ) , P C is a projection operator from H 1 into C, and the coefficients { α n } and { r n } satisfy the following conditions:
  1. (1)

    { α n } ( 0 , 1 ) , lim n α n = 0 , n = 1 α n = ;

     
  2. (2)

    lim inf n + r n > 0 , lim n | r n + 1 r n | = 0 .

     

Suppose that Ω = { p F ( T ) E P ( f ) : A p F ( S ) E P ( g ) } , then x n , u n q Ω and w n A q F ( S ) E P ( g ) , where q = P Ω h ( q ) .

Proof Let p Ω . The following inequalities are easily verified:
u n p x n p , w n A p A u n A p .
(4.11)
By Lemma 2.4,
(4.12)
From (4.10) and (4.12), we have
(4.13)
and
y n p 2 = P C ( u n + ξ B ( S w n A u n ) P C p 2 u n p + ξ B ( S w n A u n ) 2 = u n p 2 + ξ B ( S w n A u n ) 2 + 2 ξ u n p , B ( S w n A u n ) u n p 2 ξ ( 1 ξ B 2 ) S w n A u n 2 ξ T r n g A u n A u n 2 x n p 2 ξ ( 1 ξ B 2 ) S w n A u n 2 ξ T r n g A u n A u n 2 = x n p 2 ξ ( 1 ξ B 2 ) S w n A u n 2 ξ w n A u n 2 .
(4.14)
So, from (4.10)-(4.11) and (4.14), we have
y n p u n p x n p , z n p x n p .
(4.15)
We say { x n } is bounded. In fact, from (4.10) and (4.15), we have
x n + 1 p = α n ( f ( x n ) p ) + ( 1 α n ) ( z n p ) ( 1 α n ) z n p + α n f ( x n ) p ( 1 α n ) x n p + α n α x n p + α n f ( p ) p = ( 1 α n ( 1 α ) ) x n p + α n ( 1 α ) f ( p ) p 1 α ,
which implies that
x n p max { x 1 p , f ( p ) p 1 α } , n N ,
(4.16)

so { x n } is bounded. Further, { u n } , { w n } and { y n } are also bounded by (4.11).

By Lemma 2.5, from (4.10) we have
u n + 1 u n 2 = T r n + 1 f x n + 1 T r n f x n 2 u n + 1 u n 2 ( x n + 1 x n + | r n r n + 1 | r n T r n f x n x n ) 2 u n + 1 u n 2 x n + 1 x n 2 + | r n r n + 1 | r n M 1 , w n + 1 w n 2 = T r n + 1 g A u n + 1 T r n g A u n 2 w n + 1 w n 2 ( A u n + 1 A u n + | r n r n + 1 | r n T r n g A u n A u n ) 2 w n + 1 w n 2 A u n + 1 A u n 2 + | r n r n + 1 | r n M 1
(4.17)
and
y n + 1 y n 2 u n + 1 + ξ B ( S w n + 1 A u n + 1 ) u n ξ B ( S w n A u n ) 2 = u n + 1 u n + ξ B ( S w n + 1 A u n + 1 ( S w n A u n ) ) 2 = u n + 1 u n 2 + ξ B ( S w n + 1 A u n + 1 ( S w n A u n ) ) 2 + 2 ξ u n + 1 u n , B ( S w n + 1 A u n + 1 ( S w n A u n ) ) u n + 1 u n 2 + ξ 2 B 2 S w n + 1 A u n + 1 ( S w n A u n ) 2 + 2 ξ A ( u n + 1 u n ) , S w n + 1 A u n + 1 ( S w n A u n ) = u n + 1 u n 2 + ξ 2 B 2 S w n + 1 A u n + 1 ( S w n A u n ) 2 + 2 ξ A ( u n + 1 u n ) + S w n + 1 A u n + 1 ( S w n A u n ) , S w n + 1 A u n + 1 ( S w n A u n ) 2 ξ S w n + 1 A u n + 1 ( S w n A u n ) , S w n + 1 A u n + 1 ( S w n A u n ) = u n + 1 u n 2 + ξ 2 B 2 S w n + 1 A u n + 1 ( S w n A u n ) 2 + 2 ξ S w n + 1 S w n , S w n + 1 A u n + 1 ( S w n A u n ) 2 ξ S w n + 1 A u n + 1 ( S w n A u n ) 2 = u n + 1 u n 2 + ξ 2 B 2 S w n + 1 A u n + 1 ( S w n A u n ) 2 + 2 ξ 1 2 { S w n + 1 S w n 2 + S w n + 1 A u n + 1 ( S w n A u n ) 2 A u n + 1 A u n 2 } 2 ξ S w n + 1 A u n + 1 ( S w n A u n ) 2 = u n + 1 u n 2 + ξ 2 B 2 S w n + 1 A u n + 1 ( S w n A u n ) 2 + ξ { S w n + 1 S w n 2 A u n + 1 A u n 2 } ξ S w n + 1 A u n + 1 ( S w n A u n ) 2 u n + 1 u n 2 ξ ( 1 ξ B 2 ) S w n + 1 A u n + 1 ( S w n A u n ) 2 + ξ { w n + 1 w n 2 A u n + 1 A u n 2 } u n + 1 u n 2 ξ ( 1 ξ B 2 ) S w n + 1 A u n + 1 ( S w n A u n ) 2 + ξ { A u n + 1 A u n 2 + | r n r n + 1 | r n M 1 A u n + 1 A u n 2 } = u n + 1 u n 2 ξ ( 1 ξ B 2 ) S w n + 1 A u n + 1 ( S w n A u n ) 2 + ξ | r n r n + 1 | r n M 1 x n + 1 x n 2 ξ ( 1 ξ B 2 ) S w n + 1 A u n + 1 ( S w n A u n ) 2 + | r n r n + 1 | r n ( ξ M 1 + M 1 ) ,
(4.18)
where M 1 is a constant satisfying
Proving x n + 1 x n 0 as n . Setting β n = 1 ( 1 α n ) ( 1 r ) and v n = x n + 1 x n + β n x n β n , namely v n = α n f ( x n ) + ( 1 α n ) r T y n β n . Let M 2 be a constant satisfying sup n N { f ( x n + 1 ) β n + 1 , f ( x n ) β n , T y n } M 2 for all n N . Then
v n + 1 v n = α n + 1 f ( x n + 1 ) + ( 1 α n + 1 ) r T y n + 1 β n + 1 α n f ( x n ) + ( 1 α n ) r T y n β n α n + 1 f ( x n + 1 ) β n + 1 + α n f ( x n ) β n + r ( 1 α n + 1 ) T y n + 1 β n + 1 ( 1 α n ) T y n β n ( α n + 1 + α n ) M 2 + r ( 1 α n + 1 ) ( T y n + 1 T y n ) β n + 1 + ( 1 α n + 1 ) T y n β n + 1 ( 1 α n ) T y n β n ( α n + 1 + α n ) M 2 + r ( 1 α n + 1 ) y n + 1 y n β n + 1 + | ( 1 α n + 1 ) β n + 1 ( 1 α n ) β n | M 2 = ( α n + 1 + α n ) M 2 + r ( 1 α n + 1 ) y n + 1 y n β n + 1 + | ( 1 r ) ( α n α n + 1 ) + β n + 1 α n β n α n + 1 β n β n + 1 | M 2 ( α n + 1 + α n ) M 2 + r ( 1 α n + 1 ) y n + 1 y n β n + 1 + 2 α n + α n + 1 β n β n + 1 M 2 : = ρ n + r ( 1 α n + 1 ) y n + 1 y n β n + 1 .
(4.19)
From (4.18) and (4.19), we have
v n + 1 v n 2 ( ρ n + r ( 1 α n + 1 ) y n + 1 y n β n + 1 ) 2 = ρ n 2 + 2 ρ n r ( 1 α n + 1 ) y n + 1 y n β n + 1 + r 2 ( 1 α n + 1 ) 2 y n + 1 y n 2 β n + 1 2 , ρ n 2 + 2 ρ n r ( 1 α n + 1 ) y n + 1 y n β n + 1 + r 2 ( 1 α n + 1 ) 2 β n + 1 2 x n + 1 x n 2 + r 2 ( 1 α n + 1 ) 2 β n + 1 2 | r n r n + 1 | r n ( 1 + ξ ) M 1 .
(4.20)
By the conditions (1) and (2) and (4.20), we obtain
lim sup n { v n + 1 v n 2 x n + 1 x n 2 } 0 .
(4.21)
Notice v n + 1 v n 2 x n + 1 x n 2 = ( v n + 1 v n x n + 1 x n ) ( v n + 1 v n + x n + 1 x n ) , hence from (4.21) we have
lim sup n { v n + 1 v n x n + 1 x n } 0 .
(4.22)
By Lemma 2.7 and (4.22), we have lim n v n x n = 0 , which implies that
lim n x n + 1 x n = 0
(4.23)
by the definition of v n . Since x n + 1 z n 0 , together with (4.23), we have
lim n x n z n = 0 .
(4.24)
Using (4.10), (4.12) and (4.15),
x n + 1 p 2 = α n ( f ( x n ) p ) + ( 1 α n ) ( z n p ) 2 ( 1 α n ) z n p 2 + α n f ( x n ) p 2 ( 1 r ) x n p 2 + r u n p 2 + α n f ( x n ) p 2 x n p 2 r u n x n 2 + α n f ( x n ) p 2 ,
(4.25)
which yields
r u n x n 2 x n p 2 x n + 1 p 2 + α n f ( x n ) p 2 = ( x n p + x n + 1 p ) ( x n p x n + 1 p ) + α n f ( x n ) p 2 ( x n p + x n + 1 p ) x n x n + 1 + α n f ( x n ) p 2 .
(4.26)
From (4.26) we have
lim n T r n f x n x n = lim n u n x n = 0 .
(4.27)
Again, applying (4.25), (4.15) and (4.14), we have
x n + 1 p 2 ( 1 α n ) z n p 2 + α n f ( x n ) p 2 ( 1 r ) x n p 2 + r y n p 2 + α n f ( x n ) p 2 x n p 2 r ξ ( 1 ξ B 2 ) S w n A u n 2 r ξ w n A u n 2 + α n f ( x n ) p 2 ,
(4.28)
which implies that
(4.29)
From (4.29) we have
lim n T r n g A u n A u n = lim n w n A u n = 0 , lim n S w n A u n = 0
(4.30)
and
lim n S w n w n = 0 .
(4.31)
Notice y n = P C ( u n + ξ B ( S w n A u n ) ) and u n C for all n N , so
y n u n = P C ( u n + ξ B ( S w n A u n ) ) P C u n ξ B ( S w n A u n ) ξ B S w n A u n ,
so
lim n y n u n = 0 .
(4.32)
Further, from (4.27), (4.32) and (4.24), we have
lim n y n x n = 0 , lim n y n z n = 0
(4.33)
and
lim n y n T y n = 0 by (4.10), (4.24) and (4.33) .
(4.34)
Let q = P Ω f ( q ) . Choose a subsequence { x n k } such that
lim sup n f ( q ) q , x n q = lim k f ( q ) q , x n k q .
(4.35)
Since { x n } is bounded, { f ( q ) q , x n q } is bounded. Hence lim sup n f ( q ) q , x n q is a constant, namely lim n f ( q ) q , x n k q exists, which implies (4.35) is well defined. Because { x n } is bounded, { x n k } has a weak convergence subsequence which is still denoted by { x n k } . Suppose x n k x , we say x Ω . When x n k x , from (4.30), (4.32) and (4.33), we have
u n k x , y n k x , z n k x , A u n k A x , w n k A x .
(4.36)
If T x x , then by (4.34) and (4.36) and Opial’s condition, we have
lim inf k y n k x < lim inf k y n k T x lim inf k { y n k T y n k + T y n k T x } lim inf k { y n k T y n k + y n k x } = lim inf k y n k x ,
(4.37)
which is a contradiction, so T x = x and x F ( T ) . Since for each r > 0 , E P ( f ) = F ( T r f ) by Lemma 2.2, we have x F ( T r f ) . Otherwise, if there exists r > 0 such that T r f x x , then by (4.27) and Lemma 2.5 and Opial’s condition, we have
lim inf k x n k x < lim inf k x n k T r f x lim inf k { x n k T n k f x n k + T n k f x n k T r f x } = lim inf k T n k f x n k T r f x lim inf k { x n k x + | r n k r | r n k T n k f x n k x n k } = lim inf k x n k x ,
(4.38)
which is also a contradiction, so T r f x = x and x F ( T r f ) = E P ( f ) . Up to now, we have proved x F ( T ) E P ( f ) . Similarly, we can also prove A x F ( S ) E P ( g ) . Hence x Ω , because of this, we can also obtain
lim sup n f ( q ) q , x n q = lim k f ( q ) q , x n k q = f ( q ) q , x q 0 , where  q = P C f ( q ) .
(4.39)
Finally, we prove the conclusion of this theorem is right. For q = P Ω f ( q ) , from (4.10) we have
x n + 1 q 2 = α n ( h ( x n ) q ) + ( 1 α n ) ( z n q ) 2 ( 1 α n ) 2 z n q 2 + 2 α n h ( x n ) q , x n + 1 q ( 1 α n ) 2 x n q 2 + 2 α n h ( x n ) h ( q ) + h ( q ) q , x n + 1 q ( 1 α n ) 2 x n q 2 + 2 α n α x n q x n + 1 q + 2 α n h ( q ) q , x n + 1 q ( 1 α n ) 2 x n q 2 + α n α x n q 2 + α n α x n + 1 q 2 + 2 α n h ( q ) q , x n + 1 q = ( 1 2 α n ) x n q 2 + α n 2 x n q 2 + α n α x n q 2 + α n α x n + 1 q 2 + 2 α n h ( q ) q , x n + 1 q .
(4.40)
From (4.40) we have
x n + 1 q 2 ( 1 α n 2 2 α 1 α n α ) x n q 2 + α n 2 1 α n α x n q 2 + 2 α n 1 α n α h ( q ) q , x n + 1 q ,
(4.41)

by (4.41) and Lemma 2.6, we have x n q Ω . Again, from (4.27) and (4.30), we have u n q Ω and w n A q F ( S ) E P ( f ) , respectively. The proof is completed. □

Remark
  1. (1)

    In this paper, the iterative coefficient α or r can be replaced with the sequence { ζ n } if { ζ n } satisfies { ζ n } [ ϱ , ϑ ] , where ϱ , ϑ ( 0 , 1 ) ;

     
  2. (2)

    Obviously, if H 1 = H 2 in this paper, these weak and strong convergence theorems are also true;

     
  3. (3)

    In this paper, if T is a nonexpansive mapping from H 1 into H 1 and f ( x , y ) is a bi-function from H 1 × H 1 into with the conditions (A1)-(A4), S is a nonexpansive mapping from H 2 into H 2 and g ( u , v ) is a bi-function from H 2 × H 2 into with the conditions (A1)-(A4), then we may obtain a series of similar algorithms.

     

Declarations

Acknowledgements

The first author was supported by the Natural Science Foundation of Yunnan Province (2010ZC152). The second author was supported partially by grant No. NSC 101-2115-M-017-001 of the National Science Council 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 2013

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