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Weak convergence theorems for common solutions of a system of equilibrium problems and operator equations involving nonexpansive mappings

Abstract

In this paper, a monotone variational inequality, a system of equilibrium problems, and nonexpansive mappings are investigated based on an iterative algorithm. Weak convergence theorems of common solutions are established in Hilbert spaces.

AMS Subject Classification:47H05, 47H09, 47J25.

1 Introduction

Equilibrium problems which were introduced by Blum and Oettli [1] have intensively been studied. It has been shown that equilibrium problems cover fixed point problems, variational inequality problems, inclusion problems, saddle problems, complementarity problem, minimization problem, and Nash equilibrium problem; see [13] and the references therein. Equilibrium problem has emerged as an effective and powerful tool for studying a wide class of problems which arise in economics, finance, image reconstruction, ecology, transportation, network, elasticity, and optimization; see [47] and the references therein. For the existence of solutions of equilibrium problems, we refer the readers to [813] and the references therein. However, from the standpoint of real world applications, it is important not only to know the existence of solutions of equilibrium problems, but also to be able to construct an iterative algorithm to approximate their solutions. The computation of solutions is important in the study of many real world problems. For instance, in computer tomography with limited data, each piece of information implies the existence of a convex set in which the required solution lies. The problem of finding a point in the intersection of a finite of the convex sets is then of crucial interest and it cannot be directly solved. Therefore, an iterative algorithm must be used to approximate such a point. The well-known convex feasibility problem which captures applications in various disciplines such as image restoration and radiation therapy treatment planning is to find a point in the intersection of common fixed point sets of a family of nonlinear mappings; see, for example, [1419].

In this paper, a monotone variational inequality, a system of equilibrium problems, and nonexpansive mappings are investigated based on an iterative algorithm. Weak convergence theorems of common solutions are established in Hilbert spaces.

2 Preliminaries

In what follows, we always assume that H is a real Hilbert space with the inner product , and the norm , and C is a nonempty closed and convex subset of H.

Let denote the set of real numbers and F a bifunction of C×C into . Recall the bifunction equilibrium problem is to find an x such that

F(x,y)0,yC.
(2.1)

In this paper, the solution set of the equilibrium problem is denoted by EP(F), i.e.,

EP(F)= { x C : F ( x , y ) 0 , y C } .

To study the equilibrium problems (2.1), we may assume that F satisfies the following conditions:

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

(A2) F is monotone, i.e., F(x,y)+F(y,x)0 for all x,yC;

(A3) for each x,y,zC,

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

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

Let S:CC be a mapping. In this paper, we use F(S) to stand for the set of fixed points. Recall that the mapping S is said to be nonexpansive if

SxSyxy,x,yC.

If C is a bounded closed and convex subset of H, then fixed point sets of nonexpansive mappings are not empty, closed, and convex; see [20] and the references therein.

Let A:CH be a mapping. Recall that A is said to be monotone if

AxAy,xy0,x,yC.

A set-valued mapping T:H 2 H is said to be monotone if, for all x,yH, fTx and gTy imply xy,fg>0. A monotone mapping T:H 2 H is maximal if the graph G(T) of T is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping T is maximal if and only if, for any (x,f)H×H, xy,fg0 for all (y,g)G(T) implies fTx. The class of monotone operators is one of the most important classes of operators. Within the past several decades, many authors have been devoted to the studies on the existence and convergence of zero points for maximal monotone operators.

Let F(x,y)=Ax,yx, x,yC. We see that the problem (2.1) is reduced to the following classical variational inequality. Find xC such that

Ax,yx0,yC.
(2.2)

It is known that xC is a solution to (2.2) if and only if x is a fixed point of the mapping P C (IρA), where ρ>0 is a constant and I is the identity mapping.

Recently, the common solution problems have been extensively studied by many scholars; see, for example, [2133] and the references therein. In this paper, we investigate the common solution problem of a monotone variational inequality, a system of equilibrium problems, and nonexpansive mappings based on an iterative algorithm. In order to prove our main results, we need the following lemmas.

Lemma 2.1 Let C be a nonempty closed and convex subset of H. Then the following inequality holds:

x Proj C x 2 + y Proj C y 2 x y 2 ,xH,yC.

Lemma 2.2 [1, 2]

Let C be a nonempty closed convex subset of H and F:C×CR be a bifunction satisfying (A1)-(A4). Then, for any r>0 and xH, there exists zC such that

F(z,y)+ 1 r yz,zx0,yC.

Further, define

T r x= { z C : F ( z , y ) + 1 r y z , z x 0 , y C }

for all r>0 and xH. Then the following hold:

  1. (a)

    T r is single-valued;

  2. (b)

    T r is firmly nonexpansive, i.e., for any x,yH,

    T r x T r y 2 T r x T r y,xy;
  3. (c)

    F( T r )=EP(F);

  4. (d)

    EP(F) is closed and convex.

Lemma 2.3 [33]

Let A be a monotone mapping of C into H and N C v be the normal cone to C at vC, i.e.,

N C v= { w H : v u , w 0 , u C }

and define a mapping T on C by

Tv={ A v + N C v , v C , , v C .

Then T is maximal monotone and 0Tv if and only if Av,uv0 for all uC.

Lemma 2.4 [34]

Let { a n } n = 1 N be real numbers in [0,1] such that n = 1 N a n =1. Then we have the following:

i = 1 N a i x i 2 i = 1 N a i x i 2 ,

for any given bounded sequence { x n } n = 1 N in H.

Lemma 2.5 [35]

Let 0<p t n q<1 for all n1. Suppose that { x n } and { y n } are sequences in H such that

lim sup n x n d, lim sup n y n d

and

lim n t n x n + ( 1 t n ) y n =d

hold for some r0. Then lim n x n y n =0.

Lemma 2.6 [36]

Let C be a nonempty closed and convex subset of H and S:CC be a nonexpansive mapping. If { x n } is a sequence in C such that x n x and lim n x n S x n =0, then x=Sx.

Lemma 2.7 [37]

Let { a n }, { b n }, and { c n } be three nonnegative sequences satisfying the following condition:

a n + 1 (1+ b n ) a n + c n ,n n 0 ,

where n 0 is some nonnegative integer, n = 1 b n <, and n = 1 c n <. Then the limit lim n a n exists.

3 Main results

Theorem 3.1 Let C be a nonempty closed convex subset of H, S:CC be a nonexpansive mapping with a nonempty fixed point set, and A:CH be a L-Lipschitz continuous and monotone mapping. Let F m be a bifunction from C×C to which satisfies (A1)-(A4). Let N1 denote some positive integer. Assume that F:= m = 1 N EP( F m )VI(C,A)F(S) is not empty. Let { α n }, { β n }, { γ n }, { δ n , 1 }, …, { δ n , N } be real number sequences in (0,1). Let { λ n }, { r n , 1 }, …, and { r n , N } be positive real number sequences. Let { e n } be a bounded sequence in H. Let { x n } be a sequence generated in the following manner:

{ x 1 H , x n + 1 = α n x n + β n S Proj C ( m = 1 N δ n , m z n , m λ n A y n ) + γ n e n , n 1 , y n = Proj C ( m = 1 N δ n , m z n , m λ n A m = 1 N δ n , m z n , m ) ,

where z n , m is such that

F m ( z n , m ,z)+ 1 r n , m z z n , m , z n , m x n 0,zC,m{1,2,N}.

Assume that { α n }, { β n }, { γ n }, { δ n , 1 }, …, { δ n , N }, { λ n }, { r n , 1 }, …, and { r n , N } satisfy the following restrictions:

  1. (a)

    α n + β n + γ n =1,

  2. (b)

    0<a β n b<1 and n = 1 γ n <;

  3. (c)

    m = 1 N δ n , m =1 and 0<c δ n , m 1;

  4. (d)

    lim inf n r n , m >0 and d λ n e, where d,e(0,1/L).

Then the sequence { x n } weakly converges to some point x ¯ F.

Proof Put u n = Proj C ( m = 1 N δ n , m z n , m λ n A y n ) and v n = m = 1 N δ n , m z n , m . Letting pF, we see from Lemma 2.1 that

u n p 2 v n λ n A y n p 2 v n λ n A y n u n 2 = v n p 2 v n u n 2 + 2 λ n A y n , p u n = v n p 2 v n u n 2 + 2 λ n ( A y n A p , p y n + A p , p y n + A y n , y n u n ) v n p 2 v n u n 2 + 2 λ n A y n , y n u n = v n p 2 v n y n 2 y n u n 2 + 2 v n λ n A y n y n , u n y n .
(3.1)

Notice that A is L-Lipschitz continuous and y n = Proj C ( v n λ n A v n ). It follows that

v n λ n A y n y n , u n y n = v n λ n A v n y n , u n y n + λ n A v n λ n A y n , u n y n λ n L v n y n u n y n .
(3.2)

Substituting (3.2) into (3.1), we obtain that

u n p 2 v n p 2 v n y n 2 y n u n 2 + 2 λ n L v n y n u n y n v n p 2 + ( λ n 2 L 2 1 ) v n y n 2 .
(3.3)

On the other hand, we have from the restriction (c) that

v n p 2 m = 1 N δ n , m z n , m p 2 m = 1 N δ n , m z n , m p 2 m = 1 N δ n , m T r n , m x n p 2 x n p 2 .
(3.4)

Substituting (3.4) into (3.3), we obtain that

u n p 2 x n p 2 + ( λ n 2 L 2 1 ) v n y n 2 .
(3.5)

This in turn implies from the restriction (d) that

x n + 1 p 2 α n x n p 2 + β n S u n p 2 + γ n e n p 2 α n x n p 2 + β n u n p 2 + γ n e n p 2 α n x n p 2 + β n ( x n p 2 + ( λ n 2 L 2 1 ) v n y n 2 ) + γ n e n p 2 x n p 2 + β n ( λ n 2 L 2 1 ) v n y n 2 + γ n e n p 2 x n p 2 + γ n e n p 2 .
(3.6)

It follows from Lemma 2.7 that the lim n x n p exists. This in turn shows that { x n } is bounded. It follows from (3.6) that

β n ( 1 λ n 2 L 2 ) v n y n 2 x n p 2 x n + 1 p 2 + γ n e n p 2 .

This implies from the restrictions (b) and (d) that

lim n v n y n =0.
(3.7)

Notice that

y n u n = Proj C ( v n λ n A v n ) Proj C ( v n λ n A y n ) ( v n λ n A v n ) ( v n λ n A y n ) λ L v n y n .

It follows from (3.7) that

lim n y n u n =0.
(3.8)

In view of

v n u n v n y n + y n u n ,

we see from (3.7) and (3.8) that

lim n v n u n =0.
(3.9)

Notice that

z n , m p 2 = T r n , m x n T r n , m p 2 T r n , m x n T r n , m p , x n p = z n , m p , x n p = 1 2 ( z n , m p 2 + x n p 2 z n , m x n 2 ) , 1 m N .

This implies that

z n , m p 2 x n p 2 z n , m x n 2 ,1mN.
(3.10)

In view of (3.10) and v n = m = 1 N δ n , m z n , m , where m = 1 N δ n , m =1, we see from Lemma 2.4 that

v n p 2 m = 1 N δ n , m z n , m p 2 m = 1 N δ n , m ( x n p 2 z n , m x n 2 ) = x n p 2 m = 1 N δ n , m z n , m x n 2 .
(3.11)

In view of (3.3), we obtain from the restriction (d) that

x n + 1 p 2 α n x n p 2 + β n S u n p 2 + γ n e n p 2 α n x n p 2 + β n u n p 2 + γ n e n p 2 α n x n p 2 + β n v n p 2 + γ n e n p 2 x n p 2 β n m = 1 N δ n , m z n , m x n 2 + γ n e n p 2 .

It follows that

β n δ n , m z n , m x n 2 x n p 2 x n + 1 p 2 + γ n e n p 2 .

In view of the restrictions (b) and (c), we find that

lim n z n , m x n =0.
(3.12)

Since { x n } is bounded, we may assume that a subsequence { x n i } of { x n } converges weakly to ξ. It follows from (3.12) that { z n i , m } converges weakly to ξ for each 1mN. Next, we show that ξEP( F m ) for each 1mN. Since z n , m = T r n , m x n , we have

F m ( z n , m ,z)+ 1 r n , m z z n , m , z n , m x n 0,zC.

From the assumption (A2), we see that

1 r n , m z z n , m , z n , m x n F m (z, z n , m ),zC.

Replacing n by n i , we arrive at

z z n i , m , z n i , m x n i r n i , m F m (z, z n i , m ),zC.

In view of the assumption (A4), we get from (3.12) that

F m (z,ξ)0,zC.

For t m with 0< t m 1 and zC, let z t m = t m z+(1 t m )ξ for each 1mN. Since zC and ξC, we have z t m C for each 1mN. It follows that F m ( z t m ,ξ)0 for each 1mN. Notice that

0= F m ( z t m , z t m ) t m F m ( z t m ,z)+(1 t m ) F m ( z t m ,ξ) t m F m ( z t m ,z),1mN,

which yields that

F m ( z t m ,z)0,zC.

Letting t m 0 for each 1mN, we obtain from the assumption (A3) that

F m (ξ,z)0,zC.

This implies that ξEP( F m ) for each 1mN. This proves that ξ m = 1 N EP( F m ).

Next, we show that ξVI(C,A). In fact, let T be the maximal monotone mapping defined by

Tx={ A x + N C x , x C , , x C .

For any given (x,y)G(T), we have yAx N C x. So, we have xm,yAx0, for all mC. On the other hand, we have u n = Proj C ( v n λ n A y n ). We obtain that

v n λ n A y n u n , u n x0

and hence

x u n , u n v n λ n + A y n 0.

In view of the monotonicity of A, we see that

x u n i , y x u n i , A x x u n i , A x x u n i , u n i v n i λ n i + A y n i = x u n i , A x A u n i + x u n i , A u n i A y n i x u n i , u n i v n i λ n i x u n i , A u n i A y n i x u n i , u n i v n i λ n i .
(3.13)

On the other hand, we see that

v n x n m = 1 N δ n , m z n , m x n .

It follows from (3.12) that

lim n v n x n =0.
(3.14)

Notice that

u n x n u n v n + v n x n .

Combining (3.9) with (3.14), we arrive at

lim n u n x n =0.
(3.15)

This in turn implies that u n i ξ. It follows from (3.13) that xξ,y0. Notice that T is maximal monotone and hence 0Tξ. This shows from Lemma 2.3 that ξVI(C,A).

Next, we show that ξF(S). Since lim n x n p exists, we put lim n x n p=d>0. It follows that

lim n x n + 1 p= lim n ( 1 β n ) ( x n p + γ n ( e n x n ) ) + β n ( S u n p + γ n ( e n x n ) ) =d.

Notice that

S u n p + γ n ( e n x n ) S u n p + γ n e n x n u n p + γ n e n x n x n p + γ n e n x n .

This shows that

lim sup n S u n p + γ n ( e n x n ) d.

On the other hand, we have

lim sup n x n p + γ n ( e n x n ) d.

It follows from Lemma 2.5 that

lim n x n S u n =0.
(3.16)

In view of

S x n x n S x n S u n + S u n x n x n u n + S u n x n ,

we find from (3.15) and (3.16) that

lim n x n S x n =0.

This implies from Lemma 2.6 that ξF(S). This completes the proof that ξF.

Finally, we show that the whole sequence { x n } weakly converges to ξ. Let { x n j } be another subsequence of { x n } converging weakly to ξ , where ξ ξ. In the same way, we can show that ξ F. Since the space H enjoys Opial’s condition, we, therefore, obtain that

d = lim inf i x n i ξ < lim inf i x n i ξ = lim inf j x j ξ < lim inf j x j ξ = d .

This is a contradiction. Hence, ξ= ξ . This completes the proof. □

If N=1, then Theorem 3.1 is reduced to the following.

Corollary 3.2 Let C be a nonempty closed convex subset of H, S:CC be a nonexpansive mapping with a nonempty fixed point set, and A:CH be a L-Lipschitz continuous and monotone mapping. Let F be a bifunction from C×C to which satisfies (A1)-(A4). Assume that F:=EP(F)VI(C,A)F(S) is not empty. Let { α n }, { β n }, and { γ n } be real number sequences in (0,1). Let { λ n }, { r n } be positive real number sequences. Let { e n } be a bounded sequence in H. Let { x n } be a sequence generated in the following manner:

{ x 1 H , x n + 1 = α n x n + β n S Proj C ( z n λ n A y n ) + γ n e n , n 1 , y n = Proj C ( z n λ n A z n ) ,

where z n is such that

F( z n ,z)+ 1 r n z z n , z n x n 0,zC.

Assume that { α n }, { β n }, { γ n }, { δ n }, { λ n }, { r n } satisfy the following restrictions:

  1. (a)

    α n + β n + γ n =1,

  2. (b)

    0<a β n b<1 and n = 1 γ n <;

  3. (c)

    lim inf n r n >0 and c λ n d, where c,d(0,1/L).

Then the sequence { x n } weakly converges to some point x ¯ F.

If S=I, where I stands for the identity mapping, then Theorem 3.1 is reduced to the following.

Corollary 3.3 Let C be a nonempty closed convex subset of H and A:CH be a L-Lipschitz continuous and monotone mapping. Let F m be a bifunction from C×C to which satisfies (A1)-(A4). Let N1 denote some positive integer. Assume that F:= m = 1 N EP( F m )VI(C,A) is not empty. Let { α n }, { β n }, { γ n }, { δ n , 1 }, …, { δ n , N } be real number sequences in (0,1). Let { λ n }, { r n , 1 }, …, and { r n , N } be positive real number sequences. Let { e n } be a bounded sequence in H. Let { x n } be a sequence generated in the following manner:

{ x 1 H , x n + 1 = α n x n + β n Proj C ( m = 1 N δ n , m z n , m λ n A y n ) + γ n e n , n 1 , y n = Proj C ( m = 1 N δ n , m z n , m λ n A m = 1 N δ n , m z n , m ) ,

where z n , m is such that

F m ( z n , m ,z)+ 1 r n , m z z n , m , z n , m x n 0,zC,m{1,2,N}.

Assume that { α n }, { β n }, { γ n }, { δ n , 1 }, …, { δ n , N }, { λ n }, { r n , 1 }, …, and { r n , N } satisfy the following restrictions:

  1. (a)

    α n + β n + γ n =1,

  2. (b)

    0<a β n b<1 and n = 1 γ n <;

  3. (c)

    m = 1 N δ n , m =1 and 0<c δ n , m 1;

  4. (d)

    lim inf n r n , m >0 and d λ n e, where d,e(0,1/L).

Then the sequence { x n } weakly converges to some point x ¯ F.

If F m (x,y)0 for all x,yC and r n , m 1, then Theorem 3.1 is reduced to the following.

Corollary 3.4 Let C be a nonempty closed convex subset of H, S:CC be a nonexpansive mapping with a nonempty fixed point set, and A:CH be a L-Lipschitz continuous and monotone mapping. Assume that F:=VI(C,A)F(S) is not empty. Let { α n }, { β n }, and { γ n } be real number sequences in (0,1). Let { λ n } be a positive real number sequence. Let { e n } be a bounded sequence in H. Let { x n } be a sequence generated in the following manner:

{ x 1 H , x n + 1 = α n x n + β n S Proj C ( Proj C x n λ n A y n ) + γ n e n , n 1 , y n = Proj C ( Proj C x n λ n A Proj C x n ) .

Assume that { α n }, { β n }, { γ n }, and { λ n } satisfy the following restrictions:

  1. (a)

    α n + β n + γ n =1,

  2. (b)

    0<a β n b<1 and n = 1 γ n <;

  3. (c)

    c λ n d, where c,d(0,1/L).

Then the sequence { x n } weakly converges to some point x ¯ F.

4 Applications

Theorem 4.1 Let S:HH be a nonexpansive mapping with a nonempty fixed point set and A:HH be a L-Lipschitz continuous and monotone mapping. Assume that F:= A 1 (0)F(S) is not empty. Let { α n }, { β n }, and { γ n } be real number sequences in (0,1). Let { λ n } be a positive real number sequence. Let { e n } be a bounded sequence in H. Let { x n } be a sequence generated in the following manner:

x 1 H, x n + 1 = α n x n + β n S ( x n λ n A ( x n λ n A x n ) ) + γ n e n ,n1.

Assume that { α n }, { β n }, { γ n }, and { λ n } satisfy the following restrictions:

  1. (a)

    α n + β n + γ n =1,

  2. (b)

    0<a β n b<1 and n = 1 γ n <;

  3. (c)

    c λ n d, where c,d(0,1/L).

Then the sequence { x n } weakly converges to some point x ¯ F.

Proof Put F m (x,y)0 for all x,yC and r n , m 1. Notice that A 1 (0)=VI(H,A) and P H =I, we easily find from Theorem 3.1 the desired conclusion. □

Next, we consider the common zero point problem of two monotone mappings.

Theorem 4.2 Let B:H 2 H a maximal monotone mapping and A:HH be a L-Lipschitz continuous and monotone mapping. Assume that F:= A 1 (0) B 1 (0) is not empty. Let { α n }, { β n }, and { γ n } be real number sequences in (0,1). Let { λ n } be a positive real number sequence. Let { e n } be a bounded sequence in H. Let { x n } be a sequence generated in the following manner:

x 1 H, x n + 1 = α n x n + β n J r B ( x n λ n A ( x n λ n A x n ) ) + γ n e n ,n1,

where J r B stands for the resolvent of B for each r>0. Assume that { α n }, { β n }, { γ n }, and { λ n } satisfy the following restrictions:

  1. (a)

    α n + β n + γ n =1,

  2. (b)

    0<a β n b<1 and n = 1 γ n <;

  3. (c)

    c λ n d, where c,d(0,1/L).

Then the sequence { x n } weakly converges to some point x ¯ F.

Proof Put F m (x,y)0 for all x,yC and r n , m 1. Notice that A 1 (0)=VI(H,A), F( J r B )= B 1 (0), and P H =I, we easily find from Theorem 3.1 the desired conclusion. □

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Cheng, P., Zhang, A. Weak convergence theorems for common solutions of a system of equilibrium problems and operator equations involving nonexpansive mappings. J Inequal Appl 2013, 87 (2013). https://doi.org/10.1186/1029-242X-2013-87

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