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Convergence theorems for finding zero points of maximal monotone operators and equilibrium problems in Banach spaces

Abstract

In this paper, we construct a new hybrid projection method for approximating a common element of the set of zeroes of a finite family of maximal monotone operators and the set of common solutions to a system of generalized equilibrium problems in a uniformly smooth and strictly convex Banach space. We prove strong convergence theorems of the algorithm to a common element of these two sets. As application, we also apply our results to find common solutions of variational inequalities and zeroes of maximal monotone operators.

MSC:47H05, 47H09, 47H10.

1 Introduction

Let E be a Banach space with the norm and let E denote the dual space of E. Let S={xE:x=1} be the unit sphere of E. A Banach space E is said to be smooth if the limit

lim t 0 x + t y x t
(1.1)

exists for any x,yS. E is said to be uniformly smooth if the limit (1.1) is attained uniformly for (x,y) in S×S. A Banach space E is said to be strictly convex if x + y 2 <1 for all x,yE with x=y=1 and xy (see [1] for more details).

One of the major problems in the theory of monotone operators is as follows.

Find a point zE such that

0Bz,
(1.2)

where B is an operator from E into E . Such zE is called a zero point of B. We denote the set of zeroes of the operator B by B 1 0.

An operator BE× E is said to be monotone if

x y , x y 0, ( x , x ) , ( y , y ) B.

A monotone B is said to be maximal if its graph G(B)={(x, y ): y Bx} is not properly contained in the graph of any other monotone operator. If B is maximal monotone, then the solution set B 1 0 is closed and convex. The resolvent of a monotone operator B is defined by

J λ = ( J + λ B ) 1 J,λ>0.

Let C be a closed convex subset of a Banach space E, a mapping T:CC is said to be nonexpansive if

TxTyxy,x,yC.

Recall that a point xC is a fixed point of T provided Tx=x. Let E be a Banach space with dual E and let , be the pairing between E and E . The normalized duality mapping J:E 2 E is defined by

J(x)= { x E : x , x = x 2 , x = x } ,xE.

The Lyapunov functional is defined by

ϕ(x,y)= x 2 2x,Jy+ y 2 for x,yE.
(1.3)

It is obvious that

( y x ) 2 ϕ(y,x) ( y + x ) 2 ,x,yE.
(1.4)

If E is a Hilbert space, then ϕ(x,y)= x y 2 for all x,yE.

A point p in C is said to be a strongly asymptotic fixed point of T [2] if C contains a sequence { x n } which converges strongly to p such that lim n x n T x n =0. The set of strong asymptotic fixed points of T will be denoted by F ˜ (T).

A mapping T from C into itself is said to be a weak relatively nonexpansive mapping if

  1. 1.

    F(T) is nonempty;

  2. 2.

    ϕ(p,Tx)ϕ(p,x) for all xC and pF(T);

  3. 3.

    F ˜ (T)=F(T).

Kohasaka and Takahashi [3] proved that if E is a smooth strictly convex and reflexive Banach space and B is a continuous monotone operator with B 1 0, then J λ is a weak relatively nonexpansive mapping. By Takahashi [4], we know that F( J λ ) is closed and convex, where F( J λ ) is the set of fixed points of J λ .

Let B be a maximal monotone operator in a Hilbert space H. The proximal point algorithm generates, for starting x 1 =xH, a sequence { x n } in H by

x n + 1 = J λ n x n ,n1,
(1.5)

where { λ n }(0,) and J λ n = ( I + λ n B ) 1 .

Also, Rockafellar [5] proved that the sequence { x n } defined by (1.5) converges weakly to an element of B 1 0.

Let C be a nonempty closed convex subset of E and let be the set of real numbers. Let f i :C×CR be a bifunction and A i :C E be a nonlinear mapping for i=1,2,3,,N. The system of generalized equilibrium problems is as follows.

Find uC such that for all yC,

{ f 1 ( u , y ) + y u , A 1 u 0 , f 2 ( u , y ) + y u , A 2 u 0 , f N ( u , y ) + y u , A N u 0 .
(1.6)

If f i =f and A i =A in (1.6), then from the problem (1.6) we have the following generalized equilibrium problem denoted by GEP(f,A).

Find uC such that

f(u,y)+yu,Au0,yC.
(1.7)

The generalized equilibrium problems include fixed point problems, optimization problem, monotone inclusion problems, saddle point problems, variational inequality problems, minimization problems, vector equilibrium problems, Nash equilibria in noncooperative games and equilibrium problems as special cases (see, for example, [6]). Also, some solution methods have been proposed to solve the equilibrium problem (see, for example, [714]) and numerous problems in physics, optimization and economics reduce to finding a solution of problem (1.7).

Recently, Li and Su [15] introduced the hybrid iterative scheme for approximating a common solution of the equilibrium problems and the variational inequality problems in a 2-uniformly convex real Banach space which is also uniformly smooth. In 2010, Zegeye and Shahzad [16] introduced the iterative process which converges strongly to a common solution of the variational inequality problems for two monotone mappings in Banach spaces.

Quite recently, Shehu [13] introduced an iterative scheme by the hybrid method for approximating a common element of the set of zeroes of a finite family of α-inverse-strongly monotone operators and the set of common solutions of a system of generalized mixed equilibrium problems in a 2-uniformly convex real Banach space which is also uniformly smooth.

Motivated by the results of Shehu [13], we prove some strong convergence theorems for finding a common zero of a finite family of continuous monotone mappings and a solution of the system of generalized equilibrium problems in a uniformly smooth and strictly convex real Banach space with the Kadec-Klee property.

2 Preliminaries

Throughout this paper, let E be a Banach space with its dual space E . For a sequence { x n } of E and a point xE, the weak convergence of { x n } to x is denoted by x n x and the strong convergence of { x n } to x is denoted by x n x.

The normalized duality mapping J:E 2 E is defined by

J(x)= { x E : x , x = x 2 , x = x } ,xE,

where , denotes the duality pairing.

Cioranescu [17] proved the following properties:

  1. (1)

    If E is an arbitrary Banach space, then J is monotone and bounded;

  2. (2)

    If E is a strictly convex, then J is strictly monotone;

  3. (3)

    If E is a smooth, then J is single-valued and semi-continuous;

  4. (4)

    If E is uniformly smooth, then J is uniformly norm-to-norm continuous on each bounded subset of E;

  5. (5)

    If E is reflexive, smooth and strictly convex, then the normalized duality mapping J is single-valued, one-to-one and onto;

  6. (6)

    If E is a reflexive, strictly convex and smooth Banach space and J is the duality mapping from E into E , then J 1 is also single-valued, bijective and is also the duality mapping from E into E and thus J J 1 = I E and J 1 J= I E ;

  7. (7)

    If E is uniformly smooth, then E is smooth and reflexive;

  8. (8)

    E is uniformly smooth if and only if E is uniformly convex.

A Banach space E has the Kadec-Klee property [1, 17] if, for any sequence { x n }E and xE with x n x and x n x, then x n x0 as n.

Consider the functional ϕ:E×ER defined by

ϕ(x,y)= x 2 2x,Jy+ y 2 ,x,yE,
(2.1)

where J is the normalized duality mapping from E to 2 E .

It is obvious from the definition of the function ϕ that

( y x ) 2 ϕ(y,x) ( y + x ) 2 ,x,yE.
(2.2)

If E is a Hilbert space, then ϕ(y,x)= y x 2 .

Remark 2.1 If E is a reflexive, strictly convex and smooth Banach space, then, for any x,yE, ϕ(x,y)=0 if and only if x=y. It is sufficient to show that if ϕ(x,y)=0, then x=y. From (1.4) we have x=y. This implies that x,Jy= x 2 = J y 2 . From the definition of J, one has Jx=Jy. Therefore, we have x=y (see [1, 17] for more details).

Let C be a nonempty closed convex subset of a reflexive, strictly convex and smooth Banach space E. The generalized projection Π C :EC is a mapping that assigns to an arbitrary point xE the minimum point of the functional ϕ(x,y), that is, Π C x= x ¯ , where x ¯ is the solution to the minimization problem

ϕ( x ¯ ,x)= inf y C ϕ(y,x).
(2.3)

The existence and uniqueness of the operator Π C follows from the properties of the functional ϕ(y,x) and the strict monotonicity of the mapping J (see, for example, [1, 1720]). If E is a Hilbert space, then Π C becomes the metric projection of E onto C.

Example 2.2 (Qin et al. [21])

Let Π C be the generalized projection from a smooth strictly convex and reflexive Banach space E onto a nonempty closed convex subset C of E. Then Π C is a closed relatively quasi-nonexpansive mapping from E onto C with F( Π C )=C.

We also need the following lemmas for the proof of our main results.

Lemma 2.3 (Alber [19])

Let C be a nonempty closed convex subset of a smooth Banach space E and let xE. Then x 0 = Π C x if and only if

x 0 y,JxJ x 0 0,yC.

Lemma 2.4 (Alber [19])

Let E be a reflexive, strictly convex and smooth Banach space, let C be a nonempty closed convex subset of E and let xE. Then

ϕ(y, Π C x)+ϕ( Π C x,x)ϕ(y,x),yC.

Let E be a smooth strictly convex and reflexive Banach space, C be a nonempty closed convex subset of E and BE× E be a monotone operator satisfying the following:

D(B)C J 1 ( λ > 0 R ( J + λ B ) ) .

Then the resolvent J λ :CD(B) of B is defined by

J λ x= { z D ( B ) : J x J z + λ B z , x C } .

J λ is a single-valued mapping from E to D(B). For any λ>0, the Yosida approximation B λ :C E of B is defined by B λ x= J x J J λ x λ for all xC. We know that B λ xB( J λ x) for all λ>0 and xE.

Lemma 2.5 (Kohsaka and Takahashi [22])

Let E be a smooth strictly convex and reflexive Banach space, let C be a nonempty closed convex subset of E and let BE× E be a monotone operator satisfying D(B)C J 1 ( λ > 0 R(J+λB)). For any λ>0, let J λ and B λ be the resolvent and the Yosida approximation of B, respectively. Then the following hold:

  1. (1)

    ϕ(p, J λ x)+ϕ( J λ x,x)ϕ(p,x) for all xC and p B 1 0;

  2. (2)

    ( J λ x, B λ x)B for all xC;

  3. (3)

    F( J λ )= B 1 0.

Lemma 2.6 (Rockafellar [23])

Let E be a reflexive strictly convex and smooth Banach space. Then an operator BE× E is maximal monotone if and only if R(J+λB)= E for all λ>0.

For solving the equilibrium problem for a bifunction f:C×CR, 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 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.

The following result is given in Blum and Oettli [6].

Lemma 2.7 Let C be a closed convex subset of a smooth strictly convex and reflexive Banach space E and let f be a bifunction from C×C to satisfying the conditions (A1)-(A4). Then, for any r>0 and xE, there exists zC such that

f(z,y)+ 1 r yz,JzJx0,yC.

Lemma 2.8 (Takahashi and Zembayashi [24])

Let C be a closed convex subset of a uniformly smooth strictly convex and reflexive Banach space E and let f be a bifunction from C×C to satisfying the conditions (A1)-(A4). For any r>0 and xE, define a mapping T r :EC as follows:

T r x= { z C : f ( z , y ) + 1 r y z , J z J x 0 , y C } ,xC.

Then the following hold:

  1. (1)

    T r is single-valued;

  2. (2)

    T r is a firmly nonexpansive-type mapping for all x,yE, that is,

    T r x T r y,J T r xJ T r y T r x T r y,JxJy;
  3. (3)

    F( T r )=EP(f);

  4. (4)

    EP(f) is closed and convex.

Lemma 2.9 (Takahashi and Zembayashi [24])

Let C be a closed convex subset of a smooth, strictly convex and reflexive Banach space E, f be a bifunction from C×C to satisfying the conditions (A1)-(A4) and let r>0. Then, for any xE and pF( T r ),

ϕ(p, T r x)+ϕ( T r x,x)ϕ(p,x).

Lemma 2.10 Let C be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space E. Let A:C E be a continuous monotone mapping and f be a bifunction from C×C to satisfying the conditions (A1)-(A4). Then, for any r>0 and xE, there exists zC such that

f(z,y)+yz,Az+ 1 r yz,JzJx0,yC.

Proof Define a bifunction Θ:C×CR by Θ(x,y)=f(x,y)+yx,Ax for all x,yC. We show that Θ satisfies the conditions (A1)-(A4).

First, we show that Θ satisfies the condition (A1). Since

Θ(x,x)=f(x,x)+xx,Ax=0,xC,

the condition (A2) is satisfied.

Next, we show that Θ satisfies the condition (A2). Since A is a continuous monotone mapping and f satisfies the condition (A2), for any x,yC, we have

Θ ( x , y ) + Θ ( y , x ) = f ( x , y ) + f ( y , x ) + y x , A x + x y , A y 0 + y x , A x A y 0 .

So, the condition (A2) is satisfied.

Thirdly, we show that Θ satisfies the condition (A3). Since f satisfies the condition (A3) and A is a continuous monotone mapping, for any x,y,zC, we have

lim sup t 0 Θ ( x + t ( z x ) , y ) = lim sup t 0 f ( x + t ( z x ) , y ) + y ( x + t ( z x ) ) , A ( x + t ( z x ) ) lim sup t 0 f ( x + t ( z x ) , y ) + lim sup t 0 y ( x + t ( z x ) ) , A ( x + t ( z x ) ) f ( x , y ) + lim sup t 0 y ( x + t ( z x ) ) , A ( x + t ( z x ) ) = f ( x , y ) + y x , A x = Θ ( x , y ) .

The condition (A3) is satisfied.

Finally, we show that Θ satisfies the condition (A4) since yyx,Ax is convex and continuous; that is, yyx,Ax is convex and lower semi-continuous. Since yf(x,y) is convex and lower semi-continuous, yΘ(x,y) is convex and lower semi-continuous.

Therefore, Θ(x,y) satisfies the conditions (A1)-(A4). From Lemma 2.7, we can conclude that there exists zC such that

f(z,y)+yz,Az+ 1 r yz,JzJx0,yC.

This completes the proof. □

Lemma 2.11 Let C be a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach space E. Let A:C E be a continuous and monotone mapping and f be a bifunction from C×C to satisfying the conditions (A1)-(A4). Then, for any r>0 and xE, there exists zC such that

f(z,y)+yz,Az+ 1 r yz,JzJx0,yC.

Define a mapping K r :CC as follows:

K r (x)= { z C : f ( z , y ) + y z , A z + 1 r y z , J z J x 0 , y C } ,xC.
(2.4)

Then we have the following:

  1. (1)

    K r is single-valued;

  2. (2)

    K r is firmly nonexpansive, i.e., for all x,yE,

    K r x K r y,J K r xJ K r y K r x K r y,JxJy;
  3. (3)

    F( K r )=GEP(f,A);

  4. (4)

    GEP(f,A) is closed and convex;

  5. (5)

    ϕ(p, K r z)+ϕ( K r z,z)ϕ(p,z) for all pF( K r ) and zE.

Proof Define a bifunction Θ:C×CR by Θ(u,y)=f(u,y)+yu,Au for all u,yC. From Lemma 2.8, it follows that Θ satisfies the conditions (A1)-(A4). Now, we can rewrite the mapping K r :CC given in (2.4) as follows:

K r (x)= { z C : Θ ( z , y ) + 1 r y z , J z J x 0 , y C } ,xC.
(2.5)

Thus, from Lemmas 2.8 and 2.9, we obtain the conclusion. This completes the proof. □

Throughout this paper, we define a mapping K r i Θ i (x):CC by

K r i Θ i (x)= { z C : Θ i ( z , y ) + 1 r i y z , J z J x 0 , y C } ,xC,
(2.6)

where Θ i (z,y)= f i (z,y)+yz, A i z for all z,yC and i=1,2,3,,m.

3 Main results

Theorem 3.1 Let C be a nonempty closed and convex subset of a uniformly smooth and strictly convex Banach space E with the Kadec-Klee property. For any i=1,2,3,,m, let f i be a bifunction from C×C to satisfying the conditions (A1)-(A4) and let { A i } be a finite family of continuous and monotone mappings from C to E . Let B j E× E be maximal monotone operators satisfying D( B j )C and J λ j , n B j = ( J + λ j , n B j ) 1 J for all λ j , n >0 and j=1,2,,l. Assume that F:=( i = 1 m GEP( f i , A i ))( j = 1 l B j 1 0). For arbitrary x 1 C and C 1 =C, generate a sequence { x n } by

{ z n = J λ l , n B l J λ l 1 , n B l 1 J λ 1 , n B 1 x n , u n = K r m , n Θ m K r m 1 , n Θ m 1 K r 1 , n Θ 1 z n , C n + 1 = { z C n : ϕ ( z , u n ) ϕ ( z , x n ) } , x n + 1 = Π C n + 1 x 1 , n 1 ,
(3.1)

where { r i , n }[a,) for some a>0 for all i=1,2,,m and lim inf n λ j , n >0 for all j=1,2,,l. Then the sequence { x n } converges strongly to a point pF, where p= Π F x 1 .

Proof We split the proof into five steps as follows.

Step 1. We first show that C n + 1 is closed and convex for all n1. Clearly, C 1 =C is closed and convex. Suppose that C n is closed and convex for all n1. Since, for any z C n , we know that ϕ(z, u n )ϕ(z, x n ) is equivalent to the following:

2z,J x n J u n x n 2 u n 2 .

Thus C n + 1 is closed and convex for all n1.

Step 2. We show that F C n for all n1 and { x n } is well defined. Since F C 1 =C, suppose that F C n for some n1. Let qF, from Lemma 2.11 and Lemma 2.5, we have that

ϕ ( q , u n ) = ϕ ( q , K r m , n Θ m K r m 1 , n Θ m 1 K r 1 , n Θ 1 z n ) ϕ ( q , K r m 1 , n Θ m 1 K r 1 , n Θ 1 z n ) ϕ ( q , K r 1 , n Θ 1 z n ) ϕ ( q , z n ) = ϕ ( q , J λ l , n l J λ l 1 , n l 1 J λ 1 , n 1 x n ) ϕ ( q , J λ l 1 , n l 1 J λ 1 , n 1 x n ) ϕ ( q , J λ 1 , n 1 x n ) ϕ ( q , x n ) .
(3.2)

This shows that q C n + 1 , which implies that F C n + 1 . Hence F C n for all n1. This implies that the sequence { x n } is well defined.

Step 3. We show that lim n u n x n =0 and lim n J u n J x n =0. By the definition of C n + 1 with x n = Π C n x 1 and x n + 1 = Π C n + 1 x 1 C n + 1 C n , it follows that

ϕ( x n , x 1 )ϕ( x n + 1 , x 1 ),n1,
(3.3)

that is, {ϕ( x n , x 1 )} is nondecreasing. By Lemma 2.4, we get

ϕ ( x n , x 1 ) = ϕ ( Π C n x 1 , x 1 ) ϕ ( q , x 1 ) ϕ ( q , x n ) ϕ ( q , x 1 ) , q F .
(3.4)

This implies that {ϕ( x n , x 1 )} is bounded and so lim n ϕ( x n , x 1 ) exists. In particular, by (1.4), the sequence { ( x n x 1 ) 2 } is bounded. This implies { x n } is also bounded. So, { z n } and { u n } are also bounded. Since E is reflexive and C n is closed and convex, without loss of generality, we may assume that there exists p C n such that x n p.

Since x n = Π C n x 1 , we have

ϕ( x n , x 1 )ϕ(p, x 1 ),p C n .

On the other hand, since

lim inf n ϕ ( x n , x 1 ) = lim inf n { x n 2 2 x n , J x 1 + x 1 2 } p 2 2 p , J x 1 + x 1 2 = ϕ ( p , x 1 ) ,

it follows that

ϕ(p, x 1 ) lim inf n ϕ( x n , x 1 ) lim sup n ϕ( x n , x 1 )ϕ(p, x 1 ).

This implies that lim n ϕ( x n , x 1 )=ϕ(p, x 1 ). Hence we get x n p as n. In view of the Kadec-Klee property of E, we obtain

lim n x n =p.
(3.5)

Now, we claim that J u n J x n 0 as n. By the definition of Π C n x 1 , it follows that

ϕ ( x n + 1 , x n ) = ϕ ( x n + 1 , Π C n x 1 ) ϕ ( x n + 1 , x 1 ) ϕ ( Π C n x 1 , x 1 ) = ϕ ( x n + 1 , x 1 ) ϕ ( x n , x 1 ) .

Since lim n ϕ( x n , x 1 ) exists, we obtain

lim n ϕ( x n + 1 , x n )=0.
(3.6)

Since x n + 1 = Π C n + 1 x 1 C n + 1 C n and the definition of C n + 1 , we have ϕ( x n + 1 , u n )ϕ( x n + 1 , x n ). By (3.6) we also have

lim n ϕ( x n + 1 , u n )=0.
(3.7)

From (2.2) it follows that

u n p(n).
(3.8)

Since J is uniformly norm-to-norm continuous on bounded subsets of E, it follows that

J u n Jp(n).
(3.9)

This implies that {J u n } is bounded in E . Note that E is reflexive and E is also reflexive, we can assume that J u n x E . Since E is reflexive, we see that J(E)= E . Hence there exists xE such that Jx= x , and we have

ϕ ( x n + 1 , u n ) = x n + 1 2 2 x n + 1 , J u n + u n 2 = x n + 1 2 2 x n + 1 , J u n + J u n 2 .

Taking lim inf n on both sides of the equation above, in view of the weak lower semi-continuity of the norm , it follows that

0 p 2 2 p , x + x 2 = p 2 2 p , J x + J x 2 = p 2 2 p , J x + x 2 = ϕ ( p , x ) .

From Remark 2.1 we have p=x, which implies that x =Jp, and so J u n Jp E . From the Kadec-Klee property, we have that

J u n Jpas n.

Note that J 1 : E E is norm-weak-continuous, that is,

u n pas n.
(3.10)

From (3.8), (3.10) and the Kadec-Klee property of E, it follows that

lim n u n =p.
(3.11)

Since x n u n x n p+p u n , it follows that

lim n x n u n =0.
(3.12)

Since J is uniformly norm-to-norm continuous on bounded subsets of E, we obtain

lim n J u n J x n =0.
(3.13)

Step 4. We show that pF:=( i = 1 m GEP( f i , A i ))( j = 1 l B j 1 0). First, we show that p i = 1 m GEP( f i , A i ). From (3.2), (3.5) and (3.11), it follows that for any qF,

lim n ϕ(q, z n )=ϕ(q,p).
(3.14)

Denote Δ n j := J λ j , n B j J λ j 1 , n B j 1 J λ 1 , n B 1 x n for each j=1,2,,l and Δ n 0 =I. We have that z n = Δ n l x n for all n1. From Lemma 2.5(1), it follows that

ϕ ( z n , x n ) = ϕ ( Δ n l x n , x n ) ϕ ( q , x n ) ϕ ( q , Δ n l x n ) = ϕ ( q , x n ) ϕ ( q , z n ) .

Taking limit as n on both sides of the inequality, we have

lim n ϕ( z n , x n )=0.

From (2.2) it follows that ( x n z n ) 2 0. Since x n p, we have

z n p(n).
(3.15)

Since J is uniformly norm-to-norm continuous on bounded subsets of E, it follows that

J z n Jp(n).
(3.16)

This implies that {J z n } is bounded in E and E is reflexive, we can assume that J z n z E . In view of J(E)= E , there exists zE such that Jz= z , and so

ϕ ( x n , z n ) = x n 2 2 x n , J z n + z n 2 = x n 2 2 x n , J z n + J z n 2 .

Taking lim inf n on both sides of the equality above and in view of the weak lower semi-continuity of the norm , it follows that

0 p 2 2 p , z + z 2 = p 2 2 p , J z + J z 2 = p 2 2 p , J z + z 2 = ϕ ( p , z ) .

From Remark 2.1, we have p=z, which implies that z =Jp and so J z n Jp E . From (3.16) and the Kadec-Klee property of E , we have J z n Jp as n. Note that J 1 is norm-weak-continuous, that is, z n p. From (3.15) and the Kadec-Klee property of E, we have

lim n z n =p.
(3.17)

For any qF, we note that

ϕ ( q , x n ) ϕ ( q , u n ) = x n 2 u n 2 2 q , J x n J u n x n u n ( x n + u n ) + 2 q J x n J u n .

Thus it follows from x n u n 0 and J x n J u n 0 that

ϕ(q, x n )ϕ(q, u n )0(n).
(3.18)

Denote Φ n i := K r i , n Θ i K r i 1 , n Θ i 1 K r 1 , n Θ 1 for each i=1,2,,m and Φ n 0 =I. We can rewrite u n as u n = Φ n m z n . It follows that for each i=1,2,,m, we have

ϕ ( q , u n ) = ϕ ( q , Φ n m z n ) ϕ ( q , Φ n m 1 z n ) ϕ ( q , Φ n m 2 z n ) ϕ ( q , Φ n i z n ) .
(3.19)

From Lemma 2.11(5), for each i=1,2,,m, we have

ϕ ( Φ n i z n , z n ) ϕ ( q , z n ) ϕ ( q , Φ n i z n ) ϕ ( q , x n ) ϕ ( q , Φ n i z n ) ϕ ( q , x n ) ϕ ( q , u n ) .
(3.20)

From (3.18) it follows that ϕ( Φ n i z n , z n )0 as n for each i=1,2,,m, and so from (2.2), that

( Φ n i z n z n ) 2 0.

Since z n p, we also have, for each i=1,2,,m,

Φ n i z n p(n).
(3.21)

Since { Φ n i z n } is bounded for each i=1,2,,m and E is reflexive, without loss of generality, we may assume that Φ n i z n h for all i=1,2,,m. From the first step, since C n is closed and convex for each n1, it is obvious that h C n . Again, since

ϕ ( Φ n i z n , z n ) = Φ n i z n 2 2 Φ n i z n , J z n + z n 2 ,

taking lim inf n on both sides of the equality above, we have

0 h 2 2h,Jp+ p 2 =ϕ(h,p).

This implies that h=p for each i=1,2,,m, and so

Φ n i z n p.
(3.22)

From (3.21), (3.22) and the Kadec-Klee property, for each i=1,2,,m, we have

lim n Φ n i z n =p.
(3.23)

By using the triangle inequality, for each i=1,2,,m, we obtain

Φ n i z n Φ n i 1 z n Φ n i z n p + p Φ n i 1 z n .

Hence, for each i=1,2,,m, we have

lim n Φ n i z n Φ n i 1 z n =0.
(3.24)

Since { r i , n }[a,) and J is uniformly norm-to-norm continuous on bounded subsets of E, for each i=1,2,3,,m, we have

lim n J Φ n i z n J Φ n i 1 z n r i , n =0.
(3.25)

From Lemma 2.11 we have, for each i=1,2,,m,

Θ i ( Φ n i z n , y ) + 1 r i , n y Φ n i z n , J Φ n i z n J Φ n i 1 z n 0,yC,

where Θ i ( u n ,y)= f i ( u n ,y)+y u n , A i u n for all u n ,yC. From the condition (A2), it follows that for each i=1,2,,m,

1 r i , n y Φ n i z n , J Φ n i z n J Φ n i 1 z n Θ i ( y , Φ n i z n ) ,yC.

From (3.23) and (3.25), for each i=1,2,,m, we have

0 Θ i (y,p),yC.
(3.26)

For any t[0,1] and yC, let y t =ty+(1t)p. Then we get y t C. From (3.26) it follows that for each i=1,2,,m,

Θ i ( y t ,p)0, y t C.
(3.27)

By the conditions (A1) and (A4), for each i=1,2,,m, we have

0 = Θ i ( y t , y t ) t Θ i ( y t , y ) + ( 1 t ) Θ i ( y t , p ) t Θ i ( y t , y ) Θ i ( y t , y ) .
(3.28)

From the condition (A3), we get

0 Θ i ( y t ,y)= Θ i ( t y + ( 1 t ) p , y ) .

Taking t0 in the equality above, for each i=1,2,,m, we have

0= lim t 0 0 lim t 0 Θ i ( t y + ( 1 t ) p , y ) = Θ i (p,y),yC,

that is, f i (p,y)+yp, A i p0 for all yC and i=1,2,,m. This implies that pGEP( f i , A i ) for each i=1,2,,m. Therefore, p i = 1 m GEP( f i , A i ).

Next, we show that p j = 1 l B j 1 0. Let z n = Δ n l x n for each n1. For any qF, it follows that for each j=1,2,,l,

ϕ ( q , z n ) = ϕ ( q , Δ n l x n ) ϕ ( q , Δ n l 1 x n ) ϕ ( q , Δ n l 2 x n ) ϕ ( q , Δ n j x n ) .
(3.29)

By Lemma 2.5 we have, for j=1,2,3,,m,

ϕ ( Δ n j x n , x n ) ϕ ( q , x n ) ϕ ( q , Δ n j x n ) ϕ ( q , x n ) ϕ ( q , z n ) .
(3.30)

Since x n p and z n p as n, we get ϕ( Δ n j x n , x n )0 as n for j=1,2,3,,m. From (2.2) it follows that

( Δ n j x n x n ) 2 0.

Since x n p, we also have

Δ n j x n pas n,j=1,2,3,,m.
(3.31)

This implies that for each j=1,2,3,,m, { Δ n j x n } is bounded and E is reflexive, without loss of generality, we assume that Δ n j x n k. We know that C n is closed and convex for each n1, it is obvious that k C n . Again, since

ϕ ( Δ n j x n , x n ) = Δ n j x n 2 2 Δ n j x n , J x n + x n 2 ,

taking lim inf n on both sides of equality above, we have

0 k 2 2 k , J p + p 2 = ϕ ( k , p ) .

That is, k=p, j=1,2,3,,l, it follows that

Δ n j x n p.
(3.32)

From (3.31), (3.32) and the Kadec-Klee property, it follows that

lim n Δ n j x n =p,j=1,2,3,,m.
(3.33)

We also have

lim n Δ n j 1 x n =p,j=1,2,3,,m.
(3.34)

It follows that

lim n Δ n j x n Δ n j 1 x n =0,j=1,2,3,,m.
(3.35)

Since J is uniformly norm-to-norm continuous on bounded subsets of E and

lim inf n λ j , n >0

for each j=1,2,,l, we have

lim n 1 λ j , n J Δ n j x n J Δ n j 1 x n =0.

Let Δ n j x n = J λ j , n j Δ n j 1 x n for each j=1,2,,l. Then we have

lim n B λ j , n Δ n j 1 x n = lim n 1 λ j , n J Δ n j x n J Δ n j 1 x n =0.

For any (w, w )G( B j ) and ( Δ n j x n , A λ j , n Δ n j 1 x n )G( B j ) for each j=1,2,,l, it follows from the monotonicity of B j that for all n0,

w Δ n j x n , w B λ j , n Δ n j 1 x n 0.

Letting n in the inequality above, we get wp, w 0. Since B j is maximal monotone for each j=1,2,,l, we obtain p j = 1 l B j 1 0.

Step 5. We show that p= Π F x 1 . From x n = Π C n x 1 , we have J x 1 J x n , x n z0 for all z C n . Since F C n , we also have

J x 1 J x n , x n y0,yF,

and so, taking limit n, we get

J x 1 Jp,py0,yF.

Therefore, by Lemma 2.3, we can conclude that p= Π F x 1 and x n p as n. The proof is completed. □

If i=1 and j=1, we have the following.

Corollary 3.2 Let C be a nonempty closed and convex subset of a uniformly smooth and strictly convex Banach space E with the Kadec-Klee property. Let f be a bifunction from C×C to satisfying the conditions (A1)-(A4) and let A:C E be a continuous and monotone mapping. Let BE× E be a maximal monotone operator satisfying D(B)C and J λ n = ( J + λ n B ) 1 J for all λ n >0. Assume that F:=GEP(f,A) B 1 0. For arbitrary x 1 C and C 1 =C, generate a sequence { x n } by

{ z n = J λ n x n , u n = K r n z n , C n + 1 = { z C n : ϕ ( z , u n ) ϕ ( z , x n ) } , x n + 1 = Π C n + 1 x 1 , n 1 ,
(3.36)

where { r n }[a,) for some a>0 and lim inf n λ n >0. Then the sequence { x n } converges strongly to a point pF, where p= Π F x 1 .

4 Applications

In this section, we apply our result to find a common solution of the variational inequality problems and zeros of the maximal operators.

We need the following lemma for our result, which is a special case of Lemmas 2.8 and 2.9 of [24].

Lemma 4.1 (Zegeye and Shahzad [16])

Let C be a closed convex subset of a uniformly smooth, strictly convex real Banach space E. Let A:C E be a continuous monotone mapping. For any r>0 and xE, define a mapping T r :EC as follows:

T r x= { z C : y z , A z + 1 r y z , J z J x 0 , y C } ,xE.

Then we have the following:

  1. (1)

    T r is single-valued;

  2. (2)

    T r is a firmly nonexpansive-type mapping, i.e., for any x,yE,

    T r x T r y,J T r xJ T r y T r x T r y,JxJy;
  3. (3)

    F( T r )=VI(C,A);

  4. (4)

    VI(C,A) is closed and convex;

  5. (5)

    ϕ(p, T r x)+ϕ( T r x,x)ϕ(p,x) for any pF( T r ).

Theorem 4.2 Let C be a nonempty closed and convex subset of a uniformly smooth and strictly convex Banach space E with the Kadec-Klee property. For each i=1,2,,m, let { A i } be a finite family of continuous and monotone mappings C E . For r n >0 and xE, define a mapping T r i , n :EC by

T r i , n x:= { z C : y z , A i z + 1 r n y z , J z J x 0 , y C } .

Let BE× E be a maximal monotone operator satisfying D(B)C and J λ j , n B j = ( J + λ j , n B j ) 1 J for all λ>0 and j=1,2,,l. Assume that F:=( i = 1 m VI(C, A i ))( j = 1 l B j 1 0). For an initial point x 1 E with C 1 =C, define the sequence { x n } in C as follows:

{ z n = J λ l , n B l J λ l 1 , n B l 1 J λ 1 , n B 1 x n , u n = T r m , n T r m 1 , n T r 1 , n z n , C n + 1 = { z C n : ϕ ( z , u n ) ϕ ( z , x n ) } , x n + 1 = Π C n + 1 x 1 , n 1 ,
(4.1)

where { r i , n }[a,) for some a>0 for all i=1,2,,m and lim inf n λ j , n >0 for all j=1,2,,l. Then the sequence { x n } converges strongly to a point pF, where p= Π F x 1 .

Proof Taking f i ( u n ,y)=0 for all i=1,2,,m in Theorem 3.1, we can get the desired conclusion. □

By Theorem 4.2, if we set B j 0 for each j=1,2,,l, we obtain the following.

Corollary 4.3 Let C be a nonempty closed and convex subset of a uniformly smooth and strictly convex Banach space E with the Kadec-Klee property. For any i=1,2,,m, let { A i } be a finite family of continuous and monotone mappings C E . Assume that F:= i = 1 m VI(C, A i ). For an initial point x 1 E with C 1 =C, define the sequence { x n } in C as follows:

{ u n = T r m , n T r m 1 , n T r 1 , n x n , C n + 1 = { z C n : ϕ ( z , u n ) ϕ ( z , x n ) } , x n + 1 = Π C n + 1 x 1 , n 1 ,
(4.2)

where { r i , n }[a,) for some a>0 for each i=1,2,,m. Then the sequence { x n } converges strongly to a point pF, where p= Π F x 1 .

Remark 4.4 Corollary 4.3 extends and improves the result of Zegeye and Shahzad [16] to a common solution of the variational inequality problems.

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Acknowledgements

This work was supported by Thaksin University Research Fund and was also supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant Number: 2012-0008170).

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Saewan, S., Kumam, P. & Cho, Y.J. Convergence theorems for finding zero points of maximal monotone operators and equilibrium problems in Banach spaces. J Inequal Appl 2013, 247 (2013). https://doi.org/10.1186/1029-242X-2013-247

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Keywords

  • maximal monotone operators
  • hybrid projection method
  • system of generalized equilibrium problems