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Graph convergence for the H(,)-mixed mappingwith an application for solving the system of generalized variationalinclusions

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Abstract

In this paper, we investigate a class of accretive mappings called theH(,)-mixed mappingsin Banach spaces. We prove that the proximal-point mapping associated with theH(,)-mixed mapping issingle-valued and Lipschitz continuous. Some examples are given to justify thedefinition of H(,)-mixed mapping.Further, a concept of graph convergence concerned with theH(,)-mixed mapping isintroduced in Banach spaces and some equivalence theorems betweengraph-convergence and proximal-point mapping convergence for theH(,)-mixed mappingssequence are proved. As an application, we consider a system of generalizedvariational inclusions involving H(,)-mixed mappingsin real q-uniformly smooth Banach spaces. Using the proximal-pointmapping method, we prove the existence and uniqueness of solution and suggest aniterative algorithm for the system of generalized variational inclusions.Furthermore, we discuss the convergence criteria for the iterative algorithmunder some suitable conditions.

MSC: 47J19, 49J40, 49J53.

1 Introduction

Variational inclusions, as the generalization of variational inequalities, have beenwidely studied in recent years. Some of the most interesting and important problemsin the theory of variational inclusions include variational, quasi-variational,variational-like inequalities as special cases. For applications of variationalinclusions, we refer to [1]. Various kinds ofiterative methods have been studied to solve the variational inclusions. Among thesemethods, the proximal-point mapping technique for the study of variationalinclusions has been widely used by many authors. For details, we refer to[220].

In 2001, Huang and Fang [5] were the first tointroduce the generalized m-accretive mapping and give the definition ofthe proximal-point mapping for the generalized m-accretive mapping inBanach spaces. Since then a number of researchers have investigated several classesof generalized m-accretive mappings such as H-accretive,H,η-accretive,(P,η)-proximal-point,(P,η)-accretive,A-maximal relaxed accretive, (A,η)-accretive mappings.For details, we refer to [2, 3, 6, 7, 11, 14, 16, 18].

Recently, Zou and Huang [19, 20] introduced and studied H(,)-accretive mappings;Kazmi et al.[810] introduced and studied generalized H(,)-accretive mappings,H(,)-η-proximal-point mappings. Veryrecently, Li and Huang [12] studied thegraph convergence for the H(,)-accretive mappingand showed the equivalence between graph convergence and proximal-point mappingconvergence for the H(,)-accretive mappingsequence in a Banach space, and Verma [17]studied the graph convergence for an A-maximal relaxed monotone mapping andgave the equivalence between the graph convergence and the proximal-point mappingconvergence for the A-maximal relaxed monotone mapping sequence in aHilbert space. They extended the concept of graph convergence introduced andconsidered by Attouch [21].

Motivated by the research work going on in this direction, we consider a class ofaccretive mappings called H(,)-mixed mappings, anatural generalization of accretive (monotone) mappings in Banach spaces. Forrelated work, we refer to [24, 11, 14, 16, 1820]. We prove that theproximal-point mapping of the H(,)-mixed mapping issingle-valued and Lipschitz continuous and extends the concept of proximal-pointmappings associated with the H(,)-accretive mappingsto the H(,)-mixed mappings.Further, we study the graph convergence for the H(,)-mixed mappings. Wepresent an equivalence theorem between graph convergence and proximal-point mappingconvergence for the H(,)-mixed mappingsequence in Banach spaces. As an application, we consider a system of generalizedvariational inclusions involving the H(,)-mixed mappings inreal q-uniformly smooth Banach spaces. Using the proximal-point mappingmethod, we prove the existence and uniqueness of solution and suggest an iterativealgorithm for the system of generalized variational inclusions. Furthermore, wediscuss the convergence criteria of the iterative algorithm under some suitableconditions. Our results can be viewed as a generalization of some known resultsgiven in [12, 17, 1921].

2 Preliminaries

Let X be a real Banach space equipped with the norm , and let X be the topological dual space of X. Let , be the dualpair between X and X ,and let 2 X be the power set of X.

Definition 2.1[22]

For q>1,a mapping J q :X 2 X is said to be a generalized duality mapping if it is defined by

J q (x)= { f X : x , f = x q , f = x q 1 } ,xX.

In particular, J 2 is the usual normalized duality mapping on X. It is known that, ingeneral,

J q (x)= x q 1 J 2 (x)x(0)X.

If XHa real Hilbert space, then J 2 becomesan identity mapping on H.

Definition 2.2[22]

A Banach space X is called smooth if, for everyxXwith x=1,there exists a unique f X such that f=f(x)=1.

The modulus of smoothness of X is a function ρ X :[0,)[0,)defined by

ρ X (t)=sup { 1 2 ( x + y + x y ) 1 : x 1 , y t } .

Definition 2.3[22]

A Banach space X is called

  1. (i)

    uniformly smooth if

    lim t 0 ρ X ( t ) t =0;
  2. (ii)

    q-uniformly smooth, for q>1, if there exists a constant c>0 such that

    ρ X (t)c t q ,t[0,).

Note that J q is single-valued if X is uniformly smooth. Concerned with thecharacteristic inequalities in q-uniformly smooth Banach spaces, Xu[22] proved the following result.

Lemma 2.4 Let X be a real uniformly smooth Banach space. Then X is q-uniformly smooth if and only if there exists a constant c q >0such that, for allx,yX,

x + y q x q +q y , J q ( x ) + c q y q .

From Lemma 2 of Liu [13], it is easy tohave the following lemma.

Lemma 2.5 Let { a n } and { b n } be two nonnegative real sequences satisfying

a n + 1 k a n + b n

with0<k<1and b n 0. Then lim n a n =0.

Definition 2.6 Let G:XXbe a single-valued mapping. Then

  1. (i)

    G is said to be accretive if

    G ( x ) G ( y ) , J q ( x y ) 0,x,yX;
  2. (ii)

    G is said to be ξ-strongly accretive if there exists a constant ξ>0 such that

    G ( x ) G ( y ) , J q ( x y ) ξ x y q ,x,yX;
  3. (iii)

    G is said to be μ-cocoercive if there exists a constant μ>0 such that

    G ( x ) G ( y ) , J q ( x y ) μ G ( x ) G ( y ) q ,x,yX;
  4. (iv)

    G is said to be λ G -Lipschitz continuous if there exists a constant λ G >0 such that

    G ( x ) G ( y ) λ G xy,x,yX;
  5. (v)

    G is said to be α-expansive if there exists a constant α>0 such that

    G ( x ) G ( y ) αxy,x,yX;

if α=1,then it is expansive.

Definition 2.7 Let H:X×XXand A,B:XXbe three single mappings. Then

  1. (i)

    H(A,) is said to be μ-cocoercive with respect to A if there exists a constant μ>0 such that

    H ( A x , u ) H ( A y , u ) , J q ( x y ) μ A x A y q ,x,y,uX;
  2. (ii)

    H(,B) is said to be γ-relaxed accretive with respect to B if there exists a constant γ>0 such that

    H ( u , B x ) H ( u , B y ) , J q ( x y ) (γ) x y q ,x,y,uX;
  3. (iii)

    H(A,) is said to be r 1 -Lipschitz continuous with respect to A if there exists a constant r 1 >0 such that

    H ( A x , ) H ( A y , ) r 1 xy,x,yX;
  4. (iv)

    H(,B) is said to be r 2 -Lipschitz continuous with respect to B if there exists a constant r 2 >0 such that

    H ( , B x ) H ( , B y ) r 2 xy,x,yX.

Example 2.8 Let us consider the 2-uniformly smooth Banach spaceX= R 2 with the usual innerproduct. Let A,B: R 2 R 2 bedefined by

Ax= ( m x 1 m x 2 m x 1 + 2 m x 2 ) ,By= ( m y 1 + m y 2 m y 1 m y 2 )

for all scalers mR and for all x=( x 1 , x 2 ),y=( y 1 , y 2 ) R 2 .

Suppose that H: R 2 × R 2 R 2 isdefined by H(Ax,By)=Ax+By,then H(A,B) is 1 3 m -cocoercivewith respect to A and m-relaxed accretive with respect toB, and 5 m-Lipschitzcontinuous with respect to A and 2 m-Lipschitzcontinuous with respect to B.

Indeed, let for any uX,

H ( A x , u ) H ( A y , u ) , x y = A x A y , x y = ( m x 1 m x 2 , m x 1 + 2 m x 2 ) ( m y 1 m y 2 , m y 1 + 2 m y 2 ) , ( x 1 y 1 , x 2 y 2 ) = ( m ( x 1 y 1 ) m ( x 2 y 2 ) , m ( x 1 y 1 ) + 2 m ( x 2 y 2 ) ) , ( x 1 y 1 , x 2 y 2 ) = m ( x 1 y 1 ) 2 2 m ( x 1 y 1 ) ( x 2 y 2 ) + 2 m ( x 2 y 2 ) 2 , A x A y 2 = A x A y , A x A y A x A y 2 = ( ( m x 1 m x 2 , m x 1 + 2 m x 2 ) ( m y 1 m y 2 , m y 1 + 2 m y 2 ) ) , A x A y 2 = ( ( m x 1 m x 2 , m x 1 + 2 m x 2 ) ( m y 1 m y 2 , m y 1 + 2 m y 2 ) ) A x A y 2 = 2 m 2 ( x 1 y 1 ) 2 6 m 2 ( x 1 y 1 ) ( x 2 y 2 ) + 5 m 2 ( x 2 y 2 ) 2 A x A y 2 3 m 2 ( x 1 y 1 ) 2 6 m 2 ( x 1 y 1 ) ( x 2 y 2 ) + 6 m 2 ( x 2 y 2 ) 2 A x A y 2 = 3 m { m ( x 1 y 1 ) 2 2 m ( x 1 y 1 ) ( x 2 y 2 ) + 2 m ( x 2 y 2 ) 2 } A x A y 2 = 3 m { H ( A x , u ) H ( A y , u ) , x y } ,

which implies that

H ( A x , u ) H ( A y , u ) , x y 1 3 m A x A y 2 ,

that is, H(A,B) is 1 3 m -cocoercivewith respect to A.

H ( u , B x ) H ( u , B y ) , x y = B x B y , x y = B x B y , x y = ( m x 1 + m x 2 , m x 1 m x 2 ) ( m y 1 + m y 2 , m y 1 m y 2 ) , ( x 1 y 1 , x 2 y 2 ) = ( m ( x 1 y 1 ) + m ( x 2 y 2 ) , m ( x 1 y 1 ) m ( x 2 y 2 ) ) , ( x 1 y 1 , x 2 y 2 ) = m ( x 1 y 1 ) 2 m ( x 2 y 2 ) 2 = m { ( x 1 y 1 ) 2 + ( x 2 y 2 ) 2 } m x y 2 ,

which implies that

H ( u , B x ) H ( u , B y ) , x y (m) x y 2 ,

that is, H(A,B) ism-relaxed accretive with respect to B.

H ( A x , u ) H ( A y , u ) 2 = A x A y 2 = A x A y , A x A y = ( ( m x 1 m x 2 , m x 1 + 2 m x 2 ) ( m y 1 m y 2 , m y 1 + 2 m y 2 ) ) , ( ( m x 1 m x 2 , m x 1 + 2 m x 2 ) ( m y 1 m y 2 , m y 1 + 2 m y 2 ) ) = 2 m 2 ( x 1 y 1 ) 2 6 m 2 ( x 1 y 1 ) ( x 2 y 2 ) + 5 m 2 ( x 2 y 2 ) 2 5 m 2 ( x 1 y 1 ) 2 + 5 m 2 ( x 2 y 2 ) 2 ,

which implies that

H ( A x , u ) H ( A y , u ) 5 mxy,

that is, H(A,B) is 5 m-Lipschitzcontinuous with respect to A.

H ( u , B x ) H ( u , B y ) 2 = B x B y 2 = B x B y , B x B y = ( ( m x 1 + m x 2 , m x 1 m x 2 ) ( m y 1 + m y 2 , m y 1 m y 2 ) ) , ( ( m x 1 + m x 2 , m x 1 m x 2 ) ( m y 1 + m y 2 , m y 1 m y 2 ) ) = 2 m 2 ( x 1 y 1 ) 2 + 2 m 2 ( x 2 y 2 ) 2

which implies that

H ( u , B x ) H ( u , B y ) 2 mxy,

that is, H(A,B) is 2 m-Lipschitzcontinuous with respect to B.

Definition 2.9 Let η:X×XXand H,A,B:XXbe mappings. Let M:X 2 X be a set-valued mapping. Then

  1. (i)

    η is said to be τ-Lipschitz continuous if there exists a constant τ>0 such that

    η ( x , y ) τxy,x,yX;
  2. (ii)

    M is said to be accretive if

    u v , J q ( x y ) 0,x,yX,uMx,vMy;
  3. (iii)

    M is said to be μ -strongly accretive if there exists a constant μ >0 such that

    u v , J q ( x y ) μ x y q ,x,yX,uMx,vMy;
  4. (iv)

    M is said to be m-relaxed accretive if there exists a constant m>0 such that

    u v , J q ( x y ) m x y q ,x,yX,uMx,vMy;
  5. (v)

    M is said to be η-accretive if

    u v , J q ( η ( x , y ) ) 0,x,yX,uMx,vMy;
  6. (vi)

    M is said to be strictly η-accretive if M is η-accretive and equality holds if and only if x=y;

  7. (vii)

    M is said to be γ-strongly η-accretive if there exists a constant γ>0 such that

    u v , J q ( η ( x , y ) ) γ x y q ,x,yX,uMx,vMy;
  8. (viii)

    M is said to be α-relaxed η-accretive if there exists a constant α>0 such that

    u v , J q ( η ( x , y ) ) (α) x y q ,x,yX,uMx,vMy;
  9. (ix)

    M is said to be m-accretive if M is accretive and (I+ρM)(X)=X for all ρ>0, where I denotes the identity operator on X;

  10. (x)

    M is said to be generalized m-accretive if M is η-accretive and (I+ρM)(X)=X for all ρ>0;

  11. (xi)

    M is said to be H-accretive if M is accretive and (H+ρM)(X)=X for all ρ>0;

  12. (xii)

    M is said to be (H,η)-accretive if M is η-accretive and (H+ρM)(X)=X for all ρ>0;

  13. (xiii)

    M is said to be (A,η)-accretive if M is m-relaxed η-accretive and (A+ρM)(X)=X for all ρ>0.

Definition 2.10[19]

Let A,B:XX,H:X×XXbe three single-valued mappings. Let M:X 2 X be a set-valued mapping. Then M is said to be H(,)-accretivewith respect to A and B if M is accretive and(H(,)+ρM)(X)=X for allρ>0.

3 H(,)-mixedmappings

In this section, we introduce the H(,)-mixed mapping andshow some of its properties.

Definition 3.1 Let H:X×XX,A,B:XXbe three single-valued mappings. Let H(A,B) beμ-cocoercive with respect to A, γ-relaxedaccretive with respect to B. Then the set-valued mappingM:X 2 X is said be H(,)-mixed withrespect to mappings A and B if

  1. (i)

    M is m-relaxed accretive;

  2. (ii)

    (H(A,B)+ρM)(X)=X for all ρ>0.

Example 3.2 Let X, H, A, B be the same asin Example 2.8, and M: R 2 R 2 bedefined by M(x)=(3π,3 x 2 ),x=( x 1 , x 2 ) R 2 .

We claim that M is a 3-relaxed accretive mapping. Indeed, for anyx=( x 1 , x 2 ),y=( y 1 , y 2 ) R 2

M x M y , x y = ( 3 π , 3 x 2 ) ( 3 π , 3 y 2 ) , ( ( x 1 y 1 ) , ( x 2 y 2 ) ) M x M y , x y = ( 0 , 3 ( x 2 y 2 ) ) , ( ( x 1 y 1 ) , ( x 2 y 2 ) ) M x M y , x y = 3 ( x 2 y 2 ) 2 M x M y , x y 3 { ( x 1 y 1 ) 2 + ( x 2 y 2 ) 2 } M x M y , x y 3 x y 2 , M x M y , x y ( 3 ) x y 2 .

Furthermore, M is also an H(,)-mixed mapping since(H(A,B)+ρM)( R 2 )= R 2 for anyρ>0.

Proposition 3.3 Let the set-valued mappingM:X 2 X be anH(,)-mixed mappingwith respect to mappings A and B. If A is α-expansive andμ>γwithr=μ α q γ>m,then the following inequality holds:

x y , J q ( u v ) 0,(v,y)graph(M),impliesxMu.

Proof Suppose on contrary that there exists ( u 0 , x 0 )graphMsuch that

x 0 y , J q ( u 0 v ) 0,(v,y)graph(M).
(3.1)

Since M is an H(,)-mixed mapping, weknow that (H(A,B)+ρM)(X)=X holdsfor every ρ>0,and so there exists ( u 1 , x 1 )graph(M) such that

H(A u 0 ,B u 0 )+ρ x 0 =H(A u 1 ,B u 1 )+ρ x 1 X.
(3.2)

Now

ρ x ρ x 1 = H ( A u 1 , B u 1 ) H ( A u , B u ) X , ρ x ρ x 1 , J q ( u u 1 ) = H ( A u 1 , B u 1 ) H ( A u , B u 0 ) , J q ( u u 1 ) .

Setting (v,y)=( u 1 , x 1 ) in (3.1) and thenfrom the resultant (3.2) and m-relaxed accretivity of M, we obtain

m u 0 u 1 q ρ x 0 x 1 , J q ( u 0 u 1 ) = H ( A u 0 , B u 0 ) H ( A u 1 , B u 1 ) , J q ( u 0 u 1 ) = H ( A u 0 , B u 0 ) H ( A u 1 , B u 0 ) , J q ( u 0 u 1 ) H ( A u 1 , B u 0 ) H ( A u 1 , B u 1 ) , J q ( u 0 u 1 ) .
(3.3)

Since H(A,B) isμ-cocoercive with respect to A and γ-relaxedaccretive with respect to B, and A is α-expansive,thus (3.3) becomes

m u 0 u 1 q μ A u A u 1 q + γ u u 1 q μ α 2 u 0 u 1 q + γ u u 1 q ( μ α q γ ) u 0 u 1 q = r u 0 u 1 q 0 , where  r = μ α q γ ( r m ) u 0 u 1 q 0 .

It implies that u 0 = u 1 since r>m.By (3.1), we have x 0 = x 1 ,a contradiction. This completes the proof. □

Theorem 3.4 Let the set-valued mappingM:X 2 X be anH(,)-mixed mappingwith respect to mappings A and B. If A is α-expansive andμ>γwithr=μ α q γ>ρm,then ( H ( A , B ) + ρ M ) 1 is single-valued.

Proof For any given uX, letx,y ( H ( A , B ) + ρ M ) 1 (u). It follows that

H ( A x , B x ) + u ρ M x , H ( A y , B y ) + u ρ M y .

Since M is m-relaxed accretive, we have

m x y q 1 ρ H ( A x , B x ) + u ( H ( A y , B y ) + u ) , J q ( x y ) , m ρ x y q = H ( A x , B x ) H ( A y , B x ) , J q ( x y ) m ρ x y q = H ( A y , B x ) H ( A y , B y ) , J q ( x y ) ,

which is like (3.3). Hence it follows that xy0.This implies that x=y and so ( H ( A , B ) + ρ M ) 1 is single-valued. □

Definition 3.5 Let the set-valued mapping M:X 2 X be an H(,)-mixed mapping withrespect to mappings A and B. If A isα-expansive and μ>γwith r=μ α q γ>ρm,then the proximal-point mapping R ρ , M H ( , ) :XXis defined by

R ρ , M H ( , ) (u)= ( H ( A , B ) + ρ M ) 1 (u),uX.
(3.4)

Now we prove that the proximal-point mapping defined by (3.4) is Lipschitzcontinuous.

Theorem 3.6 Let the set-valued mappingM:X 2 X be anH(,)-mixed mappingwith respect to mappings A and B. If A is α-expansive andμ>γwithr=μ α q γ>ρm,then the proximal-point mapping R ρ , M H ( , ) :XXis 1 r ρ m -Lipschitzcontinuous, that is,

R ρ , M H ( , ) ( u ) R ρ , M H ( , ) ( v ) 1 r ρ m uv,u,vX.

Proof Let u and vXbe any given points in X. It follows from (3.2) that

{ R ρ , M H ( , ) ( u ) = H ( ( A , B ) + ρ M ) ) 1 ( u ) , R ρ , M H ( , ) ( v ) = H ( ( A , B ) + ρ M ) ) 1 ( v ) , 1 ρ ( u H ( A ( R ρ , M H ( , ) ( u ) ) , B ( R ρ , M H ( , ) ( u ) ) ) M ( R ρ , M H ( , ) ( u ) ) , 1 ρ ( v H ( A ( R ρ , M H ( , ) ( v ) ) , B ( R ρ , M H ( , ) ( v ) ) ) M ( R ρ , M H ( , ) ( v ) ) .

Let z 1 = R ρ , M H ( , ) (u) and z 2 = R ρ , M H ( , ) (v).

Since M is m-relaxed accretive, we have

1 ρ ( u H ( A ( z 1 ) , B ( z 1 ) ) ( v H ( A ( z 2 ) , B ( z 2 ) ) ) , J q ( z 1 z 2 ) m z 1 z 2 q , u v ( H ( A ( z 1 ) , B ( z 1 ) ) H ( A ( z 2 ) , B ( z 2 ) ) ) , J q ( z 1 z 2 ) ρ m z 1 z 2 q ,

which implies that

u v z 1 z 2 q 1 u v , J q ( z 1 z 2 ) H ( A ( z 1 ) , B ( z 1 ) ) H ( A ( z 2 ) , B ( z 2 ) ) , J q ( z 1 z 2 ) ρ m z 1 z 2 q H ( A ( z 1 ) , B ( z 1 ) ) H ( A ( z 2 ) , B ( z 1 ) ) , J q ( z 1 z 2 ) H ( A ( z 2 ) , B ( z 1 ) ) H ( A ( z 2 ) , B ( z 2 ) ) , J q ( z 1 z 2 ) ρ m z 1 z 2 q μ A ( z 1 ) A ( z 2 ) q γ z 1 z 2 q ρ m z 1 z 2 q μ α q z 1 z 2 q γ z 1 z 2 q ρ m z 1 z 2 q = ( μ α q γ ρ m ) z 1 z 2 q , = ( r ρ m ) z 1 z 2 q , where  r = μ α q γ ,

and hence

uv z 1 z 2 q 1 (rρm) z 1 z 2 q ,

that is,

R ρ , M H ( , ) ( u ) R ρ , M H ( , ) ( v ) 1 r ρ m uv,u,vX.

This completes the proof. □

4 Graph convergence for an H(,)-mixedmapping

Let M:X 2 X be a set-valued mapping. The graph of the map M is defined by

graph(M)= { ( x , y ) X × X : y M ( X ) } .

In this section we shall introduce the graph convergence for theH(,)-mixed mapping.

Definition 4.1 Let M n ,M:X 2 X be the set-valued mappings such that M, M n areH(,)-mixed mappings withrespect to the mappings A and B for n=0,1,2, .The sequence { M n } is said to begraph convergent to M, denoted by M n G M,if for every (x,y)graph(M), there exists a sequence( x n , y n )graph( M n ) such that

x n x, y n yas n.

Theorem 4.2 Let M n ,M:X 2 X be the set-valued mappings such that M, M n areH(,)-mixed mappingswith respect to the mappings A and B forn=0,1,2, .LetH(A,B)be s-Lipschitz continuous with respect to A and t-Lipschitz continuous with respect to B. If A is α-expansive andμ>γwithr=μ α q γ>ρm,then M n G Mif and only if

R ρ , M n H ( , ) (u) R ρ , M H ( , ) (u),uX,ρ>0,

where

R ρ , M n H ( , ) (u)= ( H ( A , B ) + ρ M n ) 1 (u), R ρ , M H ( , ) (u)= ( H ( A , B ) + ρ M ) 1 (u).

Proof It follows from Theorem 3.6 that R ρ , M n H ( , ) and R ρ , M H ( , ) are both 1 r ρ m -Lipschitzcontinuous.

If part: Suppose that M n G M.For any given xX,let

z n = R ρ , M n H ( , ) (x),z= R ρ , M H ( , ) (x).

Then

1 ρ [ x H ( A z , B z ) ] M(z),that is ( z , 1 ρ [ x H ( A z , B z ) ] ) graph(M).

In the light of Definition 4.1, we know that there exists a sequence( z n , y n )graph( M n ) such that

z n z, y n 1 ρ [ x H ( A z , B z ) ] as n.
(4.1)

Since y n M n ( z n ), we have

H ( A z n , B z n ) +ρ y n [ H ( A , B ) + ρ M n ] ( z n )

and so

z n = [ H ( A z n , B z n ) + ρ y n ] .

From the Lipschitz continuity of M n , we get

z n z z n z n + z n z = R ρ , M n H ( , ) ( x ) R ρ , M n H ( , ) [ H ( A z n , B z n ) + ρ y n ] + z n z 1 r ρ m x H ( A z n , B z n ) ρ y n + z n z 1 r ρ m [ x H ( A z , B z ) ρ y n + H ( A z , B z ) H ( A z n , B z n ) ] + z n z .
(4.2)

From the Lipschitz continuity of H(A,B), we have

H ( A z , B z ) H ( A z n , B z n ) H ( A z , B z ) H ( A z , B z n ) + H ( A z , B z n ) H ( A z n , B z n ) ( s + t ) z n z .
(4.3)

It follows from (4.2) and (4.3) that

z n z 1 r ρ m x H ( A z , B z ) ρ y n + [ 1 + 1 r ρ m ( s + t ) ] z n z .

By (4.1), we have

z n z 0, 1 ρ [ x H ( A z , B z ) y n ] 0

and so

z n z0as n.

Only if part: Suppose that

R ρ , M n H ( , ) R ρ , M H ( , ) ,uX,ρ>0.

For any given (x,y)graph(M), we have

H(Ax,Bx)+ρy [ H ( A , B ) + ρ M ] (x)

and so

x= R ρ , M H ( , ) [ H ( A x , B x ) + ρ y ] .

Let

x n = R ρ , M n H ( , ) [ H ( A x , B x ) + ρ y ] .

Then

1 ρ [ H ( A x , B x ) H ( A x n , B x n ) + ρ y ] M n ( x n ).

Let

y n = 1 ρ [ H ( A x , B x ) H ( A x n , B x n ) + ρ y ] .

It follows from (4.3) that

y n y 1 ρ [ H ( A x , B x ) H ( A x n , B x n ) + ρ y ] y = 1 ρ H ( A x , B x ) H ( A x n , B x n ) 1 ρ ( s + t ) x n x .
(4.4)

Since R ρ , M n H ( , ) R ρ , M H ( , ) for anyuX,we know that x n x0. Now (4.4)implies that

y n yas n,

and so M n G M.This completes the proof. □

5 An application of the H(,)-mixed mappingfor solving the system of generalized variational inclusions

Throughout the rest of the paper, unless otherwise stated, we assume that for eachi=1,2, E i is a q i -uniformlysmooth Banach space with the norm i .

Let A 1 , B 1 : X 1 X 1 , A 2 , B 2 : X 2 X 2 , N 1 , H 1 : X 1 × X 2 X 1 and N 2 , H 2 : X 1 × X 2 X 2 be nonlinear mappings. Let M 1 : X 1 2 X 1 be H 1 (,)-mixed and M 2 : X 2 2 X 2 be H 2 (,)-mixed mappings,respectively. We consider the following system of generalized variational inclusions(SGVI): Find (x,y) X 1 × X 2 such that

{ θ 1 N 1 ( x , y ) + M 1 ( x ) ; θ 2 N 2 ( x , y ) + M 2 ( y ) ,
(5.1)

where θ 1 , θ 2 are zero vectors of X 1 and X 2 ,respectively. The problem of type (5.1) was studied by Zou and Huang [20].

Definition 5.1 Let A: X 1 X 1 .A mapping N: X 1 × X 2 X 1 is said to be:

  1. (i)

    κ-strongly accretive in the first argument with respect to A if there exists a constant κ>0 such that

    N ( x 1 , y ) N ( x 2 , y ) , J q ( A ( x 1 ) A ( x 2 ) ) 1 κ x y 1 q 1 , x 1 , x 2 X 1 ,y X 2 ;
  2. (ii)

    L N -Lipschitz continuous in the first argument if there exists a constant L N >0 such that

    N ( x 1 , y ) N ( x 2 , y ) 1 L N x 1 x 2 1 q 1 , x 1 , x 2 X 1 ,y X 2 ;
  3. (iii)

    l N -Lipschitz continuous in the second argument if there exists a constant l N >0 such that

    N ( x , y 1 ) N ( x , y 2 ) 1 l N y 1 y 2 2 q 2 ,x X 1 , y 1 , y 2 X 2 .

The following lemma, which will be used in the sequel, is an immediate consequence ofthe definitions of R ρ 1 , M 1 H 1 ( , ) , R ρ 2 , M 2 H 2 ( , ) .

Lemma 5.2 For any given(x,y) X 1 × X 2 ,(x,y)is a solution of (SGVI) (5.1) if and only if(x,y)satisfies

x= R ρ 1 , M 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x ) ρ 1 N 1 ( x , y ) ] ,
(5.2)
y= R ρ 2 , M 2 H 2 ( , ) [ H 2 ( A 2 , B 2 ) ( y ) ρ 2 N 2 ( x , y ) ] ,
(5.3)

where R ρ 1 , M 1 H 1 ( , ) = ( H 1 ( A 1 , B 1 ) + ρ 1 M 1 ) 1 and R ρ 2 , M 2 H 2 ( , ) = ( H 2 ( A 2 , B 2 ) + ρ 2 M 2 ) 1 ,and ρ 1 , ρ 2 >0are constants.

Proof Consider first that an element (x,y) X 1 × X 2 is a solution to (5.1). Then it follows that

θ 1 N 1 ( x , y ) + M 1 ( x ) H 1 ( A 1 x , B 1 x ) H 1 ( A 1 x , B 1 x ) + ρ 1 N 1 ( x , y ) + ρ 1 M 1 ( x ) H 1 ( A 1 x , B 1 x ) ρ 1 N 1 ( x , y ) H 1 ( A x , B x ) + ρ 1 M 1 ( x ) x = R ρ 1 , M 1 H 1 ( , ) [ H 1 ( A 1 x , B 1 x ) ρ 1 N 1 ( x , y ) ] .

In a similar way, we can show that

y= R ρ 2 , M 2 H 2 ( , ) [ H 2 ( A 2 y , B 2 y ) ρ 2 N 2 ( x , y ) ] .

 □

A similar proof follows for the converse part:

x = R ρ 1 , M 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x ) ρ 1 N 1 ( x , y ) ] x = ( H 1 ( A 1 , B 1 ) + ρ 1 M 1 ) 1 [ H 1 ( A 1 , B 1 ) ( x ) ρ 1 N 1 ( x , y ) ] ( H 1 ( A 1 , B 1 ) + ρ 1 M 1 ) ( x ) ( H 1 ( A 1 x , B 1 x ) ) ρ 1 N 1 ( x , y ) θ 1 N 1 ( x , y ) + M 1 ( x ) .

In a similar way, we can show that

θ 2 N 2 (x,y)+ M 2 (y).

Theorem 5.3 For eachi=1,2,let X i be q i -uniformlysmooth Banach spaces, let A i , B i : X i X i be single-valued mappings. Let the set-valuedmappings M i : X i 2 X i be such that M i are H i (,)-mixed mappingswith respect to mappings A i and B i ,and A i are α i -expansiveand μ i > γ i with r i = μ i α i q i γ i > ρ i m i .Let H i : X 1 × X 2 X i be s i -Lipschitzcontinuous with respect to A i and t i -Lipschitzcontinuous with respect to B i ,and let N i : X 1 × X 2 X i be a κ i -stronglyaccretive mapping in the ith argument, L N i -Lipschitzcontinuous in the first argument and l N i -Lipschitzcontinuous in the second argument. Suppose that there are twoconstants ρ 1 , ρ 2 >0satisfying the following conditions:

{ τ 1 = a 1 + ρ 2 L 2 L N 2 < 1 ; τ 2 = a 2 + ρ 2 L 1 l N 1 < 1 ,
(5.4)

where

a 1 = L 1 [ ( 1 2 q 1 r 1 + c q 1 ( s 1 + t 1 ) q 1 ) 1 q 1 + ( 1 2 ρ 1 q 1 κ 1 + c q 1 ρ 1 q 1 L 1 q 1 ) 1 q 1 ] ; a 2 = L 2 [ ( 1 2 q 2 r 2 + c q 2 ( s 2 + t 2 ) q 2 ) 1 q 2 + ( 1 2 ρ 2 q 2 κ 2 + c q 2 ρ 2 q 2 L 2 q 2 ) 1 q 2 ] ; L 1 = 1 r 1 ρ 1 m 1 ; L 2 = 1 r 2 ρ 2 m 2 .

Then SGVI (5.1) has a unique solution(x,y) X 1 × X 2 .

Proof For i=1,2,it follows that for (x,y) X 1 × X 2 ,the proximal-point mappings R ρ 1 , M 1 H 1 ( , ) and R ρ 2 , M 2 H 2 ( , ) are 1 r 1 ρ 1 m 1 -Lipschitzcontinuous and 1 r 2 ρ 2 m 2 -Lipschitzcontinuous, respectively.

Let R: X 1 × X 2 X 1 × X 2 be defined as follows:

R(x,y)= ( P ( x , y ) , Q ( x , y ) ) ,(x,y) X 1 × X 2 ,
(5.5)

where P: X 1 × X 2 X 1 and Q: X 1 × X 2 X 2 are defined by

P(x,y)= R ρ 1 , M 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x ) ρ 1 N 1 ( x , y ) ]
(5.6)

and

Q(x,y)= R ρ 2 , M 2 H 2 ( , ) [ H 2 ( A 2 , B 2 ) ( x ) ρ 2 N 2 ( x , y ) ]
(5.7)

for ρ 1 , ρ 2 >0,respectively.

For any ( x 1 , y 1 ),( x 2 , y 2 ) X 1 × X 2 ,it follows from (5.6) and (5.7) and the Lipschitz continuity of R ρ 1 , M 1 H 1 ( , ) and R ρ 2 , M 2 H 2 ( , ) that

P ( x 1 , y 1 ) P ( x 2 , y 2 ) 1 = R ρ 1 , M 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x 1 ) ρ N 1 ( x 1 , y 1 ) ] R ρ 1 , M 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x 2 ) ρ N 1 ( x 2 , y 2 ) ] 1 Ł 1 [ H 1 ( A 1 , B 1 ) ( x 1 ) H 1 ( A 1 , B 1 ) ( x 2 ) ρ 1 ( N 1 ( x 1 , y 1 ) N 1 ( x 2 , y 1 ) ) 1 + ρ 1 N 1 ( x 2 , y 1 ) N 1 ( x 2 , y 2 ) 1 ]
(5.8)

and

Q ( x 1 , y 1 ) Q ( x 2 , y 2 ) 2 Ł 2 [ H 2 ( A 2 , B 2 ) ( y 1 ) H 2 ( A 2 , B 2 ) ( y 2 ) ρ 2 ( N 2 ( x 1 , y 1 ) N 2 ( x 1 , y 2 ) ) 2 + ρ 2 N 2 ( x 1 , y 2 ) N 2 ( x 2 , y 2 ) 2 ] .
(5.9)

Now

H 1 ( A 1 , B 1 ) ( x 1 ) H 1 ( A 1 , B 1 ) ( x 2 ) ρ 1 ( N 1 ( x 1 , y 1 ) N 1 ( x 2 , y 1 ) ) 1 = H 1 ( A 1 , B 1 ) ( x 1 ) H 1 ( A 1 , B 1 ) ( x 2 ) ( x 1 x 2 ) 1 + x 1 x 2 ρ 1 ( N 1 ( x 1 , y 1 ) N 1 ( x 2 , y 1 ) ) 1 .
(5.10)

Also,

H 2 ( A 2 , B 2 ) ( y 1 ) H 2 ( A 2 , B 2 ) ( y 2 ) ( y 1 y 2 ) 2 = H 2 ( A 2 , B 2 ) ( y 1 ) H 2 ( A 2 , B 2 ) ( y 2 ) ( y 1 y 2 ) 2 + y 1 y 2 ρ 2 ( N 2 ( x 1 , y 1 ) N 1 ( x 1 , y 2 ) ) 2 .
(5.11)

Now

H 1 ( A 1 , B 1 ) ( x 1 ) H 1 ( A 1 , B 1 ) ( x 2 ) ( x 1 x 2 ) 1 q 1 x 1 x 2 1 q 1 2 q 1 H 1 ( A 1 , B 1 ) ( x 1 ) H 1 ( A 1 , B 1 ) ( x 2 ) J q 1 ( x 1 x 2 ) 1 + c q 1 H 1 ( A 1 , B 1 ) ( x 1 ) H 1 ( A 1 , B 1 ) ( x 2 ) 1 q 1 .
(5.12)

Since M 1 is an H(,)-mixed mapping, then H 1 ( A 1 , B 1 ) is μ 1 -cocoercivewith respect to A 1 and γ 1 -relaxedaccretive with respect to B 1 , andfrom the fact that A 1 is α 1 -expansive,we can obtain

H 1 ( A 1 , B 1 ) ( x 1 ) H 1 ( A 2 , B 2 ) ( x 2 ) J q 1 ( x 1 x 2 ) 1 = H 1 ( A 1 , B 1 ) ( x 1 ) H 1 ( A 2 , B 1 ) ( x 2 ) J q 1 ( x 1 x 2 ) 1 + H 1 ( A 2 , B 1 ) ( x 1 ) H 1 ( A 2 , B 2 ) ( x 2 ) J q 1 ( x 1 x 2 ) 1 μ 1 A 1 ( x 1 ) A 1 ( x 2 ) q 1 γ 1 x 1 x 2 q 1 μ 1 α 1 q 1 x 1 x 2 q 1 γ 1 x 1 x 2 q 1 = ( μ 1 α 1 q 1 γ 1 ) x 1 x 2 q 1 .
(5.13)

Since H 1 ( A 1 , B 1 ) is s 1 -Lipschitzcontinuous with respect to A 1 and t 1 -Lipschitzcontinuous with respect to B 1 , wehave

H 1 ( A 1 , B 1 ) ( x 1 ) H 1 ( A 1 , B 1 ) ( x 2 ) ( x 1 x 2 ) 1 H 1 ( A 1 , B 1 ) ( x 1 ) H 1 ( A 2 , B 1 ) ( x 2 ) 1 + H 1 ( A 2 , B 1 ) ( x 1 ) H 1 ( A 2 , B 2 ) ( x 2 ) 1 ( s 1 + t 1 ) x 1 x 2 1 .
(5.14)

Using (5.12), (5.13) and (5.14), we have

H 1 ( A 1 , B 1 ) ( x 1 ) H 1 ( A 1 , B 1 ) ( x 2 ) ( x 1 x 2 ) 1 q 1 x 1 x 2 1 q 1 2 q 1 H 1 ( A 1 , B 1 ) ( x 1 ) H 1 ( A 1 , B 1 ) ( x 2 ) J q 1 ( x 1 x 2 ) 1 [ 1 2 q 1 r 1 + c q 1 ( s 1 + t 1 ) q 1 ] x 1 x 2 1 q 1 ,

which implies that

H 1 ( A 1 , B 1 ) ( x 1 ) H 1 ( A 1 , B 1 ) ( x 2 ) ( x 1 x 2 ) 1 [ 1 2 q 1 r 1 + c q 1 ( s 1 + t 1 ) q 1 ] 1 q 1 x 1 x 2 1 , where  r 1 = μ 1 α 1 q 1 γ 1 .
(5.15)

In the light of (5.15), we can obtain

H 2 ( A 2 , B 2 ) ( y 1 ) H 2 ( A 2 , B 2 ) ( y 2 ) ( y 1 y 2 ) 2 [ 1 2 q 2 r 2 + c q 2 ( s 2 + t 2 ) q 2 ] 1 q 2 y 1 y 2 2 , where  r 2 = μ 2 α 2 q 2 γ 2 .
(5.16)

Again, since N i is κ i -stronglyaccretive in the first argument and L N i -Lipschitzcontinuous in the first argument and l N i -Lipschitzcontinuous in the second argument, then using Lemma 2.4, we have

x 1 x 2 ρ 1 ( N 1 ( x 1 , y 1 ) N 1 ( x 2 , y 1 ) ) 1 q 1 x 1 x 2 1 q 1 2 ρ 1 q 1 N 1 ( x 1 , y 1 ) N 1 ( x 2 , y 1 ) , J q 1 ( x 1 x 2 ) 1 + ρ 1 q 1 c q 1 N 1 ( x 1 , y 1 ) N 1 ( x 2 , y 1 ) 1 q 1 ( 1 2 ρ 1 q 1 κ 1 + c q 1 ρ 1 q 1 L 1 q 1 ) x 1 x 2 1 q 1 ,

which implies that

x 1 x 2 ρ 1 ( N 1 ( x 1 , y 1 ) N 1 ( x 2 , y 1 ) ) 1 ( 1 2 ρ 1 q 1 κ 1 + c q 1 ρ 1 q 1 L 1 q 1 ) 1 q 1 x 1 x 2 1 .
(5.17)

In the light of (5.17), we have

y 1 y 2 ρ 2 ( N 2 ( x 1 , y 1 ) N 2 ( x 1 , y 2 ) ) 2 ( 1 2 ρ 2 q 2 κ 2 + c q 2 ρ 2 q 2 L 2 q 2 ) 1 q 2 y 1 y 2 2 .
(5.18)

Using (5.8), (5.15) and (5.17), we have

P ( x 1 , y 1 ) P ( x 2 , y 2 ) 1 ( L 1 [ ( 1 2 q 1 r 1 + c q 1 ( s 1 + t 1 ) q 1 ) 1 q 1 + ( 1 2 ρ 1 q 1 κ 1 + c q 1 ρ 1 q 1 L 1 q 1 ) 1 q 1 ] ) x 1 x 2 1 + L 1 ρ 1 l N 1 y 1 y 2 2 .
(5.19)

Using (5.9), (5.16) and (5.18), we have

Q ( x 1 , y 1 ) Q ( x 2 , y 2 ) 2 ( L 2 [ ( 1 2 q 2 r 2 + c q 2 ( s 2 + t 2 ) q 2 ) 1 q 2 + ( 1 2 ρ 2 q 2 κ 2 + c q 2 ρ 2 q 2 L 2 q 2 ) 1 q 2 ] ) y 1 y 2 2 + L 2 ρ 2 l N 2 x 1 x 2 1 .
(5.20)

From (5.19) and (5.20), we have

P ( x 1 , y 1 ) P ( x 2 , y 2 ) 1 + Q ( x 1 , y 1 ) Q ( x 2 , y 2 ) 2 τ 1 x 1 x 2 1 + τ 2 y 1 y 2 2 max { τ 1 , τ 2 } ( x 1 x 2 1 + y 1 y 2 2 ) ,
(5.21)

where

{ τ 1 = a 1 + ρ 2 L 2 L N 2 ; τ 2 = a 2 + ρ 2 L 1 l N 1 ,
(5.22)

and

a 1 = L 1 [ ( 1 2 q 1 r 1 + c q 1 ( s 1 + t 1 ) q 1 ) 1 q 1 + ( 1 2 ρ 1 q 1 κ 1 + c q 1 ρ 1 q 1 L 1 q 1 ) 1 q 1 ] ; a 2 = L 2 [ ( 1 2 q 2 r 2 + c q 2 ( s 2 + t 2 ) q 2 ) 1 q 2 + ( 1 2 ρ 2 q 2 κ 2 + c q 2 ρ 2 q 2 L 2 q 2 ) 1 q 2 ] ; L 1 = 1 r 1 ρ 1 m 1 ; L 2 = 1 r 2 ρ 2 m 2 .

Now define the norm on X 1 × X 2 by

( x , y ) = x 1 + y 2 ,(x,y) X 1 × X 2 .
(5.23)

We observe that ( X 1 × X 2 , ) is a Banach space.Hence it follows from (5.5), (5.21) and (5.23) that

R ( x 1 , y 1 ) R ( x 2 , y 2 ) =max{ τ 1 , τ 2 } ( x 1 , y 1 ) ( x 2 , y 2 ) .
(5.24)

Since max{ τ 1 , τ 2 }<1 by (5.2), itfollows from (5.24) that R is a contraction mapping. Hence, by the Banachcontraction principle, there exists a unique point (x,y) X 1 × X 2 such that

R(x,y)=(x,y),

which implies that

x = R ρ 1 , M 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x ) ρ 1 N 1 ( x , y ) ] , y = R ρ 2 , M 2 H 2 ( , ) [ H 2 ( A 2 , B 2 ) ( y ) ρ 2 N 2 ( x , y ) ] .

It follows from Lemma 5.2 that (x,y) is a unique solutionof SGVI (5.1). This completes the proof. □

6 Convergence of an iterative algorithm for SGVI (5.1)

Based on Lemma 5.2, we suggest and analyze the following iterative algorithm forfinding an approximate solution for SGVI (5.1).

Algorithm 6.1 For any given ( x 0 , y 0 ) X 1 × X 2 ,( x n , y n ) X 1 × X 2 by an iterative scheme

x n + 1 = R ρ 1 , M 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x n ) ρ 1 N 1 ( x n , y n ) ] ,
(6.1)
y n + 1 = R ρ 2 , M 2 H 2 ( , ) [ H 2 ( A 2 , B 2 ) ( y n ) ρ 2 N 2 ( x n , y n ) ] ,
(6.2)

where n=0,1,2,and ρ 1 , ρ 2 >0are constants.

Theorem 6.2 For eachi=1,2,let X i be q i -uniformlysmooth Banach spaces, let A i , B i : X i X i be single-valued mappings. Let the set-valuedmappings M n i , M i : X i 2 X i be such that M n i , M i are H i (,)-mixed mappingswith respect to mappings A i and B i such that M n i G M i forn=0,1,2, ,and A i is α i -expansiveand μ i > γ i with r i = μ i α i q i γ i > ρ i m i .Let H i : X 1 × X 2 X i be s i -Lipschitzcontinuous with respect to A i and t i -Lipschitzcontinuous with respect to B i ,and let N i : X 1 × X 2 X i be a κ i -stronglyaccretive mapping in the ith argument, L N i -Lipschitzcontinuous in the first argument and l N i -Lipschitzcontinuous in the second argument. Suppose that there are twoconstants ρ 1 , ρ 2 >0satisfying the following conditions:

{ τ 1 = a 1 + ρ 2 L 2 L N 2 < 1 ; τ 2 = a 2 + ρ 2 L 1 l N 1 < 1 ,
(6.3)

where

a 1 = L 1 [ ( 1 2 q 1 r 1 + c q 1 ( s 1 + t 1 ) q 1 ) 1 q 1 + ( 1 2 ρ 1 q 1 κ 1 + c q 1 ρ 1 q 1 L 1 q 1 ) 1 q 1 ] ; a 2 = L 2 [ ( 1 2 q 2 r 2 + c q 2 ( s 2 + t 2 ) q 2 ) 1 q 2 + ( 1 2 ρ 2 q 2 κ 2 + c q 2 ρ 2 q 2 L 2 q 2 ) 1 q 2 ] ; L 1 = 1 r 1 ρ 1 m 1 ; L 2 = 1 r 2 ρ 2 m 2 .

Then the approximate solution( x n , y n )generated by Algorithm  6.1 converges strongly to the uniquesolution(x,y)of SGVI (5.1).

Proof By Theorem 5.3, there exists a unique solution(x,y) X 1 × X 2 of SGVI (5.1). It follows from Algorithm 6.1 and Theorem 3.6 that

x n + 1 x = R ρ 1 , M n 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x n ) ρ 1 N 1 ( x n , y n ) ] R ρ 1 , M 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x ) ρ 1 N 1 ( x , y ) ] 1 R ρ 1 , M n 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x n ) ρ 1 N 1 ( x n , y n ) ] R ρ 1 , M n 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x ) ρ 1 N 1 ( x , y ) ] 1 + R ρ 1 , M n 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x ) ρ 1 N 1 ( x , y ) ] R ρ 1 , M 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x ) ρ 1 N 1 ( x , y ) ] 1
(6.4)

and

y n + 1 y R ρ 2 , M n 2 H 2 ( , ) [ H 2 ( A 2 , B 2 ) ( y n ) ρ 2 N 2 ( x n , y n ) ] R ρ 2 , M n 2 H 2 ( , ) [ H 2 ( A 2 , B 2 ) ( y ) ρ 2 N 2 ( x , y ) ] 2 + R ρ 2 , M n 2 H 2 ( , ) [ H 2 ( A 2 , B 2 ) ( y ) ρ 2 N 2 ( x , y ) ] R ρ 2 , M n 2 H 2 ( , ) [ H 2 ( A 2 , B 2 ) ( y ) ρ 2 N 2 ( x , y ) ] 2 .
(6.5)

By (5.8) and (5.19), we have

R ρ 1 , M n 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x n ) ρ 1 N 1 ( x n , y n ) ] R ρ 1 , M n 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x ) ρ 1 N 1 ( x , y ) ] 1 Ł 1 [ H 1 ( A 1 , B 1 ) ( x n ) H 1 ( A 1 , B 1 ) ( x ) ρ 1 ( N 1 ( x n , y n ) N 1 ( x , y n ) ) 1 + ρ 1 N 1 ( x , y n ) N 1 ( x , y ) 1 ] ( L 1 [ ( 1 2 q 1 r 1 + c q 1 ( s 1 + t 1 ) q 1 ) 1 q 1 + ( 1 2 ρ 1 q 1 κ 1 + c q 1 ρ 1 q 1 L 1 q 1 ) 1 q 1 ] ) × x n x 1 + L 1 ρ 1 l N 1 y n y 2
(6.6)

and

R ρ 2 , M n 2 H 2 ( , ) [ H 2 ( A 2 , B 2 ) ( y n ) ρ 2 N 2 ( x n , y n ) ] R ρ 2 , M n 2 H 2 ( , ) [ H 2 ( A 2 , B 2 ) ( y ) ρ 2 N 2 ( x , y ) ] 2 Ł 2 [ H 2 ( A 2 , B 2 ) ( y n ) H 2 ( A 2 , B 2 ) ( y ) ρ 2 ( N 2 ( x n , y n ) N 1 ( x n , y ) ) 2 + ρ 2 N 2 ( x n , y ) N 2 ( x , y ) 2 ] ( L 2 [ ( 1 2 q 2 r 2 + c q 2 ( s 2 + t 2 ) q 2 ) 1 q 2 + ( 1 2 ρ 2 q 2 κ 2 + c q 2 ρ 2 q 2 L 2 q 2 ) 1 q 2 ] ) × y n y 2 + L 2 ρ 2 l N 2 x n x 1 .
(6.7)

By Theorem 4.2, we have

R ρ 1 , M n 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x ) ρ 1 N 1 ( x , y ) ] R ρ 1 , M 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x ) ρ 1 N 1 ( x , y ) ] ,
(6.8)
R ρ 2 , M n 2 H 2 ( , ) [ H 2 ( A 2 , B 2 ) ( y ) ρ 2 N 2 ( x , y ) ] R ρ 2 , M 2 H 2 ( , ) [ H 1 ( A 2 , B 2 ) ( y ) ρ 2 N 2 ( x , y ) ] .
(6.9)

Let

b n 1 = R ρ 1 , M n 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x ) ρ 1 N 1 ( x , y ) ] b n 1 = R ρ 1 , M 1 H 1 ( , ) [ H 1 ( A 1 , B 1 ) ( x ) ρ 1 N 1 ( x , y ) ] 1 ,
(6.10)
b n 2 = R ρ 2 , M n 2 H 2 ( , ) [ H 2 ( A 2 , B 2 ) ( y ) ρ 2 N 2 ( x , y ) ] b n 2 = R ρ 2 , M 2 H 2 ( , ) [ H 2 ( A 2 , B 2 ) ( y ) ρ 2 N 2 ( x , y ) ] 2 .
(6.11)

From (6.4)-(6.11), we have

x n + 1 x 1 τ 1 x n x 1 + b n 1 ,
(6.12)
y n + 1 y 1 τ 2 y n y 1 + b n 2 .
(6.13)

From (6.12) and (6.13), we have

x n + 1 x 1 + y n + 1 y 2 max{ τ 1 , τ 2 } { x n x 1 + y n y 2 } +{ b n 1 + b n 2 }.
(6.14)

Since ( X 1 × X 2 , ) is a Banach spacewith the norm defined by (5.23), it follows from (5.5), (5.23) and (6.14) that

( x n + 1 , y n + 1 ) ( x , y ) = x n + 1 x 1 + y n + 1 y 2 max { τ 1 , τ 2 } ( x n , y n ) ( x , y ) + { b n 1 + b n 2 } .
(6.15)

By condition (6.3), it follows that max{ τ 1 , τ 2 }<1 andLemma 2.5 implies that ( x n + 1 , y n + 1 ) ( x , y ) 0as n.

Thus {( x n , y n )} converges strongly to the unique solution(x,y) of SGVI (5.1). Thiscompletes the proof. □

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