# Auxiliary Principle for a System of Generalized Set-Valued Nonlinear Mixed Variational-Like Inequalities in Banach Spaces

Authors: Yangqing Qiu; Xiaolin Zhan; Luchuan Ceng
Journal: Journal of Function Spaces
 (2019)

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## Abstract

In this paper, the auxiliary principle technique is extended to study a system of generalized nonlinear mixed variational-like inequalities problem for set-valued mappings without compact values in Banach spaces withp-uniformly convex bidual spaces. First, the existence of the solutions of the related auxiliary problem is proved. Then, a new iterative algorithm based on the system of auxiliary variational inequalities is constructed. Finally, both the existence of the solutions of the original problem and the convergence of the iterative sequences generated by the algorithm are proved. And we also present a numerical example to demonstrate the result. Our results improve and extend some known results.


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## Body

## 1. Introduction

Throughout the paper unless otherwise stated, letI=1,2 be an index set for each i∈I, (Ei,·i) be a real reflexive Banach space with the dual space Ei∗, u,vi be the dual pairing between u∈Ei∗ and v∈Ei, and CB(Ei∗) be the family of all nonempty bounded closed subsets of Ei∗. For each i∈I, let Ni:Ei∗×Ei∗→Ei∗, Fi:E1∗×E2∗→Ei∗, ηi:Ei×Ei→Ei be nonlinear mappings and Ai,Bi:Ei→CB(Ei∗), Ci:E1→CB(E1∗), Di:E2→CB(E2∗) be set-valued mappings. We consider the following system of generalized set-valued nonlinear mixed variational-like inequalities problem: find (x,y)∈E1×E2 and ui,vi,wi,zi such that u1∈A1x, v1∈B1x, u2∈A2y, v2∈B2y, wi∈Cix, zi∈Diy, and(1)N1u1,v1+F1w1,z1,η1h1,x1+b1x,h1-b1x,x≥0,∀h1∈E1,N2u2,v2+F2w2,z2,η2h2,y2+b2y,h2-b2y,y≥0,∀h2∈E2,where for each i∈I, the bifunction bi:Ei×Ei→R, which is not necessarily differentiable, satisfies the following properties:(i)
bi(·,·) is linear in the first argument;(ii)
bi(·,·) is bounded; that is, there exists a constant γi>0 such that (2)biui,vi≤γiuiivii,∀ui,vi∈Ei;(iii)
bi(ui,vi)-bi(ui,wi)≤bi(ui,vi-wi), ∀ui,vi,wi∈Ei;(iv)
bi(·,·) is convex in the second argument.Remark 1 (see [1]).
( 1 ) For each i∈I and arbitrary ui,vi∈Ei, property (i) implies that -bi(ui,vi)=bi(-ui,vi) and property (ii) implies that bi(-ui,vi)≤γiuiivii. Hence we have (3)biui,vi≤γiuiivii,∀ui,vi∈Ei,and (4)biui,0=bi0,vi=0,∀ui,vi∈Ei.
( 2 ) For each i∈I, it follows from properties (ii) and (iii), for all ui,vi,wi∈Ei, that we have (5)biui,vi-biui,wi≤γiuiivi-wii, and (6)biui,wi-biui,vi≤γiuiiwi-vii, and, therefore, (7)biui,vi-biui,wi≤γiuiivi-wii,∀ui,vi∈Ei. This implies that for each i∈I, bi is continuous with respect to the second argument.There are many special cases of problem (1), which can be also found in [1–11] and the references cited therein. Therefore, problem (1) is a more general and unified one.In the system of the generalized set-valued nonlinear mixed variational-like inequalities problem (1), bi is a nonlinear mapping, so the projection method cannot be applied to it. This fact motivated many authors to develop the auxiliary principle technique to study the existence of solutions of generalized mixed type variational inequalities and also to develop a large number of numerical methods for solving various variational inequalities, complementarity problems, and optimization problems. The auxiliary principle technique was first introduced by Glowinski et al. [12]. Then Ding et al. [3] extended it to study the existence and algorithm of solutions of generalized strongly nonlinear mixed variational-like inequalities in Banach spaces when N(T(·),A(·)) is a set-valued mapping. Regretfully, N(T(·),A(·)) is actually a single-valued mapping in Theorem 3.1 of [3]. In problem (1), we consider the set-valued type mappings. Very recently, auxiliary principle technique is used to solve regularized nonconvex mixed variational inequalities and strongly mixed variational-like inequalities; see [13] and [14], respectively.On the other hand, Noor [6] put forward that extending the projection methods and its variant forms for generalized set-valued mixed nonlinear variational inequalities involving the nonlinear form bi(·,·) satisfying properties (i), (ii), and (iii) is still an open problem, and this needs further research efforts.In this paper, the auxiliary principle technique is extended to study a system of generalized nonlinear mixed variational-like inequalities problem (1) for set-valued mappings without compact values in Banach spaces with p-uniformly convex bidual spaces. First, the existence of the solutions of the related auxiliary problem is proved. Then a new iterative algorithm based on the system of auxiliary variational inequalities is constructed. Finally, both the existence of the solutions of the problem (1) and the convergence of the iterative sequences generated by the algorithm are proved. And we also present a numerical example to demonstrate the result. Our results improve and extend some known results.


## 2. Preliminaries

We recall some concepts and definitions at first.LetE be a real Banach space; the modulus of convexity of E is the function δε:[0,2]→[0,1] defined by (8)δEε=inf⁡1-x+y2:x=1,y=1andx-y≥ε,0≤ε≤2.E is uniformly convex if and only if, for all ε: 0<ε≤2 such that δE(ε)>0. E is said to be p-uniformly convex, if there exists a constant c>0, such that δE(ε)≥cεp. Hilbert space Lp and lp(1<p<∞) and Sobolev space Wmp(1<p<∞) are p-uniformly convex. Hilbert space is 2-uniformly convex, while Lp,lp,Wmp are 2-uniformly convex, if 1<p≤2; they are p-uniformly convex, if p≥2. For q>1, the duality mapping Jq:E→2E∗ with gauge function φ(r)=rq-1 is defined by (9)Jqx=f∈E∗:x,f=xq,f=xq-1. In particular, J=J2 is the normalized duality mapping. It is well known that Jq is single-valued if E is smooth and Jq(x)=xq-2J(x), x≠0. The duality mappings of E∗ and E∗∗ are denoted by Jq∗ and Jq∗∗, respectively. The duality mapping Jq can be equivalently defined as the subdifferential of the gauge function φ(x)=xq/q; i.e., (10)f∈Jqx⇔(11)f∈∂φx=f∈E∗:yqq-xqq≥y-x,f,y∈E.We assume thatH(·,·) is the Hausdorff metric on CB(E) defined by (12)HA,B=maxsupx∈A⁡dx,B,supy∈B⁡dy,A,A,B∈CBE.Definition 2.
LetE be a smooth Banach space, g:E→E and N:E×E→E be single-valued mappings, and A,T:E→CB(E) be set-valued mappings. (1)
T is said to be λ-H-Lipschitz continuous, if there exists a constant λ>0 such that (13)HTu,Tv≤λu-v,∀u,v∈E,whereH(·,·) is the Hausdorff metric on CB(E);(2)
N is said to be (β,ξ)-Lipschitz continuous, if there exists a pair of constants β,ξ>0 such that (14)Nu1,v1-Nu2,v2≤βu1-u2+ξv1-v2,∀ui,vi∈E,i∈I;(3)
N is said to be α-strongly mixed accretive with set-valued mappings T, A and single-valued mapping g, if there exists a constant α>0 such that (15)Nu1,v1-Nu2,v2,Jqgx-gy≥αgx-gyq,∀x,y∈E,u1∈Tx,u2∈Ty,v1∈Ax,v2∈Ay.Remark 3.
IfE=H is a Hilbert space, g=I is an identity operator on E; then N(T(·),A(·)) is called an α-strongly mixed monotone mapping; that is, (16)Nu1,v1-Nu2,v2,x-y≥αx-y2,∀x,y∈E,u1∈Tx,u2∈Ty,v1∈Ax,v2∈Ay, which is introduced by Zeng et al. [15].Definition 4 (see [12]).
The mappingη:E×E→E is said to be ε-Lipschitz continuous, if there exists a constant ε>0 such that (17)ηx,y≤εx-y,∀x,y∈E.
In order to obtain our results, we need the following assumption.Assumption 5.
The mappingηi:Ei×Ei→Ei, i∈I, satisfies the following conditions:( 1 )   η i ( u , v ) = η i ( u , z ) + η i ( z , v ), ∀u,v∈Ei;( 2 )   η i ( · , · ) is affine in the second argument;( 3 )for each fixed ui∈Ei,vi↦ηi(ui,vi) is continuous from the weak topology to the weak topology.Remark 6.
It follows from Assumption5(1) that ηi(u,v)=-ηi(v,u) and ηi(u,u)=0,∀u,v∈Ei.We also need the following Lemmas.Lemma 7 (see [16]).
LetX be a nonempty closed convex subset of a Hausdorff linear topological space E and let ϕ,ψ:X×X→R be mappings satisfying the following conditions:
(1) ψ ( x , y ) ≤ ϕ ( x , y ) , ψ ( x , x ) ≥ 0, ∀x,y∈X;
(2)for each x∈X, ϕ(x,y) is upper semicontinuous with respect to y;
(3)for each y∈X, the set x∈X:ψx,y≤0 is convex;
(4)if there exists a nonempty compact set K⊂X and x0∈K such that (18)ψx0,y<0,∀y∈X∖K, then there exist y¯∈K such that ϕ(x,y¯)≥0,∀x∈K.Lemma 8 (see [8]).
LetE be a real Banach space with its uniformly convex dual space E∗. Assume that the modulus of convexity of E∗ satisfies δE∗(ε)≥cεq for some q≥2 and c>0. Then duality mapping Jq with gauge function φ(t)=tp-1(1/p+1/q=1) is Hölder continuous with exponent p-1; i.e., there exists a constant r>0 such that (19)Jqx-Jqy≤rx-yp-1,∀x,y∈E.


## 3. Auxiliary Problem and Algorithm

In this section, we will extend the auxiliary principle technique of Glowinski et al. [12] to study problem (1). And we will prove the existence of the solution of the system of auxiliary variational inequalities problem for (1). Then, by using the existence theorem, we will construct an iterative algorithm for problem (1).For eachi∈I, let gi:Ei→Ei∗ be a single-valued mapping. Given (x1,x2)∈E1×E2, u1∈A1x1, v1∈B1x1, u2∈A2x2, v2∈A2x2, wi∈Cix1, zi∈Dix2, we consider the following problem: find (p1,p2)∈E1×E2 such that(20)g1p1,h1-p11≥g1x1,h1-p11-ρ1N1u1,v1+F1w1,z1,η1h1,p11+ρ1b1x1,p1-b1x1,h1,g2p2,h2-p22≥g2x2,h2-p22-ρ2N2u2,v2+F2w2,z2,η2h2,p22+ρ2b2x2,p2-b2x2,h2,for all (h1,h2)∈E1×E2, where ρ1,ρ2>0 are constants. Problem (20) is called the system of auxiliary variational inequalities problem for problem (1).Theorem 9.
For eachi∈I, let gi:Ei→Ei∗ be δi-Lipschitz continuous and σi-strongly monotone, ηi:Ei×Ei→Ei be εi-Lipschitz continuous, bi:Ei×Ei→R be a function with the properties (i)-(iv), and Ni:Ei∗×Ei∗→Ei∗ and Fi:E1∗×E2∗→Ei∗ be mappings. If Assumption 5 holds, then the system of auxiliary variational inequalities problem (20) has a unique solution (x,y)∈E1×E2.Proof.
Fori∈I, define the mappings ϕi,ψi:Ei×Ei→R by (21)ϕihi,pi=gihi,hi-pii-gixi,hi-pii+ρ1Niui,vi+Fiwi,zi,ηihi,pii-ρ1bixi,pi-bixi,hi,and (22)ψihi,pi=gipi,hi-pii-gixi,hi-pii+ρ2Niui,vi+Fiwi,zi,ηihi,pii-ρ2bixi,pi-bixi,hi,respectively. We show that for each i∈I, the mappings ϕi,ψi satisfy the conditions of Lemma 7 in the weak topology. Indeed, since gi is σi-strongly monotone, it is clear that ϕi and ψi satisfy condition (1) of Lemma 7. Since ηi is εi-Lipschitz continuous, by Assumption 5, for any pi1,pi2∈Ei, we have (23)Niui,vi+Fiwi,zi,ηihi,pi1i-Niui,vi+Fiwi,zi,ηihi,pi2i=Niui,vi+Fiwi,zi,ηihi,pi1-ηihi,pi2i=Niui,vi+Fiwi,zi,ηipi1,pi2i≤Niui,vi+Fiwi,ziiηipi1,pi2i≤εiNiui,vi+Fiwi,ziipi1-pi2i. By Assumption 5(3), the function pi↦Niui,vi+Fiwi,zi,ηihi,pii is continuous and convex. It follows that the function pi↦Niui,vi+Fiwi,zi,ηihi,pii is weakly lower semicontinuous and so the function pi↦Niui,vi+Fiwi,zi,ηihi,pii is weakly upper semicontinuous. It is easy to show that, for each fixed pi∈Ei, the set hi∈Ei:ψihi,pi<0 is a convex set (we note that if pi∈Ei, the set hi∈Ei:ψihi,pi<0=∅; then this immediately implies the conclusion of Theorem 9 and hence we discuss only the case of the set hi∈Ei:ψihi,pi<0≠∅ below). Thus, conditions (2) and (3) of Lemma 7 hold.
Now, for eachi∈I, set (24)Wi=σi-1δixii+ρiεiNiui,vi+Fiwi,zii+ρiγixii,Ki=pi∈Ei:pii≤Wi. Then Ki is a weakly compact subset of Ei. For any fixed pi∈Ei∖Ki, take h0i=0∈Ki; from Assumption 5(1), the Lipschitz continuity of ηi, the strong monotonousness of gi, and Remark 1(2), we have (25)ψih0i,pi=ψi0,pi=-gipi,pii+gixi,pii+ρiNiui,vi+Fiwi,zi,ηi0,pii+ρibixi,0-bixi,pi=-gipi-gi0,pi-0i+gixi,pii+ρiNiui,vi+Fiwi,zi,ηi0,pii+ρibixi,0-bixi,pi≤-σipii2+δixiipii+ρiεiNiui,vi+Fiwi,ziipii+ρiγixiipii=-σipiipii-σi-1δixii+ρiεiNiui,vi+Fiwi,zii+ρiγixii≤0.Therefore, condition (4) of Lemma 7 holds. By Lemma 7, there exists an pi¯∈Ei such that ϕi(hi,pi¯)≥0,∀hi∈Ei, that is,(26)gihi,hi-pi¯i-gixi,hi-pi¯i+ρiNiui,vi+Fiwi,zi,ηihi,pi¯i-ρibixi,pi¯-bixi,hi≥0,∀hi∈Ei.
For arbitraryt∈(0,1) and ∀hi∈Ei, let yit=thi+(1-t)pi¯, replace hi by yit in (26); we obtain (27)0≤giyit,yit-pi¯i-gixi,yit-pi¯i+ρiNiui,vi+Fiwi,zi,ηiyit,pi¯i-ρibixi,pi¯-bixi,yit=tgiyit,hi-pi¯i-gixi,hi-pi¯i+ρiNiui,vi+Fiwi,zi,ηithi+1-tpi¯,pi¯i-ρibixi,pi¯-bixi,thi+1-tpi¯≤tgiyit,hi-pi¯i-gixi,hi-pi¯i+ρitNiui,vi+Fiwi,zi,ηihi,pi¯i+ρitbixi,hi-bixi,pi¯. Hence, we derive (28)giyit,hi-pi¯i-gixi,hi-pi¯i+ρiNiui,vi+Fiwi,zi,ηihi,pi¯i+ρibixi,hi-bixi,pi¯≥0, that is,(29)giyit,hi-pi¯i≥gixi,hi-pi¯i-ρiNiui,vi+Fiwi,zi,ηihi,pi¯i+ρibixi,pi¯-bixi,hi. Let t→0+; we have (30)gipi¯,hi-pi¯i≥gixi,hi-pi¯i-ρiNiui,vi+Fiwi,zi,ηihi,pi¯i+ρibixi,pi¯-bixi,hi. Therefore, pi¯∈Ei is a solution of the system of auxiliary variational inequalities problem (20). Now, let pi1,pi2∈Ei be any two solutions of the system of auxiliary variational inequalities problem (20). Then we have(31)gipi1-gixi,hi-pi1i≥-ρiNiui,vi+Fiwi,zi,ηihi,pi1i+ρibixi,pi1-bixi,hi,∀hi∈Ei,(32)gipi2-gixi,hi-pi2i≥-ρiNiui,vi+Fiwi,zi,ηihi,pi2i+ρibixi,pi2-bixi,hi,∀hi∈Ei,and, taking hi=pi2 in (31) and hi=pi1 in (32), then adding these two inequalities, we obtain (33)gipi1-gipi2,pi1-pi2i≤ρiNiui,vi+Fiwi,zi,ηipi2,pi1+ηipi1,pi2i=0. Since gi is σi-strongly monotone, we have (34)σipi1-pi2i2≤gipi1-gipi2,pi1-pi2i≤0. Hence, we must have pi1=pi2 and so problem (20) has a unique solution. This completes the proof.By virtue of Theorem9, we now construct an iterative algorithm for solving problem (1).For given(x0,y0)∈E1×E2, u01∈A1x0, v01∈B1x0, u02∈A2y0, v02∈B2y0, w0i∈Cix0, z0i∈Diy0, i∈I, from Theorem 9, we know that the system of auxiliary variational inequalities problem (20) has a unique solution (x1,y1)∈E1×E2; that is, (35)g1x1,h1-x11≥g1x0,h1-x11-ρ1N1u01,v01+F1w01,z01,η1h1,x11+ρ1b1x0,x1-b1x0,h1,∀h1∈E1,g2y1,h2-y12≥g2y0,h2-y12-ρ2N2u02,v02+F2w02,z02,η2h2,y12+ρ2b2y0,y1-b2y0,h2,∀h2∈E2. Since u01∈A1x0, v01∈B1x0, u02∈A2y0, v02∈B2y0, w0i∈Cix0, and z0i∈Diy0, i∈I, by Nadler’s Theorem, there exist u11∈A1x1, v11∈B1x1, u12∈A2y1, v12∈B2y1, w1i∈Cix1, and z1i∈Diy1, i∈I, such that (36)u11-u011≤1+1HA1x1,A1x0;v11-v011≤1+1HB1x1,B1x0;u12-u022≤1+1HA2y1,A2y0;v12-v022≤1+1HB2y1,B2y0;w1i-w0i1≤1+1HCix1,Cix0;z1i-z0i2≤1+1HDiy1,Diy0.Again by Theorem9, the system of auxiliary variational inequalities problem (20) has a unique solution (x2,y2)∈E1×E2; that is, (37)g1x2,h1-x21≥g1x1,h1-x21-ρ1N1u11,v11+F1w11,z11,η1h1,x21+ρ1b1x1,x2-b1x1,h1,∀h1∈E1,g2y2,h2-y22≥g2y1,h2-y22-ρ2N2u12,v12+F2w12,z12,η2h2,y22+ρ2b2y1,y2-b2y1,h2,∀h2∈E2. Since u11∈A1x1, v11∈B1x1, u12∈A2y1, v12∈B2y1, w1i∈Cix1, and z1i∈Diy1, i∈I, by Nadler’s Theorem, there exist u21∈A1x2, v21∈B1x2, u22∈A2y2, v22∈B2y2, w2i∈Cix2, and z2i∈Diy2, i∈I, such that (38)u21-u111≤1+12HA1x2,A1x1;v21-v111≤1+12HB1x2,B1x1;u22-u122≤1+12HA2y2,A2y1;v22-v122≤1+12HB2y2,B2y1;w2i-w1i1≤1+12HCix2,Cix1;z2i-z1i2≤1+12HDiy2,Diy1;by induction, we can get the iterative algorithm for solving problem (1).Algorithm 10.
For given(x0,y0)∈E1×E2, u01∈A1x0, v01∈B1x0, u02∈A2y0, v02∈B2y0, w0i∈Cix0, and z0i∈Diy0, i∈I, let ({xn},{yn})⊂E1×E2, {un1},{vn1},{wn1},{wn2}⊂E1, and {un2},{vn2},{zn1},{zn2}⊂E2 satisfy the following conditions: (39)g1xn+1,h1-xn+11≥g1xn,h1-xn+11-ρ1N1un1,vn1+F1wn1,zn1,η1h1,xn+11+ρ1b1xn,xn+1-b1xn,h1,(40)g2yn+1,h2-yn+12≥g2yn,h2-yn+12-ρ2N2un2,vn2+F2wn2,zn2,η2h2,yn+12+ρ2b2yn,yn+1-b2yn,h2,where constants ρ1,ρ2>0, ∀h1∈E1, ∀h2∈E2, and (41)un1∈A1xn,un+11-un11≤1+1n+1HA1xn+1,A1xn;vn1∈B1xn,vn+11-vn11≤1+1n+1HB1xn+1,B1xn;un2∈A2yn,un+12-un22≤1+1n+1HA2yn+1,A2yn;vn2∈B2yn,vn+12-vn22≤1+1n+1HB2yn+1,B2yn;wni∈Cixn,wn+1i-wni1≤1+1n+1HCixn+1,Cixn;zni∈Diyn,zn+1i-zni2≤1+1n+1HDiyn+1,Diyn.


## 4. Existence and Convergence Theorem

In this section, we will prove the existence of the solution of problem (1) and the convergence of the iterative sequences generated by Algorithm 10.Theorem 11.
For eachi∈I, let Ei be a real reflexive Banach space with a uniformly convex bidual space Ei∗∗; the modulus of convexity of Ei∗∗ satisfies δEi∗∗(εi)≥ciεipi for some pi≥2 and ci>0. Let Ni:Ei∗×Ei∗→Ei∗, Fi:E1∗×E2∗→Ei∗, ηi:Ei×Ei→Ei, gi:Ei→Ei∗, Ai,Bi:Ei→CB(Ei∗), Ci:E1→CB(E1∗), and Di:E2→CB(E2∗) be mappings, and let bi:Ei×Ei→R be a real valued functional. Assume that the following conditions are satisfied:
(1) N i is (αi,βi)-Lipschitz continuous and ξi-strongly mixed accretive with respect to Ai,Bi, and gi;
(2) F i is (li,ki)-H-Lipschitz continuous;
(3) A i is τi-H-Lipschitz continuous;
(4) B i is ωi-H-Lipschitz continuous;
(5) C i is λi-H-Lipschitz continuous;
(6) D i is μi-H-Lipschitz continuous;
(7) g i is σi-strongly monotone and δi-Lipschitz continuous;
(8) b i has the properties (i)-(iv);
(9) η i is εi-Lipschitz continuous and satisfies the following: there exists a constant 0<ζi≤1 such that (42)x-y-ηix,y≤1-ζix-y. If Assumption 5 holds and there exist constants ρ1,ρ2>0 such that(43)σ1-1δ11-ζ1+ρ1γ1+ρ1ε1l1λ1+ε1δ1q1-ρ1q1ξ1δ1q1+ρ12q1r1α1τ1+β1ω1q11/q1+ρ2ε2l2λ2σ2-1<1,σ2-1δ21-ζ2+ρ2γ2+ρ2ε2k2μ2+ε2δ2q2-ρ2q2ξ2δ2q2+ρ22q2r2α2τ2+β2ω2q21/q2+ρ1ε1k1μ1σ1-1<1,where r is a constant appearing in Lemma 8, then there exist (x,y)∈E1×E2, u1∈A1x, v1∈B1x, u2∈A2y, v2∈B2y, wi∈Cix, and zi∈Diy satisfying problem (1) and (44)xn→x,yn→y,uni→ui,vni→vi,wni→wi,zni→zi,as n→∞, where the sequences xn,yn,uni,vni,wni,zni are defined by Algorithm 10.Proof.
First, it follows from (39) in Algorithm 10 that for any h1∈E1,(45)g1xn,h1-xn1≥g1xn-1,h1-xn1-ρ1N1un-11,vn-11+F1wn-11,zn-11,η1h1,xn1+ρ1b1xn-1,xn-b1xn-1,h1,and(46)g1xn+1,h1-xn+11≥g1xn,h1-xn+11-ρ1N1un1,vn1+F1wn1,zn1,η1h1,xn+11+ρ1b1xn,xn+1-b1xn,h1.
Takingh1=xn+1 in (45) and h1=xn in (46), respectively, we get(47)g1xn,xn+1-xn1≥g1xn-1,xn+1-xn1-ρ1N1un-11,vn-11+F1wn-11,zn-11,η1xn+1,xn1+ρ1b1xn-1,xn-b1xn-1,xn+1,and(48)g1xn+1,xn-xn+11≥g1xn,xn-xn+11-ρ1N1un1,vn1+F1wn1,zn1,η1xn,xn+11+ρ1b1xn,xn+1-b1xn,xn.Adding (47) and (48), we obtain (49)g1xn+1-g1xn,xn-xn+11≥g1xn-g1xn-1,xn-xn+11-ρ1N1un1,vn1+F1wn1,zn1-(N1un-11,vn-11+F1wn-11,zn-11,η1xn,xn+11+ρ1b1xn-1-xn,xn-b1xn-xn-1,xn+1, and so(50)g1xn-g1xn+1,xn-xn+11≤g1xn-1-g1xn,xn-xn+11-ρ1N1un-11,vn-11+F1wn-11,zn-11-N1un1,vn1+F1wn1,zn1,η1xn,xn+11-ρ1b1xn-1-xn,xn-b1xn-xn-1,xn+1≤g1xn-1-g1xn,xn-xn+1-η1xn,xn+11+g1xn-1-g1xn-ρ1N1un-11,vn-11+F1wn-11,zn-11-N1un1,vn1+F1wn1,zn1,η1xn,xn+11+ρ1b1xn-xn-1,xn-xn+1.
By conditions(2),(7),(8),(9) and (50), we derive (51)σ1xn-xn+112≤δ11-ζ1xn-1-xn1xn-xn+11+ε1g1xn-1-g1xn-ρ1N1un-11,vn-11-N1un1,vn11xn-xn+11+ρ1ε1F1wn-11,zn-11-F1wn1,zn11xn-xn+11+ρ1γ1xn-1-xn1xn-xn+11. Therefore,(52)xn-xn+11≤1σ1δ11-ζ1xn-1-xn1+ε1σ1g1xn-1-g1xn-ρ1N1un-11,vn-11-N1un1,vn11+ρ1ε1l1σ1wn-11-wn11+ρ1ε1k1σ1zn-11-zn11+ρ1γ1σ1xn-1-xn1≤1σ1δ11-ζ1+ρ1γ1σ1xn-1-xn1+ε1σ1g1xn-1-g1xn-ρ1N1un-11,vn-11-N1un1,vn11+ρ1ε1l1σ11+1nHC1xn-1,C1xn+ρ1ε1k1σ11+1nHD1yn-1,D1yn≤1σ1δ11-ζ1+ρ1γ1+ρ1ε1l1λ11+1nxn-1-xn1+ρ1ε1k1μ1σ11+1nyn-1-yn2+ε1σ1g1xn-1-g1xn-ρ1N1un-11,vn-11-N1un1,vn11.Since E1∗∗ is p1-uniformly convex and Lemma 8 holds with r replaced by r1 and Jq replaced by Jq1∗, where Jq1∗:E1∗→E1∗∗ is duality mapping with gauge function φ(r)=rq1-1,1/p1+1/q1=1, by (50) and condition (1), we have(53)g1xn-1-g1xn-ρ1N1un-11,vn-11-N1un1,vn11q1≤g1xn-1-g1xn1q1-q1ρ1N1un-11,vn-11-N1un1,vn1,Jq1∗g1xn-1-g1xn-ρ1N1un-11,vn-11-N1un1,vn11≤δ1q1xn-1-xn1q1-q1ρ1N1un-11,vn-11-N1un1,vn1,Jq1∗g1xn-1-g1xn-ρ1N1un-11,vn-11-N1un1,vn1-Jq1∗g1xn-1-g1xn1-q1ρ1N1un-11,vn-11-N1un1,vn1,Jq1∗g1xn-1-g1xn1≤δ1q1xn-1-xn1q1+q1r1ρ12N1un-11,vn-11-N1un1,vn11q1-q1ρ1ξ1g1xn-1-g1xn1q1≤δ1q1-ρ1q1ξ1δ1q1+ρ12q1r1α1τ1+β1ω1q11+1nq1xn-1-xn1q1.
It follows from (52) and (53) that(54)xn-xn+11≤1σ1δ11-ζ1+ρ1γ1+ρ1ε1l1λ11+1n+ε1δ1q1-ρ1q1ξ1δ1q1+ρ12q1r1α1τ1+β1ω1q11+1nq11/q1xn-1-xn1+ρ1ε1k1μ1σ11+1nyn-1-yn2.
Second, it follows from (40) in Algorithm 10 that, for any ∀h2∈E2,(55)g2yn,h2-yn2≥g2yn-1,h2-yn2-ρ2N2un-12,vn-12+F2wn-12,zn-12,η2h2,yn2+ρ2b2yn-1,yn-b2yn-1,h2,and(56)g2yn+1,h2-yn+12≥g2yn,h2-yn+12-ρ2N2un2,vn2+F2wn2,zn2,η2h2,yn+12+ρ2b2yn,yn+1-b2yn,h2.
Takingh2=yn+1 in (55) and h2=yn in (56), respectively, then adding the results, we have (57)g2yn-g2yn+1,yn-yn+12≤g2yn-1-g2yn,yn-yn+1-η2yn,yn+12+g2yn-1-g2yn-ρ2N2un-12,vn-12-F2wn-12,zn-12-N2un2,vn2+F2wn2,zn2,η2yn,yn+12+ρ2b2yn-yn-1,yn-yn+1. By conditions (1)-(9), we have(58)yn-yn+12≤1σ2δ21-ζ2+ρ2γ2+ρ2ε2k2μ21+1n+ε2δ2q2-ρ2q2ξ2δ2q2+ρ22q2r2α2τ2+β2ω2q21+1nq21/q2yn-1-yn2+ρ2ε2l2λ2σ21+1nxn-1-xn1.
From (54) and (58), we have(59)xn-xn+11+yn-yn+12≤1σ1δ11-ζ1+ρ1γ1+ρ1ε1l1λ11+1n+ε1δ1q1-ρ1q1ξ1δ1q1+ρ12q1r1α1τ1+β1ω1q11+1nq11/q1+ρ2ε2l2λ2σ21+1nxn-1-xn1+1σ2δ21-ζ2+ρ2γ2+ρ2ε2k2μ21+1n+ε2δ2q2-ρ2q2ξ2δ2q2+ρ22q2r2α2τ2+β2ω2q21+1nq21/q2+ρ1ε1k1μ1σ11+1nyn-1-yn2≤max⁡θn1,θn2xn-1-xn1+yn-1-yn2.Here (60)θn1=1σ1δ11-ζ1+ρ1γ1+ρ1ε1l1λ11+1n+ε1δ1q1-ρ1q1ξ1δ1q1+ρ12q1r1α1τ1+β1ω1q11+1nq11/q1+ρ2ε2l2λ2σ21+1n,(61)θn2=1σ2δ21-ζ2+ρ2γ2+ρ2ε2k2μ21+1n+ε2δ2q2-ρ2q2ξ2δ2q2+ρ22q2r2α2τ2+β2ω2q21+1nq21/q2+ρ1ε1k1μ1σ11+1n.
Let(62)θ1=1σ1δ11-ζ1+ρ1γ1+ρ1ε1l1λ1+ε1δ1q1-ρ1q1ξ1δ1q1+ρ12q1r1α1τ1+β1ω1q11/q1+ρ2ε2l2λ2σ2,θ2=1σ2δ21-ζ2+ρ2γ2+ρ2ε2k2μ2+ε2δ2q2-ρ2q2ξ2δ2q2+ρ22q2r2α2τ2+β2ω2q21/q2+ρ1ε1k1μ1σ1.We can see that θn1→θ1,θn2→θ2, as n→∞. Now, it follows from condition (43) that max⁡θ1,θ2<1. Hence, there are a positive number θ0<1 and an integer n0≥1 such that max⁡θn1,θn2≤θ0<1, for all n≥n0. Therefore, it follows from (59) that xn,yn is a Cauchy sequence in E1×E2. Let (xn,yn)→(x,y)∈E1×E2, as n→∞. Since Ai,Bi,Ci and Di(i∈I) are H-Lipschitz continuous, by (41), we have (63)un+11-un11≤1+1n+1HA1xn+1,A1xn≤1+1n+1τ1xn+1-xn1;vn+11-vn11≤1+1n+1HB1xn+1,B1xn≤1+1n+1ω1xn+1-xn1;un+12-un22≤1+1n+1HA2yn+1,A2yn≤1+1n+1τ2yn+1-yn2;vn+12-vn22≤1+1n+1HB2yn+1,B2yn≤1+1n+1ω2yn+1-yn2;wn+1i-wni1≤1+1n+1HCixn+1,Cixn≤1+1n+1λixn+1-xn1;zn+1i-zni2≤1+1n+1HDiyn+1,Diyn≤1+1n+1μiyn+1-yn2.Therefore, uni,vni,wni and zni(i∈I) are also Cauchy sequences. Let uni→ui,vni→vi,wni→wi,zni→zi(i∈I), as n→∞. Since un1∈A1xn, we have (64)du1,A1x≤u1-un11+dun1,A1x≤u1-un11+HA1xn,A1x≤u1-un11+τ1xn-x1→0n→∞. Hence, we conclude u1∈A1x. Similarly, we can obtain u2∈A2y, v1∈B1x, v2∈B2y, wi∈Cix, zi∈Diy, i∈I.
SinceNi(uni,vni)→Ni(ui,vi), Fi(wni,zni)→Fi(wi,zi), b1(xn,xn+1)→b1(x,x), b2(yn,yn+1)→b2(y,y), g1(xn)→g1(x), g2(yn)→g2(y), as n→∞, and η1(h1,xn)→η1(h1,x) weakly in E1 and η2(h2,yn)→η2(h2,y) weakly in E2, as n→∞, we deduce from (39) and (40) that (65)g1x,h1-x1≥g1x,h1-x1-ρ1N1u1,v1+F1w1,z1,η1h1,x1+ρ1b1x,x-b1x,h1,∀h1∈E1,g2y,h2-y2≥g2y,h2-y2-ρ2N2u2,v2+F2w2,z2,η2h2,y2+ρ2b2y,y-b2y,h2,∀h2∈E2. That is, (66)N1u1,v1+F1w1,z1,η1h1,x1-b1x,x+b1x,h1,∀h1∈E1,N2u2,v2+F2w2,z2,η2h2,y2-b2y,y+b2y,h2,∀h2∈E2. This completes the proof.Remark 12.
( 1 ) Problem (1) is a very general system of variational inequalities in Banach spaces.
( 2 ) The system of auxiliary variational inequalities problem (20) is very different from the corresponding auxiliary variational inequality problems in [1, 3].
( 3 ) By using the concept of strongly mixed accretiveness and Lipschitz continuity in the sense of the Hausdorff metric, the restrictions imposed on the set-valued mapping T and the single-valued mapping N in Theorem 11 are weaker than that in [1, 3].Example 13.
LetE1=E2=R, for i=1,2, bi(x,y)=x,y, ηi(x,y)=x-y,(67)Aix=Bix=tsin⁡x-t2x,∀t∈0,1,x≤1,0,x>1,Cix=tsin⁡x,∀t∈0,1,x≤1,0,x>1,Diy=ty2,∀t∈0,1,x≤1,0,x>1,Ni(ui,vi)=1/2(ui+vi), Fi(wi,zi)=1/2(zi-2wi), where u1∈A1x, v1∈B1x, u2∈A2y, v2∈B2y, wi∈Cix,zi∈Diy. It is easy to check that these mappings satisfy the conditions of Theorem 11, and the system of variational inequalities is as follows:(68)12u1+v1+z1-2w1,h1-x+x,h1-x,x≥0,∀h1∈R,12u2+v2+z2-2w2,h2-y+y,h2-y,y≥0,∀h2∈R.
Letg1=g2=I; then Algorithm 10 can be written as(69)xn+1,h1-xn+1≥xn,h1-xn+1-ρ12un1+vn1+zn1-2wn1,h1-xn+1+ρ1xn,xn+1-xn,h1,yn+1,h2-yn+1≥yn,h2-yn+1-ρ22un2-vn2+zn2-2wn2,h2-yn+1+ρ2yn,yn+1-yn,h2.
Further, we choose(x1,y1)=(1/2,1/3), ρ1=ρ2=1/2, h1=h2=0. The numerical results presented in Table 1 and Figure 1 demonstrate problem (68).Table 1
Computational results for example 4.3.
n x n y n 1 0.500000000000000 0.333333333333333 2 -0.0833333333333333 -0.125000000000000 3 0.0312500000000000 0.0208333333333333 4 -0.00520833333333333 -0.00781250000000000 5 0.00195312500000000 0.00130208333333333 6 -0.000325520833333333 -0.000488281250000000 7 0.000122070312500000 8.13802083333333e-05 8 -2.03450520833333e-05 -3.05175781250000e-05 9 7.62939453125000e-06 5.08626302083333e-06 10 -1.27156575520833e-06 -1.90734863281250e-06Figure 1
The convergences ofxn and yn with initial value (x1,y1)=(1/2,1/3).Obviously,(0,0) is the solution of (68).



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