# On the Hadamard Well-Posedness of Generalized Mixed Variational Inequalities in Banach Spaces

Authors: Lu-Chuan Ceng; Yeong-Cheng Liou; Ching-Feng Wen; Hui-Ying Hu; Long He; Yun-Ling Cui
Journal: Journal of Mathematics
 (2021)

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## Abstract

We introduce a new concept of Hadamard well-posedness of a generalized mixed variational inequality in a Banach space. The relations between the Levitin–Polyak well-posedness and Hadamard well-posedness for a generalized mixed variational inequality are studied. The characterizations of Hadamard well-posedness for a generalized mixed variational inequality are established.


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## Body

## 1. Introduction

In [1], Tykhonov first introduced the well-posedness of a minimization problem, which means that it has a unique minimizer and every minimizing sequence converges to the unique minimizer. There are two concepts of well-posedness which are Tykhonov well-posedness [1] and Hadamard well-posedness [2].Recently, variational inequality (VI) has been extensively studied due to the facts that it has many potential applications and that it is closely related to a differentiable minimization problem. Well-posedness for a variational inequality has been then extensively investigated. See, e.g., [3–11] and the references therein.In 2013, Li and Xia [12] introduced the concept of Hadamard well-posedness of a general mixed variational inequality in Banach spaces. Under some suitable conditions, relations between Levitin–Polyak well-posedness and Hadamard well-posedness of a general mixed variational inequality were presented. They also established some characterizations of Hadamard well-posedness for a general mixed variational inequality. Very recently, some scholars still focused on the study of the well-posedness of various classes of variational inequalities, see e.g., generalized variational-hemivariational inequalities with perturbations in [13], completely generalized mixed variational inequalities in [14], noncompact generalized mixed variational inequalities in [15], generalized variational inequality with generalized mixed variational inequality constraint in [16], systems of generalized mixed quasivariational inclusion problems in [17], systems of time-dependent hemivariational inequalities in [18], and generalized hemivariational inequalities in [19].Motivated and inspired by the research work going on this field, we introduce a new concept of Hadamard well-posedness for a generalized mixed variational inequality in a Banach space. Under some suitable conditions, the relations between the Levitin–Polyak well-posedness and Hadamard well-posedness for a generalized mixed variational inequality are studied. We also establish some characterizations of Hadamard well-posedness for a generalized mixed variational inequality. Finally, we prove that under suitable conditions, the Hadamard well-posedness of a generalized mixed variational inequality is equivalent to the existence and uniqueness of its solutions. Our results improve, extend, and develop the earlier and recent ones announced by some others, e.g., Ceng and Yao [7] and Li and Xia [12, 20].


## 2. Preliminaries

LetX be a real reflexive Banach space with its dual X∗ and K be a nonempty, closed subset of X. We use the same notations in [12]. For more details about these notations and relevant definitions, please consult relevant reference; see, e.g., [12] (following [21]). Let X′ be the collection of all affine functions defined on X. It is obvious that X∗⊂X′. Let U be the collection of all nonempty set-valued mappings F:X⟶2X∗, and τX be the collection of all mappings P:X⟶2X′ such that for any x∈X, there exist F∈U and λ∈R such that(1)Px,x−y=Fx,x−y+λ,∀y∈K.For anyP1,P2∈τX, it follows that there exist Fi∈U and λi∈R,i=1,2 such that(2)P1x,x−y=F1x,x−y+λ1,∀y∈K,P2x,x−y=F2x,x−y+λ2,∀y∈K.We define(3)dP1,P2=λ1−λ2,F1=F2,1+λ1−λ2,F1≠F2.It can be routinely checked thatτX,d is a metric space. In particular, if U is the collection of all single-valued mappings F:X⟶X∗ and τX is the collection of all single-valued mappings P:X⟶X′ such that for any x∈X, there exist F∈U and λ∈R such that(4)Px,x−y=Fx,x−y+λ,∀y∈K,then the above metric space τX,d reduces to the metric space τX,d defined in [[12], p. 1619]. In this case, it is clear that the metric space τX,d is a special case of metric space Γ,d defined in [[21], p. 377].LetCX be the collection of all nonempty closed subsets of X endowed with the usual Hausdorff metric ℋ⋅,⋅, that is, for every A1,A2∈CX,(5)ℋA1,A2=maxeA1,A2,eA2,A1,where eA1,A2=supa∈A1d′a,A2 with d′a,A2=infb∈A2a−b. Let An be a sequence of nonempty closed subsets of X. We say that An converges to A in the Hausdorff metric iff ℋAn,A⟶0 as n⟶∞.LetBX be the family of all real-valued functions on X; we define(6)d1ϕ1,ϕ2=supx∈Xϕ1x−ϕ2x,where ϕ1,ϕ2∈BX; it can be routinely checked that BX,d1 is a metric space.LetM be the collection of all P,ϕ,K such that(i)
P∈τX;(ii)
ϕ∈BX;(iii)
K∈CX.Then, for anyP1,ϕ1,K1,P2,ϕ2,K2∈M, we define(7)ρP1,ϕ1,K1,P2,ϕ2,K2=dP1,P2+d1ϕ1,ϕ2+ℋK1,K2.Clearly,M,ρ is a metric space.LetF:X⟶2X∗ be a set-valued mapping, and ϕ:X⟶R∪+∞ be a proper, convex, and lower semicontinuous functional. Consider the following generalized mixed variational inequality associated with F,ϕ,K:(8)GMVIF,ϕ,K:findx∈Ksuch that for someu∈Fx,u,x−y+ϕx−ϕy≤0,∀y∈K.We denote bySF,ϕ,K the solution set of GMVIF,ϕ,K. In what follows, we first introduce new concept of Hadamard well-posedness for GMVIF,ϕ,K. It is worth mentioning that some similar ideas have also been presented in [22, 23] very recently.Definition 1.
A generalized mixed variational inequalityGMVIF,ϕ,K is called Hadamard well-posed if it has a unique solution x∗∈K, and if for every sequence of triples Pn,ϕn,Kn⊂M converging to F,ϕ,K and every sequence xn such that xn∈SPn,ϕn,Kn for each n∈N, it follows that xn⟶x∗, where K∪xn⊆Kn for all n∈N.Definition 2.
(see [20]). A sequence xn⊂X is called a LP approximating sequence for GMVIF,ϕ,K, if there exist wn∈X with wn⟶0 and 0<ϵn⟶0 such that xn+wn∈K for all n∈N, and there exists un∈Fxn such that(9)un,xn−y+ϕxn−ϕy≤ϵn,∀y∈K,n∈N.Definition 3.
(see [20]). We say that GMVIF,ϕ,K is LP well-posed if GMVIF,ϕ,K has a unique solution and every LP approximating sequence for GMVIF,ϕ,K converges strongly to the unique solution.
The product spaceCX×BX is equipped by a product metric, that is, ℋA1,A2+d1f1,f2, where A1,A2∈CX and f1,f2∈BX. Let further BC0X be the family of all real-valued continuous functions on X; it is easy to check that BC0X,d1 is a metric space and we write Q=CX×BC0X. Now, we can easily get the following lemma.Lemma 1 (see [12]).
Let the pairA,f∈Q, then the function A,f⟼infA,f is upper semicontinuous.
We consider the following gap function forGMVIF,ϕ,K:(10)gx=supy∈Kinfu∈Fxu,x−y+ϕx−ϕy,∀x∈X.Lemma 2.
The following statements hold:(i)
gx≥0,∀x∈K;(ii)
x¯∈K solves GMVIF,ϕ,K⟺gx¯=0.Proof.
For eachx∈K, we have(11)gx=supy∈Kinfu∈Fxu,x−y+ϕx−ϕy≥infu∈Fxu,x−x+ϕx−ϕx=0.
Observe that(12)x¯solves GMVIF,ϕ,K⟺∃u¯∈Fx¯s.t.u¯,x¯−y+ϕx¯−ϕy≤0,∀y∈K⟺supy∈Ku¯,x¯−y+ϕx¯−ϕy≤0⟺0≤gx¯≤supy∈Ku¯,x¯−y+ϕx¯−ϕy≤0⟺gx¯=0.
This completes the proof.
We also consider the following optimization problem:(13)OP:infx∈Kgx,with gx defined by (10). Its optimal solution set will be denoted by argminK,g and the optimal value will be denoted by infK,g, respectively.
The following Definitions4–6 can be found in [12]. However, Definition 7 is conventional.Definition 4.
A sequencexn⊂X is called an LP minimizing sequence for (OP) if there exists un∈Fxn such that(14)gnxn⟶infK,g,d′xn,K⟶0,where gnxn=supy∈K∪xnun,xn−y+ϕxn−ϕy,∀n∈N.Definition 5.
We say that (OP) is LP well-posed if and only if (OP) has a unique solution and every LP minimizing sequence for (OP) converges strongly to the unique solution.Definition 6.
A nonempty-valued functionf:X⟶R∪+∞ is said to be uniformly continuous, if for any ϵ>0, there exists δ>0 such that for all x,y∈X with x−y<δ, one has fx−fy<ε.Definition 7.
A nonempty set-valued mappingF:X⟶2X∗ is said to be monotone, if for all x,y∈X,u∈Fx and v∈Fy,(15)u−v,x−y≥0.Definition 8.
(see [7]). Let ℋ⋅,⋅ be the Hausdorff metric on the collection CBX of all nonempty, closed, and bounded subsets of X, which is defined by ℋA,B=maxeA,B,eB,A for A and B in CBX. A nonempty set-valued mapping F:X⟶2X∗ is said to be(i)
ℋ -hemicontinuous, if for any x,y∈X, the function t⟼ℋFx+ty−x,Fx from 0,1 into R+=0,+∞ is continuous at 0+;(ii)
ℋ-uniformly continuous, if for any ϵ>0, there exists δ>0 such that for all x,y∈X with x−y<δ, one has ℋFx,Fy<ϵ.
The following proposition is a special case of Lemma 2.2 in [7].Proposition 1 (see [20]).
LetK be a nonempty, closed, and convex subset of X,F:X⟶2X∗ be a nonempty compact-valued mapping which is ℋ-hemicontinuous and monotone, and ϕ:X⟶R∪+∞ be a proper and convex functional. Then, for a given x∈K, the following statements are equivalent:(i)
there existsu∈Fx such that u,x−y+ϕx−ϕy≤0,∀y∈K;(ii)
v,x−y+ϕx−ϕy≤0,∀y∈K,v∈Fy.
We can also prove the following lemma easily.Lemma 3.
LetK be a nonempty, closed subset of X,F:X⟶2X∗ be a nonempty set-valued mapping, and ϕ:X⟶R∪+∞ be a functional on X. Then, the following statements are equivalent:(i)
GMVIF,ϕ,K is LP well-posed;(ii)
(OP) is LP well-posed withgx defined by (10).Proof.
We first claim that (i)⇒ (ii). Indeed, suppose that GMVIF,ϕ,K is LP well-posed and x∗∈K is the unique solution of GMVIF,ϕ,K. By Lemma 2, x∗∈K is the unique solution of (OP). Then, we get infK,g=0. Let xn⊂X be a LP minimizing sequence for (OP). Then, there exists un∈Fxn such that(16)gnxn⟶infK,g=0,d′xn,K⟶0,where gnxn=supy∈K∪xnun,xn−y+ϕxn−ϕy,∀n∈N. So, we deduce that(17)supy∈Kun,xn−y+ϕxn−ϕy≤supy∈K∪xnun,xn−y+ϕxn−ϕy=gnxn,which immediately yields(18)limsupn⟶∞supy∈Kun,xn−y+ϕxn−ϕy≤0.
Thus, there exist0<ϵn⟶0 and un∈Fxn such that(19)supy∈Kun,xn−y+ϕxn−ϕy≤ϵn;that is,(20)un,xn−y+ϕxn−ϕy≤ϵn,∀y∈K.
Furthermore, fromd′xn,K⟶0 it follows that there exists x¯n∈K such that x¯n−xn⟶0. Putting wn=x¯n−xn, we get xn+wn=x¯n∈K with wn⟶0. Therefore, xn is a LP approximating sequence for GMVIF,ϕ,K, and hence xn⟶x∗ as n⟶∞. This means that (OP) is LP well-posed.
We show that (ii)⇒ (i). Indeed, suppose that (OP) is LP well-posed and x∗∈K is the unique solution of (OP). By Lemma 2, x∗∈K is the unique solution of GMVIF,ϕ,K. Then, we get infK,g=0. Let xn⊂X be a LP approximating sequence for GMVIF,ϕ,K. Then, there exist wn∈X with wn⟶0 and 0<ϵn⟶0 such that xn+wn∈K for all n∈N, and there exists un∈Fxn such that(21)un,xn−y+ϕxn−ϕy≤ϵn,∀y∈K,n∈N,that is,(22)supy∈Kun,xn−y+ϕxn−ϕy≤ϵn,∀n∈N.
Puttingx¯n=xn+wn for all n∈N, we get x¯n∈K for all n∈N. Then, it is easy to see that d′xn,K≤xn−x¯n=wn⟶0. Observe that for all n∈N,(23)0≤gnxn=supy∈K∪xnun,xn−y+ϕxn−ϕy=maxsupy∈Kun,xn−y+ϕxn−ϕy,0≤ϵn⟶0.
Consequently,xn is a LP minimizing sequence for (OP), and hence xn⟶x∗ as n⟶∞. This means that GMVIF,ϕ,K is LP well-posed.


## 3. Well-Posedness

In this section, we investigate the relations between Levitin–Polyak well-posedness and Hadamard well-posedness of a generalized mixed variational inequality.Theorem 1.
LetK be a nonempty, closed subset of X and F:X⟶2X∗ be a nonempty set-valued mapping. Let ϕ:X⟶R∪+∞ be a functional. Then, GMVIF,ϕ,K is LP well-posed whenever GMVIF,ϕ,K is Hadamard well-posed.Proof.
Suppose thatGMVIF,ϕ,K is Hadamard well-posed and x∗∈K is the unique solution of GMVIF,ϕ,K. Let xn⊂X be an LP approximating sequence for GMVIF,ϕ,K. Then, there exist wn∈X with wn⟶0 and 0<ϵn⟶0 such that xn+wn∈K for all n∈N, and there exists un∈Fxn such that(24)un,xn−y+ϕxn−ϕy≤ϵn,∀y∈K,that is,(25)supy∈Kun,xn−y+ϕxn−ϕy≤ϵn,∀n∈N.
So, it follows fromxn+wn∈K that there exists x¯n∈K such that xn+wn=x¯n for all n∈N. Thus, we get d′xn,K≤xn−x¯n=wn⟶0. For each n∈N,x∈X, we construct a sequence Fn,ϕn,Kn as follows:(26)Fnx,x−y=Fx,x−y−ϵn,∀y∈K,(27)ϕnx=ϕx−ϵn,and Kn=K∪xn.
It is obvious thatxn∈Kn,Fn∈τX,ϕn∈BX, and Kn∈CX. It follows from (24)–(27) that(28)Fnxn,xn−y+ϕnxn−ϕny=Fnxn,xn−y−ϵn+ϕxn−ϵn−ϕny−ϵn=Fnxn,xn−y+ϕxn−ϕny−ϵn,∀y∈K.
SinceKn=K∪xn, it follows from (24) that(29)0≤Gnxn=supy∈KninfFnxn,xn−y+ϕnxn−ϕny=supy∈KninfFnxn,xn−y+ϕxn−ϕny−ϵn=supy∈Kninfu∈Fxnu,xn−y+ϕxn−ϕny−ϵn≤supy∈Knu,xn−y+ϕxn−ϕny−ϵn≤0.
That is,Gnxn=0 for all n∈N. So, it follows from Lemma 2 that xn∈SFn,ϕn,Kn for all n∈N. From (3) and (26), we have dFn,F=ϵn⟶0. Again from (6) and (27), we have d1ϕn,ϕ⟶0. We also obtain that ℋKn,K=d′xn,K⟶0. Thus, we have ρFn,ϕn,Kn,F,ϕ,K⟶0. Since GMVIF,ϕ,K is Hadamard well-posed, we know that xn converges strongly to the unique solution x∗ of GMVIF,ϕ,K. Thus, GMVIF,ϕ,K is LP well-posed. The proof is complete.
Next, we have the following result which can be regarded as the reverse of Theorem1 under the uniform continuity of the function ϕ.Theorem 2.
LetK be a nonempty, closed subset of X and F:X⟶2X∗ be a nonempty set-valued mapping. Let ϕ:X⟶R∪+∞ be uniformly continuous on X. Then, GMVIF,ϕ,K is Hadamard well-posed whenever GMVIF,ϕ,K is LP well-posed.Proof.
Suppose thatGMVIF,ϕ,K is LP well-posed and x∗∈K is the unique solution of GMVIF,ϕ,K. Let(30)gx=supy∈KinfFx,x−y+ϕx−ϕy,∀x∈X.
SinceGMVIF,ϕ,K has the unique solution x∗∈K, by Lemma 2, we know that (OP) has the unique solution x∗∈K. That is, infK,g=0 and argminK,g=x∗. Let Fn,ϕn,Kn∈τX×BX×CX,Fn,ϕn,Kn converges to F,ϕ,K and xn∈SFn,ϕn,Kn, where K∪xn⊆Kn. So, it follows from xn∈SFn,ϕn,Kn that there exists u˜n∈Fnxn such that(31)u˜n,xn−y+ϕnxn−ϕny≤0,∀y∈Kn,which immediately yields(32)u˜n,xn−y+ϕnxn−ϕny≤0,∀y∈K.
For anyx∈X, let(33)Gnx=supy∈KninfFnx,x−y+ϕnx−ϕny,∀x∈X.
From Lemma1, it is easy to see that(i)
Gnx≥0,∀x∈Kn;(ii)
for anyx∈Kn,Gnx=0⟺x∈SFn,ϕn,Kn.
It follows from (ii) andxn∈SFn,ϕn,Kn that Gnxn=infKn,G=0. On the other hand, note that ρFn,ϕn,Kn,F,ϕ,K⟶0. Then, we deduce that dFn,F⟶0,d1ϕn,ϕ⟶0 and ℋKn,K⟶0. Since dFn,F⟶0, it follows from (3) that there exists 0<ϵn⟶0 such that for any x∈X,(34)Fnx,x−y=Fx,x−y−ϵn,∀y∈K.
In particular, we have(35)Fnxn,xn−y=Fxn,xn−y−ϵn,∀y∈K.
Fromu˜n∈Fnxn it follows that there exists un∈Fxn such that(36)u˜n,xn−y=un,xn−y−ϵn,∀y∈K.
This together with (32) leads to(37)un,xn−y+ϕnxn−ϕny≤ϵn,∀y∈K,which can be rewritten as follows:(38)un,xn−y+ϕxn−ϕy≤ϵn+ϕxn−ϕnxn−ϕy−ϕny,∀y∈K.
Next, we claim that(39)limn⟶∞ϕxn−ϕnxn−ϕy−ϕny=0,uniformly fory∈X.
As a matter of fact, for anyδ>0, since d1ϕn,ϕ=supx∈Xϕnx−ϕx⟶0, there exists an integer N≥1 such that for all n≥N,(40)supx∈Xϕnx−ϕx≤δ.
It follows that for anyx∈X,(41)ϕx−δ≤ϕnx≤ϕx+δ,n≥N.
So, for anyx,y∈X, we have(42)ϕnx−ϕny≤ϕx+δ−ϕy−δ=ϕx−ϕy+2δ,n≥N.
Meantime, we also have(43)ϕx−ϕy−2δ=ϕx−δ−ϕy+δ≤ϕnx−ϕny,n≥N.
Then, for anyx,y∈X,(44)ϕnx−ϕny−ϕx−ϕy≤2δ,n≥N.
In particular, for anyy∈X, we get(45)ϕnxn−ϕny−ϕxn−ϕy≤2δ,n≥N.
This means that (39) holds.
Finally, from (38) and (39) and 0<ϵn⟶0, we have(46)0≤gnxn=supy∈K∪xnun,xn−y+ϕxn−ϕy≤maxsupy∈Kϵn+ϕxn−ϕnxn−ϕy−ϕny,0≤ϵn+supy∈Kϕxn−ϕnxn−ϕy−ϕny⟶0.
SinceKn⟶K in the Hausdorff metric and xn∈Kn, we have d′xn,K⟶0. Thus, xn is an LP minimizing sequence for (OP). Since GMVIF,ϕ,K is LP well-posed, according to Lemma 3, we know that (OP) is LP well-posed. Therefore, xn⟶x∗ as n⟶∞. So, it follows that GMVIF,ϕ,K is Hadamard well-posed. The proof is complete.Remark 1.
Theorems1 and 2 improve, extend, and develop Theorems 3.1 and 3.2 in [12] to a great extent because the generalized mixed variational inequality considered in Theorems 1 and 2 is more general than the general mixed variational inequality considered in [[12], Theorems 3.1 and 3.2].


## 4. Metric Characterization and Conditions for Hadamard Well-Posedness

In this section, we derive the metric characterization of Hadamard well-posedness for a generalized mixed variational inequality and prove that under suitable conditions, the Hadamard well-posedness of a generalized mixed variational inequality is equivalent to the existence and uniqueness of its solutions.To characterize the Hadamard well-posedness for a generalized mixed variational inequalityGMVIF,ϕ,K, we define(47)Ωϵ=x∈X:d′x,K≤ϵ,and there existsu∈Fxsuch that∀y∈K,u,x−y+ϕx−ϕy≤ϵ,∀ϵ≥0.Theorem 3.
LetF,ϕ,K∈τX×BX×CX, K be convex, F:X⟶2X∗ be a nonempty compact-valued mapping which is ℋ-hemicontinuous and monotone, and ϕ:X⟶R∪+∞ be proper, convex, and uniformly continuous on X. Then, GMVIF,ϕ,K is Hadamard well-posed if and only if(48)Ωϵ≠∅,∀ϵ>0and diamΩϵ⟶0asϵ⟶0.Proof.
Assume thatGMVIF,ϕ,K is Hadamard well-posed. Then, GMVIF,ϕ,K has a unique solution which lies in Ωϵ for all ϵ>0. Put x0∈SF,ϕ,K. Obviously, x0∈Ωϵ for all ϵ>0. If diamΩϵ↛0 as ϵ⟶0, then for some δ>0,0<ϵn⟶0 such that for n sufficiently large,(49)diamΩϵn>δ>0.
Thus, we can find pointsxn∈Ωϵn such that(50)xn−x0>δ2.
Sincexn∈Ωϵn, we have(51)d′xn,K≤ϵn,and there exists un∈Fxn such that(52)un,xn−y+ϕxn−ϕy≤ϵn,∀y∈K.
Now, we construct a sequenceFn,ϕn,Kn as follows:(53)Fnx,x−y=Fnx,x−y−ϵn,∀y∈K,ϕnx=ϕx−ϵn,Kn=K∪xn.
It is obvious thatxn∈Kn,FnτX,ϕn∈BX, and Kn∈CX. By the similar argument to that in the proof of Theorem 1, we have xn∈SFn,ϕn,Kn. Observe that dFn,F=ϵn⟶0,d1ϕn,ϕ⟶0, and ℋKn,K=d′xn,K⟶0. Thus, we have ρFn,ϕn,Kn,F,ϕ,K⟶0. Since GMVIF,ϕ,K is Hadamard well-posed, one has xn⟶x0, a contradiction to (50).
Conversely, suppose that condition (48) holds. Let Fn,ϕn,Kn⟶F,ϕ,K, and xn∈SFn,ϕn,Kn, where K∪xn⊆Kn,n=1,2,…. So, it follows from xn∈SFn,ϕn,Kn that there exists u˜n∈Fnxn such that(54)u˜n,xn−y+ϕnxn−ϕny≤0,∀y∈Kn,which immediately yields(55)u˜n,xn−y+ϕnxn−ϕny≤0,∀y∈K.
Furthermore, note thatρFn,ϕn,Kn,F,ϕ,K⟶0. Then, we deduce that dFn,F⟶0,d1ϕn,ϕ⟶0 and ℋKn,K⟶0. Since dFn,F⟶0, it follows from (3) that there exists 0<ϵn′⟶0 such that for any x∈X,(56)Fnx,x−y=Fx,x−y−ϵn′,∀y∈K.
In particular, we have(57)Fnxn,xn−y=Fxn,xn−y−ϵn′,∀y∈K.
Fromu˜n∈Fnxn, it follows that there exists un∈Fxn such that(58)u˜n,xn−y=un,xn−y−ϵn′,∀y∈K.
This together with (55) leads to(59)un,xn−y+ϕnxn−ϕny≤ϵn′,∀y∈K,which can be rewritten as follows:(60)un,xn−y+ϕxn−ϕy≤ϵn′+ϕxn−ϕnxn−ϕy−ϕny,∀y∈K.
Repeating the same argument as that of (39) in the proof of Theorem 2, we get(61)limn⟶∞ϕxn−ϕnxn−ϕy−ϕny=0,uniformly fory∈X.
Taking into account thatKn⟶K in the Hausdorff metric and xn∈Kn, we have d′xn,K⟶0. Thus, there exists 0<ϵn∗⟶0 such that d′xn,K≤ϵn∗ and(62)supy∈Xϕxn−ϕnxn−ϕy−ϕny≤ϵn∗,∀n∈N.
Setϵn=ϵn′+ϵn∗. Then, it follows from (60) that(63)un,xn−y+ϕxn−ϕy≤ϵn,∀y∈K.
This means thatxn∈Ωϵn for all n∈N. From (48), we know that xn is a Cauchy sequence and so it converges strongly a point x¯∈K. Since F is monotone and ϕ is lower semicontinuous, it follows from (63) that for any y∈K,v∈Fy,(64)v,x¯−y+ϕx¯−ϕyliminfn⟶∞v,xn−y+ϕxn−ϕy≤liminfn⟶∞un,xn−y+ϕxn−ϕy≤liminfn⟶∞ϵn=0.
So, from Proposition1, it is easy to see that x¯ solves GMVIF,ϕ,K.
To complete the proof, we need only to prove thatGMVIF,ϕ,K has a unique solution. Assume by contradiction that GMVIF,ϕ,K has two distinct solutions x1 and x2 in K. Then, it is easy to see that x1,x2∈Ωϵ for all ϵ>0 and(65)0<x1−x2≤diamΩϵ⟶0,a contradiction to (48). The proof is complete.
Next, we prove that the Hadamard well-posedness of a generalized mixed variational inequality is equivalent to the existence and uniqueness of its solutions under suitable conditions.Theorem 4.
LetF,ϕ,K∈τX×BX×CX, K be convex, F:X⟶2X∗ be a nonempty compact-valued mapping which is ℋ -hemicontinuous and monotone, and ϕ:X⟶R∪+∞ be proper, convex, and uniformly continuous on X. Then, GMVIF,ϕ,K is Hadamard well-posed if and only if it has a unique solution.Proof.
The necessity is obvious. For the sufficiency, suppose thatGMVIF,ϕ,K has a unique solution x∗. If GMVIF,ϕ,K is not Hadamard well-posed, then there exists Fn,ϕn,Kn⊂M converging to F,ϕ,K with xn∈SFn,ϕn,Kn such that xn do not converge to x∗, where K∪xn⊆Kn,n=1,2,…. So, it follows from xn∈SFn,ϕn,Kn that there exists u˜n∈Fnxn such that(66)u˜n,xn−y+ϕnxn−ϕny≤0,∀y∈Kn,which immediately yields(67)u˜n,xn−y+ϕnxn−ϕny≤0,∀y∈K.
Furthermore, note thatρFn,ϕn,Kn,F,ϕ,K⟶0. Then, we deduce that dFn,F⟶0,d1ϕn,ϕ⟶0 and ℋKn,K⟶0. Since dFn,F⟶0, it follows from (3) that there exists 0<ϵn′⟶0 such that for any x∈X,(68)Fnx,x−y=Fx,x−y−ϵn′,∀y∈K.
In particular, we have(69)Fnxn,xn−y=Fxn,xn−y−ϵn′,∀y∈K.
Fromu˜n∈Fnxn, it follows that there exists un∈Fxn such that(70)u˜n,xn−y=un,xn−y−ϵn′,∀y∈K.
This together with (67) leads to(71)un,xn−y+ϕxn−ϕy≤ϵn′+ϕxn−ϕnxn−ϕy−ϕny,∀y∈K.
Repeating the same argument as that of (63) in the proof of Theorem 3, we obtain that there exists 0<ϵn⟶0 such that d′xn,K≤ϵn and(72)un,xn−y+ϕxn−ϕy≤ϵn,∀y∈K.
Fromd′xn,K≤ϵn<ϵn+1/n, it follows that there exists x¯n∈K such that x¯n−xn<ϵn+1/n⟶0. Putting wn=x¯n−xn, we get x¯n=wn+xn with wn⟶0.
We claim thatxn is bounded. As a matter of fact, if xn is unbounded, then x¯n is an unbounded sequence in K. Without loss of generality, we may assume that x¯n⟶+∞. Let(73)tn=1x¯n−x∗,zn=x∗+tnx¯n−x∗.
Without loss of generality, we may assume thattn∈0,1 and zn⇀z≠x∗. Then, we have for each y∈K,v∈Fy,(74)v,z−y=v,z−zn+v,zn−x∗+v,x∗−y=v,z−zn+tnv,x¯n−x∗+v,x∗−y=v,z−zn+tnv,xn+wn−x∗+v,x∗−y=v,z−zn+tnv,xn−y+1−tnv,x∗−y+tnv,wn.
Sincex∗ is the unique solution of GMVIF,ϕ,K, there exists u∗∈Fx∗ such that(75)u∗,x∗−y+ϕx∗−ϕy≤0,∀y∈K.
SinceF is monotone, we have(76)v,x∗−y≤u∗,x∗−y,v,xn−y≤un,xn−y.
It follows from (72)–(76) and the convexity of ϕ that for all v∈Fy,(77)v,z−y≤v,z−zn+tnϕy−tnϕxn+tnϵn+1−tnϕy−ϕx∗+tnv,wn=v,z−zn+ϕy−tnϕxn+1−tnϕx∗+tnϵn+tnv,wn=v,z−zn+ϕy−tnϕx¯n+1−tnϕx∗+tnϕxn−tnϕx¯n+tnϵn+tnv,wn≤v,z−zn+ϕy−ϕzn−tnϕxn−ϕx¯n+tnϵn+tnv,wn,∀y∈K.
Sinceϕ is uniformly continuous, we have(78)v,z−y≤liminfn⟶∞v,z−y+ϕy−ϕzn−tnϕxn−ϕx¯n+tnϵn+tnv,wn≤ϕy−ϕz,∀y∈K.
This together with Proposition1 implies that z solves GMVIF,ϕ,K, a contradiction. Thus, xn is bounded.
Next, we claim thatxn⟶x∗ as n⟶∞. Let xnk be any subsequence of xn such that xnk⟶x¯ as k⟶∞. Clearly, x¯∈K. It follows from (72) that(79)unk,xnk−y+ϕxnk−ϕy≤ϵnk,∀y∈K.
SinceF is monotone and ϕ is uniformly continuous, we have(80)v,x¯−y+ϕx¯−ϕy=liminfk⟶∞v,xnk−y+ϕxnk−ϕy≤liminfk⟶∞unk,xnk−y+ϕxnk−ϕy≤liminfk⟶∞ϵnk=0,∀y∈K,v∈Fy.
This together with Proposition1 implies that x¯ solves GMVIF,ϕ,K. Since GMVIF,ϕ,K has a unique solution x∗, we have x¯=x∗. Thus, xn⟶x∗, which reaches a contradiction. So, GMVIF,ϕ,K is Hadamard well-posed. The proof is complete.


## 5. Concluding Remarks

Theorems3 and 4 improve, extend, and develop Theorems 4.1 and 4.2 in [12] to a great extent because the generalized mixed variational inequality considered in Theorems 3 and 4 is more general than the general mixed variational inequality considered in ([12], Theorems 4.1 and 4.2). In addition, Theorems 3 and 4 also improve, extend, and develop Theorems 3.1 and 6.1 in [7] and Theorems 3.1 and 6.1 in [20] to a great extent because Levitin–Polyak well-posedness of a generalized mixed variational inequality is replaced by Hadamard well-posedness of a generalized mixed variational inequality.



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