# Existence Theorems for Quasivariational Inequality Problem on Proximally Smooth Sets

Authors: Jittiporn Suwannawit; Narin Petrot
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The concept of quasivariational inequality problem on proximally smooth sets is studied. Some sufficient conditions for solving the existence of solutions of  such a problem are provided; also some interesting cases are discussed. Of course, due to the significance of proximally smooth sets, the results which are presented in this paper improve and extend many important results in the literature.


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## Body

## 1. Introduction and Preliminaries

Variational inequality theory is a branch of the mathematics which is important, and it was also the inspiration for researchers to find new works, both in terms of mathematics and applications such as in economics, physical, biological, and engineering science, and other applied sciences. In 1973, Bensoussan et al. [1] introduced and studied the concept of the quasivariational inequality, which is a generalized form of the classical variational inequality that introduced by Stampacchia [2]. Later, many researchers proposed and analyzed the concept of the generalized quasivariational inequality; see, for example [3–18]. It is worth mentioning that the quasivariational inequality problem is of interest to study, since in many important problems the considered set also depends upon the solutions explicitly or implicitly. In fact, the concept of the quasi variational inequality has been applied in many fields such as in economics and transportation equilibrium, control and optimization theory, mathematical programming, and game theory.In the early period of the research, it should be pointed out that almost all the results regarding the existence and iterative schemes for solving those variational inequalities problems are being considered in the convexity setting. This is because they need the convexity assumption for guaranteeing the well definedness of the proposed iterative algorithm which depends on the projection mapping. However, in fact, the convexity assumption may not be required because it may be well defined even if the considered set is nonconvexs (e.g., when the considered set is a closed subset of a finite dimensional space or a compact subset of a Hilbert space, etc.). However, it may be from the practical point of view one may see that the nonconvex problems are more useful than convex case. Consequently, now many researchers are paying attention to many nonconvex cases.LetT:H→H be a mapping and let C:H→𝒞(H) be a set-valued mapping, where 𝒞(H) is a family of all nonempty closed subsets of H. In this paper, we are interested in the following problem: find x*∈C(x*) such that
(1)-Tx*∈NC(x*)P(x*),
where NC(x)P(x) is denoted for the proximal normal cone of C(x) of x. The problem of type (1) was introduced by Bounkhel et al. [19]. In such a paper, they proposed some iterative algorithms for finding a solution of type (1), when the considered mapping is a set-valued mapping, while, in this paper, we will provide sufficient conditions for the existence of a solution of such a problem (1). To do this, let us start by recalling some basic concepts and useful results that will be needed in this work.LetH be a real Hilbert space equipped with norm ∥·∥ and inner product 〈·,·〉. Let 2H be denoted for the class of all nonempty subsets of H and 𝒞(H) denoted for the family of all nonempty closed subsets of H.For eachK∈2H, the usual distance function on H to K is denoted by d(·,K), that is:
(2)d(u,K)=infv∈K∥u-v∥,∀u∈H.
Let K∈2H and u∈H. A point v∈K is called the closest point or the projection of u onto K if
(3)d(u,K)=∥u-v∥.
The set of all such closest points is denoted by ProjK(u), that is:
(4)ProjK(u)={v∈K:d(u,K)=∥u-v∥}.
Further, for each x∈K, the proximal normal cone to K at x is given by
(5)NKP(x)={z∈H:∃ρ>0:x∈ProjK(x+ρz)}.The following is calledthe proximal normal inequality; the proof of this characterization can be found in [20].Lemma 1 (see [19]).
LetK be a closed subset of a Hilbert space H. Then
(6)z∈NKP(x)⟺∃σ>0suchthat〈z,y-x〉≤σ∥y-x∥2,∀y∈K.We recall also [21] that the Clarke normal cone is given by
(7)N(K,x)=co¯[NKP(x)],
where co¯[S] means the closure of the convex hull of S. It is clear that one always has NKP(x)⊂N(K,x). The converse is not true in general. Note that N(K,x) is always a closed and convex cone and that NKP(x) is always a convex cone but may be nonclosed (see [20, 21]).In 1995, Clarke et al. [22] have introduced and studied a new class of nonconvex sets, which are called proximally smooth sets. This class of proximally smooth sets has played an important part in many nonconvex applications such as optimization, dynamic systems, and differential inclusions. Subsequently, the proximally smooth sets have been proposed by many researchers. In recent years, Bounkhel et al. [19], Cho et al. [23], Moudafi [24], Noor [25], Noor et al. [26], Petrot [27], and Pang et al. [28] have considered both variational inequalities and equilibrium problems in the context of proximally smooth sets. They suggested and analyzed some projection type iterative algorithms by using the prox-regular technique and auxiliary principle technique. Here, we will take the following characterization proved in [20] as the definition of proximally smooth sets. Note that the original definition was given in terms of the differentiability of the distance function (see [22, 29]).Definition 2.
For a givenr∈(0,+∞], a subset K of H is said to be uniformly prox-regular with respect tor, say, uniformlyr-prox-regular set, if for all x¯∈K and for all 0≠z∈NKP(x¯), one has
(8)〈z∥z∥,x-x¯〉≤12r∥x-x¯∥2,∀x∈K.For the case ofr=∞, the uniform r-prox-regularity K is equivalent to the convexity of K (see [22]). Moreover, it is known that the class of uniformly prox-regular sets is sufficiently large to include the class p-convex sets, C1,1 submanifolds (possibly with boundary) of H, the images under a C1,1 diffeomorphism of convex sets, and many other nonconvex sets; see [20, 29].From now on, we will denote[Cl(H)]r for the class of all uniformly r-prox-regular closed subset of H, where r∈(0,+∞] is fixed positive real number. Also, for each r∈(0,+∞], we write
(9)Kr:={x∈H;d(x,K)<r}.Remark 3.
IfT:H→H and C:H→[Cl(H)]r are mappings, then the problem of type (1) is equivalent to the following problem: find x*∈C(x*) such that
(10)〈Tx*,x-x*〉+∥Tx*∥2r∥x-x*∥2≥0,∀x∈C(x*),
see [19]. This means, in particular, that the problem (1) contains the well-known Stampacchia's variational inequality, as a special case.The following lemma which summarizes some important consequences of the uniformly prox-regularity sets is needed in the sequel. The proof of this result can be found in [22, 29].Lemma 4.
Letr∈(0,+∞] and K be a nonempty closed subset of H. If K is uniformly r-prox-regular set, then the following holds.  (i)
For allx∈Kr,ProjK(x)≠∅.(ii)
For alls∈(0,r),ProjK is Lipschitz with constant r/(r-s) on Ks.(iii)
The proximal normal cone is closed as a set-valued mapping.Remark 5.
IfK is uniformly r-prox-regular set, as a direct consequence of Lemma 4 (iii), we have N(K,x)=NKP(x).The following definition and lemma are also needed, in order to obtain our main results.Definition 6.
A set-valued mappingC:H→2H is said to be κ-Lipschitz if there exists κ>0 such that
(11)|d(y,C(x))-d(y′,C(x′))|≤∥y-y′∥+κ∥x-x′∥,∀x,x′,y,y′∈H.Lemma 7 (see [19]).
Letr∈(0,+∞] and let C:H→2H be a κ-Lipschitz set-valued mapping with uniformly r-prox-regular values then the following closedness property holds: “For any xn→x*,yn→y* and un→u* with yn∈C(xn) and un∈N(C(xn),yn), one has u*∈N(C(x*),y*).”


## 2. Main Results

In this paper, we are interested in the following classes of mappings.Definition 8.
LetT:H→H be a mapping. Then T is called  (a)
β-strongly monotone if there exists β>0 such that
(12)〈T(x)-T(y),x-y〉≥β∥x-y∥2,∀x,y∈H,(b)
ξ-Lipschitz if there exists ξ>0 such that
(13)∥T(x)-T(y)∥≤ξ∥x-y∥,∀x,y∈H.  That is, in other word, we will make the following assumption.Assumption
(
𝒜
). Let T:H→H and C:H→[Cl(H)]r be mappings.  (i)
T is a β-strongly monotone and a ξ-Lipschitz single-valued mapping;(ii)
C is a κ-Lipschitz set-valued mapping;(iii)
there isω∈[0,1) such that
(14)∥ProjC(x)(z)-ProjC(y)(z)∥≤ω∥x-y∥,∀x,y,z∈H.Remark 9.
LetH=(-∞,∞) and K=[a,b]∪[c,d], for some positive real numbers a,b,c,d with a<b<c<d. If we define C:H→2H by C(x)=m(x)+K, where m:H→H is a mapping defined by m(x)=kx for k∈R, then we see that C is a max{d-b,c-a}-Lipschitz mapping, and Assumption (𝒜)  (iii) is satisfied with a constant 2(max{d-b,c-a}). This means that Assumptions (𝒜)    (ii) and  (iii) are independent.The following remark is very useful in order to prove our results. Before seeing that, for the sake of simplicity, let us make a notation: for eachr∈(0,∞) and s∈(0,r), we will write ts∶=r/(r-s).Remark 10.
Letr,β,ξ,ω, and δ be five positive real numbers such that β∈(0,ξ), δ∈(0,rβ2-ξ2(1-(1-ω)2)/(1-ω)2) and ω∈[0,1-ξrξ2-β2/(ξ2r-βδ)). If h:[1,⋄]→[0,∞) is a function defined by
(15)h(t)=r(t-1)tδ+β2t2-ξ2[t2-(1-ω)2]tξ2
for all t∈[1,⋄], where ⋄=ξ(1-ω)/ξ2-β2, then h is an increasing continuous function on its domain. Moreover, we can check that β/ξ2 is an element of the range of h.Next, for a fixed positive real numberκ∈[0,1/(2⋄-1)), we pick a real number ζ=(1-⋄κ)/κ(⋄-1). Here, we notice that ζ∈(1,∞). Let us consider now a case when h-1(β/ξ2)∈[1,(1+ζκ)/κ(1+ζ)]. Then, by the definition of ts and h is an increasing continuous function, for s*=r(1-κ)/(1+ζκ) we have
(16)βξ2<r(ts*-1)ts*δ+β2ts*2-ξ2[ts*2-(1-ω)2]ts*ξ2,
that is,
(17)βξ2-β2ts*2-ξ2[ts*2-(1-ω)2]ts*ξ2<r(ts*-1)ts*δ.
Since r(ts*-1)/ts*=s*, this gives
(18)βξ2-β2ts*2-ξ2[ts*2-(1-ω)2]ts*ξ2<s*δ.
This allows us to take a real number ρ such that
(19)βξ2-f(ts*)<ρ<min{βξ2+f(ts*),s*δ},
where f(t)=β2t2-ξ2[t2-(1-ω)2]/tξ2.Now we are in a position to present our main results.Theorem 11.
LetT:H→H be a single-valued mapping and let C:H→[Cl(H)]r be a set-valued mapping. Assume that Assumption (𝒜) holds and the following control conditions are satisfied:(i)
δ∈(0,rβ2-ξ2(1-(1-ω)2)/(1-ω)2) and ω∈[0,1-ξrξ2-β2/(ξ2r-βδ)), where δ=sup{∥T(x)∥:x∈H};(ii)
κ∈[0,1/(2⋄-1)), where ⋄=ξ(1-ω)/ξ2-β2;(iii)
h-1(β/ξ2)∈[1,(1+ζκ)/κ(1+ζ)], where h and ζ are defined as Remark 10.
If there isx0∈H such that d(x0,C(x0))<s*-ρ∥Tx0∥, where s*=r(1-κ)/(1+ζκ) and ρ is a real number corresponding to s* which is chosen as in (19), then the problem (1) has a solution.Proof.
Firstly, we will define a sequence{xn} in H as follows: consider an element x0 in H such that d(x0,C(x0))<s*-ρ∥Tx0∥; we see that
(20)d(x0-ρTx0,C(x0))≤d(x0,C(x0))+ρ∥Tx0∥<s*.
This means x0-ρTx0∈[C(x0)]s*. Subsequently, by Lemma 4  (i), we know that ProjC(x0)[x0-ρTx0]≠∅. Let x1∈ProjC(x0)[x0-ρTx0]. So, by a choice of ρ, we see that
(21)d(x1-ρTx1,C(x1))≤d(x1,C(x1))+ρ∥Tx1∥=d(x1,C(x1))-d(x1,C(x0))+ρ∥Tx1∥≤κ∥x1-x0∥+ρ∥Tx1∥<s*+κ∥x1-x0∥.
On the other hand, we see that
(22)∥x1-x0∥≤∥x1-(x0-ρTx0)∥+∥(x0-ρTx0)-x0∥=d(x0-ρTx0,C(x0))+ρ∥Tx0∥≤d(x0,C(x0))+ρ∥Tx0∥+ρ∥Tx0∥<(s*-ρ∥Tx0∥)+ρ∥Tx0∥+ρ∥Tx0∥<2s*.
Using this one together with (21), we obtain
(23)d(x1-ρTx1,C(x1))<s*+2κs*.
Note that, by ζ>1, we have (1-κ)(1+2κ)/(1+ζκ)<1. So, since s*=r(1-κ)/(1+ζκ), by (23) we have
(24)d(x1-ρTx1,C(x1))<r(1-κ)(1+2κ)1+ζκ<r.
Hence x1-ρTx1∈[C(x1)]r(1-κ)(1+2κ)/(1+ζκ), and it follows that ProjC(x1)[x1-ρTx1]≠∅. Let x2∈ProjC(x1)[x1-ρTx1]. In the same way as obtaining (21) and (23), we see that
(25)d(x2-ρTx2,C(x2))<s*+κ∥x2-x1∥,∥x2-x1∥≤d(x1,C(x1))+2ρ∥Tx1∥=d(x1,C(x1))-d(x1,C(x0))+2ρ∥Tx1∥≤κ∥x1-x0∥+2ρ∥Tx1∥=2s*+κ∥x1-x0∥<2s*+2κs*.
By (25), we have
(26)d(x2-ρTx2,C(x2))<s*+κ(2κs*+2s*)=s*+2κs*+2κ2s*≤r(1-κ1+ζκ)(1+2κ+2κ2+⋯)=r(1-κ1+ζκ)(1+κ1-κ)<r.
By continuing this process, we can construct a sequence {xn} in H such that
(27)xn+1∈ProjC(xn)[xn-ρTxn],d(xn-ρTxn,C(xn))<s*+κ∥xn-xn-1∥,
for all n∈ℕ.
Further, let us consider(28)∥xn-xn-1∥≤∥xn-(xn-1-ρTxn-1)∥+∥(xn-1-ρTxn-1)-xn-1∥=d(xn-1-ρTxn-1,C(xn-1))+ρ∥Txn-1∥<s*+κ∥xn-1-xn-2∥+ρ∥Txn-1∥=2s*+κ∥xn-1-xn-2∥<2s*+2κs*+⋯+2κn-2+2κn-1s*=2s*(1+κ+⋯+κn-2+κn-1)=2s*(1-κn1-κ),
for each n∈ℕ. This implies
(29)d(xn-ρTxn,C(xn))<s*+2κs*(1-κn1-κ)=s*(1+2κ(1-κn1-κ))=s*(1-κ+2κ-2κn+11-κ)≤s*(1+κ1-κ)<r(1+κ1+ζκ)<r.
Write λ∶=r((1+κ)/(1+ζκ)). From the previous argument, we see that the sequence {xn} also has a property that
(30)xn-ρTxn∈[C(xn)]λ,
for all n∈ℕ.
Next, we will show that{xn} is a convergent sequence, and its limit is nothing but a solution of the problem (1).
Now, by using the Assumption (𝒜) (iii) and Lemma 4 (ii), we have
(31)∥xn+1-xn∥=∥ProjC(xn)(xn-ρTxn)-ProjC(xn-1)(xn-1-ρTxn-1)∥≤∥ProjC(xn)(xn-ρTxn)-ProjC(xn)(xn-1-ρTxn-1)∥+∥ProjC(xn)(xn-1-ρTxn-1)-ProjC(xn-1)(xn-1-ρTxn-1)∥≤tλ∥xn-ρTxn-xn-1+ρTxn-1∥+ω∥xn-xn-1∥.
Meanwhile, since T is a β-strongly monotone and ξ-Lipschitz mapping, we see that
(32)∥xn-ρTxn-xn-1+ρTxn-1∥2=∥xn-xn-1-ρ[Txn-Txn-1]∥2=∥xn-xn-1∥2-2ρ〈Txn-Txn-1,xn-xn-1〉+ρ2∥Txn-Txn-1∥2≤∥xn-xn-1∥2-2ρβ∥xn-xn-1∥2+ρ2ξ2∥xn-xn-1∥2=(1-2ρβ+ρ2ξ2)∥xn-xn-1∥2.
Let θρ=max{1-2ρβ+ρ2ξ2,((1-ω)(r-λ)/2r)2}. Observe that θρ∈[0,∞). Moreover, it follows that
(33)∥xn-ρTxn-xn-1+ρTxn-1∥≤θρ∥xn-xn-1∥.
By replacing (33) into (31), we get
(34)∥xn+1-xn∥≤tλθρ∥xn-xn-1∥+ω∥xn-xn-1∥=[tλθρ+ω]∥xn-xn-1∥.
Let γ:=tλθρ+ω. Then, by a choice of ρ, one can check that γ<1. Subsequently, by (34), we have
(35)∥xn+1-xn∥≤γn∥x1-x0∥,∀n∈ℕ.
Hence, for any m≥n, we see that
(36)∥xm-xn∥≤∑i=nm-1∥xi+1-xi∥≤∥x1-x0∥∑i=nm-1γi≤∥x1-x0∥∑i=n∞γi≤(γn1-γ)∥x1-x0∥.
This implies that {xn} is a Cauchy sequence in H, since γ<1. So, by the completeness of H, there exists x*∈H such that xn→x* as n→∞.
We now finish the proof by showing thatx* is a solution of problem (1). To do this, we will start by asserting that x*∈C(x*). Indeed, since xn+1∈C(xn) and C is a κ-Lipschitz mapping, we have
(37)d(x*,C(x*))=|d(xn+1,C(xn))-d(x*,C(x*))|≤∥xn+1-x*∥+κ∥xn-x*∥.
Thus, since xn→x*, we see that d(x*,C(x*))=0. So, by the closedness of C(x*), it guarantees that x*∈C(x*).
Next we show thatTx*∈-N(C(x*),x*). Let us observe that
(38)xn+1∈ProjC(xn)(xn-ρTxn),∀n∈ℕ
is equivalent to
(39)xn+1∈ProjC(xn)(xn+1+(xn-xn+1)-ρTxn).
And this means
(40)(xn-xn+1)-ρTxn∈N(C(xn),xn+1),∀n∈ℕ.
Thus, by using the continuity of mapping T, from Lemma 7 we see that ρTx*∈-N(C(x*),x*). This implies Tx*∈-N(C(x*),x*), as required. This means that x* is a solution of the problem (1), and the proof is completed.Remark 12.
A condition which has been proposed in the assumptions of Theorem11 is that “there isx0∈Hsuch thatd(x0,C(x0))<s*-ρ∥Tx0∥,wheres*=r(1-κ)/(1+ζκ).” Here, in view of Remark 10 together with the following facts, one may see that our choice s*=r(1-κ)/(1+ζκ) should be sharpest.  (i)
s=r(1-κ)/(1+ζκ)⇔ts=(1+ζκ)/κ(1+ζ).(ii)
The functionζ↦(1+ζκ)/κ(1+ζ) is an increasing function on its domain.(iii)
(1+ζκ)/κ(1+t)=⋄⇔ζ=(1-⋄κ)/κ(⋄-1), where ⋄=ξ(1-ω)/ξ2-β2.Remark 13.
Assume that all assumptions of Theorem11 hold. By starting with an element x0∈H such that d(x0,C(x0))<s*-ρ∥Tx0∥, a sequence {xn}⊂H which is defined by
(41)xn+1∈ProjC(xn)[xn-ρTxn]
is well defined. Moreover, it is a convergent sequence and its limit is a solution of the problem (1).Remark 14.
Recall that a set-valued mappingC:H→2H is said to be Hausdorff Lipschitz if there exists a real number κ>0 such that
(42)ℋ(C(x),C(y))≤κ∥x-y∥,∀x,y∈H,
where ℋ stands for the Hausdorff distance relative to the norm associated with the Hilbert space H, that is:
(43)ℋ(A,B)=max{supx∈Ainfy∈B∥x-y∥,supy∈Binfx∈A∥x-y∥},∀A,B∈2H.
It is easy to check that the class of Lipschitz mappings, which has been defined in Definition 6, is larger than the class of Hausdorff Lipschitz mappings. Thus, Theorem 11 can also be applied when the Assumption (𝒜) (ii) is replaced by “C is κ-Hausdorff Lipschitz set-valued mapping.”Remark 15.
LetK be a uniformly prox-regular closed subset of H. If C:H→[Cl(H)]r is defined by
(44)C(x)=K,∀x∈H,
then we see that Assumptions (𝒜) (ii) and (iii) are satisfied with a constant zero. In this case, Theorem 11 is reduced to a result which was presented by Petrot [27].It is well known that ifK is a closed convex set, then it is r-prox-regular set for every r>0 (see [22]). Using this fact, and by careful consideration of the proof of Theorem 11, one can see that control conditions (i) and (iii) of Theorem 11 can be omitted. So, we have the following results immediately.Corollary 16.
LetT:H→H be a single-valued mapping and let C:H→𝒞𝒞(H) be a set-valued mapping, where 𝒞𝒞(H) is a family of nonempty closed convex subset of H. If the Assumption (𝒜) holds and κ∈[0,1/(2⋄-1)), where ⋄=ξ(1-ω)/ξ2-β2, then the problem (1) has a solution.Finally, in view of Remark15 and Corollary 16, we can obtain the following classical result immediately.Corollary 17.
LetH be a real Hilbert space, let K be a closed convex subset of H, and let T:H→H be a single-valued mapping. If T is a strongly monotone and Lipschitz single-valued mapping, then the problem (1) has a solution.


## 3. Conclusion

In this paper, we provide some conditions for the existence theorems of the quasivariational inequality problem on a class of nonconvex sets. In fact, there are two constraints on the assumptions of considered mappingT in the main Theorem 11, that is,  (i)
T is a strongly monotone and Lipschitz mapping,(ii)
the range of mappingT is a bounded set. Hence, these lead to some natural questions in the future works for relaxing these constraints. At this point, we desire that the results presented here will be useful for those researchers, because this paper may also be extended and generalized for considering the mutlivalued and set-valued extended general variational inequalities problems.



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