AKRUR BEHERA AND PRASANTA KUMAR DAS
Received 3 October 2005; Accepted 28 February 2006
The concept ofη-invex set is explored and the concept ofT-η-invex function is intro-duced. These concepts are applied to the generalized vector variational inequality prob-lems in ordered topological vector spaces. The study of variational inequality probprob-lems is extended toH-spaces and differentiablen-manifolds. The solution of complementarity problem is also studied in the presence of fixed points or Lefschetz number.
Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.
1. Introduction
Variational inequality theory has become a rich source of inspiration in pure and applied mathematics. In recent years, classical variational inequality problem has been extended to study a wide class of problems arising in mechanics, physics, optimization and con-trol, nonlinear programming, economics, finance, regional, structural, transportation, elasticity and applied sciences, and so forth. They have been extended and generalized in different directions by using novel and innovative techniques and ideas. In this paper, we extend the study of variational inequality problems toH-spaces andn-manifolds. First we extend the concept ofη-invex sets and introduce the concept ofT-η-invex function and study their applications to the generalized vector variational inequality problems in ordered topological vector spaces.
2.η-invex set andT-η-invex function in ordered topological vector spaces
The notion of invexity was introduced by Hanson [8] as a generalization of the con-cept of convexity. Now this concon-cept is broadly used in the theory of optimization. Many authors have studied different types of convex and invex functions in different vector spaces. Suneja et al. [13] have studiedK-convex functions in finite-dimensional vector spaces. Mititelu [11] has studied the concept ofη-invex functions in the differentiable manifolds. We recall the concept ofη-invex set.
Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 78545, Pages1–18
Definition 2.1 [8]. LetX be a topological vector space andK⊂Xa nonempty subset of X.Kis said to be anη-invex set if there exists a vector functionη:K×K→Xsuch that y+tη(x,y)∈K for allx,y∈K and for allt∈(0,1). For an example ofη-invex set, see Example 2.7.
We make the following definitions for the vector functionηfor our need.
Definition 2.2 (ConditionC0). LetX be a topological vector space andK a nonempty
subset ofX. A vector functionη:K×K→Xis said to satisfy conditionC0if the following
hold:
ηx+η(x,x),x+ηx,x+η(x,x)=0,
ηx+tη(x,x),x+tη(x,x)=0
(2.1)
for allx,x∈K, for allt∈(0, 1).
Example 2.3 (ConditionC0). LetX=R. LetK=[0,∞) be any nonempty convex
sub-set ofX. Define the vector functionη:K×K→X by the ruleη(x,x)=x−xfor all x,x∈K.η(x+η(x,x),x) +η(x,x+η(x,x))=x−x−η(x,x) +x+η(x,x)−x=0
and for allt∈(0, 1),η(x+tη(x,x),x)=x−x−tη(x,x)= −tη(x,x). Thus η(x+ tη(x,x),x) +tη(x,x)=0 for allx,x∈K.
Theorem 2.4. LetK⊂X be a nonempty pointed subset of the topological vector space X. Let p:X→X be a linear projective map (p2=p). Letη:K×K→X be a vector-valued mapping defined by the ruleη(x,x)=p(x)−p(x). Thenηsatisfies condition C0.
Proof. For allx,x∈K,η(x+η(x,x),x) +η(x,x+η(x,x))=p(x)−p(x+η(x,x)) + p(x+η(x,x))−p(x)=0. If p is projective, thenη(x+tη(x,x),x)=p(x)−p(x+ tη(x,x))=p(x)−p(x)−p(tη(x,x))=−t p(η(x,x))=−t p(p(x)−p(x))=−t(p2(x)−
p2(x))= −t(p(x)−p(x))= −tη(x,x).
We use the following result of Ky Fan in our work.
Theorem 2.5 [6, Theorem 4.3.1, page 116]. LetKbe an arbitrary nonempty set in a Hous-dorfftopological vector spaceX. Let the set-valued mappingF:K→2Xbe a KKM map such that
(a)F(x) is closed for all x∈K,
(b)F(x) is compact for at least onex∈K. Then∩x∈KF(x)= ∅.
Theorem 2.6. Let X be a topological vector space and letKbe any nonemptyη-invex sub-set ofX. Let (Y,P) be an ordered topological vector space equipped with the closed convex pointed conePwith intP= ∅. LetL(X,Y) be the set of linear continuous functionals from XtoY. LetT:K→L(X,Y) andη:K×K→Xbe continuous mappings. Assume that
(a)T(x),η(x,x)∈ −int/ Pfor allx∈K,
(b) for eachu∈K, the setB(u)= {x∈K:T(u),η(x,u) ∈ −intP}is anη-invex set, (c) the mapu→ T(u),η(x,u)is continuous on the finite-dimensional subspaces (or at
least hemicontinuos),
Proof. For eachu∈K, consider the set-valued mappingF:K→2X defined by the rule F(x)= {u∈K:T(u),η(x,u)∈ −int/ P}for allx∈K.
We assert that F(x) is closed for eachx∈K. Let {un} be a sequence inF(x) such
that un→u. Sinceun∈F(x), we haveT(un),η(x,un)∈ −int/ P for allx∈K, that is,
T(un),η(x,un) ∈(Y−{−intP}) for all x∈K. SinceT andη are continuous, we have
T(un),η(x,un) → T(u),η(x,u). Since (Y−{−intP}) is a closed set,T(u),η(x,u) ∈
(Y−{−intP}) for allx∈K, that is,T(u),η(x,u)∈−int/ P for allx∈K. Thusu∈F(x) andF(x) is closed.
We claim thatFis a KKM mapping. If not there exists a finite subset{x1,x2,. . .,xn}of Ksuch thatCh({x1,x2,. . .,xn})⊂ ∪{F(x) :x∈ {x1,x2,. . .,xn}}, that is,Ch({x1,x2,. . .,xn})
⊂F(x) for anyx∈ {x1,x2,. . .,xn}, whereCh({x1,x2,. . .,xn}) denotes the convex hull of
{x1,x2,. . .,xn}. Let w∈Ch({x1,x2,. . .,xn}) such that w /∈ ∪{F(x) :x∈ {x1,x2,. . .,xn}},
that is,w /∈F(x) for any x∈ {x1,x2,. . .,xn}. Note that w∈K. Thus T(w),η(x,w) ∈
−intPfor allx∈ {x1,x2,. . .,xn}; this shows thatx∈ {x1,x2,. . .,xn} ⊂Ch({x1,x2,. . .,xn})⊂ B(w) (since every convex set is a subset of invex set). Hencew∈B(w), that is,T(w), η(w,w) ∈ −intP, which contradicts (a). ThusFis a KKM mapping. Hence byTheorem 2.6,∩{F(x) :x∈K} = ∅, that is, there existsx0∈Ksuch thatT(x0),η(x,x0)∈ −int/ P
for allx∈K.
We illustrateTheorem 2.6by an example.
Example 2.7. Let X= {si:s∈(−∞,∞)},K= {si:s∈[0,∞)},Y=R,P=[0,∞). Let η:K×K→Xbe defined byη(u,v)=u−v. LetT:K→L(X,Y) be defined byT(x)= −x for allx∈KandT(u),x =T(u)·xfor allu∈Kandx∈X.
(a) For allx∈K,T(x),η(x,x) =x(x−x)=P0.
(b) Leta,b∈B(u). We showb+tη(a,b)∈B(u). First we show thattT(u),η(a,u)+ (1−t)T(u),η(b,u) − T(u),η(b+tη(a,b),u) =0 for all a,b∈B(u) and for allt∈(0, 1) :tT(u),η(a,u)+ (1−t)T(u),η(b,u) − T(u),η(b+tη(a,b),u) = t(−u)(a−u) + (1−t)(−u)(b−u)−(−u)η(b+t(a−b),u)=(−u)(ta−tu+b− u−tb+tu)−(−u)(b+ta−tb−u)=(−u)(ta−tu+b−u−tb+tu−b−ta+ tb+u)=(−u)0=0. Hence for eachu∈K,tT(u),η(a,u) + (1−t)T(u),η(b,u)− T(u),η(b+tη(a,b),u)=0∈ −int/ Pfor alla,b∈B(u).
Asa,b∈B(u),T(u),η(a,u)∈−intP,T(u),η(b,u) ∈ −intP. Now for anyt∈(0, 1), T(u),η(b+tη(a,b),u) =tT(u),η(a,u)+ (1−t)T(u),η(b,u) ∈ −intPfor alla,b∈ B(u). Henceb+tη(a,b)∈B(u) for alla,b∈B(u) and for allt∈(0, 1).
(c) It is obvious that the mapu→ T(u),η(x,u)is continuous.
(d) Atx=u,T(u),η(x,u) =T(u)(x−u)=0 for allu∈Kshowing for at least one x=0∈K, the set{u∈K:T(u),η(x,u)∈ −/ intP}is compact.
All the conditions ofTheorem 2.6are satisfied. Hence there exists 0x0=0∈Ksolving
the problemT(x0),η(x,x0)∈ −int/ Pfor allx∈K.
We introduce the concept ofT-η-invexity.
Definition 2.8. f is said to beT-η-invex inKif f(x)−f(x)− T(x),η(x,x) ≥P0 (i.e., /
∈ −intP).
Remark 2.9. IfX=R,Y=R,K=(0, 1),T= ∇f, andP=[0,∞) and iff is differentiable, thenDefinition 2.8coincides with the definition of a differentiable convex function.
Definition 2.10. Tis said to beη-monotone if there exists a vector functionη:K×K→X such thatT(x),η(x,x)+T(x),η(x,x)∈/ intPfor allx,x∈K.
Example 2.11. LetX=R,K=R+,Y=R2,P=R2+. Let f :K→Y be defined by f(u)=
[u2
0] for allu∈Kand letT:K→L(X,Y) be defined byT(u)=[−02u] for allu∈K, where
T(u),x =T(u)·x,u∈K,x∈X. Define a vector functionη:K×K→Xbyη(u,v)= u+vfor allu,v∈K. Now for allu,v∈K, we have
f(u)−f(v)− T(v),η(u,v) = ⎡ ⎣u2
0 ⎤ ⎦− v2 0 − −2v
0 [u+v]
=
u2−v2+ 2v(u+v)
0
=
(u+v)2
0 ∈ −/ intP,
f(v)−f(u)− T(u),η(v,u) = v2 0 − u2 0 − −2u
0 [v+u]
=
v2−u2+ 2u(v+u)
0
=
(v+u)2
0 ∈ −/ intP
(2.2)
showing f isT-η-invex inK. Next we have
T(v),η(u,v)+T(u),η(v,u) =
−2v
0 [u+v] +
−2u
0 [v+u]
=
−2(u+v)
0 [u+v]
= −
2(u+v)
0 [u+v]∈/ intP
(2.3)
Example 2.12. LetX=R,K=R+,Y =R2,P=R2+. LetT:K→L(X,Y) be defined by T(u)=[−−2uu] for allu∈K, whereT(u),x =T(u)·x,u∈K,x∈X. Define a vector func-tionη:K×K→Xbyη(u,v)=u+vfor allu,v∈K. Now for allu,v∈K, we have
T(v),η(u,v)+T(u),η(v,u) =
−2v
−v (u+v) +
−2u
−u (v+u)
=
−2
−1 (u+v)2∈/ intP ∀u,v∈K,
(2.4)
showing thatTisη- monotone inK.
Proposition 2.13. LetXbe a topological vector space and let (Y,P) be an ordered topolog-ical vector space equipped with a closed pointed conePsuch that intP= ∅. LetL(X,Y) be the set of linear functional fromXtoY and letη:X×X→Xbe a vector-valued function, whereKis any subset ofX. LetT:K→L(X,Y) be an operator. Let the function f :K→Y be T-η-invex inK. ThenTisη-monotone.
Proof. Let f beT-η-invex inK, then for allx,x∈K, we havef(x)−f(x)− T(x),η(x, x)∈ −int/ P. Interchangingxandx, we have f(x)−f(x)− T(x),η(x,x)∈ −/ intP. Adding the above we get−T(x),η(x,x) − T(x),η(x,x)∈ −/ intP, that is,T(x),η(x, x)+T(x),η(x,x)∈/ intPfor allx,x∈K, showingTisη-monotone.
The converse ofProposition 2.13is not true, as the following example shows.
Example 2.14. LetX=Y =RandK=[−π/2,π/2],P=[0,∞],T= ∇f (derivative of f). Let f :K→Rbe defined by f(x)=sinxfor allx∈Kandη:K×K→Xdefined by η(x,x)=cosx−cosxfor all (x,x)∈K×K. ThenT(x),η(x,x)+T(x),η(x,x) = ∇f(x)η(x,x) +∇f(x)η(x,x)=cosx(cosx−cosx) + cosx(cosx−cosx)= −(cosx− cosx)2≤0 for allx,x∈K, showing thatTisη-monotone. But atx= −π/3 andx=π/6, we have f(x)−f(x)− T(x),η(x,x)<0, showing f is notT-η-invex.
3. A complementarity problem
In this section we study a vector complementarity problem in topological vector spaces. We use the notations of the following result.
Lemma 3.1 [5, Lemma 2.1]. Let (V,P) be an ordered topological vector space with a closed, pointed, and convex cone P with intP= ∅. Then, for ally,z∈V,
(i) y−z∈intPandy /∈intPimplyz /∈intP; (ii) y−z∈Pandy /∈intPimplyz /∈intP;
(iii)y−z∈ −intPandy /∈ −intPimplyz /∈ −intP; (iv) y−z∈ −Pandy /∈ −intPimplyz /∈ −intP.
Remark 3.2. For simplicity, we use the following terminologies: (a) y /∈ −intPif and only ify≥P0;
(b) y∈intPif and only ify >P0;
(c)y /∈intPif and only ify≤P0;
(e) y−z /∈ −intPif and only ify−z≥P0 (i.e.,y≥Pz);
(f) y−z /∈intPif and only ify−z≤P0 (i.e.,y≤Pz);
(g) y−z /∈(intP∪(−intP)) if and only ify−z=P0, (i.e.,y=Pz).
We also use the following terminologies as and when required: (A) y−z /∈ −Pandz /∈ −intPimplyy /∈ −intP;
(B) y−z /∈ −intPandz /∈ −intPimplyy /∈ −intP; (C) y−z /∈ −Pandy∈ −intPimplyz∈ −intP; (D)y−z /∈ −intPandy∈ −intPimplyz∈ −intP;
(E) y−z∈ −intPandz∈ −intPimplyy∈ −intP; (F) y /∈ −intPif and only if−y /∈intP;
(G) y /∈ −intPandz /∈ −intPimplyy+z /∈ −intP.
Definition 3.3. LetXbe a topological vector space and let (Y,P) be an ordered topological vector space equipped with a closed convex pointed coneP such that intP= ∅. LetK be any subset ofX, letL(X,Y) be the set of all linear functionals fromX toY, and let η:K×K→Xbe a vector-valued function. LetT:K→L(X,Y) be an arbitrary map.
The generalized vector variational inequality problem (GVVI) and the generalized vector complementarity problem (GVCP) are defined as follows.
GVVI: findx0∈Ksuch thatT(x0),η(x,x0)∈ −int/ Pfor allx∈K.
GVCP: findx0∈Ksuch thatT(x0),η(x,x0) ∈(Y−(intP∪(−intP))) for allx∈K.
We prove the following result concerning GVCP.
Theorem 3.4. LetK be a nonempty compact cone in a topological vector spaceXand let (Y,P) be an ordered topological vector space equipped with a convex pointed conePsuch that intP= ∅. LetL(X,Y) be the set of all linear functionals fromXtoY and letη:K×K→X be a continuous vector-valued function. LetT:K→L(X,Y) be an arbitrary continuous map and let K beη-invex.
Let the following conditions hold: (a)T(x),η(x,x) =P0 for allx∈K,
(b) for eachu∈K, the setB(u)= {x∈K:T(u)η(x,u) ∈ −intP}is anη-invex set, (c)ηsatisfies conditionC0.
Then there existsx0∈Ksuch thatT(x0),η(x,x0) ∈(Y−(intP∪(−intP))) for allx∈K. Proof. ByTheorem 2.6, there existsx0∈Ksuch thatT(x0),η(x,x0)∈ −/ intPfor allx∈ K, that is,T(x0),η(x,x0) ≥P0 for allx∈K. SinceK isη-invex cone,x0+tη(x,x0)∈K
for allx∈K. Replacingxbyx0+tη(x,x0), we getT(x0),η(x0+tη(x,x0),x0) ≥P0 for all x∈K. Thus 0≤PT(x0),η(x0+tη(x,x0),x0) = T(x0),−tη(x,x0) = −tT(x0),η(x,x0)
(by (c)) and hence−tT(x0),η(x,x0)∈ −/ intPfor allx∈K. Sincet >0, we haveT(x0), η(x,x0)∈/ intP. Hence T(x0),η(x,x0)∈ {−/ intP} ∪ {intP} for all x∈K showing
T(x0),η(x,x0) ∈(Y−(intP∪(−intP))) for allx∈K.
4. Variational inequality problems inH-spaces
H-convex sets,H-weakly convex sets, andH-KKM maps inH-spaces. In this section we establish an inequality inH-space and obtain the traditional variational and variational-type inequalities as particular cases of the newly obtained inequality. We also discuss the uniqueness of the solutions of the inequality with examples.
Several generalizations of the celebrated Ky Fan minimax inequality [7] have already appeared. This study requires the use of KKM theorem. In [1] Bardaro and Ceppitelli have explained the necessity of generalizing the reformulation of the KKM theorem for generalizing minimax inequality for functions taking values in ordered vector spaces.
In this section we prove certain results inH-spaces.
Definition 4.1 [1]. LetXbe a topological space and let{ΓA}be a given family of nonempty
contractible subsets ofX, indexed by finite subsets ofX. A pair (X,{ΓA}) is said to be an H-space ifA⊂BimpliesΓA⊂ΓB.
Let (X,{ΓA}) be anH-space. A subsetD⊂Xis said to beH-convex ifΓA⊂Dfor every
finite subsetA⊂D.
A subsetD⊂Xis said to be weaklyH-convex ifΓA∩Dis nonempty and contractible
for every finite subsetA⊂D. This is equivalent to saying that the pair (D,{ΓA∩D}) is an H-space.
A subsetK⊂Xis said to be H-compact if there exists a compact and weaklyH-convex setD⊂Xsuch thatK∪A⊂Dfor every finite subsetA⊂X.
In a givenH-space a multifunctionF:X→2X is said to be H-KKM ifΓ
A⊂ ∪{F(x) : x∈A}for every finite subsetA⊂X.
In this section we present an application of [1, Theroem 1, page 486]. In fact we estab-lish an inequality associated with the variational inequality or variational-type inequality. LetX be a topological vector space, let (Y,P) be an ordered topological vector space equipped with closed convex pointed cone with intP= ∅, and letL(X,Y) be the set of continuous linear functionals from X toY. Let the value of f ∈L(X,Y) atx∈X be denoted byf,x. Let K be a convex set inX with 0∈K and letT:K→L(X,Y) be any map. The variational inequality problem is to findx0∈Ksuch thatT(x0),v−x0∈/
−intPfor allv∈K.
Theorem 4.2. Let (X,{ΓA}) be anH-space. Assume that X is Hausdorff. Let N be a subset ofX×Xhaving the following properties.
(a) For eachx∈X, (x,x)∈N.
(b) For each fixedy∈X, the setN(y)= {x∈X: (x,y)∈N}is closed in X. (c) For eachx∈X, the setM(x)= {y∈X: (x,y)∈/ N}is H-convex.
(d) There exists a compact setL⊂X and anH-compact setW⊂X such that for each weakly H-convex setDwithW⊂D⊂X,∩y∈D({x∈X: (x,y)∈N} ∩D)⊂L. Then there existsx0∈Xsuch that{x0} ×X⊂N.
Proof. Define a set-valued mapF:X→2Xby the ruleF(y)= {x∈X: (x,y)∈N}. By (a), F(y)= ∅for eachy∈X. SinceXis Hausdorff, by (b),F(y) is compactly closed for each y∈X. We assert thatFis anH-KKM map. Suppose to the contrary thatFis not anH -KKM map .Then there exists a finite setA⊂Xsuch thatΓA⊂ ∪y∈AF(y). Thus there exists
Thusu∈M(u), that is, (u,u)∈/ N, which is a contradiction to (a). HenceFis anH-KKM map.
By (d) there exists a compact setL⊂Xand anH-compact setW⊂Xsuch that for each weaklyH-convex setDwithW⊂D⊂X, we have∩y∈D({x∈X: (x,y)∈N} ∩D)⊂ L, that is,∩y∈D(F(y)∩D)⊂L. Thus by [1, Theorem 1, page 486],∩y∈XF(y)= ∅. Hence
there existsx0∈Xsuch that{x0} ×X⊂N.
The following result is a slightly different version ofTheorem 4.2.
Theorem 4.3. Let (X,{ΓA}) be anH-space and let (Y,P) be an ordered topological vector space equipped with closed convex pointed cone with intP= ∅. LetKbe a convex set inX, with 0∈K. Assume thatXis Hausdorff. Letf :K×K→Ybe a continuous map having the following properties.
(a) For eachx∈X,f(x,x)∈ −/ intP.
(b) For each fixedv∈K, the set{x∈K:f(x,v)∈ −int/ P}is closed in X. (c) For eachx∈K, the set{v∈K:f(x,v)∈ −intP}isH-convex.
(d) There exists a compact setL⊂Xand anH-compact setW⊂X, such that for each weaklyH-convex setDwithW⊂D⊂X,∩v∈D({x∈K:f(x,v)∈ −int/ P}∩D)⊂L. Then there existsx0∈Ksuch that f(x0,v)∈ −int/ Pfor allv∈K.
Proof. LetN= {(x,v) : f(x,v)∈ −/ intP} ⊂K×K. By (a),Nis nonempty since (x,x)∈N for eachx∈K. For eachv∈K consider the setN(v)= {x∈X: (x,v)∈N} = {x∈X: f(x,v)∈ −int/ P}. By (b),N(v) is closed for eachv∈K. By (c), for eachx∈K, the set M(x)= {v∈K: (x,v)∈/ N} = {v∈K:f(x,v)∈ −intP}isH-convex . By (d), there ex-ists a compact set L⊂X and an H-compact set W⊂X such that for each weakly H -convex set D with W⊂D⊂X, we have ∩y∈D({x∈K: f(x,v)∈N} ∩D)⊂L. Thus
all the conditions ofTheorem 4.2are satisfied and hence there existsx0∈K such that
{x0} ×K⊂N, that is, (x0,v)∈Nfor allv∈K. This means there existsx0∈K such that
f(x0,v)∈ −int/ Pfor allv∈K.
Remarks 4.4. InTheorem 4.3we takeXto be a Hausdorfftopological real vector space with dualX∗,Y=RandP=[0,∞). ClearlyXis anH-space. LetKbe a nonempty con-vex subset ofX. LetT:K→X∗,η:K×K→X,θ:K×K→R,g:K→Xbe continuous functions satisfying some appropriate conditions as and when required. We consider the following cases.
Case 1. InTheorem 4.3 if we define f :K×K→Rby the rule f(x,y)= Tx,y−x, then there existsx0∈Ksuch thatTx0,y−x0 ≥0 for ally∈K, which is the variational
inequality as given in [4, Theorem 1, page 780]; also see [9, Theorem 4.32, page 116], [12, Theorem 1, page 90].
Case 2. InTheorem 4.3if we define f :K×K→Rby the rule f(x,y)= Tx,η(y,x), then there existsx0∈Ksuch thatTx0,η(y,x0) ≥0 for ally∈K, which is the
variational-type inequality as given in [2, Theorem 2.1, Theorem 2.2, page 184].
Case 3. InTheorem 4.3if we define f :K×K→Rby the rule f(x,y)= Tx,y−g(x), then there existsx0∈K such thatTx0,y−g(x0) ≥0 for ally∈K, which is the
Case 4. InTheorem 4.3if we define f :K×K→Rby the rule f(x,y)= Tx,η(y,x)+ θ(x,y), then there existsx0∈K such that Tx0,η(y,x0)+θ(x0,y)≥0 for all y∈K,
which is the variational-type inequality as given in [3, Theorem 2.1, page 346; Theorem 2.2, page 347].
The following result characterizes the uniqueness of the solution of the inequality f(x0,v)∈ −int/ Pobtained inTheorem 4.3.
Theorem 4.5. Let (X,{ΓA}) be anH-space and let (Y,P) be an ordered topological vector space equipped with closed convex pointed cone with intP= ∅. LetKbe a convex set inX, with 0∈K. Assume thatXis Hausdorff. Let f :K×K→Y be a continuous map such that
(a) f(x,v) +f(v,x)∈/ intPfor allx,v∈K, (b) f(x,v) +f(v,x)=P0 impliesx=v.
Then if the problem, findx0such that f(x0,ν) for allν∈K, is solvable, then it has a unique solution.
Proof. Letx1,x2∈K be such that f(x1,v)∈ −/ intPand f(x2,v)∈ −/ intPfor allv∈K;
putting v=x2 in the former inequality and v=x1 in the later inequality we see that f(x1,x2)∈ −/ intP and f(x2,x1)∈ −/ intP and on adding we get f(x1,x2) +f(x2,x1)∈/
−intP. This combined with inequality (a) gives f(x1,x2) +f(x2,x1)=P0. Hence by (b),
we havex1=x2.
The following examples illustrateTheorem 4.5.Example 4.6, given below, shows that fulfillment of conditions (a) and (b) does not guarantee the existence of the solution of the problem, stated inTheorem 4.5.
Example 4.6. Let X=R and define f :X×X→R by f(x,v)= −e−x|x−v|. Clearly f(x,v) + f(v,x)= −(e−x+e−v)|x−v| ≤0. Furthermore f(x,v) + f(v,x)=0 implies that x=v. It is clear that there is no x0∈K satisfying f(x0,v)= −e−x0|x0−v| ≥0 for all v∈X.
InExample 4.7the functionf :X×X→Rsatisfies conditions (a) and (b) ofTheorem 4.5and at the same time the problem stated inTheorem 4.5has a unique solution.
Example 4.7. LetX=[0,∞) and define f :X×X→R by f(x,v)= −x2|x−v|. Clearly f(x,v) + f(v,x)= −(x2+v2)|x−v| ≤0. Furthermore f(x,v) +f(v,x)=0 implies that
eitherx2+v2=0 or|x−v| =0; sincex2 andv2 are nonnegative, whenx2+v2=0, we
havex=0 andv=0 and when|x−v| =0 we have certainlyx=v. Thus conditions (a) and (b) ofTheorem 4.5hold. In this example we have a unique solutionx0=0 to the
problem ofTheorem 4.5, for f(x0,v)≥0 for allv∈X implies−x02|x0−v| ≥0 for all v∈X; since|x0−v| =0, the only solution isx0=0.
We explore some characteristics of generalized vector variational inequality problems inH-spaces.
Letη:K×K→Xbe a vector valued function and letf :K→Ybe continuous. Assume that the following conditions hold:
(a) f(x)∈ −/ intPfor allx∈K; (b) f :K→Y isT-η-invex inK;
(c) for eachu∈K the setBu= {x∈X: f(u)−f(x)∈ −intP}is eitherH-convex or empty.
Then f(x)− T(x),η(x,x)∈ −int/ Pfor allx,x∈K.
Proof. Define a set-valued mappingF:K→2Xby the ruleF(x)={x∈X:f(x)− T(x), η(x,x)∈ −int/ P}for eachx∈K. ClearlyF(x) is nonempty for eachx∈K.
It is enough to prove thatFis anH-KKM mapping. If not, then there exists a finite set A⊂K such thatΓA⊆ ∪{F(x) :x∈A}. Letz∈ΓAsuch thatz /∈ ∪{F(x) :x∈A}. Thus z /∈F(x) for allx∈A, that is, f(z)− T(z),η(x,z) ∈ −intP for allx∈A. Since f isT -η-invex inX, atz, we have f(x)−f(z)− T(z),η(x,z)∈ −/ intP for allx∈A. By (a), f(z)∈ −/ intP, so byLemma 3.1(i) we have f(x)− T(z),η(x,z)∈ −/ intPfor allx∈A. Thus f(x)− T(z),η(x,z) ∈ −intPfor allx∈A, that is,T(z),η(x,z) −f(x)∈ −intP for allx∈Aand we obtainf(z)−f(x)∈ −intPfor allx∈A. Hencex∈Bzfor allx∈A.
ThusA⊂Bz and byH-convexity ofBz,ΓA⊂Bz . Sincez∈ΓA, we have z∈Bz, that is, f(z)−f(z)∈ −intPgiving 0∈ −intP, which is a contradiction because by the pointed-ness condition ofP, 0∈P∩(−P) implies that 0∈/ (intP)∪(−intP). HenceF is anH
-KKM mapping.
Theorem 4.9. Let (X,{ΓA}) be anH-space and let (Y,P) be an ordered topological vector space equipped with a closed convex pointed cone P such that intP= ∅. LetL(X,Y) be the set of all linear functionals from X to Y and letT:K→L(X,Y) be a mapping. Let η:K×K→X be a vector-valued function and let f :K→Y be continuous. Assume that the following conditions hold:
(a) f(x)∈ −/ intPfor allx∈K; (b) f :K→Y isT-η-invex in K;
(c) for eachu∈K the setBu= {x∈X: f(u)−f(x)∈ −intP}is eitherH-convex or empty;
(d) the mappingv→f(v)− T(v),η(x,v)ofKintoYis continuous;
(e) there exists a compact setL⊂Xand anH-compactC⊂X, such that, for each weakly H-convex setDwithC⊂D⊂X,∩{x∈D:f(x)− T(x),η(x,x)∈ −int/ P} ⊂L. Then there existsx0∈Ksuch that f(x0)− T(x0),η(x,x0)∈ −/ intPfor allx∈K. Proof. As in the proof of Theorem 4.8, for eachx∈K the set-valued mappingF(x)= {x∈X:f(x)−T(x)η(x,x)∈−int/ P}is anH-KKM mapping. We show that∩{F(x) : x∈X} = ∅. By [1, Theorem 1, page 486] we need to show thatF is closed. Let{zn} ⊂ F(x) be a sequence inF(x) wherezn→z, then we show thatz∈F(x). Since the mapz→ f(z)−T(z),η(x,z)is continuous, we have f(z)−T(zn),η(x,zn) → f(z)−T(z)η(x,z).
But we havef(zn)−T(zn),η(x,zn)∈−int/ P. Thus f(zn)−T(zn),η(x,zn)∈(Y−{−intP}).
But (Y− {−intP}) is closed. Therefore f(z)− T(z),η(x,z) ∈(Y− {−intP}) giving f(z)− T(z),η(x,z)∈ −int/ P. Hencez∈F(x). Thus there existsx0∈ ∩{F(x) :x∈X},
such that f(x0)− T(x0),η(x,x0)∈ −/ intPfor allx∈K. This proves the theorem.
Theorem 4.10. Let (X,{ΓA}) be anH-space and let (Y,P) be an ordered topological vector space equipped with a closed convex pointed cone P such that intP= ∅. LetL(X,Y) be the set of all linear functionals from X to Y and letT:K→L(X,Y) be a mapping. Let η:K×K→Xbe a vector-valued function and let f :K→Y be continuous. Assume that the following conditions hold:
(a) f(x)∈ −/ intPfor allx∈K; (b) f :K→Y isT-η-invex in K;
(c) for eachu∈K the setBu= {x∈X: f(u)−f(x)∈ −intP}is eitherH-convex or empty;
(d) the mappingv→f(v)− T(v),η(x,v)ofKintoYis continuous;
(e) there exists a compact setL⊂Xand anH-compactC⊂X, such that, for each weakly H-convex setDwithC⊂D⊂X,∩{x∈D:f(x)− T(x),η(x,x)∈ −int/ P} ⊂L; (f)T(x),η(x,x) ≥P0 for allx,x∈K,x=x.
Then there existsx0∈Ksuch that
(i) f(x)−T(x),η(x0,x)∈ −int/ Pfor allx∈K;
(ii)T(x0),η(x,x0) − T(x),η(x0,x)∈ −/ intPfor allx∈K;
(iii) f(x)−f(x0)∈ −int/ Pfor allx∈K;
(iv){f(x)− T(x),η(x0,x)} − {f(x0)− T(x0),η(x,x0)}∈ −/ intPfor allx∈K.
Proof. (i)Theorem 4.9gives the existence ofx0∈K. SinceTisη-monotone (Proposition
2.13)T(x0),η(x,x0)+T(x),η(x0,x) ≤P0 for all x∈K. By (f), at x=x0, we have
−T(x),η(x0,x) ≥PT(x0),η(x,x0) ≥P0 (by (f)) for allx∈Kand hence by (a) f(x)−
T(x),η(x0,x) ≥P0 for allx∈K, that is, f(x)− T(x),η(x0,x)∈ −int/ Pfor allx∈K.
(ii) From (f), atx=x0, we haveT(x0),η(x,x0) ≥P0 for allx∈K. As in the proof
of (i)−T(x),η(x0,x) ≥P0. ThusT(x0),η(x,x0) − T(x),η(x0,x) ≥P0 for allx∈K,
that is,T(x0),η(x,x0) − T(x),η(x0,x)∈ −/ intPfor allx∈K.
(iii) By T-η-invexity of f, f(x)−f(x0)− T(x0),η(x,x0) ≥P0 for all x∈K, that
is, f(x)−f(x0)≥PT(x0),η(x,x0) ≥P0 (by (f)) for allx∈K, that is, f(x)−f(x0)∈/
−intPfor allx∈K.
(iv) Addition of (ii) and (iii) gives{f(x)− T(x),η(x0,x)} − {f(x0)− T(x0),η(x, x0)} ≥P0 for allx∈K, that is,{f(x)− T(x),η(x0,x)} − {f(x0)− T(x0),η(x,x0)}∈/
−intPfor allx∈K.
Theorem 4.11. Let (X,{ΓA}) be anH-space and let (Y,P) be an ordered topological vector space equipped with a closed convex pointed coneP such that intP= ∅. Let L(X,Y) be the set of all linear functionals from X to Y and letT:K→L(X,Y) be a mapping. Let η:K×K→Xbe a vector-valued function and let f :K→Y be continuous. Assume that the following conditions hold:
(a) f(x)∈ −/ intPfor allx∈K; (b) f :K→Y isT-η-invex inK;
(c) for eachu∈K the setBu= {x∈X: f(u)−f(x)∈ −intP}is eitherH-convex or empty;
(d) the mappingv→f(v)− T(v),η(x,v)ofKintoYis continuous;
(f)T(x),η(x,x) ≥P0 for allx,x∈K,x=x.
Then there existsx0∈Ksuch that{f(x)−T(x),η(x0,x)}−{f(x0)− T(x0),η(x,x0)}∈/
−intPfor allx∈K.
Proof. Theorem 4.9gives the existence ofx0∈K. Sincef isT-η-invex inK f(x)−f(x0)−
T(x0),η(x,x0)∈ −int/ P for allx∈K, that is, f(x)−f(x0)− T(x0),η(x,x0) ≥P0 for
allx∈K. Thus f(x)− T(x),η(x0,x) −f(x0)≥PT(x0),η(x,x0) − T(x),η(x0,x)for
all x∈K. Again since T isη-monotone, we have T(x0),η(x,x0)+T(x),η(x0,x)∈/
intP for all x∈K, that is, T(x0),η(x,x0)+T(x),η(x0,x) ≤P0 for all x∈K. Thus
−T(x0),η(x,x0) − T(x),η(x0,x) ≥P0 for allx∈K, that is,−T(x),η(x0,x)≥PT(x0), η(x,x0)for allx∈K. SinceT(x0),η(x,x0) ≥P0, we have,−T(x),η(x0,x) ≥P−T(x0), η(x,x0)for allx∈K. From the above we conclude that f(x)−T(x),η(x0,x)−f(x0)≥P
T(x0),η(x,x0) − T(x0),η(x,x0), that is, f(x)− T(x),η(x0,x) −f(x0) +T(x0),η(x, x0) ≥P T(x0),η(x,x0) ≥P0. Hence {f(x)− T(x),η(x0,x)} − {f(x0)− T(x0),η(x, x0)} ≥P0 for allx∈K, that is,{f(x)− T(x),η(x0,x)} − {f(x0)− T(x0),η(x,x0)}∈/
−intPfor allx∈K.
Theorem 4.12. Let (X,{ΓA}) be anH-space and let (Y,P) be an ordered topological vector space equipped with a closed convex pointed cone P such that intP= ∅. LetL(X,Y) be the set of all linear functionals from X to Y and letT:K→L(X,Y) be a mapping. Let η:K×K→Xbe a vector-valued function and let f :K→Y be continuous. Assume that the following conditions hold:
(a) f(x)∈ −/ intPfor allx∈K; (b) f :K→Y isT-η-invex inK;
(c) for eachu∈K the setBu= {x∈X: f(u)−f(x)∈ −intP}is eitherH-convex or empty;
(d) the mappingv→f(v)− T(v),η(x,v)ofKintoYis continuous;
(e) there exists a compact setL⊂Xand anH-compactC⊂X, such that, for each weakly H-convex setDwithC⊂D⊂X,∩{x∈D:f(x)− T(x),η(x,x)∈ −int/ P} ⊂L; (f)T(x),η(x,x) ≥P0 for allx,x∈K,x=x.
Then there existsx0∈Ksuch that f(x)−f(x0)∈ −int/ Pfor allx∈K.
Proof. Theorem 4.9gives the existence ofx0∈K. Sincef isT-η-invex inK f(x)−f(x0)−
T(x0),η(x,x0)∈ −int/ P for allx∈K, that is, f(x)−f(x0)− T(x0),η(x,x0) ≥P0 for
allx∈K. By (f), atx=x0, we haveT(x0),η(x,x0) ≥P0. Hence f(x)−f(x0)≥P0, that
is, f(x)−f(x0)∈ −int/ Pfor allx∈K.
Theorem 4.13. Let (X,{ΓA}) be anH-space and let (Y,P) be an ordered topological vector space equipped with a closed convex pointed cone P such that intP= ∅. LetL(X,Y) be the set of all linear functionals from X to Y and letT:K→L(X,Y) be a mapping. Let η:K×K→Xbe a vector-valued function and let f :K→Y be continuous. Assume that the following conditions hold:
(a) f(x)∈ −/ intPfor allx∈K; (b) f :K→Y isT-η-invex inK;
(d) the mappingv→f(v)− T(v),η(x,v)ofKintoYis continuous;
(e) there exists a compact setL⊂Xand anH-compactC⊂X, such that, for each weakly H-convex setDwithC⊂D⊂X,∩{x∈D:f(x)− T(x),η(x,x)∈ −int/ P} ⊂L; (f)T(x),η(x,x) ≥P0 for allx,x∈K,x=x;
(g)T(x),η(x,x) − T(x),η(x,x) ≥P0 for allx,x∈K,x=x.
Then there existsx0∈Ksuch that f(x)− T(x),η(x0,x)∈ −/ intPfor allx∈K.
Proof. Theorem 4.9gives the existence ofx0∈K. By condition (g), atx=x0,T(x0),η(x, x0)− T(x),η(x0,x)∈−int/ Pfor allx∈K, that is,T(x0),η(x,x0) − T(x),η(x0,x) ≥P
0 for allx∈K. Since f isT-η-invex inK, we have f(x)−f(x0)− T(x0),η(x,x0) ≥P0
for allx∈K. Adding the above two inequalities, we get f(x)−f(x0)− T(x),η(x0,x) ≥P
0 for allx∈K, that is, f(x)− T(x),η(x0,x) ≥P f(x0)≥P0 for allx∈K. Hence f(x)−
T(x),η(x0,x)∈ −int/ Pfor allx∈K.
Theorem 4.14. Let (X,{ΓA}) be anH-space and let (Y,P) be an ordered topological vector space equipped with a closed convex pointed cone P such that intP= ∅. LetL(X,Y) be the set of all linear functionals from X to Y and letT:K→L(X,Y) be a mapping. Let η:K×K→Xbe a vector-valued function and let f :K→Y be continuous. Assume that the following conditions hold:
(a) f(x)∈ −/ intPfor allx∈K; (b) f :K→Y isT-η-invex inK;
(c) for eachu∈K the setBu= {x∈X: f(u)−f(x)∈ −intP}is eitherH-convex or empty;
(d) the mappingv→f(v)− T(v),η(x,v)ofKintoYis continuous;
(e) there exists a compact setL⊂Xand anH-compactC⊂X, such that, for each weakly H-convex setDwithC⊂D⊂X,∩{x∈D:f(x)− T(x),η(x,x)∈ −int/ P} ⊂L; (f)T(x),η(x,x) ≥P0 for allx,x∈K,x=x.
Then there existsx0∈Ksuch that
(A)T(x0),η(x,x0)−T(x),η(x0,x)∈ −int/ Pfor allx∈K;
(B){f(x)−T(x), η(x0,x)} − {f(x0)− T(x0),η(x,x0)}∈ −int/ Pfor allx∈K. Proof. Theorem 4.12gives the existence ofx0∈K with f(x)−f(x0)∈ −int/ P, that is,
f(x)−f(x0)≥P0 for allx∈K.
(A) Sincef isT-η-invex inK,f(x0)−f(x)− T(x),η(x0,x)∈ −/ intPfor allx∈K,
that is, f(x0)−f(x)− T(x),η(x0,x) ≥P0 for allx∈K; adding this withf(x)− f(x0)≥P0, we get −T(x),η(x0,x) ≥P0. By condition (f) atx=x0, we have
T(x0),η(x,x0) ≥P0. HenceT(x0),η(x,x0) − T(x),η(x0,x)∈ −/ intPfor all x∈K.
(B) Addition of (A) with f(x)−f(x0)≥P0 gives (B).
5. Variational inequality problem in n-manifolds
In this section, we study the concept ofT-η-invex functions and its application in gener-alized vector variational inequality problems (in short, GVVI) on the manifolds.
ordered cone inY with intP= ∅. Letη:K×K→τX be an application and letT:K→ L(τX,τY) be the linear application.
The generalized vector variational inequality problem (GVVI)Xon theXcan be
formu-lated as follows:
(GVVI)XFindx0∈Ksuch thatT(x0),η(x,x0)∈ −/ intPfor allx∈K. Definition 5.1 [11]. Let ϕ:X→Rn be a differential vector function. dϕ
u:τ(X,u)→ τ(Rn,ϕ(u))≡Rn is called the differential of ϕ at u∈X, if dϕ
u(v)=dϕ(u)(v) for all v∈τ(X,u).
Definition 5.2 [11]. A differential vector functionϕ:X→Rnis said to be invex atu∈X
with respect toη(shortly,ϕisη-invex) if there exists an applicationη:K×K→τXsuch thatϕ(x)−ϕ(u)≥dϕu(η(x,u)) for allx∈K.
Definition 5.3. AnH-space is called anH-differentiable manifold if it is also a diff eren-tiable manifold.
Example 5.4. Ris anH-differentiable manifold.
In this section we prove some results inH-differentiable manifold (X,{ΓA}).
Theorem 5.5. Let (X,{ΓA}) be anH-differentiable manifold with the tangent bundleτX andKa nonempty closed conex subset ofX. Letϕ:X→Rnbe a linear application. LetPbe a closed, convex, and ordered pointed cone inRnwith intP= ∅. LetL(τX,Rn) denote the set of linear maps fromτX toτRn≡Rn. LetT:K→L(τX,Rn) andη:K×K→τX be an application such thatη(x,u)∈τ(X,u) for allx,u∈K. Suppose that
(a) (−ϕ) isT-η-invex onK,
(b) for eachu∈K,U(u)= {x∈X:ϕ(x)−ϕ(u)∈ −intP}is eitherH-convex or empty, (c) the applicationu→ T(u),η(x,u)ofKintoRnis continuous (or at least
hemicon-tinuous) for allx∈K,
(d) there exists a compact setL⊂X and anH-compact setC⊂X such that for each weaklyH-convex setDwithC⊂D⊂X,∩{T(u),η(x,u)∈ −/ intP:x∈D} ⊂L. Then (GVVI)K is solvable, that is, there existsx0∈K such thatT(x0),η(x,x0)∈ −/ intP for allx∈K.
Proof. LetF:K→2X be a set-valued application defined by the ruleF(x)= {u∈K:
T(u), η(x,u)∈ −/ intP}for allx∈X. We prove that ∩{F(x) :x∈K} = ∅. It can be proved by showing thatFis anH-KKM, mapping on the manifoldX. IfF is not an H-KKM then there exists a finite subsetA⊂Xsuch thatΓA⊂ ∪{F(x) :x∈A}. Assume there
existw∈ΓAsuch thatw /∈ ∪{F(x) :x∈A}. This implies thatw /∈F(x) for allx∈A, that
is,T(w),η(x,w) ∈ −intPfor allx∈A. Since (−ϕ) isT-η-invex onK,−ϕ(x) +ϕ(u) + T(u),η(x,u)∈ −/ intPfor allx,u∈K; equivalentlyϕ(x)−ϕ(u)− T(u),η(x,u)∈/ intP for allx,u∈K. Atw∈K, we haveϕ(x)−ϕ(w)− T(w),η(x,w)∈/ intP for allx∈K. Henceϕ(x)−ϕ(w)− T(w),η(x,w)∈/ intP for allx∈A(sinceAis a nonempty finite subsets ofX). Thus we getϕ(x)−ϕ(w)∈ −intPfor allx∈Aand by assumption (b) we getx∈U(w). HenceA⊂B(w). By theH-convexity ofU(w) we getΓA⊂U(w) for every
finite subsetA⊂U(w). Sincew∈ΓA,w∈U(w)⊂X. Hence 0=ϕ(w)−ϕ(w)∈ −intP
Next we prove thatF is closed. Let{yn}be a sequence inF(x) such thatyn→y. We
need to show thaty∈F(x). Since the applicationyn→ T(yn),η(x,yn)is continuous, yn→y givesT(yn),η(x,yn) → T(y),η(x,y). Also yn∈F(x) givesT(y),η(x,y)∈/
−intP for all x∈(Rn− {−intP}). Since (Rn− {−intP}) is a closed set, T(y n),η(x, yn) ∈(Rn− {−intP}) for all x∈K, showing T(y),η(x,y)∈ −/ intP for allx∈Rn.
Hencey∈F(x). By [1, Theorem 1, page 486], we get∩{F(x) :x∈K} = ∅. Hence there exists anx0∈K, such thatT(x0),η(x,x0)∈ −int/ Pfor allx∈K.
Theorem 5.6. Let (X,{ΓA}) and (Y,{ΓB}) beH-differentiable manifolds with the tangent bundleτX andτY, respectively, andK a nonempty closed convex subset ofX. LetP be a closed, convex, and pointed ordered cone Y in such that intP= ∅. LetL(τX,τY) denote the set of linear maps fromτX toτY andη:K×K→τX an application. Letϕ:X→Y be a linear application and leteu=τϕu:τX→τY be the corresponding bundle map defined by eu(v)∈τ(Y,ϕ(u))−for allv∈τ(X,u), whereτ(Y,ϕ(u))−= {w∈τ(Y,ϕ(u)) :w /∈intP}. LetT:K→L(τX,τY) be defined by the ruleT(u),v=(dϕu−eu)(v) for allv∈τ(X,u). Let
(a) (−ϕ) be a differentiableη-invex onK,
(b) for eachu∈K,U(u)= {x∈X:ϕ(x)−ϕ(u)∈ −intP}is eitherH-convex or empty, (c) the applicationu→dϕu−eu,η(x,u) ofK intoτY is continuous (or at least
hemi-continuous) for allx∈X.
(d) there exists a compact setL⊂X and anH-compact set V⊂X such that for each weakly H-compact setD withV ⊂D⊂X, ∩{dϕu−eu, η(x,u)∈ −int/ P, x∈ D} ⊂L.
Then (GVVI)K is solvable, that is, there existsx0∈K such that dϕx0−ex0,η(x,x0)∈/ −intPfor allx∈K.
Proof. First we show that (−ϕ) isT-η-invex onK. Since−ϕ:X→Y is a differentiable η-invex on K, we have (−ϕ)(x)−(−ϕ)(u)−d(−ϕ)u(η(x,u))∈ −int/ P for allx,u∈K,
that is, (−ϕ)(x)−(−ϕ)(u) +dϕu(η(x,u))∈ −/ intP and this implies that−ϕ(x) +ϕ(u)+ dϕu(η(x,u))∈−int/ P. The definition ofeu,eu(η(x,u))∈T(Y,ϕ(u))−implieseu(η(x,u))∈/
intP, that is,−eu(η(x,u))∈ −int/ P. Hence−ϕ(x) +ϕ(u) +dϕu(η(x,u))−eu(η(x,u))∈/
−intP for all x,u∈K, that is,−ϕ(x) +ϕ(u) +T(u),η(x,u)∈ −/ intP for all x,u∈K. Thus−ϕisT-η-invex onK.
Construct a set-valued mappingG:K→2Xby the ruleG(x)= {z∈K: (dϕ
z−ez)(η(x, z))∈ −int/ P}. We show thatGis an H-KKM mapping. Suppose to the contrary thatG is not an H-KKM application. Then there exists a finite subset A⊂X such thatΓA⊂
∪{G(x) :x∈A}. Letw∈ΓAbe such thatw /∈ ∪{G(x) :x∈A}. Sow /∈G(x) for allx∈A,
that is, (dϕw−ew)(η(x,w))∈ −intP for allx∈A. Since−ϕisS−η−invex onX, we
Hence 0=ϕ(w)−ϕ(w)∈ −intP gives a contradiction that 0∈ −int/ P. HenceGis an H-KKM.
Next we prove thatGis closed. Let{zn}be a sequence inG(x) such thatzn→z, then
we show thatz∈G(x). Again since the applicationzn→(dϕxn−ezn)(η(x,zn)) is finite dimensional, we have (dϕxn−ezn)(η(x,zn))→(dϕz−ezn)(η(x,z)) aszn→z. As we have assumedzn∈G(x), so that (dϕxn−ezn)(η(x,zn))∈ −/ intPfor allx∈(Y− {−intP}). But (Y− {−intP}) is a closed set; therefore, (dϕz−ezn)(η(x,z))∈(Y− {−intP}) for allx∈ X implying that (dϕz−ez)(η(x,z))∈ −int/ P for allx∈X. Hencez∈G(x). ThusG is
closed.
By [1, Theorem 1, page 486], we get∩{G(x) :x∈X} = ∅. Hence there exists anx0∈ K such that (dϕx0−ex0)(η(x,x0))∈ −/ intPfor allx∈K, that is,x0 solves the problem.
6. Complementarity problem solution using fixed point theorems in manifolds
In this section we are interested in studying the behavior of continuous functions on manifolds with particular interest in finding the solutions of complementarity problems in the presence of fixed points or coincidences. Though the complementarity problem is a classical problem, the techniques involved in this section may prove enlightening to take a brief look at some development of the problems.
LetM be a closed manifold and let f :M→M be a map. Then for eachk there is the induced homomorphism on homology with rational coefficients fk:Hk(M;Q)→ Hk(M;Q). For eachk we may choose a basis for the finite-dimensional rational vector
spaceHk(M;Q) and write fk as a matrix with respect to this basis. Denote by tr(fk) the
trace of the matrix. If we define the Lefschetz number byL(f)=k∞=0(−1)ktr(fk), then L(f) is independent of the choices involved and hence is a well-defined, rational-valued function of f andL(f) depends only on the homotopy class of f [15].
Let X be a closed, convex, and oriented Riemannian n-manifold, modeled on the Hilbert spaceH with Riemannian metric g. It is well known that the tangent bundle τ(X) can be identified with the cotangent bundleτ∗(X) by the Riemannian metric, be-causeH∗, the dual ofH, can be identified withH [10]. Ifv,w∈τx(X), then we write gx(v,w)= v,wx. LetF:X→Hbe an operator. The complementarity problem is to find x0∈Xsuch thatF(x0)∈τ∗(X) andgx0(Fx0,x0)= Fx0,x0x0=0.
Theorem 6.1. LetXbe a closed, convex, and oriented Riemannian n-manifold, modeled on the Hilbert spaceHwith Riemannian metricg and let f :X→Xbe a map with Lefschetz numberL(f). LetF:X→H be an operator. Then there exists a uniquex0∈X such that F(x0)∈τ∗(X) andgx0(Fx0,x0)= Fx0,x0x0=0.
Proof. SinceX is nonempty, closed, and convex endowed with the Riemannian metric g, for everyy∈X, there is an uniquex∈Xclosest toy−F(y). Thusx−y+F(y) ≤ z−y+F(y)for everyz∈X, that is,x,z−xx≥ y−F(y),z−xxfor allz∈X. Let f :X→Xbe defined by f(y)=y−F(y) +xfor everyy∈X, wherexis the unique ele-ment corresponding toy. Now for everyy∈X, (1X−f)(y)=1X(y)−f(y)=F(y)−x
F−1Xat the uniquex∈X. DefineG:X×I→Xby the rule
G(y,t)= ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
(1X−f)
(1−2t)y+ 2tx, 0≤t≤1 2,
(F−1X)
(2t−1)y+ 2(1−t)x, 1
2≤t≤1.
(6.1)
G(y, 0)=(1X− f)(y) andG(y, 1)=(F−1X) for each y∈X. At t=1/2, G(y, 1/2)=
(1X−f)(x)=(F−1X)(x). SoGis continuous by Pasting lemma. ThusG: (1X− f)
(F−1X). Thus, the coincidence index set of f is given by If =(1X−f)∗0X =(F−
1X)∗0X andIf =0. Since f :X→X is a mapping withL(f)=If =0, by [15,
Theo-rem 7.16, Lefschitz fixed-point theoTheo-rem], f has a fixed point. Let the fixed point bey0
inX, that is, f(y0)=y0. Letx0 be the unique element that corresponds toy0. Thus we
havex0,z−x0x0≥ y0−F(y0),z−x0x0 for allz∈X, givingx0,z−x0x0≥ f(y0)− y0,z−x0x0 for all z∈X, that is, 2x0−y0,z−x0x0≥0 for all z∈X. At y=y0, we get f(y0)=y0−F(y0) +x0, that is,x0=F(y0). We show thatF(y0)∈τ∗(X). By
defini-tion of coincidence index set, we haveIf =(1X−f)∗0X=(F−1X)∗0X=IF, which
means that f andFhave the same fixed point inX, that is,f(y0)=y0=F(y0). So, we get x0=y0. Puttingx0=y0in2F(y0)−y0,z−x0x0≥0, we getF(y0),z−y0y0≥0 for all z∈X, that is,F(y0)∈T∗(X). Again puttingz=0 andz=2y0 inF(y0),z−y0y0≥0, respectively, we getF(y0),y0y0≤0 andF(y0),y0y0≥0. HenceF(y0),y0y0=0. Theorem 6.2. Let f :Sn→Sn,n≥1, be a map of degreem=(−1)n+1. LetT:Sn→Snbe any operator. Then there exists ay0∈SnT(y0),z−y0≥0 for allz∈Sn.
Proof. SinceSn⊂Rn+1is closed, for eachy∈Sn, there exists a uniquex∈Snclosest to y−T(y), that is,x−y+T(y) ≤ z−y+T(y)for allz∈Sn. Thusx,z−x ≥ y− T(y),z−xfor allz∈Sn, that is,x−y+T(y),z−x ≥0 for allz∈Sn. Define f :Sn→ Sn by the rule f(y)=x. This is well defined since x is unique inSncorresponding to
each y∈Sn. Obviously f is a homeomorphism. Replacingxby f(y) in the inequality
x−y+T(y),z−x ≥0 we getf(y)−y+T(y),z−f(y) ≥0 for allz∈Sn
