FIXED POINT INDEX APPROACH FOR SOLUTIONS OF VARIATIONAL INEQUALITIES

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FIXED POINT INDEX APPROACH FOR SOLUTIONS OF VARIATIONAL INEQUALITIES

YISHENG LAI, YUANGUO ZHU, AND YINBING DENG Received 8 October 2004 and in revised form 18 May 2005

By using fixed point index approach for multivalued mappings, the existence of nonzero solutions for a class of generalized variational inequalities is studied in reflexive Banach space. One of the mappings concerned here is coercive or monotone and the other is set-contractive or upper semicontinuous.

1. Introduction

Since the fundamental theory of variational inequality was founded in the 1960s, the variational inequality theory with applications has made powerful progress and has become an important part of nonlinear analysis. It has been applied intensively to mechanics, dif- ferential equation, cybernetics, quantitative economics, optimization theory, nonlinear programming, and so forth (see [2]).

In virtue of minimax theorem of Ky Fan and KKM technique, variational inequalities, generalized variational inequalities, and generalized quasivariational inequalities were studied intensively in the last 20 years with topological method, variational method, semiordering method, and fixed point method [2]. However, the existence of nonzero solutions for variational inequalities, as another important topic of variational inequality theory, has been rarely discussed.

It is of theoretical and practical significance to study the existence of nonzero solutions for variational inequalities. In this paper, we will discuss the existence of nonzero solutions for a class of generalized variational inequalities for multivalued mappings by fixed point index approach in reflexive Banach space.

LetY,Zbe two topological spaces. A multivalued mappingF:Y→2^Zis called upper semicontinuous aty0∈Yif for each neighbourhoodV⊂ZofF(y0), there exists a neigh- bourhoodU ofy0such that the setF(U)⊂V. Suppose thatE1,E2are two real Banach spaces,D⊆E1. A multivalued mappingA:D→2^E² is said to bek-set-contractive onD if there exists a constantk such thatα(A(S))≤kα(S) wheneverα(S)=0,S⊆D, where αis the Kuratowski measure of noncompactness. A mappingAis called condensing on Difα(A(S))< α(S) wheneverα(S)=0,S⊆D. It is easily seen that a mappingAis condensing whenk <1. LetXbe a Banach space,X^∗its dual, and (·,·) the pair betweenX^∗

International Journal of Mathematics and Mathematical Sciences 2005:12 (2005) 1879–1887 DOI:10.1155/IJMMS.2005.1879

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andX. Suppose thatKis a closed convex subset ofXandUis an open subset ofXwith UK=U∩K= ∅. The closure and boundary ofUK relative toKare denoted byUK and

∂(UK), respectively. Assume thatT:UK→2^K is an upper semicontinuous mapping with nonempty compact convex values andTis also condensing. Ifx∈T(x) forx∈∂(UK), then the fixed point index,iK(T,U), is well defined (see [3]).

Proposition1.1 [3]. LetK be a nonempty closed convex subset of a real Banach space Xand letUbe an open subset ofX. Suppose thatT:UK→2^Kis an upper semicontinuous mapping with nonempty compact convex values andx∈T(x)forx∈∂(UK), then the index, iK(T,U), has the following properties:

(i)ifiK(T,U)=0, thenThas a fixed point;

(ii)for mappingX0with constant value{x0}, ifx0∈UK, theniK(X0,U)=1;

(iii)let U1, U2 be two open subsets of X withU1∩U2= ∅. If x∈T(x)when x∈

∂((U1)K)∪∂((U2)K), theniK(T,U1∪U2)=iK(T,U1) +iK(T,U2);

(iv)letH: [0, 1]×UK→2^K be an upper semicontinuous mapping with nonempty com- pact convex values andα(H([0, 1]×Q))< α(Q)wheneverα(Q)=0,Q⊆UK. If x∈H(t,x)for everyt∈[0, 1],x∈∂(UK), theniK(H(1,·),U)=iK(H(0,·),U).

In this paper, for mappingsA:X→X^∗andg:K→2^X^∗, we will deal with the following problem by fixed point index approach: findu∈K,u=0, andw∈g(u) such that

(Au,v−u)≥(w,v−u), ∀v∈K. (1.1)

2. Nonzero solutions when the mappingAis coercive

Suppose thatK is a subspace ofX andA:X→X^∗is a coercive and linear continuous mapping, that is, there exist constantsM,γ >0 such that

(Av,v)≥γv², AvX^∗≤Mv, ∀v∈X. (2.1)

It is well known that for any givenw∈X^∗, the variational inequality

(Au,v−u)≥(w,v−u), ∀v∈K, (2.2)

has an only solutionuinK(see [1]). Define a mapping as follows:

KA:X^∗−→K, KA(w)=u, ∀w∈X^∗, (2.3)

thenKAis a coercive and linear continuous mapping and (see [1]) KA

w1

−KA

w2≤1

γw1−w2_X_∗. (2.4)

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Theorem 2.1. Let K be a subspace of a reflexive real Banach spaceX. Suppose that A: X→X^∗is a coercive and linear continuous mapping which satisfies inequalities (2.1) and g:K→2^X^∗isβ-set-contractive and upper semicontinuous mapping with nonempty compact convex values, whereβ/γ <1. Assume

(a) lim infun→0sup_w_n_∈_g(u_n₎(wn,un)/un²< γ(un∈K);

(b)there existx0∈Kand a constantq >0such thatinfw∈g(u)(w,x0)/u> Mx0when u> q,u∈K.

Then (1.1) has a nonzero solution.

Proof. Define a mapping as follows:

KAg:K−→2^K, KAg(u)=KA

g(u), ∀u∈K. (2.5) It is easily seen thatKAg is (β/γ)-set-contractive and upper semicontinuous mapping with nonempty compact convex values by (2.4). LetK^r= {x∈K,x< r}. Assuming that there does not exist r=0 and u∈∂(K^r) such that u∈KA(g(u)) (or else u is a nonzero solution of (1.1)). We will verify thatiK(KAg,K^r)=1 for small enoughr and iK(KAg,K^R)=0 for large enoughR.

Firstly, define a mapping byH: [0, 1]×K^r→2^K,H(t,u)=tKA(g(u)). Obviously,H(t,u) is an upper semicontinuous mapping with nonempty compact convex values. We claim thatα(H([0, 1]×Q))< α(Q) wheneverα(Q)=0,Q⊆K^r. In fact, lete∈KAg(Q), then 0∈ {KAg(Q)−e}. Hence, we have

H[0, 1]×Q=

t∈[0,1]

tKAg(Q)−e+te

⊆

t∈[0,1]

tKAg(Q)−e +

t∈[0,1]

{te}

⊆

KAg(Q)−e +

t∈[0,1]

{te}.

(2.6)

Thus

αH[0, 1]×Q≤αKAg(Q) +α

t∈(0,1)

{te}

=αKAg(Q) < α(Q). (2.7)

We claim that there exists small enoughrsuch thatu∈H(t,u) for allt∈[0, 1],u∈∂(K^r).

Otherwise, there exist two sequences{tn},{un},tn∈(0, 1],un∈∂(K^r),un →0, such thatun∈H(tn,un)=tnKAg(un) orun/tn∈KAg(un), hence there existswn∈g(un) such thatun/tn=KA(wn), that is, we have

Aun

tn

,v−un

tn

≥

wn,v−un

tn

, ∀v∈K. (2.8)

Lettingv=0, we can obtain from (2.1) and (2.8) that γ≤

Aun,un un² ≤tn

wn,un un² ≤

wn,un

un² . (2.9)

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Thus lim infun→0sup_w_n_∈_g(u_n₎(wn,un)/un²≥γ, which contradicts condition (a). There- fore,

iK

KAg,K^r=iK

H(1,·),K^r=iK

H(·, 0),K^r=iK

ˆ0,K^r=1 (2.10) byProposition 1.1(ii) and (iv).

Secondly, we will verify thatiK(KAg,K^R)=0 for large enoughR. In fact, we can get from (2.1) and condition (b) that

w,x0

>Au,x0

, ∀w∈g(u), asu> q. (2.11) On the other hand, becauseg isβ-set-contractive and upper semicontinuous mapping with nonempty compact convex values, there exists a constantL >0 such thatwX^∗≤L for allw∈g(u) wheneveru ≤q,u∈K. TakeNfor large enough and f ∈X^∗so that

Mqx0+Lx0<−Nf,x0

. (2.12)

Define a mapping byH: [0, 1]×K^R→2^K,H(t,u)=KA(g(u)−tN f). ThenH(t,u) is an upper semicontinuous mapping with nonempty compact convex values. We claim that α(H([0, 1]×Q))< α(Q) wheneverα(Q)=0,Q⊆K^r. In fact,

H[0, 1]×Q=KA

t∈[0,1]

g(Q)−tN f

⊆KA

g(Q) +

t∈[0,1]

{−tN f}

, α

g(Q) +

t∈[0,1]

{−tN f}

≤αg(Q)+α

t∈[0,1]

{−Nt f}

=αg(Q)≤βα(Q).

(2.13) Thusα(H([0, 1]×Q))≤(β/γ)α(Q)< α(Q) by (2.4) andβ/γ <1. We claim that there exists large enoughRsuch thatu∈H(t,u) for allt∈[0, 1],u∈∂(K^R). Otherwise, there exist two sequences{tn},{un},tn∈[0, 1],un∈∂(K^R),un →+∞, such thatun∈H(tn, un)=KA(g(un)−tnN f), hence there existswn∈g(un) such thatun=KA(wn−tnN f), that is, we have

Aun,v−un

≥

wn−tnN f,v−un

, ∀v∈K. (2.14)

Takingv=un+x0in (2.14), we obtain from (2.1) that Mx0≥

Aun,x0

un ≥

wn,x0

un ≥ inf

wn∈g(un)

wn,x0

un , (2.15)

which contradicts condition (b). Therefore,

iKKAg,K^R=iKH(·, 0),K^R=iKH(1,·),K^R (2.16) byProposition 1.1(iv). IfiK(H(1,·),K^R)=0, then the mapping H(1,·) :K→2^K has a fixed point uin K^R byProposition 1.1(i), that is, u∈H(1,u)=KA(g(u)−N f). Thus there existsw∈g(u) such thatu=KA(w−N f), that is,

(Au,v−u)≥(w−N f,v−u), ∀v∈K. (2.17)

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Takingv=u+x0in (2.17), we get that Au,x0

− w,x0

≥ −Nf,x0

. (2.18)

That contradicts (2.11) ifu> q, henceu ≤q, then we can get from (2.1) and (2.18) that

−Nf,x0

≤Au,x0+w,x0≤Mqx0+Lx0, (2.19) but it contradicts (2.12). Therefore,iK(H(1,·),K^R)=0.

It follows from (2.10), (2.16), and Proposition 1.1(iii) that iK(KAg,K^R\K^r)= −1.

Therefore, there exists a fixed point u∈K^R\K^r which is a nonzero solution of (1.1).

Theorem 2.2. Let K be a subspace of a reflexive real Banach spaceX. Suppose that A: X→X^∗is a coercive and linear continuous mapping which satisfies inequalities (2.1) and g:K→2^X^∗isβ-set-contractive and upper semicontinuous mapping with nonempty compact convex values, whereβ/γ <1. Assume

(a) lim infun→+∞sup_w_n_∈_g(u_n₎(wn,un)/un²< γ(un∈K);

(b)there existx0∈K and an open neighbourhoodV(0)of zero point such that for any givenu∈K∩V(0)\ {0},infw∈g(u)(w,x0)/u> Mx0.

The proof ofTheorem 2.2is similar to that ofTheorem 2.1. We omit it here.

3. Nonzero solutions when the mappingAis monotone

LetA:X→X^∗be a monotone linear mapping with (Au,u)/u →+∞(asu →+∞, u∈K). It is well known that for any givenw∈X^∗, the variational inequality (2.2) has solutions inK(see [2]), thus we may define two mappings as follows:

KA:X^∗−→2^K, KA(w)=

u∈K:uis a solution of the variational inequality (2.2) , (3.1) KAg:K−→2^K, KAg(u)=KA

g(u), ∀u∈K. (3.2) Proposition3.1. LetX=Rⁿand letK⊂Xbe a nonempty closed convex set. Suppose that A:X→X^∗is a monotone hemicontinuous mapping. If for everyw∈X^∗, the variational inequality (2.2) has solutions inK, then the mappingKAin (3.1) is a monotone and upper semicontinuous mapping with nonempty compact convex values.

Proof. Letu1∈KA(w1),u2∈KA(w2). Then Aui,v−ui

≥wi,v−ui

, ∀v∈K,i=1, 2. (3.3)

It is easily obtained from above inequalities that (Au1−Au2,u1−u2)≤(w1−w2,u1−u2).

ThusKAis monotone due to the monotony ofA. Furthermore,KAis locally bounded by [3]. We claim thatKAis upper semicontinuous. Otherwise, there exists a pointw∈X^∗

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and an open setV0containingKA(w) such that for sequence{wn}converging tow, there existun∈KA(wn) such thatun∈V0. Since{un}is bounded by the locally boundedness of KA, there exists a subsequence{unk}such thatunk→u0. Obviously,u0∈K,u0∈V0. We know that monotone hemicontinuous mappingAis continuous by [3]. Lettingk→+∞ in the inequality

Aunk,v−unk

≥

wnk,v−unk

, ∀v∈K, (3.4)

yields that

Au0,v−u0

≥

w,v−u0

, ∀v∈K, (3.5)

which implies thatu0∈KA(w)⊂V0. That is a contradiction. In addition, the compact

convexity ofKA(w) is obvious.

Proposition3.2. Let K be a subspace of a reflexive real Banach spaceX. Suppose that A:X→X^∗is a monotone linear mapping with(Au,u)/u →+∞(asu →+∞,u∈K) andg:K→2^X^∗ is a mapping with nonempty convex values, thenKAg:K→2^K(3.2) is also a mapping with nonempty convex values.

Proof. Letq∈Kandu1,u2∈KAg(q). Then there existw1,w2∈g(q) such thatui∈KA(wi), i=1, 2. That is, we have

Au1,v−u1

≥

w1,v−u1

, ∀v∈K, (3.6)

Au2,v−u2

≥

w2,v−u2

, ∀v∈K. (3.7)

Substitutingv+u1−(λ1u1+λ2u2) (resp.,v+u2−(λ1u1+λ2u2)) forvin (3.6) (resp., in (3.7)), whereλ1,λ2≥0,λ1+λ2=1, we get that

λ1Au1+λ2Au2,v− 2 i=1

λiui

≥

λ1w1+λ2w2,v− 2 i=1

λiui

, ∀v∈K. (3.8)

In addition,²_i₌1λiwi∈g(q). Therefore,²_i₌1λiui∈KAg(q) which implies thatKAg is a

mapping with nonempty convex values.

We first consider the nonzero solutions of (1.1) inRⁿ.

Theorem3.3. Let K be a subspace of X=Rⁿ. Suppose thatA:X→X^∗is a monotone linear mapping with(Au,u)/u →+∞(asu →+∞,u∈K) andg:K→2^X^∗is an upper semicontinuous mapping with nonempty compact convex values. The following conditions either (a), (b) or (a), (b) are assumed to be satisfied:

(a)there exist y0∈Kand an open neighbourhoodV(0)of zero point such that for any givenu∈K∩V(0)\ {0},infw∈g(u)(Au−w,y0)>0;

(b)there existx0∈K and a constantq >0 such thatsup_w_∈_g(u)(Au−w,x0)<0when u> q,u∈K;

(a)there exist y0∈Kand an open neighbourhoodV(0)of zero point such that for any givenu∈K∩V(0)\ {0},sup_w_∈_g(u)(Au−w,y0)<0;

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(b)there existx0∈K and a constantq >0 such thatinfw∈g(u)(Au−w,x0)>0 when u> q,u∈K.

Proof. It is well known that monotone linear mapping must be semicontinuous (see [2]), hence KA:X^∗→2^K (3.1) is an upper semicontinuous mapping with nonempty compact convex values byProposition 3.1. It is easy to see from [2] thatKAg:K→2^K, (KAg)(u)=KA(g(u)),u∈K, is an upper semicontinuous mapping with nonempty com- pact values, thereforeKAgis an upper semicontinuous mapping with nonempty compact convex values byProposition 3.2.

Let K^r= {x∈K,x< r}. Similar to the proof of Theorem 2.1, we may get that iK(KAg,K^R\K^r)= −1. Therefore, there exists a fixed pointu∈K^R\K^rwhich is a non-

zero solution of (1.1).

Now, we discuss the nonzero solution of (1.1) in reflexive real Banach space.

Theorem 3.4. Let K be a subspace of a reflexive real Banach spaceX. Suppose that A: X→X^∗is a monotone linear mapping with(Au,u)/u →+∞(asu →+∞,u∈K) and g:K→2^X^∗ is an upper semicontinuous from the weak topology onXto the strong topology onX^∗, with nonempty compact convex values. The following conditions either (a), (b), (c) or (a), (b), (c) are assumed to be satisfied:

(a)there exist y0∈Kand an open neighbourhoodV(0)of zero point such that for any givenu∈K∩V(0)\ {0},infw∈g(u)(Au−w,y0)>0;

(b)there existx0∈K and a constantq >0 such thatsup_w_∈_g(u)(Au−w,x0)<0when u> q,u∈K;

(c)there existsz0∈Ksuch thatlim inf_u_α_−→^w₀sup_w_∈_g(u_α₎(Auα−w,z0)<0, whereuα∈K;

(a)there exist y0∈Kand an open neighbourhoodV(0)of zero point such that for any givenu∈K∩V(0)\ {0},sup_w_∈_g(u)(Au−w,y0)<0;

(b)there existx0∈K and a constantq >0 such thatinfw∈g(u)(Au−w,x0)>0 when u> q,u∈K.

Proof. Let F⊂X be a finite-dimensional subspace containing x0, y0, and z0. We will show that all conditions inTheorem 3.3 are satisfied on spaceF. DenoteKF=K∩F.

Let jF:F→Xbe an injective mapping andj_F^∗:X^∗→F^∗its dual mapping. DenoteAF= jF^∗(A|KF) :KF→F^∗,gF=jF^∗(g|KF) :KF→F^∗. We know thatAF=jF^∗AjF,gF=jF^∗g jF. ThenAF,gFare linear and upper semicontinuous with nonempty compact convex values, respectively. Forx1,x2,u∈KF, we have

AF x1

−AF x2

,x1−x2

= j^∗_FAx1

−j_F^∗Ax2

,x1−x2

=Ax1

−Ax2

,j_F^∗x1−x2

= Ax1

−Ax2

,x1−x2

≥0, AF(u),u

u =

jF^∗A(u),u

u =(Au,u) u .

(3.9)

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These mean thatAFis monotone with (AF(u),u)/u →+∞(asu →+∞,u∈KF). On the other hand,

w∈infgF(u)

AF(u)−w,y0

= inf

w∈j^∗g(u)

j^∗_FA(u)−w,y0

= inf

w∈g(u)

j_F^∗A(u)−j^∗_Fw,y0

= inf

w∈g(u)

A(u)−w,y0

, sup

w∈gF(u)

AF(u)−w,x0

= sup

w∈g(u)

A(u)−w,x0

.

(3.10)

Therefore, there existsuF,uF=0, andw_F∈gF(uF) from conditions (a) and (b) or (a) and (b) andTheorem 3.3such that

AF uF

,v−uF

≥

w_F,v−uF

, ∀v∈KF. (3.11)

Sincew_F∈gF(uF)=j^∗(g(uF)), there existswF∈g(uF) such thatw_F=j_F^∗(wF). Hence, AuF

,v−uF

≥

wF,v−uF

, ∀v∈KF, (3.12)

by (3.11). Suppose that conditions (a) and (b) are satisfied, takingv=uF+x0 (or else v=uF−x0), we get that (A(uF)−wF,x0)≥0. Thus sup_w_∈_g(u_F₎(A(uF)−w,x0)≥0, this conduces to a contradiction by condition (b) ifuF →+∞. Hence, there exists a con- stantM >0 such thatuF ≤Mfor all finite-dimensional subspaceFcontainingx0,y0, andz0. SinceX is reflexive andK is weakly closed, there existsu∈Ksuch that for ev- ery finite-dimensional subspaceFcontainingx0,y0, andz0,uis in the weak closure of the setVF=

F⊂F1{uF1}, whereF1is a finite-dimensional subspace inX. In fact, because VF is bounded, we know that (VF)^w (the weak closure of the set VF) is weakly compact. On the other hand, letF¹,F²,...,F^mbe finite-dimensional subspace containingx0, y0, andz0. SetF^(m):=span{F¹,F²,...,F^m}. ThenF^(m), which containsx0,y0, andz0, is a finite-dimensional subspace. Hence,^m_i₌₁VFⁱ=_m

i=1(_Fⁱ_⊂_F₁{uF1})=

F^(m)⊂F1{uF1} = ∅ and then_F(VF)^w= ∅. Now letv∈Kand letFbe a finite-dimensional subspace which containsx0,y0,z0andv. Sinceubelongs to the weak closure of the setVF=

F⊂F1{uF1}, we may find a sequence{uFα}inVFsuch thatuFα

−→w u. There exists a sequence{wFα}, wFα∈g(uFα), from (3.12) such that (A(uFα),v−uFα)≥(wFα,v−uFα). Becauseg:K→2^X^∗ is an upper semicontinuous from the weak topology onXto the strong topology onX^∗, there existw∈g(u) and a subsequence{wFβ} ⊂ {wFα}by [2,5] such that the sequence {wFβ}−→^s w (strongly converges tow). However,uFβ andwFβ satisfy the following inequality:

AuFβ

,v−uFβ

≥

wFβ,v−uFβ

. (3.13)

The monotony ofAimplies that

A(v),v−uFβ

≥

wFβ,v−uFβ

. (3.14)

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LettinguFβ

−→w uand{wFβ}−→^s wyields that Av,v−u≥

w,v−u, ∀v∈K. (3.15)

Thus

Au,v−u≥

w,v−u, ∀v∈K, (3.16) by the Minity theorem [2,4]. We claim thatu=0. OtherwiseuFβ

−→w 0. Takingv=z0+uFβ

in (3.13) yields that (A(uFβ),z0)≥(wFβ,z0). Thus

w_Fβsup∈g(u_Fβ)

AuFβ

−wFβ,z0

≥0, (3.17)

which contradicts condition (c). Therefore,uis a nonzero solution of (1.1).

Acknowledgments

This project was supported by The Natural Science Foundation of Zhejiang Province Grant no. Y104149. The authors would like to thank the anonymous referees for the help- ful suggestions.

References

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Soc.71(1965), 780–785.

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[3] P. M. Fitzpatrick and W. V. Petryshyn,Fixed point theorems and the fixed point index for multi- valued mappings in cones, J. London Math. Soc. (2)12(1975/76), no. 1, 75–85.

[4] D. Kinderlehrer and G. Stampacchia,An Introduction to Variational Inequalities and Their Ap- plications, Pure and Applied Mathematics, vol. 88, Academic Press, New York, 1980.

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Yisheng Lai: College of Statistics and Computer Science, Zhejiang Gongshang University, Hangzhou 310035, China

E-mail address:[email protected]

Yuanguo Zhu: Department of Applied Mathematics, Nanjing University of Science and Technol- ogy, Nanjing 210094, China

Yinbing Deng: Department of Mathematics, Central China Normal University, Wuhan 430079, China

10.1155/IJMMS.2005.1879