GENERALIZED NONLINEAR VARIATIONAL
INEQUALITY PROBLEMS INVOLVING
MULTIVALUED MAPPINGS
RAM U. VERMA
International
Publications, 1206
CoedDriveOrlando,
Florida 32826USA
andIstituto per la Ricerca di
Base
1-85075 Monteroduni(IS), Molise,
Italy(Received March, 1996;
RevisedMay, 1997)
The solvability of a class of
generalized
nonlinear variational inequality problems involving multivalued, strongly monotone andstrongly
Lipschitz(a
specialtype) operators,
which are closely associated with generalized nonlinear complementarity problems, is discussed.Key
words: Generalized Nonlinear Variational Inequality, Strongly LipschitzOperator,
Strongly MonotoneOperator.
AMS
subject classifications: 47H15.1. Introduction
Variational inequalities and complementarity problems play equally important roles in applied
mathematics,
physics, control theory and optimization, equilibrium theory of transportation and economics, mechanics, and engineering sciences. These pro-blems,
especially variational inequality problems, are studied in convexsets,
while complementarity problems are approached in convex cone settings leading to equiva- lences. The complementarity problem in mathematical programming is based on aspecial type ofvariational inequality in finite dimensions that has been central to the development of important algorithms. General variational inequalities can be re- duced to this special type by an application of discretization and the introduction of
Lagrange
multipliers leading to a computational approach. There are situations where computational methods for variational inequalities have anedge
over the com- plementarity.For
more details on variational inequalities, we advise the reader to refer to[1, 3-5, 14-16].
Printed in theU.S.A. ()1997by North AtlanticScience PublishingCompany 289
2. Prehminaries
Let H
be a real Hilbert space andH*
its dual with the inner product(u, v)
and norm]1
uII
foru,v
inH. Let [w,u]
denote the duality pairing between the element w inH*
and the element u inH. Let f: H*---H
be a canonical isomorphism fromH*
ontoH
definedby[w,x] (f(w),x)
for all x inH
and all w inH*.
Thus, II f II II f-111
1.Let T, U: HP(H*)
be multivalued mappings fromH
into the powersetP(H*)
of
H*. Let K
be anonempty,closed,
convexsubset ofH.
Then the problem ofdeter- mining the elements z inK,
u inT(z)
and v inU(z),
such that[u- v,
y-x] >
0 for all y inK,
iscalled the generalized nonlinear variational inequality
(GNVI)
problem.The following presents a class of generalized nonlinear complementarity
(GNC)
problems corresponding to the
GNVI
problem(2.2).
Find an element z inK,
an ele-ment u in
T(z)
and an element v inU(z)such
thatu-v is in
K*
and[u- v,x] 0,
whereK* {w
inH*:[w,z] >
0 for all z inK}.
For
T:K---,H single-valued andU--0,
theGNVI
problem(2.2)
reduces to thevariational inequality problem considered by
Yao [13]:
find an element x inK
suchthat
(x- Tx,
y-x) >
0 for all y in g.(2.4)
To
thisend,
let us recall some definitions crucial to the workat hand.Definition2.1:
An
operatorT: HP(H*)
froma Hilbert spaceH
intothe power- setP(H*)
of its dual is said to be strongly monotoneif,
for a constant r>
0 and for all x,y inH,
[u-
v,- y] >_ II -
yII
2 for all u inT(x)
and v inT(y).
Let (X, d)
be a metric space andP(X)
be the powerset ofX.
Then for anyA, B
inP(X),
we definec3(A,B) sup{d(x,y):x
is inA,
and y is inB}. (2.6) A
mappingF: X-P(X)
issaid to be an s-contractionifO(Fx, Fy) s(d(x, y))
for allx,y inX. (2.7)
Definition 2.2:
An
operatorT:HP(H*)
is said to be Lipschitz continuous if there is aconstant s>
0 such that for all x,y inH,
O(T, Ty) < II -
YII
for s>
O.Definition 2.3:
An
operatorU:HP(H*)
is said to be strongly Lipschitz if, for x,y inH
and u inU(x)
and v inU(y),
(2.9)
where k
>
0 isarbitrary.Definition2.4:
An
operatorT: H---+H
is hemicontinuous if thereal function+ z)
is continuous on
[0, 1]
for allz,
y,z inH.
Let
us consider an example ofa strongly Lipschitz operator where the constant k is slightly relaxed[13].
Example 2.5:
Let K
be a nonempty,closed,
convex subset ofa real Hilbert spaceH. Let U:K---+K
be hemicontinuousand,
for all x,y inK
and for a real numberk> -1,
(Ux Uy,
xy) _< -kllx-yll2. (2.10)
Ifwe define an operator
V" KK
byV(z) -(I- U)z
for all z inK,
thenV
is hemi- continuous andstrongly
monotone with thestrong
monotonicity constant 1+ k,
andas a
result, U
has aunique fixed point inK.
3. Auxiliary and Main Results
Before we considerour main
result,
weneed some auxiliary results.Lemma
3.1"([5]) Let K
be a nonempty,closed,
convex subsetof
a real Hilbertspace
H. Then, for
a given elementz inH, x-PK
z(x-
z,y-x) >_ O for
all y GK. (3.1)
Lemma
3.2:Let K
be a nonempty,closed,
convex subsetof H.
Then theGNVI
problem
(2.2)
has a solutioniff, for
a constant t>O,
the mappingF:H---P(H)
de-fined
byr(x) U U [PK (x- tf(u-- v))], (3.2)
T() has a
fixed
point.Proof: The proofis based on
[2,
Theorem3.2].
IfXl, uI and vI form a solution of theGNVI
problem(2.2),
then x1 is inK,
uI is inT(Xl)
and vI is inU(Xl)
suchthat
[u
I-vi,y-Xl] >_
0 for all y inK. (3.3)
This, in
turn,
implies that for aconstant t> 0,
(X
1(X
1tf(u v)),y Xl) >_
0 for all y inK. (3.4) It
follows fromLemma
3.1 thatxI
Pk(Xl tf(u
1Vl)), (3.5)
which is in
U U [PK(xl tf(ul- Vl))]- F(Xl)"
u
lT(x1)
v1U(x1)
(3.6)
That
is,
X1 is afixedpoint ofF.
Conversely, if xI is a fixed point of
F,
then there exist u1 inT(Xl)
and v1 inU(x 1)
suchthatx1
Pk(Xl tf(u
IVl) ). (3.7)
This implies that x1 isin
K,
and byLemma 3.1,
wefind(x
1(x
1tf(u
1vl)),y xl) >_
0 for all y inK. (3.8)
Since t
> 0,
it follows that[U
1 Vl, y-Xl] _0 for all y in K. (3.9)
Hence
Xl,U1 and v1 form asolution of theGNVI
problem(2.2).
Lemma
3.3:([2]) Let (X,d)
be a complete, metrically convex metric space and letF: X--P(X)
be a contraction mapping. ThenF
has afixed
point; andfor
any xo inX,
the sequence{Xn} defined
so that xn is inF(x
n_1) for
n>_ 1,
converges to afixed
pointof F
inX.
Theorem 3.4:
Let H
be a real Hilbert space andK
be a nonempty,closed,
con-vex subset
of H. Let T’H--,P(H*)
be strongly monotone and Lipschitz continuous with respective constants r>
0 and s> O. Let U’H---P(H*)
be strongly Lipschitz and Lipschitz continuous with respective constants k>_
0 and m>_
1. Then theGNVI
problem(2.2)
has a solutionfor
an arbitrary constant t such that 0<
t<
+ +
Proof: Ifwe define an
operator F: K--,P(K)
byr(x)- U U PK( x-ts(u-v))
fr all x inK,
e T()
. e
v()(3.10)
then
(by Lemma 3.2)
it would suffice to show thatF
has a fixed point.P
K is non- expanding,T
isstrongly
monotone and Lipschitz continuous, andU
isstrongly
Lip- schitz and Lipschitz continuous.Therefore,
wefindthat,
forall x,y inK,
uI inT(x),
U2 in
T(y),
V1 inU(x)
and v2 inU(y),
and
II P[K[
xtf(ul vl)]- Pk[Y- tf(u2- v2)] I[
- II II
x-xy-(tf(u
y-(tf(u
I1u2)- u2)- f tf(v (v
1Iv2) v2) II, II
2(3.11)
II -
yII I1 2t( -
yII
y,-t- f(u
t2t) -
2III f u2)- (ul
y,f(u f(v u2)-
IIu2) f v2) (vl
-[--b t2t(x- v2)II
2II f(ul
2y,f(v u2)- f(vl v2) v2)[I
2+ t2 I] f(ul u2)- f(vl v2)II u
- ]1
x yII II
2 y2tr]l II
x2t(
yII +
2)II
2tkII
yx]]
2y+ t2[O(Tx, [I
2"4-t2[ II f(ul Ty) + (9(Ux, u2)I! Uy)]
/II
2f(vl
by[6] v2)II ]2
= {
12t(r + k) + t2(s + m) 2) ]]
x-yII 2. (3.12)
From (3.11)
and(3.12),
it follows thatc9(Fx, Fy) _ L II
x-yII
for all x,y inK, (3.13)
1
where
L (1 2t(r + k) + t2(s + m)2) . Now,
under the assumptions, 0< L <
1 for all t such that 0<
t< 2(r + k)/(s + m) 2.
Since each Hilbert space is a metrically convex metric space, it follows fromLemma
3.3 thatF
has a fixed point x1 inK,
and hence xl,uI and v1form asolution totheGNVI
problem(2.2).
Theorem 3.5:
Let K
be anonempty, closed,
convex subsetof
a real Hilbert space h.Let T:H--P(H*)
be strongly monotone and Lipschitz continuous with the strongmonotonicity constant r
>
0 and Lipschitz continuity constant s>
O.Also,
letU"
H---P(H*)
be strongly Lipschitz andLipschitz continuous with strong Lipschitzity con- slant k>_
0 and Lipschitz continuity constant m>_
1. Consider the sequences{xn}
{un}
and{vn}
as generated by the iterative algorithmdefined
byXn
-t-1(1 an)x
n+ anPk(x
ntf(u
nvn) for
any xo inK (3.14)
and
for
all t such thatO<t<2(r+ k)/(s-{-m) 2,
where un is inT(Xn)
vn is inU(xn) O_a n<
1, and the series aO+a l+a2+...
is divergent. Then{Xn} {Un}
and
{Vn}
converge to inK,
-fi inH*
and inH*,
respectively, and 5, andform
a solutionof
theGNVI
problem(2.2).
When
f,T:H---,H
are the identities, U’H---,H is single-valued and an-a> O,
then Theorem 3.5 reducesto[13,
Theorem3.6].
Corollary 3.6:
Let U:H---H
be strongly Lipschitz and Lipschitz continuous with respective constants k>_
0 and m>_
1.Let
the sequence{Xn}
be generated by an iterative scheme"xn +
1PK((
1a)xn
q-aU(xn)) for
any xo inK
and 0
<
a< 2(1 + k)/(1 +
2k+ m2).
Then{xn}
converges to the uniquefixed
pointof U.
Proof of Theorem 3.5: Under the assumptions, it follows from Theorem 3.4 that 5 in
K,
inT(5)
and V inU(5)
form a solution of theGNVI
problem(2.2).
SinceP
k is nonexpansive, wehave]] Xn +
1 5(1 an)]1 xn - I[ -- an II tf(un + ft(vn - )11. (3.16)
Using the
strong
monotonicity and Lipschitz continuity ofT
and applying thestrong
Lipschitzity and Lipschitz continuity ofU,
wefind thatII :,
tf (Un -t-
tf (vn - )112
II , II = 2t<, , f(u
n)> -t- 2t<x
n", f(v
n)>
+ t II f(Un f(Vn - )11
-t-t2
[I f (Un f (Vn - )11
2II :. II
22t(r + k) II . !12 + t2(0(T,, T + 0(U., U ))2
< (1 2t(r + k))II Zn II
2+ ,2(, + m)2 II , II
2(3.17)
Now
applying(3.17)
to(3.16),
we havethatII a:, +
1- II
1<_ (1 an)II ,, II + an[1 2t(r + k) + t2(s + m)2]
7II , II
=
n(1 -(1 M)an)II . II
< H [1 -(1 M)aj] II o- II, (3.18)
j=0
1
where
0<M-(1-2t(r+k)+(s+rn)) g<l
for all such that 0<t<2(r + k)/(s + m) .
Since the series a0+
a1+
a2-t-...
diverges andM < 1,
this implies thatnlirn
j=0I-I (1- (1- M)aj)-
0and,
consequently,{n}
convergessgrongly
to.
The Lipschit continuity of
T
andU
implies that{un}
and{vn}
converge to res-pectively. This completes theproof.
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