GENERALIZED NONLINEAR VARIATIONAL INEQUALITY PROBLEMS INVOLVING MULTIVALUED MAPPINGS

The solvability of a class of generalized nonlinear variational inequality 
problems involving multivalued, strongly monotone and strongly Lipschitz 
(a special type) operators, which are closely associated with generalized 
nonlinear complementarily problems, is discussed.


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 convex sets, while complementarity problems are approached in convex cone settings leading to equiva- lences.The complementarity problem in mathematical programming is based on a special type of variational 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 an edge over the com- plementarity.For more details on variational inequalities, we advise the reader to refer to [1,[3][4][5][14][15][16].
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Prehminaries
Let H be a real Hilbert space and H* its dual with the inner product (u, v) and norm ]1 u I I for u,v in H. Let [w,u] denote the duality pairing between the element w in H* and the element u in H. Let f: H*---H be a canonical isomorphism from H* onto H defined by [w,x] (f(w),x) for all x in H and all w in H*.
Thus, I I f I I I I f-111 1.
Let T, U: HP(H*) be multivalued mappings from H into the powerset P(H*) of H*.Let K be a nonempty, closed, convex subset of H. Then the problem of deter- mining the elements z in K, u in T(z) and v in U(z), such that [u-v, y-x] > 0 for all y in K, is called 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 in K, an ele- ment u in T(z) and an element v in U(z)such that u-v is in K* and [uv,x] 0, where K* {w in H*:[w,z] > 0 for all z in K}.
For T:K---,H single-valued and U--0, the GNVI problem (2.2) reduces to the variational inequality problem considered by Yao [13]: find an element x in K such that (x-Tx, y-x) > 0 for all y in g. (2.4) To this end, let us recall some definitions crucial to the work at hand.Definition 2.1: An operator T: HP(H*) from a Hilbert space H into the power- set P(H*) of its dual is said to be strongly monotone if, for a constant r > 0 and for all x, y in H, [u-v,y] >_ I Iy I I 2 for all u in T(x) and v in T(y).
Let (X, d) be a metric space and P(X) be the powerset of X.Then for any A, B in P(X), we define c3(A,B) sup{d(x,y):x is in A, and y is in B}. (2.6) A mapping F: X-P(X) is said to be an s-contraction if O(Fx, Fy) s(d(x, y)) for all x, y in X. (2.7) Definition 2.2: An operator T:HP(H*) is said to be Lipschitz continuous if there is a constant s > 0 such that for all x, y in H, O(T, Ty) < I I -Y I I for s > O.
Definition 2.3: An operator U:HP(H*) is said to be strongly Lipschitz if, for x,y in H and u in U(x) and v in U(y), (2.9) where k > 0 is arbitrary.Definition 2.4: An operator T: H---+H is hemicontinuous if the real function + z) is continuous on [0, 1] for all z, y, z in H. Let us consider an example of a strongly Lipschitz operator where the constant k is slightly relaxed [13].
Example 2.5: Let K be a nonempty, closed, convex subset of a real Hilbert space H. Let U:K---+K be hemicontinuous and, for all x,y in K and for a real number k> -1, (Ux Uy, x y) _< -kllx-yll2.
(2.10) If we define an operator V" KK by V(z) -(I-U)z for all z in K, then V is hemi- continuous and strongly monotone with the strong monotonicity constant 1 + k, and as a result, U has a unique fixed point in K.

Auxiliary and Main Results
Before we consider our main result, we need some auxiliary results.
Lemma 3.1" ([5]) Let K be a nonempty, closed, convex subset of a real Hilbert space H.Then, for a given element z in H, x-PK z (xz,y-x) >_ O for all y G K.
Proof: The proof is based on [2, Theorem 3.2].If Xl, u I and v I form a solution of the GNVI problem (2.2), then x 1 is in K, u I is in T(Xl) and v I is in U(Xl) such that [u I -vi, y-Xl] >_ 0 for all y in K. (3.3) This, in turn, implies that for a constant t > 0, (X 1 (X 1 tf(u v)),y Xl) >_ 0 for all y in K. (3.4) It follows from Lemma 3.1 that x I Pk(Xl tf(u 1 Vl)), That is, X 1 is a fixed point of F.
Conversely, if x I is a fixed point of F, then there exist u 1 in T(Xl) and v 1 in U(x1) such that x 1 Pk(Xl tf(u I Vl)). (3.7) This implies that x 1 is in K, and by Lemma 3.1, we find (x 1 (x 1 tf(u 1 vl)),y xl) >_ 0 for all y in K. (3.8) Since t > 0, it follows that [U 1 Vl, y-Xl] _ 0 for all y in K. (3.9) Hence Xl, U 1 and v 1 form a solution of the GNVI problem (2.2).Lemma 3.3: ([2]) Let (X,d) be a complete, metrically convex metric space and let F: X--P(X) be a contraction mapping.Then F has a fixed point; and for any x o in X, the sequence {Xn} defined so that x n is in F(x n_ 1) for n >_ 1, converges to a fixed point of F in X.
Theorem 3.4: Let H be a real Hilbert space and K 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 the GNVI problem (2.2) has a solution for an arbitrary constant t such that 0 < t < + + Proof: If we define an operator F: K--,P(K) by r(x)-U U PK(x-ts(u-v)) fr all x in K, e T() .e v() (3.10) then (by Lemma 3.2) it would suffice to show that F has a fixed point.P K is non- expanding, T is strongly monotone and Lipschitz continuous, and U is strongly Lip- schitz and Lipschitz continuous.Therefore, we find that, for all x, y in K, u I in T(x), U 2 in T(y), V 1 in U(x) and v 2 in U(y), and -t-t 2 I I f(ul u2)-f(vl v2)II 2 + t2 I] f(ul u2)-f(vl v2)II u -]1 x y I I 2 2tr I I x y I I 2 2tk I I x y [I 2 "4-t2[ I I f(ul u2)I! / I I f(vl v2)II ]2 I I y ]l 2t( + )II y ]] 2 + t2[O(Tx, Ty) + (9(Ux, Uy)] 2 by [6]   = { 1 2t(r + k) + t2(s + m)2) ]] x-y I I 2. (3.12) From (3.11) and (3.12), it follows that c9(Fx, Fy) _ L I I x-y I I for all x, y in K, (3.13) 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 from Lemma 3.3 that F has a fixed point x 1 in K, and hence xl, u I and v 1 form a solution to the GNVI problem (2.2).Theorem 3.5: Let K be a nonempty, closed, convex subset of a real Hilbert space h.Let T:H--P(H*) be strongly monotone and Lipschitz continuous with the strong monotonicity constant r > 0 and Lipschitz continuity constant s > O. Also, let U" H---P(H*) be strongly Lipschitz and Lipschitz 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 algorithm defined by Xn -t-1 (1 an)x n + anPk(x n tf(u n vn) for any x o in K (3.14) and for all t such that O<t<2(r+ k)/(s-{-m) 2, where u n is in T(Xn) v n is in U(xn) O_a n< 1, and the series a O+al+a2+... is divergent.Then {Xn} {Un} and {Vn} converge to in K, -fi in H* and in H*, respectively, and 5, and form a solution of the GNVI problem (2.2).