Ordered Variational Inequalities and Ordered Complementarity Problems in Banach Lattices

and Applied Analysis 3 Lemma 3. Let (X; ≽) be a Banach lattice. Then the positive coneX is weakly closed. Proof. It is clear that the positive coneX of the Banach lattice X is convex. We have mentioned that the positive coneX of the Banach latticeX is norm closed. ApplyingMazur’s lemma (see [12] or [15]), we have in a Banach space, a convex set is norm closed if and only if it is weakly closed. This completes the proof of this lemma. Let (X; ≽X) and (U; ≽U) be two Banach lattices. A linear operator T : X → U is said to be order bounded if it maps every order bounded subset ofX to order-bounded subset of U. Let L b (X,U) denote the collection of all order bounded linear operators fromX toU.L b (X,U) is also a vector space. A natural partial order ≽ on L b (X,X) is induced by the positive cone X+ as follows: for any S, T ∈ L b (X,U), we say that S ≽T if and only if S(x) ≽T(x), for all x≽0. Then L b (X,U) is a partially ordered vector space with respect to ≽ . A linear operator T : X → U between two Banach lattices is said to be order continuous if x α 0 󳨀 → x in X implies T(x α ) 0 󳨀 → T(x) in U. It is known that all order-continuous linear operators between two Banach lattices are orderbounded linear operators. The following lemma is useful in the content of this paper. Lemma 4. Let (X; ≽X) and (U; ≽U) be two Banach lattices with U being Dedekind complete. Then one has L b (X,U) ⊆ L (X,U) . (5) Proof. A linear operator T : X → U between two Banach lattices is said to be positive whenever T(X+) ⊆ U . It is known that any positive linear operator between two Banach lattices is an order-bounded linear operator. Let L r (X,U) denote the collection of all linear operators from X to U which can be represented as a difference between two positive operators. Then as a consequence, we have L r (X,U) ⊆ L b (X,U). Furthermore, if U is Dedekind complete, then L r (X,U) = L b (X,U). From Theorem 4.3 in [14], we have that every positive linear operator between two Banach lattices is (strongly) continuous. It implies that ifU is Dedekind complete, then we have L b (X,U) = L r (X,U) ⊆ L(X,U). This lemma is proved. If (X; ≽X) and (U; ≽U) are two Banach lattices with U Dedekind complete, then for any T in L b (X,U), we have |T| ∈ L b (X,U); therefore, from Lemma 4, |T| ∈ L(X,U) holds. Hence we can define a norm ‖ ⋅ ‖ r on L b (X,U) by ‖T‖ r = ‖|T|‖, for all T ∈ L b (X,U). This norm ‖ ⋅ ‖ r is called the regular norm on L b (X,U) that satisfies the following inequality ‖T‖ ≤ ‖T‖ r , ∀T ∈ L b (X,U) . (6) By applying the Riesz-Kantorovich theorem,L b (X,U) under the regular norm and with the partial order ≽ becomes a Dedekind-complete Banach lattice. In addition, for any net {T α } inL b (X,U), we have T α ↓ 0 in L b (X,U) , iff, T α (x) ↓ 0 in U for each x ∈ X. (7) Let (X; ≽) be a Banach lattice. The norm ofX is said to be an order-continuous norm, if for any net {x α } inX, x α 0 󳨀 → 0 inX implies ‖x α ‖ → 0. A Banach lattice with order-continuous norm has many useful properties. We list some below. (1) Every Banach lattice with order-continuous norm is Dedekind complete. (2) Every reflexive Banach lattice has order-continuous norm (Nakano theorem); therefore, every reflexive Banach lattice is Dedekind complete. The class of Banach lattices with order-continuous norms is pretty large and includes many useful Banach spaces. For example, the classical L p (μ), where 1 ≤ p < ∞, are Banach lattices with order-continuous norms. The following result is a consequence of order-continuous norm. We list it as a lemma which is useful in the following sections. Lemma 5. If the norm of a Banach lattice (X; ≽X) is order continuous, then the σ-order convergence implies norm convergence, that is, x n 0 󳨀→ x implies x n 󳨀→ x, in the norm of X. (8) Proof. Suppose that {x n } is a sequence inX satisfyingx n 0 󳨀 → x. It is equivalent to x n −x 0 󳨀 → 0. Since (X; ≽X) is a Banach lattice with order-continuous norm, it implies ‖x n − x‖ → 0. This completes the proof of this lemma. 3. The Solvability of Ordered Variational Inequalities in Banach Lattices In this section, we introduce the concepts of ordered variational inequalities and ordered complementarity problems on suitable Banach lattices. Then we extend some already known solvability results about variational inequalities and complementarity problems (see [3–8, 10, 12, 13]) to the cases of ordered variational inequalities and ordered complementarity problems. Definition 6. Let (X; ≽X) and (U; ≽U) be two Banach lattices. Let C be a nonempty convex subset of X and f : C → L b (X,U) a mapping. The ordered variational inequality problem associated with C and f, denoted by VOI(C, f), is to find an x∗ ∈ C such that f (x ∗ ) (x − x ∗ ) ≽ U 0, ∀x ∈ C, (9) where, as usual, 0 denotes the origin of U. If f is linear, then the problem VOI(C, f) is called a linear ordered variational inequality problem; otherwise, it is called a nonlinear ordered variational inequality problem. 4 Abstract and Applied Analysis Definition 7. Let (X; ≽X) and (U; ≽U) be two Banach lattices. Let K be a convex cone of X and f : K → L b (X,U) a mapping. The ordered complementarity problem associated with K and f, denoted by OCP(K, f), is to find an x∗ ∈ K such that f (x ∗ ) (x ∗ ) = 0, f (x ∗ ) (x) ≽ U 0, ∀x ∈ K. (10) If f is linear, then the problem OCP(K, f) is called a linear ordered complementarity problem; otherwise, it is called a nonlinear ordered complementarity problem. For a given Banach lattice (X, ≽), let X󸀠 denote the norm dual of X, that is, X󸀠 = L(X,R). The order dual of (X, ≽) is denoted byX that is defined to beL b (X,R), where (R, ≥) is the Banach lattice of the set of real numbers with the ordinal topology and the standard order ≥, which is complete. From Garrett Birkhoff Theorem, the norm dual X󸀠 of a Banach lattice (X, ≽) coincides with its order dualX, that is, X 󸀠 = L (X,R) = L b (X,R) = X ∼ . (11) In Definitions 6 and 7, if we take (U; ≽U) = (R, ≥), then L b (X,R) = X󸀠 holds; therefore, in this case, we have f (x ∗ ) (x − x ∗ ) = ⟨f (x ∗ ) , (x − x ∗ )⟩ , (12) where ⟨⋅, ⋅⟩ is the pairing between X󸀠 and X. Hence, the ordered variational inequality VOI(C, f) and the ordered complementarity problem OCP(K, f) turn to be an ordinary variational inequality VI(C, f) and an ordinal complementarity problem CP(K, f), respectively. Thus the ordered variational inequality problems and the ordered complementarity problems in Banach lattices are generalizations of the variational inequality problems and complementarity problems in Banach spaces that extend the ranges from the real numbers to more general Banach lattices. There are close connections between variational inequality problems and complementarity problems in Banach spaces (e.g., see [3–5, 9–11]). In the next lemma, we show the similar connections between ordered variational inequality problems and ordered complementarity problems in Banach lattices. Lemma8. Let (X; ≽X) and (U; ≽U) be two Banach lattices. Let K be a convex cone of X and f : K → L b (X,U) a mapping. Then x∗ ∈ K is a solution to VOI(K, f) if and only if x∗ is a solution to OCP(K, f). Proof. It can be seen that x∗ is a solution to OCP(K, f) that implies that x∗ is a solution to VOI(K, f). Conversely, we will show that (9) implies (10). Suppose x∗ is a solution to VOI(K, f) satisfying (9). In the case, if x∗ = 0, then (10) obviously follows from (9). So we assume x∗ ̸ = 0. SinceK is a convex cone, then 2x∗ and 0.5x∗ are both inK. From (9), we have f (x ∗ ) (2x ∗ − x ∗ ) ≽ U 0, f (x ∗ ) (0.5x ∗ − x ∗ ) ≽ U 0. (13) They imply f (x ∗ ) (x ∗ ) ≽ U 0, f (x ∗ ) (− x ∗ ) ≽ U 0. (14) The last order inequality is equivalent tof(x)(x)≼0. From the antisymmetric property of ≼, we obtain f(x∗)(x∗) = 0. Sincef(x) ∈ L b (X,U), then from the linearity off(x) and by substitutingf(x)(x) = 0 into (9), it yields f(x)(x)≽0, for all x ∈ C. This lemma is proved. Definition 9. Let (X; ≽X) and (U; ≽U) be two Banach lattices. LetC be a nonempty convex subset ofX. Amappingf : C → L(X,U) is said to be linearly order comparable onCwhenever for any given x, y ∈ C and for every α ∈ (0, 1), f(αx + (1 − α)y)(x), and f(αx + (1 − α)y)(y) are ≽-comparable in U, that is, either f (αx + (1 − α) y) (x) ≽ U f (αx + (1 − α) y) (y) (15)