Extended Lorentz cones and variational inequalities on cylinders

Solutions of a variational inequality are found by giving conditions for the monotone convergence with respect to a cone of the Picard iteration corresponding to its natural map. One of these conditions is the isotonicity of the projection onto the closed convex set in the definition of the variational inequality. If the closed convex set is a cylinder and the cone is an extented Lorentz cone, then this condition can be dropped because it is automatically satisfied. The obtained result is further particularized for unbounded box constrained variational inequalities. For this case a numerical example is presented.


Introduction
In this paper we will study the solvability of variational inequalities on closed convex sets by using the isotonicity of the metric projection mapping onto these sets with respect to the partial order defined by the cone. Apparently this approach has not been considered before.
Variational inequalities are models of various important problems in physics, engineering, economics and other sciences. The classical Nash equilibrium concept can also be reformulated by using variational inequalities.
It is known that a vector is a solution of a variational inequality if and only if it is a zero of the corresponding natural mapping [1]. Hence, a variational inequality on a closed convex set is equivalent to finding the fixed points of the difference between the identity mapping and its natural mapping. The latter mapping is just the composition between the projection onto the latter closed convex set and the difference between the identity mapping and the mapping defining the variational inequality. I we could guarantee the isotonicity of the mappings in the latter composition with respect to the partial order defined by the cone, then by an iterative process we could construct an increasing sequence with respect to the partial order defined by the cone by using Picard's iteration. If we could also guarantee that this sequence is bounded from above with respect to the partial order defined by the cone, then this sequence would be convergent to the solution of the variational inequality (in fact it would be convergent to a solution of the corresponding equivalent fixed point problem). It turns out that there is a class of variational inequalities and a class of cones, that extend the Lorentz cone, for which this idea works very well. The only restriction is that the variational inequality has to be defined on a cylinder. Such problems appears in the practice. For example the unbounded box constrained variational inequalities are of this form. Based on the above idea a theorem for finding solutions of variational inequalities on a cylinder will be presented and an example will be given. Similar ideas to the above ones were presented in [2][3][4][5][6][7] for complementarity and implicit complementarity problems, but with a strong restriction on the cone defining the problem. The idea of monotone convergence for complementarity problems defined by general cones is considered first in the paper [8]. The present paper extends the results of [8] for variational inequalities.
Several other papers dealt with conditions of convergence for iterations similar to the above one, as for example [9][10][11][12][13][14][15][16][17][18][19]. However, neither of these works used the ordering defined by a cone for showing the convergence of the corresponding iterative scheme. Instead, they used as a tool the Banach fixed point theorem and assumed Kachurovskii-Minty-Browder type monotonicity (see [20][21][22][23]) and global Lipschitz properties of F . The structure of the paper is as follows: In the Preliminaries we introduce the terminology and notations use throughout this paper. In Section 3 we recall the definition and basic properties of the extended Lorentz cone, and determine all sets onto which the projection is isotone with respect to this cone. In Section 4 we will find solutions of a variational inequality by analyzing the monotone convergence with respect to a cone of the Picard iteration corresponding to its natural map. In Section 5 we will particularize these results to variational inequalities defined on cylinders, by using the extended Lorentz cone for the corresponding monotone convergence above. In this case we can drop the condition of Proposition 1 that the projection onto the closed convex set in the definition of the variational inequality is isotone with respect to the extended Lorentz cone, because this condition is automatically satisfied obtaining the more explicit result of Theorem 2. The latter result extend our results in [8] for mixed complementarity problems. In Section 6 we further particularize the results for unbounded box constrained variational inequalities. In Section 7 we will give a numerical example for this case.

Preliminaries
Denote by N the set of nonnegative integers. Let k, m, p, q ∈ N \ {0} and R m be the n-dimensional real Euclidean vector space.
Define the direct product space R p ×R q as the pair of vectors (x, u), where x ∈ R p and u ∈ R p .
Identify the vectors of R m by column vectors and consider the canonical This identification leads to the inner product in R p × R q given by The closed set K ⊂ R m will be called a cone if K ∩ (−K) = {0} and λx + µy ∈ K, whenever λ, µ ≥ 0 and x, y ∈ K. Let K ⊂ R p+q be a cone. Denote ≤ K the relation defined by x ≤ K y ⇐⇒ y − x ∈ K and call it the partial order defined by K. The relation ≤ K is reflexive, transitive, antisymmetric and compatible with the linear structure of R m in the sense that x ≤ K y implies that tx + z ≤ K ty + z, for any z ∈ R m and any t ≥ 0. Moreover, ≤ K is continuous at 0 in the sense that if x n → x when n → ∞ and 0 ≤ K x n for any n ∈ N, then 0 ≤ K x. Conversely any reflexive, transitive and antisymmetric relation in R m which is compatible with the linear structure of R m and it is continuous at 0 is defined by a cone. More specifically, =≤ K , where K = {x ∈ R m : 0 x} is a cone.
For any closed convex set C denote by P C : R m → R m the metric projection mapping onto C, that is, the mapping defined by P C (x) ∈ C and for any x ∈ R m . Since C is closed and convex, the projection mapping is well defined and by its definition it follows that for any x, y ∈ R m . It can be also shown that P C is nonexpansive (see [24]), that is, for any x, y ∈ R m . Let K ⊂ R m be a cone. The mapping F : The set Ω ⊂ R m is called K-bounded from below (K-bounded from above) if there exists a vector y ∈ R m such that y ≤ K x (x ≤ K y), for all x ∈ Ω. In this case y is called a lower K-bound (upper K-bound ) of Ω. If y ∈ Ω, then y is called the K-least element (K-greatest element) of Ω.
Let I ⊂ N be an unbounded set of nonnegative integers. The sequence The sequence {x n } n∈I is called K-bounded from below (K-bounded from above) if the set {x n : n ∈ I} is K-bounded from below (K-bounded from above).
A closed convex cone K is called regular if any K-increasing sequence which is K-bounded from above is convergent. It is easy to show that this is equivalent to the convergence of any K-decreasing sequence which is Kbounded from below. It is known (see [25]) that any cone in R m is regular.
The dual of a cone K ⊂ R m is the cone K * defined by The vectors v 1 , . . . , v k are called the generators of K.
The affine hyperplane with normal u ∈ R m \ {0} and through a ∈ R m is the set defined by An affine hyperplane H(u, a) determines two closed halfspaces H − (a, u) and and

Extended Lorentz cones
Let p, q be positive integers. For a, b ∈ R p denote a ≥ b if and only if b ≤ R p + a, that is, components of a are at least as large as the corresponding components of b. Denote by e the vector in R p with all components equal to one and by e i the canonical unit vectors of R p . In [8] we defined the following notion of an extended Lorentz cone: and showed that the dual of L is In the same paper we also showed the followings: • The extended Lorentz cone L defined by (4) is a (regular) cone.
• The cone L (or L * [26]) is a polyhedral cone if and only if q = 1.
• If q = 1, then the minimal number of generators of L is where δ denotes the Kronecker symbol. • If q = 1, then L * is a p + 1 dimensional polyhedral cone with the minimal number of generators 2p and a minimal set of generators of L * is {(e i , 1), (e i , −1) : i = 1, . . . , p}.
• If q = 1 and p > 1, then note that the number of generators of L and L * coincide if and only if they are 2 or 3-dimensional cones.
• The cone L is a subdual cone and L is self-dual if and only if p = 1, that is, L is the q + 1-dimensional Lorentz cone.
• L is a self-dual polyhedral cone if and only if p = q = 1.
In Theorem 2 of [8] we determined the L-isotone projection sets. For convenience we repeat this theorem here: where C is an arbitrary nonempty closed convex set in R q and L be the extended Lorentz cone defined by (4). Then, K is an L-isotone projection set.
2. Let p = 1, q > 1 and K ⊂ R p × R q be a nonempty closed convex set. Then, K is an L-isotone projection set if and only if K = R p × C, for some C ⊂ R q nonempty closed convex set.
3. Let p, q > 1, and where γ ℓ = (a ℓ , u ℓ ) is a unit vector. Then, K is an L-isotone projection set if and only if for each ℓ one of the following conditions hold: (a) The vector a ℓ = 0.

Variational inequalities
Let K ⊂ R m be a closed convex set and F : R m → R m be a mapping. Then, the variational inequality V I(K, F ) defined by F and K is the problem of finding an x * ∈ K such that for any y ∈ K, It is known that x * is a solution of V I(F, K) if and only if it is a fixed point of the mapping I − F nat K = P K • (I − F ), where I is the identity mapping of R m and F nat K is the natural mapping associated to V I(F, K) defined by [1]. Consider the Picard iteration If F is continuous and {x n } n∈N is convergent to x * , then by a simple limiting process in (5), it follows that x * is a fixed point of the mapping P K • (I − F ) and hence a solution of V I(F, K). Therefore, it is natural to seek convergence conditions for x n . Let us first state the following simple lemma: Let K ⊂ R m be a closed convex set, F : R m → R m be a continuous mapping and L be a cone. Consider the sequence {x n } n∈N defined by (5).
Suppose that the mappings P K and I − F are L-isotone, x 0 ≤ L x 1 , and there exists a y ∈ R m such that x n ≤ L y, for all n ∈ N sufficiently large. Then, {x n } n∈N is convergent and its limit x * is a solution of V I(F, K).
Proof. Since the mappings P K and I − F are L-isotone, the mapping x → P K • (I − F ) is also L-isotone. Then, by using (5) and a simple inductive argument, it follows that {x n } n∈N is L-increasing. Since any cone in R m is regular, {x n } n∈N is convergent and hence its limit x * is fixed point of P K • (I − F ) and therefore a solution of V I(F, K).

Remark 1.
The condition x 0 ≤ L x 1 in Lemma 1 is satisfied when , and hence by the isotonicity of P K we obtain Proposition 1. Let K ⊂ R m be a closed convex set, F : R m → R m be a continuous mapping and L be a cone. Consider the sequence {x n } n∈N defined by (5). Suppose that the mappings P K and I −F are L-isotone and x 0 ≤ L x 1 . Denote by I the identity mapping. Let Consider the following assertions: (iii) The sequence {x n } n∈N is convergent and its limit x * is a solution of V I(F, K). Moreover, x * is the L-least element of Γ.
(ii) =⇒ (iii): Suppose that Γ = ∅. Since the mappings P K and I − F are L-isotone, the mapping P K • (I − F ) is also L-isotone. Similarly to the proof of Lemma 1, it can be shown that {x n } n∈N is L-increasing. Let y ∈ Γ be arbitrary but fixed. We have y − x 0 ∈ L, that is x 0 ≤ L y. Now, suppose that x n ≤ L y.
Since the mapping P K • (I − F ) is L-isotone, x n ≤ L y implies that x n+1 = P K (x n − F (x n )) ≤ L P K (y − F (y)) ≤ L y. Thus, we have by induction that x n ≤ L y for all n ∈ N. Then, Lemma implies that {x n } n∈N is convergent and its limit x * ∈ K ∩ (x 0 + L) is a solution of V I(F, K). Since x * is a solution of V I(F, K), we have that P K (x * − F (x * )) = x * and hence x * ∈ Γ. Moreover, the relation x n ≤ L y in limit gives x * ≤ y. Therefore, x * is the smallest element of Γ with respect to the partial order defined by L.

Variational Inequality on cylinders
Let p, q be positive integers and m = p + q. By a cylinder we mean a set K = R p × C ⊂ R p × R q ≡ R m . In this section we will particularize the results of the previous section for variational inequalities on cylinders.
Lemma 2. Let K = R p ×C, where C is an arbitrary nonempty closed convex set in R q . Let G : R p × R q → R p , H : R p × R q → R q and F = (G, H) : Then, the variational inequality V I(F, K) is equivalent to the problem of finding a vector (x, u) ∈ R p × C such that for any v ∈ C.
Proof. The variational inequality V I(F, K) is equivalent to finding an (x, u) ∈ R p × C such that for any (y, v) ∈ R p × C. Let (x, u) ∈ R p × C be a solution of (7). If we choose v = u ∈ C in (7), then we get (y − x) ⊤ G(x, u) ≥ 0 for any y ∈ R p . Hence, G(x, u) = 0 and (v − u) ⊤ H(x, u) ≥ 0. Conversely, if (x, u) ∈ R p × C is a solution of (6), then it is easy to see that it is a solution of (7).
By using the notations of Lemma (2) the Picard iteration (5) can be rewritten as: Consider the partial order defined by the extended Lorentz cone (4). Then, we obtain the following proposition.
Let (x 0 , u 0 ) ∈ R p × C and consider the sequence (x n , u n ) n∈N defined by (8). Let x, y ∈ R p and u, v ∈ R q . Suppose that Let Consider the following assertions (iii) The sequence {(x n , u n )} n∈N is convergent and its limit (x * , u * ) is a solution of V I(F, K). Moreover, (x * , u * ) is the smallest element of Γ with respect to the partial order defined by L.
Proof. Let L be the extended Lorentz cone defined by (4). First observe that K ∩ (x 0 + L) = ∅. By using the definition of the extended Lorentz cone, it is easy to verify that Hence, by Proposition 1 (with m = p + q) and Lemma 2, it follows that Ω ⊂ Γ and (i) =⇒ (ii) =⇒ (iii).

Unbounded box constrained variational inequalities
Let p, q be positive integers, m = p + q and K = m ℓ=1 [a ℓ , b ℓ ] be a box, where a ℓ , b ℓ ∈ R ∪ {−∞, ∞} and a ℓ < b ℓ , for all ℓ ∈ {1, . . . , m}. The i-th entry of the projection function is (see Example 1.5.10 in [1]): and the Picard iteration (5) becomes Let L be the extended Lorentz cone defined by (4). The next proposition shows that the L-isotonicity of a box is equivalent to the box being a cylinder.
Proposition 2. Let L be the extended Lorentz cone defined by (4). Then, the projection mapping Proof. The sufficiency follows easily from item 1 of Theorem 2 in [8] (repeated in the Preliminaries as Theorem 1) For the sake of completeness we provide a proof here. Suppose that B = R p . If (x, u) ≤ L (y, v), that is, y−x ≥ v−u e, then by the nonexpansivity of P C we get which is equivalent to Hence, P K is L-isotone. Although, the necessity could also be derived from item 3 of the same theorem, it is more clear to prove this directly as follows. Suppose that P K is L-isotone. We need to prove that a i = −∞, b i = ∞, for any i = 1, . . . p. Assume to the contrary, that there exist at least one k ∈ {1, . . . , p} such that either a k or b k is a finite real number. Assume that b k is a finite real number. The case a k is a finite real number can be treated similarly. Let u and v be two different vectors in C. Then, P C (u) = u and . . , p}, where δ ik is the Kronecker symbol. Then, (x, u) ≤ L (y, v) and by (9) we have (P K (y, v)) k = (P K (x, u)) k , or equivalently (P B (y)) k = (P B (x)) k . Hence, by (10) and the L-isotonicity of P K we get which is a contradiction. Hence, the results of Theorem 2 can be particularized to the set K given by Proposition 2, with the Picard iteration taking the form (11) and the function 'mid' given by (9). In the next section we will present an example for this particularized result. 10,10]. Let L be the extended Lorentz cone defined by (4). Let f 1 (x, u) = 1/12(x 1 + u +12) and f 2 (x, u) = 1/12(x 2 + u − 7.2). Then it is easy to show that these two functions are L-monotone. Let w 1 = (1, 1, 1/6, 1/3) and w 2 = (1, 1, 1/3, 1/6) so w 1 and w 2 is in L. For any two vectors (x, u) and (y, v) in K, suppose (x, u) ≤ L (y, v),

Numerical example
Similarly, we can prove that if (x, u) ≤ L (y, v), then f 2 (y, v) − f 2 (x, u) ≥ 0. Provided that K is convex, and w 1 , w 2 ∈ L, if (x, u) ≤ L (y, v) holds, then Hence, we can easily see that: which shows that (x, u) ∈ Ω and Ω = ∅. Therefore, the conclusions of Theorem 2 will apply for our example. Now, we begin to solve the V I. Suppose that (x, u) is its solution. Since G(x, u) = 0, and where f i = f i (x, u), i = 1, 2, we have x 1 = x 2 = f 1 + f 2 . Moreover, since we get Obviously, if (x, u) ∈ ker(F ), that is, G(x, u) = 0 and H(x, u) = 0, then (x, u) is a solution of the variational inequality. The equality H(x, u) = 0 implies      By using equations (13), equations (14) become Hence, u 2 = u 1 + 4 15 . By substituting back to (15), we obtain However, the equation (15) can be transformed to By squaring both sides, we get Hence, Thus, if u 1 = 28 995 , then since u 2 = u 1 + 4 15 equation (16) will not hold. Therefore, the only solution in this case is Next, suppose that the solution (x, u) / ∈ ker(F ), that is, H = (x, u) = 0. Then, by (6) and (13), we get From the general theory of variational inequalities, it is known that u should be on the boundary of C. Let u 1 = 10. By choosing v 1 < u 1 and v 2 = u 2 and using u < 20, it can be seen that inequality (17) will not hold. Similarly, if u 2 = 10, by choosing v 2 < u 2 and v 1 = u 1 , inequality (17) will not be satisfied. If u 1 = −10, then by choosing v 1 > u 1 and v 2 = u 2 , inequality (17) will not hold. Similarly u 2 = −10 will not lead to a solution. Hence, this case will not provide a new solution. Therefore, the only solution of the variational inequality is obtained above. The Picard iteration can be completed by using any spreadsheet software. Note that since the variational inequality is box constrained, the iteration in (11) will be calculated by using the median function as it is shown in the following tables. More precisely, the initial point is given in the first row for n = 0 and the later iterations are given in the following rows, where the columns below u n 1 and u n 2 are obtained by the median of the upper bound, lowerbound and u n − H(x n , u n ). In the first table, we showed the iteration using the previous example where (x 0 , u 0 ) = (43/30, 13/30, 2, 5). In the other tables, we showed the iteration from different initial points in different directions outside of C. It can be observed that the iterations in all tables converge to the same unique solution of the variational inequality. and/or variational inequalities by using a monotone convergence. Moreover, any such result could be important in statistics as well, where the isotonicity of the projection may occur in various algorithms (see for example [27]).