On η-Upper Sign Property and Upper Sign Continuity and Their Applications in Equilibrium-Like Problems

and Applied Analysis 3 Proof. On the contrary, suppose that there exist u, V ∈ K such that f(u t , u) ≤ 0, for all t ∈ ]0, 1[, and f(u, V) < 0. By (ii) f(u, u t ) < 0 for all t ∈ ]0, 1[. Now it follows from (iii) that there exists t 0 ∈ ]0, 1[ such that 0 ≤ f (u, u t 0 ) + f (u t 0 , u) < 0, (11) which is a contradiction. This completes the proof. The following example satisfies all the assumptions of Proposition 7 while it cannot fulfill all the assumptions of Lemma 6. Example 8. LetK = [0, 1] and f : K × K → R such that f (x, y) = {{ {{ { y − 1, x = 1, 0 ≤ y ≤ 1, 󵄨󵄨󵄨󵄨x − y 󵄨󵄨󵄨󵄨 , 1 ≥ y ≥ x ≥ 0, 1, 1 > x > y > 0. (12) It is straightforward to check that f satisfies conditions of Proposition 7 and so it has upper sign property. But it is not upper sign continuous, because if we take x = 1 and y = 0, then we have f (x t = tx + (1 − t) 0, 0) = f (t, 0) = 1 ≥ 0, ∀t ∈ (0, 1) , (13) whereas f(1, 0) = −1 < 0. Hence the hypothesis of Lemma 6 cannot work. The next example shows that condition (iii) of Proposition 7 is essential. Example 9. let K = [0, 1] and f : K × K → R as f (x, y) = − 󵄨󵄨󵄨󵄨x − y 󵄨󵄨󵄨󵄨 . (14) Then f satisfies (i) and (ii) of Proposition 7 but not (iii). Note that f is upper sign continuous but does not have upper sign property. It is worth noting that Example 9 indicates that upper sign continuity cannot imply condition (iii) of Proposition 7. Now onemay ask the converse.The following example answers the question. Example 10. Let f : R ×R → R be defined by f (x, y) = {{ {{ { 0, x = y, 1, x ∈ Q (the rational numbers) , −1, x ∈ Q c (the irrational numbers) . (15) It is easy to verify that f satisfies condition (iii) while f is not upper sign continuous. 3. Application of the η-Upper Sign Property In this section we establish a link between the solution set of the equilibrium-like problem (that is, S(f,K)) and the solution set of the local Minty equilibrium-like problem (M L (f,K)). The result presented is another version of Theorem 1 of [9] without assuming convexity on the set and upper sign continuity on the map. Theorem 11. Letf : K×K → R be a bifunction with η-upper sign property and η (u, u + tη (u, V)) = tη (u, V) ∀u, V ∈ K, ∀t ∈ [0, 1] . (16) If f satisfies the following condition: f (u, V) < 0 󳨐⇒ f (u, u t = u + tη (u, V)) < 0, ∀t ∈ ]0, 1[ (17) thenM L (f,K) ⊆ S(f,K). Proof. Let u ∈ M L (f,K) and V ∈ K be arbitrary. Therefore by definition ofM L (f,K) there exists an open neighborhood U of u such that f (w, u) ≤ 0 ∀w ∈ K ∩ U. (18) Since U − u is a neighbourhood of 0, there exists t 0 ∈ ]0, 1[ such that tη(u, V) ∈ U − u for all t ∈ [0, t 0 ]. Let V = u + t 0 η(u, V) and V t = u + tη(u, V) for t ∈ [0, 1]. Then V t ∈ U ∩ K, since V t = u + tη(u, u + t 0 η(u, V)) = u + tt 0 η(u, V) and tt 0 η(u, V) ∈ U − u. Hence, by our assumption on u, f(V t , u) ≤ 0. Now from η-upper sign property of f, we have f(u, V) ≥ 0. So from (17) we conclude f(u, V) ≥ 0 and this completes the proof. Remark 12. There are many functions η : K×K → X which satisfy η(u, u + tη(u, V)) = tη(u, V). For instance, let K = R η (u, V) = {{ {{ { V − u, V ≥ 0, u ≥ 0, V − u, V ≤ 0, u ≤ 0, u, otherwise. (19) It is clear thatK is an invex set and η satisfies (17). If η(u, V) = V − u, then K will be a convex set and clearly the condition (17), that is, η (u, u + tη (u, V)) = tη (u, V) , ∀u, V ∈ K, ∀t ∈ [0, 1] , (20) is automatically satisfied. Corollary 13. If the bifunction f satisfies all the assumptions of Proposition 7 and η (u, u + tη (u, V)) = tη (u, V) ∀u, V ∈ K, ∀t ∈ [0, 1] , (21) then M L (f,K) ⊆ S(f,K). This means that the upper sign continuity property in Theorem 11 can be substituted by the conditions of Proposition 7. Remark 14. One can see that Theorem 11 and Corollary 13 extend the corresponding result given in [9] for bifunctions whose domains are not necessarily convex with weaker hypothesis. To verify this one can consider Example 8 which satisfies all the assumptions of Theorem 11 as well as Corollary 13 and M L (f,K) = {} ⊆ S(f,K) = [0, 1) while f is not upper sign continuous. Hence, even when K is a convex set, our conditions are weaker than the conditions of Lemma 3 in [9] and consequently Lemma 2.1 in [10] and Lemma 2.1 in [7]. 4 Abstract and Applied Analysis Proposition 15. Assume that f satisfies condition (ii) of Proposition 7 and, for all u, V ∈ X, we have M(f, [u, V]) ⊆ S(f, [u, V]) where [u, V] = u + tη(u, V), t ∈ [0, 1]. Then f has η-upper sign property. Proof. Assume that f does not have η-upper sign property at u ∈ U; thus, for all open neighborhoods U of u, there exists V ∈ U ∩ K such that f(u t , u) ≤ 0 for all t ∈ ]0, 1[, but f(u, V) < 0. From condition (ii) of Proposition 7, we have f(u, u t ) < 0 for all t ∈ ]0, 1[, where u t = u + tη(u, V). Clearly u ∈ M(f, [u, u t ]); therefore u ∈ S(f, [u, u t ]), which implies that f(u, u t ) ≥ 0 and a contradiction occurs. 4. Relations between Solution Set of Variational-Like Inequalities and Equilibrium-Like Problems In this section we investigate the relationship between ηupper sign property and upper sign continuity in the setting of variational inequality. Also we are going to establish a link between the solution sets of variational inequality and Minty variational inequality. Variational-like inequality problem is formulated as follows: find u ∈ K such that ∃u∗ ∈ T (u) with ⟨u∗, η (u, V)⟩ ≥ 0 ∀V ∈ K. (22) And the Minty variational-like inequality problem is formulated as follows: find u ∈ K such that ⟨V∗, η (u, V)⟩ ≥ 0 ∀V ∈ K, V∗ ∈ T (V) . (23) We denote by S(T,K) and M(T,K) the solution sets of the variational inequality problem and the Minty variational inequality problem, respectively. The set-valued map T : K → 2X ∗ is said to be ηpseudomonotone on K if, for all u, V ∈ K, ∃ u∗ ∈ T(u) such that ⟨u ∗ , η (u, V)⟩ ≥ 0 󳨐⇒ ⟨V∗, η (u, V)⟩ ≥ 0 ∀V∗ ∈ T (V) . (24) Proposition 16. Let T : K → 2X ∗ be a set-valued map with w -compact value and f T be defined as follows:


Introduction
Convexity is one of the most significant assumptions in optimization theory and plays an important role in general equilibrium theory.Hanson [1] introduced the concept of invexity as a generalization of convexity for scalar constrained optimization problems.It has been shown by Blum and Oettli [2] and Noor and Oettli [3] that variational inequalities and mathematical programming problems can be viewed as special realization of the abstract equilibrium problems.Equilibrium problems have numerous applications, including but not limited to problems in economics, game theory, finance, traffic analysis, circuit network analysis, and mechanics; see [2][3][4][5].Noor [6] introduced a new class of equilibrium problems called invex equilibrium (or equilibrium-like) problem in the setting of invexity.It has been shown that invex equilibrium problems include variational-like inequality and equilibrium problems as special cases.Hence the invex equilibrium problems cover a vast range of applications.Bianchi and Pini [7] for an equilibrium problem considered its dual equilibrium problem and then presented some results about relationship between the solution set of equilibrium problem and its dual equilibrium by using an extension of the notion of upper sign continuity introduced by Hadjisavvas [8] (in the setting of variational inequality problems) for bifunctions.
Recently, in the convex setting, Castellani and Giuli [9] introduced the notion of upper sign property.It was shown that, in the framework of variational inequalities, this notion coincides with the upper sign continuity for the setvalued map introduced by Hadjisavvas [8].However, this phenomenon is not true for bifunctions.In fact, it is not hard to construct examples of bifunctions with upper sign property which is not upper sign continuous and vice versa.Castellani and Giuli [9], by adding a suitable assumption (they called it technical assumption), proved the fact that an upper sign continuous bifunction has upper sign property.In this work, we first extend the concept of upper sign property for bifunctions whose domains are invex.Afterwards, by using two technical assumptions we highlight bifunctions with upper sign property.It is worth noting that we are going to relax the upper sign continuity assumption on the bifunction in order to establish a link between the solution sets of dual equilibrium and equilibrium problems.This result confirms that we can replace the upper sing continuity by some sufficient conditions under which a bifunction has upper sign property.Finally we present some existence results for invex equilibrium problems.The results obtained in

Preliminaries
Let  be a topological vector spaces and  * its topological dual.Let  be a nonempty subset of  and  :  ×  → R (real numbers) and (⋅, ⋅) :  ×  →  two functions.Definition 1.Let  ∈ .The set  is said to be invex at  with respect to , if The Let  :  ×  → R be a real-valued bifunction.The invex equilibrium (equilibrium-like) problem (for short, IEP) is to find  ∈  such that for all V ∈   (, V) ≥ 0.
It is well known that the IEP is closely related to the problem of finding  ∈  such that which is called Minty equilibrium-like problem (MIEP) (note that the problem was called dual equilibrium in [7,10] when  is a convex set).We designate by (, ) the solution set of IEP and by (, ) the solution set of MIEP.
If  ∈  and there exists an open neighborhood  of  such that then  is a solution of the local Minty equilibrium-like problem and the solution set of it will be denoted by   (, ).
Definition 3. We say that bifunction  :  ×  → R is -upper sign continuous, if for any , V ∈ , the following implication holds: Remark that if, for all , V ∈ , we set (, V) = V − , then the definition reduces to the upper sign continuity introduced by Bianchi and Pini [7].It is worth noting that the idea of upper sign continuity was first introduced by Hadjisavvas [8].
In what follows we extend the upper sign continuity given [8] for a set-valued mapping whose domain is not necessarily convex.
The following definition extends the definition of the upper sign continuity introduced in [8] for set-valued mappings whose domains are not necessarily convex.Definition 4. A set-valued map  :  → 2  * is said to be upper sign continuous, if for any , V ∈  and for all  ∈ ]0, 1[ we have inf where   =  + (, V).Note that if the set-valued mapping  :  → 2  * has  * -compact values and we define   :  ×  → R by then we cannot deduce -upper sign continuity of   by  and vice versa.Hence a natural question is how the definition of -upper sign continuity can be modified such that the former notion of the upper sign continuity implies the other one.The first time Castellani and Giuli in [9] answered the question by proposing a new definition for mappings whose domains are convex.In the next definition we extend their definition for mappings whose domains are invex.
Definition 5. We say that the bifunction  :  ×  → R has -upper sign property (with respect to the first variable) at  ∈  if there exists an open neighborhood  of  such that, for any V ∈  ∩ , the following implication holds: where   =  + (, V).
The following result establishes sufficient conditions under which a mapping has upper sign property.Lemma 6 (see [9,Lemma 3]).Let  be a convex set and  an equilibrium bifunction (i.e.,  :  ×  → R with (, ) = 0, for all  ∈ ) such that for all If  is upper sign continuous then it has the upper sign property.
It is a natural question: is the result of the above lemma still valid when the convexity on  and the upper sign continuity have been relaxed?The next result provides an affirmative answer.
Then  has -upper sign property.
The following example satisfies all the assumptions of Proposition 7 while it cannot fulfill all the assumptions of Lemma 6.
Example 8. Let  = [0, 1] and  :  ×  → R such that It is straightforward to check that  satisfies conditions of Proposition 7 and so it has upper sign property.But it is not upper sign continuous, because if we take  = 1 and  = 0, then we have whereas (1, 0) = −1 < 0. Hence the hypothesis of Lemma 6 cannot work.
The next example shows that condition (iii) of Proposition 7 is essential.
Then  satisfies (i) and (ii) of Proposition 7 but not (iii).Note that  is upper sign continuous but does not have upper sign property.
It is worth noting that Example 9 indicates that upper sign continuity cannot imply condition (iii) of Proposition 7. Now one may ask the converse.The following example answers the question.

(15)
It is easy to verify that  satisfies condition (iii) while  is not upper sign continuous.

Application of the 𝜂-Upper Sign Property
In this section we establish a link between the solution set of the equilibrium-like problem (that is, (, )) and the solution set of the local Minty equilibrium-like problem (  (, )).The result presented is another version of Theorem 1 of [9] without assuming convexity on the set and upper sign continuity on the map.Theorem 11.Let  : × → R be a bifunction with -upper sign property and If  satisfies the following condition: then   (, ) ⊆ (, ).
Proof.Let  ∈   (, ) and V ∈  be arbitrary.Therefore by definition of   (, ) there exists an open neighborhood  of  such that Since  −  is a neighbourhood of 0, there exists Hence, by our assumption on , (V  , ) ≤ 0. Now from -upper sign property of , we have (, V) ≥ 0. So from (17) we conclude (, V) ≥ 0 and this completes the proof.
If (, V) = V − , then  will be a convex set and clearly the condition (17), that is, is automatically satisfied.
Corollary 13.If the bifunction  satisfies all the assumptions of Proposition 7 and then   (, ) ⊆ (, ).This means that the upper sign continuity property in Theorem 11 can be substituted by the conditions of Proposition 7.
Remark 14.One can see that Theorem 11 and Corollary 13 extend the corresponding result given in [9] for bifunctions whose domains are not necessarily convex with weaker hypothesis.To verify this one can consider Example 8 which satisfies all the assumptions of Theorem 11 as well as Corollary 13 and   (, ) = {} ⊆ (, ) = [0, 1) while  is not upper sign continuous.Hence, even when  is a convex set, our conditions are weaker than the conditions of Lemma 3 in [9] and consequently Lemma 2.1 in [10] and Lemma 2.1 in [7].

Relations between Solution Set of Variational-Like Inequalities and Equilibrium-Like Problems
In this section we investigate the relationship between upper sign property and upper sign continuity in the setting of variational inequality.Also we are going to establish a link between the solution sets of variational inequality and Minty variational inequality.Variational-like inequality problem is formulated as follows: And the Minty variational-like inequality problem is formulated as follows: find  ∈  such that ⟨V * ,  (, V)⟩ ≥ 0 ∀V ∈ , V * ∈  (V) . ( We denote by (, ) and (, ) the solution sets of the variational inequality problem and the Minty variational inequality problem, respectively.
The set-valued map  :  → 2  * is said to be pseudomonotone on  if, for all , V ∈ , ∃  * ∈ () such that Proposition 16.Let  :  → 2  * be a set-valued map with  * -compact value and   be defined as follows: and since  is -upper sign continuity we have Therefore   has -upper property.Conversely, let   have -upper sign property.We want to show that  is -upper sign continuous.To see this, assume , V ∈  such that, for all  ∈ ]0, 1[, and this completes the proof.
Remark that if we assume that  is convex and take (, V) = V − ; then it is clear that − (  , ) =  (, V) , ∀ ∈ [0, 1] . ( This means that in the setting of convexity Proposition 16 collapses to the corresponding result given in [9] and moreover there are many examples of  which are not equal to the usual case (, V) = V − .For instance, see the following example.(37) It is clear that  is an invex set (in here not convex) and that  satisfies the condition −(  , ) = (, V).
set  is called an invex set with respect to , if  is invex at each  ∈ .From now on,  denotes a nonempty invex set in  with respect to .