Existence of solutions of generalized vector equilibrium problems in reflexive Banach spaces

Nonlinear Analysis 74 (2011) 2226–2234

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Existence of solutions of generalized vector equilibrium problems in reflexive Banach spaces I. Sadeqi ∗ , C.G. Alizadeh Faculty of Sciences, Sahand University of Technology, Tabriz, Iran

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Article history: Received 9 February 2010 Accepted 17 November 2010 Keywords: Generalized vector equilibrium problems Recession cones Reflexive Banach space

In this paper, a generalized vector equilibrium problem with set-valued maps defined on a reflexive Banach space is considered. By using the recession method, we first give the conditions under which the solution set is non-empty, convex and weakly compact, and then extend it to the strong generalized vector equilibrium problem. This facilitates generalizing and modifying various existence theorems. Furthermore, the topological properties of the solution set are studied and it is shown that the solution set includes some boundary points. © 2010 Elsevier Ltd. All rights reserved.

1. Introduction Let X be a real reflexive Banach space and Y be a real normed linear space. Suppose that 2Y denotes the family of all subsets of Y , and C ⊆ Y is an order cone, that is a proper, closed and convex cone such that int C ̸= ∅. Given a non-empty subset K ⊆ X and a set-valued function F : K × K → 2Y \ ∅, the problem of determining the existence of x ∈ K such that F (x, y) ∩ (−int C ) = ∅;

∀y ∈ K ,

(1.1)

has been extensively studied by many authors in recent years (see [1–8]). This problem may be written as: find x ∈ K

such that F (x, y) ⊆ Y \ (−int C ); ∀y ∈ K ,

which is called generalized vector equilibrium problem (GVEP). Moreover, the following problem, which is closely related to GVEP (1.1), is named dual generalized vector equilibrium problem (DGVEP): find x ∈ K

such that F (y, x) ∩ (int C ) = ∅; ∀y ∈ K .

(1.2)

The solution sets to GVEP and DGVEP are denoted by Ep and Ed , respectively. The strong version of GVEP is considered as follows: find x ∈ K

such that F (x, y) ⊆ C ; ∀y ∈ K ,

(1.3)

such that F (y, x) ⊆ −C ; ∀y ∈ K .

(1.4)

and its dual form: find x ∈ K

Problem (1.3) is called the strong generalized vector equilibrium problem (SGVEP) and Problem (1.4) is the dual strong generalized vector equilibrium problem (DSGVEP). The solution sets to SGVEP and DSGVEP are denoted by Esp and Esd ,

∗

Corresponding author. Tel.: +98 4123459605; fax: +98 4123224950. E-mail addresses: [email protected] (I. Sadeqi), [email protected] (C.G. Alizadeh).

0362-546X/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2010.11.027

I. Sadeqi, C.G. Alizadeh / Nonlinear Analysis 74 (2011) 2226–2234

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respectively. For more details, we refer the interested reader to [1,2,5,6,8–15]. Nowadays, many mathematicians investigate different models of GVEP and the related topics. In this article, the concept of GVEP is discussed and some results of Ansari and Flores-Bazán’s work are improved. Moreover, several alternative necessary and/or sufficient conditions are obtained, and it is proved that under these conditions, the solution sets of GVEP and SGVEP are non-empty, convex and weakly compact. Some other topological properties of the solution set are also studied. Furthermore, it is shown that for a subset K in a reflexive Banach space X , the solution set Ep contains some boundary points of K . This result has numerous applications in minimization problems and geometry of Banach spaces. 2. Preliminaries Definition 2.1 ([1]). For any weakly closed set K in X , K ∞ , so called the recession cone of K , is defined by K ∞ = {x ∈ X : ∃tn ↓ 0, ∃xn ∈ K , tn xn ⇀ x}, where ‘‘ ⇀’’ means convergence in weak topology, we define ∅∞ = ∅. If K is convex, then K ∞ is also convex and we have K ∞ = {x ∈ X : ∃x0 ∈ K , x0 + tx ∈ K , ∀t > 0}. The set K is called linearly bounded whenever K ∞ = {0}. Proposition 2.2 ([1,11]). The following statements are satisfied: (a) K1 ⊆ K2 implies that K1∞ ⊆ K2∞ ; (b) (K + x)∞ = K ∞ , for all x ∈ X ; (c) If {Ki }i∈I is any family of non-empty sets in X , then

 ∞ Ki

⊂

i∈I

i∈I

In addition, if

 (Ki )∞ .



i∈I

Ki ̸= ∅ and each Ki is closed and convex, then equality is obtained in the previous inclusion.

Let K be a non-empty convex subset of X . For a given closed convex cone P of a real normed space Y , the set-valued map F : K → 2Y \ ∅ is called: (i) P-convex, if ∀x, y ∈ K and ∀α ∈ [0, 1],

  α F (x) + (1 − α)F (y) ⊆ F α x + (1 − α)y + P ; (ii) properly P-quasiconvex, if ∀x, y ∈ K and ∀α ∈ ]0, 1[,





F (x) ⊆ F α x + (1 − α)y + P





or F (y) ⊆ F α x + (1 − α)y + P ;

(iii) strictly quasiconvex, if for all distinct x, y ∈ K and ∀α ∈ ]0, 1[,





F (y) − F (1 − α)x + α y ⊆ int P





or F (x) − F (1 − α)x + α y ⊆ int P ;

(iv) strictly quasiconcave, if (−F ) is strictly quasiconvex. (v) Weakly lower semicontinuous at x ∈ K , if for any y ∈ F (x) and for any sequence xn ∈ K weakly converging to x, there exists a sequence yn ∈ F (xn ) that converges strongly to y. The map F is weakly lower semicontinuous on K , if it is weakly lower semicontinuous at each point of K . 3. The main results on GVEP First let us consider the following hypotheses: Hypothesis (H1) ([1]). The set-valued map F : K × K → 2Y \ ∅ is such that:

(f0 ) (f1 ) (f2 ) (f3 )

F (x, x) ⊆ C ∩ (−C ) for all x ∈ K , F (x, y) ∩ (−int C ) = ∅ for all x, y ∈ K , implies F (y, x) ∩ (int C ) = ∅, the mapping F (x, ·) : K → 2Y \ ∅ is C -convex for all x ∈ K , the set {ξ ∈ [x, y] : F (ξ , y) ∩ (−int C ) = ∅} is closed for all x, y ∈ K , where [x, y] denotes the closed line segment joining x and y, (f4 ) F (x, ·) is weakly lower semicontinuous for all x ∈ K .

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If K is a bounded subset, then Problem (1.1) has a solution (see [1]). To deal with the unbounded case, it is needed to determine the behavior of F : K × K → 2Y \ ∅ along some particular directions. These directions are indicated by the following cones: R0 :=

 {ν ∈ K ∞ : 0 ∈ F (y, z + λν) + W , ∀ λ > 0, ∀z ∈ K s.t. F (y, z ) ⊆ −C } y∈K

and R1 :=

 {ν ∈ K ∞ : 0 ∈ F (y, y + λν) + W , ∀λ > 0}, y∈K

where W = Y \ (−int C ). The cones R0 and R1 have been introduced by Flores-Bazán [7] and applied in [1,11,12] for existence problems. According to assumptions (f0 ) and (f4 ), these cones are non-empty and closed. Considering the definition of R0 and R1 , the mapping F : K × K → 2Y \ ∅ must be well defined on the direction {z + λν : λ > 0}, then it should lie entirely in K . Since ν ∈ K ∞ , this occurs when K is convex. The inclusion case of the following lemma has been proved by Ansari and Flores-Bazán [1] and the conclusion is clear by Assumption (f1 ). Lemma 3.1 ([1]). Let K be a non-empty closed convex set in X . If F : K × K → 2Y \ ∅ is a set-valued map satisfying Assumptions (f0 ), (f1 ) and the mapping F (x, .) : K → 2Y \ ∅ is C -convex for all x ∈ K , then R1 =

   {ν ∈ K ∞ : 0 ∈ F (y, y + λν) + Y \ (−int C ) , ∀λ > 0} y∈K

 {ν ∈ K ∞ : F (y + λν, y) ∩ (−int C ) = ∅, ∀λ > 0}. = y∈K

Recall that, cones R0 and R1 play an important role in the existence results of the solution set of GVEP. In [11], the authors applied these cones to show that the solution set Ep is non-empty and weakly closed. In the following, we prove R0 = R1 , and then we show that the Flores-Bazán’s result is true in reflexive Banach spaces. Lemma 3.2. Let K be a non-empty closed convex subset of X . If F : K × K → 2Y \ ∅ is a set-valued map satisfying Assumptions (f0 ), (f1 ), (f2 ) and (f4 ), then R0 = R1 . Proof. Clearly, R0 ⊂ R1 . Conversely, if ν ∈ R1 , then ν ∈ K ∞ and so for all y ∈ K and all λ > 0 we have 0 ∈ F (y, y + λν) + W (W = Y \ (−int C )). By Lemma 3.1, we have F (y + λν, y) ∩ (−int C ) = ∅ for all y ∈ K and λ > 0. This implies that F (y + λν, y) ⊆ W . Assumption (f1 ) yields F (y, y + λν) ⊆ −W for all λ > 0. Now, take any z ∈ K with F (y, z ) ⊆ −C . Given y ∈ K , we put uk := y + kν ∈ K for all k ∈ N, then F (y, uk ) ⊆ −W for all k ∈ N. For any k ∈ N, if we put tk = 1k , then tk uk → ν as k → +∞. For any λ > 0 and sufficiently large k, by C -convexity of F (y, .), we have





0 ∈ F y, (1 − λtk )z + λtk uk + W . Now Assumption (f4 ) implies that 0 ∈ F (y, z + λν) + W . This means ν ∈ R0 . Therefore, R0 = R1 .



Theorem 3.3. Let K be a non-empty closed convex subset of X and F : K × K → 2Y \ ∅ be a set-valued map satisfying Assumptions (f0 ), (f1 ), (f2 ) and (f4 ). Then

(Ep )∞ ⊂ R0 = R1 ⊂

 {x ∈ K : F (x, y) ∩ (−int C ) = ∅}∞ y∈K

 ⊂ {x ∈ K : F (y, x) ∩ (int C ) = ∅}∞ . y∈K

In addition, if there exists x∗ ∈ K such that F (y, x∗ ) ⊆ −C for all y ∈ K , then (Ep )∞ = R0 = R1 . Proof. Put W = Y \ (−int C ). To prove the first inclusion, choose ν ∈ (Ep )∞ . Therefore, there exist tk ↓ 0 and uk ∈ Ep such that tk uk ⇀ ν . For an arbitrary y ∈ K , by Assumption (f1 ), we conclude F (y, uk ) ⊆ −W for all k ∈ N. Now, take any z ∈ K such that F (y, z ) ⊆ −C and fix λ > 0. C -convexity of F (y, .) implies

  (1 − λtk )F (y, z ) + λtk F (y, uk ) ⊆ F y, (1 − λtk )z + λtk uk + C , for any k sufficiently large. Thus





0 ∈ F y, (1 − λtk )z + λtk uk + W . Assumption (f4 ) implies that 0 ∈ F (y, z + λν) + W . Therefore, ν ∈ R0 .

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For the second inclusion, if ν ∈ K ∞ such that 0 ∈ F (y, y + λν) + W for all λ > 0 and all y ∈ K , then Lemma 3.1 implies that F (y + λν, y) ⊆ W for all λ > 0 and all y ∈ K . If we fix y ∈ K and put xk := y + kν ∈ K , then F (xk , y) ⊆ W for all k ∈ N. By choosing tk = 1k for k ∈ N, we have ν ∈ {x ∈ K : F (x, y) ⊆ W }∞ . Since y is arbitrary, the second inclusion is valid. The third inclusion is a consequence of Assumption (f1 ) and Proposition 2.2. Finally we show (Ep )∞ = R0 . First, it is necessary to ensure that there exists x∗ ∈ K such that F (y, x∗ ) ⊆ −C for all y ∈ K . If ν ∈ R0 , then 0 ∈ F (y, x∗ + λν) + W for all λ > 0 and all y ∈ K . Considering Remark 3.2 in [1], we get F (x∗ + λν, y) ⊆ W for all λ > 0 and all y ∈ K , which yields x∗ + λν ∈ Ep . Hence ν ∈ (Ep )∞ and R0 ⊆ (Ep )∞ . Therefore, R0 = R1 = (Ep )∞ .  Ansari and Flores-Bazán [1] proved that the solution sets of GVEP and DGVEP are identical for any non-empty closed convex and bounded subset K . In the following we extend this result for the case that K is not necessarily bounded, and then show that the generalized version of Theorem 3.2 in [1] holds. Lemma 3.4. Let K be a non-empty closed convex subset of X and F : K × K → 2Y \∅ be a set-valued map satisfying Assumptions (f0 ), (f1 ), (f2 ) and (f3 ). Then Ep = Ed .





Proof. By Assumption (f1 ), it is clear that Ep ⊆ Ed . For Ed ⊆ Ep , if x ∈ Ed , then F (y, x) ⊆ − Y \ (−int C ) for all y ∈ K . Let yt := x + t (y − x) for t ∈]0, 1[ then yt ∈ K . C -convexity of F (yt , .) implies that tF (yt , y) + (1 − t )F (yt , x) ⊆ F (yt , yt ) + C . Therefore, F (yt , y) ⊆ Y \ (−int C ). If t ↓ 0, then by Assumption (f3 ) we get F (x, y) ⊆ Y \ (−int C ) and then x ∈ Ep . Thus, Ed ⊆ Ep and hence, Ep = Ed .  By applying Lemma 3.4, the equality case of Theorem 3.3 is investigated in the following. Theorem 3.5. Let K be a non-empty closed convex subset of X and let F : K × K → 2Y \ ∅ be a set-valued map satisfying Hypothesis (H1). If the set {x ∈ K : F (y, x) ∩ (int C ) = ∅} is convex and Ep ̸= ∅, then

(Ep )∞ = R0 = R1 =

 {x ∈ K : F (x, y) ∩ (−int C ) = ∅}∞ y∈K

 {x ∈ K : F (y, x) ∩ (int C ) = ∅}∞ . = y∈K

Proof. The inclusion cases were proved in Theorem 3.3. It is sufficient consider the conclusion cases. The set {x ∈ K : F (y, x) ∩ (int C ) = ∅} is convex and by (f4 ) it is weakly closed. Using Lemma 3.4, we have

  {x ∈ K : F (y, x) ∩ (int C ) = ∅}. {x ∈ K : F (x, y) ∩ (−int C ) = ∅} = y∈K

y∈K

Hence Ed = Ep ̸= ∅. Now parts (a) and (c) of Proposition 2.2 imply that

 ∞  ∞ = {x ∈ K : F (y, x) ∩ (int C ) = ∅} {x ∈ K : F (x, y) ∩ (−int C ) = ∅} y∈K

y∈K

=



{x ∈ K : F (y, x) ∩ (int C ) = ∅}∞ .

y∈K

Therefore,

(Ep )∞ = R0 = R1 =

 {x ∈ K : F (x, y) ∩ (−int C ) = ∅}∞ y∈K

=

 {x ∈ K : F (y, x) ∩ (int C ) = ∅}∞ .  y∈K

Now we consider the following theorem to present the conditions under which the solution set of an unbounded set K becomes weakly compact. Theorem 3.6. Let K be a non-empty closed convex set in X and F : K ×K → 2Y \∅ be a set-valued map satisfying Hypothesis (H1). If there exists x∗ ∈ K such that F (y, x∗ ) ⊆ −C for all y ∈ K , then the following statements are equivalent: (a) Ep is a non-empty and weakly compact. (b) ∃r > 0, ∀x ∈ K \ Kr , ∃y ∈ Kr : F (x, y) ̸⊆ Y \ (−int C ), where Kr = {x ∈ K : ‖x‖ ≤ r } ̸= ∅.

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Proof. (a) ⇒ (b). For any r > 0 suppose that there exists x ∈ K \ Kr such that for all y ∈ Kr , F (x, y) ⊆ W = Y \ (−int C ). Since Ep is bounded, fix s > Supx∈Ep ‖x‖ + 1. Then there exists x ∈ K \ Ks such that F (x, y) ⊆ W for all y ∈ Ks . Consider α ∈]0, 1[ and set z = x + α(x∗ − x), then we have s − 1 ≤ ‖z ‖ < s. It is claimed that F (z , y) ⊆ W for all y ∈ K with ‖y‖ ≤ s. C -convexity of F (y, .) implies that

α F (y, x∗ ) + (1 − α)F (y, x) ⊆ F (y, z ) + C . Thus, 0 ∈ F (y, z ) + W for all y ∈ Ks . Based on Remark 3.2 in [1] we have F (z , y) ⊆ W for all y ∈ K with ‖y‖ ≤ s. It can easily be shown that F (z , y) ⊆ W for all y ∈ K . Hence z ∈ Ep . On the other hand ‖z ‖ ≥ s − 1 > Supx∈Ep ‖x‖, which contradicts to z ∈ Ep . Therefore, the non-emptiness and compactness of Ep implies (b). (b) ⇒ (a). Put W := Y \ (−int C ). Since there exists x∗ ∈ K such that F (y, x∗ ) ⊆ −W for all y ∈ K , so we have 0 ∈ F (y, x∗ ) + W . By Remark 3.2 in [1] we get F (x∗ , y) ⊆ W for all y ∈ K . Then x∗ ∈ Ep , which means Ep is non-empty. Now we prove the weak closedness of Ep . If {xn } is a sequence in Ep such that xn ⇀ x ∈ K , then F (xn , y) ⊆ W for all y ∈ K and all n ∈ N. By Assumptions (f1 ) and (f4 ), 0 ∈ F (y, x) + W , for all y ∈ K . According to Remark 3.2 in [1], we conclude F (x, y) ⊆ W for all y ∈ K . Therefore, x ∈ Ep and hence Ep is weakly closed. It only remains to prove that Ep is bounded. If not, up to a subsequence, we get ‖xk ‖ → ∞ and so there exists n0 ∈ N such that for all k > n0 , xk ∈ K \ Kr . On the other hand, by applying (b) there exists yk ∈ Kr such that F (xk , yk ) ̸⊆ W , this contradicts to the assumption and then Ep is bounded.  Remark 3.7. In [12], another version of Theorem 3.6 was discussed on Euclidean spaces. The non-emptiness of the solution set has been considered in [12] where the given proof is too long. Since the provided assumption that ‘‘there exists x∗ ∈ K such that F (y, x∗ ) ⊆ −C for all y ∈ K ’’ clearly implies non-emptiness of the solution set so no long proof is needed. Note that since the notion of convexity plays an essential role in most areas of mathematical analysis, it is important to make some conditions on set-valued maps and the subset K of X such that the solution set becomes convex. Ansari and Flores-Bazán [1] proved the convexity of Ep when K ⊆ R. Now, we discuss the convexity of Ep in reflexive Banach spaces. Theorem 3.8. Let K be a non-empty closed convex subset of X and F : K × K → 2Y \ ∅ be a set-valued map satisfying Hypothesis (H1). If there exists x∗ ∈ K such that F (y, x∗ ) ⊆ −C for all y ∈ K and the condition (c) ∃r > 0, ∀x ∈ K \ Kr , ∃y ∈ Kr : F (x, y) ̸⊆ Y \ (−int C ), where Kr = {x ∈ K : ‖x‖ ≤ r } ̸= ∅, is satisfied, then Ep is non-empty and linearly bounded. In addition, if we have

∀x, y ∈ K ,

F (x, y) ∩ (−int C ) = ∅ H⇒ F (y, x) ⊆ −C ,

(3.1)

then Ep is convex. Proof. Put W := Y \ (−int C ). The proof of non-emptiness of Ep is the same as that of the previous theorem. It needs to be noted that Condition (c ) implies that R0 = {0}. If not, there exists ν ∈ R0 such that ν ̸= 0. Then, in particular, we have x∗ + λν ∈ K \ Kr for all λ > 0 sufficiently large. By considering Condition (c ) we obtain

∃ξλ ∈ Kr such that F (x∗ + λν, ξλ ) ̸⊆ W .

(3.2)

On the other hand, ν ∈ R0 implies that 0 ∈ F (y, z + λν) + W for all λ > 0 and all y, z ∈ K with F (y, z ) ⊆ −C . Therefore, F (z + λν, y) ⊆ W . Since F (y, x∗ ) ⊆ −C and ξλ ∈ K , we get F (x∗ + λν, ξλ ) ⊆ W which contradicts (3.2). Therefore, R0 = {0}. Considering Theorem 3.3, Ep∞ = {0} and then Ep is linearly bounded. To prove the convexity of Ep , take x1 , x2 ∈ Ep , then

∀y ∈ K ;

F (x1 , y) ⊆ W ,

F (x2 , y) ⊆ W .

By applying (3.1), we have the following

∀y ∈ K ;

F (y, x1 ) ⊆ −C

F (y, x2 ) ⊆ −C .









Now, Fix y ∈ K and α ∈]0, 1[. Since F (y, .) is C -convex so 0 ∈ F y, α x1 +(1−α)x2 +W and then F α x1 +(1−α)x2 , y ⊆ W . Thus α x1 + (1 − α)x2 ∈ Ep and Ep is convex.



Theorem 3.9. Let K be a non-empty closed convex subset of X and F : K × K → 2Y \ ∅ be a set-valued map such that F (., y) : K → 2Y \ ∅ is strictly quasiconcave. Then Ep is convex. Proof. If x1 , x2 ∈ Ep with x1 ̸= x2 , then F (x1 , y) ⊆ W and F (x2 , y) ⊆ W for all y ∈ K , where W = Y \ (−int C ). For any y ∈ K , the strictly quasiconcaveness of F (., y) implies

     −F (x1 , y) − −F (1 − α)x1 + α x2 , y ⊆ int C ,

(3.3)

     −F (x2 , y) − −F (1 − α)x1 + α x2 , y ⊆ int C .

(3.4)

or

I. Sadeqi, C.G. Alizadeh / Nonlinear Analysis 74 (2011) 2226–2234

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2231



Without loss of generality assume that the inclusion (3.3) is satisfied. If u ∈ F (1 − α)x1 + α x2 , y and v ∈ F (x1 , y), then we have v − u ∈ −int C . On the other hand F (x1 , y) ⊆ Y \ (−int C ), then v ̸∈ −int C . We claim that u ̸∈ −int C . If u ∈−int C , then v = v − u + u ∈ −int C − int C ⊆ −int C . It means v ∈ −int C , which contradicts to v ̸∈ −int C . Hence F (1 − α)x1 + α x2 , y convex.

⊆ Y \ (−int C ) and so (1 − α)x1 + α x2 ∈ Ep . This means Ep is



Based on the maximum and minimum principles, most of the smooth functions attain their extremum on some boundary points of a convex set. So, determining, whether the solution set includes some boundary points or not is important. Lemma 3.10. Let K be a non-empty closed convex set in X and F : K ×K → 2Y \∅ be a set-valued map satisfying Hypothesis (H1). If

(f5 ) for any sequence {xk } ∈ K with ‖xk ‖ → ∞, there is n0 ∈ N, u ∈ K such that F (xk , u) ⊆ −C for all k ≥ n0 , then the solution of the problem find x ∈ Kk

such that F (x, y) ⊆ Y \ (−int C ); ∀y ∈ Kk ,

(3.5)

is also a solution of Problem (1.1), where Kk = {x ∈ K : ‖x‖ ≤ k}. Indeed, Ep is non-empty. Proof. From Theorem 3.1 in [1], we conclude that Problem (3.5) has a solution, say xk ∈ Kk for all k ∈ N. If ‖xk ‖ < k for some k ∈ N, then xk is a solution of Problem (1.1) [1]. Now consider the case that ‖xk ‖ = k for all k ∈ N. It may be assumed, up to subsequence, ‖xk ‖ → +∞. By Assumption (f5 ), there exist n0 ∈ N and u ∈ K such that F (xk , u) ⊆ −C for all k ≥ n0 . Since ‖xk ‖ → +∞, there exists, n1 ∈ N such that for all k ≥ n1 , the inequality ‖u‖ < ‖xk ‖ holds. Considering n2 = max{n0 , n1 }, we have ‖u‖ < ‖xk ‖ = k for all k ≥ n2 . It is shown that F (xk , y) ⊆ W for all y ∈ K with ‖y‖ > k. Since ‖u‖ < ‖xk ‖ = k, for k ≥ n2 we can find z ∈]u, y[ such that ‖z ‖ < k. By C -convexity of F (xk , .), we have F (xk , y) ⊆ W . Therefore, each solution to Problem (3.5) is also a solution to Problem (1.1). Hence Ep is non-empty.  Now, we are ready to prove the main theorem. Theorem 3.11. Let K be a non-empty closed convex subset of X not containing the origin. Let F : K × K → 2Y \ ∅ be a setvalued map satisfying Hypothesis (H1) and Assumption (f5 ). Then, there exist some boundary points of K , which are solutions to Problem (1.1). Proof. Let α = inf {r : Kr ̸= ∅}, it is shown that Kα ̸= ∅, i.e., there exists x0 ∈ K such that ‖x0 ‖ ≤ α . Minimum principle states that there exists a sequence {rn } converging to α . Since Krn ̸= ∅, there exists xn ∈ Krn with ‖xn ‖ ≤ rn and so limn→∞ ‖xn ‖ ≤ α . On the other hand, we have ‖xn ‖ ≥ α , and then limn→∞ ‖xn ‖ = α . Since K ⊆ X is non-empty, closed and convex and X is a reflexive Banach space, then by Corollary 9.9.2 in [16], there exists x0 ∈ K such that ‖x0 ‖ = infx∈K ‖x‖. This yields ‖x0 ‖ = infx∈K ‖x‖ ≤ ‖xn ‖ for all n ∈ N. Then ‖x0 ‖ ≤ limn→∞ ‖xn ‖ = α and ‖x0 ‖ ≤ α , where x0 ∈ K . Thus x0 ∈ Kα and so Kα ̸= ∅. Since α = inf {r : Kr ̸= ∅}, we get ‖x0 ‖ = α . By applying Theorem 3.1 in [1], we conclude that Problem (3.5) has a solution. If x ∈ Kα is a solution to Problem (3.5), then according to Lemma 3.10, x will also be a solution to Problem (1.1).   Note that x is a boundary point of K . Since x ∈ K , so x ∈ K . Considering the sequence xn = x 1 − 1n , it follows that

       x 1 − 1  ≤ ‖x‖ 1 −   n 

    = ‖x‖ 1 − 1 < α. n n

1

Thus, xn ∈ K c , because ‖x‖ ≥ α , for all x ∈ K . On the other hand, xn → x; hence x ∈ K c and x is a boundary point of K . Therefore, Ep ∩ ∂ K ̸= ∅. This means that there exists an element in the boundary point of K , which is a solution to Problem (1.1).  4. Some new results on SGVEP In this section we consider the following hypothesis. Hypothesis (H2) ([1]). The set-valued map F : K × K → 2Y \ ∅ is such that

(f0 ) (f1′ ) (f2′ ) (f3′ ) (f4′ )

F (x, x) ⊆ C ∩ (−C ) for all x ∈ K , F (x, y) ⊆ C for all x, y ∈ K , implies F (y, x) ⊆ −C , mapping F (x, .) : K → 2Y \ ∅ is properly C -quasiconvex for all x ∈ K , the set {ξ ∈ [x, y] : F (ξ , y) ⊆ C } is closed for all x, y ∈ K , where [x, y] denotes the closed line segment joining x and y, F (x, .) is weakly lower semicontinuous on K for all x ∈ K .

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Remark 4.1. Let K be a non-empty closed convex subset of X , and F : K × K → 2Y \ ∅ a set-valued map satisfying Assumptions (f0 ), (f1′ ) and (f3′ ) such that for all x ∈ K , F (x, .) is C -convex. Then Esp = Esd . Indeed, if x ∈ Esd then F (y, x) ⊆ −C , for all y ∈ K . Considering yt := x + t (y − x) for t ∈ ]0, 1[, C -convexity of F (yt , .) along with F (yt , x) ⊆ −C , implies F (yt , y) ⊆ C . Now if t ↓ 0, then by Assumption (f3′ ), we conclude F (x, y) ⊆ C and therefore, x ∈ Esp . This means that Esd ⊆ Esp . It can easily be shown that Esp is convex. When the set K is unbounded, we need the following cones. R′0 :=

 {ν ∈ K ∞ : 0 ∈ F (y, z + λν) + C , ∀λ > 0, ∀z ∈ K s.t. F (y, z ) ⊆ −C } y∈K

and R′1 :=

 {ν ∈ K ∞ : 0 ∈ F (y, y + λν) + C , ∀λ > 0}. y∈K

Recall that the cones R′0 and R′1 have already been introduced in [1], and some efficient results have been given. By applying them, the non-emptiness and weak compactness of the solution set are proved. Next we show that R′0 = R′1 . Then, all results obtained for R′1 in [1] hold for R′0 and vice versa. Lemma 4.2. Let K be a non-empty closed convex set, and F : K × K → 2Y \ ∅ be a set-valued map satisfying Assumptions (f0 ) and (f1′ ) such that for all x ∈ K , F (x, .) : K → 2Y \ ∅ is C -convex. Then R′1 =



{ν ∈ K ∞ : 0 ∈ F (y, y + λν) + C , ∀λ > 0} =

 {ν ∈ K ∞ : F (y + λν, y) ∩ (Y \ C ) = ∅, ∀λ > 0}. y∈K

y∈K

Proof. If ν ∈ R′1 , then ν ∈ K ∞ such that 0 ∈ F (y, y + λν) + C for all y ∈ K and all λ > 0. Since F (y + λν, .) is C -convex, then for any y ∈ K and λ > 0, we obtain 1 2

F (y + λν, y + λν + λν) +

1 2

F (y + λν, y) ⊂ F (y + λν, y + λν) + C ⊆ C .

Hence F (y + λν, y) ⊂ C , and then ν ∈ straightforward. 



y∈K

{ν ∈ K ∞ : F (y + λν, y) ∩ (Y \ C ) = ∅, ∀λ > 0}. The converse is

Theorem 4.3. Let K be a non-empty closed convex subset of X , and F : K × K → 2Y \ ∅ be a set-valued map satisfying Assumptions (f0 ), (f1′ ) and (f4′ ) such that for all x ∈ K , F (x, .) : K → 2Y \ ∅ is C -convex. Then R′0 = R′1 . Proof. It is clear that R′0 ⊂ R′1 . Conversely, if ν ∈ R′1 then Lemma 4.2 and Assumption (f1′ ) imply that F (y, y + λν) ⊆ −C . In addition, take any z ∈ K such that F (y, z ) ⊆ −C . For any fixed y ∈ K , put xk := y + kν ∈ K for k ∈ N. Then F (y, xk ) ⊆ −C for all k ∈ N. By choosing tk = 1k for k ∈ N, we get tk xk → ν as k → +∞. For any fixed λ > 0 and k sufficiently large, C -convexity of F (y, .) implies

  (1 − λtk )F (y, z ) + (λtk )F (y, xk ) ⊆ F y, (1 − λtk )z + λtk xk + C , hence we conclude





0 ∈ F y, (1 − λtk )z + λtk xk + C . Assumption (f4′ ) implies that 0 ∈ F (y, z + λν) + C and so ν ∈ R′0 . Therefore, R′0 = R′1 .



Theorem 4.4. Let K be a non-empty closed convex subset of X and F : K × K → 2Y \ ∅ be a set-valued map satisfying Assumptions (f0 ), (f1′ ) and (f4′ ) such that for all x ∈ K , F (x, .) : K → 2Y \ ∅ is C -convex. Then

(Esp )∞ ⊂ R′0 ⊂ R′1 ⊂

 {x ∈ K : F (x, y) ⊆ C }∞ y∈K

⊂

 {x ∈ K : F (y, x) ⊆ −C }∞ . y∈K

In addition, if there exists x∗ ∈ K such that F (y, x∗ ) ̸⊆ inclusions.





Y \ (−int C ) for all y ∈ K , then we have equalities, instead of

I. Sadeqi, C.G. Alizadeh / Nonlinear Analysis 74 (2011) 2226–2234

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Proof. Let ν ∈ (Esp )∞ , then there exists tk ↓ 0 and xk ∈ Esp such that tk xk ⇀ ν . By applying Assumption (f1′ ), we get F (y, xk ) ⊆ −C for all k ∈ N. Take any z ∈ K such that F (y, z ) ⊆ −C . Fixing any λ > 0, for k sufficiently large, C -convexity of F (y, .) implies

  (1 − λtk )F (y, z ) + (λtk )F (y, xk ) ⊆ F y, (1 − λtk )z + λtk xk + C . Hence





0 ∈ F y, (1 − λtk )z + λtk xk + C . From Assumption (f4′ ), we have ν ∈ R′0 . Then the first inclusion is proved. The second inclusion is straightforward. To prove the third inclusion, let ν ∈ K ∞ such that 0 ∈ F (y, y + λν) + C for all λ > 0 and all y ∈ K . By Lemma 4.2, we have F (y + λν, y) ⊆ C . For any fixed y ∈ K , set xk := y + kν ∈ K for all k ∈ N. Then F (xk , y) ⊆ C for all k ∈ N. If we choose tk = 1k for k ∈ N, then ν ∈ {x ∈ K : F (x, y) ⊆ C }∞ , and therefore, the proof of the third inclusion is complete. The fourth inclusion is an immediate consequence of Assumption (f1′ ) and Proposition 2.2. Now we prove the last part of the theorem. There exists x∗ ∈ K such that F (y, x∗ ) ⊆ −int C ⊆ −C for all y ∈ K and so 0 ∈ F (y, x∗ ) + C . We claim that F (x∗ , y) ⊆ C for all y ∈ k. Hence, x∗ ∈ Esp and Esp ̸= ∅. It is easy to see that the set {x ∈ K : F (y, x) ⊆ −C } is convex by C -convexity of F (y, .) and weakly closed by Assumption (f4′ ). Moreover, according to Remark 4.1, we have ∅ ̸= Esp = Esd = ∩y∈K {x ∈ K : F (y, x) ⊆ −C }. Therefore, based on parts (a) and (c) of Proposition 2.2, we obtain

(Esp )∞ = R′0 = R′1 =



{x ∈ K : F (x, y) ⊆ C }∞

y∈K

=



{x ∈ K : F (y, x) ⊆ −C }∞ . 

y∈K

By Remark 4.2 in [1], the authors assumed R′1 ⊆ −R′1 to provide Condition (∗)′ , in the case that Y is finite dimensional. However, it can easily be shown that the latter remains true when Y is not necessarily a finite dimensional space. By applying Condition (f5′ ) in [1], they have also proved that Esp is non-empty and weakly compact. In the following, the same result is proved by dropping (f5′ ). Theorem 4.5. Let K be a non-empty closed convex set in X , and F : K × K → 2Y \ ∅ be a set-valued map satisfying Hypothesis (H2) such that for all x ∈ K , F (x, .) : K → 2Y \ ∅ is C -convex. Then, the following assertions are equivalent: (a) Esp is non-empty and weakly compact. (b) ∃r > 0, ∀x ∈ K \ Kr , ∃y ∈ Kr : F (x, y) ̸⊆ C , where Kr = {x ∈ K : ‖x‖ ≤ r } ̸= ∅. Proof. (a) ⇒ (b). Let for all r > 0, there exists x ∈ K \ Kr such that F (x, y) ⊆ C for all y ∈ Kr . Since Esp is bounded, fix s > Supx∈Esp ‖x‖ + 1. There exists x ∈ K \ Ks such that for all y ∈ Ks , F (x, y) ⊆ C . If x ∈ Esp ̸= ∅, then by Assumption (f1′ ) we have F (y, x) ⊆ −C for all y ∈ K . Take any α ∈ ]0, 1[ and z = x + α(x − x), then s − 1 ≤ ‖z ‖ < s. We show that F (z , y) ⊆ C for all y ∈ Ks . For any y ∈ Ks , C -convexity of F (y, .) implies

α F (y, x) + (1 − α)F (y, x) ⊆ F (y, z ) + C . Thus we conclude 0 ∈ F (y, z ) + C for all y ∈ Ks . Considering Remark 4.1 in [1], we get F (z , y) ⊆ C for all y ∈ Ks . Now, we prove that F (z , y) ⊆ C for all y ∈ K . Let there exists y ∈ K \ Ks satisfying F (z , y) ̸⊆ C . Put z = α y + (1 − α)z for α ∈ ]0, 1[ such that z ∈ Ks . Since F (z , .) is C -convex we derive

α F (z , y) + (1 − α)F (z , z ) ⊆ F (z , z ) + C . Assumption (f0 ) and F (z , z ) ⊆ C imply F (z , y) ⊆ C , which is a contradiction. Hence z ∈ Esp . On the other hand, ‖z ‖ ≥ s − 1 > Supx∈Esp ‖x‖ which is impossible. Therefore, (a) implies (b). (b) ⇒ (a). Let us consider the problem find x ∈ Kn

such that F (x, y) ⊆ C ; ∀y ∈ Kn .

(4.1)

Using Lemma 4.1 in [1], we conclude that Problem (4.1) admits a solution, say xn ∈ Kn for all n ∈ N. Assumption (b) follows that such a sequence is bounded. Therefore, there exists x ∈ K such that xn ⇀ x. By fixing y ∈ K , ∃n0 ∈ N such that ‖y‖ < n0 and so y ∈ Kn0 . It is clear that F (xn , y) ⊆ C for all n ≥ n0 . By Conditions (f1′ ) and (f4′ ) we have F (y, x) ⊆ −C . This proves that x is a solution to Problem (1.3). Hence, Esp ̸= ∅. Condition (b) follows that Esp is bounded. In fact, if not, up to subsequence, we have ‖xn ‖ → ∞. Then xn ∈ K \ Kr and by (b), there exists yn ∈ Kr such that F (xn , yn ) ̸⊆ C . This is impossible, since xn ∈ Esp . Further, the weak closedness of Esp is consequence of (f1′ ) and (f4′ ). Then Esp is weakly compact. 

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Corollary 4.6. Let K ⊆ X be a non-empty closed convex set and F : K × K → 2Y \ ∅ be a set-valued map satisfying Hypothesis (H2). Let F (x, .) : K → 2Y \ ∅ be C -convex for all x ∈ K . If ∃r > 0, ∀x ∈ K \ Kr , ∃y ∈ Kr : F (x, y) ̸⊆ C , where Kr = {x ∈ K : ‖x‖ ≤ r } ̸= ∅, then Esp is the closed convex hull of the set of its extreme points. Proof. Remark 4.1 and Theorem 4.4 imply that Esp is a weakly compact and convex set. By Krein–Milman Theorem [17], Esp is the closed convex hull of the set of its extreme points.  Acknowledgements The authors would like to thank the Editor and the anonymous referees for bringing up very good points to be considered in the revision. Also we would like to thank Prof. K. Ghanbari for valuable comments. References [1] Q.H. Ansari, F. Flores-Bazán, Recession methods for generalized vector equilibrium problems, J. Math. Anal. Appl. 321 (2006) 132–146. [2] Q.H. Ansari, I.V. Konnov, J.C. Yao, On generalized vector equilibrium problems, Nonlinear Anal. 47 (2001) 543–554. [3] Q.H. Ansari, I.V. Konnov, J.C. Yao, Existence of a solution and variational principles for vector equilibrium problems, J. Optim. Theory Appl. 110 (2001) 481–492. [4] Q.H. Ansari, I.V. Konnov, J.C. Yao, Characterizations of solutions for vector equilibrium problems, J. Optim. Theory Appl. 113 (2002) 435–447. [5] Q.H. Ansari, A.H. Siddiqi, S.Y. Wu, Existence and duality of generalized vector equilibrium problems, J. Math. Anal. Appl. 259 (2001) 115–126. [6] Q.H. Ansari, J.C. Yao, An existence result for the generalized vector equilibrium problem, Appl. Math. Lett. 12 (1999) 53–56. [7] F. Flores-Bazán, Ideal, weakly efficient solutions for vector optimization problems, Math. Program. 93 (2002) 453–475. [8] I.V. Konnov, J.C. Yao, Existence of solutions for generalized vector equilibrium problems, J. Math. Anal. Appl. 233 (1999) 328–335. [9] Q.H. Ansari, W. Oettli, D. Schläger, A generalization of vectorial equilibria, Math. Methods Oper. Res. 46 (1997) 147–152. [10] M. Bianchi, N. Hadjisavvas, S. Schaible, Vector equilibrium problems with generalized monotone bifunctions, J. Optim. Theory Appl. 92 (1997) 527–542. [11] F. Flores-Bazán, F. Flores-Bazán, Vector equilibrium problems under asymptotic analysis, J. Global Optim. 26 (2003) 141–166. [12] G.M. Lee, I.J. Bu, On vector equilibrium problems with multifunctions, Taiwanese J. Math. 10 (2006) 399–407. [13] B.S. Mordukhovich, Variational Analysis and Generalized Differentiation, I: Basic Theory, in: Grundlehren Series (Fundamental Principles of Mathematical Sciences), vol. 330, Springer, Berlin, 2006. [14] B.S. Mordukhovich, Variational Analysis and Generalized Differentiation, II: Applications, in: Grundlehren Series (Fundamental Principles of Mathematical Sciences), vol. 331, Springer, Berlin, 2006. [15] W. Oettli, D. Schläger, Existence of equilibria for monotone multivalued mappings, Math. Methods Oper. Res. 48 (1998) 219–228. [16] R. Larsen, Functional Analysis, Macel Dekker, Inc., New York, 1973. [17] W. Rudin, Functional Analysis, McGraw-Hill, 1990.