Optimality Conditions of Set-Valued Optimization Problem Involving Relative Algebraic Interior in Ordered Linear Spaces

Optimality conditions of set-valued optimization problem involving relative algebraic interior in ordered linear spaces

Zhi-Ang Zhoua, Xin-Min Yangb and Jian-Wen Pengc1

aDepartment of Applied Mathematics, Chongqing University of Technology, Chongqing

400054, P.R. China; bSchool of Mathematics, Chongqing Normal University,

Chongqing 400047, P.R. China; cSchool of Mathematics, Chongqing

Normal University, Chongqing 400047, P.R. China

In this paper, ﬁrstly, a generalized subconvexlike set-valued map involving the relative algebraic interior is introduced in ordered linear spaces. Secondly, some properties of a generalized subconvexlike set-valued map are investigated. Finally, the optimality conditions of set-valued optimization problem are established.

Keywords Relative algebraic interior · Generalized cone subconvexlikeness · Set-valued map· Separation property· Optimality condition

AMS 2010 Subject Classiﬁcations: 90C26, 90C29, 90C30

1 Introduction

Recently, many authors have been interested in vector optimization problems with set-valued maps. For example, Rong and Wu [11] gave characterizations of super eﬃciency for vector

1Corresponding author. E-mail: [email protected]

1 optimization with cone convexlike set-valued maps. Li [2] established an alternative theorem for cone subconvexlike set-valued maps and Kuhn-Tucker conditions for vector optimization problems with cone subconvexlike set-valued maps. Yang et al. [12] established an alternative theorem for the generalized cone subconvexlike set-valued maps. Yang et al. [13] established an alternative theorem for the nearly cone subconvexlike set-valued maps. We note that the results in [2] and [11-13] are established in linear topological spaces. It is well-known that linear spaces are wider than linear topological spaces. Hence, it is natural to consider the following interesting and meaningful problem: How to generalize those results in [2, 12, 13] from linear topological spaces to linear spaces. Li [1] generalized those results in [2] from linear topological spaces to linear spaces. Huang and Li [3] also generalized the results obtained in

[1] from cone subconvexlikeness case to generalized cone-subconvexlikeness case. Note that the results in [1] and [3] were established under the condition that the algebraic interior of ordered cone C denoted by cor(C) is nonempty. However, in some optimization problems, it is possible that cor(C) = ∅. In order to overcome this ﬂaw, the authors in [15-16] introduced the notion of relative algebraic interiors in linear spaces. It is worth noting that the algebraic interior of a set C is the subset of the relative algebraic interior of C. Thus, the notion of the relative algebraic interior generalizes that of the algebraic interior. Ad´anand Novo [5-

8] investigated weak or proper eﬃciency of vector optimization problems with generalized convex set-valued maps involving relative algebraic interior and vector closure of C in linear spaces. Hern´andezet al. [9] introduced a cone subconvexlike set-valued map and established optimality conditions and Lagrangian multiplier rule involving relative algebraic interior of C in linear spaces.

This paper is organized as follows. In section 2, we give some preliminaries. In section

3, the new notion of generalized cone subconvexlike set-valued map involving the relative

2 algebraic interior of C is introduced, and some properties of the generalized cone subconvexlike set-valued maps are investigated. In section 4, a separation property is obtained, and the optimality conditions of set-valued optimization problem are established. The results in this paper generalize some known results in some literature.

2 Preliminaries

Let X be a nonempty set, and let Y and Z be two ordered linear spaces. Let 0 stand for the zero element of every space. Let K be a nonempty subset in Y. The generated cone of K is deﬁned as coneK = {λa|a ∈ K, λ ≥ 0}. A cone K ⊆ Y is said to be pointed if

K ∩ (−K) = {0}. A cone K ⊆ Y is said to be nontrivial if K ≠ {0} and K ≠ Y.

Let Y ∗ and Z∗ stand for algebraic dual spaces in Y and Z, respectively. From now on, let C and D be nontrivial pointed convex cones in Y and Z, respectively. The algebraic dual cone C+ and strictly algebraic dual cone C+i of C are deﬁned as

C+ = {y∗ ∈ Y ∗|⟨y, y∗⟩ > 0, ∀y ∈ C} and

C+i = {y∗ ∈ Y ∗|⟨y, y∗⟩ > 0, ∀y ∈ C \{0}}, where ⟨y, y∗⟩ denotes the value of the linear functional y∗ at the point y. The meanings of

D+ and D+i are similar.

Let K be a nonempty subset in Y . We denote by aff(K), span(K) and L(K) = span(K −

K) the aﬃne hull, linear hull and generated linear subspace of K, respectively.

Deﬁnition 2.1 [14] Let K be a nonempty subset in Y . The algebraic interior of K is the set

cor(K) = {k ∈ K|∀v ∈ Y, ∃λ0 > 0, ∀λ ∈ [0, λ0], k + λv ∈ K}.

3 Deﬁnition 2.2 [15,16] Let K be a nonempty subset in Y . The relative algebraic interior of K is the set

icr(K) = {k ∈ K|∀v ∈ L(K), ∃λ0 > 0, ∀λ ∈ [0, λ0], k + λv ∈ K}.

LEMMA 2.1 Let K be a nonempty subset in Y . Then

aff(K) = x + L(K), ∀x ∈ K. (1)

Proof Firstly, we will show that

aff(K) ⊆ x + L(K), ∀x ∈ K. (2)

∑n Let y ∈ aff(K). Then, there exist ki ∈ K, αi ∈ R with αi = 1 such that i=1

∑n y = αiki. i=1

Thus, we obtain

∑n ∑n ∑n y = x + y − x = x + αiki − αix = x + αi(ki − x) ∈ x + span(K − K) = x + L(K). i=1 i=1 i=1

Therefore, (2) holds.

Finally, we will show that

x + L(K) ⊆ aff(K), ∀x ∈ K. (3)

Let z ∈ x + L(K). Then, there exist xi, yi ∈ K and λi ∈ R(i = 1, ··· , n) such that

∑n ∑n ∑n z = x + λi(xi − yi) = x + λixi − λiyi. (4) i=1 i=1 i=1

Clearly, ∑n ∑n 1 + λi − λi = 1. (5) i=1 i=1

4 It follows from (4) and (5) that z ∈ aff(K). Therefore, (3) holds. It follows from (2) and (3) that (1) holds.

Remark 2.1 It follows from Lemma 2.1 that

icr(K) = {k ∈ K|∀v ∈ aff(K) − k, ∃λ0 > 0, ∀λ ∈ [0, λ0], k + λv ∈ K}.

Remark 2.2 It follows from Lemma 2.1 that if 0 ∈ K ⊆ Y, then

icr(K) = {k ∈ K|∀v ∈ aff(K), ∃λ0 > 0, ∀λ ∈ [0, λ0], k + λv ∈ K}.

Remark 2.3 If K be a nontrivial pointed cone in Y then 0 ∈/ icr(K). In fact, if 0 ∈ icr(K), then it follows from Remark 2.2 that

aff(K) = K. (6)

Since 0 ∈ K, it is clear that

aff(K) = span(K). (7)

By (6) and (7), we have

span(K) = K. (8)

Since K is nontrivial, by (8), there exists a nonzero k ∈ K such that −k ∈ K, which contradicts that K is pointed.

Remark 2.4 It is easy to check that icr(K) is a convex set and icr(K) ∪ {0} is a convex cone if K is a convex cone.

LEMMA 2.2 Let K be a convex cone in Y. Then

K + icr(K) = icr(K). (9)

Proof Clearly, (9) holds if icr(K) = ∅.

5 Now, let k1 ∈ K and k2 ∈ icr(K). k2 ∈ icr(K) implies that k2 ∈ K, ∀v ∈ L(K), ∃λ0 >

0, ∀λ ∈ [0, λ0], we have k2 + λv ∈ K. Since K is a convex cone, we obtain

(k1 + k2) + λv = k1 + (k2 + λv) ∈ K + K ⊆ K,

which implies k1 + k2 ∈ icr(K). Therefore,

K + icr(K) ⊆ icr(K). (10)

Since 0 ∈ K, we have

icr(K) ⊆ K + icr(K). (11)

According to (10) and (11), (9) holds.

LEMMA 2.3 Let K1 and K2 be two nonempty subsets in Y . Then

aff(K1 + K2) = aff(K1) + aff(K2). (12)

Proof Clearly,

aff(K1 + K2) ⊆ aff(K1) + aff(K2). (13)

Now, we will show that

aff(K1) + aff(K2) ⊆ aff(K1 + K2). (14)

∈ i ∈ j ∈ i ∈ j ∈ Let y aff(K1) + aff(K2). Then, there exist k1 K1, k2 K2, λ1 R, λ2 R(i =

∑m1 ∑m2 ··· ··· i j 1, m1; j = 1, m2) with λ1 = 1 and λ2 = 1 such that i=1 j=1

∑m1 ∑m2 ∑m1 ∑m2 i i j j i j y = λ1k1 + λ2k2 = λij(k1 + k2), i=1 j=1 i=1 j=1

i j ··· ··· where λij = λ1λ2(i = 1, , m1; j = 1, , m2).

6 ∑m1 ∑m2 i j ∈ ··· ··· It is easy to check that λij = 1 and k1 +k2 K1 +K2(i = 1, , m1; j = 1, , m2). i=1 j=1

Hence, y ∈ aff(K1+K2). Thus, (14) holds. By (13) and (14), (12) holds.

LEMMA 2.4 Let K be a nonempty subset in Y . Then

(a) aff(K) = k + aff(K − K), ∀k ∈ K.

If K is convex and icr(K) ≠ ∅, then

(b) icr(icr(K)) = icr(K);

Proof (b) and (c) can be found in [5] and [9], respectively. We will only prove (a). Clearly, it follows from that Lemma 2.3 that

aff(K) − k ⊆ aff(K) − aff(K) ⊆ aff(K − K), ∀k ∈ K, i.e.,

aff(K) ⊆ k + aff(K − K), ∀k ∈ K. (15)

Finally, we will show that

k + aff(K − K) ⊆ aff(K), ∀k ∈ K. (16)

By Lemma 2.3, we only need to show that

k + aff(K) − aff(K) ⊆ aff(K), ∀k ∈ K. (17)

∈ − i ∈ j ∈ i ∈ j ∈ Let y k + aff(K) aff(K). Then, there exist k1 K, k2 K, λ1 R, λ2 R(i =

∑m1 ∑m2 ··· ··· i j 1, , m1; j = 1, , m2) with λ1 = 1 and λ2 = 1 such that i=1 j=1

∑m1 ∑m2 i i − j j y = k + λ1k1 λ2k2. (18) i=1 j=1

7 Clearly, ∑m1 ∑m2 i − j 1 + λ1 λ2 = 1. (19) i=1 j=1 By (18) and (19), y ∈ aff(K). Thus, (17) holds. Hence, (16) holds. By (15) and (16), we obtain

aff(K) = k + aff(K − K), ∀k ∈ K.

LEMMA 2.5 Let K1 and K2 be two nonempty subsets in Y . Then

aff(K1 × K2) = aff(K1) × aff(K2). (20)

Proof Clearly,

aff(K1 × K2) = aff(K1 × {0} + {0} × K2). (21)

It follows from Lemma 2.3 that

aff(K1 × {0} + {0} × K2) = aff(K1 × {0}) + aff({0} × K2)

= aff(K1) × {0} + {0} × aff(K2)

= aff(K1) × aff(K2) (22)

By (21) and (22), (20) holds.

LEMMA 2.6 Let K1 and K2 be two nontrivial convex cone in Y and Z. If icr(K1) ≠ ∅ and icr(K2) ≠ ∅, then

icr(K1 × K2) = icr(K1) × icr(K2). (23)

Proof Firstly, we will show that

icr(K1) × icr(K2) ⊆ icr(K1 × K2). (24)

Let (k1, k2) ∈ icr(K1) × icr(K2). Clearly, (k1, k2) ∈ K1 × K2. Let (v1, v2) ∈ aff(K1 × K2) −

(k1, k2). By Lemma 2.5, (v1, v2) ∈ (aff(K1) − k1) × (aff(K2) − k2). Since k1 ∈ icr(K1), for

8 ∈ − 1 v1 aff(K1) k1, there exists ε0 > 0 such that

∈ ∀ ∈ 1 k1 + εv1 K1, ε [0, ε0]. (25)

∈ ∈ − 2 Since k2 icr(K2), for v2 aff(K2) k2, there exists ε0 > 0 such that

∈ ∀ ∈ 2 k2 + εv2 K2,, ε [0, ε0]. (26)

3 { 1 2} 3 Choose ε0 = min ε0, ε0 . It follows from (25) and (26) that there exists ε0 > 0 such that

∈ × ∀ ∈ 3 (k1, k2) + ε(v1, v2) K1 K2, ε [0, ε0], which implies

(k1, k2) ∈ icr(K1 × K2).

Thus, (24) holds.

Finally, we will show that

icr(K1 × K2) ⊆ icr(K1) × icr(K2). (27)

Since icr(K1) ≠ ∅ and icr(K2) ≠ ∅, it follows from (24) that icr(K1 × K2) ≠ ∅. Let (k1, k2) ∈ icr(K1 × K2). Clearly, (k1, k2) ∈ K1 × K2. Let (v1, v2) ∈ (aff(K1) − k1) × (aff(K2) − k2).

Clearly, (v1, v2) ∈ aff(K1 × K2) − (k1, k2). Therefore, there exists ε0 > 0 such that

(k1, k2) + ε(v1, v2) ⊆ K1 × K2, ∀ε ∈ [0, ε0], which implies

k1 + εv1 ⊆ K1 (28) and

k2 + εv2 ⊆ K2. (29)

9 By (28) and (29), we have

(k1, k2) ∈ icr(K1) × icr(K2).

Therefore, (27) holds. By (24) and (27), (23) holds.

LEMMA 2.7 [15,16] Let K be a convex set with icr(K) ≠ ∅ in Y. If 0 ∈/ icr(K), then there exists y∗ ∈ Y ∗ \{0} such that

⟨k, y∗⟩ ≥ 0, ∀k ∈ K.

3 Generalized subconvexlike set-valued map

In this section, ﬁrstly, we will introduce several classes of generalized convex set-valued maps.

Secondly, we will discuss their relationships. Finally, we will obtain some properties of the generalized cone subconvexlike set-valued maps.

From now on, let A be a nonempty subset in X, and let F : A → 2Y be a set-valued map. ∪ Write F (A) = F (x). We suppose that icr(C) × icr(D) ≠ ∅. x∈A Y Deﬁnition 3.1 [11] A set-valued map F : A → 2 is called C-convexlike if, ∀x1, x2 ∈ A and

∀λ ∈ (0, 1),

λF (x1) + (1 − λ)F (x2) ⊆ F (A) + C,

Remark 3.1 It follows from [11] that F : A → 2Y is C-convexlike if and only if F (A) + C is a convex set.

Deﬁnition 3.2 [9] A set-valued map F : A → 2Y is called C-subconvexlike if, ∃c ∈ icr(C) such that, ∀x1, x2 ∈ A, ∀λ ∈ (0, 1), ∀ε > 0,

εc + λF (x1) + (1 − λ)F (x2) ⊆ F (A) + C.

Remark 3.2 Proposition 3.2 of [9] shows that F : A → 2Y is C-subconvexlike if and only if

10 F (A) + icr(C) is a convex set.

Deﬁnition 3.3 A set-valued map F : A → 2Y is called generalized C-subconvexlike if cone(F (A)) + icr(C) is a convex set.

THEOREM 3.1 If the set-valued map F : A → 2Y is C-convexlike, then F is C-subconvexlike.

Proof It follows from the C-convexlikeness of F that F (A) + C is a convex set. We need to prove that F (A) + icr(C) is a convex set. Indeed, let mi ∈ F (A) + icr(C)(i = 1, 2), λ ∈ (0, 1).

Then, there exist xi ∈ A, ci ∈ icr(C)(i = 1, 2) such that

mi = yi + ci, yi ∈ F (xi), i = 1, 2.

By Remark 3.1 and Lemma 2.2,

λm1 + (1 − λ)m2 = λ(y1 + c1) + (1 − λ)(y2 + c2) = [λy1 + (1 − λ)y2] + [λc1 + (1 − λ)c2]

∈ (F (A) + C) + icr(C) = F (A) + (C + icr(C)) = F (A) + icr(C)

Remark 3.3 The following example shows that a C-subconvexlike set-valued map may not be

C-convexlike. Thus, the C-subconvexlikeness is a generalization of C-convexlikeness.

2 Example 3.1 Let X = Y = R ,C = {(y1, 0)|y1 ≥ 0} and A = {(1, 0), (0, 2)}. The set-valued map F : A → 2Y is deﬁned as follows:

F (1, 0) = {(y1, y2)|1 < y1 ≤ 2, 0 ≤ y2 ≤ 1} ∪ {(1, 0), (1, 1)};

F (0, 2) = {(y1, y2)|1 < y1 ≤ 2, 1 ≤ y2 ≤ 2} ∪ {(1, 2), (1, 1)}.

Clearly, F (A) + icr(C) is a convex set, and F (A) + C is not a convex set. Therefore, F is

C-subconvexlike. However, F is not C-convexlike.

THEOREM 3.2 If the set-valued map F : A → 2Y is C-subconvexlike, then F is generalized

C-subconvexlike.

Proof It follows from the C-subconvexlikeness of F that F (A)+icr(C) is a convex set. We need

11 to prove that cone(F (A)) + icr(C) is a convex set. Indeed, let mi ∈ cone(F (A)) + icr(C)(i =

1, 2), λ ∈ (0, 1). Then, there exist ρi ≥ 0, xi ∈ A, ci ∈ icr(C)(i = 1, 2) such that

mi = ρiyi + ci, yi ∈ F (xi), i = 1, 2.

Case 1. ρ1 = 0 or ρ2 = 0. Clearly, λm1 + (1 − λ)m2 ∈ cone(F (A)) + icr(C).

Case 2. ρ1 > 0 and ρ2 > 0. Since F (A) + icr(C) is a convex set, we have

1 1 λm1 + (1 − λ)m2 = λ(ρ1y1 + c1) + (1 − λ)(ρ2y2 + c2) = λρ1(y1 + c1) + (1 − λ)ρ2(y2 + c2) ρ1 ρ2

λρ1 1 (1−λ)ρ2 1 = [λρ1 + (1 − λ)ρ2]{ (y1 + c1) + (y2 + c2)} λρ1+(1−λ)ρ2 ρ1 λρ1+(1−λ)ρ2 ρ2

∈ [λρ1 + (1 − λ)ρ2](F (A) + icr(C)) ⊆ cone(F (A)) + icr(C).

Case 1 and Case 2 show that cone(F (A)) + icr(C) is a convex set.

Remark 3.4 The following example shows that a generalized C-subconvexlike set-valued map may be not C-subconvexlike. Thus, generalized C-subconvexlikeness is a generalization of

C-subconvexlikeness.

2 Example 3.2 Let X = Y = R ,C = {(y1, 0)|y1 ≥ 0} and A = {(1, 0), (0, 2)}. The set-valued map F : A → 2Y is deﬁned as follows:

F (1, 0) = {(y1, y2)|y2 ≤ −y1 + 2, y1 ≥ 0, y2 ≥ 1};

F (0, 2) = {(y1, y2)|y2 ≤ −y1 + 2, y1 ≥ 1, y2 ≥ 0}.

Clearly, cone(F (A))+icr(C) is a convex set, and F (A)+icr(C) is not a convex set. Therefore,

F is generalized C-subconvexlike. However, F is not C-subconvexlike.

Next, we give some equivalent properties of generalized C-subconvexlike set-valued maps.

THEOREM 3.3 A set-valued map F : A → 2Y is generalized C-subconvexlike if and only if,

∀c ∈ icr(C), ∀x1, x2 ∈ A, ∀λ ∈ (0, 1),

c + λF (x1) + (1 − λ)F (x2) ⊆ cone(F (A)) + icr(C). (30)

12 Proof Necessity. Let c ∈ icr(C), x1, x2 ∈ A, y1 ∈ F (x1), y2 ∈ F (x2) and λ ∈ (0, 1). Clearly,

y1 + c ∈ cone(F (A)) + icr(C) (31) and

y2 + c ∈ cone(F (A)) + icr(C). (32)

Since F is generalized C-subconvexlike, it follows from (31) and (32) that

c + λy1 + (1 − λ)y2 = λ(y1 + c) + (1 − λ)(y2 + c) ∈ cone(F (A)) + icr(C).

Therefore, (30) holds.

Suﬃciency. Let mi ∈ cone(F (A)) + icr(C)(i = 1, 2), λ ∈ (0, 1). Then, there exist ρi ≥

0, xi ∈ A, ci ∈ icr(C)(i = 1, 2) such that

mi = ρiyi + ci, yi ∈ F (xi), i = 1, 2.

Case 1. ρ1 = 0 or ρ2 = 0. Clearly, λm1 + (1 − λ)m2 ∈ cone(F (A)) + icr(C).

Case 2. ρ1 > 0 and ρ2 > 0. Since F (A) + icr(C) is a convex set, we have

λm1 + (1 − λ)m2 = λ(ρ1y1 + c1) + (1 − λ)(ρ2y2 + c2) = [λc1 + (1 − λ)c2] + [λρ1y1 + (1 − λ)ρ2y2]

1 λρ1 (1 − λ)ρ2 = [λρ1 +(1−λ)ρ2]{ [λc1 +(1−λ)c2]+ y1 + y2}. λρ1 + (1 − λ)ρ2 λρ1 + (1 − λ)ρ2 λρ1 + (1 − λ)ρ2 (33)

Clearly, 1 [λc1 + (1 − λ)c2] ∈ icr(C), λρ1 + (1 − λ)ρ2 λρ 1 ∈ (0, 1), λρ1 + (1 − λ)ρ2 (1 − λ)ρ 2 ∈ (0, 1) λρ1 + (1 − λ)ρ2

13 and λρ (1 − λ)ρ 1 + 2 = 1. λρ1 + (1 − λ)ρ2 λρ1 + (1 − λ)ρ2 By (30) and (33), we obtain

λm1 + (1 − λ)m2 ∈ [λρ1 + (1 − λ)ρ2](cone(F (A)) + icr(C)) ⊆ cone(F (A)) + icr(C).

Case 1 and Case 2 shows that cone(F (A))+icr(C) is a convex set. Therefore, F is generalized

C-subconvexlike.

THEOREM 3.4 The following statements are equivalent:

(a) F : A → 2Y is generalized C-subconvexlike;

(b) ∀c ∈ icr(C), ∀x1, x2 ∈ A, ∀λ ∈ (0, 1),

c + λF (x1) + (1 − λ)F (x2) ⊆ cone(F (A)) + icr(C);

′ (c) ∃c ∈ icr(C), ∀x1, x2 ∈ A, ∀λ ∈ (0, 1), ∀ε > 0,

′ εc + λF (x1) + (1 − λ)F (x2) ⊆ cone(F (A)) + C;

′′ (d) ∀x1, x2 ∈ A, ∀λ ∈ (0, 1), ∃c ∈ C, ∀ε > 0,

′′ εc + λF (x1) + (1 − λ)F (x2) ⊆ cone(F (A)) + C.

Proof By Theorem 3.3, (a) ⇔ (b). The implications (b) ⇒ (c) ⇒ (d) are clear. Therefore, we

′′ need to prove that (d) ⇒ (b). Let c ∈ icr(C), x1, x2 ∈ A and λ ∈ (0, 1). Then, ∃c ∈ C, ∀ε > 0,

′′ εc + λF (x1) + (1 − λ)F (x2) ⊆ cone(F (A)) + C. (34)

Since c ∈ icr(C) = icr(icr(C)), by Lemma 2.4 (a) and (c), for −c′′ = 0 − c′′ ∈ C − C ⊆ aff(C − C) = aff(C) − c = aff(icr(C)) − c, there exists λ0 > 0 such that

′′ c + λ0(−c ) ∈ icr(C).

14 ′′ ′′ Write c = c + λ0(−c ). Clearly, c = c + λ0c . By (34) and Lemma 2.2, we obtain

′′ ′′ c+λF (x1)+(1−λ)F (x2) = (c+λ0c )+λF (x1)+(1−λ)F (x2) = [λ0c +λF (x1)+(1−λ)F (x2)]+c

⊆ cone(F (A)) + C + c ⊆ cone(F (A)) + C + icr(C) = cone(F (A)) + icr(C).

4 Optimality conditions

In order to establish optimality conditions for vector optimization problems with set-valued maps, we need to present a separation property for a generalized C-subconvexlike set-valued map. Now, we consider the following two systems:

System 1: There exists x0 ∈ A such that F (x0) ∩ (−icr(C)) ≠ ∅;

System 2: There exists y∗ ∈ C+ \{0} such that ⟨y, y∗⟩ ≥ 0, ∀y ∈ F (A).

THEOREM 4.1 (i) Suppose that F : A → 2Y is generalized C-subconvexlike and icr(coneF (A)+ icr(C)) ≠ ∅. If System 1 has no solutions, then System 2 has a solution.

(ii) If y∗ ∈ C+i is a solution of System 2, then System 1 has no solutions.

Proof (i) Firstly, we assert that 0 ∈/ cone(F (A))+icr(C). Otherwise, there exist x0 ∈ A, α ≥ 0 such that 0 ∈ αF (x0) + icr(C).

Case 1. If α = 0, then 0 ∈ icr(C), which contradicts 0 ∈/ icr(C) (see Remark 2.3).

∈ − ∈ 1 ⊆ Case 2. If α > 0, there exists y0 F (x0) such that y0 α icr(C) icr(C), which contradicts that System 1 has no solutions.

Thus, we obtain

0 ∈/ cone(F (A)) + icr(C).

15 Since F is generalized C-subconvexlike, cone(F (A)) + icr(C) is a convex set. By Lemma 2.7, there exists y∗ ∈ Y ∗ \{0} such that

⟨y, y∗⟩ ≥ 0, ∀y ∈ cone(F (A)) + icr(C).

So,

⟨αF (x) + c, y∗⟩ ≥ 0, ∀x ∈ A, ∀c ∈ icr(C), ∀α ≥ 0. (35)

Letting α = 0 in (35), we obtain

⟨c, y∗⟩ ≥ 0, ∀c ∈ icr(C). (36)

We will show that y∗ ∈ C+. Otherwise, there exists c′ ∈ C such that ⟨c′, y∗⟩ < 0. Hence,

⟨θc′, y∗⟩ < 0, ∀θ > 0. By Lemma 2.2, we have

θc′ + c ∈ icr(C), ∀θ > 0, ∀c ∈ icr(C). (37)

It follows from (36) and (37) that

θ⟨c′, y∗⟩ + ⟨c, y∗⟩ ≥ 0, ∀θ > 0, ∀c ∈ icr(C). (38)

∗ − ⟨c,y ⟩ ≥ ∗ ∈ + We note that (38) does not hold when θ > ⟨c′,y∗⟩ 0. Therefore, y C . Letting α = 1 in (35), we have

⟨F (x) + c, y∗⟩ ≥ 0, ∀x ∈ A, ∀c ∈ icr(C).

Letting c0 ∈ icr(C) and λn > 0 with lim λn = 0, we have n→∞

∗ ⟨F (x) + λnc0, y ⟩ ≥ 0, ∀x ∈ A, ∀n ∈ N. (39)

It follows from (39) that ⟨F (x), y∗⟩ ≥ 0, ∀x ∈ A.

16 (ii) Since y∗ ∈ C+i is a solution of System 2, we have

⟨y, y∗⟩ ≥ 0, ∀y ∈ F (A). (40)

Now, we suppose that System 1 has a solution. Then, there exists x0 ∈ A such that F (x0) ∩

(−icr(C)) ≠ ∅. Thus, there exists y0 ∈ F (x0) such that −y0 ∈ icr(C). Clearly, −y0 ≠ 0. So, we

∗ have ⟨y0, y ⟩ < 0, which contradicts (40).

Remark 4.1 In (i), the condition icr(cone(F (A)) + icr(C)) ≠ ∅ can be replaced by the condition aff(C)) = aff(cone(F (A)) + icr(C)). In fact, when aff(C) = aff(cone(F (A)) + icr(C)), icr(C) ⊆ icr(cone(F (A)) + icr(C)). Since icr(C) ≠ ∅, icr(cone(F (A)) + icr(C)) ≠ ∅.

However, the following example shows that the condition icr(cone(F (A)) + icr(C)) ≠ ∅ is weaker than the condition aff(C) = aff(cone(F (A)) + icr(C)).

Example 4.1 In Example 3.2, it is clear that icr(cone(F (A))+icr(C)) = {(y1, y2)|y1 > 0, y2 ≥

2 0} ̸= ∅. However, aff(C) = {(y1, y2)|y1 ∈ R, y2 = 0} ̸= R = aff(cone(F (A)) + icr(C)).

Remark 4.2 Since cone(F (A)) + icr(C) is a nonempty convex subset in Y, The condition icr(cone(F (A)) + icr(C)) ≠ ∅ can be deleted if Y is a ﬁnite-dimensional space.

Remark 4.3 Theorem 4.1 improves Theorem 2.1 of [1], Theorem 3.5 of [5] and Theorem 3.9 of [9].

Let F : A → 2Y and G : A → 2Z be two set-valued maps. Now, we consider the following vector optimization problem with set-valued maps:    min F (x) (VP)  s.t. − G(x) ∩ D ≠ ∅.

The feasible set of (VP) is denoted by S = {x ∈ A| − G(x) ∩ D ≠ ∅}. Now, we deﬁne

W Min(F (A),C) = {y0 ∈ F (A)|(y0 − F (A)) ∩ icr(C) = ∅}.

17 Deﬁnition 4.1 A point x0 is called a weakly eﬃcient solution of (VP) if there exists x0 ∈ S such that F (x0) ∩ W Min(F (S),C) ≠ ∅. A point pair (x0, y0) is called a weak minimizer of

(VP) if y0 ∈ F (x0) ∩ W Min(F (S),C).

Let I(x) = F (x) × G(x), ∀x ∈ A. It is clear that I is a set-valued map from A to Y × Z, where Y × Z is a linear space with nontrivial pointed convex cone C × D. The algebraic dual space of Y × Z is Y ∗ × Z∗, and the algebraic dual cone of C × D is C+ × D+.

By Deﬁnition 3.3, we say that the set-valued map I : A → 2Y ×Z is generalized C × D- subconvexlike if cone(I(A)) + icr(C × D) is a convex set in Y × Z.

Now we present a necessary optimality condition for (VP) as follows:

THEOREM 4.2 Suppose that the following conditions hold:

(i) (x0, y0) is a weak minimizer of (VP);

Y ×Z (ii) I1 : A → 2 is generalized C ×D-subconvexlike, where I1(x) = (F (x)−y0)×G(x), ∀x ∈

(iii) icr(cone(I1(A)) + icr(C × D)) ≠ ∅.

Then, there exists (y∗, z∗) ∈ C+ × D+ with (y∗, z∗) ≠ (0, 0) such that

∗ ∗ ∗ inf (⟨F (x), y ⟩ + ⟨G(x), z ⟩) = ⟨y0, y ⟩, x∈A

∗ inf⟨G(x0), z ⟩ = 0.

Proof According to Deﬁnition 4.1, we have

(y0 − F (S)) ∩ icr(C) = ∅. (41)

Clearly, I1(x) = I(x) − (y0, 0), ∀x ∈ A. We assert that

−I1(x) ∩ icr(C × D) = ∅, ∀x ∈ A. (42)

18 Otherwise, there exists x ∈ A such that

−I1(x) ∩ icr(C × D) ≠ ∅.

By Lemma 2.6, icr(C × D) = icr(C) × icr(D). Therefore,

−I1(x) ∩ (icr(C) × icr(D)) ≠ ∅. (43)

By (43), we obtain

(y0 − F (x)) ∩ icr(C) ≠ ∅ (44) and

−G(x) ∩ icr(D) ≠ ∅. (45)

It follows from (45) that x ∈ S. Thus, by (44), we have

(y0 − F (S)) ∩ icr(C) ≠ ∅, which contradicts (41). Therefore, (42) holds.

By Theorem 4.1, there exists (y∗, z∗) ∈ C+ × D+ with (y∗, z∗) ≠ (0, 0) such that

∗ ∗ ⟨I1(x), (y , z )⟩ ≥ 0, ∀x ∈ A, i.e.,

∗ ∗ ∗ ⟨F (x), y ⟩ + ⟨G(x), z ⟩ ≥ ⟨y0, y ⟩, ∀x ∈ A. (46)

∗ + Since x0 ∈ S, there exists p ∈ G(x0) such that −p ∈ D. Because z ∈ D , we obtain

∗ ⟨p, z ⟩ ≤ 0. Taking x = x0 in (46), we get

∗ ∗ ∗ ⟨y0, y ⟩ + ⟨p, z ⟩ ≥ ⟨y0, y ⟩

It follows that ⟨p, z∗⟩ ≥ 0. So, ⟨p, z∗⟩ = 0. Thus, we have

∗ ∗ ∗ ⟨y0, y ⟩ ∈ ⟨F (x0), y ⟩ + ⟨G(x0), z ⟩. (47)

19 Therefore, it follows from (46) and (47) that

∗ ∗ ∗ inf (⟨F (x), y ⟩ + ⟨G(x), z ⟩) = ⟨y0, y ⟩. x∈A

Taking again x = x0 in (46), we obtain

∗ ∗ ∗ ⟨y0, y ⟩ + ⟨G(x0), z ⟩ ≥ ⟨y0, y ⟩.

∗ ∗ So, ⟨G(x0), z ⟩ ≥ 0. We have shown that there exists p ∈ G(x0) such that ⟨p, z ⟩ = 0. Thus, we have

∗ inf⟨G(x0), z ⟩ = 0.

The following example shows that the conditions of Theorem 4.2 can be satisﬁed.

2 Example 4.2 Let X = Y = Z = R ,C = D = {(y1, 0)|y1 ≥ 0} and A = {(1, 0), (1, 2)}. The set-valued map F : X → 2Y on A is deﬁned as follows:

F (1, 0) = {(y1, y2)|y1 = 1, 0 ≤ y2 ≤ 1},F (1, 2) = {(y1, y2)|y1 > 1, 0 ≤ y2 ≤ −y1 + 2}.

The set-valued map G : X → 2Y on A is deﬁned as follows:

G(1, 0) = {(y1, y2)|y1 ≤ 0, 0 ≤ y2 ≤ y1 + 1},G(1, 2) = {(y1, y2)|y1 ≥ −1, y1 + 1 ≤ y2 ≤ 1}.

Let x0 = (1, 0) and y0 = (1, 0) ∈ F (x0). Clearly, all conditions of Theorem 4.2 are satisﬁed.

∗ ∗ ∗ ∗ Therefore, there exist y : ⟨(y1, y2), y ⟩ = y1 + y2 and z : ⟨(y1, y2), z ⟩ = −y1 + y2 such that

∗ ∗ ∗ inf (⟨F (x), y ⟩ + ⟨G(x), z ⟩) = ⟨y0, y ⟩ x∈A and

∗ inf⟨G(x0), z ⟩ = 0.

Remark 4.4 Theorem 4.2 improves Theorem 3.2 of [1], Theorem 3.1 of [3] and Theorem 4.2 of [9].

20 We also show a suﬃcient optimality condition for (VP) as follows:

THEOREM 4.3 Suppose that the following conditions hold:

(i) x0 ∈ S;

∗ ∗ +i + (ii) there exist y0 ∈ F (x0) and (y , z ) ∈ C × D such that

∗ ∗ ∗ inf (⟨F (x), y ⟩ + ⟨G(x), z ⟩) ≥ ⟨y0, y ⟩. x∈A

Then (x0, y0) is a weak minimizer of (VP).

Proof By condition (ii), we have

∗ ∗ ⟨F (x) − y0, y ⟩ + ⟨G(x), z ⟩ ≥ 0, ∀x ∈ A. (48)

We assert (x0, y0) is a weak minimizer of (VP). Otherwise, there exists x ∈ S such that

(y0 −F (x))∩icr(C) ≠ ∅. Therefore, there exists y ∈ F (x) such that y0 −y ∈ icr(C) ⊆ C \{0}.

Thus, we obtain

∗ ⟨y − y0, y ⟩ < 0. (49)

Since x ∈ S, there exists q ∈ G(x) such that −q ∈ D. Hence,

⟨q, z∗⟩ ≤ 0. (50)

Adding (49) to (50), we have

∗ ∗ ⟨y − y0, y ⟩ + ⟨q, z ⟩ < 0, which contradicts (48). Therefore, (x0, y0) is a weak minimizer of (VP).

The following example shows that the conditions of Theorem 4.3 can be satisﬁed.

2 Example 4.3 Let X = Y = Z = R ,C = D = {(y1, 0)|y1 ≥ 0} and A = {(1, 0), (1, 2)}. The set-valued map F : X → 2Y on A is deﬁned as follows:

F (1, 0) = {(y1, y2)|y1 ≥ 1, y1 ≤ y2 ≤ 2},F (1, 2) = {(y1, y2)|y1 ≤ 2, 1 ≤ y2 ≤ y1}.

21 The set-valued map G : X → 2Y on A is deﬁned as follows:

G(1, 0) = {(y1, y2)| − 1 ≤ y1 ≤ 0, y2 = 0},G(1, 2) = {(y1, y2)| − 1 ≤ y1 ≤ 0, 0 ≤ y2 ≤ 1}.

∗ ∗ Let x0 = (1, 0), y0 = (1, 1) ∈ F (x0), ⟨(y1, y2), y ⟩ = y1 + y2 and ⟨(y1, y2), z ⟩ = −y1. Clearly, all conditions of Theorem 4.3 are satisﬁed. Therefore, (x0, y0) is a weak minimizer of (VP).

Acknowledgements

This work was supported by the National Nature Science Foundation of China (Grant No.

10831009) and the Natural Science Foundation of Chongqing (Grant No. CSTC, 2009BB8240). References

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