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

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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 gen- eralized 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 = fλaja 2 K; λ ≥ 0g: A cone K ⊆ Y is said to be pointed if K \ (−K) = f0g: A cone K ⊆ Y is said to be nontrivial if K =6 f0g and K =6 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+ = fy∗ 2 Y ∗jhy; y∗i > 0; 8y 2 Cg and C+i = fy∗ 2 Y ∗jhy; y∗i > 0; 8y 2 C n f0gg; where hy; y∗i 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) = fk 2 Kj8v 2 Y; 9λ0 > 0; 8λ 2 [0; λ0]; k + λv 2 Kg: 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) = fk 2 Kj8v 2 L(K); 9λ0 > 0; 8λ 2 [0; λ0]; k + λv 2 Kg: LEMMA 2.1 Let K be a nonempty subset in Y . Then aff(K) = x + L(K); 8x 2 K: (1) Proof Firstly, we will show that aff(K) ⊆ x + L(K); 8x 2 K: (2) Pn Let y 2 aff(K): Then, there exist ki 2 K; αi 2 R with αi = 1 such that i=1 Xn y = αiki: i=1 Thus, we obtain Xn Xn Xn y = x + y − x = x + αiki − αix = x + αi(ki − x) 2 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); 8x 2 K: (3) Let z 2 x + L(K): Then, there exist xi; yi 2 K and λi 2 R(i = 1; ··· ; n) such that Xn Xn Xn z = x + λi(xi − yi) = x + λixi − λiyi: (4) i=1 i=1 i=1 Clearly, Xn Xn 1 + λi − λi = 1: (5) i=1 i=1 4 It follows from (4) and (5) that z 2 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) = fk 2 Kj8v 2 aff(K) − k; 9λ0 > 0; 8λ 2 [0; λ0]; k + λv 2 Kg: Remark 2.2 It follows from Lemma 2.1 that if 0 2 K ⊆ Y; then icr(K) = fk 2 Kj8v 2 aff(K); 9λ0 > 0; 8λ 2 [0; λ0]; k + λv 2 Kg: Remark 2.3 If K be a nontrivial pointed cone in Y then 0 2= icr(K): In fact, if 0 2 icr(K); then it follows from Remark 2.2 that aff(K) = K: (6) Since 0 2 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 2 K such that −k 2 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) [ f0g 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 2 K and k2 2 icr(K): k2 2 icr(K) implies that k2 2 K; 8v 2 L(K); 9λ0 > 0; 8λ 2 [0; λ0]; we have k2 + λv 2 K: Since K is a convex cone, we obtain (k1 + k2) + λv = k1 + (k2 + λv) 2 K + K ⊆ K; which implies k1 + k2 2 icr(K): Therefore, K + icr(K) ⊆ icr(K): (10) Since 0 2 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) 2 i 2 j 2 i 2 j 2 Let y aff(K1) + aff(K2): Then, there exist k1 K1; k2 K2; λ1 R; λ2 R(i = Pm1 Pm2 ··· ··· i j 1; m1; j = 1; m2) with λ1 = 1 and λ2 = 1 such that i=1 j=1 Xm1 Xm2 Xm1 Xm2 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 Pm1 Pm2 i j 2 ··· ··· 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 2 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); 8k 2 K: If K is convex and icr(K) =6 ;; then (b) icr(icr(K)) = icr(K); (c) aff(icr(K)) = aff(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); 8k 2 K; i.e., aff(K) ⊆ k + aff(K − K); 8k 2 K: (15) Finally, we will show that k + aff(K − K) ⊆ aff(K); 8k 2 K: (16) By Lemma 2.3, we only need to show that k + aff(K) − aff(K) ⊆ aff(K); 8k 2 K: (17) 2 − i 2 j 2 i 2 j 2 Let y k + aff(K) aff(K): Then, there exist k1 K; k2 K; λ1 R; λ2 R(i = Pm1 Pm2 ··· ··· i j 1; ; m1; j = 1; ; m2) with λ1 = 1 and λ2 = 1 such that i=1 j=1 Xm1 Xm2 i i − j j y = k + λ1k1 λ2k2: (18) i=1 j=1 7 Clearly, Xm1 Xm2 i − j 1 + λ1 λ2 = 1: (19) i=1 j=1 By (18) and (19), y 2 aff(K): Thus, (17) holds.
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