Lagrangian Duality for Multiobjective Programming Problems in Lexicographic Order

Abstract

This paper deals with a constraint multiobjective programming problem and its dual problem in the lexicographic order. We establish some duality theorems and present several existence results of a Lagrange multiplier and a lexicographic saddle point theorem. Meanwhile, we study the relations between the lexicographic saddle point and the lexicographic solution to a multiobjective programming problem.

1. Introduction

Duality assertions are very important in optimization researches from the theoretical as well as from the numerical point of view. Duality theorems in mathematical programming establish typical connections between a constrained minimization problem and a constrained maximization problem. The relationship is such that the existence of a solution to one of these problems ensures the existence of a solution to other, both having the same optimal value. In the past centuries, many authors have studied the duality problems of vector optimization problems; see, for example, [1–11] and reference therein.

On the other hand, it is well known that partial order plays an important role in multiobjective optimization theory. But partial efficient points are usually large, so that one needs certain additional rules to reduce them. One of the possible approaches is to utilize the lexicographic order, which is introduced by the lexicographic cone. The main reason is that the lexicographic order is a total ordering and it can overcome the shortcoming that not all points can be compared in partial order. The lexicographic order has been investigated in connection with its applications in optimization and decision making theory; see, for example, [12–19] and references therein. However, the lexicographic cone is neither open nor closed. Note that a lot of results for vector optimization problems are gotten under the assumption that the ordering cone is open or closed. Therefore, it is valuable to investigate multiobjective optimization problems in the lexicographic order. Konnov [12] first discussed the vector equilibrium problems in lexicographic order. Recently, Li et al. [13] studied the minimax inequality problem and have obtained minmax theorems in the lexicographic order.

The rest of the paper is organized as follows. In Section 2, we first recall some definitions and their properties. In Section 3, with respect to the lexicographic order, we first establish weak duality theorem and the Lagrangian multiplier rules for a multiobjective programming. In Section 4, we investigate a lexicographic saddle point of a vector-valued Lagrangian function and discuss the connection between the lexicographic saddle point and the lexicographic solution to a multiobjective programming problem.

2. Preliminaries and Notations

Throughout this paper, unless otherwise specified, let 𝑋 be an arbitrarily nonempty subset of a topology space 𝑌 and ℝ^ℓ𝑛-dimensional Euclidean space. Let ℝ^ℓ₊∶={𝑥∈ℝ^ℓ∶𝑥_𝑖≥0,∀𝑖}. Let ℒ(ℝ^𝑚,ℝ^𝑛) be the space of continuous linear operator from ℝ^𝑚 to ℝ^𝑛 and ℒ⁺(ℝ^𝑚,ℝ^𝑛)∶={Λ∈ℒ(ℝ^𝑚,ℝ^𝑛)∶Λ(ℝ^𝑚₊)⊂ℝ^𝑛₊}. By 0_ℓ denote the zero vector of ℝ^ℓ.

For convenience, we set 𝐼_𝑛={1,2,…,𝑛}. As usual, for any 𝑣∈ℝ^𝑛 and 𝑖∈𝐼_𝑛, 𝑣_𝑖 will denote the 𝑖th coordinate of 𝑣 with respect to the canonical basis {𝑒₁,𝑒₂,…,𝑒_𝑛}. The lexicographic cone of ℝ^𝑛 is defined as the set of all vectors whose first nonzero coordinate (if any) is positive: 𝐶_lex={0}∪𝑣∈𝑅^𝑛∶∃𝑖∈𝐼_𝑛suchthat𝑣_𝑖>0;∄𝑗∈𝐼_𝑛,𝑗<𝑖suchthat𝑣_𝑗≠0. (1)

Note that the lexicographic cone 𝐶_lex is neither closed nor open. However, it is convex, pointed and 𝐶_lex∪(−𝐶_lex)=ℝ^𝑛. Thus the binary relation defined for any 𝑢,𝑣∈ℝ^𝑛 by 𝑢≤_lex𝑣⟺𝑢∈𝑣−𝐶_lex (2) is a total order on ℝ^𝑛 (i.e., it is reflexive, transitive, antisymmetric and any two vectors are comparable). The binary relation induced by 𝐶_lex is called a lexicographic order.

Now we recall the definitions of efficient points of a nonempty subset in the lexicographic order.

Definition 1 (see [13, 14]). Let 𝑍⊂ℝ^𝑛 be a nonempty subset. (i) A point 𝑧₀∈𝑍 is called a lexicographic maximal point of 𝑍 if 𝑍⊂𝑧₀−𝐶_lex; that is, for any 𝑧∈𝑍,𝑧≤_lex𝑧₀; by max_lex𝑍, we denote the set of all lexicographic maximal points of 𝑍.(ii) A point 𝑧₀∈𝑍 is called a lexicographic minimal point of 𝑍 if 𝑍⊂𝑧₀+𝐶_lex; that is, for any 𝑧∈𝑍,𝑧₀≤_lex𝑧; by min_lex𝑍, we denote the set of all lexicographic minimal points of 𝑍.

Obviously, if max_lex𝑍≠∅, max_lex𝑍 is a single point set and so is min_lex𝑍.

Lemma 2 (see [13]). If 𝑍⊂ℝ^𝑛 is a nonempty compact set, then min_𝑙𝑒𝑥𝑍≠∅ and max_𝑙𝑒𝑥𝑍≠∅.

Proposition 3. If 𝐶,𝐷⊂ℝ^𝑛 are two nonempty compact subsets, then (i)min_𝑙𝑒𝑥(𝐶+𝐷)=min_𝑙𝑒𝑥(𝐶)+min_𝑙𝑒𝑥(𝐷),(ii)max_𝑙𝑒𝑥(𝐶+𝐷)=max_𝑙𝑒𝑥(𝐶)+max_𝑙𝑒𝑥(𝐷).

Proof. It suffices to verify (i) since (ii) is evident by max_lex(𝐶)=−min_lex(−𝐶). Indeed, let 𝑥=min_lex(𝐶) and 𝑦=min_lex(𝐷). Then 𝑥∈𝐶 and 𝑥≤_lex𝑥,forall𝑥∈𝐶; 𝑦∈𝐷 and 𝑦≤_lex𝑦,forall𝑦∈𝐷. Hence, 𝑥+𝑦∈𝐶+𝐷 and 𝑥+𝑦≤_lex𝑥+𝑦forall𝑥∈𝐶, and 𝑦∈𝐷. Namely, {𝑥+𝑦}=min_lex(𝐶+𝐷) and the proof is complete.

Definition 4. A vector-valued mapping 𝑔∶𝑋→ℝ^ℓ is called lower semicontinuous if, for any 𝑧∈ℝ^ℓ, the set {𝑥∈𝑋∶𝑔(𝑥)∈𝑧−ℝ^ℓ₊} is closed in X.

Definition 5. A vector-valued mapping 𝑔∶𝑋→ℝ^ℓ is called convex on 𝑌 if and only if, for any 𝑥₁,𝑥₂∈𝑋, 𝑡∈[0,1], and 𝑖=1,2,…,ℓ, 𝑔_𝑖(𝑡𝑥₁+(1−𝑡)𝑥₂)≤𝑡𝑔_𝑖(𝑥₁)+(1−𝑡)𝑔_𝑖(𝑥₂).

Consider the following multiobjective programming problem: (MP)min_lex𝑓(𝑥)=𝑓₁(𝑥),𝑓₂(𝑥),…,𝑓_𝑛(𝑥)^𝑇s.t.𝑔(𝑥)∈−𝑅^𝑚₊,𝑥∈𝑋, (3) where 𝑓_𝑖∶𝑋→ℝ(𝑖=1,2,…,𝑛) and 𝑔(𝑥)=(𝑔₁(𝑥),𝑔₂(𝑥),…,𝑔_𝑚(𝑥))^𝑇, and 𝑔_𝑗∶𝑋→ℝ(𝑗=1,2,…,𝑚).

Let 𝐸 denote the set of all feasible points for the multiobjective optimization problem (MP); that is, 𝐸=𝑥∈𝑋∶𝑔(𝑥)∈−𝑅^𝑚₊. (4) And let 𝐸_𝑖 denote the set of all minimal points for 𝑓_𝑖 on 𝐸; that is, 𝐸_𝑖=𝑥∈𝐸∶𝑓_𝑖(𝑥)=min_𝑦∈𝐸𝑓_𝑖𝑦,for𝑖=1,2,…,𝑛. (5) Then, we write 𝑓(𝐸)=⋃_𝑥∈𝐸𝑓(𝑥). Throughout this paper, we always assume that 𝐸≠∅ and 𝐸_𝑖≠∅(𝑖=1,2,…,𝑛).

A vector-valued Lagrangian function for (MP) is defined from 𝐸×𝑅^𝑛×𝑚→𝑅^𝑛 by 𝐿(𝑥,Λ)=𝑓(𝑥)+Λ𝑔(𝑥), (6) where Λ∈ℒ⁺(𝑅^𝑚,𝑅^𝑛); that is, Λ_𝑛×𝑚=(𝜆₁,𝜆₂,…,𝜆_𝑛)^𝑇; 𝜆_𝑖=(𝜆_𝑖1,𝜆_𝑖2,…,𝜆_𝑖𝑚)∈𝑅^𝑚₊,𝑖=1,2,…,𝑛.

Then the Lagrangian dual problem is defined as the following multiobjective programming problem: (DMP)max_lex𝜙(Λ)s.t.Λ∈ℒ⁺𝑅^𝑚,𝑅^𝑛, (7) where 𝜙(Λ)=min_lex{𝐿(𝑥,Λ)∶𝑥∈𝑋}.

Definition 6. (i) A point 𝑥∈𝐸 is said to be a lexicographic minimal solution to (MP) if 𝑥∈𝐸 and 𝑓(𝑥)≤_lex𝑓(𝑥),forall𝑥∈𝐸. By min_lex𝐸andmin_lex𝑓(𝐸) denote the set of all lexicographic minimal solutions and the lexicographic minimal value to (MP), respectively.
(ii) A point 𝑥∈𝐸 is said to be a strong minimal solution to (MP) if 𝑥∈𝐸 and 𝑓(𝑥)∈𝑓(𝑥)−𝑅^𝑛₊,forall𝑥∈𝐸; that is, 𝑓_𝑖(𝑥)≤𝑓_𝑖(𝑥),and𝑥∈𝐸,and𝑖=1,2,…,𝑛. The set of all strong minimal solutions to (MP) is denoted by min_𝑅^𝑛₊𝑓(𝐸).

Remark 7. (1) From [14], it is easy to verify that (i) of Definition 6 is equivalent and refers to the following concept: a point 𝑥∈𝐸 is said to be a lexicographic minimal solution to (MP) if 𝑥∈𝐸 and {𝑓(𝑥)−𝐶_lex⧵{0}}∩𝑓(𝐸)=∅.
(2) If 𝑥∈𝐸 is a strong solution to (MP), then 𝑥 is a lexicographic solution to (MP). However, the converse generally is not valid.

3. Duality Results

The following result shows that, with respect to lexicographic order, the objective value of any feasible point to the dual problem (DMP) is less than or equal to the objective value of any feasible point to the primal problem (MP).

Theorem 8 (weak duality theorem). Consider the primal problem (MP) given by (3) and its Lagrangian dual problem (DMP) given by (7). Let 𝑥 be a feasible point of (MP); that is, 𝑥∈𝑋and𝑔(𝑥)∈−𝑅^𝑚₊. Also, let Λ be a feasible point of (𝐷𝑀𝑃); that is, Λ∈ℒ⁺(𝑅^𝑚,𝑅^𝑛). Then 𝑓(𝑥)≥_𝑙𝑒𝑥𝜙(Λ). (8)

Proof. Since 𝑥∈𝑋,𝑔(𝑥)∈−𝑅^𝑚₊, and Λ∈ℒ⁺(𝑅^𝑚,𝑅^𝑛), we obtain that Λ𝑔(𝑥)∈−𝑅^𝑛₊⊂−𝐶_lex; that is, Λ𝑔(𝑥)≤_lex0_𝑛. Then, it follows from the definition of 𝜙(Λ) that 𝜙(Λ)=min_lex𝑓(𝑥)(𝑥)∶≤+Λ𝑔𝑥∈𝑋_lex𝑓(𝑥)+Λ𝑔(𝑥)≤_lex𝑓(𝑥), (9) and the proof is complete.

As a corollary of the previous theorem, we have the following result.

Corollary 9. Assume that max_𝑙𝑒𝑥{𝜙(Λ)∶Λ∈𝐿⁺(𝑅^𝑛,𝑅^𝑚)≠∅. Then, min_𝑙𝑒𝑥𝑓(𝑥)∶𝑔(𝑥)∈−𝑅^𝑚₊≥,𝑥∈𝑋_𝑙𝑒𝑥max_𝑙𝑒𝑥𝜙(Λ)∶Λ∈𝐿⁺𝑅^𝑛,𝑅^𝑚. (10)

Before stating the Lagrangian multiplier rule to (MP) in lexicographic sense, we need the following result.

Lemma 10. Assume that ⋂^𝑛_𝑖=1𝐸_𝑖≠∅. A point 𝑥∈𝐸 is a lexicographic solution for (MP) if and only if 𝑥∈𝐸 is a strong solution for (MP).

Proof. If 𝑥 is a lexicographic solution for (MP), then 𝑓(𝑥)∈min_lex𝑓(𝐸). Assume that ⋂^𝑛_𝑖=1𝐸_𝑖≠∅. Then each 𝑥∈⋂^𝑛_𝑖=1𝐸_𝑖 is a strong solution for (MP), which implies that 𝑓(𝑥)∈min_lex𝑓(𝐸). Since ∈min_lex𝑓(𝐸) is singleton, we have 𝑓(𝑥)=𝑓(𝑥). That is, 𝑥 is a strong solution for (MP).
Conversely, suppose that 𝑥 is a lexicographic solution for (MP). Since ⋂^𝑛_𝑖=1𝐸_𝑖≠∅, there exists 𝑥∈𝐸 which is a strong solution for (MP). By induction, we can show that 𝑓(𝑥)=𝑓(𝑥) which means that 𝑥 is a strong solution for (MP). Let 𝐷_𝑘={𝑥∈𝐷_𝑘−1∶𝑓_𝑘(𝑥)≤𝑓_𝑘(𝑦),∀𝑦∈𝐷_𝑘−1} for 𝑘=1,2,…,𝑛 and 𝐷₀=𝐸. Obviously, 𝐷_𝑛⊂𝐷_𝑛−1⊂⋯⊂𝐷₀. Noting that 𝑥 is a lexicographic solution for (MP), we have 𝑥∈𝐷_𝑛. Then, 𝑥∈𝐷₁; that is, 𝑓₁(𝑥)≥𝑓₁(𝑥),∀𝑥∈𝐷₀. (11) Taking 𝑥=𝑥 in the above inequality, we get 𝑓₁(𝑥)≥𝑓₁(𝑥). (12) On the other hand, 𝑓₁(𝑥)≥𝑓₁(𝑥) since 𝑥∈𝐷₀ is a strong solution and 𝑥∈𝑋_𝑛⊂𝐷₀. Therefore, 𝑓₁(𝑥)=𝑓₁(𝑥) and 𝑥∈𝐷₁. Suppose that 𝑓_𝑖(𝑥)=𝑓_𝑖(𝑥) for 𝑖=1,2,…,𝑛−1. Obviously, 𝑥∈𝐷_𝑛−1. Similarly, we can verify that 𝑓_𝑛(𝑥)=𝑓_𝑛(𝑥) and 𝑥∈𝐷_𝑛. The proof is complete.

Theorem 11 (Lagrangian multiplier rule). Suppose that the following conditions are satisfied: (i)𝑋 is a nonempty convex subset of 𝑅^ℓ;(ii)𝑓∶𝑋→𝑅^𝑛 is a 𝑅^𝑛₊-convex vector function; that is, 𝑓_𝑖∶𝑋→𝑅, 𝑖=1,2,…,𝑛, are convex functions;(iii)𝑔∶𝑋→𝑅^𝑚 is a 𝑅^𝑚₊-convex vector function; that is, 𝑔_𝑗∶𝑋→𝑅,𝑗=1,2,…,𝑚, are convex functions;(iv) the Slater constraint qualification is satisfied; that is, there exists 𝑥₀∈𝑋 such that 𝑔(𝑥₀)∈−int𝑅^𝑚₊, where int𝑅^𝑚₊ is the topology interior of 𝑅^𝑚₊;(v)⋂^𝑛_𝑖=1𝐸_𝑖≠∅. If 𝑥 is a lexicographic solution to (MP), then there exists Λ∈ℒ⁺(𝑅^𝑚,𝑅^𝑛) such that 𝑓(𝑥)=min_𝑙𝑒𝑥𝑓(𝑥)+Λ𝑔(𝑥)∶𝑥∈𝐸,Λ𝑔(𝑥)=0_𝑛. (13)

Proof. Let 𝑥 be a lexicographic solution of the problem (MP). By Lemma 10, 𝑥 is a strong solution for (MP). Then, for any 𝑥∈𝐸∶={𝑥∈𝑋∶𝑔(𝑥)∈−𝑅^𝑚₊}, 𝑓₁(𝑥)≥𝑓₁(𝑥),𝑓₂(𝑥)≥𝑓₂(𝑥),⋮𝑓_𝑛(𝑥)≥𝑓_𝑛(𝑥). (14)
Then, the inequality system 𝑓₁(𝑥)<𝑓₁(𝑥),𝑔(𝑥)∈−𝑅^𝑚₊ for some 𝑥∈𝑋 has no solution. Define the following set: 𝑆₁∶=𝑝,𝑞∈𝑅×𝑅^𝑚∶𝑓₁(𝑥)−𝑓₁(𝑥)<𝑝,𝑔(𝑥)∈𝑞−𝑅^𝑚₊,forsome𝑥∈𝑋. (15) The set 𝑆₁ is convex subset since 𝑋,𝑓₁, and 𝑔 are convex. Since the above inequality system has no solution, we have (0,0_𝑚)∉𝑆₁. By the standard separation theorem, there exists a nonzero vector (𝜇₁,𝜆₁) such that 𝜇₁𝑓₁(𝑥)−𝑓₁(𝑥)+𝜆^𝑇₁𝑔(𝑥)≥0,∀𝑥∈𝑋. (16) That is, 𝜇₁𝑓₁(𝑥)+𝜆^𝑇₁𝑔(𝑥)≥𝜇₁𝑓₁(𝑥),∀𝑥∈𝑋. (17)
Now, fix an 𝑥∈𝑋. From the definition of 𝑆₁, we have that 𝑝 and 𝑞 can be arbitrarily large. In order to satisfy (17), we must have (𝜇₁,𝜆₁)∈𝑅₊×𝑅^𝑚₊. We will next show that 𝜇₁>0. On the contrary, suppose that 𝜇₁=0. By the Slater constraint qualification, there exists 𝑥₀∈𝑋 such that 𝑔(𝑥₀)∈−int𝑅^𝑚₊. Then, letting 𝑥=𝑥₀ in (17), we have 𝜆₁𝑔(𝑥₀)≥0. But, since 𝑔(𝑥₀)∈−int𝑅^𝑚₊ and 𝜆₁∈𝑅^𝑚₊,𝜆₁𝑔(𝑥₀)≥0 is only possible if 𝜆₁=0_𝑚. Thus, it has been shown that (𝜇₁, 𝜆₁)=(0, 0_𝑚), which is a contradiction.
After dividing (17) by 𝜇₁ and denoting 𝜆₁=𝜆₁/𝜇₁, we obtain 𝑓₁(𝑥)+𝜆^𝑇₁𝑔(𝑥)≥𝑓₁(𝑥),∀𝑥∈𝑋. (18) From (18), letting 𝑥=𝑥, we get 𝜆^𝑇₁𝑔(𝑥)≥0. Since 𝑔(𝑥)∈−𝑅^𝑚₊ and 𝜆₁∈𝑅^𝑚₊, we obtain 𝜆^𝑇₁𝑔(𝑥)=0.
Similarly, we can show that there exist 𝜆₂,𝜆₃,…,𝜆_𝑛∈𝑅^𝑚₊ such that, for any 𝑥∈𝑋, 𝑓₂(𝑥)+𝜆^𝑇₂𝑔(𝑥)≥𝑓₂(𝑥),𝜆^𝑇₂𝑔(𝑥)𝑓=0,₃(𝑥)+𝜆^𝑇₃𝑔(𝑥)≥𝑓₃(𝑥),𝜆^𝑇₃𝑔(𝑥)⋮𝑓=0,_𝑛(𝑥)+𝜆^𝑇_𝑛𝑔(𝑥)≥𝑓_𝑛(𝑥),𝜆^𝑇_𝑛𝑔(𝑥)=0. (19) Set Λ=(𝜆₁,𝜆₂,…,𝜆_𝑛)^𝑇. Then, it follows from 𝑅^𝑛₊⊂𝐶_lex, (18), and (19) that 𝑓(𝑥)+Λ𝑔(𝑥)∈𝑓(𝑥)+𝑅^𝑛₊∈𝑓(𝑥)+𝐶_lex,∀𝑥∈𝑋,Λ𝑔(𝑥)=0_𝑛. (20) Namely, 𝑓(𝑥)+Λ𝑔(𝑥)≥_lex𝑓(𝑥),∀𝑥∈𝑋,Λ𝑔(𝑥)=0_𝑛. (21)(22)
Thus, (21) and weak duality theorem (Theorem 8) together yield that 𝑓(𝑥)=min_lex𝑓(𝑥)+Λ𝑔(𝑥)∶𝑥∈𝐸. (23) And the proof is complete.

Now, we give an example to illustrate that Theorem 11 is applicable.

Example 12. Let 𝑋=[−2, 2]⊂𝑅; 𝑔∶𝑋→𝑅 is defined by 𝑔(𝑥)=𝑥²−1 and 𝑓=(𝑓₁, 𝑓₂)^𝑇∶𝑋→𝑅² is defined by 𝑓₁(𝑥)=𝑓−2𝑥+1,if−2≤𝑥<−1,3,if−1≤𝑥≤1,2𝑥+1,if1<𝑥≤2,₂(𝑥)=(𝑥−2)²+1. (24)
Consider the following convex multiobjective programming problem: min_lex{𝑓(𝑥)∶𝑔(𝑥)≤0,𝑥∈𝑋}. Direct computation shows that 𝐸={𝑥∈𝑥∶𝑔(𝑥)≤0}=[−1,1], 𝐸₁=[−1,1], and 𝐸₂={1}. Obviously, 𝑥=1∈𝐸₁∩𝐸₂ is a lexicographic solution of the multiobjective programming problem. Therefore, Theorem 11 is applicable. Indeed, by directly computing, there exists Λ=(0,1)^𝑇 such that min_lex𝑓(𝑥)+Λ𝑔(𝑥)∶𝑥∈𝑋=𝑓(𝑥)=(3,−1)^𝑇,Λ𝑔(𝑥)=0₂. (25)

From Theorem 11, we get a sufficient condition for the existence of Lagrangian multiplier rule for the problem (MP). But this is not a necessary condition, as the following example shows.

Example 13. Let 𝑋⊂𝑅, and 𝑔(𝑥) is given as in Example 12. Let 𝑓=(𝑓₁,𝑓₂)∶𝑋→𝑅² be defined by 𝑓₁(𝑥)=(𝑥−2)²𝑓+1,₂(𝑥)=𝑥. (26) Clearly, the following problem: min_lex𝑓(𝑥)∶𝑔(𝑥)≤0,𝑥∈𝑋, (27) is a convex multiobjective programming. It follows from direct computation that 𝐸=[−1,1]𝐸₁={1}, and 𝐸₂={−1}, and 𝑥=1 is a unique lexicographic solution for (MP). Thus, the assumption (v) of Theorem 11 is not satisfied. However, we can verify that there exists Λ=(1,0)^𝑇 such that, for any 𝑥∈𝐸, min_lex𝑓(𝑥)+Λ𝑔(𝑥)∶𝑥∈𝑋=𝑓(𝑥)=(2,1)^𝑇,Λ𝑔(𝑥)=0₂. (28)

In order to obtain another sufficient condition for the existence of Lagrangian multiplier rule, we consider the following assumptions.

Theorem 14 (Lagrangian multiplier rule). Assume that all conditions of Theorem 11 are satisfied except for the hypothesis (v) of Theorem 11 which is replaced by the following hypothesis:   (v^') 𝑓₁∶𝑋→𝑅 is a strict convex function; that is, for any 𝑥₁,𝑥₂∈𝑋 with 𝑥₁≠𝑥₂ and 𝜃∈(0,1), 𝑓₁(𝜃𝑥₁+(1−𝜃)𝑥₂)<𝜃𝑓₁(𝑥₁)+(1−𝜃)𝑓₁(𝑥₂).
If 𝑥 is a lexicographic solution to (MP), then there exists Λ∈𝐿⁺(𝑅^𝑚,𝑅^𝑛) such that 𝑓(𝑥)=min_𝑙𝑒𝑥𝑓(𝑥)+Λ𝑔(𝑥)∶𝑥∈𝐸,Λ𝑔(𝑥)=0_𝑛. (29)

Proof. Since 𝑋 is convex and 𝑔(𝑥) is 𝑅^𝑚₊-convex, the set 𝐸={𝑥∈𝑋∶𝑔(𝑥)∈−𝑅^𝑚₊} is convex. Noting that 𝑓₁(𝑥) is strict convex, we obtain 𝐸₁={𝑥∈𝐸∶𝑓₁(𝑥)=min𝑓₁(𝐸)} is singleton. Obviously, 𝑥∈𝐸₁={𝑥} is a lexicographic solution to (MP). Following the same arguments as in Theorem 11, there exists 𝜆∈𝑅^𝑚₊ such that 𝑓₁(𝑥)=min{𝑓₁(𝑥)+𝜆^𝑇𝑔(𝑥)∶𝑥∈𝑋}, 𝜆^𝑇𝑔(𝑥)=0. Since 𝑔(𝑥) is 𝑅^𝑚₊-convex and 𝜆∈𝑅^𝑚₊, we obtain that 𝜆^𝑇𝑔(𝑥) is a convex function (𝑅⁺-convex function). By hypothesis (v^′), 𝑓₁(𝑥)+𝜆^𝑇𝑔(𝑥) is a strict convex function. Thus, the solution to min{𝑓₁(𝑥)+𝜆^𝑇𝑔(𝑥)∶𝑥∈𝑋} is also singleton. Let 𝜆₁=𝜆,𝜆₂=⋯=𝜆_𝑛=0_𝑚 in Theorem 11. Therefore, 𝑥 is also a lexicographic solution to min_lex{𝑓(𝑥)+Λ^𝑇𝑔(𝑥)∶𝑥∈𝑋} and Λ𝑔(𝑥)=0_𝑛. It completes the proof.

Remark 15. The hypothesis (v^′) in Theorem 14 is redundant if 𝑓 is a real-valued function. Under this case, the hypothesis (v) in Theorem 11 always holds.

From Theorems 8 and 11, we have the following strong duality theorem. The following result shows that, under suitable assumptions, there is no duality gap between the primal and dual lexicographic optimal objective function values.

Theorem 16 (strong duality theorem). Assume that max_𝑙𝑒𝑥{𝜙(Λ)∶Λ∈𝐿⁺(𝑅^𝑛,𝑅^𝑚)≠∅. If all conditions of Theorem 11 or Theorem 14 are satisfied, then min_𝑙𝑒𝑥𝑓(𝑥)∶𝑔(𝑥)∈−𝑅^𝑚₊,𝑥∈𝑋=max_𝑙𝑒𝑥𝜙(Λ)∶Λ∈𝐿⁺𝑅^𝑛,𝑅^𝑚. (30) And there exists Λ such that Λ𝑔(𝑥)=0_𝑛, where 𝑥∈{𝑥∈𝑋∶𝑓(𝑥)=min_𝑙𝑒𝑥𝑓(𝐸)}.

The following example is to illustrate that there is no duality gap if all conditions of Theorem 16 are satisfied.

Example 17. Let 𝑋⊂𝑅,𝑔(𝑥) is given as in Example 12. Let 𝑓(𝑥) be defined by 𝑓₁(𝑥)=(𝑥−2)²𝑓+1,₂(𝑥)=−𝑥. (31)
Direct computation shows that min_lex𝑓(𝑥)∶𝑔(𝑥)∈−𝑅^𝑚₊,𝑥∈𝑋=max_lex𝜙(Λ)∶Λ∈𝐿⁺𝑅^𝑛,𝑅^𝑚=(2,−1)^𝑇=𝑓(1). (32) And there exists Λ=(1,1/2)^𝑇or(1,0)^𝑇 such that Λ𝑔(1)=0₂.

4. Lexicographic Saddle Point

Now, we introduce the notion of lexicographic saddle point for the vector-valued Lagrangian map 𝐿(⋅,⋅) in terms of lexicographic order and give some optimality conditions.

Definition 18. A pair (𝑥,Λ)∈𝐸×ℒ(𝑅^𝑛,𝑅^𝑚) is said to be a lexicographic saddle point for the vector-valued Lagrangian function 𝐿(𝑥,Λ) if, for all 𝑥∈𝐸 and Λ∈ℒ⁺(𝑅^𝑚,𝑅^𝑛), 𝐿(𝑥,Λ)≤_lex𝐿𝑥,Λ≤_lex𝐿𝑥,Λ. (33)

By Theorem 11 or Theorem 14, we directly have the following result.

Theorem 19. Suppose that all conditions of Theorem 11 or Theorem 14 are satisfied. If 𝑥 is a lexicographic solution to (MP), then there exists Λ∈ℒ⁺(𝑅^𝑚,𝑅^𝑛) such that (𝑥,Λ) is a lexicographic saddle point of the vector Lagrangian function 𝐿(𝑥,Λ).

Thus, we have verified the existence of a lexicographic saddle point for the vector-valued Lagrangian function 𝐿(𝑥,Λ) under the appropriate conditions. We conclude this result by showing that the saddle point condition is sufficient for optimality for problem (MP).

Theorem 20. If (𝑥,Λ)∈𝐸×ℒ(𝑅^𝑚,𝑅^𝑛) is a lexicographic saddle point for the vector-valued Lagrangian function 𝐿(𝑥,Λ), then 𝑥 is a lexicographic solution for (MP), and Λ𝑔(𝑥)=0_𝑛.

Proof. Assume that (𝑥,Λ) is a lexicographic saddle point of 𝐿(𝑥,Λ). Then, from the first inequality of (33), we get 𝑓(𝑥)+Λ𝑔(𝑥)≥_lex𝑓(𝑥)+Λ𝑔(𝑥),∀Λ∈ℒ⁺𝑅^𝑚,𝑅^𝑛. (34) This implies that Λ𝑔(𝑥)≥_lexΛ𝑔(𝑥),∀Λ∈ℒ⁺𝑅^𝑚,𝑅^𝑛. (35) We claim that 𝑔(𝑥)∈−𝑅^𝑚₊. Otherwise, we suppose that 𝑔(𝑥)∉−𝑅^𝑚₊. Since Λ∈ℒ⁺(𝑅^𝑚,𝑅^𝑛) can be taken arbitrarily to be large, Λ𝑔(𝑥) can be large enough, which contradicts (35).
Therefore, we also get Λ𝑔(𝑥)∈−𝑅^𝑛₊⊂−𝐶_lex (36) since 𝑔(𝑥)∈−𝑅^𝑚₊ and Λ∈ℒ⁺(𝑅^𝑚,𝑅^𝑛). This implies that Λ𝑔(𝑥)≤_lex0_𝑛. (37)
On the other hand, by taking Λ=0 in (35), we can get Λ𝑔(𝑥)≥_lex0_𝑛. (38) Noting the reflexivity of the lexicographic order ≤_lex, (37) and (38) together yields Λ𝑔(𝑥)=0_𝑛. (39)
Since 𝑔(𝑥)∈−𝑅^𝑚₊, 𝑥 is a feasible point of (MP). Let 𝑥 be any feasible point of (MP); that is, 𝑔(𝑥)∈−𝑅^𝑚₊. Then, it follows from the second inequality of (33) and (39) that 𝑓(𝑥)=𝑓(𝑥)+Λ𝑔(𝑥)≤_lex𝑓(𝑥)+Λ𝑔(𝑥)≤_lex𝑓(𝑥), (40) which implies that 𝑥 is a lexicographic solution to (MP).

Acknowledgments

The authors would like to express their deep gratitude to the anonymous referee and the associate editor for their valuable comments and suggestions which helped improve the paper. This research was partially supported by the National Natural Science Foundation of China (Grant no. 11201509).