Một cách tiếp cận mới cho khoảng cách đối ngẫu bằng không của bài toán tối ưu véctơ sử dụng tập đặc trưng

 , a contradiction. 
 We now show that . Indeed, take Hence, ̅ and we get 
arbitrarily . For any , one has 
 ̅ ̅ , and hence, by (5) ̅ . This contradicts the fact that 
 ̅ . Consequently, 
 ̅ ̅ ̅ 
 ̅ . 
 Secondly, we next claim that 
which implies 
 ̅ . For this purpose, we take 
 arbitrarily ̃ and show that ̅ ̃ , or 
 ̅ ̅ ̅ 
 equivalently, ̅ ̃ . 
 As ̅ and 
 Letting , one gains . 
 ̃
 ̅ ̅, there is ̃ such that 
Consequently, . 
 ̅ ̃ ̃, or equivalently, 
 We proceed to show that ̅ 
 ̃
Indeed, pick . Since , it follows ̅ ̃ (6) 
that . Let ̅ defined by 
 As ̃ , there exists ̃ 
̅ 
 . Then, it is easy to check that 
 such that 
 and ̅ . 
 ̃ ( ̃ ) 
 Take . As 
 from (5), we have 
 ̅ Moreover, by the convex assumption, 
 ̃
and hence, with the help of ̅, is a 
 convex set of (Nguyen Dinh et al., 2019, 
 ̅
 ̅ Remark 4.1). Hence, the convex separation 
or equivalently, theorem (Rudin, 1991, Theorem 3.4) ensures 
 the existence of satisfying 
 ̅ ̅ 
 ̃ 
 So, there is such that 
 ̃ 
 ̅ ̅ 
or equivalently, So, according to Nguyen Dinh et al. 
 (2019, Lemma 3.3), one gets and 
 ̅ 
 ⟨ ̅ ⟩ ̃ ( ̃ ) 
 (7) 
 As , the last inequality entails 
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Natural Sciences issue 
 Take now . Then, there is ̂
 hand, since ̅ ̅ there is 
 ̃ such that 
 ̂
 such that ̅ , and consequently, 
 ̃ ̃ (8) 
 ̅ ̂ . 
 It is worth noting that ̃ , one 
gets from (8) that We get a contradiction, and hence, 
 ̃ ( ̃ ) ̃ ̃ (9) ̅ for all . 
 Since , it follows from (6), (7), Lastly, it follows from Steps , 
and (9) that ̅ ̃ ̃ and the definition of weak supremum 
 that ̅ . The proof 
 ̃
 ̃ ̃ ̃ , is complete. 
 ̃ , and ̃ ̃ ̃ . 
From these inequalities, Remark 4. According to the proof of 
 Proposition 5, we see that if all the 
 ̅ ̃ ̅ ̃ ̃ (10) 
 assumptions of this proposition hold then one 
 ̃ ̃ also has 
(recall that as and ). 
Note that (10) holds for any . This 4. Zero duality gap for vector 
 ̃ optimization problem 
means that ̅ is strictly separated 
 ̃
from , and consequently, ̅ Consider the pair of primal-dual problems 
(see Zalinescu, 2002, Theorem 1.1.7). and as in the previous section. 
 Lastly, we have just shown that Definition 1. We say that has weak 
 zero duality gap if 
 ̅ . So, ̅ . 
 and that has a strong zero duality gap if 
 Take ̅ , we will prove that . 
 ̅ . 
 Theorem 1. Assume that is -convex, 
 Firstly, take ̃ such that ̃ ̅. that is -convex, and that is a convex 
Then, as ̅ one has ̃ . We subset of . Then, the following statements 
now apply the argument in Step again, 
 are equivalent: 
with ̅ replaced by ̃ to obtain ̃ , 
or in the other words, there is such { } 
that ̃ . { } 
 Secondly, prove that ̅ for all 
 for some and , 
 . Suppose, contrary to our claim, that 
there is ̂ such that ̅ ̂. Then, has a weak zero duality gap. 
there is ̂ such that ̅ ̂ ̂. Hence, Proof. [(i) (ii)] Assume that there are 
 ̂ and ̅ ̂ ̅ ̂ ̂. Letting 
 and satisfying 
 ̅ ̂ and ̂ play the roles of ̅ ̃ and ̃ 
 { } 
(respectively) in Step and using the same 
argument as in this step, one gets ̅ ̂ { } (11) 
which also means ̅ ̂ . On the other Let 
 { } 
 { } 
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 Dong Thap University Journal of Science, Vol. 9, No. 5, 2020, 03-16 
(see Proposition 3). Then, according to Lemma In brief, we have just proved that 
2, one has which, which also 
together with Proposition 4, yields means that . 
 . 
 [(ii) (i)] Assume that there is 
 We now prove that . . Pick arbitrarily 
With the help of Lemma 2 and Proposition 5, 
 . We now prove that 
we begin by proving 
 { } { } 
 { } 
(see Proposition 3). { } . (13) 
Set { } It is easy to see that 
 As , one has { } { } 
 (12) and that { } is a closed 
 Three following cases are possible: set. So, the inclusion “ ” in (13) holds trivially. 
 For the converse inclusion, take arbitrarily 
 Case 1. . Then, (12) yields ̃
 { } we 
 . will prove that 
 Case 2. . Then, one has 
 ̃
 { } . 
 , or equivalently, 
 ̃ ̃
 { } . This accounts As we have , 
 ̃ { } 
for { } , and then, which implies that 
 On the other hand, it holds 
by (11), one gets { } 
 (see Proposition 5), and hence, 
which yields . So, 
 { } (see Lemma 2). 
 . 
 So, one gets ̃ which yields 
 Case 3. . We claim that 
Conversely, by (12), suppose that . ( ̃ ) (14) 
Then, there is such that , 
or equivalently, Note that, one also has . 
 So, for each , it follows from (14) and 
 { } . 
This, together with (11), leads to the definition of infimum that the existence of 
 ̃ 
 such that ( ) , and 
 { } 
and hence, consequently, ( ̃ ) 
 (see Proposition 3) which yields 
( * +) { } 
 ( ̃ ) { } . 
 } 
 As ( ̃ ) ̃ 
(as * + is a neighborhood of 
 ). Consequently, there is we obtain 
 ̃
 such that which { } . 
yields . This contradicts the The proof is complete. 
fact that { }. We now recall the qualification condition 
 So, . (Nguyen Dinh et al., 2020) 
 { } { } 
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Natural Sciences issue 
 We now study the results on a strong zero On the other hand, by the weak duality 
duality gap between the problem (VP) and its (see Remark 3), one has 
Lagrange dual problems, which are established which, together with (15), gives 
under the condition without using and is achieved, taking 
Farkas-type results while the such ones were (Lohne, 2011, Corollary 1.48) into account. 
established in Nguyen Dinh et al., 2020, where 
the authors have used Farkas-type results for [ ] Assume that holds, we 
vector optimization under the condition will prove that holds. It is clear that 
 to obtain the ones (see Nguyen Dinh et 
 { } { } (16) 
al., 2020, Theorem 6.1). We will show that it is 
possible to obtain the ones by using the convex So, we only need to show that the 
separation theorem (through the use of converse inclusion of holds. Take 
Proposition 5 given in the previous section). ̅ . Then, one has ̅ . 
The important point to note here is the use of 
 Assume that { }. Then, in the 
the convex separation theorem to establish the 
 light of Proposition 4, one has { } 
Farkas-type results for vector optimization in 
 which also means that (see 
Nguyen Dinh et al., 2020 while the convex 
separation theorem to calculate the supremum Remark 1). Observing that , 
 consequently, . This entails ̅ , or 
value of in this paper. 
 equivalently, ̅ showing that 
 Theorem 2. Assume that { }. ̅ { } . 
Assume further that is -convex, that is - 
convex, and that is convex. Then, the Assume that { }. Then, as 
following statements are equivalent: holds, from Propositions 4 and 5, 
 { }. 
 holds, By the decomposition 
 has a strong zero duality gap. 
 (see Proposition 1) and the fact that 
 Proof. [ ] Assume that holds. 
Since is continuous, we have (see Remark 2), one 
 gets . So, there are 
 ( { } ) { } ̅ ̅
 and such that ̅ . 
 As holds, it follows from Proposition 
 Pick . For each , one has 
3 that . Recall that are 
nonempty subset of (by the definition of ̅
 ̅ 
 and Proposition 3 . So, 
 { } and , and then, This, together with the fact that 
Proposition 2 shows that yields the existence 
Noting that (Nguyen of sequence { } such that 
Dinh et al., 2017, Proposition 2.1(iv)). Hence, ̅ for all . 
 Combining this 
 Then, ̅ 
with the fact that and 
 , we get (see Proposition 3) which is equivalent to 
 ̅ { } . Here, note 
As we have 
 that ̅ ̅ , we obtain 
 (15) 
 ̅ { } , which is desired. 
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 Dong Thap University Journal of Science, Vol. 9, No. 5, 2020, 03-16 
 Remark 5. It is worth mentioning that 
when we take , K , , 
 { } 
 the problem collapses 
to the problem in Pham Duy Khanh et al. { } 
(2019). Then, the result on strong duality for the { } { } 
problem in Pham Duy Khanh et al. { } 
(2019, Theorem 4.3) follows from Theorem 2. 
 { }, and 
 Let us now introduce the second 
qualification condition, saying that is closed 
regarding the set { } concretely, 
 { } { } { } { } 
 Theorem 3. Assume that the problem { } 
 is feasible and { }. It is clear that the converse implication in 
Assume further that is -convex, that is - Theorem 3 does not hold. 
convex, and that is a convex set of . If the 
 5. A special case: Linear programming 
condition holds then the problem 
has a strong zero duality gap. In the this section, as an illustrate example 
 for the results established above, we consider a 
 Proof. According to Proposition 4 and 
 special case of the problem (VP), that is the 
Proposition 5, we have 
 linear programming: 
and . As holds, 
one finds that . Consequently, one has 
 . where , , and . 
 The following example shows that the Observing that the problem collapses to 
converse implication in Theorem 3 does not hold. the problem when we take , 
 K , , . 
 Example 1 Let 
 Then, the corresponding characterizing set of 
 and . Let 
 is 
and be such that 
and for all . It is easy to { } 
see that is -convex, that is -convex, and The qualification condition now is 
 is convex. In this case, we have 
 { } { } 
 ⋃ Recall that the Lagrange dual problem of 
 denoted by is 
 ⋃ 
 It is worth mentioning that the problem 
 By some calculations, we obtain is a special case of the linear 
 programming problem (IP) in Anderson (1983) 
 { } { } { 
 and the problem (ILP) in Pham Duy Khanh et 
 } 
 al. (2019) where . The duality for the 
 { } { } { problem was considered in Anderson 
 } (1983) under the closedness conditions. 
 Recently, Pham Duy Khanh et al. (2019) had 
 On the other hand, 
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Natural Sciences issue 
studied the duality for the problem under (i) holds, 
some necessary and sufficient conditions. 
 (ii) 
 We now introduce a new type of dual 
 Proof. Firstly, by Proposition 6, is 
problem of called the sequential dual 
 equivalent to The 
problem as follows: 
 conclusion now follows from Theorem 2. 
 [ ] We next introduce a sufficient condition, 
 which ensures the fulfillment of the condition 
 , and then, leads to the results on zero 
 The relations between the values of the duality gap for the pairs 
problem and its dual problems are given 
by the following proposition. 
 Proposition 7. Assume that there are 
 Proposition 6. It holds: 
 and such that 
 . 
 (14) 
 Proof. 
 (15) 
 Prove that : It is 
easy to see that Then, holds. 
 { } Proof. It is sufficient to prove that 
 { } { } . To do 
 { } 
 this, take . We will show that 
where { } . Indeed, since 
 { , it follows that there exists a net 
 } such that 
 { (16) 
 } 
 . (17) 
 Obviously, . So, . 
 Prove that : Take Assume that there are and 
 such that (14) and (15) holds. This, together 
 and such that 
 with (17), leads to the fact that 
Then, for all 
 , and hence, 
 for all . 
 , 
 Since and , it follows from 
 or, . 
 the above inequality that 
 The desired inequality follows from the 
 This, together with the last one of (15) and 
definition of the problems and . 
 (14), one gets 
 The next result extends (Pham Duy Khanh 
et al., 2019, Theorem 4.3) in the case when 
taking 
 Corollary 3. The following statements are 
equivalent: or equivalently, . From this and the 
 first one of (15), we obtain 
14 
 Dong Thap University Journal of Science, Vol. 9, No. 5, 2020, 03-16 
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