AWP-04

We consider two partitions over the space of linear semi-infinite programming parameters with a fixed index set and bounded coefficients (the functions of the constraints are bounded). The first one is the primal-dual partition inspired by consistency and boundedness of the optimal value of the linear semi-infinite optimization problems. The second one is a refinement of the primal-dual partition that arises considering the boundedness of the optimal set. These two partitions have been studied in the continuous case, this is, the set of indices is a compact infinite compact Hausdorff topological space and the functions defining the constraints are continuous. In this work, we present an extension of this case. We study same topological properties of the cells generated by the primal-dual partitions and characterize their interior. Through examples, we show that the results characterizing the sets of the partitions in the continuous case are neither necessary nor sufficient in both refinements. In addition, a sufficient condition for the boundedness of the optimal set of the dual problem has been presented. .


Introduction
We associate with each triplet π ∈  =    Above and henceforth, ℝ + () denotes the set of nonnegative general finite sequences, that is, functions :  → ℝ + satisfying that   = 0 for all  ∈  except maybe for a finite number of indices.In ℝ + () we consider the norms  ∞ and  1 .
As both primal and dual problems are defined with the same data ,  and , these are represented by the triplet  ∶= (, , ).The parameters space  is defined as the set of all triplets  with  and  fixed, equipped with the pseudometrics  ∶  ×  → [0, ∞], defined by where, π  = (  ,   ,   ) ∈ ,  = 1, 2 and ‖•‖ ∞ represents the uniform norm.
The partitions where  is an infinite compact Hausdorff topological space and the functions  and  are continuous, have been analyzed in [1], [5] and [6].In particular, the last reference deals with the consistence and boundedness of the optimal value of the problems and these properties define the primaldual partition.The interior of the sets generated through the partition is also studied.In [8], partitions corresponding to an arbitrary index set  and arbitrary functions  and  are considered.In the present paper a refinement of the primal-dual partition is presented.Trough the article we shall consider only bounded linear semi-infinite optimization problems.The new sets of the partition arise from the boundedness of the optimal set of the optimization problems.This work extends the study of the primaldual partition and its refinement to the case of bounded coefficients.
The new results are presented in Sections 3 and 4. In Section 3, we characterize the interior of the sets that are generated by the primal-dual partition and we show that the characterization is like that one obtained in the continuous case.In Section 4 we show that the conditions that characterize the sets generated by the refinement are neither necessary nor sufficient.In addition, we present a condition that implies the boundedness of the optimal set of the dual problem.

PRELIMINARY
This section begins with the notations to follow in the rest of the work.We denote by ℝ + the set of positive real numbers, and by ℝ ++ the set of positive real numbers where the zero is not included.In the -dimensional space ℝ  endowed with the Euclidean norm,  ′ stands for the transpose of the vector column , the null vector will be denoted by   .If  is a set of any topological space,   and   MOL2NET, 2018, 4, http://sciforum.net/conference/mol2net-04/awp-04 3 will denote the interior and closure, respectively.Given a nonempty set  ⊂ ℝ  ,   and   will denote its convex and conical hull, respectively.
Remember that  satisfies the Slater condition if there exist  ̅ ∈ ℝ  such that,   ′  ̅ >   for all  ∈ .
Also,  satisfies the strong Slater condition if there are  > 0 and  ̅ ∈ ℝ  , such that,   ′  ̅ ≥   +  for all  ∈ .From the definition we have that every parameter that satisfies the strong Slater condition is consistent and satisfies the Slater condition.However, the opposite is not true in general (see Example 4).In [4,Theorem]  We conclude this section with the characterization of the sets   ,  = 1, … , 6 where ,  and  play a crucial role, look at [6].The next theorem, proved in [8], holds for the general linear semi-infinite optimization, hence in the particular case when  and  are bounded, as well.

Primal-dual stability
In [6], the following theorem presents the characterization of the interior of the sets generated by the primal-dual partition in the continuous case.Only continuous perturbations are considered.(iv)    = ∅ for  = 4, 5, 6.
The theorem, obtained in [8], shows the results which characterize the interior of the sets generated by the primal-dual partition in the general case.Arbitrary perturbations are considered.(v)    = ∅ for  = 5, 6.
The following theorem shows that the characterization of the interior of the sets that are generated with the primal-dual partition, in the case of bounded coefficients, is like the continuous case.However, in the new case Slater condition is replaced by strong Slater condition, because in the case of bounded coefficient there are parameters that satisfy the Slater condition, but not the strong Slater condition (see Example 4).Note that in the case of bounded coefficients,  ∈   3 ,  ≠ ℝ  which is impossible, because otherwise (  , 1) ′ ∈  + ( ) (see Theorem 3.2), and in this case,  + ( ) = { +1 }.
In the same way we get that   4 = ∅.

First refined primal-dual partition
A refinement of the primal-dual partition follows from classifying the bounded primal and dual problems in two categories.The first one, is formed by solvable problems with bounded optimal set (S).The second one, includes unsolvable problems and those that have unbounded optimal set (N).The refinement is called refined primal-dual partition and it is shown in Table 2.
Table 2 In the refinement,  The other sets are the same as in the primal-dual partition.
The condition that characterizes the set Π   and which is presented in Lemma 2.1 is true in the case of bounded coefficients.In the following example,  1 ∈  1 and  1 satisfies the Slater condition.However, the dual problem associated with the parameter  1 is unsolvable.Hence, we show that the condition that characterizes the set Π   fails in the case of bounded coefficients.
Example 1.Let  = [0, 1] and  = 2.We define  1 ∶= ( 1 ,  1 ,  1 ) such that,   1 ≔ (, 1) ′ for all and ′ . In [7] it is shown that  1 ∈  1 and  1 satisfies the Salter condition, but the dual problem is not solvable.In the above example, the parameter has unsolvable dual problem, which means that the optimal set of the dual problem is empty, and in particular it is bounded.In addition, the parameter satisfies the strong Slater condition, in fact, for  =  In this way, we have that Λ * is bounded with the norm  ∞ .
In the case when  ≠   we shall only demonstrate that the infimum is strictly positive.Proof.Suppose that  {‖‖ 1 :  ∈  * } = 0.
Then there is {  } in  * such that, Since   ∈  * , for each , we have The above equality implies  =   , which is a contradiction.
With the examples below we will show that the conditions that characterize the sets generated by the refined primal-dual partition, in the continuous case, do not hold in the case of bounded coefficients.
We consider again the parameter of Example 1, and we show that  ∈   and that  satisfying the strong Slater condition are not sufficient conditions for  ∈  1 1 .
The following example shows that if  ∈  1 1 , it is not necessary condition that  ∈   and that  satisfy the strong Slater condition.
The dual problem is: The function is the only feasible solution and it is optimal.We conclude that  3 is solvable with bounded optimal set.This way, we have a parameter  3 ∈  1 1 , where  3 ∈   3 , but  3 does not satisfy the strong Slater condition.
In the particular case  =   we present a sufficient condition which implies that a given parameter  belongs to the set  The feasible set is shown in Figure 1.The problem is solvable with unbounded optimal set.On the other hand, it is true that   The function  = { 1,   = 1 0,  0 <  < 1, is the only feasible solution and it is also optimal.It implies that  4 is solvable and it has bounded optimal set.This way,  4 ∈  1  3 and although  4 ∈  4 \  4 .we have that  4 does not satisfy the strong Slater condition.
With the following example we show that the conditions  ∈ \  and  satisfying the strong Slater condition are not sufficient for  ∈  1 3 .

Figure 1 .
Figure 1.Feasible set of  4 .From the above figure, we have, ∈ (0, 1]}, which implies that 4 does not satisfy the strong Slater condition.The figure2shows us that 4

𝑆 𝐷 if and only if 𝒄 ∈ 𝑀 and 𝜋 satisfies the Slater condition.
it is shown that a parameter satisfies the strong Slater condition if and only if  +1 ≔ (  , 0) ′ ∉  , where  ∶=  {(  ,   ) ′ ∶  ∈ }.It is worth mentioning that in the continuous case strong Slater and Slater conditions coincide.
Since  * ⊆  and, for all ,   is in  * , we have From the equality above we have that for each , (,   ()) ′ is a positive linear combination of {(  ,   ) ′ ,  ∈ }.From the previous Lemma we have, for all , If Λ * is bounded with the norm  1 , then it is bounded with the norm  ∞ too.In fact, if there is  ∈ ℝ ++ such that, ‖‖ 1 ≤  for all  ∈  * , then .The last lemma shows that all linear positive combination can be represented by  + 1 elements and that the sums of the coefficients are equal.The next theorem proves the conjecture.Theorem 4.5.Let  a parameter, with || ≥  + 2. If  ∈  and  satisfies the strong Slater condition, then  * is bounded with respect to the norm  1 .Proof.As  ∈ , then  ∈    .In addition, as  satisfies the strong Slater condition,  ∈    .This means that,  ∈  1 , in particular,  ∈    .Now, if  * = ∅, the result is obvious.Consider that  * ≠ ∅ and  * is not bounded.Let {  } in  * ′ } are bounded, if  → ∞ in (1), it follows that this means that the strong Slater condition fails in this case, which contradicts the hypothesis.Corollary 4.6.If Λ * is not bounded with the norm  1 and || ≥  + 2, then  does not satisfy the strong Slater condition.Observation 4.7.
Proof.The first part of the last proof shows that  * is bounded,  * ≠ ∅ and   () = 0. Now, suppose there is  ∈  * such that, ∑    ∈  > 0. If  > 0, we have  ∈  * .In fact, since  ∈  * and  = ) ,  ∈  }, this means that  does not satisfy the strong Slater condition, which is a contradiction.Thus ∑    ∈  = 0Since  was arbitrary, we conclude that  * = { ≡ 0}.Now, we present another proof of the above theorem.
1  1.The proof follows from Lemma 2.1 and Theorem 4.9.Let  = (, , ) be a parameter with  = 0  and || ≥  + 2. If  ∈   and  satisfies the strong Slater condition, then  ∈  11.As a consequence of the above corollary, we have that the feasible set of the system {   ≥   :  ∈ } (when || ≥  + 2) is nonempty and bounded, if 0  ∈   {  :  ∈ } and there are  > 0 and MOL2NET, 2018, 4, http://sciforum.net/conference/mol2net-04/awp-0411  ̅ ∈ ℝ  such that,    ̅ ≥   +  for all  ∈ .In fact, if the conditions are true and we consider the parameter  = (, ,   ), then  ∈  1 1 .In particular,  * is nonempty and bounded.Since in this case  =  * the result is immediate.Let  = (, , ) be a parameter with  ≡ 0 and || ≥  + 2. If  ∈   and  satisfies the strong Slater condition, then  ∈  1 1 .Proof.If  ∈  and  satisfies the strong Slater condition, then, by Theorem 4.5,  * is bounded.Now, since  ∈  we have that  ≠ ∅.Furthermore, every feasible solution of the dual problem is optimal because  ≡ 0. In this way, we have  * ≠ ∅.This implies that  ∈    .It only remains to prove that  ∈    , but this one is equivalent to  ∈  , if the parameter  has consistent primal problem ([4, Corollary 9.3.1]).As  ∈  , we will only show that  has consistent primal problem.But it is consistent because, by hypothesis,  satisfies the strong Slater condition. without the strong Slater condition are not sufficient conditions for  to belong to  1 2 .In fact, (0, 1) ′ ∉   3 because the primal problem is consistent, also  3 ∈   3 and  3 does not satisfy the strong Salter condition, but  3 ∉  1 2 .The parameter  2 of Example 2, also shows that (0, 1) ′ ∉  ,  ∈   and  without the strong Slater condition are not necessary conditions for the belonging of  to  1 2 .In fact,  2 ∈  1 2 , (0, 1) ′ ∉   2 and  2 ∈   2 , but  2 satisfies the Slater condition.With the next example we demonstrate that the condition presented in Theorem 4.2 (iii) is not valid in the case of bounded coefficients.In particular, it shows that  ∈  \   and  with the strong Slater condition are not necessary conditions for  ∈  1 3 .