Solvability and Primal-dual Partitions of the Space of Continuous Linear Semi-infinite Optimization Problems

Barragán, Abraham; Hernández, Lidia; Todorov, Maxim; Barragán, Abraham; Hernández, Lidia; Todorov, Maxim

doi:10.13053/cys-22-2-2943

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Computación y Sistemas

versión On-line ISSN 2007-9737versión impresa ISSN 1405-5546

Comp. y Sist. vol.22 no.2 Ciudad de México abr./jun. 2018 Epub 21-Ene-2021

https://doi.org/10.13053/cys-22-2-2943

Articles of the Thematic Issue

Solvability and Primal-dual Partitions of the Space of Continuous Linear Semi-infinite Optimization Problems

Abraham Barragán¹

Lidia Hernández¹

Maxim Todorov²

^¹ BUAP, Puebla, Mexico.

^² UDLAP, Puebla, Mexico.

Abstract:

Different partitions of the parameter space of all linear semi-infinite programming problems with a fixed compact set of indices and continuous right and left hand side coefficients have been considered in this paper. The optimization problems are classified in a different manner, e.g., consistent and inconsistent, solvable (with bounded optimal value and nonempty optimal set), unsolvable (with bounded optimal value and empty optimal set) and unbounded (with infinite optimal value). The classification we propose generates a partition of the parameter space, called second general primal-dual partition. We characterize each cell of the partition by means of necessary and sufficient, and in some cases only necessary or sufficient conditions, assuring that the pair of problems (primal and dual), belongs to that cell. In addition, we show non emptiness of each cell of the partition and with plenty of examples we demonstrate that some of the conditions are only necessary or sufficient. Finally, we investigate various questions of stability of the presented partition.

Keywords: Linear semi-infinite programming; parameter space of continuous problems; primal-dual partition; stability properties

1 Introduction

For a given infinite compact Hausdorff topological space $T$ and $n∈ℕ$ , we associate with each triple $π=(a,b,c)∈Π:=C(T)n×C(T)×ℝn$ , where $C(T)$ is the Banach space of all continuous functions over the compact $T$ , a continuous linear semi-infinite programming problem:

$P:min⁡x∈ℝnc′x,s.t.a′tx≥bt, t∈T,$

and its corresponding Haar’s dual problem:

$D:max⁡λ∈ℝ+(T)∑t∈Tλtbts.t.∑t∈Tλtat=c,$

where $ℝ+(T)$ is the set of nonnegative functions $λ$ , and $λ:T→ℝ+$ such that $λt≠0$ for at most a finite number of indices belonging to $T$ .

Among the recent applications of continuous linear semi-infinite programming (LSIP), let us mention that the primal problem $P$ arises in functional approximation [⁵, ⁶], finance [¹²], Bayesian statistics [¹⁴], and the design of telecommunications networks [⁴, ¹³, ¹⁶], whereas the dual $D$ has been used in robust Bayesian analysis [³] and optimization under uncertainty [¹].

We denote the feasible (optimal) set of $P$ and $D$ by $F(F*)$ and $Λ(Λ*)$ , respectively. In the space of parameters $Π$ we consider the topology of the uniform convergence generated by the following extended distance: for given $πi=(ai,bi,ci)∈Π$ , $i=1,2$ , the distance between $π1$ and $π2$ is:

$d(π1,π2):=max⁡{∥c1−c2∥∞,max⁡t∈T‖(at1bt1)−(at2bt2)‖∞}.$

By $ΠCP$ , $ΠICP$ , $ΠBP$ and $ΠUBP$ , ( $ΠCD$ , $ΠICD$ , $ΠBD$ and $ΠUBD$ ) we denote the sets of parameters providing primal (dual) consistent, inconsistent, bounded, (with finite optimal value), and unbounded, (with unbounded optimal value) problem, respectively.

In ordinary linear programming (LP), if the primal problem is solvable then the dual problem is also solvable and the optimal values of both problems coincide. In LSIP these properties fail in general. The continuity property of $π=(a,b,c)$ ensures nice theoretical properties (e.g., in the duality context) and has computational implications (for instance, continuity guarantees the convergence of LSIP discretization algorithms). In particular, Goberna, Lopez, Todorov, Ochoa and Vera de Serio, among others, have investigated conditions under which the primal-dual pair in LSIP satisfies some of the above mentioned properties (see [⁷, ¹⁰, ¹¹, ¹⁵]).

In [¹¹], Goberna and Todorov have considered the consistency of the primal and dual problems, and have presented a characterization of the sets of the primal partition { $ΠICP$ , $ΠBP$ , $ΠUBP$ } the sets of the dual partition { $ΠICD$ , $ΠBD$ , $ΠUBD$ } and the sets (or states) of the primal-dual partition, formed by the intersections of the corresponding states of the primal and dual partitions. The stability properties of the different states have been studied, as well.

In [¹⁰], Goberna and Todorov divided the set of parameters with bounded primal (dual) problem $ΠBP$ ( $ΠBD$ ) into sets of parameters that have solvable primal (dual), problem with bounded optimal set $ΠSP$ ( $ΠSD$ ) and a set of parameters that have unsolvable primal (dual), problem or unbounded optimal set $ΠNP$ ( $ΠND$ ). This generates what we call a first general primal partition { $ΠICP$ , $ΠSP$ , $ΠNP$ , $ΠUBP$ }, a first general dual partition { $ΠICD$ , $ΠSD$ , $ΠND$ , $ΠUBD$ }, and a first general primal-dual partition. In the same article, Goberna and Todorov characterized, by means of necessary and sufficient conditions, when a given parameter belongs to a certain state of the above partitions.

They have also studied several topological and stability properties of each cell of the partitions. Later on, Ochoa and Vera de Serio reconsidered the characterizations presented in [¹¹], and investigated the stability of the states in the general case [¹⁵], i.e., without continuity properties of the functions involved in LSIP problems.

In this paper we present a more natural primal-dual partition in continuous LSIP than the one given in [¹⁰]. The new partition is generated by dividing the set of bounded primal (dual) problems $ΠBP$ ( $ΠBD$ ) in two: the set of parameters that have solvable primal (dual) problem $ΠSP$ ( $ΠSD$ ), and the set of parameters that have unsolvable primal (dual) problem $ΠnP$ ( $ΠnD$ ). Formally, we consider the partitions { $ΠICP$ , $ΠsP$ , $ΠnP$ , $ΠUBP$ } ({ $ΠICD$ , $ΠsD$ , $ΠnD$ , $ΠUBD$ }) of $Π$ from the primal (dual) perspective, and the primal-dual { $ΠsP∩ΠsD$ , $ΠnP∩ΠsD$ , $ΠsP∩ΠnD$ , $ΠnP∩ΠnD$ } of the set of parameters with primal and dual bounded problems.

We have obtained some necessary and some sufficient conditions showing that a certain element of the space of parameters belongs to certain subset generated by the second general primal-dual partition. Intersecting the nonempty pairs of the new general primal and dual partitions we obtain the second general primal-dual partition. By means of suitable examples we demonstrate that each subset of the second general primal-dual partition is nonempty. Only in a few cases we have succeeded to find necessary and sufficient conditions. If it is not the case, we provide counterexamples showing that given conditions are only necessary or sufficient. Finally, we investigate several topological and stability properties of the cells in the second general primal-dual partition.

This paper is organized as follows. In Section 2, we introduce the necessary notations, recall the basic results on LSIP which are frequently used throughout this article, and summarize the conditions presented in [¹⁰] and [¹¹] that characterize the states of the primal-dual partition and the first general primal-dual partition. Section 2.1 provides new conditions which are either necessary of sufficient guaranteeing that a given parameter belongs to a certain element of the second general primal-dual partition. Finally, we study some topological properties of the states of the second general primal-dual partition in Section 2.1.1. More precisely, we prove that the interior of certain cells is nonempty and provide some density results.

The results of this paper are partially announced without proofs in [²].

2 Preliminaries

Let us introduce the necessary notations for this paper. The symbol $0n$ denotes the null-vector in $ℝn$ , the $j$ -th element of the canonical basis of $ℝn$ is $ej$ . Given a nonempty set $X⊂ℝn$ , $conv X$ and $cone X$ are the convex and the canonical hull of $X$ , respectively ( $cone ∅={0n}$ ). If $X$ is a convex set, $dim⁡X(dim⁡∅=−1)$ denotes its dimension. From the topological side, if $X$ is a subset of any topological space, $int⁡X$ , $cl X$ and $bd X$ represent the interior, the closure and the boundary of $X$ , respectively.

We recall some concepts and basic results on LSIP, we shall use (all the proofs and references can be found in [⁹] and [¹⁷]). We associate with each triple $π=(a,b,c)$ the first and second moment cones of $π$ :

$M=cone{at:t∈T},$

and

$N=cone{(atbt):t∈T},$

as well as its characteristic cone

$K=cone{(atbt):t∈T;(0n−1)}.$

The Existence Theorem establishes that $P$ is consistent if and only if $(0n,1)′∉cl N$ . In such a case, the non-homogeneous Farkas Lemma establishes that the inequality $c′x≥d$ holds for all $x∈F$ , if and only if $(c,d)∈cl K$ . For the dual problem, $D$ is consistent if and only if $c∈M$ .

When various triples are simultaneously considered, they and their associated feasible, optimal, etc. sets will be distinguished by means of superscripts or subscripts: $πi$ , $Pi$ , $Di$ , $Fi$ , $Fi*$ , $Λi$ , $Λi*$ , $Mi$ , $Ni$ , $Ki$ .

We denote by $υP(π)$ and $υD(π)$ the optimal value of $P$ and $D$ , defining as usual:

$υP(π)=+∞ and υD(π)=−∞$

respectively, when the corresponding problem is inconsistent.

If we consider different primal-dual states of the LSIP problems, since $P$ and $D$ can be either inconsistent ( $IC$ ), bounded ( $B$ ), or unbounded ( $UB$ ), crossing both criteria, we get nine possible duality states, which are reduced to six by the Weak Duality Theorem: $υD(π)≤υP(π)$ . The primal-dual partition is presented in Table 1 (according to the Duality Theorem, the duality states 5 and 6 are impossible in LP [⁸], Proposition 4.2):

Table 1 Primal-dual partition

$D\P$	$IC$	$B$	$UB$
$IC$	$Π4$	$Π5$	$Π2$
$B$	$Π6$	$Π1$
$UB$	$Π3$

where $Π1=ΠBP∩ΠBD$ , $Π2=ΠUBP∩ΠICD$ , $Π3=ΠICP∩ΠUBD$ , $Π4=ΠICP∩ΠICD$ , $Π5=ΠBP∩ΠICD$ , and $Π6=ΠICP∩ΠBD$ .

The next Lemma describes the characterization of the duality states $Πi$ , $i=1,…,6$ in terms of $M$ , $N$ , and $K$ . This characterization appears in [¹¹].

Lemma 2.1. The following assertions hold:

(i) $π∈Π1$ if and only if $(0n,1)′∉cl N$ and $c∈M$ .

(ii) $π∈Π2$ if and only if $(0n,1)′∉cl N$ and $({c}×ℝ)∩cl N=∅$ .

(iii) $π∈Π3$ if and only if ${c}×ℝ⊆K$ .

(iv) $π∈Π4$ if and only if $(0n,1)′∈cl N$ and $c∉M$ .

(v) $π∈Π5$ if and only if $c∉M$ , $(0n,1)′∉cl N$ and $({c}×ℝ)∩cl N≠∅$ .

(vi) $π∈Π6$ if and only if $(0n,1)′∈cl N$ , $c∈M$ and ${c}×ℝ⊆K$ .

Corollary 2.2. [[⁹], Corollary 9.3.1] Given a consistent problem $P$ in LSIP, the next statements are equivalent.

(i) $F*$ is nonempty and bounded;

(ii) $c∈int⁡M$ .

The next results are valid in continuous LSIP, where the Slater Condition plays a crucial role. Recall that $π=(a,b,c)$ satisfies the Slater Condition if there exists $x¯∈ℝn$ such that $a′tx¯>bt$ , for all $t∈T$ . $π$ satisfies the Slater Condition if and only if $π∈int⁡ΠCP$ (This result can be found in [⁹]).

Theorem 2.3. [[⁹], Theorem 9.8] If $P$ is a consistent LSIP problem with consistent dual problem $D$ , then the next statements are equivalent.

(i) $Λ*$ is nonempty and bounded;

(ii) $π$ satisfies the Slater Condition.

We will use the next characterization of $ΠSP$ and $ΠSD$ .

Lemma 2.4. [[¹⁰], Lemma 2.2]

(i) $π∈ΠSP$ if and only if $(0n1)∉cl N$ and $c∈int⁡M$ .

(ii) $π∈ΠSD$ if and only if $c∈M$ and $π$ satisfies the Slater Condition.

If we consider the parameter space $Π$ , $P$ and $D$ can be either consistent, inconsistent, bounded or unbounded. Now, if in addition to the boundedness we consider the solvability, the bounded problems can be either solvable with optimal set nonempty and bounded ( $S$ ) or unsolvable ( $N$ ). In the latter case we include the problems that have optimal set unbounded. With this classification we obtain the first general primal partition { $ΠICP$ , $ΠSP$ , $ΠNP$ , $ΠUBP$ } and the first general dual partition { $ΠICD$ , $ΠSD$ , $ΠND$ , $ΠUBD$ } of the optimization problems space. Crossing both partitions, we get the first general primal-dual partition, which is presented in Table 2:

Table 2 First general primal-dual partition

$D\P$			$IC$			$UB$
				$B$
				$S$	$N$
$IC$			$Π4$		$Π5$	$Π2$

	$B$	$S$		$Π11$	$Π13$
	$B$	$N$	$Π6$	$Π12$	$Π14$

$UB$			$Π3$

where $Π11=ΠSP∩ΠSD$ , $Π12=ΠSP∩ΠND$ , $Π13=ΠNP∩ΠSD$ and $Π14=ΠNP∩ΠND$ .

In [¹⁰], Goberna and Todorov showed that $ΠSP∩ΠICD$ and $ΠICP∩ΠSD$ are empty sets, for this reason the corresponding boxes do not appear numbered. The next Theorem confirms that $Π1i$ , $i=1,…,4$ are nonempty.

Theorem 2.5. [[¹⁰], Theorem 3.1] $Π1i≠∅$ , $i=1,…,4$ .

The states $Π1i≠∅$ , $i=1,…,4$ , in continuous LSIP, are characterized by Goberna and Todorov in the next Theorem.

Theorem 2.6. [[¹⁰], Theorem 3.3]

(i) $π∈Π11$ if and only if $c∈int⁡M$ and $π$ satisfies the Slater Condition.

(ii) $π∈Π12$ if and only if $(0n1)∉cl N$ , $c∈int⁡M$ and $π$ does not satisfy the Slater Condition.

(iii) $π∈Π13$ if and only if $c∈M\int⁡M$ and $π$ satisfies the Slater Condition.

(iv) $π∈Π14$ if and only if $(0n1)∉cl N$ , $c∈M\int⁡M$ and $π$ does not satisfy the Slater Condition.

2.1 Second Refined Primal-dual Partition

In this section we present a refinement of Table 1, different to the refinement of Goberna and Todorov, presented in Table 2. To do this, we separate the parameter set with bounded primal problem $ΠBP$ , into two parameter sets. The first one, with solvable primal problem $ΠsP$ and the other with unsolvable primal problem $ΠnP$ . The same classification is made with respect to the dual problem. Having in mind the new notations, we get the second general primal partition { $ΠICP$ , $ΠsP$ , $ΠnP$ , $ΠUBP$ } and the second general dual partition { $ΠICD$ , $ΠsD$ , $ΠnD$ , $ΠUBD$ } of the parameter space. Crossing both partitions we obtain the second general primal-dual partition. The possible duality states in continuous linear optimization are enumerated in Table 3:

Table 3 Possible duality states in continuous linear optimization

$D\P$			$IC$			$UB$
				$B$
				$s$	$n$
$IC$			$Π4$	$Π^51$	$Π^52$	$Π2$

	$B$	$s$	$Π^61$	$Π^11$	$Π^13$
	$B$	$n$	$Π^62$	$Π^12$	$Π^14$

$UB$			$Π3$

where $Π^11=ΠsP∩ΠsD$ , $Π^12=ΠsP∩ΠnD$ , $Π^13=ΠnP∩ΠsD$ , $Π^14=ΠnP∩ΠnD$ , $Π^51=ΠsP∩ΠICD$ , $Π^52=ΠnP∩ΠICD$ , $Π^61=ΠICP∩ΠsD$ and $Π^62=ΠICP∩ΠnD$ .

Lemma 2.7. $Π^11⊇Π11$ and $Π^14⊆Π14$ .

Theorem 2.8. Let $π∈Π$ with primal and dual problems be consistent and bounded. The following assertions are true:

(i) If $c∈int⁡M$ and $π$ satisfies Slater Condition, then $π∈Π^11$ ;

(ii) $π∈Π^12$ , then $π$ does not satisfy the Slater Condition;

(iii) If $π∈Π^13$ , then $c∈M\int⁡M$ ;

(iv) If $π∈Π^14$ , then $c∈M\int⁡M$ and $π$ does not satisfy the Slater Condition.

Proof. i) Suppose that $c∈int⁡M$ and $π$ satisfies the Slater Condition. First, if $c∈int⁡M$ , then $F*≠∅$ and $F*$ is bounded [Corollary 2.2]. On the other hand, if $π$ satisfies the Slater Condition, then $Λ*≠∅$ and $Λ*$ is bounded [Theorem 2.3]. Therefore, if $c∈int⁡M$ and $π$ satisfies the Slater Condition, then $π∈Π^11$ .

ii) if $π∈Π^12$ , then $Λ*=∅$ , by Theorem 2.3 $π$ does not satisfy the Slater condition.

iii) If $π∈Π^13$ , then $F*=∅$ , by Corollary 2.2 $c∉int⁡M$ , in addition from hypothesis $c∈M$ . So, we conclude that $c∈M\int⁡M$ .

iv) If $π∈Π^14$ , then $F*=∅$ and $Λ*=∅$ this implies $c∈M\int⁡M$ and $π$ does not satisfy the Slater Condition.

With the following examples, we show that the above conditions, are only sufficient or necessary, respectively. The examples also show that $Π^1i≠∅$ for $i=1,2,3,4$ , which justifies the previous partition and Theorem. In ordinary linear programming we have $Π^12=Π^13=Π^14=∅$ , according to the Duality Theorem [[⁸], Theorem 4.4].

In the Example 2.9, $π1∈Π^12$ and $π1$ does not satisfy the Slater Condition, while in the Example 2.10, $ℝ2$ does not satisfy the Slater Condition and $π2∉Π^12$ . This shows that the condition (ii), stated in Theorem 2.8, is not a sufficient one.

Example 2.9. Consider the optimization problem in $ℝ2$

$P1:min⁡x∈ℝ2x2s.t. x1+rx2≥−r2,r∈[0,1], −x1+sx2≥−s2,s∈[0,1].$

If $r=0=s$ , then $x1≥0$ and $−x1≥0$ , we conclude that $x1=0$ . Now, as $x1=0$ , if $r,s∈(0,1]$ then $rx2≥−r2$ y $sx2≥−s2$ , i.e., $x2≥−r$ y $x2≥−s$ , it follows that $x2≥0$ . Therefore:

$F1={(0x2)∈ℝ2:x2≥0},υP(π1)=0,$

and $F1*={(00)}$ . As $dim⁡F1=1$ , $F1⊂ℝ2$ and $π1$ is continuous, we have that $π1$ does not satisfy the Slater Condition. InFigure 1, we show $F1$ :

The dual problem of $P1$ is:

$D1:max⁡λ,γ∈ℝ([0,1])(∑r∈[0,1]−λrr2+∑s∈[0,1]−γss2),$

Fig. 1 Feasible set of $P1$

$s.t.∑r∈[0,1]λr(1r)+∑s∈[0,1]γs(−1s)=(01),$

which is equivalent to:

$−(min⁡λ,γ∈ℝ+([0,1])(∑r∈[0,1]λrr2+∑s∈[0,1]γss2)), s.t.∑r∈[0,1]λr(1r)+∑s∈[0,1]γs(−1s)=(01).$

If:

$υ1:min⁡λ,γ∈ℝ+([0,1]){∑r∈[0,1]λrr2+∑s∈[0,1]γss2|,∑r∈[0,1]λr(1r)+∑s∈[0,1]γs(−1s)=(01)},$

we have $0≤υ1$ . Now, if $t0∈(0,1]$ , then $θ¯0=(λ0;γ0)$ is in $Λ1$ , if and only if:

$λt00(1t0)+λt00(−1t0)=(01),$

where $λt00=γt00$ and $λr0=0=γs0$ for all $r,s∈[0,1]\{t0}$ . l.e., $λt00$ has to satisfy the equality:

$1=λt00t0+λt00t0.$

So, we conclude that $θ¯0∈Λ1$ , if and only if $λt00=12t0$ . Then:

$υ1≤∑r∈[0,1]λr0r2+∑s∈[0,1]γs0s2=t022t0+t022t0=t0.$ (1)

If $t0$ approaches to zero in(1), we have $υ1≤0$ . Therefore $υD(π1)=0$ . On the other hand, $υD(π1)=0$ , if and only if $λrr2=0=λss2$ , for all $(r,s)∈[0,1]×[0,1]$ . Then, the only possible optimal solutions have the form $θ¯1=(λ1;γ1)$ , where $λ01,γ01∈ℝ+$ and $λr1=0=γs1$ for all $r,s∈(0,1]$ , but $θ¯1∉Λ1$ , because:

$λ01(10)+γ01(−10)=(01),$

is impossible. Therefore $Λ1*=∅$ . All these show that $π1∈Π^12$ and $Π^12≠∅$ .

Example 2.10. We consider, the following problem:

$P2:min⁡x∈ℝ2x2s.t. x1≥0, −x1≥0, x2≥0.$

InFigure 2, we show the feasible set of $P2$ .

Fig. 2 Feasible set of $P2$

We observe that $P2$ is consistent, $υP(π2)=0$ and $F2*={(00)}$ . In addition, as $dim⁡F2=1$ and $F2⊂ℝ2$ , we have that $π2$ does not satisfy the Slater Condition.

The dual problem of $P2$ is:

$D2:max⁡λ1,λ2,λ3≥0(λ10+λ20+λ30),s.t. λ1(01)+λ2(0−1)+λ3(10)=(10).$

From the problem it follows that $υD(π2)=0$ . Now, $λ¯=(λ1,λ2,λ3)′$ with $λ1,λ2,λ3≥0$ is in $Λ2=Λ2*$ , if and only if:

$λ1(10)+λ2(−10)+λ3(01)=(01),$

i.e., $0=λ1−λ2$ and $1=λ3$ , or equivalently $λ1=λ2$ and $1=λ3$ . Then:

$Λ2=Λ2*={(λ1,λ1,1)′∈ℝ3:λ1≥0}.$

Therefore, $π2∉Π^12.$ .

In the Example 2.11, $π3∈Π^13$ and $c3∈M3\int⁡M3$ . While, in the Example 2.12, $π4∉Π^13$ and $c4∈M4\int⁡M4$ . This shows that the condition (iii), stated in Theorem 2.8, is not a sufficient condition.

Example 2.11. Consider in $ℝ2$ the problem:

$P3: min⁡x∈ℝ2x1,s.t. x1+t2x2≥2t, t∈[0,1].$

InFigure 3, we show the feasible set of $P3$ :

Fig. 3 Feasible set of $P3$

$π3$ satisfies the Slater Condition and $(22)$ is a Slater point. In fact $1+t2>t$ for all $t∈[0,1]$ if and only if $2+2t2>2t$ for all $t∈[0,1]$ , so, $P3$ is consistent. InFigure 3, we observe that $υP(π3)=0$ but $F3*=∅$ , i.e., $P3$ is not solvable. Moreover, inFigure 4, we show that $c3∈M3\int⁡M3$ .

Fig. 4 First moment cone of $P3$

The dual problem of $P3$ is:

$D3:max⁡λ∈ℝ([0,1])∑t∈[0,1]λt2t,s.t. ∑t∈[0,1]λt(1t2)=(10).$

The function $λ∈ℝ+([0,1])$ defined as:

$λt:={1,if t=0,0,if t∈(0,1],$

is a feasible solution for the problem $D3$ with $υD(π3)=0$ . It follows that $D3$ is solvable. Therefore, $π3∈Π^13$ and $Π^13≠∅$ .

Example 2.12. The primal problem $P4$ is formulated as follows:

$P4: min⁡x∈ℝ2x1, s.t. x1+t2x2≥0, t∈[0,1].$

InFigure 5, we show the feasible set of $P4$ .

Fig. 5 Feasible set of $P4$

Since $π4$ satisfies the Slater Condition, then $P4$ is consistent. We observe that $υP(π4)=0$ and $F4*={0}×ℝ+$ , therefore $P4$ is solvable. On the other hand, as in the above example, $c4∈M4\int⁡M4$ .

The dual problem of $P4$ is now:

$D4:max⁡λ∈ℝ([0,1])∑t∈[0,1]λt0,s.t. ∑t∈[0,1]λt(1t2)=(10).$

Again, we have that the function $λ∈ℝ([0,1])$ defined as:

$λt:={1,if t=0,0,if t∈(0,1],$

is a feasible solution of the problem $D4$ with $υD(π4)=0$ . Thus, we conclude that $D4$ is solvable.

Observation 2.13. With the previous example we also show that $Π^11∉∅$ . In the same example $π4∈Π^11$ and $c4∉int⁡M$ . On the other hand, in the Example 2.10 $π2∈Π^11$ and $π2$ does not satisfy the Slater Condition. This means that both $c∈int⁡M$ and the Slater Condition are not necessary conditions for $π∈Π^11$ .

In the next Example 2.14, $π5∈Π^14$ , $c5∈M5\int⁡M5$ and $π5$ does not satisfy the Slater Condition. Later on, in the Example 2.15, $π6∈Π^11$ , $(0n1)∉cl N6$ , $c6∈M6\int⁡M6$ and $π6$ does not satisfy Slater Condition. This shows two things. First the condition (i) in Theorem 2.8 is not a necessary condition, and second the condition (iv), stated in Theorem 2.8, is not a sufficient one.

Example 2.14. Consider in $ℝ3$ the primal problem, with $α>0$ :

$P5: min⁡x∈ℝ3(x1+αx3),s.t. x1+t2x2≥2t, t∈[0,1], sx3≥−s2, s∈[0,1], −rx3≥−r2, r∈[0,1].$

If $s, r∈(0,1]$ then $x3≥−s$ and $x3≤r$ , which means that $x3=0$ . So, $dim⁡F5≤2$ . We can look at $F5$ in $ℝ2$ , as the feasible set of $P3$ . Therefore, $υP(π5)=0$ and $F5*=∅$ . In addition, as $F5=2$ and $F5⊂ℝ3$ , we have that $π5$ does not satisfy the Slater condition. On the other hand, $c5∈M5\int⁡M5$ . In fact:

$(10α)=1(100)+α(001),$

and, for all $ε>0$ :

$(1−ε2α)∉M5,$

and:

$‖10α−1−ε2α‖2=ε2<ε.$

Therefore, $c5∈M5\int⁡M5$ .

The dual problem $P5$ is:

$D5:max⁡λ,β,γ∈ℝ+([0,1])(∑t∈[0,1]λt2t+∑s∈[0,1]βs(−s2)+∑r∈[0,1]γr(−r2)),s.t. ∑t∈[0,1]λt(1t20)+∑s∈[0,1]βs(00s)+∑r∈[0,1]γr(00−r)=(10α).$

This is equivalent to:

$−(min⁡γ,β,γ∈ℝ+([0,1])(−(∑t∈[0,1]λt2t)+∑s∈[0,1]βss2+∑r∈[0,1]γrr2)),s.t. ∑t∈[0,1]λt(1t20)+∑s∈[0,1]βs(00s)+∑r∈[0,1]γr(00−r)=(10α).$

From the above equality system, we have that:

$(10)=∑t∈[0,1]λt(1t2) where λ∈ℝ+([0,1]).$

The only solution of the equation above is $λ0∈ℝ+([0,1])$ with $λ00=1$ and $λt0=0$ , for all $t∈(0,1]$ . Then, feasible points of $D5$ have the form:

$θ¯=(λ0;β;γ),$

where $β,γ∈ℝ+([0,1])$ . If we evaluate the objective function of the last problem at the points that have the above form, the problem will be reduced to:

$−(min⁡β,γ∈ℝ+([0,1])(∑s∈[0,1]βss2+∑r∈[0,1]γrr2)), s.t. ∑s∈[0,1]βss−∑r∈[0,1]γrr=α.$ (2)

From(2)it follows that:

$υ5: min⁡β,γ∈ℝ+([0,1]){∑s∈[0,1]βss2+∑r∈[0,1]γrr2|,∑s∈[0,1]βss−∑r∈[0,1]γrr=α}≥0.$

Note that for each $i∈(0,1]$ :

$θ0:=(β0;γ0),$

where $β0,γ0∈ℝ+([0,1])$ , $βi0=αi$ , $βs0=0$ for all $s∈[0,1]\{i}$ and $γr0=0$ for all $r∈[0,1]$ , is a feasible point of(2). Moreover, if we evaluate $∑s∈[0,1]βss2+∑r∈[0,1]γrr2$ in $θ0$ , we have that:

$υ5≤∑s∈[0,1]βs0s2+∑r∈[0,1]γr0r2=αi.$ (3)

If $i→0$ in(3), we have that $υ5≤0$ . Therefore $υD(π5)=0$ . On the other hand, if a feasible point of the problem $D5$ is an optimal solution, the objective function evaluated at this point must satisfy:

$∑s∈[0,1]βss2+∑r∈[0,1]γrr2=0.$

Then, the possible optimal solutions of problem $D5$ have the following form: $θ¯0=(λ0;β1;γ1)$ , where $λ0,β1,γ1∈ℝ+([0,1])$ , $λ00=1$ , $λt0=0$ for all $t∈(0,1]$ , $β01∈ℝ+$ , $βs1=0$ for all $s∈(0,1]$ , $γ01∈ℝ+$ and $γr1=0$ for all $r∈(0,1]$ . But $θ¯0∉Λ5$ , because it implies $α=0$ , which is a contradiction with the hypothesis $α>0$ . Therefore, $Λ5*=∅$ .

Example 2.15. Consider the following problem in: $ℝ2$

$P6:min⁡x1,x∈ℝ2s.t. x1≥0, −x1≥0, x2≥0.$

The feasible set of $P6$ is the same as the feasible set of the Example 2.10. Then $υP(π6)=0$ and $F6*={0}×ℝ+$ in addition, since $dim⁡F6=1$ and $F6⊂ℝ2$ , we have that $π6$ does not satisfy the Slater Condition. On the other hand, inFigure 6, we show that $c6∈M6\int⁡M6$ .

Fig. 6 First moment cone of $P6$

The dual problem of $P6$ is:

$D6:max⁡λ1,λ2,λ3≥0(λ10+λ20+λ30),s.t. (10)=λ1(10)+λ2(−10)+λ3(01).$

From the problem it follows that, $υD(π6)=0$ . Now:

$λ¯=(λ1λ2λ3)∈Λ6=Λ6*,$

if and only if:

$(10)=λ1(10)+λ2(−10)+λ3(01),$

i.e., $1=λ1−λ2$ and $0=λ3$ , or equivalently, $λ1=1+λ2$ and $λ3=0$ . Then:

$Λ6=Λ6*={(1+λ2λ20)∈ℝ3:λ2≥0},$

which implies that $Λ6*≠∅$ . Therefore, $π6∉Π^14$ .

Remember that in the first refined primal-dual partition, the sets $ΠSP∩ΠICD$ and $ΠICP∩ΠSP$ are empty. However, we will show that the sets $ΠsP∩ΠICD=Π^51$ and $ΠICP∩ΠsD=Π^61$ , in the second general primal-dual partition, are nonempty. We shall present some necessary conditions for the fact that a given parameter $π$ belongs to the state $Π^52=ΠnP∩ΠICD$ and $Π^62=ΠICP∩ΠnD$ , respectively. Using also the definitions of the states, we shall characterize the cells $Π^51$ and $Π^61$ .

Proposition 2.16. $π∈Π^51$ , if and only if $c∉M$ and $F*$ is not bounded.

Proof. Suppose that $π∈Π^51$ and $c∈M$ or $F*$ is bounded. First, if $c∈M$ , we have the contradiction $Λ≠∅$ . Second, if $F*$ is bounded, then $π∈ΠSP∩ΠICD$ and again we get to a contradiction. On the other hand, if $c∉M$ and $F*$ is not bounded, then $Λ=∅$ and $F*≠∅$ . Therefore, $π∈Π^51$ .

Now, $Π^5i⊂Π5$ for $i=1,2$ . Then, for the Lemma 2.1:

$c∉M, (0n,1)′∉cl N and ({c}×ℝ)∩cl N≠∅,$

is a necessary condition for $π$ belongs to $Π^5i$ for $i=1,2$ , respectively.

Proposition 2.17. $π∈Π^61$ , if and only if $(0n1)∈cl N$ and $Λ*$ is not bounded.

Proof. Suppose that $π∈Π^61$ and $(0n1)∉cl N$ or $Λ*$ is bounded. First, if $(0n1)∉cl N$ , we have the contradiction $F≠∅$ . Second, if $Λ*$ is bounded, then $π∈ΠICP∩ΠSD$ and also we get again to a contradiction. On the other hand, if $(0n1)∈cl N$ and $Λ*$ is not bounded, then $F=∅$ and $Λ*≠∅$ . Therefore, $π∈Π^61$ .

Again, $Π^6i⊂Π6$ for $i=1,2$ . Then, by the Lemma 2.1:

$(0n,1)′∈cl N, c∈M and {c}×ℝ⊆K,$

is a necessary condition for $π$ belongs to $Π^6i$ for $i=1,2$ , respectively.

With the following examples, we will show that the conditions (v) and (vi), presented in Lemma 2.1, are only necessary. We will also show that in continuous LSIP $Π^ji≠∅$ for $i=1,2$ and $j=5,6$ . However, in ordinary linear programming all these sets are empty [[⁸], Proposition 4.2].

Example 2.18. Consider, in $ℝ2$ , the optimization problem:

$P7: min⁡x∈ℝ2x2,s.t. t2x1+tx2≥t, t∈[0,1]$

The feasible set of $P7$ is presented in theFigure 7.

Fig. 7 Feasible set of $P7$

Figure 7, shows that $P7$ is consistent and bounded, with an optimal value $υP(π7)=1$ , and an optimal set:

$F7*={(x1x2)∈ℝ2:x1≥0 and x2=1}.$

From the previous equality it follows that $F7*$ is unbounded. The cone $M7$ of $P7$ , is shown in theFigure 8.

As $c7=(01)$ , we have that $c7∈cl M7\M7$ , then the dual problem $D7$ is inconsistent. We conclude that $π7∈Π^51$ , therefore $Π^51≠∅$ .

Fig. 8 First moment cone of $P7$

Example 2.19. Consider now, the next problem in $ℝ2$ :

$P8: min⁡x∈ℝ2x1,s.t. tx1+t3x2≥t2, t∈[0,1].$

The feasible set of $P8$ is shown in theFigure 9.

Fig. 9 Feasible set of $P8$

Figure 9, shows that $P8$ is consistent and bounded, with optimal value $υP(π8)=0$ , however $F*=∅$ , i.e., $P8$ is unsolvable. The cone $M8$ of $P8$ , is presented in theFigure 10:

As $c8=(10)$ , we have that $c8∈cl M8\M8$ , whereby the dual problem $D8$ is inconsistent. So, whe observe that $π8∈Π^52$ , therefore $Π^52≠∅$ .

Fig. 10 First moment cone of $P8$

Observation 2.20. In the Example 2.19, $c8∉M8$ , $(0n,1)′∉cl N8$ ( $P8$ is consistent), also $({c8}×ℝ)∩cl N8≠∅$ ( $P8$ is bounded), but $π8∉Π^51$ . It shows that $c∉M$ , $(0n,1)′∉cl N$ and $({c}×ℝ)∩cl N≠∅$ is not a sufficient condition for $π∈Π^51$ . On the other hand, in the Example 2.18, $c7∉M7$ , $(0n,1)′∉cl N7$ , and $({c7}×ℝ)∩cl N7≠∅$ , but $π7∉Π^52$ . It shows that $c∉M$ , $(0n,1)′∉cl N$ and $({c}×ℝ)∩cl N≠∅$ is not a sufficient condition for $π∈Π^52$ , as well.

Example 2.21. We now study the following problem in $ℝ2$ :

$P9: min⁡x∈ℝ2x1, s.t. x1+t2x2≥2t, t∈[0,1], −x1≥0.$

We observe that for each $t∈(0,1]$ :

$[12t(1t22t)+12t(−100)]=(0t21),$ (4)

is an element of $N9$ . If $t→0$ in(4), we have that $(02,1)′∈cl N9$ . Therefore $P9$ is inconsistent.

The dual problem of $P9$ is:

$D9: max⁡λ∈ℝ+([0,1]),γ∈ℝ+∑t∈[0,1]λt2t,s.t. ∑t∈[0,1]λt(1t2)+γ(−10)=(10).$

we have that $θ¯=(λ0;γ)$ , where $γ∈ℝ+$ and $λ0∈ℝ+([0,1])$ defined as:

$λt0:={1+γ,if t=0,0,if t∈(0,1],$

is a feasible point of $D9$ . Then $υD(π9)=0$ and $Λ*$ is unbounded. This means that $π9∈Π^61$ . Therefore $Π^61≠∅$ .

Example 2.22. Consider, in $ℝ2$ , the following problem:

$P10: min⁡x∈ℝ2x2,s.t. t2x1≥t, t∈[0,1] sx2≥−s2, s∈[0,1]$

We observed that for each $t∈(0,1]$ :

$[1t(t20t)]=(t01),$ (5)

is an element of $N10$ . If $t→0$ in(5), we observe that $(02,1)′∈cl N10$ . Therefore, $P10$ is inconsistent.

The dual problem of $P10$ is:

$D10:max⁡λ,γ∈ℝ+([0,1])(∑t∈[0,1]λtt+∑s∈[0,1]γs(−s2)), s.t. ∑t∈[0,1]λt(t20)+∑s∈[0,1]γs(0s)=(01).$

From the system above, we have that:

$0=∑t∈[0,1]λtt2 con λ∈ℝ+([0,1]),$

whose solutions have the form $λ0∈ℝ+([0,1])$ with $λ00∈ℝ+$ and $λt0=0$ for all $t∈(0,1]$ . Then, the feasible points of $D10$ look like:

$θ¯10=(λ0;γ),$

where $γ∈ℝ+([0,1])$ . If we evaluate the objective function of dual problem in points that have the above form, the problem is reduced to:

$max⁡γ∈ℝ+([0,1])∑s∈[0,1]γs(−s2),s.t. ∑s∈[0,1]γss=1,$

which is equivalent to:

$−min⁡γ∈ℝ+([0,1])∑s∈[0,1]γss2, s.t. ∑s∈[0,1]γss=1.$

we have that:

$υ10:=min⁡γ∈ℝ+([0,1]){∑s∈[0,1]γss2|∑s∈[0,1]γss=1}≥0.$

Now, if $s0∈(0,1]$ , then $γ0∈ℝ+([0,1])$ satisfies:

$∑s∈[0,1]γs0s=1,$

where $γs00=1s0$ and $γs0=0$ for all $s∈[0,1]\{s0}$ . Hence:

$υ10≤∑s∈[0,1]γs0s2=1s0s02=s0.$ (6)

If $s0→0$ in(6), we have $υ10≤0$ . Therefore, $υD(π10)=0$ . On the other hand, $υD(π10)=0$ , if and only if:

$∑t∈[0,1]λt0t=∑s∈[0,1]γss2, for all θ¯10=(λ0;γ)∈Λ10,$

but, if $θ¯10∈Λ10$ ,

$0=∑t∈[0,1]λt0t,$

it follows that:

$0=∑s∈[0,1]γss2 for all θ¯10=(λ0;γ)∈Λ10.$

So, the possible optimal solutions of $D10$ , have the form $θ¯100=(λ0;γ1)$ , where $λ0,γ1∈ℝ+([0,1])$ , $λ00,γ01∈ℝ+$ and $λt0=0=γs1$ for all $t,s∈(0,1]$ . If $θ¯100∈Λ10$ , then:

$(01)=λ00(00)+γ01(00),$

which is impossible. Therefore, $θ¯100∉Λ10$ , whereby $Λ10*=∅$ . So, we conclude that $π10∈Π^62$ , and thus $Π^62≠∅$ .

Observation 2.23. We see in Example 2.22 that $(0n,1)′∈cl N10$ ( $P10$ is inconsistent), $c10∈M10$ ( $θ¯10=(λ0;γ0)∈Λ10$ , i.e, $D10$ is consistent) and ${c10}×ℝ⊆K10$ ( $υD(π10)=0$ , i.e, $D10$ is bounded), but $π10∉Π^61$ . This shows that $(0n,1)′∈cl N$ , $c∈M$ and ${c}×ℝ⊆K$ is not a sufficient condition for $π∈Π^61$ . On the other hand, in the Example 2.21, $(0n,1)′∈cl N9$ , $c9∈M9$ and ${c9}×ℝ⊆K9$ , but $π9∉Π^62$ . It demonstrates that $(0n,1)′∈cl N$ , $c∈M$ and ${c}×ℝ⊆K$ is not a sufficient condition for the following statement $π∈Π^62$ .

2.1.1 Some Topological Properties of the Sets Generated by the Second General Primal-dual Partition

In [¹¹], Goberna and Todorov presented the characterization of the interior of the sets $Π2$ , $Π3$ and $Π4$ . They also studied the density properties and the interior of these and other sets of the first general primal-dual partition. In this section, we shall study some topological properties of the sets $Π^11$ , $Π^12$ , $Π^13$ , $Π^14$ , $Π^51$ , $Π^52$ , $Π^61$ and $Π^62$ . In particular, we investigate the interior and some density properties of the mentioned sets of the new second general primal-dual partition.

Theorem 2.24. $Π^11$ is dense in $Π1$ .

Proof. Since $Π11⊆Π^11$ and $Π11$ is dense in $Π1$ [[¹⁰], Theorem 3.3], hence $Π^11$ is dense in $Π1$ .

Theorem 2.25. The sets $Π^1i$ , $i=1,2,3,4$ are neither closed nor open.

Proof. Since these sets are cones with the null triplet belonging to $Π^11$ , only $Π^11$ could be closed:

$Π^11$ is not closed. In fact, consider the sequence ${πr}$ in $Π^11$ , where $πr:=(1re1,1,e1)$ , obviously:

$lim⁡r→∞πr=(0n,1,e1),$

but $(0n,1,e1)∈Π4$ . Therefore, $Π^11$ is not closed.

$Π^11$ is not open. Indeed, consider $π:=(0n,0,0n)$ in $Π^11$ . Let $r>0$ , we define $πr:=(0n,0,r2e1)$ . It is easy to see that, for all $r>0$ , $0,πr∈Π2$ and $d(π,πr)=r2<r$ . This implies that $Π^11$ is not open.

We shall prove that the sets $Π^1i$ , $i=2,3,4$ are not open. Since $Π^1i⊂Π1$ , then $int⁡Π^1i⊂int⁡Π1$ , but $int⁡Π1=Π11$ [[¹⁰], Theorem 2], as $Π11⊂Π^11$ , so it follows that $int⁡Π^1i⊂Π^11$ . However $Π^1i∩Π^11=∅$ , and, the above inclusion is only possible if $int⁡Π^1i=∅$ . Since $Π^1i≠∅$ it follows that $Π^1i$ is not open for every $i=2,3,4$ .

Corollary 2.26. $int⁡Π^1i$ , $i=2,3,4$ are empty.

Theorem 2.27. $int⁡Π^11≠∅$ .

Proof. Since $Π11⊂Π^11$ then $int⁡Π11⊂int⁡Π^11$ .

but:

$int⁡Π11=Π11$

and:

$Π11≠∅$

we conclude that $int⁡Π^11≠∅$ .

Theorem 2.28. $int⁡Π^11$ is dense in $Π1$ .

Proof. Since $Π11⊂Π^11⊂Π1$ , then:

$int⁡Π11⊂int⁡Π^11⊂int⁡Π1$

it follows:

$int⁡Π11¯⊂int⁡Π^11¯⊂int⁡Π1¯.$

As $int⁡Π11$ is dense on $Π1$ [[¹⁰], Theorem 3.3] and $int⁡Π1$ is dense on $Π1$ [[¹¹], Theorem 2], we have that $int⁡Π^11¯=Π1$ . Therefore, $int⁡Π^11$ is dense in $Π1$ .

Theorem 2.29. $int⁡Π^11=Π11$ .

Proof. Since $Π11$ is open [[¹⁰], Theorem 3.3], it follows $Π11⊂int⁡Π^11$ . We will only show that $int⁡Π^11\Π11=∅$ . Suppose the contrary, i.e., $int⁡Π^11∩(Π11)c≠∅$ . Then, there exists $π∈int⁡Π^11$ , such that $π∉Π11$ , so it follows that $c∉int⁡M$ or $π$ does not satisfy the Slater Condition [Theorem 2.6].

First, $c∉int⁡M$ implies that $c∈M\int⁡M$ because, by hypothesis $π∈int⁡Π^11$ , but:

$int⁡Π^11⊂Π^11⊂Π1.$

In addition, $c∉int⁡M$ implies $M≠ℝn$ . From the above two implications, we conclude that, there exists a sequence ${cr}$ from $ℝn\M$ such that:

$lim⁡r→∞cr=c.$

We define $πr:=(a,b,cr)$ . The sequence ${πr}$ is in $ΠICD$ and satisfies:

$lim⁡r→∞πr=π.$

This is a contradiction, because, by hypothesis, $π∈int⁡Π^11$ .

Second, if $π$ does not satisfy the Slater Condition, then $π∉int⁡ΠCP$ . Now, by hypothesis $π∈int⁡Π^11$ , we have $π∈ΠCP$ . So, if $π$ does not satisfy the Slater Condition and the hypothesis is true, $π∈bd ΠCP$ . Therefore there exists a sequence ${πr}$ on $ΠICP$ such that:

$lim⁡r→∞πr=π.$

This is a contradiction, because $π∈int⁡Π^11$ .

It follows that $int⁡Π^11\Π11=∅$ , or equivalently $int⁡Π^11=Π11$ .

Observation 2.30. Since $int⁡Π5=∅$ and $int⁡Π6=∅$ [[¹¹], Theorem 2], we conclude that $int⁡Π^ji=∅$ for $i=1,2$ and $j=5,6$ .

We have proved that all parameters in $Π1$ can be approached by parameters in $Π^11$ . In addition, we have shown that the sets $Π^11$ , $Π^12$ , $Π^13$ and $Π^14$ are neither closed nor open, and that the interior of the sets $Π^12$ , $Π^13$ and $Π^14$ are empty. The characterization of $Π^11$ follows from the equality $Π^11=Π11$ which was also proved in this section.

3 Conclusion and Future Work

To conclude, we would like to mention that the lack of necessary and sufficient conditions, characterizing the majority of the states of the second general primal-dual partition has, in some sense, its justification. Namely, there are no necessary and sufficient conditions for the solvability, neither for the primal nor for the dual linear semi-infinite optimization problems. On the contrary, for the solvability, considered in the first general primal-dual partition, the conditions could be found in lemma 2.4. Anyway, finding such necessary and sufficient conditions is still a challenging problem. Another open question is how will apply the developed theory in this article removing the continuity in the LSIP.

Acknowledgements

The authors are grateful to the referees and the editor for their constructive comments and helpful suggestions which have contributed to the final preparation of the paper. This research was supported by CONACYT, Mexico, Sistema Nacional de Investigadores, Mexico, and partially supported by MINECO of Spain and ERDF of EU, Grant MTM2014-59179-C2-1-P.

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Received: January 30, 2017; Accepted: April 21, 2017

Corresponding author is Abraham Barragán. ab_beni211@hotmail.com, lhernan@fcfm.buap.mx, maxim.todorov@udlap.mx.

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