EXISTENCE OF SOLUTIONS OF MINIMIZATION PROBLEMS WITH AN INCREASING COST FUNCTION AND POROSITY

We consider the minimization problem f (x)→min, x ∈ K , where K is a closed subset of an ordered Banach spaceX and f belongs to a space of increasing lower semicontinuous functions on K . In our previous work, we showed that the complement of the set of all functions f , for which the corresponding minimization problem has a solution, is of the first category. In the present paper we show that this complement is also a σ-porous set.


Introduction
The study of a generic existence of solutions in optimization has recently been a rapidly growing area of research (see [1,2,3,4,5,6,8,9,10,12,13,14,15] and the references mentioned there).Instead of considering the existence of solutions for a single cost function, we study it for a space of all such cost functions equipped with an appropriate complete uniformity and show that a solution exists for most of these functions.Namely, we show that in the space of functions, there exists a subset which is a countable intersection of open everywhere dense sets such that for each cost function in this subset, the corresponding minimization problem has a unique solution.This approach allows us to establish the existence of solutions of minimization problems without restrictive assumptions on the functions and on their domains.
Let K be a nonempty closed subset of a Banach ordered space (X, • ,≥).A function f : In this paper, we study the existence of a solution of the minimization problem f (x) −→ min, x ∈ K, (1.2) where f : K → R 1 ∪ {+∞} is an increasing lower semicontinuous function.In [10,12], it was established the generic existence of solutions of problem (1.2) for certain classes of increasing lower semicontinuous functions f .Note that the perturbations which are usually used to obtain a generic existence result are not suitable for these classes since they break the monotonicity.In [10], we proposed the new kind of perturbations which allowed us to establish the generic existence of solutions for certain classes of increasing lower semicontinuous functions.In the present paper, we show that the complement of the set of all functions f , for which the corresponding minimization problem has a solution, is not only of the first category but also σ-porous.
Before we continue, we briefly recall the concept of porosity [2,4].As a matter of fact, several different notions of porosity have been used in the literature.In the present paper, we will use porosity with respect to a pair of metrics, a concept which was introduced in [15].
When (Y, d) is a metric space, we denote by B d (y,r) the closed ball of center y ∈ Y and radius r > 0. Assume that Y is a nonempty set and d 1 ,d 2 : Y × Y → [0,∞) are two metrics which satisfy d 1 (x, y) ≤ d 2 (x, y) for all x, y ∈ Y .A subset E ⊂ Y is called porous in Y with respect to the pair (d 1 ,d 2 ) (or just porous in Y if the pair of metrics is understood) if there exist α ∈ (0,1) and r 0 > 0 such that for each r ∈ (0,r 0 ] and each y ∈ Y , there exists z ∈ Y for which d 2 (z, y) ≤ r and B d1 (z,αr) ∩ E = ∅.A subset of the space Y is called σ-porous in Y with respect to (d 1 ,d 2 ) (or just σ-porous in Y if the pair of metrics is understood) if it is a countable union of porous (with respect to (d 1 ,d 2 )) subsets of Y .Note that if d 1 = d 2 , then by [15,Proposition 1.1] our definitions reduce to those in [2,4].We use porosity with respect to a pair of metrics because in applications a space is usually endowed with a pair of metrics and one of them is weaker than the other.Note that porosity of a set with respect to one of these two metrics does not imply its porosity with respect to the other metric.However, it is shown in [15 , then E is porous in Y with respect to any metric which is weaker than d 2 and stronger than d 1 .
We obtain our main results as a realization of a general variational principle which is established in Section 3.

Well-posedness of optimization problems with increasing cost functions
In this paper, we use the following notations and definitions.Let (X, • ,≥) be a Banach ordered space and X + = {x ∈ X : x ≥ 0} the cone of its positive elements.Assume that X + is a closed convex cone such that x ≤ y for each x, y ∈ X + satisfying x ≤ y.We assume that the cone X + has the following property: i=1 converges.The property (A) is well known in the theory of ordered Banach spaces (see, e.g., [7,10,11]).Recall that the cone X + has the property (A) if the space X is reflexive.The property (A) also holds for the cone of nonnegative functions (with respect to usual order relation) in the space L 1 of all integrable on a measure space functions.
Assume that K is a closed subset of X.
Assume that Ꮽ is a nonempty set and d w ,d s : ) for all a,b ∈ Ꮽ.We assume that the metric space (Ꮽ,d s ) is complete.The topology induced in Ꮽ by the metric d s is called the strong topology and the topology induced in Ꮽ by the metric d w is called the weak topology.
We assume that with every a ∈ Ꮽ a lower semicontinuous function Let a ∈ Ꮽ.We say that the minimization problem for f a on K is strongly well posed with respect to (Ꮽ,d w ) if the following assertions hold: (1) the infimum inf( f a ) is finite and attained at a point x (a) ∈ K such that for each x ∈ K satisfying f a (x) = inf( f a ), the inequality x ≤ x (a) holds; (2) for any > 0, there exist δ > 0 and a neighborhood U of a in Ꮽ with the weak topology such that for each b ∈ U, inf( f b ) is finite; and if | < and there is u ∈ X such that u < and x ≤ x (a) + u.
For each integer n ≥ 1, denote by Ꮽ n the set of all a ∈ Ꮽ which have the following property: (P1) there exist x ∈ K and positive numbers r, η, and c such that and for each b and there is u ∈ X such that u ≤ 1/n and z ≤ x + u.
Proposition 2.1.Assume that a ∈ ∩ ∞ n=1 Ꮽ n .Then the minimization problem for f a on K is strongly well posed with respect to (Ꮽ,d w ).
Proof.By (P1) for each integer n ≥ 1, there exist x n ∈ K, r n > 0, η n > 0, and c n > 0 such that and the following property holds: We may assume without loss of generality that for all integers n ≥ 1, There exists a strictly increasing sequence of natural numbers Let n ≥ 1 be an integer.Inequality (2.3) implies that (2.7) By (2.7), (2.5), and the definition of c 1 , It follows from (2.7), (P2) (see (2.4)), and the definitions of x kn and η kn that there exists u n ∈ X such that (2.10) Set Clearly, the sequence {y n } ∞ n=1 is well defined.By (2.11) and (2.9), for each integer n ≥ 1, (2.16) These inequalities and (2.14) imply that x ≤ x (a) .By (2.15) and the property (P2), for all integers n ≥ 1, (2.17) Let > 0. Choose a natural number m for which Assume that b ∈ B w (a,r km /2), x ∈ K, and By (2.19) and the definitions of η km , r km , and x km (see the property (P2)), It follows from (2.21) and (2.18) that Inequalities (2.20), (2.17), and (2.18) imply that This completes the proof of Proposition 2.1.
An element x ∈ K is called minimal if for each y ∈ K satisfying y ≤ x, the equality x = y is true.Denote by K min the set of all minimal elements of K.
For each integer n ≥ 1, denote by Ꮽn the set of all a ∈ Ꮽ which has the property (P1) with x ∈ K min .
Analogously to the proof of Proposition 2.1, we can prove the following result.
Proposition 2.2.Assume that the set K min is a closed subset of the Banach space X and a ∈ ∩ ∞ n=1 Ꮽn .Then the minimization problem for f a on K is strongly well posed with respect to (Ꮽ,d w ) and inf( f a ) is attained at a unique point.
In the proof of Proposition 2.2, we choose x n ∈ K min , n = 1,2,....This implies that inf( f a ) is attained at the unique point x (a) ∈ K min (see (2.13)).
Remark 2.3.Note that assertion (1) in the definition of a strongly well-posed minimization problem for f a can be represented in the following way: inf( f a ) is finite and the set has the largest element.
We construct an example of an increasing function h for which the set argmin(h) is not a singleton and has the largest element.Define a continuous increasing function Let n be a natural number and consider the Euclidean space R n .Let It is easy to see that h is a continuous increasing function: and the set is not a singleton and has the largest element (1/2,...,1/2).
Remark 2.4.The following example shows that in some cases the sets Ꮽ n can be empty.Let where f a = 0 for any x ≤ a and f a (x) > 0 for any x > a.It is easy to see that the set Ꮽ n is empty for any natural number n.

Variational principles
We use the notations and definitions introduced in Section 2. The following are the basic hypotheses about the functions: (H1) for each a ∈ Ꮽ, inf( f a ) is finite; (H2) for each > 0 and each integer m ≥ 1, there exist numbers δ > 0 and r 0 > 0 such that the following property holds: (P3) for each a ∈ Ꮽ satisfying inf( f a ) ≤ m and each r ∈ (0,r 0 ], there exist ā ∈ Ꮽ, x ∈ K, and d > 0 such that and if x ∈ K satisfies then x ≤ d and there exists u ∈ X for which u ≤ and x ≤ x + u; (H3) for each integer m ≥ 1, there exist α ∈ (0,1) and r 0 > 0 such that for each r ∈ (0,r 0 ], each a 1 ,a 2 ∈ Ꮽ satisfying d w (a 1 ,a 2 ) ≤ αr, and each Assume that (H1), (H2), and (H3) hold.Then there exists a set and for each a ∈ Ᏺ, the minimization problem for f a on K is strongly well posed with respect to (Ꮽ,d w ).
Proof.Recall that for each integer n ≥ 1, Ꮽ n is the set of all a ∈ Ꮽ which has the property (P1).By Proposition 2.1, in order to prove the theorem, it is sufficient to show that the set Ꮽ \ Ꮽ n is σ-porous in Ꮽ with respect to (d w ,d s ) for any integer n ≥ 1.Then the theorem is true with Let n ≥ 1 be an integer.We will show that the set Ꮽ \ Ꮽ n is σ-porous in Ꮽ with respect to (d w ,d s ).To meet this goal, it is sufficient to show that for each integer m ≥ 1, the set is porous in Ꮽ with respect to (d w ,d s ).Let m ≥ 1 be an integer.By (H3), there exist such that for each r ∈ (0,r 1 ], each a 1 ,a 2 ∈ Ꮽ satisfying d w (a 1 ,a 2 ) ≤ α 1 r, and each By (H2), there exist α 2 ,r 2 ∈ (0,1) such that the following property holds: (P4) for each a ∈ Ꮽ satisfying inf( f a ) ≤ m + 2 and each r ∈ (0,r 2 ], there exist ā ∈ Ꮽ, x ∈ K, and d > 0 such that and if x ∈ K satisfies f ā(x) ≤ inf( f ā) + 4rα 2 , then x ≤ d and there exists u ∈ X for which u ≤ (2n) −1 and x ≤ x + u.Choose Let a ∈ Ꮽ and r ∈ (0, r].There are two cases Assume that (3.8) holds.We will show that for each ξ ∈ B dw (a, r), the inequality inf( f ξ ) > m is valid.Assume the contrary.Then there exists ξ ∈ Ꮽ such that There exists y ∈ K such that a contradiction (see (3.8)).Therefore and by (3.3), Thus, we have shown that (3.8) implies (3.15).Assume that (3.9) holds.Then there exists a 1 ∈ Ꮽ such that By the definitions of α 2 , r 2 , the property (P4), (3.16), and (3.7), there exist ā ∈ Ꮽ, and that the following property holds: Since these inequalities hold for any x ∈ K satisfying (3.21), the relation (3.20) implies that Moreover, since (3.21) holds with x = x (see (3.17)), we obtain that Thus, the following property holds: The property (P6) implies that  We also use the following hypotheses about the functions: (H4) for each > 0 and each integer m ≥ 1, there exist numbers δ > 0 and r 0 > 0 such that the following property holds: (P7) for each a ∈ Ꮽ satisfying inf( f a ) ≤ m and each r ∈ (0,r 0 ], there exist ā ∈ Ꮽ, x ∈ K min , and d > 0 such that (3.1) is true; and if x ∈ K satisfies (3.2), then x ≤ d and there exists u ∈ X for which u ≤ , x ≤ x + u.
Theorem 3.2.Assume that (H1), (H3), and (H4) hold and K min is a closed subset of the Banach space X.Then there exists a set Ᏺ ⊂ Ꮽ such that the complement Ꮽ \ Ᏺ is σ-porous in Ꮽ with respect to (d w ,d s ) and that for each a ∈ Ᏺ the following assertions hold: (1) the minimization problem for f a on K is strongly well posed with respect to (Ꮽ,d w ), ( 2) the infimum inf( f a ) is attained at a unique point.
We can prove Theorem 3.2 analogously to the proof of Theorem 3.1.Recall that for each integer n ≥ 1, Ãn is the set of all a ∈ Ꮽ which have the property (P1) with x ∈ K min .Set Ᏺ = ∩ ∞ n=1 Ãn .By Proposition 2.2 for each a ∈ Ᏺ, assertions (1) and ( 2) hold.Therefore, in order to prove Theorem 3.2, it is sufficient to show that for each integer n ≥ 1, the set Ꮽ \ Ꮽn is σ-porous in Ꮽ with respect to (d w ,d s ).We can show this fact analogously to the proof of Theorem 3.1.

Spaces of increasing functions
In the sequel, we use the functional λ : X → R 1 defined by λ(x) = inf y : y ≥ x , x ∈ X. (4.1) The function λ has the following properties (see [10,Proposition 6.1]): (i) the function λ is sublinear.Namely, Clearly, |λ(x) − λ(y)| ≤ x − y for each x, y ∈ X. Denote by ᏹ the set of all increasing lower semicontinuos bounded-frombelow functions f : It is not difficult to see that the metric space (ᏹ,d s ) is complete.Denote by ᏹ v the set of all finite-valued functions f ∈ ᏹ and by ᏹ c the set of all finite-valued continuous functions f ∈ ᏹ.Clearly, ᏹ v and ᏹ c are closed subsets of the metric space (ᏹ,d s ).We say that the set K has property (C) if K min is a closed subset of K and for each x ∈ K, there is y ∈ K min such that y ≤ x.
Denote by ᏹ g the set of all f ∈ ᏹ such that f (x) → ∞ as x → ∞.Clearly, ᏹ g is a closed subset of the metric space (ᏹ,d s ).Set ᏹ gc = ᏹ g ∩ ᏹ c and ᏹ gv = ᏹ g ∩ ᏹ v .
It is easy to see that Remark 4.1.Let K = X + and define Clearly, Theorem 4.2.Assume that Ꮽ is either ᏹ g , ᏹ gv , or ᏹ gc and that f a = a for all a ∈ Ꮽ.Then there exists a set Ᏺ ⊂ Ꮽ such that the complement Ꮽ \ Ᏺ is σ-porous in Ꮽ with respect to (d s ,d s ) and that for each f ∈ Ᏺ the minimization problem for f on K is strongly well posed with respect to (Ꮽ,d s ).If K has the property (C), then for each f ∈ Ᏺ, inf( f ) is attained at a unique point.
Proof.By Theorems 3.1 and 3.2, we need to show that (H1), (H2), and (H3) hold and that the property (C) implies (H4).Clearly, (H1) holds.For each f ,g ∈ ᏹ, we have that and that if d s ( f ,g) ≤ 1/2, then ds ( f ,g) ≤ 2d s ( f ,g).Combined with (4.7), this property implies (H3).We will show that (H2) holds and that the property (C) implies (H4).Let f ∈ Ꮽ, ∈ (0,1), and If K has the property (C), then we assume that x is a minimal element of K. Define Let x ∈ K and f (x) ≤ inf( f ) + r/8.Then by (4.9) and (4.8), By (4.1), there exists u ∈ X such that x ≤ x + u and u < .Since f (y) → ∞ as y → ∞, we obtain that x ≤ d, where d > 0 is a constant which depends only on f .Thus, (H2) is true and if K has the property (C), then (H4) holds.Theorem 4.2 is proved.
Theorem 4.3.Assume that there exists z ∈ X such that z ≤ x for all x ∈ K, that a space Ꮽ is either ᏹ, ᏹ v , or ᏹ c , and that f a = a for all a ∈ Ꮽ.Then there exists a set and that for each f ∈ Ᏺ, the minimization problem for f on K is strongly well posed with respect to (Ꮽ,d s ).If K has the property (C), then for each f ∈ Ᏺ, inf( f ) is attained at a unique point.
Proof.We can prove Theorem 4.3 analogously to the proof of Theorem 4.2.The existence of a constant d is obtained in the following manner.Let x ∈ K, u ∈ X, x ≤ x + u, and u < .Then where d = 2 z + x + .
Denote by ᏹ + the set of all f ∈ ᏹ such that f (x) ≥ 0 for all x ∈ K. Clearly, ᏹ + is a closed subset of the metric space (ᏹ,d s ).Define It is not difficult to see that the metrtic space (ᏹ + ,d w ) is complete and that ᏹ + v , ᏹ + c , ᏹ + g , ᏹ + gv , and ᏹ + gc are closed subsets of (ᏹ + ,d w ).Clearly, d w ( f ,g) ≤ d s ( f ,g) for all f ,g ∈ ᏹ + .Theorem 4.4.Assume that one of the following cases holds: (1) the space Ꮽ is either ᏹ + g , ᏹ + gv , or ᏹ + gc ; (2) there is z ∈ X such that z ≤ x for all x ∈ K and that Ꮽ is either ᏹ + , ᏹ + v , or ᏹ + c .Let f a = a for all a ∈ Ꮽ.Then there exists a set Ᏺ ⊂ Ꮽ such that Ꮽ \ Ᏺ is σporous in Ꮽ with respect to (d w ,d s ) and that for each f ∈ Ᏺ, the minimization problem for f on K is strongly well posed with respect to (Ꮽ,d w ).If K has the property (C), then for each f ∈ Ᏺ, inf( f ) is attained at a unique point.
Proof.By Theorems 3.1 and 3.2, we need to show that (H1), (H2), and (H3) hold and that the property (C) implies (H4).Clearly, (H1) holds.Analogously to the proofs of Theorems 4.2 and 4.3, we can show that (H2) is true.Therefore, in order to prove Theorem 4.4, it is sufficient to show that (H3) holds.
Let m ≥ 1 be an integer.Choose α ∈ (0,1) such that We may assume without loss of generality that Fix a number c 0 > 0 and denote by ᏹ (co) the set of all convex functions f ∈ ᏹ such that f (x) ≥ c 0 x for all x ∈ K. Clearly, ᏹ (co) is a closed subset of the metric space (ᏹ + ,d w ).Set ᏹ are closed subsets of the metric space (ᏹ + ,d w ).
Theorem 4.5.Assume that a space Ꮽ is either ᏹ co , ᏹ (co) v , or ᏹ (co) c and that f a = a for all a ∈ Ꮽ.Then there exists a set Ᏺ ⊂ Ꮽ such that the complement Ꮽ \ Ᏺ is σporous in Ꮽ with respect to (d w ,d w ) and that for each f ∈ Ᏺ, the minimization problem for f on K is strongly well posed with respect to (Ꮽ,d w ).If the set K has the property (C), then for each f ∈ Ᏺ, inf( f ) is obtained at a unique point.Proof.By Theorems 3.1 and 3.2, we need to show that (H1), (H2), and (H3) hold and that the property (C) implies (H4).Clearly, (H1) holds.Analogously to the proof of Theorem 4.4, we can show that (H3) holds.Now, we show that (H2) holds and that the property (C) implies (H4).Let f ∈ Ꮽ, ∈ (0,1), and m ≥ 1 an integer.Choose a natural number m 0 and a number r 0 > 0 such that Choose a positive number If K has the property (C), we assume that x is a minimal element of K. Define (4.32) Therefore, there exists u ∈ X such that u < 4 −1 and x ≤ x + u.Hence, (H2) is true and if K has the property (C), then (H4) holds.This completes the proof of Theorem 4.5.
A function f : Denote by ᏹ (qu) the set of all quasiconvex functions f ∈ ᏹ.Clearly, ᏹ (qu) is a closed subset of the metric space (ᏹ,d s ).Set gc = ᏹ (qu) ∩ ᏹ gc .
(4.33) Theorem 4.6.Assume that one of the following cases holds: (1) a space Ꮽ is either ᏹ gv , or ᏹ (qu) gc ; (2) there is z ∈ X such that z ≤ x for all x ∈ K and a space Ꮽ is either ᏹ (qu) , ᏹ .
Let f a = a for all a ∈ Ꮽ.Then there exists a set Ᏺ ⊂ Ꮽ such that the complement Ꮽ \ Ᏺ is σ-porous in Ꮽ with respect to (d s ,d s ) and that for each f ∈ Ᏺ, the minimization problem for f on K is strongly well posed with respect to (Ꮽ,d s ).If K has the property (C), then for each f ∈ Ᏺ, inf( f ) is attained at a unique point.Proof.By Theorems 3.1 and 3.2, we need to show that (H1), (H2), and (H3) hold and that the property (C) implies (H4).Clearly, (H1) holds.Analogously to the proof of Theorem 4.2, we can show that (H3) is true (see (4.7)).Now, we show that (H2) holds and that the property (C) implies (H4).
Let f ∈ Ꮽ, ∈ (0,1], and r ∈ (0 Clearly, f ∈ Ꮽ, In the first case, we choose d > 0 such that f (y) > inf( f ) + 1 for all y ∈ K satisfying y > d and obtain that x ≤ d.In the second case, z ≤ x ≤ x + u, x ≤ z + x − z ≤ z + x + u − z ≤ 2 z + x + u ≤ 2 z| + x + 1, (4.40) and x ≤ d := 2 z + x + 1.Therefore, in both cases, (H2) is true and if the set K has the property (C), then (H4) holds.Theorem 4.6 is proved.Theorem 4.7.Let Ꮽ be defined as in Theorem 4.6 and let Ꮽ + be the set of all f ∈ Ꮽ such that f (x) ≥ 0 for all x ∈ K. Assume that f a = a for all a ∈ Ꮽ + .Then the metric spaces (Ꮽ + ,d s ) and (Ꮽ + ,d w ) are complete and there exists a set Ᏺ ⊂ Ꮽ + such that the complement Ꮽ + \ Ᏺ is σ-porous in Ꮽ + with respect to (d w ,d s ) and that for each f ∈ Ᏺ, the minimization problem for f on K is strongly well posed with respect to (Ꮽ + ,d w ).If the set K has the property (C), then for each f ∈ Ᏺ, inf( f ) is attained at a unique point.
Proof.By Theorems 3.1 and 3.2, we need to show that (H1), (H2), and (H3) hold and that the property (C) implies (H4).Clearly, (H1) holds.Analogously to the proof of Theorem 4.6, we can show that (H2) is true and that the property (C) implies (H4).We can prove (H3) as in the proof of Theorem 4.4.