Well-Posedness and Porosity for Symmetric Optimization Problems

In the present work, we investigate a collection of symmetric minimization problems, which is identified with a complete metric space of lower semi-continuous and bounded from below functions. In our recent paper, we showed that for a generic objective function, the corresponding symmetric optimization problem possesses two solutions. In this paper, we strengthen this result using a porosity notion. We investigate the collection of all functions such that the corresponding optimization problem is well-posed and prove that its complement is a σ-porous set.


Introduction
In this paper, we study a class of symmetric minimization problems, which was studied recently in our paper [1]. The results of [1] and of the present paper have prototypes in [2,3], where some minimization problems arising in crystallography were considered. It was shown in [2,3] that a typical symmetric minimization problem possesses exactly two minimizers, and every minimizing sequence converges to them in some natural sense. In [1], we extend the results of [2,3] for a sufficiently large class of symmetric minimization problems by showing that for a generic objective function, the corresponding symmetric optimization problem possesses two solutions. In this paper, we strengthen this result using a porosity notion. We investigate the collection of all functions such that the corresponding optimization problem is well-posed and prove that its complement is a σ-porous set.
More precisely, we study an optimization problem where X is a complete metric space and g is a lower semi-continuous and bounded from below function. It is well-known that the above problem possesses a minimizer when the space X is compact or when the objective function f possesses a growth property and all bounded subsets of the space X satisfy certain compactness assumptions. Without such assumptions, the existence problem becomes more difficult. This difficulty is overcome by applying the Baire category approach, which was used for many mathematical problems [4][5][6][7][8][9].
Namely, it is known that the minimization problem stated above can be solved for a generic objective function [8][9][10]. More precisely, there is a collection F in a complete metric space of objective functions, which is a countable intersection of open and everywhere dense sets such that for every objective function f ∈ F , the corresponding minimization problem has a unique solution, which is a limit of every minimizing sequence. See [9], which contains this result and its several extensions and modifications. Note that the generic approach in nonlinear analysis is used in [11][12][13][14][15], generic solvability of best approximation problems are discussed in [4,11,13], while generic existence of fixed points for nonlinear operators is established in [7,12,13].
In our recent paper [1] the goal was to establish a generic solvability of optimization problems with symmetry. These results have applications in crystallography [2,3]. In this paper, we strengthen this result using a porosity notion. We investigate the set of all functions for which the corresponding minimization problem is well-posed and show that its complement is a σ-porous set.

The Main Result
We begin this section recalling the following notion of porosity [3,4,7,9,12,13]. Suppose that (Y, d) is a complete metric space and define We say that a set E ⊂ Y is porous with respect to d (or just porous if the metric is understood) if there are a real number α ∈ (0, 1] and a positive number r 0 such that for every positive number r ≤ r 0 and every point y ∈ Y there is a point z ∈ Y such that We say that a set in the complete metric space Y is σ-porous with respect to d (or just σ-porous if the metric is understood) if this set is a countable union of porous (with respect to d) subsets of Y.
For every function h : Suppose that (X, ρ) is a complete metric space. For every z ∈ X and every positive ∆ put For every z ∈ X and every subset D = ∅ of the space X, define Denote by M l the collection of all functions f : X → R 1 ∪ {∞}, which are bounded from below, lower semi-continuous, and which are not identical infinity. For each h 1 , Note that by convention, is a complete metric. We denote by M c the collection of all continuous finite-valued functions f : X → R 1 which are bounded from below. Clearly, M c is a closed set in the complete metric space (M l , d). We endow the space M c with the metric d too.
Suppose that T : X → X is a continuous operator such that We denote by M l,T the collection of all functions f ∈ M l for which Evidently, M l,T and M c,T are closed subsets of the complete metric space M l . We endow them with the same metric d too.
We investigate the optimization problem where the objective function f ∈ M l,T . Given f ∈ M l,T , we say that the problem of minimization for f on X is well-posed with respect to (M l , d) if the following properties are true: There exists x f ∈ X, which satisfies and for every > 0 there are an open neighborhood U of f in M l and a positive number δ such that if a function g ∈ U and if a point z ∈ X satisfies g(z) ≤ inf(g) + δ, then This notion has an analog in the optimization theory [9], where the set of minimizers is a singleton. Here, since the problem is symmetric, the set of minimizers contains two points in general.
The next theorem is our sole main result.

Theorem 1.
Suppose that A is either M l,T or M c,T . Then, there is a set B ⊂ A such that its complement A \ B is σ-porous in the metric space (A, d) and that for every function f ∈ B the minimization problem for f on the space X is well-posed with respect to (M l , d).
Proof. Let r ∈ (0, 1], f , g ∈ M l satisfy and let x ∈ X be given. By (2) and (3), In view of (1) and the equation above,

Lemma 2.
Suppose that f ∈ M l,T , ∈ (0, 1), r ∈ (0, 1]. Then there aref ∈ M l,T andx ∈ X such thatf ∈ M c,T if f ∈ M c,T , and that for each y ∈ X, which satisfies Define a functionf ∈ M l as follows: Clearly,f ∈ M l,T , andf ∈ M c,T if f ∈ M c,T and (4) is true. Let y ∈ X and (5) hold. By (5) and (6),

Proof of Theorem 1
For every integer n ≥ 1 let A n be the collection of all functions f ∈ A such that: (i) there are a pointx ∈ X and δ > 0 such that if z ∈ X and f (z) ≤ inf( f ) + δ, then the inequality ρ(x, {z, T(z)}) ≤ 1/n is valid.
Let a natural number n be given. We claim that the set A \ A n is porous. By Lemma 1, for every positive number r ≤ 1, each f , g ∈ M l satisfying d( f , g) ≤ 4 −1 r and each x ∈ X, |g(x) − f (x)| ≤ r.
By Lemma 2 applied with = (2n) −1 , the following property is valid: (ii) for each function f ∈ A and every positive number r ≤ 1, there existf ∈ A and x ∈ X such thatd ( f ,f ) ≤ r/4 (8) and that for each y ∈ X satisfyinḡ is valid.
Let f ∈ A and a positive number r ≤r be given. By property (ii), there existf ∈ A andx ∈ X such thatd ( f ,f ) ≤ r/4 (12) and that the next property is true: (iii) for every point y ∈ X satisfying (9), Equation (10) is true.
Thus g ∈ A n by definition. Together with (14), this implies that Thus, the set A \ A n is σ-porous. Then the set By (20), for every integer n ≥ 1, there are x n ∈ X and δ n > 0 such that the following property is valid: (iv) if a point z ∈ X satisfies the inequality f (z) ≤ inf( f ) + δ n , then the equation Let a natural number n be given. By (21) and property (iv), for every large enough positive integer i, Since n is an arbitrary positive integer, there is a sub-sequence {z i p } ∞ p=1 such that at least one of the sequences {z i p } ∞ p=1 and {T(z i p )} ∞ p=1 converges. Since T is continuous and T 2 is the identity operator, they both converge and By (21), (23) and the lower semi-continuity of f , Applying property (iv) with z i = x f for every natural number i, we obtain that Let ξ ∈ X be such that By (26) and property (iv) applied with z i = ξ for every integer i ≥ 1 we obtain that ρ(x n , {ξ, T(ξ)}) ≤ n −1 for every natural number n.
Equations (25) and (27) imply that Since n is an arbitrary positive integer, we conclude that and at least one of the following equalities is true: Let > 0. Fix a natural number n such that Property (iv) and (25) imply that for every z ∈ X which satisfies the inequality we have ρ(x n , {z, T(z)}) ≤ 1/n (30) and min{ρ(z, x f ), ρ(T(z), x f ), ρ(z, T(x f )), ρ(T(z), T(x f ))} ≤ 2n −1 .
Let a function g ∈ M l satisfyd (g, f ) ≤ δ (33) and let a point z ∈ X be such that g(z) ≤ inf(g) + δ.
Thus, (28) holds and for each function g ∈ M l which satisfies (33) and every point z ∈ X satisfying (34) Equation (36) holds. By Equations (33) and (34), we have Thus, the minimization problem for f on X is well-posed with respect to (M l , d) for all f ∈ ∩ ∞ n=1 A n . Theorem 1 is proved.