A POROSITY RESULT IN CONVEX MINIMIZATION

We study the minimization problem f (x)→min, x ∈ C, where f belongs to a complete metric space of convex functions and the set C is a countable intersection of a decreasing sequence of closed convex sets Ci in a reflexive Banach space. Let be the set of all f ∈ for which the solutions of the minimization problem over the set Ci converge strongly as i→∞ to the solution over the set C. In our recent work we show that the set contains an everywhere dense Gδ subset of . In this paper, we show that the complement \ is not only of the first Baire category but also a σ-porous set.


Introduction
Let X be a reflexive Banach space with the norm • and let where C i+1 ⊂ C i for each i = 1,2,... and where each C i is a closed convex subset of X.We study the minimization problem In this paper, we study the set of all functions f ∈ ᏹ for which the solutions of the minimization problem over the sets C i converge strongly as i → ∞ to the solution of the minimization problem over the set C ∞ .We show that the complement of this set is not only of the first Baire category but also a σ-porous set.
We now recall the concept of porosity [5,6,7,12].Let (Y ,d) be a complete metric space.We denote by B d (y,r) the closed ball of center y ∈ Y and radius r > 0. A subset E ⊂ Y is called porous in (Y ,d) 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 B d (z,αr) ⊂ B d (y,r) \ E. (1.3) Hence every ball under a certain size includes a smaller ball of fixed proportional size that is contained in the complement of E.
Remark 1.1.In the above definition of porosity it is known that the point y can be assumed to belong to E.
Other notions of porosity have been used in the literature [16,17].We use the rather strong notion which appears in [5,6,7,12].
Since porous sets are nowhere dense, all σ-porous sets are of the first category.That is, each σ-porous set can be expressed as a countable union of nowhere dense subsets.If Y is a finite-dimensional Euclidean space, then all σ-porous sets are of Lebesgue measure zero.In fact, the class of σ-porous sets in such a space is smaller than the class of sets which have measure zero and are of the first category [16].Furthermore every complete metric space without isolated points contains a closed nowhere dense set which is not σ-porous [17].
To differentiate between porous and nowhere dense sets note that if E ⊂ Y is nowhere dense, y ∈ Y , and r > 0, then there is a point z ∈ Y and a number s > 0 such that B d (z,s) ⊂ B d (y,r) \ E. If, however, E is also porous, then for small enough r we can choose s = αr, where α ∈ (0,1) is a constant which depends only on E.

The main result
We use the convention that ∞ − ∞ = 0 and ∞/∞ = 1.Let X be a reflexive Banach space with the norm • and let where C i+1 ⊂ C i for each i = 1,2,... and where each C i is a closed convex subset of X.Let ϕ : Denote by ᏹ the set of all convex lower semicontinuous functions f : C 1 → R 1 ∪ {∞} which are not identically infinity on C ∞ and satisfy For each f ∈ ᏹ and each nonempty set It is well known that for each f ∈ ᏹ and each i ∈ {1, 2,...} ∪ {∞}, the following minimization problem: Denote by ᏹ 1 the set of all finite-valued functions f ∈ ᏹ and by ᏹ 2 the set of all finite-valued continuous functions f ∈ ᏹ.Next we endow the set ᏹ with a metric d.For each f , g ∈ ᏹ and each m ∈ N, we first set and then define (see the convention at the start of this section).We adopt the convention that the supremum of the empty set is zero.Clearly (ᏹ,d) is a complete metric space.It is also not difficult to see that the collection of sets where m ∈ N and δ > 0, is a base for the uniformity generated by the metric d.Evidently ᏹ 1 and ᏹ 2 are closed subsets of the metric space (ᏹ,d).In the sequel we assign to all these spaces the same metric d.
In [8] for a function f ∈ ᏹ 2 we studied the convergence of solutions to the problem (P then this convergence property was established by Semple [14].A similar result was also obtained for certain Banach spaces and the distance function by Israel Jr. and Reich [10]. In [8] we showed that the convergence property holds for most functions f ∈ ᏹ 2 .More precisely, we considered the metric space (ᏹ 2 ,d) with where a 1 and a 2 are positive numbers, and showed that there exists a subset of ᏹ 2 which is a countable intersection of open everywhere dense sets such that for each function belonging to this subset the convergence property holds.Note that this result is true for reflexive Banach spaces but not necessarily for nonreflexive Banach spaces.For more information consider [8, Examples 1 and 2].In this paper for the spaces ᏹ, ᏹ 1 , and ᏹ 2 we show that the complements of subsets of functions which have the convergence property are not only of the first Baire category but are also σ-porous sets.We will establish the following result.
Theorem 2.1.Let Ꮽ be either ᏹ, ᏹ 1 , or ᏹ 2 .There exists a set Ᏺ ⊂ Ꮽ such that the complement Ꮽ \ Ᏺ is σ-porous in (Ꮽ,d) and such that for each f ∈ Ᏺ the following properties hold.
(P 1 ) There exists a unique point the relation y − x f ≤ also holds.

Auxiliary results
We begin with a lemma.
The next proposition introduces a useful property.
Proof of Proposition 3.2.Property (P 1 ) holds by virtue of the first assumption.For each i = 1,2,..., we choose x i ∈ C i with f (x i ) = inf( f ;C i ).The truth of property (P 4 ) clearly implies the truth of (P 2 ).We will show that property (P 3 ) also holds.Choose > 0. By property (P 4 ) there exist δ ∈ (0,1/2) and p ∈ N such that if it follows that z − x f ≤ .Hence property (P 3 ) is also true.
The proof is also similar and is omitted.

Proof of the main result
It is convenient to split the proof into several smaller parts.We use the notation Ꮽ to denote either ᏹ, ᏹ 1 , or ᏹ 2 .
Lemma 4.1.For each n ∈ N let Ᏺ n denote the set of all f ∈ Ꮽ with the following property.
for all n ∈ N. Hence x f = lim n→∞ x n .Therefore x f is the unique minimizer of f on C ∞ .Let > 0 and n ∈ N be such that n > 2/ .For each i ∈ {p n , p n + 1,...} ∪ {∞} and each Hence y − x f ≤ .Thus property (P 4 ) is valid and hence also properties (P 1 ), (P 2 ), and (P 3 ).
Remark 4.2.To complete the proof of Theorem 2.1, we need to show that is defined for each γ > 0 and all Proof of Lemma 4.3.Clearly f γ ∈ Ꮽ.We will estimate d( f γ , f ).Since it follows from property (Q 2 ) that x f ≤ m.For each k = 1,2,... we have and hence we deduce that y ≤ m and hence also that Since this holds for any such y, we obtain We can now deduce that f γ (z and hence
We can now complete the proof of the main result.
Proof of Theorem 2.1.In Remark 4.2, we observed that the proof of Theorem 2.1 would be complete if we could show that for each m,n ∈ N the set E m \ Ᏺ n is porous in (Ꮽ,d).Let f ∈ E m \ Ᏺ n .Choose any real number r with 0 < r ≤ 1 and choose real numbers γ = γ(r) and θ = θ(r,m,n) as in Lemma 4.

4 .
If we define α = α(m,n) by the formula θ = αr and for each r with 0 < r ≤ 1 we can see from Lemma 4.4 thatg ∈ Ꮽ | d g, f γ ≤ αr ⊂ g ∈ Ꮽ | d(g, f ) ≤ r ∩ Ᏺ n .(4.14)Hence each sufficiently small ball B d ( f ,r) ⊂ (Ꮽ,d) centred at a point f ∈ E m \ Ᏺ n contains a smaller ball B d ( f γ ,αr) of fixed proportional radius centred at the point f γ and lying entirely within Ᏺ n .Hence E m \ Ᏺ n is porous in (Ꮽ,d).This completes the proof.