A convex-valued selection theorem with a non-separable Banach space

Pascal Gourdel; Nadia Mâagli

doi:10.1515/anona-2016-0053

Article Open Access

A convex-valued selection theorem with a non-separable Banach space

Pascal Gourdel and Nadia Mâagli

Published/Copyright: July 2, 2016

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From the journal Advances in Nonlinear Analysis Volume 7 Issue 2

Abstract

In the spirit of Michael’s selection theorem [6, Theorem 3.1”’], we consider a nonempty convex-valued lower semicontinuous correspondence φ:X→2Y. We prove that if φ has either closed or finite-dimensional images, then there admits a continuous single-valued selection, where X is a metric space and Y is a Banach space. We provide a geometric and constructive proof of our main result based on the concept of peeling introduced in this paper.

Keywords: Barycentric coordinates; continuous selections; lower semicontinuous correspondence; closed-valued correspondence; finite-dimensional convex values; separable Banach spaces

MSC 2010: 54C60; 54C65; 54D65

1 Introduction

The area of continuous selections is closely associated with the publication by Ernest Michael of two fundamental papers [10]. It is important to notice that the axiom of choice ensures the existence of a selection for any nonempty family of subsets of X (see [9]). Yet, the axiom of choice does not guarantee the continuity of the selection. Michael’s studies are more concerned about continuous selections for correspondences φ:X→2Y. He guarantees the continuity under specific structures on X (paracompact spaces, perfectly normal spaces, collectionwise normal spaces, etc.) and on Y (Banach spaces, separable Banach spaces, Fréchet spaces, etc.) [5]. Without any doubt, the most known selection theorems are: closed-convex valued, compact-valued, zero-dimensional and finite-dimensional theorems [6, 7, 8].

The closed-convex valued theorem is considered as one of the most famous of Michael’s contributions in the continuous selection theory for correspondences. This theorem gives sufficient conditions for the existence of a continuous selection with the paracompact domain: paracompactness of the domain is a necessary condition for the existence of continuous selections of lower semicontinuous correspondences into Banach spaces with convex closed values [9].

However, despite their importance, all the theorems mentioned above were obtained for closed-valued correspondences. One of the selection theorems obtained by Michael in order to relax the closeness restriction is the convex-valued selection theorem [6, Theorem 3.1”’]. The result was obtained by an alternative assumption on X (Perfect normality), a separability assumption on Y and an additional assumption involving three alternative conditions on the images. Besides, Michael shows that when Y=ℝ, then perfect normality is a necessary and sufficient condition in order to get a continuous selection of any convex-valued lower semicontinuous correspondence. The proof of the convex-valued selection theorem is based on the existence of a dense family of selections. The technique is quite involved and exploits the characterization of perfect normality of X and separability of Y.

An interesting question is the following: is it possible to relax the separability of Y? To answer this question, Michael provided in his paper a counterexample [6, Example 6.3] showing that the separability of Y can not be omitted. Even though, the correspondence satisfies one of the three conditions, Michael established an overall conclusion. One question arises naturally: is it possible to omit the separability of Y when the images satisfy one of the two remaining conditions? This study aims to prove that the answer is affirmative.

The paper is organized as follows: in Section 2, we begin with some definitions and results which will be very useful in the sequel. Section 3 is dedicated to recall the two Michael’s selection theorems that will be used later: the closed-convex valued and the convex-valued theorems. In Section 4, we first state a partial result when the dimension of the images is finite and constant. Then, we introduce and motivate the concept of peeling. Finally, we state the general case and Sections 5 and 6 provide the proofs of our results.

2 Preliminaries and notations

We start by introducing some notations which will be useful throughout this paper.

2.1 Notations

Let Y be a normed space and C⊂Y. We shall denote

by C¯ the closure of C in Y,
by co⁡(C) the convex hull of C and by aff⁡(C) the affine space of C,
by dima⁡(C)=dim⁡aff⁡(C) the dimension of C which is by definition the dimension of aff⁡(C),
if C is finite-dimensional^[1], then ri⁡(C) the relative interior of C in aff⁡(C) is given by
ri⁡(C)={x∈C∣there exists a neighborhood ⁢Vx⁢ of ⁢x⁢ such that ⁢Vx∩aff⁡(C)⊂C},
by BC⁢(a,r):=B⁢(a,r)∩aff⁡(C), where B⁢(a,r) is the open ball of radius r>0 centered at a point a∈X and B¯C⁢(a,r):=B¯⁢(a,r)∩aff⁡(C), where B¯⁢(a,r) is the closed ball of radius r>0 centered at a point a∈X,
by Si-1⁢(0,1):={x=(x1,…,xi)∈ℝi∣∥x∥=1} the unit (i-1)-sphere of ℝi embedded with the euclidean norm,
by (Ya⁢ip)p∈ℕ the set of affinely independent families of (Yp)p∈ℕ.

We recall that if {x0,x1,…,xi} is a set of i+1 affinely independent points of Y. We call an i-simplex the convex hull of {x0,x1,…,xi} given by

Si={z∈Y|z=∑k=0iαk⁢xk,αk≥0,∑k=0iαk=1}=co⁡(x0,…,xi).

2.2 Classical definitions

We go on with formal definitions and related terms of correspondences. Let us consider nonempty topological spaces X and Y.

Definition 2.1.

Let φ:X→2Y be a correspondence, B⊂Y. We define by

φ+⁢(B)={x∈X∣φ⁢(x)⊂B},φ-⁢(B)={x∈X∣φ⁢(x)∩B≠∅}.

Definition 2.2.

Let φ:X→2Y be any correspondence. A correspondence ψ:X→2Y satisfying ψ⁢(x)⊂φ⁢(x), for each x∈X, is called a selection of φ. In particular, if ψ is single-valued (associated to some function f:X→Y), then f is a single-valued selection when f⁢(x)∈φ⁢(x), for each x∈X.

We recall some alternative characteristics of lower semicontinuous correspondences.

Definition 2.3 ([3]).

Let φ:X→2Y be a correspondence. We say that φ is lower semicontinuous (abbreviated to lsc) if one of the equivalent conditions is satisfied.

For all open sets V⊂Y, we have that φ-⁢(V) is open.
For all closed sets V⊂Y, we have that φ+⁢(V) is closed.

In the case of metric spaces, an alternative characterization is given by the following proposition.

Proposition 2.4 ([3]).

Let X and Y be metric spaces and φ:X→2Y a correspondence. We have that φ is lsc on X if and only if for all x∈X, for all convergent sequences (xn)n∈N to x, and for all y∈φ⁢(x), there exists a sequence (yn)n≥n0 in Y such that yn→y and for all n≥n0, we have yn∈φ⁢(xn).

3 Michael’s selection theorems (1956)

Let us first recall one of the main selection theorems: the closed-convex valued selection theorem.

Theorem 3.1 (Closed-convex valued selection theorem).

Let X be a paracompact space, Y a Banach space and φ:X→2Y an lsc correspondence with nonempty closed convex values. Then φ admits a continuous single-valued selection.

Before stating the next theorem, we recall^[2] that a topological space is perfectly normal if it is normal and every closed subset is a G-delta subset (Gδ).

The following Michael selection theorem dedicated for non-closed valued correspondences is much more difficult to prove. The assumption on Y is reinforced by adding the separability. We recall that perfect normality does neither imply paracompactness nor the converse.

Theorem 3.2 (Convex-valued selection theorem).

Let X be a perfectly normal space, Y a separable Banach space and φ:X→Y an lsc correspondence with nonempty convex values. If for any x∈X, φ⁢(x) is either finite-dimensional, or closed, or has an interior point, then φ admits a continuous single-valued selection.

Note that as explained before, in his paper [6, Example 6.3], Michael provided the following counterexample showing that the assumption of separability of Y can not be omitted in Theorem 3.2.

Example 3.3.

There exists an lsc correspondence φ from the closed unit interval X to the nonempty, open, convex subsets of a Banach space Y for which there exists no selection.

Indeed, let X be the closed unit interval [0,1] and let

Y=ℓ1(X)={y:X→ℝ|∑x∈X|y(x)|<+∞}.

Michael showed that the correspondence φ:X→2Y given by φ⁢(x)={y∈Y∣y⁢(x)>0} has open values, consequently, images have an interior point but Michael proved that there does not exist a continuous selection. The case where the correspondence is either finite-dimensional or closed valued still remains to be dealt with. In order to provide an answer, we now state the main results of this paper.

4 The results

We start by recalling that if X is a metric space, then it is both paracompact and perfectly normal. In many applications, both paracompactness and perfect normality aspects are ensured by the metric character. Therefore, throughout this section, we assume that (X,d) is a metric space. In addition, let (Y,∥⋅∥) be a Banach space. We recall that the relative interior of a convex set C is a convex set of same dimension and that ri⁡(C¯)=ri⁡(C) and ri⁡(C)¯=C¯. In the first instance, in order to prove the main result, we focus on the case of constant (finite-)dimensional images. In addition, compared with Theorem 3.2 of Michael, we suppose first that X is a metric space and omit the separability of Y. We denote by Di the set Di:={x∈X∣dima⁡φ⁢(x)=i}. Then we state the following theorem.

Theorem 4.1.

Let φ:X→2Y be an lsc correspondence with nonempty convex values. Then, for any i∈N, the restriction of ri⁡(φ) to Di admits a continuous single-valued selection hi:Di→Y. In addition, if i>0, then there exists a continuous function βi:Di→]0,+∞[ such that for any x∈Di, we have B¯φ⁢(x)⁢(hi⁢(x),βi⁢(x))⊂ri⁡(φ⁢(x)).

Once we have Theorem 4.1, we will be able to prove our main result given by

Theorem 4.2.

Let X be a metric space and Y a Banach space. Let φ:X→2Y be an lsc correspondence with nonempty convex values. If for any x∈X, φ⁢(x) is either finite-dimensional or closed, then φ admits a continuous single-valued selection.

Note that the property of φ being either closed or finite-dimensional valued is not inherited by co⁡(φ). Consequently, we can not directly convert Theorem 4.2 in terms of the convex hull. Yet, first, a direct consequence of both Theorem 4.2 and Theorem 3.1 is the following.

Corollary 4.3.

Let X be a metric space and Y a Banach space. Let φ:X→2Y be an lsc correspondence with nonempty values. Then co¯⁢(φ⁢(x)) admits a continuous single-valued selection.

Secondly, we can also deduce from Theorem 4.2 the following result.

Corollary 4.4.

Let X be a metric space and Y a Banach space. Let φ:X→2Y be an lsc correspondence with nonempty values. If for any x∈X, φ⁢(x) is either finite-dimensional or closed convex, then co⁡(φ) admits a continuous single-valued selection.

It is also worth noting that under the conditions of Theorem 4.2, we may have an lsc correspondence with both closed and finite-dimensional values. This is made clear in the following example.

Example 4.5.

Let Y be an infinite-dimensional Banach space and X=[0,1]. Consider (e0,…,en,…) some linearly independent normed family of Y. Let φ:ℝ→2Y defined by

φ⁢(x)={{0}if ⁢x∈{0,1},ri⁡(co⁡(0,e0))if ⁢x∈[12,1),ri⁡(co⁡(0,e0,e1))if ⁢x∈[13,12),⋮Yif ⁢x∈Xc.

Remark that φ is lsc on ]0,1[ since it is locally increasing. In other terms, for any x∈X, there exists Vx such that for all x′∈Vx, we have φ⁢(x)⊂φ⁢(x′).

Besides, φ is lsc at the point x=0. It suffices to remark that for ε>0, we have φ-⁢(B⁢(0,ε))=ℝ. Indeed, since for any x∈ℝ, 0∈φ¯⁢(x), then B⁢(0,ε)∩φ⁢(x)≠∅. The same argument is used for x=1. Therefore, we conclude that φ is lsc.

By Theorem 4.2, we can conclude that φ admits a continuous selection. It should also be noted that we can even build an explicit selection.

The proofs of Theorem 4.1 and Theorem 4.2 are postponed until Sections 5 and 6 respectively. The proof of Theorem 4.1 is based on the concept of “peeling” that we will introduce and motivate here.

Definition 4.6.

Let C be a nonempty finite-dimensional subset of Y. We say that C′ is a peeling of C of parameter ρ≥0 if

C′=Γ(C,ρ):={y∈C such that B¯C(y,ρ)⊂ri(C)}.

In order to gain some geometric intuition, the concept is illustrated by Figure 1.

Figure 1

Peeling concept.

Figure 2

Graphic illustration.

Definition 4.7.

Let η be a non-negative real-valued function defined on X, and φ a correspondence from X to Y. We will say that the correspondence φη:X→2Y is a peeling of φ of parameter η if for each x∈X, we have φη⁢(x)=Γ⁢(φ⁢(x),η⁢(x)).

The motivation of the peeling concept is given by the next proposition (whose proof is postponed until the next section) where we show that when the dimension is constant, continuous peeling of an lsc correspondence is also a (possibly empty) lsc correspondence. This proposition is a key argument for the proof of Theorem 4.1.

Proposition 4.8.

Let φ:X→2Y be an lsc correspondence with nonempty convex values. If there exists an i∈N* such that for any x∈X, dima⁡φ⁢(x)=i, then the continuity of the function η implies the lower semicontinuity of φη.

Remark 1.

The following simple example shows that the above proposition is no more valid if the dimension of φ is equal to zero. Let φ:ℝ→2ℝ, defined by φ⁢(x)={0} and η⁢(x)=|x|. Obviously φ is lsc and η is continuous, but φη:ℝ→2ℝ is not lsc since φη⁢(0)={0} while for x≠0, we have φη⁢(x)=∅.

Remark 2.

Modifying slightly the previous example, we also show that the above proposition does not hold true if the dimension of φ is not constant. Let us consider the case where X=ℝ, Y=ℝ2, and φ:X→2Y, is the lsc correspondence defined by

φ⁢(x)={(y1,y2)∈ℝ2∣y1≥0⁢ and ⁢y2≥tan⁡(2⁢arctan⁡(|x|))⁢y1},if ⁢x≠0⁢ and ⁢φ⁢(0)=[0,1/2]×{0}.

Using the same η⁢(x)=|x|, we obtain that when x≠0, φη⁢(x) is a translation of φ⁢(x). More precisely,

φη⁢(x)={(1,|x|)}+φ⁢(x)

(see Figure 2). In particular, d⁢(φ⁢(0),φη⁢(x))≥1/2, which allows us to conclude that φη is not lsc.

5 Proof of Theorem 4.1

In Section 5.1, we first present elementary results about the “peeling” of a set. Section 5.2 is dedicated to prove some affine geometry results used to prove Proposition 4.8 in Section 5.3. Finally, we deduce Theorem 4.1 from this proposition in the last subsection.

5.1 Elementary results on a set “peeling”

In this subsection, C is a finite-dimensional set.

Definition 5.1.

We define^[3] the internal radius of a finite-dimensional set C by

α⁢(C):=sup⁡{ρ∈ℝ+∣there exists ⁢y∈C⁢ such that ⁢B¯C⁢(y,ρ)⊂ri⁡(C)}.

Lemma 5.2.

Let C be a nonempty convex set. Then one has Γ⁢(C,0)=ri⁡(C). Yet, if α⁢(C) is finite, then we have Γ⁢(C,α⁢(C))=∅.

Proof.

The equality on Γ⁢(C,0) is a simple consequence of the definition. Let us prove by contradiction that Γ⁢(C,α⁢(C))=∅. Indeed, if y∈Γ⁢(C,α⁢(C)), we have B¯C⁢(y,α⁢(C))⊂ri⁡(C). By a compactness argument on the circle of center y and radius α⁢(C) and the openness of r⁢i⁢(C) in aff(C), we can prove the existence of some ε>0 such that B¯C⁢(y,α⁢(C)+ε)⊂ri⁡(C). ∎

We first establish that a peeling of a convex set remains convex. In addition, we can characterize the nonemptiness.

Lemma 5.3.

Let C be a nonempty convex set and ρ∈[0,+∞[. Then, the set Γ⁢(C,ρ) is convex. In addition, the set Γ⁢(C,ρ) is nonempty if and only if ρ<α⁢(C).

Proof.

First, we have that Γ⁢(C,ρ) is convex. Let x1,x2∈Γ⁢(C,ρ) and λ∈[0,1]. We claim that

B¯C⁢((λ⁢x1+(1-λ)⁢x2),ρ)⊂ri⁡(C).

By triangle inequality, it is easy to see that

B¯⁢(λ⁢x1+(1-λ)⁢x2,ρ)=λ⁢B¯⁢(x1,ρ)+(1-λ)⁢B¯⁢(x2,ρ).

Yet, since ri⁡(C) is convex, then we have

(λ⁢B¯⁢(x1,ρ)+(1-λ)⁢B¯⁢(x2,ρ))∩aff⁡(C)⊂λ⁢ri⁡(C)+(1-λ)⁢ri⁡(C)=ri⁡(C),

which establishes the result. Now, remark that if Γ⁢(C,ρ)≠∅, then by definition of Γ, we have ρ≤α⁢(C) and in view of Lemma 5.2, ρ<α⁢(C). It remains to prove the converse. Using the definition of α, there exists yρ∈C such that B¯C⁢(yρ,ρ)⊂ri⁡(C). Therefore, yρ∈Γ⁢(C,ρ). ∎

Lemma 5.4.

Let C be a convex set and ρ1,ρ2 non-negative real numbers such that ρ1<ρ2. Then we have Γ¯⁢(C,ρ2)⊂Γ⁢(C,ρ1)⊂ri⁡(C).

Proof.

Let y¯∈Γ¯⁢(C,ρ2). Since ε=ρ2-ρ1>0, there exists y∈Γ⁢(C,ρ2)∩B⁢(y¯,ε). Consequently, by triangle inequality, B¯C⁢(y¯,ρ1)⊂B¯C⁢(y,ρ2), and therefore y∈Γ⁢(C,ρ1). Finally, it comes from the definition that Γ⁢(C,ρ1)⊂ri⁡(C). ∎

5.2 Affine geometry

Next, we will use known results about linear independence in order to raise a series of results about affine independence and barycentric coordinates.

Lemma 5.5.

The set Ya⁢ii+1 is an open subset of Yi+1.

Proof.

Let (z0,…,zi)∈Yaii+1. In order to get a contradiction, we suppose that for all r>0, and for all k∈{0,…,i}, there exists a zk′∈B⁢(zk,r) such that (z0′,…,zi′) are affinely dependent. In particular, for all p∈ℕ* and for all k∈{0,…,i}, there exists zk,p′∈B⁢(zk,1/p) such that (z0,p′,…,zi,p′) are affinely dependent. Thus, the family of vectors

(v1p,…,vip)=(z1,p′-z0,p′,…,zi,p′-z0,p′)

is linearly dependent. Hence, there exists a λp=(λ1p,…,λip)∈ℝi∖{0} such that ∑k=1iλkp⁢vkp=0. We can normalize by letting μp=λp/∥λp∥∈Si-1⁢(0,1). By a compactness argument, the sequence μp admits a convergent subsequence μφ⁢(p) to μ¯∈Si-1⁢(0,1). Since for all k∈{0,…,i}, there exists a zk,p′∈B⁢(zk,1/p), then the sequence (zk,p′)p∈ℕ* converges to zk. Therefore, we have

∑k=1iμφ⁢(p)⁢vkφ⁢(p)=∑k=1i(λkφ⁢(p)/∥λφ⁢(p)∥)⁢vkφ⁢(p)=0→∑k=1iμ¯k⁢(zk-z0).

That is, ∑k=1iμ¯k⁢(zk-z0)=0. Since, by the starting assumption, (z1-z0,…,zi-z0) is linearly independent, then we obtain μ¯=0, which is absurd. ∎

We recall that if yn=(y0n,…,yin)∈Ya⁢ii+1, then every point zn of aff⁡(yn) has a unique representation

zn=∑k=0iλkn⁢ykn,λn=(λ0n,…,λin)∈ℝi+1,∑k=0iλkn=1,

where λ0n,…,λin are the barycentric coordinates of the point zn relative to (y0n,…,yin). Using the previous notations and adopting obvious ones for the limits, our next result is formulated in the following way, where the assumption y¯∈Ya⁢ii+1 can not be omitted.

Lemma 5.6.

Let yn∈Ya⁢ii+1 tending to y¯∈Ya⁢ii+1. Let zn∈aff⁡(yn). Hence, we have:

If zn is bounded, then λn is bounded and zn has a cluster point in aff⁡(y¯).
If zn converges to z¯, then λn converges to λ¯.

Proof.

We start by proving assertion (1). We denote by wn=zn-y0n. Since ∑k=0iλkn=1, it follows that wn=∑k=0iλkn⁢(ykn-y0n). Hence, wn=∑k=1iλkn⁢(ykn-y0n). We denote by λ~n=(λ1n,…,λin)∈ℝi, where the first component is omitted. First, we will prove by contradiction that λ~n is bounded. Assume the contrary. Then there exists a subsequence λ~ψ⁢(n) of λ~n such that ∥λ~ψ⁢(n)∥ diverges to infinity. Since zn is bounded, it follows that wψ⁢(n)/∥λ~ψ⁢(n)∥→0. Moreover, by normalizing, the sequence μψ⁢(n)=λ~ψ⁢(n)/∥λ~ψ⁢(n)∥ belongs to the compact set Si-1⁢(0,1). Therefore, the sequence μψ⁢(n) admits a convergent subsequence μψ⁢(φ⁢(n)) converging to μ¯ in Si-1⁢(0,1). Thus, we obtain that

wψ⁢(φ⁢(n))∥λ~ψ⁢(φ⁢(n))∥→∑k=1iμ¯k⁢(yk¯-y0¯).

By uniqueness of the limit, we deduce that ∑k=1iμ¯k⁢(yk¯-y0¯)=0. Since (y1¯-y0¯,…,yi¯-y0¯) is independent, we obtain μ¯=0, which is impossible. Now it suffices to remark that since λ0n=1-∑k=1iλkn, the boundedness of λ~n implies the one of λ0n and therefore of the whole vector λn.

It remains to check that zn has a cluster point. We have already established that λn is bounded. Therefore, there admits a convergent subsequence λψ⁢(n) converging to λ¯. Hence,

zψ⁢(n)=∑k=0iλkψ⁢(n)⁢ykψ⁢(n)→z¯=∑k=0iλ¯k⁢y¯k.

That is, z¯ is a cluster point of zn. Moreover, observe that since ∑k=0iλkψ⁢(n)=1 converges to ∑k=0iλ¯k, we have z¯∈aff⁡(y¯), which establishes the result.

Now, we are able to prove the second assertion. By assertion (1), we know that λn is bounded. By [1], in order to prove that λn converges to λ¯, we claim that the bounded sequence λn has a unique cluster point. Using the same notations, let a∈F, where F is the set of cluster points of λn. Then there exists a subsequence λφ⁢(n) of λn such that λφ⁢(n)→a. Consequently,

wφ⁢(n)=∑k=1iλkφ⁢(n)⁢(ykφ⁢(n)-y0φ⁢(n))→∑k=1iak⁢(y¯k-y¯0).

Alternatively, we have that wφ⁢(n)=zφ⁢(n)-y0φ⁢(n) converges to

z¯-y¯0=∑k=0iλk⁢y¯k-y¯0=∑k=1iλk⁢(y¯k-y¯0),

since ∑k=0iλk=1. By uniqueness of the limit and given that (y1¯-y0¯,…,yi¯-y0¯) is independent, we obtain that ak=λ¯k, for all k∈{1,…,i}. Therefore, we have a unique cluster point. Consequently, λkn→λ¯k, for all k∈{1,…,i}. In addition, for the first component, we obtain λ0n=1-∑k=1nλkn→1-∑k=1nλ¯k=λ¯0. Thus, for all k∈{0,…,i}, we have λkn→λ¯k. This completes the proof. ∎

5.3 Proof of Proposition 4.8

The key argument we will use is based on the following lemma. Note that Tan and Yuan [11] have a similar but weaker result, since they only treated the case of a finite-dimensional set Y when the images have a nonempty interior which avoids to introduce the relative interior.

Lemma 5.7 (Fundamental Lemma).

Under the assumptions of Theorem 4.1, let x¯∈Di, γ>0 and y¯∈Γ⁢(φ⁢(x¯),γ). Then, for any ε∈]0,γ[, there exists a neighborhood V of x¯ such that for every x∈V∩Di, we have that Γ⁢(φ⁢(x),γ-ε)∩B⁢(y¯,ε)≠∅.

Proof.

First, let us fix some ε∈]0,γ[. In order to simplify the notations, we will assume within this proof that X=Di. We will start by proving the following claim:

Claim 1.

There exists a neighborhood V1 of x¯, such that for all x∈V1, we have B¯φ⁢(x)⁢(y¯,γ-ε/3)⊂φ⁢(x).

To prove the claim, let us denote by r the positive quantity r=γ-ε/3. In order to get a contradiction, assume that for all n>0, there exists an xn∈B⁢(x¯,1/n), and there exists a zn∈B¯φ⁢(xn)⁢(y¯,r) such that zn∉φ⁢(xn). Since for all x¯∈X, we have dima⁡φ⁢(x¯)=i, there exists (y^0,…,y^i)∈Ya⁢ii+1∩φ⁢(x¯). Moreover, we have xn→x¯ and φ is lsc. Therefore, for n sufficiently large, there exists ykn→y^k such that ykn∈φ⁢(xn), for all k∈{0,…,i}. Using Lemma 5.5, for n large enough, we obtain that dima(y0n,…,yin)=i.

Besides, we have zn∈aff⁡(φ⁢(xn))=aff⁡(y0n,…,yin). Since zn∈B¯⁢(y¯,r), applying the first assertion of Lemma 5.6, we conclude that zn has a cluster point z¯ in aff⁡(φ⁢(x¯))=aff⁡(y^0,…,y^i). Furthermore, since zn∈B¯⁢(y¯,r), we have z¯∈B¯⁢(y¯,r)⊂B⁢(y¯,γ). It follows that z¯ belongs to ri⁡(B¯φ⁢(x¯)⁢(y¯,γ)).

Hence, in view of the dimension of B¯φ⁢(x¯)⁢(y¯,γ), there exists an i-simplex Si=co⁡(U¯0,…,U¯i) contained in B¯φ⁢(x¯)⁢(y¯,γ) such that z¯∈ri⁡(Si). We can write the affine decomposition z¯=∑k=0iμ¯k⁢U¯k, for some μ∈ℝi+1 such that ∑k=0iμ¯k=1 and μ¯k>0. In particular, in view of the assumption of the lemma, U¯k⊂B¯φ⁢(x¯)⁢(y¯,γ)⊂ri⁡(φ⁢(x¯)). Using again that xn→x¯ and the lower semicontinuity of φ, we obtain that there exists a sequence Ukn→U¯k such that Ukn∈φ⁢(xn), for all k∈{0,…,i} and n large enough. Then we consider μn=(μ0n,…,μin), the affine coordinates of zn in (U0n,…,Uin). Since zn has a cluster point z¯, there exists a subsequence zψ⁢(n) of zn converging to z¯. By assertion (2) of Lemma 5.6, we have μkψ⁢(n)→μ¯k, for all k∈{0,…,i}. However, μ¯k>0, then μkψ⁢(n)≥0, for n large enough. Consequently, zψ⁢(n) is a convex combination of Ukψ⁢(n). By the convexity of φ⁢(xψ⁢(n)), we conclude that zψ⁢(n)∈φ⁢(xψ⁢(n)), which is a contradiction. This proves Claim 1.

Note that an obvious consequence of Claim 1 is that for all x∈V1, we have

B¯φ⁢(x)⁢(y¯,γ-2⁢ε3)⊂ri⁡(φ⁢(x)).

Now, since ε>0, we have y¯∈φ⁢(x¯)∩B⁢(y¯,ε/3). From the lower semicontinuity of φ, there exists a neighborhood V2 of x¯, such that for all x∈V2, we have the existence of some y∈φ⁢(x)∩B⁢(y¯,ε/3). Therefore, for any x∈V=V1∩V2, we have

B¯φ⁢(x)⁢(y,γ-ε)⊂B¯φ⁢(x)⁢(y¯,γ-2⁢ε3)⊂ri⁡(φ⁢(x)).

Thus, we finish the proof. ∎

Note that Proposition 4.8 was already stated without proof in Section 4. We are now ready to prove it.

Proof of Proposition 4.8.

Let us consider an open set O and x¯∈φ-⁢(O). This means that φη⁢(x¯)∩O is nonempty and contains some y¯. Since O is an open set containing y¯, we can first remark that there exists r>0 such that B¯⁢(y¯,r)⊂O.

On one hand, by the condition on η, letting γ=α⁢(x¯)-η⁢(x¯)3>0 and ηi=η⁢(x¯)+i⁢γ, for any i={0,…,3}, we have

η⁢(x¯)=η0<η1<η2<η3=α⁢(x¯).

By the definition of α, there exists z¯∈Γ⁢(φ⁢(x¯),η2). Applying the Fundamental Lemma, we obtain that there exists a neighborhood V1 of x¯ such that for any x∈V1, there exists z∈Γ⁢(φ⁢(x),η1)∩B⁢(z¯,γ).

On the other hand, we will distinguish two cases depending whether at a point x, there is or is not a significant peeling. Let us denote by M=1+2γ⁢(∥y¯-z¯∥+γ); we will consider the positive number ε¯=min⁡(r/M,γ).

First case: η⁢(x¯)>0.

Let us denote ε=min⁡(ε¯,η⁢(x¯)/2). Since y¯∈φη⁢(x¯)=Γ⁢(φ⁢(x¯),η⁢(x¯)), then once again, by the Fundamental Lemma, we have the existence of a neighborhood V2 of x¯ such that for any x∈V2, there exists y∈Γ⁢(φ⁢(x),η⁢(x¯)-ε)∩B⁢(y¯,ε). Now, let us consider λ=2⁢εγ+ε∈]0,1] and yλ=(1-λ)⁢y+λ⁢z.

Second case: η⁢(x¯)=0.

In this case, φη⁢(x¯)=ri⁡(φ⁢(x¯)). Let ε=ε¯. Since φ is lsc, there exists a neighborhood V2 of x¯ such that for any x∈V2, there exists y∈φ⁢(x)∩B⁢(y¯,ε) satisfying

B¯φ⁢(x)⁢(y,0)⊂ri⁡(φ⁢(x)).

Now, let us consider λ=εγ∈]0,1] and yλ=(1-λ)⁢y+λ⁢z.

In both cases, since the set φ⁢(x) is convex, in view of our choice of λ, by a simple computation, we can prove that if x∈V1∩V2, then yλ∈Γ⁢(φ⁢(x),η⁢(x¯)+ε).

Note that in both cases λ≤2⁢ε/γ. The following computation will show that our choice of ε implies yλ∈O:

∥yλ-y¯∥≤(1-λ)⁢∥y-y¯∥+λ⁢∥z-y¯∥

≤∥y-y¯∥+λ⁢(∥z-z¯∥+∥z¯-y¯∥)

≤ε+2⁢εγ⁢(∥z-z¯∥+∥z¯-y¯∥)

≤ε+2⁢εγ⁢(γ+∥y¯-z¯∥)=ε⁢M≤r.

Finally, by the continuity of η, there exists a neighborhood V3 of x¯ such that for any x∈V3, we have η⁢(x¯)-ε<η⁢(x)<η⁢(x¯)+ε. Summarizing the previous results, for all x∈V1∩V2∩V3, there exists yλ∈O such that B¯φ⁢(x)⁢(yλ,η⁢(x))⊂B¯φ⁢(x)⁢(yλ,η⁢(x¯)+ε)⊂ri⁡(φ⁢(x)), which means that yλ∈φη⁢(x)∩O and establishes the result. ∎

5.4 Proof of Theorem 4.1

Using the above lemma 5.7, we are able to prove the following result on the regularity of the internal radius of an lsc correspondence.

Lemma 5.8.

Let us consider φ:X→2Y such that there exists some positive integer i such that for all x∈X, dima⁡φ⁢(x)=i. Then α∘φ:X→R+∩{+∞} is a lower semicontinuous function.

Proof.

Let us first recall that

α⁢(φ⁢(x)):=sup⁡{ρ∈ℝ+∣there exists ⁢y∈φ⁢(x)⁢ such that ⁢B¯φ⁢(x)⁢(y,ρ)⊂ri⁡(φ⁢(x))}.

Let us fix x¯ in X, and ρ<α⁢(φ⁢(x¯)). We can consider γ such that ρ<γ<α⁢(φ⁢(x¯)). By the definition of α, there exists y∈Γ⁢(φ⁢(x¯),γ). Now, by the Fundamental Lemma, there exists a neighborhood V of x¯ such that for all x∈V, Γ⁢(φ⁢(x),ρ)≠∅. Using Lemma 5.3, we obtain that for all x in V, α⁢(φ⁢(x))>ρ, which establishes the result. ∎

An immediate consequence is the following result.

Corollary 5.9.

For all positive integers i, there exists ηi:Di→R+ continuous such that 0<ηi⁢(x)<α⁢(φ⁢(x)).

Proof.

Since on Di, α∘φ is a lower semicontinuous function, the correspondence given by

Λ:Di→2ℝ,x↦]0,α(φ(x))[

is lsc. Applying Michael’s Selection Theorem 3.2, we deduce that there exists a single-valued continuous selection ηi of Λ. ∎

Proof of Theorem 4.1.

If i=0, then φ is reduced to a singleton on D0. Moreover, on D0, the correspondences φ and ri⁡(φ) coincide. Therefore, there exists a mapping h0:D0→Y such that for all x∈D0, φ⁢(x)={h0⁢(x)}. Since φ is lsc on D0, it is well known that h0 is a continuous function.

If i>0, let us first apply Corollary 5.9 in order to get a continuous function ηi. Consequently, by Proposition 4.8, we have for any i∈ℕ*, that φηi is lsc on Di. By a classical result, this implies that φ¯ηi is also lsc on Di. Moreover, in view of the double inequality satisfied by ηi, we can apply Lemma 5.3, in order to state that φ¯ηi is a correspondence with nonempty convex-closed values. Therefore, applying Theorem 3.1 of Michael gives a continuous single-valued selection hi of φ¯ηi. That is for any x∈Di, hi⁢(x)∈Γ¯⁢(φ⁢(x),ηi⁢(x)). Since ηi⁢(x)>0, we can apply Lemma 5.4 in order to show that hi⁢(x)∈Γ⁢(φ⁢(x),ηi⁢(x)/2). Now, for all x∈Di, let βi⁢(x):=ηi⁢(x)/2; then the previous condition can be rewritten as B¯φ⁢(x)⁢(hi⁢(x),βi⁢(x))⊂ri⁡(φ⁢(x)), which proves the result. ∎

It is worth noting that the idea of peeling should be distinguished from the approximation method introduced by Cellina [2]. Indeed, mainly Cellina’s method consists of approximating an upper semicontinuous correspondence φ by a lower semicontinuous one.^[4] In addition, unlike the peeling concept which can be seen as an “inside approximation” (the approximated set is a subset of the original one), the approximation of Cellina is an “outside one” ( the original set is a subset of the approximated one). As it is well known (see [2]), applying selection theorems to Cellina approximations is used to deduce Kakutani’s fixed point theorem.

We are now ready to prove the main result of this paper.

6 Proof of Theorem 4.2

6.1 Notations and preliminaries

We denote

by D∞={x∈X∣φ⁢(x)⁢ has infinite dimension},
by D≤i=D0∪D1∪⋯∪Di,
by D≥i=(Di∪Di+1∪⋯)∪D∞.

As in the previous section, we begin by listing a series of parallel results which will be used in order to prove Theorem 4.2. The first one is a classical result (see for example [4]).

Lemma 6.1.

Let C be a convex set such that ri⁡(C) is nonempty. Then, for any α such that 0<α<1, we have (1-α)⁢A¯+α⁢ri⁡(A)⊂ri⁡(A).

Lemma 6.2.

The set D≥i is an open subset of X.

Proof.

Let x∈D≥i. Then there exists j≥i such that x∈Dj. Therefore, since dima⁡φ⁢(x)=j, then we have Ya⁢ij+1∩aff⁡(φj+1⁢(x))≠∅ where φj+1⁢(x) is the cartesian product φj+1⁢(x)=φ⁢(x)×⋯×φ⁢(x). Since the lower semicontinuity of φ implies that φj+1 is also lsc, in view of the openness of Ya⁢ij+1, there exists a neighborhood Vx of x such that for any x′∈Vx, Ya⁢ij+1∩aff⁡(φj+1⁢(x′))≠∅. This implies that dima⁡φ⁢(x′)≥j. That is, Vx⊂D≥j. Yet, since D≥j⊂D≥i, we finish the proof. ∎

Lemma 6.3.

The set D≤(i-1) is closed in D≤i.

Proof.

In view of the partition D≤i=D≤i-1∪Di, in order to prove the result, it suffices to prove that Di is an open set in D≤i. Using the previous remark, we already know that D≥i is an open set of X. Yet, since Di=D≥i∩D≤i, the result is established. ∎

Lemma 6.4.

Let X and Y be two topological spaces and F a closed subset of X. Suppose that φ:X→2Y is an lsc correspondence and f:F→Y is a continuous single-valued selection of φ|F. Then, the correspondence ψ given by

ψ⁢(x)={{f⁢(x)}if ⁢x∈F,φ⁢(x)if ⁢x∉F,

is also lsc.

Proof.

Let V be a closed subset of Y. We have

ψ+⁢(V)={x∈F∣ψ⁢(x)⊂V}∪{x∈X∖F∣ψ⁢(x)⊂V}={x∈F∣f⁢(x)∈V}∪{x∈X∖F∣φ⁢(x)⊂V}.

Since f is a selection of φ|F, we deduce that

{x∈X∣φ⁢(x)⊂V}={x∈X∖F∣φ⁢(x)⊂V}∪{x∈F∣φ⁢(x)⊂V}

={x∈X∖F∣φ⁢(x)⊂V}∪{x∈F∣f⁢(x)∈φ⁢(x)⊂V}.

Therefore,

ψ+⁢(V)={x∈F∣f⁢(x)∈V}∪{x∈X∣φ⁢(x)⊂V}=f-1⁢(V)∪φ+⁢(V).

Since φ is lsc, the set φ+⁢(V) is closed. Moreover, since V is a closed subset of Y and f is continuous, it follows that f-1⁢(V) is a closed subset of F. Now, since F is closed in X, the set f-1⁢(V) is closed in X. Hence, ψ+⁢(V) is closed, as required. ∎

6.2 Proof of Theorem 4.2

The proof of Theorem 4.2 is outlined in three steps as follows:

For any k∈ℕ, we have that ri⁡(φ) admits a continuous selection h≤k on D≤k.
For any k∈ℕ, there exists jk:X→Y such that
1. jk is a continuous selection of φ¯ on X.
2. jk is a continuous selection of ri⁡(φ) on D≤k.
There exists a continuous selection f of φ on X.

Proof of Step 1.

Let us apply for any k∈ℕ, Theorem 4.1 in order to get the existence of a continuous single-valued function hk defined on Dk such that for all x∈Dk, hk⁢(x)∈ri⁡(φk⁢(x)). Let Pn be the following heredity property: the restriction of ri⁡(φ) to D≤n admits a continuous selection h≤n.

For n=0, it suffices to notice that D≤0=D0. Therefore, we can let h≤0:=h0 which is a continuous selection of ri⁡(φ). Thus, P0 is true.

Let n≥1. Suppose that Pn-1 holds true and let us prove that Pn is true. By the heredity hypothesis, we have that ri⁡(φ) admits a continuous selection h≤(n-1) on D≤(n-1). We will introduce an auxiliary mapping φ~n defined on D≤n, which is lsc and such that the graph is contained in the graph of φ, by taking

φ~n:D≤n→2Y defined by φ~n⁢(x)={{h≤(n-1)⁢(x)}if ⁢x∈D≤n-1,φ¯⁢(x)if ⁢x∈Dn.

By Lemma 6.3 and Lemma 6.4, we conclude that φ~n is lsc. Moreover, φ~n has closed-convex values. By Michael’s selection Theorem 3.1, there exists gn continuous such that gn⁢(x)∈φ~n⁢(x)⊂φ¯⁢(x), for every x∈D≤n. In the following, we will construct a continuous single-valued selection h≤n of ri⁡(φ) on D≤n. We consider the continuous application λ:Dn→]0,1[ given by

λ⁢(x)=min⁡(d⁢(x,D≤n-1),1)2+∥hn⁢(x)∥.

Then we define h≤n:D≤n→Y given by

h≤n⁢(x)={gn⁢(x)if ⁢x∈D≤n-1,(1-λ⁢(x))⁢gn⁢(x)+λ⁢(x)⁢hn⁢(x)if ⁢x∈Dn.

Using the heredity property, we know that h≤(n-1) is a selection of ri⁡(φ) on D≤(n-1). On the other hand, on Dn, the function h≤n is a strict convex combination of gn∈φ¯ and hn∈ri⁡(φ), thus by Lemma 6.1, h≤n∈ri⁡(φ). Therefore, we conclude that h≤n is a selection of ri⁡(φ) on D≤n. It remains to check the continuity of h≤n. It is clear that the restriction of h≤n on Dn (respectively on D≤n-1) is continuous. Since Dn is open in D≤n, it suffices to consider the case of a sequence xk∈Dn such that xk→x¯∈D≤n-1. It is obvious that since d⁢(xk,D≤n-1) tends to zero, λ⁢(xk) tends to zero when xk tends to x¯. Besides, we have that ∥hn⁢(x)∥/(2+∥hn⁢(x)∥) is bounded. That is, λ⁢(xk)⁢hn⁢(xk) tends to zero when xk tends to x¯. Therefore, combining the previous remarks and the continuity of gn allows us to conclude that h≤n⁢(xk) tends to gn⁢(x¯)=h≤n⁢(x¯), which establishes the result. ∎

Proof of Step 2.

Using Lemma 6.2, we have already proved that D≥i is an open set of X. Consequently, D≤i-1 is closed in X. Using again Lemma 6.3, we can define an lsc function Tk:X→2Y given by

Tk⁢(x)={{h≤k⁢(x)}if ⁢x∈D≤k,φ¯⁢(x)if ⁢x∈D≥k+1.

By Michael’s Selection Theorem 3.1, there exists jk continuous such that for all x∈X, jk⁢(x)∈Tk⁢(x)⊂φ¯⁢(x). That is, jk is a selection of φ¯. On the other hand, for any x∈D≤k, we have jk⁢(x)∈Tk⁢(x)={h≤k⁢(x)}, which finishes the proof of Step 2. ∎

Proof of Step 3.

We can write X as a partition between X1 and D∞, where X1=⋃k∈ℕDk. Under the hypothesis of Theorem 4.2, we have that D∞ is a subset of X2, where X2={x∈X∣φ⁢(x)⁢ has closed values}.

Now, in the spirit of Michael’s proof, for any k∈ℕ*, we define

fk⁢(x)=λk⁢(x)⁢jk⁢(x)+(1-λk⁢(x))⁢j0⁢(x),where ⁢λk⁢(x)=1max⁡(1,∥jk⁢(x)-j0⁢(x)∥).

It is easy to check in view of Lemma 6.1 that for each k∈ℕ*, fk is also a selection of φ¯ on X satisfying for any x∈D≤k, fk⁢(x)∈ri⁡(φ⁢(x)). In addition, we have that fk⁢(x) is bounded since fk⁢(x) can be written as fk⁢(x)=j0⁢(x)+λk⁢(x)⁢(jk⁢(x)-j0⁢(x)) and our choice of λk⁢(x) ensures that ∥fk⁢(x)∥≤∥j0⁢(x)∥+1.

We can now define for any k∈ℕ*,

f~n⁢(x)=∑k=1n12k⁢fk⁢(x),f⁢(x)=∑k∈ℕ*12k⁢fk⁢(x).

First, we claim that f is continuous at any x¯∈X. As a consequence of the continuity of j0, the set

Wx¯={x∈X∣∥j0⁢(x)∥≤∥j0⁢(x¯)∥+1}

is an open neighborhood of x¯. Indeed, each fk is continuous and the series f~n converges uniformly to f on Wx¯, since

∥f⁢(x)-f~n⁢(x)∥≤∑k=n+1∞12k⁢∥fk⁢(x)∥≤(∥j0⁢(x¯)∥+2)⁢∑k=n+1∞12k.

Moreover, we have f⁢(x)∈φ¯⁢(x) for any x∈X, since f is an “infinite convex combination”^[5] of elements of φ¯⁢(x).

Now, we claim that for all x∈X, f⁢(x) is an element of φ⁢(x). We have to distinguish three cases.

If x∈D0, then f⁢(x)=j0⁢(x) since φ⁢(x) is a singleton.
If x∈X1∖D0, then there exists k0⁢(x)>0 such that x∈Dk0⁢(x)⊂D≤k0⁢(x). Let us remark that f⁢(x) can be written as
f⁢(x)=μ⁢fk0⁢(x)⁢(x)+(1-μ)⁢∑k≠k0⁢(x)fk⁢(x)2k⁢(1-μ),
where μ=12k0⁢(x). Using once again Footnote 6.2, we easily check that ∑k≠k0⁢(x)fk⁢(x)2k⁢(1-μ) belongs to φ¯⁢(x). Therefore, by Lemma 6.1, f⁢(x) is an interior point of φ⁢(x), then a selection of φ, which finishes the proof.
If x∈D∞, then x∈X2. That is, φ¯⁢(x)=φ⁢(x). Since we have already established that f⁢(x)∈φ¯⁢(x) for all x∈X, the result is immediate.∎

Dedicated to the memory of Monique Florenzano.

Acknowledgements

We thank Philippe Bich, Jean-Marc Bonnisseau, Souhail Chebbi and an anonymous referee for their useful comments.

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Received: 2016-2-25

Revised: 2016-4-29

Accepted: 2016-5-7

Published Online: 2016-7-2

Published in Print: 2018-5-1

This article is distributed under the terms of the Creative Commons Attribution Non-Commercial License, which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.

Articles in the same Issue

https://doi.org/10.1515/anona-2016-0053

Keywords for this article

Barycentric coordinates; continuous selections; lower semicontinuous correspondence; closed-valued correspondence; finite-dimensional convex values; separable Banach spaces

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