On the Core of a Low Dimensional Set-Valued Mapping
aa r X i v : . [ m a t h . F A ] F e b On the Core of a Low Dimensional Set-Valued Mapping
By P avel S hvartsman Department of Mathematics, Technion - Israel Institute of Technology,32000 Haifa, Israele-mail: [email protected]
Abstract
Let M = ( M , ρ ) be a metric space and let X be a Banach space. Let F be a set-valued mapping from M into the family K m ( X ) of all compact convex subsets of X of dimension at most m . The main result in our recent joint paper [16] with CharlesFe ff erman (which is referred to as a “Finiteness Principle for Lipschitz selections”)provides e ffi cient conditions for the existence of a Lipschitz selection of F , i.e., aLipschitz mapping f : M → X such that f ( x ) ∈ F ( x ) for every x ∈ M .We give new alternative proofs of this result in two special cases. When m = X = R , and when m = X . Both ofthese proofs make use of a simple reiteration formula for the “core” of a set-valuedmapping F , i.e., for a mapping G : M → K m ( X ) which is Lipschitz with respect tothe Hausdor ff distance, and such that G ( x ) ⊂ F ( x ) for all x ∈ M .
1. Introduction
Let M = ( M , ρ ) be a pseudometric space , i.e., suppose that the “distance function” ρ : M × M → [0 , + ∞ ] satisfies ρ ( x , x ) = ρ ( x , y ) = ρ ( y , x ), and ρ ( x , y ) ≤ ρ ( x , z ) + ρ ( z , y )for all x , y , z ∈ M . Note that ρ ( x , y ) = x , y , and ρ ( x , y ) may be + ∞ .Let ( X , k · k ) be a real Banach space. Given a non-negative integer m we let K m ( X )denote the family of all non-empty compact convex subsets K ⊂ X of dimension at most m . (We say that a convex subset of X has dimension at most m if it is contained in ana ffi ne subspace of X of dimension at most m .) We let K ( X ) = S {K m ( X ) : m = , , ... } denote the family of all non-empty compact convex finite-dimensional subsets of X .By Lip( M , X ) we denote the space of all Lipschitz mappings from M to X equippedwith the Lipschitz seminorm k f k Lip( M , X ) = inf { λ > k f ( x ) − f ( y ) k ≤ λ ρ ( x , y ) for all x , y ∈ M } . In this paper we study the following problem.
Math Subject Classification:
Key Words and Phrases:
Set-valued mapping, Lipschitz selection, Helly’s theorem, the core of a set-valuedmapping, Hausdor ff distance, balanced refinement.This research was supported by Grant No 2014055 from the United States-Israel Binational ScienceFoundation (BSF). roblem 1.1. Suppose that we are given a set-valued mapping F which to each point x ∈ M assigns a set F ( x ) ∈ K m ( X ). A selection of F is a map f : M → X such that f ( x ) ∈ F ( x ) for all x ∈ M .We want to know whether there exists a selection f of F in the space Lip( M , X ). Suchan f is called a Lipschitz selection of the set-valued mapping F : M → K m ( X ).If a Lipschitz selection f exists, then we ask how small we can take its Lipschitzseminorm .The following result provides e ffi cient conditions for the existence of a Lipschitz se-lection of an arbitrary set-valued mapping from a pseudometric space into the family K m ( X ). We refer to it as a “Finiteness Principle for Lipschitz selections”, or simply as a“Finiteness Principle”. Theorem 1.2. (Fe ff erman,Shvartsman [16]) Fix m ≥ . Let ( M , ρ ) be a pseudometricspace, and let F : M → K m ( X ) for a Banach space X. LetN ( m , X ) = ℓ ( m , X ) where ℓ ( m , X ) = min { m + , dim X } . (1.1) Suppose that for every subset M ′ ⊂ M consisting of at most N = N ( m , X ) points,the restriction F | M ′ of F to M ′ has a Lipschitz selection f M ′ with Lipschitz seminorm k f M ′ k Lip( M ′ , X ) ≤ . Then F has a Lipschitz selection f with Lipschitz seminorm k f k Lip( M , X ) ≤ γ (1.2) where γ = γ ( m ) is a positive constant depending only m. There is an extensive literature devoted to various versions of Finiteness Principles forLipschitz selections and related topics. We refer the reader to the papers [1, 2, 4, 14–16,19–21, 23–26] and references therein for numerous results in this direction.We note that the “finiteness number” N ( m , X ) in Theorem 1.2 is optimal; see [24].For the case of the trivial distance function ρ ≡
0, Theorem 1.2 agrees with the classi-cal Helly’s Theorem [9], except that the optimal finiteness constant for ρ ≡ n ( m , X ) = ℓ ( m , X ) + = min { m + , dim X + } in place of N ( m , X ) = ℓ ( m , X ) . Thus, Theorem 1.2 may be regarded as a generalization of Helly’s Theorem.Our interest in Helly-type criteria for the existence of Lipschitz selections was ini-tially motivated by some intriguing close connections of this problem with the classicalWhitney extension problem [28], namely, the problem of characterizing those functionsdefined on a closed subset, say E ⊂ R n , which are the restrictions to E of C m -smoothfunctions on R n . We refer the reader to the papers [5–7,10–13,26] and references thereinfor numerous results and techniques concerning this topic.One of the main ingredients of the proof of Theorem 1.2 is the construction of a specialset-valued mapping G : M → K m ( X ) introduced in [16] which we call a “core” of theset-valued mapping F . In fact each core is associated with a positive constant. Here arethe relevant definitions. 2 efinition 1.3. Let γ be a positive constant, and let F : M → K m ( X ) be a set-valuedmapping. A set-valued mapping G : M → K m ( X ) is said to be a γ -core of F if(i). G ( x ) ⊂ F ( x ) for all x ∈ M ;(ii). G is γ -Lipschitz with respect to Hausdor ff distance, i.e.,d H ( G ( x ) , G ( y )) ≤ γ ρ ( x , y ) for all x , y ∈ M . We refer to a map G as a core of F if G is a γ -core of F for some γ > ff distance d H ( A , B ) between two non-empty bounded subsets A and B of X is defined byd H ( A , B ) = inf { r > A + B X (0 , r ) ⊃ B and B + B X (0 , r ) ⊃ A } . (1.3)Here and throughout this paper, for each x ∈ X and r >
0, we use the standard notation B X ( x , r ) for the closed ball in X with center x and radius r . We also let B X = B X (0 , X , and we write rB X to denote the ball B X (0 , r ).In Definition 1.3 m can be any non-negative integer not exceeding the dimension ofthe Banach space X . It can happen that a core G : M → K m ( X ) of a given set-valuedmapping F : M → K m ( X ) in fact maps M into the smaller collection K m ′ ( X ) for someinteger m ′ ∈ [0 , m ). The next claim shows that the existence of some core G : M →K m ( X ) for F implies the existence of a (possibly di ff erent) core which maps M into K ( X ). Since K ( X ) is identified with X , that core is simply a Lipschitz selection of F . Claim 1.4. ( [16, Section 5]) Let γ be a positive constant, let m be a non-negative integer,and let G : M → K m ( X ) be a γ -core of a set-valued mapping F : M → K m ( X ) for someBanach space X. Then F has a Lipschitz selection f : M →
X with k f k Lip( M , X ) ≤ C γ where C = C ( m ) is a constant depending only on m. In [16] we showed that this claim follows from Definition 1.3 and the existence of theso-called “Steiner-type point” map St : K m ( X ) → X [25]. See formula (1.13).In [16] given a set-valued mapping F : M → K m ( X ) satisfying the hypothesis ofTheorem 1.2, we constructed a γ -core G of F with a positive constant γ depending onlyon m . We produced the core G using a rather delicate and complicated procedure whosemain ingredients are families of Basic Convex Sets associated with F , metric spaces withbounded Nagata dimension , ideas and methods of work [14] related to the case M = R n ,and Lipschitz selections on finite metric trees. See [16] for more details.In the present paper we suggest and discuss a di ff erent new geometrical method forproducing a core of a set-valued mapping. Its main ingredient is the so-called balancedrefinement of a set-valued mapping which we define as follows. Definition 1.5.
Let λ ≥
0, let ( M , ρ ) be a pseudometric space, let X be a Banach space,and let F : M → K m ( X ) be a set-valued mapping for some non-negative integer m . Foreach x ∈ M we consider the subset of F ( x ) defined by BR [ F : λ ; ρ ]( x ) = \ z ∈M (cid:2) F ( z ) + λ ρ ( x , z ) B X (cid:3) . We refer to the set-valued mapping BR [ F : λ ; ρ ] : M → K m ( X ) ∪{∅} as the λ -balancedrefinement of the mapping F . 3e note that any Lipschitz selection f of a set-valued mapping F : M → K m ( X ) with k f k Lip( M , X ) ≤ λ is also a Lipschitz selection of the λ -balanced refinement of F , i.e., f ( x ) ∈ BR [ F : λ ; ρ ]( x ) for all x ∈ M . Various geometrical parameters of the set BR [ F : λ ; ρ ]( x ) (such as diameter and width,etc.) may turn out to be smaller than the same parameters for the set F ( x ) which containsit. When attempting to find Lipschitz selections of F it may turn out to be convenient forour purposes to search for them in the more “concentrated” setting provided by the sets BR [ F : λ ; ρ ]( x ). One can take this approach still further by searching in even smallersets which can be obtained from consecutive iterations of balanced refinements of F , i.e.from the set functions which we describe in the following definition. Definition 1.6.
Let ℓ be a positive integer, and let ~λ = { λ k : 1 ≤ k ≤ ℓ } be a finite se-quence of ℓ non-negative numbers λ k . We set F [0] = F , and, for every x ∈ M and integer k ∈ [0 , ℓ − F [ k + ( x ) = BR [ F [ k ] : λ k + ; ρ ]( x ) = \ z ∈M h F [ k ] ( z ) + λ k + ρ ( x , z ) B X i . (1.4)We refer to the set-valued mapping F [ k ] : M → K m ( X ) ∪ {∅} , k ∈ [1 , ℓ ], as the k-thorder ( ~λ, ρ ) -balanced refinement of F .Clearly, F [ k + ( x ) ⊂ F [ k ] ( x ) on M for every k ∈ [0 , ℓ − . (1.5)(Put z = x in the right hand side of (1.4).) Remark 1.7.
Of course, for each integer k ∈ [1 , ℓ ] the set F [ k ] ( x ) also depends on thesequence ~λ = { λ k : 1 ≤ k ≤ ℓ } , on the pseudometric space M = ( M , ρ ) and the Banachspace X . However, in all places where we use F [ k ] ’s, these objects, i.e., ~λ , M and X , areclear from the context. Therefore, in these cases, we omit any mention of ~λ , M and X inthe notation of F [ k ] ’s. ⊳ We formulate the following
Conjecture 1.8.
Let ( M , ρ ) be a pseudometric space, and let X be a Banach space. Letm be a fixed positive integer and (as in the formula (1.1) of Theorem 1.2) let N ( m , X ) denote the “finiteness number” N ( m , X ) = ℓ where ℓ = ℓ ( m , X ) = min { m + , dim X } .There exist a constant γ ≥ and a sequence ~λ = { λ k : 1 ≤ k ≤ ℓ } of ℓ numbers λ k allsatisfying λ k ≥ such that the following holds:Let F : M → K m ( X ) be a set-valued mapping such that, for every subset M ′ ⊂ M with M ′ ≤ N ( m , X ) , the restriction F | M ′ of F to M ′ has a Lipschitz selection f M ′ : M ′ → X with Lipschitz seminorm k f M ′ k Lip( M ′ , X ) ≤ .Then the ℓ − th order balanced refinement of the mapping F, namely the set-valuedmapping F [ ℓ ] : M → K m ( X ) is a γ -core of F.Here F [ ℓ ] is defined as in Definition 1.6 using the particular sequence ~λ . m = X =
2, or (ii) m = X is an arbitrary Banach space. Note that in both of these cases the above mentionedfiniteness number N ( m , X ) equals 4. Theorem 1.9.
Let M = ( M , ρ ) be a pseudometric space, and let X be a two dimensionalBanach space. Let m = so that the number ℓ ( m , X ) = . In this case Conjecture 1.8holds for every λ , λ and γ such that λ ≥ e ( M , X ) , λ ≥ λ , γ ≥ λ (3 λ + λ ) / ( λ − λ ) . (1.6) Here e ( M , X ) denotes the Lipschitz extension constant of X with respect to M . (SeeDefinition 3.1.)Thus, the following statement is true: Let F : M → K ( X ) be a set-valued mappingfrom a pseudometric space ( M , ρ ) into the family K ( X ) of all non-empty convex compactsubsets of X. Given x ∈ M letF [1] ( x ) = \ z ∈M (cid:2) F ( z ) + λ ρ ( x , z ) B X (cid:3) , F [2] ( x ) = \ z ∈M h F [1] ( z ) + λ ρ ( x , z ) B X i . (1.7) Suppose that for every subset M ′ ⊂ M with M ′ ≤ , the restriction F | M ′ of F to M ′ has a Lipschitz selection with Lipschitz seminorm at most .Then for every λ , λ and γ satisfying (1.6) the setF [2] ( x ) , ∅ for every x ∈ M . (1.8) Furthermore, d H ( F [2] ( x ) , F [2] ( y )) ≤ γ ρ ( x , y ) for every x , y ∈ M . (1.9) If X is a Euclidean two dimensional space, (1.8) and (1.9) hold when (1.6) is replacedby the weaker requirements that λ ≥ e ( M , X ) , λ ≥ λ , γ ≥ λ (cid:26) + λ / (cid:16) λ − λ (cid:17) (cid:27) . (1.10)In particular, in Section we show that the mapping F [2] satisfies (1.8) and (1.9) when-ever X is an arbitrary two dimensional Banach space and λ = / λ = γ = X is also Euclidean, then one can set λ = /π , λ = /π and γ =
38. Furthermore, weprove that if M is a subset of a Euclidean space E , ρ is the Euclidean metric in E , and X is a two dimensional Euclidean space, then properties (1.8) and (1.9) hold for λ = λ =
3, and γ = X = ℓ ∞ , i.e., for R equipped with the norm k x k = max {| x | , | x |} , x = ( x , x ).More specifically, we show that in this case properties (1.8) and (1.9) hold whenever λ ≥ λ ≥ λ , and γ ≥ λ (3 λ + λ ) / ( λ − λ ). In particular, these properties hold for λ = λ = γ = K ( X ) of all bounded closed line segments of an arbitrary Banach space X . 5 heorem 1.10. Let ( M , ρ ) be a pseudometric space. Let m = and let X be a Banachspace with dim X > ; thus, ℓ ( m , X ) = , see (1.1). In this case Conjecture 1.8 holds forevery λ , λ and γ such that λ ≥ , λ ≥ λ , γ ≥ λ (3 λ + λ ) / ( λ − λ ) . (1.11) Thus, the following statement is true: Let F : M → K ( X ) be a set-valued mappingsuch that for every subset M ′ ⊂ M with M ′ ≤ , the restriction F | M ′ of F to M ′ has aLipschitz selection with Lipschitz seminorm at most .Let F [2] be the mapping defined by (1.7). Then properties (1.8) and (1.9) hold when-ever λ , λ and γ satisfy (1.11). In particular, one can set λ = , λ = and γ = .If X is a Euclidean space, the same statement is also true whenever, instead of (1.11), λ , λ and γ satisfy the weaker condition λ ≥ , λ ≥ λ , γ ≥ λ + λ / (cid:16) λ − λ (cid:17) . (1.12) In particular, in this case, (1.8) and (1.9) hold whenever λ = , λ = and γ = . In Section 5.1 we note that Conjecture 1.8 also holds for a one dimensional space X and m =
1. In this case the statement of the conjecture is true for every λ ≥ γ ≥ F satisfying thehypothesis of this theorem, the mapping F [2] determined by (1.7) with λ = / λ = γ -core of F with γ = F [2] corresponding to the parameters λ = λ = F satisfying the conditions of this theorem.We note that the proofs of Theorem 1.9 and Theorem 1.10 rely on Helly’s IntersectionTheorem and a series of auxiliary results about neighborhoods of intersections of convexsets. See Section . Remark 1.11.
Let us compare Conjecture 1.8 (and Theorems 1.9 and 1.10) with theFiniteness Principle (FP) formulated in Theorem 1.2. First we note that FP is invari-ant with respect to the transition to an equivalent norm on X , while the statement ofConjecture 1.8 is not.To express this more precisely, let k · k and k · k be two equivalent norms on X , i.e.,suppose that for some α ≥ /α ) k · k ≤ k · k ≤ α k · k holds.Clearly, if FP holds for ( X , k · k ) then it immediately holds also for ( X , k · k ) (with theconstant α γ in (1.2) instead of γ ). However the validity of Conjecture 1.8 for the norm k · k does not imply its validity for an equivalent norm k · k on X (at least we do notsee any obvious way for obtaining such an implication). For example, the validity ofConjecture 1.8 in ℓ n ∞ (i.e., R n equipped with the uniform norm) does not automaticallyimply its validity in the space ℓ n (i.e., R n with the Euclidean norm).We also note the following: in a certain sense, the result of Theorem 1.9 is “stronger”than Theorem 1.2 (i.e., FP for the case of a two dimensional Banach space X ). Indeed,in this case, the hypotheses of FP and Theorem 1.9 coincide. Moreover, Theorem 1.96nsures that the set-valued mapping F [2] is a core of F . This property of F [2] implies, viaarguments in [16] that the function f ( x ) = St ( F [2] ) ( x ) , x ∈ M , (1.13)is a Lipschitz selection of F . Here St : K m ( X ) → X is the Steiner-type point map [25].Thus, FP (in the two dimensional case) follows immediately from Theorem 1.9. How-ever, it is absolutely unclear how the statement of Theorem 1.9 can be deduced from FP.I would like to thank Charles Fe ff erman who kindly drew my attention to this interestingfact. ⊳ Let us reformulate Conjecture 1.8 in a way which does not require the use of thenotion of a core of a set-valued mapping . We recall that the mapping F [ ℓ ] : M → K m ( X )which appears in Conjecture 1.8 is a γ -core of F ifd H ( F [ ℓ ] ( x ) , F [ ℓ ] ( y )) ≤ γ ρ ( x , y ) for all x , y ∈ M . See part (ii) of Definition 1.3. Hence, given x ∈ M , F [ ℓ ] ( x ) ⊂ F [ ℓ ] ( y ) + γ ρ ( x , y ) B X for every y ∈ M . (1.14)We also recall that F [ ℓ + ( x ) = BR [ F [ ℓ ] : γ ; ρ ]( x ) = \ y ∈M h F [ ℓ ] ( y ) + γ ρ ( x , y ) B X i . See (1.4). This and (1.14) imply the inclusion F [ ℓ + ( x ) ⊃ F [ ℓ ] ( x ), x ∈ M . On the otherhand, (1.5) tells us that F [ ℓ + ( x ) ⊂ F [ ℓ ] ( x ) proving that F [ ℓ + = F [ ℓ ] on M .These observations enable us to reformulate Conjecture 1.8 as follows. Conjecture 1.12.
Let ( M , ρ ) be a pseudometric space, and let X be a Banach space. Letm be a fixed positive integer and let ℓ = ℓ ( m , X ) , see(1.1).There exists a sequence ~λ = { λ k : 1 ≤ k ≤ ℓ + } of ℓ + numbers λ k all satisfying λ k ≥ such that, for every set-valued mapping F : M → K m ( X ) satisfying the hypothesis ofthe Finiteness Principle (Theorem 1.2), the family { F [ k ] : k = , ..., ℓ + } of set-valuedmappings constructed by formula (1.4) has the following property:F [ ℓ ] ( x ) , ∅ and F [ ℓ + ( x ) = F [ ℓ ] ( x ) for all x ∈ M . (1.15)We refer to (1.15) as a Stabilization Property of balanced refinements.Thus, Theorem 1.9 and Theorem 1.10 tell us that a Stabilization Property of balancedrefinements holds whenever dim X = m = X is an arbitrary). More specifi-cally, Theorem 1.9 shows that if m = X =
2, Conjecture 1.12 holds with ℓ = ~λ = { / , , } .In other words, in this case, F [2] ( x ) , ∅ for each x ∈ M and F [3] = F [2] on M . In turn,Theorem 1.10 states that the same property holds whenever X is an arbitrary Banachspace, m =
1, and ~λ = { , , } .Readers might find it helpful to also consult a much more detailed version of thispaper posted on the arXiv [27]. 7 cknowledgements. I am very thankful to Michael Cwikel for useful suggestionsand remarks. I am also very grateful to Charles Fe ff erman for stimulating discussionsand valuable advice.The results of this paper were presented at the 12th Whitney Problems Workshop,August 2019, the University of Texas at Austin, TX. I am very thankful to all participantsof that workshop for valuable conversations and useful remarks.
2. Neighborhoods of intersections of convex sets in a Banach space
We first need to fix some notation.Let ( X , k · k ) be a Banach space, and let B X be the unit ball in X . Let A and B benon-empty subsets of X . We let A + B = { a + b : a ∈ A , b ∈ B } denote the Minkowskisum of these sets. Given λ ≥
0, by λ A we denote the set λ A = { λ a : a ∈ A } .Sometimes, for a given set M , we will be looking simultaneously at two distinctpseudometrics on M , say ρ and δ . In this case we will speak of a ρ -Lipschitz selectionand ρ -Lipschitz seminorm, or a δ -Lipschitz selection and δ -Lipschitz seminorm to makeclear which pseudometric we are using. Furthermore, given a mapping f : M → X we will write k f k Lip(( M ; ρ ) , X ) or k f k Lip(( M ; δ ) , X ) to denote the Lipschitz seminorm of f withrespect to the pseudometric ρ or δ respectively.For each finite set S , we let S denote the number of elements of S .There are two known results, which provide essential tools for proving the main resultsof the paper. Here is the first of them. Theorem 2.1.
Let X be a Banach space, and let C ⊂ X be a convex set. Let a ∈ X andlet r ≥ . Suppose that C ∩ B X ( a , r ) , ∅ . (2.1) Then for every s > and L > C ∩ B X ( a , Lr ) + θ ( L ) s B X ⊃ ( C + sB X ) ∩ B X ( a , Lr + s ) (2.2) where θ ( L ) = (3 L + / ( L − . (2.3) If X is a Euclidean space then (2.2) holds with θ ( L ) = + L / √ L − . For the case of an arbitrary Banach space X , Theorem 2.1 was proved by Przesławskiand Rybinski [19, p. 279]. For the case of a Euclidean space X see Przesławski, Yost [21,Theorem 4]. See also [27, Section 2]. For similar results we refer the reader to [1], [2, p.369] and [4, p. 26].The second of these known results is the classical Helly Intersection Theorem for twodimensional Banach spaces. It can be formulated as follows:8 heorem 2.2.
Let K be a collection of convex closed subsets of a two dimensionalBanach space X. Suppose that K is finite or at least one member of the family K isbounded.If every subfamily of K consisting of at most three elements has a common point thenthere exists a point common to all of the family K . We conclude this section by stating and proving one more result which will be a thirdmain tool for proving our main theorems. As we shall see, its proof makes use of thepreceding two theorems.
Proposition 2.3.
Let X be a two dimensional Banach space. Let C , C , C ⊂ X be convexsubsets, and let r > . Suppose thatC ∩ C ∩ ( C + rB X ) , ∅ . (2.4) Then for every L > and every ε > the following inclusion ( C ∩ C + LrB X ) ∩ C + θ ( L ) ε B X ⊃ [ C ∩ C + ( Lr + ε ) B X ] ∩ [( C + rB X ) ∩ C + ε B X ] ∩ [( C + rB X ) ∩ C + ε B X ] holds. The function θ in the above inclusion is as defined in Theorem 2.1. I.e., for anarbitrary Banach space X that inclusion holds for θ ( L ) = (3 L + / ( L − , and if X isEuclidean it also holds for θ ( L ) = + L / √ L − .Proof. Suppose that a ∈ [ C ∩ C + ( Lr + ε ) B X ] ∩ [( C + rB X ) ∩ C + ε B X ] ∩ [( C + rB X ) ∩ C + ε B X ](2.5)and prove that a ∈ ( C ∩ C + LrB X ) ∩ C + θ ( L ) ε B X . (2.6)First, let us show that C ∩ C ∩ ( C + rB X ) ∩ B X ( a , Lr + ε ) , ∅ . (2.7)Helly’s Theorem 2.2 tells us that this statement holds provided any three sets in theleft hand side of (2.7) have a common point. Note that, thanks to (2.4), this is true for C , C and C + rB X . Also property (2.5) tells us that a ∈ C ∩ C + ( Lr + ε ) B X , so that C ∩ C ∩ B X ( a , Lr + ε ) , ∅ .Let us prove that C ∩ ( C + rB X ) ∩ B X ( a , Lr + ε ) , ∅ . (2.8)By (2.5), the point a ∈ ( C + rB X ) ∩ C + ε B X . Therefore, there exist elements b ∈ C and b ∈ C such that k b − b k ≤ r and k a − b k ≤ ε . In particular, b ∈ C ∩ ( C + rB X ).Furthermore, k a − b k ≤ k a − b k + k b − b k ≤ ε + r ≤ ε + Lr
9o that b ∈ B X ( a , Lr + ε ). Hence, b ∈ C ∩ ( C + rB X ) ∩ B X ( a , Lr + ε ) proving (2.8). Ina similar way we show that C ∩ ( C + rB X ) ∩ B X ( a , Lr + ε ) , ∅ .Thus, (2.7) holds proving the existence of a point x ∈ X such that x ∈ C ∩ C ∩ ( C + rB X ) ∩ B X ( a , Lr + ε ) . (2.9)In particular, x ∈ C + rB X so that B X ( x , r ) ∩ C , ∅ proving that condition (2.1) ofTheorem 2.1 holds. This theorem tells us that B X ( x , Lr ) ∩ C + θ ( L ) ε B X ⊃ ( C + ε B X ) ∩ B X ( x , Lr + ε ) . (2.10)From (2.5) and (2.9) we learn that a ∈ C + ε B X and a ∈ B X ( x , Lr + ε ). Thus, the point a belongs to the set ( C + ε B X ) ∩ B X ( x , Lr + ε ). Therefore, by (2.10), a ∈ B X ( x , Lr ) ∩ C + θ ( L ) ε B X = ( x + LrB X ) ∩ C + θ ( L ) ε B X . But x ∈ C ∩ C , see (2.9), and the required inclusion (2.6) follows.The proof of the proposition is complete. (cid:4)
3. The main theorem for two dimensional Banach spaces
In this section we prove Theorem 1.9First, let us recall the notion of the Lipschitz extension constant e ( M , X ) which we usein the formulation of this theorem. Definition 3.1.
Let M = ( M , ρ ) be a pseudometric space, and let X be a Banach space.We define the Lipschitz extension constant e ( M , X ) of X with respect to M as the infimumof the constants λ > M ′ ⊂ M , and every Lipschitz mapping f : M ′ → X , there exists a Lipschitz extension ˜ f : M → X of f to all of M such that k ˜ f k Lip( M , X ) ≤ λ k f k Lip( M ′ , X ) . Remark 3.2.
Recall several results about Lipschitz extension constants which we use inthis paper. In particular, thanks to the McShane-Whitney extension theorem, e ( M , R ) = M = ( M , ρ ). Hence, e ( M , ℓ ∞ ) = e ( M , X ) ≤ / X is an arbitrary two di-mensional Banach space . See also [3]. Furthermore, e ( M , X ) ≤ /π whenever X is anarbitrary two dimensional Euclidean space . See [22] and [17].We also note that, thanks to Kirszbraun’s extension theorem [18], e ( M , X ) = X is a Euclidean space, M is a subset of a Euclidean space E , and ρ is the Eu-clidean metric in E . ⊳ Proof of Theorem 1.9.
Let M = ( M , ρ ) be a pseudometric space, and let X be a twodimensional Banach space. Let F : M → K ( X ) be a set-valued mapping satisfying thehypothesis of Theorem 1.9. This enables us to make the following Assumption 3.3.
For every subset M ′ ⊂ M with M ′ ≤ , the restriction F | M ′ of F to M ′ has a ρ -Lipschitz selection f M ′ : M ′ → X with k f M ′ k Lip(( M ′ , ρ ) , X ) ≤ .
10e fix constants L and α satisfying the following inequalities: L ≥ , α ≥ e ( M , X ) . (3.1)Then we introduce a new pseudometric on M defined byd( x , y ) = α ρ ( x , y ) , x , y ∈ M . (3.2)This definition, Definition 3.1 and the inequality α ≥ e ( M , X ) imply the following Claim 3.4.
Let M ′ ⊂ M , and let f : M ′ → X be a ρ -Lipschitz mapping on M ′ .There exists a d -Lipschitz extension ˜ f : M →
X of f to all of M with k ˜ f k Lip(( M , d) , X ) ≤k f k Lip(( M ′ , ρ ) , X ) . We introduce set-valued mappings F [1] ( x ) = \ z ∈M [ F ( z ) + d( x , z ) B X ] , x ∈ M , (3.3)and F [2] ( x ) = \ z ∈M h F [1] ( z ) + L d( x , z ) B X i , x ∈ M . (3.4)Thus, F [1] and F [2] are the first and the second order ( { , L } , d)-balanced refinementsof F respectively. See Definition 1.6.Our aim is to show that, if L and α satisfy inequality (3.1), then (i) F [2] ( x ) , ∅ on M ,and (ii) the mapping F [2] is d-Lipschitz with respect to the Hausdor ff distance. We prove(i) and (ii) in Proposition 3.8 and Proposition 3.9 respectively.We begin with the property (i). Its proof relies on a series of auxiliary lemmas. Lemma 3.5.
Let X be a two dimensional Banach space, and let
K ⊂ K ( X ) be a collec-tion of convex compact subsets of X with non-empty intersection. Let B ⊂ X be a convexclosed subset symmetric with respect to . Then \ K ∈ K K + B = \ K , K ′ ∈ K n (cid:16) K ∩ K ′ (cid:17) + B o . (3.5) Proof.
Obviously, the right hand side of (3.5) contains its left hand side. Let us provethe converse statement. Fix a point x ∈ \ K , K ′ ∈ K n (cid:16) K ∩ K ′ (cid:17) + B o (3.6)and prove that x ∈ ∩ { K : K ∈ K } + B . We know that B is symmetric with respect 0 sothat − B = B . Therefore, x ∈ ∩ { K : K ∈ K } + B if and only if B ( x ) \ \ K ∈ K K , ∅ where B ( x ) = x + B . (3.7)11et S = K ∩ { B ( x ) } . Helly’s intersection Theorem 2.2 tells us that property (3.7) holdsprovided ∩ { K : K ∈ S ′ } , ∅ for every subfamily S ′ ⊂ S consisting of at most threeelements. Clearly, this is true if B ( x ) < S ′ because there exists a point common to all ofthe sets from K .Suppose that B ( x ) ∈ S ′ . Then S ′ = { B ( x ) , K , K ′ } for some K , K ′ ∈ K . Then, thanks to(3.6), x ∈ ( K ∩ K ′ ) + B proving that B ( x ) ∩ K ∩ K ′ , ∅ .Thus, (3.7) holds, and the proof of the lemma is complete. (cid:4) Lemma 3.6.
For each x ∈ M the set F [1] ( x ) ∈ K ( X ) , i.e., F [1] ( x ) is a non-empty convexcompact subset of X. Furthermore, for every x , z ∈ M we haveF [1] ( z ) + L d( x , z ) B X = \ y ′ , y ′′ ∈M n [ F ( y ′ ) + d( z , y ′ ) B X ] ∩ [ F ( y ′′ ) + d( z , y ′′ ) B X ] + L d( x , z ) B X o . Proof.
Let x ∈ M . Clearly, F [1] ( x ) is a convex bounded subset of X . See (3.3).Prove that F [1] ( x ) , ∅ . Indeed, formula (3.3) and Helly’s Theorem 2.2 tell us that F [1] ( x ) , ∅ provided[ F ( z ) + d( x , z ) B X ] ∩ [ F ( z ) + d( x , z ) B X ] ∩ [ F ( z ) + d( x , z ) B X ] , ∅ (3.8)for every z , z , z ∈ M .To prove this property, we introduce a set M ′ = { x , z , z , z } and apply Assumption3.3 to M ′ . By this assumption, there exists a ρ -Lipschitz selection f M ′ : M ′ → X of F with ρ -Lipschitz seminorm k f M ′ k Lip(( M ′ , ρ ) , X ) ≤
1. In particular, f M ′ ( z i ) ∈ F ( z i ) and k f M ′ ( x ) − f M ′ ( z i ) k ≤ ρ ( x , z i ) ≤ αρ ( x , z i ) = d( x , z i ) for every i = , , . See (3.2). These properties of f M ′ tell us that f M ′ ( x ) belongs to the left hand side of(3.8). Thus, (3.8) holds for arbitrary z i ∈ M , i = , ,
3, proving that F [1] ( x ) , ∅ .Finally, the second statement of the lemma immediately follows from this property,Lemma 3.5 and formula (3.3). The proof of the lemma is complete. (cid:4) Lemma 3.7.
For every x ∈ M the set F [2] ( x ) admits the following representation:F [2] ( x ) = \ u , u ′ , u ′′ ∈M n [ F ( u ′ ) + d( u ′ , u ) B X ] ∩ [ F ( u ′′ ) + d( u ′′ , u ) B X ] + L d( u , x ) B X o . Proof.
The lemma is immediate from (3.4) and Lemma 3.6. (cid:4)
Given x , u , u ′ , u ′′ ∈ M we set T x ( u , u ′ , u ′′ ) = [ F ( u ′ ) + d( u ′ , u ) B X ] ∩ [ F ( u ′′ ) + d( u ′′ , u ) B X ] + L d( u , x ) B X . (3.9)In these settings, Lemma 3.7 reformulates as follows: F [2] ( x ) = \ u , u ′ , u ′′ ∈M T x ( u , u ′ , u ′′ ) . (3.10) Proposition 3.8.
For every x ∈ M the set F [2] ( x ) is non-empty. roof. Formula (3.10) and Helly’s Theorem 2.2 tell us that F [2] ( x ) , ∅ provided forevery choice of elements u i , u ′ i , u ′′ i ∈ M , i = , ,
3, we have T x ( u , u ′ , u ′′ ) ∩ T x ( u , u ′ , u ′′ ) ∩ T x ( u , u ′ , u ′′ ) , ∅ . (3.11)We set r i = d( x , u i ), i = , ,
3. Without loss of generality, we may assume that r ≤ r ≤ r . For each i ∈ { , , } we also set G ( u ′ i ) = F ( u ′ i ) + d( u ′ i , u i ) B X and G ( u ′′ i ) = F ( u ′′ i ) + d( u ′′ i , u i ) B X . (3.12)We will prove that there exist points y i ∈ X , i = , ,
3, such that y i ∈ G ( u ′ i ) ∩ G ( u ′′ i ) for every i = , , , (3.13)and k y − y k ≤ r + r and k y − y k ≤ r + r + r . (3.14)Let us see that the existence of the points y i with these properties implies (3.11).Indeed, let us set z = y + τ ( y − y ) = y + (1 − τ )( y − y )with τ = r / ( r + r + r ). Then, thanks to (3.14), k y − z k = τ k y − y k ≤ r r + r + r · ( r + r + r ) = r , and k y − z k = (1 − τ ) k y − y k ≤ r + r r + r + r · ( r + r + r ) = r + r . Hence, k y − z k ≤ k y − y k + k y − z k ≤ r + r + r = r + r . Recall that r i = d( x , u i ) and r ≤ r ≤ r . From this and above inequalities, we have k z − y i k ≤ r i = x , u i ) , i = , , . (3.15)Let us prove that z ∈ T x ( u i , u ′ i , u ′′ i ) for each i ∈ { , , } . In fact, we know that L ≥ y i ∈ G ( u ′ i ) ∩ G ( u ′′ i ), so that, thanks to (3.15), (3.12)and (3.9), z ∈ G ( u ′ i ) ∩ G ( u ′′ i ) + x , u i ) B X ⊂ G ( u ′ i ) ∩ G ( u ′′ i ) + L d( x , u i ) B X = T x ( u i , u ′ i , u ′′ i )proving (3.11).Thus, our aim is to prove the existence of points y i satisfying (3.13) and (3.14). Wewill do this in three steps. STEP 1 . We introduce sets W i ⊂ X , i = , ...,
4, defined by W = G ( u ′ ) , W = G ( u ′′ ) , W = G ( u ′ ) ∩ G ( u ′′ ) + ( r + r ) B X , (3.16)13nd W = G ( u ′ ) ∩ G ( u ′′ ) + ( r + r + r ) B X . (3.17)Obviously, there exist the points y i satisfying (3.13) and (3.14) whenever W ∩ W ∩ W ∩ W , ∅ . (3.18)By Helly’s Theorem 2.2, this property holds provided any three members of the familyof sets { W , W , W , W } have a common point. STEP 2.
Prove that W ∩ W ∩ W , ∅ .To see this, we set V = G ( u ′ ) + ( r + r ) B X , V = G ( u ′ ) , V = G ( u ′′ ) , (3.19)and V = G ( u ′ ) ∩ G ( u ′′ ) + ( r + r ) B X . (3.20)Let us show that if V ∩ V ∩ V ∩ V , ∅ , (3.21)then W ∩ W ∩ W is non-empty as well. Indeed, this property, definitions (3.19) and(3.20) imply the existence of points z ∈ G ( u ′ ), z ∈ G ( u ′ ) ∩ G ( u ′′ ), z ∈ G ( u ′ ) ∩ G ( u ′′ )such that k z − z k ≤ r + r and k z − z k ≤ r + r . Hence, k z − z k ≤ k z − z k + k z − z k ≤ ( r + r ) + ( r + r ) = r + r + r . Thus, thanks to (3.16) and (3.17), the point z belongs to W ∩ W ∩ W proving thatthis set is non-empty.Let us prove (3.21). Helly’s Theorem 2.2 tells us that (3.21) holds whenever everythree members of the family V = { V , V , V , V } have a common point.Let us verify this property. First, let us show that V ∩ V ∩ V , ∅ . (3.22)Let M = { u ′ , u ′ , u ′ , u ′′ } . Thanks to Assumption 3.3, there exists a ρ -Lipschitz map-ping f M : M → X with k f M k Lip(( M , ρ ) , X ) ≤ f M ( u ′ ) ∈ F ( u ′ ) , f M ( u ′ ) ∈ F ( u ′ ) , f M ( u ′ ) ∈ F ( u ′ ) , and f M ( u ′′ ) ∈ F ( u ′′ ) . Claim 3.4 tells us that there exists a d-Lipschitz mapping ˜ f : M → X with d-Lipschitzseminorm k ˜ f k Lip(( M , d) , X ) ≤ k f M k Lip(( M , ρ ) , X ) ≤ f | M = f M . Prove that˜ f ( u ) ∈ V ∩ V ∩ V . (3.23)We know that ˜ f ( u ′ ) = f M ( u ′ ) ∈ F ( u ′ ) and k ˜ f ( u ′ ) − ˜ f ( u ) k ≤ d( u ′ , u ). Hence,˜ f ( u ) ∈ F ( u ′ ) + d( u ′ , u ) B X = G ( u ′ ) = V .
14n the same way we prove that ˜ f ( u ) ∈ G ( u ′ ).We also know that k ˜ f ( u ) − ˜ f ( u ) k ≤ d( u , u ). Hence, ˜ f ( u ) ∈ G ( u ′ ) + d( u , u ) B X . By the triangle inequality,d( u , u ) ≤ d( u , x ) + d( x , u ) = r + r so that ˜ f ( u ) ∈ G ( u ′ ) + ( r + r ) B X = V .It remains to show that ˜ f ( u ) ∈ V . We know that˜ f ( u ′ ) = f M ( u ′ ) ∈ F ( u ′ ) , ˜ f ( u ′′ ) = f M ( u ′′ ) ∈ F ( u ′′ )and k ˜ f ( u ) − ˜ f ( u ′ ) k ≤ d( u , u ′ ) , k ˜ f ( u ) − ˜ f ( u ′′ ) k ≤ d( u , u ′′ ) . Hence, ˜ f ( u ) ∈ [ F ( u ′ ) + d( u ′ , u ) B X ] ∩ [ F ( u ′′ ) + d( u ′′ , u ) B X ] = G ( u ′ ) ∩ G ( u ′′ ) . Furthermore, k ˜ f ( u ) − ˜ f ( u ) k ≤ d( u , u ). These properties of ˜ f ( u ) and the triangleinequality d( u , u ) ≤ d( u , x ) + d( x , u ) = r + r imply the following:˜ f ( u ) ∈ G ( u ′ ) ∩ G ( u ′′ ) + d( u , u ) B X ⊂ G ( u ′ ) ∩ G ( u ′′ ) + ( r + r ) B X = V . Thus, ˜ f ( u ) ∈ V ∩ V ∩ V proving (3.22).In the same fashion we show that V ∩ V ∩ V , ∅ .Next, prove that V ∩ V ∩ V = G ( u ′ ) ∩ G ( u ′′ ) ∩ [ G ( u ′ ) ∩ G ( u ′′ ) + ( r + r ) B X ] , ∅ . (3.24)Following the scheme of the proof of (3.22), we introduce a set M = { u ′ , u ′′ , u ′ , u ′′ } .Assumption 3.3 provides the existence of a ρ -Lipschitz selection f M : M → X of therestriction F | M with k f M k Lip(( M , ρ ) , X ) ≤
1. In turn, Claim 3.4 tells us that there exists ad-Lipschitz mapping ˜ f : M → X with k ˜ f k Lip(( M , d) , X ) ≤ k f M k Lip(( M , ρ ) , X ) ≤ f | M = f M .Then, following the scheme of the proof of (3.23), we show that ˜ f ( u ) ∈ V ∩ V ∩ V proving the required property (3.24).Finally, following the same approach, we prove that V ∩ V ∩ V = [ G ( u ′ ) + ( r + r ) B X ] ∩ G ( u ′ ) ∩ G ( u ′′ ) , ∅ . (3.25)More specifically, me introduce a set M = { u ′ , u ′′ , u ′ , u ′′ } . Assumption 3.3 guar-antees the existence of a ρ -Lipschitz selection f M : M → X of the restriction F | M with k f M k Lip(( M , ρ ) , X ) ≤
1. Claim 3.4 provides the existence of a d-Lipschitz mapping˜ f : M → X with k ˜ f k Lip(( M , d) , X ) ≤ k f M k Lip(( M , ρ ) , X ) ≤ f | M = f M . Finally,following the proof of (3.23), we show that ˜ f ( u ) ∈ V ∩ V ∩ V completing the proofof (3.25).Thus, (3.21) is proven, so that W ∩ W ∩ W , ∅ . STEP 3 . First, using a similar approach, we show that W ∩ W ∩ W , ∅ .15ext, we prove W ∩ W ∩ W = G ( u ′ ) ∩ G ( u ′′ ) ∩ [ G ( u ′ ) ∩ G ( u ′′ ) + ( r + r + r ) B X ] , ∅ . (3.26)To see this, we set M = { u ′ , u ′′ , u ′ , u ′′ } . By Assumption 3.3, there exists a ρ -Lipschitzmapping f M : M → X with k f M k Lip(( M , ρ ) , X ) ≤ f M ( u ) ∈ F ( u ) for every u ∈ M . Claim 3.4 enables us to extend f M to a d-Lipschitz mapping ˜ f : M → X withd-Lipschitz seminorm k ˜ f k Lip(( M , d) , X ) ≤ k f M k Lip(( M , ρ ) , X ) ≤
1. The reader can easily seethat from this inequality and definitions (3.16), (3.17), we have ˜ f ( u ) ∈ W ∩ W ∩ W .Thus, (3.26) holds.In a similar way we prove that W ∩ W ∩ W = G ( u ′ ) ∩ G ( u ′′ ) ∩ [ G ( u ′ ) ∩ G ( u ′′ ) + ( r + r ) B X ] , ∅ . (3.27)More specifically, we set M = { u ′ , u ′′ , u ′ , u ′′ } . By Assumption 3.3, there exists amapping f M : M → X with k f M k Lip(( M , ρ ) , X ) ≤ f M ( u ) ∈ F ( u ) for every u ∈M . We again use Claim 3.4 to extend f M to a d-Lipschitz mapping ˜ f : M → X withd-Lipschitz seminorm k ˜ f k Lip(( M , d) , X ) ≤ k f M k Lip(( M , ρ ) , X ) ≤
1. One can readily see that,thanks to this inequality and definition (3.16), the point ˜ f ( u ) belongs to W ∩ W ∩ W ,and the required property (3.27) follows.The proof of the proposition is complete. (cid:4) Proposition 3.9.
For every x , y ∈ M the following inequality d H ( F [2] ( x ) , F [2] ( y )) ≤ γ ( L ) d( x , y ) (3.28) holds. Here γ ( L ) = L θ ( L ) where θ ( L ) is the constant from Theorem 2.1.Proof. Let x , y ∈ M . Formula (3.10) tells us that F [2] ( x ) = \ u , u ′ , u ′′ ∈M T x ( u , u ′ , u ′′ ) and F [2] ( y ) = \ u , u ′ , u ′′ ∈M T y ( u , u ′ , u ′′ ) . (3.29)Let τ = γ ( L ) d( x , y ). Representation (3.29), Lemma 3.5 and Proposition 3.8 implythe following formula: F [2] ( x ) + τ B X = \ { T x ( u , u ′ , u ′′ ) ∩ T x ( v , v ′ , v ′′ ) + τ B X } . (3.30)Here the first intersection in the right hand side of (3.30) is taken over all elements u , u ′ , u ′′ , v , v ′ , v ′′ ∈ M .Given u , u ′ , u ′′ , v , v ′ , v ′′ ∈ M , prove that A = T x ( u , u ′ , u ′′ ) ∩ T x ( v , v ′ , v ′′ ) + τ B X ⊃ F [2] ( y ) . (3.31)To establish this fact, we introduce the following sets: C = F ( u ′ ) + d( u ′ , u ) B X , C = F ( u ′′ ) + d( u ′′ , u ) B X , C = T x ( v , v ′ , v ′′ ) . (3.32)16et ε = L θ ( L ) d( x , y ) and r = d( x , u ) . (3.33)Then τ = γ ( L ) d( x , y ) = θ ( L ) ε , and A = T x ( u , u ′ , u ′′ ) ∩ T x ( v , v ′ , v ′′ ) + τ B X = ( C ∩ C + LrB X ) ∩ C + θ ( L ) ε B X . We want to apply Proposition 2.3 to the set A . To do this, we have to verify condition(2.4) of this proposition, i.e., to show that C ∩ C ∩ ( C + rB X ) , ∅ . (3.34)Let M ′ = { u ′ , u ′′ , v ′ , v ′′ } . Thanks to Assumption 3.3, there exists a ρ -Lipschitz se-lection f M ′ of the restriction F | M ′ with k f M ′ k Lip(( M ′ , ρ ) , X ) ≤
1. Claim 3.4 enables us toextend f M ′ to a d-Lipschitz mapping ˜ f : M → X defined on all of M with d-Lipschitzseminorm k ˜ f k Lip(( M , d) , X ) ≤ k f M ′ k Lip(( M ′ , ρ ) , X ) ≤ . In particular, ˜ f ( u ′ ) = f M ′ ( u ′ ) ∈ F ( u ′ ), ˜ f ( u ′′ ) = f M ′ ( u ′′ ) ∈ F ( u ′′ ), k ˜ f ( u ′ ) − ˜ f ( u ) k ≤ d( u ′ , u ) , k ˜ f ( u ′′ ) − ˜ f ( u ) k ≤ d( u ′′ , u ) k ˜ f ( x ) − ˜ f ( u ) k ≤ d( x , u ) = r . and k ˜ f ( x ) − ˜ f ( u ) k ≤ d( x , u ) = r . (3.35)These properties of f and definition (3.32) imply that ˜ f ( u ) ∈ C ∩ C .In a similar way we show that ˜ f ( x ) ∈ T x ( v , v ′ , v ′′ ) = C , see (3.32) and (3.9). Fromthis and (3.35), we have ˜ f ( u ) ∈ C + rB X . Thus, C ∩ C ∩ ( C + rB X ) ∋ ˜ f ( u ) proving(3.34).We see that property (2.4) of Proposition 2.3 holds. This proposition tells us that A = ( C ∩ C + LrB X ) ∩ C + θ ( L ) ε B X ⊃ [ C ∩ C + ( Lr + ε ) B X ] ∩ [( C + rB X ) ∩ C + ε B X ] ∩ [( C + rB X ) ∩ C + ε B X ] = S ∩ S ∩ S . Prove that S i ⊃ F [2] ( y ) for every i = , , . (3.36)We begin with the set S = C ∩ C + ( Lr + ε ) B X . Thus, S = { F ( u ′ ) + d( u ′ , u ) B X } ∩ { F ( u ′′ ) + d( u ′′ , u ) B X } + ( L d( u , x ) + L θ ( L ) d( x , y )) B X . See (3.32). By the triangle inequality,d( u , x ) + θ ( L ) d( x , y ) ≥ d( u , x ) + d( x , y ) ≥ d( u , y ) . S ⊃ { F ( u ′ ) + d( u ′ , u ) B X } ∩ { F ( u ′′ ) + d( u ′′ , u ) B X } + L d( u , y ) B X = T y ( u , u ′ , u ′′ ) . But T y ( u , u ′ , u ′′ ) ⊃ F [2] ( y ), see (3.29), so that S ⊃ F [2] ( y ).We turn to the proof of the inclusion S ⊃ F [2] ( y ). Note that S is defined by S = ( C + rB X ) ∩ C + ε B X . (3.37)By the triangle inequality, C + rB X = F ( u ′ ) + d( u ′ , u ) B X + d( u , x ) B X ⊃ F ( u ′ ) + d( u ′ , x ) B X . (3.38)Let e C = F ( u ′ ) + d( u ′ , x ) B X , e C = F ( v ′ ) + d( v ′ , v ) B X , e C = F ( v ′′ ) + d( v ′′ , v ) B X , (3.39)and let ˜ r = d( v , x ) . (3.40)In these settings C = T x ( v , v ′ , v ′′ ) = e C ∩ e C + L ˜ rB X . Let e A = ( e C ∩ e C + L ˜ rB X ) ∩ e C + ε B X . (3.41)Then, thanks to (3.37) and (3.38), S ⊃ { F ( u ′ ) + d( u ′ , x ) B X } ∩ C + ε B X = ( e C ∩ e C + L ˜ rB X ) ∩ e C + ε B X = e A . (3.42)Prove that e A ⊃ F [2] ( y ). We will do this by applying Proposition 2.3 to the set e A . Butfirst we must check the hypothesis of this proposition, i.e., to show that e C ∩ e C ∩ ( e C + ˜ rB X ) , ∅ . (3.43)To establish this property, we set b M = { u ′ , v ′ , v ′′ } and apply Assumption 3.3 to b M .This assumption tells us that the restriction F | b M has a ρ -Lipschitz selection f b M : b M → X with k f b M k Lip(( b M , ρ ) , X ) ≤
1. By Claim 3.4, there exists a d-Lipschitz mapping ˜ f : M → X with d-Lipschitz seminorm k ˜ f k Lip(( M , d) , X ) ≤ k f b M k Lip(( b M , ρ ) , X ) ≤ f | b M = f b M .In particular,˜ f ( u ′ ) = f b M ( u ′ ) ∈ F ( u ′ ) , ˜ f ( v ′ ) = f b M ( v ′ ) ∈ F ( v ′ ) , ˜ f ( v ′′ ) = f b M ( v ′′ ) ∈ F ( v ′′ ) . Futhermore, k ˜ f ( x ) − ˜ f ( u ′ ) k ≤ d( x , u ′ ), k ˜ f ( v ′ ) − ˜ f ( v ) k ≤ d( v ′ , v ) , k ˜ f ( v ′′ ) − ˜ f ( v ) k ≤ d( v ′′ , v ) and k ˜ f ( x ) − ˜ f ( v ) k ≤ d( x , v ) . Combining these properties of ˜ f with definitions (3.39) and (3.40), we conclude that e C ∩ e C ∩ ( e C + ˜ rB X ) ∋ ˜ f ( v ) proving (3.43).18e recall that ε = L θ ( L ) d( x , y ), see (3.33), so that e A = ( e C ∩ e C + L ˜ rB X ) ∩ e C + L θ ( L ) d( x , y ) B X , (see (3.41)).We apply Proposition 2.3 to e A and obtain the following: e A ⊃ { e C ∩ e C + ( L ˜ r + L d( x , y )) B X } ∩ { ( e C + ˜ rB X ) ∩ e C + L d( x , y ) B X } ∩ { ( e C + ˜ rB X ) ∩ e C + L d( x , y ) B X } = e S ∩ e S ∩ e S . Prove that e S i ⊃ F [2] ( y ) for every i = , ,
3. First, let us show that e S = e C ∩ e C + ( L ˜ r + L d( x , y )) B X ⊃ F [2] ( y ) . (3.44)By (3.40) and the triangle inequality, ˜ r + d( x , y ) = d( v , x ) + d( x , y ) ≥ d( v , y ) so that e S ⊃ e C ∩ e C + L d( v , y ) B X = { F ( v ′ ) + d( v ′ , v ) B X } ∩ { F ( v ′′ ) + d( v ′′ , v ) B X } + L d( v , y ) B X = T y ( v , v ′ , v ′′ ) . See (3.39) and (3.9). This inclusion and (3.29) imply (3.44).Next, let us show that e S = ( e C + ˜ rB X ) ∩ e C + L d( x , y ) B X ⊃ F [2] ( y )) . (3.45)Thanks to (3.39), (3.40) and the triangle inequality, e C + ˜ rB X = F ( v ′ ) + d( v ′ , v ) B X + d( v , x ) B X ⊃ F ( v ′ ) + d( v ′ , x ) B X so that e S ⊃ { F ( v ′ ) + d( v ′ , x ) B X } ∩ { F ( u ′ ) + d( u ′ , x ) B X } + L d( x , y ) B X = T y ( x , u ′ , v ′ ) . See (3.9). From this and (3.29), we have e S ⊃ T y ( x , u ′ , v ′ ) ⊃ F [2] ( y ), proving (3.45).In the same way we show that e S = ( e C + ˜ rB X ) ∩ e C + L d( x , y ) B X ⊃ T y ( x , u ′ , v ′′ ) ⊃ F [2] ( y ) . Combining this with (3.44) and (3.45), we obtain the required inclusion e S i ⊃ F [2] ( y )for every i = , ,
3. In turn, this proves that e A ⊃ F [2] ( y ) because e A ⊃ e S ∩ e S ∩ e S ⊃ F [2] ( y ) . We know that S ⊃ e A , see (3.42), so that S ⊃ F [2] ( y ). In the same fashion we showthat S = ( C + rB X ) ∩ C + L ε B X ⊃ F [2] ( y )19roving (3.36). Hence, A ⊃ S ∩ S ∩ S ⊃ F [2] ( y ) so that (3.31) holds.Now, from (3.30) and (3.31) it follows that F [2] ( x ) + γ ( L ) d( x , y ) B X = F [2] ( x ) + τ B X ⊃ F [2] ( y ) . By interchanging the roles of elements x and y in this inclusion we obtain the inclusion F [2] ( y ) + γ ( L ) d( x , y ) B X ⊃ F [2] ( x ). These two inclusions imply inequality (3.28) provingthe proposition. (cid:4) We are in a position to complete the proof of Theorem 1.9.Let λ , λ and γ be parameters satisfying (1.6). Thus, λ ≥ e ( M , X ) and λ ≥ λ .We set α = λ , L = λ /λ . In this case, L and α satisfy (3.1). We also recall that d = αρ = λ ρ , see (3.2). In these settings, the mappings F [1] and F [2] defined by formulae(3.3) and (3.4) are the first and the second order ( { λ , λ } , ρ )-balanced refinements of F respectively. See Definition 1.6.Proposition 3.8 tells us that, under these conditions, F [2] ( x ) , ∅ on M . In turn,Proposition 3.9 states thatd H ( F [2] ( x ) , F [2] ( y )) ≤ γ ( L ) d( x , y ) for all x , y ∈ M . Recall that γ ( L ) = L · θ ( L ) where θ = θ ( L ) = (3 L + / ( L − θ ( L ) = (3 λ + λ ) / ( λ − λ ) so thatd H ( F [2] ( x ) , F [2] ( y )) ≤ γ ( L ) d( x , y ) ≤ L · θ ( L ) d( x , y )) = λ (3 λ + λ ) ( λ − λ ) ρ ( x , y ) . Combining this inequality with the third inequality in (1.6), we obtain (1.9) provingTheorem 1.9 for the parameters λ , λ and γ satisfying (1.6).In particular, one can set λ = / λ =
4, and γ = e ( M , X ) ≤ λ = /
3, see Remark 3.2, so that λ , λ and γ satisfy (1.6).Next, let X be a two dimensional Euclidean space, and let λ , λ and γ satisfy (1.10). Inthis case, we prove (1.8) and (1.9) by replacing in the proof of Theorem 1.9 the function θ = θ ( L ) defined by (2.3) with the function θ ( L ) = + L / √ L −
1. We leave the detailsto the interested reader. See also [27, Section 3].In particular, we can set λ = /π , λ = /π , γ =
38. Indeed, in this case, e ( M , X ) ≤ /π , see Remark 3.2, which implies (1.10) for these values of parameters λ , λ and γ .Finally, suppose that X is a Euclidean space, M is a subset of a Euclidean space E ,and ρ is the Euclidean metric in E . In this case e ( M , X ) =
1, see Remark 3.2, so that onecan set λ = λ = γ =
25. Clearly, in this case the inequalities (1.10) hold.The proof of Theorem 1.9 is complete. (cid:4)
4. Balanced refinements of line segments in a Banach space
In this section we prove Theorem 1.10.Let ( M , ρ ) be a pseudometric space, and let ( X , k · k ) be a Banach space. We assumethat dim X >
1. Let us recall that K ( X ) is the family of all non-empty convex compacts20n X of dimension at most 1 (i.e., the family of all points and all bounded closed linesegments in X ).In this section we need the following version of Helly’s Theorem. Theorem 4.1.
Let K be a collection of closed convex subsets of X containing a setK ∈ K ( X ) . If K has a common point with any two members of K , then there exists apoint common to all of the collection K .Proof. We introduce a family e K = { K ∩ K : K ∈ K } , and apply to e K the onedimensional Helly’s Theorem. (See part (a) of Lemma 5.2.) (cid:4) We also need the following variant of Proposition 2.3 for the family K ( X ). Proposition 4.2.
Let X be a Banach space, and let r ≥ . Let C , C , C ⊂ X be convexclosed subsets, and let C ∈ K ( X ) . Suppose thatC ∩ C ∩ ( C + rB X ) , ∅ . (4.1) Then for every L > and every ε > the following inclusion ( C ∩ C + LrB X ) ∩ C + θ ( L ) ε B X ⊃ [ C ∩ C + ( Lr + ε ) B X ] ∩ [( C + rB X ) ∩ C + ε B X ] holds. Here θ ( L ) is the same as in Theorem 2.1, i.e., θ ( L ) = (3 L + / ( L − for anarbitrary X, and θ ( L ) = + L / √ L − whenever X is a Euclidean space.Proof. For the detailed proof of the proposition we refer the reader to [27]. This proofis a slight modification of the proof of Proposition 2.3, in which we use Theorem 4.1(i.e., the one dimensional version of the Helly Theorem) rather than Theorem 2.2 (i.e.,the two dimensional Helly’s Theorem). (cid:4)
We recall that dim X > N (1 , X ) = min { , dim X } = F : M → K ( X ) be a set-valued mapping. We suppose that F satisfies the hypothesisof Theorem 1.10, i.e., that the following assumption is true. Assumption 4.3.
For every subset M ′ ⊂ M with M ′ ≤ the restriction F | M ′ of F to M ′ has a Lipschitz selection f M ′ : M ′ → X with k f k Lip( M ′ , X ) ≤ . Let ~λ = { λ , λ } , and let F [1] and F [2] be the first and the second order ( ~λ, ρ )-balancedrefinements of F . See Definition 1.6. Our aim is to show that if λ ≥ , λ ≥ λ , γ ≥ λ (3 λ + λ ) / ( λ − λ ) , (4.2)then the set-valued mapping F [2] is a γ -core of F (with respect to ρ ), i.e., F [2] ( x ) , ∅ for every x ∈ M , and d H ( F [2] ( x ) , F [2] ( y )) ≤ γρ ( x , y ) for all x , y ∈ M . To prove this, we set L = λ /λ and introduce a new pseudometric on M defined byd( x , y ) = λ ρ ( x , y ) , x , y ∈ M . L ≥ ρ ≤ d on M . (4.3)We also note that in these settings, for every x ∈ M , F [1] ( x ) = \ z ∈M [ F ( z ) + d( x , z ) B X ] and F [2] ( x ) = \ z ∈M h F [1] ( z ) + L d( x , z ) B X i . (4.4)Next, we need the following analog of Lemma 3.5. Lemma 4.4.
Let K be a collection of convex closed subsets of X containing a set K ∈K ( X ) . Suppose that ∩ { K : K ∈ K } , ∅ . Then for every r ≥ , we have \ K ∈ K K + rB X = \ K ∈ K n h K ∩ K i + rB X o . Proof.
Let e K = { K ∩ K : K ∈ K } . Clearly, e K ⊂ K ( X ). It is also clear that thestatement of the lemma is equivalent to the equality \ e K ∈ e K e K + rB X = \ e K ∈ e K n e K + rB X o provided ∩ { e K : e K ∈ K } , ∅ . We prove this equality by a slight modification of theproof of Lemma 3.5. More specifically, we obtain the result by using in that proof Helly’sTheorem 4.1 instead of Theorem 2.2. We leave the details to the interested reader. (cid:4) Lemma 4.5.
For every x ∈ M the set F [1] ( x ) belongs to the family K ( X ) . Moreover, forevery x , z ∈ M , we haveF [1] ( z ) + L d( x , z ) B X = \ v ∈M n [ F ( v ) + d( z , v ) B X ] ∩ F ( z ) + L d( x , z ) B X o . (4.5) Proof.
Let K = { F ( z ) + d( z , x ) B X : z ∈ M} . Clearly, K is a family of bounded closedconvex subsets of X containing the set F ( x ) ∈ K ( X ). Theorem 4.1 tells us that the set F [1] ( x ) = ∩ { K : K ∈ K } is non-empty whenever for every z ′ , z ′′ ∈ M the set E ( x , z ′ , z ′′ ) = F ( x ) ∩ [ F ( z ′ ) + d( z ′ , x ) B X ] ∩ [ F ( z ′′ ) + d( z ′′ , x ) B X ] , ∅ . (4.6)Fix z ′ , z ′′ ∈ M and set M ′ = { x , z ′ , z ′′ } . By Assumption 4.3, there exists a function f M ′ : M ′ → X satisfying the following conditions: f M ′ ( x ) ∈ F ( x ), f M ′ ( z ′ ) ∈ F ( z ′ ), f M ′ ( z ′′ ) ∈ F ( z ′′ ), k f M ′ ( z ′ ) − f M ′ ( x ) k ≤ ρ ( z ′ , x ) ≤ d( z ′ , x ) , and k f M ′ ( z ′′ ) − f M ′ ( x ) k ≤ ρ ( z ′′ , x ) ≤ d( z ′′ , x ) . See (4.3). Then f M ′ ( x ) ∈ E ( x , z ′ , z ′′ ) so that (4.6) holds. Hence, F [1] ( x ) , ∅ proving that F [1] ( x ) ∈ K ( X ).It remains to note that equality (4.5) is immediate from (4.4) and Lemma 4.4.The proof of the lemma is complete. (cid:4) emma 4.6. For every x ∈ M , we have equalityF [2] ( x ) = \ u , u ′ ∈M n [ F ( u ′ ) + d( u ′ , u ) B X ] ∩ F ( u ) + L d( u , x ) B X o holds.Proof. The statement of the lemma is immediate from (4.4) and Lemma 4.5. (cid:4)
Given x , u , u ′ ∈ M we put e T x ( u , u ′ ) = [ F ( u ′ ) + d( u ′ , u ) B X ] ∩ F ( u ) + L d( u , x ) B X . (4.7)Now, Lemma 4.6 reformulates as follows: F [2] ( x ) = \ u , u ′ ∈M e T x ( u , u ′ ) . (4.8) Proposition 4.7.
For every x ∈ M the set F [2] ( x ) is non-empty.Proof. Recall that F ( x ) ∈ K ( X ). Furthermore, by (4.7), F ( x ) = e T x ( x , x ). Therefore,by (4.8) and Helly’s Theorem 4.1, the set F [2] ( x ) , ∅ whenever for every u i , u ′ i ∈ M , i = ,
2, we have F ( x ) ∩ e T x ( u , u ′ ) ∩ e T x ( u , u ′ ) , ∅ . (4.9)We recall that e T x ( u i , u ′ i ) = [ F ( u ′ i ) + d( u ′ i , u i ) B X ] ∩ F ( u i ) + L d( u i , x ) B X , i = , . See (4.7). (4.10)Fix elements u , u ′ , u , u ′ ∈ M and prove that property (4.9) holds.First we note that, without loss of generality, one may assume that ρ ( u , x ) ≥ ρ ( u , x ).Next, we introduce the following sets: G = F ( u ) , G = F ( u ′ ) + ρ ( u , u ′ ) B X , G = F ( x ) + ρ ( u , x ) B X , (4.11)and G = [ F ( u ′ ) + ρ ( u ′ , u ) B X ] ∩ F ( u ) + ρ ( u , u , ) B X . (4.12)Prove that \ i = G i , ∅ . (4.13)We know that G = F ( u ) ∈ K ( X ) so that, by Theorem 4.1, (4.13) holds provided G ∩ G i ∩ G j , ∅ for every 2 ≤ i , j ≤ , i , j . (4.14)First prove that G ∩ G ∩ G , ∅ . Let M = { u ′ , u , x } . By Assumption 4.3, thereexists a mapping f : M → X with the following properties: f ( x ) ∈ F ( x ), f ( u ) ∈ F ( u ), f ( u ′ ) ∈ F ( u ′ ), k f ( u ) − f ( x ) k ≤ ρ ( u , x ) and k f ( u ) − f ( u ′ ) k ≤ ρ ( u , u ′ ) . f and definition (4.11) tell us that f ( u ) ∈ G ∩ G ∩ G provingthat the sets G , G and G have a common point.Prove that G ∩ G ∩ G , ∅ . Let M = { u ′ , u , u ′ , u } . Using Assumption 4.3, weproduce a mapping f : M → X such that f ( u i ) ∈ F ( u i ), f ( u ′ i ) ∈ F ( u ′ i ) for every i = , k f ( u ) − f ( u ′ ) k ≤ ρ ( u , u ′ ), k f ( u ) − f ( u ) k ≤ ρ ( u , u ) and k f ( u ) − f ( u ′ ) k ≤ ρ ( u , u ′ ) . These properties of f and (4.11), (4.12) yield f ( u ) ∈ G ∩ G ∩ G proving that thesets G , G and G have a common point.In the same way we show that G ∩ G ∩ G , ∅ . (We set M = { u ′ , u , x , u } ,produce a corresponding function f : M → X and show that f ( u ) ∈ G ∩ G ∩ G .)Thus, (4.14) holds, and the proof of property (4.13) is complete.Let f M = { u ′ , u , x , u , u ′ } . Property (4.13) and definitions (4.11), (4.12) imply theexistence of a mapping g : f M → X with the following properties: g ( v ) ∈ F ( v ) on f M , k g ( u ) − g ( u ′ ) k ≤ ρ ( u , u ′ ) , k g ( u ) − g ( u ) k ≤ ρ ( u , u ) , (4.15)and k g ( u ) − g ( u ′ ) k ≤ ρ ( u , u ′ ) , k g ( u ) − g ( x ) k ≤ ρ ( u , x ) . (4.16)We establish (4.9) by showing that g ( x ) ∈ F ( x ) ∩ e T x ( u , u ′ ) ∩ e T x ( u , u ′ ) . (4.17)Indeed, from the above properties of g it follows that g ( x ) ∈ F ( x ). We also know that g ( u ) ∈ F ( u ), g ( u ′ ) ∈ F ( u ′ ). Thanks to (4.15), (4.16) and (4.3), k g ( u ) − g ( u ′ ) k ≤ ρ ( u , u ′ ) ≤ d( u , u ′ ) and k g ( u ) − g ( x ) k ≤ ρ ( u , x ) ≤ L d( u , x ) . From these properties of g and definition (4.10), we have g ( x ) ∈ [ F ( u ′ ) + d( u ′ , u ) B X ] ∩ F ( u ) + L d( u , x ) B X = e T x ( u , u ′ ) . It remains to show that g ( x ) ∈ e T x ( u , u ′ ). In fact, as we know, g ( x ) ∈ F ( x ) , g ( u ) ∈ F ( u ) , and g ( u ′ ) ∈ F ( u ′ ) . (4.18)Furthermore, thanks to (4.15) and (4.3), k g ( u ) − g ( u ′ ) k ≤ ρ ( u , u ′ ) ≤ d( u , u ′ ) . (4.19)Let us estimate k g ( u ) − g ( x ) k . From (4.15), (4.16) and the triangle inequality, we have k g ( u ) − g ( x ) k ≤ k g ( u ) − g ( u ) k + k g ( u ) − g ( x ) k ≤ ρ ( u , u ) + ρ ( u , x ) ≤ ρ ( u , x ) + ρ ( x , u ) + ρ ( u , x ) = ρ ( u , x ) + ρ ( x , u ) . ρ ( u , x ) ≥ ρ ( u , x ). This and (4.3) yield k g ( u ) − g ( x ) k ≤ ρ ( u , x ) ≤ L d( u , x ) . This inequality, (4.18), (4.19) and (4.10) imply the required property g ( x ) ∈ e T x ( u , u ′ )proving (4.17).The proof of the proposition is complete. (cid:4) As in Section , we again set γ = γ ( L ) = L θ ( L ) . Proposition 4.8.
For every x , y ∈ M the following inequality d H ( F [2] ( x ) , F [2] ( y )) ≤ γ ( L ) d( x , y ) holds.Proof. Let x , y ∈ M . By (4.8), F [2] ( x ) = \ u , u ′ ∈M e T x ( u , u ′ ) and F [2] ( y ) = \ u , u ′ ∈M e T y ( u , u ′ ) . (4.20)Recall that e T x ( u , u ′ ) = [ F ( u ′ ) + d( u ′ , u ) B X ] ∩ F ( u ) + L d( u , x ) B X . (4.21)We know that F [2] ( x ) , ∅ , see Proposition 4.7, and e T x ( x , x ) = F ( x ) ∈ K ( X ). Theseproperties, (4.20) and Lemma 4.4 tell us that F [2] ( x ) + γ ( L ) d( x , y ) B X = \ u , u ′ ∈M n e T x ( u , u ′ ) ∩ F ( x ) + γ ( L ) d( x , y ) B X o . (4.22)We fix u , u ′ ∈ M and introduce a set e A = e T x ( u , u ′ ) ∩ F ( x ) + γ ( L ) d( x , y ) B X . We alsointroduce sets C = F ( u ) , C = F ( u ′ ) + d( u ′ , u ) B X , and C = F ( x ) . (4.23)Let ε = L d( x , y ) and r = d( x , u ) . (4.24)In these settings, γ ( L ) d( x , y ) = θ ( L ) ε and e A = e T x ( u , u ′ ) ∩ F ( x ) + γ ( L ) d( x , y ) B X = ( C ∩ C + LrB X ) ∩ C + θ ( L ) ε B X . Let us apply Proposition 4.2 to the set e A . To do this, we have to verify condition (4.1),i.e., to show that C ∩ C ∩ ( C + rB X ) , ∅ . (4.25)25et f M = { x , u , u ′ } . Thanks to Assumption 4.3, there exists a ρ -Lipschitz selection f f M of the restriction F | f M with k f f M k Lip(( f M ; ρ ) , X ) ≤
1. Thus, f f M ( u ′ ) ∈ F ( u ′ ), f f M ( u ) ∈ F ( u ), f f M ( x ) ∈ F ( x ), k f f M ( u ′ ) − f f M ( u ) k ≤ ρ ( u ′ , u ) and k f f M ( x ) − f f M ( u ) k ≤ ρ ( x , u ) . Let us prove that f f M ( u ) ∈ C ∩ C ∩ ( C + rB X ) . (4.26)Indeed, f f M ( u ) ∈ F ( u ) = C , see (4.23). Furthermore, f f M ( u ′ ) ∈ F ( u ′ ) and, thanks to(4.3), ρ ≤ d on M . Hence, k f f M ( u ′ ) − f f M ( u ) k ≤ ρ ( u ′ , u ) ≤ d( u ′ , u )proving that f f M ( u ) ∈ C , see (4.23). Finally, by (4.23) and (4.24), f f M ( x ) ∈ F ( x ) = C and k f f M ( x ) − f f M ( u ) k ≤ ρ ( x , u ) ≤ d( x , u ) = r , so that f f M ( u ) ∈ C + rB X . Thus, (4.26) is true so that property (4.25) holds. Furthermore, C = F ( u ) ∈ K ( X ), sothat all conditions of the hypothesis of Proposition 4.2 are satisfied. By this proposition, e A = ( C ∩ C + LrB X ) ∩ C + θ ( L ) ε B X ⊃ [ C ∩ C + ( Lr + ε ) B X ] ∩ [( C + rB X ) ∩ C + ε B X ] = e S ∩ e S . Prove that e S i ⊃ F [2] ( y ) for every i = , e S = C ∩ C + ( Lr + ε ) B X . Thanks to (4.23) and (4.24), e S = { F ( u ′ ) + d( u ′ , u ) B X } ∩ F ( u ) + ( L d( u , x ) + L d( x , y )) B X . By the triangle inequality, ρ ( u , x ) + ρ ( x , y ) ≥ ρ ( u , y ) so that e S ⊃ [ F ( u ′ ) + d( u ′ , u ) B X ] ∩ F ( u ) + L d( u , y ) B X = e T y ( u , u ′ ) , see (4.21) . But, by (4.20), e T y ( u , u ′ ) ⊃ F [2] ( y ) which implies the required inclusion e S ⊃ F [2] ( y ).We turn to the set e S = ( C + rB X ) ∩ C + ε B X . By (4.7), (4.23) and (4.24), e S = [ F ( u ) + d( u , x ) B X ] ∩ F ( x ) + L d( x , y ) B X = T y ( u , x ) . Thanks to (4.20), e T y ( u , x ) ⊃ F [2] ( y ) proving that e S ⊃ F [2] ( y ).Thus, e A = e T x ( u , u ′ ) ∩ F ( x ) + γ ( L ) d( x , y ) B X ⊃ e S ∩ e S ⊃ F [2] ( y ) for every u , u ′ ∈ M . From this and representation (4.22), we have F [2] ( x ) + γ ( L ) d( x , y ) B X ⊃ F [2] ( y ) .
26y interchanging the roles of x and y we obtain also F [2] ( y ) + γ ( L ) d( x , y ) B X ⊃ F [2] ( x ) . These two inclusions and (1.3) imply the statement of the proposition. (cid:4)
We complete the proof of Theorem 1.10 as follows. We fix λ , λ and γ satisfyinginequalities (1.11). Then, by Proposition 4.7, the set F [2] ( x ) , ∅ for every x ∈ M .In turn, Proposition 4.8 tells us that d H ( F [2] ( x ) , F [2] ( y )) ≤ γ ( L ) d( x , y ) on M . Werecall that L = λ /λ , d = λ ρ , γ ( L ) = L θ ( L ) and θ ( L ) = (3 L + / ( L − H ( F [2] ( x ) , F [2] ( y )) ≤ γ ( L ) d( x , y ) = L L + L − ! d( x , y ) = λ (3 λ + λ )( λ − λ ) ρ ( x , y ) . This inequality and (1.11) imply (1.9).Thus, (1.8) and (1.9) hold provided λ , λ and γ satisfy inequalities (1.11). In partic-ular, we can set λ = λ = γ = λ (3 λ + λ ) / ( λ − λ ) = X be a Euclidean space, and let λ , λ and γ be the parameters satisfyinginequalities (1.12). In this case, we replace in the above calculations the constant θ ( L ) = (3 L + / ( L −
1) with θ ( L ) = + L / √ L −
1. This leads us to the required estimated H ( F [2] ( x ) , F [2] ( y )) ≤ (cid:26) λ + λ / (cid:16) λ − λ (cid:17) (cid:27) ρ ( x , y ) ≤ γ ρ ( x , y )proving that (1.8) and (1.9) hold for λ , λ and γ satisfying (1.12).The proof of Theorem 1.10 is complete. (cid:4)
5. The main theorem in ℓ ∞ . X = RProposition 5.1.
Let ( M , ρ ) be a pseudometric space. Let m = and let X = R ; thus, ℓ = ℓ ( m , X ) = , see (1.1). In this case Conjecture 1.8 holds for every λ ≥ and γ ≥ .Thus, the following statement is true: Let F be a set-valued mapping from M into thefamily K ( R ) of all closed bounded intervals in R . Suppose that for every x , y ∈ M thereexist points g ( x ) ∈ F ( x ) and g ( y ) ∈ F ( y ) such that | g ( x ) − g ( y ) | ≤ ρ ( x , y ) .Let F [1] ( x ) , x ∈ M , be the λ -balanced refinement of the mapping F, i.e., the setF [1] ( x ) = \ z ∈M (cid:2) F ( z ) + λ ρ ( x , z ) I (cid:3) where I = [ − , . Then F [1] ( x ) , ∅ for every x ∈ M , and d H ( F [1] ( x ) , F [1] ( y )) ≤ γ ρ ( x , y ) for all x , y ∈ M . For a detailed proof of the proposition we refer the reader to [27, Section 5].Here we only note that Proposition 5.1 easily follows from the one dimensional HellyTheorem and a formula for a neighborhood of the intersection of intervals in R . Weformulate these statements in the following27 emma 5.2. Let
K ⊂ K ( R ) be a collection of closed bounded intervals in R .(a) (Helly’s Theorem in R .) If the intersection of every two intervals from K is non-empty, then there exists a point in R common to all of the family K .(b) Suppose that ∩ { K : K ∈ K } , ∅ . Then for every r ≥ we have \ K ∈ K K + rI = \ K ∈ K { K + rI } . Proof of part (b).
In Lemma 3.5 we proved an analog of property (b) for R . Weprove part (b) by replacing in that proof the Helly Theorem 2.2 with the one dimensionalHelly Theorem formulated in part (a) of the present lemma. We leave the details to theinterested reader. (cid:4) Let us fix some additional notation. We let R ( R ) denote the family of all boundedclosed rectangles in R with sides parallel to the coordinate axes . We refer to every Π ∈ R ( R ) as a “rectangle” . By Ox and Ox we denote the coordinate axes in R . Definition 5.3.
Let S be a non-empty bounded convex closed subset in R . We let H [ S ]denote the smallest (with respect to inclusion) rectangle containing S . Thus, H [ S ] = ∩ { Π : Π ∈ R ( R ) , Π ⊃ S } . We refer to H [ S ] as the “rectangular hull“ of the set S .Note the following useful representation of the rectangular hull which easily followsfrom Definition 5.3: H [ S ] = ( S + Ox ) ∩ ( S + Ox ) . (5.1)In the next section we need the following auxiliary Lemma 5.4.
Let K , K ⊂ K ( R ) be two convex compacts in R with non-empty inter-section, and let Π be a rectangular with center . ThenK ∩ K + Π = ( K + Π ) ∩ ( K + Π ) ∩ H [( K ∩ K ) + Π ] . (5.2) Proof.
Obviously, the right hand side of (5.2) contains its left hand side.Let us prove the converse statement. Fix a point x ∈ ( K + Π ) ∩ ( K + Π ) ∩ H [ K ∩ K + Π ] . (5.3)Our aim is to prove that x ∈ K ∩ K + Π . Clearly, this property holds if and only if( x + Π ) ∩ K ∩ K , ∅ . It is also clear that x + Π = Π ( x ) ∩ Π ( x ) where Π i ( x ) = x + Ox i + Π , i = , . x ∈ K ∩ K + Π provided K ∩ K ∩ Π ( x ) ∩ Π ( x ) , ∅ . By Theorem 2.2, the family of sets { K , K , Π ( x ) , Π ( x ) } has a common point when-ever any three members of this family have a non-empty intersection.Prove that it is true for x satisfying (5.3). Indeed, x ∈ K i + Π and Π = − Π , so that K i ∩ Π ( x ) ∩ Π ( x ) = K i ∩ ( x + Π ) , ∅ , i = , . Next, thanks to (5.1) and (5.3), for every i = , x ∈ H [( K ∩ K ) + Π ] ⊂ ( K ∩ K ) + Π + Ox i . Hence, K ∩ K ∩ Π i ( x ) , ∅ , i = ,
2, and the proof of the lemma is complete. (cid:4) ℓ ∞ In this section we refine the result of Theorem 1.9 for the space X = ℓ ∞ . Theorem 5.5.
In the settings of Theorem 1.9 properties (1.8) and (1.9) hold providedX = ℓ ∞ , λ ≥ , λ ≥ λ , and γ ≥ λ (3 λ + λ ) / ( λ − λ ) . (5.4) In particular, (1.8) and (1.9) hold whenever λ = , λ = and γ = .Proof. We mainly follow the scheme of the proof of Theorem 1.9 given in Section .Let F : M → K ( R ) be a set-valued mapping satisfying the hypothesis of Theorem5.5. This enables us to make the following Assumption 5.6.
For every M ′ ⊂ M , M ′ ≤ , the restriction F | M ′ of F to M ′ has a ρ -Lipschitz selection f M ′ : M ′ → ℓ ∞ with ρ -Lipschitz seminorm k f M ′ k Lip(( M ′ , ρ ) ,ℓ ∞ ) ≤ . Let Q = B X , be the unit ball of the Banach space X = ℓ ∞ , i.e., Q = [ − , . Given r ≥
0, let rQ = [ − r , r ] .We fix a constant L ≥ α ≥
1, and introduce a pseudometric d( x , y ) = αρ ( x , y ), x , y ∈ M . We let F [1] and F [2] denote the first and the second order ( { , L } , d)-balanced refinements of F respectively. See Definition 1.6. Thus, F [1] ( x ) = \ z ∈M [ F ( z ) + d( x , z ) Q ] and F [2] ( x ) = \ z ∈M h F [1] ( z ) + L d( x , z ) Q i , x ∈ M . We also recall that e ( M , ℓ ∞ ) =
1. In this case, Lemma 3.6 and Proposition 3.8 tell usthat F [1] ( x ) , ∅ and F [2] ( x ) , ∅ for every x ∈ M . Thus, our aim is to show that for every α ≥ L ≥
3, and every x , y ∈ M the following inequalityd H ( F [2] ( x ) , F [2] ( y )) ≤ ˜ γ ( L ) d( x , y ) (5.5)holds. Here ˜ γ ( L ) = L θ ( L ) where θ ( L ) = (3 L + / ( L − F [2] ( x ) = \ u , u ′ , u ′′ ∈M T x ( u , u ′ , u ′′ ) (5.6)where, given u , u ′ , u ′′ ∈ M , T x ( u , u ′ , u ′′ ) = [ F ( u ′ ) + d( u ′ , u ) Q ] ∩ [ F ( u ′′ ) + d( u ′′ , u ) Q ] + L d( u , x ) Q . (5.7)The next lemma is a refinement of the formula (3.30) for the special case of X = ℓ ∞ . Lemma 5.7.
For every x ∈ M and every rectangle Π ∈ R ( R ) with center , we haveF [2] ( x ) + Π = \ v , u , u ′ , u ′′ ∈M { ( T x ( u , u ′ , u ′′ ) ∩ ( F ( v ) + d( x , v ) Q )) + Π } . (5.8) Proof.
Representation (3.10) and Lemma 3.5 tell us that F [2] ( x ) + Π = \ { T x ( u , u ′ , u ′′ ) ∩ T x ( u , u ′ , u ′′ ) + Π } (5.9)where the intersection is taken over all u i , u ′ i , u ′′ i ∈ M , i = ,
2. Note also that, by (5.7), F ( v ) + d( x , v ) Q = T x ( x , v , v ). From this and (5.9) it follows that the right hand side of(5.8) contains its left hand side.Prove the converse statement. Fix a ∈ \ v , u , u ′ , u ′′ ∈M { ( T x ( u , u ′ , u ′′ ) ∩ ( F ( v ) + d( x , v ) Q )) + Π } (5.10)and show that a ∈ F [2] ( x ) + Π . In view of formula (5.9), it su ffi ce to prove that for every u , u ′ , u ′′ , u , u ′ , u ′′ ∈ M the point a belongs to the set A defined by A = T x ( u , u ′ , u ′′ ) ∩ T x ( u , u ′ , u ′′ ) + Π . (5.11)To see this, given i ∈ { , } we introduce the following sets: Q i = L d( u i , x ) Q , K ′ i = F ( u ′ i ) + d( u , u ′ i ) Q and K ′′ i = F ( u ′′ i ) + d( u , u ′′ i ) Q . (5.12)In these settings, T x ( u i , u ′ i , u ′′ i ) = K ′ i ∩ K ′′ i + Q i , i = ,
2. See (5.7). Therefore, thanksto Lemma 5.4, T x ( u i , u ′ i , u ′′ i ) = ( K ′ i + Q i ) ∩ ( K ′′ i + Q i ) ∩ H [ T x ( u i , u ′ i , u ′′ i )] , i = , . (5.13)Now, let us introduce the following families of sets: K + = { K ′ i + Q i , K ′′ i + Q i : i = , } , K ++ = {H [ T x ( u i , u ′ i , u ′′ i )] : i = , } , K = K + ∪ K ++ . Then, thanks to (5.11) and (5.13), A = ∩ { K : K ∈ K } + Π . From this formula for A andLemma 3.5, we have A = ∩ { K ∩ K ′ + Π : K , K ′ ∈ K } .30hus, our aim is to show that a ∈ K ∩ K ′ + Π for every K , K ′ ∈ K . Let us note that,by (5.12), for every i = , K ′ i + Q i = F ( u ′ i ) + d( u i , u ′ i ) Q + L d( u i , x ) Q ⊃ F ( u ′ i ) + (d( u i , u ′ i ) + d( u i , x )) Q so that, thanks to the triangle inequality and (5.7), K ′ i + Q i ⊃ F ( u ′ i ) + d( u ′ i , x ) Q = T x ( x , u ′ i , u ′ i ) . (5.14)In the same way we prove that K ′′ i + Q i ⊃ F ( u ′′ i ) + d( u ′′ i , x ) Q = T x ( x , u ′′ i , u ′′ i ) , i = , . (5.15)Furthermore, we know that H [ T x ( u i , u ′ i , u ′′ i )] ⊃ T x ( u i , u ′ i , u ′′ i ) , i = , . (5.16)On the other hand, property (5.10) tells us that a ∈ T x ( u , u ′ , u ′′ ) ∩ ( F ( v ) + d( x , v ) Q )) + Π for every u , u ′ , u ′′ , v ∈ M . (5.17)Combining this property with (5.14), (5.15) and (5.16), we conclude that a ∈ K ∩ K ′ + Π whenever either K ∈ K + , K ′ ∈ K ++ or K , K ′ ∈ K + . It remains to prove that a ∈ H ∩ H + Π where H i = H [ T x ( u i , u ′ i , u ′′ i )] , i = , . (5.18)It is immediate from Lemma 5.2, part (b), that H ∩ H + Π = ( H + Π ) ∩ ( H + Π ).Hence, H ∩ H + Π = {H [ T x ( u , u ′ , u ′′ )] + Π } ∩ {H [ T x ( u , u ′ , u ′′ )] + Π } so that, by (5.16), H ∩ H + Π ⊃ { T x ( u , u ′ , u ′′ ) + Π } ∩ { T x ( u , u ′ , u ′′ ) + Π } . (5.19)But, thanks to (5.17), a ∈ T x ( u i , u ′ i , u ′′ i ) + Π , i = ,
2. Combining this property with(5.19), we obtain the required property (5.18) completing the proof of the lemma. (cid:4)
We are in a position to prove inequality (5.5). Our proof will follow the scheme of theproof of Proposition 3.9.Let x , y ∈ M , and let τ = ˜ γ ( L ) d( x , y ). (Recall that ˜ γ ( L ) = L θ ( L ) and d = αρ .)Lemma 5.7 tells us that F [2] ( x ) + τ Q = \ v , u , u ′ , u ′′ ∈M { ( T x ( u , u ′ , u ′′ ) ∩ ( F ( v ) + d( x , v ) Q )) + τ Q } . (5.20)31et us fix elements u , u ′ , u ′′ , v ∈ M and set A = ( T x ( u , u ′ , u ′′ ) ∩ ( F ( v ) + d( x , v ) Q )) + τ Q . (5.21)Prove that A ⊃ F [2] ( y ). Let C = F ( u ′ ) + d( u ′ , u ) Q , C = F ( u ′′ ) + d( u ′′ , u ) Q X , C = F ( v ) + d( x , v ) Q , (5.22)and let ε = L d( x , y ) and r = d( u , x ). Then τ = ˜ γ ( L ) d( x , y ) = L θ ( L ) d( x , y ) = θ ( L ) ε, and A = { [( F ( u ′ ) + d( u ′ , u ) Q ) ∩ ( F ( u ′′ ) + d( u ′′ , u ) Q )] ∩ ( F ( v ) + d( x , v ) Q ) + L d( u , x ) Q = ( C ∩ C + LrQ ) ∩ C + θ ( L ) ε Q . Let us apply Proposition 2.3 to the sets C , C and C defined by (5.22). To do this, wehave to verify condition (2.4) of that proposition, i.e., to prove that C ∩ C ∩ ( C + rQ ) , ∅ . (5.23)Let M ′ = { u , u ′ , v } . Then, by Assumption 5.6, there exists a ρ -Lipschitz selection f M ′ : M ′ → ℓ ∞ of the restriction F | M ′ with k f M ′ k Lip(( M ′ , ρ ) ,ℓ ∞ ) ≤ e ( M , ℓ ∞ ) = = αρ ≥ ρ , the mapping f M ′ : M ′ → ℓ ∞ can be extendedto a d-Lipschitz mapping ˜ f : M → ℓ ∞ defined on all of M with d-Lipschitz seminorm k ˜ f k Lip(( M , d) ,ℓ ∞ ) ≤ k f M ′ k Lip(( M ′ , ρ ) ,ℓ ∞ ) ≤ . Thus, ˜ f ( u ′ ) = f M ′ ( u ′ ) ∈ F ( u ′ ), ˜ f ( u ′′ ) = f M ′ ( u ′′ ) ∈ F ( u ′′ ), ˜ f ( v ) = f M ′ ( v ) ∈ F ( v ), k ˜ f ( u ′ ) − ˜ f ( u ) k ≤ d( u ′ , u ) , k ˜ f ( u ′′ ) − ˜ f ( u ) k ≤ d( u ′′ , u )and k ˜ f ( x ) − ˜ f ( u ) k ≤ d( u , x ) = r , k ˜ f ( x ) − ˜ f ( v ) k ≤ d( v , x ) . Hence, ˜ f ( u ) ∈ C ∩ C and ˜ f ( x ) ∈ C , so that C ∩ C ∩ ( C + rQ ) ∋ ˜ f ( u ) proving (5.23).This enables us to apply Proposition 2.3 to the sets C , C and C . By this proposition, A = ( C ∩ C + LrQ ) ∩ C + θ ( L ) ε Q ⊃ [ C ∩ C + ( Lr + ε ) Q ] ∩ [( C + rQ ) ∩ C + ε Q ] ∩ [( C + rQ ) ∩ C + ε Q ] = S ∩ S ∩ S . Prove that S i ⊃ F [2] ( y ) for every i = , ,
3. We begin with the set S = C ∩ C + ( Lr + ε ) Q = { F ( u ′ ) + d( u ′ , u ) Q } ∩ { F ( u ′′ ) + d( u ′′ , u ) Q } + ( L d( u , x ) + L d( x , y )) Q . u , x ) + d( x , y ) ≥ d( u , y ) so that S ⊃ { F ( u ′ ) + d( u ′ , u ) B X } ∩ { F ( u ′′ ) + d( u ′′ , u ) Q } + L d( u , y ) B X = T y ( u , u ′ , u ′′ ) . But, by (5.6), T y ( u , u ′ , u ′′ ) ⊃ F [2] ( y ) proving the required inclusion S ⊃ F [2] ( y ).Prove that S ⊃ F [2] ( y ). We have S = ( C + rQ ) ∩ C + ε Q = { ( F ( u ′ ) + d( u ′ , u ) Q ) + d( x , u ) Q } ∩ { F ( v ) + d( x , v ) Q } + L d( x , y ) Q . Therefore, by (5.6) and the triangle inequality, S ⊃ { ( F ( u ′ ) + d( u ′ , x ) Q } ∩ { F ( v ) + d( x , v ) Q } + L d( x , y ) Q = T y ( x , u ′ , v ) ⊃ F [2] ( y ) . In the same way we show that S ⊃ F [2] ( y ). Hence, A ⊃ S ∩ S ∩ S ⊃ F [2] ( y ).Combining this with (5.20) and (5.21), we conclude that F [2] ( x ) + τ Q ⊃ F [2] ( y ). Byinterchanging the roles of x and y we obtain also F [2] ( y ) + τ Q ⊃ F [2] ( x ). These twoinclusions imply inequalityd H ( F [2] ( x ) , F [2] ( y )) ≤ τ = ˜ γ ( L ) d( x , y ) = α ˜ γ ( L ) ρ ( x , y )proving (1.9) with γ = α L (3 L + / ( L − α = λ and L = λ /λ . This together with (5.4) provides inequalities L ≥ α ≥ γ ≥ λ (3 λ + λ ) / ( λ − λ ) completing the proof ofTheorem 5.5. (cid:4) References [1] Z. Artstein, Extension of Lipschitz selections and an application to di ff erentialinclusions, Nonlinear Anal. 16 (1991) 701–704.[2] J.-P. Aubin, H. Frankowska, Set-valued analysis, Systems & Control: Founda-tions & Applications, 2. Birkhauser Boston, 1990.[3] G. Basso, Computation of maximal projection constants, J. Funct. Anal. 277(2019) 3560–3585.[4] Y. Benyamini, J. Lindenstrauss, Geometric nonlinear functional analysis, Vol.1, in: American Mathematical Society Colloquium Publications, 48. AmericanMathematical Society, Providence, RI, 2000. xii +
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