Non-expansive bijections, uniformities and polyhedral faces
NNON-EXPANSIVE BIJECTIONS, UNIFORMITIES ANDPOLYHEDRAL FACES
CARLOS ANGOSTO, VLADIMIR KADETS, AND OLESIA ZAVARZINA
Dedicated to the memory of Bernardo Cascales
Abstract.
We extend the result of B. Cascales at al. aboutexpand-contract plasticity of the unit ball of strictly convex Banachspace to those spaces whose unit ball is the union of all its finite-dimensional polyhedral extreme subsets. We also extend the def-inition of expand-contract plasticity to uniform spaces and gener-alize the theorem on expand-contract plasticity of totally boundedmetric spaces to this new setting. Introduction
Let E , E be metric spaces. A map F : E → E is called non-expansive (resp. non-contractive) if d ( F ( x ) , F ( y )) (cid:54) d ( x, y ) (resp. d ( F ( x ) , F ( y )) (cid:62) d ( x, y )) for all x, y ∈ E . A metric space E is called expand-contract plastic (or simply, an EC-space) if every non-expansivebijection from E onto itself is an isometry.[5, Satz IV] or [10, Theorem 1.1] imply that every totally boundedmetric space is an EC-space, but there are examples of EC-spaces thatare not totally bounded (and even unbounded).In general bounded closed subsets of infinite-dimensional Banachspaces are not EC-spaces, see [2, Example 2.7]. It is not known whetherit is true that for every Banach space X its unit ball B X is an EC-space. There is no known counterexample and there are some knownpartial positive results: finite-dimesional spaces (the unit ball is com-pact), strictly convex Banach spaces (see [2, Theorem 2.6]) or (cid:96) -sum ofstrictly convex Banach spaces (see [7, Theorem 3.1]). A more generalproblem is studied in [12]: Date : July 26, 2018.2010
Mathematics Subject Classification.
Key words and phrases. non-expansive map; unit ball; expand-contract plasticspace.The research of the first author is supported by MINECO grant MTM2014-57838-C2-1-P and Fundaci´on S´eneca, Regi´on de Murcia grant 19368/PI/14. Theresearch of the second author is done in frames of Ukrainian Ministry of Scienceand Education Research Program 0118U002036, and it was partially supported bySpanish MINECO/FEDER projects MTM2015-65020-P and MTM2017-83262-C2-2-P and Fundaci´on S´eneca, Regi´on de Murcia grant 19368/PI/14. The third authoris partially supported by a grant from Akhiezer’s Fund, 2018. a r X i v : . [ m a t h . F A ] A ug ANGOSTO, KADETS, AND ZAVARZINA
Problem 1.1.
Let X and Y be Banach spaces and F : B X → B Y be abijective non-expansive map. Is F an isometry? There are positive answers when Y is (cid:96) , a finite-dimensional Banachspace or a strictly convex Banach space (see [12]).The unit sphere of a strictly convex space consists of its extremepoints. The main result of our paper is Theorem 4.11, in which we sub-stitute extreme points by finite-dimensional polyhedral extreme sub-sets. Namely, we demonstrate that if X , Y are Banach spaces, F : B X → B Y is a bijective non-expansive map and S Y is the union of allits finite-dimensional polyhedral extreme subsets, then F is an isome-try.Let us briefly explain the structure of the paper. In Section 2 weextend the results about EC-spaces in totally bounded metric spacesto totally bounded uniform spaces, see Theorem 2.3 and Lemma 2.4.In Section 3 we recollect some known results about bijective non-expansive maps between unit balls that we will need in the sequel.The goal of Section 4 is to demonstrate the main result. On the waywe collect as much as possible information about preimages under abijective non-expansive map of finite-dimensional faces of the unit ball.Using this information we obtain positive answers for the Problem 1.1for the case when X is strictly convex (Theorem 4.2) and for the casewhen S Y is the union of all its finite-dimensional polyhedral extremesubsets (Theorem 4.11).We dedicate this paper to the memory of Bernardo Cascales , whopassed away in April, 2018. From the very beginning our activity re-lated to EC-spaces was motivated by Bernardo’s interest to the subject.It was his idea to search for a definition of EC-spaces that could be ap-plicable to topological vector spaces and to uniform spaces. It wasBernardo who communicated to us the Ellis’ result [4] and the wayhow this result was used by Isaac Namioka [11] in his elegant demon-stration of EC-plasticity of compact metric spaces. We were planningto start a joint project, but. . .2.
Non-expansive maps and uniformities
The aim of this section is to extend the results about EC-spaces inmetric spaces to uniform spaces. We denote by ( E, U ) a uniform spaceand by B a basis of the uniformity. For A ⊂ E , U, V ⊂ E × E , and u, v, w ∈ E we denote: A − = { ( u, v ) : ( v, u ) ∈ A } , ( u, v ) ◦ ( v, w ) = ( u, w ); U ◦ V = { ( u, v ) : there is w ∈ E such that ( u, w ) ∈ U and ( w, v ) ∈ V } ; ON-EXPANSIVE BIJECTIONS, UNIFORMITIES AND FACES 3 U [ A ] = { u ∈ E such that there is v ∈ A with ( u, v ) ∈ U } . The uniform space ( E, U ) is called totally bounded if for every U ∈ B there is a finite subset (cid:101) E ⊂ E such that E = U [ (cid:101) E ].Let us recall the following definitions that were introduced in [3] andextend the concepts of non-expansive, non-contractive and isometricmaps to uniform spaces. Definition 2.1.
Let ( E, U ) be a uniform space, B a basis of the uni-formity and F : E → E a map. We say that F is non-contractive forthe basis B if for every V ∈ B (2.1) ( F ( x ) , F ( y )) ∈ V ⇒ ( x, y ) ∈ V We say that F is non-expansive for the basis B if for every V ∈ B ( x, y ) ∈ V ⇒ ( F ( x ) , F ( y )) ∈ V. We say that F is an isobasism for the basis B if for every V ∈ B ( F ( x ) , F ( y )) ∈ V ⇔ ( x, y ) ∈ V. For unexplained standard definitions and terminology we refer to [8,Chapter 6].The next proposition will be used in the proof of Theorem 2.3.
Proposition 2.2 (Ellis [4]) . Let K be a compact space, let S ⊂ C ( K, K ) be a semigroup for the composition, and let Σ := S ⊂ K K . The follow-ing are equivalent: (1) each member of Σ is onto, (2) each member of Σ is one to one, (3) Σ is a group and id : K → K is the identity element of thegroup. Theorem 2.3.
Let K be a compact Hausdorff uniform space, B a basisfor the uniformity made of open sets in K × K . If F : K → K is anon-contractive bijection for the basis B , then F is an isobasism for thebasis B .Proof. The demonstration follows the idea of Namioka’s unpublishedproof of EC-plasticity of compact metric spaces [11], and is presentedhere with his kind permission.Observe that since F is non-contractive, then F − is non-expansiveand then ( x, y ) ∈ V ⇒ ( F − ( x ) , F − ( y )) ∈ V, so F − is a continuous bijection between compact spaces and then F is a continuous function. Consider the semigroup S = { F n : n ∈ N } ⊂ C ( K, K )and let G ∈ Σ = S be the pointwise closure of S . Choose a net ( G i ) i ∈ I in S that converges to G . Let x, y ∈ K and V ∈ B be such that ANGOSTO, KADETS, AND ZAVARZINA ( G ( x ) , G ( y )) ∈ V . There is j ∈ I such that ( G j ( x ) , G j ( y )) ∈ V . Let n ∈ N be such that G j = F n . Then since F is non-contractive we havethat( F n ( x ) , F n ( y )) ∈ V ⇒ ( F n − ( x ) , F n − ( y )) ∈ V ⇒ · · · ⇒ ( x, y ) ∈ V. We have proved that(2.2) ( G ( x ) , G ( y )) ∈ V ⇒ ( x, y ) ∈ V for every G ∈ Σ, x, y ∈ K and V ∈ B . Then we have that G is one toone. By Proposition 2.2, Σ is a group so F − ∈ Σ and then by (2.2)we have that ( F − ( x ) , F − ( y )) ∈ V ⇒ ( x, y ) ∈ V for every x, y ∈ K and V ∈ B , so F − is non-contractive and then F isan isobasism for the basis B . (cid:3) We know that every totally bounded metric space is an EC-space.The above theorem generalizes this result for uniformities when thespace is compact. We can use some ideas of [10] to get the followingresults for uniformities in totally bounded spaces.
Lemma 2.4.
Let ( E, U ) be a totally bounded uniform space, B a basisfor the uniformity in E × E and F : E → E a non-contractive bijectionfor B . Then F satisfies that for every V ∈ B ( x, y ) ∈ V ⇒ for each W ∈ U there is k ∈ N such that ( F k ( x ) , F k ( y )) ∈ W ◦ V ◦ W. (2.3) Proof.
Choose x, y ∈ E and V ∈ B with ( x, y ) ∈ V . Choose W ∈ U , W (cid:48) ∈ B a subset of W , Z ∈ B such that Z ◦ Z ⊂ W (cid:48) and U ∈ B suchthat U ⊂ Z ∩ Z − . Since E is totally bounded there is a finite set (cid:101) E ⊂ E such that E = U [ (cid:101) E ]. Then there is a infinite set M ⊂ N and z , z ∈ (cid:101) E such that { F n ( x ) : n ∈ M } ⊂ U [ z ] and { F n ( y ) : n ∈ M } ⊂ U [ z ].Pick n, m ∈ M with m > n and let k = m − n . Then( F m ( x ) , F n ( x )) = ( F m ( x ) , z ) ◦ ( z , F n ( x )) ∈ U ◦ U − ⊂ Z ◦ Z ⊂ W (cid:48) . Then by (2.1) we have that ( F k ( x ) , x ) ∈ W (cid:48) ⊂ W . Analogously( y, F k ( y )) ∈ W . Then( F k ( x ) , F k ( y )) = ( F k ( x ) , x ) ◦ ( x, y ) ◦ ( y, F k ( y )) ∈ W ◦ V ◦ W. (cid:3) Corollary 2.5.
Let ( E, U ) be a totally bounded uniform space, B abasis for the uniformity and F : E → E a non-contractive bijection for B . Then F satisfies that (2.4) ( x, y ) ∈ V ⇒ ( F ( x ) , F ( y )) ∈ V for every V ∈ B . ON-EXPANSIVE BIJECTIONS, UNIFORMITIES AND FACES 5
Proof.
Choose x, y ∈ E and V ∈ B with ( x, y ) ∈ V . By Lemma 2.4 wehave that for each W ∈ B there is k ∈ N such that ( F k ( x ) , F k ( y )) ∈ W ◦ V ◦ W . Then since F k is a bijection, we can choose w, z ∈ E suchthat ( F k ( x ) , F k ( w )) ∈ W , ( F k ( w ) , F k ( z )) ∈ V and ( F k ( z ) , F k ( y )) ∈ W . Since F is a non-contractive map we have that ( F ( x ) , F ( w )) ∈ W ,( F ( w ) , F ( z )) ∈ V and ( F ( z ) , F ( y )) ∈ W so ( F ( x ) , F ( y )) ∈ W ◦ V ◦ W and then ( F ( x ) , F ( y )) ∈ (cid:92) W ∈B W ◦ V ◦ W = V . (cid:3)
Let (
E, d ) be a metric space, if we denote U ε = { ( x, y ) : d ( x, y ) < ε } then B = { U ε : ε > } is a basis of the uniformity and F : E → E isnon-expansive, non-contractive or an isometry for the metric d if andonly if F is non-expansive, non-contractive or an isobasism for the basisof the uniformity B . Then Corollary 2.5 implies the following result: Corollary 2.6 ([5, Satz IV]) . Let ( E, d ) be a totally bounded metricspace and F : E → E a bijective non-contractive (or non-expansive)map. Then F is an isometry. Corollary 2.7.
Let X be a topological vector space, A ⊂ X a totallybounded set and B a basis of closed neighborhoods of 0. Let F : A → A be a bijection such that for every x, y ∈ A and V ∈ B F ( x ) − F ( y ) ∈ V ⇒ x − y ∈ V. Then f satisfies that for every x, y ∈ A and V ∈ B x − y ∈ V ⇒ F ( x ) − F ( y ) ∈ V. Proof.
This result follows from Corollary 2.5 applied to the set A andthe basis for a uniformity { U V : V ∈ B} where U V = { ( x, y ) : x − y ∈ V } for each V ∈ B . (cid:3) The following result is a reformulation of the last corollary:
Corollary 2.8.
Let X be a topological vector space, A ⊂ X a totallybounded set and B a basis of closed neighborhoods of 0. Let F : A → A be a bijection. If there is x, y ∈ A and V ∈ B such that x − y ∈ V and F ( x ) − F ( y ) / ∈ V, then there is z, w ∈ A and W ∈ B such that F ( z ) − F ( w ) ∈ W and z − w / ∈ W. Notation and auxiliary statements for Banach spaces
In this short section we fix the necessary notation and recollect someknown results that we will need in the sequel. Below the letters X and Y always stand for real Banach spaces. We denote by S X and B X theunit sphere and the closed unit ball of X respectively. For a convex ANGOSTO, KADETS, AND ZAVARZINA set A ⊂ X denote by ext( A ) the set of extreme points of A ; that is, x ∈ ext( A ) if x ∈ A and for every y ∈ X \ { } either x + y (cid:54)∈ A or x − y (cid:54)∈ A . Recall that X is called strictly convex if all elements of S X are extreme points of B X , or in other words, S X does not contain non-trivial line segments. Strict convexity of X is equivalent to the stricttriangle inequality (cid:107) x + y (cid:107) < (cid:107) x (cid:107) + (cid:107) y (cid:107) holding for all pairs of vectors x, y ∈ X that do not have the same direction. For subsets A, B ⊂ X we use the standard notation A + B = { x + y : x ∈ A, y ∈ B } and aA = { ax : x ∈ A } . Proposition 3.1 (P. Mankiewicz’s [9]) . If A ⊂ X and B ⊂ Y areconvex subsets with non-empty interior, then every bijective isometry F : A → B can be extended to a bijective affine isometry (cid:101) F : X → Y . Taking into account that in the case of A , B being the unit ballsevery isometry maps 0 to 0, this result implies that every bijectiveisometry F : B X → B Y is the restriction of a linear isometry from X onto Y . Proposition 3.2 (Brower’s invariance of domain principle [1]) . Let U be an open subset of R n and f : U → R n be an injective continuousmap, then f ( U ) is open in R n . Proposition 3.3 ([6, Proposition 4]) . Let X be a finite-dimensionalnormed space and V be a subset of B X with the following two properties: V is homeomorphic to B X and V ⊃ S X . Then V = B X . The remaining results of this section listed below appeared first in[2] for the particular case of X = Y . The generalizations to the case oftwo different spaces were made in [12] and [7].The following theorem appears in [12, Theorem 2.1] and it can bedemonstrated repeating the proof of [2, Theorem 2.3] almost word toword (see [13, Theorem 2.3] for details). Theorem 3.4.
Let F : B X → B Y be a bijective non-expansive (briefly,a BnE) map. In the above notations the following hold. (1) F (0) = 0 . (2) F − ( S Y ) ⊂ S X . (3) If F ( x ) is an extreme point of B Y , then x is also an extremepoint of B X , F ( ax ) = aF ( x ) for all a ∈ [ − , .Moreover, if Y is strictly convex, then (i) F maps S X bijectively onto S Y ; (ii) F ( ax ) = aF ( x ) for all x ∈ S X and a ∈ [ − , . Lemma 3.5 ([12, Lemma 2.3]) . Let F : B X → B Y be a BnE mapsuch that F ( S X ) = S Y . Let V ⊂ S X be the subset of all those v ∈ S X that F ( av ) = aF ( v ) for all a ∈ [ − , . Denote A = { tx : x ∈ V, t ∈ [ − , } , then F | A is a bijective isometry between A and F ( A ) . ON-EXPANSIVE BIJECTIONS, UNIFORMITIES AND FACES 7
Lemma 3.6 ([7, Lemma 2.9]) . Let F : B X → B Y be a BnE mapsuch that for every v ∈ F − ( S Y ) and every t ∈ [ − , the condition F ( tv ) = tF ( v ) holds true. Then F is an isometry. Proposition 3.7 ([12, Theorem 3.1]) . Let F : B X → B Y be a BnEmap. If Y is strictly convex, then F is an isometry. Let us list some more definitions. • An extreme subset of a set B ⊂ X is a subset C ⊂ B with theproperty ∀ y ,y ∈ B ∀ α ∈ (0 , ( αy + (1 − α ) y ∈ C ) = ⇒ ( y , y ∈ C ) . • The generating subspace of a convex set C is span( C − C ). • The dimension of a convex set C is the dimension of its gener-ating subspace. • For a convex set B ⊂ X we will say that a point x ∈ B is n -extreme if for any ( n + 1)-dimensional subspace E ⊂ X andany ε > e ∈ S E , such that x + εe / ∈ B . • For n ∈ N a point x of the convex set B is called sharp n -extremein B if it is n -extreme and is not ( n − B is an extreme subset of B . Nevertheless, we mostly deal with convexsets and convex extreme subsets. Observe also that being 0-extremepoint and being extreme point of B in the usual sense are equivalent.Every n -extreme point of B is also ( n + 1)-extreme point of B . Every n -dimensional convex extreme subset C of a convex set B consists of n -extreme points of B and contains a sharp n -extreme point. If E isthe generating subspace of the n -dimensional convex extreme subset C ⊂ B , then x ∈ C is a sharp n -extreme point of B if and only if x belongs to the relative interior of C in the affine subspace x + E = C + E .For a convex set C ⊂ X with generating subspase E by ∂C we denotethe relative boundary of C in C + E .Evidently, in a normed space for collinear vectors x, y looking in thesame direction (codirected vectors) we have(3.1) (cid:107) x + y (cid:107) = (cid:107) x (cid:107) + (cid:107) y (cid:107) . In spaces that are not strictly convex the converse statement is nottrue, which motivates the following definition.
Definition 3.8.
Elements x, y ∈ X are said to be quasi-codirected , ifthey satisfy (3.1).By the triangle inequality, in order to verify (3.1) it is sufficient tocheck (cid:107) x + y (cid:107) (cid:62) (cid:107) x (cid:107) + (cid:107) y (cid:107) . The next lemma is well-known, but thisis the example of a fact which is much easier to demonstrate than tofind out when and who observed it first (cid:44) ANGOSTO, KADETS, AND ZAVARZINA
Lemma 3.9. If x, y ∈ X are quasi-codirected, then for every a, b > the elements ax and by are quasi-codirected as well.Proof. By symmetry we may assume a (cid:62) b . Then, (cid:107) ax + by (cid:107) = (cid:107) a ( x + y ) − ( a − b ) y (cid:107) (cid:62) a (cid:107) x + y (cid:107) − ( a − b ) (cid:107) y (cid:107) = a (cid:107) x (cid:107) + b (cid:107) y (cid:107) . (cid:3) Geometrically speaking x, y ∈ S X are quasi-codirected, if the wholesegment [ x, y ] := { tx + (1 − t ) y : t ∈ [0 , } lies on the unit sphere. If C ⊂ S X is convex, then every two elementsof C are quasi-codirected.4. Non-expansive maps and finite-dimensional faces
The aim of this section is, in the setting of Section 3 and usingsome similar ideas, to obtain as much as possible information aboutpreimages of finite-dimensional faces of the unit ball. The main resultis Theorem 4.11 that gives a positive answer for the Problem 1.1 when S Y is the union of all its finite-dimensional polyhedral extreme subsets.Let us start with a very simple observation. Lemma 4.1.
Let X , Y be Banach spaces, F : B X → B Y be a BnEmap, and y , y ∈ S Y be quasi-codirected. Then, (1) F − ( y ) is quasi-codirected with − F − ( − y ) , so (2) if F − ( − y ) = − F − ( y ) , then F − ( y ) is quasi-codirected with F − ( y ) . (3) In particular if y is an extreme point of B Y , then F − ( y ) isquasi-codirected with F − ( y ) .Proof. (cid:13)(cid:13) F − ( y ) + (cid:0) − F − ( − y ) (cid:1)(cid:13)(cid:13) = (cid:13)(cid:13) F − ( y ) − F − ( − y ) (cid:13)(cid:13) (cid:62) (cid:107) y − ( − y ) (cid:107) = (cid:107) y + y (cid:107) = 2 . (cid:3) The above lemma readily implies the following natural counterpartto Proposition 3.7.
Theorem 4.2.
Let X , Y be Banach spaces, X be strictly convex and F : B X → B Y be a BnE map. Then F is an isometry.Proof. According to Proposition 3.7 it is sufficient to demonstrate that Y is strictly convex. Assume to the contrary that S Y contains a non-void segment [ y , y ] := { ty + (1 − t ) y : t ∈ [0 , } . Since X isstrictly convex, the only element of S X quasi-codirected with F − ( y )is F − ( y ) itself. But, according to (1) of Lemma 4.1 all elements − F − ( − y t ), where y t := ty + (1 − t ) y , t ∈ [0 , F − ( y ). This contradiction completes the proof. (cid:3) ON-EXPANSIVE BIJECTIONS, UNIFORMITIES AND FACES 9
Let Y be a Banach space, y , y ∈ S Y be quasi-codirected. Denote D ( y , y ) := ( y + B Y ) ∩ ( − y + B Y )= { y ∈ Y : (cid:107) y − y (cid:107) (cid:54) (cid:107) y + y (cid:107) (cid:54) } (4.1) = (cid:8) y ∈ Y : (cid:107) y − y (cid:107) = (cid:107) y + y (cid:107) = 1 (cid:9) . Some evident properties of D ( y , y ) are listed below without proof. Lemma 4.3.
Let Y be a Banach space, y , y ∈ S Y be quasi-codirected.Then • D ( y , y ) is a convex closed subset of Y . • ∈ D ( y , y ) . • tD ( y , y ) ⊂ D ( y , y ) for every t ∈ [0 , . • D ( y , y ) ⊂ B Y , consequently • D ( y , y ) ⊂ D ( y , y ) ∩ B Y . Lemma 4.4.
Let Y be a Banach space, y , y ∈ S Y be quasi-codirected,and h ∈ Y be such that y ± h ∈ S Y . Then (4.2) (cid:110)
12 ( y − y ) ± h (cid:111) ⊂ D ( y , y ) . In particular, substituting y = y we obtain ± h ∈ D ( y , y ) . Substituting h = 0 we obtain
12 ( y − y ) ∈ D ( y , y ) , which implies that for all t ∈ [0 , / t ( y − y ) ∈ D ( y , y ) . Proof.
We have to verify two inequalities: (cid:13)(cid:13)(cid:13)(cid:13)
12 ( y − y ) ± h + y (cid:13)(cid:13)(cid:13)(cid:13) (cid:54) (cid:13)(cid:13)(cid:13)(cid:13)
12 ( y − y ) ± h − y (cid:13)(cid:13)(cid:13)(cid:13) (cid:54) . Each of them reduces to the same inequality (cid:13)(cid:13)(cid:13)(cid:13)
12 ( y + y ) ± h (cid:13)(cid:13)(cid:13)(cid:13) (cid:54) . Let us demonstrate this: (cid:107) ( y + y ) ± h (cid:107) = (cid:107) y + ( y ± h ) (cid:107) (cid:54) (cid:107) y (cid:107) + (cid:107) y ± h (cid:107) = 2. (cid:3) Lemma 4.5.
Let Y be a Banach space, C ⊂ S Y be a convex extremesubset, and E be the generating subspace of C . Then D ( y , y ) ⊂ E for every y , y ∈ C . Proof.
Let y ∈ D ( y , y ). Then, y − y, y + y ∈ B Y and12 (cid:0) ( y − y ) + ( y + y ) (cid:1) = 12 ( y + y ) ∈ C. Consequently, by the definition of extreme subset, y + y ∈ C , so y = ( y + y ) − y ∈ C − C ⊂ E . (cid:3) Lemma 4.6.
Let X , Y be Banach spaces, F : B X → B Y be a BnEmap, y , y ∈ S Y be quasi-codirected, x = F − ( y ) ∈ S X , x = − F − ( − y ) ∈ S X . Then F ( D ( x , x ) ∩ B X ) ⊂ D ( y , y ) ∩ B Y . In particular, F (cid:0) D ( x , x ) (cid:1) ⊂ D ( y , y ) ∩ B Y .Proof. According to (1) of Lemma 4.1, x and x are quasi-codirected,so the set D ( x , x ) is well-defined. Consider arbitrary x ∈ D ( x , x ) ∩ B X . We have (cid:107) x − x (cid:107) (cid:54) (cid:107) ( − x ) − x (cid:107) = (cid:107) x + x (cid:107) (cid:54)
1, so (cid:107) F ( x ) − F ( x ) (cid:107) (cid:54) (cid:107) F ( − x ) − F ( x ) (cid:107) (cid:54)
1. In other words, (cid:107) y − F ( x ) (cid:107) (cid:54) (cid:107) ( − y ) − F ( x ) (cid:107) = (cid:107) y + F ( x ) (cid:107) (cid:54)
1, which meansthat F ( x ) ∈ D ( y , y ). (cid:3) Lemma 4.7.
Let X , Y be Banach spaces, F : B X → B Y be a BnEmap, n ∈ N , and C ⊂ S Y be an n -dimensional convex extreme subset.Then for every y ∈ C its preimage x = F − ( y ) ∈ S X is an n -extremepoint of B X .Proof. Denote x = − F − ( − y ) ∈ S X . Assume that x is not n -extreme point of B X . Then, according to the definition, there existan ( n + 1)-dimensional subspace E ⊂ X and an ε > x + εB E ⊂ S X . According to Lemma 4.412 ( x − x ) + εB E ⊂ D ( x , x ) . The above inclusion implies that D ( x , x ) contains an ( n + 1)-dimensional ball. Then Lemma 4.6 implies that D ( y , y ) containsa homeomorphic copy of ( n + 1)-dimensional ball, which is impossibleby Lemma 4.5. (cid:3) Note, that under conditions of the previous lemma x may be also m -extreme point for some m < n . Now we are coming to the mostimportant and at the same time most difficult result of the paper. Theorem 4.8.
Let X , Y be Banach spaces, F : B X → B Y be a BnEmap, then for every n ∈ N the preimage of any n -dimensional convexpolyhedral extreme subset C ⊂ S Y is an n -dimensional convex poly-hedral extreme subset of S X . Moreover, the equality − F − ( − C ) = F − ( C ) holds true.Proof. We will use the induction in n . The initial case of n = 0 (i.e., ofextreme points) is covered by the assertion (3) of Theorem 3.4. Let us ON-EXPANSIVE BIJECTIONS, UNIFORMITIES AND FACES 11 assume that the theorem is demonstrated for extreme subsets of dimen-sion smaller than n , and let us demonstrate it for a given n -dimensionalpolyhedral extreme subset C ⊂ S Y . Denote E the generating subspaceof C , dim E = n . The boundary ∂C of polyhedron C consists of finiteunion of its convex ( n − A := F − ( ∂C ) also consists of finiteunion of some convex ( n − S X . Consequently, A is an extreme subset of B X . Also, A is com-pact, and F | A performs a homeomorphism between A and F ( A ) = ∂C .Let y ∈ C \ ∂C be an arbitrary point. Denote x = F − ( y ). Since y is quasi-codirected with every point y ∈ ∂C , x is quasi-codirectedwith the corresponding x = − F − ( − y ) ∈ S X . By the inclusion (4.3)and Lemmas 4.6 and 4.5 F ( t ( x − x )) ∈ F (cid:18) D ( x , x ) (cid:19) ⊂ D ( y , y ) ⊂ E for all t ∈ (cid:2) , (cid:3) . By the inductive hypothesis, when y runs through ∂C the corresponding x runs through the whole A . So, denoting (cid:101) A = (cid:2) , (cid:3) ( x − A ) = (cid:8) t ( x − x ) : t ∈ (cid:2) , (cid:3) , x ∈ A (cid:9) we obtain(4.4) F (cid:0) (cid:101) A (cid:1) ⊂ E. Let us demonstrate that the segments (cid:0) , (cid:3) ( x − x ) with different x ∈ A are pairwise disjoint. We will argue by contradiction. Let twosegments of the form (cid:0) , (cid:3) ( x − (cid:98) x ), (cid:0) , (cid:3) ( x − (cid:101) x ) with (cid:98) x , (cid:101) x ∈ A , (cid:98) x (cid:54) = (cid:101) x intersect at some point y . Then the corresponding closedsegments (cid:2) , (cid:3) ( x − (cid:98) x ), (cid:2) , (cid:3) ( x − (cid:101) x ) intersect in two points (0 and y ), so either they coincide or one segment contains the other one. Thatis, ( x − (cid:98) x ) and ( x − (cid:101) x ) are codirected. There are two cases:( x − (cid:98) x ) = λ ( x − (cid:101) x ) or ( x − (cid:101) x ) = λ ( x − (cid:98) x ) with some 0 < λ <
1. We will discussthe first one, the second one is analogous. We get (cid:98) x = λ (cid:101) x + (1 − λ ) x ,so these three points are on the same segment and (cid:98) x lies between x and (cid:101) x . Since A is extreme subset, we get x ∈ A , which contradictsthe fact y / ∈ ∂C .The set ( x − A ) is homeomorphic to the unit sphere of R n . Let usshow, that (cid:101) A is homeomorphic to the unit ball of R n , with 0 mappedto 0. Let S n and B n denote the unit sphere and the unit ball of R n respectively, and h : S n → ( x − A ) be a homeomorphism. One maydefine the mapping H : B n → (cid:101) A as H ( x ) = (cid:40) , when x = 0 (cid:107) x (cid:107) h (cid:16) x (cid:107) x (cid:107) (cid:17) , when x ∈ B n \{ } . Obviously, this mapping is bijective and continuous at 0. We are go-ing to show that H is continuous at all points. Let us consider some sequence { x n } ∞ n =1 in B n converging to an x ∈ B n \ { } , that islim n →∞ x n = x (cid:54) = 0 . Thenlim n →∞ H ( x n ) = lim n →∞ (cid:107) x n (cid:107) h (cid:18) x n (cid:107) x n (cid:107) (cid:19) = 14 lim n →∞ (cid:107) x n (cid:107) lim n →∞ h (cid:18) x n (cid:107) x n (cid:107) (cid:19) = 14 (cid:107) x (cid:107) h (cid:18) lim n →∞ x n (cid:107) x n (cid:107) (cid:19) = 14 (cid:107) x (cid:107) h (cid:18) x (cid:107) x (cid:107) (cid:19) = H ( x ) . So, H is a bijective continuous map from compact B n to Hausdorfspace, thus H is a homeomorphism.Consequently, F (cid:0) (cid:101) A (cid:1) ⊂ E is homeomorphic to the unit ball of R n ,with 0 being a relative (in E ) interior point of F (cid:0) (cid:101) A (cid:1) .Consider now any point (cid:101) y ∈ C \ ∂C , (cid:101) y (cid:54) = y , such that the cor-responding (cid:101) x = − F − (cid:0) − (cid:101) y (cid:1) is not equal to x . By the same reasonas before, the segment F (cid:0)(cid:2) , (cid:3) ( x − (cid:101) x ) (cid:1) ⊂ D ( y , (cid:101) y ) ⊂ E . The set F (cid:0)(cid:2) , (cid:3) ( x − (cid:101) x ) (cid:1) is a continuous curve in E connecting F ( ( x − (cid:101) x ))with 0, which is an interior point of F (cid:0) (cid:101) A (cid:1) . So there is a t ∈ (cid:0) , (cid:3) suchthat F (cid:0) t ( x − (cid:101) x ) (cid:1) ∈ F (cid:0) (cid:101) A (cid:1) , that is t ( x − (cid:101) x ) ∈ (cid:101) A . This means thatfor some t ∈ (cid:0) , (cid:3) and some x ∈ A we have t ( x − (cid:101) x ) = t ( x − x ).In other words, there is an α > x − (cid:101) x = α ( x − x ) . Let us demonstrate that α <
1. Indeed, if α (cid:62)
1, the above formulawould give the representation x = (cid:18) − α (cid:19) x + 1 α (cid:101) x of x ∈ A as a convex combination of x , (cid:101) x ∈ S X \ A , which contradictsthe fact that A is extreme in S X .Since α <
1, the formula (4.5) gives the representation (cid:101) x = (1 − α ) x + αx of (cid:101) x as a convex combination of x and some x ∈ A .If we consider the BnE mapping G : B X → B Y defined as G ( x ) = − F ( − x ), all the above reasoning is applicable for G as well, becauseby the inductive hypothesis G − ( ∂C ) = F − ( ∂C ) = A . Since (cid:101) x = G − (cid:0)(cid:101) y (cid:1) and x = − G − ( − y ) the roles of these elements for G inter-change, and we deduce that also x is a convex combination of (cid:101) x andsome x ∈ A . So, we obtain the following properties of sets F − ( C \ ∂C )and G − ( C \ ∂C ): Properties. (i) For every u ∈ F − ( C \ ∂C ) G − ( C \ ∂C ) ⊂ { tx + (1 − t ) u : t ∈ [0 , , x ∈ A } . ON-EXPANSIVE BIJECTIONS, UNIFORMITIES AND FACES 13 (ii) For every v ∈ G − ( C \ ∂C ) F − ( C \ ∂C ) ⊂ { tx + (1 − t ) v : t ∈ [0 , , x ∈ A } . (iii) For every u ∈ F − ( C \ ∂C ) and every v ∈ G − ( C \ ∂C ), u (cid:54) = v there are (unique) elements w, z ∈ A such that [ u, v ] ⊂ [ w, z ].Properties (i) and (ii) imply that F − ( C ) and G − ( C ) lie in somefinite-dimensional subspace of X . Since both these sets are boundedand closed, they are compacts. Continuous mappings F and G mapcorresponding compacts F − ( C ) and G − ( C ) to C bijectively, so both F − ( C ) and G − ( C ) are homeomorphic to C , i.e. homeomorphic tothe unit ball of R n . Since the set { tx + (1 − t ) u : t ∈ [0 , , x ∈ A } fora fixed u is also homeomorphic to the unit ball of R n and A correspondsto the unit sphere and belongs to both { tx + (1 − t ) u : t ∈ [0 , , x ∈ A } and G − ( C ), the inclusion (i) and Proposition 3.3 imply that(i)’ for every u ∈ F − ( C \ ∂C ) G − ( C ) = { tx + (1 − t ) u : t ∈ [0 , , x ∈ A } , and by the same reason(ii)’ For every v ∈ G − ( C \ ∂C ) F − ( C ) = { tx + (1 − t ) v : t ∈ [0 , , x ∈ A } . In particular, from (i)’ it follows that every u ∈ F − ( C \ ∂C ) belongsto G − ( C ), so F − ( C ) ⊂ G − ( C ), and (ii)’ implies the inverse inclusion G − ( C ) ⊂ F − ( C ), so G − ( C ) = F − ( C ) . Coming back to the already used inclusion (4.3) and Lemmas 4.6and 4.5 we obtain that for all x , x ∈ F − ( C ) F (cid:18)
14 ( x − x ) (cid:19) ∈ E, in other words(4.6) F (cid:18)
14 ( F − ( C ) − F − ( C )) (cid:19) ⊂ E. Recall, that by the inductive hypothesis, A = F − ( ∂C ) consists offinite union of some convex ( n − (cid:102) W i , i = 1 , . . . , N which are preimages of corresponding partsof ∂C . Let us fix some v ∈ F − ( C \ ∂C ). Denote W i = (cid:110) tx + (1 − t ) v : t ∈ [0 , , x ∈ (cid:102) W i (cid:111) . These W i are n -dimensional convex polyhedrons, and, according to(ii)’, F − ( C ) = N (cid:91) i =1 W i . We state that all polyhedrons W i (and also their union F − ( C )) aresituated in one and the same n -dimensional affine subspace (cid:101) E .To this end, consider the generating subspaces Z i = span( W i − W i )of W i and let us demonstrate that all of Z i are equal one to another, i.e.all of them are equal to some n -dimensional linear subspace Z . Then (cid:101) E = v + Z will be the n -dimensional affine subspace (cid:101) E we are lookingfor.Let us argue “ad absurdum”. Assume that Z i (cid:54) = Z j for some i (cid:54) = j .Then Z i + Z j has dimension strictly greater than n , anddim( W i − W j ) = dim(span(( W i − W j ) − ( W i − W j ))) = dim( Z i + Z j ) > n. Taking into account that W i − W j ⊂ ( F − ( C ) − F − ( C )) the dimensionof F − ( C ) − F − ( C ) is strictly greater than n , which makes the inclusion(4.6) impossible.It remains to demonstrate that F − ( C ) is convex and is an extremesubset. For the convexity let us show that F − ( C ) = B X ∩ (cid:101) E . Wehave already known, that F − ( C ) ⊂ B X ∩ (cid:101) E . Let us show the inverseinclusion. Again we will argue by contradiction. Suppose there is apoint z ∈ ( B X ∩ (cid:101) E ) \ F − ( C ). We may fix some v ∈ F − ( C \ ∂C )and consider the segment [ z, v ]. As we already remarked, F − ( C ) ishomeomorphic to C and hence to B n , that is, v lies in the relativeinterior of F − ( C ) in (cid:101) E . So, the segment [ z, v ] must intersect A = F − ( ∂C ) in some point. In other words, there is λ ∈ (0 ,
1) such that λz + (1 − λ ) v ∈ A , which contradicts the fact, that A is an extremesubset in B X . (cid:3) Theorem 4.9.
Let X , Y be Banach spaces, F : B X → B Y be aBnE map, then for every n -dimensional convex polyhedral extreme sub-set C ⊂ S Y the following equality holds true: F (conv(0 , F − ( C ))) =conv(0 , C ) .Proof. We will carry out the proof by induction in n . For n = 0 (i.e.,when C is extreme point) the required equality may be obtained fromthe assertion (3) of Theorem 3.4. Suppose our theorem is proved forall extreme subsets of dimension smaller than n , and let us show thesame for a given n -dimensional polyhedral extreme subset C ⊂ S Y .Consider x ∈ F − ( C \ ∂C ) and α ∈ (0 , F is non-expansivewe have(4.7) (cid:107) F ( αx ) (cid:107) (cid:54) (cid:107) αx (cid:107) , and (cid:107) F ( x ) − F ( αx ) (cid:107) (cid:54) (cid:107) x − αx (cid:107) . Also 1 = (cid:107) F ( x ) (cid:107) (cid:54) (cid:107) F ( αx ) (cid:107) + (cid:107) F ( x ) − F ( αx ) (cid:107) (cid:54) (cid:107) αx (cid:107) + (cid:107) x − αx (cid:107) = 1 . (4.8) ON-EXPANSIVE BIJECTIONS, UNIFORMITIES AND FACES 15
That is why (cid:107) F ( αx ) (cid:107) + (cid:107) F ( x ) − F ( αx ) (cid:107) = 1 . So one may write F ( x ) as a convex combination F ( x ) = (cid:107) F ( αx ) (cid:107) F ( αx ) (cid:107) F ( αx ) (cid:107) + (cid:107) F ( x ) − F ( αx ) (cid:107) F ( x ) − F ( αx ) (cid:107) F ( x ) − F ( αx ) (cid:107) . Since F ( x ) ∈ C and C is extreme subset in B X we get F ( αx ) (cid:107) F ( αx ) (cid:107) ∈ C and F ( x ) − F ( αx ) (cid:107) F ( x ) − F ( αx ) (cid:107) ∈ C . So, F ( αx ) = (cid:107) F ( αx ) (cid:107) F ( αx ) (cid:107) F ( αx ) (cid:107) ∈ conv( F ( αx ) (cid:107) F ( αx ) (cid:107) , ⊂ conv(0 , C ) and thus F (conv(0 , F − ( C ))) ⊂ conv(0 , C ). By the in-ductive hypothesis F (conv(0 , A )) = conv(0 , ∂C ) and ∂ conv(0 , C ) ⊂ F (conv(0 , F − ( C ))). Besides, conv(0 , F − ( C )) is homeomorphic to B n +1 and ∂ conv(0 , C ) is homeomorphic to S n +1 . In this way Proposi-tion 3.3 implies the statement of the theorem. (cid:3) Lemma 4.10.
Let X , Y be Banach spaces, F : B X → B Y be a BnEmap, then (cid:107) F ( αx ) (cid:107) = (cid:107) αx (cid:107) = α for all x ∈ F − ( S Y ) , α ∈ [0 , .Proof. Since F is non-expansive, we may use inequalities (4.7) and(4.8). The inequality (4.8) implies (cid:107) F ( αx ) (cid:107) + (cid:107) F ( x ) − F ( αx ) (cid:107) = (cid:107) αx (cid:107) + (cid:107) x − αx (cid:107) , and application of (4.7) concludes the proof. (cid:3) Theorem 4.11.
Let X , Y be Banach spaces, F : B X → B Y be aBnE map and S Y be the union of all its finite-dimensional polyhedralextreme subsets. Then F is an isometry.Proof. Let us first show, that F ( S X ) = S Y . Since(4.9) S Y = (cid:91) i ∈ I C i , where C i are finite-dimensional polyhedral extreme subsets of S Y and I is some index set, one may deduce B Y = (cid:91) i ∈ I conv(0 , C i ) . Due to bijectivity of F , theorem 4.9 implies B X = (cid:91) i ∈ I conv(0 , F − ( C i )) . Consequently, there is no other norm-one points in B X except for pointsfrom F − ( C i ), and we get S X = (cid:91) i ∈ I F − ( C i ) = F − ( S Y ) . To prove that F is an isometry we will use lemmas 3.6 and 3.5. We aregoing to show for the set V from lemma 3.5 that(4.10) F − ( C ) ⊂ V for every n -dimensional polyhedral extreme subset C of S Y . To dothat, we will use induction by dimension. For 0-dimensional sets, i.e.extreme points, the statement we need follows from item (3) of theorem3.4. Now suppose that the inclusion is proved for all ( n − n . Con-sider some n -dimensional extreme subset C in S Y . For every pair x, y ∈ F − ( C ) there are u, v ∈ F − ( ∂C ) such that x = λu + (1 − λ ) v and y = µu + (1 − µ ) v, λ, µ ∈ (0 , λ > µ . Since ∂C consists of ( n − (cid:107) u − v (cid:107) = (cid:107) F ( u ) − F ( v ) (cid:107) . Since F is non-expansive, (cid:107) u − v (cid:107) = (cid:107) F ( u ) − F ( v ) (cid:107) (cid:54) (cid:107) F ( u ) − F ( x ) (cid:107) + (cid:107) F ( x ) − F ( y ) (cid:107) + (cid:107) F ( y ) − F ( v ) (cid:107) (cid:54) (cid:107) u − x (cid:107) + (cid:107) x − y (cid:107) + (cid:107) y − v (cid:107) = (1 − λ ) (cid:107) u − v (cid:107) + ( λ − µ ) (cid:107) u − v (cid:107) + µ (cid:107) u − v (cid:107) = (cid:107) u − v (cid:107) . So we get (cid:107) F ( u ) − F ( x ) (cid:107) = (cid:107) u − x (cid:107) , (cid:107) F ( y ) − F ( v ) (cid:107) = (cid:107) y − v (cid:107) , (cid:107) F ( x ) − F ( y ) (cid:107) = (cid:107) x − y (cid:107) . Thus, F is bijective isometry between F − ( C ) and C and Proposition 3.1 implies that F is affine on F − ( C ). Lemma4.10 together with Theorem 4.9 give the equality F ( αF − ( C )) = αC for α ∈ [0 , α ∈ [ − , F is bijective isometry between αF − ( C ) and αC , so F is affine on αF − ( C ). We are going to show that F ( αx ) = αF ( x ) for all x ∈ F − ( C ), α ∈ [ − , x ∈ F − ( C ) isof the form x = λu + (1 − λ ) v , where u, v ∈ F − ( ∂C ) and λ ∈ (0 , F ( αx ) = F ( λαu + (1 − λ ) αv ) = λF ( αu ) + (1 − λ ) F ( αv ) , because F is affine on αF − ( C ). By the inductive hypothesis F ( αu ) = αF ( u ), F ( αv ) = αF ( v ), so F ( αx ) = λαF ( u ) + (1 − λ ) αF ( v ) = α ( λF ( u ) + (1 − λ ) F ( v )) . It remains to use the fact that F is affine on F − ( C ) to conclude that F ( αx ) = αF ( λu + (1 − λ ) v ) = αF ( x ) . So, the required inclusion (4.10) is demonstrated. At last, (4.9) and thewritten above imply that for every v ∈ F − ( S Y ) and every t ∈ [ − , F ( tv ) = tF ( v ). So, the application of lemma 3.6 completes the proofof the theorem. (cid:3) References [1] Brouwer L.E.J.
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Departamento de Matem´atica Aplicada y Estad´ıstica Uni-versidad Polit´ecnica de CartagenaORCID:
E-mail address : [email protected] (Kadets) School of Mathematics and Informatics, V.N. Karazin KharkivNational University, 61022 Kharkiv, UkraineORCID:
E-mail address : [email protected] (Zavarzina) School of Mathematics and Informatics, V.N. KarazinKharkiv National University, 61022 Kharkiv, UkraineORCID:
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