NONLINEAR CLASSIFICATION OF BANACH SPACES. A Dissertation NIRINA LOVASOA RANDRIANARIVONY

Transcription

1 NONLINEAR CLASSIFICATION OF BANACH SPACES A Dissertation by NIRINA LOVASOA RANDRIANARIVONY Submitted to the Office of Graduate Studies of Texas A&M University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY August 2005 Major Subject: Mathematics

2 NONLINEAR CLASSIFICATION OF BANACH SPACES A Dissertation by NIRINA LOVASOA RANDRIANARIVONY Submitted to the Office of Graduate Studies of Texas A&M University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Approved by: Chair of Committee, Committee Members, Head of Department: William B. Johnson Harold Boas N. Sivakumar Jerald Caton Albert Boggess August 2005 Major Subject: Mathematics

3 iii ABSTRACT Nonlinear Classification of Banach Spaces. (August 2005) Nirina Lovasoa Randrianarivony, Licence de Mathématiques; Maîtrise de Mathématiques, Université d Antananarivo Madagascar Chair of Advisory Committee: Dr. William B. Johnson We study the geometric classification of Banach spaces via Lipschitz, uniformly continuous, and coarse mappings. We prove that a Banach space which is uniformly homeomorphic to a linear quotient of l p is itself a linear quotient of l p when p<2. We show that a Banach space which is Lipschitz universal for all separable metric spaces cannot be asymptotically uniformly convex. Next we consider coarse embedding maps as defined by Gromov, and show that l p cannot coarsely embed into a Hilbert space when p>2. We then build upon the method of this proof to show that a quasi- Banach space coarsely embeds into a Hilbert space if and only if it is isomorphic to a subspace of L 0 (µ) for some probability space (Ω, B,µ).

4 For my family iv

5 v ACKNOWLEDGEMENTS I thank Dr. William B. Johnson for his patience, professional guidance and the mathematical expertise and dedication that he exemplified throughout the course my study. Thanks also to all those who have served on my committee: Dr. David Larson, Dr. Emil Straube, Dr. Harold Boas, Dr. N. Sivakumar, and Dr. Jerald Caton. I wish to thank my colleagues at the Mathematics Department for their help and encouragement throughout my study, my examinations, and the writing of my dissertation. I want to extend my gratitude especially to the many friends I have in the community, they have helped me in many ways to overcome my difficulties throughout my stay here. Thanks to Dr. Gérard Rambolamanana, my teacher and friend from the Université d Antananarivo, who has helped me get this far. Thanks to my family and to my son for their patience, encouragement and love. And finally, thanks to the Holy Spirit without whom I am nothing.

6 vi TABLE OF CONTENTS CHAPTER Page I INTRODUCTION II BANACH SPACES UNIFORMLY HOMEOMORPHIC TO A QUO- TIENT OF l p (1 <p<2) Uniform homeomorphism Quotient mappings Results from the linear theory Approximate metric midpoints Main result III BANACH SPACES CONTAINING A LIPSCHITZ COPY OF c Modulus of asymptotic uniform convexity Modulus of asymptotic uniform smoothness Properties of the asymptotic moduli Asymptotic moduli and approximate midpoints Main result IV COARSE EMBEDDING INTO A HILBERT SPACE Coarse embeddings Negative definite kernels and related notions Extensions Positive definite functions on l p, p> Main result V CHARACTERIZATION OF SPACES THAT COARSELY EMBED INTO A HILBERT SPACE Quasi-Banach spaces Positive definite functions and the space L 0 (µ) Main result VI CONCLUSION REFERENCES VITA

7 1 CHAPTER I INTRODUCTION How rigid is the linear structure of a Banach space? That is the question we set to investigate in this study. Banach spaces are considered as metric spaces, and considered between them are maps which control the distances between points in a uniform manner. The question is to study which aspects of the linear geometry of these spaces have to be transported by such maps from one of the Banach spaces into the other. Such questions have been widely studied. We mention in particular the Mazur- Ulam Theorem [MaU] which says that if the map considered is a surjective isometry, then all of the aspects of the linear geometry of one of the Banach spaces are transported into the other, namely the two Banach spaces are linearly isometric. In the present study, the control we impose on a map f : X Y between two Banach spaces X and Y is not quite as strict as an isometry. We relax a bit by allowing (uniform) lower and upper bounds on the distances. Namely, we request that for any x, y X we have: ϕ 1 ( x y ) f(x) f(y) ϕ 2 ( x y ), where ϕ 1,ϕ 2 :[0, ) [0, ) are nondecreasing functions. The study divides into two parts. In Chapters II and III, we focus our attention to controlling small distances. Namely we request that ϕ 1 (t) > 0whent>0, and The journal model is Geometric and Functional Analysis.

8 2 that ϕ 2 (t) 0ast 0. The maps that have such properties are exactly the uniform and/or Lipschitz embeddings. In Chapter II, we study Banach spaces that are uniformly homeomorphic to a linear quotient of l p when 1 <p<2. Such spaces are of course uniform quotients of l p as well (see for the formal definition). The case p>2 has been studied. Namely, in [GoKL2], Godefroy, Kalton and Lancien show that if a Banach space X is uniformly homeomorphic to a quotient of l p when p>2, then X has to be a linear quotient of l p as well. Johnson, Lindenstrauss, Preiss and Schechtman in [JoLPS] show that l 2 cannot be a Lipschitz quotient of l p when p>2. The techniques for these were respectively the conservation of the convex Szlenk index under a uniform homeomorphism, and the ɛ-fréchet differentiability of a Lipschitz map from l p to l 2 when p>2. In our study of the case p<2, we employ the technique introduced by Enflo in his proof that l 1 and L 1 are not uniformly homeomorphic, namely we use a comparison of the sizes of approximate metric midpoints. Using this technique, we get that a Banach space which is uniformly homeomorphic to a linear quotient of l p must itself be a linear quotient of l p when 1 <p<2. In Chapter III, we refine the Enflo technique by relating the sizes of approximate metric midpoints with the moduli introduced by Milman in [Mi]: the modulus of asymptotic uniform convexity δ X and the modulus of asymptotic uniform smoothness ρ X of a Banach space X. Through this refinement, the Enflo technique gives us an estimate of the form δ Y (t) Cρ X (Ct) when the Banach space X Lipschitz embeds into the Banach space Y. This enables us to give one characteristic of Banach spaces that contain a Lipschitz copy of every separable metric space. In Chapters IV and V, we change focus and control large distances instead. Now our requirement on the control that the map f : X Y makes on distances

9 3 is that ϕ 1 (t) as t. These maps were introduced by Gromov [Gr1] and are called coarse embeddings. They were introduced in order to study groups as geometric objects. Finitely generated groups are considered as metric spaces under the word distance. In relation to algebraic topology, Yu [Y] proved that a metric space with bounded geometry that coarsely embeds into a Hilbert space satisfies the coarse geometric Novikov conjecture. Later, Kasparov and Yu [KaY] strengthened this result by showing that it is enough for the metric space in question to admit a coarse embedding into a uniformly convex Banach space. Whether this was really a strengthening was not very clear though, because it is not apparent at first sight that there are uniformly convex Banach spaces that do not coarsely embed into a Hilbert space. This inspires the study we do in Chapters IV and V. In Chapter IV, we show that the uniformly convex Banach space l p (p>2) does not admit a coarse embedding into a Hilbert space. In Chapter V, we give an actual characterization of all quasi-banach spaces that coarsely embed into a Hilbert space. The techniques used here rely on the classical work of Schoenberg [S] regarding the theory of positive and negative definite kernels; and on the work of Aharoni, Maurey and Mityagin [AMM] regarding the study of uniform embeddings into a Hilbert space.

10 4 CHAPTER II BANACH SPACES UNIFORMLY HOMEOMORPHIC TO A QUOTIENT OF l p (1 <p<2) Let us recall the definition and the main property of a uniform homeomorphism. 2.1 Uniform homeomorphism Definition Amapfis called a uniform homeomorphism between two metric spaces X and Y if it is bijective, uniformly continuous, and its inverse is also uniformly continuous. If f and f 1 are actually Lipschitz maps, then we say f is a Lipschitz homeomorphism or a Lipschitz isomorphism. AmapΩ f called the modulus of uniform continuity of f gives a way to quantify the uniform continuity of f. It is defined for r [0, ) by Ω f (r)=sup{d(f(x),f(y)),d(x, y) r}. The map f is uniformly continuous if Ω f (r) 0whenr 0. When the function is Lipschitz, its modulus of uniform continuity has a linear form, i.e. there is a constant C>0sothatΩ f (r) Cr for all r 0. The Lipschitz behavior of a map is closer to a linear behavior than the uniform behavior is, so it is natural to Lipschitz-ize a uniformly continuous map. The following lemma (see e.g. [BeL, page 18]), which does this in an explicit manner, will be of importance to us throughout this work: Lemma (Lipschitz for large distances property). Let f : X Y be a uniformly continuous map. If the metric space X is metrically convex, then f is

11 5 Lipschitz for large distances, i.e. c>0, L< such that d X (x, y) c d Y (f(x),f(y)) Ld X (x, y). Proof: Recall that a metric space X is called metrically convex if for all x, y in X, and for all n N, one can find in X points x 0 = x, x 1,x 2,,x n = y such that d(x i,x i+1 ) d(x, y)/n. Call Ω f the modulus of uniform continuity of f, andletc>0. Let x, y X with d(x, y) c, andcallnthe smallest integer bigger than or equal to d(x, y)/c. Since X is metrically convex, we can find in X points x 0 = x, x 1,x 2,,x n =y such that d(x i,x i+1 ) c for all i =0,1,2,,n 1. Then by the triangle inequality in Y,wehave: n 1 d(f(x),f(y)) d(f(x i ),f(x i+1 )) i=0 n 1 Ω f (c) k=0 So we can take L =2Ω f (c)/c. =Ω f (c)n 2d(x, y) Ω f (c) c Remark A Banach space is convex, and hence metrically convex, so it follows from Lemma that a uniform homeomorphism f from a Banach space X onto a subset of another Banach space Y is Lipschitz for large distances. However, f(x) need not be a convex subset of Y, and hence we cannot always assume that f 1 is Lipschitz for large distances as well. Unless imposed by assumption, the only case where f 1 is guaranteed to be Lipschitz for large distances is when f is surjective, i.e. the two Banach spaces are uniformly homeomorphic.

12 6 2.2 Quotient mappings Definition A Banach space Y is called a linear quotient of a Banach space X if there is a linear bounded map T : X Y that is surjective. Recall the Open Mapping Theorem, which gives a very distinctive property of linear quotient mappings. Theorem (Open Mapping Theorem). Let X and Y be Banach spaces, and let T : X Y be a linear bounded map. If T is surjective, then there exists a constant c (0, ) such that: 1 c B Y T (B X ), (where B X and B Y represent the unit ball of X and Y respectively). In other words, linear quotient mappings are uniformly open in the sense that the image of any ball of a given radius centered at a point x contains a ball of proportional radius centered at T (x). This observation leads to the following definition for general nonlinear mappings. Definition Let X and Y be Banach spaces. A map f : X Y is called a uniform quotient map ([BJLPS], [J]) if it is uniformly continuous, surjective, and uniformly open in the sense that one can find a nondecreasing map ω :[0, ) [0, ) such that ω(r) > 0whenr>0, and x X, B(f(x),ω(r)) f(b(x, r)). ω is called the modulus of co-uniformity of f. If f is Lipschitz, and ω is such that ω(r) r c for some constant c independent of x and r, then fis called a Lipschitz quotient map, and the smallest constant c

13 7 which makes the inequality true for all r [0, ) is called its co-lipschitz constant [Gr2]. Let us remark that if X is a separable Banach space, and Y is a uniform quotient of X, theny is separable. In fact, all we need for this is the continuity and the surjectivity of the uniform quotient map. A uniform quotient f map between Banach spaces is also co-lipschitz for large distances [BJLPS]. Roughly speaking, it means that if the map were invertible, its inverse would have been Lipschitz for large distances. Formally, it means the following: Lemma (Co-Lipschitz for large distances property). Let f be a uniform quotient map between two Banach spaces X and Y. Then f is co-lipschitz for large distances, i.e. c>0, L (0, ) such that r c x X, B ( f(x), r ) f(b(x, r)). L Proof: Let f : X Y be a uniform quotient, with modulus of co-uniformity ω. Let r c, letx X,andlety B(f(x),ω(r)). Divide the segment [f(x),y]inton ω(r) segments y 0 = f(x),y 1,,y n = y with y i y i 1 ω(c)andn=. Now, ω(c) by the co-uniform condition on f, we can find points x 0 = x, x 1,,x n X such that f(x i )=y i and x i x i 1 cforevery i =1,,n. Using the triangle inequality in X,wethenseethat x x n nc 2 c ω(c) ω(r)andf(x n)=y.denotingl=2 c ω(c), we see that for every x X and for every r c, B(f(x), ω(r)) f (B(x, Lω(r))). Now if we write R = Lω(r), we see that R > 2c implies that r > c and hence ( B f(x), R ) f (B(x, R)). L

14 8 [BJLPS]): Now we can make use of an ultrapower argument to get the following (see Proposition Let X and Y be Banach spaces, and assume Y is a uniform quotient of X. Let U be a free ultrafilter on N. Then the ultrapower (Y ) U is a Lipschitz quotient of the ultrapower (X) U. Proof (see e.g. [BeL, page 273]): Let f : X Y be the uniform quotient map. We know that f is Lipschitz and co-lipschitz for large distances, so we can find a constant L (0, ) such that: B ( f(x), r ) f (B(x, r)) B (f(x),lr) L for every x X and every r 1. Then the maps f n (x) = f(nx) also satisfy the n same condition for every r 1. By taking limits along the ultrafilter U, weseethat n their ultraproduct f :(X) U (Y) U defined by f(x 1,x 2, )=(f 1 (x 1 ),f 2 (x 2 ), ) satisfies it for every r>0. Recall that a Banach space X is called uniformly convex if for every ɛ>0, one can find δ>0such that if x and y are two elements of the unit sphere of X with x y ɛ,then x+y 2 1 δ. Bates, Johnson, Lindenstrauss, Preiss and Schechtman [BJLPS] improved the previous proposition in the case of uniform quotients of a uniformly convex Banach space (see also [BeL, Theorem 11.18]): Theorem Assume Y is a Banach space that is a uniform quotient of a uniformly convex Banach space X. Then there is a constant c>0such that for each finite dimensional subspace F Y, there is a subspace F 1 X such that d(f, F 1 ) c. In particular, Y is isomorphic to a linear quotient of some ultrapower of X. (Here d(, ) represents the Banach-Mazur distance between Banach spaces.)

15 9 Idea of the proof: First, the previous proposition gives us a Lipschitz quotient map between ultrapowers. Now, since a Banach space and its ultrapowers have the same finite-dimensional subspaces, we can just reduce to f being a Lipschitz quotient map between X and Y. Bates, Johnson, Lindenstrauss, Preiss and Schechtman introduce and study a property of pairs of Banach spaces that they call UAAP (Uniform Approximation by Affine Property). A pair (X, Y ) is said to have the UAAP if for every ɛ>0there is a constant c>0such that for every ball B X with radius r, and for every Lipschitz map f : B Y with Lipschitz constant L, one can find a ball B 1 B (not necessarily cocentric to B) with radius r 1 >cr, and a continuous affine map g : X Y such that f(x) g(x) ɛr 1 L for every x B 1. Then they show that if X is a uniformly convex Banach space, and G is a finite-dimensional Banach space, then (X, G) has the UAAP. Now, for each finite dimensional subspace F of Y, they consider the finite dimensional space G = Y/F where F is the anihilator of F in Y. They apply the UAAP property of (X, G) to the composition map Q f where f : X Y is the given Lipschitz quotient map and Q : Y G is the canonical quotient map. The affine approximation of Q f that they get gives a linear map T : X G, andthen the Lipschitz and co-lipschitz conditions on Q f dualize and give that the adjoint map T : F X is an isomorphism into. As a corollary to this theorem, we have the following [BJLPS]: Corollary (Uniform quotients of l p ). Let X be a Banach space that is a uniform quotient of l p (1 <p< ). Then X is a linear quotient of L p [0, 1]. Proof: Recall that l p is uniformly convex when 1 <p<, hence we can apply

16 10 Theorem to get that X is a linear quotient of an ultrapower of l p.butsuchan ultrapower is isometric to L p (µ) for some measure µ [H]. Since X is separable, X is then a linear quotient of L p [0, 1]. To summarize what we have learned so far, some problems in the classification of uniform quotients of a Banach space can be transformed into linear problems. Since the linear classification of Banach spaces has been very extensively studied, this technique then enables us to use the wealth that the linear theory gives us. Some results that are of importance to us will be presented in the following section. 2.3 Results from the linear theory The following is a result of Johnson and Odell [JoO]: Theorem (Subspaces of L q that embed into l q ). Let 2 <q<, and let X be isomorphic to a subspace of L q.ifl 2 is not isomorphic to a subspace of X, then X embeds into l q. Another result that we will use is the Maurey Extension Theorem [M]: Theorem (Maurey Extension Theorem). Let X be a Banach space with type 2, and let Y beasubspaceofx.lethbe a Hilbert space. Then for every bounded linear operator T : Y H, there exists an operator S : X H which extends T and which satisfies S C T, where C is the type-2 constant of X. In particular, every subspace of X which is isomorphic to a Hilbert space is complemented (i.e. is the range of a bounded linear projection). The last result that we will use in this chapter is the following theorem of Johnson and Zippin [JoZ]:

17 11 Theorem If Y is a linear quotient of l p, then Y embeds into an l p -sum of finite-dimensional spaces. 2.4 Approximate metric midpoints In this section, we are going to present an aspect of the geometry of Banach spaces considered as metric spaces. Definition Given two points x and y in a Banach space X, andgivenδ 0, we define the metric midpoint of x and y with error δ by { } Mid(x, y, δ) = z X, max( x z, y z ) (1 + δ) x y 2. The following proposition gives a way to quantify the sizes of approximate midpoints in l p -type spaces ([BeL, Proposition 10.10]): Proposition Let 1 p<, and let X =( k F k) p be an l p -sum of the finite dimensional spaces (F k ) k. Then for every x X and every 0 <δ<1, 1. there is a finite-dimensional subspace F of X such that Mid(x, x, δ) F +2(pδ) 1 p x BX, 2. there is a finite-codimensional subspace Y of X such that Y δ 1 p x BX Mid(x, x, δ). Proof: Let 0 <δ<1, let x X, andletɛ>0. We can uniquely write x = x n, with x n F n for every n and x p = x n p. N Let N be such that n=1 x n p (1 ɛ) x p, and hence n=1 n>n Set F = span{f n,n =1,,N} and Y = span{f n,n > N}. And for every y X, write y uniquely as y = y F + y Y with y F F and y Y Y. n=1 x n p ɛ x p.

18 12 1. Let y = y F +y Y Mid(x, x, δ). By virtue of the triangle inequality, we cannot both have x F + y F < x F and x F y F < x F. So let us assume witout loss of generality that x F + y F x F. Since x + y (1 + δ) x, weget: (1 + δ) p x p x+y p = x F +y F p + x Y +y Y p x F p + x Y +y Y p (1 ɛ) x p + x Y + y Y p. This implies that x Y + y Y ((1 + δ) p (1 ɛ)) 1 p x and hence y Y ) (((1 + δ) p (1 ɛ)) 1 1 p + ɛ p x. Now, since 0 <δ<1, we have that (1 + δ) p 1+2 p 1 pδ, and hence we have ) (((1 + δ) p (1 ɛ)) 1 1 p + ɛ p ( 2 p 1 pδ + ɛ ) 1 p + ɛ 1 p, andthisissmallerthan2(pδ) 1 p if ɛ is chosen to be small enough. 2. Let y Y where Y is the finite-codimensional subspace described above. Assume y δ 1 p x.wehave: x F ±y p = x F p + y p x F p +δ x p. Letusconsiderfirstthecasep= 1. From the above we have x F ± y x F +δ x,

19 13 and hence by triangle inequality, x ± y x F +δ x + x Y =(1+δ) x. For the case p>1, let us use the trivial upper bound that x F x to get So x F ± y p (1 + δ) x p x F ± y (1 + δ) 1 p x. When ɛ is small enough, we have x ± y (1 + δ) 1 p x + xy ) ((1 + δ) 1 1 p + ɛ p x. ) ((1 + δ) 1 1 p + ɛ p (1 + δ). The idea of comparing the sizes of approximate midpoints in the study of the uniform classification of Banach spaces is due to Enflo (as is mentioned in [BeL, page 254]) when he proved that L 1 and l 1 are not uniformly homeomorphic. The main ingredient in being able to do this comparison is expressed in the following lemma (see [BeL, Lemma 10.11]). Lemma (Stretching Lemma). Let X and Y be Banach spaces and let f : X Y be uniformly continuous. Call K d the Lipschitz constant of f for distances larger than d, i.e. f(x) f(y) K d x y for every x, y X such that x y d. Define K = lim d K d. Assume K > 0. Then for every 0 <δ<1/2and for every D>0there are points x, y X with x y Dsuch that f(mid(x, y, δ)) Mid(f(x), f(y), 5δ).

20 14 Proof: Let δ (0, 1/2), and let d>0belarge enough so that K d 4 1+δ.Letx, y K d X with x y dbe such that f(x) f(y) K d x y,andletz Mid(x, y, δ). 1+δ x y Then by the triangle inequality we have that x z (1 δ) d and hence: 2 4 f(x) f(z) Kd x z 4 x y Kd(1 + δ) 4 2 K d 4 2 f(x) f(y) (1 + δ) K d 2 3 f(x) f(y) (1 + δ) 2 f(x) f(y) (1 + 5δ). 2 The previous lemma says that pairs of points that stretch the function f, i.e. at which f almost attains its Lipschitz constant, are such that their approximate midpoints are sent into approximate midpoints of their images, and hence by comparing the sizes of approximate midpoints we get the following result [JoLS]: Proposition Let 1 p<r<+. Suppose that X is an l r -sum of the finite-dimensional spaces (E n ) n, and Y is an l p -sum of the finite-dimensional spaces (F n ) n. Then it is impossible that there be a uniform embedding of X into Y whose inverse is colipschitz for large distances. Proof: Say j : X Y is a uniform embedding. Then we can find positive constants K and L such that whenever x y 1wehave: L x y j(x) j(y) K x y Fix 0 <δ<1/2andchoosedsuch that δ 1/r d>2. By the Stretching Lemma (Lemma 2.4.3), and via possible translations, we can find a point x X with x d

21 15 for which j(mid(x, x, δ)) Mid(j(x), j(x), 5δ). By the second part of Proposition for X we can find an infinite sequence (y k ) k Mid(x, x, δ) such that y k y l 1 2 δ1/r x. Thus y k y l 1by the choice of d, and j(y k ) j(y l ) L y k y l L 2 δ1/r x. The first part of Proposition for Y, and the finite-dimensionality of F give that there are k l with j(y k ) j(y l ) 5(5pδ) 1/p j(x) 5(5pδ) 1/p K x. Thus 5(5pδ) 1/p K L 2 δ1/r, which is impossible when δ is small enough because p<r. We now come to the main result of this chapter. 2.5 Main result Theorem Let 1 <p<2, and let Y be a linear quotient of l p and u : Y X be a uniform homeomorphism. Then X is also a linear quotient of l p. Proof: By the result of [BJLPS] stated in Corollary 2.2.7, we know that X is a linear quotient of L p, and so in particular it is reflexive. By dualizing the result in Theorem 2.3.1, we only need to show that l 2 is not a linear quotient of X. Bywayof contradiction, let us assume that is the case, and let Q : X l 2 be a linear quotient map. By taking adjoints, we get that l 2 embeds into X which then embeds into L p where 1 p + 1 p = 1 (so in particular p > 2). We can use Maurey s Extension Theorem (Theorem 2.3.2) to get that l 2 is complemented in X by a projection say P : X l 2. By taking the adjoint of P again, we get that l 2 embeds into X. On the other hand, Y is a linear quotient of l p, so by Theorem Y embeds linearly into an l p -sum of finite-dimensional spaces, say by an embedding I : Y

22 16 ( k F k) p. So I u 1 P is a uniform embedding of l 2 into ( k F k) p which is co-lipschitz for large distances, and we know from Proposition that this is impossible since p<2.

23 17 CHAPTER III BANACH SPACES CONTAINING A LIPSCHITZ COPY OF c 0 In this chapter, we are going to generalize the method we used in previous chapter via the use of two moduli introduced by Milman in 1971 [Mi]. These moduli were later investigated by Johnson, Lindenstrauss, Preiss and Schechtman in [JoLPS] and they called them moduli of asymptotic uniform convexity and of asymptotic uniform smoothness. We will recall the definitions and properties of these moduli in the first half of the chapter. In the second half, we will tie them to the method presented in the previous chapter. Throughout, we will restrict our attention to real Banach spaces. 3.1 Modulus of asymptotic uniform convexity Definition Let X be a Banach space. The modulus of asymptotic uniform convexity of X is the map δ X defined on the interval [0, 1] by δ X (t) = inf x S X sup Y cof(x) inf x + y 1, y Y y t where S X is the unit sphere of X and cof(x) is the family of all finite codimensional subspaces of X. The Banach space X is called asymptotically uniformly convex if δ X (t) > 0 for all 0 <t 1. This notion generalizes that of uniform convexity as is pointed out by the next Proposition [JoLPS]: Proposition If X is uniformly convex, then X is asymptotically uniformly convex.

24 18 Proof: Let 0 <t 1 and fix ɛ>0. Denote by δ X the modulus of uniform convexity of X, i.e. δ X (t) = inf 1 x,y B X x y t Since X is uniformly convex, we have δ X (t) > 0. x + y 2 Let x S X,letx be a norming functional for x, i.e. x S X and x (x) = x. Denote Y =ker(x ), and denote. δ X (x, Y, t) = inf x + y 1. y Y y t It suffices to show that δ X (t) δ X (x, Y, t)+ɛ. First of all, let us notice that δ X (x, Y, t) = inf x+y 1. y Y y =t In fact, if we fix y S Y and denote ϕ(t) = x+ty 1, then since Y =ker(x )we see that ϕ(t) x (x + ty) 1 = 0, i.e. ϕ attains its minimum at t =0. Sinceϕ is a convex function and ϕ(t) as t,wenoticethatϕis a nondecreasing function. Now, let y 0 Y be such that y 0 = t and 1 + δ X (x, Y, t)+ɛ> x+y 0.Call a= x+y 0 x+y 0 and b = a y 0.Then a b =t, a =1,and ( ) 1 b= x+y 0 x+ 1 1 ( y 0 ) is a convex combination of two elements of x+y 0 B X,so b 1. So we have:

25 19 a + b δ X (t) 1 ( 2 ) a + b 1 x x + y δ X (x, Y, t)+ɛ δ X (x, Y, t)+ɛ. So uniformly convex spaces are asymptotically uniformly convex. Notice however that the notion of asymptotic uniform convexity is a strict generalization of that of uniform convexity as we shall see next. Proposition (Modulus of AUC of l 1 ). For every t [0, 1], δ l1 (t) =t. Proof: As in the previous proof, notice that we have for any Banach space X δ X (t) = inf x S X sup Y cof(x) inf x + y 1, y Y y =t and hence by the triangle inequality we always have δ X (t) t. Conversely, let x =(x n ) n l 1 with x = x n =1. Fixɛ>0andletN be such that N x n > 1 ɛ. Denote by Y 0 the finite codimensional subspace of l 1 n=1 consisting of those vectors supported only on {n N,n > N},andlety Y 0 with y t.wehave: n=1

26 20 x + y = N x n + n=1 1 ɛ + n=n+1 n=n+1 1 2ɛ+ y 1 2ɛ+t. Hencewehave inf x + y 1 t 2ɛ. y Y 0 y t x n + y n ( y n x n ) Taking supremum over all finite codimensional subspaces, and then infimum over all unit vectors, we then get δ l1 (t) t 2ɛ. Since ɛ is arbitrary, we have the result. As we can see, l 1 is a nonreflexive Banach space which is asymptotically uniformly convex. Moreover, since for any Banach space X we have δ X (t) t, wesee that l 1 attains the best modulus of asymptotic uniform convexity. So we can think of l 1 as the most asymptotically uniformly convex space. 3.2 Modulus of asymptotic uniform smoothness Definition The modulus of asymptotic uniform smoothness of a Banach space X is the map ρ X defined on the interval [0, 1] by ρ X (t) = sup x S X inf Y cof(x) sup x + y 1, y Y y t where again S X is the unit sphere of X and cof(x) is the family of all finite codimensional subspaces of X. The Banach space X is called asymptotically uniformly smooth if ρ X(t) t as t 0. 0

27 21 Proposition If X is uniformly smooth, then X is asymptotically uniformly smooth. Proof: Recall that the modulus of uniform smoothness ρ X of X is x + y + x y ρ X (t) = sup 1, x =1 2 y t and that ρ X(t) 0ast 0sinceXis uniformly smooth. t Let x S X,letx S X be a norming functional for x and set Y 0 =ker(x ). For each y Y 0 with y twe have because x y x (x y)=1. 1 x + y +1 ( x + y 1) = x+y + x y 1 2 ρ X (t) Hence by taking supremum over all y Y 0 with y t, and then infimum over all finite codimensional subspaces, and then supremum over all x S X,weget 1 2 ρ X(t) ρ X (t). On the other hand, for every y Y 0 we have x + y 1 x (x+y) 1=0, so for any finite codimensional subspace Y of X we have Hencewealsohave and thus ρ X (t) 0. sup y Y Y 0 y t x + y 1 0. sup x + y 1 0, y Y y t Proposition (Modulus of AUS of c 0 ). The Banach space c 0 is asymptotically uniformly flat, namely ρ c0 (1) = 0.

28 22 Proof: From the previous proof, we always have for any Banach space X that ρ X (t) 0 for any t [0, 1]. Conversely, let x =(x n ) n c 0 with x =sup{ x n,n 1} =1. Fix0< ɛ<1, and let N > 1 be such that x n <ɛfor all n>n. Hence in particular max{ x n, 1 n N} =1. CallY 0 the finite codimensional subspace of c 0 consisting of those vectors supported on {n N,n>N},andlety Y 0 be such that y 1. We have: x + y =sup x n +y n n =max{sup x n, sup x n + y n } n N n>n max{1, sup ( x n + y n )} n>n max{1, sup x n +sup y n } n>n n>n max{1, 1+ɛ} 1+ɛ. Taking supremum, infimum, supremum as is now customary, we get that ρ c0 (1) ɛ. 3.3 Properties of the asymptotic moduli The following are properties of the moduli δ X and ρ X that come directly from their definitions [JoLPS]: Proposition For a Banach space X, 1. δ X and ρ X are nondecreasing. 2. for each t [0, 1], 0 δ X (t) ρ X (t) t.

29 23 3. if Y is an infinite-dimensional subspace of X, then δ X δ Y and ρ Y ρ X. Proof: 1. Directly from the definitions. 2. We have practically seen in the previous proofs that δ X (t) 0andρ X (t) t. For the middle inequality, let x S X and denote δ X,x (t) = ρ X,x (t) = sup Y cof(x) inf Y cof(x) inf x + y 1, y Y y t sup x + y 1. y Y y t Let ɛ>0, and let Y 0 and Y 1 be finite codimensional subspaces such that inf x + y 1>δ X,x (t) ɛ, y Y 0 y t sup x + y 1<ρ X,x (t)+ɛ. y Y 1 y t Then Y 0 Y 1 is also a finite codimensional subspace of X and inf y Y 0 Y 1 y =t In this last inequality, we have and similarly x + y 1 RHS sup y Y 0 Y 1 y t sup y Y 0 Y 1 y =t x + y 1 sup x + y 1 y Y 1 y t <ρ X,x (t)+ɛ, LHS > δ X,x (t) ɛ. x + y 1. Hencewealwayshaveδ X,x (t) ρ X,x (t) and the result follows immediately from here.

30 24 3. For a subspace Y of X, we have{z Y,Z cof(x)} cof(y ). Hence: δ Y (t) = inf y S Y inf y S Y inf y S Y sup W cof(y ) sup Z cof(x) sup Z cof(x) inf y + w 1 w W w t inf y + z 1 z Z Y z t inf y + z 1 z Z z t inf x S X =δ X (t). sup Z cof(x) inf x + z 1 z Z z t In a similar manner we also get ρ Y (t) ρ X (t). Remark From the previous Proposition we see that if Y is a subspace of X, then we have δ Y (t) ρ X (t). The purpose of this chapter is to show that we still have such an inequality up to some constant if X is just assumed to contain a Lipschitz copy of Y. 3.4 Asymptotic moduli and approximate midpoints In this section, we will relate the sizes of the approximate metric midpoints of points in the Banach space X with the moduli of asymptotic uniform convexity and smoothness of X. Proposition (Modulus of AUS and approximate midpoints). Let X be a Banach space. Let t [0, 1], letɛ>0and let x X. Then one can find a finite codimensional subspace Y of X such that t x B Y Mid(x, x, ρ X (t)+ɛ).

31 25 Proof: Without loss of generality, we can assume that x =1. Sinceɛ>0, we have: so in particular, sup u S X inf Y cof(x) sup u + y < 1+ρ X (t)+ɛ, y Y y t inf Y cof(x) sup x + y < 1+ρ X (t)+ɛ. y Y y t Hence we can find a finite codimensional subspace Y 0 of X such that sup x + y < 1+ρ X (t)+ɛ, y Y 0 y t or in other words we have for every y tb Y0 x + y < 1+ρ X (t)+ɛ. By the symmetry of the ball B Y0,wealsothenhave x y <1+ρ X (t)+ɛ. This means that tb Y0 Mid(x, x, ρ X (t)+ɛ). Proposition (Modulus of AUC and approximate midpoints). Let X be a real Banach space that is asymptotically uniformly convex. Let t (0, 1], let 0<ɛ<δ X (t)and let x X. Then one can find a compact subset K of X such that Mid(x, x, δ X (t) ɛ) K +3t x B X. Proof: Again without loss of generality, we may assume that x =1. Sinceɛ>0, we have: inf u S X sup Y cof(x) inf u + y > 1+δ X (t) ɛ, y Y y t

32 26 so in particular, sup Y cof(x) inf x + y > 1+δ X (t) ɛ. y Y y t Hence we can find a finite codimensional subspace Y 0 of X such that inf x + y > 1+δ X (t) ɛ, y Y 0 y t or in other words we have for every y Y 0 with y t, x+y >1+δ X (t) ɛ. Since y Y 0 and y t,wealsothenhave x y >1+δ X (t) ɛ. By contraposition, this means that if y Y 0 and at least one of x+y and x y is less than or equal to 1 + δ X (t) ɛ, then y <t. We then have the weaker statement: Y 0 Mid(x, x, δ X (t) ɛ) tb X. Denote C =Mid(x, x, δ X (t) ɛ). Then since ɛ<δ X (t), we notice that C is a convex closed symmetric bounded subset of X that contains a neighborhood of 0. So since we are working in a real Banach space, the Minkowski functional µ C of C defines an equivalent norm on X whose unit ball is C. In particular, (X, µ C )isa Banach space. We have an easy lemma (see e.g. [JoLPS]): Lemma Let X be a Banach space, and let Y be a finite codimensional subspace of X. Then one can find a compact subset K of X such that B X K +3B Y.

33 27 Proof: Denote by Q : X X/Y the canonical quotient map, and let ɛ = 1 4. We have 2+ɛ 1 ɛ =3. SinceX/Y is finite dimensional, we can find an ɛ-net (e i) i=1,,n of B X/Y. We can choose e i < 1 for each i {1,,n}, and so can choose points x i B X such that Q(x i )=e i.sete=span{x i,i=1,,n}. Let x B X, and pick i 0 {1,,n} such that Q(x) e i0 <ɛ, i.e. d(x x i0,y)<ɛ.picky Ysuch that x x i0 y <ɛ.then y ɛ+ x x i0 2+ɛ. As a result, we have written x as an element of B E +(2+ɛ)B Y +ɛb X. Now we have so by induction we get for every k 1 i.e. for every k 1 B X B E +(2+ɛ)B Y +ɛb X, B X (1 + ɛ + +ɛ k )(B E +(2+ɛ)B Y )+ɛ k+1 B X, B X Write x B X as x = s k + r k with s k ( ) 1 (B E +(2+ɛ)B Y )+ɛ k+1 B X. 1 ɛ ( 1 ) 1 ɛ So r k 0 and hence s k x. Since B E is compact, ( ) 1 closed and so x (B E +(2+ɛ)B Y ). ( 1 ɛ 1 Set K = 1 ɛ (B E +(2+ɛ)B Y )andr k ɛ k+1 B X. ( ) 1 (B E +(2+ɛ)B Y )is 1 ɛ ) B E and we have our result B X K +3B Y. Continuation of the proof of 3.4.2: Applying Lemma to the Banach space (X, µ C ) and the finite codimensional subspace Y 0, we can find a compact subset K of X such that B X,µC K +3B Y0,µ C. But B X,µC = C and B Y0,µ C = Y 0 C. And since Y 0 C = Y 0 Mid(x, x, δ X (t) ɛ) tb X,

34 28 we get that Mid(x, x, δ X (t) ɛ) K +3tB X. 3.5 Main result Theorem Let X and Y be real Banach spaces, and let f : X Y beauniform embedding which is co-lipschitz for large distances. Assume that Y is asymptotically uniformly convex. Then there exists a constant C 1 such that for each t (0, 1/C] we have: δ Y (t) 10ρ X (Ct). Proof: Let L be such that we have for all x, x X with x x 1: 1 L x x f(x) f(x ) L x x. Let t (0, 1], and notice that 0 < δ Y (t) 10 < 1. Use the Stretching Lemma 2 (Lemma 2.4.3) to find points x and x in X with x x aslargeaswepleasesuch that f ( ( Mid x, x, δ )) ( Y (t) Mid f(x),f(x ), δ ) Y(t) Via possible translations, we can assume without loss of generality that x = x and f(x )= f(x). Use ɛ = δ Y (t) in Proposition to produce a compact subset K of Y such 2 that ( Mid f(x),f(x ), δ ) Y(t) K+3t f(x) B Y, 2 i.e. ( Mid f(x),f(x ), δ ) Y(t) K +3tL x B Y. 2

35 29 Now, assume that we can find an ɛ>0 such that ρ X (28L 2 t)+ɛ< δ Y(t) 10. Then Mid ( x, x, ρ X (28L 2 t)+ɛ ) ( Mid x, x, δ ) Y (t), 10 and using Proposition we can then find a finite codimensional subspace X 0 of X such that f(28tl 2 x B X0 ) K +3tL x B Y. Now, since X 0 is infinite dimensional, we can find an infinite sequence (x n ) n in 28tL 2 x B X0 such that x n x m > 14tL 2 x for every n m. Since x was chosen to be large enough, we can use the co-lipschitz condition on f for distances larger than 1 to get for every n m f(x n ) f(x m ) > 14tL x. Write f (x n )asf(x n )=k n +y n where k n K and y n 3tL x. Then since K is compact, we can assume (by possibly passing to a further subsequence) that for every n m k n k m < 7tL x. Hence 6tL x y n + y m y n y m f(x n ) f(x m ) k n k m 14tL x 7tL x.

36 30 This is a contradiction, and hence we must have δ Y (t) 10 ρ X (28L 2 t). So we can take C =28L 2 and this finishes the proof. Corollary Assume that c 0 Lipschitz embeds into a real Banach space Y. Then Y cannot be asymptotically uniformly convex under any renorming. Remark A separable Banach space Y is called a Lipschitz universal space for separable metric spaces if every separable metric space admits a Lipschitz embedding into Y. It is well known that the Banach space C[0, 1] is a linear and therefore Lipschitz universal space. It was proved by Aharoni [A] that c 0 is Lipschitz universal for separable metric spaces. As a result, when we study a Banach space Y that is Lipschitz universal for separable metric spaces, it is enough to assume that c 0 Lipschitz embeds into Y.

37 31 CHAPTER IV COARSE EMBEDDING INTO A HILBERT SPACE In these last two chapters, we are going to concentrate on nonlinear maps that give a control on the distances when the distances are large. These maps are called coarse embeddings. The formal definition is given below (see [Gr1, 7.G]): 4.1 Coarse embeddings Definition A (not necessarily continuous) map f between two metric spaces (X, d) and(y,δ) is called a coarse embedding if there exist two non-decreasing functions ϕ 1 :[0, ) [0, ) andϕ 2 :[0, ) [0, ) such that 1. ϕ 1 (d(x, y)) δ(f(x),f(y)) ϕ 2 (d(x, y)) 2. ϕ 1 (t) as t. We will be interested mainly in coarse embeddings of Banach spaces into a Hilbert space. One of the main ingredients we need for this study is Schoenberg s classical work on positive definite functions [S]. We will recall the definitions and the results of most importance to our study in the next section. This overview can also be seen in [BeL, page 185] with slightly different proofs. Again, all Banach spaces and Hilbert spaces considered in this chapter are real. 4.2 Negative definite kernels and related notions Throughout this section, X will be an additive group, e.g. a Banach space. Definition A map K:X X Ris called a positive definite kernel on X if x, y X, K(x, y) =K(y, x),

38 32 and n c 1,c 2,,c n R, x 1,x 2,,x n X, c i c j K(x i,x j ) 0. i,j=1 Amapf :X Ris called a positive definite function on X if the map (x, y) f(x y) is a positive definite kernel. A positive definite kernel K is called normalized if K(x, x) = 1 for each x X. Similarly, a positive definite function f is normalized if f(0) = 1. Definition A map N:X X Ris called a negative definite kernel on X if x, y X, N(x, y) =N(y, x), and x 1,x 2,,x n X, c 1,c 2,,c n R, n n c i =0 c i c j N(x i,x j ) 0. i=1 i,j=1 Amapg:X Ris called a negative definite function on X if the map (x, y) g(x y) is a negative definite kernel. A negative definite kernel N is called normalized if N(x, x) = 0 for each x X, and a negative definite function g is normalized if g(0) = 0. Examples Some examples of negative and positive definite kernels are the following: The constant kernel (x, y) 1 is both a positive and a negative definite kernel. Let H be a Hilbert space. Let K and N be the kernels defined on H by (x, y) K(x, y) = x, y and (x, y) N(x, y) = x y 2. Then K is a positive definite kernel, and N is a negative definite kernel. In fact:

39 33 1. Let c 1,c 2, c n R.Wehave: n n c i c j K(x i,x j )= c i c j x i,x j i,j=1 i,j=1 n n = c i x i, c j x j i=1 j=1 n 2 = c i x i 0. i=1 2. Now assume n c i =0. Wehave: i=1 n n c i c j N(x i,x j )= c i c j x i x j 2 i,j=1 i,j=1 n ( = c i c j xi 2 + x j 2 2 x i,x j ) i,j=1 ( n )( n ) ( n )( n ) = c i x i 2 c j + c j x j 2 c i i=1 j=1 j=1 i=1 n 2 c i c j x i,x j i,j=1 =0+0 2 n c i c j K(x i,x j ) i,j=1 0. One of the most important of Schoenberg s results that we will use here is the one showing that there are no other examples. More precisely, we have the following theorem. Theorem (Schoenberg s Theorem). 1. K is a positive definite kernel on X if and only if there exists a Hilbert space H and a map T : X H such that K(x, y) = T(x),T(y) for every x, y X.

40 34 2. N is a normalized negative definite kernel on X if and only if there exists a Hilbert space H and a map T : X H such that N(x, y) = T(x) T(y) 2 for every x, y X. Proof: 1. Let V K be the linear vector space of all finitely supported real-valued maps on X, and define on V K f,g = f(x)g(y)k(x, y). x,y X Then, is an inner product on the quotient V K /N K where N K = {f V K, f,f =0}. Let H K be the completion of V K /N K,thenH K is a Hilbert space. Define T K : X H K by T K (x) being the equivalence class of the map δ x where δ x (y) =1ifx=yand δ x (y) =0ifx y.wehave: K(x, y) = δ x (u)δ y (v)k(u, v) u,v X = δ x,δ y = T K (x),t K (y). 2. Now let V N be the linear vector space of all finitely supported real-valued maps f on X that satisfy x X f(x) = 0, and define on V N f,g = 1 2 f(x)g(y)n(x, y). x,y X Then, is a semi-inner product. Let N N = {f V N, f,f =0},andletH N be the completion of V N /N N.ThenH N is a Hilbert space.

41 35 Define T N : X H N by T N (x) being the equivalence class of the map δ x where δ x is defined as above. We have for every x and y: δ x,δ y = 1 δ x (u)δ y (v)n(u, v) 2 = 1 N(x, y). 2 So δ x 2 = δ y 2 =0sinceN is normalized, and hence u,v T N (x) T N (y) 2 = δ x δ y 2 = 2 δ x,δ y =N(x, y). The relationship between negative and positive definite kernels is expressed in the following lemma: Lemma (Schoenberg s Lemma). The kernel N is negative definite if and only if the kernel e tn is positive definite for every t>0. Proof: Assume that e tn is positive definite for every t>0. Then e tn is negative definite. Since the constant kernel (x, y) 1 is negative definite, we also have that 1 e tn is negative definite. Since we have t 1 e tn N = lim, t 0 t we get that N is a negative definite kernel. For the converse, we can assume without loss of generality that t = 1 2, i.e we want to show that if N is negative definite then e N 2 is positive definite. By Schoenberg s Theorem, write N(x, y) = T(x) T(y) 2 where T is a map from X to some Hilbert space H. So N(x, y) = T(x) 2 + T(y) 2 2 T(x),T(y).

42 36 {1,,n}: Let x 1,x 2,,x n X and let c 1,c 2,,c n R. We have for every i, j c i c j e 1 2 N(x i,x j ) = c i e 1 2 T (x i) 2 c j e 1 2 T (x j) 2 e T (x i),t (x j ). So by setting for every i, c i = c ie 1 2 T (x i) 2, it suffices to prove that the kernel (x, y) e T (x),t (y) is positive definite. Writing this in power series, it then suffices to prove that (x, y) T (x),t(y) k is positive definite for every k 0. This we prove by induction. Assume that for k 2, the kernel (x, y) T (x),t(y) k 1 is posistive definite. Let x 1,x 2,,x n X and let c 1,c 2,,c n R. Since the kernel (x, y) T (x),t(y) is positive definite, the matrix ( T (x i ),T(x j ) ) i,j=1,,n is a positive definite matrix. Hence we can find a matrix (a ij ) i,j such that for each i and j, we have: n T (x i ),T(x j ) = a il a jl, and n n n c i c j T (x i ),T(x j ) k = c i a il c j a jl T(x i ),T(x j ) k 1. i,j=1 l=1 i,j=1 l=1 And this is nonnegative because the inner sum is always nonnegative from the induction hypothesis. One last property that we will use is the following: Proposition Let N be a negative definite kernel on X satisfying N(x, y) 0 for every x, y X. Then for every 0 <λ<1, the kernel N λ is also negative definite. Proof: To see this we use the classical identity x λ 1 e tx = c λ dt, 0 t λ+1

43 37 where c λ is the constant ( c λ = 0 ) 1 e u 1 dt. u λ+1 (Make the change of variable u = tx.) So then N λ 1 e tn = c λ dt, 0 t λ+1 and the integrand is negative definite by applying Schoenberg s Lemma to N. Hence N λ is also negative definite. 4.3 Extensions In this section, we present another ingredient of importance to our study in this chapter, and that is the problem of extension. We have a metric space X and we have a map f : S Y defined on a subset S of X into some metric space Y. This map f is assumed to be Lipschitz or α-hölder for some λ, and the question is to study the structure of X and/or Y under which such a map can always be extended to the whole space X, the extension being also Lipschitz or α-hölder respectively (with the same α). Such problems have been widely studied, one of the latest results being that of Naor, Peres, Schramm, Sheffield [NPSS] about extending Lipschitz maps from a subset of l p (p>2) to a Hilbert space. The result that we will use in this chapter is the following [WW, last statement of Theorem 19.1]: Theorem Let X beametricspace,andlet0<α 1.Letf:S Hbe an 2 α-hölder map from a subset S of X toahilbertspaceh, i.e. there exists a constant C such that for every x, y S one has: f(x) f(y) Cd(x, y) α.

44 38 Then one can find a map f : X Y such that f extends f and satisfying for every x, y X: f(x) f(y) Cd(x, y) α. 4.4 Positive definite functions on l p, p>2 Definition Let X be a Banach space with a normalized Schauder basis (e n ) n. Assume that (e n ) n is 1-symmetric, i.e. for any choice of signs (θ n ) n { 1,+1} and any choice of permutation σ : N N, n θ n a n e σ(n) X = n a n e n X. A function g : X R is called symmetric if it satisfies for any choice of signs (θ n ) n { 1,+1} and any choice of permutation σ : N N the equality: ( ) ( ) g θ n a n e σ(n) = g a n e n. n n Aharoni, Maurey and Mityagin proved the following theorem [AMM] which characterizes the symmetric continuous positive definite functions on a Banach space X with a symmetric basis behaving like that of l p, p>2. Theorem Let X be a Banach space with a symmetric basis (e i ) i 1. Assume that lim inf n e 1 + e 2 + +e n n =0. Then every symmetric continuous positive definite function on X is constant. 4.5 Main result The following is our main result:

45 39 Theorem Suppose that a Banach space X has a normalized symmetric basis n (e n ) n and that lim inf 1 n n 2 e i =0. Then X does not coarsely embed into a Hilbert i=1 space. We present the proof in five steps. Step (Reducing to the α-hölder case). Let f : X H be a coarse embedding satisfying 1. ϕ 1 ( x y ) f(x) f(y) ϕ 2 ( x y ) 2. ϕ 1 (t) as t. Our first claim is that we do not lose generality by assuming that ϕ 2 (t) =t α with 0 <α 1 2. To prove this claim, note first that (x, y) f(x) f(y) 2 is a negative definite kernel on X. This follows from Schoenberg s Theorem (Theorem 4.2.4). So by Proposition 4.2.6, for any 0 <α<1, N(x, y) = f(x) f(y) 2α is also negative definite kernel which is normalized. As a result, Schoenberg s Theorem again allows us to find a Hilbert space H α and a function f α : X H α such that N(x, y) = f α (x) f α (y) 2. On the other hand, since X, being a normed space, is (metrically) convex, the original function f : X H is Lipschitz for large distances (same proof as 2.1.2). Consequently, without loss of generality, we can assume by rescaling that we have the following for x y 1: f(x) f(y) H x y and (ϕ 1 ( x y )) α f α (x) f α (y) Hα x y α.

46 40 Now, let S be a 1-net in X (i.e. S is a maximal 1-separated subset of X). The restriction of f α to S is α-hölder, so if 0 <α 1 2, then we can extend f α to an α-hölder map f α defined on the whole of X by applying Theorem 4.3.1: f α : X H α x S, fα (x) =f α (x) and x, y X, f α (x) f α (y) Hα x y α. Now, write ϕ 1 (t) =inf{ f α (x) f α (y), x y t}, and let us make sure that ϕ 1 (t) as t. To this end, let x, y X and find x S,y S S such that x x S < 1and y y S <1. We have: f α (x) f α (y) f α (x S ) f α (y S ) f α (x) f α (x S ) f α (y) f α (y S ) = f α (x S ) f α (y S ) f α (x) f α (x S ) f α (y) f α (y S ) (ϕ 1 ( x S y S )) α x x S α y y S α (ϕ 1 ( x S y S )) α 2 as x y. This finishes the proof of our reduction to the case where f is α-hölder and thus uniformly continuous. So from now on we will assume that our coarse embedding

47 41 is a map f : X H satisfying the following for all x, y X: ϕ 1 ( x y ) f(x) f(y) x y α where ϕ 1 (t) as t. Step (Controlling the growth of the negative definite kernel). Set N(x, y) = f(x) f(y) 2. Then N is a normalized (i.e. N(x, x) = 0) negative definite kernel on X. Now if we write φ 1 (t) =(ϕ 1 (t)) 2 and φ 2 (t) =t 2α,thenN satisfies: φ 1 ( x y ) N(x, y) φ 2 ( x y ), φ 1 (t) as t. Step (Reducing to a negative definite function). The argument in this paragraph comes from [AMM, Lemma 3.5.]. Let µ be an invariant mean on the bounded functions on X (see e.g. [BeL]). We can think of µ as a finitely additive translation invariant positive measure for which µ(x) = 1. Define: g(x)= N(y+x, y) dµ(y) Then we have the following for g: X g is well-defined because the map y N(y + x, y) is bounded for each x X, g(0) = N(y,y) dµ(y)=0, X

48 42 For scalars (c i ) 1 i n satisfying n n c i =0,wehave: i=1 i,j=1 c i c j g(x i x j )= i,j c i c j N(y + x i x j,y)dµ(y) X n = c i c j N(y + x i,y+x j )dµ(y) i,j=1 X ( n ) = c i c j N(y + x i,y+x j ) dµ(y) X i,j=1 = ( 0) dµ(y) X 0. This is because µ is translation invariant, and N is negative definite. This shows that g is a negative definite function on X. Finally, since dµ(y) = 1, we have: X φ 1 ( x ) g(x) φ 2 ( x ). In summary, we have found a negative definite function g on X which satisfies g(0) = 0 and φ 1 ( x ) g(x) φ 2 ( x ), where φ 1 (t) as t. Step (Reducing to a continuous symmetric negative definite function). Let (e n ) n be the normalized symmetric basis for X. This means that for any choice of signs (θ n ) n { 1,+1} and any choice of permutation σ : N N, n θ n a n e σ(n) X = n a n e n X.