There is no finitely isometric Krivine's theorem
aa r X i v : . [ m a t h . F A ] A ug There is no finitely isometric Krivine’s theorem
James Kilbane and Mikhail I. OstrovskiiAugust 7, 2017
Abstract.
We prove that for every p ∈ (1 , ∞ ), p = 2, there exist a Banach space X isomorphic to ℓ p and a finite subset U in ℓ p , such that U is not isometric to a subsetof X . This result shows that the finite isometric version of the Krivine theorem(which would be a strengthening of the Krivine theorem (1976)) does not hold. Keywords: isometric embedding, isomorphism of Banach spaces, Krivine theorem, Orlicz space, modularspace
Primary: 46B03; Secondary: 30L05, 46B07, 46B85
One of the most fundamental results on the structure of the general infinite-dimensio-nal Banach spaces is the following theorem of Dvoretzky.
Theorem 1.1 (Dvoretzky [Dvo61]) . For each infinite-dimensional Banach space X ,each n ∈ N , and each ε > , there is an n -dimensional subspace X n ⊂ X and anisomorphism T : X n → ℓ n such that || T || · || T − || ≤ ε . It is well known that (1+ ε ) cannot be replaced by 1 in this theorem. This follows,for example, from the fact that the unit ball of any finite-dimensional subspace in c is a polytope. The fact that ℓ p does not contain all of ℓ n isometrically, unless p isan even integer, was proved in [DJP98].Recently embeddability of finite metric spaces into Banach spaces became a veryimportant direction in the Banach space theory. One of the main reasons for thisis the discovery that such embeddings have important algorithmic applications, see[LLR95, AR98]. Low-distortion embedding of finite metric spaces into Banach spacesbecame a very powerful toolkit for designing efficient algorithms, the interestedreader can find more information in surveys such as [Ind01, IM04, Lin02, Mat02,Nao10, Ost13, WS11].In this connection it is worthwhile to observe that Theorem 1.1 can be derivedfrom the following seemingly weaker theorem. Our terminology follows [Ost13]. Theorem 1.2 (Finite Dvoretzky Theorem) . For each infinite-dimensional Banachspace X , each finite subset F ⊂ ℓ , and each ε > , there is a bilipschitz embeddingof F into X with distortion at most (1 + ε ) . roof of (Theorem 1.2) ⇒ (Theorem 1.1) . We are going to use ultrapowers of Banachspaces (see [Ost13, Section 2.2]). We need to show that if the conclusion of Theorem1.2 holds for a Banach space X , then there exists an ultrapower of X containing anisometric copy of ℓ . This can be done as follows (this is a slightly modified versionof [Ost13, Proposition 2.33]).Denote by J the set of all finite subsets of ℓ containing 0. Consider the set I = J × (0 ,
1) as an ordered set: ( j , ε ) (cid:23) ( j , ε ) if and only if j ⊇ j and ε ≤ ε .Consider an ultrafilter U on I containing the filter generated by sets of the form { ( j, ε ) : ( j, ε ) (cid:23) ( j , ε ) } , where j ∈ J , ε ∈ (0 , j, ε ) ∈ J × (0 ,
1) thereis a map T ( j,ε ) : j → X with distortion ≤ ε satisfying T (0) = 0.It remains to observe that the maps z ( T ( j,ε ) ( z ) if z ∈ j z / ∈ j (parameterized by pairs ( j, ε ) ∈ I ) induce an isometric embedding of ℓ into X U .An important difference between Theorem 1.1 and Theorem 1.2 is that there areno known examples showing the necessity of the + ε in Theorem 1.2. The examplesabove showing the necessity of the + ε in Theorem 1.1 do not serve as examples inthe finite case. In fact, Fr´echet [Fre10] (see also [Ost13, Proposition 1.17]) provedthat each n -element set embeds isometrically into ℓ n ∞ , and thus, into c . Ball [Bal90]proved that each n -element subset of L p embeds isometrically into ℓ ( n ) p . Since, as iswell known [JL01, p. 16] ℓ embeds isometrically into L p [0 ,
1] for every p , it followsthat for X = ℓ p the statement of finite Dvoretzky theorem remains true if we replace(1 + ε ) by 1.In this connection, the second-named author asked whether the result which canbe called “finite isometric Dvoretzky theorem” is true for all infinite-dimensionalBanach spaces X , that is, Problem 1.3 ([Ost15a]) . Does there exist a finite subset F of ℓ and an infinite-dimensional Banach space X such that F does not admit an isometric embeddinginto X ? This problem remains open. In this paper we show that the result which couldbe called “finite isometric Krivine theorem” does not hold for any p ∈ [1 , ∞ ], p = 2.More precisely, we answer in the negative, for every p ∈ (1 , ∞ ), p = 2, the followingproblem suggested in [Ost15b]: Problem 1.4 ([Ost15b]) . Let Y be a Banach space isomorphic to ℓ p , < p < ∞ .Is it true that any finite subset of ℓ p is isometric to some finite subset of Y ? To justify the term “finite isometric Krivine theorem” let us recall the followinglandmark result of Krivine. heorem 1.5 (Krivine [Kri76]) . For each p ∈ [1 , ∞ ] , each Banach space X isomor-phic to ℓ p , each n ∈ N , and each ε > , there is an n -dimensional subspace X n ⊂ X and an isomorphism T : X n → ℓ np such that || T || · || T − || ≤ ε . The cases p = 1 and p = ∞ were not included in Problem 1.4 because Bill Johnsonhad already described examples in these cases in his answer to [Kil15]. The examplesare the following: both ℓ and ℓ ∞ are isomorphic to strictly convex spaces. On theother hand, both ℓ and ℓ ∞ contain quadruples of points a, b, c, d such that b = c and both b and c are metric midpoints between a and d . It is easy to see that suchquadruples do not exist in strictly convex Banach spaces.It is worth mentioning that although we prove that the answer to Problem 1.4 isnegative, there exist “many” subsets of the unit sphere of ℓ p for which the result ispositive, see the paper [Kil17+] of the first-named author for precise statement.Our main result is the following (we denote by { e i } the unit vector basis of ℓ p ): Theorem 1.6. (a)
For each < p < there exist a Banach space X isomorphic to ℓ p such that the set U := { e , e , − e , − e , } , considered as a subset of ℓ p , does notembed isometrically into X . (b) For each < p < ∞ there exist a Banach space X isomorphic to ℓ p such thatthe set V := {± − /p ( e + e ) , ± − /p ( e − e ) , } , considered as a subset of ℓ p , doesnot embed isometrically into X .Remark . The coefficient 2 − /p in the statement of (b) is needed to make thevectors normalized (this will be convenient), of course the result holds without thiscoefficient.The main technical tools we will use in the proof of Theorem 1.6 are the Clarksoninequalities. In the following theorem, if q ∈ (1 , ∞ ) we set q ′ to be the so-calledconjugate index of q , defined by q + q ′ = 1. We recall the following (see [Gar07,Theorem 9.7.2] for generalized Clarkson inequalities): Theorem 1.8 ([Cla36, Theorem 2]) . Suppose that x, y ∈ ℓ p , where < p < ∞ , and r = min( p, p ′ ) . Then, (1) 2( k x k r ′ p + k y k r ′ p ) ≤ k x + y k r ′ p + k x − y k r ′ p ≤ r ′ − ( k x k r ′ p + k y k r ′ p )(2) 2 r − ( k x k rp + k y k rp ) ≤ k x + y k rp + k x − y k rp ≤ k x k rp + k y k rp ) .Remark . The following remark is for the reader who knows the definition ofthe
James constant of a Banach space, which is defined as J ( X ) = sup { min( k x + y k , k x − y k ) : x, y ∈ S X } . One can unify parts (a) and (b) of Theorem 1.6 in termsof the following: Theorem 1.10.
For each p = 2 there is a Banach space X such that X is isomorphicto ℓ p , J ( X ) = J ( ℓ p ) , but the supremum in the definition of the James constant isnot attained. Proving Theorem 1.10 one can use some of the results of [RR02] and [Yan07]. Tomake our argument as elementary and self-contained we prefer to present a directargument in terms of the metric spaces U and V . The case p ∈ (1 , We show that in this case we can choose X to be an Orlicz sequence space ℓ M fora suitably chosen function M ( t ). Let us recall the definition of an Orlicz sequencespace. Definition 2.1.
Let M : [0 , ∞ ) → [0 , ∞ ) be a continuous, non-decreasing andconvex function such that M (0) = 0 and lim t →∞ M ( t ) = ∞ . We define the sequencespace ℓ M to be the collection of sequences x = ( x , x , . . . ) such that P M ( | x n | /ρ ) < ∞ for some ρ and define the norm k x k M to be k x k M = inf ( ρ > ∞ X i =1 M (cid:18) | x i | ρ (cid:19) ≤ ) We refer to [LT77, Section 4.a] for basic properties of Orlicz sequence spaces,however, our proof will require very little of this theory to understand.Let p ∈ (1 , r ∈ ( p,
2) and let M ( t ) = t p + t r . We show that thecorresponding Orlicz space ℓ M has all of the desired properties. The fact that ℓ M isisomorphic to ℓ p follows immediately from [LT77, Proposition 4.a.5].Assume that ℓ M does not have the second property, that is, assume that U admitsan isometric embedding f into ℓ M . Without loss of generality we may assume that f (0) = 0. Denote f ( e ) by x and f ( e ) by y . It is easy to see that, since ℓ M is astrictly convex space [RR91, Chapter VII], we have f ( − e ) = − x and f ( − e ) = − y .We have || x || = || y || = 1. So we need to get a contradiction by showing that it isnot possible that both of the vectors: x + y and x − y have norm 2 p in ℓ M .Since x, y ∈ S ℓ M , we have P ∞ i =1 | x i | p + P ∞ i =1 | x i | r = 1 and P ∞ i =1 | y i | p + P ∞ i =1 | y i | r =1. We can write this as k x k pp + k x k rr = 1 and k y k pp + k y k rr = 1, where by k · k p wedenote the norm of a sequence in ℓ p . Adding these equalities we get2 = k x k pp + k y k pp + k x k rr + k y k rr . By the Clarkson inequality ((2) of Theorem 1.8), we get4 ≥ k x + y k pp + k x − y k pp + k x + y k rr + k x − y k rr (1)Denote u = k x + y k M and v = k x − y k M , and note that k x + y k pp u p + k x + y k rr u r = 1and k x − y k pp v p + k x − y k rr v r = 1 . Suppose that u = v = 2 /p , i.e., the embedding described above is isometric. Weget that 12 k x + y k pp + 12 r/p k x + y k rr = 1 nd 12 k x − y k pp + 12 r/p k x − y k rr = 1 . Doubling and adding gives k x + y k pp + k x − y k pp + 2 − rp k x + y k rr + 2 − rp k x − y k rr = 4Since 2 − r/p <
1, we get a contradiction with (1). p ∈ (2 , ∞ ) This case is more difficult. The reason is the following: to get a counterexample inthe above we “perturbed” ℓ p slightly “in the direction of ℓ ”. While for p < t r to the Orlicz function corresponding to ℓ p , this is nolonger possible for p > t p + t r with r < p is isomorphicto ℓ r , and not to ℓ p ), for this reason we have to consider more complicated, so-called modular spaces .Let us recall the definition of modular spaces: Definition 3.1.
For each i ∈ N , let M i : [0 , ∞ ) → [0 , ∞ ) be a continuous, convex,non-decreasing and convex function such that M (0) = 0 and lim t →∞ M ( t ) = ∞ .Then the modular sequence space ℓ { M i } is the Banach space of all sequences x = { x i } ∞ i =1 with P ∞ i =1 M i ( | x i | /ρ ) < ∞ for some ρ >
0, equipped with the norm || x || = inf ( ρ > ∞ X i =1 M i (cid:18) | x i | ρ (cid:19) ≤ ) . We refer to [LT77, Section 4.d] for basic information on modular sequence spaces.Let p >
2. We introduce a sequence { M i } ∞ i =1 of functions given by M i ( t ) = t p + t p i where 2 < p i < p for any i , and the sequence p i converges to p rapidly enough,so that the obtained modular space ℓ { M i } is isomorphic to ℓ p . To see that this isachievable, we recall the following criterion from [LT77, p. 167]: if M i and N i are twocollections of functions having the properties in Definition 3.1, then ℓ { M i } and ℓ { N i } are isomorphic with the identity map being an isomorphism if there exist numbers K > t i ≥ i = 1 , , . . . , and an integer i so that(a) K − N i ( t ) ≤ M i ( t ) ≤ KN i ( t ) for all i ≥ i and t ≥ t i .(b) P ∞ i =1 N i ( t i ) < ∞ .We are going to apply this criterion with N i ( t ) = t p for every i ∈ N . We choose t i > P ∞ i =1 t pi < ∞ (so (b) is satisfied).Finally, we let i = 1 and choose the sequence { p i } ∈ (2 , p ) so rapidly approaching p ,that the condition (a) is satisfied with K = 3. It is easy to see that this is possible. e wish to use an argument similar to the argument in Section 2. To do this wewill need to first prove the strict convexity of ℓ { M i } . To show this, suppose that wepick two distinct elements u, v ∈ S ℓ { Mi } . This means that ∞ X i =1 ( | u i | p + | u i | p i ) = || u || pp + ∞ X i =1 | u i | p i = 1 (2)and ∞ X i =1 ( | v i | p + | v i | p i ) = || v || pp + ∞ X i =1 | v i | p i = 1 (3)Adding these together, and using the Clarkson inequality ((1) of Theorem 1.8),we get 2 − p ( k u + v k pp + k u − v k pp ) + ∞ X i =1 − p i ( | u i + v i | p i + | u i − v i | p i ) ≤ k u + v k <
2, assume that k u + v k = 2, that is,2 − p k u + v k pp + ∞ X i =1 − p i | u i + v i | p i = 1 . Multiplying by 2 and comparing with Equation (4) is a contradiction, therefore k u + v k < V does not isometrically embed into ℓ { M i } .Observe that the distance in ℓ p between any of the vector ± − /p ( e + e ) andany of the vectors ± − /p ( e − e ) is equal to 2 − p . Assume that V admits anisometric embedding f into ℓ { M i } . Without loss of generality we may assume that f (0) = 0. Set x = f (2 − /p ( e + e )) and y = f (2 − /p ( e − e )). By the strictconvexity we get − x = f ( − − /p ( e + e )) and − y = f ( − − /p ( e − e )). We have || x || { M i } = || y || { M i } = 1. To complete the proof it suffices to show that || x − y || { M i } = || x + y || { M i } = 2 − p leads to a contradiction. This gives us that k x + y k pp p − + ∞ X i =1 | x + y | p i p i (1 − p ) = 1and k x − y k pp p − + ∞ X i =1 | x − y | p i p i (1 − p ) = 1Adding and rearranging we get − p ( k x + y k pp + k x − y k pp ) + ∞ X i =1 p i ( p − ( | x i + y i | p i + | x i − y i | p i ) = 2 (5)Since 2 < p i < p , and therefore 1 − p i > p i ( p − S ℓ { Mi } , so we are free to set u = x and v = y ) and (5)contradict each other. The second-named author gratefully acknowledges the support by National ScienceFoundation DMS-1700176. [AR98] Y. Aumann, Y. Rabani, An O (log k ) approximate min-cut max-flow theorem and approxi-mation algorithm, SIAM J. Comput. , (1998), no. 1, 291–301.[Bal90] K. Ball, Isometric embedding in ℓ p -spaces, European J. Combin. , (1990), no. 4, 305–311.[Cla36] J. A. Clarkson, Uniformly convex spaces, Trans. Amer. Math. Soc. , (1936), no. 3, 396–414.[DJP98] F. Delbaen, H. Jarchow, A. Pe lczy´nski, Subspaces of L p isometric to subspaces of l p . Posi-tivity (1998), no. 4, 339–367.[Dvo61] A. Dvoretzky, Some results on convex bodies and Banach spaces, in: Proc. Internat. Sym-pos. Linear Spaces (Jerusalem, 1960), pp. 123–160, Jerusalem Academic Press, Jerusalem;Pergamon, Oxford, 1961; Announcement: A theorem on convex bodies and applications toBanach spaces,
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Acta Math. Sin. (Engl.Ser.) (2007), no. 5, 827–846. (J.K.) Department of Pure Maths and Mathematical Statistics, University ofCambridge, Cambridge, CB3 0WB, UK E-mail address : [email protected] (M.O.) Department of Mathematics and Computer Science, St. John’s Univer-sity, 8000 Utopia Parkway, Queens, NY 11439, USA E-mail address : [email protected]@stjohns.edu