aa r X i v : . [ m a t h . M G ] A ug UNIFORM NONEXTENDABILITY FROM NETS
ASSAF NAOR
Abstract.
It is shown that there exist Banach spaces
X, Y , a 1-net N of X and a Lipschitz function f : N → Y such that every F : X → Y that extends f is not uniformly continuous. Introduction
A metric space (
X, d X ) is said to embed uniformly into a metric space ( Y, d Y ) if there exists aninjection f : X → Y such that both f and f − are uniformly continuous. ( X, d X ) is said [5] to embedcoarsely into ( Y, d Y ) if there exists f : X → Y and nondecreasing functions α, β : [0 , ∞ ) → [0 , ∞ )with lim t →∞ α ( t ) = ∞ such that α ( d X ( x, y )) d Y ( f ( x ) , f ( y )) β ( d X ( x, y )) for every x, y ∈ X .While making no attempt to survey the very large literature on these topics, we only indicate herethat in addition to their intrinsic geometric interest, uniform and coarse embeddings have importantapplications in areas ranging from functional analysis [4] to group theory and topology [17], andtheoretical computer science [2].In the context of embeddings of Banach spaces, the literature suggests that uniform and coarseembeddings are closely related, despite dealing with infinitesimal and large-scale structures, respec-tively. Specifically, by [1, 6, 15] a Banach space X embeds uniformly into a Hilbert space if and onlyif it embeds coarsely into a Hilbert space. Also, certain obstructions work equally well [7, 11, 9]for ruling out both uniform and coarse embeddings of Banach spaces. Despite this, it remainsunknown whether or not the existence of a coarse embedding of a Banach space X into a Banachspace Y implies that X also embeds uniformly into Y . The analogous question with the roles ofcoarse and uniform embeddings interchanged is open as well. The only available negative result inthis context treats uniform and coarse equivalences rather than embeddings: Kalton [8] proved theexistence of two Banach spaces X, Y that are coarsely equivalent but not uniformly equivalent.Recent work of Rosendal [16] yields progress towards the above questions. It implies that if X and Y are Banach spaces such that X embeds uniformly into Y , then X also embeds coarsely into ℓ p ( Y ) for every p >
1. As for the deduction of uniform embeddability from coarse embeddability,Rosendal’s work [16] implies that if X and Y are Banach spaces with the property that for every1-net N of X , every Lipschitz function f : N → Y admits an extension F : X → Y that is uniformlycontinuous, then the existence of a coarse embedding of X into Y implies that X embeds uniformlyinto ℓ p ( Y ) for every p >
1. Rosendal therefore asked [16] whether or not every pair of Banachspaces
X, Y has this (seemingly weak) extension property. Here we show that this is not the case.
Theorem 1.
There exist two Banach spaces ( X, k · k X ) and ( Y, k · k Y ) , a -net N of X anda Lipschitz function f : N → Y such that every F : X → Y that extends f is not uniformlycontinuous. Moreover, any F : X → Y that is uniformly continuous satisfies sup x ∈ N k F ( x ) − f ( x ) k Y = ∞ . (1)It remains an interesting open question to understand those pairs of Banach spaces X, Y forwhich Rosendal’s question has a positive answer, i.e., to prove theorems asserting the existence of
Supported by NSF grant CCF-0832795, BSF grant 2010021, the Packard Foundation and the Simons Foundation. uniformly continuous extension of any Y -valued Lipschitz function that is defined on a 1-net N of X . If the initial function f : N → Y is assumed to be H¨older with sufficiently small exponentrather than Lipschitz (which is a more stringent requirement since the minimum positive distancein N is at least 1), then such an extension result holds true provided that Y is superreflexive. Thisfollows from deep work of Ball [3] (see [12] for the precise statement that we need here). Indeed,if Y is superreflexive then by the work of Pisier [14] we know that Y admits an equivalent normwhose modulus of uniform convexity has power-type q for some q ∈ [2 , ∞ ). Therefore, by [3, 12], Y has metric Markov cotype q . Since any metric raised to the power 1 /q has Markov type q , itfollows from [3, 12] that every 1 /q -H¨older function from a subset of X into Y can be extended to a1 /q -H¨older function defined on all X . The role of superreflexivity here is only through the finitenessof the metric Markov cotype of Y , so by [12] similar statements hold true when Y is a q -barycentricmetric space (in particular, if Y is a Hadamard space then this holds true with q = 2). Theseconsiderations, however, do not address Rosendal’s question, where the initial function f : N → Y is only assumed to be Lipschitz. At the same time, an inspection of the proof below reveals thatin Theorem 1 we can ensure for every α ∈ (0 ,
1) that f is α -H¨older, so in general the uniformlycontinuous extension problem for H¨older functions on nets in Banach spaces has a negative answer.2. Proof of Theorem 1
The proof below is a variant of the argument in Section 5 of [13], which itself uses an averagingidea that is inspired by Lemma 6 in [10].For p > M p : ℓ → ℓ p be the Mazur map, i.e., for every x ∈ ℓ and j ∈ N , M p ( x ) j def = | x j | p sign( x j ) . Since p >
2, it is elementary to check that every u, v ∈ R satisfy (cid:12)(cid:12)(cid:12) | u | p sign( u ) − | v | p sign( v ) (cid:12)(cid:12)(cid:12) p p − | u − v | . Consequently, for every x, y ∈ ℓ we have. k M p ( x ) − M p ( y ) k p − p k x − y k p k x − y k p . (2)Denote X def = (cid:16) ∞ M p =2 ℓ (cid:17) ∞ and Y def = (cid:16) ∞ M p =2 ℓ p (cid:17) ∞ . (3)Fix a 1-net M of ℓ and denote N = Q ∞ p =2 M . Then (by the definition of X ) N is a 1-net of X .Define f : N → Y by setting for every ( x p ) ∞ p =2 ∈ X , f (( x p ) ∞ p =2 ) def = ( M p ( x p )) ∞ p =2 . Since the minimum distance in N is at least 1, it follows from (2) (and the definitions of X and Y )that f is 2-Lipschitz.Suppose for the purpose of obtaining a contradiction that there exists F : X → Y that isuniformly continuous and satisfies γ def = sup x ∈ N k F ( x ) − f ( x ) k Y < ∞ . (4)Let ω : [0 , ∞ ) → [0 , ∞ ) be the modulus of uniform continuity of F . Thus ω is nondecreasing andlim s → ω ( s ) = 0. Write F = ( F p ) ∞ p =2 , where for every integer p > F p : ℓ → ℓ p also as modulus of continuity that is bounded from above by ω (by the definitions of X and Y ). By (4)and the definition of f , for every y ∈ M we have k F p ( y ) − M p ( y ) k p γ . Hence, since M is a 1-net,sup x ∈ ℓ k F p ( x ) − M p ( x ) k p sup x ∈ ℓ inf y ∈ M (cid:0) k F p ( x ) − F p ( y ) k p + k F p ( y ) − M p ( y ) k p + k M p ( y ) − M p ( x ) k p (cid:1) (2) sup x ∈ ℓ inf y ∈ M (cid:18) ω ( k x − y k ) + γ + 2 k y − x k p (cid:19) ω (1) + γ + 2 . (5)In what follows, for every n ∈ N we let J n : ℓ n → ℓ be the canonical embedding, i.e., J n ( x , . . . , x n ) = ( x , . . . , x n , , . . . ). Also, we let Q n : ℓ p → ℓ np be the canonical projection, i.e., Q n (( x j ) ∞ j =1 ) = ( x , . . . , x n ). Given n ∈ N , we identify a permutation π ∈ S n with its associatedpermutation matrix, i.e., πx = ( x π − (1) , . . . , x π − ( n ) ) for every x ∈ R n . Similarly, we identify ε ∈ {− , } n with the corresponding diagonal matrix, i.e., εx = ( ε x , . . . , ε n x n ) for every x ∈ R n .Fix two integers p, n ∈ N with p >
2. Define G np : ℓ n → ℓ np by ∀ x ∈ ℓ n , G np ( x ) def = 12 n n ! X π ∈ S n X ε ∈{− , } n ( επ ) − Q n ◦ F p ◦ J n ( επx ) . (6)Then, because { επ : ( ε, π ) ∈ {− , } n × S n } forms a group of linear operators on R n , we have G np ( επx ) = επG np ( x ) for every ( ε, π ) ∈ {− , } n × S n and x ∈ ℓ n . Since for every A ⊆ { , . . . , n } and t ∈ R we have π ( t A ) = t A whenever π ∈ S n fixes A , it follows that there exist α ( t, A ) , β ( t, A ) ∈ R such that G np ( t A ) = α ( t, A ) A + β ( t, A ) { ,...,n } r A . Since ( A − { ,...,n } r A )( t A ) = t A , it followsthat β ( t, A ) = − β ( t, A ), thus G np ( t A ) = α ( t, A ) A . Finally, since for every A, B ⊆ { , . . . , n } ofthe same cardinality there exists π ∈ S n with π ( t A ) = t B , we conclude that α ( t, A ) depends onlyon the cardinality of A . In other words, there exists a sequence { α k ( t ) } nk =0 ⊆ R such that ∀ A ⊆ { , . . . , n } , ∀ t ∈ R , G np ( t A ) = α | A | ( t ) A . (7)Since Q n ◦ M p ◦ J n ( επx ) = επQ n ◦ M p ◦ J n ( x ) for every x ∈ ℓ n and ( ε, π ) ∈ {− , } n × S n ,sup x ∈ ℓ n (cid:13)(cid:13) G np ( x ) − Q n ◦ M p ◦ J n ( x ) (cid:13)(cid:13) p (6) = sup x ∈ ℓ n (cid:13)(cid:13)(cid:13) n n ! X π ∈ S n X ε ∈{− , } n ( επ ) − Q n ◦ ( F p − M p ) ◦ J n ( επx ) (cid:13)(cid:13)(cid:13) p (5) ω (1) + γ + 2 . In particular, for every k ∈ { , . . . , n − } and t ∈ (0 , ∞ ) we have (cid:13)(cid:13)(cid:13) G np (cid:0) t { n − k +1 ,...,n } (cid:1) − t p { n − k +1 ,...,n } (cid:13)(cid:13)(cid:13) p = (cid:13)(cid:13) G np (cid:0) t { n − k +1 ,...,n } (cid:1) − Q n ◦ M p ◦ J n ◦ (cid:0) t { n − k +1 ,...,n } (cid:1)(cid:13)(cid:13) p ω (1) + γ + 2 . and (cid:13)(cid:13)(cid:13) G np (cid:0) t { ,...,k } (cid:1) − t p { ,...,k } (cid:13)(cid:13)(cid:13) p = (cid:13)(cid:13) G np (cid:0) t { ,...,k } (cid:1) − Q n ◦ M p ◦ J n ◦ (cid:0) t { ,...,k } (cid:1)(cid:13)(cid:13) p ω (1) + γ + 2 . Consequently, assuming from now on that 2 k n + 1 we have t p (2 k ) p = (cid:13)(cid:13)(cid:13) t p (cid:0) { n − k +1 ,...,n } − { ,...,k } (cid:1)(cid:13)(cid:13)(cid:13) p (cid:13)(cid:13) G np (cid:0) t { n − k +1 ,...,n } (cid:1) − G np (cid:0) t { ,...,k } (cid:1)(cid:13)(cid:13) p + (cid:13)(cid:13)(cid:13) G np (cid:0) t { n − k +1 ,...,n } (cid:1) − t p { n − k +1 ,...,n } (cid:13)(cid:13)(cid:13) p + (cid:13)(cid:13)(cid:13) G np (cid:0) t { ,...,k } (cid:1) − t p { ,...,k } (cid:13)(cid:13)(cid:13) p (cid:13)(cid:13) G np (cid:0) t { n − k +1 ,...,n } (cid:1) − G np (cid:0) t { ,...,k } (cid:1)(cid:13)(cid:13) p + 2 ω (1) + 2 γ + 4 . (8) t the same time, since G np is obtained by averaging compositions of Q n ◦ F p ◦ J n : ℓ n → ℓ np withisometries (of both the source space and the target space), the modulus of uniform continuity of G np is bounded from above by ω . Hence, (cid:13)(cid:13) G np (cid:0) t { n − k +1 ,...,n } (cid:1) − G np (cid:0) t { ,...,k } (cid:1)(cid:13)(cid:13) p (7) = (cid:13)(cid:13) α k ( t ) (cid:0) { n − k +1 ,...,n } − { ,...,k } (cid:1)(cid:13)(cid:13) p = | α k ( t ) | (2 k ) p = k p (cid:13)(cid:13) α k ( t ) (cid:0) { ,...,k } − { ,...,k +1 } (cid:1)(cid:13)(cid:13) p (7) = k p (cid:13)(cid:13) G np (cid:0) t { ,...,k } (cid:1) − G np (cid:0) t { ,...,k +1 } (cid:1)(cid:13)(cid:13) p k p ω ( √ t ) . In combination with (8), this yields the following estimate, which holds for every p, k ∈ N with p > t ∈ (0 , ∞ ). t p (2 k ) p k p ω ( √ t ) + 2 ω (1) + 2 γ + 4 . (9)Suppose that 0 < t < / ( √ e ) and choose p = (cid:24) log (cid:18) t (cid:19)(cid:25) > k = (cid:18) ω (1) + 2 γ + 4 ω ( √ t ) (cid:19) (cid:16) t (cid:17) . These choices ensure that k p ω ( √ t ) > ω (1) + 2 γ + 4 and (2 t ) p > /e , so (9) implies that ω ( √ t ) > (2 t ) p > e . Thus lim inf s → ω ( s ) >
0, a contradiction. (cid:3)
Remark 2.
In order to obtain an example of separable Banach spaces ( X ′ , k · k X ′ ) and ( Y ′ , k · k Y ′ )that satisfy the conclusion of Theorem 1, replace the ℓ ∞ products in (3) by c products, i.e., define X ′ def = (cid:16) ∞ M p =2 ℓ (cid:17) c and Y ′ def = (cid:16) ∞ M p =2 ℓ p (cid:17) c . If the initial 1-net M ⊆ ℓ is chosen so that 0 ∈ M , then the set N ′ def = X ′ ∩ Q ∞ p =2 M is a 1-net in X ′ and the above proof of Theorem 1 goes through in this modified setting without any other change. Acknowledgements.
I am grateful to Bill Johnson for suggesting that I consider Rosendal’squestion and several helpful discussions on these topics. I also thank Christian Rosendal for sharinghis work [16] and helpful comments, in particular noting that the proof of Theorem 1 also yields (1).
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Mathematics Department, Princeton University, Fine Hall, Washington Road, Princeton, NJ 08544-1000, USA.
E-mail address : [email protected]@math.princeton.edu