Lipschitz Embeddings of Metric Spaces into c_0
aa r X i v : . [ m a t h . F A ] D ec LIPSCHITZ EMBEDDINGS OF METRIC SPACES INTO c F. Baudier and R. DevilleAbstract. let M be a separable metric space. We say that f = ( f n ) : M → c is agood- λ -embedding if, whenever x, y ∈ M , x = y implies d ( x, y ) ≤ k f ( x ) − f ( y ) k and,for each n , Lip ( f n ) < λ , where Lip ( f n ) denotes the Lipschitz constant of f n . We provethat there exists a good- λ -embedding from M into c if and only if M satisfies an internalproperty called π ( λ ). As a consequence, we obtain that for any separable metric space M ,there exists a good-2-embedding from M into c . These statements slightly extend formerresults obtained by N. Kalton and G. Lancien, with simplified proofs.
1) Introduction.
First, let us recall that if f is a mapping between the metric spaces ( M, d ) and (
N, δ ),the Lipschitz constant
Lip ( f ) is the infimum of all λ such that for all ( x, y ) ∈ M , δ ( f ( x ) , f ( y )) ≤ λd ( x, y ).Let ( M, d ) be a separable metric space and λ ≥
1. We say that f : M → c is a λ -embedding if, whenever x, y ∈ M , then :(1) d ( x, y ) ≤ k f ( x ) − f ( y ) k ≤ λd ( x, y ) . Let us denote f = ( f n ) and, for each n , E n = { ( x, y ) ∈ M × M ; d ( x, y ) ≤ | f n ( x ) − f n ( y ) |} .Whenever x, y ∈ M , we have k f ( x ) − f ( y ) k = max n | f n ( x ) − f n ( y ) | . Hence f is a λ -embedding if for each n , Lip ( f n ) ≤ λ and M × M = S n E n .I. Aharoni [1] proved that for any separable metric space M , there exists a λ -embeddingfrom M into c for any λ >
6, and that there is no λ -embedding from ℓ into c if λ < λ -embedding fromany separable metric space M into c for any λ >
3. Later on, J. Pelant [4] obtained thesame result with λ = 3. It was also observed that there is no λ -embedding from ℓ into c +0 if λ <
3. All these authors actually constructed λ -embeddings into the positive cone c +0 of c . Finally N. Kalton and G. Lancien [3] proved that for any separable metric space M ,there exists a 2-embedding from M into c , and this result is optimal (consider M = ℓ ).We say that f : M → c is a strict- λ -embedding if, whenever x, y ∈ M and x = y , then :(2) d ( x, y ) < k f ( x ) − f ( y ) k < λd ( x, y ) . We say that f = ( f n ) : M → c is a good- λ -embedding if, for each n , Lip ( f n ) < λ , and M × M = S n E n . Proposition 1.1.
Assume f : M → c is a good- λ -embedding. Then, there exists g : M → c which is a strict and good- λ -embedding. roof. Let λ n < λ be such that f n : M → RI is λ n -Lipschitz continuous, and let us define g = ( g n ) : M → c such that for each n , g n = α n f n with 1 < α n < α n λ n < λ .Clearly, g is still a good- λ -embedding. If x = y , since the sequences ( f n ( x )) and ( f n ( y ))tend to zero, the sequence (cid:0) g n ( x ) − g n ( y ) (cid:1) also converges to 0 and there exists n suchthat k g ( x ) − g ( y ) k = | g n ( x ) − g n ( y ) | ≤ α n λ n d ( x, y ) < λd ( x, y )Since k f ( x ) − f ( y ) k ≤ k g ( x ) − g ( y ) k , this implies that k f ( x ) − f ( y ) k < λd ( x, y ). On theother hand, let m be such that k f ( x ) − f ( y ) k = | f m ( x ) − f m ( y ) | . We have d ( x, y ) ≤ k f ( x ) − f ( y ) k = | f m ( x ) − f m ( y ) | < | α m f m ( x ) − α m f m ( y ) | ≤ k g ( x ) − g ( y ) k . Therefore, g is also a strict- λ -embedding.Our purpose is to prove that for every separable metric space, one can construct a strict-2-embedding from M into c . We introduce also a property π ( λ ) of a metric space, slightlyweaker than a property introduced by N. Kalton and G. Lancien, and we prove that if1 < λ ≤
2, a separable metric space M admits a good- λ -embedding into c if and only ifit has the property π ( λ ).
2) Necessary condition for the existence of good- λ -embedding into c . Let (
M, d ) be a metric space and E be a non empty subset of M × M . We denote π ( E ) = { x ∈ M ; ∃ y ∈ M, ( x, y ) ∈ E } , π ( E ) = { y ∈ M ; ∃ x ∈ M, ( x, y ) ∈ E } theprojections of E , and π ( E ) = π ( E ) × π ( E ) the smallest rectangle containing E . Wealso define the gap of E by δ ( E ) := inf { d ( x, y ); ( x, y ) ∈ E } and the diameter of E by diam ( E ) = sup { d ( x, y ); ( x, y ) ∈ E } . These notions are not quite standard, and requiresome comments. Let us denote ∆ := { ( x, x ); x ∈ M } the diagonal of M × M , and let usendow the set M × M with the metric d (cid:0) ( x, y ) , ( x ′ , y ′ ) (cid:1) = d ( x, x ′ ) + d ( y, y ′ ). The distancefrom a point ( y, z ) ∈ M × M to ∆ is d (cid:0) ( y, z ) , ∆ (cid:1) = inf { d (cid:0) ( y, z ) , ( x, x ) (cid:1) ; ( x, x ) ∈ ∆ } and it is easy to check that d (cid:0) ( y, z ) , ∆ (cid:1) = d ( y, z ). Consequently, if ∅ 6 = E ⊂ M × M , thesmallest distance from a point of E to ∆ is the quantity d ( E, ∆) = inf { d (cid:0) ( y, z ) , ∆ (cid:1) ; ( y, z ) ∈ E } = δ ( E )On the other hand, the largest distance from a point of E to ∆ is D ( E, ∆) = sup { d (cid:0) ( y, z ) , ∆ (cid:1) ; ( y, z ) ∈ E } = diam ( E )Whenever E is of the form U × V , then δ ( E ) = inf { d ( x, y ); x ∈ U, y ∈ V } is the gapbetween U and V , and diam ( E ) = sup { d ( x, y ); x ∈ U, y ∈ V } . Thus, if U = V , diam ( E )is the usual diameter of U . 2 act 2.1. Let E be a bounded subset of M × M , F be a finite dimensional normed vectorspace, let P : M → F be such that Lip ( P ) ≤ λ and d ( x, y ) ≤ k P ( x ) − P ( y ) k for each ( x, y ) ∈ E , and let ε > . Then, there exists a finite partition { E , · · · , E N } of E so thatfor each n, diam ( E n ) < λδ ( π ( E n )) + ε. Proof.
The set P ( π ( E ) ∪ π ( E )) ⊂ F is bounded as E is bounded and P , π and π are Lipschitz. Hence we can find a finite partition of this set into subsets F j of diameter < ε/
4. The sets E j,k = ( P − ( F j ) × P − ( F k )) ∩ E which are non empty form a partitionof E . If ( x, y ) ∈ E j,k and ( u, v ) ∈ π ( E j,k ), then k P ( x ) − P ( y ) k ≤ k P ( x ) − P ( u ) k + k P ( u ) − P ( v ) k + k P ( v ) − P ( y ) k ≤ ε/ λd ( u, v )Thus d ( x, y ) ≤ λd ( u, v ) + ε/
2. The result follows by taking the infimum over all ( u, v ) ∈ π ( E j,k ), the supremum over all ( x, y ) ∈ E j,k , and by relabeling the sets E j,k . Definition 2.2.
A metric space (
M, d ) has property π ( λ ) if, for any balls B and B ofradii r and r and for any non empty subset E of B × B satisfying δ ( E ) > λ ( r + r ),there exists a partition { E , · · · , E N } of E , such thatfor each n, diam ( E n ) < λδ ( π ( E n ))We say that ( M, d ) has the property weak- π ( λ ) if the conclusion is replaced by the weakerconclusion : there exists non empty closed subsets F , · · · , F N covering E such thatfor each n, r + r < δ ( π ( F n ))this conclusion is indeed weaker : if F n is the closure of E n in M × M , then λ ( r + r ) <δ ( E ) ≤ diam ( E n ) < λδ ( π ( E n )) = λδ ( π ( F n )). It is also easy to see that if λ < µ and if M has π ( λ ), then M has π ( µ ), and that if M has at least 2 elements, M never has π (1). Proposition 2.3.
1) Assume that there is a good- λ -embedding from M into c . Then M has property π ( λ ) .2) If ( M, d ) λ -embeds into c , then M has property weak- π ( λ ) . Proof.
Let f : M → c be a λ -embedding. If ( e i ) is the unit vector basis of c , then f ( x ) = + ∞ P i =0 f i ( x ) e i . Let B and B be balls of radii r and r and of centers a and a ,and E ⊂ B × B such that δ ( E ) > λ ( r + r ). We claim that the function E ∋ ( x, y ) f ( x ) − f ( y ) k depends on finitely many coordinates, i. e. there exists i ∈ NI such that,if P ( x ) = i P i =0 f i ( x ) e i then k f ( x ) − f ( y ) k = k P ( x ) − P ( y ) k .Fix ε > ε < δ ( E ) − λ ( r + r ). We choose i such that, if Q = f − P , then k Q ( a ) − Q ( a ) k < ε . If ( x, y ) ∈ E , then k Q ( x ) − Q ( y ) k ≤ k Q ( x ) − Q ( a ) k + k Q ( a ) − Q ( a ) k + k Q ( a ) − Q ( y ) k < λ ( r + r ) + ε < δ ( E ) ≤ d ( x, y ) . d ( x, y ) ≤ k f ( x ) − f ( y ) k = max {k Q ( x ) − Q ( y ) k , k P ( x ) − P ( y ) k} = k P ( x ) − P ( y ) k .This proves our claim. Since Lip ( P ) ≤ λ , Fact 2.1 implies the existence of a a partition { E , · · · , E N } of E such that for all n , diam ( E n ) < λδ ( π ( E n )) + ε . Since we also have λ ( r + r ) + ε ≤ diam ( E n ), we have r + r < δ ( π ( E n )), so if F n is the norm closure of E n in E , r + r < δ ( π ( F n ))When f is a good- λ -embedding, the mapping P is µ -Lipschitz continuous for some µ < λ , sowe can asume that for all n , diam ( E n ) < µδ ( π ( E n ))+ α , where α = min { ε, ( λ − µ )( r + r ) } .This still implies r + r < δ ( π ( E n )). Finally, diam ( E n ) < µδ ( π ( E n )) + ( λ − µ )( r + r ) < λδ ( π ( E n )) . Corollary 2.4 (see [3]).
Let X be a Banach space. If there exists u ∈ S X and aninfinite dimensional subspace Y of X such that inf {k u + y k ; y ∈ S Y } > λ , then there is no λ -embedding from X into c . Proof. If E = { ( u + y, − u − y ); y ∈ S Y } ⊂ B ( u, × B ( − u, E satisfies δ ( E ) > λ .Assume there exists a λ -embedding from M into c . Then X has the weak- π ( λ ) property, sothere exists closed subsets F , · · · , F N of E covering E such that for each n , δ ( π ( F n )) > A n = { y ∈ S Y ; ( u + y, − u − y ) ∈ F n } is closed and A ∪ · · · ∪ A N = S Y .Since dim ( Y ) > N , the Borsuk-Ulam thoeorem yields the existence of y ∈ S Y and n suchthat { y, − y } ⊂ A n . Hence ( u + y, − u + y ) ∈ π ( F n ) and so δ ( π ( F n )) ≤
2, which is absurd.
Example 2.5.
There is no λ -embedding from ℓ p into c for any λ < /p . In particular, ℓ is a metric space which does not λ -embed into c with λ < . (If u = e and Y = { y =( y i ) ∈ ℓ p ; y = 0 } , then k u + y k = 2 /p for all y ∈ S Y ).
3) Examples of spaces with property π ( λ ) .Example 3.1. A metric space such that the bounded subsets of M are totally bounded hasproperty π (1 + ε ) for all ε > E into subsets E n of small d -diameter). Example 3.2. If ( M, d ) is a metric space, then ( M, d ) has property π (2) .Therefore, property π ( λ ) is of interest only if < λ ≤ . Proof.
Let E ⊂ B × B , with B et B balls of radii r ≥ r , and assume ε := δ ( E ) − r + r ) >
0. Let a = δ ( E ) < a < · · · < a N − < diam ( E ) < a N so that for all 1 ≤ n ≤ N , a n − a n − < ε . Define, for 1 ≤ n ≤ N , E n = (cid:8) ( x, y ) ∈ E ; a n − ≤ d ( x, y ) < a n (cid:9) . Thus δ ( E n ) + ε > diam ( E n ). If ( u, v ) ∈ π ( E n ), one can find v ′ ∈ B such that ( u, v ′ ) ∈ E n .Moreover v, v ′ ∈ B , hence : diam ( E n ) < δ ( E n ) − r + r ) ≤ d ( u, v ′ ) − d ( v ′ , v ) ≤ d ( u, v )Taking the infimum over all ( u, v ) ∈ π ( E n ), we get diam ( E n ) < δ ( π ( E n )).4 xample 3.3. If ( X n ) is a sequence of finite dimensional Banach spaces, then ( ⊕ X n ) p has property π (2 /p ) . Proof.
Let E ⊂ B ( a , r ) × B ( a , r ) such that α = δ ( E ) p − r + r ) p >
0. We select ε > r + r + ε ) p < δ ( E ) p / − α/ t + ε ) p − α/ < t p whenever0 ≤ t ≤ diam ( E ). If x ∈ ( ⊕ X n ) p , then x = ∞ P n =1 x n with x n ∈ X n for each n . Define P, Q : ( ⊕ X n ) p → ( ⊕ X n ) p by P ( ∞ P i =0 x i ) = i P i =0 x i and Q = I − P , where i is such that k Qa − Qa k < ε . According to Fact 2.1, since P is an operator of norm 1 with valuesin a finite dimensional subspace of ( ⊕ X n ) p , we can find relatively closed subsets E n of E covering E such that, for all n , if ( x, y ) ∈ E n , then k P x − P y k ≤ δ ( π ( E n )) + ε . Onthe other hand, k Qx − Qy k ≤ k Qx − Qa k + k Qy − Qa k + k Qa − Qa k ≤ r + r + ε .Moreover, k x − y k p = k P x − P y k p + k Qx − Qy k p , so diam ( E n ) p ≤ (cid:0) δ ( π ( E n )) + ε (cid:1) p + ( r + r + ε ) p < (cid:0) δ ( π ( E n )) + ε (cid:1) p + diam ( E n ) p / − α/ diam ( E n ) p < (cid:0) δ ( π ( E n )) + ε (cid:1) p − α/ < δ ( π ( E n )) p . Remark 3.4.
In the definition of property π ( λ ), if a and a are centers of B and B ,we can assume that ( λ − r + r ) < d ( a , a ) ≤ λ + 1 λ − r + r )Indeed, if E = ∅ , then, for each ( x, y ) ∈ E , λ ( r + r ) < δ ( E ) ≤ d ( x, y ) ≤ d ( a , a ) +( r + r ), which proves the first inequality. If λ +1 λ − ( r + r ) < d ( a , a ), the conclusion ofproperty π ( λ ) is always true if we take N = 1 and E = E , since then diam ( E ) ≤ d ( a , a ) + ( r + r ) < λ (cid:0) d ( a , a ) − ( r + r ) (cid:1) ≤ λδ ( π ( E ))
4) Constructing strict- λ -embeddings into c .Theorem 4.1. Let ( M, d ) be a separable metric space and < λ ≤ . If M has property π ( λ ) , there exists a λ -embedding from M into c which is strict and good. Theorem 4.1 and Proposition 2.3 show that there exists a good- λ -embedding from M into c if and only if M has property π ( λ ). We do not know any internal characterization ofseparable metric spaces that admit a λ -embedding into c . Corollary 4.2.
Let ( M, d ) be a metric space such that the bounded subsets of M aretotally bounded. For all ε > , there exists a (1 + ε ) -embedding from M into c . Corollary 4.3.
Every separable metric space strictly- -embeds into c . ℓ does not λ -embed into c whenever λ < Corollary 4.4. If ( X n ) is a sequence of finite dimensional Banach spaces, then ( ⊕ X n ) p strictly- /p -embeds into c . This result is optimal since we have seen that there is no λ -embedding from ℓ p into c with λ < /p . We now turn to the proof of Theorem 4;1. We need some further notations. If( M, d ) is a metric space , x ∈ M and U, V ⊂ M , the distance from x to U is d ( x, U ) = δ ( { x } × U ) and the gap between U and V is δ ( U, V ) = δ ( U × V ). The coordinates of theembedding from M into c are of the following type : Lemma 4.5.
Let ( M, d ) be a metric space, U, V, F three non empty subsets of M and ε ≥ . There exists f : M → RI , -Lipschitz, such that :1) For all x ∈ F , | f ( x ) | ≤ ε .2) For all ( x, y ) ∈ U × V , f ( x ) − f ( y ) = min (cid:8) δ ( U, V ) , δ ( U, F ) + δ ( V, F ) + 2 ε (cid:9) . Lemma 4.6.
Let < λ ≤ , ( M, d ) be a metric space with property π ( λ ) , F ⊂ G be finitesubsets of M and < α < β . We set : A ( F, β ) = (cid:8) ( x, y ) ∈ M × M ; λ (cid:0) d ( x, F ) + d ( y, F ) + β (cid:1) ≤ d ( x, y ) (cid:9) Then there exists a finite partition { E , · · · , E N } of A ( G, α ) \ A ( F, β ) such that, if we denote π ( E n ) = U n × V n thenfor each n diam ( E n ) < λ min (cid:8) δ ( U n , V n ) , δ ( U n , F ) + δ ( V n , F ) + 2 β (cid:9) . Proof of Theorem 4.1.
The goal is to construct a sequence ( f n ) of 1-Lipschitz continuous functions satisfying, forevery x ∈ M , lim n →∞ f n ( x ) = 0, and a partition { E n ; n ∈ NI } of { ( x, y ) ∈ M × M ; x = y } ,so that for each n , the function ( x, y ) → f n ( x ) − f n ( y ) is equal to some constant c n on E n and diam ( E n ) < λc n . The required strict and good- λ -embedding is then f = ( λ n f n )where λ n < λ is chosen so that diam ( E n ) < λ n c n .Let ( a k ) be a dense sequence of distinct points of M , F k = { a , · · · , a k } , and ( ε k ) be a de-creasing sequence of real numbers converging to 0. We set ∆ k = A ( F k +1 , ε k +1 ) \ A ( F k , ε k ).The sets ∆ k form a partition of { ( x, y ) ∈ M × M ; x = y } . Indeed, if x, y ∈ M , x = y and if σ k = λ (cid:0) d ( x, F k ) + d ( y, F k ) + ε k (cid:1) , then 0 < d ( x, y ) < σ , ( σ k ) is decreasing and lim k →∞ σ k = 0,so there exists a unique k such that σ k +1 ≤ d ( x, y ) < σ k , which means ( x, y ) ∈ ∆ k .By Lemma 4.6, there exists integers 0 = n < n < · · · < n k < · · · and subsets E n of M × M such that for all k , { E n ; n k < n ≤ n k +1 } is a partition of ∆ k , and, whenever n k < n ≤ n k +1 then diam ( E n ) < λ min (cid:8) δ ( U n , V n ) , δ ( U n , F k ) + δ ( V n , F k ) + 2 ε k (cid:9) . π ( E n ) = U n × V n . In particular, { E n ; n ∈ NI } is a partition of { ( x, y ) ∈ M × M ; x = y } . By Lemma 4.5, there are 1-Lipschitz functions f n : M → RI so that1) if x ∈ F k and n k < n ≤ n k +1 , then | f n ( x ) | ≤ ε k ,2) if n k < n ≤ n k +1 and ( x, y ) ∈ U n × V n , then f n ( x ) − f n ( y ) = c n := min (cid:8) δ ( U n , V n ) , δ ( U n , F k ) + δ ( V n , F k ) + 2 ε k (cid:9) . and so diam ( E n ) < λc n .If x ∈ M , let us show that lim n →∞ f n ( x ) = 0. If ε >
0, we fix j such that d ( x, a j ) < ε/
2, then k ≥ j such that ε k < ε/
2. Since the functions f n are 1-Lipschitz continuous, if n ≥ n k ,then | f n ( x ) | ≤ d ( x, a j ) + | f n ( a j ) | < ε/ ε k < ε . Proof of Lemma 4.5.
We fix s, t such that − δ ( V, F ) − ε ≤ s ≤ ≤ t ≤ δ ( U, F ) + ε and t − s = min (cid:8) δ ( U, V ) , δ ( U, F ) + δ ( V, F ) + 2 ε (cid:9) , and we set f ( x ) := min (cid:8) d ( x, U ) + t, d ( x, V ) + s, d ( x, F ) + ε (cid:9) The function f is 1-Lipschitz continuous as the infimum of 1-Lipschitz continuous functions.If x ∈ U , f ( x ) = min (cid:8) t, d ( x, V ) + s, d ( x, F ) + ε (cid:9) = t because d ( x, V ) + s ≥ δ ( U, V ) + s ≥ t and d ( x, F ) + ε ≥ δ ( U, F ) + ε ≥ t . If y ∈ V , f ( y ) = min (cid:8) d ( y, U ) + t, s, d ( x, F ) + ε (cid:9) = s because s ≤
0. Therefore, if x ∈ U and y ∈ V , then f ( x ) − f ( y ) = t − s , which proves 2).Finally, if x ∈ F , then f ( x ) = min (cid:8) d ( x, U ) + t, d ( x, V ) + s, ε (cid:9) ≤ ε . On the other hand d ( x, U ) + t ≥ d ( x, V ) + s ≥ δ ( V, F ) + s ≥ − ε , so f ( x ) ≥ − ε , which proves 1). Proof of Lemma 4.6.
Set ∆ := A ( G, α ) \ A ( F, β ). There is a bounded subset B of M such that ∆ ⊂ B × B ,because λ > G is bounded and λ (cid:0) d ( x, G ) + d ( y, G ) (cid:1) ≤ d ( x, y ) ≤ d ( x, G ) + d ( y, G ) + diam ( G ) . whenever ( x, y ) ∈ A ( G, α ). Thus, there is a partition { B , B , · · · , B m } of the boundedset B such that for all j , if x, x ′ ∈ B j and a ∈ G , then | d ( x, a ) − d ( x ′ , a ) | ≤ α/
5, and so,for all x ∈ B j , for all a ∈ G, d ( x, a ) < d ( B j , a ) + α/ . Since G is finite, there exists a j ∈ G such that d ( B j , a j ) = δ ( B j , G ), and so B j ⊂ B ( a j , r j ),where r j = δ ( B j , G ) + α/
4. The subsets E jk = ∆ ∩ B j × B k of ∆ form a partition of ∆, E jk ⊂ B ( a j , r j ) × B ( a k , r k ), and, if ( x, y ) ∈ E jk : d ( x, y ) ≥ λ (cid:0) d ( x, G ) + d ( y, G ) + α (cid:1) ≥ λ (cid:0) δ ( B j , G ) + δ ( B k , G ) + α (cid:1) = λ ( r j + r k + α/ > λ ( r j + r k ) . δ ( E jk ) > λ ( r j + r k ). According to property π ( λ ) applied to each E jk , there exists afinite partition { E , · · · , E N } of ∆ such that, diam ( E n ) < λδ ( π ( E n )) = λδ ( U n , V n ) , where π ( E n ) = U n × V n . Moreover, if j, k, n are such that E n ⊂ B j × B k and if ( x, y ) ∈ E n ,then d ( x, y ) ≤ λ (cid:0) d ( x, F ) + d ( y, F ) + β (cid:1) ≤ λ (cid:0) δ ( B j , F ) + α/ δ ( B k , F ) + α/ β (cid:1) ≤ λ (cid:0) δ ( U n , F ) + δ ( V n , F ) + α/ β (cid:1) . hence diam ( E n ) ≤ λ (cid:0) δ ( U n , F ) + δ ( V n , F ) + α/ β (cid:1) < λ (cid:0) δ ( U n , F ) + δ ( V n , F ) + 2 β (cid:1) .
5) Some consequences.
Observe that a metric space has property π ( λ ) (resp. weak- π ( λ )) if and only if its boundedsubsets have it. In particular a Banach space has property π ( λ ) (resp. weak- π ( λ )) if andonly if its unit ball has it. Since the property “there exists a good- λ -embedding from M into c ” is equivant to the property “ M has π ( λ )”, we can state : Proposition 5.1.
Assume that ( M, d ) is a separable metric space and that for each ball B of M , there is a good- λ -embedding from B into c . Then there is a strict and good- λ -embedding from M into c . In particular, if X is a Banach space and if there exists a good- λ -embedding from its closedunit ball into c , then there exists a good- λ -embedding from X into c . The followingextension result is obvious. Proposition 5.2.
Assume that ( M, d ) is a separable metric space and that N is a densesubset of M . If there is a good- λ -embedding from N into c , then there is a good- λ -embedding from M into c . Remark 5.3.
In Definition 2.2, we didn’t specify if the balls were closed or open. Wecan define two different properties, π ( λ ) with closed balls and π ( λ ) with open balls. Thesetwo properties are equivalent! Indeed, the proof of Proposition 2.3 shows that if there isa good- λ -embedding from M into c , then M has property π ( λ ) with closed balls, whichin turn implies that M has property π ( λ ) with open balls. On the othe hand, the proof ofTheorem 4.1 shows that if M has property π ( λ ) with open balls, then there is a good- λ -embedding from M into c . This proves that property π ( λ ) with open balls is equivalentto property π ( λ ) with closed balls.N. Kalton and G. Lancien introduced the following definition : A metric space ( M, d ) hasΠ( λ ) if, for every µ > λ , there exists ν > µ such that for every closed balls B and B r and r , the exists subsets U , · · · U N , V , · · · , V n of M such that thesets U n × V n are a covering of E µ := { ( x, y ) ∈ B × B ; d ( x, y ) > µ ( r + r ) } and,for all n, λδ ( U n , V n ) ≥ ν ( r + r ) Lemma 5.4.
Property Π( λ ) implies property π ( λ ) . We do not know if the converse is true. Let us notice that N. Kalton and G. Lancien provedthat if a separable metric space satisfies property Π( λ ), then there exists f : M → c suchthat, for all x, y ∈ M , x = y , we have d ( x, y ) < k f ( x ) − f ( y ) k ≤ λd ( x, y )which is weaker than the condition f is a strict and good- λ -embedding. Theorem 1 improvetheir result since our hypothesis, M has π ( λ ), is weaker, and our conclusion, f is a strict andgood- λ -embedding, is stronger. Moreover, our condition π ( λ ) is a necessary and sufficientcondition for the existence of a good- λ -embedding. Proof of Lemma 5.4.
Let us assume that (
M, d ) has property Π( λ ). Let E ⊂ B × B suchthat δ ( E ) > λ ( r + r ). We fix µ > λ such that δ ( E ) > µ ( r + r ). Then E ⊂ E µ . Let ν > µ be given by property Π( λ ). Let 1 = a < a < · · · < a K be a sequence such that diam ( E µ ) = a K µ ( r + r ) and a k +1 a k < νµ whenever 1 ≤ k < K . We denote E k := { ( x, y ) ∈ B × B ; a k µ ( r + r ) < d ( x, y ) ≤ µ ( r + r ) a k +1 } . The E k ’s form a covering of E µ . Let B k and B k be the closed balls of the same center as B and B and of radius a k r and a k r respectively. Obviously, E k ⊂ B k × B k . Applyingproperty Π( λ ) for each k , We can find subsets U k, , · · · , U k,N k , V k, , · · · , V k,N k of M suchthat the sets U k,n × V k,n for 1 ≤ n ≤ N k form a covering of E k andfor all n, λδ ( U k,n , V k,n ) ≥ ν ( a k r + a k r ) > µa k +1 ( r + r )We can assume in addition that for each n , the sets U k,n × V k,n are pairwise disjoint(because a finite union of products can always be written as a finite union of pairwisedisjoint products). If we denote E k,n = E ∩ E k ∩ ( U k,n × V k,n ), the E k,n ’s form a partitionof E . Moreover, π ( E k,n ) ⊂ U k,n × V k,n , and the above inequality impliesfor all n, λδ ( π ( E k,n )) > µa k +1 ( r + r ) ≥ diam ( E k ) ≥ diam ( E k,n ) . We have proved property π ( λ ).
6) Strict- λ -embeddings into c +0 . Here, c +0 denotes the positive cone of c . Observe that : u, v ∈ c +0 ⇒ k u − v k ≤ max {k u k , k v k} . λ -embedding into c +0 follows from the following property π + ( λ ). Definition 6.1.
A metric space (
M, d ) has property π + ( λ ) (with λ >
1) if,a) Whenever B and B are balls of positive radii r et r and E is a subset of B × B such that δ ( E ) > λ max( r , r ), there exists a finite partition { E , · · · , E N } of E satisfyingfor each n, diam ( E n ) < λδ ( π ( E n ))b) There exists θ < λ and ϕ : M → [0 , + ∞ [ such that | ϕ ( x ) − ϕ ( y ) | ≤ d ( x, y ) ≤ θ max (cid:0) ϕ ( x ) , ϕ ( y ) (cid:1) for all x, y ∈ M. The function ϕ is called a control function. Remark 6.2.
1) It is easy to see that it is enough to check a) whenever r = r (= r ).2) If λ >
2, the function ϕ ( x ) = d ( x, a ) is a control function (take θ = 2). Therefore, themetric space M has property π + ( λ ) if and only if the bounded subsets of M have property π + ( λ ). In particular a Banach space X has property π + ( λ ) (with λ >
2) if and only if itsunit ball has property π + ( λ ).3) If M is bounded, then, for any λ >
1, the function ϕ : M → [0 , + ∞ [ given by ϕ ( x ) = d ( x, a ) + diam ( M ) satisfies condition b) of property π + ( λ ) (take θ = 1). Proposition 6.3.
1) If there is a λ -embedding from ( M, d ) into c +0 , then M has property π + ( µ ) for all µ > λ .2) Assume that M is bounded or that λ > . If there is a good- λ -embedding from M into c , then M has property π ( λ ) . Proof.
Let B = B ( a , r ) and B = B ( a , r ). Let E ⊂ B × B such that λr + ε < δ ( E )for some ε >
0. Let f : M → c be a λ -embedding given by f ( x ) = + ∞ P i =0 f i ( x ) e i . We denote P ( x ) = i P i =0 f i ( x ) e i and Q = f − P , where i is such that max {k Q ( a ) k , k Q ( a ) k} < ε . If( x, y ) ∈ E , then k Q ( x ) − Q ( y ) k ≤ max {k Q ( x ) k , k Q ( y ) k}≤ max (cid:8) k Q ( x ) − Q ( a ) k + k Q ( a ) k , k Q ( y ) − Q ( a ) k + k Q ( a ) k (cid:9) ≤ λr + ε < δ ( E ) ≤ d ( x, y ) ≤ k f ( x ) − f ( y ) k . Thus, k f ( x ) − f ( y ) k = k P ( x ) − P ( y ) k . Following the lines of the proof of Proposition 2.3we get, for any µ > λ , a partition { E , · · · , E N } of E such that for each n , diam ( E n ) ≤ µδ ( π ( E n )). Condition a) can be checked and the E n ’s satisfy also δ ( π ( E n )) > r . Moreover,if ϕ ( x ) = k f ( x ) k /λ , then | ϕ ( x ) − ϕ ( y ) | ≤ d ( x, y ) ≤ λ max (cid:0) ϕ ( x ) , ϕ ( y ) (cid:1) for all x, y ∈ M .This proves that ϕ is a control function of π + ( µ ) because λ < µ .The proof of 2) also follows the lines of the corresponding case of Proposition 2.3, and herewe do not have to worry about the existence of a control function by Remark 6.2.10 orollary 6.4. Let X be a Banach space. If there exists u ∈ S X and an infinite dimen-sional subspace Y of X such that inf {k u + 2 y k ; y ∈ S Y } > λ , then there is no λ -embeddingfrom M into c +0 . Proof. If E = { ( u + 2 y, − u − y ); y ∈ S Y } ⊂ B ( u, × B ( − u, E satisfies δ ( E ) > λ .If there is a λ -embedding from X into c +0 , then, by Proposition 6.3, there is a partition { E , · · · , E N } of E such that for each n , δ ( π ( E n )) >
2. If F n is the norm closure of E n ,then { F , · · · , F N } is a covering of E and we still have δ ( π ( F n )) > δ ( π ( F n )) ≤
2, which is absurd.
Example 6.5.
The metric space ℓ does not λ -embed into c +0 whenever λ < . (If u = e and Y = { y = ( y i ) ∈ ℓ ; y = 0 } , then k u + 2 y k = 3 for each y ∈ S Y ). The space ℓ p does not λ -embed into c +0 whenever λ < (1 + 2 p ) /p (since in this case, forall y ∈ S Y , k u + 2 y k = (1 + 2 p ) /p ).The balls of positive radius of ℓ p do not λ -embed into c +0 whenever λ < (1 + 2 p ) /p . Onthe other hand, it follows from Corollary 4.4 that the balls of positive radius of ℓ p embedin c if λ = 2 /p < (1 + 2 p ) /p . Indeed, assume that for some λ < (1 + 2 p ) /p , a ball B of positive radius of ℓ p λ -embedsinto c +0 . We can assume that B is the unit ball of ℓ p . According to Proposition 6.3, B has property π + ( µ ) for every µ > λ , and by Remark 6.2 2), ℓ p has property π + ( µ ) forevery µ > λ , and from Theorem 6.9 below, ℓ p µ -embeds into c +0 for every µ > λ . But thiscontradicts the fact that ℓ p does not µ -embed into c +0 whenever µ < (1 + 2 p ) /p . Example 6.6.
A compact metric space M has property π + ( λ ) for all λ > . A metricspace M such that its bounded subsets are totally bounded has property π + ( λ ) for all λ > ,but may fail property π + (2) . A metric space M such that its bounded subsets are totally bounded satisfies conditiona) of property π + ( λ ) for all λ >
1, and any metric space satisfies condition b) of property π + ( λ ) for all λ >
2. Moreover, if λ > M is compact, then M is bounded and sosatisfies condition b).The bounded subsets of the set Z of integers are finite. Let ϕ : Z → [0 , + ∞ [ such that | ϕ ( x ) − ϕ ( y ) | ≤ | x − y | for all x, y ∈ Z . If x n = ( − n n , then | x n − x n +1 | = 2 n + 1 and ϕ ( x n ) ≤ ϕ (0) + n . Consequently, if θ <
2, then | x n − x n +1 | > θ max (cid:8) ϕ ( x n ) , ϕ ( x n +1 ) (cid:9) for n large enough. Thus Z do not admit any control function ϕ for property π + (2). Example 6.7.
Each metric space M has property π + (3) . Proof.
Let B and B be balls of radius r and E ⊂ B × B such that ε := δ ( E ) − r > { E , · · · , E N } of E such that foreach n , δ ( E n ) + ε > diam ( E n ). If ( u, v ) ∈ π ( E n ), there is v ′ ∈ B so that ( u, v ′ ) ∈ E n .Moreover v, v ′ ∈ B , hence :3 d ( u, v ) ≥ d ( u, v ′ ) − d ( v ′ , v ) ≥ δ ( E n ) − r ≥ δ ( E n ) + 2 ε > diam ( E n )Taking the infimum over all ( u, v ) ∈ π ( E n ), we get 3 δ ( π ( E n )) > diam ( E n ).11 xample 6.8. ℓ p ( NI ) has property π + (cid:0) (1 + 2 p ) /p (cid:1) . Proof.
Let E ⊂ B ( a , r ) × B ( a , r ) such that α = δ ( E ) p − (1 + 2 p ) r p >
0. We choose ε > r + ε ) p < p p δ ( E ) p − α/ t + ε ) p − α/ < t p si 0 ≤ t ≤ diam ( E ).Let ( e i ) be the canonical basis of ℓ p , P, Q : ℓ p → ℓ p be defined by P ( ∞ P i =0 x i e i ) = i P i =0 x i e i and Q = I − P , where i is chosen so that k Qa − Qa k < ε . Since Lip ( P ) = 1 and P has its values in a finite dimensional space, Fact 2.1 implies the existence of a partition { E , · · · , E N } of E such that, for each n , if ( x, y ) ∈ E n , then k P x − P y k ≤ δ ( π ( E n )) + ε .On the other hand, k Qx − Qy k ≤ k Qx − Qa k + k Qy − Qa k + k Qa − Qa k ≤ r + ε .Hence diam ( E n ) p ≤ (cid:0) δ ( π ( E n )) + ε (cid:1) p + (2 r + ε ) p ≤ (cid:0) δ ( π ( E n )) + ε (cid:1) p + 2 p p δ ( E ) p − α/ diam ( E n ) p ≤ (1 + 2 p ) (cid:0) δ ( π ( E n )) + ε (cid:1) p − (1 + 2 p ) α/ < (1 + 2 p ) δ ( π ( E n )) p . Theorem 6.9.
If the separable metric space ( M, d ) has property π + ( λ ) with < λ ≤ ,then there exists f : M → c +0 such that for all x, y ∈ M , x = y , we have : d ( x, y ) < k f ( x ) − f ( y ) k < λd ( x, y ) . Corollary 6.10.
Let ( M, d ) be a separable metric space. Then there exists f : M → c +0 such that, for all x, y ∈ M , x = y , we have : d ( x, y ) < k f ( x ) − f ( y ) k < d ( x, y ) . This result is optimal since we observed that there is no λ -embedding from ℓ into c +0 with λ < Corollary 6.11.
There exists f : ℓ p → c +0 such that, for all x, y ∈ M , x = y , we have : d ( x, y ) < k f ( x ) − f ( y ) k < (1 + 2 p ) /p d ( x, y ) . This result is optimal since we observed that there is no λ -embedding from ℓ p into c +0 whenever λ < (1 + 2 p ) /p . Corollary 6.12. If ( M, d ) is a compact space and ε > , there exists f : M → c +0 suchthat, for all x, y ∈ M , x = y , we have : d ( x, y ) < k f ( x ) − f ( y ) k < (1 + ε ) d ( x, y ) . If ( M, d ) is a metric space such that its bounded subsets are totally bounded, there exists f : M → c +0 such that, for all x, y ∈ M , x = y , we have : d ( x, y ) < k f ( x ) − f ( y ) k < (2 + ε ) d ( x, y ) . The following result shows that we cannot replace 2 + ε by 2 in the above statement.12 roposition 6.13. There exists a separable metric space M such that, for any λ > , Mλ -embeds into c but there is no -embedding from M into c +0 . Proof.
Let ( e n ) be the canonical basis of ℓ ( NI ) and F p := { pe k , e + pe k ; 1 ≤ k ≤ p } . Wedefine M = { , e } ∪ + ∞ S p =1 F p ⊂ ℓ ( NI ). The bounded sets of M are finite, hence totallybounded, so, by Corollary 2, for any λ > M λ -embeds into c .Assume now that there exists f = ( f n ) : M → c +0 such that, for all x, y ∈ M , k x − y k ≤ k f ( x ) − f ( y ) k ∞ ≤ k x − y k Let us denote C = max {k f (0) k ∞ , k f ( e ) k ∞ } , fix n ≥ n > n , onehas f n (0) < f n ( e ) <
1, and finally fix p > C/ ϕ : { , · · · , p } → { , } n defined by ϕ ( k ) = (cid:0) [0 ,C +1] ( f n ( pe k )) (cid:1) n ≤ n is injective. Thisleads to a contradiction if we also have p > n .If n ≥ ≤ k ≤ p , then f n ( pe k ) ≤ | f n ( pe k ) − f n (0) | + f n (0) and f n ( pe k + e ) ≤ | f n ( pe k + e ) − f n ( e ) | + f n ( e )so(1) f n ( pe k ) ≤ p + C and f n ( e + pe k ) ≤ p + C. Whenever n > n , we have a better estimate: f n ( pe k ) < p + 1 and f n ( e + pe k ) < p + 1So, if n > n , | f n ( pe k ) − f n ( e + pe k ) | ≤ max { f n ( pe k ) , f n ( e + pe k ) } < p + 1On the other hand, if 1 ≤ k = ℓ ≤ p , we have2 p + 1 = k e + pe k − pe ℓ k ≤ k f ( e + pe k ) − f ( pe ℓ ) k ∞ , hence, there exists n ≤ n such that | f n ( e + pe k ) − f n ( pe ℓ ) | ≥ p + 1, and using the factthat | f n ( e + pe k ) − f n ( pe k ) | ≤
2, we obtain(2) | f n ( pe k ) − f n ( pe ℓ ) | ≥ p − f n ( pe k ) ≤ C + 1 and f n ( pe ℓ ) ≥ p −
1, or f n ( pe ℓ ) ≤ C + 1 and f n ( pe k ) ≥ p −
1, hence 1I [0 ,C +1] ( f n ( pe k )) = 1I [0 ,C +1] ( f n ( pe ℓ )), and ϕ is injective.The proof of Theorem 6.9 is analogous to the proof of Theorem 4.1 and relies on thefollowing two lemmas (analogous to Lemmas 4.5 and 4.6).13 emma 6.14. Let ( M, d ) be a metric space, U, V, F non empty bounded subsets of M and ε ≥ . There exists f : M → RI + , such that Lip ( f ) ≤ and :1) For all x ∈ F , f ( x ) ≤ ε ,2) For all ( x, y ) ∈ U × V , f ( x ) − f ( y ) = min (cid:8) δ ( U, V ) , max( δ ( U, F ) , δ ( V, F )) + ε (cid:9) . Proof.
Indeed, if δ ( V, F ) ≤ δ ( U, F ) and if we put t = min( δ ( U, V ) , δ ( U, F )+ ε ), the function f defined by f ( x ) = max( t − d ( x, U ) ,
0) satisfies Lemma 6.14.
Lemma 6.15.
Let ( M, d ) be a metric space with property π + ( λ ) with < λ ≤ , F ⊂ G be finite subsets of M and < α < β , we set : A + ( G, α ) = (cid:8) ( x, y ) ∈ M × M ; d ( x, y ) ≥ λ (cid:0) max( d ( x, G ) , d ( y, G )) + α (cid:1)(cid:9) Then there exists a finite partition { E , · · · , E N } of A + ( G, α ) \ A + ( F, β ) such that, if wedenote π ( E n ) = U n × V n , thenfor each n, diam ( E n ) < λ min (cid:8) δ ( U n , V n ) , max( δ ( U n , F ) , δ ( V n , F )) + 2 β (cid:9) . Proof.
Denote K = sup {| ϕ ( a ) | ; a ∈ G } , where ϕ is the control function. For all x ∈ M ,we have ϕ ( x ) ≤ d ( x, G ) + K . If ( x, y ) ∈ A + ( G, α ), then λ max (cid:0) d ( x, G ) , d ( y, G ) (cid:1) ≤ d ( x, y ) ≤ θ max (cid:0) ϕ ( x ) , ϕ ( y ) (cid:1) ≤ θ max (cid:0) d ( x, G ) , d ( y, G ) (cid:1) + θK. Since G is bounded and λ > θ , we can find B ⊂ M bounded such that A + ( G, α ) ⊂ B × B .The rest of the proof follows the lines of Lemma 4.6 (show that δ ( E jk ) > max { r j , r k } ).For the proof of Theorem 6.9, choose ε > ϕ ( a ), which implies, for all x, y ∈ M , d ( x, y ) ≤ λ max( ϕ ( x ) , ϕ ( y )) < λ (cid:0) max( d ( x, a ) , d ( y, a )) + ε (cid:1) := σ . Remark 6.16.
Property π ( λ ) characterizes the existence of a good- λ -embedding into c ,but we do not know if property π + ( λ ) characterizes the existence of a good- λ -embeddinginto c +0 . We do not know of any internal characterization of the existence of a good- λ -embedding into c +0 , or of the existence of a λ -embedding into c +0 . However, it seems verylikely that if M is bounded, or if 2 < λ ≤
3, then the existence of a good- λ -embeddinginto c +0 is equivalent to the fact that M has property π + ( λ ). We wish to thank anonymous referees for their useful comments on the presentation of thispaper.
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