Lipschitz-Free Spaces over Cantor Sets and Approximation Properties

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Introduction
Lipschitz-free spaces have become an active research area in Banach space theory in recent years.For a metric space (M, d) and a point x 0 ∈ M we define the Banach space Lip 0 (M, x 0 ) consisting of all real-valued Lipschitz functions f on M that vanish at x 0 , equipped with the norm ||f || := sup x,y∈M,x =y |f (x) − f (y)| d(x, y) .
For any x ∈ M we define the bounded linear functional δ x ∈ Lip 0 (M, x 0 ) * by δ x (f ) = f (x), f ∈ Lip 0 (M, x 0 ).The closed linear span of the set {δ x : x ∈ M } is called the Lipschitz-free space F(M, x 0 ).It is well-known that Lip 0 (M, x 0 ) is isometric to the dual space of F(M, x 0 ) and that the Banach space structure of both spaces does not depend on the choice of the base point x 0 .We will hereafter write Lip 0 (M ) and F(M ) without specifying the base point, and call F(M ) simply the free space over M .In the book [20], Weaver provides a comprehensive introduction to Lipschitz and Lipschitz-free spaces.In it, the latter are called Arens-Eells spaces and are denoted by AE(M ).
In particular, we mention that if M is a sufficiently 'thin' totally disconnected metric space then F(M ) has the MAP.For example, if M is a countable proper metric space (i.e.where all closed balls are compact), then F(M ) has the MAP [4].Also, as a corollary of [20,Corollary 4.39], F(M ) has the MAP when M is compact and uniformly disconnected (this notion is defined at the beginning of Section 2).Moreover, as a corollary of [15,Proposition 2.3] and [1,Theorem B], F(M ) has the MAP when M is a subset of a finite-dimensional normed space and is purely 1-unrectifiable, which is equivalent to the condition that M contains no bi-Lipschitz image of a compact subset of R of positive measure.On the other hand, in [12], G. Godefroy and N. Ozawa constructed a compact convex subset C of a separable Banach space such that F(C) fails the AP.Also, by a result in [13], there exists a metric space M homeomorphic to the Cantor space such that F(M ) fails the AP.
Given a compact metric space M with metric d and a Lipschitz function f : M → R, we say that f is locally flat [20,Definition 4.1] if for every x ∈ M , lim →0 Lip(f | B (x) ) = 0, where B (x) = {y ∈ M : d(y, x) < }.The Little Lipschitz space lip 0 (M ) is the subspace of Lip 0 (M ) consisting of all locally flat Lipschitz functions.By [20,Corollary 4.5], lip 0 (M ) is a Banach space.It often happens that lip 0 (M ) = {0}, for example when M is a connected smooth submanifold of R N .We say that lip 0 (M ) separates points uniformly [20,Definition 4.10] if there exists a ∈ (0, 1] such that for every x, y ∈ M there is f ∈ lip 0 (M ) such that Lip(f ) ≤ 1 and |f (x) − f (y)| = ad(x, y).By [1, Theorem A, Theorem B], lip 0 (M ) separates points uniformly if and only if lip 0 (M ) is an isometric predual to F(M ), which holds if and only if M is purely 1unrectifiable.If M is compact and purely 1-unrectifiable, then by [2, Proposition 3.5], if F(M ) has the MAP then lip 0 (M ) has the MAP.But also, by the remarks after [9, Problem 6.5], if lip 0 (M ) has the MAP then F(M ) has the MAP as well.
In [9], G. Godefroy surveys various aspects of the theory of free spaces, including the lifting property for separable Banach spaces, approximation properties of free spaces, and norm attainment of Lipschitz functions and operators.Regarding the bounded approximation property of the free space over a compact set M , he states a very useful criterion (see the end of Section 2) in terms of 'almost-extension' operators from Lipschitz spaces over finite subsets of M to Lip 0 (M ).In the last section he states a number of open problems, several of which concern approximation properties of free spaces.
Let K = 2 N be the Cantor set, equipped with the usual (product) topology.Define M ⊆ C(K 2 ) to be the space of all metrics d on K compatible with its topology.We equip M with the metric induced by the usual (supremum) norm of C(K 2 ).The set M is a G δ subset of C(K 2 ) (a proof is provided at the beginning of Section 2), and is therefore a Polish (i.e.separable and completely metrisable) space.For d ∈ M we write Lip 0 (K, d) and F(K, d) for the Lipschitz space and free space over the metric space (K, d), respectively.We define the following subsets of M: In [9,Problem 6.6] Godefroy asks what the topological nature of the set A f is (it is nonempty by [13,Corollary 2.2]).Also, more precisely, he asks whether the set A f is residual in M.
In this paper, we investigate the topological properties the subsets of M defined above.Section 2 concerns notation and preliminary results.In Section 3 we prove that the set A 1 is a residual F σδ set in M, and that A f is a dense meager set.Furthermore we prove that P is a dense G δ set and M\P is dense.As a corollary we obtain that the set of metrics d for which F(M ) is a dual space to lip 0 (M ) and both F(M ) and lip 0 (M ) have the MAP is residual.Finally, in Section 4, we construct a family (d α ) of metrics on K of size continuum, such that Lip 0 (K, d α ) and Lip 0 (K, d β ) are not isomorphic as algebras for α = β.This should be compared with [13, Corollary 2.3], wherein it is shown that there exists an family (d α ) of metrics on K of size ℵ 1 , such that F(K, d α ) is not isomorphic to F(K, d β ) for α = β.For d ∈ M, K 1 , K 2 ⊆ K, and x ∈ K we put D(K 1 , K 2 ) = sup{d(x, y) : x ∈ K 1 , y ∈ K 2 }, and D(x, K 1 ) = D({x}, K 1 ).We write D(K 1 ) or diam d (K 1 ) for D(K 1 , K 1 ).If S is a nonempty set then we call a finite family {S 1 , . . ., S n } of nonempty subsets of S a partition of S if S i ∩ S j = ∅ whenever i = j and n i=1 S i = S.If X is a Banach space then B X denotes the closed unit ball of X.If Y is another Banach space then X Y means that X is isomorphic to Y .

Notation and preliminary results
We will first provide a short proof of the fact that M is a G δ set in C(K 2 ).Let µ be the canonical metric on K: and for n ∈ N, define The following is a crucial lemma that allows us to make small changes to a given metric on K in a very flexible way.
Lemma 2.1.Suppose that d ∈ M, > 0, δ ∈ (0, 1] and K is a nonempty clopen subset of K. Let {K 1 , . . ., K n } be an arbitrary partition of K into clopen sets satisfying D(K i ) < 2 , and let e 1 , . . ., e n ∈ M be arbitrary.Then there exists and let e 1 , . . ., e n ∈ M be arbitrary.As each K i is homeomorphic to K, we can find a metric d i on K i compatible with its topology such that Now define We will now show that ∼ d is a metric on K.It is clearly symmetric and satisfies ∼ d(x, y) = 0 if and only if x = y.To show the triangle inequality, pick x, y, z ∈ K.If x, y, z ∈ K \ K or x, y, z ∈ K i for some i then the triangle inequality follows from the triangle inequality for the metric d or d i , respectively.We now consider the remaining cases: Case 1: x, y ∈ K \ K and z ∈ K i for some i.
We have By compactness, there exists z ∈ K i such that We can similarly show Similarly, and similarly by (2.1).The last assertion of the lemma follows from (2.2).
Remark 2.2.Note that the metric ∼ d satisfies ∼ d(x, y) ≥ d(x, y) whenever x, y ∈ K and x, y do not both belong to K i for any i = 1, . . ., n.
Corollary 2.3.Let d ∈ M, > 0, and K ⊆ K be a clopen subset.Then there exists a partition {K 1 , . . ., K n } of K consisting of clopen sets such that, for any arbitrary e 1 , . . ., e n ∈ M, there exists a metric Proof.Let d ∈ M, > 0 and K be given.Using the compactness of K and the fact that each point in K has a local base for its topology consisting of clopen neighbourhoods, we can find a cover C 1 , . . ., C p of K such that C i is clopen and D(C i ) < 2 for each i = 1, . . ., p. Now inductively define K 1 = C 1 and K i+1 = C i+1 \ i j=1 K j .By dismissing any empty K i we obtain a partition {K 1 , . . ., K n } of K (where n ≤ p), consisting of clopen sets satisfying D(K i ) < 2 for each i = 1, . . ., n.Now if e 1 , . . ., e n ∈ M are arbitrary, the corollary follows from an application of Lemma 2.1 with δ = 1.
We will also need the following results.
Lemma 2.4.Suppose that d ∈ M and (d n ) n ⊆ M are such that d n → d uniformly.Let (M, ρ) be a compact metric space, L > 0 and h n : K → M be functions such that Lip dn,ρ (h n ) ≤ L for each n ∈ N. Then there exists a subsequence (h ni ) i of (h n ) n and a function h if h n is surjective for all n, then h is surjective.
Proof.We will show that the functions (h n ) n are d-ρ-equicontinuous.Pick > 0, and pick n ∈ N such that ||d k − d|| ∞ < 2L for k ≥ n.Then for k ≥ n and x, y ∈ K such that d(x, y) < 2L , Therefore (h n ) n is equicontiuous on (K, d).By the Arzelà-Ascoli theorem, there exists a continuous h : K → M and a subsequence ( Hence Lip d,ρ (h) ≤ L.
If J > 0 and h n is bilipschitz with Lip ρ,dn (h −1 n ) ≤ J for all n ∈ N, then, similarly as previously we can show that h is bilipschitz with Lip ρ,d (h −1 ) ≤ J. Now assume the h n are surjective.Pick y ∈ M , and let > 0. Choose i ∈ N such that ||h − h ki || ∞ < .If h ki (x) = y then |h(x) − y| < .As was arbitrary, y is a limit point of h(K).Since h(K) is compact, y ∈ h(K) and h is surjective.Proposition 2.5.Let d ∈ M, x 0 be the base point of (K, d) and {K 1 , . . ., K n } be a partition of K consisting of clopen sets.
Proof.Let x 0 be the base point of each K i as well.We define the bounded linear operator , which shows that T is an isomorphism.It is not hard to see that T is the adjoint of the operator T * : (the spaces F(K i , d)) are seen as subspaces of F(K, d), by [20,Theorem 3.7]).Therefore T * is the required isomorphism.
Theorem 2.6 (Grothendieck).If X is a separable dual Banach space with the AP then X has the MAP.
A proof of the previous theorem can be found in e.g.[16,Theorem 1.e.15].Finally, in this section we give a useful criterion, due to Godefroy, for the λ-BAP of free spaces over compact metric spaces in terms of 'almost-extension' operators [9, Theorem 3.2] (also [8,Theorem 1]).We will need one of the several equivalent conditions for the λ-BAP stated in [9, Theorem 3.2].For a finite subset M ⊆ M of a compact metric space M , and > 0, we say that M is -dense in M if for every y ∈ M there is an x ∈ M such that d(y, x) < .Note that if (M n ) n is an increasing sequence (with respect to inclusion) of finite subsets of M such that ∞ n=1 M n is dense in M , then there exists a sequence ( n ) n of positive numbers tending to 0 such that M n is n -dense in M for all n ∈ N.
Theorem 2.7.Let M be a compact metric space and (M n ) n be a sequence of finite n -dense subsets of M with lim n→∞ n = 0. Then F(M ) has the λ-BAP if and only if there is a sequence (T n ) n of operators T n : Lip 0 (M n ) → Lip 0 (M ) such that ||T n || ≤ λ for all n and

Topological properties of A λ , A f and P
In this section we give our main results.We will first prove that A λ is an F σδ set in M for any λ ≥ 1. Fix a dense sequence of distinct elements (x n ) ∞ n=1 in K.For n ∈ N, define A n = {x 1 , . . ., x n }, and for n ∈ N, λ ≥ 1 and > 0, define k=1 is a bounded sequence.Therefore, by compactness, (S k ) k has a subsequence (S kj ) j which converges to an operator P m : Lip m=n+1 has a w * -convergent subsequence.As the w * -topology on bounded subsets of Lip 0 (K, d) coincides with the topology of pointwise convergence [20,Theorem 2.37], we can obtain a subsequence (P mj ) ∞ j=1 of (P m ) ∞ m=n+1 such that P mj (f i ) converges pointwise to some T (f i ) ∈ λB Lip 0 (K,d) for all i ∈ {1, . . ., n − 1}.We extend T by linearity to Lip 0 (A n , d).
For each p ∈ N and each i = {1, . . ., n − 1}, we have Therefore, by linearity, T (f )(x p ) = lim j→∞ P mj (f )(x p ) for each f ∈ Lip 0 (A n , d).Now let p, l ∈ N, p < l.Then x p , x l ∈ A mj whenever m j ≥ l, hence From the fact that (x p ) ∞ p=1 is dense in K follows that Lip d (T (f )) ≤ λ Lip d (f ), and so ||T || ≤ λ.Also, it is not hard to see that sup n, and B λ n, is closed.Remark 3.2.In the second part of the previous proof, extending P m (f i ) by McShane's extension theorem is done for convenience rather than necessity.Alternatively, we can define T f only on the set {x n : n ∈ N} and then extend to K using the density of {x n : n ∈ N}.
Proof.According to Theorem 2.7 with M = K, M n = A n , n ∈ N, F(K, d) has the λ-BAP if and only if there exists a sequence (α n ) n of positive real numbers converging to 0 such that d ∈ B λ n,αn for all n ∈ N.This is equivalent to Now the statement follows from Proposition 3.1.
We next prove that A 1 is residual in M. For a ∈ 2 <N (that is, a is a finite sequence of 0s and 1s) and n ∈ N, let l(a) denote the length of a, and define R n = {a ∈ 2 <N : l(a) = n}, and C a = {x ∈ K : x| {1,2,...,l(a)} = a}.Each C a is a clopen subset of K with diam µ (C a ) = 2 −l(a)−1 , and, for each n ∈ N, K is the disjoint union of the sets {C a : a ∈ R n }.For each a ∈ 2 <N , let r a = (a(1), . . ., a(l(a)), 0, 0, . ..) ∈ K.For d ∈ M and n ∈ N, we define ).Also define and apply Lemma 2.1 to d, , δ, K = K, the partition {C a : a ∈ R n } of K , and e i = µ for all i.We obtain a metric where the last inequality holds by (3.1).As a, b ∈ R n , a = b were arbitrary, we have χ Theorem 3.6.The set A 1 is a residual F σδ set in M.
Proof.By Proposition 3.3 it suffices to prove that A 1 is residual, and by Proposition 3.5 it suffices to prove If d is in the set on the left-hand side, then there exists an increasing sequence of natural numbers ni for all i, by Lemma 3.4.As the set {r a : a ∈ 2 <N } is dense in K, an application of Theorem 2.7 with M = K, M i = {r a : a ∈ R ni }, and operators Proof.The fact that A f is meager follows from Theorem 3.6.To show it is dense, let d ∈ M and > 0. By [13, Corollary 2.2] there exists d ∈ A f .According to Corollary 2.3 with K = K and e i = d for all i, there exists where K i = K i ∪ {x 0 } and x 0 is the base point of K. Since proportional metric spaces have isometrically isomorphic free spaces, F(K 1 , We will prove that V m,k is closed.Let (d n ) n∈N ⊆ V m,k converge uniformly to d ∈ M, and let K n and h n : K n → R be the compact set and bilipschitz function, respectively, associated to d n for each n ∈ N. By taking a subsequence, we can assume that the K n converge to a compact set K ⊆ K (in the compact-open topology of R).We extend each function h n to a function (again denoted by h n ) on K by [20, Theorem 1.33], while preserving its Lipschitz constant with respect to d n .We can assume, by translation, that h n (0, 0, . ..) = 0 for each n.As the diameters of (K, d n ) are uniformly bounded, the sets h n (K) are all contained in some fixed bounded interval.By Lemma 2.4, there exists a subsequence (h ni ) i converging uniformly to a function h : K → R with Lip d (h) ≤ m.Now let x, y ∈ K be arbitrary and let (x i ) i , (y i ) i ⊆ K converge to x and y, respectively, and be such that x i , y i ∈ K ni for each i ∈ N. Let > 0 and pick i such that As was arbitrary, we get |h(x) − h(y)| ≥ m −1 d(x, y).This means that h satisfies the bilipschitz condition in the definition of V m,k on the set K .
To show that ∆(h(K )) ≥ k −1 , pick > 0 and set As was arbitrary, we get that ∆(h(K )) ≥ k −1 .Therefore d ∈ V m,k and so V m,k is closed.As M \ P = m,k∈N V m,k , we have that M \ P is F σ , and so P is G δ .Proposition 3.9.The set P is dense in M.
Proof.Let d ∈ M and > 0 be arbitrary.It is not hard to see that (K, µ) is uniformly disconnected.By [20, Corollary 4.39 (ii)], F(K, µ) is a dual space.Then by [1, Theorem B], (K, µ) is purely 1-unrectifiable.According to Corollary 2.3 applied to d, , K = K and e i = µ for all i, there exists a partition {K 1 , . . ., K n } of K consisting of clopen sets, and a metric

Lipschitz equivalence classes
In this section we consider subsets of M consisting of Lipschitz-equivalent metrics, and describe some of their basic topological properties.For any d ∈ M, we consider the 'Lipschitz equivalence class' of metrics E d = {d ∈ M : there exists a surjection h : K → K and n > 0 such that n −1 d(x, y) ≤ d (h(x), h(y)) ≤ nd(x, y) for all x, y ∈ K}.For a bounded metric space M , we define Lip(M ) (resp.Lip(M, C)) to be the space of all real (resp.complex)-valued Lipschitz functions on M equipped with the norm ||f || = Lip(f ) + ||f || ∞ (the Lipschitz constant of a complex-valued function is defined by the same supremum as for a real-valued function).Under pointwise multiplication of functions, Lip(M ) and Lip(M, C) are Banach algebras.Note that by [20,Lemma 1.28], Lip(M, C) is isomorphic to Lip(M ) ⊕ Lip(M ) as a Banach space.We can also view Lip 0 (M ) as an algebra under pointwise multiplication, however the Lipschitz constant is not submultiplicative.Instead, it satisfies Lip(f g) ≤ ||f || ∞ Lip(g) + ||g|| ∞ Lip(f ), which implies ||f g|| ≤ 2diam(M )||f ||||g||.Therefore the product is continuous, which implies that there is an equivalent norm which is submultiplicative.Proof.Suppose that d 1 , d 2 ∈ M are such that Lip 0 (K, d 1 ) and Lip 0 (K, d 2 ) are isomorphic as algebras.If h : Lip 0 (K, d 1 ) → Lip 0 (K, d 2 ) is the isomorphism, then the map ∼ h : Lip(K, d 1 ) → Lip(K, d 2 ), given by ∼ h(λ1 K + f ) = λ1 K + h(f ), is an algebra isomorphism, where λ ∈ R, 1 K is the constant function on K equal to 1, and f ∈ Lip 0 (K, d 1 ).Then the map ĥ : Lip((K, d 1 ), C) → Lip((K, d 2 ), C), given by ĥ(f + ig) = ∼ h(f ) + i ∼ h(g), is again an algebra isomorphism.Finally, by [18,Theorem 5.1], there exists a bilipschitz surjection from (K, d 1 ) to (K, d 2 ), and the corollary follows from Proposition 4.1.
We do not know (in ZFC) whether there exists a family (d α ) ⊆ M of size continuum such that F(K, d α ) is not linearly isomorphic to F(K, d β ) for α = β.

For a metric
space (A, d), x ∈ A and r > 0 we write B d r (x) = {y ∈ A : d(x, y) < r}.If C ⊆ A then write B d r (C) = {y ∈ A : d(y, C) < r}, where d(y, C) = inf{d(y, x) : x ∈ C}.For a real-valued Lipschitz function f on (A, d), Lip d (f ) denotes the Lipschitz constant of f with respect to d.If (B, e) is another metric space and f : A → B then the Lipschitz constant of f is denoted by Lip d,e (f ).We call (A, d) and (B, e) proportional if there exists a surjection f : A → B and c > 0 such that d(x, y) = ce(f (x), f (y)) for all x, y ∈ A. The space (A, d) is called uniformly disconnected [20, Proposition 4.12] if there exists r ∈ (0, 1] such that for any distinct x, y ∈ A there are complementary clopen sets C, D ⊆ A such that x ∈ C, y ∈ D and d(C, D) ≥ rd(x, y), where d(C, D) = inf{d(z, t) : z ∈ C, t ∈ D}.
The set A is clearly closed and hence G δ in C(K 2 ), and the sets B n are open by compactness.Therefore the set A ∩ ∞ n=1 B n is G δ .Pick any d ∈ A ∩ ∞ n=1 B n and observe that d is a metric on K. Let T d be the topology on K induced by d.Since for any x ∈ K, the function d(x, •) is continuous on K, B d r (x) is open in the topology of K for any r > 0. Thus T d is a coarser topology than the one on K. Therefore any closed subset C of K is compact in T d , and is therefore closed, because T d is Hausdorff.Thus T d agrees with the topology on K, so d ∈ M. As d was arbitrary,

∼d)Proposition 3 . 8 .
fails the AP.As F(K 1 , ∼ d) has codimension 1 in F(K 1 , ∼ d), we have that F(K 1 , ∼ d) fails the AP as well.Hence F(K, ∼ d) fails the AP.We now state and prove some properties of the set P of metrics d for which (K, d) is purely 1-unrectifiable.Denote by ∆ the Lebesgue measure on R. The set P is G δ in M. Proof.For m, k ∈ N define V m,k = {d ∈ M : there exists a compact K ⊆ K and h : K → R, such that m −1 d(x, y) ≤ |h(x) − h(y)| ≤ md(x, y) for all x, y ∈ K , and ∆(h(K )) ≥ k −1 }.

∼d)
is also purely 1-unrectifiable.Proposition 3.10.The set M \ P is dense in M. Proof.Let d ∈ M and > 0 be arbitrary.Let d ∈ M be such that (K, d ) is isometric to a Cantor subset of R of positive measure.Then clearly (K, d ) is not purely 1-unrectifiable.According to Corollary 2.3 with K = K and e i = d for all i, there exists ∼ d ∈ M such that ||d− ∼ d|| ∞ < and (K, ∼ d) contains a clopen subset proportional to (K, d ).Then (K, ∼ d) is not purely 1-unrectifiable.Corollary 3.11.The set of metrics d for which F(K, d) is the dual space of lip 0 (K, d) and both lip 0 (M ) and F(M ) have the MAP is residual in M. Proof.Let G be the set of metrics in question.Clearly G ⊆ A 1 , and by [1, Theorem B], G ⊆ P. Also, if d ∈ P ∩ A 1 then by [2, Proposition 3.5], lip 0 (K, d) has the MAP, so d ∈ G. Therefore G = P ∩ A 1 and the corollary follows from Proposition 3.8, Proposition 3.9, and Theorem 3.6.

Proposition 4 . 2 .
For each d ∈ M, E d is a dense meager F σ set in M.Proof.Fix d ∈ M. If d ∈ M and > 0 are arbitrary, we will show that there exists a metric ∼ d ∈ E d which is -close to d.We apply Corollary 2.3 to d, K = K, and , to get a partition K 1 , . . ., K n of K consisting of clopen sets.Now if e i ∈ M is such that (K, e i ) is isometric to (K i , d) for i = 1, . . ., n, then by the same corollary we obtain a metric∼ d ∈ M, such that || d − ∼ d|| ∞ < , and (K i , ∼ d) is proportional to (K, e i ), and hence to (K i , d), for all i.It is not hard to see that then∼ d ∈ E d .Therefore E d is dense in M.Now for n ∈ N, consider the setE n d = {d ∈ M : there exists a surjection h : K → K such that n −1 d(x, y) ≤ d (h(x), h(y)) ≤ nd(x, y) for all x, y ∈ K}.Suppose that (d k ) ∞ k=1 is a sequence in E n d converging to d 0 ∈ M. Let h k : K → Kbe the surjection associated with d k , for each k ∈ N. By Lemma 2.4, there exists a surjective function h : K → K satisfying the condition in the definition of E n d with d 0 for d .This shows that d 0 ∈ E n d and so E n d is closed.Since E d = n∈N E n d , E d is F σ .Moreover, E n d is nowhere dense because if e ∈ E d then E e is dense in M and E e ∩ E d = ∅.Therefore E d is meager.