Lipschitz quotients from metric trees and from Banach spaces

containing ℓ_{1}

William B. Johnson^{∗}, Joram Lindenstrauss^{†}, David Preiss and Gideon Schechtman^{†}

1 Introduction

A Lipschitz map f between the metric spaces X and Y is called a Lipschitz quotient map if there is a C > 0 (the smallest such C, the co-Lipschitz constant, is denoted coLip(f), while Lip(f) denotes the Lipschitz constant of f) so that for every x ∈ X and r > 0, fB_{X}(x, r) ⊃ B_{Y} (f(x), r/C). Thus Lipschitz quotient maps are surjective maps which by definition have the property ensured by the open mapping theorem for surjective linear operators between Banach spaces.

If there is a Lipschitz quotient mapping f from X onto Y with Lip(f) · coLip(f) ≤ C, we say that Y is a C-Lipschitz quotient of X. If Y is a C-Lipschitz quotient of X for some C, we say that Y is a Lipschitz quotient of X.

Lipschitz quotient maps were introduced and studied in [4]. Among other results the following proposition was proved in [4] (see also [5], Theorem 11.18). Proposition 1. Let X be a superreflexive Banach space and let the Banach space Y be a Lipschitz quotient of X. Then Y^{∗} is crudely finitely representable in X^{∗}.

The conclusion of the Proposition means that there is a constant λ so that for every finite dimensional (linear) subspace F of Y^{∗} there is a finite dimensional subspace E of X^{∗} so that d(E, F) ≤ λ where d denotes here the Banach Mazur distance. An equivalent way to express the conclusion of Proposition 1 is that Y^{∗} is linearly isomorphic to a subspace of an ultraproduct of X^{∗}. An immediate consequence of Proposition 1 is that any Banach space Y which is a Lipschitz quotient of a Hilbert space must be linearly isomorphic to a Hilbert space.

Proposition 1 was proved in [4] by using a special result on approximating Lipschitz functions from a superreflexive Banach space into finite dimensional Banach spaces (the so-called UAAP, see also [5], Proposition 11.16). It was proved in [4] that this approximation property actually characterizes superreflexive spaces and thus this proof of Proposition 1 does not work beyond the class of superreflexive spaces.

∗Supported in part by NSF DMS-9900185 and the U.S.-Israel Binational Science Foundation.

†Supported in part by the U.S.-Israel Binational Science Foundation and participants in the NSF Workshop in Linear Analysis and Probability at Texas A&M University MR subject classification: 46T99, 54E40.

In [12] it was shown that if X is asymptotically uniformly smooth (AUS) then every Lipschitz map from X to a finite dimensional space has for every ε > 0 points of ε-Fréchet differentiability. We do not intend to recall here the definitions of AUS or of ε-Fréchet differentiability. We just mention that the class of spaces having an equivalent AUS norm is much richer than that of superreflexive spaces and contains, for example, the non-reflexive space c_{0}. On the other hand any space X with an AUS norm is an Asplund space (any separable subspace of X has a separable dual). The existence of points of ε-Fréchet differentiability for every ε > 0 provides an approximation of Lipschitz functions by affine functions which also suffices for proving the conclusion of Proposition 1. Thus Proposition 1 is valid also if X has an AUS norm. In [12] the result on existence of points of ε-Fréchet differentiability and thus Proposition 1 is proved also for some spaces X which do not admit an equivalent AUS norm (for example, every X which is a C(K) space with K compact countable; unless such X is isomorphic to c_{0} it does not have an equivalent AUS norm).

It is possible that the ε-Fréchet differentiability result (and thus Proposition 1) holds for every Asplund space X. The ε-Fréchet differentiability result is known to be false for every non-Asplund space (see e.g. [5], Proposition 4.12, or Chapter III of [7]). In this connection we mention also that in [4] it is proved that a Banach space which is a Lipschitz quotient of an Asplund space is itself an Asplund space.

All the results mentioned so far leave open the question whether the conclusion of Proposition 1 is valid for an arbitrary Banach space X. Recall that in the formally dual setting of Lipschitz embeddings the dual version of Proposition 1 is valid for an arbitrary Banach space X; i.e., if X Lipschitz embeds in Y then X is crudely finitely representable in Y. This fact is due to Ribe [17] (see also [5], Corollary 7.10).

We prove in this paper that Proposition 1 may fail to hold for many common Banach spaces. We do this by proving in Theorem 1 that if X is any separable Banach space containing ℓ_{1} then there is a Lipschitz quotient map from X onto any separable Banach space Y. In particular, there is a Lipschitz quotient map from C[0, 1] onto ℓ_{1}; in fact, known results in the linear theory reduce the general theorem to this special case. This provides also the first known examples of pairs of separable Banach spaces X and Y so that there is a Lipschitz quotient map from X to Y but no such linear quotient map.

Theorem 1 can be considered in a way as a dual result to a theorem of Aharoni [1] (see also [3] or [5], Theorem 7.11) that every separable metric space is Lipschitz equivalent to a subset of c_{0}. There is however no obvious connection between the proof of Theorem 1 and the known proofs of Aharoni’s theorem. The proof of Theorem 1 depends strongly on the special structure of ℓ_{1}.

Theorem 1 actually holds for a general complete separable metrically convex metric space Y.

Recall that a metric space Y is metrically convex provided that for each pair {x, y} of points in Y and 0 < λ < 1 there exists a point z in Y so that d(x, z) = λd(x, y) and d(y, z) = (1 − λ)d(x, y). In other words, for each x and y in Y there is a geodesic arc from x to y (i.e., an arc which is the range of an isometry from an interval). It is not simply for the sake of generalization that we mention metrically convex spaces. The Lipschitz quotient maps f constructed in the proof of Theorem 1 factor through a special metrically convex metric space which we call an ℓ_{1} tree. This factorization forces the mapping f to have the property that whenever the Gâteaux derivative f′(x) of f exists at a point x ∈ X then the rank of f′(x) is ≤ 1 (see Proposition 5 below). Note that if X has an AUS norm and 1 < dim Y < ∞ such a situation is impossible by the results in [12] on ε-Fréchet differentiability mentioned above.

An ℓ_{1} tree is a special kind of metric tree. Metric trees are used in geometry, metric topology, and group theory, but, as far as we know, this is the first time they have been used in the geometry of Banach spaces.

There is a weakening of the notion of Lipschitz quotient map which has been studied by David and Semmes [6]. Say that a Lipschitz map f from the metric space X into the metric space Y is ball non collapsing provided there is a c > 0 so that for every r > 0, the image under f of any ball of radius r contains a ball of radius cr. If f is a Lipschitz map from a metric space X into ℝ^{n}, say that f is measure non collapsing provided there is a c > 0 so that for every r > 0, the image under f of any ball of radius r has Lebesgue measure at least cr^{n}. David and Semmes [6] showed that that if X is finite dimensional and f : X → ℝ^{n} is measure non collapsing then f is ball non collapsing. In [4] this was extended to the case where X is any superreflexive Banach space. In Theorem 4 we show that if X is any Banach space which contains a subspace isomorphic to ℓ_{1}, then there is a measure non collapsing map from X onto the plane whose range has empty interior (and hence the mapping is not ball non collapsing).

In Section 3 we prove non separable versions of Theorem 1. Theorem 5 implies in particular that if Y is a Banach space (or more generally a complete metrically convex space) whose density character is at most that of the continuum then there is a Lipschitz quotient map from ℓ_{∞} onto Y.

2 Proofs of the separable results

We begin by recalling the definition of a metric tree. A metric space X is a metric tree provided it is complete, metrically convex, and there is a unique arc (which then by metric convexity must be a geodesic arc) joining each pair of points in X. For an introduction to metric trees see [10]. However, except for the proof of Proposition 5 we do not use the theory of metric trees. Instead we work with a different notion of ‘separable metric tree’ which we denote by SMT and a different notion of ‘metric tree’ which we denote by MT. In Section 4 we further discuss metric trees and observe that a metric space is an SMT if and only if it is a separable metric tree and that a metric space is a MT if and only if it is a metric tree.

The concept of SMT is defined by describing how to construct one. In order to explain the construction, we need (a special case of) the concept of the ℓ_{1}-union of two metric spaces. Suppose that (X, d_{X}) and (Y, d_{Y}) are two metric spaces whose intersection is a single point, p. We define the ℓ_{1} union X ∪_{1} Y of X and Y to be (X ∪ Y, d), where the metric d agrees with d_{X} on X, d agrees with d_{Y} on Y, and if x ∈ X, y ∈ Y, then d(x, y) is defined to be d_{X}(x, p) + d_{Y} (p, y). We can now describe the construction of an SMT. Let I_{1} be a closed interval or a closed ray and define T_{1} := I_{1}. The metric space T_{1} is the first approximation to our SMT. Having defined T_{n}, let I_{n+1} be a closed interval or a closed ray whose intersection with T_{n} is an end point, p_{n}, of I_{n}, and define T_{n+1} := T_{n} ∪_{1} I_{n+1}. The completion, T, of ∪_{n=1}^{∞} T_{n} is an SMT. If each I_{n} is a ray with end point p_{n−1} for n > 1 and the set {p_{n}}_{n=1}^{∞} of nodal points is dense in T, then we call T an ‘ℓ_{1} tree’ and say that {I_{n}}_{n=1}^{∞}, {T_{n}}_{n=1}^{∞} describe an allowed construction of T. (Of course, typically there are many different allowed constructions of a given SMT.)

The relevance of SMTs to Lipschitz quotients is made clear by Proposition 2. Proposition 2. Let T be an ℓ_{1} tree. Then every separable, complete, metrically convex metric space is a 1-Lipschitz quotient of T.

Proof. Let {I_{n}}_{n=1}^{∞}, {T_{n}}_{n=1}^{∞} describe an allowed construction of T and let the nodes be {p_{n}}_{n=1}^{∞}. Let Y be a separable, complete, metrically convex metric space. We build the desired Lipschitz quotient map by defining it on T_{n} by induction. That we can do this is clear from the following. Suppose you have a 1-Lipschitz map f : T_{n} → Y, and y is taken from some countable dense subset Y_{0} of Y. Extend f to T_{n+1} by mapping I_{n+1} to a geodesic arc [f(p_{n}), y] which joins f(p_{n}) to y; f is an isometry on {z ∈ I_{n+1} : d(p_{n}, z) ≤ d(f(p_{n}), y)} and f maps points on I_{n+1} whose distance to p_{n} is larger than d(f(p_{n}), y) to y. This makes f act like a Lipschitz quotient at p_{n} relative to [f(p_{n}), y]. Since the nodal points are dense in T, it is evident that a judicious selection of the points from Y_{0} will produce a 1-Lipschitz quotient mapping. For example, order Y_{0} as a sequence and for each node p pick a sequence of nodes {p_{k_{n}}} tending to p so that the sequences corresponding to different p-s are disjoint. Then, at the k_{n} + 1 step, extend f to T_{k_{n}+1} by mapping I_{k_{n}+1} to a geodesic arc which joins f(p_{k_{n}}) to the n-th element of Y_{0}. (We use the fact that to build a 1-Lipschitz quotient mapping from a complete metric space X onto a complete metric space Y it is enough to construct a function f from a dense subset X_{0} of X into Y so that for each x in X_{0} and for a dense set of r in ℝ^{+}, the closure in Y of f[B_{X}(x, r)] ∩ X_{0} is B_{Y} (f(x), r). See, for example, the argument for Proposition 3.15 in [4].)

For the proof of Theorem 1 we use the fact that an SMT is a 1-absolute Lipschitz retract. A metric space X is called a λ-absolute Lipschitz retract (where 1 ≤ λ < ∞) provided that whenever X is isometrically contained in a metric space Y, there is a retraction, f, from Y onto X with Lip(f) ≤ λ. A metric space X is a 1-absolute Lipschitz retract if and only if X is metrically convex and every collection of mutually intersecting closed balls in X have a common point (see e.g. [5], Proposition 1.4).

An isomorphic version of Lemma 1 was proved in a more general context in [13]. Lemma 1. Assume that X and Y are 1-absolute Lipschitz retracts which intersect in a single point, p. Then X ∪_{1} Y is also a 1-absolute Lipschitz retract.

Proof. It is evident that X ∪_{1} Y is metrically convex. We check that any collection of mutually intersecting closed balls in X ∪_{1} Y have a point in common. Let

{B(x_{α}, r_{α}), x_{α} ∈ X, α ∈ Γ_{1}} ∪ {B(y_{β}, ρ_{β}), y_{β} ∈ Y, β ∈ Γ_{2}}

be such a collection in X ∪ Y. We may clearly assume that both Γ_{1} and Γ_{2} are non-empty. By assumption we have for every α ∈ Γ_{1} and β ∈ Γ_{2}

r_{α} + ρ_{β} ≥ d(x_{α}, p) + d(y_{β}, p).

If r_{α} ≥ d(x_{α}, p) for every α ∈ Γ_{1} and ρ_{β} ≥ d(y_{β}, p) for every β ∈ Γ_{2} then p is a point common to all the balls in the collection. Otherwise there is, say, an α_{0} ∈ Γ_{1} so that r_{α_{0}} < d(x_{α_{0}}, p). Then R = inf_{β∈Γ_{2}}(ρ_{β} − d(y_{β}, p)) > 0. The collection of balls {B(x_{α}, r_{α}), α ∈ Γ_{1}} ∪ {B(p, R)} in X will be mutually intersecting and thus have a point x in common. It is trivial to check that this point x is a common point to the given collection of balls in X ∪_{1} Y.

That every metric tree is a 1-absolute Lipschitz retract is basic to the theory of metric trees. We now show, independently of this fact, that the same holds for SMTs. Corollary 1. Let T be an SMT. Then T is a 1-absolute Lipschitz retract.

Proof. Let {I_{n}}_{n=1}^{∞}, {T_{n}}_{n=1}^{∞} describe an allowed construction of T and assume that T is isometrically contained in a metric space S. By Lemma 1 and induction, for each n there is a nonexpansive retraction P_{n} from S onto T_{n}. It is natural to try to get a cluster point of the mappings P_{n} in the space of functions from S to T under the topology of pointwise convergence. Since, except in some degenerate situations, T is not locally compact, this presents a problem. However, the space T can be isometrically embedded into ℓ_{1} in an obvious way as a weak^{∗} closed subset. (Map I_{1} isometrically to an interval or ray in the direction e_{1}. Having defined the mapping, f, on T_{n}, extend f to T_{n+1} by mapping I_{n+1} isometrically into f(p_{n}) + ℝ^{+}e_{n+1}.) Any cluster point of {P_{n}}_{n=1}^{∞} in the space of functions from S to T under the topology of weak^{∗}-pointwise convergence is a nonexpansive retraction from S onto T.

Let Δ be the Cantor set {−1, 1}ℕ and for each n let r_{n} be the nth coordinate projection on Δ. In the space C(Δ), the sequence {r_{n}}_{n=1}^{∞} is isometrically equivalent to the unit vector basis of ℓ_{1}. For n = 1, 2, …, let E_{n} be the functions in C(Δ) which depend only on the first n coordinates.

Notice that if x is in E_{n} and m > n then for all real t, ∥x + tr_{m}∥ = ∥x∥ + |t|. In other words, if I is a ray in the direction of r_{m} emanating from a point p in E_{n}, then, in C(Δ), the set E_{n} ∪ I is an ℓ_{1} union of E_{n} and I. That {r_{n}}_{n=1}^{∞} acts like the ℓ_{1} basis over C(Δ) is the key to proving Proposition 3. Proposition 3. Every SMT is a 1-Lipschitz quotient of C(Δ).

Proof. Let {I_{n}}_{n=1}^{∞}, {T_{n}}_{n=1}^{∞} describe an allowed construction of the SMT T. The desired Lipschitz quotient map is defined by induction by defining at step n the map on E_{n}. Let {x_{n}}_{n=1}^{∞} be a dense sequence in ∪_{n=1}^{∞} E_{n} so that for each n, x_{n} is in E_{n}, and let Y_{0} be a subset of ∪_{n=1}^{∞} T_{n} such that Y_{0} ∩T_{n} is dense in T_{n} for each n. Suppose that we have a 1-Lipschitz mapping, f, from E_{n} into T_{n} and y is in Y_{0} ∩ T_{n+1}. We can extend f to a 1-Lipschitz map, f̃, from E_{n} ∪ (x_{n} + ℝ^{+}r_{n+1}) by mapping the set {x_{n} + te_{n+1} : 0 ≤ t ≤ d(f(x_{n}), y)} isometrically onto the geodesic arc from f(x_{n}) to y and setting f̃(x_{n} + te_{n+1}) = y for t > d(f(x_{n}), y)}. Then, since T_{n+1} is a 1-absolute Lipschitz retract by Corollary 1, we can extend f̃ to a 1-Lipschitz map from E_{n+1} into T_{n+1}. Since the sequence {x_{n}}_{n=1}^{∞} is dense in C(Δ), it is evident that a judicious selection of the points from Y_{0} will produce a 1-Lipschitz quotient mapping from C(Δ) onto T.

Corollary 2 is an immediate consequence of Propositions 3 and 2. Corollary 2. If Y is a separable, complete, metrically convex metric space, then Y is a 1-Lipschitz quotient of C(Δ). Remark. Corollary 2 can be proved without using Proposition 2 and in fact without mentioning the concept of SMT. One defines the Lipschitz quotient on E_{n} by induction as in the proof of Proposition 3. To extend the 1-Lipschitz mapping f̃: E_{n} ∪ (x_{n} + ℝ^{+}r_{n+1}) → Y to a 1-Lipschitz mapping from E_{n+1} into Y, one uses the fact that E_{n} ∪ (x_{n} + ℝ^{+}r_{n+1}) is a 1-absolute Lipschitz retract, which follows from Lemma 1 because the space E_{n} is isometrically isomorphic to ℓ_{∞}^{n} and thus is a 1-absolute Lipschitz retract. Theorem 1. Let X be a separable Banach space which contains a subspace isomorphic to ℓ_{1} and let ε > 0. Then every separable, complete, metrically convex metric space is a (1 + ε)-Lipschitz quotient of X.

Proof. By a theorem of James [11] X contains a subspace (1 + ε)-isomorphic to ℓ_{1}. Pełczyński [15],

[16] proved that this implies that C(Δ) is (1 + ε)-isomorphic to a linear quotient of X. Indeed, let T be a (1 + ε) linear quotient from the subspace of X (1 + ε)-isomorphic to ℓ_{1} onto C(Δ). Extend T to a map S, with norm ∥T ∥, from X to L_{∞}(Δ) and let Y denote the (separable) closure of the range of S. The main result of [16] implies that C(Δ) contains a space isometric to C(Δ) which is norm one complemented in Y. Denote the projection by P. Then P S is a (1 + ε)-linear quotient map.

Theorem 1 follows from this and Corollary 2.

An alternative way to prove Theorem 1 without referring to [15] is sketched in the proof of Theorem 5 below.

Recall that a metric space Y is C-quasiconvex provided that any two points x and y in Y can be joined by an arc whose length is at most Cd(x, y). Non essential modifications of the proofs given above yield Theorem 2. Theorem 2. Let X be a separable Banach space which contains a subspace isomorphic to ℓ_{1}, let ε > 0, and let 1 ≤ C < ∞. Then every separable, complete, C-quasiconvex metric space is a (1 + ε)C-Lipschitz quotient of X.

We do not know if a Banach space X with the property that every separable, complete, Cquasiconvex metric space is a Lipschitz quotient of X must contain an isomorphic image of ℓ_{1}. However the next proposition says that such an X must contain a metric space eqivalent to an ℓ_{1} tree. Proposition 4. If X is a Banach space that has an ℓ_{1} tree as a Lipschitz quotient space then X contains a subset which is Lipschitz equivalent to an ℓ_{1} tree.

Proof. By [4], since X admits a non-Asplund Banach space as a Lipschitz quotient, X is non-Asplund. Let Y be any separable subspace of X whose dual is non separable. Then by a theorem of Stegall ([18] or [8] p. 194), there is a bounded linear operator from Y to C(Δ) such that the image of the unit ball of Y contains a Haar system in C(Δ). Recall that in this context a Haar system is a collection of indicator functions of sets which form a dyadic tree under the order of inclusion. Passing to a subsequence of the sets we get a collection of sets {An̄} indexed by all the finite sequences of integers n̄ = (n_{1}, …, n_{k}) and such that An̄ ⊃ Am̄ whenever n̄ is an initial segment of m̄ and A_{(n_{1},...,n_{k},r)} ∩ A_{(n_{1},...,n_{k},s)} = ∅ whenever r ≠ s. Let {r_{i}} be a dense sequence in [0, ∞). For t̄ = (r_{i_{1}}, …, r_{i_{k}}, s) with s ∈ [0, ∞) let ut̄ = r_{i_{1}}1_{A_{∅}} + r_{i_{2}}1_{A_{i_{1}}} + ⋯ + r_{i_{k}}1_{A_{(i_{1},...,i_{k−1})}} + s1_{A_{(i_{1},...,i_{k})}}, where 1_{A} denotes the indicator function of the set A. It is not difficult to check that the closure in C(Δ) of the collection of all such ut̄ is 2-Lipschitz equivalent to an ℓ_{1} tree (with nodes of the form r_{i_{1}}1_{A_{∅}} + r_{i_{2}}1_{A_{i_{1}}} + ⋯ + r_{i_{k}}1_{A_{(i_{1},...,i_{k−1})}}). For each n̄ let xn̄ be an element in the unit ball of Y which is mapped by Stegall’s map to 1_{An̄} and extend this discrete map (of {xn̄} to {1_{An̄}}) linearly, denoting the inverse image of ut̄ by xt̄. The set {xt̄} is easily seen to be Lipschitz equivalent to an ℓ_{1} tree.

Remark. We do not know if every Banach space X containing (Lipschitz equivalent image of) an ℓ_{1} tree has the property that every separable, complete, metrically convex metric space is a Lipschitz quotient of X.

It is known, and more or less clear from the unique arc property, that every connected subset of a metric tree is contractible. Using this fact it is very simple to prove Proposition 5.

Proposition 5. Let f be a Lipschitz mapping from the Banach space X into the Banach space Y. Assume that f factors continuously through a metric tree; i.e., there is a metric tree T and continuous functions g : X → T and h : T → Y so that hg = f. If f is Gâteaux differentiable at some point x in X, then the Gâteaux derivative f′(x) of f at x has rank at most one.

Proof. Assume, without loss of generality, that x = 0 and f(0) = 0. Suppose, for contradiction, that f′(0) has rank at least two. By composing with a linear projection onto a two dimensional subspace of Y we can assume that Y is two dimensional and that f′(0) has rank two. By restricting to a suitable two dimensional subspace of X, we may further assume that X is two dimensional. Finally, by composing with suitable isomorphisms from R^{2} to X and from Y to R^{2}, we may assume that X = Y = R^{2} and f′(0) is the identity mapping on R^{2}.

Now that we are in the finite dimensional setting we know that f′(0) is a Fréchet derivative because f is a Lipschitz function (this is the only place where the Lipschitz assumption on f is used). So there exists r > 0 so that if x is in R^{2} and |x| = r, then |x − f(x)| < r/2 (recall that f(0) = 0 and f′(0) is the identity). In particular, zero is not in f[rS^{1}]. However, g[rS^{1}], hence also f[rS^{1}] = hg[rS^{1}], is contractible in itself. This implies the absurdity that rS^{1} is also contractible in R^{2} ∼ {0}.

Using the fact, to be explained in Section 4, that every SMT is a metric tree, we get from Propositions 2, 3, and 5 that: Theorem 3. Let X be a separable Banach space which contains a subspace isomorphic to ℓ_{1} and let Y be any separable Banach space. There there is a Lipschitz quotient map from X onto Y so that whenever the Gâteaux derivative of f exists, it has rank at most one.

It is well known (see e.g. [5], p. 155) that if Y has the Radon Nikodym property then any Lipschitz function from a separable space into Y is Gâteaux differentiable outside a Gauss null set.

One can prove Theorem 3 without referring to the theory of metric trees. One needs only to prove that any connected subset of an SMT is contractible and use the proof of Proposition 5 to prove an analogous result for SMTs. The direct proof of the contractability of connected sets in an SMT is not hard but quite lengthy so we prefer not to include it here.

We end this section with an application of Theorem 1 to concepts studied by David and Semmes [6]. Theorem 4. Let X be a separable Banach space which has a subspace isomorphic to ℓ_{1}. Then there is a measure non collapsing mapping f from X into R^{2} whose range is a closed set with empty interior.

Proof. First note that if {B_{n}}_{n=1}^{∞} is a sequence of non overlapping open balls in R^{2} then R^{2} ∼ ∪_{n=1}^{∞} B_{n} is π-quasiconvex. To prove this it is sufficient to check that for each N = 1, 2, …, R^{2} ∼ ∪_{n=1}^{N} B_{n} is π-quasiconvex. To prove this latter statement, draw the line segment between two points

in R^{2} ∼ ∪_{n=1}^{N} B_{n}. If the segment intersects B_{n} with n ≤ N, then it intersects \overline{B_{n}} in a chord. Replace the chord by the arc on the boundary of B_{n} cut by the chord, which has length at most π times the length of the chord.

Now let {x_{n}}_{n=1}^{∞} be dense in the plane. We choose by recursion open balls B_{n} centered at x_{n} so that for each n, B_{n} does not intersect any B_{i} with i < n and with the convention that B_{n} is the “empty ball” exactly when x_{n} is in the closure of ∪_{i=1}^{n} B_{i}. The balls are chosen so that for

each N and each x in R^{2} ∼ ∪_{n=1}^{N} B_{n}, the measure of B(x, r) ∼ ∪_{n=1}^{N} B_{n} is larger than \frac{π}{2} r^{2}. The first ball can be any ball centered at x_{1}. Assume that B_{1}, B_{2}, …, B_{n} have been chosen to satisfy the stated conditions. In the non trivial case that D := d(x_{n+1}, ∪_{n=1}^{N} B_{n}) > 0, we need to choose ε > 0

sufficiently small so that if we set B_{n+1} := B(x_{n+1}, ε), then ∪_{n=1}^{N+1} B_{n} satisfies the desired conditions.

Now balls with radius at most D/4 cannot intersect both B(x_{n+1}, D/2) and ∪_{n=1}^{N} B_{n}. So as long as we choose ε < D/2, we only have to worry about balls with radius r at least D/4. For balls in that range it is evident that for every x in R^{2} ∼ ∪_{n=1}^{N} B_{n}, the measure of B(x, r) ∼ ∪_{n=1}^{N} B_{n} is larger than \frac{π}{2} r^{2} by an amount δ > 0 independent of x and r ≥ D/2. So we can take care of such balls by choosing ε > 0 so that πε^{2} is less than δ. This completes the inductive construction. It is clear that if x is in R^{2} ∼ ∪_{n=1}^{∞} B_{n}, then for every r > 0 the measure of B(x, r) ∼ ∪_{n=1}^{∞} B_{n} is at least (π/2)r^{2}.

Now the set R^{2} ∼ ∪_{n=1}^{∞} B_{n} is π-quasiconvex and complete, so by Theorem 2 there is a Lipschitz quotient mapping f from X onto R^{2} ∼ ∪_{n=1}^{∞} B_{n}. But then f, when considered as map into R^{2}, is measure non collapsing, and of course the range of f is closed and has empty interior.

3 The nonseparable case

We prove in this section non separable versions of Theorem 1. First we discuss the notion of MT. The concept of MT is defined by describing how an MT is constructed. The construction is a transfinite version of the construction of an SMT. Let ℵ be a cardinal number and identify ℵ with the set of all ordinals whose cardinality is less than ℵ. Let I_{1} be a closed interval or a closed ray and define T_{1} := I_{1}. Having defined T_{α} for all α < β, where β < ℵ, let I_{β} be a closed interval or a closed ray whose intersection with the completion T̃_{β} of ∪_{α<β} T_{α} is an end point, p_{β}, of I_{β}. Define T_{β} to be the ℓ_{1} union of T̃_{β} and I_{β}. The completion, T, of ∪_{α<ℵ} T_{α} is an MT. If each I_{α} is a ray with end point p_{α} for α > 1 and for every non empty open subset U of T the set {α < ℵ : p_{α} ∈ U} has cardinality ℵ, then we call T an ℓ_{1}(ℵ) tree and say that {I_{α}}_{α<ℵ}, {T_{α}}_{α<ℵ} describe an allowed construction of T. It is more or less clear that there are ℓ_{1}(ℵ) trees for each infinite cardinal ℵ.

In Section 4 we explain why a metric space is an MT if and only if it is a metric tree. From the general theory of metric trees we can then conclude that every MT is a 1-absolute Lipschitz retract.

Alternatively, this can be proved directly: Let {I_{α}}_{α<ℵ}, {T_{α}}_{α<ℵ} describe an allowed construction of the tree T and show by transfinite induction, using an argument similar to that in the proof of Corollary 1, that each T_{α} is a 1-absolute Lipschitz retract.

By replacing the induction argument in Proposition 2 by a transfinite induction argument, we get a non separable version: Proposition 6. Let T be an ℓ_{1}(ℵ) tree. Then every complete, metrically convex metric space whose density character is at most ℵ is a 1-Lipschitz quotient of T.

There is also a non separable version of Proposition 3. However, since Pełczyński’s theorem [15] does not have a non separable analogue, there are limited applications of the non separable extension of Proposition 3. As a replacement in the non separable setting, we have Proposition 7. Let Δ_{ℵ} be the group {−1, 1}^{ℵ} (where ℵ is an uncountable cardinal) and let μ be the normalized Haar measure on the Borel subsets of Δ_{ℵ}; i. e., μ is the product measure ν^{ℵ} where ν is normalized counting measure on {−1, 1}. For each α < ℵ, let r_{α} be the projection of Δ_{ℵ} onto its α-th coordinate. Then {r_{α}}_{α<ℵ} is ‘ℓ_{1}(ℵ) over L_{∞}(Δ_{ℵ}, μ)’ in an obvious sense. Specifically, if X is a subspace of L_{∞}(Δ_{ℵ}, μ) with density character strictly less than ℵ, then (since every function in L_{∞}(Δ_{ℵ}, μ) depends on only countably many coordinates) there is a subset Δ_{0} of Δ_{ℵ} of cardinality strictly less than ℵ so that each function in x depends only on the coordinates in Δ_{0}. It follows that if α is in the (non empty) set Δ_{ℵ} ∼ Δ_{0} and x is in X, then ∥x + tr_{α}∥ = ∥x∥ + |t| for all real t. Thus transfinite induction and the same considerations used in the proof of Proposition 3 yield: Proposition 7. Let X be a subspace of L_{∞}(Δ_{ℵ}, μ) (where ℵ is an uncountable cardinal) with density character ℵ and assume that {r_{α}}_{α<ℵ} ⊂ X. Then every MT of density character at most ℵ is a 1-Lipschitz quotient of X.

If ℵ^{ℵ_{0}} = ℵ, then the density character of L_{∞}(Δ_{ℵ}, μ) itself is ℵ and hence L_{∞}(Δ_{ℵ}, μ) is isometrically isomorphic to a quotient space of ℓ_{1}(ℵ). Now if X is a Banach space which contains a subspace isomorphic to ℓ_{1}(ℵ) then for each ε > 0, X contains a subspace 1 + ε-isomorphic to ℓ_{1}(ℵ) by the non separable version of a theorem of James [11] (this requires only that ℵ be infinite). Since L_{∞}(Δ_{ℵ}, μ) is injective, it follows that L_{∞}(Δ_{ℵ}, μ) is 1 + ε-isomorphic to a linear quotient of X for every ε > 0. Therefore, from Propositions 7 and 6 we get our first non separable version of Theorem 1. Theorem 5. Let X be a Banach space which contains a subspace isomorphic to ℓ_{1}(ℵ) and let ε > 0. Assume that ℵ = ℵ^{ℵ_{0}}. Then every complete, metrically convex metric space of density character at most ℵ is a (1 + ε)-Lipschitz quotient of X.

In case ℵ < ℵ^{ℵ_{0}} we can prove a weaker version of Theorem 5: Theorem 6. Let X be a Banach space of density character ℵ which contains a subspace isomorphic to ℓ_{1}(ℵ) and let ε > 0. Then every complete, metrically convex metric space of density character at most ℵ is a (1 + ε)-Lipschitz quotient of X.

Proof. Let {x_{α}}_{α<ℵ} be unit vectors in X which are 1+ε-equivalent to the unit vector basis of ℓ_{1}(ℵ). Since L_{∞}(Δ_{ℵ}, μ) is injective, the mapping x_{α} → r_{α} extends to a linear operator L of norm at most 1 + ε from X into L_{∞}(Δ_{ℵ}, μ). The range of this linear operator has density character at most ℵ and L is an isomorphism from the closed space of {x_{α}}_{α<ℵ} onto the closed span of {r_{α}}_{α<ℵ}, so the same considerations as in Proposition 7 yield that every ℓ_{1}(ℵ) tree is a 1 + ε-Lipschitz quotient of X. Now apply Proposition 6.

Remark. 1. It is well known that ℓ_{∞} has a subspace isometric to ℓ_{1}(2^{ℵ_{0}}). Indeed, ℓ_{∞} ⊃ C(0, 1)^{∗} (like the dual of any separable space) and the atomic measures in C(0, 1)^{∗} are isometric to ℓ_{1}(2^{ℵ_{0}}). Hence by Theorem 5 any complete metrically convex metric space of density character ≤ 2^{ℵ_{0}} is an image of ℓ_{∞} by a map whose Lipschitz and co-Lipschitz constants are 1. In particular there is a Lipschitz quotient map from ℓ_{∞} onto c_{0} (the question of the validity of this result was raised in several places, see e.g. [5], p. 279).

2. It is a formal consequence of the preceding remark that there is retraction from ℓ_{∞} onto c_{0} that is a Lipschitz quotient map. To see this, let P be a Lipschitz retraction from ℓ_{∞} onto c_{0}; for a proof of the existence of such a mapping see [5], p. 14. Let f be a Lipschitz quotient mapping from ℓ_{∞} onto c_{0} with f(0) = 0. Define a function g on ℓ_{∞} ⊕ ℓ_{∞} by g(x, y) := (P (x) + f(y), 0). The map g is easily seen to be a retraction and Lipschitz quotient map from ℓ_{∞} ⊕ ℓ_{∞}(≡ ℓ_{∞}) onto its subspace c_{0} ⊕ {0}. By the subspace homogeneity of ℓ_{∞} (see [14]), there is also a retraction from ℓ_{∞} onto c_{0} that is a Lipschitz quotient map.

3. In the literature there are several non-obvious criteria for deciding for which uncountable cardinal ℵ does ℓ_{1}(ℵ) embed isomorphically into a given Banach space X. See [2] and its references. These criteria depend often on the axioms one uses in set theory. From these criteria and Theorem 5 one may deduce some results on the existence of non-linear quotient maps between suitable spaces. We do not enter into this here.

4 Metric trees

Recall that a complete metric space X is a metric tree (also called an R-tree) provided that for any two points x and y in X, there is a unique arc joining x to y, and this arc is a geodesic arc. See [10] for an overview of the study of metric trees and [9] for statements and sometimes proofs of the most basic properties of metric trees. References to original sources are contained in these two papers.

One particularly elegant characterization of metric trees and subsets thereof is given by the four point condition. A metric space X is said to satisfy the four point condition provided that for each set x, y, w, z of four points in X, two of the three numbers d(x, y) + d(w, z), d(x, w) + d(y, z), d(x, z)+d(y, w) are the same, and that number is at least as large as the third. The characterizations are that

1. A metric space embeds isometrically into some metric tree if and only if it satisfies the four point condition.

2. A metric space is a metric tree if and only if it is complete, connected, and satisfies the four point condition.

This characterization of metric trees makes Proposition 8 very easy to prove. Proposition 8. 1. A metric space T is an SMT if and only if T is a separable metric tree.

2. A metric space T is an MT if and only if T is a metric tree.

Proof. Assume that T is an SMT and suppose that {I_{n}}_{n=1}^{∞}, {T_{n}}_{n=1}^{∞} describe an allowed construction of T. It is easy to check by induction that each T_{n} satisfies the four point condition and hence so does ∪_{n=1}^{∞} T_{n}, whence so does T because the four point condition obviously passes to completions. Furthermore, ∪_{n=1}^{∞} T_{n} is arcwise connected, hence its completion T is connected. Thus T is a metric tree by the characterization (2) above. Of course, an SMT is separable, so this completes the proof of the “only if” part of (1). To go the other way, assume that T is a separable metric tree and let {x_{n}}_{n=1}^{∞} be a countable dense subset of T. We need to build subsets {I_{n}}_{n=1}^{∞}, {T_{n}}_{n=1}^{∞} of T which describe an allowed construction of T. Let I_{1} = T_{1} = {x_{1}}. Having defined T_{n}, let Ĩ_{n+1} be the geodesic arc from x_{1} to x_{n}, let x̃_{n} be the last point on this arc which is in T_{n} (set x̃_{n} = x_{n} if x_{n} is in T_{n}), let I_{n+1} be the geodesic arc from x̃_{n} to x_{n}, and set T_{n+1} = T_{n} ∪ I_{n+1}. It is easy to check that this works.

The second statement is proved similarly.

References

[1] I. Aharoni, Every separable metric space is Lipschitz equivalent to a subset of c_{0}, Israel J. Math. 19 (1974), 284–291.

[2] S. Argyros, J. Bourgain and T. Zachariades, A remark on the isomorphic embeddability of ℓ_{1}(Γ), Studia Math. 78 (1984), 77–91.

[3] P. Assouad, Remarques sur un article de Israel Aharoni sur les prolongements Lipschitziens dans c_{0}, Israel J. Math. 31 (1978), 97–100.

[4] S. M. Bates, W. B. Johnson, J. Lindenstrauss, D. Preiss and G. Schechtman, Affine approximation of Lipschitz functions and nonlinear quotients, Geom. Funct. Anal. 9 (1999), 1092–1127.

[5] Y. Benyamini and J. Lindenstrauss, Geometric nonlinear functional analysis, Vol. 1 Colloquium Publication Amer. Math. Soc., Volume 48 (2000).

[6] G. David and S. Semmes, Regular mappings between dimensions, Publ. Mat. 44 (2000), 369– 417.

[7] R. Deville, G. Godefroy and V. Zizler, Smoothness and renormings in Banach spaces, Pitman Monographs, Vol. 64 Longman Scientific and Technical Publisher (1993).

[8] J. Diestel and J. J. Uhl, Vector measures, Mathematical Surveys, No. 15. AMS, Providence, R.I. (1977).

[9] Andreas W. M. Dress, Trees, tight extensions of metric spaces, and the cohomological dimension of certain groups: a note on combinatorial properties of metric spaces, Adv. in Math. 53 (1984), no. 3, 321–402.

[10] A. Dress, V. Moulton, and W. Terhalle, T -theory: an overview, Europ. J. Combinatorics 17 (1996), 161–175.

[11] R. C. James, Uniformly non-square Banach spaces, Ann. of Math. (2) 80 (1964), 542–550.

[12] W. B. Johnson, J. Lindenstrauss, D. Preiss and G. Schechtman, Almost Fréchet differentiability of Lipschitz mappings between infinite-dimensional Banach spaces, Proc. London Math. Soc., to appear.

[13] N. V. Khue and N. T. Nhu, Lipschitz extensions and Lipschitz retractions in metric spaces, Colloquium Math. 45 (1981), 245–250.

[14] J. Lindenstrauss and H. P. Rosenthal, Automorphisms in c_{0}, ℓ_{1} and m, Israel J. Math. 7 (1969), 227–239.

[15] A. Pełczyński, On Banach spaces containing L_{1}(μ). Studia Math. 30 (1968), 231–246.

[16] A. Pełczyński, On C(S)-subspaces of separable Banach spaces. Studia Math. 31 (1968), 513– 522.

[17] M. Ribe, On uniformly homeomorphic normed spaces. Ark. Mat. 14 (1976), 237–244.

[18] C. Stegall, The Radon-Nikodym property in conjugate Banach spaces. Trans. Amer. Math. Soc. 206 (1975), 213–223.

W.B. Johnson  J. Lindenstrauss
Department of Mathematics  Institute of Mathematics
Texas A&M University  Hebrew University of Jerusalem
College Station, TX 77843 U.S.A. Jerusalem, Israel

johnson@math.tamu.edu         joram@math.huji.ac.il

D. Preiss  G. Schechtman
Department of Mathematics  Department of Mathematics
University College London  Weizmann Institute of Science
London, Great Britain  Rehovot, Israel

dp@math.ucl.ac.uk             gideon@wisdom.weizmann.ac.il