Lipschitz-free spaces over ultrametric spaces

We prove that the Lipschitz-free space over a separable ultrametric space has a monotone Schauder basis and is isomorphic to $\ell_1$. This extends results of A. Dalet using an alternative approach.

x ∈ M }. It is known that its dual space is isometrically isomorphic to Lip 0 (M ). We refer to [15] and [7] for an introduction to Lipschitz-free spaces and its basic properties.
The study of the linear structure of Lipschitz-free spaces over metric spaces has become an active field of study, see e.g. [1,2,[6][7][8][9]11]. In this note we are interested in the structure of Lipschitz-free spaces over ultrametric spaces. Let us recall that a metric space (M, d) is said to be ultrametric if for every x, y, z ∈ M , we have d(x, z) ≤ max{d(x, y), d(y, z)}.
A. Dalet [2] proved, among other things, that the Lipschitz-free space over a separable proper ultrametric space has the metric approximation property and is isomorphic to ℓ 1 . We improve the result and show the following. Theorem 1. The Lipschitz-free space over a separable ultrametric space has a monotone Schauder basis.
Theorem 2. The Lipschitz-free space over a separable ultrametric space is isomorphic to ℓ 1 .
The improvement is that we do not assume the ultrametric space to be proper. Moreover, in Theorem 1 we get a stronger conclusion. Our proofs do not follow the lines of the proofs from [2] and so they can be viewed as an alternative approach to the the above mentioned results of A. Dalet as well. In the final section we collect few corollaries of our results and suggest some open problems.
Before coming to the proofs, let us recall some basic results. One of the main properties of the Lipschitz-free spaces is the following universality property that provides a connection between the Lipschitz maps in metric spaces and linear maps in Banach spaces; see [7,Lemma 2.5]. Note that Lemma 2.5 in [7] is formulated only for the case when M is a Banach space; however, a similar proof works also in the more general setting of Lemma 3. Moreover, it is possible to prove Lemma 3 directly in a similar way as [7, Lemma 2.2] -it is enough to replace Lip 0 (Y ) by Y * in its proof. Note that it is straightforward to check that for Lipschitz maps L : M → N ⊂ F(N ) and S:N → P ⊂ F(P ) with L(0 M ) = 0 N and S(0 N ) = 0 P we have SL = S L. Hence, as an immediate consequence of Lemma 3 we get the following facts, which we will use later.
Let (M, d) be an ultrametric space. We will often use the following property of ultrametric spaces which is easy to prove. For x, y, z ∈ M , if d(x, y) = d(y, z) then d(x, z) = max{d(x, y), d(y, z)}.

Monotone Schauder basis
The purpose of this section is to prove Theorem 1. Proof. Since r n (M ) ⊂ r n+1 (M ) for every n ∈ N, we have r n+1 • r n = r n . By Lemma 3, there are projections P n : F(M ) → F(M ) with P n ≤ 1, P n (F(M )) = span{δ M (s k ); k ≤ n} and P n • P m = P min{n,m} for every n, m ∈ N. Obviously, dim P n (F(M )) = n. Since n∈N P n (F(M )) is dense in F(M ), we have P n (x) → x for every x ∈ F(M ). Now, it is a folklore fact [5,Lemma 4.7] that such a system of projections gives us a monotone Schauder basis on F(M ).
Proof of Theorem 1. Let (M, d) be a separable ultrametric space and fix a one-to-one sequence (s n ) n∈N of points from M with 0 M = s 1 and {s n ; n ∈ N} = M . For every n ∈ N, put S n := {s k ; k ≤ n}. We will find a sequence of retractions (r n ) n∈N satisfying the assumptions of Lemma 6. Fix n ∈ N. First, we put I n (x) := {k ∈ N; k ≤ n and dist(x, S n ) = d(x, s k )}. We denote by i n (x) the minimal natural number from I n (x). Finally, we define r n : M → S n by Now, we will verify that the sequence (r n ) n∈N meets the requirements (i) and (ii) from Lemma 6. First, observe the following.
Proof. Fix x, y ∈ M with d(x, y) < dist(x, S n ). In order to see that (1) holds, we show dist(y, S n ) = dist(x, S n ). Indeed, Thus, in order to get a contradiction let us assume dist(y, S n ) < dist(x, S n ). Then In order to see that r n is a 1-Lipschitz mapping, we need to verify If d(x, y) < max{dist(x, S n ), dist(y, S n )} we get from (1) and the symmetry of the situation that i n (x) = i n (y) and (2) is obvious.
and (2) holds. It remains to show that, for every n ∈ N, we have r n • r n+1 = r n . Fix n ∈ N and x ∈ M . Then it follows from the definitions above that either i n+1 (x) = i n (x) or i n+1 (x) = n + 1. In both cases we will get i n (s i n+1 (x) ) = i n (x). Indeed, this is trivial in the first case. Assume i n+1 (x) = n + 1. Then and we are done.

Isomorphism with ℓ 1
The purpose of this section is to prove Theorem 2. First, let us recall the notion of R-trees and its link with Lipschitz-free spaces and ultrametric spaces.
Definition 7. Let (T, d) be a metric space such that, for every x, y ∈ T , there exists a unique isometry φ x,y : [0, d(x, y)] → T with φ x,y (0) = x and φ x,y (d(x, y)) = y. Then we say that T is an R-tree and we define the segment [x, y] by [x, y] := φ x,y ([0, d(x, y)]).
Moreover, we say that v ∈ T is a branching point of T if there are three points We denote by Br(T ) the set of branching points of T .
The link with Lipschitz-free spaces is contained in the following result, which has been proved by Godard in [6,Corollary 3.4]. Note that the definition of an R-tree and of a branching point above is not exactly as in [6], but it is equivalent to it; see [4,Chapter 3].
It belongs to a folklore fact that every ultrametric space embeds into an R-tree. The shortest way of proving this statement is probably to show that an ultrametric space satisfies the "four-point condition" and then use the well-known fact that every metric space which satisfies this condition isometrically embeds into an R-tree, see e.g. [4,Theorem 3.38].
In the following we will combine those two links and show that the Lipschitz-free space over a separable ultrametric space is isomorphic to ℓ 1 . First, we observe that it is enough to consider only "2 n -valued ultrametric spaces".
Definition 9. A metric is said to be 2 n -valued if the only values assumed by the metric are 2 n , n ∈ Z. Fact 10. Any ultrametric space is 2-bi-Lipschitz equivalent to a 2 n -valued ultrametric space.
Next, we show that the embedding of an ultrametric space into R-tree may be done in such a way that it satisfies certain additional conditions, see Proposition 12. In order to find an R-tree into which our ultrametric spaces embeds, we will follow the ideas from [4,Theorem 3.38], where it is proved that any metric space satisfying the "four-point condition" embeds isometrically into an R-tree. However, our space is an ultrametric space, so the construction will be done in an easier way. Once the construction is done, we will show that the additional conditions mentioned above are satisfied.
We begin with the following Lemma, which is inspired by [4,Lemma 3.5].
We will show that τ (i) = φ x,y (i). Put , depending on how u and v are arranged in ρ. It follows that either u = φ x,y (i) or v = φ x,y (i).
Let us consider the case when σ 2 ∩ ρ = {φ x,y (i)}. Then, by (i) and (ii), we have φ As i ∈ (0, d(x, y)) was arbitrary, we have that τ = φ x,y . Hence, isometries φ x,y are unique and (M, d) is R-tree.  Observe that, if d(m, n)/2 > i and d(m, n)/2 > j, we have ρ( m, i , n, j ) = (d(m, n)/2 − i) + (d(m, n)/2 − j). Otherwise, ρ( m, i , n, j ) = |i − j|. It is straightforward to check that ρ is well defined metric on T . Obviously, M ∋ m → m, 0 is an isometric embedding of M into T and T is separable whenever M is. In order to see that (T, ρ) is R-tree we will find, for every x, y ∈ T , isometry φ x,y : [0, ρ(x, y)] → T in such a way that the family (φ x,y ) x,y∈T satisfies the assumptions of Lemma 11. Fix x, y ∈ T . There are m, n ∈ M and i, j ∈ [0, ∞) with x = m, i and y = n, j . We distinguish the following cases: • If j ≤ i and i ≥ d(m, n)/2, we put φ x,y (t) := n, ρ(x, y) + j − t , t ∈ [0, ρ(x, y)].
• If j ≤ i and i < d(m, n)/2, we put . Considering all the possible cases, it is straightforward to check that the family (φ x,y ) x,y∈T satisfies the assumptions of Lemma 11; hence, (T, d) is R-tree and φ x,y are the unique isometries from the definition of an R-tree. "⊂" Fix v ∈ Br(t) and let x k = m k , i k ∈ T \ {v}, k ∈ {1, 2, 3} be the points witnessing the fact that v ∈ Br(T ). There cannot be a point x ∈ M such that for every k ∈ {1, 2, 3} we would have x k = x, i k , as otherwise, we would have that all the points lie on a common line segment. We distinguish two cases: • Two points, let us say x 1 , x 2 , lie on a common branch, i.e. there exists x ∈ M such that x k = x, i k for k ∈ {1, 2}. We will show that in this case v = x, d(x, m 3 )/2 . We may without loss of generality assume that i 1 < i 2 . Notice that Note that so far we have not used the fact that (M, d) is 2 n -valued. Thus, the embedding as described above works for an arbitrary ultrametric space. In order to prove (ii) and (iii) we will use the assumption that (M, d) is 2 n -valued. From now on we will not distinguish between m ∈ M and m, 0 , its isometric copy in T . Find n 0 such that i > d(m, n 0 )/2 = sup{d(m, n)/2; n ∈ M, i > d(m, n)/2} ≥ d(m, m)/2 = 0; note that such an n 0 exists, because and the set {2 n ; n ∈ Z} does not have any positive cluster point. If, for every n ∈ M \ {m}, i > d(m, n)/2, we put ε := min{i − d(m, n 0 )/2, i/2}. Otherwise, we find n 1 with i < d(m, n 1 )/2 = inf{d(m, n)/2; n ∈ M, i < d(m, n)/2} and we put ε := min{i − d(m, n 0 )/2, d(m, n 1 )/2 − i, i/2}. In any case straightforward computations show that B(x, ǫ) = { m, j ; |j − i| < ε} and B(x, ε) ∩ (Br(T ) ∪ M ) = ∅.
As the distances between points in Br(T ) ∪ M cannot be irrational numbers, it follows that Br(T ) ∪ M = Br(T ) ∪ M does not contain any segment [x, y] for x = y. Hence, it remains to prove (iii). For every v ∈ Br(T ) find some m v , n v ∈ M with v = m v , d(m v , n v )/2 . We define the retraction r : Br(T ) ∪ M → M as follows: if a ∈ Br(T ).
Obviously, r • r = r. It remains to show that r is 4-Lipschitz. If a, b ∈ M then obviously ρ(r(a), r(b)) = ρ(a, b). Fix a ∈ M and b ∈ Br(T ). Then ρ(r(a), r(b)) = d(a, m b ). Hence, the estimation of the Lipschitz constant follows from the following Claim.  2ρ( a, 0 , b).
Proof. First, let us prove that Otherwise, Now, the following computation proves the Claim: Proof. As M is 2 n -valued, there are m, n, k ∈ Z with d(m a , m b ) = 2 m , d(m a , n a ) = 2 n and d(m b , n b ) = 2 k . Interchanging the roles of a, b we may without loss of generality assume that n ≥ k. Now, we will show that Indeed, if m > n we have 1 = 2(1 − 2 −1 ) ≤ 2(1 − 2 n−m ) and If m ≤ n and n > k, then we have 1 = 2(1 − 2 −1 ) ≤ 2(1 − 2 k−n ) and We have verified that r : Br(T ) ∪ M → M is a 4-Lipschitz retraction, which proves (iii). This completes the proof of the Proposition. Now, it is straightforward to use the above and prove Theorem 2.
Proof of Theorem 2. Let M be a separable ultrametric space. By Fact 10 and Fact 5, there is a 2 n -valued separable ultrametric space N such that F(M ) is isomorphic to F(N ). By Proposition 12 and Fact 4, there is a separable R-tree T such that F(N ) is isometric to a complemented subspace of F(Br(T ) ∪ N ) and Br(T ) ∪ N does not contain any segment. By Proposition 8, F(Br(T ) ∪ N ) is isometric to ℓ 1 . Thus, F(M ) is isomorphic to a complemented subspace of ℓ 1 . It is a well-known result of Pe lczyński, see e.g. [5,Corollary 4.48], that this is possible only if F(M ) is isomorphic to ℓ 1 .
Remark 13. Note that in the proof of Theorem 2 we could also use the result of Matoušek [12] to see that F(N ) is isometric to a complemented subspace of F(Br(T ) ∪ N ) (because, by [12], there is a linear extension operator from Lip 0 (N ) to Lip 0 (T ) ⊃ Lip 0 (Br(T ) ∪ N )). However, we decided to prove the existence of a retraction instead as it gives us deeper insight into the situation. In this case, the linear extension operator is just Lip 0 (N ) ∋ f → f • r, where r : Br(T ) ∪ N → N is the retraction from (iii) in Proposition 12.

Final Remarks and Open problems
Remark 14. Note that in Fact 10 and Proposition 12 we used the constant 2; however, the same proof works for any constant λ > 1 . Hence, a careful inspection of the proof of Theorem 2 shows us, that for any λ > 1, any separable ultrametric space is λ-isomorphic to a λ-complemented subspace of ℓ 1 . This motivates the following question. Question 1. Is the Lipschitz-free space over a separable ultrametric space isometric to ℓ 1 ?
It is proved in [3,Proposition 15.7] that every "uniformly disconnected" metric space is Lipschitz equivalent to an ultrametric space. We also refer to [3,Definition 15.1] for a precise definition of uniform disconnectedness and for examples. Hence, the following corollary follows from Theorem 2 and Fact 5. Every uniformly disconnected is totally disconnected; however, there exists a countable totally disconnected compact space K which is not uniformly disconnected -consider, for example, K = {1/n; n ∈ N} ∪ {0}. However, A. Dalet proved in [1] that F(K) is a dual space and has MAP whenever K is countable compact. This suggests the following question which has already been asked by G. Godefroy.

Question 2. Does the Lipschitz-free space over a totally disconnected compact metric space have the BAP? Is it a dual space?
Our last observation concerns linearly rigidity. R. Holmes in [10] proved that the Urysohn universal metric space admits (up to isometry) a unique linearly dense isometric embedding into a Banach space. In other words, any isometric embedding of the Urysohn space with a distinguished point into a Banach space X that maps the distinguished point on 0 X extends to a linear isometric embedding of the free space over the Urysohn space F(U) into X. J. Melleray, F. Petrov and A. Vershik in [13] investigated this property of the Urysohn space further and found other metric spaces with this property. They call them linearly rigid metric spaces. This is another corollary of Theorem 2.

Corollary 16. No separable ultrametric space is linearly rigid.
Proof. By [14,Theorem 1], any separable ultrametric space isometrically embeds into c 0 . If it were linearly rigid the embedding would extend to an embedding of the free space of this ultrametric space. However, by Theorem 2 this free space is isomorphic to ℓ 1 , while it is well known that c 0 does not contain a copy of ℓ 1 , a contradiction.