Metrical characterization of super-reflexivity and linear type of Banach
  spaces
METRICAL CHARACTERIZATION OF SUPER-REFLEXIVITY AND
LINEAR TYPE OF BANACH SPACES
FLORENT BAUDIER†
Abstract. We prove that a Banach space X is not super-reflexive if and only if the
hyperbolic infinite tree embeds metrically into X. We improve one implication of
J.Bourgain’s result who gave a metrical characterization of super-reflexivity in Banach
spaces in terms of uniforms embeddings of the finite trees. A characterization of the
linear type for Banach spaces is given using the embedding of the infinite tree equipped
with the metrics dp induced by the ℓp norms.
1. Introduction and Notation
We fix some notation and recall basic results.
Let (M,d) and (N, δ) be two metric spaces and an injective map f : M → N . Following
[11], we define the distortion of f to be
dist(f) := ‖f‖Lip‖f
−1‖Lip = sup
x 6=y∈M
δ(f(x), f(y))
d(x, y)
. sup
x 6=y∈M
d(x, y)
δ(f(x), f(y))
If dist(f) is finite, we say that f is a metric embedding, or simply an embedding of M into
And if there exists an embedding f from M into N , with dist(f) ≤ C, we use the notation
→֒ N .
Denote Ω0 = {∅}, the root of the tree. Let Ωn = {−1, 1}
n, Tn =
i=0 Ωi and
n=0 Tn. Thus Tn is the finite tree with n levels and T the infinite tree.
For ε, ε′ ∈ T , we note ε ≤ ε′ if ε′ is an extension of ε.
Denote |ε| the length of ε; i.e the numbers of nodes of ε. We define the hyperbolic dis-
tance between ε and ε′ by ρ(ε, ε′) = |ε|+ |ε′|−2|δ|, where δ is the greatest common ancestor
of ε and ε′. The metric on Tn, is the restriction of ρ.
For a Banach space X , we denote BX its closed unit ball, and X
∗ its dual space.
T embeds isometrically into ℓ1(N) in a trivial way. Actually, let (eε)ε∈T be the canonical
basis of ℓ1(T ) (T is countable), then the embedding is given by ε 7→
s≤ε es.
Laboratoire de Mathématiques, UMR 6623
Université de Franche-Comté,
25030 Besançon, cedex - France
†florent.baudier@math.univ-fcomte.fr
2000 Mathematics Subject Classification. (46B20) (51F99)
http://arxiv.org/abs/0704.1955v1
2 FLORENT BAUDIER†
Aharoni proved in [1] that every separable metric space embeds into c0, so T does.
The main result of this article is an improvement of Bourgain’s metrical characterization
of super-reflexivity. Bourgain proved in [2] that X is not super-reflexive if and only if the
finite trees Tn uniformly embed into X (i.e with embedding constants independent of n).
Obviously if T embeds into X then the T ′ns embed uniformly into X and X is not super-
reflexive, but if X is not super-reflexive we did not know whether the infinite tree T embeds
into X . In this paper, we prove that it is indeed the case :
Theorem 1.1. Let X be a non super-reflexive Banach space, then (T, ρ) embeds into X.
The proof of the direct part of Bourgain’s Theorem essentially uses James’ characteriza-
tion of super-reflexivity (see [7]) and an enumeration of the finite trees Tn. We recall James’
Theorem :
Theorem 1.2 (James). Let 0 < θ < 1 and X a non super-reflexive Banach space, then :
∀ n ∈ N, ∃ x1, x2, . . . , xn ∈ BX , ∃ x
2, . . . , x
n ∈ BX∗ s.t :
x∗k(xj) = θ ∀k < j
x∗k(xj) = 0 ∀k ≥ j
2. Metrical characterization of super-reflexivity
The main obstruction to the embedding of T into any non-super-reflexive Banach space
X is the finiteness of the sequences in James’ characterization. How, with a sequence of
Bourgain’s type embedding, can we construct a single embedding from T into X ?
In [13], Ribe shows in particular, that
2 lpn and (
2 lpn)
l1 are uniformly homeo-
morphic, where (pn)n is a sequence of numbers such that pn > 1, and pn tends to 1. But T
embeds into l1, hence via the uniform homeomorphism T embeds into
2 lpn . However T
does not embed into any lpn(they are super-reflexive).
The problem solved in the next theorem, inspired in part by Ribe’s proof, is to construct
a subspace with a Schauder decomposition
Fn where T2n+1 embeds into Fn and to repast
properly the embeddings in order to obtain the desired embedding.
Proof of Theorem 1.1 : Let (εi)i≥0, a sequence of positive real numbers such that∏
i≥0(1 + εi) ≤ 2, and fix 0 < θ < 1. Let kn = 2
2n+1+1 − 1.
First we construct inductively a sequence (Fn)n≥0 of subspaces of X , which is a Schauder
finite dimensional decomposition of a subspace of X s.t the projection from
i=0 Fi onto⊕p
i=0 Fi, with kernel
i=p+1 Fi (with p < q) is of norm at most
i=p (1+εi), and sequences
xn,1, xn,2, . . . , xn,kn ∈ BFn
x∗n,1, x
n,2, . . . , x
∈ BX∗
s.t :
x∗n,k(xn,j) = θ ∀k < j
x∗n,k(xn,j) = 0 ∀k ≥ j.
Metrical characterization of super-reflexivity and linear type of Banach spaces 3
Denote Φn : Tn → {1, 2, . . . , 2
n+1− 1} the enumeration of Tn following the lexicographic
order. It is an enumeration of Tn such that any pair of segments in Tn starting at incompa-
rable nodes (with respect to the tree ordering ≤) are mapped inside disjoint intervals.
Let Ψn = Φ2n+1 and Γn = T2n+1.
X is non super-reflexive, hence from James’ Theorem :
∃ x0,1, x0,2, . . . , x0,7 ∈ BX , ∃ x
0,1, x
0,2, . . . , x
0,7 ∈ BX∗ s.t :
x∗0,k(x0,j) = θ ∀k < j
x∗0,k(x0,j) = 0 ∀k ≥ j.
Γ0 = T2 embeds into X via the embedding f0(ε) =
s≤ε x0,Ψ0(s) (see [2]).
Let F0 = Span{x0,1, . . . , x0,7}, then dim(F0) < ∞.
Suppose that F0, . . . , Fp, and
xp,1, xp,2, . . . , xp,kp ∈ BFp
x∗p,1, x
p,2, . . . , x
∈ BX∗
verifying the required conditions, are constructed for all p ≤ n.
We apply Mazur’s Lemma (see [9] page 4) to the finite dimensional subspace
i=0 Fi of X .
Thus there exists Yn ⊂ X such that dim(X/Yn) < ∞ and :
‖x‖ ≤ (1 + εn)‖x+ y‖, ∀(x, y) ∈
Fi × Yn
But Yn is of finite codimension in X , hence is not super-reflexive.
From James’ Theorem and Hahn-Banach Theorem:
∃ xn+1,1, xn+1,2, . . . , xn+1,kn+1 ∈ BYn ,
∃ x∗n+1,1, x
n+1,2, . . . , x
n+1,kn+1
∈ BX∗ ,
s.t :
x∗n+1,k(xn+1,j) = θ ∀k < j
x∗n+1,k(xn+1,j) = 0 ∀k ≥ j.
Γn+1 embeds into Yn via the embedding fn+1(ε) =
s≤ε xn+1,Ψn+1(s) .
Let Fn+1 = Span{xn+1,j ; 1 ≤ j ≤ kn+1}, then dim(Fn+1) < ∞, which achieves the induc-
tion.
Now define the following projections :
Let, Pn the projection from Span(
i=0 Fi) onto F0
· · ·
Fn with kernel Span(
i=n+1 Fi).
It is easy to show that ‖Pn‖ ≤
i=n(1 + εi) ≤ 2.
We denote now Π0 = P0 and Πn = Pn − Pn−1 for n ≥ 1. We have that ‖Πn‖ ≤ 4.
4 FLORENT BAUDIER†
From Bourgain’s construction, for all n :
ρ(ε, ε′) ≤ ‖fn(ε)− fn(ε
′)‖ ≤ ρ(ε, ε′),
where fn denotes the Bourgain’s type embedding from Γn in Fn, i.e fn(ε) =
s≤ε xn,Ψn(s).
Note that :
∀ n, ∀ ε ∈ Γn ‖fn(ε)‖ ≤ |ε|.
Now we define our embedding.
f : T → Y = Span(
i=0 Fi) ⊂ X
ε 7→ λfn(ε) + (1− λ)fn+1(ε) , if 2
n ≤ |ε| ≤ 2n+1
where,
2n+1 − |ε|
f(∅) = 0.
We will prove that :
(2) ∀ε, ε′ ∈ T
ρ(ε, ε′) ≤ ‖f(ε)− f(ε′)‖ ≤ 9ρ(ε, ε′).
Remark 2.1 We have θ
|ε| ≤ ‖f(ε)‖ ≤ |ε|.
First of all, we show that f is 9−Lipschitz.
We can suppose that 0 < |ε| ≤ |ε′| w.r.t remark 2.1.
If |ε| ≤ 1
|ε′| then :
ρ(ε, ε′) ≥ |ε′| − |ε| ≥
|ε|+ |ε′|
Hence,
‖f(ε)− f(ε′)‖ ≤ 3ρ(ε, ε′).
|ε′| < |ε| ≤ |ε′|, we have two different cases to consider.
1) if 2n ≤ |ε| ≤ |ε′| < 2n+1.
Then, let
2n+1 − |ε|
and λ′ =
2n+1 − |ε′|
‖f(ε)− f(ε′)‖ = ‖λfn(ε)− λ
′fn(ε
′) + (1 − λ)fn+1(ε)− (1− λ
′)fn+1(ε
≤ λ‖fn(ε)− fn(ε
′)‖+ |λ− λ′|(‖fn(ε
′)‖+ ‖fn+1(ε
′)‖) + (1 − λ)‖fn+1(ε)− fn+1(ε
≤ ρ(ε, ε′) + 2ρ(ε, ε′) + 2ρ(ε, ε′)
≤ 5ρ(ε, ε′),
Metrical characterization of super-reflexivity and linear type of Banach spaces 5
because ‖fn(ε
′)‖ < 2n+1, ‖fn+1(ε
′)‖ < 2n+1 and,
|λ− λ′| =
|ε′| − |ε|
ρ(ε, ε′)
2) if 2n ≤ |ε| ≤ 2n+1 ≤ |ε′| < 2n+2.
Then, let
2n+1 − |ε|
and λ′ =
2n+2 − |ε′|
‖f(ε)− f(ε′)‖ = ‖λfn(ε) + (1− λ)fn+1(ε)− λ
′fn+1(ε
′)− (1− λ′)fn+2(ε
≤ λ(‖fn(ε)‖ + ‖fn+1(ε)‖) + (1− λ
′)(‖fn+1(ε
′)‖+ ‖fn+2(ε
′)‖) + ‖fn+1(ε)− fn+1(ε
≤ ρ(ε, ε′) + 2λ|ε|+ 2(1− λ′)|ε′|
≤ 9ρ(ε, ε′),
because,
ρ(ε, ε′)
, so λ|ε| ≤ 2ρ(ε, ε′).
Similarly
1− λ′ =
|ε′| − 2n+1
ρ(ε, ε′)
and (1− λ′)|ε′| ≤ 2ρ(ε, ε′).
Finally, f is 9-Lipschitz.
Now we deal with the minoration.
In our next discussion, whenever |ε| (respectively |ε′|) will belong to [2n, 2n+1), for some
integer n, we shall denote
2n+1 − |ε|
(respectively λ′ =
2n+1 − |ε′|
We can suppose that ε is smaller than ε′ in the lexicographic order. Denote δ the greatest
common ancestor of ε and ε′. And let d = |ε| − |δ| (respectively d′ = |ε′| − |δ|).
1) if 2n ≤ |ε|, |ε′| ≤ 2n+1.
We have,
x∗n,Ψn(δ)Πn(f(ε)− f(ε
′)) = θ(λd− λ′d′)
x∗n+1,Ψn+1(δ)Πn+1(f(ε)− f(ε
′)) = θ((1 − λ)d− (1 − λ′)d′).
Hence,
‖f(ε)− f(ε′)‖ ≥
θ(d− d′)
−x∗n,Ψn(ε)Πn(f(ε)− f(ε
′)) = θλ′d′
−x∗n+1,Ψn+1(ε)Πn+1(f(ε)− f(ε
′)) = θ(1− λ′)d′.
6 FLORENT BAUDIER†
‖f(ε)− f(ε′)‖ ≥
Finally if we distinguish the cases d
≤ d′, and d′ < d
we obtain :
‖f(ε)− f(ε′)‖ ≥
θ(d + d′)
ρ(ε, ε′).
2) if 2n ≤ |ε| ≤ 2n+1 ≤ 2q+1 ≤ |ε′| ≤ 2q+2,
or 2n ≤ |ε′| ≤ 2n+1 ≤ 2q+1 ≤ |ε| ≤ 2q+2.
If n < q,
|x∗q+1,Ψq+1(δ)Πq+1(f(ε)− f(ε
′)) + x∗q+2,Ψq+2(δ)Πq+2(f(ε)− f(ε
′))| = θMax(d, d′)
Hence,
‖f(ε)− f(ε′)‖ ≥
ρ(ε, ε′).
If n = q and |ε| ≤ |ε′|,
|x∗n+1,Ψn+1(ε)Πn+1(f(ε)− f(ε
′)) + x∗n+2,Ψn+2(δ)Πn+2(f(ε)− f(ε
′))| ≥ θd′.
‖f(ε)− f(ε′)‖ ≥
ρ(ε, ε′).
If n = q and |ε′| < |ε|,
x∗n+1,Ψn+1(δ)Πn+1(f(ε)−f(ε
′))−x∗n+1,Ψn+1(ε)Πn+1(f(ε)−f(ε
′))+x∗n+2,Ψn+2(δ)Πn+2(f(ε)−f(ε
′)) = θd.
Hence,
‖f(ε)− f(ε′)‖ ≥
ρ(ε, ε′).
Finally T
→֒ X .
Corollary 2.2. X is non super-reflexive if and only if (T, ρ) embeds into X.
Proof : It follows clearly from Bourgain’s result [2] and Theorem 1.1. �
Metrical characterization of super-reflexivity and linear type of Banach spaces 7
3. Metric characterization of the linear type
First we identify canonicaly {−1, 1}n with Kn = {−1, 1}
k>n{0}.
Let p ∈ [1,∞).
Then we define an other metric on T =
Kn as follows :
∀ ε, ε′ ∈ T ,
dp(ε, ε
′) = (
|εi − ε
The length of ε ∈ T can be viewed as |ε| = (dp(ε, 0))
The norm ‖.‖p on ℓp coincides with dp for the elements of T .
We recall now two classical definitions :
Let X and Y be two Banach spaces. If X and Y are linearly isomorphic, the Banach-
Mazur distance between X and Y , denoted by dBM (X,Y ), is the infimum of ‖T ‖ ‖T
over all linear isomorphisms T from X onto Y .
For p ∈ [1,∞], we say that a Banach space X uniformly contains the ℓnp ’s if there is a
constant C ≥ 1 such that for every integer n, X admits an n-dimensional subspace Y so
that dBM (ℓ
p , Y ) ≤ C.
We state and prove now the following result.
Theorem 3.1. Let p ∈ [1,∞).
If X uniformly contains the ℓnp ’s then (T, dp) embeds into X.
Proof : We first recall a fundamental result due to Krivine (for 1 < p < ∞ in [8]) and
James (for p = 1 and ∞ in [7]).
Theorem 3.2 (James-Krivine). Let p ∈ [1,∞] and X be a Banach space uniformly con-
taining the ℓnp ’s. Then, for any finite codimensional subspace Y of X, any ǫ > 0 and any
n ∈ N, there exists a subspace F of Y such that dBM (ℓ
p , F ) < 1 + ǫ.
Using Theorem 3.2 together with the fact that each ℓnp is finite dimensional, we can build
inductively finite dimensional subspaces (Fn)
n=0 of X and (Rn)
n=0 so that for every n ≥ 0,
Rn is a linear isomorphism from ℓ
p onto Fn satisfying
∀u ∈ ℓnp
‖u‖ ≤ ‖Rnu‖ ≤ ‖u‖
and also such that (Fn)
n=0 is a Schauder finite dimensional decomposition of its closed
linear span Z. More precisely, if Pn is the projection from Z onto F0 ⊕ ...⊕ Fn with kernel
Span (
i=n+1 Fi), we will assume as we may, that ‖Pn‖ ≤ 2. We denote now Π0 = P0 and
Πn = Pn − Pn−1 for n ≥ 1. We have that ‖Πn‖ ≤ 4.
We now consider ϕn : Tn → ℓ
p defined by
∀ε ∈ Tn, ϕn(ε) =
εiei,
where (ei) is the canonical basis of ℓ
p . The map ϕn is clearly an isometric embedding of Tn
into ℓnp .
8 FLORENT BAUDIER†
Then we set :
∀ε ∈ Tn, fn(ε) = Rn(ϕn(ε)) ∈ Fn.
Finally we construct a map f : T → X as follows :
f : T → X
ε 7→ λfm(ε) + (1− λ)fm+1(ε) , if 2
m ≤ |ε| < 2m+1,
where,
2m+1 − |ε|
Remark 3.3 We have 1
p ≤ ‖f(ε)‖ ≤ |ε|
Like in the proof of Theorem 1.1 ,we prove that f is 9-Lipschitz using exactly the same
computations.
We shall now prove that f−1 is Lipschitz. We consider ε, ε′ ∈ T and assume again that
0 < |ε| ≤ |ε′|. We need to study two different cases. Again, whenever |ε| (respectively |ε′|)
will belong to [2m, 2m+1), for some integer m, we shall denote
2m+1 − |ε|
(respectively λ′ =
2m+1 − |ε′|
1) if 2m ≤ |ε|, |ε′| < 2m+1.
dp(ε, ε
′) ≤ ‖λ
i=1 εiei − λ
∑|ε′|
i=1 ε
iei‖p + ‖(1− λ)
i=1 εiei − (1− λ
∑|ε′|
i=1 ε
iei‖p
≤ 2‖Πm(f(ε)− f(ε
′))‖ + 2‖Πm+1(f(ε)− f(ε
≤ 16‖f(ε)− f(ε′)‖.
2) if 2m ≤ |ε| ≤ 2m+1 ≤ 2q+1 ≤ |ε′| < 2q+2.
if m < q,
dp(ε, ε
′) ≤ 2dp(ε
′, 0)
≤ 2((1− λ′)dp(ε
′, 0) + λ′dp(ε
′, 0))
≤ 2(2‖Πq+2(f(ε)− f(ε
′))‖+ 2‖Πm+1(f(ε)− f(ε
′))‖)
≤ 32‖f(ε)− f(ε′)‖.
Metrical characterization of super-reflexivity and linear type of Banach spaces 9
if m = q,
dp(ε, ε
′) ≤ λdp(ε, 0) + ‖(1− λ)
i=1 εiei − λ
∑|ε′|
i=1 ε
iei‖p + (1 − λ
′)dp(ε
′, 0)
≤ 2‖Πm(f(ε)− f(ε
′))‖+ 2‖Πm+1(f(ε)− f(ε
′))‖ + 2‖Πm+2(f(ε)− f(ε
≤ 24‖f(ε)− f(ε′)‖.
Finally we obtain that f−1 is 32-Lipschitz, and T
→֒ X .
In the sequel a Banach space X is said to have type p > 0 if there exists a constant T < ∞
such that for every n and every x1, . . . , xn ∈ X ,
εjxj‖
≤ T p
where the expectation Eε is with respect to a uniform choice of signs ε1, . . . , εn ∈ {−1, 1}
The set of p’s for which X contains ℓnp ’s uniformly is closely related to the type of X
according to the following result due to Maurey, Pisier [10] and Krivine [8], which clarifies
the meaning of these notions.
Theorem 3.4 (Maurey-Pisier-Krivine). Let X be an infinite-dimensional Banach space.
pX = sup{p ; X is of type p},
Then X contains ℓnp ’s uniformly for p = pX.
Equivalently, we have
pX = inf{p ; X contains ℓ
p’s uniformly}.
We deduce from Theorem 3.1 two corollaries.
Corollary 3.5. Let X a Banach space and 1 ≤ p < 2.
The following assertions are equivalent :
i) pX ≤ p.
ii) X uniformly contains the ℓnp ’s.
iii) the (Tn, dp)’s uniformly embed into X .
iv) (T, dp) embeds into X.
Proof : ii) implies i) is obvious.
i) implies ii) is due to Theorem 3.2 and the work of Bretagnolle, Dacunha-Castelle and
Krivine [4].
For the equivalence between ii) and iii) see the work of Bourgain, Milman and Wolfson
[3] and Krivine [8].
iv) implies iii) is obvious.
And ii) implies iv) is Theorem 3.1.
10 FLORENT BAUDIER†
Corollary 3.6. Let X be an infinite dimensional Banach space, then (T, d2) embeds into
Proof : This corollary is a consequence of the Dvoretsky’s Theorem [6] and Theorem 3.1.
References
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http://arxiv.org/abs/math/0506201
	1. Introduction and Notation
	2. Metrical characterization of super-reflexivity
	3. Metric characterization of the linear type
	References
