Lipschitz quotients from metric trees and from Banach spaces

containing

William B Johnson

Department of Mathematics Texas AM University Col lege Station TX

USA

Email johnsonmathtamuedu

Joram Lindenstrauss

Institute of Mathematics Hebrew University of Jerusalem

Email jorammathhujiacil

David Preiss

Department of Mathematics University Col lege London London Great Britain

Email dpmathuclacukj

and

Gideon Schechtman

Department of Mathematics Weizmann Institute of Science Rehovot Israel

Email gideonwisdomweizmannacil

INTRODUCTION

ALipschitz map f between the metric spaces X and Y is called a Lips

chitz quotient map if there is a C the smallest such C the coLipschitz

constant is denoted coLipf while Lipf denotes the Lipschitz constant

of f so that for every x X and r fB x r B f xrC

X Y

Thus Lipschitz quotient maps are surjective maps whichby denition have

the prop erty ensured by the op en mapping theorem for surjective linear

op erators b etween Banach spaces

WB Johnson was supp orted in part by NSF DMS and the USIsrael

Binational Science Foundation J Lindenstrauss and G Schechtman were supp orted in

part by the USIsrael Binational Science Foundation and were participants in the NSF

Workshop in Linear Analysis and ProbabilityatTexas AM University

JOHNSON LINDENSTRAUSS PREISS AND SCHECHTMAN

If there is a Lipschitz quotient mapping f from X onto Y with Lipf

coLipf C wesaythat Y is a C Lipschitz quotientofX If Y is a C

Lipschitz quotientof X for some C wesaythat Y is a Lipschitz quotient

of X

Lipschitz quotient maps were intro duced and studied in Among

other results the following prop osition was proved in seealso The

orem

Proposition Let X be a superreexive and let the

Banach space Y bea Lipschitz quotient of X Then Y is crudely nitely

representable in X

The conclusion of the Prop osition means that there is a constant so

that for every nite dimensional linear subspace F of Y there is a nite

dimensional subspace E of X so that dE F where d denotes here

the Banach Mazur distance An equivalentway to express the conclusion

of Prop osition is that Y is linearly isomorphic to a subspace of an

ultrapro duct of X An immediate consequence of Prop osition is that

any Banachspace Y which is a Lipschitz quotient of a Hilb ert space must

b e linearly isomorphic to a Hilb ert space

Prop osition was proved in by using a sp ecial result on approxi

mating Lipschitz functions from a sup erreexive Banach space into nite

dimensional Banach spaces the socalled UAAP see also Prop osi

tion It was proved in that this approximation prop erty actually

characterizes sup erreexive spaces and thus this pro of of Prop osition

do es not work b eyond the class of sup erreexive spaces

In itwas shown that if X is asymptotically uniformly smo oth AUS

then every Lipschitz map from X to a nite dimensional space has for

every points of Frechet dierentiability We do not intend to

recall here the denitions of AUS or of Frechet dierentiability Wejust

mention that the class of spaces ha ving an equivalentAUS norm is much

richer than that of sup erreexive spaces and contains for example the non

reexive space c On the other hand anyspaceX with an AUS norm is

an Asplund space any separable subspace of X has a separable dual The

existence of p oints of Frechet dierentiabilityforevery provides an

approximation of Lipschitz functions by ane functions which also suces

for proving the conclusion of Prop osition Thus Prop osition is valid

also if X has an AUS norm In the result on existence of points of

Frechet dierentiabilityandthus Prop osition is proved also for some

spaces X which do not admit an equivalentAUS norm for example every

table unless such X is X which is a C K space with K compact coun

isomorphic to c it do es not have an equivalentAUS norm

LIPSCHITZ QUOTIENTS FROM METRIC TREES

It is p ossible that the Frechet dierentiability result and thus Prop o

sition holds for every Asplund space X The Frechet dierentiability

result is known to b e false for every nonAsplund space see eg Prop o

sition or Chapter I I I of In this connection wemention also that

in itisproved that a Banach space which is a Lipschitz quotientofan

Asplund space is itself an Asplund space

All the results mentioned so far leave op en the question whether the

conclusion of Prop osition is valid for an arbitrary Banach space X

Recall that in the formally dual setting of Lipschitz emb eddings the dual

version of Prop osition is valid for an arbitrary Banachspace X ie

if X Lipschitz emb eds in Y then X is crudely nitely representable in Y

This fact is due to Rib e seealso Corollary

We prove in this pap er that Prop osition may fail to hold for many

common Banach spaces Wedothisbyproving in Theorem that if X is

any separable Banach space containing then there is a Lipschitz quotient



map from X onto any separable BanachspaceY In particular there is a

Lipschitz quotient map from C onto in fact known results in the



linear theory reduce the general theorem to this sp ecial case This provides

also the rst known examples of pairs of separable Banach spaces X and Y

so that there is a Lipschitz quotient map from X to Y butnosuch linear

quotient map

Theorem can b e considered in a way as a dual result to a theorem of

Aharoni see also or Theorem that every separable metric

space is Lipschitz equivalent to a subset of c There is however no obvious

connection between the pro of of Theorem and the known pro ofs of

Aharonis theorem The pro of of Theorem dep ends strongly on the

sp ecial structure of



Theorem actually holds for a general complete separable metrically

convex metric space Y Recall that a metric space Y is metrical ly convex

provided that for eachpairfx y g of p oints in Y and there exists

a p oint z in Y so that dx z dx y and dy z dx y In

other words for each x and y in Y there is a geo desic arc from x to y ie

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 conv ex spaces

The Lipschitz quotient maps f constructed in the pro of of Theorem

factor through a sp ecial metrically convex metric space which wecall an

tree This factorization forces the mapping f to have the prop erty that





whenever the Gateaux derivative f xoff exists at a p oint x X then



the rank of f x is see Prop osition b elow Note that if X has

an AUS norm and dim Y such a situation is imp ossible bythe

results in onFrechet dierentiability mentioned ab ove

JOHNSON LINDENSTRAUSS PREISS AND SCHECHTMAN

An tree is a sp ecial kind of metric tree Metric trees are used in



geometry metric top ology and group theory but as far as weknow this

is the rst time they have b een used in the geometry of Banach spaces

There is a weakening of the notion of Lipschitz quotient map whichhas

b een studied byDavid and Semmes Say that a Lipschitz map f from

the metric space X into the metric space Y is bal l non col lapsing provided

there is a c so that for every r 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

n

space X into R say that f is measure non col lapsing provided there is

a csothat for ev ery r the image under f of anyball of radius

n

r has Leb esgue measure at least cr David and Semmes showed that

n

that if X is nite dimensional and f X R is measure non collapsing

then f is ball non collapsing In this was extended to the case where X

is any sup erreexive Banachspace In Theorem weshow that if X is

anyBanach space which contains a subspace isomorphic to then there



is a measure non collapsing map from X onto the plane whose range has

emptyinterior and hence the mapping is not ball non collapsing

In Section we prove non separable versions of Theorem Theo

rem 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 onto



Y

PROOFS OF THE SEPARABLE RESULTS

We b egin by recalling the denition 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 convexitymust be a geo desic arc

joining each pair of p oints in X For an intro duction to metric trees see

However except for the pro of of Prop osition wedonotusethe

theory of metric trees Instead we dene and use a concept which we

denote by SMT which is sucient for our purp oses Later in Section

we show that this notion actually coincides with the notion of separable

metric tree Similarlywe dene a notion of MT sucient for our purp oses

and in Section show that it coincides with the notion of a metric tree

The concept of SMT is dened by describing how to construct one In

order to explain the construction we need a sp ecial case of the concept

of the union of two metric spaces Supp ose that X d and Y d are

 X Y

two metric spaces whose intersection is a single p oint p We dene the



union X Y of X and Y to b e X Y d where the metric d agrees with



x y is d on X d agrees with d on Y and if x X y Y then d

X Y

dened to b e d x pd p y We can now describ e the construction of

X Y

an SMT Let I b e a closed interval or a closed ray and dene T I The

  

LIPSCHITZ QUOTIENTS FROM METRIC TREES

metric space T is the rst approximation to our SMT Having dened T

 n

let I b e a closed interval or a closed ray whose intersection with T is

n n

an end p oint p ofI and dene T T I The completion

n n n n  n



T of T is an SMT If each I is a ray with end p oint p for n

n n n

n



and the set fp g of no dal p oints is dense in T then we call T an

n 

n

 

tree and say that fI g fT g describ e an allowed construction of

n n

n n

T Of course typically there are many dierent allowed constructions of

agiven SMT

The relevance of SMTs to Lipschitz quotients is made clear by Prop osi

tion

Proposition Let T bean tree Then every separable complete



metrical ly convex metric spaceisaLipschitz quotient of T

 

Proof Let fI g fT g describ e an allowed construction of T and

n n

n n



let the no des b e fp g Let Y b e a separable complete metrically convex

n

n

metric space We build the desired Lipsc hitz quotient map by dening it on

T by induction That we can do this is clear from the following Supp ose

n

you have a Lipschitz map f T Y and y is taken from some countable

n

dense subset Y of Y Extend f to T by mapping I to a geo desic arc

n n

f p y whichjoinsf p toy f is an isometry on fz I dp z

n n n n

df p yg and f maps p oints on I whose distance to p is larger than

n n n

p yto y This makes f act like a Lipschitz quotientatp relative df

n n

to f p y Since the no dal points are dense in T it is evident that a

n

judicious selection of the p oints from Y will pro duce a Lipschitz quotient

mapping For example order Y as a sequence and for eachnode p picka

sequence of no des fp g tending to p so that the sequences corresp onding

k

n

to dierent ps are disjoint Then at the k step extend f to T

n k 

n

by mapping I to a geo desic arc which joins f p th element tothen

k  k

n n

of Y We use the fact that to build a 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 of X into Y so that for each



x in X and for a dense set of r in R the closure in Y of f B x r X

X

is B f xr See for example the argument for Prop osition in

Y

For the pro of of Theorem we use the fact that an SMT is a absolute

Lipschitz retract A metric space X is called a absolute Lipschitz retract

where provided that whenever X is isometrically contained in

a metric space Y there is a retraction f fromY onto X with Lipf

A metric space X is a absolute Lipschitz retract if and only if X is

metrically convex and every collection of mutually intersecting closed balls

in X have a common p oint see eg Prop osition

JOHNSON LINDENSTRAUSS PREISS AND SCHECHTMAN

An isomorphic version of Lemma was proved in a more general con

text in

Lemma Assume that X and Y are absolute Lipschitz retracts

which intersect in a single point p Then X Y is also a absolute



Lipschitz retract

Proof It is evident that X Y is metrically convex We check that



any collection of mutually intersecting closed balls in X Y haveapoint



in common Let

fB x r x X gfB y y Y g

 

b e such a collection in X Y Wemay clearly assume that b oth and



are nonempty By assumption wehave for every and

  

r dx pdy p

If r dx p for every and dy p for every then p

 

is a p oint common to all the balls in the collection Otherwise there is say

an so that r dx p Then R inf dy p The







collection of balls fB x r g fB p R g in X will b e mutually



intersecting and thus havea point x in common It is trivial to check that

this p oint x is a common p oint to the given collection of balls in X Y



That every metric tree is a absolute Lipschitz retract is basic to the

theory of metric trees Wenowshow indep endently of this fact that the

same holds for SMTs

Corollary Let T be an SMT Then T is a absolute Lipschitz

retract

 

Proof Let fI g fT g describ e an allowed construction of T

n n

n n

and assume that T is isometrically contained in a metric space S By

Lemma and induction for each n there is a nonexpansive retraction

P from S onto T It is natural to try to get a cluster point of the

n n

mappings P in the space of functions from S to T under the top ology

n

of pointwise convergence Since except in some degenerate situations

T is not lo cally compact this presents a problem However the space

T can be isometrically embedded into in an obvious way as a weak



closed subset Map I isometrically to an interval or ray in the direc



e Having dened the mapping f on T extend f to T by tion

 n n



mapping I isometrically into f p R e Any cluster point of

n n n

LIPSCHITZ QUOTIENTS FROM METRIC TREES



fP g in the space of functions from S to T under the top ology of

n

n

weak p ointwise convergence is a nonexpansive retraction from S onto T

N

Let be the Cantor set f g and for each n let r be the nth

n



co ordinate pro jection on In the space C the sequence fr g is

n

n

isometrically equivalent to the unit vector basis of For n let



E b e the functions in C which dep end only on the rst n co ordinates

n

Notice that if x is in E and mnthen for all real t kx tr k kxk jtj

n m

In other words if I isaray in the direction of r emanating from a p oint

m

p in E then in C the set E I is an union of E and I That

n n  n



fr g acts likethe basis over C is the key to proving Prop osition

n 

n

Proposition Every SMT is a Lipschitz quotient of C

 

Proof Let fI g fT g describ e an allowed construction of the

n n

n n

SMT T The desired Lipschitz quotient map is dened by induction by



dening at step n the map on E Let fx g be a dense sequence in

n n

n

 

E so that for each n x is in E and let Y b e a subset of T such

n n n n

n n

that Y T is dense in T for each n Supp ose that wehave a Lipschitz

n n

mapping f from E into T and y is in Y T We can extend f

n n n





to a Lipschitz map f from E x R r by mapping the set

n n n

fx te t df x yg isometrically onto the geo desic arc from

n n n

f x toy and setting f x te y for tdf x yg Then since

n n n n

T is a absolute Lipschitz retract by Corollary we can extend f

n



to a Lipschitz map from E into T Since the sequence fx g

n n n

n 

is dense in C it is evident that a judicious selection of the points

from Y will pro duce a Lipschitz quotient mapping from C onto T

Corollary is an immediate consequence of Prop ositions and

Corollary If Y is a separable complete metrical ly convex metric

space then Y is a Lipschitz quotient of C

Remark Corollary can b e proved without using Prop osition

and in fact without mentioning the concept of SMT One denes the Lips

chitz quotientonE by induction as in the pro of of Prop osition Toex

n





tend the Lipschitz mapping f E x R r Y to a Lipschitz

n n n





mapping from E into Y one uses the fact that E x R r

n n n n

b ecause is a absolute Lipschitz retract which follows from Lemma

n

the space E is isometrically isomorphic to and thus is a absolute

n



Lipschitz retract

JOHNSON LINDENSTRAUSS PREISS AND SCHECHTMAN

Theorem Let X be a separable Banach space which contains a

subspace isomorphic to and let Then every separable complete



metrical ly convex metric spaceisa Lipschitz quotient of X

Proof By a theorem of James X contains a subspace

isomorphic to Pelczynski proved that this implies that C



is isomorphic to a linear quotientofX Indeed let T be a

linear quotient from the subspace of X isomorphic to onto C



Extend T to a map S with norm kT k from X to L and let Y



denote the separable closure of the range of S The main result of

implies that C contains a space isometric to C whichisnormone

complemented in Y Denote the pro jection by P Then PS isa

linear quotient map

Theorem follows from this and Corollary

An alternative way to prove Theorem without referring to is

sketched in the pro of of Theorem b elow

Recall that a metric space Y is C quasiconvex provided that any two

points x and y in Y can b e joined by an arc whose length is at most Cdx y

Non essential mo dications of the pro ofs given ab ove yield Theorem

Theorem Let X be a separable Banach space which contains a

subspace isomorphic to let and let C Then every



separable complete C quasiconvex metric space is a C Lipschitz

quotient of X

We do not know if a Banach space X with the prop erty that every

separable complete C quasiconvex metric space is a Lipschitz quotientof

X must contain an isomorphic image of However the next prop osition



says that suchanX must contain a metric space eqivalenttoan tree



Proposition If X is a Banach space that has an treeasaLips



chitz quotient spacethenX contains a subset which is Lipschitz equivalent

to an tree



Proof By since X admits a nonAsplund Banach space as a Lip

schitz quotient X is nonAsplund Let Y be any separable subspace of

X whose dual is non separable Then by a theorem of Stegall or

p there is a b ounded linear op erator from Y to C suchthatthe

image of the unit ball of Y contains a Haar system in C Recall that

in this context a Haar system is a collection of indicator functions of sets

LIPSCHITZ QUOTIENTS FROM METRIC TREES

whichformadyadic tree under the order of inclusion Passing to a subse

quence of the sets we get a collection of sets fA g indexed by all the nite

n

sequences of integers n n n and such that A A whenever

 k n m

n is an initial segmentof m and A A whenever

n n r n n s

 

k k

r s Let fr g b e a dense sequence in For t r r swith

i i i



k

s r r r s let u



i A i A i A A

t

  i

k

i i  i i 



 

k  k

where denotes the indicator function of the set A It is not di

A

cult to check that the closure in C of the collection of all such u



t

is Lipschitz equivalent to an tree with no des of the form r

A  i



r For each n let x be an element in the r

i A n A i

i 

k

i i 





k 

unit ball of Y whichismappedby Stegalls map to and extend this

A

n

discrete map of fx g to f g linearly denoting the inverse image of u



n A

t

n

by x The set fx g is easily seen to b e Lipschitz equivalenttoan tree

 



t t

Remark We do not know if every Banach space X containing

Lipschitz equivalent image of an tree has the prop erty 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 prop erty that

every connected subset of a metric tree is contractible Using this fact it is

very simple to prove Prop osition

Proposition 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 ie there is a metric tree T and continuous functions g X

T and h T Y so that hg f If f is Gateaux dierentiable at some



point x in X then the Gateaux derivative f x of f at x has rank at most

one

Proof Assume without loss of generality that x and f



Supp ose for contradiction that f has rank at least two By comp osing

with a linear pro jection onto a two dimensional subspace of Y we can as



sume that Y is two dimensional and that f has rank two By restricting

to a suitable two dimensional subspace of X wemay further assume that

X is two dimensional Finally by comp osing with suitable isomorphisms

  

from R to X and from Y to R wemay assume that X Y R and

 

f is the identity mapping on R



Now that we are in the nite dimensional setting we know that f

is a Frechet derivative b ecause f is a Lipschitz function this is the only

place where the Lipschitz assumption on f is used So there exists r



so that if x is in R and jxj r then jx f xj r recall that

JOHNSON LINDENSTRAUSS PREISS AND SCHECHTMAN

 

f and f is the identity In particular zero is not in f rS

  

However g rS hence also f rS hg rS is contractible in itself

 

This implies the absurditythat rS is also contractible in R fg

Using the fact to b e explained in Section that every SMT is a metric

tree we get from Prop ositions and that

Theorem Let X be a separable Banach space which contains a

subspace isomorphic to and let Y beanyseparable Banach space There



there is a Lipschitz quotient map from X onto Y so that whenever the

Gateaux derivative of f exists it has rank at most one

It is well known see eg p that if Y has the Radon Nikodym

prop erty then any Lipschitz function from a separable space into Y is

Gateaux dierentiable outside a Gauss null set

One can prove Theorem 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 pro of of Prop osition to prove an analogous

result for SMTs The direct pro of of the contractability of connected sets

in an SMT is not hard but quite lengthysowe prefer not to include it here

We end this section with an application of Theorem to concepts

studied byDavid and Semmes

Theorem Let X bea separable Banach space which has a subspace

isomorphic to Then thereisa measurenoncol lapsing mapping f from





X into R whose range is a closed set with empty interior



Proof First note that if fB g is a sequence of non overlapping op en

n

n



 

balls in R then R B is quasiconvex Toprove this it is sucient

n

n

N



to check that for each N R B is quasiconvex To

n

n

prove this latter statement draw the line segmentbetween two p oints in

N



R B If the segmentintersects B with n N then it intersects

n n

n

B in a chord Replace the chord by the arc on the b oundary of B cut by

n n

the chord which has length at most times the length of the chord



be dense in the plane We cho ose by recursion op en Now let fx g

n

n

balls B centered at x so that for each n B do es not intersect any B

n n n i

with in and with the convention that B is the empty ball exactly

n

n

when x is in the closure of B The balls are chosen so that for each

n i

i

N N



the measure of B x r B is larger N and each x in R B

n n

n n



than r The rst ball can be any ball centered at x Assume that

 

LIPSCHITZ QUOTIENTS FROM METRIC TREES

B B B have been chosen to satisfy the stated conditions In

  n

N

the non trivial case that D dx B we need to cho ose

n n

n

N 

suciently small so that if wesetB B x then B

n n n

n

satises the desired conditions Now balls with radius at most D cannot

N

intersect b oth B x D and B So as long as wecho ose D

n n

n

w e only haveto worry ab out balls with radius r at least D For balls

N



in that range it is evident that for every x in R B the measure

n

n

N



r by an amount indep endent of B x r B is larger than

n



n

of x and r D So wecan take care of such balls by cho osing



so that is less than This completes the inductive construction It





is clear that if x is in R B then for every r the measure of

n

n





B x r B is at least r

n

n





Now the set R B is quasiconvex and complete so by The

n

n



orem there is a Lipschitz quotient mapping f from X onto R





B But then f when considered as map into R is measure non

n

n

collapsing and of course the range of f is closed and has emptyinterior

THE NONSEPARABLE CASE

We provein this section non separable versions of Theorem First

we discuss the notion of MT The concept of MT is dened by describing

how an MT is constructed The construction is a transnite version of

the construction of an SMT Let be a cardinal number and identify

with the set of all ordinals whose cardinalityis less than Let I be a



closed interval or a closed ray and dene T I Having dened T for

 

all where let I b e a closed interval or a closed ray whose

T of T is an end point p of I intersection with the completion

Dene T to b e the union of T and I The completion T of T





is an MT If each I is aray with end point p for and for every

non empty op en subset U of T the set f p U g has cardinality

then wecall T an tree and say that fI g fT g describ e

  

an allowed construction of T It is more or less clear that there are



trees for each innite cardinal

In Section 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

JOHNSON LINDENSTRAUSS PREISS AND SCHECHTMAN

that every MT is a absolute Lipschitz retract Alternatively this can b e

proved directly Let fI g fT g describ e an allowed construction

 

of the tree T and showby transnite induction using an argument similar

to that in the pro of of Corollary that each T is a absolute Lipschitz

retract

By replacing the induction argument in Prop osition by a transnite

induction argument we get a non separable version

Proposition Let T be an tree Then every complete met



rical ly convex metric space whose density character is at most is a

Lipschitz quotient of T

There is also a non separable version of Prop osition However since

Pelczynskis theorem do es not have a non separable analogue there are

limited applications of the non separable extension of Prop osition As

a replacement in the non separable setting wehave Prop osition Let



b e the group f g where is an uncountable cardinal and let



b e the normalized Haar measure on the Borel subsets of ie is the





pro duct measure where is normalized counting measure on f g

For each let r be the pro jection of onto its th co ordinate



Then fr g is over L in an obvious sense Sp ecically

   

if X is a subspace of L with densitycharacter strictly less than

 

then since every function in L dep ends on only countably many

 

co ordinates there is a subset of of cardinality strictly less than



so that each function in x dep ends only on the co ordinates in It

follows that if is in the non empty set and x is in X then



kx tr k kxk jtj for all real t Thus transnite induction and the same

considerations used in the pro of of Prop osition yield

Proposition Let X be a subspace of L where is an

 

uncountable cardinal with density character and assume that fr g



X Then every MT of density character at most is a Lipschitz quotient

of X



If then the densitycharacter of L itself is and hence

 

L is isometrically isomorphic to a quotientspaceof Nowif

  

X is a Banach space whichcontains a subspace isomorphic to then



for each X contains a subspace isomorphic to by the



non separable version of a theorem of James this requires only that

is injective it follows that L is be innite Since L

   

isomorphic to a linear quotientofX for every Therefore from

Prop ositions and we get our rst non separable version of Theorem

LIPSCHITZ QUOTIENTS FROM METRIC TREES

Theorem Let X be a Banach space which contains a subspace iso



morphic to and let Assume that Then every com



plete metrical ly convex metric space of density character at most is a

Lipschitz quotient of X



we can provea weaker version of Theorem In case

Theorem Let X be a Banach space of density character which

contains a subspace isomorphic to and let Then every com



plete metrical ly convex metric space of density character at most is a

Lipschitz quotient of X

Proof Let fx g b e unit vectors in X which are equivalentto



the unit vector basis of Since L is injective the mapping

  

x r extends to a linear op erator L of norm at most from X into

L The range of this linear op erator has densitycharacter at most

 

and L is an isomorphism from the closed space of fx g onto the closed



span of fr g so the same considerations as in Prop osition yield that



every tree is a Lipschitz quotientofX Now apply Prop osition



Remark It is well known that has a subspace isometric to





Indeed C like the dual of any separable space and

 



Hence byTheo the atomic measures in C are isometric to



rem any complete metrically convex metric space of densitycharacter



is an image of by a map whose Lipschitz and coLipschitz con



stants are In particular there is a Lipschitz quotient map from onto



c the question of the validityofthisresultwas raised in several places

see eg p

It is a formal consequence of the preceding remark that there is retraction

from onto c that is a Lipschitz quotient map To see this let P be



a Lipschitz retraction from onto c for a pro of of the existence of



such a mapping see p Let f be a Lipschitz quotient mapping

from onto c with f Dene a function g on by

  

g x y P xf y The map g is easily seen to b e a retraction and

By Lipschitz quotient map from onto its subspace c fg

  

the subspace homogeneityof see there is also a retraction from



onto c that is a Lipschitz quotient map



In the literature there are several nonobvious criteria for deciding for

which uncountable cardinal do es emb ed isomorphically into a given



Banachspace X See and its references These criteria dep end often

on the axioms one uses in set theory From these criteria and Theorem

JOHNSON LINDENSTRAUSS PREISS AND SCHECHTMAN

one may deduce some results on the existence of nonlinear quotientmaps

between suitable spaces Wedonotenter into this here

METRIC TREES

Recall that a complete metric space X is a metric tree also called an

Rtree provided that for anytwo p oints x and y in X there is a unique

arc joining x to y and this arc is a geo desic arc See for an overview

of the study of metric trees and for statements and sometimes pro ofs

of the most basic prop erties of metric trees References to original sources

are contained in these two pap ers

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 p oints in X twoofthethreenumb ers dx y dw z dx w dy z

dx z dy w are the same and that numb er is at least as large as the

third The characterizations are that

A metric space emb eds isometrically into some metric tree if and only

if it satises the four p oint condition

A metric space is a metric tree if and only if it is complete connected

and satises the four p oint condition

This characterization of metric trees makes Prop osition very easy to

prove

Proposition

A metric space T is an SMT if and only if T is a separable metric tree

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

 

Proof Assume that T is an SMT and supp ose that fI g fT g

n n

n n

describ e an allowed construction of T It is easy to checkby induction that



each T satises the four p oint condition and hence so do es T whence

n n

n

so do es T b ecause the four p oint condition obviously passes to comple



tions Furthermore T is arcwise connected hence its completion T

n

n

is connected Thus T is a metric tree by the characterization ab ove

Of course an SMT is separable so this completes the pro of of the only

if part of To go the other way assume that T is a separable metric



We need to build tree and let fx g be a countable dense subset of T

n

n

 

subsets fI g fT g of T which describ e an allowed construction of

n n

n n

LIPSCHITZ QUOTIENTS FROM METRIC TREES

T Let I T fx g Having dened T let I be the geo desic arc

   n n

from x to x let x be the last point on this arc which is in T set

 n n n

x x if x is in T let I b e the geo desic arc fromx to x and set

n n n n n n n

T T I It is easy to check that this works

n n n

The second statementisproved similarly

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Israel J Math 

S Argyros J Bourgain and T Zachariades A remark on the isomorphic embed

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Collo quium Publication Amer Math So c Volume

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Adv in Math  no

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Pro c London Math So c to app ear

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Math 

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