Nagata-Assouad dimension via Lipschitz extensions

0.2 MB·English

by  N. Brodskiy

NAGATA-ASSOUAD DIMENSION VIA LIPSCHITZ
6 EXTENSIONS
0
0
2 N.BRODSKIY,J. DYDAK,J. HIGES, ANDA. MITRA

n
a
Abstract. Inthefirstpartofthepaperweshowhowtorelateseveral
J
dimensiontheories(asymptoticdimensionwithHigsonproperty,asymp-
0
totic dimension of Gromov, and capacity dimension of Buyalo [7]) to
1
Nagata-Assouad dimension. This is done by applying two functors on

the Lipschitz category of metric spaces: microscopic and macroscopic.
]
G In the second part we identify (among spaces of finite Nagata-Assouad
dimension) spaces of Nagata-Assouad dimension at most n as those for
M
which the n-sphere Sn is a Lipschitz extensor. Large scale and small
. scale analogs of that result are given.
h
t
a
m
[

1
v Contents
6
2 1. Introduction 1
2 2. Microscopic and macroscopic Nagata-Assouad dimensions 3
1
3. Spheres as Lipschitz extensors 6
0
4. Lipschitz extensions and Nagata-Assouad dimension 10
6
0 5. Coarsely equivalent metrics and Nagata-Assouad dimension 13
h/ References 14
t
a
m
:
v
1. Introduction
i
X
The large-scale geometry of metric spaces has been the subject of intense
r
a research in the last 15 years. For a comprehensive account of the area see
Gromov’s paper [12]. Thereare two large-scale concepts related to the topic
ofour paper: asymptotic dimension andasymptotic dimension of linear type
(Dranishnikov and Zarichnyi[11] refer to thelatter asasymptotic dimension
with Higson property). Asymptotic dimension is an invariant of the coarse
Date: January 10, 2006.
1991 Mathematics Subject Classification. Primary: 54F45, 54C55, Secondary: 54E35,
18B30, 54D35, 54D40, 20H15.
Key words and phrases. Asymptotic dimension, coarse category, Higson property,
Lebesquenumber,Lipschitz extensors, Nagata dimension.
The second-named author was partially supported by Grant No.2004047 from the
UnitedStates-Israel Binational Science Foundation (BSF), Jerusalem, Israel.
1
2 N.BRODSKIY,J.DYDAK,J.HIGES,ANDA.MITRA
category of Roe [19]. Asymptotic dimension of linear type is preserved by
bi-Lipschitz functions, so its natural place is in the Lipschitz category.
Notice that our notion of a Lipschitz function is a bit different from that
used in [13]. Namely, we allow Lipschitz constant to be smaller than 1.
Thus, a Lipschitz function f : (X,d ) (Y,d ) satisfies d (f(x),f(y))
X Y Y
→ ≤
λ d (x,y) for some λ 0 and all x,y X. The infimum of all possible λ
X
· ≥ ∈
is denoted by Lip(f). f is called bi-Lipschitz if there are constants µ,λ> 0
such that µ d (x,y) d (f(x),f(y)) λ d (x,y) for all x,y X.
X Y X
· ≤ ≤ · ∈
In [14] a variation of asymptotic dimension is considered. That invariant
of bi-Lipschitz functions was introduced and named Nagata dimension by
Assouad [1] as it is closely related to a theorem of Nagata characterizing
the topological dimension of metrizable spaces (cf. [[16], Thm. 5] or [[17],
p. 138]). In our paper we refer to it as Nagata-Assouad dimension. In
contrast to the asymptotic dimension, the Nagata dimension of a metric
space is in general not preserved under quasi-isometries, but it is still a
bi-Lipschitz invariant and, as it turns out, even a quasisymmetry invariant
(see [14]). The class of metric spaces with finite Nagata dimension includes
all doubling spaces, metric trees, euclidean buildings, and homogeneous or
pinched negatively curved Hadamard manifolds as shown in [14]. One of
main results of [14] relates to theorems of Assouad [2] and Dranishnikov
[10]. Namely, Theorem 1.3 of [14] states: Every metric space with Nagata-
Assouad dimension at most n admits a quasisymmetric embedding into the
product of n+1 metric trees.
The result from [14] of major importance to us is Theorem 1.4:
Theorem 1.1. Suppose that X, Y are metric spaces, dim X n < ,
NA
≤ ∞
and Y is complete. If Y is Lipschitz m-connected for m = 0,1,... ,n 1,
−
then the pair (X,Y) has the Lipschitz extension property.
Recall that (X,Y) has the Lipschitz extension property if there is a con-
stant C > 0 such that for any Lipschitz map f : A Y, A any subset of X,
→
there is a Lipschitz extension g : X Y of f such that Lip(g) C Lip(f).
→ ≤ ·
We call Y a Lipschitz extensor of X in such a case. Y is Lipschitz m-
connected if there is a constant C > 0 such that any Lipschitz function
m
f : Sm Y extends over the (m+1)-ball Bm+1 to g : Bm+1 Y so that
→ →
Lip(g) C Lip(f).
m
≤ ·
One of the main themes of our paper is characterizing Nagata-Assouad
dimension via Lipschitz extensions. In Section 4 we characterize spaces of
Nagata-Assouad dimension at most n (among all spaces of finite Nagata-
Assouad dimension) as those for which n-sphere Sn is a Lipschitz extensor.
In the case of dimension zero the assumption of dim (X) being finite can
NA
be dropped.
Recently, S.Buyalo[7]introducedthecapacity dimensionofametricspace
and proved many analogs of results obtained by U. Lang and T. Schlichen-
maier [14] for Nagata-Assouad dimension. It is clear that capacity dimen-
sion is the small scale version of Nagata-Assouad dimension. In Section 2
NAGATA-ASSOUAD DIMENSION VIA LIPSCHITZ EXTENSIONS 3
we formalize that observation by introducing microscopic and macroscopic
functors on the Lipschitz category. That way many results from Section
3 of [7] can be deduced formally from [14]. Also, the main result of [7]
(the asymptotic dimension of a visual hyperbolic space X is bounded by 1
plus the capacity dimension of the visual boundary of X) is really about
Nagata-Assouad dimension of the visual boundary of X - see 2.12.
2. Microscopic and macroscopic Nagata-Assouad dimensions
Definition 2.1. A metric space X is said to be of Nagata-Assouad dimen-
sion at most n (notation: dim (X) n) if there is C > 0 such that for all
NA
≤
n+1
r > 0 there is a cover = of X so that each is r-disjoint and the
i i
U U U
i=1
diameter of elements of isSbounded by C r.
U ·
Nagata-Assouad dimension can be characterized in many ways (see [14]):
Proposition 2.2. For a metric space (X,d) the following conditions are
equivalent:
(1) dim (X) n.
NA
≤
(2) There is a constant C > 0 such that for any r > 0 there is a cover
1
of X of multiplicity at most n+1, of mesh at most C r, and of
r 1
U ·
Lebesque number at least r.
(3) There is a constant C > 0 such that for any r > 0 there is a cover
2
of mesh at most C r such that each r-ball B(x,r) intersects at
r 2
V ·
most n+1 elements of .
r
V
If one replaces all r > 0 in 2.1 by r sufficiently small, then one gets the
concept of capacity dimension of Buyalo [7] for which he proved analog of
2.2. If one replaces all r > 0 in 2.1 by r sufficiently large, then one gets
the concept of asymptotic dimension of linear type or asymptotic dimension
with Higson property. In this section we will show how to introduce those
dimensions formally from Nagata-Assouad dimension. That way small scale
version and large scale version of 2.2 are in fact consequences of 2.2.
Given a metric space (X,d) and ǫ > 0 we consider the metric max(d,ǫ)
on X. Needles to say, the formula should not be read literally, only in the
case of x= y. Similarly, we consider the metric min(d,ǫ).
6
A metric space (X,d) is called discrete if (X,d) is δ-discrete for some
δ > 0. That means d(x,y) > δ for all x =y.
6
Lemma 2.3. Any discrete metric space (X,d) is bi-Lipschitz equivalent to
(X,max(d,ǫ)) for all ǫ > 0.
Proof. Suppose (X,d) is δ-discrete. Notice the identity map
id: (X,max(d,ǫ)) (X,d) is 1-Lipschitz and its inverse is (1+ ǫ)-Lipschitz
as d(x,y) max(d→(x,y),ǫ)) (1+ ǫ) d(x,y) for all x= y Xδ. (cid:4)
≤ ≤ δ · 6 ∈
Corollary2.4. Foranymetricspace (X,d)andǫ,δ > 0thespace (X,max(d,ǫ))
is bi-Lipschitz equivalent to (X,max(d,δ)).
4 N.BRODSKIY,J.DYDAK,J.HIGES,ANDA.MITRA
Proof. Assume ǫ > δ and notice max(d ,ǫ) = d , where d = max(d,a).
δ ǫ a
Use 2.3. (cid:4)
Since Nagata-Assouad dimension is an invariant of the Lipschitz category
(see [14] for a stronger result for quasisymmetric embeddings), one gets the
following.
Corollary 2.5. For any metric space (X,d) the Nagata-Assouad dimension
of (X,max(d,ǫ)) does not depend on ǫ > 0.
Given a metric space (X,d) one can disregard its microscopic features
by considering the space (X,max(d,1)). The macroscopic Nagata-Assouad
dimension of (X,d) is defined as dim (X,max(d,1)).
NA
Corollary 2.6. If (X,d) is a discrete metric space, then its macroscopic
Nagata-AssouaddimensionequalstheNagata-Assouaddimensiondim (X,d)
NA
of (X,d).
Lemma2.7. The macroscopic Nagata-Assouad dimension of ametric space
X is at most n 0 if and only if there is C > 0 such that for sufficiently
≥
n+1
large r > 0 there is a cover = of X so that each is r-disjoint
i i
U U U
i=1
and the diameter of elements of Sis bounded by C r.
U ·
Proof. Suppose X has a constant C > 0 such that for all r > M, where
n+1
M > 0, thereis acover = of X sothateach is r-disjointandthe
r i i
U U U
i=1
diameterof elements of r isbSoundedbyC r. LetdM = max(d,M). Notice
U ·
the cover , r M, consisting of one-point sets is r-disjoint in (X,d ).
r M
U ≤
Sincecovers , r > M, have the same desired properties in (X,d )as they
r M
U
do in (X,d), dim (X,d ) n.
NA M
≤
n+1
If dim (X,d ) n, then for each r > 1 there is a a cover = of
NA 1 r i
≤ U U
i=1
X so that each i is r-disjoint (in (X,d1)) and the diameter of elemSents of
U
is bounded by C r in (X,d ). Notice that is also r-disjoint in (X,d)
r 1 i
Uand the diameter of·elements of is boundedUby (C +1) r in (X,d). (cid:4)
r
U ·
In view of definition of the Higson property in [9] one has the following
consequence of 2.7.
Corollary 2.8. If (X,d) is a metric space, then the macroscopic Nagata-
Assouad dimension of (X,d) is at most n if and only if asdim(X) n with
≤
the Higson property.
Inthereminderof thissection wewilldualize theabove resultsfromlarge
scale/macroscopic category to small scale/microscopic category.
Lemma 2.9. Any bounded metric space (X,d) is bi-Lipschitz equivalent to
(X,min(d,ǫ)) for all ǫ > 0.
Proof. Suppose (X,d) is δ-bounded. Notice the identity map id :
(X,min(d,ǫ)) (X,d) is (1 + ǫ)-Lipschitz and its inverse is 1-Lipschitz
→ δ
as min(d(x,y),ǫ) d(x,y) (1+ δ) min(d(x,y),ǫ)) for all x= y X. (cid:4)
≤ ≤ ǫ · 6 ∈
NAGATA-ASSOUAD DIMENSION VIA LIPSCHITZ EXTENSIONS 5
Corollary2.10. Foranymetricspace(X,d)andǫ,δ > 0thespace (X,min(d,ǫ))
is bi-Lipschitz equivalent to (X,min(d,δ)).
Proof. Assume ǫ < δ and notice min(d ,ǫ) = d , where d = min(d,a).
δ ǫ a
Use 2.9. (cid:4)
SinceNagata-Assouad dimensionisaninvariantof theLipschitzcategory,
one gets the following.
Corollary 2.11. For any metric space (X,d) the Nagata-Assouad dimen-
sion of (X,min(d,ǫ)) does not depend on ǫ > 0.
Given a metric space (X,d) one can disregard its macroscopic features
by considering the space (X,min(d,1)). The microscopic Nagata-Assouad
dimension of (X,d) is defined as dim (X,min(d,1)).
NA
Corollary 2.12. If (X,d) is a bounded metric space, then its microscopic
Nagata-AssouaddimensionequalstheNagata-Assouaddimensiondim (X,d)
NA
of (X,d).
Lemma 2.13. The microscopic Nagata-Assouad dimension of a metric
space X is at most n if and only if there is C > 0 such that for sufficiently
n+1
small r > 0 there is a cover = of X so that each is r-disjoint
i i
U U U
i=1
and the diameter of elements of iSs bounded by C r.
U ·
Proof. Suppose X has a constant C > 0 such that for all r < M, where
n+1
M > 0, there is a cover = of X so that each is r-disjoint and
r i i
U U U
i=1
the diameter of elements of rSis bounded by C r. Let dM = min(d,M).
U ·
Notice the cover , r M, consisting of the whole X has diameter at
r
U ≥
most (C +1) r in (X,d ). Since covers , r < M, have the same desired
M r
· U
properties in (X,d ) as they do in (X,d), dim (X,d ) n.
M NA M
≤
n+1
If dim (X,d ) n, then for each r < 1 there is a a cover = of
NA 1 r i
≤ U U
i=1
X so that each i is r-disjoint (in (X,d1)) and the diameter of elemSents of
U
is bounded by C r in (X,d ). Notice that is also r-disjoint in (X,d)
r 1 i
Uand the diameter of·elements of is boundedUby C r in (X,d). (cid:4)
r
U ·
Since the capacity dimension of Buyalo [7] can be characterized by the
condition appearing in 2.13, one derives the following.
Corollary 2.14. If (X,d) is a metric space, then the microscopic Nagata-
Assouddimension of (X,d) isatmost nif and only ifthe capacity dimension
of X is at most n.
In [14] it is shown that if X = A B, then
∪
dim (X) = max(dim (A),dim (B)).
NA NA NA
Corollary 2.15. Let D(Y) stand for either the capacity dimension of Y or
the asymptotic dimension of linear type of Y. If X = A B, then
∪
D(X) = max(D(A),D(B)).
6 N.BRODSKIY,J.DYDAK,J.HIGES,ANDA.MITRA
Proof. Switch to either max(d,1) or min(d,1) metrics. (cid:4)
3. Spheres as Lipschitz extensors
A metric space E is a Lipschitz extensor of X if there is a constant C > 0
such that any λ-Lipschitz function f : A E, A a subset of X, extends to
a C λ-Lipschitz function f˜:X E. →
· →
The purpose of this section is to find necessary and sufficient conditions
for a sphere Sm to be a Lipschitz extensor of X. This is done by comparing
existence of Lipschitz extensions in a finite range of Lipschitz constants to
existence of Lebesque refinements in a finite range of Lebesque constants
(see 3.1, 3.3, and 3.5).
Given a cover = Us s∈S of a metric space (X,d) there is a natural
U { }
family of functions fs s∈S associated to : fs(x) := dist(x,X Us). To
{ } U \
simplify matters by the local Lebesque number LU(x) of at x we mean
U
sup f (x) s S
s
{ | ∈ }
and by the (global) Lebesque number L( ) of we mean
U U
inf LU(x) x X .
{ | ∈ }
We are interested in covers with positive Lebesque number. For those the
local multiplicity mU(x) can be defined as 1+ T(x), where T(x) = s S
| | { ∈ |
f (x) > 0 and the global multiplicity m( ) can be defined as
s
} U
sup mU(x) x X .
{ | ∈ }
If the multiplicity m( ) is finite, then has a natural partition of unity
U U
φs s∈S associated to it:
{ }
f (x)
s
φ (x) = .
s
f (x)
t
t∈S
P
That partition can be considered as a barycentric map φ : X ( ) from
→ N U
X tothenerveof . Sinceeachf is1-Lipschitz, f (x)is2m( )-Lipschitz
s t
U t∈S U
and each φ is 2m(U)-Lipschitz (use the fact thPat u is max(Lip(u),Lip(v))-
s L(U) u+v inf(u+v)
4m(U)2
Lipschitz). Therefore φ : X ( ) is -Lipschitz. See [3] and [8] for
→ N U L(U)
more details and better estimates of Lipschitz constants.
Most estimates in this paper work well for both the l and l metrics on
1 2
Rn and simplicial complexes.
Inanalogytoλ-Lipschitzfunctionsweintroducetheconceptofr-Lebesque
cover . That is simply a shortcut to r L( ).
U ≤ U
Proposition 3.1. Suppose X is metric, m 0, C > 0, and λ > λ > 0.
2 1
≥
If any λ-Lipschitz function f : A Sm, A a subset of X and λ < λ < λ ,
1 2
extends to a C λ-Lipschitz funct→ion f˜: X Sm, then t = 1
· → 4C(m+2)2(m+1)
has the property that any finite r-Lebesque cover = U ,... ,U of X
0 m+1
U { }
NAGATA-ASSOUAD DIMENSION VIA LIPSCHITZ EXTENSIONS 7
admits a refinement so that is t r-Lebesque and the multiplicity of
V 4(m+V2)2 · 4(m+2)2 V
is at most m+1 provided < r < .
λ2 λ1
Proof. Assume 4(m+2)2 < r < 4(m+2)2 and is r-Lebesque. Therefore
λ2 λ1 U
λ < 4(m+2)2 < λ . Consider a barycentric map φ : X ( ) = ∆m+1.
1 r 2 → N U
Notice Lip(φ) 4(m+2)2. There is g : X ∂∆m+1 such that Lip(g)
≤ r → ≤
4C(m+2)2 and g(x) = φ(x) for all x X so that φ(x) ∂∆m+1. Consider
r ∈ ∈
V = x X g (x) > 0 . Notice = V i=m+1 is of multiplicity at most
i { ∈ | i } V { i}i=0
m+1. Alsox V impliesx U ,so refines . Givenx X thereisisuch
i i
that g (x) ∈1 . If d(x,y∈) < V r U, then g (x∈) g (y) < 1
i ≥ m+1 4C(m+2)2(m+1) | i − i | m+1
and g (y) > 0. Thus, the ball at x of radius r is contained in
i 4C(m+2)2(m+1)
one element of . That proves is t r-Lebesque, where t = 1 .
V V · 4C(m+2)2(m+1)
(cid:4)
InthispaperweconsiderthespaceRn witheitherthel -metricd (x,y) =
1 1
n n
x y or the l -metric d (x,y) = (x y )2. The inequalities
i i 2 2 i i
| − | −
i=1 si=1
P P
d (x,y) n max x y n d (x,y)
1 i i 2
≤ · {| − |} ≤ ·
d (x,y) √n max x y √n d (x,y)
2 i i 1
≤ · {| − |} ≤ ·
showthattheidentity map (Rn,d ) (Rn,d )is n-Lipschitzanditsinverse
2 1
→
is √n-Lipschitz.
Lemma 3.2. Let be a closed convex subset of the space (Rn,d ), j =
j
△
1 or 2. For any metric space X any λ-Lipschitz map f: A of a sub-
space A X can be extended to a n2 λ-Lipschitz map f˜: X→ △ .
⊂ · → △
Proof. Fix an orthogonal coordinate system in (Rn,d ). Given a λ-
2
Lipschitz map f: A (Rn,d ), every coordinate map f : A R is λ-
2 i
→ →
Lipschitz and can be extended to a λ-Lipschitz map fˆ: X R [20]. These
i
→
coordinate extensions define the map fˆ: X Rn which is λ√n-Lipschitz.
Clearly,thenearestpointretractionr△: (Rn,→d2) is1-Lipschitz. There-
→ △
fore the composition f˜= r△ fˆis √nλ-Lipschitz.
Since the nearest point retr◦action r△: (Rn,d1) may be multivalued,
→ △
we proceed as follows. The composition id f: A ( ,d ) ( ,d ) is
1 2
◦ → △ → △
√nλ-Lipschitz and admits a nλ-Lipschitz extension f˜ : X ( ,d ) by the
2 2
→ △
first part of the proof. Then the composition f˜= id f˜ : X ( ,d )
2 2
( ,d ) is n2λ-Lipschitz. ◦ → △ →(cid:4)
1
△
The idea behind the proof of the next proposition is best understood if
one thinks of maps from X to an (m+1)-simplex ∆m+1 as a partition of
unity. Since we want to create a map to its boundary Sm = ∂∆m+1, a
geometrical tool is the radial projection r which we splice in the form of
(1 β) r+β φ with a partition of unity φ coming from a covering of X of
− · ·
multiplicity at most m+1.
8 N.BRODSKIY,J.DYDAK,J.HIGES,ANDA.MITRA
Proposition 3.3. Suppose X is metric, m 0, t > 0, and r > r > 0.
2 1
≥
There is s > 0 such that if any finite r-Lebesque cover = U ,... ,U
0 m+1
U { }
of X, where r < r < r , admits an t r-Lebesque refinement satisfying
1 2
· V
m( ) m+1, then any λ-Lipschitz function f : A Sm, A X, admits a
V ≤ → ⊂
C λ-Lipschitz extension f˜: X Sm provided 1 < λ < 1
· → 12sr2(m+2) 12sr1(m+2)
and C = 50(m+2)2s+ 150s2(m+2)5.
t
Proof. s > 0 is chosen so that given a λ-Lipschitz f : A ∆m+1 one
→
can extend it to an s λ-Lipschitz g :X ∆m+1 (see 3.2).
· →
Suppose 1 < λ < 1 and f : A ∂∆m+1 is λ-Lipschitz.
12sr2(m+2) 12sr1(m+2) →
Extend it to an s λ-Lipschitz g :X ∆m+1. Let α :X [0,1] be defined
· → →
as α(x) = (m+2) min g (x) 0 i m+1 . Notice Lip(α) (m+2)s λ.
i
· { | ≤ ≤ } ≤ ·
Let β :[0,1] [0,1] be defined by β(z) = 3z 1 on [1/3,2/3], β(z) = 0 for
→ −
z 1/3 and β(z) = 1 for z 2/3. Notice Lip(β) 3.
≤ ≥ ≤
α(x)
Put U = x X g (x) > or α(x) > 2/3 and notice L( ) r =
i { ∈ | i m+2 } U ≥
1 as follows:
12sλ(m+2)
Case 1: x X and α(x) > 3/4. Now, for any y X with d(x,y) <
∈ ∈
1 one has α(x) α(y) 1/12, so α(y) 2/3 is not possible. Thus,
12sλ(m+2) − ≤ ≤
in that case the ball B(x, 1 ) is contained in all U .
12sλ(m+2) i
Case 2: α(x) 3/4. There is i so that g (x) 1 . Since ψ = g
≤ i ≥ m+2 i i −
α is 2sλ-Lipschitz, for any y X satisfying d(x,y) < 1 one has
m+2 ∈ 8sλ(m+2)
ψ (x) ψ (y) < 1 and ψ (y) > 0 as ψ (x) 1 .
i − i 4(m+2) i i ≥ 4(m+2)
Thus is r-Lebesque and r < r < r . Shrink each U to V so that
1 2 i i
m( ) mU +1andL( ) t . Thebarycentric map φ: X ∂∆m+1
V ≤ V ≥ 12sλ(m+2) →
4(m+2)2 48sλ(m+2)3
corresponding to has Lip(φ) .
V ≤ L(V) ≤ t
m+1 m+1
α(x) 1−β(α(x))
Define h(x) = (g (x) ) e + β(α(x)) φ (x) e .
i − m+2 · 1−α(x) · i · i · i
i=0 i=0
To show Lip(h) PC λ we will use the following obsePrvations.
≤ ·
(1) If u,v : X [0,M], then Lip(u v) M (Lip(u)+Lip(v)).
→ · ≤ ·
(2) In addition, if v :X [k,M] and k > 0, then
→
u Lip(u)+Lip(v)
Lip( ) M .
v ≤ · k2
1−β(α(x))
(3) v(x) = 1 α(x) 1/3 if > 0.
− ≥ 1−α(x)
m+1
Therefore Lip( β(α(x)) φ (x) e ) 3 Lip(α) (m +2) Lip(φ)
i i
· · ≤ · · · ≤
i=0
3(m+2) (m+2)Ps λ 48sλ(m+2)3 150s2λ(m+2)5. Also, Lip(1−β(α(x)))
· · · t ≤ t 1−α(x) ≤
m+1
α(x) 1−β(α(x))
9 4 Lip(α) 36(m+2)sλ, so Lip( (g (x) ) ) (m+
· · ≤ i − m+2 · 1−α(x) ≤
i=0
2) (2sλ+36(m+2)sλ) 50(m+2)2sPλ and C = 50(m+2)2s+ 150s2(m+2)5
· ≤ t
works.
NAGATA-ASSOUAD DIMENSION VIA LIPSCHITZ EXTENSIONS 9
It remains to show h(X) ∂∆m+1 and hA = f. hA = f follows from
⊂ | |
the fact α(x) = 0 if x A. It is clear h(x) ∂∆m+1 if either β(α(x)) = 0
∈ ∈
or β(α(x)) = 1, so assume 0 < β(α(x)) < 1. In that case φ (x) > 0 implies
i
g (x) α(x) > 0, so the only possibility for h(x) to miss ∂∆m+1 is when
i − m+2
g (x) α(x) > 0 for all i which is not possible. (cid:4)
i − m+2
Propositions 3.1 and 3.4 imply the following
Corollary 3.4. If X is metric and m 0, then the following conditions
≥
are equivalent:
a. Sm is a Lipschitz extensor of X.
b. There is t > 0 such that any finite cover = U ,... ,U of X
0 m+1
U { }
admits a refinement so that L( ) tL( ) and the multiplicity of
V V ≥ U
is at most m+1.
V
Proposition 3.5. Suppose X is a metric space, n 0, 1 > t > 0, and
≥
r > r > 0. If every r-Lebesque cover = U ,... ,U of X, where
2 1 0 n+1
U { }
r < r < r , admits a 4t r-Lebesque refinement satisfying m( ) n+1,
1 2
· V V ≤
then any s-Lebesquecover = W ,... ,W of X, where 4r < s < 4r ,
0 n+2 1 2
W { }
admits a t s-Lebesque refinement of multiplicity at most n+2.
· V
Proof. Suppose 4r < s < 4r . First, let us show that any s/2-Lebesque
1 2
cover = U i=n+1 of A X consisting of n+2elements has a refinement
U { i}i=0 ⊂ ′
such that L( ) t s and m( ) n+1. Define U = U (X A) for
V V ≥ ′· V ≤ i i ∪ \
i n+1. Notice L( ) s/4. Indeed, if x X, then B(x,s/4) A is either
≤ U ≥ ∈ ∩
empty or is contained in B(y,s/2) for some y A. Since B(y,s/2) A U
i
′ ∈ ′ ∩ ⊂
for some i n+1, B(y,s/2) U and B(x,s/4) U . There is a cover
≤ ′ ⊂ i ⊂ i W
of X such that refines , L( ) 4t s/4, and m( ) n+1. Putting
W U W ≥ · W ≤
= finishes the task.
A
V W|
Suppose = W ,... ,W is an s-Lebesquecover of X. LetAbethe
0 n+2
W { }
union of balls B(x,s/2) such that B(x,s) is not contained in W . Define
n+2
U = W A for i n+1 and observe L( ) s/2 for = U i=n+1 as a
i i∩ ≤ U ≥ U { i}i=0
cover of A. Indeed, if x A, then there is y X such that B(y,s) is not
∈ ∈
contained in W and x B(y,s/2). Therefore, B(y,s) W for some
n+2 i
∈ ⊂
i n+1 which means B(x,s/2) A B(y,s) A W A= U .
i i
≤ ∩ ⊂ ∩ ⊂ ∩
Shrinkeach U toV so thattheintersection of all V is empty and L( )
i i i
′ ′ ′V ≥
t s. Define W = V for i n+1 and W = W . The cover is of
· i i ≤ n+2 ′ n+2 W
multiplicityatmostn+2. WewanttoshowL( ) t s. IfB(x,s) W ,
n+2
W ≥ · ⊂
we are done. Otherwise B(x,s/2) A and there is i n + 1 such that
B(x,t s) V in which case B(x,t⊂s) W′. ≤ (cid:4)
· ⊂ i · ⊂ i
Corollary 3.6. Suppose X is a metric space and n 0. If Sn is a Lipschitz
≥
extensor of X, then so is Sn+1.
Proof. By3.4thereist > 0suchthatanycover ofX consistingofn+2
U
elements has a refinement satisfying L( ) t L( ) and m( ) n+1.
V V ≥ · U V ≤
Use λ very large and λ very small. We may assume t < 1, so applying 3.5
2 1
and 3.4 completes the proof. (cid:4)
10 N.BRODSKIY,J.DYDAK,J.HIGES,ANDA.MITRA
Definition 3.7. AmetricspaceE isalarge scale Lipschitz extensor(respec-
tively, a small scale Lipschitz extensor) of X if there are constants C,M > 0
such that any λ-Lipschitz function f : A E, A a subset of X, extends to
a C λ-Lipschitz function f˜ : X E fo→r all λ < M (respectively, for all
· →
λ> M).
Using 3.1, 3.3, and 3.5 one proves easily the following large/small scale
analogs of 3.4.
Corollary 3.8. If X is metric and m 0, then the following conditions
≥
are equivalent:
a. Sm is a large scale Lipschitz extensor (respectively, a small scale
Lipschitz extensor) of X.
b. There are constants t,M > 0 such that any finite r-Lebesque cover
= U ,... ,U of X, where r > M (respectively, r < M),
0 m+1
U { }
admits a t r-Lebesque refinement of multiplicity at most m+1.
· V
As in 3.6 one proves its large/small scale analogs:
Corollary 3.9. Suppose X is a metric space and n 0. If Sn is a large
≥
scale Lipschitz extensor (respectively, a small scale Lipschitz extensor) of X,
then so is Sn+1.
4. Lipschitz extensions and Nagata-Assouad dimension
Theorem 4.1. Suppose X is a metric space of finite Nagata-Assouad di-
mension. If n 0, then the following conditions are equivalent:
≥
a. Sn is a Lipschitz extensor of X.
b. dim (X) n.
NA
≤
Proof. The direction (b) = (a) follows from 3.4 as follows. Given a
⇒
cover of X of Lebesque number L( ) > 0 pick of mesh( ) < L( )
U U V V U
so that L( ) > L( )/C and m( ) m+1. Notice that refines and
V U V ≤ V U
L( ) > L( )/C.
V U
(a) = (b). Without loss of generality (in view of 3.6), we may assume
⇒
dim (X) n+1. k > 0 is chosen so that given a λ-Lipschitz f : A
NA
∂∆n+1 one≤can extend it to an k λ-Lipschitz g : X ∂∆n+1. Let c >→1
· →
be a constant such that for any r > 0 there is a cover of X of mesh at
U
most c r, Lebesque number at least r, that can be expressed as a union
·
n+2
so that each is r-disjoint. For such a cover pick a barycentric map
i i
U U
i=1
fS: X ( ) to the nerve of such that Lip(f) 4(n+2)2. Given an
(n+1)-→simNpleUx ∆ in ( ) we looUk at f f−1(∂∆) and e≤xtendrit over f−1(∆)
to obtain g :f−1(∆N) U∂∆ of Lipschit|z number at most k 4(n+2)2. Paste
∆ → · r
all g together to g : X ( ). Our goal it to estimate the mesh and
∆
Lebesque number of g−1(s→t(vN)),Uv a vertex of ( ).
The mesh of g−1(st(v)) is at most twice tNhaUt of f−1(st(v)) . Indeed,
{ } { }
if g(x) st(w) and g(x) = f(x), then f(x) must belong to the interior of a
∈ 6