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ON THE EXTENSION OF MUCKENHOUPT WEIGHTS IN METRIC SPACES

arXiv:2012
...
CA] 25 Oct 2021

EMMA-KAROLIINA KURKI AND CARLOS MUDARRA
Abstract
...
We give a complete and self-contained proof of this theorem generalized
into metric measure spaces supporting a doubling measure
...


1
...
In addition to Ap weights being ubiquitous in harmonic analysis,
weighted norm inequalities have applications in the study of regularity of certain partial differential
equations
...

Our main result is the following theorem that provides an abstract starting point for the investigation of extensions
...
Wolff
...
An outline of the Euclidean proof can be
found in [9], Theorem 5
...
However, the metric setting brings about technical challenges that are
not present in the Euclidean case
...
1
...
Then, for 1 < p < ∞, the following statements are equivalent
...
e
...

(1)
w
dµ
sup
µ(B) B∩E
µ(B) B∩E w1+ε
B⊂X
B ball

In addition, whenever p = 1, the condition (ii) takes the following form: There exists a constant
C > 0 such that
Z
1
w1+ε dµ ≤ C ess inf w1+ε
B∩E
µ(B) B∩E
(E-K
...
) Aalto University, Department of Mathematics and Systems Analysis, P
...
BOX 11100,
FI-00076 Aalto, Finland
(C
...
) Aalto University, Department of Mathematics and Systems Analysis, P
...
BOX 11100, FI00076 Aalto, Finland
E-mail addresses: emma-karoliina
...
fi, carlos
...
mudarra@jyu
...
mudarra@aalto
...

Date: October 26, 2021
...
30L99, 42B25, 42B37
...
Muckenhoupt weight, metric measure space, doubling condition
...
E
...
Kurki was funded by a young researcher’s grant from the Emil Aaltonen Foundation
...

Mudarra acknowledges financial support from the Academy of Finland
...

1

for every ball B ⊂ X
...
13) for measurable sets in a metric space supporting a doubling measure
...
We have chosen to call these
classes induced Ap weights; see Definition 2
...
It is not obvious at the outset whether all
properties of globally defined weights hold true for this class as well
...
In particular, we need to show that the
restricted maximal function is bounded on Lp (E, w) when the weight w belongs to the induced Aq
class for some q < p (Theorem 2
...
The proofs of this theorem in the whole space are based either
on Calderón-Zygmund decompositions on cubes (when X = Rn ) or on Vitali-type coverings of the
distributional sets of the maximal function
...
e
...

The reader might wonder why we need to assume the Muckenhoupt-type condition (1) for w1+ǫ
instead of simply stating the corresponding condition for w
...
12)
...
11)
...
However, it is unclear whether the induced classes of weights satisfy a
RHI, since it is yet again impossible to control the measures of the relative balls B ∩ E in terms of
those of B
...
This technicality destroys our ability to compute
the averaged integrals that would lead to the RHI
...
Muckenhoupt weights in
particular are discussed in [12] and [28]
...
For
recent results concerning reverse Hölder inequalities for A∞ weights or strong Ap weights, as well
as versions of the Gehring lemma in various spaces, see the articles [1–3, 7, 8, 19, 20, 22–24, 26]
...
1
...
One might ask what are the
subsets E and weights w that satisfy (1) and consequently possess an extension to the entire space
...
Holden [17], working in Rn , has verified (1) for weights in Ap (E) under additional geometric
assumptions on the set E
...
1
...
In particular, Lemma 4
...
In [17], this lemma is used to recover the Euclidean equivalent of Theorem
1
...

2
...
In addition, we
assume that the nontrivial Borel regular measure µ satisfies the doubling condition: there exists a
constant Cd = Cd (µ) > 1 only depending on µ such that
0 < µ (2B) ≤ Cd µ (B) < ∞
2

(2)

for all balls B ⊂ X
...
In particular, we assume
that every ball in X has positive and finite measure
...
This implies that X is
locally compact and proper, which in turn means that every closed and bounded subset of X is
compact
...
Observe that in general,
the center and the radius of a ball B are not uniquely determined by B as a set
...

For any two nonnegative numbers A and B, if there exists a constant C ∈ (0, ∞) such that A ≤
CB, we write A
...
Furthermore, we write A ≈ B whenever there exist constants C1 , C2 ∈ (0, ∞)
such that C1 A ≤ B ≤ C2 A
...

Whenever E ⊂ X is a measurable subset and the function f is integrable on every compact subset
of E we say that f is locally integrable on E, denoted f ∈ L1loc (E)
...
[28] In practice, we
assume w to be positive almost
everywhere in E
...

For the purposes of Theorem 2
...

weights on a subset, which we denote by A

Definition 2
...
On a metric space X, let E ⊂ X be a measurable subset with µ(E) > 0
...
If 1 < p < ∞, we say that w ∈ A
!p−1

  1

Z
Z
1
1 p−1
1
w dµ
dµ
< ∞
...
We denote by JwK1 the infimum of the C > 0 for which the inequality (4)
holds
...
Whenever E = X, the above classes coincide with Muckenhoupt weights as usually defined
...
Notice that it is not possible
to reduce (3) to the Ap (X) condition e
...
by replacing w with XE w
...
2
...
Whenever E ⊂ X is a measurable
set and x ∈ E, we also define a maximal function relative to the set E by
Z
1
|f | dµ
...
These correspond to well-known results for Ap weights in Rn
...

Throughout the rest of this section (X, d, µ) will denote a complete metric measure space, with the
measure µ satisfying doubling condition (2) and thus all the properties mentioned at the beginning
of the section
...
502
...
3
...
If w ∈ A
mE w(x) ≤ JwK1 w(x) for almost every x ∈ E
...
The first inequality is a consequence of the Lebesgue differentiation theorem for µ
...
4
...
We aim to show that µ(A) = 0
...
Define a countable collection of
balls F = {B(zk , q) : k ≥ 1, q ∈ Q+ }
...

JwK1 w(x) < (1 − δ)
µ(B) B∩E

For every ε ∈ (0, 1) denote Bε = B(z, (1 − ε)r)
...
If ε ∈ (0, 1) is small enough so that x ∈ Bε and µ (B \ Bε ) ≤ η,
then (B\Bε )∩E w ≤ δ B∩E w
...
The
R triangle inequality gives the inclusions
Bε ⊂ B ′ ⊂ B, implying that x ∈ B ′ and (B\B ′ )∩E w ≤ δ B∩E w
...

B ∩E
µ(B) B∩E
µ(B ′ ) B ′ ∩E
We have shown that w(x) < ess inf B ′ ∩E w, which means that xSbelongs to the set DB ′ = {y ∈
B ′ ∩ E : w(y) < ess inf B ′ ∩E w}, where µ(DB ′ ) = 0
...


We remark that the above proposition remains true if the maximal function mE is defined by
taking a supremum over closed balls instead
...

The next lemma follows from Proposition 2
...
13 below
...
4
...
Here
wXE is the function in X that coincides with w on E and vanishes outside E
...
It is immediate that wXE ∈ L1loc (X) because B∩E w is finite for every ball B ⊂ X
...
3 implies that M (wXE )(x) = mE w(x) < ∞ for a
...
x ∈ E
...
Defining
A = {y ∈ E : mE w(y) ≤ JwK1 w(y) < ∞},
4

Proposition 2
...
For a
...

x ∈ X \ A, the Lebesgue differentiation theorem states that
Z
1
lim
wXA dµ = wXA (x) = 0,
B∋x µ(B) B
r(B)→0

and thus there exists a radius rx > 0 such that
Z
1
wXA dµ ≤ 1
...
For such a ball B it clearly holds that d (x, A) ≤ 2r(B), and consequently
r(B) ≥ max{rx , d (x, A)/2}
...
Denoting by z the center of B, we have
d (y0 , z) ≤ d(y0 , x) + d(x, z) < max{rx , 2 d (x, A)} + r(B) ≤ 4r(B) + r(B) = 5r(B),
and hence y0 ∈ 5B
...

′
µ(B) B
µ(5B) 5B
B ′ ∋y0 µ(B ) B ′ ∩A
We conclude that M (wXA )(x) < ∞ for almost every x ∈ X
...

ep (E), and
Lemma 2
...
Let E ⊂ X be a measurable set with µ(E) > 0
...
Also, if q ≥ 1, v ∈ A
q
y
eq (E) with v δ
ep (E) ⊂ A
eq (E) for every
and δ ∈ [0, 1], then v δ ∈ A
≤ JvKδ
...


Proof
...
Let A ⊂ X be measurable, 0 ≤ s ≤ 1, and
h ∈ L1 (A)
...

A

A

1)−1

Since the exponents δ and δ(p − 1)(q −
are in [0, 1], we can apply (5) to obtain
δ
Z
Z
v δ dµ ≤ µ(B ∩ E)1−δ
v dµ ,
B∩E

Z

B∩E

(6)

B∩E



1
vδ



1
q−1

  δ(p−1)
1 (q−1)(p−1)
1− δ(p−1)
q−1
dµ =
dµ ≤ µ(B ∩ E)
v
B∩E
Z

! δ(p−1)
  1
q−1
1 p−1
dµ

...
Then (6) and (7) lead to
!q−1
  1
Z
Z
q−1
1
1
dµ
v δ dµ
µ(B)q B∩E
vδ
B∩E
!δ(p−1)
  1
δ Z
Z
1 p−1
µ(B)q−δp
≤
v dµ
dµ
≤ JvKδp ,
µ(B)q
v
B∩E
B∩E
5

e1 (E), and q > 1, using first (5)
which proves the statement for p > 1
...
For q = 1, the result follows immediately from (5)
...
6
...
If 1 ≤ q < ∞, v ∈ A
g ∈ Lq (E, v), then for every ball B ⊂ X we have

q
Z
Z
1
v(B ∩ E)
|g|q v dµ
...
We may and do assume g ≥ 0
...

µ(B) B∩E
µ(B)q B∩E
v
v(B
∩
E)
B∩E
B∩E
e1 (E) (4)
...
The
following lemma provides a version for the maximal function relative to a subset
...

Proposition 2
...
Let E ⊂ X be a measurable set with µ(E) > 0
...
Then
v∈A
Z
−q
v ({x ∈ E : mE f (x) > t}) ≤ Ct
|f |q v dµ,
E

where the constant C only depends on q, JvKq , and the doubling constant Cd (µ)
...
We may assume f ≥ 0
...
Once we show the estimate
R
for mE , the statement will follow by the monotone convergence theorem
...

The
set
E
is
contained
in
x
t
x∈Et Bx ∩ E
...
Now let us write
Z
Z
XZ
v dµ
...
6 with B = 5Bj and g = f XBj ∩E to deduce that the sum (8) is
smaller than
!−q
!−q
Z
Z
XZ
XZ
1
1
q
q
f v dµ
g v dµ
g dµ
f dµ
= JvKq

...




We next show a strong-type estimate for the maximal function mE restricted to E, in the space
Lp (E, v), provided v is an induced Muckenhoupt weight of a higher class
...
8
...
Then
Z
Z
p
|f |p v dµ,
(mE f ) v dµ ≤ C
E

E

where the constant C depends only on p, q, JvKq , and the doubling constant Cd (µ)
...
For simplicity, we again assume that f ≥ 0, and proceed to write
f = f X{f >t/2} + f X{f ≤t/2} = ft + f X{f ≤t/2}
...
Combining
this observation with Cavalieri’s principle for the measure v dµ, and then using Proposition 2
...


Z

{x∈E:f (x)>t/2}

f (x)p v(x) dµ(x),
E



The following factorization theorem will be one of the main ingredients in the proof of Theorem
1
...

Proposition 2
...
Let E ⊂ X be a measurable set with µ(E) > 0, p > 1, and v a weight on E such
ep (E) for some r > 1
...

that v r ∈ A
2

Proof
...
5, v =
eq (E) with JvK ≤ Jv r K1/r
...
Applying
p

p

eq (E) and
again Lemma 2
...
Notice that q1 < p and q2 < p
...
8 applied first with v and q1 , and then with
2
E

v
′
p
p
−1/(p−1)
a bounded operator both in L (E, v) and L E, v
, with norms bounded by constants
depending only on r, p, and Jv r Kp
...

T f = v m E v p f p′
7

This is a bounded operator in Lp (E), which can be verified by applying Proposition 2
...

v p |f |p v p−1 dµ =
|f |p dµ,
m E v p f p′
dµ =
E

E

E

E

Lp (E)

E

Z 
Z
Z
 1 p
 1 p
1
−
−
v p mE v p f
mE v p f v dµ(x)
...

dµ =
E

P∞

−k

E

T kf
...
The operator T is subadditive since p/p′ ≥ 1,
so
∞
∞
X
X
Tη ≤
(2c)−k T k+1 f =
(2c)1−k T k f ≤ 2cη
...

m E v2 = m E v η ≤ v
v m E v p η p′
′

In the case 1 < p < 2, we instead factorize v 1−p = v1 v21−p as above, and raise this equation to
the power 1/(1 − p′ )
...

We will not be needing the reverse statement
...
10
...
e
...
Then, the weight g (M f )ε = w
belongs to A1 (X)
...
Since g, g−1 ∈ L∞ (X), it is enough to show that for every ball B ⊂ X and x ∈ B
Z
1
(M f )ε dµ ≤ C(M f )(x)ε ,
µ(B) B

(9)

where the constant C depends on ε and the doubling constant Cd
...
Then, owing to subadditivity of the maximal function, we have
(M f )ε ≤ (M f1 )ε + (M f2 )ε
...
Beginning with (M f1 )ε , by Cavalieri’s principle we have
Z
Z ∞
1
1
ε
(M f1 )(y) dµ(y) =
εtε−1 µ ({y ∈ B : M f1 (y) > t}) dt
µ(B) B
µ(B) 0

Z a
Z ∞
1
=
···
...

µ(B) 0
µ(B) 0
8

(10)

As for the second, Proposition 2
...

≤
εt
·
µ(B) a
t
1−ε
µ(B) 4B
X
R
We choose a = µ(B)−1 4B |f | dµ and combine the two parts
...
Then (10)
becomes

ε 

Z
Z
1
1
C(µ)ε
ε
(M f1 )(y) dµ(y) ≤
|f (y)| dµ(y)
1+
µ(B) B
µ(B) 4B
1−ε
ε


Z
µ(4B) 1
C(µ)ε
|f (y)| dµ(y)
= 1+
1−ε
µ(B) µ(4B) 4B


Z
ε
1
|f (y)| dµ(y) ≤ C(M f )(x)ε ,
≤ C(µ, ε)
µ(4B) 4B
where we have used the fact that µ satisfies the doubling condition (2)
...
Let x, y ∈ B = B(z, r) and let B ′ = B(z ′ , r ′ ) be another ball containing y
...
We claim that r ≤ r ′
...
Using that r ≤ r ′ , we have for any q ∈ B
d(q, z ′ ) ≤ d(q, y) + d(y, z ′ ) ≤ 2r ′ + r ′ = 3r ′ ,
which shows that B ⊂ 4B ′
...

µ(B ′ ) B ′
µ(4B ′ ) 4B ′
B∋x µ(B) B
R
In the case B ′ ⊂ 4B, we have that B ′ |f2 | dµ = 0, and the preceding estimate trivially holds
...




In order to show that (i) implies (ii) in Theorem 1
...
This is Lemma 2
...

Proposition 2
...
Let 1 ≤ p < ∞, and w ∈ Ap (X)
...

(11)
µ(B) B
µ(B) B
For a proof of Proposition 2
...
15
...
12
...
There exists an ε > 0 such that w1+ε ∈ Ap (X)
...
Let ε > 0 be such that w satisfies the reverse Hölder inequality (11) with δ = ε
...

As for p > 1 we start by observing that, as a consequence of Jensen’s inequality, if a weight v
satisfies (11) for some δ > 0, then v safisfies the same inequality for every 0 < δ′ ≤ δ
...
As a
′

consequence, we obtain that both w and w1−p satisfy a reverse Hölder inequality (11) for ε > 0
small enough
...




We are now ready to prove our main result, Theorem 1
...

Theorem 2
...
Let X be a complete metric space with a doubling measure, E ⊂ X a measurable
set with µ(E) > 0, and w a weight on E
...

(i) There exists a weight W ∈ Ap (X) such that W = w a
...
on E;
ep (E)
...
The implication (i) ⇒ (ii) follows from Lemma 2
...
Because W ∈ Ap (X) for a given
1 ≤ p < ∞, there exists an ε > 0 such that W 1+ε ∈ Ap (X)
...
Then, for all
balls B ⊂ X,
p−1


Z
Z
1+ε
1
1
w1+ε dµ
w 1−p dµ
µ(B) B∩E
µ(B) B∩E


p−1
Z
Z
1+ε
1
1
=
W 1+ε dµ
W 1−p dµ
µ(B) B∩E
µ(B) B∩E


p−1
Z
Z
1+ε
1
1
1+ε
1−p
W
dµ
W
≤ C
...

B
B∩E
B∩E
µ(B) B∩E
µ(B) B
ε

Next, let us prove (ii) ⇒ (i)
...
Consider first the case
ep (E), it is clear that v satisfies the hypothesis of Proposition 2
...
Because w1+ε ∈ A
1−p
e1 (E)
...
These are weights in A1 (X) as per Lemma 2
...
10
...
2
...
3)
...
Defining g = g1δ g2
that g, g −1 ∈ L∞ (X), g > 0, and

δ
g(x)V1 (x)V2 (x)1−p = v1 (x)v2 (x)1−p = v(x)δ = w(x)

for almost every x ∈ E
...
e
...

Finally, if p = 1, we reproduce the above argument taking v1 as v and discarding the weight
v2
...
Balls and chains

The aim of this section is to collect several preparatory results concerning balls in a metric space
with a doubling measure
...
In particular, Lemma 3
...
4 in the next section, which in turn is an integral part of Holden’s
argument in [17]
...
8, as we
could not locate one in the literature
...
To cite an example of geodesic
spaces relevant to partial differential equations, Corollary 8
...
16 in [16] states that a complete,
doubling metric space that supports a Poincaré inequality admits a geodesic metric that is bilipschitz
equivalent to the underlying metric, with constant depending on the doubling constant of the
measure and the data of the Poincaré inequality
...
A rectifiable curve γ : [a, b] → X satisfying ℓ(γ) = d(γ(a), γ(b)) is
called a geodesic on X
...

We will invoke the following well-known property of geodesics: if [a′ , b′ ] ⊂ [a, b], the subarc γ|[a′ ,b′ ]
of the geodesic γ : [a, b] → X is a geodesic too
...

Slightly abusing notation, we write γ|[x ,x ] to mean γ|[t ,t ] whenever γ(ti ) = xi , i = 1, 2
...
Also, when using the notation A ≈ B or
A
...

We begin by showing two lemmas in metric geometry for future reference
...

Lemma 3
...
Let X be a geodesic space, and B, B ′ any two balls in X
...

rad(B ′ ) and that B ′ contains the center of B
...

11

Proof
...
In the first
place, assume that d (z, z ′ ) ≤ 12 rad(B)
...

In this case, define B ′′ as the ball centered at z ′ and of radius a4 rad(B)
...
On the other hand, for any x ∈ B ′′ we can write
d (x, z) ≤ d (x, z ′ ) + d (z ′ , z) ≤

a
4

rad(B) +

1
2

rad(B) <

1
4

rad(B) + 21 rad(B) < rad(B),

which shows that B ′′ ⊂ B, and we also have rad(B ′′ ) = a4 rad(B) ≈ rad(B)
...
Let γ be a continuous curve joining z and z ′ with
ℓ(γ) = d (z, z ′ )
...
Let q ∈ γ be the midpoint between z and p, that is, d (z, q) = d (q, p) = 21 d (z, p)
...
For any x ∈ B ′′ we have
d (x, z) ≤ d (x, q) + d(q, z) ≤

1
2

d (z, q) + d (q, z) < 2 d (z, q) = d(z, p) =

1
2

rad(B)
...
To verify that B ′′ ⊂ B ′ , notice first that d(z, q) + d (q, z ′ ) = d (z, z ′ ) as
subarcs of the geodesic γ
...
Finally, because d (z, p) =
rad(B ′′ ) =

1
2

d (z, q) =

1
4

d (z, p) =

1
8

1
2

rad(B), we have

rad(B),

which completes the proof of the lemma
...
2
...
rad(B) for
some p ∈ B, p′ ∈ B ′
...

Proof
...
For any x ∈ B ′ we have
d (x, zB ) ≤ d (x, p′ ) + d (p′ , p) + d (p, zB ) ≤ 2 rad(B ′ ) + 2 d (B, B ′ ) + rad(B)
...

As a result, there exists a constant 1 ≤ λ < ∞ such that d(x, zB ) ≤ λ rad(B) for every q ∈ B ′ ,
which means that B ′ ⊂ λB and therefore µ(B ′ ) ≤ µ(λB)
...
µ(B) because the measure
is doubling, so µ(B ′ )
...
Reversing the roles of B and B ′ gives the inequality in the other
direction
...
8 in [14], whose proof is based on ideas from
[27, Lemma 2] and [6, Theorem 1
...
See also [29, Lemma 5] and [12, Lemma 1
...
3]
...
3
...
Then there exists a collection
W(D) = {Bk = B(xk , rk )}k of balls with the following properties:
S
S
(i) the balls {B(xk , rk /4)}k are pairwise disjoint and k Bk = k 2Bk = D;
(ii) 2 rad Bk ≤ d (x, X \ D) ≤ 6 rad Bk for every x ∈ 2Bk ;
(iii) for each B ∈ W(D), there are at most N = N (Cd ) < ∞ balls in W(D) that intersect B
...
We have
(iv) 14 rad B1 ≤ rad B2 ≤ 4 rad B1 ;
(v) µ(B1 ) ≈ µ(B2 )
...
8], but they are direct consequences of (ii) and Lemma 3
...
The next lemma pertains to balls whose radius is comparable to
their distance from the boundary
...

12

Lemma 3
...
Let D ⊂ X be open and proper, and B ⊂ D a ball such that rad(B) ≈ d (B, X \ D)
...
If B ′ ∈ W(D) and B ′ contains the center of B,
then there exists a ball B ′′ ⊂ B ∩ B ′ such that rad(B ′′ ) ≈ rad(B) and µ(B ′′ ) ≈ µ(B)
...
(i) Let y ∈ B ∩ B ′
...

Similarly, we obtain rad(B ′ ) & rad(B)
...
Let us write B =
{Bi }N
i=1
...
2 µ(Bi ) ≈ µ(B)
for every i
...


i=1



The Bi being Whitney balls, the collection { 41 Bi }i is pairwise disjoint
...
Since we obviously have the inclusion N
i=1 4 Bi ⊆ λB, we may write
!
N
N
N
X
X
[
1
1
µ(
µ(B) = N µ(B)
...

(iii) We know from (i) above that r(B) ≈ r(B ′ )
...
1 provides a ball B ′′ ⊂ B ∩ B ′ such that
r(B ′′ ) ≈ r(B) ≈ r(B ′ )
...
2
...

Definition 3
...
Let D ⊂ X be a domain, k ∈ {0, 1, 2,
...
, k
...
, Bk )
is a (Whitney) chain joining B0 to Bk , if Bj ∩ Bj−1 6= ∅ for every j ∈ {1,
...
In this case, we
say that k is the length of the chain C(B0 , Bk )
...
Because there is no possibility of confusion, we drop the subscript D
from now on
...


Definition 3
...
Let X be a geodesic space
...
The
quantity kD satisfies the axioms of a metric on D × D
...

γ|[y ,y ] d (y, ∂D)
1

13

2

If E1 , E2 are subsets of D, we define kD (E1 , E2 ) = inf x1 ∈E1 , x2 ∈E2 kD (x1 , x2 )
...

It is easy to see that the quasihyperbolic diameter of any Whitney-like ball is bounded, which is
the content of the following lemma
...
7
...
If B ⊂ D is
a ball such that d(B, ∂D) ≈ rad(B), then k(x, y) ≤ C for any two points x, y ∈ B
...
Let z denote the center of B, and let γ ⊂ B be a rectifiable curve connecting z and x such
that ℓ(γ|[z,x] ) = d(z, x)
...

=
≤ C
...




The next lemma establishes an equivalence between shortest Whitney chains and quasihyperbolic
distance
...
For a detailed proof of the
corresponding lemma in Rn , see Proposition 6
...
Notice that if the space X is geodesic and
D ⊂ X is a proper subset, the distance functions d (·, ∂D) and d (·, X \ D) coincide over D
...
3 and 3
...

Lemma 3
...
Assume further that X is a geodesic space
...
Then e
k(B1 , B2 ) ≈ k(x1 , x2 )
...
In the
Proof
...
Suppose now x1 and
x2 are distinct points
...
k(x1 , x2 )
...
Of all the Whitney balls
containing z, we choose the one with the smallest radius, say, B = B(x, r)
...
It is clear that Bz ⊂ 2B, and thus Bz is contained in D with
d (Bz , ∂D) ≥ d (2B, ∂D) ≥ r by virtue of Lemma 3
...
Also, by the properties of the Whitney
decomposition (Lemma 3
...

Let γz be a subarc of γ ∩ Bz passing through z and of maximal length
...
Whenever γ is not entirely contained in Bz , by the continuity of γ, there exists a
point q ∈ γz such that d(q, z) > r/2
...
In the case γ ⊂ Bz ,
by the properties of the Whitney decomposition there exists a constant 0 < c < 1 such that
ℓ(γz ) = ℓ(γ) ≥ d(x1 , x2 ) ≥ cr1
...
4 (i) gives r ≈ r1 and consequently
ℓ(γz ) ≥ C1 r
...

(12)
d(y,
∂D)
r
+
d
(z,
∂D)
9r
γz
Next, we cover the geodesic γ by balls {Bzi }i , with the points {zi }i ⊂ γ chosen so that every point
is contained in at most two balls Bzi
...
For any z ∈ γ, Lemma 3
...
Now let M1 be the minimal number of Whitney balls needed to cover i Bzi ,
and denote this collection by F
...
Also, we have that # F = M1 and, by minimality, for every B ∈ F there is at
14

S
least one i such that B ∩ Bzi 6= ∅
...

{B ∈ WD : B ∩ Bzi 6= ∅} ≤
M1 = # F ≤ #
i=1

i=1

We obtain Cm ≥ M1 ≥ M
...

d
(y,
∂D)
2
d
(y,
∂D)
2
2C
γ
i=1 γi


As for the inequality in the other direction, take the the shortest chain C = B 1 , B 2 ,
...
For every j ∈ {1,
...
We have k(xj , pj ) ≤ C and k(xj+1 , pj ) ≤ C owing to Lemma 3
...
Using the
triangle inequality repeatedly, we obtain
k(x1 , x2 ) = k(x1 , xM ) ≤

M
−1
X
j=1

whereby the statement is proven
...
M = e
k(B1 , B2 ),



4
...
e
...
Holden
[17] gives certain sufficient conditions for extension domains in Rn
...
In this
final section we adapt [17, Lemma 2] into the metric setting, resulting in Lemma 4
...
In
Holden’s Euclidean argument, [17, Lemma 2] is used to estimate integrals over each cube in a
dyadic decomposition of Q ∩ E in terms of integrals over cubes arising from Holden’s assumptions
that enjoy additional good properties
...
These are intimately related to Muckenhoupt
weights: whenever a weight w belongs to Ap , then log w is of bounded mean oscillation
...
Peter W
...
Vodop’yanov and Greshnov
[29] extended Jones’ characterization to metric spaces supporting a doubling measure
...

For Muckenhoupt weights, the question remains open
...

For the purposes of this section we need to introduce classical Ap weights defined on a subset
...
1
...

Definition 4
...
Let D ⊂ X be a nonempty open subset in a metric space X, and 1 < p < ∞
...
e
...

(13)
dµ
[w]p = sup
µ(B) B
B⊂D µ(B) B
15

The supremum is taken over all balls B ⊂ D
...

(14)
B
µ(B) B

We denote by [w]1 the infimum of the C > 0 for which the inequality (14) holds
...
This
property follows from statement (ii) of the next lemma, that collects some estimates for weights on
balls and chains
...

Lemma 4
...
Let D ⊂ X be an open proper subset, and w ∈ Ap (D) with 1 ≤ p < ∞
...

µ(B) B
µ(B) B
(ii) If B is a ball in D and E ⊂ B is a measurable subset with µ(E) > 0, then
 Z

Z
µ(B) p
w dµ
...

w(B)
µ(B)
(iv) Assume further that D is a domain
...


Proof
...
Now, the inequality (i) fol-

lows from the Ap condition (Definition 4
...
Indeed, notice that the function t 7→ exp −t(p − 1)−1
is convex, and apply Jensen’s inequality:


p−1
Z
Z
1
1
− 1
w dµ
exp (log w) p−1 dµ
[w]p ≥
µ(B) B
µ(B) B


− 1 (p−1)

Z
Z
p−1
1
1

...
1):

−(p−1)
Z
Z
1
1
− 1
w dµ ≤ [w]p
w p−1 dµ
µ(B) B
µ(B) B1
−(p−1)
−(p−1) 

Z
1
1
µ(E)
− p−1
dµ
w
≤ [w]p
µ(B)
µ(E) E


Z
µ(B) p−1 1
≤ [w]p
w dµ,
µ(E)
µ(E) E
16

where the last estimate follows from Hölder’s inequality
...

B
E
µ(B) B
µ(E) E
For a proof of the A∞ condition (iii) we refer to [28], Theorem I
...
There, the weights are
globally defined in X, but the proof for weights in Ap (D) is exactly the same
...
11), using it and the classical Hölder inequality we have
Z

E

w dµ ≤ µ(E)

δ
1+δ

Z

w
B

1+δ

dµ



1
1+δ

≤C



µ(E)
µ(B)



δ
1+δ

Z

w dµ
B

for all balls B ⊂ D and all measurable subsets E ⊂ B
...
1 we only consider balls that are entirely
contained in D
...
77
...
Let Bj = B(pj , rj ) and Bj+1 = B(pj+1 , rj+1 ) be two consecutive balls
in the chain C(B1 , B2 )
...
To begin with, we show that there is a constant C
such that
Z
Z
w dµ ≤ C

Bj+1

w dµ
...
For any z ∈ X, we have
d (z, pj ) ≤ d (z, pj+1 ) + d (pj , y) + d (y, pj+1 ) < d (z, pj+1 ) + rj + 18 rj+1
...
3, we have
d (z, pj ) < 18 rj+1 + rj + 18 rj+1 ≤

1
4

· 4rj + rj = 2rj ,

which shows that 18 Bj+1 ⊂ 2Bj ⊂ D as guaranteed by (i) of Lemma 3
...
Using (ii) of the current
lemma and the fact that µ is doubling, we may write
!p Z
Z
Z
µ(Bj+1 )
w dµ
...
[w]p
w dµ
1
1
µ( 18 Bj+1 )
Bj+1
B
B
8 j+1
8 j+1
 Z

Z
Z
µ(2Bj ) p
w dµ,
w dµ
...
[w]p
≤ [w]p
µ(Bj )
Bj
Bj
2Bj
which proves (15)
...
The balls Bj+1
= B(y, 18 rj+1 ) and 81 Bj+1 have the
same radius and d (y, pj+1 ) ≈ 81 rj+1
...
2 and the doubling condition (2), it holds
that


∗
µ(Bj+1
) ≈ µ 18 Bj+1 ≈ µ(Bj+1 )
...
3 (iv), and the fact that y ∈ Bj ∩ Bj+1 , we easily obtain for
∗
a z ∈ Bj+1
d(z, pj+1 ) ≤ d (z, y) + d(y, pj+1 ) ≤ 18 rj+1 + rj+1 < 2rj+1 ,
d (z, pj ) ≤ d (z, y) + d(y, pj ) ≤ 81 rj+1 + rj ≤ 32 rj < 2rj ,
17

∗
which implies that Bj+1
⊂ 2Bj+1 ∩ 2Bj
...
[w]p
w dµ ≤
w dµ
...
[w]p
≤ [w]p
w dµ
...
[w]p
µ(Bj )
2Bj
Bj
Bj
R
R
Thus, in any case we have Bj+1 w dµ
...
Reversing the roles of Bj and Bj+1 , we obtain
R
R
Bj w dµ ≈ Bj+1 w dµ, where the constants involved depend on p, the doubling constant, and [w]p
...
3 (v), there exists a constant C1 such that
Z
Z
1
1
w dµ ≤ C1
w dµ
...

µ(B1 ) B1
µ(B2 ) B2
Choose C = log C1 to get the desired expression
...
3
...
Statement (ii) is a “reverse A∞ condition” that follows immediately from the
Ap condition
...

≤ C(w)
µ(B)
w(B)
Lemma 4
...
Assume further that X is a geodesic space and let D ⊂ X be a domain, w ∈ Ap (D)
with 1 ≤ p < ∞, C a constant possibly depending on Cd , and B1 , B2 ⊂ D balls satisfying
(i) d (Bi , ∂D) ≈ rad(Bi ), i = 1, 2,
(ii) k(B1 , B2 ) ≤ C
...

Proof
...
Lemma 3
...

Furthermore, it holds that k(z1′ , z2′ ) ≤ C
...
Using the triangle inequality for k we have
k(z1′ , z2′ ) ≤ k(z1′ , z1 ) + k(z1 , x1 ) + k(x1 , x2 ) + k(x2 , z2 ) + k(z2 , z2′ )
...
By Lemma 3
...

18

k(B1′ , B2′ )
...
With these remarks,
k(B1′ , B2′ ) and thus e
Also, by Lemma 3
...
2 (ii) allows us to estimate


Z
Z
1
µ(B1 ) p−1 1
w dµ
(16)
w dµ
...

w dµ
µ(B1′′ ) µ(B1′ ) B1′
Z
1
w dµ
(17)

...

w dµ
(18)
µ(B2′ ) B2′


Z
µ(B2′ ) p−1 1
w dµ
(19)

...

w dµ
(20)
µ(B2′′ ) B2′′
Z
1
w dµ
...

µ(B2 ) B2
Line (16) follows from the fact that the measure w dµ is doubling, while (18) and (19) are Lemma
4
...
On lines
(17) and (20) we used the fact that µ(Bi′′ ) ≈ µ(Bi′ ) ≈ µ(Bi )
...
, B N = B2′ is the shortest Whitney chain connecting B1′ and B2′ , we
have that N
...
Since each pair of consecutive balls (B j−1 , B j ) in the
chain has nonempty intersection, we have rad(B j−1 ) ≈ rad(B j ) by Lemma 3
...
, N, because N
...
Moreover, if
pj ∈ B j−1 ∩ B j , the triangle inequality gives
d (p0 , pN ) ≤

N
X

d (pj−1 , pj ) ≤

N
X

2 rad(Bj−1 )
...


j=1

j=1

It follows from Lemma 3
...
We conclude
that
Z
Z
w dµ
w dµ
...




References
[1] T
...
Anderson, D
...
Moen, Extrapolation in the scale of generalized reverse Hölder weights,
Rev
...
Complut
...
2, 263–286
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C
...
Hytönen, and O
...
Geom
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27 (2017), no
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[3] P
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Bortz, M
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Saari, Non-local Gehring lemmas in spaces of homogeneous type and
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Björn and J
...
17, European Mathematical Society (EMS), Zürich, 2011
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Butaev and G
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[7] D
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M
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Pérez, Extrapolation from A∞ weights and applications, J
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Cruz-Uribe, K
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Rodney, Matrix Ap weights, degenerate Sobolev spaces, and mappings of finite
distortion, J
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4, 2797–2830
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García-Cuerva and J
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Rubio de Francia, Weighted norm inequalities and related topics, North-Holland
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Osgood, Uniform domains and the quasihyperbolic metric, J
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[12] I
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Gogatishvili, V
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249, Springer, New
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Grafakos, L
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Maldonado, and D
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497 (2014), 121 pp
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Heinonen, Lectures on analysis on metric spaces, Universitext, Springer-Verlag, New York, 2001
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Heinonen, P
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Shanmugalingam, and J
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Tyson, Sobolev spaces on metric measure spaces: an
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Pérez, Sharp weighted bounds involving A∞ , Anal
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Rela, Sharp reverse Hölder property for A∞ weights on spaces of homogeneous
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