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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
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We give a complete and self-contained proof of this theorem generalized
into metric measure spaces supporting a doubling measure
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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
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Our main result is the following theorem that provides an abstract starting point for the investigation of extensions
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Wolff
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An outline of the Euclidean proof can be
found in [9], Theorem 5
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However, the metric setting brings about technical challenges that are
not present in the Euclidean case
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1
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Then, for 1 < p < ∞, the following statements are equivalent
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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
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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
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fi, carlos
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mudarra@jyu
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mudarra@aalto
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Date: October 26, 2021
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30L99, 42B25, 42B37
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Muckenhoupt weight, metric measure space, doubling condition
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E
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Kurki was funded by a young researcher’s grant from the Emil Aaltonen Foundation
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Mudarra acknowledges financial support from the Academy of Finland
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1

for every ball B ⊂ X
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13) for measurable sets in a metric space supporting a doubling measure
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We have chosen to call these
classes induced Ap weights; see Definition 2
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It is not obvious at the outset whether all
properties of globally defined weights hold true for this class as well
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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
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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+Ç«
instead of simply stating the corresponding condition for w
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12)
...
11)
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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
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This technicality destroys our ability to compute
the averaged integrals that would lead to the RHI
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Muckenhoupt weights in
particular are discussed in [12] and [28]
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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
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1
...
In particular, Lemma 4
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In [17], this lemma is used to recover the Euclidean equivalent of Theorem
1
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2
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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
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In particular, we assume
that every ball in X has positive and finite measure
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This implies that X is
locally compact and proper, which in turn means that every closed and bounded subset of X is
compact
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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
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Furthermore, we write A ≈ B whenever there exist constants C1 , C2 ∈ (0, ∞)
such that C1 A ≤ B ≤ C2 A
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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)
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[28] In practice, we
assume w to be positive almost
everywhere in E
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For the purposes of Theorem 2
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weights on a subset, which we denote by A

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

  1

Z
Z
1
1 p−1
1
w dµ
dµ
< ∞
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We denote by JwK1 the infimum of the C > 0 for which the inequality (4)
holds
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Whenever E = X, the above classes coincide with Muckenhoupt weights as usually defined
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Notice that it is not possible
to reduce (3) to the Ap (X) condition e
...
by replacing w with XE w
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2
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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
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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
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If w ∈ A
mE w(x) ≤ JwK1 w(x) for almost every x ∈ E
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The first inequality is a consequence of the Lebesgue differentiation theorem for µ
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4
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We aim to show that µ(A) = 0
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Define a countable collection of
balls F = {B(zk , q) : k ≥ 1, q ∈ Q+ }
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JwK1 w(x) < (1 − δ)
µ(B) B∩E

For every ε ∈ (0, 1) denote Bε = B(z, (1 − ε)r)
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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
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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
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The next lemma follows from Proposition 2
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13 below
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4
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Here
wXE is the function in X that coincides with w on E and vanishes outside E
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It is immediate that wXE ∈ L1loc (X) because B∩E w is finite for every ball B ⊂ X
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3 implies that M (wXE )(x) = mE w(x) < ∞ for a
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x ∈ E
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Defining
A = {y ∈ E : mE w(y) ≤ JwK1 w(y) < ∞},
4

Proposition 2
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For a
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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
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For such a ball B it clearly holds that d (x, A) ≤ 2r(B), and consequently
r(B) ≥ max{rx , d (x, A)/2}
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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
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′
µ(B) B
µ(5B) 5B
B ′ ∋y0 µ(B ) B ′ ∩A
We conclude that M (wXA )(x) < ∞ for almost every x ∈ X
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ep (E), and
Lemma 2
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Let E ⊂ X be a measurable set with µ(E) > 0
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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
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Let A ⊂ X be measurable, 0 ≤ s ≤ 1, and
h ∈ L1 (A)
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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
...
Applying
p

p

eq (E) and
again Lemma 2
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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′ )
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We will not be needing the reverse statement
...
10
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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
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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
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≤
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 â‰

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