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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
...
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+ǫ
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
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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
which, recalling that w ∈ A
Z
Z
1
1
JwK1 w(x) < (1 − δ)
w dµ <
w dµ ≤ JwK1 ess′ inf w
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Hence A ⊂ B ′ ∈F DB ′ , which is a countable
union of sets of measure zero
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The proof is similar, except that the absolute continuity
of the Lebesgue integral is not needed
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3, and is needed in the proof of Theorem 2
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e1 (E), the function
Lemma 2
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For a measurable set E ⊂ X with µ(E) > 0 and a weight w ∈ A
wXE is in L1loc (X) and its maximal function M (wXE ) is finite at almost every point of X
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R
Proof
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As for
the second statement, Proposition 2
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e
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It
remains to verify that M (wXE ) < ∞ on X \ E
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3 shows that µ(E \ A) = 0, and therefore M (wXE ) = M (wXA ) on X
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e
...

sup
B∋x µ(B) B
r(B)≤rx

For almost every x ∈
/ A, we estimate the averages over balls B such that x ∈ B, r(B) > rx ,
and B ∩ A 6= ∅
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Also, there exists a point y0 = y0 (x) ∈ A such that d (x, y0 ) <
max{rx , 2d(x, A)}
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Using the doubling condition for µ and the definition of A, we obtain
Z
Z
Z
1
1
1
wXA dµ ≤ C(Cd )
wXA dµ ≤ C(Cd ) sup
w dµ ≤ C(Cd ) JwK1 w(y0 )
...




In the following two technical lemmas, we will not be using the fact that the measure is doubling
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5
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If p, q > 1, v ∈ A
q δy
δ
−1
δ
e
e1 (E),
0 ≤ δ ≤ min{1, (q − 1)(p − 1) }, then v ∈ Aq (E) with v q ≤ JvKp
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In particular, A
1

q

1 ≤ p ≤ q
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We will use the following basic estimate
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Then it follows from Hölder’s inequality that
s
Z
Z
s
1−s
(5)
h dµ ≤ µ(A)
h dµ
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(7)
v
B∩E

Z

Since the exponents 1 − δ and (q − 1) − δ(p − 1) are nonnegative, we have µ(B ∩ E)1−δ ≤ µ(B)1−δ
−1
−1
and µ(B ∩ E)1−δ(p−1)(q−1) ≤ µ(B)1−δ(p−1)(q−1)
...
In the case δ ∈ [0, 1], v ∈ A
e1 (E) we can write
and then the definition of A
!q−1
δ
  1
Z
Z
Z
q−1
µ(B ∩ E)q−1
1
1
µ(B ∩ E)1−δ
δ
v
dµ
v
dµ
dµ
≤
µ(B)q B∩E
vδ
µ(B)q
ess inf B∩E v δ
B∩E
B∩E


µ(B ∩ E) q−δ
δ
≤ JvKδ1 ,
≤ JvK1
µ(B)
where we have used the fact that δ ≤ 1
...




eq (E) and
Lemma 2
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Let E ⊂ X be a measurable set with µ(E) > 0
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|g| dµ ≤ JvKq
µ(B) B∩E
B∩E
Proof
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In the case q > 1, applying Hölder’s inequality we readily
obtain
!q−1

q
  1
Z
Z
Z
Z
JvKq
1
1
1 q−1
q
dµ
≤
g v dµ
g dµ ≤
g q v dµ
...

When q = 1, the assertion follows immediately from the definition of A



The Hardy-Littlewood maximal function is well known to satisfy a weak type inequality
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Notice that by
letting E = X, we recover the classical result
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7
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Furthermore, let 1 ≤ q < ∞,
eq (E), f ∈ Lq (E, v), and t > 0
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Proof
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We restrict the supremum defining the maximal function to balls with
radius no greater than R, and denote the resulting function by mR
E f
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R
R For every x ∈ Et = {x ∈ E : mE f (x) > t}, there
S is a ball Bx ∋ x such that rad(Bx ) ≤ R and
f
dµ
>
tµ(B
)
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Since the space X is separable, by
Bx ∩E
the Vitali S
covering lemma
S we can find a disjoint sequence of balls {Bj }j belonging to this collection
such that x∈Et Bx ⊂ j 5Bj
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(8)
v dµ ≤
v dµ ≤ S
Et

j (5Bj ∩E)

j

5Bj ∩E

For each j, we apply Lemma 2
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JvKq
µ(5Bj ) 5Bj ∩E
µ(5Bj ) Bj ∩E
Bj ∩E
5Bj ∩E
j

j

6

By the choice of the balls Bx , this in turn is smaller than


Z
Z
XZ
tµ(Bj ) −q
q
−q
q
−q
f v dµ
JvKq
≤ JvKq C(q, Cd )t
f v dµ ≤ Ct
f q v dµ,
S
µ(5Bj )
Bj ∩E
B
∩E
E
j j
j

where we have applied the doubling property of µ
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eq (E), and
Proposition 2
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Let E ⊂ X be a measurable set with µ(E) > 0, 1 ≤ q < p, v ∈ A
p
f ∈ L (E, v)
...

Proof
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Using the subadditivity of the maximal function mE , we have that mE f ≤ mE (ft )+t/2, from which
it is clear that the set {x ∈ E : mE f (x) > t} is contained in {x ∈ E : mE ft (x) > t/2}
...
7 for
q and ft , we arrive at
Z ∞
Z
tp−1 v ({x ∈ E : mE ft (x) > t/2}) dt
(mE f )(x)p v(x) dµ(x) ≤ p
0
E
Z ∞
Z ∞
Z
Z
p−q−1
p−q−1
q
t
f (x)q v(x) dµ(x) dt
t
ft (x) v(x) dµ(x) dt = C
≤C
=C

Z

0

f (x)q v(x)
E

0

E

Z

2f (x)

tp−q−1 dt dµ(x) ≤ C

0

where C depends on p, q, JvKq , and Cd (µ)
...
1
...
9
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Then there exist weights v1 , v2 ∈ A
e1 (E) such that v = v1 v 1−p
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Writing q1 = r −1 (p − 1) + 1, we have 1 < q1 < p and, by virtue of Lemma 2
...
Also, by the hypothesis, the weight (v −r )1/(p−1) belongs
(v r )

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