Table Of ContentThe Gehring lemma in metric spaces
8
0 Outi Elina Maasalo
0
2 21th March 2006∗
n
a
J
5
Abstract
1
We present a proof for theGehring lemma in a metric measure space
] endowed with a doubling measure. As an application we show the self–
A
improving property of Muckenhouptweights.
C
.
h
t 1 Introduction
a
m
Thefollowingself–improvingpropertyofthereverseHo¨lderinequalityisaresult
[
due to Gehring [6]. Assume that f is a non–negativelocallyintegrablefunction
3 and 1<p<∞. If there is a constant c such that the inequality
v
6 1/p
1 fpdx ≤c fdx (1.1)
9 (cid:18)ZB (cid:19) ZB
3
. holds for all balls B of Rn, then there exists ε>0 such that
4
0
1/p+ε
7 fp+εdx ≤c fdx (1.2)
0
(cid:18)ZB (cid:19) ZB
:
v
for some other constant c. The theorem remains true also in metric spaces.
i
X However,the proof seems to be slightly difficult to find in the literature.
r ThesubjecthasbeenstudiedforexamplebyFiorenza[5]aswellasD’Apuzzo
a
andSbordone[4],[14]. Gianazza[7]showsthatifafunctionsatisfies(1.1),then
there exists ε>0 such that
1/p+ε
fp+εdµ ≤c fdµ (1.3)
(cid:18)ZX (cid:19) ZX
forsomeconstantc. Theresultisobtainedinaspaceofhomogeneoustype,with
the assumption that 0 < µ(X) < ∞. In this paper, our purpose is to present
a transparent proof for a version of the Gehring lemma in a metric space that
has the annular decay property and that supports a doubling measure.
∗Revised15thJan2008
1
Also Kinnunen examines various minimal, maximal and reverse Ho¨lder in-
equalities in [11] and [12]. Stro¨mberg and Torchinsky prove Gehring’s result
under the additional assumption that the measure of a ball depends continu-
ously on its radius, see [15]. Zatorska–Goldstein [16] proves a version of the
lemma, where on the right–hand side there is a ball with a bigger radius.
We present a proof of the Gehring lemma in a doubling metric measure
space. In addition, we assume that the space has the annular decay property,
see Section 2. This is true for example in length spaces, i.e. metric spaces in
which the distance between any pair of points is equal to the infimum of the
length of rectifiable paths joining them. Our method is classicaland intends to
be as transparent as possible. In particular, we obtain the result for balls in
the sense of (1.2) in the metric setting instead of (1.3). The proof is based on
a Caldero´n–Zygmundtype argument.
The Gehring lemma has a number of possible applications. As an example
we show that the Muckenhoupt class is an open ended condition. The proof is
classicalandholdswithouttheassumptionofannulardecayproperty. Moreover,
the Gehring lemma can be applied for example to prove higher integrability of
the volume derivative, also known as the Jacobian, of a quasisymmetric map-
ping, see [10].
2 General Assumptions
Let (X,d,µ) be a metric measure space equipped with a Borelregular measure
µ such that the measure of every nonempty open set is positive and that the
measure of every bounded set is finite.
Our notation is standard. We assume that a ball B in X comes always
with a fixed centre and radius, i.e. B = B(x,r) = {y ∈ X: d(x,y) < r} with
0<r <∞. We denote
1
u = udµ= udµ,
B
µ(B)
ZB ZB
and when there is no possibility for confusion we denote kB the ball B(x,kr).
We assume in addition that µ is doubling i.e. there exists a constant c such
d
that
µ(B(x,2r))≤c µ(B(x,r))
d
for all balls B in X. We refer to this property by calling (X,d,µ) a doubling
metricmeasurespaceanddenoteitbrieflyX. Thisisdifferentfromtheconcept
of doubling space. The latter is a property of the metric space (X,d), where all
ballscanbecoveredbyaconstantnumberofballswithradiushalfoftheradius
of the original ball. A doubling metric measure space is always doubling as a
metric space.
Agoodreferenceforthebasicpropertiesofadoublingmetricmeasurespace
is [9]. In particular, we will need two elementary facts. Consider a ball con-
taining disjoint balls such that their radii are bounded below. In a doubling
2
space the number of these balls is bounded. Secondly, the doubling property of
µ implies that for all pairs of radii 0<r ≤R the inequality
µ(B(x,R)) R Q
≤c
d
µ(B(x,r)) r
(cid:18) (cid:19)
holds true for all x ∈ X. Here Q = log c is called the doubling dimension of
2 d
(X,d,µ).
Given0<α≤1andametricspace(X,d,µ)withadoublingµ,wesaythat
thespacesatisfiesthe α–annular decay property ifthere existsaconstantc≥1,
such that
µ(B(x,r)\B(x,(1−δ)r))≤cδαµ(B(x,r))
for all x ∈ X, r > 0 and 0 < δ < 1. We omit α in the notation if we do not
care about its value. See [1] for further information on spaces that satisfy the
annular decay property.
Throughout the paper, constants are denoted c and they may not be the
same everywhere. However, if not otherwise mentioned, they depend only on
fixed constants such as those associated with the structure of the space, the
doubling constant etc.
3 Gehring lemma
The following theorem is our main result.
Theorem 3.1 (Gehring lemma). Consider (X,d,µ), where µ is doubling. Let
1 < p < ∞ and f ∈ L1 (X) be non–negative. If there exists a constant c such
loc
that f satisfies the reverse Ho¨lder inequality
1/p
fpdµ ≤c fdµ (3.1)
(cid:18)ZB (cid:19) ZB
for all balls B of X, then there exists q >p such that
1/q
fqdµ ≤c fdµ (3.2)
q
(cid:18)ZB (cid:19) Z2B
for all balls B of X. The constant c as well as q depend only on the doubling
q
constant, p, and on the constant in (3.1).
In metric spaces with a priori more geometrical structure this can be im-
proved.
Corollary 3.2. In addition to the assumptions in Theorem 3.1, suppose X
satisfies the annular decay property. Then the measure induced by f is doubling
and (3.1) implies
1/q
fqdµ ≤c fdµ (3.3)
q
(cid:18)ZB (cid:19) ZB
for all balls B of X. The constant c as well as q depend only on the doubling
q
constant, p, and on the constant in (3.1).
3
The following is a standard iteration lemma, see [8].
Lemma 3.3. Let Z : [R ,R ] ⊂ R → [0,∞) be a bounded non–negative func-
1 2
tion. Suppose that for all ρ,r such that R ≤ρ<r ≤R
1 2
Z(ρ)≤ A(r−ρ)−α+B(r−ρ)−β +C +θZ(r) (3.4)
holds true for some con(cid:0)stants A,B,C ≥0, α>β >0(cid:1)and 0≤θ <1. Then
Z(R )≤c(α,θ) A(R −R )−α+B(R −R )−β +C . (3.5)
1 2 1 2 1
Lemma 3.3 is needed in(cid:0)the proof of our first key lemma: (cid:1)
Lemma 3.4. Let R > 0, q > 1, k > 1 and f ∈ Lq (X) non–negative. There
loc
exists ε>0 such that, if for all 0<r ≤R and for a constant c
q
fqdµ≤ε fqdµ+c fdµ (3.6)
ZB(x,r) ZB(x,kr) ZB(x,kr) !
holds, then
q
fqdµ≤c fdµ . (3.7)
ZB(x,R) ZB(x,2R) !
Theconstantin (3.7)dependsonk,onthedoublingconstantandontheconstant
in (3.6).
Proof. Fix R > 0 and choose r,ρ > 0 such that R ≤ ρ < r ≤ 2R. Set
r˜=(r−ρ)/k. Now
B(x,ρ)⊂ B(y,r˜/5)
y∈B[(x,ρ)
and by the Vitali covering theorem there exist disjoint balls {B(x ,r˜/5)}∞
i i=1
such that x ∈B(x,ρ) and
i
B(x,ρ)⊂ B(x ,r˜).
i
i
[
These balls can be chosen so that
χ ≤M (3.8)
B(xi,kr˜)
i
X
for some constant M < ∞. This follows from the doubling property of the
space. Indeed, assume that y belongs to N balls B(x ,kr˜). Clearly
i
B(x ,kr˜)⊂B(y,2kr˜)⊂B(y,2R).
i
Remember that r˜ and R are fixed and choose K = 20R/r˜. Now there are N
disjoint balls with radius r˜/5 ≥ 2R/K included in a fixed ball B(y,2R). Since
the space is doubling, we must have N ≤M(K). The inequality (3.8) follows.
4
Observethenthatbythedoublingpropertyandtheconstructionoftheballs
{B(x ,r˜)} we have
i i
µ(B(x ,r˜))≤c µ(B(x ,r˜/5))=cµ(∪ B(x ,r˜/5))
i i i i
i i
X X
Q
r
≤cµ(B(x,r)) ≤c µ(B(x,ρ)).
ρ
(cid:18) (cid:19)
On the other hand, B(x,ρ)⊂B(x ,2kρ), so that
i
Q
2kρ
µ(B(x,ρ))≤µ(B(x ,2kρ))≤c µ(B(x ,r˜))
i i
r˜
(cid:18) (cid:19)
Q
ρ
=c µ(B(x ,r˜)).
i
r−ρ
(cid:18) (cid:19)
Combining these two inequalities implies
r −Q
µ(B(x,ρ))≥c µ(B(x ,r˜))
i
ρ
(cid:18) (cid:19) i
X
r −Q ρ −Q
≥c µ(B(x,ρ)).
ρ r−ρ
(cid:18) (cid:19) (cid:18) (cid:19) i
X
And as a consequence
r Q ρ Q
#{B(x ,r˜)}≤c ,
i
ρ r−ρ
(cid:18) (cid:19) (cid:18) (cid:19)
Q
i.e. the number ofballs B(x ,r˜)is atmostc r/(r−ρ) ,where c depends only
i
on k, on the doubling constant and Q=log c .
(cid:0)2 d (cid:1)
Observe that (3.6) holds true for r˜, so that
µ(B(x ,r˜))
fqdµ≤ε i fqdµ
µ(B(x ,kr˜))
ZB(xi,r˜) i ZB(xi,kr˜)
q
µ(B(x ,r˜))
i
+c fdµ
µ(B(x ,kr˜))q
i ZB(xi,kr˜) !
q
≤ε fqdµ+cµ(B(x ,r˜))1−q fdµ . (3.9)
i
ZB(xi,kr˜) ZB(xi,kr˜) !
We note that
µ(B(x,r)) µ(B(x ,2r)) 2r Q r Q
i
≤ ≤c ≤c ,
d
µ(B(x ,r˜)) µ(B(x ,r˜)) r˜ r−ρ
i i (cid:18) (cid:19) (cid:18) (cid:19)
from which it follows that
Q(q−1)
r
µ(B(x ,r˜))1−q ≤c µ(B(x,r))1−q.
i
r−ρ
(cid:18) (cid:19)
5
Together with (3.9) this implies
fqdµ≤ε fqdµ
ZB(xi,r˜) ZB(xi,kr˜)
Q(q−1) q
r
+c µ(B(x,r))1−q fdµ . (3.10)
r−ρ
(cid:18) (cid:19) ZB(xi,kr˜) !
Since B(x,ρ)⊂∪ B(x ,r˜), summing over i in (3.10) gives
i i
fqdµ≤ fqdµ
ZB(x,ρ) i ZB(xi,r˜)
X
≤ε fqdµ
i ZB(xi,kr˜)
X
r Q(q−1) q
+c µ(B(x,r))1−q fdµ
r−ρ
(cid:18) (cid:19) i ZB(xi,kr˜) !
X
≤εM fqdµ
ZB(x,r)
Q(q−1) Q q
r r
+c µ(B(x,r))1−q fdµ
r−ρ r−ρ
(cid:18) (cid:19) (cid:18) (cid:19) ZB(x,r) !
Qq q
r
=εM fqdµ+c µ(B(x,r))1−q fdµ .
r−ρ
ZB(x,r) (cid:18) (cid:19) ZB(x,r) !
Finally, remember that R≤ρ<r ≤2R, so that
fqdµ≤εM fqdµ
ZB(x,ρ) ZB(x,r)
q
+cRQq(r−ρ)−Qqµ(B(x,r))1−q fdµ
ZB(x,r) !
and furthermore
fqdµ≤εc fqdµ
ZB(x,ρ) ZB(x,r)
q
+cRQq(r−ρ)−Qq fdµ . (3.11)
ZB(x,2R) !
We are able to iterate this. In Lemma 3.3 set
Z(ρ):= fqdµ,
ZB(x,ρ)
6
so that Z is bounded on [R,2R]. Set also R =R, R =2R, α=Qq and
1 2
q
A=cRQq fdµ ≥0,
ZB(x,2R) !
where c is the constant in (3.11). Putting θ = cε and choosing ε so small that
cε < 1, (3.11) satisfies the assumptions of Lemma 3.3 with B = C = 0. This
yields Z(R)≤cA(2R−R)−Qq, that is
q
|f|qdµ≤cRQq(cR−R)−Qq fdµ
ZB(x,R) ZB(x,2R) !
q
=c fdµ .
ZB(x,2R) !
In the following, we consider the Hardy–Littlewood maximal function re-
stricted to a fixed ball 100B , that is
0
Mf(x)= sup |f|dµ.
B∋x ZB
B⊂100B0
Clearlythecoefficient100canbereplacedbyanyothersufficientlybigconstant.
The role of this constantis setting a playgroundlargeenough to assurethat all
balls we are dealing with stay inside this fixed ball.
Lemma 3.5. Let f be a non–negative function in L1 (X) and suppose that is
loc
satisfies the reverse Ho¨lder inequality (3.1). Then for all balls B in X
0
fpdµ≤cλpµ({x∈100B : Mf(x)>λ}), (3.12)
0
Z{x∈B0:Mf(x)>λ}
for all λ > essinf Mf with some constant depending only on p, the doubling
B0
constant and on the constant in (3.1).
Proof. LetusfixaballB withradiusr >0. Wedenote{x∈X : Mf(x)>λ}
0 0
briefly by {Mf > λ}. Let λ > essinf Mf. Now there exists x ∈ B so that
B0 0
Mf(x)≤λ. This implies that B ∩{Mf ≤λ}6=∅. For every x∈B ∩{Mf >
0 0
λ}, set
r =dist(x,100B \{Mf >λ}),
x 0
so that B(x,r )⊂100B . Remark that the radii r are uniformly bounded by
x 0 x
2r .
0
In the consequence of the Vitali covering theorem there are disjoint balls
{B(x ,r )}∞ such that
i xi i=1
B ∩{Mf >λ}⊂ 5B ,
0 i
i
[
7
where we denote B = B(x ,r ). Both B ⊂ 100B and 5B ⊂ 100B for all
i i i i 0 i 0
i=1,2,..., so they are still balls of (X,d). Furthermore, 5B ∩{Mf ≤λ}=6 ∅
i
for all i=1,2,... so that
fdµ≤λ (3.13)
Z5Bi
for all i =1,2,.... We can now estimate the integral on the left side in (3.12).
A standard estimation shows that
fpdµ≤ fpdµ≤ fpdµ
ZB0∩{Mf>λ} Z∪i5Bi i Z5Bi
X
p
= µ(5B ) fpdµ≤cp µ(5B ) fdµ
i i
i Z5Bi i (cid:18)Z5Bi (cid:19)
X X
≤cpλp µ(5B ),
i
i
X
where the second last inequality follows from the reverse Ho¨lder inequality and
the last from (3.13). Since µ is doubling and the balls B are disjoint we get
i
µ(5B )≤c µ(B )=cµ(∪ B ).
i i i i
i i
X X
By the definition of B ⊂100B ∩{Mf >λ} for all i=1,2,.... Therefore
i 0
fpdµ≤cλpµ(∪ B )≤cλpµ(100B ∩{Mf >λ})
i i 0
ZB0∩{Mf>λ}
for all λ>essinf Mf.
B0
Remark. Note that essinf Mf 6=∞.
B0
Indeed,inthewellknownweaktypeestimateforlocallyintegrablefunctions
c
µ(B ∩{Mf >λ})≤ fdµ,
0
λ
Z100B0
the right–hand side tends to zero when λ → ∞. The constant c depends only
on the doubling constant c . We can thus choose 0<λ <∞ so that
d 0
c 1
fdµ≤ µ(B ).
0
λ 2
0 Z100B0
As a consequence,
µ(B ∩{Mf ≤λ })=µ(B )−µ(B ∩{Mf >λ })
0 0 0 0 0
c 1
≥µ(B )− fdµ≥ µ(B ).
0 0
λ 2
0 Z100B0
8
This leads to essinf Mf ≤λ , for if essinf Mf >λ , then Mf(x)>λ for
B0 0 B0 0 0
almost every x∈B . This impossible since
0
1
µ(B ∩{Mf ≤λ })≥ µ(B ).
0 0 0
2
Forthereader’sconvenience,wepresenthereonetechnicalpartofourproof
as a separate lemma.
Lemma 3.6. Let 1<q <∞ and let f ∈Lq (X) be non–negative. Suppose in
loc
addition that f satisfies the reverse Ho¨lder inequality. Then for every ball B
0
in X and for all 1<p<q
q−p
fqdµ≤cαqµ(100B ∩{Mf >α})+c (Mf)qdµ,
0
q
ZB0∩{Mf>α} Z100B0
(3.14)
where α = essinf Mf and c depends on p, the doubling constant and on the
B0
constant in (3.1).
Proof. Fix a ball B in X. Let α = essinf Mf, so that Mf ≥ α µ–a.e. on
0 B0
B . Set dν =fpdµ. Now
0
fqdµ= fq−pfpdµ≤ (Mf)q−pdν.
ZB0∩{Mf>α} ZB0∩{Mf>α} Z{Mf>α}
However, for every positive measure and measurable non–negative function g
and a measurable set E, we have
∞
gpdν =p λp−1ν({x∈E : g(x)>λ})dλ
ZE Z0
for all 0<p<∞. This implies
∞
fqdµ≤(q−p) λq−p−1ν(B ∩{Mf >α}∩{Mf >λ})dλ
0
ZB0∩{Mf>α} Z0
α
=(q−p) λq−p−1ν(B ∩{Mf >α})dλ
0
Z0
∞
+(q−p) λq−p−1ν(B ∩{Mf >λ})dλ.
0
Zα
Replacing dν =fpdµ and integrating over λ, we get
fqdµ≤ αq−pfpdµ
ZB0∩{Mf>α} ZB0∩{Mf>α}
∞
+(q−p) λq−p−1 fpdµdλ.
Zα ZB0∩{Mf>λ}
9
WecannowuseLemma3.5forbothintegralsontheright–handsideandobtain
fqdµ≤cαqµ(100B ∩{Mf >α})
0
ZB0∩{Mf>α}
∞
+c(q−p) λq−1µ(100B ∩{Mf >λ})dλ.
0
Zα
Then by changing the order of integration we arrive at
fqdµ≤cαqµ(100B ∩{Mf >α})
0
ZB0∩{Mf>α}
∞
+c(q−p) λq−1 dµdλ
Zα Z100B0∩{Mf>λ}
=cαqµ(100B ∩{Mf >α})
0
Mf
+c(q−p) λq−1dλdµ,
Z100B0Zα
from which by integrating over α we conclude that
fqdµ≤cαqµ(100B ∩{Mf >α})
0
ZB0∩{Mf>α}
q−p
+c (Mf)q−αq dµ
q
Z100B0
≤cαqµ(100B ∩{M(cid:0) f >α}) (cid:1)
0
q−p
+c (Mf)qdµ.
q
Z100B0
Proof of the Gehring lemma. Consider a fixed ball B . Set α = essinf Mf
0 B0
and let q > p be an arbitrary real number for the moment. We divide the
integral of fq over B into two parts:
0
fqdµ= fqdµ+ fqdµ. (3.15)
ZB0 ZB0∩{Mf>α} ZB0∩{Mf≤α}
The second integral in (3.15) is easier to estimate, and we have
fqdµ≤ (Mf)qdµ≤αqµ(100B ∩{Mf ≤α}).
0
ZB0∩{Mf≤α} ZB0∩{Mf≤α}
It would be tempting to use Lemma 3.6 to the first integral in (3.15), but this
would require f ∈ Lq (X). Unfortunately, that is exactly what we need to
loc
prove. The function f is assumed to be locally integrable and by the reverse
10
