Given a doubling measure $\mu$ on $R^d$, it is a classical result of harmonic
analysis that Calderon-Zygmund operators which are bounded in $L^2(\mu)$ are
also of weak type (1,1). Recently it has been shown that the same result holds
if one substitutes the doubling condition on $\mu$ by a mild growth condition
on $\mu$. In this paper another proof of this result is given. The proof is
very close in spirit to the classical argument for doubling measures and it is
based on a new Calderon-Zygmund decomposition adapted to the non doubling
situation.Comment: 10 page

Tolsa, Xavier

English

arXiv

Chalmers Publication Library

A proof of the weak (1,1) inequality for singular integrals with non-doubling measures based on a Calderón-Zygmund decomposition

Chalmers Research

A proof of the weak (1,1) inequality for singular integrals with non-doubling measures based on a Calder\uf3n-Zygmund decomposition

Given a doubling measure $\mu$ on ${\mathbb R}^d$, it is a classical result of harmonic analysis that Calderón-Zygmund operators which are bounded in $L^2(\mu)$ are also of weak type (1,1). Recently it has been shown that the same result holds if one substitutes the doubling condition on $\mu$ by a mild growth condition on $\mu$. In this paper another proof of this result is given. The proof is very close in spirit to the classical argument for doubling measures and it is based on a new Calderón-Zygmund decomposition adapted to the non doubling situation

Tolsa, X.

Revistes Catalanes amb Accés Obert

Publ. Mat. 45 (2001), 163–174
A PROOF OF THE WEAK (1,1) INEQUALITY FOR
SINGULAR INTEGRALS WITH NON DOUBLING
MEASURES BASED ON A CALDERO´N-ZYGMUND
DECOMPOSITION
Xavier Tolsa
Abstract
Given a doubling measure µ on Rd, it is a classical result of
harmonic analysis that Caldero´n-Zygmund operators which are
bounded in L2(µ) are also of weak type (1, 1). Recently it has
been shown that the same result holds if one substitutes the dou-
bling condition on µ by a mild growth condition on µ. In this
paper another proof of this result is given. The proof is very close
in spirit to the classical argument for doubling measures and it is
based on a new Caldero´n-Zygmund decomposition adapted to the
non doubling situation.
1. Introduction
Let µ be a positive Radon measure on Rd satisfying the growth con-
dition
µ(B(x, r)) ≤ C0 rn for all x ∈ Rd, r > 0,(1.1)
where n is some ﬁxed number with 0 < n ≤ d. We do not assume that
µ is doubling [µ is said to be doubling if there exists some constant C
such that µ(B(x, 2r)) ≤ C µ(B(x, r)) for all x ∈ supp(µ), r > 0]. Let
us remark that the doubling condition on the underlying measure µ
on Rd is an essential assumption in most results of classical Caldero´n-
Zygmund theory. However, recently it has been shown that a big part
of the classical theory remains valid if the doubling assumption on µ is
2000 Mathematics Subject Classiﬁcation. 42B20.
Key words. Caldero´n-Zygmund operators, non doubling measures, non homogeneous
spaces, weak estimates.
Supported by a postdoctoral grant from the European Comission for the TMR Net-
work “Harmonic Analysis”. Also partially supported by grants DGICYT PB96-1183
and CIRIT 1998-SGR00052 (Spain).
164 X. Tolsa
substituted by the size condition (1.1) (see for example the references
cited at the end of the paper).
In this note we will prove that Caldero´n-Zygmund operators (CZO’s)
which are bounded in L2(µ) are also of weak type (1, 1), as in the usual
doubling situation. This result has already been proved in [To1] in
the particular case of the Cauchy integral operator, and by Nazarov,
Treil and Volberg [NTV2] in the general case. The proof that we will
show here is diﬀerent from the one of [NTV2] (and also from the one
of [To1], of course) and it is closer in spirit to the classical proof of the
corresponding result for doubling measures. The basic tool for the proof
is a decomposition of Caldero´n-Zygmund type for functions in L1(µ)
obtained in [To4].
Our purpose in writing this paper is not only to obtain another proof
in the non doubling situation of the basic result that CZO’s bounded in
L2(µ) are of weak type (1, 1), but to show that the Caldero´n-Zygmund
decompositon of [To4] is a good substitute of its classical doubling ver-
sion.
Let us introduce some notation and deﬁnitions. A kernel k(·, ·) from
L1loc((R
d×Rd) \ {(x, y) : x = y}) is called a Caldero´n-Zygmund kernel if
1. |k(x, y)| ≤ C|x− y|n ,
2. there exists 0 < δ ≤ 1 such that
|k(x, y)− k(x′, y)|+ |k(y, x)− k(y, x′)| ≤ C |x− x
′|δ
|x− y|n+δ
if |x− x′| ≤ |x− y|/2.
Throughout all the paper we will assume that µ is a Radon measure on
R
d satisfying (1.1). The CZO associated to the kernel k(·, ·) and the
measure µ is deﬁned (at least, formally) as
Tf(x) =
∫
k(x, y) f(y) dµ(y).
The above integral may be not convergent for many functions f because
k(x, y) may have a singularity for x = y. For this reason, one introduces
the truncated operators Tε, ε > 0:
Tεf(x) =
∫
|x−y|>ε
k(x, y) f(y) dµ(y),
and then one says that T is bounded in Lp(µ) if the operators Tε are
bounded in Lp(µ) uniformly on ε > 0. It is said that T is bounded from
Weak (1,1) Inequality with Non Doubling Measures 165
L1(µ) into L1,∞(µ) (or of weak type (1, 1)) if
µ{x : |Tεf(x)| > λ} ≤ C
‖f‖L1(µ)
λ
for all f ∈ L1(µ), uniformly on ε > 0. Also, T is bounded from M(C)
(the space of complex Radon measures) into L1,∞(µ) if
µ{x : |Tεν(x)| > λ} ≤ C ‖ν‖
λ
for all ν ∈ M(C), uniformly on ε > 0. In the last inequality, Tεν(x)
stands for
∫
|x−y|>ε k(x, y) dν(y) and ‖ν‖ ≡ |ν|(Rd).
The result that we will prove in this note is the following.
Theorem 1.1. Let µ be a Radon measure on Rd satisfying the growth
condition (1.1). If T is a Caldero´n-Zygmund operator which is bounded
in L2(µ), then it is also bounded from M(C) into L1,∞(µ). In particular,
it is of weak type (1, 1).
2. The proof
First we will introduce some additional notation and terminology. As
usual, the letter C will denote a constant which may change its value
from one occurrence to another. Constants with subscripts, such as C0,
do not change in diﬀerent occurrences.
By a cube Q ⊂ Rd we mean a closed cube with sides parallel to the
axes. We denote its side length by (Q) and its center by xQ. Given
α > 1 and β > αn, we say that Q is (α, β)-doubling if µ(αQ) ≤ β µ(Q),
where αQ is the cube concentric with Q with side length α (Q). For
deﬁniteness, if α and β are not speciﬁed, by a doubling cube we mean a
(2, 2d+1)-doubling cube.
Before proving Theorem 1.1 we state some remarks about the exis-
tence of doubling cubes.
Remark 2.1. Because µ satisﬁes the growth condition (1.1), there are a
lot of “big” doubling cubes. To be precise, given any point x ∈ supp(µ)
and c > 0, there exists some (α, β)-doubling cube Q centered at x with
l(Q) ≥ c. This follows easily from (1.1) and the fact that β > αn.
Indeed, if there are no doubling cubes centered at x with l(Q) ≥ c, then
µ(αmQ) > βmµ(Q) for each m, and letting m → ∞ one sees that (1.1)
cannot hold.
Remark 2.2. There are a lot of “small” doubling cubes too: if β >
αd, then for µ-a.e. x ∈ Rd there exists a sequence of (α, β)-doubling
166 X. Tolsa
cubes {Qk}k centered at x with (Qk) → 0 as k → ∞. This is a prop-
erty that any Radon measure on Rd satisﬁes (the growth condition (1.1)
is not necessary in this argument). The proof is an easy exercise on
geometric measure theory that is left for the reader.
Observe that, by the Lebesgue diﬀerentiation theorem, for µ-almost
all x ∈ Rd one can ﬁnd a sequence of (2, 2d+1)-doubling cubes {Qk}k
centered at x with (Qk) → 0 such that
lim
k→∞
1
µ(Qk)
∫
Qk
f dµ = f(x).
As a consequence, for any ﬁxed λ > 0, for µ-almost all x ∈ Rd such that
|f(x)| > λ, there exists a sequence of cubes {Qk}k centered at x with
(Qk) → 0 such that
lim sup
k→∞
1
µ(2Qk)
∫
Qk
|f | dµ > λ
2d+1
.
In the following lemma we will prove an easy but essential estimate
which will be used below. This result has already appeared in previous
works ([DM], [NTV2]) and it plays a basic role in [To2] and [To4] too.
Lemma 2.3. If Q ⊂ R are concentric cubes such that there are no
(α, β)-doubling cubes (with β > αn) of the form αkQ, k ≥ 0, with Q ⊂
αkQ ⊂ R, then, ∫
R\Q
1
|x− xQ|n dµ(x) ≤ C1,
where C1 depends only on α, β, n, d and C0.
Proof: Let N be the least integer such that R ⊂ αNQ. For 0 ≤ k ≤ N
we have µ(αkQ) ≤ µ(αNQ)/βN−k. Then,
∫
R\Q
1
|x− xQ|n dµ(x) ≤
N∑
k=1
∫
αkQ\αk−1Q
1
|x− xQ|n dµ(x)
≤ C
N∑
k=1
µ(αkQ)
(αkQ)n
≤ C
N∑
k=1
βk−N µ(αNQ)
α(k−N)n (αNQ)n
≤ C µ(α
NQ)
(αNQ)n
∞∑
j=0
(
αn
β
)j
≤ C.
Weak (1,1) Inequality with Non Doubling Measures 167
The Caldero´n-Zygmund decomposition mentioned above has been ob-
tained in Lemma 7.3 of [To4] and in that paper it has been used to show
that if a linear operator is bounded from a suitable space of type H1 into
L1(µ) and from L∞(µ) into a space of type BMO , then it is bounded
in Lp(µ). We will use a slight variant of this decompositon to prove
Theorem 1.1. Let us state the result that we need in detail.
Lemma 2.4 (Caldero´n-Zygmund decomposition). Assume that µ satis-
ﬁes (1.1). For any f ∈ L1(µ) and any λ > 0 (with λ > 2d+1‖f‖L1(µ)/‖µ‖
if ‖µ‖ <∞) we have:
(a) There exists a family of almost disjoint cubes {Qi}i (that is,∑
i χQi ≤ C) such that
1
µ(2Qi)
∫
Qi
|f | dµ > λ
2d+1
,(2.1)
1
µ(2ηQi)
∫
ηQi
|f | dµ ≤ λ
2d+1
for η > 2,(2.2)
|f | ≤ λ a.e. (µ) on Rd \⋃iQi.(2.3)
(b) For each i, let Ri be a (6, 6n+1)-doubling cube concentric with Qi,
with l(Ri) > 4l(Qi) and denote wi =
χQi∑
k
χQk
. Then, there exists a
family of functions ϕi with supp(ϕi) ⊂ Ri and with constant sign
satisfying ∫
ϕi dµ =
∫
Qi
f wi dµ,(2.4)
∑
i
|ϕi| ≤ B λ(2.5)
(where B is some constant), and
‖ϕi‖L∞(µ)µ(Ri) ≤ C
∫
Qi
|f | dµ.(2.6)
Let us remark that other related decompositons with non doubling
measures have been obtained in [NTV2] and [MMNO]. However, these
results are not suitable for our purposes.
Although the proof of the lemma can be found in [To4], for the
reader’s convenience we have included it in the last section of the present
paper.
168 X. Tolsa
Proof of Theorem 1.1: We will show that T is of weak type (1, 1). By
similar arguments, one gets that T is bounded from M(C) into L1,∞(µ).
In this case, one has to use a version of the Caldero´n-Zygmund decom-
position in the lemma above suitable for complex measures (see the end
of the proof for more details).
Let f ∈ L1(µ) and λ > 0. It is straightforward to check that we may
assume λ > 2d+1‖f‖L1(µ)/‖µ‖. Let {Qi}i be the almost disjoint family
of cubes of Lemma 2.4. Let Ri be the smallest (6, 6n+1)-doubling cube
of the form 6kQi, k ≥ 1. Then we can write f = g + b, with
g = f χ
Rd\
⋃
i
Qi
+
∑
i
ϕi
and
b =
∑
i
bi :=
∑
i
(wi f − ϕi) ,
where the functions ϕi satisfy (2.4), (2.5), (2.6) and wi =
χQi∑
k
χQk
.
By (2.1) we have
µ
(⋃
i
2Qi
)
≤ C
λ
∑
i
∫
Qi
|f | dµ ≤ C
λ
∫
|f | dµ.
So we have to show that
µ
{
x ∈ Rd \
⋃
i
2Qi : |Tεf(x)| > λ
}
≤ C
λ
∫
|f | dµ.(2.7)
Since
∫
bi dµ = 0, supp(bi) ⊂ Ri and ‖bi‖L1(µ) ≤ C
∫
Qi
|f | dµ, using con-
dition 2 in the deﬁnition of a Caldero´n-Zygmund kernel (which implies
Ho¨rmander’s condition), we get∫
Rd\2Ri
|Tεbi| dµ ≤ C
∫
|bi| dµ ≤ C
∫
Qi
|f | dµ.
Let us see that ∫
2Ri\2Qi
|Tεbi| dµ ≤ C
∫
Qi
|f | dµ(2.8)
Weak (1,1) Inequality with Non Doubling Measures 169
too. On the one hand, by (2.6) and using the L2(µ) boundedness of T
and that Ri is (6, 6n+1)-doubling we get
∫
2Ri
|Tεϕi| dµ ≤
(∫
2Ri
|Tεϕi|2 dµ
)1/2
µ(2Ri)1/2
≤ C
(∫
|ϕi|2 dµ
)1/2
µ(Ri)1/2
≤ C
∫
Qi
|f | dµ.
On the other hand, since supp(wif) ⊂ Qi, if x ∈ 2Ri \ 2Qi, then
|Tε(wi f)(x)| ≤ C
∫
Qi
|f | dµ/|x− xQi |n, and so∫
2Ri\2Qi
|Tε(wi f)| dµ ≤ C
∫
2Ri\2Qi
1
|x− xQi |n
dµ(x)×
∫
Qi
|f | dµ.
By Lemma 2.3, the ﬁrst integral on the right hand side is bounded by
some constant independent ofQi andRi, since there are no (6, 6n+1)-dou-
bling cubes of the form 6kQi between 6Qi and Ri. Therefore, (2.8) holds.
Then we have∫
Rd\
⋃
k
2Qk
|Tεb| dµ ≤
∑
i
∫
Rd\
⋃
k
2Qk
|Tεbi| dµ
≤ C
∑
i
∫
Qi
|f | dµ ≤ C
∫
|f | dµ.
Therefore,
µ
{
x ∈ Rd \
⋃
i
2Qi : |Tεb(x)| > λ
}
≤ C
λ
∫
|f | dµ.(2.9)
The corresponding integral for the function g is easier to estimate. Tak-
ing into account that |g| ≤ C λ, we get
µ
{
x ∈ Rd \
⋃
i
2Qi : |Tεg(x)| > λ
}
≤ C
λ2
∫
|g|2 dµ
≤ C
λ
∫
|g| dµ.
(2.10)
170 X. Tolsa
Also, we have∫
|g| dµ ≤
∫
Rd\
⋃
i
Qi
|f | dµ+
∑
i
∫
|ϕi| dµ
≤
∫
|f | dµ+
∑
i
∫
Qi
|f | dµ ≤ C
∫
|f | dµ.
Now, by (2.9) and (2.10) we get (2.7).
If we want to show that T is bounded from M(C) into L1,∞(µ), then
in Lemma 2.4 and in the arguments above f dµ must be substituted
by dν, with ν ∈ M(C), and |f | dµ by d|ν|. Also, condition (2.3) of
Lemma 2.4 should be stated as “On Rd\⋃iQi, ν is absolutely continuous
with respect to µ, that is ν = f dν, and moreover |f(x)| ≤ λ a.e. (µ)
x ∈ Rd\⋃iQi”. With other minor changes, the arguments and estimates
above work in this situation too.
3. Proof of Lemma 2.4
(a) Taking into account Remark 2.2, for µ-almost all x ∈ Rd such
that |f(x)| > λ, there exists some cube Qx satisfying
1
µ(2Qx)
∫
Qx
|f | dµ > λ
2d+1
(3.1)
and such that if Q′x is centered at x with l(Q
′
x) > 2l(Qx), then
1
µ(2Q′x)
∫
Q′x
|f | dµ ≤ λ
2d+1
.
Now we can apply Besicovich’s covering theorem (see Remark 3.1 below)
to get an almost disjoint subfamily of cubes {Qi}i ⊂ {Qx}x satisfying
(2.1), (2.2) and (2.3).
(b) Assume ﬁrst that the family of cubes {Qi}i is ﬁnite. Then we
may suppose that this family of cubes is ordered in such a way that
the sizes of the cubes Ri are non decreasing (i.e. l(Ri+1) ≥ l(Ri)). The
functions ϕi that we will construct will be of the form ϕi = αi χAi , with
αi ∈ R and Ai ⊂ Ri. We set A1 = R1 and ϕ1 = α1 χR1 , where the
constant α1 is chosen so that
∫
Q1
f w1 dµ =
∫
ϕ1 dµ.
Suppose that ϕ1, . . . , ϕk−1 have been constructed, satisfy (2.4) and
k−1∑
i=1
|ϕi| ≤ B λ,
where B is some constant which will be ﬁxed below.
Weak (1,1) Inequality with Non Doubling Measures 171
Let Rs1 , . . . , Rsm be the subfamily of R1, . . . , Rk−1 such that Rsj ∩
Rk = ∅. As l(Rsj ) ≤ l(Rk) (because of the non decreasing sizes of Ri),
we have Rsj ⊂ 3Rk. Taking into account that for i = 1, . . . , k − 1∫
|ϕi| dµ ≤
∫
Qi
|f | dµ
by (2.4), and using that Rk is (6, 6n+1)-doubling and (2.2), we get
∑
j
∫
|ϕsj | dµ ≤
∑
j
∫
Qsj
|f | dµ
≤ C
∫
3Rk
|f | dµ ≤ Cλµ(6Rk) ≤ C2λµ(Rk).
Therefore,
µ
{∑
j |ϕsj | > 2C2λ
}
≤ µ(Rk)
2
.
So we set
Ak = Rk ∩
{∑
j |ϕsj | ≤ 2C2λ
}
,
and then µ(Ak) ≥ µ(Rk)/2.
The constant αk is chosen so that for ϕk = αk χAk we have
∫
ϕk dµ =∫
Qk
f wk dµ. Then we obtain
|αk| ≤ 1
µ(Ak)
∫
Qk
|f | dµ ≤ 2
µ(Rk)
∫
1
2Rk
|f | dµ ≤ C3λ
(this calculation also applies to k = 1). Thus,
|ϕk|+
∑
j
|ϕsj | ≤ (2C2 + C3)λ.
If we choose B = 2C2 + C3, (2.5) follows.
Now it is easy to check that (2.6) also holds. Indeed we have
‖ϕi‖L∞(µ) µ(Ri) ≤ C |αi|µ(Ai) = C
∣∣∣∣
∫
Qi
f wi dµ
∣∣∣∣ ≤ C
∫
Qi
|f | dµ.
172 X. Tolsa
Suppose now that the collection of cubes {Qi}i is not ﬁnite. For each
ﬁxed N we consider the family of cubes {Qi}1≤i≤N . Then, as above, we
construct functions ϕN1 , . . . , ϕ
N
N with supp(ϕ
N
i ) ⊂ Ri satisfying∫
ϕNi dµ =
∫
Qi
f wi dµ,
N∑
i=1
|ϕNi | ≤ B λ
and
‖ϕNi ‖L∞(µ) µ(Ri) ≤ C
∫
Qi
|f | dµ.
Notice that the sign of ϕNi equals the sign of
∫
f wi dµ and so it does
not depend on N .
Then there is a subsequence {ϕk1}k∈I1 which is convergent in the
weak ∗ topology of L∞(µ) to some function ϕ1 ∈ L∞(µ). Now we can
consider a subsequence {ϕk2}k∈I2 with I2 ⊂ I1 which is also convergent in
the weak ∗ topology of L∞(µ) to some function ϕ2 ∈ L∞(µ). In general,
for each j we consider a subsequence {ϕkj }k∈Ij with Ij ⊂ Ij−1 that con-
verges in the weak ∗ topology of L∞(µ) to some function ϕj ∈ L∞(µ). It
is easily checked that the functions ϕj satisfy the required properties.
Remark 3.1. Recall that Besicovich’s covering theorem asserts that if
Ω ⊂ Rd is a bounded set and for each x ∈ Ω there is a cube Qx centered
at x, then there exists a family of cubes {Qxi}i with ﬁnite overlap, that
is
∑
i χQi ≤ C, which covers Ω.
In (a) of the preceeding proof we have applied Besicovich’s covering
theorem to Ω = {x : |f(x)| > λ}. However this set may be unbounded,
and the boundedness property is a necessary assumption in Besicovich’s
theorem (example: take Ω = [0,+∞) ⊂ R and consider Qx = [0, 2x] for
all x ∈ Ω).
We can solve this problem using diﬀerent arguments. One possibility
is to consider for each r > 0 the set Ωr = {x : |x| ≤ r, |f(x)| > λ} and
to apply Besicovich’s covering theorem to Ωr. With the same arguments
as above, we can decompose f = g+ b, with |g| ≤ λ only on Ωr and b as
above. Then the proof of Theorem 1.1 can be modiﬁed to show that for
any ﬁxed constants λ,R > 0 one has
µ{x ∈ B(0, R) : |Tεf(x)| > λ} ≤ C
‖f‖L1(µ)
λ
.
Weak (1,1) Inequality with Non Doubling Measures 173
However we prefer the following solution. We are interested in showing
that the Caldero´n-Zygmund decomposition of Lemma 2.4 works also
without assuming Ω = {x : |f(x)| > λ} bounded. Let us sketch the
argument. Consider a cube Q0 centered at 0 big enough so that
2d+1 ‖f‖L1(µ)/µ(Q0) < λ.
So for any cube Q containing Q0 we will have
2d+1 ‖f‖L1(µ)/µ(Q) < λ.(3.2)
For m ≥ 0 we set Qm :=
(
5
4
)m
Q0. For each m we can apply Besicovich’s
covering theorem to the annulus Qm \Qm−1 (we take Q−1 := ∅), with
cubes Qx centered at x ∈ supp(µ) ∩ (Qm \Qm−1) as in (a) of the proof
above, satisfying (3.1).
In this argument we have to be careful with the overlapping among
the cubes belonging to coverings of diﬀerent annuli. Indeed, there exist
some ﬁxed constants N and N ′ such that if m ≥ N ′, for x ∈ supp(µ) ∩
(Qm \Qm−1) we have
Qx ⊂ Qm+N \Qm−N .(3.3)
Otherwise, it is easily seen that (Qx) > 34(Qm), choosing N big enough.
It follows that Q0 ⊂ 2Qx since (Q0)  (Qm) for N ′ big enough too.
This cannot happen because then 2Qx satisﬁes (3.2), which contradicts
(3.1).
Because of (3.3), the covering made up of squares belonging to the
Besicovich coverings of diﬀerent annuli Qm \ Qm−1, m ≥ 0, will have
ﬁnite overlap.
Notice that in this argument, it is essential the fact that in (3.1) we
are not dividing by µ(Qx), but by µ(2Qx).
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[NTV3] F. Nazarov, S. Treil and A. Volberg, Tb-theorem on
non-homogeneous spaces, Preprint (1999).
[NTV4] F. Nazarov, S. Treil and A. Volberg, Accretive Tb-sys-
tems on non-homogeneous spaces, Preprint (1999).
[OP] J. Orobitg and C. Pe´rez, Ap weights for non doubling
measures in Rn and applications, Trans. Amer. Math. Soc.
(to appear).
[To1] X. Tolsa, L2-boundedness of the Cauchy integral opera-
tor for continuous measures, Duke Math. J. 98(2) (1999),
269–304.
[To2] X. Tolsa, Cotlar’s inequality without the doubling condition
and existence of principal values for the Cauchy integral of
measures, J. Reine Angew. Math. 502 (1998), 199–235.
[To3] X. Tolsa, A T (1) theorem for non doubling measures with
atoms, Proc. London Math. Soc. 82 (2001), 195–228.
[To4] X. Tolsa, BMO , H1 and Caldero´n-zygmund operators for
non doubling measures, Math. Ann. (to appear).
[Ve] J. Verdera, On the T (1)-theorem for the Cauchy integral,
Ark. Mat. 38 (2000), 183–199.
Department of Mathematics
Chalmers
412 96 Go¨teborg
Sweden
Current address:
De´partement de Mathe´matique
Universite´ de Paris-Sud
Baˆt. 425
91405 Orsay Cedex
France
E-mail address: xavier.tolsa@math.u-psud.fr
Primera versio´ rebuda el 2 de maig de 2000,
darrera versio´ rebuda el 17 de novembre de 2000.


A proof of the weak (1,1) inequality for singular integrals with non doubling measures based on a Calderón-Zygmund decomposition

Given a doubling measure µ on Rd, it is a classical result of harmonic analysis that Calderón-Zygmund operators which are bounded in L2(µ) are also of weak type (1, 1). Recently it has been shown that the same result holds if one substitutes the doubling condition on µ by a mild growth condition on µ. In this paper another proof of this result is given. The proof is very close in spirit to the classical argument for doubling measures and it is based on a new Calderón-Zygmund decomposition adapted to the non doubling situation

Tolsa Domènech, Xavier

Diposit Digital de Documents de la UAB

Publ. Mat. 45 (2001), 163–174A PROOF OF THE WEAK (1,1) INEQUALITY FORSINGULAR INTEGRALS WITH NON DOUBLINGMEASURES BASED ON A CALDERO´N-ZYGMUNDDECOMPOSITIONXavier TolsaAbstractGiven a doubling measure µ on Rd, it is a classical result ofharmonic analysis that Caldero´n-Zygmund operators which arebounded in L2(µ) are also of weak type (1, 1). Recently it hasbeen shown that the same result holds if one substitutes the dou-bling condition on µ by a mild growth condition on µ. In thispaper another proof of this result is given. The proof is very closein spirit to the classical argument for doubling measures and it isbased on a new Caldero´n-Zygmund decomposition adapted to thenon doubling situation.1. IntroductionLet µ be a positive Radon measure on Rd satisfying the growth con-ditionµ(B(x, r)) ≤ C0 rn for all x ∈ Rd, r > 0,(1.1)where n is some fixed number with 0 < n ≤ d. We do not assume thatµ is doubling [µ is said to be doubling if there exists some constant Csuch that µ(B(x, 2r)) ≤ C µ(B(x, r)) for all x ∈ supp(µ), r > 0]. Letus remark that the doubling condition on the underlying measure µon Rd is an essential assumption in most results of classical Caldero´n-Zygmund theory. However, recently it has been shown that a big partof the classical theory remains valid if the doubling assumption on µ is2000 Mathematics Subject Classification. 42B20.Key words. Caldero´n-Zygmund operators, non doubling measures, non homogeneousspaces, weak estimates.Supported by a postdoctoral grant from the European Comission for the TMR Net-work “Harmonic Analysis”. Also partially supported by grants DGICYT PB96-1183and CIRIT 1998-SGR00052 (Spain).164 X. Tolsasubstituted by the size condition (1.1) (see for example the referencescited at the end of the paper).In this note we will prove that Caldero´n-Zygmund operators (CZO’s)which are bounded in L2(µ) are also of weak type (1, 1), as in the usualdoubling situation. This result has already been proved in [To1] inthe particular case of the Cauchy integral operator, and by Nazarov,Treil and Volberg [NTV2] in the general case. The proof that we willshow here is different from the one of [NTV2] (and also from the oneof [To1], of course) and it is closer in spirit to the classical proof of thecorresponding result for doubling measures. The basic tool for the proofis a decomposition of Caldero´n-Zygmund type for functions in L1(µ)obtained in [To4].Our purpose in writing this paper is not only to obtain another proofin the non doubling situation of the basic result that CZO’s bounded inL2(µ) are of weak type (1, 1), but to show that the Caldero´n-Zygmunddecompositon of [To4] is a good substitute of its classical doubling ver-sion.Let us introduce some notation and definitions. A kernel k(·, ·) fromL1loc((Rd×Rd) \ {(x, y) : x = y}) is called a Caldero´n-Zygmund kernel if1. |k(x, y)| ≤ C|x− y|n ,2. there exists 0 < δ ≤ 1 such that|k(x, y)− k(x′, y)|+ |k(y, x)− k(y, x′)| ≤ C |x− x′|δ|x− y|n+δif |x− x′| ≤ |x− y|/2.Throughout all the paper we will assume that µ is a Radon measure onRd satisfying (1.1). The CZO associated to the kernel k(·, ·) and themeasure µ is defined (at least, formally) asTf(x) =∫k(x, y) f(y) dµ(y).The above integral may be not convergent for many functions f becausek(x, y) may have a singularity for x = y. For this reason, one introducesthe truncated operators Tε, ε > 0:Tεf(x) =∫|x−y|>εk(x, y) f(y) dµ(y),and then one says that T is bounded in Lp(µ) if the operators Tε arebounded in Lp(µ) uniformly on ε > 0. It is said that T is bounded fromWeak (1,1) Inequality with Non Doubling Measures 165L1(µ) into L1,∞(µ) (or of weak type (1, 1)) ifµ{x : |Tεf(x)| > λ} ≤ C‖f‖L1(µ)λfor all f ∈ L1(µ), uniformly on ε > 0. Also, T is bounded from M(C)(the space of complex Radon measures) into L1,∞(µ) ifµ{x : |Tεν(x)| > λ} ≤ C ‖ν‖λfor all ν ∈ M(C), uniformly on ε > 0. In the last inequality, Tεν(x)stands for∫|x−y|>ε k(x, y) dν(y) and ‖ν‖ ≡ |ν|(Rd).The result that we will prove in this note is the following.Theorem 1.1. Let µ be a Radon measure on Rd satisfying the growthcondition (1.1). If T is a Caldero´n-Zygmund operator which is boundedin L2(µ), then it is also bounded from M(C) into L1,∞(µ). In particular,it is of weak type (1, 1).2. The proofFirst we will introduce some additional notation and terminology. Asusual, the letter C will denote a constant which may change its valuefrom one occurrence to another. Constants with subscripts, such as C0,do not change in different occurrences.By a cube Q ⊂ Rd we mean a closed cube with sides parallel to theaxes. We denote its side length by (Q) and its center by xQ. Givenα > 1 and β > αn, we say that Q is (α, β)-doubling if µ(αQ) ≤ β µ(Q),where αQ is the cube concentric with Q with side length α (Q). Fordefiniteness, if α and β are not specified, by a doubling cube we mean a(2, 2d+1)-doubling cube.Before proving Theorem 1.1 we state some remarks about the exis-tence of doubling cubes.Remark 2.1. Because µ satisfies the growth condition (1.1), there are alot of “big” doubling cubes. To be precise, given any point x ∈ supp(µ)and c > 0, there exists some (α, β)-doubling cube Q centered at x withl(Q) ≥ c. This follows easily from (1.1) and the fact that β > αn.Indeed, if there are no doubling cubes centered at x with l(Q) ≥ c, thenµ(αmQ) > βmµ(Q) for each m, and letting m → ∞ one sees that (1.1)cannot hold.Remark 2.2. There are a lot of “small” doubling cubes too: if β >αd, then for µ-a.e. x ∈ Rd there exists a sequence of (α, β)-doubling166 X. Tolsacubes {Qk}k centered at x with (Qk) → 0 as k → ∞. This is a prop-erty that any Radon measure on Rd satisfies (the growth condition (1.1)is not necessary in this argument). The proof is an easy exercise ongeometric measure theory that is left for the reader.Observe that, by the Lebesgue differentiation theorem, for µ-almostall x ∈ Rd one can find a sequence of (2, 2d+1)-doubling cubes {Qk}kcentered at x with (Qk) → 0 such thatlimk→∞1µ(Qk)∫Qkf dµ = f(x).As a consequence, for any fixed λ > 0, for µ-almost all x ∈ Rd such that|f(x)| > λ, there exists a sequence of cubes {Qk}k centered at x with(Qk) → 0 such thatlim supk→∞1µ(2Qk)∫Qk|f | dµ > λ2d+1.In the following lemma we will prove an easy but essential estimatewhich will be used below. This result has already appeared in previousworks ([DM], [NTV2]) and it plays a basic role in [To2] and [To4] too.Lemma 2.3. If Q ⊂ R are concentric cubes such that there are no(α, β)-doubling cubes (with β > αn) of the form αkQ, k ≥ 0, with Q ⊂αkQ ⊂ R, then, ∫R\Q1|x− xQ|n dµ(x) ≤ C1,where C1 depends only on α, β, n, d and C0.Proof: Let N be the least integer such that R ⊂ αNQ. For 0 ≤ k ≤ Nwe have µ(αkQ) ≤ µ(αNQ)/βN−k. Then,∫R\Q1|x− xQ|n dµ(x) ≤N∑k=1∫αkQ\αk−1Q1|x− xQ|n dµ(x)≤ CN∑k=1µ(αkQ)(αkQ)n≤ CN∑k=1βk−N µ(αNQ)α(k−N)n (αNQ)n≤ C µ(αNQ)(αNQ)n∞∑j=0(αnβ)j≤ C.Weak (1,1) Inequality with Non Doubling Measures 167The Caldero´n-Zygmund decomposition mentioned above has been ob-tained in Lemma 7.3 of [To4] and in that paper it has been used to showthat if a linear operator is bounded from a suitable space of type H1 intoL1(µ) and from L∞(µ) into a space of type BMO , then it is boundedin Lp(µ). We will use a slight variant of this decompositon to proveTheorem 1.1. Let us state the result that we need in detail.Lemma 2.4 (Caldero´n-Zygmund decomposition). Assume that µ satis-fies (1.1). For any f ∈ L1(µ) and any λ > 0 (with λ > 2d+1‖f‖L1(µ)/‖µ‖if ‖µ‖ <∞) we have:(a) There exists a family of almost disjoint cubes {Qi}i (that is,∑i χQi ≤ C) such that1µ(2Qi)∫Qi|f | dµ > λ2d+1,(2.1)1µ(2ηQi)∫ηQi|f | dµ ≤ λ2d+1for η > 2,(2.2)|f | ≤ λ a.e. (µ) on Rd \⋃iQi.(2.3)(b) For each i, let Ri be a (6, 6n+1)-doubling cube concentric with Qi,with l(Ri) > 4l(Qi) and denote wi =χQi∑kχQk. Then, there exists afamily of functions ϕi with supp(ϕi) ⊂ Ri and with constant signsatisfying ∫ϕi dµ =∫Qif wi dµ,(2.4)∑i|ϕi| ≤ B λ(2.5)(where B is some constant), and‖ϕi‖L∞(µ)µ(Ri) ≤ C∫Qi|f | dµ.(2.6)Let us remark that other related decompositons with non doublingmeasures have been obtained in [NTV2] and [MMNO]. However, theseresults are not suitable for our purposes.Although the proof of the lemma can be found in [To4], for thereader’s convenience we have included it in the last section of the presentpaper.168 X. TolsaProof of Theorem 1.1: We will show that T is of weak type (1, 1). Bysimilar arguments, one gets that T is bounded from M(C) into L1,∞(µ).In this case, one has to use a version of the Caldero´n-Zygmund decom-position in the lemma above suitable for complex measures (see the endof the proof for more details).Let f ∈ L1(µ) and λ > 0. It is straightforward to check that we mayassume λ > 2d+1‖f‖L1(µ)/‖µ‖. Let {Qi}i be the almost disjoint familyof cubes of Lemma 2.4. Let Ri be the smallest (6, 6n+1)-doubling cubeof the form 6kQi, k ≥ 1. Then we can write f = g + b, withg = f χRd\⋃iQi+∑iϕiandb =∑ibi :=∑i(wi f − ϕi) ,where the functions ϕi satisfy (2.4), (2.5), (2.6) and wi =χQi∑kχQk.By (2.1) we haveµ(⋃i2Qi)≤ Cλ∑i∫Qi|f | dµ ≤ Cλ∫|f | dµ.So we have to show thatµ{x ∈ Rd \⋃i2Qi : |Tεf(x)| > λ}≤ Cλ∫|f | dµ.(2.7)Since∫bi dµ = 0, supp(bi) ⊂ Ri and ‖bi‖L1(µ) ≤ C∫Qi|f | dµ, using con-dition 2 in the definition of a Caldero´n-Zygmund kernel (which impliesHo¨rmander’s condition), we get∫Rd\2Ri|Tεbi| dµ ≤ C∫|bi| dµ ≤ C∫Qi|f | dµ.Let us see that ∫2Ri\2Qi|Tεbi| dµ ≤ C∫Qi|f | dµ(2.8)Weak (1,1) Inequality with Non Doubling Measures 169too. On the one hand, by (2.6) and using the L2(µ) boundedness of Tand that Ri is (6, 6n+1)-doubling we get∫2Ri|Tεϕi| dµ ≤(∫2Ri|Tεϕi|2 dµ)1/2µ(2Ri)1/2≤ C(∫|ϕi|2 dµ)1/2µ(Ri)1/2≤ C∫Qi|f | dµ.On the other hand, since supp(wif) ⊂ Qi, if x ∈ 2Ri \ 2Qi, then|Tε(wi f)(x)| ≤ C∫Qi|f | dµ/|x− xQi |n, and so∫2Ri\2Qi|Tε(wi f)| dµ ≤ C∫2Ri\2Qi1|x− xQi |ndµ(x)×∫Qi|f | dµ.By Lemma 2.3, the first integral on the right hand side is bounded bysome constant independent ofQi andRi, since there are no (6, 6n+1)-dou-bling cubes of the form 6kQi between 6Qi and Ri. Therefore, (2.8) holds.Then we have∫Rd\⋃k2Qk|Tεb| dµ ≤∑i∫Rd\⋃k2Qk|Tεbi| dµ≤ C∑i∫Qi|f | dµ ≤ C∫|f | dµ.Therefore,µ{x ∈ Rd \⋃i2Qi : |Tεb(x)| > λ}≤ Cλ∫|f | dµ.(2.9)The corresponding integral for the function g is easier to estimate. Tak-ing into account that |g| ≤ C λ, we getµ{x ∈ Rd \⋃i2Qi : |Tεg(x)| > λ}≤ Cλ2∫|g|2 dµ≤ Cλ∫|g| dµ.(2.10)170 X. TolsaAlso, we have∫|g| dµ ≤∫Rd\⋃iQi|f | dµ+∑i∫|ϕi| dµ≤∫|f | dµ+∑i∫Qi|f | dµ ≤ C∫|f | dµ.Now, by (2.9) and (2.10) we get (2.7).If we want to show that T is bounded from M(C) into L1,∞(µ), thenin Lemma 2.4 and in the arguments above f dµ must be substitutedby dν, with ν ∈ M(C), and |f | dµ by d|ν|. Also, condition (2.3) ofLemma 2.4 should be stated as “On Rd\⋃iQi, ν is absolutely continuouswith respect to µ, that is ν = f dν, and moreover |f(x)| ≤ λ a.e. (µ)x ∈ Rd\⋃iQi”. With other minor changes, the arguments and estimatesabove work in this situation too.3. Proof of Lemma 2.4(a) Taking into account Remark 2.2, for µ-almost all x ∈ Rd suchthat |f(x)| > λ, there exists some cube Qx satisfying1µ(2Qx)∫Qx|f | dµ > λ2d+1(3.1)and such that if Q′x is centered at x with l(Q′x) > 2l(Qx), then1µ(2Q′x)∫Q′x|f | dµ ≤ λ2d+1.Now we can apply Besicovich’s covering theorem (see Remark 3.1 below)to get an almost disjoint subfamily of cubes {Qi}i ⊂ {Qx}x satisfying(2.1), (2.2) and (2.3).(b) Assume first that the family of cubes {Qi}i is finite. Then wemay suppose that this family of cubes is ordered in such a way thatthe sizes of the cubes Ri are non decreasing (i.e. l(Ri+1) ≥ l(Ri)). Thefunctions ϕi that we will construct will be of the form ϕi = αi χAi , withαi ∈ R and Ai ⊂ Ri. We set A1 = R1 and ϕ1 = α1 χR1 , where theconstant α1 is chosen so that∫Q1f w1 dµ =∫ϕ1 dµ.Suppose that ϕ1, . . . , ϕk−1 have been constructed, satisfy (2.4) andk−1∑i=1|ϕi| ≤ B λ,where B is some constant which will be fixed below.Weak (1,1) Inequality with Non Doubling Measures 171Let Rs1 , . . . , Rsm be the subfamily of R1, . . . , Rk−1 such that Rsj ∩Rk = ∅. As l(Rsj ) ≤ l(Rk) (because of the non decreasing sizes of Ri),we have Rsj ⊂ 3Rk. Taking into account that for i = 1, . . . , k − 1∫|ϕi| dµ ≤∫Qi|f | dµby (2.4), and using that Rk is (6, 6n+1)-doubling and (2.2), we get∑j∫|ϕsj | dµ ≤∑j∫Qsj|f | dµ≤ C∫3Rk|f | dµ ≤ Cλµ(6Rk) ≤ C2λµ(Rk).Therefore,µ{∑j |ϕsj | > 2C2λ}≤ µ(Rk)2.So we setAk = Rk ∩{∑j |ϕsj | ≤ 2C2λ},and then µ(Ak) ≥ µ(Rk)/2.The constant αk is chosen so that for ϕk = αk χAk we have∫ϕk dµ =∫Qkf wk dµ. Then we obtain|αk| ≤ 1µ(Ak)∫Qk|f | dµ ≤ 2µ(Rk)∫12Rk|f | dµ ≤ C3λ(this calculation also applies to k = 1). Thus,|ϕk|+∑j|ϕsj | ≤ (2C2 + C3)λ.If we choose B = 2C2 + C3, (2.5) follows.Now it is easy to check that (2.6) also holds. Indeed we have‖ϕi‖L∞(µ) µ(Ri) ≤ C |αi|µ(Ai) = C∣∣∣∣∫Qif wi dµ∣∣∣∣ ≤ C∫Qi|f | dµ.172 X. TolsaSuppose now that the collection of cubes {Qi}i is not finite. For eachfixed N we consider the family of cubes {Qi}1≤i≤N . Then, as above, weconstruct functions ϕN1 , . . . , ϕNN with supp(ϕNi ) ⊂ Ri satisfying∫ϕNi dµ =∫Qif wi dµ,N∑i=1|ϕNi | ≤ B λand‖ϕNi ‖L∞(µ) µ(Ri) ≤ C∫Qi|f | dµ.Notice that the sign of ϕNi equals the sign of∫f wi dµ and so it doesnot depend on N .Then there is a subsequence {ϕk1}k∈I1 which is convergent in theweak ∗ topology of L∞(µ) to some function ϕ1 ∈ L∞(µ). Now we canconsider a subsequence {ϕk2}k∈I2 with I2 ⊂ I1 which is also convergent inthe weak ∗ topology of L∞(µ) to some function ϕ2 ∈ L∞(µ). In general,for each j we consider a subsequence {ϕkj }k∈Ij with Ij ⊂ Ij−1 that con-verges in the weak ∗ topology of L∞(µ) to some function ϕj ∈ L∞(µ). Itis easily checked that the functions ϕj satisfy the required properties.Remark 3.1. Recall that Besicovich’s covering theorem asserts that ifΩ ⊂ Rd is a bounded set and for each x ∈ Ω there is a cube Qx centeredat x, then there exists a family of cubes {Qxi}i with finite overlap, thatis∑i χQi ≤ C, which covers Ω.In (a) of the preceeding proof we have applied Besicovich’s coveringtheorem to Ω = {x : |f(x)| > λ}. However this set may be unbounded,and the boundedness property is a necessary assumption in Besicovich’stheorem (example: take Ω = [0,+∞) ⊂ R and consider Qx = [0, 2x] forall x ∈ Ω).We can solve this problem using different arguments. One possibilityis to consider for each r > 0 the set Ωr = {x : |x| ≤ r, |f(x)| > λ} andto apply Besicovich’s covering theorem to Ωr. With the same argumentsas above, we can decompose f = g+ b, with |g| ≤ λ only on Ωr and b asabove. Then the proof of Theorem 1.1 can be modified to show that forany fixed constants λ,R > 0 one hasµ{x ∈ B(0, R) : |Tεf(x)| > λ} ≤ C‖f‖L1(µ)λ.Weak (1,1) Inequality with Non Doubling Measures 173However we prefer the following solution. We are interested in showingthat the Caldero´n-Zygmund decomposition of Lemma 2.4 works alsowithout assuming Ω = {x : |f(x)| > λ} bounded. Let us sketch theargument. Consider a cube Q0 centered at 0 big enough so that2d+1 ‖f‖L1(µ)/µ(Q0) < λ.So for any cube Q containing Q0 we will have2d+1 ‖f‖L1(µ)/µ(Q) < λ.(3.2)For m ≥ 0 we set Qm :=(54)mQ0. For each m we can apply Besicovich’scovering theorem to the annulus Qm \Qm−1 (we take Q−1 := ∅), withcubes Qx centered at x ∈ supp(µ) ∩ (Qm \Qm−1) as in (a) of the proofabove, satisfying (3.1).In this argument we have to be careful with the overlapping amongthe cubes belonging to coverings of different annuli. Indeed, there existsome fixed constants N and N ′ such that if m ≥ N ′, for x ∈ supp(µ) ∩(Qm \Qm−1) we haveQx ⊂ Qm+N \Qm−N .(3.3)Otherwise, it is easily seen that (Qx) > 34(Qm), choosing N big enough.It follows that Q0 ⊂ 2Qx since (Q0)  (Qm) for N ′ big enough too.This cannot happen because then 2Qx satisfies (3.2), which contradicts(3.1).Because of (3.3), the covering made up of squares belonging to theBesicovich coverings of different annuli Qm \ Qm−1, m ≥ 0, will havefinite overlap.Notice that in this argument, it is essential the fact that in (3.1) weare not dividing by µ(Qx), but by µ(2Qx).References[DM] G. David and P. Mattila, Removable sets for Lipschitzharmonic functions in the plane, Rev. Mat. Iberoamericana16(1) (2000), 137–215.[GM] J. Garc´ıa Cuerva and J. M. Martell, Weighted inequal-ities and vector-valued Caldero´n-Zygmund operators on non-homogeneous spaces, Publ. Mat. 44(2) (2000), 613–640.[MMNO] J. Mateu, P. Mattila, A. Nicolau and J. Orobitg,BMO for nondoubling measures, Duke Math. J. 102(3)(2000), 533–565.174 X. Tolsa[NTV1] F. Nazarov, S. Treil and A. Volberg, Cauchy integraland Caldero´n-Zygmund operators on nonhomogeneous spaces,Internat. Math. Res. Notices 15 (1997), 703–726.[NTV2] F. Nazarov, S. Treil and A. Volberg, Weak type esti-mates and Cotlar inequalities for Caldero´n-Zygmund opera-tors on nonhomogeneous spaces, Internat. Math. Res. Notices9 (1998), 463–487.[NTV3] F. Nazarov, S. Treil and A. Volberg, Tb-theorem onnon-homogeneous spaces, Preprint (1999).[NTV4] F. Nazarov, S. Treil and A. Volberg, Accretive Tb-sys-tems on non-homogeneous spaces, Preprint (1999).[OP] J. Orobitg and C. Pe´rez, Ap weights for non doublingmeasures in Rn and applications, Trans. Amer. Math. Soc.(to appear).[To1] X. Tolsa, L2-boundedness of the Cauchy integral opera-tor for continuous measures, Duke Math. J. 98(2) (1999),269–304.[To2] X. Tolsa, Cotlar’s inequality without the doubling conditionand existence of principal values for the Cauchy integral ofmeasures, J. Reine Angew. Math. 502 (1998), 199–235.[To3] X. Tolsa, A T (1) theorem for non doubling measures withatoms, Proc. London Math. Soc. 82 (2001), 195–228.[To4] X. Tolsa, BMO , H1 and Caldero´n-zygmund operators fornon doubling measures, Math. Ann. (to appear).[Ve] J. Verdera, On the T (1)-theorem for the Cauchy integral,Ark. Mat. 38 (2000), 183–199.Department of MathematicsChalmers412 96 Go¨teborgSwedenCurrent address:De´partement de Mathe´matiqueUniversite´ de Paris-SudBaˆt. 42591405 Orsay CedexFranceE-mail address: xavier.tolsa@math.u-psud.frPrimera versio´ rebuda el 2 de maig de 2000,darrera versio´ rebuda el 17 de novembre de 2000.

A proof of the weak (1,1) inequality for singular integrals with non doubling measures based on a Calderon-Zygmund decomposition

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