WEIGHTED NORM INEQUALITIES FOR SINGULAR INTEGRAL

WEIGHTED NORM INEQUALITIES FOR SINGULAR INTEGRAL
OPERATORS
BAE JUN PARK
Abstract. This is just a note for my study and most contents are written based on
the textbook ”Weighted Norm Inequalities and Related Topics” by J. Garcia-cuerva and
Ruibio De Francia. It may have some mistakes or typos.
Contents
1. B.M.O. and some inequalities
2. Calderón-Zygmund Kernel and Operators
3. Weighted norm inequality for 1 < p < ∞
4. Alternative proof for Theorem 3.13
5. Weighted norm inequality for p = ∞
6. Relation between weights and B.M.O.
7. Weighted Norm inequalities for CZO(α)-First
8. Orlicz Spaces (based in [1])
9. Weighted Norm inequalities for CZO(α)- Second
References
1
5
8
20
21
28
30
34
43
45
1. B.M.O. and some inequalities
Theorem 1.1. (Calderón-Zygmund Decomposition) Let f ∈ L1 (Rd ) and t > 0. Then there
exist functions g and b on Rd such that
(1)
f = g + b;
(2)
g 1 ≤ f 1 and g ∞ ≤ 2d t
L
L
L
(3)
b=
X
bj
j
where each bj is supported in a dyadic cube Qj . Furthermore, the cubes Qk and Qj
are disjoint when j 6= j.
(4)
Z
bj (x)dx = 0
Qj
(5)
bj 1 ≤ 2d+1 t|Qj |
L
(6)
X
j
1 |Qj | ≤ f L1
t
1
2
BAE JUN PARK
Remark.
(1) This decomposition is called the Calderón-Zygmund decomposition of f
at height t. The function g is called the ”good function” since it is both integrable
and bounded. The function b is called ”bad function” since it contains the singular
part of f , but it is carefully chosen to have mean value zero.
(2)

f (x) Z
1
g(x) =
f (y)dy

|Qj | Qj
x∈
S
j
Qj
c
x ∈ Qj
(3) For each j,
1
bj = f −
|Qj |
Z
f dx χQj
Qj
and
f =g
on
[
c
Qj .
j
Let M be the Hardy-Littlewood Maximal operator(uncentered Maximal operators using
cubes) defined by
Z
1
|f (y)|dy
Mf (x) = sup
Q3x |Q| Q
and denote by Mk the operator M iterated k times.
M is a very nice tool in harmonic analysis and it can be used for Lp estimates for many
operators when 1 < p < ∞ or Lp (lq ) when 1 < p, q < ∞.
For the full range of p and q, we could apply the maximal operator Mt , defined by
1 Z
1/t
Mt f (x) = sup
|f (y)|t dy
Q3x |Q| Q
for 0 < t < p, q. Here the supremum is over any cubes Q containing x.
Now for any cubes Q, define the mean of f over Q as
Z
1
f (x)dx
fQ =
|Q| Q
and define the sharp maximal operator M# by
Z
1
M# f (x) = sup
|f (y) − fQ |dy.
Q3x |Q| Q
When M# f is L∞ function, we are saying that f ∈ BM O. That is,
BM O = {f ∈ L1loc : M# f ∈ L∞ }
and
# f M f ∞ .
=
BM O
L
Remark. (Some properties of BMO)
(1) · BM O is not a norm. The problem is that if f BM O = 0, this does not imply
that f = 0,
but that f is a constant a.e.. Moreoever, every constant function c
satisfies cBM O = 0.
(2) By the above reason, f and f + c have the same BM O norms whenever c is a
constant. In the sequel, we keepin mind that elements of BMO whose difference is
a constant are identified. Then · BM O is a seminorm.
(3) L∞ ⊂ BM O.
WEIGHTED NORM INEQUALITIES
3
(4) It is clear that
M# f (x) ≤ CMf (x)
for some C > 0.
Lemma 1.2.
(1) 12 f BM O ≤ supQ inf a∈C
1
|Q|
f .
|f
(y)
−
a|dy
≤
Q
BM O
R
(2) M# |f |(x) ≤ 2M# f (x).
Now we consider a maximal operator introduced by Peetre.
For r > 0 and σ > 0, set
|u(x + y)|
Mσ,r u(x) = sup
.
σ
y∈Rd (1 + r|y|)
One has the majorization
Mσ,r u(x) . Mt u(x)
(1.1)
provided that u ∈ S 0 whose Fourier transform is supported in {ξ : |ξ| ≤ r}.
for all σ ≥
These estimates imply the following maximal inequality.
d
t,
Lemma 1.3. (Maximal inequality)
Suppose 0 max d/p, d/q .
|Mσ,r uk |q q p . L
k
L
k
k
Lemma 1.4. If f is a function satisfying
Mf ∈ Lp0
for some 0 < p0 < ∞, then for any p0 ≤ p < ∞,
Z
Z
p
(Mf (x)) dx ≤ C
Rd
(M# f (x))p dx
Rd
for some C > 0.
Proof. We may assume that f ≥ 0 because Mf = M(|f |) and M# (|f |) ≤ 2M# f . Let
t > 0 and suppose Q is a cube satisfying fQ > t. For every x ∈ Q,
Z
1
t<
f (y)dy ≤ Mf (x).
|Q| Q
Thus,
Z
p
1
1
Mf (x) 0 dx ≤
kMf kp0
t ≤
|Q| Q
|Q|
If Q1 ( Q2 ( . . . are increasing dyadic cubes with fQk > t, then the family {Qk }K is
necessarily finite since |Qk | is boundded uniformly in k. Thus each dyadic cube Q satisfying
fQ > t will be contained in a maximal one. Let {Qj } be the family of those maximal dyadic
cubes. Then each of them satisfies
Z
1
t<
(1.2)
f (y)dy ≤ 2d t.
|Q| Q
p0
Since this family depends on t let’s denote it by {Qt,j }j . Observe that
[
(1.3)
for a.e. x ∈
/
Qt,j , f (x) ≤ t.
j
and
(1.4)
if t < s,
then each Qs,j ⊂ Qtk
for some k.
4
BAE JUN PARK
Given t > 0 we fix Q0 := Qt/2d+1 ,j0 and take A > 0.
If Q0 ⊂ x : M# f (x) > t/A , then
X
|Qt,j | ≤ |Q0 | ≤ x : M# f (x) > t/A {j:Qt,j ⊂Q0 }
because {Qt,j }j are non-overlapping.
If Q0 6⊂ x : M# f (x) > t/A , then
Z
1
|f (y) − fQ0 |dy ≤ t/A.
|Q0 | Q0
R
( If not, we have |Q10 | Q0 f (y) − fQ0 dy > t/A and then for all x ∈ Q0 M# f (x) > t/A,
which implies Q0 ⊂ x : M# f (x) > t/A - contradiction! ) Furthermore, fQ0 ≤ t/2 by
(1.2) and thus
Z
X
X
f (y) − fQ dy
t/2|Qt,j | ≤
0
{j:Qt,j ⊂Q0 } Qt,j
{j:Qt,j ⊂Q0 }
Z
≤
Q0 f (y)−fQ0 dy
≤ t/A|Q0 .|
By combining these two cases, we have
X
X
Qt/2d+1 ,k .
|Qt,j | ≤ {x : M# f (x) > t/A} + 2/A
j
Let’s call α(t) =
k
x : Mf (x) > t . Then observe that
j |Qt,j | and β(t) =
P
α(t) ≤ β(t),
X
Q3 d . α t/C
β(t) ≤
t/4 ,j
(1.5)
(1.6)
j
for some C, and
α(t) ≤ x : M# f (x) > t/A + 2/Aα t/2d+1 .
(1.7)
By (1.6) for N > 0
Z
IN :=
N
pt
p−1
0
≤
N
Z
dt ≤
Z
p−p0
pp−1
0 N
pt
0
p−1
β(t)dt ≤
p−p0
pp−1
0 N
Z
N
p0 tp0 −1 β(t)dt
0
p
Mf (x) 0 dx < ∞.
Rd
And by (1.7)
Z
IN
N
≤
pt
p−1 x : M# f (x) > t/A dt + 2/A
0
Z
pt
p−1 x : M# f (x) > t/A dt + C/A
0
≤
ptp−1 α t/2d+1 dt
0
N
=
Z
N
Z
Z
N/2d+1
0
N
ptp−1 x : M# f (x) > t/A dt + (C/A)IN .
0
By taking A = 2C we obtain the estimate
Z N
IN ≤ 2
ptp−1 x : M# f (x) > t/A dt
0
ptp−1 α(t)dt
WEIGHTED NORM INEQUALITIES
because IN < ∞.
Letting N → ∞
Z ∞
pt
p−1
Z
α(t)dt ≤ 2
5
∞
ptp−1 x : M# f (x) > t/A dt.
0
0
By combining this with (1.6), we finally get
Z
Z ∞
p
Mf (x) dx =
ptp−1 β(t)dt
Rd
0
Z ∞
.
ptp−1 α t/C dt
Z0 ∞
ptp−1 α t dt
≈
Z0 ∞
ptp−1 x : M# f (x) > t/A dt
.
Z0 ∞
≈
ptp−1 x : M# f (x) > t dt
Z0
p
=
M# f (x) dx.
Rd
Remark. For 1 < p < ∞
f ∈ Lp ⇔ Mf ∈ Lp ⇔ M# f ∈ Lp .
since the boundedness of M, Lemma 1.4 and the fact that M# f ≤ 2Mf ≤ 2f .
For p = ∞
Mf ∈ L∞ ⇔ f ∈ L∞ ,
but
M# f ∈ L∞ ⇔ f ∈ BM O.
Lemma 1.5. (Interpolation theorem)[5]
(1) Let T be a linear operator bounded in Lp0 for some 1 < p0 < ∞. Suppose T carries
L∞ to BM O boundedly. Then for any p0 < p < ∞, T is bounded in Lp .
(2) Let T be a linear operator bounded in Lp0 for some 1 < p0 < ∞. Suppose T carries
H 1 to L1 boundedly. Then for any 1 < p < p0 , T is bounded in Lp .
2. Calderón-Zygmund Kernel and Operators
We denote by 4 = {(x, x) : x ∈ Rd } the diagonal of Rd × Rd .
Definition 1. (Calderón-Zygmund Kernel or Standard Kernel)
Let 0 < α ≤ 1. A Calderón-Zygmund Kernel (or Standard Kernel) of order α is a
continuous function K : 4c → C such that there exists a constant C > 0 such that
(1) For all (x, y) ∈ 4c ,
|K(x, y)| ≤
C
,
|x − y|d
(2) For all x, y, z ∈ Rd satisfying |y − z| ≤ 12 |x − y| when x 6= y,
|y − z| α
1
|K(x, y) − K(x, z)| ≤ C
,
|x − y| |x − y|d
6
BAE JUN PARK
(3) For all x, y, z ∈ Rd satisfying |y − z| ≤ 12 |x − y| when x 6= y,
|K(x, y) − K(z, y)| ≤ C
|x − z| α
|x − y|
1
.
|x − y|d
We say that K ∈ CZK(α) if the above conditions are satisfied and then define K α to
be the infimum of implicit constants C in (1)-(3). Note that the adjoint kernel K ∗ (x, y) =
K(y, x) of K ∈ CZK(α) is also in CZK(α)
Definition 2. (Singular integral operator) T is called a singular integral operator corresponding to the Calderón-Zygmund kernel K if
Z
Z Z
g(x)T (f )(x)dx =
g(x)K(x, y)f (y)dydx
where f and g are smooth and have disjoint supports. Such operator need not to be
bounded on Lp .
Definition 3. (Calderón-Zygmund Operator) A Calderón-Zygmund Operator of order α
is a singular integral operator T that is associated to a K ∈ CZK(α) and bounded in L2 .
We define CZO(α) to be the collection of all Calderón-Zygmund operators of order α.
Note that every T ∈ CZO(α) is also bounded on all Lp with 1 1 there exists a constant Cs > 0 such that
M# (T f )(x) ≤ Cs Ms f (x).
Proof. Due to Lemma 1.2 it suffices to find a constant a such that for given x and a cube
Q containing x
Z
1
|T f (y) − a|dy ≤ CMs f (x).
|Q| Q
Write f as
f = f χ2Q + f χ(2Q)c := f1 + f2 .
and choose a = T f2 (x) = T (f χ(2Q)c )(x). Then
≤
Z
1
|T f (y) − T (f2 )|dy
|Q| Q
Z
Z
1
1
|T f1 (y)|dy +
|T f2 (y) − T f2 (x)|dy.
|Q| Q
|Q| Q
Since T is bounded in Ls we have
Z
1 Z
1/s
1
|T f1 (y)|dy ≤
|T f1 (y)|s dy
|Q| Q
|Q| Rd
1 Z
1/s
.
|f1 (y)|s dy
|Q| 2Q
. Ms f (x).
WEIGHTED NORM INEQUALITIES
7
Now use the property of CZK(δ) to get
Z
Z Z
1
1
|T f2 (y) − T f2 (x)|dy ≤
|K(y, z) − K(x, z)||f (z)|dzdy
|Q| Q
|Q| Q (2Q)c
Z Z
1
|y − x|δ
.
|f (z)|dzdy
|Q| Q (2Q)c |x − z|d+δ
∞ Z
X
|f (z)|
dz
. l(Q)δ
d+δ
k l(Q)<|x−z|<2k+1 l(Q) |x − z|
2
k=−1
Z
∞
X
1
δ
. l(Q)
|f (z)|dz
(2k l(Q))d+δ |y−z|<2k+1 l(Q)
k=−1
. Mf (x)
. Ms f (x).
Define a truncated operators
Z
T f (x) =
K(x, y)f (y)dy
|x−y|>
for f ∈ Lp , 1 ≤ p < ∞ and one naturally wishes to have a more explicit definition of T f as
T f (x) = lim T f (x).
→0
Thus we are interested in uniform estimates for T and we are going to see that we can
actually estimate the maximal operator
(2.1)
T ∗ f (x) = sup T f (x).
>0
Lemma 2.2. Let T ∈ CZO(δ) and f ∈ L∞
c so that the integral
Z
T f (x) =
K(x, y)f (y)dy
|x−y|>
exists for every > 0. Then for all 1 < t < ∞ we have
Z
1
|T f (y) − T f (x)|dy
d |x−y|</2
Z Z
1
≤ Ct Mt f (x) + d
|K(x, z) − K(y, z)||f (z)|dydz
2|x−y|<<|z−x|
. Ct Mt f (x)
where Ct is independent of and f .
Proof. For x ∈ Rd set
f1 := f · χ{y:|x−y|<}
and
f2 := f − f1 .
Then
≤
Z
1
|T f (y) − T f (x)|dy
d |x−y|</2
Z
Z
1
1
|T
f
(y)|dy
+
|T f2 (y) − T f (x)|dy
1
d |x−y|</2
d |x−y|</2
8
BAE JUN PARK
By Hölder’s inequality, the first one is less than
−d/t T f1 t ≤ Ct Mt f (x)
L
We see that the second term does not exceed
Z Z
1
(2.2)
|K(x, z) − K(y, z)||f (z)|dydz,
d
2|x−y|<<|z−x|
and
Z
≤
≤
≤
K(x, z) − K(y, z)|f (z)|dz
|x−z|>2|x−y|
∞
XZ
j
j+1 |x−y|
j=1 2 |x−y|<|x−z|≤2
∞ Z
X
j
j+1 |x−y|
j=1 2 |x−y|<|x−z|≤2
Z
∞
X
1
2d
j=1
2jα (2j+1 |x − y|)d
K(x, z) − K(y, z)|f (z)|dz
|x − y|α
|f (z)|dz
|x − z|d+α
|f (z)|dz
|x−z|≤2j+1 |x−y|
. Mf (x) ≤ Mt f (x),
which completes the proof.
Theorem 2.3. Let T ∈ CZO(α) and 1 ≤ p < ∞. Suppose f ∈ Lp . Then
T ∗ f (x) ≤ Cq Mt f (x) + CM(T f )(x)
(2.3)
for 1 < t < ∞.
Proof. By Lemma 2.2,
Ct Mt f (x) ≥
≥
≥
Z
1
T f (y) − T f (x)dy
d
|x−y|</2
Z
Z
1
T f (x)dy − 1
T f (y)dy
d
d
|x−y|</2
|x−y|</2
1 T f (x) − M(T f )(x) .
2d
Thus,
T f (x) ≤ Ct Mt f (x) + CM(T f )(x).
Taking supremum over > 0 we are done.
Remark.
(1) T ∈ CZO(α) if and only if T ∗ ∈ CZO(α).
(2) The map T 7→ K, where T ∈ CZO(α) and K ∈ CZK(α) the associated kernel, is
not injective. Consequently, one cannot define a CZO(α) uniquely given a kernel
K ∈ CZO(α).
3. Weighted norm inequality for 1 < p < ∞
Definition 4. Let µ and ν be two measures and T be an operator.
(1) When T is bounded from Lp (µ) to Lq (ν), then we are saying that T is of strong type
(p, q) w.r.t. (ν, µ). That is,
Z
Z
1
1
q
q
|f (x)|p dµ p .
|T f (x)| dν .
Rd
Rd
WEIGHTED NORM INEQUALITIES
9
(2) When
1
sup |t ν {x : |T f (x)| > t} q | ≤ C f Lp (µ) ,
t>0
we are saying that T is of weak type (p, q) w.r.t. (ν, µ).
Lemma 3.1. (Interpolation theorem) Let dµ and dν be two measures and suppose 1 ≤ p0 <
p1 ≤ ∞. Suppose T is an operator mapping functions in Lp0 (µ) + Lp1 (ν) to ν-measurable
functions satisfying the following properties;
(1) for f1 , f2 ∈ Lp0 (µ) + Lp1 (µ),
|T (f1 + f2 )(x)| ≤ |T f1 (x)| + |T f2 (x)| ν-a.e.;
(2) T is of weak type (p0 , p0 ) w.r.t. (ν, µ);
(3) T is of weak type (p1 , p1 ) w.r.t. (ν, µ).
Then for p0 0 such that
Z
Z
(3.1)
(M(f )(x))p φ(x)dx ≤ Cp
|f (x)|p M(φ)(x)dx
Rd
Rd
Moreover, there exists C1 > 0 such that
Proof. Suppose that {x : M(φ)(x) < ∞} > 0, otherwise it is trivial. Define two measures
dµ and dν as
dµ(x) = M(φ)(x)dx
and
dν(x) = φ(x)dx.
Then (3.1) means
Z
Z
p
(M(f )(x)) dν(x) ≤ Cp
(3.2)
Rd
|f (x)|p dµ(x).
Rd
Lp (ν).
That is, M is a bounded operator from Lp (µ) to
Now we apply interpolation theory.
First, assume p = ∞. If M(φ)(x) = 0 for some x., then φ(x) = 0 a.e. and thus
dν
= 0.
Then we are done. Therefore we suppose that M(φ)(x) > 0 for all x. Let α > f L∞ (µ) .
Then
Z
M(φ)(x)dx = 0 → |f (x)| ≤ α a.e. → M(f )(x) ≤ α a.e. → M(f )L∞ (ν) ≤ α
{|f |>α}
because
Z
ν({M(f ) > α}) =
φ(x)dx = 0.
{Mf >α}
Thus we have
M(f ) ∞ ≤ f ∞ ,
L (ν)
L (µ)
which means that M is of weak type (∞, ∞) w.r.t. (ν, µ).
Now consider the case p = 1. We need to prove that
Z
Z
1
(3.3)
φ(x)dx ≤ C
|f (x)|M(φ)(x)dx
t Rd
{x:Mf (x)>t}
We may assume that f ≥ 0 and then we can find a sequence of nonnegative integrable
functions fj such that
f1 ≤ f2 · · · % f a.e.
10
BAE JUN PARK
and
{x : M(f )(x) > t} =
[
{x : M(fj )(x) > t}.
j
This allows us to assume f ∈ L1 (Rd ). For given t > 0 there exists a family of nonoverlapping cubes {Qj } such that for each j,
Z
t
1
t
<
f (x)dx ≤ d
d
|Qj | Qj
4
2
and
[
{x : M(f )(x) > t} ⊂
Q3j .
j
Since
|Q3j | = 3d |Qj | <
3d · 4d
t
Z
f (x)dx,
Qj
we have
Z
φ(y)dy ≤
{x:M(f )(x)>t}
XZ
j
φ(y)dy
Q3j
Z
X 1 Z
3d · 4d
≤
f (x)dx
φ(y)dy
t
|Q3j | Q3j
Qj
j
Z
3d · 4d X
≤
f (x)M(φ)(x)dx
t
Qj
j
Z
1
f (x)M(φ)(x)dx
.
t Rd
and this implies that M is of weak type of (1, 1) w.r.t. (ν, µ). Then lemma 3.1 yields (3.2).
Any measurable functions which are positive almost everywhere are called ”weight functions”.
Definition 5. (Weight function spaces) Let 1 0}
1 Z
1 Z
p−1
Ap = (u, w) : sup
u(x)dx
(w(x))−1/(p−1) dx
≤ C for some C > 0
|Q| Q
Q:cubes |Q| Q
and the smallest such a constant C is called Ap constant for (u, w).
1 Z
p−1 1 Z
Ap -constant for (u, w) = inf
u(x)dx
(w(x))−1/(p−1) dx
Q
|Q| Q
|Q| Q
Here we are considering just weight functions u and w.
Theorem 3.3. Let 1 ≤ p < ∞. Then TFAE.
(1) M is of weak type (p, p) w.r.t. (u, w).
(2) There exists a constant C > 0 such that for f ≥ 0 and a cube Q,
Z
(fQ )p u(Q) ≤ C
(f (x))p w(x)dx
Q
where u(Q) :=
(3) (u, w) ∈ Ap .
R
Q u(x)dx
WEIGHTED NORM INEQUALITIES
11
Note that (2) implies that
|S| p
(3.4)
|Q|
u(Q) ≤ Cw(S).
for any cubes Q and any measurable subset S of Q. (Put f (x) = χS (x))
Proof. First we shall prove that (1) implies (2). Suppose that M is of weak type (p, p)
w.r.t. (u, w). Thus, there exists C > 0 such that
Z
1
u M(f )(x) > t ≤ C p
|f (x)|p w(x)dx.
t Rd
We may assume f ≥ 0. Choose a cube Q satisfying fQ > 0 ( If not possible, then f ≡ 0
a.e. ). For any 0 < t < fQ and x ∈ Q,
t < fQ ≤ M(f χQ )(x) a.e.
Thus,
Q ⊂ M(f χQ ) > t
and
u(Q) ≤ u
1
M(f χQ ) > t ≤ C p
t
Z
|f (x)|p w(x)dx.
Q
This implies that
tp u(Q) ≤ C
Z
|f (x)|p w(x)dx
Q
and by taking supremum we get
(fQ )p u(Q) ≤ C
Z
|f (x)|p w(x)dx.
Q
Now ssume p = 1 and prove that
(2) ⇒ (3) ⇒ (1).
By (3.4) we see
1
1
u(Q) ≤ C
w(S).
|Q|
|S|
Fix Q and let
a > ess.infQ w := sup b ∈ Q : |{x : w(x) < b}| = 0
and
Sa = {x ∈ Q : w(x) < a}.
Note that |Sa | > 0. Then
u(Q)
w(Sa )
≤C
≤ Ca
|Q|
|Sa |
and this gives that
(3.5)
u(Q)
≤ Cess.infQ w ≤ Cw(x) a.e. x ∈ Q,
|Q|
which establish (3).
Suppose (u, w) ∈ A1 . Then By definition Mu(X) ≤ Cw(X) a.e. Applying (3.3),
Z
1
u({Mf (x) > t}) ≤ C
f (x)M(u)(x)dx
t d
Z R
1
.
f (x)w(x)dx
t Rd
12
BAE JUN PARK
and thus (1) follows.
Now assume 1 1/j} for j = 1, 2, . . . . On each Sj f is bounded and thus
Z
f (x)dx < ∞.
Sj
It follows from (2) that
1 Z
|Q|
f (x)dx
Sj
p
1 Z
p
u(Q) =
f (x)χSj (x)dx u(Q)
|Q| Q
Z
.
f (x)p w(x)dx
Sj
Z
=
f (x)dx,
Sj
and thus
Z
p 1
1 Z
1
f (x)dx
f (x)dx.
u(Q) .
|Q| Sj
|Q|
|Q| Sj
This gives
1 Z
p−1 1
(w(x))−1/p−1 dx
u(Q) . 1
|Q| Sj
|Q|
S∞
Since S1 ⊂ S2 ⊂ . . . , j=1 Sj = {x ∈ Q : w(x) > 0}, and w(x) > 0 a.e. we have
∞
[
Sj = |Q|,
j=1
which implies
p−1 1
1 Z
(w(x))−1/p−1 dx
u(Q) . 1.
|Q| Q
|Q|
This proves that (3) implies (2).
Next assume (2) holds and we prove (3). Let Q be any cubes and suppose f ≥ 0. By
Hölder’s inequality
Z
1/p
−1/p
1
fQ =
f (x) w(x)
w(x)
dx
|Q| Q
1 Z
1/p 1 Z
(p−1)/p
p
≤
f (x) w(x)dx
w(x)−1/(p−1) dx
.
|Q| Q
|Q| Q
and thus
Z
1 Z
p−1
p
u(Q) (fQ ) u(Q) ≤
f (x) w(x)dx
w(x)−1/(p−1) dx
|Q|
|Q| Q
Q
Z
p
≤ C
f (x) w(x)dx.
p
Q
Now show that (2) implies (1). Suppose f ∈ Lp (w) and f ≥ 0. Then
fk (x) := f (x)χQ(0,k) (x)
WEIGHTED NORM INEQUALITIES
13
belongs to L1 (w) by Hölder’s inequality. Let Et := {x ∈ Rd : M(fk )(x) > t}. Then there
exists a family of non-overlapping cubes {Qj } such that
[
Et ⊂
Q3j
j
and
t
1
<
d
|Qj |
4
From (2) we obtain
1 Z
|Q3j |
Q3j
fk (x)χQj (x)dx
Z
fk (x)dx ≤
Qj
p
u(Q3j )
t
.
2d
Z
≤C
Q3j
p
fk (x)χQj (x) dx
and thus
1 Z
−p Z
f
(x)dx
(fk (x))p w(x)dx
k
3
|Qj | Qj
Qj
1 Z
−p Z
p
≤ 3dp
fk (x)dx
fk (x) w(x)dx
|Qj | Qj
Qj
Z
dp
dp
3 4
p
fk (x) w(x)dx.
≤
tp
Qj
u(Q3j ) .
Summing over j
u(Et ) ≤
X
u(Q3j )
j
Z
p
3dp 4dp X
fk (x) w(x)dx
p
t
Qj
j
Z
p
1
f (x) w(x)dx
p
t Rd
≤ C
.
Since the above inequality is independent of k, we conclude that
Z
p
1
d
u {x ∈ R : M(f )(x) > t} ≤ C p
f (x) w(x)dx.
t Rd
We are saying that
w ∈ Ap
when (w, w) ∈ Ap .
Lemma 3.4. Let w ∈ L1loc (Rd ) be positive a.e.. Then TFAE.
(1) w ∈ A1 ;
(2) For any cubes Q,
Z
1
w(x)dx ess.supQ (w−1 ) ≤ C
|Q| Q
Proof. By (3.5) w ∈ A1 is equivalent to
w(Q)
ess.supQ w−1 < C
|Q|
for any cube Q.
14
BAE JUN PARK
Theorem 3.5. Let µ be a positive
γ Borel measure such that Mµ(x) < ∞ for a.e. and let
0 < γ < 1.Then w(x) := Mµ(x) is an A1 weight with a constant depending only on γ
and d.
Proof. Let Q be a fixed cube. We shall show
Z
1
(3.6)
w(x)dx . w(x)
|Q| Q
e = Q3 , the 3-dilated of Q. We write µ = µ1 + µ2 with
uniformly in Q for a.e.. Let Q
e Then it is clear that
µ1 = χQe µ1 , which is the restriction of µ to Q.
γ
γ
≤ Mµ1 (x) + Mµ2 (x) .
γ
It suffices to show that the averages of each Mµi (x) over Q are bounded by Cw(x) for
any x ∈ Q with C depending only on γ and d. First we see that
Z
Z ∞
γ
1
1
γtγ−1 {x ∈ Q : Mµ1 (x) > t}dt
Mµ1 (x) dx =
|Q| Q
|Q| 0
Z ∞
Z
1 R
:=
+
|Q| 0
R
Mµ(x)
γ
for R > 0 to be chosen later. Clearly the first part is less than |Q|Rγ . By the weak
boundedness of the Hardy Littlewood maximal operator, the second one is bounded by
Z ∞
Z ∞
γ
γ−1 1
C
γt
kµ1 kdt = Ckµ1 k
γtγ−2 dt = C
Rγ−1 kµ1 k
t
1−γ
R
R
Taking R =
kµ1 k
|Q|
1
|Q|
we obtain
Z
Q
µ(Q)
kµ k γ
e γ
γ
1
≤C
Mµ1 (x) dx ≤ C
≤ Cw(x).
e
|Q|
|Q|
For the term corresponding to µ2 observe that for any x, y ∈ Q
Mµ2 (x) ≤ CMµ2 (y)
(3.7)
because µ2 lives far away from Q. To be precise, let Q0 be a cube containing x and meeting
e Then Q ⊂ (Q0 )3 . We have
Rd \ Q.
1
|Q0 |
3d
dµ2 ≤
|(Q0 )3 |
Q0
Z
Z
(Q0 )3
dµ2 ≤ 3d Mµ2 (y),
which implies (3.7). Then
Z
γ
γ
1
Mµ2 (x) dx ≤ C Mµ2 (y) = Cw(y)
|Q| Q
for all y ∈ Q.
Theorem 3.6. Let 1 < p < ∞ and w0 , w1 ∈ A1 . Then
1−p
w(x) := w0 (x) w1 (x)
∈ Ap .
WEIGHTED NORM INEQUALITIES
15
Proof.
1 Z
1 Z
−1/(p−1) p−1
1−p
w0 (x)w1 (x) dx
w0 (x)w1 (x)1−p
dx
|Q| Q
|Q| Q
1 Z
p−1
1 Z
w0 (x)w1 (x)1−p dx
w0 (x)−1/(p−1) w1 (x)dx
=
|Q| Q
|Q| Q
Z
Z
p−1
1
−1 p−1 1
−1
≤ ess. sup (w1 )
w0 (x)dx ess. sup (w0 )
w1 (x)dx
|Q| Q
|Q| Q
Q
Q
Z
Z
p−1
1
=
ess. sup (w0−1 )
w0 (x)dx ess. sup (w1−1 )
w1 (x)dx
|Q| Q
Q
Q
Q
. C
by Lemma 3.4.
Note that the converse also holds in Theorem 3.6, which is called ”Factorization theorem”. That is, every Ap weight w is actually of the form
1−p
w(x) = w0 (x) w1 (x)
for some w0 , w1 ∈ A1 .
Corollary 3.7.
(1) |x|m ∈ A1 if and only if −d < m ≤ 0.
(2) |x|m ∈ Ap if and only if −d < m < d(p − 1).
Proof. The first statement is immediate from Theorem 3.12.
For the second one suppose that for −d < a, b ≤ 0, |x|a and |x|b belongs to A1 . Then
Theorem 3.6 yields that
|x|a+b(1−p) ∈ Ap .
Since −d < a + b(1 − p) < d(p − 1) we have |x|m ∈ Ap if −d < m < d(p − 1).
Now suppose |x|m ∈ Ap . Then |x|m , |x|−m/(p−1) ∈ L1lock , which implies −d < m and
−d < −m/(p − 1). Thus −d < m < d(p − 1).
Lemma 3.8. Let 1 < p < ∞. Then
w ∈ Ap
is equivalent to
w−1/(p−1)∈Ap0
where
1
p
+
1
p0
= 1.
Proof.
w ∈ Ap
1 Z
1 Z
−1/(p−1) p−1
≤C
⇔ sup
w(x)dx
w(x)
dx
|Q| Q
Q:cubes |Q| Q
Z
1 Z
−1/(p−1) −1/(p0 −1) 1
−1/(p−1) p−1
⇔ sup
w(x)
dx
w(x)
dx
≤C
|Q| Q
Q:cubes |Q| Q
1 Z
p0 −1 1 Z
0
−1/p−1 −1/(p0 −1)
⇔ sup
((w(x))
)
dx
(w(x))−1/(p−1) dx ≤ C p −1
|Q| Q
Q:cubes |Q| Q
⇔ w−1/(p−1) ∈ Ap0
16
BAE JUN PARK
Lemma 3.9. Let 1 < p < ∞ and 0 < α < 1. Suppose w ∈ Ap . Then there exists 0 < β < 1,
depending on α, such that for any measurable subset A of Q with |A| ≤ α|Q|,
w(A) ≤ βw(Q).
Proof. Suppose w ∈ Ap and |A| ≤ α|Q| and put S = Q \ A. Then by (3.4)
|S| p
w(Q) ≤ Cw(S)
|Q|
and this means that
|Q| − |A| p
|Q|
w(Q) ≤ Cw(S).
Hence
(1 − α)p w(Q) ≤ C(w(Q) − w(A))
and
w(A) ≤
We are done by putting β =
C − (1 − α)p
w(Q).
C
C−(1−α)p
.
C
Lemma 3.10. (Reverse Hölder inequality) Let 1 0, depending only on p and Ap constant for w, such that for any cube Q,
Z
Z
1/(1+)
1
1
1+
(w(x)) dx
≤C
w(x)dx
|Q| Q
|Q| Q
with a constant C, independent of Q.
Proof. Suppose 0 < α < 1. Fix a cube Q and let
λ0 = wQ =
1
w(Q)
|Q|
and λk =
2 d k
α
λ0 .
Observe that
λ0 < λ1 < λ2 < . . . .
For each λk consider the maximal dyadic subcubes of Q over which the average of w is
greater than λk . Let {Qk,j } be such maximal dyadic subcubes and let
Dk =
∞
[
Qk,j .
j=1
Then for each j we have
λk < wQk,j < 2d λk
and for a.e. x ∈
S
j
Qk,j
c
w(x) ≤ λk .
Since λk+1 > λk each Qk+1,j is contained in Qk,i for some i and thus
Dk+1 ⊂ Dk .
WEIGHTED NORM INEQUALITIES
17
Then by definition
2d λ k ≥
=
1
|Qk,i |
1
|Qk,i |
Z
w(x)dx
Qk,i
T
Dk+1
Z
X
w(x)dx
Qk+1,j
Qk+1,j ⊂Qk,i
X
1
>
λk+1
|Qk+1,j |
|Qk,i |
Qk+1,j ⊂Qk,i
T
|Qk,i Dk+1 |
= λk+1
|Qk,i |
and this gives
T
|Qk,j Dk+1 |
2d λ k
=α<1
<
|Qk,i |
λk+1
for each k. By Lemma 3.9 there exists 0 < β < 1,independent of k, such that
\
w(Qk,i Dk+1 ) ≤ βw(Qk,i ).
Therefore
w(Dk+1 ) =
X
w(Qk,i
\
Dk+1 ) ≤ β
i
X
w(Qk,i ) = βw(Dk )
i
and by iteration we have
w(Dk ) ≤ β k w(D0 ).
Since
|Dk+1 | =
X
|Qk,i
\
Dk+1 | ≤ α
i
X
|Qk,i | = α|Dk |,
i
we obtain
|Dk | ≤ αk |D0 |.
Now
Z
w(x)
1+
Z
1+
w(x)
dx +
dx =
Q
Q\D0
Z
1+
w(x)
dx +
=
Q\D0
Z
1+
w(x)
dx
D0
∞
XZ
Q\D0
∞ Z
X
k=0
= (λ0 ) w(Q \ D0 ) +
∞
X
w(x)
1+
dx
Dk \Dk+1
k=0
(λ0 ) w(x)dx +
≤
Z
(λk+1 ) w(x)dx
Dk \Dk+1
(λk+1 ) w(Dk \ Dk+1 )
k=0
= (λ0 ) w(Q \ D0 ) + 2d /α
∞
X
k=0
because
λk+1 =
2d k+1
2d λ k
= ··· =
λ0 .
α
α
By choosing > 0 so small that
2d /α β < 1,
k
(2d /α) β w(D0 )
18
BAE JUN PARK
it follows that
Z
w(x)
1+
dx ≤ C(λ0 ) w(Q)
Q
= C(wQ ) w(Q),
which is equivalent to
1
|Q|
Z
1+
(w(x))
Q
1 Z
1+
dx ≤ C
w(x)dx
.
|Q| Q
Corollary 3.11. Let 1 < p < ∞ and w ∈ Ap . Then there exists q ∈ (1, p) such that
w ∈ Aq .
That is,
Ap ⊂
[
Aq .
1<q 0 and C > 0 such that for any cubes Q
Z
1 Z
−(1+)/(p−1) 1/(1+)
1
w(x)
(w(x))−1/(p−1) dx.
dx
≤C
|Q| Q
|Q| Q
Since
1+
p−1
>
1
p−1 ,
there exists q ∈ (1, p) such that
1+
1
=
.
p−1
q−1
Then
1 Z
1 Z
q−1
w(x)dx
(w(x))−1/(q−1) dx
|Q| Q
|Q| Q
(p−1)/(1+)
1 Z
1 Z
=
w(x)dx
(w(x))−(1+)(p−1) dx
|Q| Q
|Q| Q
1 Z
1 Z
p−1
≤ C
w(x)dx
(w(x))−1/(p−1) dx
|Q| Q
|Q| Q
. 1
Theorem 3.12. Let 0 < w(x) < ∞ a.e. Then
w ∈ A1
if and only if
w(x) = k(x) Mf (x)
γ
where k(x) ≥ 0, k, k −1 ∈ L∞ and f is locally integrable and 0 < γ < 1.
WEIGHTED NORM INEQUALITIES
19
Proof. ”if” direction is an immediate consequence of Lemma 3.5.
Suppose w ∈ A1 . Then by Theorem 3.10 there exists > 0 such that
Z
1/(1+)
1 Z
1
1+
≤ C
w(x) dx
w(x)dx
|Q| Q
|Q| Q
≤ CMw(x) ≤ Cw(x)
for a.e. x ∈ Q. This implies that
w(x) ≤ M1 w(x) ≤ Cw(x)
for a.e. x ∈ Rd . Let
w(x)
,
M1+ w(x)
and γ = 1/(1 + ). Then clearly
1
≤ k(x) ≤ 1
a.e.,
C
k(x) :=
and put f (x) = w1+
f ∈ L1loc ,
and
γ
w(x) = k(x) Mf (x) .
Theorem 3.13. Let 1 0 such that for any f ≥ 0 and any cubes Q,
Z
Z
p
1
f (x)dx w(Q) ≤ C
(f (x))p w(x)dx.
|Q| Q
Q
(3) w ∈ Ap .
(4) M is of strong type (p, p) w.r.t. w.
Proof. By Theorem 3.3
(1) ⇔ (2) ⇔ (3)
and we already know
(4) ⇒ (1).
Therefore it suffices to show
(3) ⇒ (4).
Let w ∈ Ap . By Lemma 3.11 w ∈ Aq for some 1 < q α}) > 0}
L (w)
≤ sup {α : {x ∈ Rd : Mf (x) > α} > 0}
= Mf L∞
. f L∞
= sup {α : {x ∈ Rd : f (x) > α} > 0}
≤ sup {α : w({x ∈ Rd : f (x) > α}) > 0}
= f L∞ (w) .
Here, we use the fact that
w(x) > 0 a.e.
20
BAE JUN PARK
and
w(A) > 0 ⇒ |A| > 0.
By using Lemma 3.1 M is of strong type (p, p) w.r.t. w.
4. Alternative proof for Theorem 3.13
The proof of Muckenhoupt depends on Reverse Hölder’s inequality and was not totally
elementary. Christ and Fefferman [2] provided an extremely elementary proof of Theorem
3.13 without Reverse Hölder’s inequality.
Theorem 4.1. Let 1 2d . For each k ∈ Z let {Qk,j }j
be the Calderon-Zygmund cubes of f at height ak . That is,
Qk,j : nonoverlapping cubes,
Z
1
ak <
|f (x)|dx ≤ 2d ak for all j,
|Qk,j | Qk,j
[
{x : Mf (x) > ak } =
Qk,j ≡ Ωk ,
j
k
|f (x)| ≤ a a.e. on (Ωk )c .
Observe that Ωk contains Ωk+1 and each Qk+1,s completely belongs to Qk,j if the intersection
of Qk,j and Qk+1,s is empty. Let Ek,j := Qk,j \ Ωk+1 and σ := w−1/(p−1) ∈ Ap0 . It is trivial
S
that Ek,j are nonoverlapping and k,j Ek,j = Rd . Thus
Z
(Mf (x))p w(X)dx
d
R
XZ
(Mf (x))p w(x)dx
=
k,j
. ap
Ek,j
X
k,j
(4.1)
= ap
X
k,j
1
|Qk,j |
Z
|f (y)|dy
p
1
σ(Qk,j )
Observe that
\
Ωk+1 Qk,j =
Z
Qk,j
p
w(E ) σ(Q ) p |f (y)|
k,j
k,j
σ(y)dy σ(Ek,j )
.
σ(y)
σ(Ek,j ) |Qk,j |
X
|Qk+1,s | ≤
Qk+1,s ⊂Qk,j
=
w(Ek,j )
Qk,j
1
ak+1
X
1
Qk+1,s ⊂Qk,j
ak+1
Z
Z
|f (x)|dx ≤
Qk,j ∩Ωk+1
1
ak+1
and this gives
|Ek,j |
2d
≥1− .
|Qk,j |
a
Furthermore, by Lemma 3.9
σ ∈ Ap0 ⇒
σ(Ek,j )
≥α
σ(Qk,j )
Z
|f (x)|dx
Qk+1,s
|f (x)|dx ≤
Qk,j
2d
|Qk,j |
a
WEIGHTED NORM INEQUALITIES
21
for some α. Therefore
w(Ek,j ) σ(Qk,j ) p
=
σ(Ek,j ) |Qk,j |
≤
w(Ek,j ) σ(Qk,j ) σ(Qk,j ) p−1
σ(Ek,j ) |Qk,j |
|Qk,j |
1 w(Ek,j ) σ(Qk,j ) p−1
.1
α |Qk,j |
|Qk,j |
by the definition of Ap . So far we proved
Z
p
X 1
|f (y)|
(4.1) .
σ(y)dy σ(Ek,j ).
σ(Qk,j ) Qk,j σ(y)
k,j
Since
1
σ(Qk,j )
Z
f (y)
Qk,j
p
1
f p
(x) ,
σ(y)dy ≤ inf Mσ
x∈Ek,j
σ(y)
σ
our estimate is controlled by
Z XZ
f p
f p
inf Mσ
Mσ
dσ(x)
(x) σ(x)dx ≤
y∈Ek,j
σ
σ
Rd
k,j Ek,j
Z
.
|f (x)|p σ(x)1−p dx
Rd
Z
=
|f (x)|p w(x)dx.
Rd
The last inequality is from the fact that Mσ is bounded in Lp (σ) because σ is a regular
measure with doubling condition.
5. Weighted norm inequality for p = ∞
In this section we discuss the weight space Ap when p = ∞, and the relationship of
{Ap }1≤p≤∞ .
Definition 6. We say that w ∈ W∞ if there exist δ > 0 and C > 0 such that for all cubes
Q, and measurable subset A of Q,
|A| δ
w(A)
.
≤C
w(Q)
|Q|
Lemma 5.1. Let w ∈ Ap for some 1 ≤ p < ∞. Then
w ∈ A∞ .
Remark. Lemma 5.1 gives
[
Ap ⊂ A∞
1≤p<∞
and is more powerful version of Lemma 3.9.
Proof. By Lemma 3.10 there exists > 0 such that
Z
1 Z
1/(1+)
1
(w(x))1+ dx
.
w(x)dx.
|Q| Q
|Q| Q
22
BAE JUN PARK
Let A be a measurable subset of Q. Then
Z
w(A) =
w(x)dx
A
1/(1+)
Z
/(1+)
(w(x))1+ dx
≤ |A|
A
1 Z
1/(1+)
1/(1+)
/(1+)
= |Q|
|A|
(w(x))1+ dx
|Q| Q
A /(1+)
.
w(Q).
Q
Thus,
w(A) |A| /(1+)
.
.
w(Q)
|Q|
Definition 7. Let µ1 and µ2 be doubling measures. We say that for any cubes Q and any
measurable subset A of Q,
µ1 is comparable to µ2
if there exist 0 < α, β < 1 such that
µ2 (A)
µ1 (A)
≤ α implies
≤β
µ2 (Q)
µ1 (Q)
Lemma 5.2. TFAE.
(1) There exist δ > 0 and C > 0 such that for all cubes Q and all measurable subset A
of Q,
µ (A) δ
µ2 (A)
1
≤C
.
µ2 (Q)
µ1 (Q)
(2) µ2 is comparable to µ1 .
(3) µ1 is comparable to µ2 .
(4) There exists a locally integrable function w w.r.t. µ1 such that
dµ2 (x) = w(x)dµ1 (x)
and
(5.1)
1
µ1 (Q)
Z
1/(1+)
1+
w(x)
dµ1 (x)
.
Q
1
µ1 (Q)
Z
w(x)dµ1 (x)
Q
for some > 0.
Remark. Lemma 5.2 means
w ∈ A∞
⇔ w(A) . w(Q) implies |A| . |Q|
⇔ |A| . |Q| implies w(A) . w(Q)
for any cubes Q and any measurable subset A of Q.
(2) ⇔ (3) implies that Comparability of measures is an equivalence relation.
Proof. (1) ⇒ (2) is trivial by definition.
(2) ⇒ (3). Suppose µ2 is comparable to µ1 . That is, there exist 0 < α, β < 1 such that
µ2 (A)
µ1 (A)
µ2 (A)
1−β
µ2 (Q) > β implies µ1 (Q) > α. If µ2 (Q) ≤ 2 < 1 − β, then
µ2 (Q \ A)
µ2 (A)
=1−
>β
µ2 (Q)
µ2 (Q)
WEIGHTED NORM INEQUALITIES
23
and thus
µ1 (A)
< 1 − α.
µ1 (Q)
0
So µ1 is comparable to µ2 with constants α0 = 1−β
2 and β = 1 − α.
By symmetry it is clear (3) implies (2).
Now show (2) ⇒ (4). Suppose µ2 is comparable to µ1 with constants α and β. Then our
claim is that if µ1 (E) = 0 then µ2 (E) = 0. That is ,
(5.2)
µ2 << µ1 .
Suppose that µ1 (E) = 0 and µ2 (E) > 0. Then there exists an open set Ω such that E ⊂ Ω
and µ2 (Ω) < β1 µ2 (E) because µ2 is regular. Decompose Ω as
[
Ω=
Qj
j
where Qj is non-overlapping cubes. For each j
\
0 = µ1 (Qj
E) ≤ αµ1 (Q).
By our assumption,
µ2 (Qj
\
E) ≤ βµ2 (Qj )
and thus we have
µ2 (E) ≤ βµ2 (Ω) < µ2 (E),
which is a contradiction. So (5.2) follows. By Radon-Nikodym theorem there exists a
locally integrable function w w.r.t. µ1 such that
dµ2 (x) = w(x)dµ1 (x).
Now we prove (5.1). Fix a cube Q and let
λ0 =
µ2 (Q)
µ1 (Q)
and λk =
2d k
λ0 .
α
Observe that
λ0 < λ1 < λ2 < . . .
and
α=
2d λk
.
λk+1
For each λk , consider the maximal dyadic subcubes of Q over which the average of w w.r.t.
µ1 is greater λk . Let {Qk,j }j be a family of such maximal dyadic subcubes and let
Dk :=
∞
[
Qk,j .
j=1
Then for each j
λk <
and for a.e. x ∈
S
j
Qk,j
µ2 (Qk,j )
< 2d λk
µ1 (Qk,j )
c
w(x) ≤ λk .
24
BAE JUN PARK
Since λk+1 > λk , each Qk+1,j is contained in Qk,i for some i. Thus, Dk+1 ⊂ Dk . Then we
have
d
2 λk ≥
=
=
T
µ2 (Qk,j Dk+1 )
µ1 (Qk,i )
Z
1
w(x)dµ1 (x)
µ1 (Qk,i ) Qk,i T Dk+!
Z
X
1
w(x)dµ1 (x)
µ1 (Qk,j )
Qk+1,j
Qk+1,j ⊂Qk,i
=
1
µ1 (Qk,i )
>
1
µ1 (Qk,i )
X
µ2 (Qk+1,j )
Qk+1,j ⊂Qk,j
X
λk+1 µ1 (Qk+1,j )
Qk+1,j ⊂Qk,i
T
µ1 (Qk,i Dk+1 )
= λk+1
µ1 (Qk,i )
Hence for each k
T
µ1 (Qk,i Dk+1 )
≤ α.
µ1 (Qk,i )
Since µ2 is comparable to µ1 , there exists β > 0 such that
µ2 (Qk,i
\
Dk+1 ) ≤ βµ2 (Qk,i ).
Combining these results,
µ2 (Dk+1 ) =
X
µ2 (Qk,i
\
Dk+1 )
i
≤ β
X
µ2 (Qk,i )
i
= βµ2 (Dk ).
By iteration we have proved
µ2 (Dk ) ≤ β k µ2 (D0 ).
(5.3)
Furthermore, since
µ1 (Dk+1 ) =
X
i
µ1 (Qk,i
\
Dk+1 ) ≤ α
X
i
we get by iteration
µ1 (Dk ) ≤ αk µ1 (D0 ).
µ1 (Qk,i ) = αµ1 (Dk ),
WEIGHTED NORM INEQUALITIES
25
By applying (5.3)
Z
(w(x))1+ dµ2 (x)
Q
Z
1+
(w(x))
=
dµ1 (x) +
Q\D0
Z
λ0 w(x)dµ1 (x) +
≤
Q\D0
∞ Z
X
k=0
∞
XZ
k=0
= λ0 µ2 (Q \ D0 ) +
∞
X
(w(x))1+ dµ1 (x)
Dk \Dk+1
(λk+1 ) w(x)dµ1 (x)
Dk \Dk+1
(λk+1 ) µ2 (Dk \ Dk+1 )
k=0
∞
2d X 2d k
≤ (λ0 ) µ2 (Q \ D0 ) +
β) µ2 (D0 ) .
(
α
α
k=0
Choose > 0 sufficiently small that
2 d α
β < 1.
Finally we obtain
Z
(w(x))1+ dµ1 (x) ≤ C(λ0 ) µ2 (Q)
Q
= C
µ2 (Q) µ2 (Q)
µ1 (Q)
and this ends the proof of (5.1).
Next let us prove (4) ⇒ (1). The key idea is Hölder’s inequality and the Reverse Hölder’s
inequality. Suppose that
Z
1 Z
1/(1+)
1+
1
w(x)
dµ1 (x)
w(x)dµ1 (x) < ∞
.
µ1 (Q) Q
µ1 (Q) Q
and
dµ2 (x) = w(x)dµ1 (x).
Let A be any measurable sets in Q. Then
Z
µ2 (A) =
w(x)dµ1 (x)
A
Z
1/(1+)
/(1+)
≤
(w(x))1+ dµ1 (x)
µ1 (A)
A
1 Z
1/1+
1/(1+)
/(1+)
1+
=
(w(x)) dµ1 (x)
µ1 (Q)
µ1 (A)
µ1 (Q) A
Z
1/(1+)
/1+
1
≤ C
w(x)dµ1 (x) µ1 (Q)
µ1 (A)
µ1 (Q) Q
µ (A) 1/(1+)
1
= Cµ2 (Q)
µ1 (Q)
Now it is done by putting δ =
1
1+ .
Theorem 5.3. Let w ∈ L1loc (Rd ) be positive a.e..Then TFAE.
(1) w ∈ Ap for some 1 ≤ p < ∞.
26
BAE JUN PARK
(2) There exist 0 < α, β < 1 such that
|E| ≤ α|Q| ⇒ w(E) ≤ βw(Q)
where E is a measurable subset of Q.
(3) There exist > 0 and C > 0 such that for every cube Q,
Z
1 Z
1+ 1/(1+)
1
≤C
w(x)dx.
w(x)
dx
|Q| Q
|Q| Q
(4) w ∈ A∞ .
Proof. We already know (1) ⇒ (3) ⇒ (2) ⇒ (4). Thus it suffices to show (4) ⇒ (1). Let
w ∈ A∞ . By Lemma 5.2 dx = (w(x))−1 w(x)dx and w(x)dx are comparable and thus there
exist η > 0 and C > 0 such that
Z
1 Z
1/(1+η)
−(1+η)
−1
1
≤ C
w(x)
w(x)dx
w(x) w(x)dx
w(Q) Q
w(Q) Q
= C
|Q|
.
w(Q)
Then we have
1
w(Q)
Z
Q
−η 1/η
|Q|(1+η)/η
w(x)
dx
≤ C
w(Q)(1+η)/η
which implies
1 Z
1/η 1 Z
−η
(w(x)) dx
w(x)dx . 1.
|Q| Q
|Q| Q
Due to Theorem 5.3,
A∞ =
[
Ap .
1≤p<∞
Lemma 5.4. Let w ∈ L1loc (Rd ) be positive a.e.. Then TFAE.
(1) There exist 0 < α, β < 1 such that for every cube Q,
{x ∈ Q : w(x) ≤ αwQ } ≤ β|Q|;
(2) w ∈ A∞ ;
(3) There exists C > 0 such that for every cube Q,
Z
1 Z
1
w(x)dx ≤ Cexp
log w(x)dx .
|Q| Q
|Q| Q
Proof. (1) ⇒ (2). Let Q be a cube and E be a measurable subset of Q. Assume
w(E)
≤γ
w(Q)
for some γ. Split
E = E1
[
˙
E2
where
E1 = {x ∈ E : w(x) > αwQ }
and
E2 = {x ∈ E : w(x) ≤ αwQ }.
Then by assumption,
|E2 | ≤ β|Q|.
WEIGHTED NORM INEQUALITIES
27
Furthermore we have
1
|E1 | ≤
αwQ
Z
w(x)dx =
E
|Q| w(E)
γ
≤ |Q|
α w(Q)
α
Putting together these two estimates
|E| ≤ (β +
Choose γ so small that β +
γ
α
γ
)|Q|.
α
< 1 and put δ = β + αγ . Since
w(E)
|E|
≤γ⇒
≤ δ,
w(Q)
|Q|
by Lemma 5.2 dx and w(x)dx are comparable and the reverse is also true, which is equivalent to w ∈ A∞ .
(2) ⇒ (3). Let w ∈ A∞ . Then w ∈ Ap for some 1 ≤ p < ∞. Now for all p < q < ∞,
w ∈ Aq because Ap ⊂ Aq . That is,
Z
Z
1
q−1
1
− 1
w(x)dx
(w(x)) q−1 dx
≤ C < ∞.
|Q| Q
|Q| Q
Since
lim
q→∞
1
|Q|
Z
1
− q−1
(w(x))
Q
dx
q−1
Z
1
dx q−1
(w(x)−1 ) q−1
q→∞
|Q|
Q
−1 = lim w Lr (Q, dx )
r→0
|Q|
Z
1
= exp
log w(x)−1 dx
|Q| Q
Z
1
= exp −
log w(x)dx ,
|Q| Q
=
lim
we conclude (3).
R
(3) ⇒ (1). Let Q be a cube and we may assume Q log w(x)dx = 0 ( otherwise, divide
w(x) by some constants ). Thus,
w(Q)
≤C
|Q|
With λ > 0 still undetermined,
x ∈ Q : w(x) ≤ λ = x ∈ Q : log (1 + 1/w(x)) ≥ log (1 + 1/λ) Z
1
1
≤
log (1 +
)dx
log (1 + 1/λ) Q
w(x)
Z
1
=
log (w(x) + 1)dx
log (1 + 1/λ) Q
1
≤
w(Q)
log (1 + 1/λ)
1
.
|Q|
log (1 + 1/λ)
. |Q|
28
BAE JUN PARK
for λ small enough because log(1 + w) ≤ w. For α sufficiently small,
x ∈ Q : w(x) ≤ αwQ ≤ x ∈ Q : w(x) ≤ αC ≤ x ∈ Q : w(x) ≤ λ ≤
1
|Q|.
2
Remark. (Relation between A0p s) Let 1 < q < p < ∞. Then
A1 ⊂ Aq ⊂ Ap ⊂ A∞
(5.4)
and
[
Ap =
[
Aq , A∞ =
Aq .
1≤q<∞
1≤q<p
Another observation is that
lim Ap = A1
p&1
and
lim Ap = A∞
p%∞
The meaning of the limits is that Ap -constant goes to A1 -constant and A∞ -constant. The
first limit is from that
Z
Z
q dx 1/q
dx p−1
lim
w(x)−1/(p−1)
= lim
w(x)−1
q→∞
p&1
|Q|
|Q|
Q
Q
−1 = lim w q dx
q→∞
L (Q; |Q| )
= w−1 L∞ (Q; dx )
|Q|
= ess.supQ w
−1
where the second limit is from Lemma
5.4.
We conclude that for w ∈ A1 , wAp is a continuous function in p and
lim wAp = wA∞ .
p→∞
6. Relation between weights and B.M.O.
Theorem 6.1. Let ϕ be a locally integrable real valued function on Rd and 1 < p < ∞.
(1) eϕ ∈ Ap if and only if
Z
1
eϕ(x)−ϕQ dx . 1
|Q| Q
and
1
|Q|
Z
e−(ϕ(x)−ϕQ )/(p−1) dx . 1.
Q
(2) eϕ ∈ A∞ if and only if
1
|Q|
Z
eϕ(x)−ϕQ dx . 1
Q
(3) w ∈ Ap if and only if
w, w−1/(p−1) ∈ A∞ .
WEIGHTED NORM INEQUALITIES
29
Proof. (1) Suppose eϕ ∈ Ap .
Z
Z
1
1
eϕ(x)−ϕQ dx = e−ϕQ
eϕ(x) dx
|Q| Q
|Q| Q
Z
−ϕQ /(p−1) p−1 1
eϕ(x) dx
= (e
)
|Q| Q
1 Z
p−1 1 Z
−ϕ(x)/(p−1)
≤
e
dx
eϕ(x) dx . 1
|Q| Q
|Q| Q
by Jensen’s inequality and our hypothesis.
Moreover,
Z
Z
1
−(ϕ(x)−ϕQ )/(p−1)
ϕQ 1/(p−1) 1
e
dx = e
e−ϕ(x)/(p−1) dx
|Q| Q
|Q| Q
1 Z
1/(p−1) 1 Z
≤
eϕ(x) dx
e−ϕ(x)/(p−1) dx . 1
|Q| Q
|Q| Q
by Jensen’s inequality and our hypothesis
The converse is trivial.
(2) By Lemma 5.4,
ϕ
e ∈ A∞ ⇔
⇔
Z
1
eϕ(x) dx . eϕQ
|Q| Q
Z
1
eϕ(x)−ϕQ dx . 1.
|Q| Q
(3) is immediate from (1) and (2).
Rd .
Corollary 6.2. Let ϕ be locally integrable real valued function on
Then
d
only if there exists a constant C > 0 such that for every cube Q in R
Z
1
e|ϕ(x)−ϕQ | dx . 1.
|Q| Q
Proof. Immediate result with p = 2 from Theorem 6.1 (1).
eϕ
∈ A2 if and
Corollary 6.3. If w ∈ A∞ , then log w ∈ B.M.O.
Proof. Let w ∈ A∞ and write w = eϕ . If w ∈ A2 , then
kϕkB.M.O. = kM# ϕk∞
Z
1
≤ sup
e|ϕ(x)−ϕQ | dx . 1
|Q|
Q
Q
S
Now suppose w ∈ A∞ . Then w ∈ Ap for some 1 ≤ p < ∞ since A∞ = 1≤p<∞ Ap .
When p ≤ 2, w ∈ A2 so we are done. When p > 2, w−1/(p−1) ∈ Ap0 ⊂ A2 and thus
1
− p−1
log w ∈ B.M.O.
Lemma 6.4. (Generalization of Lemma 1.4 with Weight) Let w ∈ A∞ . If f is a function
satisfying
Mf ∈ Lp0 (w)
for some 0 < p0 < ∞, then for every p0 ≤ p < ∞,
Z
Z
p
p
Mf (x) w(x)dx ≤ C
M# f (x) w(x)dx
Rd
for some C > 0.
Rd
30
BAE JUN PARK
7. Weighted Norm inequalities for CZO(α)-First
Theorem 7.1. Let T ∈ CZO(α).
(1) For 1 < p < ∞, if w ∈ Ap , then
T is of strong type (p, p) w.r.t. w.
(2) If w ∈ A1 , then
T is of weak type (1, 1) w.r.t. w.
p
∞
Proof. (1) Let w ∈ Ap . Since L∞
c (w) is dense in L (w), we may prove it with f ∈ Lc . By
Lemma 3.11 there exists s > 1 such that w ∈ Ap/s . Then
Z
Z
p
(M(T f )(x))p w(x)dx
|T f (x)| w(x)dx ≤
d
d
R
ZR
(M# (T f )(x))p w(x)dx
.
Rd
Z
(Ms f (x))p w(x)dx
.
d
ZR
|f (x)|p w(x)dx.
.
Rd
where the second inequality is from Lemma 6.4, the third from Lemma 2.1, and the last
from Theorem 3.13 with A ∈ A p . However, in order to apply Lemma 6.4, we need the
s
p0
p
condition that M(T
f ) ∈ L (w)for some 1 ≤ p0 ≤ p. It suffices to show M(T f ) ∈ L (w)
and since M(T f )Lp (w) . T f Lp (w) , it is done when we have
T f p
(7.1)
< ∞.
L (w)
Thus our goal is to prove (7.1). Suppose that Supp(f ) ⊂ Q. By Lemma 3.10, there exists
> 0 such that
Z
1/(1+)
1 Z
1
1+
(w(x)) dx
.
w(x)dx.
|Q| Q
|Q| Q
Now by Hölder’s inequality
Z
Z
T f (x)p w(x)dx ≤
1+
(w(x))
dx
1/(1+) Z
2Q
2Q
|T f (x)|p(1+)/ dx
/(1+)
.
2Q
The first integral is bounded by a constant, depending on Q and w, due to (3.4). The
second one is also finite because T is bounded in Lq for any 1 < q < ∞.
For another part, we use the property of CZK(α). Then
Z
Z
Z
p
T f (x)p w(x)dx ≤
|f (y)||K(x, y)|dy w(x)dx
(2Q)c
(2Q)c
.
(7.2)
Q
p
|f (y)|
dy
w(x)dx
d
(2Q)c
Q |x − y|
Z
p
1
. f L∞
w(x)dx
dp
(2Q)c |x|
∞ Z
p X
w(x)
. f L∞
dx
dp
k+1
k
(2
Q)\(2 Q) |x|
Z
Z
k=1
∞
p X
. f L∞
k=1
1
2kdp l(Q)kdp
w 2k+1 Q
WEIGHTED NORM INEQUALITIES
31
By Lemma 3.11 there exists 1 < q < p such that w ∈ Aq , and by (3.4)
|Q| q
≤ Cw(Q),
w(2k+1 Q) k+1
|2 Q|
which gives
w(2k+1 Q) . 2kdq w(Q).
Finally,
∞
p w(Q) X
|T f (x)| w(x)dx . f L∞
2−kd(p−q) < ∞.
|Q|p
(2Q)c
Z
p
k=1
(2) The proof is very similar to the proof of weak (1,1) boundedness of CZO(α) without
weights. Thus, the key idea is Calderón-Zygmund decomposition of f . We want to prove
that
Z
1
d
w({x ∈ R : |T f (x)| > t}) ≤ C
|f (x)|w(x)dx.
t Rd
Decompose f as
f =g+b
where g and b are ”good” function and ”bad” function at height t. Then it suffice to
estimate
w({x ∈ Rd : |T g(x)| > t})
and
w({x ∈ Rd : |T b(x)| > t}).
Since w ∈ A1 , it is also in A2 . By (1),
d
w({x ∈ R : |T g(x)| > t}) ≤
.
.
Z
1
|T g(x)|2 w(x)dx
t2 Rd
Z
1
|g(x)|2 w(x)dx
t2 Rd
Z
2d
|g(x)|w(x)dx
t Rd
Let
S {Qjc} be the family of cubes for Calderón-Zygmund decomposition. Then f = g on
( j Qj ) . Thus we need to prove that for each Qj
Z
Z
|g(x)|w(x)dx .
|f (y)|w(y)dy,
Qj
Qj
but this is true because
Z
Z
|g(x)|w(x)dx ≤
Qj
Qj
1
|Qj |
Z
Z
|f (y)|
=
Qj
|f (y)|dyw(x)dx
Qj
w(Qj )
dy
|Qj |
Z
≤
|f (y)|Mw(y)dy
Qj
Z
|f (y)|w(y)dy.
.
Qj
32
BAE JUN PARK
For a bad function b,
[
w {x ∈ Rd : |T b(x) > t|} ≤ w {x ∈
(2Qj ) : |T b(x) > t|}
j
+w {x ∈ Rd \
[
(2Qj ) : |T b(x) > t|}
j
and the first term is dominated by
Z
[
X
X
X w(Qj )
w(Qj )
1X
w
(2Qj ) ≤
w(2Qj ) .
w(Qj ) .
|Qj | .
|f (y)|
dy
|Qj |
t
|Qj |
Qj
j
j
j
j
j
Z
Z
1X
1
|f (y)|w(y)dy.
.
|f (y)|w(y)dy .
t
t Rd
Qj
j
Let cj be the center of Qj . Then
[
w {x ∈ Rd \ (2Qj ) : |T b(x) > t|}
j
.
.
.
.
.
.
.
.
Z
1X
t
j
1
t
(2Qj )c
XZ
j
T bj (x)w(x)dx
Z
|K(x, y) − K(x, cj )||bj (y)|dydx
(2Qj )c
Qj
|y − cj |δ
w(x)dx
dy
d+δ
Qj
(2Qj )c |x − cj |
j
Z
∞ Z
X
1X
1
w(x)dxdy
|bj (y)|l(Qj )δ
t
k+1 Q )\(2k Q ) |x − cj |
Q
(2
j
j
j
j
k=−1
Z
∞
X
X
1
1
1
δ
|bj (y)|l(Qj )
Mw(y)dy
kδ
t
2 l(Qj )δ
Qj
j
k=−1
Z
1
|b(y)|w(y)dy
t Rd
Z
1
|f (y)| + |g(y)| w(y)dy
t Rd
Z
1
|f (y)|w(y)dy.
t Rd
1X
t
Z
Z
|bj (y)|
If T is a regular singular integral operator with kernel K and f ∈ Lp (w) with w ∈ Ap
and 1 0, can be defined acting directly on
f by means of the formula
Z
T f (x) =
K(x, y)f (y)dy.
|x−y|>
Indeed, this integral converges absolutely, since
Z
|K(x, y)||f (y)|dy
|x−y|>
≤
Z
Rd
|f (y)|p w(y)dy
1/p Z
|x−y|>
0
0
|K(x, y)|p w(y)−p /p dy
1/p0
WEIGHTED NORM INEQUALITIES
33
and it is clear that the right hand side is finite since the second factor is bounded by a
constant times
1/p0
Z
0
,
w(x − y)1/(p−1) |y|−dp dy
|y|>
which is finite because w(x − y)−1/(p−1) is an Ap0 as a function y and we apply the idea of
(7.2).
Of course after Theorem 7.1, T f may be defined by density. However it is natural to ask
whether a more explicit definition of T f as a limit of T f as → 0 is possible.
Theorem 7.2. Let T ∈ CZO(α), 1 < p < ∞ and w ∈ Ap . Then the corresponding
maximal operator T ∗ arising from the truncated operators T is bounded in Lp (w). More
precisely, there is a constant C depending on T and w, such that for every f ∈ Lp (w)
Z
Z
p
∗
T f (x) w(x)dx ≤ C
|f (x)|p w(x)dx.
Rd
Rd
Proof. First, observe that for all 1 < t < ∞ if w ∈ Ap , then Lt (w) ⊂ L1loc (Rd ). This is
because
Z
Z
|f (x)|dx =
|f (x)|w(x)1/p w(x)−1/p dx
Q
Q
Z
1/t Z
1/t0
0
≤
|f (x)|t w(x)dx
w(x)−t /t dx
Q
Q
Z
(t−1)/t
= kf kLt (w)
w(x)−1/(t−1) dx
<∞
Q
by w ∈ At .
For any 1 < q < p, w ∈ Ap/q by (5.4). Let f ∈ Lp (w). Then |f |q ∈ Lp/q (w) ⊂ L1loc (Rd )
by the above argument with t = p/q > 1. Therefore f ∈ Lqloc (Rd ) for all 1 < q < ∞.
By Theorem 7.2,
T ∗ f (x) ≤ Cq Mq f (x) + CM(T f )(x).
Then
Z
1/p
∗
T f (x)p w(x)dx
Rd
≤
Cq
Z
1/p
Z
p
Mq f (x) w(x)dx
+C
Rd
1/p
p
M(T f )(x) w(x)dx
Rd
:= I + II
By Theorem 3.13(or Theorem 4.1) with w ∈ Ap/q ,
Z
1/p
p/q
I =
M(|f |q )(x)
w(x)dx
Rd
Z
1/p
≤ C
|f (x)|p w(x)dx
Rd
By Theorem 3.13 and Theorem 7.1,
Z 1/p
T f (x)p w(x)dx
II ≤ C
Rd
Z
1/p
≤ C
|f (x)|p w(x)dx
.
Rd
34
BAE JUN PARK
Theorem 7.3. [3] Let T ∈ CZO(α) and u be a positive function. Then for any > 0 and
1<p<∞
(1)
Z
Z
T f (x)p u(x)dx ≤ C,p
|f (x)|p M1+ u(x)dx.
Rd
Rd
(2)
Z
Rd
∗
T f (x)p u(x)dx ≤ C,p
Z
|f (x)|p M1+ u(x)dx.
Rd
for arbitrary functions f .
Proof. We may assume u1+ is locally integrable because M1+ u(x) = ∞ for all x if not.
By Theorem 3.12, M1+ u ∈ A1 ⊂ Ap . Thus,
Z
Z
T f (x)p u(x)dx ≤
T f (x)p M1+ u(x)dx
Rd
Rd
Z
|f (x)|p M1+ u(x)dx
≤ C,p
Rd
where the last inequality is from Theorem 7.1. Also,
Z
Z
∗
∗
T f (x)p M1+ u(x)dx
T f (x)p u(x)dx ≤
Rd
Rd
Z
≤ C,p
|f (x)|p M1+ u(x)dx
Rd
by Theorem 7.2.
8. Orlicz Spaces (based in [1])
Definition 8. (Young’s function) Let φ : [0, ∞) → [0, ∞) be increasing and left-continuous
with φ(0) = 0. Suppose φ(x) > 0 for x > 0. Then the function Φ defined by
Z s
Φ(s) =
φ(u)du
0≤s
0
is said to be a Young’s function.
Definition 9. (Complementary Young’s function) Let Φ be a Young’s function. Let
ψ(v) := inf {u : φ(u) ≥ v}
0 ≤ v.
Then the function
Z
Ψ(t) :=
t
ψ(v)dv
0≤t
0
is called the complementary Young’s function of Φ. The function ψ is called the leftcontinuous inverse of φ.
Note that the function ψ is increasing, left-continuous, ψ(0) = 0, and ψ(t) > 0 for t > 0.
Thus, Ψ is indeed a Young’s function. Also clearly
φ(u) = inf{v : ψ(v) ≥ u},
which implies Φ is the complementary Young’s function of Ψ. In conclusion, we may refer
to Φ and Ψ simply as complementary Young’s functions.
WEIGHTED NORM INEQUALITIES
35
Theorem 8.1. Let Φ and Ψ be complementary Young’s functions. Then
st ≤ Φ(s) + Ψ(t)
for 0 ≤ s, t < ∞.
Definition 10. Let Φ be a Young’s function. Then define the right-containuous inverse
Φ−1 of Φ to be
Φ−1 (t) = sup{s : Φ(x) ≤ t}
for 0 ≤ t < ∞.
Note that s = Φ−1 (t) if and only if t = Φ(s) for a Young’s function Φ.
Lemma 8.2. Let Φ and Ψ be complementary Young’s functions. Then
(8.1)
w ≤ Φ−1 (w)Ψ−1 (w) ≤ 2w
for 0 ≤ w < ∞.
Proof. It suffice to establish (8.1) for 0 < w < ∞ because Φ−1 and Ψ−1 are right continuous
on [0, ∞). Set s = Φ−1 (w) and t = Ψ−1 (w). Then Φ(s) = w = Ψ(t) and the second
inequality follows from Theorem 8.1. The first estimate will follow from the inequality
n φ(u) ψ(v) o
(8.2)
,
≤1
min
v
u
for all u, v > 0. To establish (8.2), simply note that if φ(u) ≥ v, then ψ(v) ≤ u by the
definition of ψ. Thus,
w = min Φ(s), Ψ(t) ≤ min{sφ(s), tψ(t)} ≤ st.
Let Φ and Ψ be complementary Young’s functions. Suppose Φ(1) = 1 and Φ satisfies the
doubling condition
(8.3)
Φ(2t) ≤ CΦ(t)
t>k
for some C > 0 and k ≥ 0. Then the doubling condition implies that
(8.4)
Φ(t) ≈ tφ(t)
t>0
and thus the mapping t 7→ Φ(t)/t(≈ φ(t)) is increasing.
We want to define a scale of maximal-type operators w 7→ MΦ w such that
(8.5)
Mw(x) ≤ MΦ w(x) ≤ Mr w(x)
for each x ∈ Rd . To define MΦ we introduce, for each cube Q, the Φ-average of a function
f over Q by means of the following Luxemburg norm:
Z
n
|f (y)| o
1
kf kΦ,Q := inf λ > 0 :
Φ
dy ≤ 1 .
|Q| Q
λ
We define a maximal operator
MΦ f (x) = sup kf kΦ,Q
x∈Q
and
Mdyad
f (x) = sup kf kΦ,Q
Φ
x∈Q∈D
where f is a locally integrable function.
Clearly for nonnegative functions f and g, if f ≤ g then
kf kΦ,Q ≤ kgkΦ,Q .
and note that k · kΦ,Q is indeed a norm:
36
BAE JUN PARK
(1) For a constant c kcf kΦ,Q = |c|kf kΦ,Q ;
(2) Let u = kf kΦ,Q and v = kgkΦ,Q . Then observe that
Z
f (y) 1
Φ
dy ≤ 1
|Q| Q
u
and
1
|Q|
Z
Φ
g(y) Q
v
dy ≤ 1
because Φ is a continuous and increasing function. By using a convexity of Φ,
Z
f (y) + g(y) 1
Φ
dy
|Q| Q
u+v
Z
Z
f (y) g(y) 1
1
v
u
Φ
Φ
dy +
dy
≤
u + v |Q| Q
u
u + v |Q| Q
v
u
v
≤
+
=1
u+v u+v
, which implies
kf + gkΦ,Q ≤ kf kΦ,Q + kgkΦ,Q .
Lemma 8.3. (Hölder’s inequality) Let Φ and Ψ be complementary Young’s functions. Let
f and g be locally integrable functions. Then for y ∈ Rd
M(f g)(y) ≤ 2MΦ f (y)MΨ g(y).
Proof. By Theorem 8.1,
f (y) g(y) f (y) g(y)
≤Φ
+Ψ
kf kΦ,Q kgkΨ,Q
kf kΦ,Q
kgkΨ,Q
and by taking expectation on Q
Z
1
1
f (y)g(y)dy
kf kΦ,Q kgkΨ,Q |Q| Q
Z
Z
f (y) g(y) 1
1
≤
Φ
dy +
Φ
dy
|Q| Q
kf kΦ,Q
|Q| Q
kgkΨ,Q
≤ 2.
Lemma 8.4. Let f be a non-negative bounded function with compact support. For each
t > 0, let
Ωt = y ∈ Rd : MΦ f (y) > t
and
Dt = y ∈ Rd : Mdyad
f (y) > t .
Φ
Suppose Ωt is not empty. Then there is a family {Qj } of non-overlapping maximal dyadic
cubes satisfying
[
(8.6)
Dt/4d =
Qj ,
j
(8.7)
Ωt ⊂
[
3Qj ,
j
(8.8)
t/4d < kf kΦ,Qj ≤ t/2d ,
WEIGHTED NORM INEQUALITIES
37
and
(8.9)
Ωt ≤ C
Z
Φ
f (y) {y∈Rd :f (y)>t/2}
t
dy.
Proof. The proof is a simple adaptation of the proof of Calderón-Zygmund cube decomposition. Suppose the support of f is contained in a compact set K. Since f (x) ≤ kf kL∞ χK ,
for each cube Q
kf kΦ,Q ≤ kf kL∞ χK Φ,Q .
By the definition of k · kΦ,Q ,
kχK kΦ,Q
(8.10)
Z
χ (y) 1
K
Φ
= inf λ > 0 :
dy ≤ 1
|Q| Q
λ
1 |Q ∩ K|
= inf λ > 0 : Φ
≤1
λ
|Q|
|Q| 1
≤
= inf λ > 0 : Φ
λ
|Q ∩ K|
|Q| i−1
h
= Φ−1
.
|Q ∩ K|
Thus we have
h
kf kΦ,Q ≤ kf kL∞ Φ−1
|Q| i−1
→0
|Q ∩ K|
as |Q| → ∞. Therefore if there are any dyadic cubes Q with kf kΦ,Q > t then they are
contained in cubes of this type which are maximal with respect to inclusion. Let Ct = Pj
fj
be the family of the dyadic maximal non-overlapping cubes satisfying t < kf kΦ,Pj . Let P
be the only dyadic cube containing Pj with side length twice that of Pj . Then
t < kf kΦ,Pj ≤ 2d kf kΦ,Pfj
where the second inequality follows since
f (y) 1 f (y) Φ
≤ dΦ
,
λ
2
λ/2d
which is from the fact that Φ(t)/t is an increasing function. By the maximality, we have
(8.11)
t < kf kΦ,Pj ≤ 2d t.
Observe that
(8.12)
{y ∈ Rd : Mdyad
f (y) > t} =
Φ
[
Pj .
j
Now let x ∈ Ωt . By definition of Ωt , there is a cube(may not be a dyadic cube) R
containing x such that t < kf kΦ,R . Let k be the unique integer such that 2−(k+1)d < |R| ≤
2−kd . Then there are at most 2d cubes of side length 2−k meeting the interior of R, say
{Ji : i = 1, 2, · · · 2d }. At least one of such cubes, say J1 , satisfies
t/2d < f χJ1 Φ,R
38
BAE JUN PARK
P d
P2d
(If not, kf kΦ,R = 2i=1 f χJi Φ,R ≤
i=1 kf χJi kΦ,R ≤ t, which is a contradiction).
d
Clearly |R| ≤ |J1 | ≤ 2 |R|. Then
Z
f (y) o
n
1
Φ
kf kΦ,J1 = inf λ > 0 :
dy ≤ 1
|J1 | J1
λ
Z
n
o
f (y) 1
Φ
≥ inf λ > 0 :
dy ≤ 1
|J1 | J1 ∩R
λ
Z
n
o
1
f (y)χJ1 (y) 1
≥
Φ
inf λ > 0 :
dy ≤ 1
|R| R
λ
2d
1
t
=
(8.13)
kf χJ1 kΦ,R > d
d
2
4
because Φ(t)/t is a increasing function.
By letting Ct/4d = {Qj }, (8.6) and (8.8) holds by (8.12) and (8.11). By (8.13) J1 ⊂ Qk
P
for some k and then R ⊂ 3J1 ⊂ 3Qk . This gives (8.7). Clearly |Ωt | ≤ C j |Qj | and
Z
f (y) 1
Φ
dy > 1,
|Qj | Qj
t/4d
and thus
|Ωt | ≤ C
XZ
Φ
Qj
j
XZ
4d f (y) t
f (y) dy
dy
t
j
Z
f (y) .
Φ
(8.14)
dy
t
Rd
where the second inequality is from the doubling condition of Φ.
Let f1 (x) = f (x) if f (x) > t/2 and vanishes otherwise and let f2 = f − f1 . It is clear
that
MΦ f (x) ≤ MΦ f1 (x) + MΦ f2 (x) ≤ MΦ f1 (x) + t/2.
Apply (8.14) and doubling condition to get
|Ωt | ≤ {y ∈ Rd : MΦ f1 (y) > t/2}
Z
f (y) 1
.
Φ
dy
t/2
d
R
Z
f (y) 1
.
Φ
dy
t
Rd
Z
f (y) =
Φ
dy.
t
|{y∈Rd :f (y)>t/2}|
.
Φ
Qj
Definition 11. Let 1 < p < ∞. We say that a doubling Young’s function Φ satisfies the
Bp -condition if there is a positive constant c such that
Z ∞
Z ∞ p0 p−1 dt
Φ(t) dt
t
≈
<∞
p
t t
Ψ(t)
t
c
c
where Ψ is the complementary Young’s function of Φ.
Theorem 8.5. [8] Let 1 < p < ∞. Suppose that Φ is a doubling Young’s function. Then
the followings are equivalent.
WEIGHTED NORM INEQUALITIES
(1) Φ ∈ Bp ;
(2) there is a constant c such that
Z
Z
p
MΦ f (y) dy ≤ c
Rd
39
|f (y)|p dy
Rd
for a function f ;
(3) there is a constant c such that
Z
Z
p
MΦ f (y) w(y)dy ≤ c
|f (y)|p Mw(y)dy
Rd
Rd
for a function f and a weight w;
(4) there is a constant c such that
Z
Z
p
w(y)
Mw(y)
p dy ≤ c
Mf (y)
|f (y)|p
dy
1/p
u(y)
MΨ (u )(y)
Rd
Rd
for a function f and weights w and u.
Proof. Assume (1). To prove (2) we shall use the classical approach.
Z
Z ∞ p
dt
MΦ f (y) dy = p
tp y ∈ Rd : MΦ f (y) > t t
Rd
0
Z
Z ∞
f (y) dt
Φ
.
tp
dy
t
t
{y:f (y)>t/2}
0
Z Z 2f (y)
f (y) dt
=
tp Φ
dy
t
t
d
R
0
Z Z ∞
f (y) s
(f (y))p
Φ(s) 2
dsdy
=
p
s
s f (y)
d
R
1/2
Z
Z ∞ Φ(s) ds =
f (y)p
dy
p s
1/2 s
Rd
Z
.
f (y)p dy
Rd
2d and
Now assume (2). Fix a constant a >
for each k let
Ωk = y ∈ Rd : MΦ f (y) > ak
and
f (y) > ak /4d .
Dk = y ∈ Rd : Mdyad
Φ
By Lemma 8.4 there exist a family of maximal non-overlapping dyadic cubes {Qk,j } for
which
[
(8.15)
Ωk ⊂
Qk,j ,
j
(8.16)
Dk =
[
Qk,j ,
j
and
(8.17)
ak /4d < kf kΦ,Qk,j ≤ ak /2d .
Let Ek,j = Qk,j − Qk,j ∩ Dk+1 . Then our claim is that {Ek,j }k,j is a disjoint family of sets
which satisfy
d
Qk,j ∩ Dk+1 < 2 |Qk,j |
(8.18)
a
40
BAE JUN PARK
and
1
|Ek,j |.
1 − 2d /a
|Qk,j | <
(8.19)
The family is clearly disjoint. Due to (8.17) we have
1
1<
|Qk,j |
(8.20)
Z
Φ
Qk,j
f (y) dy
ak /4d
and
1
|Qk,j |
Z
Φ
Qk,j
f (y) dy ≤ 1
ak /2d
Thus,
|Qk,j ∩ Dk+1 |
|Qk,j |
=
=
X |Qk,j ∩ Qk+1,i |
|Qk,j |
i
X
|Qk+1,i |
|Qk,j |
Z
f (y) 1
Φ k+1 d dy
|Qk,j | Qk+1,i
a /4
Z
f (y) 1
1
Φ
dy
|Qk,j | Qk+1,i a/2d
ak /2d
Qk+1,i ⊂Qk,j
X
<
Qk+1,i ⊂Qk,j
X
≤
Qk+1,i ⊂Qk,j
≤
2d 1
a |Qk,j |
Z
Φ
Qk,j
f (y) dy
ak /2d
d
≤ 2 /a
where the first inequality is from (8.20), the second one is from the fact that the mapping
1
t 7→ Φ(t)
t is increasing and a/2d , and the last one follows from (8.21), which implies (8.18)
and (8.19) is a consequence of (8.18).
WEIGHTED NORM INEQUALITIES
Therefore
Z
p
MΦ f (y) w(y)dy =
Rd
XZ
MΦ f (y)p w(y)dy
Ωk \Ωk+1
k
≤
41
XZ
a(k+1)p w(y)dy
Ωk \Ωk+!
k
≤ ap
X
p
X
akp w(Ωk )
k
≤ a
akp w(3Qk,j )
k,j
≤ 3d 4dp ap
X
= 3d 4dp ap
X
kf kpΦ,Qk,j
k,j
k,j
<
≤
≤
≤
.
3d 4dp ap
w(3Qk,j )
|Qk,j |
|3Qk,j |
w(3Q ) 1/p p
k,j
|Qk,j |f ·
|3Qk,j |
Φ,Qk,j
w(3Q ) 1/p p
k,j
|E
|
k,j f ·
|3Qk,j |
1 − 2d /a
Φ,Qk,j
k,j
Z
1/p 3d 4dp ap X
p
f
·
Mw
dy
1 − 2d /a
Φ,Qk,j
k,j Ek,j
Z
1/p p
3d 4dp ap X
MΦ f · Mw
(y) dy
d
1 − 2 /a
k,j Ek,j
Z
1/p p
3d 4dp ap
MΦ f · Mw
(y) dy
d
1 − 2 /a Rd
Z
f (y)p Mw(y)dy
X
Rd
f
Now assume that (3) holds. By setting g = u1/p
and h = u1/p , (4) is equivalent to
Z
Z
w(y)
p dy .
M(gh)(y)p
g(y)p Mw(y)dy
d
d
M
h(y)
R
R
Ψ
By Lemma 8.3
M(gh)(y) ≤ 2MΦ g(y)MΨ h(y),
which implies the desired result.
To prove that (4) implies (1) we let w = 1 to obtain
Z
Z
1
1
Mf (y)p
dy
.
f (y)p
dy
1/p )(y)p
u(y)
d
d
M
(u
Ψ
R
R
e be the dilate of K by a factor of 2.
Let K be the unit cube centered at the origin and K
Choose f = u = χK . Then
Z
1
1 &
MχK (y)p
dy
MΨ χK (y)p
Rd
Z
1
≥
MχK (y)p
dy
MΨ χK (y)p
ec
K
42
BAE JUN PARK
ec
For y ∈ K
MχK (y) ∼
1
|y|d
and
MΨ χK (y ∼
1
Ψ−1 (|y|d )
by (8.10). Thus
Z
1
Ψ−1 (|y|d )p dy
e c |y|dp
K
Z ∞
1 −1 p
Ψ (s) ds
≈
sp
c
Z ∞ p0 p−1 dt
t
≈
Ψ(t)
t
c
1 &
by change of variables Ψ−1 (s) = t and the fact that Ψ(t) ≈ tψ(y).
Theorem 8.6. [7] Let 1 0. Then there exists a constant C > 0 such that for each weight w
Z
(8.22)
Rd
T f (y)p w(y)dy ≤ C
Z
|f (y)|p MA w(y)dy.
Rd
Proof. Assume MA w is finite almost everywhere. Let Ψ(t) = A(tp ). Then Ψ is also a
doubling Young’s function. Let Φ be the complementary Young’s function of Ψ. By using
change of variables, the condition (8.21) is equivalent to
Φ ∈ Bp 0 .
By Theorem 8.5,
Z
Mf (y)
p0
v(y)
MΨ u(y)
Rd
Z
p0 dy ≤ C
f (y)p
Rd
0
Mv(y)
dy.
u(y)
for any positive functions f, u, v. Setting v ≡ 1, u = w1/p and using the fact that MΨ u =
1/p
MA (up )
, this would be
Z
p0
Mf (y)
(8.23)
Rd
1−p0
MA w(y)
dy ≤ C
Z
Rd
0
0
f (y)p w(y)1−p dy
WEIGHTED NORM INEQUALITIES
43
Now let T ∗ be the adjoint operator of T . Then T ∗ ∈ CZO(α) and
Z 1/p
T f (y)p w(y)dy
= T f · w1/p Lp
Rd
=
sup T f · w1/p , g 0 ≤1
Lp
kgk
=
sup
f, T ∗ (g · w1/p ) 0 ≤1
Lp
kgk
≤
Z
1/p
|f (y)|p MA w(y)dy
Rd
Z 1/p0
0
0
T ∗ (g · w1/p )(y)p MA w(y) −p /p dy
× sup
0 ≤1
Lp
kgk
Rd
Thus, (8.22) holds if we show
Z
Z
∗
0
0
T f (y)p MA w(y) 1−p dy ≤ C
(8.24)
Rd
0
0
|f (y)|p w(y)1−p dy
Rd
By using the same argument in Theorem 3.5, for 0 < δ < 1
δ
MA w ∈ A1 .
Since u1−r ∈ Ar for any u ∈ A1 , we have
1−p0 (p0 −1)/(r−1) 1−r
MA w
= MA w
∈ Ar ⊂ A∞
for r > p0 . By Coifman and Fefferman’s inequality and (8.23), we have
Z
Z
0
∗
0
p 0
1−p0
T f (y)p MA w(y) 1−p dy ≤ Cp
Mf (y)
MA w(y)
dy
d
Rd
ZR
0
0
≤ Cp
f (y)p w(y)1−p dy,
Rd
which is (8.24).
Theorem 8.7. Let 1 0 let
A (t) := t log (1 + t) .
Then there exists a constant C > 0 such that for each weight w and all λ > 0 we have
Z
C
d
(8.25)
|f (y)|MA w(y)dy
w y ∈ R : |T f (y)| > λ ≤
λ Rd
Proof. Refer to [7].
9. Weighted Norm inequalities for CZO(α)- Second
Theorem 9.1. [7] Let T ∈ CZO(α) and 1 0
such that for each weight w
Z
Z
p
(9.1)
T f (x) w(x)dx ≤ C
|f (x)|p M[p]+1 w(x)dx
Rd
Rd
where [p] is the integer part of p. Furthermore, the result is sharp since it does not hold for
M[p] .
44
BAE JUN PARK
Proof. Let us assume that M[p]+1 w is finite almost everywhere since otherwise it is trivial.
Let
[p]
A(t) = t log(1 + t) .
Then A(t) is a doubling Young’s function satisfying (8.21). By Theorem 8.6,
1/p
Z Z
1/p
T f (y)p w(y)dy
|f (y)|p MA w(y)dy
≤C
Rd
Rd
Then (9.1) follows if we prove the pointwise inequality
MA w(x) ≤ CM[p]+1 w(x)
and it suffices to show for each cube Q
kf kA,Q
(9.2)
C
≤
|Q|
Z
M[p] w(x)dx.
Q
By assumption the right-hand side average is finite and by homogeneity we may assume
that it is equal to one. Then by the definition of k · kA,Q we need to prove that
Z
[p]
1
w(y) log(1 + w(y)) dy ≤ C
|Q| Q
But this is a consequence of iterating the following inequality of Stein [9]:
Z
Z
k
k−1
(9.3)
w(y) log(1 + w(y)) dy ≤ C
Mw(y) log(1 + Mw(y))
dy.
Q
Q
To show the sharpness we need to show that for arbitrary 1 < p < ∞ the inequality
Z
Z
T f (x)p w(x)dx ≤ C
|f (x)|p M[p] w(x)dx
Rd
Rd
is false in general. We consider the Hilbert transform
Z
f (y)
dy.
Hf (x) = p.v.
x
−y
R
By using the same argument to get (8.24), we may show that the inequality
Z
Z
p0
−(p0 −1)
−(p0 −1)
0
[p]
Hf (x) M w(x)
(9.4)
dx ≤ C
|f (x)|p w(x)
dx
R
R
is false. Let f = w = χ(−1,1) . By a standard computation for k = 1, 2, . . .
k−1
log (1 + |x|)
k
M f (x) ≈
|x|
for |x| > e and for large x > e observe that Hχ(−1,1) (x) & 1/x. Thus the left hand side of
(9.4) diverges to infinity while the right hand side is bounded by a constant.
Theorem 9.2. [7] Let T ∈ CZO(α). Then there exists a constant C > 0 such that for
each weight w and for all λ > 0
Z
C
d w {x ∈ R : T f (x) > λ} ≤
|f (x)|M2 w(x)dx.
λ Rd
Proof. By Theorem 8.7 it suffices to show
MA w(x) . M2 w(x).
By definition we need to prove that for each Q
Z
C
kf kA ,Q ≤
Mw(x)dx.
|Q| Q
WEIGHTED NORM INEQUALITIES
45
As in the proof of Theorem 9.1 we may assume the right hand side is equal to 1 and show
Z
1
|w(y)| log (1 + |w(y)|) dy . 1,
|Q| Q
but this is true by (9.3) and so we are done.
References
[1] C. Bennett, R. Sharpley Interpolation of operators, Academic Press, New York (1988).
[2] M. Christ, R. Fefferman A note on weighted norm inequalities for the Hardy-Littlewood maximal operator, Proc. Amer. Math. Soc. 87 (1983) 447-448.
[3] A. Córdoba, C. Fefferman A weighted norm inequality for singular integrals, Studia Math. 57 (1976)
97-101.
[4] C. Fefferman, E. M. Stein Some maximal inequalities, Amer. J. Math. 93 (1971) 107-115.
[5] C. Fefferman, E. M. Stein, H p spaces of several variables, Acta Math. 129 (1972) 137-193.
[6] J. Garcia-Cuerva, J. L. Rubio De Francia Weighted norm inequalities and related topics, North-Holland
Mathematical Studies 116 (1985).
[7] C. Pérez Weighted norm inequalities for singular integral operators, J. London Math. Soc. (2) 49(2)
(1994) 296-308.
[8] C. Pérez On sufficient conditions for the boundedness of the Hardy-Littlewood maximal operator between
weighted Lp -spaces with different weights, Proc. London Math. Soc. (3) 71 (1995) 135-157.
[9] E. M. Stein Note on the class L log L, Studia Math. 32 (1969) 305-310.
Department of Mathematics, University of Wisconsin - Madison, , U.S.A.
E-mail address: [email protected]

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