Tài liệu Đề tài " Uniform bounds for the bilinear Hilbert transforms, I " ppt

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892 LOUKAS GRAFAKOS AND XIAOCHUN LI
pends only on the exponents p
1
,p
2
and is independent of any other parameter,
in particular of the parameter m. Finally N will denote a large (but fixed)
integer whose value may be chosen appropriately at different times.
Acknowledgments. The authors would like to thank M. Lacey for many
helpful discussions during a visit at the Georgia Institute of Technology. They
are grateful to C. Thiele for his inspirational work [10] and for some help-
ful remarks. They also thank the referee for pointing out an oversight in a
construction in the first version of this article.
2. The decomposition of the bilinear operator T
m
We begin with a decomposition of the half plane η<2
m
ξ on the ξ-η plane.
We can write the characteristic function of the half plane η<2
m
ξ as a union
of rectangles of size 2
−k
× 2
−k+m
as in Figure 1. Precisely, for k, l ∈ Z we set
J
(1)
1
(k, l)=[2
−k
(2l), 2
−k
(2l + 1)],J
(1)
2
(k, l)=[2
−k+m
(2l − 2), 2
−k+m
(2l − 1)],
J
(2)
1
(k, l)=[2
−k
(2l +1), 2
−k
(2l + 2)],J
(2)
2
(k, l)=[2
−k+m
(2l − 2), 2
−k+m
(2l − 1)],
J
(3)
1
(k, l)=[2
−k
(2l +1), 2
−k
(2l + 2)],J
(3)
2
(k, l)=[2
−k+m
(2l − 1), 2
−k+m
(2l)].
We call the rectangles J
(r)
1
(k, l) × J
(r)
2
(k, l) of type r, r ∈{1, 2, 3}.Itis
easy to see that
1
η<2
m
ξ
=

k∈Z

l∈Z

1
J
(1)
1
(k,l)
(ξ)1
J
(1)
2
(k,l)
(η)
+1
J
(2)
1
(k,l)
(ξ)1
J
(2)
2
(k,l)
(η)+1
J
(3)
1
(k,l)
(ξ)1
J
(3)
2
(k,l)
(η)

,
which provides a (nonsmooth) partition of unity of the half-plane η<2
m
ξ.
Next we pick a smooth partition of unity {Ψ
(r)
k,l
(ξ,η)}
k,l,r
of the half-plane
η<2
m
ξ with each Ψ
(r)
k,l
supported only in a small enlargement of the rectangle
J
(r)
1
(k, l) × J
(r)
2
(k, l) and satisfying standard derivative estimates. Since the
functions Ψ
(r)
k,l
(ξ,η) are not of tensor type, (i.e. products of functions of ξ and
functions of η) we apply the Fourier series method of Coifman and Meyer [4,
pp. 55–57] to write
Ψ
(r)
k,l
(ξ,η)=

n∈Z
2
C(n)(Φ
(r)
1,k,l,n
)(ξ)(Φ
(r)
2,k,l,n
)(η)
where |C(n)|≤C
M
(1 + |n|
2
)
−M
for all M>0(n =(n
1
,n
2
), |n|
2
= n
2
1
+ n
2
2
)
and the functions Φ
(r)
1,k,l,n
and Φ
(r)
2,k,l,n
are Schwartz and satisfy:
UNIFORM BOUNDS FOR THE BILINEAR HILBERT TRANSFORMS, I
893
rectangle size
2
−k
× 2
−k+m
ξ
rectangle
tof ype 2
rectangle
tof ype 1
rectangle
tof ype 3
η
η =2
m
ξ − 2
m−k 1
η =2
m
ξ
θ = arctan(2
−m
)
+
Figure 1: The decomposition of the plane η<2
m
ξ.
(2.1)
|D
α
((Φ
(r)
1,k,l,n
))|≤C
α
(1 + |n|)
α
2
αk
, supp (Φ
(r)
1,k,l,n
)⊂(1+2
−2L
)J
(r)
1
(k, l),
(Φ
(r)
1,k,l,n
)(ξ)=e
2πin
1
2
k
(ξ−c(J
(r)
1
(k,l)))
on (1 − 2
−2L
)J
(r)
1
(k, l),
(2.2)
|D
α
((Φ
(r)
2,k,l,n
))|≤C
α
(1 + |n|)
α
2
α(k−m)
, supp (Φ
(r)
2,k,l,n
)⊂(1+2
−2L
)J
(r)
2
(k, l),
(Φ
(r)
2,k,l,n
)(η)=e
2πin
2
2
k−m
(η−c(J
(r)
2
(k,l)))
on (1 − 2
−2L
)J
(r)
2
(k, l),
for all nonnegative integers α and all r ∈{1, 2, 3}. In the sequel, for notational
convenience, we will drop the dependence of these functions on r and we will
concentrate on the case n =(n
1
,n
2
)=(0, 0). In the cases n = 0, the polyno-
mial appearance of |n| in the estimates will be controlled by the rapid decay
894 LOUKAS GRAFAKOS AND XIAOCHUN LI
of C(n), while the exponential functions in (2.1) and (2.2) can be thought of
as almost “constant” locally (such as when n
1
= n
2
= 0), and thus a small
adjustment of the case n =(0, 0) will yield the case for general n in Z
2
.
Based on these remarks, we may set Φ
j,k,l
=Φ
j,k,l,0
and it will be sufficient
to prove the uniform (in m) boundedness of the operator T
0
m
defined by
T
0
m
(f
1
,f
2
)(x)=

k∈Z

l∈Z

R

R

f
1
(ξ)

f
2
(η)e
2πi(ξ+η)x

Φ
1,k,l
(ξ)

Φ
2,k,l
(η)dξdη .
(2.3)
The representation of T
0
m
into a sum of products of functions of ξ and η will be
crucial in its study. If follows from (2.1) and (2.2) that there exist the following
size estimates for the functions Φ
1,k,l
and Φ
2,k,l
.
|Φ
1,k,l
(x)|≤C
N
2
−k
(1+2
−k
|x|)
−N
,(2.4)
|Φ
2,k,l
(x)|≤C
N
2
−k+m
(1+2
−k+m
|x|)
−N
(2.5)
for any N ∈ Z
+
. The next lemma is also a consequence of (2.1) and (2.2).
Lemma 1. For al l N ∈ Z
+
, there exists C
N
> 0 such that for all f ∈
S(R),

l∈Z
|(f ∗ Φ
1,k,l
)(x)|
2
≤ C
N

|f(y)|
2
2
−k
(1+2
−k
|x − y|)
N
dy,(2.6)

l∈Z
|(f ∗ Φ
2,k,l
)(x)|
2
≤ C
N

|f(y)|
2
2
−k+m
(1+2
−k+m
|x − y|)
N
dy,(2.7)
where C
N
is independent of m.
Proof. To prove the lemma we first observe that whenever Φ
l
∈S
has Fourier transform supported in the interval [2l − 3, 2l + 3] and satisfies
sup
l
D
α

Φ
l

∞
≤ C
α
for all sufficiently large integers α, then we have

l∈Z
|(f ∗ Φ
l
)(x)|
2
≤ C
N

R
|f(y)|
2
(1 + |x − y|)
N
dy.(2.8)
Once (2.8) is established, we apply it to the function Φ
l
(x)=2
k
Φ
1,k,l
(2
k
x),
which by (2.1) satisfies |D
α

Φ
l
(ξ)|≤C
α
, to obtain (2.6). Similarly, applying
(2.8) to the function Φ
l
(x)=2
k−m
Φ
2,k,l
(2
k−m
x), which by (2.2) also satisfies
|D
α

Φ
l
(ξ)|≤C
α
, we obtain (2.7).
UNIFORM BOUNDS FOR THE BILINEAR HILBERT TRANSFORMS, I
895
By a simple translation, it will suffice to prove (2.8) when x = 0. Then
we have

l∈Z
|(f ∗ Φ
l
)(0)|
2
=

l∈Z






[2l−3,2l+3]

f(−·)
(1+4π
2
|·|
2
)
N

∧
(y)

(I − ∆)
N

Φ
l

(y) dy





2
≤

l∈Z

[2l−3,2l+3]





f(−·)
(1+4π
2
|·|
2
)
N

∧
(y)




2
dy

|(I − ∆)
N

Φ
l
(y)|
2
dy
≤ C
N

R
|f(−y)|
2
(1+4π
2
|y|
2
)
N
dy ≤ C
N

R
|f(y)|
2
(1 + |y|)
N
dy.
3. The truncated trilinear form
Let ψ be a nonnegative Schwartz function such that

ψ is supported in
[−1, 1] and satisfies

ψ(0) = 1. Let ψ
k
(x)=2
−k
ψ(2
−k
x). For E ⊂ R and k ∈ Z
define
E
k
= {x ∈ E : dist(x, E
c
) ≥ 2
k
},(3.1)
ψ
1,k
(x)=(1
(E
k
)
c
∗ ψ
k
)(x), and ψ
2,k
(x)=ψ
3,k
(x)=ψ
1,k−m
(x).(3.2)
Note that ψ
1,k
, ψ
2,k
, and ψ
3,k
depend on the set E but we will suppress this
dependence for notational convenience, since we will be working with a fixed
set E. Also note that the functions ψ
2,k
and ψ
3,k
depend on m, but this
dependence will also be suppressed in our notation. The crucial thing is that
all of our estimates will be independent of m. Define
Λ
E
(f
1
,f
2
,f
3
)=

k∈Z

l∈Z

3

j=1
ψ
j,k
(x)(f
j
∗ Φ
j,k,l
)(x)dx(3.3)
where for any α ≥ 0, Φ
3,k,l
depends on Φ
1,k,l
and Φ
2,k,l
and is chosen so that
it satisfies
|D
α

Φ
3,k,l
|≤C2
α(k−m)
, supp

Φ
3,k,l
⊂ (1+2
−2L
)J
(r)
3
(k, l), and(3.4)

Φ
3,k,l
=1 on J
(r)
3
(k, l)=−(1 + 2
−2L
)J
(r)
1
(k, l) − (1+2
−2L
)J
(r)
2
(k, l),
for all nonnegative integers α. (The number r in (3.4) is the type of the
rectangle in which the Fourier transforms of Φ
1,k,l
and Φ
2,k,l
are supported.)
One easily obtains the size estimate
|Φ
3,k,l
(x)|≤C2
−k+m
(1+2
−k+m
|x|)
−N
.(3.5)
896 LOUKAS GRAFAKOS AND XIAOCHUN LI
Because of the assumption on the indices p
1
,p
2
, there exists a 2 <p
3
< ∞
such that
1
p
1
+
1
p
2
+
1
p
3
> 1. Fix such a p
3
throughout the rest of the paper. The
following two lemmas reduce matters to the truncated trilinear form (3.3).
Lemma 2. Let 2 <p
1
,p
2
,p
3
< ∞,
1
p
1
+
1
p
2
+
1
p
3
> 1, and f
j

p
j
=1for
f
j
∈S and j ∈{1, 2, 3}. Define
E =
3

j=1
{x ∈ R : M
p
j
(Mf
j
)(x) > 2}.
Then for some constant C independent of m and f
1
,f
2
,f
3
,
|Λ
E
(f
1
,f
2
,f
3
)|≤C.
Lemma 2 will be proved in the next sections. Now, we have
Lemma 3. Lemma 2 implies (1.2).
Proof. To prove (1.2), it will be sufficient to prove that for all λ>0,
|{x : |T
0
m
(f
1
,f
2
)(x)| >λ}| ≤ Cλ
−
p
1
p
2
p
1
+p
2
whenever f
1

p
1
= f
2

p
2
= 1. By linearity and scaling invariance, it suffices
to show that
|{x : |T
0
m
(f
1
,f
2
)(x)| > 2}| ≤ C.(3.6)
Let E =

2
j=1
{x ∈ R : M
p
j
(Mf
j
)(x) > 1}. Since |E|≤C, it will be enough
to show that
|{x ∈ E
c
: |T
0
m
(f
1
,f
2
)(x)| > 2}| ≤ C.(3.7)
Let G = E
c

{|T
0
m
(f
1
,f
2
)| > 2}, and assuming |G|≥1 (otherwise there is
nothing to prove) choose f
3
∈S with f
3

L
∞
(E
c
)
≤ 1, supp f
3
⊂ E
c
, and




f
3
−
1
G
|G|
1/p
3
T
0
m
(f
1
,f
2
)
|T
0
m
(f
1
,f
2
)|




p
3
≤ min{1, T
0
m
(f
1
,f
2
)
−1
p

3
}.
Note that for the f
3
chosen we have f
3

p
3
≤ 2 and thus the set
{x ∈ R : M
p
3
(Mf
3
)(x) > 2} is empty. Now define
Λ(f
1
,f
2
,f
3
)=

k∈Z

l∈Z

3

j=1
(f
j
∗ Φ
j,k,l
)(x)dx.(3.8)
Then by Lemma 2 it follows that
|G|
1/p

3
≤

T
0
m
(f
1
,f
2
),
1
G
|G|
1/p
3
T
0
m
(f
1
,f
2
)
|T
0
m
(f
1
,f
2
)|

≤|Λ(f
1
,f
2
,f
3
)−Λ
E
(f
1
,f
2
,f
3
)|+C.
UNIFORM BOUNDS FOR THE BILINEAR HILBERT TRANSFORMS, I
897
Therefore, to prove (3.7), we only need to show that
|Λ(f
1
,f
2
,f
3
) − Λ
E
(f
1
,f
2
,f
3
)|≤C(3.9)
whenever f
3

L
∞
(E
c
)
≤ 1 and supp f
3
⊂ E
c
. To prove (3.9) note that
|Λ(f
1
,f
2
,f
3
) − Λ
E
(f
1
,f
2
,f
3
)|≤







k∈Z

l∈Z

(1 −
3

j=1
ψ
j,k
(x))
3

j=1
(f
j
∗ Φ
j,k,l
)(x)dx






.
(3.10)
But recall that ψ
2,k
= ψ
3,k
, hence
|1 −
3

j=1
ψ
j,k
(x)|≤|1 − ψ
1,k
(x)| +2|1 − ψ
2,k
(x)|.
Thus the expression on the right in (3.10) is at most equal to the sum of the
following two quantities

k∈Z

|1 − ψ
1,k
(x)|
2

j=1


l
|f
j
∗ Φ
j,k,l
(x)|
2

1
2
sup
l
|f
3
∗ Φ
3,k,l
(x)|dx,(3.11)
2

k∈Z

|1 − ψ
2,k
(x)|
2

j=1


l
|f
j
∗ Φ
j,k,l
(x)|
2

1
2
sup
l
|f
3
∗ Φ
3,k,l
(x)|dx.(3.12)
Using (2.6) and the fact that p
1
> 2, for any point z
0
∈ E
c
, we obtain
(

l∈Z
|f
1
∗ Φ
1,k,l
(x)|
2
)
1
2
≤ C


|f
1
(y)|
p
1
2
−k
(1+2
−k
|x − y|)
N
dy

1
p
1
≤ C

1+2
−k
dist(x, E
c
)

.
Similarly, using (2.7) and the fact that p
2
> 2 we obtain
(

l∈Z
|f
2
∗ Φ
2,k,l
(x)|
2
)
1
2
≤ C

1+2
−k+m
dist(x, E
c
)

.
By (3.5) and the facts that f
3

L
∞
(E
c
)
≤ 1 and supp f
3
⊂ E
c
,
|f
3
∗ Φ
3,k,l
(x)|≤C
N

1+2
−k+m
dist(x, E
c
)

−N
(3.13)
for all N>0. Therefore, (3.11) can be estimated by
C

k

E
k
2
−k
(1+2
−k
|x − y|)
N
dy
1
(1+2
−k+m
dist(x, E
c
))
N−2
dx
≤ C

E

k∈Z
2
k
≤dist(y,E
c
)
1
(1+2
−k
dist(y, E
c
))
N−2
dy ≤ C|E|≤C.
Similar reasoning works for (3.12). This completes the proof of (3.9) and
therefore of Lemma 3.
898 LOUKAS GRAFAKOS AND XIAOCHUN LI
We now set up some notation. For k,n ∈ Z, define I
k,n
=[2
k
n, 2
k
(n + 1)]
and let
φ
1,k,n
(x) =(1
I
k,n
∗ ψ
k
)(x),
φ
j,k,n
(x) =(1
I
k,n
∗ ψ
k−m
)(x), when j ∈{2, 3}.
(3.14)
Next, we can write
Λ
E
(f
1
,f
2
,f
3
)=

k∈Z

l∈Z

3

j=1


n∈Z
φ
j,k,n
(x)ψ
j,k
(x)(f
j
∗ Φ
j,k,l
)(x)

dx.
(3.15)
For an integer r with 0 ≤ r<L, let Z
r
= { ∈ Z :  = κL+r for some κ ∈ Z}.
Also for S ⊂ Z
r
× Z × Z
r
we let S
k,l
= {n ∈ Z :(k, n, l) ∈ S} and define
Λ
E,S
(f
1
,f
2
,f
3
)=

k∈Z
r

l∈Z
r

3

j=1



n∈S
k,l
φ
j,k,n
(x)ψ
j,k
(x)(f
j
∗ Φ
j,k,l
)(x)


dx.
(3.16)
For simplicity we will only consider the case where m ∈ Z
0
. The argument
below can be suitably adjusted to the case where m has a different remainder
when divided by L. We will therefore concentrate on proving Lemma 2 for the
expression Λ
E,S
(f
1
,f
2
,f
3
) when m ∈ Z
0
. To achieve this goal, we introduce
the grid structure.
Definition 1. A set of intervals G is called a grid if the condition below
holds:
for J,J

∈G,ifJ∩ J

= ∅, then J ⊂ J

or J

⊂ J.(3.17)
If a grid G satisfies the additional condition:
for J,J

∈G,ifJ J

, then 5J ⊂ J

,(3.18)
then it will be called a central grid.
Given S ⊂ Z
r
× Z × Z
r
and s =(k, n, l) ∈ S we set I
s
= I
k,n
. For each
function Φ
j,k,l
and each n ∈ Z we define a family of intervals ω
j,s
, s =(k, n, l)
∈ S so that conditions (3.19)–(3.25) below hold: Say that

Φ
1,k,l
(ξ)

Φ
2,k,l
(η)is
supported in a small neighborhood of a rectangle of type r = 1. Then we
define ω
j,s
such that
|c(ω
1,s
) − 2
−k
(2l +
1
2
)|≤5 · 2
−L
2
−k
,(3.19)


c(ω
2,s
) − 2
−k+m
(2l −
3
2
)


≤ 5 · 2
−L
2
−k+m
and ω
2,s
= ω
3,s
,(3.20)
supp

Φ
j,k,l
⊂ ω
j,s
for j ∈{1, 2},(3.21)
UNIFORM BOUNDS FOR THE BILINEAR HILBERT TRANSFORMS, I
899
supp

Φ
3,k,l
⊂ [−(1 + 2
−m
)a, −(1 + 2
−m
)b), where [a, b)=ω
3,s
,(3.22)
(1+2
−2L
)2
−k
≤|ω
1,s
|≤(1+10· 2
−L
)2
−k
,(3.23)
(1+2
−2L
)2
−k+m
≤|ω
j,s
|≤(1+2· 2
−2L
)(1+5· 2
−L
)2
−k+m
for j ∈{2, 3},
(3.24)
{ω
j,s
}
s∈S
is a central grid, for j ∈{1, 2, 3}.(3.25)
These properties are trivially adjusted when

Φ
1,k,l
(ξ)

Φ
2,k,l
(η) is supported in
a small neighborhood, a rectangle of type r =2orr =3.
As in [7], we prove the existence of ω
j,s
by induction. If S is nonempty, pick
s
0
=(k, n, l) ∈ S such that k is minimal and define S

= S\{s
0
}. By induction,
we may assume that for any s

∈ S

there exists a ω
j,s

so that the collection
of all such intervals satisfies (3.19)–(3.25). Now we try to define ω
j,s
0
so that
(3.19)–(3.25) still hold. Let [a
1
,b
1
) be an interval with length (1 + 2
−2L
)2
−k
which contains supp

Φ
1,k,l
. And for j =2, 3, let [a
j
,b
j
) be an interval with
length (1 + 2· 2
−2L
)2
−k+m
which contains supp

Φ
j,k,l
. By (3.22), we know that
supp

Φ
3,k,l
⊂ [−(1+2
−m
)a
3
, −(1+2
−m
)b
3
). Define ω
j,s
0
,1
as the union of [a
j
,b
j
)
and all intervals 5ω
j,s

with s

∈ S

which satisfy dist(ω
j,s

, [a
j
,b
j
)) ≤ 2|ω
j,s

|
and ω
j,s

be the next smaller interval in S. Inductively we define ω
j,s
0
,l
for
l ≥ 1. Let ω
j,s
0
=

l≥1
ω
j,s
0
,l
. It is easy to verify conditions (3.19)–(3.25) for
ω
j,s
0
. This completes the proof of the existence of a grid structure.
Furthermore, we have the following geometric picture for ω
j,s
.
Lemma 4. For s, s

∈ S and ω
j,s
= ω
j,s

, the following properties hold:
(1) If ω
1,s
⊂ ω
1,s

, then ω
j,s

<ω
j,s
and
1
2
|ω
j,s

| < dist(ω
j,s
,ω
j,s

) < 2|ω
2,s

|
for j =2, 3.
(2) If ω
j,s
⊂ ω
j,s

for j =2, 3, then ω
1,s
<ω
1,s

and
1
8
|ω
1,s

| < dist(ω
1,s
,ω
1,s

)
< 2|ω
1,s

|.
Proof. For simplicity, let us assume that the ω
j,s
are associated with
rectangles of type 1.
ω
1,s
=[2
−k
(2l − L
−1
), 2
−k
(2l +1+L
−1
)],
ω
1,s

=[2
−k

(2l

− L
−1
), 2
−k

(2l

+1+L
−1
)],
ω
2,s
=[2
−k+m
(2l − 2 − 2L
−1
), 2
−k+m
(2l − 1+2L
−1
)],
and ω
2,s

=[2
−k

+m
(2l

− 2 − 2L
−1
), 2
−k

+m
(2l

− 1+2L
−1
)].
For (1), note that if ω
1,s
 ω
1,s

, we have 2
−k

(2l

− L
−1
) < 2
−k
(2l − L
−1
) <
2
−k
(2l + L
−1
+1)< 2
−k

(2l

+ L
−1
+ 1). Thus, 2l

< 2
k

−k
(2l − L
−1
)+L
−1
and 2l + L
−1
+1 < 2
k−k

(2l

+ L
−1
+ 1). By this, we have 2
−k

+m−1
<
2
−k+m
(2l − 2L
−1
− 2) − 2
−k

+m
(2l

− 1+2L
−1
) < 2
−k

+m+1
, which proves (1).
We omit the proof of (2) since it is similar.
900 LOUKAS GRAFAKOS AND XIAOCHUN LI
As in [10] we give the following definition.
Definition 2. A subset S of Z
r
×Z×Z
r
is called convex if for all s, s

∈ S,
s

∈ Z
r
× Z × Z
r
, j ∈{1, 2} with I
s
⊂ I
s

⊂ I
s

and ω
j,s

⊂ ω
j,s

⊂ ω
j,s
,we
have s

∈ S.
It is sufficient to prove bounds on Λ
E,S
for all finite convex sets S of triples
of integers, provided the bound is independent of S and of course m.
4. The selection of the trees
Definition 3. Fix T ⊂ S and t ∈ T. If for any s ∈ T,wehaveI
s
⊂ I
t
and ω
j,s
⊃ ω
j,t
, then we call T a tree of type j with top t.Now,T is called a
maximal tree of type j ∈{1, 2} with top t in S if there does not exist a larger
tree of type j with the same top strictly containing T . Let T be a maximal tree
of type j ∈{1, 2} with top t in S, and i ∈{1, 2}, i = j. Denote the maximal
tree of type i with top t in S by

T .
Lemma 5. Let S ⊂ Z
r
× Z × Z
r
be a convex set and T ⊂ S be a maximal
tree of type j ∈{1, 2} with top t in S. Then T is a convex set.
Proof. Let s, s

∈ T , s

∈ Z
r
× Z × Z
r
, i ∈{1, 2} with I
s
⊂ I
s

⊂ I
s

and
ω
i,s

⊂ ω
i,s

⊂ ω
i,s
. Then s

∈ S by the convexity of S. Since s = s

, it follows
from Lemma 4 that i = j. Using that I
s

⊂ I
s

⊂ I
t
, ω
j,t
⊂ ω
j,s

⊂ ω
j,s

, and
the maximality of T , we obtain that s

∈ T , hence the convexity of T follows.
Lemma 6. Let S ⊂ Z
r
× Z × Z
r
be a convex set and T be a maximal tree
of type j ∈{1, 2} with top t in S. Then S\(T


T ) is convex.
Proof. Assume that S\(T


T ) is not convex. Then there exist s, s

∈
S\(T


T ), s

∈ T


T , i ∈{1, 2} with I
s
⊂ I
s

⊂ I
s

and ω
i,s

⊂ ω
i,s

⊂ ω
i,s
.
If s

∈ T, then I
s

⊂ I
t
and ω
j,t
⊂ ω
j,s

. Since s is not in T , we have i = j.
By Lemma 4, we have dist(ω
i,t
,ω
i,s

) < 2|ω
j,s

|. Since 5ω
i,s

⊂ ω
i,s
we have
ω
i,t
⊂ ω
i,s
.Thuss ∈

T , which is a contradiction.
For a given subset T of S we define T
k,l
to be the set
{n ∈ Z :(k, n, l) ∈ T }.
If T is a tree of type j for j ∈{1, 2, 3} and k ∈ Z
r
, then there is at most
one l ∈ Z
r
such that T
k,l
= ∅. If such an l exists, then let T
k
= T
k,l
and
Φ
j,k,T
=Φ
j,k,l
. Otherwise, let T
k
= ∅ and Φ
j,k,T
= 0. For brevity, we write
(k, n) ∈ T if and only if there exists an l ∈ Z
r
with (k, n, l) ∈ T.Thus
UNIFORM BOUNDS FOR THE BILINEAR HILBERT TRANSFORMS, I
901
identifying trees with sets of pairs of integers, we will use this identification
throughout.
Therefore, if (k, n, l) ∈ T , we can write ω
j,k,n,l
= ω
j,k,l
= ω
j,k,T
, and
Λ
E,T
(f
1
,f
2
,f
3
)=

k∈Z
r

3

j=1


n∈T
k
φ
j,k,n
(x)ψ
j,k
(x)(f
j
∗ Φ
j,k,T
)(x)

dx.(4.1)
Let t =(k
T
,n
T
,l
T
) be the top of T . We write I
T
= I
k
T
,n
T
and ω
j,T
= ω
j,k
T
,T
.
For a tree T of type 2 (or 3) with top t and k ∈ Z
r
, define θ
+
j,k,T
and θ
−
j,k,T
by

θ
+
j,k,T
(ξ)=(Φ
j,k−L,T
− Φ
j,k,T
)
∧
(ξ)1
ξ≥α
j
c(w
j,t
)
(ξ),

θ
−
j,k,T
(ξ)=(Φ
j,k−L,T
− Φ
j,k,T
)
∧
(ξ)1
ξ≤α
j
c(w
j,t
)
(ξ),
where α
j
=1ifj = 2 and α
j
=1+2
−m
,ifj = 3. Let ψ
∗
(x) = (1 + x
2
)
−N
.In
accordance with the definitions of φ
j,k,n
and ψ
j,k
we define the functions
ψ
∗
1,k
(x)=(1
(E
k
)
c
∗ ψ
∗
k
)(x),ψ
∗
j,k
(x)=ψ
∗
1,k−m
(x), when j ∈{2, 3}.
(4.2)
φ
∗
1,k,n
(x)=(1
I
k,n
∗ ψ
∗
k
)(x),φ
∗
j,k,n
(x)=(1
I
k,n
∗ ψ
∗
k−m
)(x), when j ∈{2, 3}.
(4.3)
Let ∆
k
be the set of all connected components of E
k
\E
k+L
. Obviously ∆
k
is
a set of intervals. Observe that if J ∈ ∆
k
, then 2
k
≤|J| < 2
k+L
, and

k
∆
k
is
a set of pairwise disjoint intervals. Define
∆
k,T
= {J ∈ ∆
k
: J ⊂ I
k+m+L,n
, for some (k + m + L, n) ∈ T },
and for J ∈ ∆
k,T
define
ρ
k,J
(x)=1
J
∗ ψ
∗
k
(x), where ψ
∗
k
(x)=2
−k
ψ
∗
(2
−k
x).(4.4)
Throughout this paper fix 0 <η≤ L
−1


3
j=1
1
p
j
− 1

min
j∈{1,2,3}
{
1
p
j
}
and let
H =
3

j=1
{(1,j,1), (2, 1, 1), (3, 1, 1)}


5

ν=2
{(2, 2,ν), (2, 3,ν), (3, 2,ν), (3, 3,ν)}

.