THE T 1 THEOREM
V. CHOUSIONIS AND X. TOLSA
Introduction
These are the notes of a short course given by X. Tolsa at the Universitat Autònoma de Barcelona between November and December of 2012. The notes have been
typed by V. Chousionis.
We present a “dyadic proof” of the classical T 1 theorem of David and Journé. We
have preferred to state the T 1 theorem in terms of the boundedness over characteristic functions of cubes rather than in terms of the typical conditions T 1, T ∗ 1 ∈ BMO,
which is the usual approach in the literature. Nevertheless, no attempt at originality
is claimed.
The notation used below is standard. We use the shortcuts Lp or C ∞ for Lp (Rn )
or C ∞ (Rn ), as well as other analogous terminology. As usual, the letter C denotes
some constant that may change its value at different occurrences, and typically
depends on some absolute constants and other fixed parameters.
1. Statement and reductions
A kernel K : Rn × Rn \ {(x, y) : x = y} → R is called Calderón-Zygmund
standard, or simply standard, if there exist constants C, δ > 0 such that for all
distinct x, y ∈ Rn and for all x′ such that |x − x′ | ≤ |x−y|
,
2
C
(1) |K(x, y)| ≤ |x−y|n
′ δ
|x−x |
(2) |K(x, y) − K(x′ , y)| + |K(y, x) + K(y, x′ )| ≤ C |x−y|
n+δ .
From now on K will always denote a standard kernel.
Given a function f ∈ C0∞ we define
Z
T f (x) = K(x, y)f (y)dy for x ∈
/ supp(f ).
We then say that T is Calderón-Zygmund operator (CZO) with kernel K(·, ·). If
f ∈ Lp , for some 1 < p < ∞, we define the truncated singular integrals Tε associated
with the standard kernel K by
Z
Tε (f )(x) =
K(x, y)f (y)dy.
|x−y|>ε
The operator T is said to be bounded in L2 if there exists some constant C not
depending on ε such that
kTε (f )k2 ≤ Ckf k2
1
2
V. CHOUSIONIS AND X. TOLSA
for all f ∈ L2 . The definition of the L2 boundedness of T in terms of the truncated
operators Tε is convenient because it avoids the delicate question of the existence of
principal values for T f .
By T ∗ we denote the CZO associated with the kernel K̃(x, y) = K(y, x), furthermore it holds that, for f, g ∈ C0∞ ,
hTε∗ f, gi = hf, Tε gi.
The following theorem provides checkable criteria for the L2 -boundedness of CZO’s.
Theorem 1 (David-Journé). Let T be a CZO as above. Then T is bounded in L2
if and only if there exists some constant C > 0 such that
(1) supε>0 kTε χQ k2 ≤ Cm(Q)1/2
(2) supε>0 kTε∗ χQ k2 ≤ Cm(Q)1/2
for all cubes Q.
The first step consists of regularizing the kernel. We define a function ϕ : R →
R, ϕ ∈ C ∞ , 0 ≤ ϕ ≤ 1, such that ϕ = 0 in [−1/2, 1/2] and ϕ = 1 in R \ [−1, 1].
Given the function ϕ we regularize the kernel K at level ε by
|x − y|
e
Kε (x, y) = K(x, y)ϕ
.
ε
e ε (x, y) = K(x, y) when |x − y| > ε and K
e ε (x, y) = 0 when |x − y| ≤ ε .
Notice that K
2
e is also a standard kernel with constants not depending on
It follows easily that K
eε
ε. Next we consider the operators associated to K
Z
e ε (x, y)f (y)dy.
Teε f (x) = K
Notice that, for Kε (x, y) = K(x, y)χ{r:|r|>ε}(|x − y|),
Z
e ε (x, y)||f (y)|dy
|Tε f (x) − Teε f (x)| ≤
|Kε (x, y) − K
ε
<|x−y|<ε
2
C
≤
(ε/2)n
Z
|f (y)|dy ≤ CMf (x)
|x−y|<ε
where M denotes the usual Hardy-Littlewood maximal operator. Since M is bounded
in Lp for all 1 < p ≤ ∞ we deduce that Teε is bounded in Lp , uniformly on ε, if and
only if the operators Tε are bounded in Lp uniformly on ε.
In the same manner, for R > 2ε, we can define the kernels
|x
−
y|
|x
−
y|
e ε,R (x, y) = ϕ
1−ϕ
K(x, y),
K
ε
R
and the corresponding operators
Teε,R f (x) =
Z
e ε,R (x, y)f (y)dy.
K
THE T 1 THEOREM
3
Since R > 2ε it follows that Teε,R = Teε − TeR therefore if Teε : Lp → Lp is bounded
for all ε > 0 then Teε,R is bounded as well for all positive R, ε such that R > 2ε.
Furthermore if Teε,R is bounded in L2 uniformly on ε and R then Teε is bounded
uniformly on ε. To see this let f ∈ L2 with compact support and notice that the
functions in L2 with compact support are dense in L2 . Then
Z
Z
e
e
e ε (x, y)f (y)dy,
Tε f (x) =
Kε (x, y)f (y)dy +
K
|x−y|≤R
|x−y|>R
and the second term tends to 0 as R → ∞. Therefore we can choose R > 0 large
enough such that
kTeε (f )k2 ≤ Ckf k2 .
Hence by density this operator uniquely extends to a bounded operator in L2 with
the same bounds. Now note that for f ∈ L2 all x ∈ Rn , by Hölder’s inequality,
1/2
Z
1/2 Z
Z
2
2
e
e
|Kε (x, y)||f (y)|dy ≤
|Kε (x, y)| dy
|f (y)|
≤ C,
e ε (x, y)|2 is integrable at infinity. Therefore by density we deduce that
because |K
there exists a sequence (by passing to a subsequence if necessary) of fn ∈ L2 with
compact support such that fn → f in Lp and fn → f a.e. Hence by the dominated
convergence theorem,
Teε (fn )(x) → Teε (f )(x)
and hence Teε is bounded in L2 uniformly on ε.
Therefore, if
kTε χQ k2 ≤ Cm(Q)1/2 and kTε∗ χQ k2 ≤ Cm(Q)1/2
for all cubes Q uniformly on ε, we also have that
∗
kTeε,R χQ k2 ≤ C ′ m(Q)1/2 and kTeε,R
χQ k2 ≤ C ′ m(Q)1/2
for all cubes Q uniformly on ε and R. The proof will be complete if we show that
the above condition implies that Teε,R : L2 → L2 is bounded uniformly on ε and R.
Notice that Teε,R : Lp → Lp is bounded for 1 ≤ p ≤ ∞ with norm depending on ε
and R. For f ∈ L∞ ,
Z
Z
e
e ε,R (x, y)|dy
|Tε,R (f )(x)| ≤ |Kε,R (x, y)||f (y)|dy ≤ kf k∞ |K
Z
1
≤ Ckf k∞
dy
n
ε<|x−y|<R |x − y|
n
R
≤ Ckf k∞
,
ε
and analogously for the L1 bound. Hence it is enough to prove the following.
4
V. CHOUSIONIS AND X. TOLSA
Theorem 2 (Reduced restatement of Theorem 1). Let the CZO T : Lp → Lp ,
bounded for some 1 ≤ p ≤ ∞, and associated with a kernel K such that kKk∞ ≤ C0
and K(x, y) = 0 whenever |x − y| ≥ R for some R > 0. If furthermore
(1) kT χQ k2 ≤ C1 m(Q)1/2 and
(2) kT ∗ χQ k2 ≤ C1 m(Q)1/2
for all cubes Q, then kT kL2 →L2 ≤ C2 with C2 depending on C1 but independent of
C0 and R.
2. The operators ∆Q
Let D denote the family of dyadic cubes of Rn . For Q ∈ D and f ∈ L1loc we define
(
0
if x ∈
/ Q,
∆Q f (x) =
mP f − mQ f if x ∈ P and P is a son of Q,
where mQ f is the average of f on Q. The following proposition gathers several
elementary but useful properties of the operators ∆Q .
Proposition 3. For Q, R ∈ D and f ∈ L1loc ,
(1) supp∆Q f ⊂ Q,
(2) R∆Q is constant in every son of Q,
(3) ∆Q f = 0,
(4) h∆Q f, ∆R gi = 0 whenever Q 6= R,
(5) ∆Q ◦ ∆Q = ∆Q .
(6) ∆Q : L2 → L2 is bounded and ∆∗Q = ∆Q .
For simplicity from now on we will denote k · k2 := k · k.
P
Proposition 4. If f ∈ L2 then f = Q∈D ∆Q f , the convergence is unconditional
P
in L2 and moreover kf k2 = Q∈D k∆Q f k2 .
Proof. The essential step consists in showing that
X
k∆Q f k2 ≤ kf k2 .
(1)
Q∈D
P
P
To see this, let F ⊂ D be finite and set g = f − Q∈F ∆Q f . Then g ⊥ Q∈F ∆Q f
because by (4),(5) and (6) of Proposition 3:
X
X
X
X
hf −
∆Q f,
∆Q f i =
hf, ∆Q f i −
h∆Q f, ∆Q f i = 0.
Q∈F
Hence kf k2 = kgk2 +
Q∈F
P
Q∈F
Q∈F
Q∈F
k∆Q f k2 and
X
k∆Q f k2 ≤ kf k2 .
Q∈F
Therefore since this holds for any finite subset F ⊂ D, (1) follows.
THE T 1 THEOREM
5
P
Now we order D = {Q1 , Q2 , . . . } and we have to prove that limm→∞ m
i=1 ∆Qi f =
2
f in L . We first see that the sequence is convergent because
2
n
n
X
X
k∆Qi f k2 → 0
∆Qi f =
i=m
i=m
P
2
as n → ∞ since i k∆Qi f k < ∞. Let A = ∪k∈N Ak where Ak are the classes of
functions with compact support such that they are constant on some dyadic cube
in Dk and furthermore they have zero mean on each orthant. Then we notice that
the result is true for the class of functions A and furthermore A is dense in L2 .
Let ε >P
0. Then for allPf ∈ L2 , there exists
Pamfunction g ∈ A such that kf −gk < ε.
m
m
We write i=1 ∆Qi f = i=1 ∆Qi (f − g) + i=1 ∆Qi g and we deduce using (1) that
2
m
m
m
X
X
X
∆Qi (g − f )
∆Qi g − g + kg − f k + ∆Qi f − f ≤ i=1
i=1
i=1
!1/2
m
m
X
X
∆Qi g − g + kg − f k +
k∆Qi (g − f )k2
=
i=1
i=1
m
X
∆Qi g − g + 2kg − f k
≤
i=1
m
X
∆Qi g − g ,
≤ 2ε + i=1
and the proof is complete after letting m → ∞.
3. Proof of the T 1 theorem
We will prove that
P
P
|hT f, gi| ≤ Ckf kkgk,
(2)
for f = Q∈F1 ∆Q f, g = Q∈F2 ∆Q g for finite F1 , F2 ⊂ D and a constant C not
depending on f or g. Then by Proposition 4, (2) holds for all f, g ∈ L2 and Theorem
2 follows, taking into account that kT kL2 →L2 < ∞ by assumption.
We have
X
hT ∆Q f, ∆R gi.
hT f, gi =
Q∈F1 ,R∈F2
We then observe the following two easy facts: first,
k∆Q f k2 ≈ k∆Q f k2∞ m(Q)
(3)
kT (∆Q f )k ≤ Ck∆Q f k.
(4)
and, second,
6
V. CHOUSIONIS AND X. TOLSA
n
To see (3), suppose that Q = ∪2i=1 Pi where Pi are the sons of Q and let ci =
mPi f − mQ f . Then,
2
k∆Q f k =
Z
n
2
|∆Q f | =
2
X
2n
m(Pi )c2i
i=1
m(Q) X 2
=
c,
2n i=1 i
and because k∆Q f k∞ = max ci , we get
1
m(Q)k∆Q f k2∞ ≤ k∆Q f k2 ≤ m(Q)k∆Q f k2∞ .
n
2
Using assumption (1) of Theorem 2, (4) follows analogously.
In the following we prove an auxiliary lemma which deals with the case when two
cubes are far each other.
Lemma
5. Let two functions ϕQ , ψR be such that suppϕQ ⊂ Q, suppψR ⊂ R, ϕQ , ψR ∈
R
1
L , ϕQ = 0 and d(Q, suppψR ) ≥ l(Q). Then,
(
l(Q)δ
C d(Q,suppψ
δ+n kϕQ k1 kψR k1
R)
|hT ϕQ, ψR i| ≤
(5)
CkϕQ k1 kψR k∞ .
R
Proof. Let xQ be the center of Q. Using that K is a standard kernel and ϕQ = 0
we get
Z Z
hT ϕQ , ψR i = K(x,
y)ϕ
(x)ψ
(y)dx
dy
Q
R
Z Z
(K(x, y) − K(xQ , y))ϕQ (x)ψR (y)dx dy =
ZZ
|x − xQ |δ
≤C
|ϕQ (x)||ψR (y)|dxdy
x∈Q,
|x − y|n+δ
y∈suppψR
δ
≤C
l(Q)
kϕQ k1 kψR k1 .
d(Q, suppψR )δ+n
To complete the proof of the lemma notice that for x ∈ Q and y ∈ R,
|xQ − y| ≤ |x − y| + l(Q) ≤ |x − y| + d(Q, suppψR ) ≤ 2|x − y|.
Therefore,
ZZ
|x − xQ |δ
|ϕQ (x)||ψR (y)|dxdy
x∈Q, |x − y|n+δ
y∈suppψR
Z
δ
≤ Cl(Q) kϕQ k1 kψR k∞
1
dy
n+δ
|xQ −y|>l(Q) |xQ − y|
≤ CkϕQ k1 kψR k∞ .
THE T 1 THEOREM
7
R
1
−δ
. This follows by
The last inequality follows because |xQ−y|>l(Q) |xQ −y|
n+δ d . l(Q)
integrating on annuli, for instance,
Z
∞ Z
X
1
1
dy
≤
dy
n+δ
n+δ
i+1 l(Q))\B(x ,2i l(Q)) |xQ − y|
B(x
,2
|xQ −y|>l(Q) |xQ − y|
Q
Q
i=0
i
∞
∞ X
(l(Q)2i+1 )n
2n X 1
≤
=
i )n+δ
(l(Q)2
l(Q)δ i=0 2δ
i=0
≤ Cl(Q)−δ .
Now we do the following splitting:
X
hT ∆Q , ∆R gi =
Q∈F1 ,R∈F2
X
hT ∆Q , ∆R gi
Q∈F1 ,R∈F2 ,
l(Q)≤l(R)
+
X
hT ∆Q , ∆R gi
Q∈F1 ,R∈F2 ,
l(Q)>l(R)
= S1 + S2 .
We are going to bound S1 , the boundedness of S2 follows analogously. We consider
the following three cases for Q, R ∈ D such that l(Q) ≤ l(R):
(1) d(∪i ∂Pi , Q) ≤ l(Q)γ l(R)1−γ where the Pi ’s are the sons of R,
(2) d(Q, R) ≥ l(Q)γ l(R)1−γ ,
(3) d(Q, ∪i ∂Pi ) ≥ l(Q)l(R)1−γ and Q ( R,
δ
where γ = 2(δ+n)
.
We will use the notation (Q, R) ∈ (1) if Q, R satisfy (1) and so on. We have
X
X
X
hT ∆Q f, ∆R gi
S1 ≤
|hT ∆Q f, ∆R gi| +
|hT ∆Q f, ∆R gi| + (Q,R)∈(1)
(Q,R)∈(2)
(Q,R)∈(3)
and we will bound each of the three previous terms separately.
3.1. Estimates for (Q, R) ∈ (1). First observe that
n/2
l(Q)
|hT ∆Q f, ∆R gi| ≤ k∆Q f kk∆R gk
.
l(R)
To see this, let
ψR1 = ∆R gχ3Qc and ψR2 = ∆R gχ3Q .
(6)
8
V. CHOUSIONIS AND X. TOLSA
Since d(suppψR1 , Q) ≥ l(Q), by Lemma 5, Hölder’s inequality and (3) we obtain
|hT ∆Q f, ψR1 i| ≤ Ck∆Q f k1 kψR1 k∞
≤ Ck∆Q f k1 k∆R gk∞
≤ Ck∆Q f km(Q)1/2 k∆R gkm(R)−1/2
n/2
l(Q)
.
= Ck∆Q f kk∆R gk
l(R)
Furthermore, by (3), Hölder’s inequality and (4)
Z
2
|hT ∆Q f, ψR i| ≤ Ck∆R gk∞
|T ∆Q f |
3Q
≤ Ck∆R gkm(R)−1/ 2 m(3Q)1/2 kT ∆Q f k
≤ Ck∆R gkm(R)−1/2 m(Q)1/2 k∆Q f k
n/2
l(Q)
= Ck∆Q f kk∆R gk
.
l(R)
Therefore (6) is proved.
Lemma 6. We have
X
(Q,R)∈(1)
l(Q)
k∆Q f kk∆R gk
l(R)
n/2
Proof. By Cauchy-Schwartz,
n/2 X
X
l(Q)
≤
k∆Q f k
k∆Q f kk∆R gk
l(R)
Q∈D
(Q,R)∈(1)
≤
≤ Ckf kkgk.

!1/2 
X
X


k∆Q f k2
Q∈D

≤ kf k 
X
Q∈D


X
Q∈D


X
R:(Q,R)∈(1)
l(Q)
k∆R gk
l(R)
R:(Q,R)∈(1)
Q∈D
Thus we only need to show that
X
k∆R gk


l(Q)
l(R)
X
n/2
2 1/2

X
k∆R gk
R:(Q,R)∈(1)
k∆R gk
R:(Q,R)∈(1)
n/2
≤ kgk.
l(Q)
l(R)
n/2
l(Q)
l(R)
2 1/2

(7)
.
n/2
2 1/2

THE T 1 THEOREM
We start with the following trick: we rewrite
n/2
X
X
l(Q)
=
k∆R gk
l(R)
R:(Q,R)∈(1)
R:(Q,R)∈(1)
9
l(Q)
k∆R gk
l(R)
n−ε
2
l(Q)
l(R)
ε2
where ε is some small number to be chosen later. Now we apply Cauchy-Schwartz
and we get

 1/2

n/2 2
X
X
l(Q)



k∆R gk
l(R)
Q∈D
R:(Q,R)∈(1)

≤
X
Q∈D


X
k∆R gk2
R:(Q,R)∈(1)
l(Q)
l(R)
n−ε

X

R:(Q,R)∈(1)
l(Q)
l(R)
ε
1/2

.
For (Q, R) ∈ (1) we have that d(∪i ∂Pi , Q) ≤ l(R), where the Pi ’s are the sons of
R. Now notice that for a fixed cube Q ∈ D, and k ∈ N,
♯{R ∈ D : (Q, R) ∈ (1) and l(R) = 2k l(Q)} ≤ C
where C only depends on n, the dimension of the space. If RQ is the dyadic cube
which contains Q and has length 2k l(Q), then C is smaller than (3n − 1)2 which is
the cardinality of the set of the neighbors of all neighbors of RQ . Thus,
ε
X
X
X l(Q) ε X
l(Q)
2−kε ≤ C.
=
=C
l(R)
l(R)
k∈N
k∈N
R:(Q,R)∈(1)
R:(Q,R)∈(1),
l(R)=2k l(Q)
Furthermore, by Fubini,



n−ε
X
X
X
l(Q)

=
k∆R gk2
k∆R gk2 
l(R)
Q∈D
R∈D
R:(Q,R)∈(1)
X
Q:(Q,R)∈(1)
l(Q)
l(R)
n−ε

.
Hence in order to prove (7) and thus settle the proof, we only need to show that


X l(Q) n−ε
 ≤ C.

(8)
l(R)
Q:(Q,R)∈(1)
To prove this we split again:
X l(Q) n−ε X
=
l(R)
X
k∈N Q:(Q,R)∈(1),
l(Q)=2−k l(R)
Q:(Q,R)∈(1)
=
X
k∈N
2
kε
X
l(Q)
l(R)
Q:(Q,R)∈(1),
l(Q)=2−k l(R)
n l(Q)
l(R)
l(R)
l(Q)
n
.
ε
10
V. CHOUSIONIS AND X. TOLSA
Let dk = 2 · 2−kγ l(R), and for A ⊂ Rn denote
Ndk (A) = {x : there is a ∈ A such that d(a, y) ≤ dk }.
Then for R ∈ D fixed
∪{Q ∈ D : (Q, R) ∈ (1), l(Q) = 2−k l(R)} ⊂ Ndk (∪i ∂Pi ),
where Pi are the sons of R. Therefore,
n X
X
X
l(Q)
kε
=
2kε
2
l(R)
k∈N
k∈N
Q:(Q,R)∈(1),
l(Q)=2−k l(R)
≤
X
2kε
k∈N
≤C
X
k∈N
≤C
X
X
Q:(Q,R)∈(1),
l(Q)=2−k l(R)
m(Q)
m(R)
m(Ndk (∪i ∂Pi ))
m(R)
2kε
dk l(R)n−1
l(R)n
2k(ε−γ) .
k∈N
Hence if we choose ε = γ/2 the proof of (8) is complete and we are done.
Combining (6) and Lemma 6 we derive the desired estimate
X
|hT ∆Q f, ∆R gi| ≤ Ckf kkgk.
(Q,R)∈(1)
3.2. Estimates for (Q, R) ∈ (2). In this subsection we are going to show that
P
an auxiliary lemma. As usual if
(Q,R)∈(2) |hT ∆Q f, ∆R gi| ≤ kf kkgk. We start with
P
2
2
I is an index set, ℓ (I) = {(xi )i∈I : xi ∈ R and
i∈I xi < ∞}.
Lemma 7 (Shur’s lemma). Let I be some set of indices, and for each i ∈ I a number
wi > 0. Suppose that, for some constant a ≥ 0, the matrix {Ti,j }i,j∈I satisfies
X
|Ti,j | wj ≤ a wi for each i
j
and
X
|Ti,j | wi ≤ a wj
for each j.
i
Then, the matrix {Ti,j }i,j∈I defines a bounded operator in ℓ2 (I) with norm ≤ a.
P
Proof. Let {xi }i∈I ∈ ℓ2 (I), and set yi =
j∈I Ti,j xj . We intend to show that
P
P
2
2
2
i |yi | ≤ a
i |xi | . So we write
1/2 −1/2
|Ti,j ||xj | = |Ti,j |1/2 wj
|Ti,j |1/2 wj
|xj | ,
THE T 1 THEOREM
11
and then by Cauchy-Schwartz,
X
X
X
2
−1
2
|Ti,j | wj−1 |xj |2 .
|yi| ≤
|Ti,j | wj
|Ti,j | wj |xj | ≤ a wi
j∈I
j∈I
j∈I
Summing on i and interchanging the sums, we get
X
X
XX
|xj |2 .
|yi|2 ≤ a
|Ti,j | wi wj−1 |xj |2 ≤ a2
i
j
i
j
For Q, R ∈ (2) we have that d(Q, ∂R) ≥ l(Q) hence by Lemma 5 and CauchySchwartz we have
l(Q)δ
k∆Q f k1 k∆R gk1
d(Q, R)δ+n
l(Q)δ
≤C
m(Q)1/2 m(R)1/2 k∆Q f kk∆R gk.
δ+n
d(Q, R)
|hT ∆Q f, ∆R gi| ≤ C
We denote D(Q, R) = l(Q) + l(R) + d(Q, R). It then follows that
l(Q)δ/2 l(R)δ/2
l(Q)δ
≤
C
.
d(Q, R)n+δ
D(Q, R)n+δ
(9)
To prove (9) we consider two cases. If d(Q, R) ≥ l(R) then D(Q, R) ≤ 3d(Q, R)
and furthermore l(Q)δ ≤ l(Q)δ/2 l(R)δ/2 hence (9) follows. If d(Q, R) ≤ l(R) then
D(Q, R) ≤ 3l(R) and recalling that for (Q, R) ∈ (2), d(Q, R) ≥ l(Q)γ l(R)1−γ ,
l(Q)δ
l(Q)δ
l(Q)δ−γ(n+δ) l(R)γ(n+δ)
≤
=
d(Q, R)n+δ
l(Q)γ(n+δ) l(R)(1−γ)(n+δ)
l(R)(n+δ)
l(Q)δ/2 l(R)δ/2
≤C
,
D(Q, R)n+δ
as γ =
δ
.
2(n+δ)
Therefore,
|hT ∆Q f, ∆R gi| ≤ C
l(Q)δ/2 l(R)δ/2
m(Q)1/2 m(R)1/2 k∆Q f kk∆R gk.
D(Q, R)δ+n
(10)
Our goal now is to apply Schur’s Lemma. To this end, consider the matrix (TQ,R )(Q,R)∈(2)
defined by
TQ,R
l(Q)δ/2 l(R)δ/2
=
m(Q)1/2 m(R)1/2 .
δ+n
D(Q, R)
12
V. CHOUSIONIS AND X. TOLSA
Applying Cauchy-Schwartz in ℓ2 (I), where I is the index set associated to the set
{(Q, R) ∈ (2)} we get,
X
(Q,R)∈(2)

TQ,R k∆Q f kk∆R gk ≤ 
X
(Q,R)∈(2)

1/2 
(TQ,R k∆Q f k)2
X
≤ kgk 
(Q,R)∈(2)

X
(Q,R)∈(2)
1/2
(TQ,R k∆Q f k)2 
1/2
k∆R gk2
.
(11)
Hence we only need to show that
X
(TQ,R k∆Q f k)2 ≤ C
(Q,R)∈(2)
X
k∆Q f k2 .
(12)
(Q,R)∈(2)
For Q ∈ D let wQ = m(Q)1/2 , by Schur’s lemma in order to prove (12) it is enough
to show that
X
TQ,R wR ≤ CwQ
(13)
X
TQ,R wQ ≤ CwR .
(14)
R:(Q,R)∈(2)
and
Q:(Q,R)∈(2)
For (13),
X
R:(Q,R)∈(2)
TQ,R wR ≤
X
δ
X
k∈N R:l(R)=2k l(Q)
= l(Q)n/2
X
k∈N
δ
l(Q)δ/2 2k 2 l(Q)δ/2
l(Q)n/2 l(R)n
D(Q, R)δ+n
2k 2 l(Q)δ
X
R:l(R)=2k l(Q)
l(R)n
.
D(Q, R)n+δ
THE T 1 THEOREM
13
Let xQ be the center of Q. Then for x ∈ R, |xQ − x| + l(R) ≤ C(l(Q) + l(R) +
d(Q, R)) = CD(Q, R) and we obtain
Z
X
X
l(R)n
1
=
dx
n+δ
n+δ
D(Q, R)
R D(Q, R)
k
k
R:l(R)=2 l(Q)
R:l(R)=2 l(Q)
Z
X
1
dx
≤C
n+δ
R (|xQ − x| + l(R))
R:l(R)=2k l(Q)
Z
1
dx
≤C
(|xQ − x| + 2k l(Q))n+δ
Z
1
dx
=C
k
n+δ
|x−xQ |≤2k l(Q) (|xQ − x| + 2 l(Q))
Z
1
+
dx
k
n+δ
|x−xQ |>2k l(Q) (|xQ − x| + 2 l(Q))
= C(I1 + I2 ).
For I1 ,
I1 ≤
Z
|x−xQ |≤2k l(Q)
(2k l(Q))n
1
dx
≤
= (2k l(Q))−δ ,
(2k l(Q))n+δ
(2k l(Q))n+δ
and for I2 as usual after splitting the set |xQ − x| > 2k l(Q) in annuli and integrating
we get
Z
C
1
dx ≤ k
.
I2 ≤
n+δ
(2 l(Q))δ
|x−xQ |>2k l(Q) |xQ − x|
Therefore,
X
C
l(R)n
≤ k
,
n+δ
D(Q, R)
(2 l(Q))δ
k
R:l(R)=2 l(Q)
and by (3.2),
X
TQ,R wR ≤ Cl(Q)n/2
X
δ
2−k 2 ≤ Cl(Q)n/2 .
k∈N
R:(Q,R)∈(2)
The proof of (14) follows in a similar manner and it is omitted.
Estimates for (Q, R) ∈ (3). In this subsection we are going to show that
3.3.
P
(Q,R)∈(3) hT ∆Q f, ∆R gi ≤ kf kkgk. To this end we need the following discrete
version of the famous embedding theorem of Carleson.
Theorem 8. [Carleson’s Embedding Theorem] Let σ be a Radon measure on Rd .
Let D be the dyadic lattice from Rd and let {aQ }Q∈D be a family of non negative
numbers. Suppose that for every cube R ∈ D we have
X
aQ ≤ c2 σ(R).
(15)
Q∈D:Q⊂R
14
V. CHOUSIONIS AND X. TOLSA
Then, every family of non negative numbers {wQ }Q∈D satisfies
Z
X
wQ aQ ≤ c2 sup wQ dσ(x).
In particular, if f ∈ L2 (σ),
X
where hf iσ,Q =
(16)
Q∋x
Q∈D
|hf iσ,Q |2 aQ ≤ c c2 kf k2L2(σ) ,
(17)
Q∈D
R
Q
f dσ/σ(Q) and c is an absolute constant.
Proof. To prove (16), consider the characteristic function defined by χ(Q, t) = 1 if
0 < t < wQ , and 0 otherwise. Then,
Z ∞X
X Z ∞
X
wQ aQ =
χ(Q, t) aQ dt =
χ(Q, t) aQ dt.
(18)
Q∈D
0
Q∈D
0
For each t > 0, let
Ωt =
[
Q∈D
Q.
Q∈D: wQ >t
Notice that if χ(Q, t) = 1, then Q ⊂ Ωt , and thus
X
X
χ(Q, t) aQ ≤
Q∈D
aQ
Q∈D: Q⊂Ωt
For m ≥ 1, let Im ⊂ D be the subfamily of the cubes Q ⊂ Ωt such that ℓ(Q) ≤ 2m ,
and let Jm ⊂ Im be the subfamily of maximal cubes from Im . Then we have
X X
X
X
aQ .
aQ = lim
aQ = lim
m→∞
Q∈D: Q⊂Ωt
Q∈Im
m→∞
R∈Jm Q⊂R
By the assumption (15), for all m ≥ 1,
X
X X
σ(R) ≤ c2 σ(Ωt ),
aQ ≤ c2
R∈Jm Q⊂R
and so
X
R∈Jm
χ(Q, t) aQ ≤ c2 σ(Ωt ).
Q∈D
Since Ωt coincides with {x ∈ Rd : w ∗ (x) > t}, where w ∗ (x) = supQ∋x wQ , by (18)
we obtain
Z ∞
Z
X
wQ aQ ≤ c2
σ(Ωt ) = c2 w ∗ (x) dσ(x).
0
Q∈D
To prove the estimate (17), we take wQ = |hf iσ,Q |2 , and then from (16) we deduce
X
|hf iσ,Q |2 aQ ≤ c2 kMσ,d f k2L2 (σ) ,
Q∈D
THE T 1 THEOREM
15
where Mσ,d is the Hardy-Littlewood maximal dyadic operator with respect to σ.
From the L2 (σ) boundedness (with absolute constants) of this operator, we obtain
(17).
We now observe that if RQ is the son of R that contains Q, ∆R g is constant in Q
and
∆R g|RQ = mRQ g − mR g := cR,Q (g).
Therefore we write,
∆R g = χR\RQ ∆R g + cR,Q (g)χRQ = χR\RQ ∆R g + cR,Q (g) · 1 − cR,Q (g)χRcQ .
Hence,
X
X
≤
|hT ∆Q f, χR\RQ ∆R gi|
hT
∆
f,
∆
gi
Q
R (Q,R)∈(3)
(Q,R)∈(3)
X
hT ∆Q f, cR,Q gi
+
|hT ∆Q f, cR,Q gχRcQ gi| + (Q,R)∈(3)
(Q,R)∈(3)
X
= A1 + A2 + A3 .
(19)
We will finish the proof of Theorem 2 by showing that Ai ≤ Ckf kkgk for i = 1, 2, 3.
The following lemma deals with A1 .
Lemma 9. We have
A1 =
X
|h∆Q f, T ∗ (χR\RQ ∆R g)i| ≤ Ckf kkgk.
(Q,R)∈(3)
Proof. If ψR = χR\RQ ∆R g then
d(Q, suppψR ) ≥ l(Q)γ l(R)1−γ .
Reasoning as in the proof of (10) we get that,
|hT ∆Q f, χR\RQ ∆R gi| ≤ C
l(Q)δ/2 l(R)δ/2
m(Q)1/2 m(R)1/2 k∆Q f k2 k∆R gk2 ,
δ+n
D(Q, R)
and following the same proof as in case (2) we get that,
X
|h∆Q f, T ∗ (χR\RQ ∆R g)i| ≤ Ckf kkgk.
(Q,R)∈(3)
We know turn our attention to A2 .
Lemma 10. We have
A2 =
X
(Q,R)∈(3)
|cR,Q (g)hT ∆Qf, χRcQ i| ≤ Ckf kkgk.
16
V. CHOUSIONIS AND X. TOLSA
c
Proof. UsingR(3) of Proposition 3 we have that for x ∈ RQ
and yQ denoting the
center of Q, K(x, yQ )∆Q f (y)dy = 0, since K is a standard kernel,
Z
|T ∆Q f (x)| = (K(x, y) − K(x, yQ ))∆Q f (y)dy Z
≤
|K(x, y) − K(x, yQ )||∆Q f (y)|dy
Q
≤C
Z
Q
(20)
|y − yQ |δ
|∆Q f (y)|dy
|x − yQ |n+δ
≤ Ck∆Q f k1
l(Q)δ
.
|x − yQ |n+δ
c
c
Therefore, as |x − yQ | ≥ d(RQ
, Q) ≥ l(Q)γ l(R)1−γ for x ∈ RQ
,
|hT ∆Q f, χRcQ i| ≤ Ck∆Q f k1
≤ Ck∆Q f k1
Z
RcQ
l(Q)δ
dx
|x − yQ |n+δ
Z
|x−yQ |≥l(Q)γ l(R)1−γ
l(Q)δ
dx
|x − yQ |n+δ
l(Q)δ
≤ Ck∆Q f k1
l(Q)γδ l(R)(1−γ)δ
δ(1−γ)
l(Q)
1/2
,
≤ Ck∆Q f km(Q)
l(R)
where we used integration on annuli and Hölder’s inequality. Furthermore by (3),
|cR,Q (g)| ≤ k∆R gk∞ ≈
k∆R gk
.
m(R)1/2
Thus,
l(Q)
|cR,Q (g)hT ∆Qf, χRcQ i| ≤ Ck∆Q f kk∆R gk
l(R)
δ(1−γ) m(Q)
m(R)
1/2
,
THE T 1 THEOREM
and so, for ε = δ(1 − γ)
X
|cR,Q (g)hT ∆Q f, χRcQ i|
17
(Q,R)∈(3)
δ(1−γ) 1/2
m(Q)
l(Q)
≤C
k∆Q f kk∆R gk
l(R)
m(R)
(Q,R)∈(3)
δ(1−γ) 1/2
X
X
l(Q)
m(Q)
≤C
k∆R gk
k∆Q f k
l(R)
m(R)
R∈D
Q:(Q,R)∈(3)
!2 1/2
!1/2 
δ(1−γ)
1/2
X
X X
m(Q)
l(Q)


≤C
k∆R gk2
k∆Q f k
l(R)
m(R)
R∈D
R∈D Q:Q⊂R
X

= Ckgk 
X
R∈D
≤ Ckgk
X
R∈D
X
k∆Q f k
2
Q:Q⊂R
"
X
k∆Q f k
Q:Q⊂R
l(Q)
l(R)
ε/2 l(Q)
l(R)
l(Q)
l(R)
ε/2 m(Q)
m(R)
1/2 !2
1/2

#!1/2
ε # " X ε
l(Q) m(Q)
,
l(R) m(R)
Q:Q⊂R
where we used twice Cauchy-Schwartz. For a cube R ∈ D let Dk (R) = {Q ∈ D :
Q ⊂ R, l(Q) = 2−k l(R)}. We then notice that,
X l(Q) ε m(Q) X X l(Q) ε m(Q)
=
l(R)
m(R)
l(R) m(R)
Q:Q⊂R
k∈N
Q∈Dk (R)
=
X
k∈N
2−kε
X
Q∈Dk (R)
m(Q) X −kε
2
≤ C.
=
m(R) k∈N
Therefore by the previous two estimates and Fubini,
X
|cR,Q (g)hT ∆Q f, χRcQ i| ≤ Ckgk
X X
R∈D Q:Q⊂R
(Q,R)∈(3)
k∆Q f k2
l(Q)
l(R)
ε !1/2
!1/2
X l(Q) ε
= Ckgk
k∆Q f k2
l(R)
Q∈D
R:R⊃Q
X
≤ Ckf kkgk.
The last inequality follows, because as previously
X l(Q) ε X X l(Q) ε X 1 k
=
=
≤ C.
l(R)
l(R)
2ε
R∈D:R⊃Q
k∈N R:R⊃Q,
k∈N
l(R)=2k l(Q)
18
V. CHOUSIONIS AND X. TOLSA
Given a cube Q ∈ D let (Ri )i∈N be the increasing sequence of cubes in D which
son of Ri which contains Q and
strictly contain Q. Recall that RiQ denotes the P
notice that R1Q = Q, Ri+1 Q = Ri and since g = R∈F2 ∆R f and F2 is a finite set
of cubes, there exists some i0 and some F2′ ⊂ F2 such that
suppg ∩ Ri0 = ∪R∈F2′ R.
Hence, by Proposition 3,
Z
g=
Ri0
X Z
R∈F2′
∆R g = 0
R
and mRi0 g = 0. Furthermore,
X
X
X
cR,Q (g) =
cRi ,Q g =
(mRiQ g − mRi g)
R)Q
i∈N
i∈N
By the previous observations this sum is telescopic and since mRi0 = 0 we get
X
cR,Q (g) = mQ g.
R)Q
Now we can estimate:
X
hT ∆Q f, cR,Q (g)1i =
(Q,R):R)Q
=
X
(Q,R):R)Q
X
Q∈D
=
cR,Q (g)hT ∆Qf, 1i
X
X
!
cR,Q (g) hT ∆Q f, 1i
R)Q
mQ (g)h∆Q f, T ∗ 1i.
Q∈D
Furthermore,
X
cR,Q (g)hT ∆Q f, 1i =
(Q,R)∈(3)
=
X
cR,Q (g)hT ∆Qf, 1i −
(Q,R):R)Q
X
X
cR,Q (g)hT ∆Q f, 1i
(Q,R):R)Q,
(Q,R)∈(3)
/
mQ (g)h∆Q f, T ∗ 1i −
(Q,R):R)Q
X
cR,Q (g)hT ∆Qf, 1i
(Q,R):R)Q,
(Q,R)∈(3)
/
= S3 − S4 .
(21)
Therefore |A3 | ≤ |S3 | + |S4 |, and in order to finish the proof we only need to bound
the terms S3 and S4 .
Lemma 11. For S4 as in (21) we have
|S4 | ≤ Ckf kkgk.
THE T 1 THEOREM
19
Proof. We have,
|S4 | ≤
X
|cR,Q (g)||hT ∆Qf, 1i|
X
|cR,Q (g)|kT ∆Qf k1 .
(Q,R):R)Q,
d(Q,R)≤l(Q)γ l(R)1−γ
≤
(Q,R):R)Q,
d(Q,R)≤l(Q)γ l(R)1−γ
We write kT ∆Q f k1 = k(T ∆Q f )χ3Q k1 + k(T ∆Q f )χ(3Q)c k1 , as in (20) we have that
for x ∈ (3Q)c and yQ being the center of Q,
|T ∆Q f (x)| ≤ Ck∆Q f k1
l(Q)δ
,
|x − yQ |n+δ
By integrating on annuli and using Hölder’s inequality we conclude that
Z
l(Q)δ
k(T ∆Q f )χ(3Q)c k1 ≤ Ck∆Q f k1
dx ≤ Ck∆Q f km(Q)1/2 .
n+δ
|x
−
y
|
Q
|x−yQ |>l(Q)
Again by Hölder’s inequality and (4),
k(T ∆Q f )χ3Q k1 ≤ CkT ∆Q f km(Q)1/2 ≤ Ck∆Q f km(Q)1/2 .
On the other hand, by (3),
k∆R gk
.
m(R)1/2
|cR,Q (g)| ≤ k∆R gk∞ ≈
So by all the previous estimates and Cauchy-Schwartz,
1/2
X
m(Q)
|S4 | ≤ C
k∆R gkk∆Q f k
m(R)
(Q,R)∈(3)
!1/2
!1/2
X
X
≤C
k∆Q f k2
k∆R gk2
Q∈D
R∈D
= Ckf kkgk.
Finally we need to bound |S3 |. By (5) and (6) of Proposition 3,
X
X
S3 =
mQ gh∆Q f, T ∗ 1i =
mQ ghf, ∆Q (T ∗ 1)i
Q∈D
=
X
Q∈F1
Q∈D
mQ ghf, ∆Q (T ∗ 1)i = hf,
X
Q∈F1
mQ g∆Q (T ∗ 1)i.
20
V. CHOUSIONIS AND X. TOLSA
Notice that the functions gQ = mQ g∆Q (T ∗ 1) are orthogonal, and thus
2
X
X
mQ g∆Q (T ∗ 1)
mQ g∆Q (T ∗ 1)i ≤ kf k2 |S3 | ≤ hf,
Q∈F1
Q∈F1
X
|mQ g|2k∆Q (T ∗ 1)k2 .
= kf k2
Q∈F1
The following lemma settles the case for S3 .
Lemma 12. We have
X
|mQ g|2k∆Q T ∗ 1k2 ≤ Ckgk2.
(22)
Q∈D
Proof. By Theorem 8 it suffices to prove that
X
k∆Q T ∗ 1k2 ≤ Cm(R)
(23)
∆Q ((T ∗ 1 − mR (T ∗ 1))χR ) = 0.
(24)
Q⊂R,Q∈D
for all R ∈ D.
For Q, R ∈ D, R ( Q and then
To see this let QR be the son of Q which contains R. Then, for h = (T ∗ 1 −
mR (T ∗ 1))χR
Z
Z
h=
T ∗ 1 − m(R)mR (T ∗ 1) = 0,
R
hence for x ∈ R
∆Q h(x) = mQR h − mQ h = 0.
Now by Proposition 4, since (T ∗ 1−mR (T ∗ 1))χR ∈ L2 , using the fact that ∆Q (mQ (T ∗ 1)) =
0 and (24), we get
X
X
k∆Q T ∗ 1k2 =
k∆Q ((T ∗ 1 − mR (T ∗ 1))χR )k2
Q∈D,Q⊂R
Q∈D,Q⊂R
=
X
k∆Q ((T ∗ 1 − mR (T ∗ 1))χR )k2
Q∈D
= k(T ∗ 1 − mR (T ∗ 1))χR k2 .
Thus in order to complete the proof it is enough to show that for all R ∈ D,
Z
1
|T ∗ 1 − mR (T ∗ 1)|2 ≤ C.
m(R) R
This is equivalent to saying that T ∗ ∈ BMOd2 , the dyadic BMO space.
Lemma 13. If kT ∗ χ2R k2 ≤ Cm(R),
Z
1
|T ∗ 1 − mR (T ∗ 1)|2 ≤ C.
m(R) R
(25)
THE T 1 THEOREM
21
Proof. We have
T ∗ 1 − mR (T ∗ 1) = T ∗ χ2R − mR (T ∗ χ2R ) + T ∗ χ(2R)c − mR (T ∗ χ(2R)c )
= A + B.
We first deal with A:
Z
Z
2
|A| =
R
|T ∗ χ2R − mR (T ∗ χ2R )|2
R
. kT ∗ χ2R k2 + m(R)|mR (T ∗ χ2R )|2
≤ Cm(R) + m(R)|mR (T ∗ χ2R )|2 .
But by Hölder’s inequality,
∗
2
|mR (T χ2R )| ≤
1
m(R)
1
m(R)
≤
Z
∗
|T χ2R |
R
Z
2
|T ∗ χ2R |2
1
kT ∗ χ2R k2
m(R)
1
≤C
m(R) = C.
m(R)
1/2
m(R)1/2
!2
≤C
Hence,
Now we will estimate
R
R
Z
|A|2 ≤ Cm(R).
R
2
|B| . The first step is to show that for all x1 , x2 ∈ R,
|T ∗ χ(2R)c (x1 ) − T ∗ χ(2R)c (x2 )| ≤ C.
To see this,
∗
∗
|T χ(2R)c (x1 ) − T χ(2R)c (x2 )| ≤
Z
|K(y, x1 ) − K(y, x2)|dy
(2R)c
≤C
Z
≤C
l(R)δ
= C,
l(R)δ
|x1 − x2 |
dy
n+δ
(2R)c |x1 − y|
Z
1
δ
dy
≤ Cl(R)
n+δ
(2R)c |x1 − y|
where in the last inequality we used integration in annuli as usual.
(26)
22
V. CHOUSIONIS AND X. TOLSA
Now,
Z
2
Z
Z ∗
1
∗
c (x2 )dx2 dx1
c (x1 ) −
T
χ
T
χ
|B| =
(2R)
(2R)
m(R) R
R
R
2
Z
Z
Z
1
1
∗
∗
c (x1 )dx2 −
c (x2 )dx2 dx1
=
T
χ
T
χ
(2R)
(2R)
m(R) R
R m(R) R
2
Z
Z
1 ∗
∗
c (x1 ) − T χ(2R)c (x2 ))dx2 dx1
(T
χ
=
(2R)
2 R m(R)
ZR
Z
1
≤
|T ∗ χ(2R)c (x1 ) − T ∗ χ(2R)c (x2 )|2 m(R)dx1
2
R m(R)
R
≤ Cm(R).
2
For the last two inequalities we used Hölder’s inequality and (26). This finishes the
proof of Lemma 13 and the proof of Lemma 12 as well. Hence the proof of Theorem
2 is completed.
Remark 14. Expressions of the form
Πh g =
X
mQ g∆Q h
Q∈D
P
are called paraproducts of g associated to h. In our case Πg = Q∈D mQ g∆Q (T ∗ 1)
is the paraproduct associated to T ∗ (1). Using Lemma 12 and orthogonality it follows
that kΠgk ≤ Ckgk.